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\begin{document}
\title[Navier-Stokes equations in Lorentz spaces]
{The Navier-Stokes equations in nonendpoint borderline Lorentz spaces}
\author[Nguyen Cong Phuc]
{Nguyen Cong Phuc}
\address{Department of Mathematics,
Louisiana State University,
303 Lockett Hall, Baton Rouge, LA 70803, USA.}
\email{pcnguyen@math.lsu.edu}
\begin{abstract} It is shown both locally and globally that $L_t^{\infty}(L_x^{3,q})$ solutions to the three-dimensional Navier-Stokes equations are regular provided $q\not=\infty$. Here $L_x^{3,q}$, $0<q\leq\infty$, is an increasing scale of Lorentz spaces containing $L^3_x$. Thus the result provides an improvement of a result by Escauriaza, Seregin and {\v S}ver\'ak ((Russian) Uspekhi Mat. Nauk {\bf 58} (2003), 3--44; translation in Russian Math. Surveys {\bf 58} (2003), 211--250), which treated the case $q=3$. A new local energy bound and a new $\epsilon$-regularity criterion are combined with the backward uniqueness theory of parabolic equations to obtain the result. A weak-strong uniqueness of Leray-Hopf weak solutions in $L_t^{\infty}(L_x^{3,q})$, $q\not=\infty$, is also obtained as a consequence.
\end{abstract}
\maketitle
\section{Introduction}
This paper addresses certain regularity and uniqueness criteria for the three-dimensional Navier-Stokes equations
\begin{equation}\label{NSEqu}
\partial_t u -\Delta u + {\rm div}\, u\otimes u +\nabla p=0, \quad {\rm div}\, u=0,
\end{equation}
where $u=u(x,t)=(u_1(x,t), u_2(x,t), u_3(x,t))\in \RR^3$ and $p=p(x,t)\in \RR$, with $x\in \RR^3$ and $t\geq 0$. The initial condition associated to \eqref{NSEqu} is given by
\begin{equation}\label{IniC}
u(x,0)=a(x), \quad x\in \RR^3.
\end{equation}
Equations \eqref{NSEqu}--\eqref{IniC} describes the motion an incompressible fluid in three spatial dimensions
with unit viscosity and zero external force. Here $u$ and $p$ are referred to as the fluid velocity and pressure, respectively.
From the classical works of Leray \cite{Ler} and Hopf \cite{Hop}, it is known that for any divergence-free vector field $a\in L^2(\RR^3)$
there exists at least one weak solution to the Cauchy problem
\eqref{NSEqu}--\eqref{IniC} in $\RR^3\times (0,\infty)$. Such a solution is now called {\it Leray-Hopf weak solution} whose precise definition will be given next. Let $\dot{C}_0^\infty$ denote the space of all divergence-free infinitely
differentiable vector fields with compact support in $\RR^3$. Let $\dot{J}$ be the closure of $\dot{C}_0^\infty$
in $L^2(\RR^3)$, and $\dot{J}^{1,2}$ be the closure of the same set with respect to the Dirichlet integral.
\begin{definition} A Leray-Hopf weak solution of the Cauchy problem \eqref{NSEqu}--\eqref{IniC} in $Q_\infty:=\RR^3\times (0, \infty)$ is a vector field
$u: Q_\infty \rightarrow \RR^3$ such that:
{\rm (i)} $u\in L^\infty(0,\infty; \dot{J})\cap L^2(0,\infty; \dot{J}^{1,2})$;
{\rm (ii)} The function $t\rightarrow \int_{\RR^3} u(x,t) w(x) dx$ is continuous on $[0,\infty)$ for any $w\in L^2(\RR^3)$;
{\rm (iii)} For any $w\in \dot{C}_0^\infty(Q_\infty)$ there holds
$$\int_{Q_\infty} (-u \cdot \partial_t w -u\otimes u:\nabla w +\nabla u:\nabla w) dx dt=0;$$
{\rm (iv)} The energy inequality
$$ \int_{\RR^3} |u(x,t)|^2 dx + 2 \int_{0}^{t}\int_{\RR^3} |\nabla u|^2 dxds \leq \int_{\RR^3} |a(x)|^2 dx$$
holds for all $t\in [0, \infty)$, and
$$\norm{u(\cdot,t)-a(\cdot)}_{L^2(\RR^3)}\rightarrow 0 \quad {\rm as~} t\rightarrow 0^{+}.$$
\end{definition}
As of now the problems of uniqueness and regularity of Leray
Hopf weak solutions are still open. Only some partial results are known. The partial uniqueness result of Prodi \cite{Pro} and Serrin \cite{Ser2}, and the partial smoothness result of Ladyzhenskaya \cite{Lad} can be summarized in the following theorem.
\begin{theorem} \label{PSL} Suppose that $a\in \dot{J}$ and $u$, $u_1$ are two Leray-Hopf weak solutions to the Cauchy problem \eqref{NSEqu}--\eqref{IniC}.
If $u\in L^s(0,T; L^p(\RR^3))$ for some $T>0$, where
$$\frac{3}{p} + \frac{2}{s}=1, \qquad p\in (3, \infty],$$
then $u=u_1$ in $Q_T:=\RR^3\times (0, T)$ and, moreover, $u$ is smooth on $\RR^3\times (0, T]$.
\end{theorem}
Here recall that the condition $u\in L^s(0,T; L^p(\RR^3))$ means that
$$\norm{u}_{L^s(0,T; L^p(\RR^3))}:=\Big(\int_{0}^{T}\norm{u(\cdot,t)}^s_{L^p(\RR^n)} dt\Big)^{\frac{1}{s}}<+\infty \qquad {\rm if~} s\in[1,\infty),$$
and
$$\norm{u}_{L^s(0,T; L^p(\RR^3))}:=\esssup_{t\in(0,T)}\norm{u(\cdot,t)}_{L^p(\RR^n)}<+\infty \qquad {\rm if~} s=\infty.$$
It is obvious that, when $s=p$, $L^p(0,T; L^p(\RR^3))= L^p(Q_T)$. In general,
if X is a Banach space with norm $\norm{\cdot}_{X}$, then
$L^s(a,b; X)$, $a<b$, means the usual Banach space of measurable X-valued functions $f(t)$ on $(a,b)$ such that the norm
\begin{equation}\label{norms}
\norm{u}_{L^s(a,b; X)}:=\Big(\int_{0}^{T}\norm{f(t)}^s_{X} dt\Big)^{\frac{1}{s}}<+\infty
\end{equation}
for $s\in [1, \infty)$, and with the usual modification of \eqref{norms} in the case $s=\infty$.
The endpoint case $p=3$ and $s=\infty$, which is not covered by Theorem \ref{PSL}, was considered harder and has been settled by Escauriaza-Seregin-{\v S}ver\'ak
in the interesting work \cite{ESS1}:
\begin{theorem}\label{global-ESS} Let $a\in \dot{J}\cap L^3(\RR^3)$.
Suppose that $u$ is a Leray-Hopf weak solution of the Cauchy problem
\eqref{NSEqu}--\eqref{IniC}, and u satisfies the additional condition
\begin{equation}\label{PSLcrit}
u\in L^\infty(0,T; L^3(\RR^3))
\end{equation}
for some $T>0$. Then $u\in L^5(Q_T)$ and hence it is unique and smooth on $\RR^3\times(0,T]$.
\end{theorem}
We remark that the condition $a\in L^3(\RR^3)$ in the above theorem is superfluous as it can be deduce from condition \eqref{PSLcrit}.
A basic consequence of Theorem \ref{global-ESS} is that if a Leray-Hopf weak solution $u$ develops a singularity at a first finite time $t_0>0$
then there necessarily holds
\begin{equation}\label{L3blowup}
\limsup_{t\uparrow t_0} \norm{u(\cdot,t)}_{L^3(\RR^3)}=+\infty.
\end{equation}
An improvement of this necessary condition of potential blow up can be found in the recent work \cite{Sere}. See also the papers \cite{GKP, KeKo}
for another approach to regularity using certain profile decompositions.
It should be noticed that the uniqueness of $u$ under condition \eqref{PSLcrit} had been known earlier (see \cite{KS}).
Moreover, local versions of the corresponding partial regularity results are also available (see \cite{Ser1}, \cite{Stru}, and \cite{ESS1}).
In particular, the local regularity result of \cite{ESS1} reads as follows.
\begin{theorem} \label{localregularity-ESS} Suppose that the pair of functions $(u,p)$ satisfies
the Navier-Stokes equations \eqref{NSEqu} in $Q_1(0,0)=B_1(0)\times(-1,0)$ in the sense of distributions and has the following properties:
\begin{equation}\label{u-reg}
u\in L^\infty(-1,0; L^2(B_1)) \cap L^2(-1,0; W^{1, 2}(B_1))
\end{equation}
and
\begin{equation*}
p\in L^{3/2}(-1, 0; L^{3/2}(B_{1})).
\end{equation*}
Suppose further that
\begin{equation*}
u\in L^\infty(-1,0; L^{3}(B_1)).
\end{equation*}
Then the velocity function $u$ is H\"older continuous on $\overline{Q}_{1/2}(0,0)$.
\end{theorem}
The main goal of this paper is to improve Theorems \ref{global-ESS} and \ref{localregularity-ESS} by means of Lorentz spaces.
Given a measurable set $\Om\subset\RR^3$, recall that the Lorentz space
$L^{p,q}(\Om)$, with $p\in(0, \infty), q\in(0,\infty]$, is the set of
measurable functions $g$ on $\Omega$ such that the quasinorm $\|g\|_{L^{p, q}(\Omega)}$ is finite. Here we define
\begin{equation*}
\|g\|_{L^{p, q}(\Omega)}:=\left\{ \begin{array}{lcr}
\displaystyle \left(p \int_{0}^{\infty}\alpha^{q} |\{x\in\Om: |g(x)|>\alpha\}|^{\frac{q}{p}} \frac{d\alpha}{\alpha}
\right)^{\frac{1}{q}} \text{~if~} q<\infty, \\
\, \ \displaystyle \sup_{\alpha >0} \,\alpha \, |\{x\in \Om: |g(x)|>\alpha\}|^{\frac{1}{p} } \text{~if~} q=\infty.
\end{array}\right.
\end{equation*}
The space $L^{p,\infty}(\Om)$ is often referred to as the Marcinkiewicz or weak $L^p$ space. It is known that $L^{p,p}(\Om)=L^p(\Om)$ and $L^{p,q_1}(\Om) \subset L^{p,q_2}(\Om)$ whenever $q_1\leq q_2$.
On the other hand, if $|\Om|$ is finite then $L^{p,q}(\Om)\subset L^r(\Om)$ for any $0<q\leq\infty$ and $0<r<p$. Moreover,
$$\norm{g}_{L^{r}(\Om)}\leq |\Om|^{\frac{1}{r}-\frac{1}{p}} \norm{g}_{L^{p,q}(\Om)}.$$
Lorentz spaces can be used to capture logarithmic singularities. For example, in $\RR^3$, for any $\beta>0$ we have
\begin{equation}\label{loglorentz}
|x|^{-1} | \log (|x|/2)|^{-\beta} \in L^{3,q}(B_1(0)) \quad {\rm if~and~only~if~} q>\frac{1}{\beta}.
\end{equation}
Note that the inequality in \eqref{loglorentz} is strict.
Of course, in the case $\beta=0$, the function $|x|^{-1}$ belongs to the Marcinkiewicz space $L^{3,\infty}(\RR^3)$.
To the best of our knowledge, a criterion of local regularity for the Navier-Stokes equations in $L^\infty(-1,0; L^{3,\infty}(B_1))$
is still unknown. See \cite{KK, LT} for some partial results, which require a smallness condition. See also \cite{Soh, Tak} for some nonendpoint
related results. The first result of this paper provides instead a regularity condition in terms of the borderline space
$L^\infty(-1,0; L^{3,q}(B_1))$ for {\it any} $q\in (0,\infty)$, and thus excluding only the endpoint case $q=\infty$.
\begin{theorem} \label{localregularity} Suppose that the pair of functions $(u,p)$ satisfies
the Navier-Stokes equations \eqref{NSEqu} in $Q_1(0,0)=B_1(0)\times(-1,0)$ in the sense of distributions such that \eqref{u-reg} holds
and
\begin{equation}\label{p-assum}
p\in L^{2}(-1, 0; L^{1}(B_{1})).
\end{equation}
Suppose further that
\begin{equation}\label{serrinlorentz}
u\in L^\infty(-1,0; L^{3,q}(B_1))
\end{equation}
for some $q\in (3,\infty)$. Then the velocity function $u$ is H\"older continuous on $\overline{Q}_{1/2}(0,0)$.
\end{theorem}
It is worth mentioning that even the regularity at $(0,0)$ is still unknown for solutions $u$ satisfying the pointwise bound
$$|u(x,t)|\leq C\, |x|^{-1} $$
for a.e. $(x,t)\in Q_1(0,0)$. A regularity result under this condition is known only for axially symmetric solutions (see \cite{SS2} and also \cite{CSTY1, CSTY2}). On the other hand, in view of \eqref{loglorentz}, Theorem \ref{localregularity} yields the regularity of $u$ under a logarithmic `bump' condition
$$|u(x,t)|\leq C\, |x|^{-1} | \log (|x|/2)|^{-\beta} $$
for any $\beta>0$.
In fact, it is possible to obtain regularity under a weaker pointwise bound condition on the solution. In this case
Theorem \ref{localregularity} is no longer applicable.
\begin{theorem}\label{nearMarc} Suppose that the pair of functions $(u,p)$ satisfies
the Navier-Stokes equations \eqref{NSEqu} in $Q_1(0,0)$ in the sense of distributions such that \eqref{u-reg} holds, and
$$p\in L^{3/2}(-1, 0; L^{1}(B_{1})).$$
Suppose further that for a.e. $(x,t)\, \in Q_1(0,0)$, there holds
\begin{equation}\label{LinftyX}
|u(x,t)|\leq f(t)|x|^{-1} g(x)
\end{equation}
for nonnegative functions $f\in L^\infty((-1,0))$ and $g\in L^{\infty}(B_1(0))$ such that $\lim_{x\rightarrow 0} g(x)=0$. Then $u$ is H\"older continuous on $\overline{Q}_{1/2}(0,0)$.
\end{theorem}
On the other hand, Theorem \ref{localregularity} can be used to deduce the following uniqueness and global regularity results, which give an
improvement of Theorem \ref{global-ESS}.
\begin{theorem}\label{global-ESS-lorentz} Let $a\in \dot{J}$.
Suppose that $u$ is a Leray-Hopf weak solution of the Cauchy problem
\eqref{NSEqu}--\eqref{IniC}, and it satisfies the additional condition
\begin{equation}\label{PSLcrit-lorentz}
u\in L^\infty(0,T; L^{3,q}(\RR^3))
\end{equation}
for some $q\in (3,\infty)$ and $T>0$. Then $u$ is smooth on $\RR^3\times(0,T]$. Moreover, if in addition $a\in L^3(\RR^3)$ then $u\in L^5(Q_T)$ and hence it is unique in $Q_T$ (in the sense of weak-strong uniqueness as in Theorem \ref{PSL}).
\end{theorem}
Theorem \ref{global-ESS-lorentz} implies that the necessary condition of potential blow up \eqref{L3blowup} can now be improved by replacing the $L^3$ norm
with any smaller $L^{3, q}$ quasi-norm provided $q\not=\infty$. We should mention that this kind of potential blow up criterion has recently been
extended in \cite{GKP2} to the norms of Besov spaces $\dot{B}_{q, p}^{-1+3/p}(\RR^3)$, $3<p,q<\infty$, using profile decompositions in the framework of ``strong" solutions. See also \cite{CP} for an earlier related result. In such a setting of strong solutions, the blow up criterion of \cite{GKP2}
is more general than ours since for $q>3$,
$$L^{3,q}(\RR^3)\subset \dot{B}_{q, q}^{-1+3/q}(\RR^3).$$
However, our blow up criterion here is obtained for Leray-Hopf weak solutions.
Moreover, using instead the local regularity criterion, Theorem \ref{localregularity}, we see that if $(x_0,t_0)$ is a singular
point of a Leray-Hopf weak solution $u$ then there holds
$$\limsup_{t\uparrow t_0} \norm{u(\cdot,t)}_{L^{3,q}(B_{\delta}(x_0))}=+\infty$$
for any $\delta>0$ and $q\not=\infty$. Note that for a Leray-Hopf weak solution $u$, the associated pressure $p$ can be chosen so that
$p\in L^2(0,\infty; L^{3/2}(\RR^3))$ since $|u|^2$ belongs to the same space (see, e.g., \cite{SS}).
Our approach to Theorems \ref{localregularity} and \ref{global-ESS-lorentz} is influenced by the above mentioned work of Escauriaza-Seregin-{\v S}ver\'ak
\cite{ESS1}, which reduces the regularity matter to the backward uniqueness problem for parabolic equations
with variable lower-order terms. A key ingredient, which makes our results stronger than that
of \cite{ESS1}, is a new $\epsilon$-regularity criterion for {\it suitable weak solutions} to the Navier-Stokes equations (see Proposition \ref{epsilon1}).
See Definition \ref{SWS} below for the notion of suitable weak solutions.
In turn, this kind of $\epsilon$-regularity criterion is a consequence of a new bound for some scaling invariant energy quantities (see Corollary \ref{ABcontrol2}). Moreover, this new energy bound is also essential in a blow-up procedure needed in the proof of Theorem \ref{localregularity}.
It provides a certain compactness result and thus yields a non-trivial `ancient solution' (see Proposition \ref{limit-infty}), another important
ingredient in the proof of Theorem \ref{localregularity}.
On the other hand, the proof of Theorem \ref{nearMarc} is simple. It requires only an $\epsilon$-regularity criterion of Seregin and {\v S}ver\'ak in \cite[Lemma 3.3]{SS}.
\section{ Preliminaries and local energy estimates}
Throughout the paper we use the following notations for balls and parabolic cylinders:
$$B_r(x)=\{y\in \RR^3: |x-y|<r\}, \quad x\in\RR^3, \, r>0,$$
and
$$Q_r(z)=B_r(x)\times (t-r^2, t) \quad {\rm with~} z=(x,t).$$
The following scaling invariant quantities will be employed:
\begin{eqnarray*}
A(z_0, r)=A(u, z_0, r)&=& \sup_{t_0-r^2\leq t\leq t_0} r^{-1}\int_{B_r(x_0)} |u(x,t)|^2 dx,\\
B(z_0, r)=B(u, z_0, r)&=& r^{-1}\int_{Q_r(x_0)} |\nabla u(x,t)|^2 dxdt,\\
C(z_0, r)=C(u, z_0, r)&=& r^{-3}\int_{t_0-r^2}^{t_0} \norm{u}_{L^{\frac{12}{5}}(B_r(x_0))}^4 dt,\\
C_1(z_0, r)=C_1(u, z_0, r)&=& r^{-2}\int_{t_0-r^2}^{t_0} \norm{u}_{L^{3}(B_r(x_0))}^3 dt,\\
D(z_0, r)=D(p,z_0, r)&=& r^{-3}\int_{t_0-r^2}^{t_0} \norm{p}_{L^{\frac{6}{5}}(B_r(x_0))}^2 dt,\\
D_1(z_0, r)=D_1(p, z_0, r)&=& r^{-2}\int_{t_0-r^2}^{t_0} \norm{p}_{L^{\frac{3}{2}}(B_r(x_0))}^{3/2} dt.\\
\end{eqnarray*}
To analyze local properties of solutions, it is often useful to use the notion of suitable weak solutions. Such a notion of weak solutions was introduced in Caffarelli-Kohn-Nirenberg \cite{CKN} following the work of Scheffer \cite{Sche1}--\cite{Sche4}. Here we use the version introduced by Lin in \cite{Lin}.
\begin{definition}\label{SWS} Let $\om$ be an open set in $\RR^3$ and let $-\infty<a< b< \infty$. We say that a pair $(u, p)$ is a suitable weak solution
to the Navier-Stokes equations in $Q=\om\times (a, b)$ if the following conditions hold:
{\rm (i)} $u\in L^\infty(a, b; L^2(\om))\cap L^2(a, b; W^{1,\, 2}(\om)) {\rm ~and~} p\in L^{3/2}(\om\times (a, b));$
{\rm (ii)} $(u,p)$ satisfies the Navier-Stokes equations in the sense of distributions. That is,
$$\int_{a}^{b}\int_{\om} \left\{-u \, \psi_{t} + \nabla u : \nabla \psi - (u\otimes u):\nabla \psi - p \, {\rm div}\, \psi \right\} dxdt =0$$
for all vector fields $\psi\in C_0^{\infty}(\om\times(a,b); \RR^3)$, and
$$\int_{\om\times \{t\}} u(x,t)\cdot \nabla \phi(x) \, dx=0$$
for a.e. $t\in (a, b)$ and all real valued functions $\phi\in C_0^{\infty}(\om)$;
{\rm (iii)} $(u,p)$ satisfies the local generalized energy inequality
\begin{eqnarray*}
\lefteqn{ \int_{\om}|u(x, t)|^2 \phi(x,t) dx + 2\int_{a}^t\int_{\om} |\nabla u|^2 \phi(x, s) dxds} \\
&\leq& \int_{a}^t\int_{\om} |u|^2 (\phi_t +\Delta \phi) dx ds + \int_{a}^t\int_{\om}(|u|^2 + 2p)u\cdot \nabla \phi dx ds.
\end{eqnarray*}
for a.e. $t\in (a, b)$ and any nonnegative function $\phi \in C_0^{\infty}(\RR^3\times\RR)$ vanishing in a neighborhood of the parabolic boundary
$\partial'Q=\om\times\{t=a\} \cup \partial\om\times [a, b]$.
\end{definition}
A proof of the following lemma can be found in \cite[Lemma 6.1]{Giu}.
\begin{lemma}\label{Giusti-lem}
Let $I(s)$ be a bounded nonnegative function in the interval $[R_1, R_2]$. Assume that for every $s, \rho\in [R_1, R_2]$ and $s<\rho$ we have
$$I(s)\leq [A(\rho-s)^{-\alpha} +B(\rho-s)^{-\beta} +C] +\theta I(\rho)$$
with $A, B, C\geq 0$, $\alpha>\beta>0$ and $\theta\in [0,1)$. Then there holds
$$I(R_1)\leq c(\alpha, \theta) [A(R_2-R_1)^{-\alpha} +B(R_2-R_1)^{-\beta} +C].$$
\end{lemma}
In the next lemma $L^{-1, \, 2}(B_r(x_0))$ stands for the dual of the Sobolev space $W^{1, \, 2}_0(B_r(x_0))$. The latter is defined as the completion of $C_0^{\infty}(B_r(x_0))$ under the Dirichlet norm
$$\norm{\varphi}_{W^{1, \, 2}_0(B_r(x_0))}=\Big(\int_{B_r(x_0)} |\nabla \varphi|^2 dx \Big)^{1/2}.$$
\begin{lemma}\label{ABcontrol0} Suppose that $(u,p)$ is a suitable weak solution
to the Navier-Stokes equations in $Q=\om\times (a, b)$. Let $z_0=(x_0, t_0)$ and $r>0$ be such that $Q_r(z_0)\subset Q$. Then there holds
\begin{eqnarray*}
A(z_0, r/2) + B(z_0, r/2)&\leq& C \left[r^{-3} \int_{t_0-r^2}^{t_0} \norm{|u|^2}^2_{L^{-1, \, 2}(B_r(x_0))} dt\right]^{1/2}\nonumber\\
&& + \, C r^{-3} \int_{t_0-r^2}^{t_0} \norm{|u|^2 +2p}^2_{L^{-1, \, 2}(B_r(x_0))} dt.\nonumber
\end{eqnarray*}
\end{lemma}
\begin{proof}
For $z_0=(x_0, t_0)$ and $r>0$ such that $Q_r(z_0)\subset Q$, we consider the cylinders $$Q_s(z_0)=B_s(x_0)\times (t_0-s^2, t_0)\subset Q_\rho(z_0)=B_\rho(x_0)\times (t_0-\rho^2, t_0),$$
where $r/2\leq s<\rho\leq r$.
Let $\phi(x,t)=\eta_1(x)\eta_2(t)$ where $\eta_1\in C_0^{\infty}(B_\rho(x_0))$, $0\leq \eta_1\leq 1$ in $\RR^n$, $\eta_1\equiv 1$ on $B_s(x_0)$, and $$|\nabla^{\alpha} \eta_1|\leq \frac{c}{(\rho-s)^{|\alpha|}}$$ for all multi-indices $\alpha$ with $|\alpha|\leq 3$. The function $\eta_2(t)$ is chosen so that
$\eta_2\in C_0^{\infty}(t_0-\rho^2, t_0+\rho^2)$, $0\leq\eta_2\leq 1$ in $\RR$, $\eta_2(t)\equiv 1$ for $t\in [t_0-s^2, t_0+s^2]$, and $$|\eta'_2(t)|\leq \frac{c}{\rho^2-s^2}\leq \frac{c}{r(\rho-s)}.$$
Then
$$|\nabla \phi_t|\leq \frac{c}{r(\rho-s)^2}\leq \frac{c}{(\rho-s)^3},\quad |\nabla \Delta \phi|\leq \frac{c}{(\rho-s)^3},$$
$$| \nabla^2 \phi|\leq \frac{c}{(\rho-s)^2},\quad |\nabla \phi|\leq \frac{c}{\rho-s}.$$
We next define
$$I(s)=I_1(s) +I_2(s),$$
where
$$I_1(s)=\sup_{t_0-s^2\leq t\leq t_0}\int_{B_s(x_0)}|u(x, t)|^2 dx=s\, A(z_0,s)$$
and
$$I_2(s)=\int_{t_0-s^2}^{t_0}\int_{B_s(x_0)}|\nabla u(x, t)|^2 dx dt=s\, B(z_0,s).$$
Using $\phi$ as a test function in the generalized energy inequality we find
\begin{eqnarray}\label{Isenergy}
\lefteqn{I(s)}\\
&\leq& \int_{t_0-\rho^2}^{t_0} \norm{|u|^2}_{L^{-1, \, 2}(B_\rho(x_0))}\norm{\nabla \phi_t +\nabla \Delta \phi}_{L^2(B_{\rho}(x_0))} dt \nonumber\\
&& + \int_{t_0-\rho^2}^{t_0} \Big\{\norm{|u|^2 + 2p}_{L^{-1, \, 2}(B_\rho(x_0))}\times\nonumber\\
&& \qquad \qquad\qquad\times\norm{\nabla u\cdot \nabla \phi + u\cdot \nabla^2\phi}_{L^2(B_{\rho}(x_0))}\Big\} dt\nonumber \\
&=:& J_1+ J_2.\nonumber
\end{eqnarray}
By the choice of test function we have
\begin{eqnarray}\label{firstterm}
J_1&\leq& C \frac{\rho^{3/2}}{(\rho-s)^3} \int_{t_0-\rho^2}^{t_0} \norm{|u|^2}_{L^{-1, \, 2}(B_\rho(x_0))} dt\\
&\leq& C \frac{\rho^{5/2}}{(\rho-s)^3} \left[ \int_{t_0-\rho^2}^{t_0} \norm{|u|^2}^2_{L^{-1, \, 2}(B_\rho(x_0))} dt\right]^{1/2}.\nonumber
\end{eqnarray}
Also,
\begin{eqnarray*}
J_2 &\leq& C \int_{t_0-\rho^2}^{t_0} \Big\{ \norm{|u|^2+2p}_{L^{-1, \, 2}(B_\rho(x_0))} \times\\
&& \qquad \qquad\qquad \times \Big[\frac{\norm{\nabla u}_{L^2(B_\rho(x_0))}}{\rho-s} +
\frac{\norm{u}_{L^2(B_\rho(x_0))}}{(\rho-s)^2}\Big] \Big\}dt,
\end{eqnarray*}
and thus by H\"older's inequality we get
\begin{eqnarray}\label{secondterm}
J_2 &\leq& \frac{C}{\rho-s} \left[\int_{t_0-\rho^2}^{t_0} \norm{|u|^2 +2p}^2_{L^{-1, \, 2}(B_\rho(x_0))} dt\right]^{1/2} I_{2}(\rho)^{1/2} \\
&&+ \frac{C \rho}{(\rho-s)^2} \left[\int_{t_0-\rho^2}^{t_0} \norm{|u|^2 +2p}^2_{L^{-1, \, 2}(B_\rho(x_0))} dt\right]^{1/2} I_{1}(\rho)^{1/2}.\nonumber
\end{eqnarray}
Combining inequalities \eqref{Isenergy}--\eqref{secondterm} and using $\rho\leq r$ we arrive at
\begin{eqnarray*}
I(s) &\leq& \frac{C r^{5/2}}{(\rho-s)^3} \left[ \int_{t_0-\rho^2}^{t_0} \norm{|u|^2}^2_{L^{-1, \, 2}(B_\rho(x_0))} dt\right]^{1/2}+\\
&& +\frac{C}{\rho-s} \left[\int_{t_0-\rho^2}^{t_0} \norm{|u|^2 +2p}^2_{L^{-1, \, 2}(B_\rho(x_0))} dt\right]^{1/2} I_{2}(\rho)^{1/2} +\\
&&+ \frac{C r}{(\rho-s)^2} \left[\int_{t_0-\rho^2}^{t_0} \norm{|u|^2 +2p}^2_{L^{-1, \, 2}(B_\rho(x_0))} dt\right]^{1/2} I_{1}(\rho)^{1/2}.
\end{eqnarray*}
By Young's inequality this yields
\begin{eqnarray*}
I(s) &\leq& \frac{C r^{5/2}}{(\rho-s)^3} \left[ \int_{t_0-\rho^2}^{t_0} \norm{|u|^2}^2_{L^{-1, \, 2}(B_\rho(x_0))} dt\right]^{1/2}\\
&& +\left\{\frac{C}{(\rho-s)^2} +\frac{C r^2}{(\rho-s)^4} \right\}\int_{t_0-\rho^2}^{t_0} \norm{|u|^2+2p}^2_{L^{-1, \, 2}(B_\rho(x_0))} dt \\
&& + \frac{1}{2}I(\rho),
\end{eqnarray*}
which implies in particular that
\begin{eqnarray*}
I(s) &\leq& \frac{C r^{5/2}}{(\rho-s)^3} \left[ \int_{t_0-r^2}^{t_0} \norm{|u|^2}^2_{L^{-1, \, 2}(B_r(x_0))} dt\right]^{1/2}\\
&& +\frac{C r^2}{(\rho-s)^4} \int_{t_0-r^2}^{t_0} \norm{|u|^2+2p}^2_{L^{-1, \, 2}(B_r(x_0))} dt + \frac{1}{2}I(\rho).
\end{eqnarray*}
Since this holds for all $r/2\leq s<\rho\leq r$ by Lemma \ref{Giusti-lem} we find
\begin{eqnarray*}
I(r/2) &\leq& C r^{-1/2} \left[ \int_{t_0-r^2}^{t_0} \norm{|u|^2}^2_{L^{-1, \, 2}(B_r(x_0))} dt\right]^{1/2}\\
&& + \, C r^{-2} \int_{t_0-r^2}^{t_0} \norm{|u|^2 +2p}^2_{L^{-1, \, 2}(B_r(x_0))} dt.
\end{eqnarray*}
Thus
\begin{eqnarray*}
A(z_0, r/2) + B(z_0, r/2)&\leq& C \left[r^{-3} \int_{t_0-r^2}^{t_0} \norm{|u|^2}^2_{L^{-1, \, 2}(B_r(x_0))} dt\right]^{1/2}\nonumber\\
&& + \, C r^{-3} \int_{t_0-r^2}^{t_0} \norm{|u|^2 +2p}^2_{L^{-1, \, 2}(B_r(x_0))} dt
\end{eqnarray*}
as desired.
\end{proof}
Note that for $f\in L^{6/5}(B_r(x_0))$ and for $\varphi\in C_0^{\infty}(B_r(x_0))$ we have
\begin{eqnarray*}
\left |\int_{B_r(x_0)} \varphi(x) f(x)dx \right | &\leq& C \int_{B_r(x_0)} \left[\int_{B_r(x_0)}\frac{|\nabla \varphi(y)|}{|x-y|^{2}} dy\right]
|f(x)| dx\\
&=& C \int_{B_r(x_0)}|\nabla \varphi(y)| \left[\int_{B_r(x_0)}\frac{ |f(x)|dx}{|x-y|^{2}}\right]
dy\\
&\leq& C \norm{\nabla \varphi}_{L^2(B_r(x_0))} \norm{ {\rm\bf I}_1(\chi_{B_r(x_0)}|f|)}_{L^2(B_r(x_0))}.
\end{eqnarray*}
Here ${\rm\bf I}_1$ is the first order Riesz's potential defined by
$${\rm\bf I}_1(\mu)(x)= c \int_{\RR^3}\frac{d\mu(y)}{|x-y|^{2}}, \qquad x\in \RR^3, $$
for a nonnegative locally finite measure $\mu$ in $\RR^3$.
Thus we find
\begin{equation}\label{dualnorm}
\norm{f}_{L^{-1,\, 2}(B_r(x_0))} \leq C \norm{ {\rm\bf I}_1(\chi_{B_r(x_0)}|f|)}_{L^2(B_r(x_0))}\leq C\norm{f}_{L^{\frac{6}{5}}(B_r(x_0))}
\end{equation}
by the embedding property of Riesz's potentials.
Using \eqref{dualnorm} we obtain the following important consequence of Lemma \ref{ABcontrol0}.
\begin{corollary} \label{ABcontrol2} Suppose that $(u,p)$ is a suitable weak solution
to the Navier-Stokes equations in $Q=\om\times (a, b)$. Let $z_0=(x_0, t_0)$ and $r>0$ be such that $Q_r(z_0)\subset Q$. Then there holds
\begin{equation*}
A(z_0, r/2) + B(z_0, r/2)\leq C[ C(z_0, r)^{1/2} + C(z_0, r) + D(z_0, r)].
\end{equation*}
\end{corollary}
\section{$\epsilon$-regularity criteria}
As demonstrated in \cite{ESS1}, the proof of Theorem \ref{localregularity-ESS} above relies heavily on the following $\epsilon$-regularity criterion for suitable weak solutions to the Navier-Stokes equations (see \cite[Lemma 2.2]{ESS1}, see also \cite{CKN, LS, NRS}).
\begin{proposition}\label{C1D1ep0}
There exist positive constants $\epsilon_0$ and $C_k$, $k=0, 1, 2, \dots$, such that the following holds. Suppose that the pair
$(u,p)$ is a suitable solution to the Navier-Stokes equations in $Q_1(z_0)$ and satisfies the smallness condition
\begin{equation*}
C_1(u,z_0,1) + D_1(p,z_0,1) \leq \epsilon_0.
\end{equation*}
Then $\nabla^{k} u$ is H\"older continuous on $\overline{Q}_{1/2}(z_0)$ for any integer $k\geq 0$, and
\begin{equation*}
\max_{z\in \overline{Q}_{1/2}(z_0)} |\nabla^{k} u(z)| \leq C_k.
\end{equation*}
\end{proposition}
To prove Theorem \ref{localregularity} we use instead a new different version of $\epsilon$-regularity criterion.
\begin{proposition}\label{epsilon1}
There exist positive constants $\epsilon_1$ and $C_k$, $k=0, 1, 2, \dots$, such that the following holds. Suppose that the pair
$(u,p)$ is a suitable solution to the Navier-Stokes equations in $Q_8(z_0)$ and satisfies the smallness condition
\begin{equation}\label{smallnessCD}
C(u,z_0,8) + D(p,z_0,8) \leq \epsilon_1.
\end{equation}
Then $\nabla^{k} u$ is H\"older continuous on $\overline{Q}_{1/2}(z_0)$ for any integer $k\geq 0$, and
\begin{equation*}
\max_{z\in \overline{Q}_{1/2}(z_0)} |\nabla^{k} u(z)| \leq C_k.
\end{equation*}
\end{proposition}
The proof of Proposition \ref{epsilon1} will be given at the end of this section. It requires the following two preliminary results. The first one is by now a well-known lemma that can be found in \cite[Lemma 2.1]{Lin}.
\begin{lemma}\label{boundC1}
Suppose that $(u,p)$ is a suitable weak solution
to the Navier-Stokes equations in $Q=\om\times (a, b)$.
Let $z_0=(x_0, t_0)$ and let $\rho>0$ be such that $Q_{\rho}(z_0)\subset Q=\om\times (a, b)$. For any $r\in (0, \rho]$ we have
$$C_1(z_0,r)\leq C \Big(\frac{\rho}{r} \Big)^3 A(z_0,\rho)^{3/4} B(z_0,\rho)^{3/4} + C \Big(\frac{r}{\rho} \Big)^3 A(z_0,\rho)^{3/2}.$$
\end{lemma}
In what follows we shall use the following notation to denote the spatial average of a function $f$ over a ball $B_r(x_0)$:
$$[f]_{x_0, r}:=\frac{1}{|B_r(x_0)|}\int_{B_r(x_0)} f(x) \, dx.$$
\begin{lemma}\label{Dbar}
Suppose that $(u,p)$ is a suitable weak solution
to the Navier-Stokes equations in $Q=\om\times (a, b)$.
Let $z_0=(x_0, t_0)$ and let $\rho>0$ be such that $Q_{\rho}(z_0)\subset Q=\om\times (a, b)$. For any $r\in (0, \rho/4]$ we have
\begin{equation*}
D_1(z_0, r)\leq C \Big(\frac{\rho}{r}\Big)^{3/2} A(z_0,\rho)^{3/4} B(z_0,\rho)^{3/4} + C \Big(\frac{r}{\rho}\Big)^{3/2} D(z_0,\rho)^{3/4}.
\end{equation*}
\end{lemma}
\begin{proof} Let $h_{x_0, \rho}=h_{x_0, \rho}(\cdot, t)$ be a function on $B_\rho(x_0)$ for a.e. $t$ such that
$$h_{x_0, \rho}=p-\tilde{p}_{x_0, \rho}\quad {\rm in~} B_\rho(x_0),$$
where $\tilde{p}_{x_0, \rho}$ is defined by
$$\tilde{p}_{x_0, \rho}=R_iR_j[(u_i-[u_i]_{x_0,\rho})(u_j-[u_j]_{x_0,\rho})\chi_{B_\rho(x_0)}].$$
Here $R_i=D_i(-\Delta)^{-\frac{1}{2}}$, $i=1,2,3$, is the $i$-th Riesz transform. Note that for any $\varphi\in C_0^{\infty}(B_\rho(x_0))$, we have
\begin{eqnarray*}
-\int_{B_\rho(x_0)} \tilde{p}_{x_0, \rho}\Delta \varphi dx&=&\int_{B_{\rho}(x_0)}(u_i-[u_i]_{x_0,\rho})(u_j-[u_j]_{x_0,\rho}) D_{ij}\varphi\, dx\\
&=&\int_{B}u_i u_j D_{ij}\varphi\, dx,
\end{eqnarray*}
which follows from the properties $-R_i R_j(\Delta \varphi)=D_{ij}\varphi$ and ${\rm div}\, u=0$.
Thus, as $p$ also solves
$$-\Delta p={\rm div}\, {\rm div} (u\otimes u)$$
in the distributional sense, we see that $h_{x_0, \rho}$ is harmonic in $B_{\rho}(x_0)$ for a.e. $t$.
With this decomposition of the pressure $p$, we have
\begin{eqnarray*}
\int_{B_r(x_0)} |p(x,t)|^{3/2} dx &=& C \int_{B_r(x_0)} |\tilde{p}_{x_0,\rho}+h_{x_0,\rho}|^{3/2} dx\\
&\leq& C \int_{B_\rho(x_0)} |\tilde{p}_{x_0, \rho}|^{3/2} dx + C\int_{B_r(x_0)} |h_{x_0, \rho}|^{3/2} dx.
\end{eqnarray*}
Next, as $h_{x_0,\rho}$ is harmonic in $B_{\rho}(x_0)$, for $r\in (0, \rho/4]$ there holds
\begin{eqnarray*}
\Big(\fint_{B_r(x_0)} |h_{x_0, \rho}|^{3/2} dx\Big)^{2/3}&\leq& \Big(\fint_{B_{r}(x_0)} |h_{x_0, \rho}|^{2} dx\Big)^{1/2}\\
&\leq& C\, \Big(\fint_{B_{\rho/4}(x_0)} |h_{x_0, \rho}|^{2} dx\Big)^{1/2}\\
&\leq& C \Big(\fint_{B_{\rho/2}(x_0)} |h_{x_0, \rho}|^{6/5} dx\Big)^{5/6}.
\end{eqnarray*}
This gives
\begin{eqnarray*}
\int_{B_r(x_0)} |p(x,t)|^{3/2} dx &\leq& C \int_{B_\rho(x_0)} |\tilde{p}_{x_0, \rho}|^{3/2} dx\\
&& +\, C \frac{r^3}{\rho^{15/4}}\Big(\int_{B_{\rho/2}(x_0)} |h_{x_0, \rho}|^{6/5} dx\Big)^{5/4}.
\end{eqnarray*}
Thus using $h_{x_0,\rho}=p-\tilde{p}_{x_0,\rho}$ again we find
\begin{eqnarray*}
\int_{B_r(x_0)} |p(x,t)|^{3/2} dx &\leq& C \int_{B_\rho(x_0)} |\tilde{p}_{x_0, \rho}|^{3/2} dx\\
&& +\, C \frac{r^3}{\rho^{15/4}}\Big(\int_{B_{\rho}(x_0)} |\tilde{p}_{x_0, \rho}|^{6/5} dx\Big)^{5/4}\\
&& +\, C \frac{r^3}{\rho^{15/4}}\Big(\int_{B_{\rho}(x_0)} |p|^{6/5} dx\Big)^{5/4}.
\end{eqnarray*}
By H\"older's inequality this yields
\begin{eqnarray}\label{pptil}
\int_{B_r(x_0)} |p(x,t)|^{3/2} dx &\leq& C\Big[1+\Big(\frac{r}{\rho}\Big)^3\Big] \int_{B_\rho(x_0)} |\tilde{p}_{x_0, \rho}|^{3/2} dx\\
&& +\, C \frac{r^3}{\rho^{15/4}}\Big(\int_{B_{\rho}(x_0)} |p|^{6/5} dx\Big)^{5/4}.\nonumber
\end{eqnarray}
On the other hand, by the Calder\'on-Zygmund estimate and a Sobolev interpolation inequality (see, e.g., (1.1) of \cite{LS}) we find
\begin{eqnarray}\label{ptil1}
\lefteqn{\int_{B_\rho(x_0)}|\tilde{p}_{x_0,\rho}|^{3/2} dx\leq C \int_{B_\rho(x_0)}| u-[u]_{x_0,\rho}|^{3} dx}\\
&\leq& C \Big(\int_{B_\rho(x_0)}| \nabla u|^{2} dx\Big)^{3/4} \Big(\int_{B_\rho(x_0)}|u-[u]_{x_0,\rho}|^{2} dx\Big)^{3/4}\nonumber\\
&\leq& C \Big(\int_{B_\rho(x_0)}| \nabla u|^{2} dx\Big)^{3/4} \Big(\int_{B_\rho(x_0)}|u|^{2} dx\Big)^{3/4},\nonumber
\end{eqnarray}
where we used the bound
$$\int_{B_r(x_0)}|u-[u]_{x_0,r}|^{2} dx\leq \int_{B_r(x_0)}|u|^{2} dx$$
in the last inequality.
Combining \eqref{pptil}, \eqref{ptil1} and using $r/\rho\leq 1/4$ we have
\begin{eqnarray*}
\int_{B_r(x_0)} |p(x,t)|^{3/2} dx&\leq& C \Big(\int_{B_\rho(x_0)}| \nabla u|^{2} dx\Big)^{3/4} \Big(\int_{B_\rho(x_0)}|u|^{2} dx\Big)^{3/4}\\
&& +\, C \frac{r^3}{\rho^{15/4}}\Big(\int_{B_{\rho}(x_0)} |p|^{6/5} dx\Big)^{5/4}.
\end{eqnarray*}
Integrating the last bound with respect to $dt/r^2$ over the interval $(t_0-r^2, t_0)$ and using H\"older's inequality we obtain
\begin{eqnarray*}
D_1(z_0, r) &\leq& C \Big(\frac{\rho}{r}\Big)^{3/2} A(z_0,\rho)^{3/4} B(z_0,\rho)^{3/4} + C \Big(\frac{r}{\rho}\Big)^{3/2} D(z_0,\rho)^{3/4}
\end{eqnarray*}
as desired.
\end{proof}
We are now ready to prove Proposition \ref{epsilon1}.
\begin{proof}[{\bf Proof of Proposition \ref{epsilon1}}]
By Lemma \ref{boundC1} and Corollary \ref{ABcontrol2} we have
\begin{eqnarray*}
C_1(z_0,1) &\leq& C A(z_0,1)^{3/4} B(z_0,1)^{3/4} + C A(z_0,1)^{3/2}\\
&\leq& C [A(z_0,1) + B(z_0,1)]^{3/2}\\
&\leq& C [C(z_0,2)^{1/2}+ C(z_0,2) + D(z_0,2)]^{3/2}.
\end{eqnarray*}
Thus using \eqref{smallnessCD} we find
\begin{eqnarray}\label{C1ep1}
C_1(z_0,1) &\leq& C (\epsilon_1^{1/2}+ \epsilon_1)^{3/2}.
\end{eqnarray}
On the other hand, using Lemma \ref{Dbar} with $r=1$ and $\rho=4$ there holds
\begin{equation*}
D_1(z_0, 1) \leq C [A(z_0,4) + B(z_0,4)]^{3/2}+ C D(z_0,4)^{3/4},
\end{equation*}
which by Corollary \ref{ABcontrol2} and \eqref{smallnessCD} yields
\begin{eqnarray*}
D_1(z_0, 1) &\leq& C [C(z_0,8)^{1/2} + C(z_0,8)+ D(z_0,8)]^{3/2}+ C D(z_0,8)^{3/4}\\
&\leq& C \epsilon_1^{3/2}+ C \epsilon_1^{3/4}.
\end{eqnarray*}
Now choosing $\epsilon_1$ sufficiently small in \eqref{C1ep1} and the last bound, we can make
\begin{equation*}
C_1(z_0,1) + D_1(z_0,1) \leq \epsilon_0,
\end{equation*}
and thus Lemma \ref{C1D1ep0} implies the desired regularity result.
\end{proof}
\section{Proof of Theorems \ref{localregularity} and \ref{nearMarc}}
This section is devoted to the proof of Theorems \ref{localregularity} and \ref{nearMarc}. We shall need the following lemma.
\begin{lemma}\label{weaktosuitable} Suppose that the pair of functions $(u,p)$ satisfies
the Navier-Stokes equations in $Q_1(0,0)=B_1(0)\times(-1,0)$ in the sense of distributions and has the properties \eqref{u-reg}, \eqref{p-assum}, and \eqref{serrinlorentz} for some $q\in (3,\infty]$.
Then $(u,p)$ forms a suitable solution to the Navier-Stokes equations in $Q_{5/6}$ with a generalized energy {\it equality},
$u\in L^4(Q)$, and $p\in L^2(Q_{5/6})$. Moreover,
the inequality
\begin{equation}\label{foralltbound}
\norm{u(\cdot, t)}_{L^{3, q}(B_{3/4})}\leq \norm{u}_{L^{\infty}(-(3/4)^2,0; L^{3, q}(B_{3/4}))}
\end{equation}
holds for \emph{all} $t\in [-(3/4)^2, 0]$, and the function
$$t\rightarrow \int_{B_{3/4}} u(x,t) w(x) dx$$
is continuous on $[-(3/4)^2, 0]$ for any $w\in L^{3/2, q/(q-1)}(B_{3/4})$. Here it is understood as usual that $q/(q-1)=1$ in the case $q=\infty$.
\end{lemma}
\begin{proof}
By Sobolev inequality we have $u\in L^2(-1,0; L^6(B_1)),$
which using \eqref{serrinlorentz} and the interpolative inequality
$$\norm{u(\cdot, t)}_{L^4(B_1)} \leq C \norm{u(\cdot,t)}_{L^{3,q}(B_1)}^{\frac{1}{2}} \norm{u(\cdot,t)}_{L^6(B_1)}^{\frac{1}{2}}$$
yields
\begin{equation}\label{uinLfour}
u\in L^4(Q).
\end{equation}
Thus by H\"older's inequality the nonlinear term
\begin{equation}\label{non-reg}
u\cdot\nabla u \in L^{4/3}(Q).
\end{equation}
As above, we have a decomposition
$$p=\tilde{p}+h,$$
where $\tilde{p}=R_iR_j[(u_i u_j)\chi_{B_1}]$, and $h$ is harmonic in $B_{1}$. By Calder\'on-Zygmund estimate we have
\begin{equation}\label{ptilbound}
\norm{\tilde{p}}_{L^{2}(-1, 0; L^{2}(B_1))}\leq C \norm{u}_{L^{4}(-1, 0; L^{4}(B_1))}^2=C \norm{u}^2_{L^4(Q)},
\end{equation}
and by harmonicity and assumption \eqref{p-assum} there holds
\begin{eqnarray}\label{hbound}
\lefteqn{\norm{h}_{L^{2}(-1, 0; L^{\infty}(B_{5/6}))}\leq C \norm{h}_{L^{2}(-1, 0; L^{1}(B_{1}))}}\\
&=& C \norm{p-\tilde{p}}_{L^{2}(-1, 0; L^{1}(B_{1}))}\nonumber\\
&\leq& C \Big[\norm{p}_{L^{2}(-1, 0; L^{1}(B_{1}))}+ \norm{u}^2_{L^4(Q)}\Big].\nonumber
\end{eqnarray}
Estimates \eqref{ptilbound}--\eqref{hbound} imply in particular that the pressure
\begin{equation}\label{p-reg}
p\in L^2(Q_{5/6}).
\end{equation}
Using the inclusions \eqref{u-reg}, \eqref{uinLfour}, \eqref{non-reg}, \eqref{p-reg}, and the local interior regularity of non-stationary Stokes systems we eventually find
$$\int_{Q_{3/4}} (|u|^4 + |\partial_t u|^{4/3} + |\nabla^2 u|^{4/3} +|\nabla p|^{4/3}) dxdt <+\infty.$$
It then follows that
$$u\in C(-[(3/4)^2, 0]; L^{4/3}(B_{3/4}))$$
and thus the function
$$g_{\varphi}(t):=\int_{B_{3/4}} u(x,t) \varphi(x) dx$$
is continuous on $[-(3/4)^2, 0]$ for any $\varphi\in C_0^{\infty}(B_{3/4})$. This yields
$$\Big|\int_{B_{3/4}} u(x,t) \varphi(x) dx\Big| \leq C \norm{\varphi}_{L^{3/2, q/(q-1)}(B_{3/4})} \norm{u}_{L^{\infty}(-(3/4)^2,0; L^{3, q}(B_{3/4}))}$$
for {\it any} $t\in [-(3/4)^2, 0]$ and any $\varphi\in C_0^{\infty}(B_{3/4})$. Thus by the density of $C_0^{\infty}(B_{3/4})$ in $L^{3/2, q/(q-1)}(B_{3/4})$
we see that
$$\norm{u(\cdot, t)}_{L^{3, q}(B_{3/4})}\leq C \norm{u}_{L^{\infty}(-(3/4)^2,0; L^{3, q}(B_{3/4}))}$$
for any $t\in [-(3/4)^2, 0]$. Then it can be seen, again by density, that the function $g_{\varphi}(t)$ above is actually continuous on $[-(3/4)^2, 0]$ for any
$\varphi\in L^{3/2, q/(q-1)}(B_{3/4})$.
Finally, using \eqref{uinLfour} and a standard mollification in $\RR^{3+1}$ combined with a truncation in time of test functions, we obtain the local generalized energy equality
in $Q_{5/6}$.
\end{proof}
We now proceed with the proof of Theorem \ref{localregularity}.
\vspace{.15in}
\noindent {\bf Proof of Theorem \ref{localregularity}.} Henceforth, let the hypothesis of Theorem \ref{localregularity} be enforced. Notice that by Lemma \ref{weaktosuitable} $(u,p)$ forms a suitable weak solution to the Navier-Stokes equations in $Q_{5/6}(0,0)$.
As in \cite{ESS1}, the proof of Theorem \ref{localregularity} goes by a contradiction.
Suppose that $z_0=(x_0, t_0)\in \overline{Q}_{1/2}(0,0)$ is a singular point. By definition, this means that there exists
no neighborhood $\mathcal{N}$ of $z_0$ such that $u$ has a H\"older continuous representative on $\mathcal{N}\cap B_1(0)\times(-1,0])$.
By Lemma 3.3 of \cite{SS}, there exist $c_0>0$ and a sequence of numbers $\epsilon_k\in (0, 1)$
such that $\epsilon_k\rightarrow 0$ as $k\rightarrow +\infty$ and
\begin{equation}\label{singcond}
A(z_0, \epsilon_k) = \sup_{t_0-\epsilon_k^2\leq s\leq t_0}\frac{1}{\epsilon_k}\int_{B(x_0,\epsilon_k)}|u(x,s)|^2 dx \geq c_0
\end{equation}
for any $k\in \NN$. Moreover, by Lemma \ref{weaktosuitable} we have in particular
\begin{equation}\label{att0}
u(\cdot, t_0)\in L^{3,q}(B_{3/4}(0)).
\end{equation}
Recall that we can decompose
$$p=\tilde{p}+h,$$
where $h$ is harmonic in $B_{1}$, and
$\tilde{p}= R_iR_j[(u_i u_j)\chi_{B_1}]$.
For each $Q=\om\times (a,b)$, where $\om\Subset\RR^3$ and $-\infty<a<b\leq 0$, we
choose a large $k_0=k_0(Q) \geq 1$ so that for any $k\geq k_0$ there hold
the implications
$$x\in \om \Longrightarrow x_0+\epsilon_k x\in B_{2/3},$$
and
$$t\in(a,b) \Longrightarrow t_0+\epsilon_k^2 t\in (-(2/3)^2, 0),$$
where the sequence $\{\epsilon_k\}$ is as in \eqref{singcond}.
Given such a $Q=\om\times (a,b)$, let us set
$$u_k(x, t)=\epsilon_k u(x_0+\epsilon_k x, t_0 +\epsilon_k^2 t),\quad p_k(x, t)=\epsilon_k^2 p(x_0+\epsilon_k x, t_0 +\epsilon_k^2 t),$$
and
$${\tilde p}_{k}(x, t)=\epsilon_k^2 \, \tilde{p}(x_0+\epsilon_k x, t_0 +\epsilon_k^2 t),\quad h_{k}(x, t)=\epsilon_k^2\, h(x_0+\epsilon_k x, t_0 +\epsilon_k^2 t)$$
for any $(x,t)\in Q$ and $k\geq k_0(Q)$.
The following proposition provides a non-trivial {\it ancient solution} (see \cite{Sere2} for this notion) that is essential in the proof of Theorem
\ref{localregularity}.
\begin{proposition}\label{limit-infty} {\rm (i)} There exist subsequence of $(u_k,p_k)$, still denoted by $(u_k,p_k)$, and a pair of functions
\begin{equation}\label{uinfpinf}
(u_\infty, p_\infty)\in L^{\infty}(-\infty,0; L^{3,q}(\RR^3)\times L^{\infty}(-\infty,0; L^{3/2,q/2}(\RR^3),
\end{equation}
with ${\rm div}\, u_\infty=0$ in $\RR^3\times (-\infty, 0)$, such that
\begin{equation}\label{Ccons}
u_k \rightarrow u_\infty \quad {\rm in} \quad C([a,b]; L^s(\om)),
\end{equation}
\begin{equation*}
\tilde{p}_k \rightarrow p_\infty \quad {\rm weakly^{*} ~ in} \quad L^{\infty}(a,b; L^{3/2,q/2}(\om),
\end{equation*}
for any $s\in(1,3)$, and any $\om\Subset
\RR^3$, $-\infty<a<b\leq0$.
\noindent {\rm (ii)} Moreover, for any $Q=\om\times (a,b)$ with $\om\Subset\RR^3$, $-\infty<a<b\leq 0$,
$$|u_\infty|^2, \nabla u_\infty\in L^2(Q),\quad \partial_t u_\infty, \nabla^2 u_\infty, \nabla p_\infty\in L^{4/3}(Q),$$
and
$(u_\infty, p_\infty)$ forms a suitable weak solution of the Navier-Stokes equations in any such $Q$.
\noindent {\rm (iii)} Additionally, $u_\infty$ satisfies the lower bound
\begin{equation}\label{contradic-goal}
\sup_{t\in[-1,0]} \int_{B_1(0)} |u_\infty(x,t)|^2 dx \geq c_0,
\end{equation}
where $c_0>0$ is the constant in \eqref{singcond}.
\end{proposition}
\begin{proof}
For each $Q=\om\times (a,b)$, where $\om\Subset\RR^3$, $-\infty<a<b\leq 0$,
and for every $t\in [a,b]$ we have
\begin{equation}\label{ukboundinfty}
\norm{u_k(\cdot, t)}_{L^{3, q}(\om)}\leq \norm{u(\cdot, t_0+\epsilon_k^2 t)}_{L^{3, q}(B_{3/4})}\leq
\norm{u}_{L^{\infty}(-1, 0; L^{3, q}(B_1))}.
\end{equation}
By Calder\'on-Zygmund estimate, for a.e. $t\in (a,b)$ there holds
\begin{eqnarray}\label{ptil-k-bound}
\norm{{\tilde p}_k(\cdot, t)}_{L^{3/2, q/2}(\om)}&\leq& \norm{{\tilde p}(\cdot, t_0+\epsilon_k^2 t)}_{L^{3/2, q/2}(B_{3/4})}\\
&\leq& \esssup_{t'\in(-(3/4)^2,0)}\norm{{\tilde p}(\cdot, t')}_{L^{3/2, q/2}(B_{3/4})} \nonumber\\
&\leq& C \norm{u}_{L^{\infty}(-1, 0; L^{3, q}(B_1))}^2. \nonumber
\end{eqnarray}
On the other hand, by harmonicity we have
\begin{eqnarray*}
\int_{a}^{b}\sup_{x\in\om} |h_k(x,t)|^{2} dt&\leq& \epsilon_k^2 \int_{-(3/4)^2}^{0}\sup_{x\in\om} |h(x_0+\epsilon_k x,s)|^{2} ds\nonumber\\
&\leq& \epsilon_k^2 \norm{h}^{2}_{L^{2}(-1, 0; L^{\infty}(B_{3/4}))} \nonumber\\
&\leq& C\, \epsilon_k^2 \norm{h}^{2}_{L^{2}(-1, 0; L^{1}(B_{5/6}))} \nonumber
\end{eqnarray*}
provided $k\geq k_0(Q)$. Thus again by Calder\'on-Zygmund estimate we find
\begin{eqnarray}\label{h-k-bound}
\lefteqn{\int_{a}^{b}\sup_{x\in\om} |h_k(x,t)|^{2} dt}\\
&\leq& C\, \epsilon_k^2 \norm{p-\tilde{p}}^{2}_{L^{2}(-1, 0; L^{1}(B_{5/6}))} \nonumber\\
&\leq& C \,\epsilon_k^2 \Big[\norm{p}_{L^{2}(-1, 0; L^{1}(B_{1}))}^2+ \norm{u}_{L^{\infty}(-1, 0; L^{3, q}(B_1))}^4\Big].\nonumber
\end{eqnarray}
Using the last estimates for ${\tilde p}_{k}$ and $h_k$ and H\"older's inequality we have the following uniform bound for $p_k$:
\begin{eqnarray}\label{pkbound}
\lefteqn{\norm{p_k}_{L^{2}(a, b; L^{6/5}(\om))}\leq \norm{{\tilde p}_k}_{L^{2}(a, b; L^{6/5}(\om))} + \norm{h_k}_{L^{2}(a, b; L^{6/5}(\om))}}\\
&\leq& C(Q)\Big[ \norm{p}_{L^{2}(-1, 0; L^{1}(B_{1}))}+ \norm{u}_{L^{\infty}(-1, 0; L^{3, q}(B_1))}^2 \Big] \nonumber
\end{eqnarray}
for any $k\geq k_0(Q)$. Here the constant $C(Q)$ is independent of such $k$.
With regard to $u_k$, with $k\geq k_0(Q)$, we have
\begin{eqnarray}\label{ukbound}
\norm{u_k}_{L^{4}(a, b; L^{12/5}(\om))} &\leq& C(Q) \norm{u_k}_{L^{\infty}(a, b; L^{3, q}(\om))}\\
&\leq& C(Q) \norm{u}_{L^{\infty}(-1, 0; L^{3, q}(B_1))}. \nonumber
\end{eqnarray}
For each $\varphi\in C_0^{\infty}(\RR^3\times \RR)$ that vanishes in a neighborhood of the parabolic boundary
$\partial'Q=\om\times\{t=a\} \cup \partial\om\times [a, b]$, we define
$$\varphi_k(x,t)=\epsilon_k^{-1}\varphi(\epsilon_k^{-1}(x-x_0), \epsilon_k^{-2}(t-t_0)).$$
Then with $k\geq k_0(Q)$ we see that $\varphi_k$ vanishes in a neighborhood of the parabolic boundary of $Q_{3/4}(0,0)$.
Using $\varphi_k$ as a test function in the generalized energy equality for $(u, p)$ at
$t=t_0+ \epsilon_k^2 \tau$ with a.e. $\tau\in (a, b)$ we find
\begin{eqnarray*}
\lefteqn{\int_{B_{3/4}}|u(x, t)|^2 \varphi_k(x,t) dx + 2\int_{-(3/4)^2}^t\int_{B_{3/4}} |\nabla u|^2 \varphi_k(x, s) dxds} \\
&=& \int_{-(3/4)^2}^t\int_{B_{3/4}} |u|^2 (\partial_t\phi_k +\Delta \phi_k) dx ds \\
&& \quad +\, \int_{-(3/4)^2}^t\int_{B_{3/4}}(|u|^2 + 2p)u\cdot \nabla \varphi_k dx ds.
\end{eqnarray*}
Hence by making a change of variables we obtain
\begin{eqnarray*}
\lefteqn{ \int_{\om}|u_k(y, \tau)|^2 \varphi(y,\tau) dy + 2\int_{a}^{\tau}\int_{\om} |\nabla u_k|^2 \varphi(y, s') dyds'} \\
&=& \int_{a}^{\tau}\int_{\om} |u_k|^2 (\phi_t +\Delta \phi) dy ds' + \int_{a}^{\tau}\int_{\om}(|u_k|^2 + 2p_k)u_k\cdot \nabla \varphi dy ds'
\end{eqnarray*}
for a.e. $\tau \in (a,b)$.
Thus each $u_k$ is a suitable solution in $Q$ for any $Q=\om\times (a,b)$, with $\om\Subset\RR^3$ and $-\infty<a<b\leq 0$, and any $k\geq k_0(Q)$.
Then, given such a $Q$, it follows from Corollary \ref{ABcontrol2} and inequalities \eqref{pkbound}--\eqref{ukbound} (applied to an appropriate enlargement of $Q$) that
\begin{equation}\label{ABkbound}
\norm{u_k}_{L^{\infty}(a, b; L^{2}(\om))} + \norm{\nabla u_k}_{L^{2}(a, b; L^{2}(\om))}\leq C(Q)
\end{equation}
for all sufficiently large $k$ depending only on $Q$.
Using \eqref{ABkbound} and Sobolev inequality we have
\begin{equation*}
\norm{u_k}_{L^{2}(a, b; L^{6}(\om))} \leq C(Q),
\end{equation*}
which by \eqref{ukboundinfty}, interpolation, and H\"older's inequality gives
\begin{equation}\label{uknonlinearterm}
\norm{u_k}_{L^4(\om\times (a,b))} + \norm{u_k\cdot\nabla u_k}_{L^{4/3}(\om\times (a,b))}\leq C.
\end{equation}
From the bounds \eqref{ptil-k-bound} and \eqref{h-k-bound}
for $\tilde{p}_k$ and $h_k$ we also have
\begin{equation}\label{pk3halfs}
\norm{p_k}_{L^{s}(\om\times(a,b))} \leq C(Q, s) \norm{p_k}_{L^{2}(a,b; L^{3/2, q/2}(\om))} \leq C
\end{equation}
for any $s\in (0, 3/2)$.
Using \eqref{ABkbound}--\eqref{pk3halfs}, it follows from the local interior regularity of solutions to non-stationary Stokes equations
we find
\begin{equation}\label{partterm}
\norm{\partial_t u_k}_{L^{4/3}(\om\times (a,b))} + \norm{\nabla^2 u_k}_{L^{4/3}(\om\times (a,b))} + \norm{ \nabla p_k}_{L^{4/3}(\om\times (a,b))}\leq C
\end{equation}
for all sufficiently large $k$ depending only on $Q$.
At this point, using \eqref{ukboundinfty}--\eqref{ptil-k-bound} and a diagonal process we may assume that
\begin{equation*}
u_k \rightarrow u_\infty \quad {\rm weakly^{*} ~ in} \quad L^{\infty}(a,b; L^{3,q}(\om)
\end{equation*}
\begin{equation*}
\tilde{p}_k \rightarrow p_\infty \quad {\rm weakly^{*} ~ in} \quad L^{\infty}(a,b; L^{3/2,q/2}(\om),
\end{equation*}
for a pair of functions $(u_\infty, p_\infty)$ satisfying \eqref{uinfpinf},
with ${\rm div}\, u_\infty=0$ in $\RR^3\times (-\infty, 0)$.
Estimates \eqref{ABkbound} and \eqref{partterm} now yield
\begin{equation}\label{con4thirds}
u_k \rightarrow u_\infty \quad {\rm in} \quad C([a,b]; L^{4/3}(\om)).
\end{equation}
For any $s\in (1, 3)$, the uniform bound \eqref{ukboundinfty}, and the interpolation inequality
\begin{eqnarray*}
\lefteqn{\norm{u_k(\cdot, t) -u_k(\cdot, t')}_{L^s(\om)}}\\
&\leq& C(s) \norm{u_k(\cdot, t) -u_k(\cdot, t')}_{L^{4/3}(\om)}^{\frac{12}{5}\left(\frac{1}{s}-\frac{1}{3}\right)} \norm{u_k(\cdot, t) -u_k(\cdot, t')}_{L^{3,q}(\om)}^{\frac{12}{5}\left(\frac{3}{4}-\frac{1}{s}\right)}
\end{eqnarray*}
imply that each $u_k\in C([a,b]; L^s(\om)$. Thus by using \eqref{con4thirds} and interpolating we obtain
\eqref{Ccons} for any $s\in (1, 3)$. This completes the proof of ${\rm (i)}$.
On the other hand, by \eqref{h-k-bound} we have
\begin{equation*}
h_k \rightarrow 0 \quad {\rm strongly ~ in} \quad L^{2}(a,b; L^{\infty}(\om),
\end{equation*}
for any $-\infty<a<b\leq 0$ and $\om\Subset\RR^3$, and thus in the limit $(u_\infty, p_\infty)$ satisfies the Navier-Stokes equations in the sense of
distributions in $\om\times (a,b)$. Now ${\rm (ii)}$ follows from ${\rm (i)}$, \eqref{ABkbound} and \eqref{partterm} via an argument as in the proof of Lemma \ref{weaktosuitable}.
Finally, note that by \eqref{singcond} and a change of variables we have
$$\sup_{-1\leq t\leq 0}\int_{B(0,1)}|u_k(x,t)|^2 dx=\sup_{t_0-\epsilon_k^2\leq s\leq t_0}\frac{1}{\epsilon_k}\int_{B(x_0,\epsilon_k)}|u(y,s)|^2 dy \geq c_0.$$
Thus using \eqref{Ccons} with $s=2$ we obtain \eqref{contradic-goal}, which proves ${\rm (iii)}$.
\end{proof}
We now continue with the proof of Theorem \ref{localregularity}. By ${\rm (i)}$ of Proposition \ref{limit-infty}, we have
$$\int_{-M}^{0} (\norm{u_\infty(\cdot, t)}_{L^{3,q}(\RR^3)}^{4} + \norm{p_\infty(\cdot, t)}_{L^{3/2,q/2}(\RR^3)}^{2})dt<+\infty$$
for any real number $M>0$. Note that for a.e. $t$,
$$\norm{u_\infty(\cdot, t)}_{L^{3,q}(\RR^3\setminus \overline{B}_R(0))}^{4} + \norm{p_\infty(\cdot, t)}_{L^{3/2,q/2}(\RR^3\setminus \overline{B}_R(0))}^{2}
\rightarrow 0$$
as $R\rightarrow +\infty$. We thus have
$$\int_{-M}^{0} (\norm{u_\infty(\cdot, t)}_{L^{3,q}(\RR^3\setminus \overline{B}_R(0))}^{4} + \norm{p_\infty(\cdot, t)}_{L^{3/2,q/2}(\RR^3\setminus \overline{B}_R(0))}^{2})dt\rightarrow 0$$
as $R\rightarrow +\infty$. This yields that for any $M>200$ there exists $N=N(\epsilon_1, M)>10$ such that
$$\int_{-M}^{0} (\norm{u_\infty(\cdot, t)}_{L^{3,q}(\RR^3\setminus \overline{B}_N(0))}^{4} + \norm{p_\infty(\cdot, t)}_{L^{3/2,q/2}(\RR^3\setminus \overline{B}_N(0))}^{2})dt\leq \epsilon_2,$$
where
\begin{equation}\label{epsilon2}
\epsilon_2 =8^2|B_1(0)|^{-1/3}\epsilon_1
\end{equation}
with $\epsilon_1$ being as found in Proposition \ref{epsilon1}.
We now fix such numbers $M$ and $N$ and consider any $z_1$ that
$$z_1=(x_1,t_1)\in (\RR^3\setminus \overline{B}_{2N}(0))\times (-M/2, 0].$$
Then there holds
$$Q_8(z_1)=B_8(x_1)\times (t_1-8^2, t_1)\subset (\RR^3\setminus \overline{B}_{N}(0))\times (-M, 0],$$
and hence
\begin{equation}\label{up3}
\int_{t_1-8^2}^{t_1} (\norm{u_\infty(\cdot, t)}_{L^{3,q}(B_8(x_1))}^{4} + \norm{p_\infty(\cdot, t)}_{L^{3/2,q/2}(B_8(x_1))}^{2})dt\leq \epsilon_2.
\end{equation}
Since
\begin{eqnarray*}
\lefteqn{\norm{u_\infty(\cdot, t)}_{L^{12/5}(B_8(x_1))}^{4} + \norm{p_\infty(\cdot, t)}_{L^{6/5}(B_8(x_1))}^{2}}\\
&\leq& |B_8(x_1)|^{1/3} \left\{\norm{u_\infty(\cdot, t)}_{L^{3,q}(B_8(x_1))}^{4}+ \norm{p_\infty(\cdot, t)}_{L^{3/2,q/2}(B_8(x_1))}^{2}\right\}
\end{eqnarray*}
we see from \eqref{epsilon2}--\eqref{up3} that
\begin{equation}\label{CDz1}
C(u_\infty,z_1,8) + D(p_\infty,z_1,8) \leq \epsilon_1.
\end{equation}
The smallness property \eqref{CDz1} and Proposition \ref{epsilon1} now yield that $\nabla^k u_\infty$, $k=0,1,2, \dots$, is H\"older continuous
on $(\RR^3\setminus \overline{B}_{2N}(0))\times (-M/2, 0]$, and
\begin{equation}\label{reg-bound}
\max_{z\in \overline{Q}_{1/2}(z_1)} |\nabla^{k} u_\infty(z)| \leq C_k.
\end{equation}
Let $\om_\infty={\rm curl}\, u_\infty$ be the vorticity of $u_\infty$. Then $\om_\infty$ satisfies the equation
$$\partial_t \om_\infty- \Delta \om_\infty + (u_{\infty} \cdot \nabla) \om_\infty - (\om_{\infty}\cdot \nabla) u_\infty=0$$
on the set $(\RR^3\setminus \overline{B}_{4N}(0))\times (-M/4, 0]$, which by \eqref{reg-bound} gives
\begin{equation}\label{vorticbound1}
|\partial_t \om_\infty- \Delta \om_\infty|\leq C(|\om_\infty| + |\nabla \om_\infty|)
\end{equation}
with
\begin{equation}\label{vorticbound2}
|\om_\infty| \leq C<+\infty
\end{equation}
on the set $(\RR^3\setminus \overline{B}_{4N}(0))\times (-M/4, 0]$, for a universal constant $C>0$.
We now claim that
\begin{equation}\label{zerooutside}
\om_\infty=0 \quad {\rm on~} (\RR^3\setminus \overline{B}_{4N}(0))\times (-M/4, 0].
\end{equation}
To see this, by applying the backward uniqueness theorem (see \cite[Theorem 5.1]{ESS1} and \cite{ESS2}) and the bounds
\eqref{vorticbound1}--\eqref{vorticbound2}, it is enough to show that
\begin{equation}\label{vorzeroat0}
\om_\infty(y,0)=0 \quad {\rm for ~ all~} y\in \RR^3\setminus \overline{B}_{4N}(0)).
\end{equation}
Note that for any $y\in \RR^3$ we have
\begin{eqnarray*}
\lefteqn{\int_{B_1(y)} |u_\infty(x, 0)| dx}\\
&\leq& \int_{B_1(y)} |u_\infty(x, 0)-u_k(x,0)| dx + \int_{B_1(y)} |u_k(x, 0)| dx\\
&\leq& \int_{B_1(y)} |u_\infty(x, 0)-u_k(x,0)| dx + |B_1(0)|^{\frac{2}{3}}\norm{u_k(\cdot,0)}_{L^{3,q}(B_1(y))}\\
&\leq& \norm{u_\infty-u_k}_{C([-M/4, 0]; L^1(B_1(y)))} + |B_1(0)|^{\frac{2}{3}}\norm{u(\cdot,t_0)}_{L^{3,q}(B_{\epsilon_k}(x_0+\epsilon_k y))}.
\end{eqnarray*}
Thus sending $k\rightarrow +\infty$ we see that
$$\int_{B_1(y)} |u_\infty(x, 0)| dx=0 $$
for all $y\in \RR^3$, which yields \eqref{vorzeroat0} as desired. Here we have used ${\rm (i)}$ of Proposition \ref{limit-infty} and \eqref{att0}.
At this point using \eqref{zerooutside} combined with the argument on pages 227--229 of \cite{ESS1}, which ultimately employs
the theory of unique continuation for parabolic inequalities, we see that in fact
$$\om_\infty(\cdot,t)=0\quad {\rm in~ the~ whole~} \RR^3$$
for a.e. $t\in (-M/4, 0)$. Thus $u_\infty(\cdot,t)$ is globally harmonic and by a Liouville theorem it follows that $u_\infty(\cdot,t)=0$ for a.e.
$t\in (-M/4, 0)$. This leads to a contradiction to the lower bound \eqref{contradic-goal} and hence completes the proof of Theorem \ref{localregularity}.
We next prove Theorem \ref{nearMarc}.
\begin{proof}[{\bf Proof of Theorem \ref{nearMarc}}] Arguing as in the proof of Lemma
\ref{weaktosuitable} we see that $(u,p)$ forms a suitable solution to the Navier-Stokes equations in $Q_{5/6}$.
Suppose that $(x_0,t_0)\in \overline{Q}_{1/2}(0,0)$ is a singular point.
Then we must have that $x_0=0$.
By Lemma 3.3 of \cite{SS}, there exist $c_0>0$ and a sequence of numbers $\epsilon_k\in (0, 1/8)$
such that $\epsilon_k\rightarrow 0$ as $k\rightarrow +\infty$ and
\begin{equation*}
\sup_{t_0-\epsilon_k^2\leq s\leq t_0}\frac{1}{\epsilon_k}\int_{B(0,\epsilon_k)}|u(x,s)|^2 dx \geq c_0.
\end{equation*}
Using a change of variables and the condition \eqref{LinftyX}, we then have
\begin{eqnarray*}
0<c_0 &\leq& \sup_{-1\leq t\leq 0} \int_{B(0,1)}|\epsilon_k u(\epsilon_k y, t_0+\epsilon_k^2 t)|^2 dy\\
&\leq& \norm{f}^2_{L^\infty((-1,0))} \int_{B_1(0)} |y|^{-2} g(\epsilon_k y)^2 dy.
\end{eqnarray*}
By the property of $g$, this is impossible to hold for all $k\in \NN$, and thus the proof of Theorem \ref{nearMarc} is complete.
\end{proof}
\section{Proof of Theorems \ref{global-ESS-lorentz}}
We shall prove Theorems \ref{global-ESS-lorentz} in this section. First observe that under the hypothesis of Theorem \ref{global-ESS-lorentz}, we have
\begin{eqnarray*}
\norm{u(\cdot, t)}_{L^4(\RR^3)} &\leq& C \norm{u(\cdot,t)}_{L^{3,q}(\RR^3)}^{\frac{1}{2}} \norm{u(\cdot,t)}_{L^6(\RR^3)}^{\frac{1}{2}}\\
&\leq& C \norm{u(\cdot,t)}_{L^{3,q}(\RR^3)}^{\frac{1}{2}} \norm{\nabla u(\cdot,t)}_{L^2(\RR^3)}^{\frac{1}{2}}.
\end{eqnarray*}
Thus,
\begin{equation}\label{L4}
u\in L^4(Q_T) \quad {\rm and} \quad u\cdot\nabla u\in L^{4/3}(Q_T),
\end{equation}
where the latter follows from H\"older's inequality. Using these inclusions, the coercive estimates (see \cite{GS, MS, Sol}) and the uniqueness theorem (see \cite{Lad1}) for the Stokes problem, we can introduce the {\it associated pressure} $p$ such that
$$\partial_t u, \nabla^2 u, \nabla p\in L^{4/3}(\RR^3\times (\delta, T))$$
for any $\delta \in (0, T)$. Moreover, it follows from the pressure equation and the global condition \eqref{PSLcrit-lorentz} that
\begin{equation}\label{presurelorentz}
p\in L^\infty(0, T; L^{3/2, q/2}(\RR^3)).
\end{equation}
Arguing as in the proof of Lemma \ref{weaktosuitable} we see that $(u,p)$ forms a suitable weak solution in any bounded cylinder of $Q_T$. By
\eqref{PSLcrit-lorentz} and \eqref{presurelorentz}, we have
\begin{equation}\label{T-for-decay}
\int_{0}^{T} (\norm{u(\cdot, t)}_{L^{3,q}(\RR^3)}^{4} + \norm{p(\cdot, t)}_{L^{3/2,q/2}(\RR^3)}^{2})dt\leq C(T)<+\infty.
\end{equation}
We next fix a $\delta\in (0, T)$ and set $r_0=\sqrt{\delta/128}$. Then using \eqref{T-for-decay} we find a large number $R=R(T,\delta)>0$ such that
\begin{equation*}
C(u, z_0, 8r_0) + D(p, z_0, 8r_0)\leq \epsilon_1
\end{equation*}
for any $z_0=(x_0,t_0)\in \RR^3\setminus B_R(0)\times [\delta, T]$. Thus by scaling and Proposition \ref{epsilon1}, there holds
$$\sup_{\RR^3\setminus B_R(0)\times[\delta, T]} |u|\leq C(\delta).$$
On the other hand, for any $z_0=(x_0,t_0)\in \overline{B}_R(0)\times [\delta, T]$ and with $r_0$ as above, $(u,p)$ is a suitable solution in
$Q_{r_0}(z_0)$. Thus by scaling, Theorem \ref{localregularity}, and the compactness of $\overline{B}_R(0)\times [\delta, T]$, we have
$$\sup_{\overline{B}_R(0)\times[\delta, T]} |u|\leq C(\delta).$$
Combining the last two bounds we obtain
\begin{equation*}
\sup_{\RR^3\times[\delta, T]} |u|\leq C(\delta)<+\infty
\end{equation*}
which holds for any $\delta\in(0,T)$. Thus $u$ is smooth on $\RR^3\times (0,T]$, and using $u\in L^4(Q_T)$ (see \eqref{L4}) and interpolation we see that $u\in L^5(\RR^3\times(\delta,T))$
for any $\delta\in (0, T)$.
On the other hand, if $a\in \dot{J}\cap L^3$ then by local strong solvability and weak-strong uniqueness $u\in L^5(\RR^3\times(0,\delta_0))$ for some $\delta_0>0$ (see, e.g., \cite[Theorem 7.4]{ESS1} and \cite{Kat}).
Thus we conclude that $u\in L^5(\RR^3\times(0,T))$ and hence by Theorem \ref{PSL} it is unique in $Q_T$.
\begin{center} {\sc Acknowledgements}
\end{center}
The author wishes to acknowledge the support from the Hausdorff Research Institute for Mathematics (Bonn, Germany), where this
work has been finalized. Part of this work was also done during a visit to the Institut Mittag-Leffler (Djursholm,
Sweden). | 193,478 |
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Cobra Kai Season 4 - The Loop
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Contents
Overview[]
Network: Discovery Channel
First Episode: October 1, 1996
Last Episode: May 2005
Status: Cancelled
Cast:
- Various Investigators officials and Law Enforcement.
History[]
Details: The New Detectives is a show about murders in the United States and how forensic scientists used to convict murderers of of their crimes.
Links:
Episodes[]
The New Detectives has the following cases in common with Unsolved Mysteries:
- Dr. Emily Craig
- David Davis and Shannon Davis
- John Hawkins
- Marie Hilley
- Joan, Michelle, and Christe Rogers
- Tanya Smith
- Darlie, Devon and Damon Routier
- Tracy Jo Shine
- O.J. Simpson
- Linda Sobek
- Aimee Willard
Other Cases[]
- Cassidy Senter (Featured on The New Detectives only) | 24,092 |
A Versatile Tree Set!This is my first tree set and I'm not sure why I've waited so long to get one! These trees are cool!
This one is from a team challenge card sketch. These are the colors I see out my window in Maine when a snowstorm is moving out and the sunset breaks through in a pretty pink! Powder Pink to be exact, and the trees are of course shaded spruce!
This is from the catalog, it's one I want to make, I'll update this blog when I do to show you my version! | 170,551 |
Having problems with your HP Pavilion dv6000z.
Answers :
Make sure you are not using DMA/UDMA in your internal DVD Burner
Are you suue that the drive on the pc is a burner?
Repair Help & Product Troubleshooting for HP Pavilion dv6000 | 96,291 |
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TITLE: Evaluating $\sum\limits_{k=0}^n \frac{(-1)^k (4 n - 2 k)!}{k! (2 n - k)! (2 n - 2 k)! (2 n + 2 r + 1 - 2 k)}$
QUESTION [1 upvotes]: I'm trying to prove an property of Legendre polynomial, namely $\int_{-1}^1x^{2r}P_{2n}(x)dx=\frac{2^{2n+1}(2r)!(r+n)!}{(2r+2n+1)!(r-n)!}$
I'm required to use the general formula for Legendre polynomial,
which is $\displaystyle P_{n}(x) = \sum_{r=0}^{[n/2]} \frac{(-1)^r (2n-2r)!}{2^n r! (n-r)! (n-2 r)!} x^{n-2 r}.$
So basically I need to show (I guess it may be possible using orthogonality, but not sure how to make it that way.)
$\displaystyle \sum_{k=0}^n \frac{(-1)^k (4 n - 2 k)!}{k! (2 n - k)! (2 n - 2 k)! (2 n + 2 r + 1 - 2 k)} = \frac{2^{4n}(2r)!(r+n)!}{(2r+2n+1)!(r-n)!}
$
When I plug in the sum expression into Wolfram alpha, it makes an output looks similar to RHS, so I guess there must be an algorithm to calculate such kind of sum.
Can anybody help me?
REPLY [1 votes]: We may safely assume $r\geq n$: otherwise we may write $x^{2r}$ as a linear combination of Legendre polynomials $P_k(x)$ with $k<2n$ and the outcome is zero by orthogonality. After this, in order to prove the first identity it is easier to exploit Rodrigues' formula and integration by parts:
$$\begin{eqnarray*} \int_{-1}^{1}x^{2r}P_{2n}(x)\,dx &=& \frac{1}{2^{2n}(2n)!}\int_{-1}^{1}x^{2r}\frac{d^{2n}}{dx^{2n}}(x^2-1)^{2n}\,dx\\(\text{IBP})\qquad&=&\frac{1}{2^{2n}(2n)!}\int_{-1}^{1}\frac{d^{2n}}{dx^{2n}}x^{2r}\cdot(x^2-1)^{2n}\,dx\\&=&\frac{1}{2^{2n}(2n)!}\int_{-1}^{1}\frac{(2r)!}{(2r-2n)!}x^{2r-2n}(x^2-1)^{2n}\,dx\\(\text{symmetry})\qquad&=&\frac{2}{4^n}\binom{2r}{2n}\int_{0}^{1}x^{2r-2n}(1-x^2)^{2n}\,dx\\(x^2\mapsto u)\qquad&=&\frac{1}{4^n}\binom{2r}{2n}\int_{0}^{1}u^{r-n-1/2}(1-u)^{2n}\,du\\(\text{Beta function})\qquad&=&\frac{\Gamma(2r+1)\Gamma(r-n+1/2)}{4^n\Gamma(2n-2r+1)\Gamma(r+n+3/2)}\end{eqnarray*}$$
Legendre's duplication formula converts the last expression into the wanted one, proving the combinatorial identity $\sum_{k=0}^{n}(\ldots)=(\ldots)$ as a consequence.
An algorithmic way for tackling combinatorial identities involving the product of (not too many) binomial coefficients is given by the Wilf-Zeilberger method. | 5,772 |
\begin{document}
\maketitle
\begin{abstract}
The hexabasic book is the cone of the 1-dimensional skeleton
of the union of two tetrahedra glued along a common face.
The universal 3-dimensional polyhedron $\up$
is the product of a segment and the hexabasic book.
We show that any 2-dimensional link in 4-space is isotopic
to a surface in $\up$.
The proof is based on a representation of surfaces in 4-space
by marked graphs, links with double intersections in 3-space.
We construct a finitely presented semigroup whose central elements
uniquely encode all isotopy classes of 2-dimensional links.
\end{abstract}
\section{Introduction}
\subsection{Brief summary}
\noindent
\smallskip
This is a research on the interface between
geometric topology, singularity theory and semigroups.
A 2-link is a closed 2-dimensional surface in 4-dimensional space $\R^4$.
We study 2-links up to isotopy that is a smooth deformation of
the ambient 4-dimensional space.
We prove that any 2-link is isotopic to a surface embedded
into the universal 3-dimensional polyhedron $\up$.
We also reduce the isotopy classification of 2-links in 4-space
to a word problem in a finitely presented semigroup.
\medskip
\subsection{The universal polyhedron containing 2-dimensional links}
\noindent
\smallskip
First we define the universal 3-dimensional polyhedron $\up$.
\medskip
\noindent
{\bf Definition 1.1.}
The \emph{theta} graph $\tg$ consists of 3 edges connecting 2 vertices.
The \emph{circled} theta graph $\ct$ is $\tg\cup S^1$, where
the circle $S^1$ meets each edge of $\tg$ in one point, see Fig.~1.
Then $\ct$ is the 1-dimensional skeleton of two tetrahedra glued
along a common face.
The \emph{hexabasic book} $\hb$ is the cone of $\ct$.
Being embedded in 3-space, the book $\hb$ divides
a neighbourhood of the central vertex into 6 parts.
The \emph{universal} 3-dimensional polyhedron
is $\up=\hb\times[-1,1]$.
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig1.eps}
\caption{The theta graph $\tg$, circled theta graph $\ct$, book $\hb$}
\end{figure}
We will work in the smooth category, i.e. all diffeomorphisms are $C^{\infty}$-smooth.
We will make necessary comments on similar constructions in the PL case.
\medskip
\noindent
{\bf Definition 1.2.}
An \emph{embedding} is a diffeomorphism onto its image.
A \emph{2-link} is a closed (possibly disconnected or
non-orientable) smooth surface $S$ embedded into $\R^4$.
An \emph{isotopy} between 2-links $S$ and $S'$ is
a continuous family of diffeomorphisms $F^u:\R^4\to\R^4$,
$u\in[0,1]$, such that $F^0=\id_{\R^4}$, $F^1(S)=S'$.
\medskip
Fix the 4th coordinate $t$ in 4-space $\R^3\times \R$.
Then a 2-link in $\R^3\times\R$ can be studied in terms of
its \emph{cross-sections} $S_t=S\cap(\R^3\times\{t\})$ \cite{FM}.
Any 2-link can be isotopically deformed to
a surface $S\subset\R^3\times[-1,1]$ such that the projection
$\pr:S\to[-1,1]$ has distinct non-degenerate critical values.
A general cross-section $S_t$ is a classical link in $\R^3\times\{t\}$, while
a cross-section containing a saddle is a link with a double point.
When $t$ passes through a saddle,
the cross-section $S_t=S\cap(\R^3\times\{t\})$ changes
by the Morse modification in the left picture of Fig.~2.
\smallskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig2.eps}
\caption{Resolving a singular point and a band in $\R^3$}
\end{figure}
A PL analogue of the smooth approach is to decompose
a 2-link $S\subset\R^3\times[-1,1]$ into handles located
in different sections $\R^3\times\{t_j\}$.
The 1-handles of $S$ will be represented by bands that have
a distinguished core and are attached to a classical link in 3-space.
Any attached band can be retracted to a singular point
marked by a bridge encoding the core of the band.
The cross-sections of $S$ below and above
every 1-handle locally look like the right picture of Fig.~2.
\noindent
\medskip
\subsection{Main results}
\noindent
\smallskip
\noindent
The hexabasic book $\hb$ is closely related to the
\emph{3-page book} $\tb$, the cone of the \emph{theta} graph $\tg$
consisting of 3 edges connecting 2 vertices, see Fig.~7.
The \emph{binding} segment of $\tb$ is the cone of the 2 vertices of $\tg$.
From another point of view, the 3-page book $\tb$ can be considered as $\R\times T$,
where $T$ is the \emph{triod} consisting of 3 edges connecting
the central vertex $O$ to other 3 vertices, here the binding axis $\al$ is $\R\times O$.
The hexabasic book $\hb$ is obtained from $\tb$ by adding
3 half-disks whose 6 boundary radii are attached to
the 3 edges of $\{0\}\times T$, see Fig.~1.
\medskip
\noindent
{\bf Theorem~1.3}.
Any 2-dimensional link $S\subset\R^4$ is isotopic to a surface embedded
into the universal 3-dimensional polyhedron $\up=\hb\times[-1,1]$.
\medskip
The key idea of Theorem~1.3 is to put a given $S$ surface in general position
and consider its cross-sections $S_t$ through saddles of $\pr:S\to[-1,1]$,
see Claim~2.3.
Such a cross-section $S_t$ is a link with exactly one singular point, so
$S_t$ can be embedded into the 3-page book $\tb$ using
the technique of 3-page embeddings developed in \cite{Kur,KV},
see Proposition~3.2.
Both resolutions of the singular point of $S_t$ can be realised in $\tb$,
i.e. the embedding extends to a regular neighbourhood of $S_t$ in $S$.
It remains to embed the complement of the regular neighbourhoods of all saddles
into $\hb\times[-1,1]$ realising any isotopy of classical links in $\hb$,
see Lemma~3.4.
\medskip
We will develop a 1-dimensional calculus for 2-links as follows.
Any 2-link $S$ in general position in $\R^3\times[-1,1]$
can be represented by a banded link $BL$ whose bands are
associated to the saddles of $\pr:S\to[-1,1]$, see Proposition~2.6(i).
Retracting each band to a point, we get a marked graph whose
singular points are marked by bridges encoding the cores of bands.
There is a complete set of moves on marked graphs generating
any isotopy of 2-links in 4-space, see Proposition~4.2.
Any marked graph can be embedded into the 3-page book $\tb$
and can be encoded by a word in the alphabet of 15 letters.
The moves on marked graphs are translated into relations on words,
which leads to the universal semigroup $\SL$ of 2-links in 4-space.
\smallskip
Introduce the universal semigroup $\SL$ generated by
the letters $a_i,b_i,c_i,d_i,x_i$ subject to relations (1)-(8),
where $i\in\Z_3=\{0,1,2\}$, e.g. $0-1=2\pmod{3}$.
\medskip
\noindent
(1) $d_0d_1d_2=1$, $\quad b_id_i=1=d_ib_i$;
\smallskip
\noindent
(2) $a_i=a_{i+1}d_{i-1}$, $\quad b_i=a_{i-1}c_{i+1}$,
$\quad c_i=b_{i-1}c_{i+1}$, $\quad d_i=a_{i+1}c_{i-1}$;
\smallskip
\noindent
(3) $uv=vu$, $u\in\{a_ib_i, d_ic_i, b_{i-1}d_id_{i-1}b_i, d_ix_ib_i\}$,
$v\in\{a_{i+1}, b_{i+1}, c_{i+1}, b_id_{i+1}d_i, x_{i+1}\}$;
\smallskip
\noindent
(4) $x_{i-1}=b_{i+1}x_id_{i+1}$,
$\quad b_i x_i b_i=a_i (b_i x_i b_i) c_i$,
$\quad d_i x_i d_i=a_i (d_i x_i d_i) c_i$;
\smallskip
\noindent
(5) $(d_i x_i b_i)d_i^2d_{i+1}^2d_{i-1}^2=d_i^2d_{i+1}^2d_{i-1}^2(d_i x_i b_i)$;
\smallskip
\noindent
(6) $a_i x_i=a_i$, $a_i b_i x_i d_i c_i =1$;
\smallskip
\noindent
(7) $d_i x_i b_i c_i x_i=b_i x_i d_i c_i x_i$;
\smallskip
\noindent
(8)
$w_i d_{i+1} d_i^2 d_{i-1} a_{i+1} b_{i+1} x_i b_i d_{i+1} b_i^2 b_{i+1} d_i^2=
w_i b_{i-1} b_i a_i b_{i+1} a_{i+1} d_i^2 c_{i-1} b_i x_i b_i$,\\
where $w_i=a_i b_i x_i b_i c_i$.
\bigskip
One of the 6 relations $b_id_i=1=d_ib_i$ is superfluous and can be deduced
from the remaining relations in (1).
Moreover, the commutativity of $d_ic_i$ with $a_{i+1},b_{i+1}$
follows from the other relations in (3), see more details in \cite{Kur}.
So the semigroup $\SL$ is generated by 15 letters and 96 relations.
\bigskip
\noindent
{\bf Theorem~1.4}.
Any 2-link $S\subset\R^4$ is encoded by an element
$w_S\in\SL$ in such a way that 2-links $S,S'$ are isotopic
if and only if their encoding elements $w_S$ and $w_{S'}$
are equal in $\SL$.
An element $w\in\SL$ encodes a 2-link if and only if $w$ is central in $\SL$.
\medskip
\noindent
\emph{Outline.}
In section~2 one represents 2-links in 4-space
by banded links and marked graphs in 3-space.
Theorems~1.3 and 1.4 are proved in sections~3 and 4, respectively.
Banded links are more convenient for deriving a complete set of moves
generating any isotopy of 2-links.
Marked graphs will be used to prove our main results on embedding
and encoding 2-links up to isotopy.
\smallskip
\noindent
{\bf Acknowledgement.}
The authors thank S.~Carter, F.~Tari for useful discussions.
\medskip
\section{Representing 2-links by banded links and marked graphs}
\subsection{Critical level embeddings of 2-links in 4-space}
\noindent
\smallskip
\noindent
Here we describe the PL approach where a 2-link is isotopically deformed
to a nice embedding with handles at different levels.
The smooth version of crucial Claim~2.3(ii) is a standard statement
on general position proved in Appendix.
\medskip
\noindent
{\bf Definition 2.1.}
A \emph{handle} of dimension $n$ and index $k$ is $D^k\times D^{n-k}$.
A \emph{handle decomposition} of a manifold $M^n$ is a sequence
of submanifolds $M_0\subset M_1\subset\dots\subset M_l=M$, where
$M_0$ is a disjoint union of $n$-dimensional disks,
each $M_{i+1}$ is obtained from $M_i$
by adding a handle of some index $k_i$.
Formally, one can express $M_{i+1}=M_i\cup_{\ph_i}(D^{k_i}\times D^{n-k_i})$,
where $\ph_i:\bd D^{k_i}\times D^{n-k_i}\to\bd M_i$ is an embedding.
If before and after each handle addition one inserts a \emph{collar},
the product of the attaching area and a segment, then
one gets a \emph{collared handle} decomposition \cite[p.~416]{KL}.
\medskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig3.eps}
\caption{A critical level embedded torus and its banded link $BL\subset\R^3$}
\end{figure}
\smallskip
\noindent
A 2-link with a collared handle decomposition
can be nicely embedded in $\R^4$.
The left picture of Fig.~3 shows a similar embedding,
where the standard 2-torus in $\R^3$
has the collared handle decomposition
consisting of 4 handles and 3 collars:
\smallskip
\noindent
1) the lowest handle is a 0-handle (a disk) at the level $t=-1$;
\smallskip
\noindent
2) the 2 intermediate handles are 1-handles (bands) at the levels $t=\pm 1/2$;
\smallskip
\noindent
3) the highest handle is a 2-handle (a disk) at the level $t=+1$.
\bigskip
\noindent
{\bf Definition 2.2.}
A \emph{critical level} PL embedding is a PL embedding of
a 2-link $S\subset\R^3\times[-1,1]$ with
a collared handle decomposition satisfying (i), (ii) \cite[p.~417]{KL}:
\smallskip
\noindent
(i) the handles are in different sections $\R^3\times\{t_j\}$,
where $-1<t_1<\dots<t_n<1$;
\smallskip
\noindent
(ii) each collar between adjacent handles of $S$ is embedded as
the direct product $A\times[t_j,t_{j+1}]\subset\R^3\times[t_j,t_{j+1}]$,
where $A\subset\R^3$ is the attaching area of the handles.
\medskip
\noindent
A smooth embedding $S\subset\R^3\times[-1,1]$
is called a smooth \emph{critical level} embedding
if the projection $\pr:S\to[-1,1]$ has all its critical points
in different sections $\R^3\times\{t_j\}$.
This is a general position assumption.
\medskip
\noindent
{\bf Claim~2.3.} (i) \cite[Theorem~1, p.~420]{KL}
Any 2-dimensional PL link in 4-space is isotopic to the image of
a critical level PL embedding $S\subset\R^3\times[-1,1]$.
\smallskip
\noindent
(ii) Any smooth 2-link is smoothly isotopic to
a surface $S\subset\R^3\times[-1,1]$ such that all critical points of
$\pr:S\to[-1,1]$ are non-degenerate and have distinct values.
\qed
\medskip
We will use the smooth version of Lemma~2.3(ii), which will be
deduced from the transversality theorem of Thom in Appendix.
Claim~2.3(i) is worth keeping in mind when one associates
a banded link to a 2-link in Proposition~2.6(i).
\smallskip
\subsection{Representing 2-links in 4-space by banded links in 3-space}
\noindent
\smallskip
\noindent
We define banded links, links with bands, which will represent 2-links in 4-space.
\medskip
\noindent
{\bf Definition 2.4.}
A \emph{banded} link is a collection of circles and bands in $\R^3$ such that
\smallskip
\noindent
(i) the circles and bands are non-oriented and non-self-intersecting;
\smallskip
\noindent
(ii) the circles and bands are disjoint except for each band having
a pair of opposite sides \emph{attached} to disjoint arcs in the circles,
the other sides are called \emph{free}.
\smallskip
\noindent
In every band we mark its \emph{core},
an arc connecting its attached opposite sides, see Fig.~3.
Banded links are considered up to isotopy of $\R^3$.
The bands of a banded link will represent 1-handles of a 2-link.
In every band $B$ of a banded link $BL$ consider the opposite free sides
not connected by the core of $B$.
Replace $B$ by its free sides, the resulting usual non-oriented link in $\R^3$
is called the \emph{positive} resolution $BL_+$ of the banded link $BL$, see Fig.~3.
Similarly define the \emph{negative} resolution $BL_-$ replacing
every band $B$ by the opposite attached sides connected by the core of $B$.
A banded link $BL$ is \emph{admissible}, if
both resolutions $BL_{\pm}$ are trivial links.
\medskip
If a PL 2-link $S\subset\R^3\times[-1,1]$
has all its 1-handles in the zero section $\R^3\times\{t=0\}$,
then the cross-section $S_0=S\cap(\R^3\times\{t=0\})$ is a banded link.
We will use much weaker assumptions and construct
a banded link for any critical level embedding.
Proposition~2.6 leads to a calculus for 2-links in Proposition~4.2 and
provides a function from the set of 2-links to the set of admissible banded links.
\medskip
\noindent
{\bf Definition 2.5.}
Given a 2-dimensional surface $S$, consider the space of all smooth functions
$f:S\to\R^4$ with the Whitney topology, see Appendix.
The space $\CS$ of all 2-links $S\subset\R^4$ has the induced topology.
Points in $\CS$
will be classified using the projection $\pr:S\to\R$ to the 4th coordinate $t$.
A 2-link $S\in\CS$ is called
\smallskip
\noindent
$\bu$
\emph{generic} if all critical points of $\pr$
are non-degenerate and have distinct values;
\smallskip
\noindent
$\bu$
an \emph{$A_1^+ A_1^+$-singularity} if $S$ fails to be generic because of
2 non-degenerate extrema of $\pr:S\to\R$ that have the same value;
\smallskip
\noindent
$\bu$
an \emph{$A_1^+ A_1^-$ -singularity} if $S$ fails to be generic because of
a non-degenerate saddle and extremum of $\pr:S\to\R$ that have the same value;
\smallskip
\noindent
$\bu$
an \emph{$A_1^- A_1^-$ -singularity} if $S$ fails to be generic because of
2 non-degenerate saddles of $\pr:S\to\R$ that have the same value;
\smallskip
\noindent
$\bu$
an \emph{$A_2$-singularity} if $S$ fails to be generic because of
a singularity of $\pr:S\to\R$ having the form $\pr(x,y)=x^2-y^3$
in local coordinates $x,y$.
\medskip
The sign in the notation above is the sign of the determinant
$\pr_{xx}\pr_{yy}-\pr_{xy}^2$ of the Jacobi matrix of 2nd order derivatives
at a critical point.
Denote by $\Si_{++},\Si_{+-},\Si_{--}$ and $\Si_{2}$ the \emph{subspaces}
of the corresponding singularities in the space $\CS$.
Introduce the \emph{singular} subspace $\Si=\Si_{++}\cup\Si_{+-}\cup\Si_{--}\cup\Si_2$.
An isotopy of 2-links can be considered as a path in $\CS$.
In Proposition~2.6 we consider paths
nicely meeting the singular subspace $\Si$.
\medskip
\noindent
{\bf Proposition 2.6.}
(i) To any a critical level embedding $S\subset\R^3\times[-1,1]$
we associate a banded link $BL$ well-defined up to
the slide/swim moves in Fig.~4.
\smallskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig4.eps}
\caption{Cup/cap moves and slide/swim moves of banded links}
\end{figure}
\noindent
(ii) If 2-links $S, S'$ are isotopic through generic 2-links,
then the associated banded links $BL,BL'$ are related
by the slide/swim moves in Fig.~4.
\smallskip
\noindent
(iii) If 2-links $S, S'$ are isotopic through generic 2-links
and one of $A_1^+A_1^+,A_1^+A_1^-,A_1^-A_1^-$-singularities,
then $BL,BL'$ are related by the slide/swim moves in Fig.~4.
\smallskip
\noindent
(iv) If 2-links $S, S'$ are isotopic through generic 2-links
and exactly one $A_2$-singularity, then
$BL,BL'$ are related
by the cap/cup and slide/swim moves in Fig.~4.
\medskip
\noindent
\emph{Proof}.
(i)
The lowest critical point of a generic 2-link $S$ with respect to $\pr:S\to[-1,1]$
at $t=t_1$ is a minimum, so the cross-section $S_{t_1+\ep}$
is a trivial knot for some $\ep>0$.
The section $S_{t_1+\ep}$ is a prototype of a future banded link $BL$,
which will be located in a fixed copy of $\R^3$.
The key idea in constructing $BL$ is to watch
the current cross-section $S_t=S\cap(\R^3\times\{t\})$
simultaneously adding bands and trivial knots corresponding to
new saddles and minima, respectively.
The left column of Fig.~5 contains cross-sections $S_t$ for different values of $t$.
The right column shows successive stages of constructing $BL$
whose final form is the top right.
\smallskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig5.eps}
\caption{Cross-sections and a banded link of the spun 2-knot of the trefoil}
\end{figure}
While $t$ is increasing, we isotopically deform
the current banded link $BL\subset\R^3$
following $S_t=S\cap(\R^3\times\{t\})$, see Fig.~5.
The existing bands of $BL$ can be deformed
to avoid intersections with the rest of $BL$.
For each new minimum of $S$ in $\R\times\{t_j\}$, add a trivial knot
from $S_{t_j+\ep}$ to the current banded link $BL\subset\R^3$.
\smallskip
For each new saddle of $S$, attach a small band $B$ to $BL$.
The band $B$ has 2 opposite sides attached
to branches of the previous link $BL$.
While $t$ passes the critical value, the attached sides of $B$ are retracted
to a point and are replaced by the free sides of $B$.
The band $B$ can not meet the attached sides of other bands of $BL$
since these sides are not included into the current cross-section of $S$.
So there are only 2 cases when the new link
with bands does not satisfy Definition~2.4.
\medskip
\noindent
(a) One (or two) of the attached sides of $B$ may meet
a free side of another band $B'$ of $BL$, see the upper picture of Fig.~6.
Then slide $B$ along the free side of $B'$ in any of the two directions
so that in the end the attached side of $B$ does not meet $B'$.
\medskip
\noindent
(b) The band $B$ intersects the interior of another band $B'$ of $BL$,
see the lower picture of Fig.~6.
Then $B$ swims through any of the attached sides of $B'$, so $B,B'$ fall apart.
The band $B$ can not swim through the free sides of $B'$
as they belong to the current cross-section of $S$.
For each new 2-handle (a maximum), we keep the corresponding trivial knot
of $BL$, although it disappears from $S_t=S\cap(\R^3\times\{t\})$.
\medskip
After we have passed all critical values of $\pr:S\to\R$,
the associated banded link $BL\subset\R^3$ has been constructed.
\medskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig6.eps}
\caption{A band slides or swims to remove an intersection}
\end{figure}
\medskip
\noindent
(ii) The construction above is not affected by an isotopy of $S$
keeping the order of critical points of $\pr:S\to[-1,1]$.
Indeed all cross-sections $S_t$ are replaced by isotopic links,
so the resulting banded link is isotopic to the original one
provided that we remove intersections of bands in Fig.~6 in the same way.
\medskip
\noindent
(iii)
The given isotopy of $S$ is a smooth path passing through
one of $A_1^+ A_1^+, A_1^+ A_1^-, A_1^- A_1^-$-singularities in the space $\CS$ of 2-links.
For $A_1^+ A_1^+$ or $A_1^+ A_1^-$,
an extremum and another singularity swap their heights, so
we add a new trivial knot (passing a minimum) or
keep an existing trivial knot (passing a maximum) that does not affect the other singularity.
For an $A_2$-singularity, two saddles of $S$ swap their heights,
so we add 2 bands to $BL$ in the reverse order.
Consider the critical moment when both saddles are
in the same section $\R^3\times\{t_j\}$.
If the associated bands do not intersect each other,
then the new banded link is isotopic to the original one.
In (i) we listed the only cases (a), (b) when one band
may intersect another, which led to the moves in Fig.~6
so the banded links are equivalent through
the slide/swim moves.
\medskip
\noindent
(iv)
If an isotopy of $S$ passes through an $A_2$-singularity, then
around this moment a non-degenerate saddle and extremum appear in a 2-link,
see Claim~4.3(iv).
In the case of a minimum, one adds a trivial knot to the current banded link $BL$
and a band attached to the trivial knot and to an existing branch of $BL$
as shown in the cup move of Fig.~4.
In the case of a maximum, one adds a band attached by both sides to
a branch of the current banded link $BL$ as shown in the cap move of Fig.~4.
Recall that we keep the trivial knot when $t$ passes a maximum.
The leftmost and rightmost columns of Fig.~4 describe
projections of 2-links to $\R^3$ around singular moments.
The 4th axis of $\R^3\times\R$ projects to the vertical axis of $\R^3$.
\qed
\medskip
Conversely, any admissible banded link will give rise to
a 2-link in 4-space, see Lemma~2.8.
One can describe all moves of banded links
generating any isotopy of 2-links in 4-space.
Banded links were called \emph{knots with bands} in \cite{Swe}.
\subsection{Representing 2-links in 4-space by marked graphs in 3-space}
\noindent
\smallskip
\noindent
Theorem~1.4 is easier to prove representing 2-links by marked graphs,
which are singular links with bridges at singular points.
\medskip
\noindent
{\bf Definition 2.7.}
After deformation retracting each band of a banded link $BL$ to a point,
we get a \emph{singular} link \cite{KV}, a collection of closed curves
with finitely many double transversal intersections, see Fig.~2, 4.
The core of each retracted band defines a bridge at the singular point,
a straight arc in a small plane neighbourhood of each singular point.
We consider the resulting \emph{marked} graph $MG$ up to isotopy in $\R^3$
keeping a neighbourhood of each singular point in a (moving) plane.
\medskip
\noindent
In the smooth approach, the zero section $S\cap(\R^3\times\{0\})$
containing all saddles of $\pr:S\to[-1,1]$ is a marked graph
whose bridges show how to resolve the singular points
for $t>0$ (along bridges) and $t<0$ (across bridges), see Fig.~2, 3.
An abstract marked graph $MG$, i.e. a singular link with bridges,
can be converted into a banded link $BL$ replacing each bridge
by a small rectangle whose core coincides with the bridge.
So there is a 1-1 correspondence between banded links and marked graphs.
Lemma~2.8 provides a unique function from the set of admissible banded links
to the set of 2-links, which is the inverse of the function from Proposition~2.6.
\medskip
\noindent
{\bf Lemma 2.8.}
Any admissible banded link $BL\subset\R^3$
gives rise to a 2-link $S\subset\R^4$ that
can be represented by $BL$ as in Proposition~2.6(i).
\smallskip
\noindent
\emph{Proof.}
Take the marked graph $MG\subset\R^3$
associated to the given banded link.
Isotopically deform $MG$ in such a way that neighbourhoods
of all singular points of $MG$ are contained
in a single hyperplane of $\R^3\times\{0\}$.
\smallskip
Resolving the singular points along the bridges for $t>0$
and across the bridges for $t<0$, extend the embedding
$MG\subset\R^3\times\{0\}$ to a surface $S'\subset\R^3\times[-\ep,\ep]$
for some $\ep>0$, such that the boundary $\bd S'$ consists of
trivial links in $\R^3\times\{\pm\ep\}$.
\smallskip
Since both sections $S'_{\pm\ep}=S'\cap(\R^3\times\{t=\pm\ep\})$ are unlinks,
one can find isotopies $\ph_t^{\pm}:\R^3\to\R^3$, $t\in[\ep,1-\ep]$, such that
each $\ph^{\pm}_{1-\ep}(S'_{\pm\ep})$ is
a collection of small disjoint circles in a plane.
The isotopies $\ph_t^{\pm}$ define the embedding of a 2-link $S$ without
small disks into $\R^3\times[\ep-1,1-\ep]$, one disk for each component of $\bd S$.
Attaching a disc to each boundary circle gives
a closed surface $S\subset\R^3\times[-1,1]$.
\smallskip
The zero section $S\cap(\R^3\times\{0\})$ is the original marked graph $MG$.
A small isotopy deformation makes $S$ generic.
The construction of Proposition~2.6(i) gives a banded link
equivalent to $MG$ as all bands may be chosen small and non-intersecting.
\qed
\medskip
\section{Three-page embeddings of marked graphs}
\subsection{Any marked graph can be embedded into the 3-page book}
\noindent
\smallskip
\noindent
Recall that the 3-page book is $\tb=\R\times T$, where
$T$ is the triod consisting of 3 edges $E_0,E_1,E_2$
joining the vertex $O$ to the other 3 vertices.
The line $\al=\R\times O$ is said to be the \emph{binding} axis,
$P_i=\R\times E_i$ are called the \emph{pages}, $i=0,1,2$.
\medskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig7.eps}
\caption{The encoding letters for 3-page embeddings of marked graphs}
\end{figure}
\noindent
{\bf Definition 3.1.}
An embedding of a marked graph $G$ into the 3-page book $\tb$
is called a \emph{3-page} embedding, if conditions (i)-(v) hold:
\smallskip
\noindent
(i) the intersection $G\cap\al$ of $G$ and the binding axis $\al$ is a finite set of points;
\smallskip
\noindent
(ii) the arcs at every point of $G\cap\al$ lie in 2 pages $P_i,P_j$,
$i\neq j$, see Fig.~7;
\smallskip
\noindent
(iii) all singular points of $G$ lie in $\al$,
a neighbourhood of each singular point lies\\
\hspace*{7mm}
in a broken plane consisting of two pages
and looks locally like a cross $\times$;
\smallskip
\noindent
(iv) the bridge at each singular point lies in the binding axis $\al$;
\medskip
\noindent
(v) every connected component of $G\cap P_i$ is projected
monotonically to $\al$.
\bigskip
\noindent
The arcs in the page $P_2$ are dashed in Fig.~7, 8.
All classical and singular links can be embedded into $\tb$
in the sense of Definition~3.1, see Fig.~8.
\medskip
The pictures in each vertical column of Fig.~7 are obtained
from each other by rotation around $\al$.
The rotation corresponds to the shift $i\mapsto i+1$ of indices,
$i\in\Z_3=\{0,1,2\}$.
A 3-page embedding can be encoded by a word in the alphabet
of 15 letters describing the local behaviour of $G$
near the intersection points $G\cap\al$, see Fig.~7.
The 3-page embedding in Fig.~8 is encoded by
$w_G=a_0 a_1 (b_2b_0b_1)^2 d_0 a_1
(x_1 b_1)^2 c_1 d_1 b_0 (d_1 d_0 d_2)^2 c_1 c_0$.
So a 3-page embedding of the marked graph $G_S$ of a 2-link $S$
is a 1-dimensional representation of $S\subset\R^4$.
\medskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig8.eps}
\caption{A 3-page embedding of the marked graph from Fig.~9}
\end{figure}
We give a proof of the embedding result from \cite{KV},
because this construction plays an important role
in further considerations.
\medskip
\noindent
{\bf Proposition~3.2.} \cite{KV}
Any marked graph $G\subset\R^3$ is isotopic to
a 3-page embedding $G\subset\tb$ in the sense of Definition~3.1.
\medskip
\noindent
\emph{Proof}.
Consider a plane diagram $D$ of $G\subset\R^3$ in general position
with finitely many double crossings.
At each crossing in the diagram $D$ mark a small overcrossing arc.
Recall that, at each singular point of $G$, there is a marked bridge
transversally intersecting both branches of $G$ passing through the singular point.
\smallskip
In the plane containing the diagram $D$, draw a continuous path $\al$ such that
\smallskip
\noindent
(1) the path $\al$ passes through each marked arc and bridge exactly once;
\smallskip
\noindent
(2) $\al$ transversally intersects the rest of $D$,
the endpoints of $\al$ are away from $D$.
\medskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig9.eps}
\caption{How to construct a 3-page embedding of a marked graph}
\end{figure}
Isotopically deform the plane containing $D$ in such a way that $\al$ becomes
a straight line containing all marked arcs and bridges of $D$.
Denote the upper half-plane and lower half-plane of $\R^2-\al$
by $P_0$ and $P_2$, respectively.
Notice that a neighbourhood of each singular point
looks like a cross $\times$ with a centre in the axis $\al$, see Fig.~9.
\smallskip
Attach the third half-plane $P_1$ to $\al$ and
push all marked arcs into $P_1$, see Fig.~8.
If both (say) upper arcs at some singular point $v\in G$
go to points on one side of the point $v\in\al$,
then make an additional couple of crossings in the intersection $\al\cap D$ like
in Reidemeister move II, see Fig.~13.
For instance, in the embedding $a_2 b_2 x_2$ both upper arcs go to the right,
see the lower right picture of Fig.~15 \cite{KV}.
Then the intersection $G\cap P_i$
is a finite collection of disjoint arcs, which can be made monotonic
with respect to the projection $\tb\to\al$, $i=0,1,2$.
\qed
\smallskip
\subsection{Any isotopy of links can be realised in the hexabasic book}
\noindent
\smallskip
The following lemma is a key stone of the 3-page approach to knot theory.
\medskip
\noindent
{\bf Lemma 3.3.} \cite{Kur}
Any isotopy of 3-page embeddings of classical links is decomposed
into finitely moves in Fig.~10 and theirs images under
$i\mapsto i+1$, $i\in\Z_3$.
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig10.eps}
\caption{Finitely many moves generating any isotopy of classical links}
\end{figure}
\medskip
The algebraic form of the moves in Fig.~10 is below,
$i\in\Z_3=\{0,1,2\}$, see \cite{Kur}:
\smallskip
\noindent
(1) $d_0d_1d_2=1$, $b_id_i=1=d_ib_i$;
\smallskip
\noindent
(2) $a_i=a_{i+1}d_{i-1}$, $b_i=a_{i-1}c_{i+1}$,
$c_i=b_{i-1}c_{i+1}$, $d_i=a_{i+1}c_{i-1}$;
\smallskip
\noindent
(3) $uv=vu$, where $u\in\{a_ib_i,d_ic_i,b_{i-1}d_id_{i-1}b_i\}$,
$v\in\{a_{i+1},b_{i+1},c_{i+1},b_id_{i+1}d_i\}$.
\medskip
Lemma~3.4 is the crucial step in Theorem~1.3.
\medskip
\noindent
{\bf Lemma 3.4.}
The moves in Fig.~10 are realised in the hexabasic book $\hb$.
\smallskip
\noindent
\emph{Proof.} All the moves in Fig.~10, apart from the commutativity
of $a_i,b_i,c_i,b_{i-1}d_id_{i-1}$ with $b_{i+1}d_{i-1}d_{i+1}b_{i-1}$,
can be realised in the 3-page book $\tb$.
For instance, the relation $b_2d_2=1$ is realised
by compressing the slice between the 2 intersection points
and removing the resulting point from $\al$.
The other relations are realised in $\hb$, see a geometric realisation
of $(b_1d_2d_1b_2)a_0=a_0(b_1d_2d_1b_2)$ in Fig.~11.
\qed
\medskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig11.eps}
\caption{Realising a commutative relation in the hexabasic book $\hb$}
\end{figure}
\subsection{Any 2-link is isotopic to a surface in the universal polyhedron}
\noindent
\smallskip
Here we finish the proof of Theorem~1.3.
\medskip
\noindent
\emph{Proof of Theorem~1.3}.
By Lemma~2.3 any 2-link in 4-space is isotopic to
a surface $S\subset\R^3\times[-1,1]$ having
all maxima, minima and saddles in
different sections $\R^3\times\{t_j\}$
for some $-1<t_1<\dots<t_n<1$.
In Step~1 we embed each cross-section $S_{t_j}$ into the 3-page book.
In Step~2 we extend this embedding to a regular neighbourhood of $S_{t_j}$.
In Step~3 we embed the complement of the neighbourhoods into $\hb\times[-1,1]$.
\smallskip
\noindent
\emph{Step 1}.
Choose $\ep>0$ such that the closed $\ep$-neighbourhoods
$N_{\ep}(S_{t_j})$ of $S_{t_j}$ in $S$ are disjoint and each of them
contains exactly one critical point of $\pr:S\to[-1,1]$, $j=1,\dots,n$.
Then the boundaries $\bd N_{\ep}(S_{t_j})$ are classical links.
By Proposition~3.2 there is an isotopy $f^u_j:\R^3\times\{t_j\}\to\R^3\times\{t_j\}$,
$u\in[0,1]$, moving $S_{t_j}$ into $\tb$, i.e. $f^0_j=\id_{\R^3}$,
$f^1_j(S_{t_j})\subset\tb\times\{t_j\}$ is a 3-page embedding.
Take smooth functions $g_j:[t_j-\ep,t_j+\ep]\to[0,1]$
such that $g_j(t_j)=1$ and $g_j(t_j\pm\ep)=0$.
Extend $f_u^j$ to
$$F^u_j:\R^3\times[t_j-\ep,t_j+\ep]\to\R^3\times[t_j-\ep,t_j+\ep],
\quad u\in[0,1],$$
$$F^u_j(x,t)=(f^{ug_j(t)}_j(x),t), \mbox{ where }x\in\R^3, t\in[t_j-\ep,t_j+\ep].$$
Then $F^u_j=f^u_j$ for $t=t_j$ and $F^u_j=\id$ for $t=t_j\pm\ep$.
Hence $\bd N_{\ep}(S_{t_j})$ are pointwise fixed and we may
combine $F^u_j$ together to form a smooth isotopy
$F^u:\R^3\times[-1,1]\to\R^3\times[-1,1]$ moving
each $S_{t_j}$ into $\tb\times\{t_j\}$.
Denote the resulting surface by $S'$.
\medskip
\noindent
\emph{Step 2}.
If a singular cross-section $S'_{t_j}$ has a double intersection, then
both positive and negative resolutions of $S'_{t_j}$ can be embedded into $\tb$.
Indeed the positive and negative resolutions of the singular point $x_i$
are encoded by $1$ and $c_ia_i$, respectively, see Fig.~12.
Given an encoding word $w_j$ of $S'_{t_j}\subset\tb$,
the positive resolution of $S'_{t_j}$ is encoded by $w_j$ after removing
the letter $x_i$ representing the double point of $S'_{t_j}$.
\smallskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig12.eps}
\caption{Resolving a singular point in the 3-page book $\tb$}
\end{figure}
The argument below with the sign $\pm$ covers 2 cases
when either $+$ or $-$ is taken in all formulae.
If $S'_{t_j}$ contains a maximum or minimum,
$S'_{t_j\pm\ep/2}$ can be embedded into $\tb$.
So there are isotopies
$h^u_{\pm j}:\R^3\times\{t_j\pm\ep/2\}\to\R^3\times\{t_j\pm\ep/2\}$, $u\in[0,1]$,
moving each $S'_{t_j\pm\ep/2}$ into $\tb\times\{t_j\pm\ep/2\}$.
Take smooth functions $\ti g_{j}:[t_j-\ep,t_j+\ep]\to[0,1]$ such that
$\ti g_j(t_j)=0=\ti g_j(t_j\pm\ep)$ and $\ti g_j(t_j\pm\ep/2)=1$.
Extend $h^u_{\pm j}$ to
$$H^u_{j}:\R^3\times[t_j-\ep,t_j+\ep]\to\R^3\times[t_j-\ep,t_j+\ep],
\quad u\in[0,1],$$
$$H^u_{j}(x,t)=(h^{u\ti g_j(t)}_{\pm j}(x),t) \mbox{ for } x\in\R^3,\;
t \mbox{ between } t_j \mbox{ and } t_j\pm\ep.$$
Then $H^u_j=h^u_{\pm j}$ for $t=t_j\pm\ep/2$ and
$H^u_j=\id$ for $t=t_j$, $t=t_j\pm\ep$.
Hence $S'_{t_j}$ and $\bd N_{\ep}(S'_{t_j})$ are pointwise fixed and
we may combine $H^u_{j}$ together to form a smooth isotopy
$H^u:\R^3\times[-1,1]\to\R^3\times[-1,1]$ moving
each $N_{\ep/2}(S'_{t_j})$ into $\tb\times[t_j-\ep/2,t_j+\ep/2]$.
Denote the resulting surface by $S''$.
\medskip
\noindent
\emph{Step 3}.
The cross-sections $S''_{t_j+\ep/2}$ and $S''_{t_{j+1}-\ep/2}$
are isotopic classical links, $j=1,\dots,n-1$.
By Lemma~3.3 and Proposition~3.4 any isotopy of classical links
can be realised in $\hb$.
Then the layers $S''\cap(\R^3\times[t_j+\ep/2,t_{j+1}-\ep/2])$
can be replaced by an isotopy of links in $\hb\times[t_j+\ep/2,t_{j+1}-\ep/2]$.
It remains to extend the embedding to the neighbourhoods
of the lowest minimum and highest maximum of $S''$
shrinking their boundaries in $\hb$.
So the final surface is embedded into $\hb\times[-1,1]$.
\qed
\medskip
\section{The universal semigroup of 2-dimensional links}
\subsection{Local moves of marked graphs generate any isotopy of 2-links}
\noindent
\smallskip
Here we derive a complete set of moves of banded links and marked graphs,
that generate any isotopy of 2-links in 4-space.
Marked graphs can be represented by plane diagrams with
small straight arcs denoting bridges over singular points, see Fig.~2, 9.
In particular, the cyclic order of edges at each singular point is invariant.
\medskip
\noindent
{\bf Lemma~4.1} \cite{Kau}
Marked graphs are isotopic in $\R^3$ if and only if
their plane diagrams can be obtained from each other
by finitely many Reidemeister moves in Fig.~13,
where all symmetric images of the moves should be considered.
\medskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig13.eps}
\caption{Reidemeister moves for rigid isotopy of marked graphs}
\end{figure}
The moves in Fig.~13 are \emph{local} in the sense,
that a small disk in the left part of each move is replaced by
another small disk in the right part of the move, while
the rest of the diagram remains unchanged.
The singular points in moves IV and V of Fig.~13
can be equipped with arbitrary corresponding bridges.
The proof is a direct application of the transversality theorem of Thom
similarly to a proof of the Reidemeister theorem
for plane diagrams of classical links, see \cite[section~2]{FK}.
\medskip
\noindent
{\bf Proposition~4.2.}
Marked graphs represent isotopic 2-links in 4-space
if and only if they can be obtained from each other
by finitely many moves in Fig.~14.
\medskip
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig14.eps}
\caption{Moves of marked graphs generating isotopy of 2-links}
\end{figure}
Symmetric images of the moves in Fig.~14 are skipped as
they can be reduced to the standard moves using an isotopy in $\R^3$.
Proposition~4.2 was conjectured by K.~Yoshikawa in \cite{Yos}.
F.~Swenton \cite{Swe} claimed a proof of Proposition~4.2
using banded links and the equivalent moves in Fig.~4.
M.~Saito wrote in his review for the MathSciNet:
`It is claimed that this set of moves is equivalent to Yoshikawa's moves.
It might be beneficial of some more detailed accounts, for example,
those for the above claim, are discussed further and
presented elsewhere in the literature'.
The authors were asked by S.~Carter to fill in these details,
so we give a more detailed proof of Proposition~4.2 for banded links.
Recall that the singular subspace $\Si$ of the space $\CS$
of 2-links was introduced in Definition~2.5.
The following result will be formally deduced in Appendix using
the transversality theorem of Thom.
\medskip
\noindent
{\bf Claim~4.3.}
(i) The closure of the subspace $\Si$ has codimension~1 in the space $\CS$.
\smallskip
\noindent
(ii)
The complement of the closure $\overline{\Si}$ in $\CS$ consists of generic 2-links.
\smallskip
\noindent
(iii)
Any isotopy of 2-links can be deformed in such a way that
all intermediate 2-links are generic apart from
finitely many singularities of Definition~2.5.
\smallskip
\noindent
(iv)
If an isotopy passes through an $A_2$-singularity, then
a non-degenerate saddle and extremum collide and disappear
as shown in the top picture of Fig.~18.
\medskip
\noindent
Claims~4.3(i,ii) say that any point of $\CS$ can be removed from $\Si$
by a small perturbation, i.e. a 2-link can be made generic, which implies Claim~2.3(ii).
Claim~4.3(iii) says that the singularities of Definition~2.5 are the only singularities
occuring in any isotopy of 2-links in general position.
\medskip
\noindent
\emph{Proof} of Proposition~4.2.
By Claim~4.3(iii) any isotopy of 2-links can be deformed into a smooth path
transversal to the subspace $\Si\subset\CS$.
When the path passes through one of the singularities, the associated
banded link changes according to Proposition 2.6(iii),(iv),
which led to the moves in Fig.~4 as required.
\qed
\medskip
\subsection{A 1-dimensional encoding 2-links up to isotopy in 4-space}
\noindent
\smallskip
Here we reduce the isotopy classification of 2-links in 4-space
to a word problem in the finitely presented semigroup $\SL$,
the universal semigroup of 2-links.
Recall that moves (1)-(8) on 3-page embeddings were defined in subsection~1.3.
Theorem~1.4 follows from the following generalisation of
Lemma~3.3 to singular links.
\medskip
\noindent
{\bf Proposition~4.4.} \cite{KV}
Consider the semigroup $\SK$ generated by $a_i,b_i,c_i,d_i,x_i$,
$i\in\Z_3$, subject to relations (1)-(5) from subsection~1.3.
Then any singular link $G\subset\R^3$ is encoded by an element
$w_G\in\SK$ in such a way that singular links $G,G'$ are isotopic
if and only if their encoding elements $w_G$ and $w_{G'}$
are equal in $\SK$.
An element $w\in\SK$ encodes a singular link
if and only if $w$ is central in $\SK$.
\medskip
\noindent
\emph{Proof of Theorem~1.4.}
Any 2-link can be represented by its marked graph $G$ whose 3-page embedding
is encoded by a word in the letters $a_i,b_i,c_i,d_i,x_i$, $i\in\Z_3$,
as described before Proposition~3.2.
All encoding elements form the centre of $\SL$ as the same result
holds for the universal semigroup $\SK$ of singular links, i.e. relations
(1)-(5) imply that any encoding element commutes with the generators.
\smallskip
The remaining part of Theorem~1.4 states that two 3-page embeddings
of marked graphs represent isotopic 2-links in 4-space if and only if
they can be related by algebraic moves (1)--(8) in subsection~1.3.
By Lemmas~3.3-3.4 and Proposition 4.2 it suffices to realise
moves VI, VII, VIII in Fig.~14 by 3-page embeddings.
\smallskip
In moves VI, VII, VIII a small disk in the left part
is replaced by another small disk in the right part.
Similarly to the construction of a 3-page embedding, choose a path $\al$
passing through overcrossing arcs and bridges at singular points, see Fig.~15, 16, 17.
Deform the diagrams in such a way that $\al$ becomes a straight line and
push all overcrossing arcs into the half-plane $P_1$, all bridges remain in $\al$.
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig15.eps}
\caption{Realising moves VI of Fig.~14 in terms of 3-page embeddings}
\end{figure}
In Fig.~15 moves VI are encoded by $a_1 x_1=a_1$ and
$a_1 b_1 x_1 d_1 c_1=1$ equivalent to (6) for $i=1$.
We made additional intersections of $\al$ with the diagram to decompose the resulting
embedding into local 3-page embeddings from Fig.~7.
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig16.eps}
\caption{Realising move VII of Fig.~14 in terms of 3-page embeddings}
\end{figure}
In Fig.~16 move VII is encoded by
$d_1 x_1 b_1 c_1 x_1=b_1 x_1 d_1 c_1 x_1$,
which is (7) for $i=1$.
Numbers 1, 2, 3, 4, 5, 6 denote arcs going out of the small disk replaced by move VII,
e.g. the path $\al$ starts between arcs 1, 4 and ends between arcs 3, 6.
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig17.eps}
\caption{Realising move VIII of Fig.~14 in terms of 3-page embeddings}
\end{figure}
In Fig.~17 move VIII is encoded by
$$(a_1 b_1 x_1 b_1 c_1) d_2 d_1 (b_2 d_2)
d_1 d_0 a_2 b_2 x_1 b_1 d_2 b_1 (b_2 d_2) b_1 b_2 d_1^2=$$
$$(a_1 b_1 x_1 b_1 c_1) b_0 b_1 (b_2 d_2) a_1 b_2 a_2 d_1 (b_2 d_2) d_1 c_0 b_1 x_1 b_1,$$
which is equivalent to (8) for $i=1$ after removing $b_2 d_2=1$ by relation~(1).
The relations for other $i\in\Z_3$ were added
to make the presentation symmetric.
\qed
\section*{Appendix: the multi-jet transversality theorem of Thom}
\smallskip
Here we follow \cite[sections I.2, I.8]{AVG}.
Let $\xi,\eta:M\to N$ be smooth maps between
finite dimensional manifolds
with Riemannian metrics $\rho_M,\rho_N$, respectively.
\medskip
\noindent
{\bf Definition~A.1.}
The maps $\xi$ and $\eta$ have the tangency of \emph{order} $k$
at a point $z\in M$ if $k$ is the maximal integer such that
$\rho_N(\xi(w),\eta(w))/(\rho_M(z,w))^k\to 0$ as $w\in M$ tends to $z$, e.g.
the curve $\xi(w)=w^{k+1}$ has the tangency of order $k$ with $\eta(w)=0$.
\smallskip
The $l$-tuple $k$-\emph{jet} of the map $\xi$ at
$(z_1,\dots,z_l)\in M^l$ is the equivalence class
of smooth maps $\eta:M\to N$ up to tangency of order $k$
at the points $z_1,\dots,z_l\in M$, e.g.
the 1-tuple 1-jet $j^1_{[1]}\xi(z)$ of a map $\xi:\R\to\R$
is determined by $z,\xi(z),\dot \xi(z)$.
\medskip
Denote by $J^k_{[l]}(M,N)$ the space of all $l$-tuple $k$-jets of
smooth maps $\xi:M\to N$ for all $(z_1,\dots,z_l)\in M^l$.
Let $(x_1,\dots,x_m)$ and $(y_1,\dots,y_n)$ be
local coordinates in $M$ and $N$, respectively.
If $\xi$ is defined locally by
$y_j=\xi_j(x_1,\dots,x_m)$, $j=1,\dots,n$, then
the $l$-tuple $k$-\emph{jet} of $\xi$ at $(z_1,\dots,z_l)$
is determined by $l$ arrays of the data below
$$\{x_1,\dots,x_m\};\quad
\{y_1,\dots,y_n\};\quad
\left\{ \dfrac{ \bd\xi_j }{ \bd x_i } \right\};\quad
\; \ldots \;
\left\{ \dfrac{ \bd^{k}\xi_j }{ \bd x_{i_1}\dots x_{i_s} } \right\},
i_1+\dots +i_s=k.$$
The quantities above define local coordinates
in $J^k_{[l]}(M,N)$.
The $l$-tuple $k$-\emph{jet} $j^k_{[l]}\xi$ of
a smooth map $\xi:M\to N$ can be considered as
the map $j^k_{[l]}\xi:M^l\to J^k_{[l]}(M,N)$,
namely $(z_1,\dots,z_l)$ goes to
the $l$-tuple $k$-jet of $\xi$ at $(z_1,\dots,z_l)$.
\medskip
The manifold $J^k_{[l]}(M,N)$ is finite dimensional, e.g.
$J^0_{[l]}(M,N)=(M\times N)^l$,
$$\dim J^1_{[l]}(M,N)=(m+n+mn)l, \;
\dim J^2_{[l]}(M,N)=(m+n+mn+\frac{m(m+1)}{2}n)l.$$
\smallskip
\noindent
{\bf Definition~A.2.}
Take an open set $W\subset J^k_{[l]}(M,N)$.
The set of smooth maps $f:M\to N$ with $l$-tuple $k$-jets from $W$
is \emph{open}.
These sets for all open $W\subset J^k_{[l]}(M,N)$ over all $k,l$
form a basis of the \emph{Whitney} topology in $C^{\infty}(M,N)$.
The space $\CS$ of all 2-links $S\subset\R^4$ inherits
the \emph{Whitney} topology from $C^{\infty}(S,\R^4)$.
So two maps are close in the Whitney topology
if they are close with all derivatives.
\medskip
\noindent
{\bf Definition A.3.}
Let $M$ be a finite dimensional smooth manifold.
A subspace $\La\subset M$ is called
\emph{a stratified space} if $\La$ is the union
of disjoint smooth submanifolds $\La^i$ (\emph{strata})
such that the boundary of each stratum
is a finite union of strata of less dimensions.
Let $N$ be a finite dimensional manifold.
A smooth map $\xi:M\to N$ is \emph{transversal} to
a smooth submanifold $U\subset N$ if
the spaces $\xi_*(T_zM)$ and $T_{\xi(z)}U$ generate $T_{\xi(z)}N$
for each $z\in M$.
A smooth map is $\eta:M\to V$ \emph{transversal} to
a stratified space $\La\subset V$
if the the map $\eta$ is transversal to each stratum of $\La$.
\medskip
\noindent
Briefly Theorem~A.4 says that any map
can be approximated by `a nice map'.
\medskip
\noindent
{\bf Theorem A.4.}
(Multi-jet \emph{transversality} theorem of Thom,
see \cite[section~I.2]{AVG})\\
Let $M,N$ be compact smooth manifolds,
$\La\subset J^k_{[l]}(M,N)$ be a stratified space.
Given a smooth map $\xi:M\to N$, there is
a smooth map $\eta:M\to N$ such that
\smallskip
$\bu$
the map $\eta$ is arbitrarily close to $\xi$
with respect to the Whitney topology;
\smallskip
$\bu$
the $l$-tuple $k$-jet
$j^k_{[l]}\eta:M^l\to J^k_{[l]}(M,N)$
is transversal to $\La\subset J^k_{[l]}(M,N)$.
\qed
\medskip
\noindent
\emph{Proof of Claim~4.3}.
(i)
For any critical point of $\pr:S\to\R$, fix local coordinates $(x,y)\in S$
such that the derivatives $\pr_x=\pr_y=0$.
The closures of the subspaces $\overline{\Si_{++}\cup\Si_{+-}\cup\Si_{--}}$
and $\bar\Si_2$ from Definition~2.5 can be mapped onto the subspaces
of the finite-dimensional spaces $J_{[2]}^1(S,\R)$ and $J_{[1]}^2(S,\R)$
given by the equations $\pr(x_1,y_1)=\pr(x_2,y_2)$ and
$\pr_{xx}\pr_{yy}-\pr_{xy}^2=0$, respectively.
The resulting subspaces of jets have codimension~1
as preimages of 0 under smooth functions, e.g. the image of $\bar\Si_2$
in $J_{[1]}^2(S,\R)$ is $(\pr_{xx}\pr_{yy}-\pr_{xy}^2)^{-1}(0)$.
Hence the closures $\overline{\Si_{++}\cup\Si_{+-}\cup\Si_{--}}$ and
$\bar\Si_2$ have codimension~1 in the space $\CS$ of 2-links.
\smallskip
\noindent
(ii)
If a 2-link is not generic, then either some critical points of
the projection $\pr:S\to\R$ are degenerate or have the same value.
The singularities of Definition~2.5 are all multi local codimension~1
singularities of smooth functions $\R^2\to\R$, see \cite{AVG}.
\smallskip
\noindent
(iii)
By Theorem A.4 any smooth isotopy of 2-links is a path in $\CS$
and can be made transversal to the singular subspace $\bar\Si$,
which has codimension~1 by (i), hence the new path
will contain only finitely many isolated singularities of Definition~2.5.
\smallskip
\noindent
(iv)
The normal form of an $A_2$-singularity of a function $\R^2\to\R$
is $\pr(x,y)=x^2-y^3$, i.e. the projection $\pr:S\to\R$ has
the form above in suitable local coordinates $(x,y)\in S$.
A 2-link $S$, its cross-sections around the singularity and
the graph of $y^3$ look like the middle pictures of Fig.~18.
The versal deformation of an $A_2$-singularity is $\pr(x,y;\ep)=x^2-y^3+\ep y$
\cite{AVG}, i.e. any smooth deformation of $\pr(x,y)$ can be expressed as
$f_1(x,y;\ep)\cdot\pr(f_2(x,y;\ep),f_3(x,y;\ep);f_4(\ep))$,
where $f_1,f_2,f_3,f_4$ are smooth, $f_1(0,0;0)\neq 0$,
$f_2(x,y;0)\equiv x$, $f_3(x,y;0)\equiv y$ and $f_4(0)=0$.
\begin{figure}[!h]
\includegraphics[scale=1.0]{fig18.eps}
\caption{Transformation of a 2-link near an $A_2$-singularity}
\end{figure}
\smallskip
For $\ep<0$, a 2-link $S$, its cross-sections around the singularity
and the graph of $y^3-\ep y$ look like the left pictures of Fig.~18.
For $\ep>0$, a 2-link $S$, its cross-sections around the singularity
and the graph of $y^3-\ep y$ look like the right pictures of Fig.~18.
For instance, 2-links for $\ep>0$ have a non-degenerate saddle at
$x=0$, $y=\sqrt{\ep/3}$ and a local extremum at $x=0$, $y=-\sqrt{\ep/3}$.
\qed
\medskip | 128,001 |
If you need someone to help you feel better, let them know.
If there is an imbalance in your relationship, talk about it.
Just because you love someone who is suffering doesn't mean you have to let them hurt you.
If you think someone looks nice, tell them and maybe put a smile on their face.
Cat-calling does not fall under that umbrella. Be respectful.
Just because you give a compliment does not mean someone is obligated to take it.
If you're feeling sad, let it out.
BUT: Be smart about who you vent to.
It's okay if you forget.
Do not apologise for your feelings. | 32,864 |
Oct. 3, 2005
Weekly Press Conference Audio
Good afternoon everyone. I want to recap our game from last week before talking about Cal. We had probably not as good a game as we anticipated last week coming off a bye week. Getting some things done over the bye week, we felt pretty good going into the game. Unfortunately we didn't have those things click, particularly in the first half of the game. We did respond in the second half, getting some things together both defensively and offensively. The good thing that we can build on from this particular game was the fact that this was the first situation we've been in where we were behind almost the whole game. We were down 17-7 in the fourth quarter, but we found a way to get ourselves back in the game and win the football game in the end. We had a great fourth quarter rally that you all know about. I thought our defense, in the last two series before Washington's Hail Mary series, played very good defense. We needed them to make some stops and they did that to get the ball back for our offense to give them some opportunities to put some points on the board. So we did rise up to the occasion given the circumstances. That is something we haven't done in the past and it is something that we are really going to build from.
Cal is going to be a tremendous challenge for us. They are very talented in all three phases of football. They are very good on offense. They are leading the conference in rushing right now. They have some good backs and their quarterback is playing very well. They have an experienced offensive line. Those things create a lot of continuity on their offense and momentum as they come into town here this week. Defensively, they are very active, very aggressive, and put a lot of pressure on the quarterback. It's difficult to run the ball against them. They also do some great things on pass coverage. And then on special teams they have been very sound and very solid. They are a top-10 team and they are playing like a top-10 team right now. So our challenge this week is to shore up our fundamentals and get ourselves back to playing the caliber of football that we're capable of playing. There will be no problem getting up for this particular game with both teams being undefeated. So there is a lot at stake for both programs going into our conference season.
For coach Dorrell's complete press conference, please click on the audio link at the top of this page. | 8,672 |
\begin{document}
\title{Effective descent morphisms of regular epimorphisms}
\author{Tomas Everaert}
\address{
Vakgroep Wiskunde \\
Vrije Universiteit Brussel \\
Department of Mathematics \\
Pleinlaan 2\\
1050 Brussel \\
Belgium.
}
\email{teveraer@vub.ac.be}
\date{\today}
\maketitle
\begin{abstract}
Let $\Ac$ be a regular category with pushouts of regular epimorphisms by regular epimorphism and $\Reg(\Ac)$ the category of regular epimorphisms in $\Ac$. We prove that every regular epimorphism in $\Reg(\Ac)$ is an effective descent morphism if, and only if, $\Reg(\Ac)$ is a regular category. Then, moreover, every regular epimorphism in $\Ac$ is an effective descent morphism. This is the case, for instance, when $\Ac$ is either exact Goursat, or ideal determined, or is a category of topological Mal'tsev algebras, or is the category of $n$-fold regular epimorphisms in any of the three previous cases, for any $n\geq 1$.
\end{abstract}
\pagestyle{myheadings}{\markboth{TOMAS EVERAERT},\markright{}}
\section{Introduction}
A useful way of weakening the notion of Barr exactness, for a regular category, is to require that every regular epimorphism is an effective descent morphism, which assures the effectiveness of a certain class of equivalence relations, rather than of all. This weaker condition turns out to be strong enough for many purposes: indeed, the authors of \cite{Janelidze-Sobral-Goursat} had good reasons to say that a regular category satisfying this condition is ``almost Barr exact''!
In the present article, we are interested in the category $\Reg(\Ac)$ of regular epimorphisms in a regular category $\Ac$. This category is usually not exact. For instance, if $\Ac$ is a non-trivial abelian category, then $\Reg(\Ac)$ is (equivalent to) the category of short exact sequences in $\Ac$, which is well known not to be abelian, while it is obviously additive---hence it can not be exact by ``Tierney's equation''. However, there are many examples of categories $\Ac$ for which every regular epimorphism in $\Reg(\Ac)$ is an effective descent morphism. Indeed, this is the case, for instance, for any abelian category and, much more generally, for any exact Goursat category $\Ac$, as was shown in \cite{Janelidze-Sobral-Goursat}. It was pointed out in \cite{Janelidze-Sobral-Goursat}, however, that the exact Goursat condition is most likely too strong, even if $\Ac$ is assumed to be a variety. This is confirmed in the present article. In particular, we show, for a regular category $\Ac$ with pushouts of regular epimorphisms by regular epimorphisms that, in order to have that every regular epimorphism in $\Reg(\Ac)$ is an effective descent morphism, it is both necessary and sufficient that also $\Reg(\Ac)$ is a regular category. This condition turns out to be satisfied not only in the exact Goursat case, but also when $\Ac$ is ideal determined, or is a category of topological Mal'tsev algebras, or is the category of $n$-fold regular epimorphisms in any of the three previous cases, for any $n\geq 1$. In particular, we find that in each of these cases the category $\Ac$ is itself ``almost exact''.
Let us begin by recalling the definition of an effective descent morphism---for instance from \cite{Janelidze-Sobral-Tholen}, but note that the monadic description recalled here goes back to B\'enabou and Roubaud's article \cite{Benabou-Roubaud}; for the reformulation in terms of discrete fibrations, see also \cite{Janelidze-Tholen2}.
Let $\Ac$ be a category with pullbacks. If $B$ is an object of $\Ac$, then we write $(\Ac\downarrow B)$ for the slice category over $B$.
If $p\colon E\to B$ is a morphism in $\Ac$, we write $p^*\colon (\Ac\downarrow B)\to (\Ac\downarrow E)$ for the induced ``change of base'' functor given by pulling back along $p$.
\begin{definition}
An effective (global) descent morphism in a category with pullbacks $\Ac$ is a morphism $p\colon E\to B$ such that $p^*\colon (\Ac\downarrow B)\to (\Ac\downarrow E)$ is monadic.
\end{definition}
Note that a left adjoint for $p^*\colon (\Ac\downarrow B)\to (\Ac\downarrow E)$ exists for any morphism $p\colon E\to B$, and is given by composition with $p$. We denote it $\Sigma_p$, and write $T^p=p^*\circ \Sigma_p$ for the corresponding monad on $(\Ac\downarrow E)$. Writing $(\Ac\downarrow E)^{T^p}$ for the corresponding category of (Eilenberg-Moore) algebras, we obtain the following commutative triangle of functors, where $U^{T^p}$ is the forgetful functor and $K^{T^p}$ the comparison functor:
\[
\xymatrix{
(\Ac\downarrow B) \ar[d]_{K^{T^p}} \ar[r]^{p^*} & (\Ac\downarrow E)\\
(\Ac\downarrow E)^{T^p} \ar[ru]_{U^{T^p}} &}
\]
Thus $p\colon E\to B$ is an effective descent morphism if and only if $K^{T^p}$ is a category equivalence. Note that when $K^{T^p}$ is merely full and faithful, one says that $p$ is a \emph{descent morphism}.
There is an equivalent way of describing the above diagram, via discrete fibrations. Recall that a discrete fibration of equivalence relations in $\Ac$ is a (downward) morphism
\[
\xymatrix{
S \ar@{}[rd]|<<<{\pullback} \ar@<0.5 ex>[r] \ar@<-0.5 ex>[r] \ar[d] & A \ar[d]\\
R \ar@<0.5 ex>[r] \ar@<-0.5 ex>[r] & E }
\]
of equivalence relations, such that the commutative square involving the second projections (hence, also the square involving the first projections) is a pullback. Let $p\colon E\to B$ be a morphism in $\Ac$. Write $Eq(p)$ for the equivalence relation $\xymatrix{E\times_B E\ar@<0.5 ex>[r] \ar@<-0.5 ex>[r] & E}$ (i.e. the kernel pair of $p$) and $\DiscFib (Eq(p))$ for the category of discrete fibrations over $Eq(p)$, with the obvious morphisms. It was proved in \cite{Janelidze-Tholen2} (but see also \cite{Janelidze-Sobral-Tholen}) that for any $p\colon E\to B$ the category of algebras $(\Ac\downarrow E)^{T^p}$ is equivalent to the category $\DiscFib (Eq(p))$ of discrete fibrations over the kernel pair of $p$, and the commutative diagram above becomes:
\[
\xymatrix{
(\Ac\downarrow B) \ar[d]_{K^{p}} \ar[r]^{p^*} & (\Ac\downarrow E)\\
\DiscFib (Eq(p)) \ar[ru]_{U^{p}} &}
\]
Here $K^p$ sends a morphism $f\colon A\to B$ to the discrete fibration displayed in the left hand side of the diagram below, obtained by first pulling back $f$ along $p$, and next taking kernel pairs:
\[
\xymatrix{
E\times_B E\times_B A \ar@{}[rd]|<<{\pullback} \ar[d] \ar@<0.5 ex>[r] \ar@<-0.5 ex>[r] & E\times_BA \ar@{}[rd]|<<{\pullback} \ar[r] \ar[d] & A \ar[d]^f\\
E\times_B E \ar@<0.5 ex>[r] \ar@<-0.5 ex>[r] & E \ar[r]_p & B}
\]
and $U^p$ is the obvious forgetful functor.
We shall need the following lemma, which can be found, for instance, in \cite{Janelidze-Sobral-Tholen}. Recall that an equivalence relation in a category is \emph{effective} if it is the kernel pair of some morphism. A category is \emph{regular} if it is finitely complete, with pullback-stable regular epimorphisms and coequalisers of effective equivalence relations. It is \emph{Barr exact} if, moreover, every internal equivalence relation is effective.
\begin{lemma}\begin{enumerate}\label{known}
\item
In a finitely complete category, a descent morphism is the same as a pullback-stable regular epimorphism.
\item
In a regular category, a regular epimorphism $p\colon E\to B$ is an effective descent morphism if and only if for any discrete fibration
\[
\xymatrix{
R \ar@{}[rd]|<<<{\pullback} \ar[d] \ar@<0.5 ex>[r]^{\pi_1} \ar@<-0.5 ex>[r]_{\pi_2} & A \ar[d] \\
E\times_B E \ar@<0.5 ex>[r] \ar@<-0.5 ex>[r] & E}
\]
over the kernel pair of $p$, the equivalence relation $(\pi_1, \pi_2)$ is effective.
\end{enumerate}
\end{lemma}
Note that the second part of this lemma immediately implies the well-known fact that in an exact category every regular epimorphism is an effective descent morphism.
\section{Main results}
Throughout this section, we shall assume that $\Ac$ is a regular category with pushouts of regular epimorphisms by regular epimorphisms. Then, in particular, every morphism in $\Ac$ factors (essentially uniquely) as a regular epimorphism followed by a monomorphism. Let us denote by $\Reg(\Ac)$ the full subcategory of the arrow category $\Ac^2$ with as objects all regular epimorphisms in $\Ac$. Thus, a morphism $(a\colon A'\to A)\to (b\colon B'\to B)$ is a pair $(f',f)$ of morphisms in $\Ac$ such that $b\circ f'=f\circ a$. Like $\Ac$, the category $\Reg(\Ac)$ is finitely complete: limits are given by the regular epi part of the regular epi-mono factorisation of the degreewise pullback. An effective equivalence relation in $\Reg(\Ac)$ is the same as a graph
\[
\xymatrix{
R' \ar@<0.5 ex>[r]^{\pi_1'} \ar@<-0.5 ex>[r]_{\pi_2'} \ar[d]_r & E' \ar[d]^e\\
R \ar@<0.5 ex>[r]^{\pi_1} \ar@<-0.5 ex>[r]_{\pi_2} & E}
\]
in $\Reg(\Ac)$ such that $(\pi_1',\pi_2')$ is an effective equivalence relation in $\Ac$ and $\pi_1$ and $\pi_2$ are jointly monic. A regular epimorphism in $\Reg(\Ac)$ is the same as a pushout square of regular epimorphisms in $\Ac$. Notice that a regular category $\Ac$ admits pushouts of regular epimorphisms by regular epimorphisms if and only if $\Reg(\Ac)$ admits coequalisers of effective equivalence relations.
By Lemma \ref{known}.1, any effective descent morphism in $\Reg(\Ac)$ is necessarily a regular epimorphism. We would like to know when we have the converse. Certainly, this can only happen if every regular epimorphism \emph{in $\Ac$} is an effective descent morphism, since for any regular epimorphism $p
\colon E\to B$ in $\Ac$ a category equivalence $K^{(p,p)}\colon (\Ac^2\downarrow 1_B)\to \DiscFib(Eq(p,p))$ will restrict to an equivalence $K^p\colon (\Ac\downarrow B)\to \DiscFib(Eq(p))$. Moreover, by Lemma \ref{known}.1, $\Reg(\Ac)$ will need to have pullback-stable regular epimorphisms, which means---since it is finitely complete and has coequalisers of effective equivalence relations---that $\Reg(\Ac)$ is regular.
Somewhat surprisingly, perhaps, these conditions turn out to be sufficient. Indeed, we have:
\begin{theorem}\label{maintheorem}
For a regular category $\Ac$ with pushouts of regular epimorphisms by regular epimorphisms, in which every regular epimorphism is an effective descent morphism, the following conditions are equivalent:
\begin{enumerate}
\item
Every regular epimorphism in $\Reg(\Ac)$ is an effective descent morphism;
\item
$\Reg(\Ac)$ is a regular category.
\end{enumerate}
\end{theorem}
\begin{proof}
We only need to prove that 2. implies 1. so let us assume that $\Reg(\Ac)$ is regular. To prove that condition 1. holds, it suffices, by Lemma \ref{known}.2, to show for every discrete fibration of equivalence relations in $\Reg(\Ac)$, as in the diagram
\begin{equation}\label{dfcube}\vcenter{
\xymatrix{
& S \ar@<0.5 ex>[rr]^{\overline{\pi}_1} \ar@<-0.5 ex>[rr]_{\overline{\pi}_2} \ar@{.>}[dd]_>>>>>g && A \ar[dd]^f \\
S' \ar@<0.5 ex>[rr]^>>>>>>>>>{\overline{\pi}_1'} \ar@<-0.5 ex>[rr]_>>>>>>>>>{\overline{\pi}_2'} \ar[dd] \ar[ru] && A' \ar[dd] \ar[ru]\\
& R \ar@<0.5 ex>@{.>}[rr]^<<<<<<<<<{\pi_1} \ar@<-0.5 ex>@{.>}[rr]_<<<<<<<<<{\pi_2} && E\\
R' \ar@<0.5 ex>[rr]^{\pi_1'} \ar@<-0.5 ex>[rr]_{\pi_2'} \ar@{.>}[ru] && E' \ar[ru]}}
\end{equation}
that whenever the bottom equivalence relation is effective, the top one is effective as well. First of all recall that the bottom equivalence relation being effective means that $(\pi_1',\pi_2')$ is an effective equivalence relation in $\Ac$ and $\pi_1$ and $\pi_2$ are jointly monic. The cube being a discrete fibration in $\Reg(\Ac)$ means that the front square is a discrete fibration in $\Ac$ and, in the back square, $\overline{\pi}_2$ and $g$ are jointly monic.
Since $(\pi_1',\pi_2')$ is the kernel pair of an effective descent morphism, by assumption, we have, by Lemma \ref{known}.2 that $(\overline{\pi}_1',\overline{\pi}_2')$ is an effective equivalence relation in $\Ac$, and it remains to be shown that $\overline{\pi}_1$ and $\overline{\pi}_2$ are jointly monic. However, this follows from the assumption that $\overline{\pi}_2$ and $g$ are jointly monic, as well as $\pi_1$ and $\pi_2$.
\end{proof}
Here's another surprise, or at least it was a surprise to us: the assumption that every regular epimorphism in $\Ac$ is an effective descent morphism is superfluous in the statement of the above theorem---it follows at once from the regularity of $\Reg(\Ac)$! To see this, we will need the following lemma.
Recall that a diagram
\[
\xymatrix{
A'' \ar@<0.5 ex>[r]^{a_1} \ar@<-0.5 ex>[r]_{a_2} & A' \ar[r]^a& A}
\]
is called a \emph{fork} if $a\circ a_1=a\circ a_2$.
\begin{lemma}\label{George}
Let $\Ac$ be a regular category such that $\Reg(\Ac)$ is regular as well.
Consider a morphism
\begin{equation}\label{forks}\vcenter{
\xymatrix{
E'' \ar@{}[rd]|<<{\pullback} \ar@<0.5 ex>[r]^>>>>{e_1} \ar@<-0.5 ex>[r]_>>>>{e_2} \ar[d] & E' \ar@{}[rd]|<<{\pullback} \ar[d] \ar[r]^{e} & E \ar[d]\\
B'' \ar@<0.5 ex>[r]^{b_1} \ar@<-0.5 ex>[r]_{b_2} & B' \ar[r]_b & B}}
\end{equation}
of forks in $\Ac$ such that the right hand square as well as the left hand squares are pullbacks. Assume, moreover, that the graph $(b_1,b_2)$ (hence also the graph $(e_1,e_2)$) is reflexive. If $b$ is the coequaliser of $b_1$ and $b_2$, then $e$ is the coequaliser of $e_1$ and $e_2$.
\end{lemma}
\begin{proof}
The given diagram induces a commutative cube
\[
\xymatrix{
& E' \ar[rr]^{e} \ar@{.>}[dd] && E \ar[dd] \\
E'' \ar[rr]^>>>>>>>>>{e_2} \ar[dd] \ar[ru]^{e_1} && E' \ar[dd] \ar[ru]_{e}\\
& B' \ar@{.>}[rr]^<<<<<<<<<{} && B\\
B'' \ar[rr]_{b_2} \ar@{.>}[ru]^>>>>>>{b_1} && B' \ar[ru]_b}
\]
in which the bottom square is a pushout since $b$ is the coequaliser of $b_1$ and $b_2$, and because $b_1$ and $b_2$ have a common splitting. Since both front and back squares are pullbacks, and because each of the backward pointing morphisms is a regular epimorphism, the cube is a pullback in $\Reg(\Ac)$. As $\Reg(\Ac)$ was assumed to be regular, it follows that the top square is a pushout as well, so that $e$ is the coequaliser of $e_1$ and $e_2$, as desired.
\end{proof}
Note that although we do not assume that arbitrary coequalisers exist in $\Ac$, coequalisers in the slice categories $(\Ac\downarrow B)$ are always coequalisers in $\Ac$: indeed, the existence of binary products implies that the ``domain'' functors $(\Ac\downarrow B)\to \Ac$ have a right adjoint. The previous lemma is then easily seen to imply, for any morphism $p\colon E\to B$, that the ``change of base'' functor $p^*\colon (\Ac\downarrow B)\to (\Ac\downarrow E)$ preserves coequalisers of reflexive graphs. (In fact, one can easily proof that the lemma is equivalent to this property---see also Remark \ref{admissibility}.) When $p$ is a regular epimorphism, then, moreover, $p^*$ reflects isomorphisms, as follows from Lemma \ref{known}.1. Since $p^*$ has a left adjoint $\Sigma_p$, we can then conclude, via the ``reflexive form'' of Beck's Monadicity Theorem, that $p^*$ is monadic.
Thus we have proved:
\begin{theorem}\label{new}
If $\Ac$ is a regular category such that $\Reg(\Ac)$ is regular as well, then every regular epimorphism in $\Ac$ is an effective descent morphism.
\end{theorem}
Combining Theorems \ref{maintheorem} and \ref{new} we obtain:
\begin{corollary}\label{corollary}
For a regular category $\Ac$ with pushouts of regular epimorphisms by regular epimorphisms, the following conditions are equivalent:
\begin{enumerate}
\item
Every regular epimorphism in $\Reg(\Ac)$ is an effective descent morphism;
\item
$\Reg(\Ac)$ is a regular category.
\end{enumerate}
\end{corollary}
\section{Examples}
The results of the previous section suggest to investigate which regular categories $\Ac$ have the property that $\Reg(\Ac)$ is regular. We have the following examples:
\subsection{Exact Goursat categories}\label{ExGoursat}
Recall from \cite[Theorem 6.8]{Carboni-Kelly-Pedicchio} that a regular category is \emph{Goursat} if and only if for every (downward) morphism
\[
\xymatrix{
R' \ar@<0.5 ex>[r]^{\pi_1'} \ar@<-0.5 ex>[r]_{\pi_2'} \ar[d]_r & E' \ar[d]^e\\
R \ar@<0.5 ex>[r]^{\pi_1} \ar@<-0.5 ex>[r]_{\pi_2} & E}
\]
of relations in $\Ac$ with $r$ and $e$ regular epimorphisms, $(\pi_1,\pi_2)$ is an equivalence relation as soon as so is $(\pi_1',\pi_2')$. In universal algebra, Goursat varieties are usually called \emph{$3$-permutable varieties}, as they are characterised by the property $RSR=SRS$ for every two congruences $R$ and $S$ on any algebra $A$. Exact Goursat categories are easily seen to admit pushouts of regular epimorphisms by regular epimorphisms. Moreover, given the latter property, an exact category is Goursat if and only if it satisfies the following condition:
\begin{itemize}
\item[($\mathsf{G}$)]
\emph{For any morphism
\begin{equation}\label{Goursat}\vcenter{
\xymatrix{
R' \ar@<0.5 ex>[r] \ar@<-0.5 ex>[r] \ar[d]_r & E' \ar[d]^e \ar[r] & B' \ar[d]^b\\
R \ar@<0.5 ex>[r] \ar@<-0.5 ex>[r] & E \ar[r] & B}}
\end{equation}
of exact forks (which means that both rows consist of a regular epimorphism together with its kernel pair) with, moreover, $e$ and $b$ regular epimorphisms, one has that the right hand square is a pushout if and only if $r$ is a regular epimorphism.}
\end{itemize}
Notice that the ``if'' part is true in any category. In fact, in order to conclude that the right hand square in the diagram is a pushout it is sufficient that $r$ is an epimorphism.
Thus we have that a commutative square of regular epimorphisms in an exact Goursat category $\Ac$ is a regular epimorphism in $\Reg(\Ac)$ if and only if the induced morphism between the kernel pairs is a regular epimorphism. The regularity of $\Reg(\Ac)$ is now easily deduced from that of $\Ac$. Indeed, consider a pullback square
\begin{equation}\label{cube}\vcenter{
\xymatrix{
& P \ar[rr]^{} \ar@{.>}[dd] && A \ar[dd] \\
P' \ar[rr]^>>>>>>>>>{} \ar[dd] \ar[ru] && A' \ar[dd] \ar[ru]\\
& E \ar@{.>}[rr]^<<<<<<<<<{} && B\\
E' \ar[rr]^{} \ar@{.>}[ru] && B' \ar[ru]}}
\end{equation}
in $\Reg(\Ac)$ and assume that the bottom morphism is a regular epimorphism: a pushout of regular epimorphisms in $\Ac$. By taking kernel pairs we obtain a pullback square
\begin{equation}\label{inducedpb}\vcenter{
\xymatrix{
P'\times_P P' \ar[r] \ar[d] & A'\times_AA' \ar[d]\\
E'\times_EE' \ar[r] & B'\times_BB'}}
\end{equation}
in $\Ac$ in which, by condition ($\mathsf{G}$), the bottom morphism is a regular epimorphism. Hence, since $\Ac$ is a regular category, the top morphism is a regular epimorphism as well, and this implies, again by condition ($\mathsf{G}$), that the top side of the cube \eqref{cube} is a pushout of regular epimorphisms, as desired. In fact, it is clear that the following weaker condition is sufficient for $\Reg(\Ac)$ to be regular:
\begin{itemize}
\item[($\mathsf{G}^-$)]
\emph{For any morphism \eqref{Goursat} of exact forks with, moreover, $e$ and $b$ regular epimorphisms, the right hand square is a pushout if and only if $r$ is a pullback-stable epimorphism.}
\end{itemize}
Thus we conclude that if $\Ac$ is a regular category with pushouts of regular epimorphisms by regular epimorphisms, which satisfies the above condition ($\mathsf{G}^-$), then $\Reg(\Ac)$ is regular as well. Then, by Corollary \ref{corollary}, every regular epimorphism in $\Reg(\Ac)$ is an effective descent morphism. In particular, we find that this is the case if $\Ac$ is an exact Goursat category, a result which was already obtained in \cite{Janelidze-Sobral-Goursat}, via a different argument.
\subsection{Ideal determined categories}\label{ExIDT}
Recall from \cite{Janelidze-Marki-Tholen-Ursini} that a pointed finitely complete and finitely cocomplete regular category is \emph{ideal determined} if every regular epimorphism is normal (that is, it is the cokernel of its kernel) and if for any commutative square with $k$ and $e$ regular epimorphisms, and $\kappa$ and $\kappa'$ monomorphisms,
\[
\xymatrix{
K' \ar[r]^{\kappa'} \ar[d]_k & E' \ar[d]^e\\
K \ar[r]_{\kappa} & E}
\]
if $\kappa'$ is normal (that is, it is the kernel of its cokernel) then $\kappa$ is normal as well. (Note that a pointed variety is ideal determined if it is ideal determined in the sense of \cite{Gumm-Ursini}.) This stability condition for normal monomorphisms is easily seen to be equivalent to the following ``normalised'' version of property ($\mathsf{G}$)---see also \cite{MM-NC}:
\begin{itemize}
\item[($\mathsf{Id}$)] \emph{For any commutative diagram
\begin{equation}\label{ideal}\vcenter{
\xymatrix{
K' \ar[r] \ar[d]_k & E' \ar[r] \ar[d]^e & B'\ar[d]^b\\
K \ar[r] & E \ar[r] & B}}
\end{equation}
with (short) exact rows (which means that both rows consist of a regular epimorphism together with its kernel) with, moreover, $e$ and $b$ regular epimorphisms,
the right hand square is a pushout if and only if $k$ is a regular epimorphism.}
\end{itemize}
As with condition ($\mathsf{G}$), the ``if'' part holds more generally: it is true in any pointed category in which every regular epimorphism is normal. Also, it is sufficient that $k$ is an epimorphism in order to conclude that the right hand side square in the diagram is a pushout.
Thus we have that a commutative square of regular epimorphisms in an ideal determined category $\Ac$ is a regular epimorphism in $\Reg(\Ac)$ if and only if the restriction to the kernels is a regular epimorphism. The arguments from the exact Goursat case are then easily adapted (simply take kernels instead of kernel pairs) in order to deduce the regularity of $\Reg(\Ac)$. And again, a weaker condition suffices:
\begin{itemize}
\item[($\mathsf{Id}^-$)] \emph{For any commutative diagram \eqref{ideal} with exact rows with, moreover, $e$ and $b$ regular epimorphisms, the right hand square is a pushout if and only if $k$ is a pullback-stable epimorphism.}
\end{itemize}
Thus we see that if $\Ac$ is a pointed regular category with pushouts of regular epimorphisms by regular epimorphisms, which satisfies the above condition ($\mathsf{Id}^-$), then $\Reg(\Ac)$ is regular as well. Then, by Corollary \ref{corollary}, every regular epimorphism in $\Reg(\Ac)$ is an effective descent morphism, and the same is true for the regular epimorphisms in $\Ac$, by Theorem \ref{new}. In particular, we find that these results hold for an ideal determined category $\Ac$. Note that for an ideal determined category $\Ac$ already the result that every regular epimorphism in $\Ac$ itself is an effective descent morphism is new, as far as we know. Note also that, in this case, $\Reg(\Ac)$ is equivalent to the category of short exact sequences in $\Ac$, since every regular epimorphism is normal.
\subsection{Topological Mal'tsev algebras}
Recall that an algebraic theory $\mathbb{T}$ is a \emph{Mal'tsev} theory if it contains a ternary operation $p$ satisfying $p(x,y,y)=x$ and $p(x,x,y)=y$. The varieties $\Set^{\mathbb{T}}$ of $\mathbb{T}$-algebras for such theories $\mathbb{T}$ are exactly the congruence permutable ones---such that $RS=SR$ for any two congruences $R$ and $S$ on a same $\mathbb{T}$-algebra---and are often called \emph{Mal'tsev varieties}. Hence, every Mal'tsev variety $\Set^{\mathbb{T}}$ is Goursat. In particular, by Example \ref{ExGoursat}, $\Reg(\Set^{\mathbb{T}})$ is a regular category and every regular epimorphism in $\Reg(\Set^{\mathbb{T}})$ is an effective descent morphism, for any Mal'tsev theory $\mathbb{T}$.
Now, let us replace sets by topological spaces. More precisely, we consider categories $\Top^{\mathbb{T}}$ of topological $\mathbb{T}$-algebras for Mal'tsev theories $\mathbb{T}$. Contrary to the varieties case, the categories $\Top^{\mathbb{T}}$ are not Barr exact. However, they are well known to be regular, since regular epimorphisms are open surjections. Using the fact that the forgetful functor $\Top^{\mathbb{T}}\to \Set^{\mathbb{T}}$ preserves both limits and colimits, and reflects epimorphisms, it is then easily deduced from the Goursat condition ($\mathsf{G}$) for $\Set^{\mathbb{T}}$ that $\Top^{\mathbb{T}}$ satisfies the weaker condition ($\mathsf{G}^-$), for any Mal'tsev theory $\mathbb{T}$. Hence, by Example \ref{ExGoursat}, for a category of topological Mal'tsev algebras $\Top^{\mathbb{T}}$ the category $\Reg(\Top^{\mathbb{T}})$ is regular, and every regular epimorphism in it is an effective descent morphism.
\subsection{$n$-Fold regular epimorphisms}
For a category $\Ac$, let us put $\Reg^1\!(\Ac)=\Reg(\Ac)$ and define, inductively, for $n\geq 2$, the categories of \emph{$n$-fold regular epimorphisms} in $\Ac$ by $\Reg^n\!(\Ac)=\Reg(\Reg^{n-1}\!(\Ac))$. In any of the above considered examples, the category $\Reg^n\!(\Ac)$ is not only regular for $n=1$, but for \emph{any} $n\geq 1$. Indeed, if $\Ac$ is a regular category which admits pushouts of regular epimorphisms by regular epimorphisms and satisfies condition $(\mathsf{G}^-)$, then $\Reg(\Ac)$ has the same properties: the regularity follows from Example \ref{ExGoursat}, pushouts of regular epimorphisms by regular epimorphisms are degreewise pushouts in $\Ac$, and condition $(\mathsf{G}^-)$ follows from the corresponding condition on $\Ac$. Indeed, for the latter, recall that the ``if'' part of $(\mathsf{G}^-)$ is true in an arbitrary category, and notice that a morphism
\begin{equation}\label{square}\vcenter{
\xymatrix{
E' \ar[d] \ar[r]^{p'} & B' \ar[d]\\
E \ar[r]_p \ar[r] & B}}
\end{equation}
in $\Reg(\Ac)$ is a pullback-stable epimorphism as soon as $p'$ is a pullback-stable epimorphism in $\Ac$.
Similarly, if $\Ac$ is a regular category in which every regular epimorphism is normal, which admits pushouts of regular epimorphisms by regular epimorphisms and satisfies condition $(\mathsf{Id}^-)$, then $\Reg(\Ac)$ has the same properties. It follows, in both cases, that the category $\Reg^n\!(\Ac)$ is regular for any $n\geq 1$.
In fact, we have the following, more general, property:
\begin{proposition}\label{goesup}
If $\Ac$ is a regular category such that also $\Reg(\Ac)$ is regular, then, for any $n\geq 2$, $\Reg^n\!(\Ac)$ is regular as well.
\end{proposition}
\begin{proof}
Of course, it suffices to prove that $\Reg^2\!(\Ac)$ is regular.
First of all note that $\Reg^2\!(\Ac)$ has coequalisers of effective equivalence relations, since $\Ac$, hence also $\Reg(\Ac)$, has pushouts of regular epimorphisms by regular epimorphisms. To see that $\Reg^2\!(\Ac)$ has pullback-stable regular epimorphisms, consider the functor
\[
\dom\colon \Reg^2\!(\Ac)\to \Reg(\Ac)
\]
which sends a double regular epimorphism $(e\colon E'\to E)\to (b\colon B'\to B)$ to its domain $e$. Notice that $\dom$ preserves pullbacks. Moreover, $\dom$ preserves and reflects regular epimorphisms, as easily follows from the fact that a regular epimorphism in $\Reg^2\!(\Ac)$ is the same as a commutative cube in $\Ac$ of regular epimorphisms such that each of the sides is a pushout. Hence, the regularity of $\Reg^2\!(\Ac)$ follows from that of $\Reg(\Ac)$.
\end{proof}
Combining the above proposition and Corollary \ref{corollary}, we find that if $\Ac$ is a regular category such that $\Reg(\Ac)$ is regular as well, then we have for any $n\geq 1$ that $\Reg^n\!(\Ac)$ is regular, and that every regular epimorphism in $\Reg^n\!(\Ac)$ is an effective descent morphism.
Remark that when $\Ac$ is an exact Mal'tsev category ($\Ac$ is exact and $RS=SR$ for any two equivalence relations $R$ and $S$ on a same object of $\Ac$) then by a result in \cite{Carboni-Kelly-Pedicchio} a pushout of regular epimorphisms is the same as a \emph{double extension} (a notion from ``higher dimensional'' Galois theory---see, for instance, \cite{Janelidze:Double,EGVdL}): a commutative square \eqref{square} of regular epimorphisms, such that the factorisation $E'\to E\times_BB'$ to the pullback is a regular epimorphism as well.
Notice that Proposition \ref{goesup} together with Theorem \ref{new} provide an alternative proof for Corollary \ref{corollary}.
\section{Remarks}
\subsection{} The relations between the various conditions considered above---conditions ($\mathsf{G}$), ($\mathsf{G}^-$), ($\mathsf{Id}$), ($\mathsf{Id}^-$) and the condition that $\Reg(\Ac)$ is a regular category---are still to be better understood. However, we do know the following:
-Any of these conditions holds in any semi-abelian category \cite{Janelidze-Marki-Tholen}. Hence, we obtain as examples any variety of groups, rings, Lie algebras, and, more generally, any variety of $\Omega$-groups; the variety of loops; the variety of Heyting semi-lattices; any abelian category; \dots.
-There exist ideal determined varieties that are not $3$-permutable (see \cite{Barbour-Raftery}), hence ($\mathsf{Id}$) does not imply ($\mathsf{G}$). In particular, this means that $3$-permutability is not a necessary condition on a variety $\Ac$ in order for every regular epimorphism in $\Reg(\Ac)$ to be an effective descent morphism. This answers a question posed in \cite{Janelidze-Sobral-Goursat}.
-The examples of topological Mal'tsev algebras and of $n$-fold regular epimorphisms show that the weaker conditions ($\mathsf{G}^-$) and ($\mathsf{Id}^-$) are strictly weaker than ($\mathsf{G}$) and ($\mathsf{Id}$).
Let us also remark that we do not know, at present, any example of a regular category whose category of regular epimorphisms is regular, which does not satisfy either condition ($\mathsf{G}^-$) or ($\mathsf{Id}^-$).
\subsection{}\label{admissibility}
As mentioned earlier, Lemma \ref{George} precisely says the following, for $\Ac$ a regular category such that $\Reg(\Ac)$ is regular as well:
\emph{For any morphism $p\colon E\to B$ in $\Ac$, the ``change of base'' functor $p^*\colon (\Ac\downarrow B)\to (\Ac\downarrow E)$ preserves coequalisers of reflexive graphs.}
Here's another way of reformulating this same property. Recall from \cite{CHK} that a reflector $I\colon \Ac\to\Xc$ into a full subcategory $\Xc$ of a category $\Ac$ is \emph{semi-left exact} if it preserves any pullback square
\[
\xymatrix{
P \ar@{}[rd]|<<{\pullback}\ar[r] \ar[d] & X \ar[d]\\
B \ar[r]_-{\eta_B} & I(B)}
\]
of a unit $\eta_B$ along a morphism in $\Xc$. Clearly, this means that, for every such pullback square, the morphism $P\to X$ coincides, up to isomorphism, with the unit $\eta_P\colon P\to I(P)$. Note that semi-left exactness is the same as \emph{admissibility} in the sense of categorical Galois theory (see \cite{CJKP}).
Now, let $\Ac$ be a regular category such that $\Reg(\Ac)$ is regular as well, and $\RG(\Ac)$ the category of reflexive graphs in $\Ac$.
If we assume, moreover, that $\Ac$ admits coequalisers of reflexive graphs, then Lemma \ref{George} can also be reformulated as follows:
\emph{The functor $\pi_0\colon \RG(\Ac)\to \Ac$, which sends a reflexive graph to its coequaliser, is semi-left exact.}
\subsection{}
For an object $B$ of a category $\Ac$, write $\Pt(B)$ for the category defined as follows: an object of $\Pt(B)$ is a triple $(A,f,s)$, where $A$ is an object of $\Ac$ and $f\colon A\to B$ and $s\colon B\to A$ are morphisms in $\Ac$ such that $f\circ s=1_B$; a morphism $(A,f,s)\to (C,g,t)$ in $\Pt(B)$ is a morphism $h\colon A\to C$ in $\Ac$ such that $g\circ h=f$ and $h\circ s=t$.
When $\Ac$ has pullbacks of split epimorphisms, any morphism $p\colon E\to B$ in $\Ac$ induces a ``change of base'' functor $p^*\colon \Pt(B)\to \Pt(E)$ given by pulling back along $p$. If, moreover, $\Ac$ admits pushouts of split monomorphisms, then any such $p^*$ has a left adjoint (see \cite{Bourn-Janelidze:Semidirect}). Recall from \cite{Bourn1991} that $\Ac$ is called \emph{protomodular} if $p^*$ reflects isomorphisms for every morphism $p$.
Now, let $\Ac$ be a regular category such that $\Reg(\Ac)$ is regular as well, and assume that $\Ac$ admits coequalisers of reflexive graphs. Then, for any object $B$ of $\Ac$, coequalisers of reflexive graphs in $\Pt(B)$ are necessarily coequalisers in $\Ac$, and we can use Lemma \ref{George} to prove that they are preserved by $p^*\colon \Pt(B)\to \Pt(E)$, for every morphism $p\colon E\to B$ in $\Ac$. Hence, using the ``reflexive form'' of Beck's Monadicity Theorem we find, for any regular protomodular category $\Ac$ with coequalisers of reflexive graphs and pushouts of split monomorphisms such that $\Reg(\Ac)$ is regular as well, that the functor $p^*\colon \Pt(B)\to \Pt(E)$ is monadic for any morphism $p\colon E\to B$ in $\Ac$. This means, according to \cite{Bourn-Janelidze:Semidirect}, that $\Ac$ is a \emph{category with semidirect products}.
\subsection{}
Let $\Ac$ be a regular category and $\Mon(\Ac)$ the full subcategory of the arrow category $\Ac^2$ with as objects all monomorphisms in $\Ac$. Like $\Ac$, the category $\Mon(\Ac)$ is finitely complete: limits in $\Mon(\Ac)$ are degreewise limits in $\Ac$. Colimits, if they exist, are given by the mono part of the regular epi-mono factorisation of the degreewise colimit. A regular epimorphism in $\Mon(\Ac)$ is the same as a degreewise regular epimorphism in $\Ac$. In particular, we have that $\Mon(\Ac)$ is a regular category.
Now let $\cod\colon \Mon(\Ac)\to\Ac$ be the functor which sends a monomorphism $a\colon A'\to A$ to its codomain $A$. Then $\cod$ preserves pullbacks, pushouts and regular epi-mono factorisations, and reflects pushouts of regular epimorphisms in the following sense: any commutative square of regular epimorphisms in $\Mon(\Ac)$ that is sent to a pushout in $\Ac$ is necessarily a pushout itself. It follows that whenever $\Reg(\Ac)$ is regular, $\Reg(\Mon(\Ac))$ is regular as well. Hence, categories of monomorphisms provide another class of examples of regular categories whose category of regular epimorphisms is regular as well. In particular, we find that $\Reg(\Mon(\Ac))$ is regular for any semi-abelian category $\Ac$. In this case $\Mon(\Ac)$ is also protomodular and finitely complete, so that by the previous remark $\Mon(\Ac)$ admits semi-direct products, a result which was already obtained in \cite{Ferreira-Sobral}.
\subsection{} Let $\Ac$ be a regular category. By an \emph{extension} we simply mean a regular epimorphism in $\Ac$; a \emph{double extension} is, as recalled above, a commutative square of extensions such that the induced factorisation to the pullback is an extension as well. Let us write $\Ext(\Ac)$ and $\Ext^2\!(\Ac)$ for the categories of extensions and of double extensions, respectively (where we have that $\Ext(\Ac)=\Reg(\Ac)$, of course). One can then, inductively, define \emph{$n$-fold extensions} for $n\geq 3$, as those commutative squares of ($n-1$)-fold extensions such that the induced factorisation to the pullback (in the category $\Ext^{n-2}\!(\Ac)$) is an ($n-1$)-fold extension. It was noted in \cite{EGVdL} that every $n$-fold extension (for $n\geq 2$) is an effective descent morphism in $\Ext^{n-1}\!(\Ac)$, basically because pullbacks along $n$-fold extensions are computed degreewise in the exact category $\Ac$.
As mentioned earlier, if $\Ac$ is an exact Mal'tsev category, then a regular epimorphism in $\Ext(\Ac)=\Reg(\Ac)$ is the same as a double extension, by a result in \cite{Carboni-Kelly-Pedicchio}. In fact, it is shown in \cite{Carboni-Kelly-Pedicchio} that this property characterises the exact Mal'tsev categories $\Ac$ among the regular ones. Hence, we have, in $\Ac$, that the effective descent morphisms are exactly the extensions, since $\Ac$ is exact, and, in $\Ext(\Ac)$, exactly the double extensions, by Theorem \ref{maintheorem} and the result in \cite{Carboni-Kelly-Pedicchio}. This might lead one to expect that the effective descent morphisms in $\Ext^2(\Ac)$ are exactly the three-fold extensions. However, this turns out not to be the case, in general, and happens only when the category $\Ac$ is \emph{arithmetical} in the sense of \cite{Pedicchio2} (for instance, when $\Ac$ is the variety of Boolean rings, or of Heyting algebras). Indeed, by Theorem \ref{maintheorem}, the effective descent morphisms in $\Ext^2(\Ac)$ are exactly the pushout squares of regular epimorphisms in $\Ext(\Ac)$, so that, by the above mentioned result in \cite{Carboni-Kelly-Pedicchio}, we would have that $\Ext(\Ac)$ is exact Mal'tsev which, by a result in \cite{Bourn2001b} is the case if and only if $\Ac$ is an arithmetical category.
\\ \\
\emph{Acknowledgement.} Many thanks to George Janelidze for pointing out that in the assumptions of Theorem \ref{maintheorem} the ``almost exactness'' of the category $\Ac$ is superfluous.
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\providecommand{\MRhref}[2]{
\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2} | 13,450 |
Online COP26 Climate Justice Journalism Fellowship
This fall 20 young journalists will get a chance to report virtually on the UN Climate Talks. Young journalists interested in climate change reporting are invited to apply for a month-long virtual Climate Justice Journalism Fellowship. Participants will receive a €250 stipend and take part in various workshops. The deadline is on Monday 20th September 2021.
Copy and paste this URL into your WordPress site to embed
Copy and paste this code into your site to embed | 82,383 |
Hamilton168 (1 review written)
We have owned a red Francis Francis X1 since 2003. I use it for making ground coffee. Except for an internal leak repaired for a modest fee by a local shop we have had few problems and the coffee is excellent with good crema. I should say that the red enamel has flaked off around the edges of the drip tray where coffee acids accumulate, and the thermometer has ceased to function, although the thermostat still works. In the event that the machine finally gives up the ghost I would seriously… Read Full Review
Written… Read Full Review
Written on: 19/01/2012 by geordie42 (1 review written)
I have had a number of coffee machines over the years inclduing 3 kenwood retro which all were good machines made good latte's espresso... but largely died after 18 or so months. So i thought i'll spend 3x as much for a better quality machine that will hopefully last x3 as long. I'm not sure I could argue this machine is even close to as good as the Kenwood. The bad points are numerous and to give you a few: The machine is ~ 8 months old i make 4-5 coffees per day and it now leaks like no ones… Read Full Review
Written on: 28/03/2010 by Gill 111 (1 review written)
This is a great machine. I have owned three espresso makers before, and this is the absolute best one. It is being said that it is expensive, due to its looks. I would say that if you compare to machines at the same price level, this one does as well compete on quality… Read Full Review
Written on: 22/12/2009 by Treesy (1 review written)
I have owned this machine for almost 2 years and it is the best coffee maker I have ever had. Yes, it's expensive, but it consistently produces excellent coffee. I have tried numerous cheaper machines and none heat the water to a sufficient temperature to extract the full flavour of the coffee. The machine is heavy but for the job it is doing, it needs to be. so far I have had absolutely no problems with it and I love the retro look of it. My only critisism is that it is quite difficult to… Read Full Review
Written on: 20/11/2009
I bought the Francis Francis! X1 around 7 months ago, and as yet I haven't managed to force a decent cup of coffee out of it. I have tried the Illy beans which are recommended for it, plus many others bought from individual roasters. I have tested these identical beans for flavour in a caffetiere, with perfect results. I am well used to coffee machines, so I'm fairly confident that the problem is with the machine and not with me!
The machine is difficult to use - it is heavy and you have to… Read Full Review
Written on: 05/07/2009 by rosindale19 (1 review written)
I have owned this FrancisFrancis! X1 for 2 years and had no problems. Good with a grind, better for the E.S.E pods. Impressive retro style look… Read Full Review
Written on: 09/05/2009
I had the FrancisFrancis X1 for nine months when it developed a fault in the steam arm and a large piece of enamel flaked off the drawer front. EuroFoodBrands of Sywell, Northampstonshire are the UK importers and the machine had to be sent back to them under the terms of the 2 year warranty. They offered to collect but only on the understanding that the machine would not be insured during transit and if lost or damaged in transit this would be my responsibility. I was not happy about this and… Read Full Review
Written on: 15/10/2008
I have had many coffee machines in my life, and got the FrancisFrancis! X1 as a birthday present. In my opinion it under performs those a third of its price. It just doesn't work properly and mine has also just broken after only 6 months. The seal no longer works, so water spills over the bit that holds the coffee instead of forcing it through at high pressure. Coffe now full of grains and weak. Very dissapointed.Do not buy this machine… Read Full Review
Written on: 20/07/2008
We used to have a Kenwood retro espresso machine and it started rusting under the coating as well as the pump breaking within one year. The francis francis x1 is powder coated which seems more durable. It is very quick with reheating between cups of coffee and cools down quickly from steam function to coffee. It does leak a little from the steam spout when brewing coffee but I keep a glass under it… Read Full Review
Written on: 28/04/2008 by biff480 (1 review written)
I've had my stylish FrancisFrancis! X1 for over two years now and it's never given a moment of trouble. Since I've had it I've given up all but espresso, because it makes it so well. It was, I believe, the first machine developed to use coffee pods and it makes a great espresso with them. However, though can also get a normal coffee holder so that you can put your own grind into it. Heats up very fast and makes great coffee every time, strong but mellow, with great crema… Read Full Review
Written on: 19/02/2009
This is for the person who needed instructions to make an espresso on the Francis X1. You begin by filling it with water, putting all switches in the up position and plugging it in. Next, put the left switch down and wait for the green light to go on. In the mean time, put coffee in the portafilta and put it on the machine. You can also get your cup ready and if your going to froth milk, get that ready. When the green light goes on, have your cup ready and put the second switch from the left down. Once you have the amount of coffee you want in the cup, put the same switch back up. To froth, leave the first switch down, the second up and put the third down. When the green light goes on again open the steam on the right and froth until your heart's content. When you're done, turn the steam off and put the switch back up. To turn off the machine all switches should be up. That's about it.
As rated by real users
" I couldn't live without it "
Written by Natasha_Shy
" Durable machine that makes excellent coffee "
" Our experience identical to above. we had an earlier AEG model tha... "
" Excellent Coffee Products. "
" A kávéfőzők királya! "
433693_Bearbug's Response to biff480's Review
Written on: 20/01/2009
I've got the instructions to this and nowhere does it actually say how to use it. Can anyone give a basic rundown? I'm just trying to make an espresso... all switches down in front.. mine could be broken. | 49,753 |
Shares of OLED specialist Universal Display (NASDAQ: OLED ) skyrocketed nearly 16% during Thursday's trading after the company reported its fourth-quarter and full-year 2012 earnings.
So why the optimism?
Though Universal Display's quarterly earnings per share of $0.12 slightly missed analyst estimates that called for $0.13, its revenue beat expectations of $26.4 million after rising 51% from the year-ago period to $28.1 million, reassuring analysts that last quarter's miss was likely a short-term issue.
Inventory also rose to $11 million by the end of the year, up from $9.5 million in the third quarter. When we remember last quarter's barrage of inventory questions from concerned analysts, perhaps its unsurprising that CFO Sid Rosenblatt was quick to point out the company considers "the buildup of inventory an excellent investment of our cash because it improves our ability to quickly fill customer orders and significantly lands our customer goodwill."
In addition, making good on its previously announced (link opens PDF) share repurchase plan, Universal Display used $5.2 million of its available $50 million to repurchase 206,000 outstanding shares. As I suggested in November, it's good to see the company increasing shareholders' slice of the pie after it incurred significant dilution in 2011 to raise capital for acquiring Fuji's portfolio of 1,200 OLED patents.
Fortunately, the Fuji acquisition cost much less than management originally expected, which left the company flush with cash and nowhere to use the excess. Now, with its future prospects looking bright and with shares currently trading significantly below 2011's $46-per-share offering price, this buyback is increasingly looking like a great long-term move.
A guiding light
With regard to guidance, Universal Display was cautious to note OLED adoption is still in its early stages, "where many variables can have a material effect on growth." Even still, the company expects to see $110 to $125 million in 2013 revenue, depending on a few factors including how long it takes Samsung to add production capacity and introduce new OLED products, how quickly those products incorporate Universal's green technology -- remembering this was the concern voiced by Piper Jaffray analysts a few days ago -- and how fast the young OLED television and lighting markets grow.
Even though guidance represents a solid revenue increase ranging from 32% to 50% over 2012, analysts at Cowen wasted no time voicing their approval and saying the numbers were probably conservative given the industry's wider growth potential and Universal Display's collaboration with its Asian supply chain -- a likely reference to Universal Display's cooperative deal with manufacturing specialist Duksan Hi-Metal announced last September.
Considering the pop in shares of Universal Display today, it seems the rest of the market agrees with Cowen, especially when we consider management's comments that electronics giants Samsung and LG (NYSE: LPL ) will spend 73% and 54% of their 2013 capital expenditure budgets on OLED development, respectively. AU Optronics is also reportedly set to mass-produce OLED displays for cell phones in the first half of this year, and Sony and Panasonic joined forces last year to develop their own large-screen OLED televisions.
For its part, however, LG increasingly appears to be the next huge proponent of Universal Display's technology; part of its spending will include $655 million to build a new OLED TV production line with a monthly capacity of 156,000 55-inch screens, with plans to begin mass production by the first half of next year. Additionally, as I mentioned a few days ago, LG is also pushing hard to commercialize both rigid and flexible OLED lighting solutions which, after OLED television achieves mass adoption, will likely be Universal Display's next big catalyst as other manufacturers continue to move forward with the technology.
Out of the lab, into the fray
Apart from smartphones, television, and lighting, Universal Display CEO Steve Abramson also reminded investors of another potential growth area for the company: encapsulation technology.
Though Universal Display actually unveiled its novel encapsulation techniques in 2011, as adoption of flexible displays and lighting continues to increase, so, too, will the necessity for manufacturers to find effective encapsulation methods for sealing sensitive display materials from environmental elements. As a result, encapsulation could become a substantial source of revenue for Universal Display down the road. According to Abramson during the conference call, the company is currently "discussing potential commercialization roadmaps with a number of partners and [is] in the process of developing [its] encapsulation technology business model."
What now?
Patience is key to Foolishly investing in a high-growth company like Universal Display. Though Samsung remains its single largest customer, the rest of the tech world is finally beginning to take note of the advantages of using OLED. When that happens, and if long-term shareholders can keep their wits about them amid the volatility, I'm convinced long-term shareholders of Universal Display will be handsomely rewarded in due time.. | 170,917 |
CHLAMYDIA GONORRHOEA TEST KIT FACTS
CHLAMYDIA GONORRHOEA TEST KIT FACTS
About the chlamydia gonorrhoea test kit
Our chlamydia gonorrhoea home test is an accurate laboratory test for genital chlamydia and genital gonorrhoea. The test for men requires you to collect a urine sample.
You should wait for at least two weeks after you think you were exposed to chlamydia or gonorrhoea before doing this test. That way the test is most likely to detect these infections.
If you think you could have got chlamydia or gonorrhoea within the last two weeks, you can buy the test now but don’t collect and return your sample until two weeks have gone by.
How does the test work?
Use the kit we send you to collect a urine sample. Send the sample to our partner laboratory. When the lab sends us your test result we’ll share it with you in your secure account.
How do I take a urine sample?
Follow our simple instructions to pee into the pot we provide. Put it in the packaging we include and send it to our partner laboratory.
Is the test kit accurate?
Yes. The test that the laboratory runs is the same test that’s routinely run in the NHS.
If you got chlamydia or gonorrhoea in the last two weeks, the test is unlikely to detect your infection.
If you start to show symptoms before you take a test or before you get your results you must visit your GP or health clinic.
What happens if I test positive?
If you test positive just for chlamydia, we can offer you treatment through the Boots Online Doctor Service..
What are the symptoms?
Many men with genital chlamydia or gonorrhoea don’t have symptoms. If you do, they are often similar for all three infections.
Symptoms include unusual discharge from the penis or pain or a burning sensation when peeing.
What happens if I already have symptoms?
If you’ve got any symptoms of an STI, | 227,499 |
Oh shitting hell, this amazing day just got better: our buddies over at Lazy Oaf revealed the two exclusive shoes they're releasing in collaboration with ever-epic brand Jeffrey Campbell. As you'd predict, they're not for the faint-hearted but for the typically bold Lazy Oaf girl in all of us and range in price from £90 to £150.
As with the first batch of their Underground creepers (which sold out in record time) these are out online only in very limited numbers. So if you just got your results back and need a congratulatory gift, you better get asking/demanding (and if you're just rich then here's to owning another pair of great shoes). The LO online store is also stocking other Jeffrey Campbells which you can click over to see here.
Go forth and obliterate your bank accounts, etc.
1 comment:
Ohhhmygod, I love you for posting this. | 299,264 |
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TIME: 13:00 – 14:00, 8th May 2017
VENUE: The Henry Jackson Society, 26th Floor
Millbank Tower, 21-24 Millbank, London, SW1P 4QP
SPEAKER
Charles Clover
Journalist and author of Black Wind, White Snow
For a summary of this event click here
Black Wind, White Snow: The Rise of Russia’s New Nationalism provides a fascinating study of the root motivations behind the political activities and philosophies of Putin’s government. The book strives to offer an explanation of the rationale behind Russia’s geopolitical actions and ambitions and delves into the largely unknown Russian philosophy – Eurasianism – that underpins seismic current events in Russia’s foreign policy.
The Henry Jackson Society is delighted to invite you to an event with Charles Clover, journalist and author who will discuss the ideological underpinnings of President Putin’s Russia. His new book traces the rise of Eurasianism – an obscure theory of Russian national identity based on ethnicity and geography – from its roots in the writings of White Russian exiles in 1920s Europe, through the Gulag, and to the dissolution of the Soviet Union in 1991. Today, Eurasianism is represented in the actions of Russia’s ruling elite, from the annexation of Crimea to the rise of anti-Western paranoia and imperialist rhetoric. Based on extensive research and interviews with close advisors of President Putin, Clover will guide you through this fascinating story of how Russia became what it is and where it is going.
*We will be selling copies of Charles Clover’s Black Wind, White Snow on the day for the special price of £10 (RRP £12.99) cash only. Please bring the correct amount of change.*
Charles Clover currently writes from Beijing for the Financial Times having previously worked as the Moscow bureau chief for the FT from 2008 to 2013. He is an award-winning journalist, achieving the Foreign Reporter of the year award from the British Press Awards in 2011. | 216,166 |
Burns Night Reading and Tasks
Yesterday, a special day was celebrated in Scotland, as it is on January 25th every year.
Burns Night is celebrated in Scotland to commemorate one of the country's most well-known poets. To go alongside this, today's reading comprehension is based on Burns Night and some of the traditions that take place.
Below, there is also a tartan design sheet so that you can create your own tartan pattern. Perhaps you could research your family name to see if you already have a tartan design! There is also a traditional shortbread recipe to have a go at baking at home. Remember, it is written in law that any baking that takes place at home must be sampled by Mr Hyde.
Have a bonnie time!! | 325,367 |
Got great photos of Lewiston & want to share your civic pride?
LEWISTON: Got great photos of Lewiston & want to share your civic pride?
In preparation for a new more user-friendly municipal web site, the City of Lewiston's Web Site Steering Committee is seeking professional & amateur photos of various aspects of Lewiston. The deadline for submitting photos is March 3, 2011.
It is expected that 15-20 photos will be selected by the Steering Committee and then submitted to CivicPlus, the web firm that Lewiston is working with to design & build the new site. Photos that are utilized for the site will have online photo credits.
Pictures submitted must be at least 480 x 640 pixels in size, and the pictures should be 72 dpi, preferably in jpg format.
Those who wish to submit a photo by March 3rd may mail or deliver them to the Web Site Steering Committee, c/o Dottie, City Administrator's Office, 27 Pine Street, Lewiston, ME, 04240. | 34,431 |
Alaafin Of Oyo Gifts Wife, Omowumi A New House
Goals �: One of Alaafin of Oyo, Lamidi Adeyemi’s Queens, Queen Omowumi, pours encomium on him, as she acquires a new home
Shared with Caption ….. | 120,340 |
Oct last …
November 18, 2010 at 11:00 am
468492 Commentshttp%3A%2F%2Fherit.ag%2F1fqskVtBoomers+Age+of+Appetites+Leaves+Americans+with+Tough+Policy+and+Personal+Choices2010-11-18+16%3A00%3A16Chuck+Donovanhttp%3A%2F%2F
“The Greatest Generation … stormed beaches in places like Normandy and Okinawa,” says today’s lead editorial in USA Today. “Their children, by contrast, stormed places like Woodstock. For the Baby Boomers — people born from 1946 to 1964 — the prosperity their parents built was never good enough. In later … … More
July 7, 2008 at 5:07 pm
9452 Commentshttp%3A%2F%2Fherit.ag%2FGPK2TOA+Funny+Bit+of+Entitlements+Nonsense+from+the+U.S.+Navy2008-07-07+21%3A07%3A36J.D.+Foster%2C+Ph.D.http%3A%2F%2F | 25,594 |
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ifm electronicProduction, Manufacturing Senior Specialist / Project Manager
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PKS Software GmbHInformation Technology, Telecommunications Specialist
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Coperion Waeschle GmbH & Co. KGProcurement, Materials, Logistics Senior Specialist / Project Manager
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We're sorry, Geo Mug is no longer available Why not try one of these instead? Personalised 'Great Beard' Man Mug by Oakdene Designs £12.50$21.45€14.98$16.01 'Never Known Without My Phone' Mug by Bespoke Verse £12$20.59€14.38$15.37 Sawdust Is Man Glitter Mug by Bread & Jam £9.50$16.30€11.38$12.17 ‘Great Beard' Man Mug by Oakdene Designs £10.75$18.44€12.88$13.77 Pinch to zoom Geo Mug by Joanna Corney £10$17.16€11.98$12.81 This item is no longer available free gift wrap Favourite Favourite View your favourites we're sorry we're sorry, This item is no longer available buy a gift voucher related categories home > dining room > tableware > mugs by Joanna Corney Location: East Sussex, GB read seller's profile see seller's complete range more items from this seller Geo Tea Towel by Joanna Corney £11$18.87€13.18$14.09 Geo Tea Cosy by Joanna Corney £25$42.90€29.96$32.03 Sunbeam Cushion by Joanna Corney £45$77.22€53.93$57.66 Starlings Lampshade by Joanna Corney £45$77.22€53.93$57.66 Beach Hut Wallpaper by Joanna Corney £95$163.03€113.85$121.73 Walk Along The Laines Print by Joanna Corney £22$37.75€26.36$28.19 view all you may also like PRODUCT DETAILS DELIVERY RETURNS Made in Britain free gift wrap ask seller a question save for later A glossy white ceramic mug, featuring geometric motifs.59 This product is dispatched to you by Joanna Cor Please see notonthehighstreet.com's returns policy.
keep in touch | 154,065 |
TITLE: Find eigenvalues of the matrix
QUESTION [6 upvotes]: Find eigenvalues of the $(n+1) \times (n+1)$-matrix
$$ \left( \begin {array}{cccccccc} 0&0&0&0&0&0&-n&0\\ 0
&0&0&0&0&-(n-1)&0&1\\ 0&0&0&0&-(n-2)&0&2&0
\\ 0&0&0&\ldots&0&3&0&0\\ 0&0&-3&0&\ldots&0
&0&0\\ 0&-2&0&n-2&0&0&0&0\\ -1&0&n-1&0
&0&0&0&0\\ 0&n&0&0&0&0&0&0\end {array} \right)
$$
I have tried for some small $n$ by hand and have got that for $n=3$ the eigenvalues are $1,-1,3,-3$, for $n=4$ eigenvalues are $0,2,-2,4,-4$. So I am conjectured than for arbitrary $n$ the eigenvalues are $n, n-2, n-4, \ldots, -n+2,-n.$
How to prove it?
REPLY [3 votes]: Call your matrix $B_{n+1}$. Denote the Kac matrix of the same size by
$$
A_{n+1}=\pmatrix{
0&n\\
1&0&n-1\\
&2&\ddots&\ddots\\
& &\ddots&\ddots&\ddots\\
& & &\ddots&0 &1\\
& & & & n &0}.
$$
It is known that the spectrum of the Kac matrix is given by
$$
\sigma(A_{n+1})=\{-n,\,-n+2,\,-n+4,\ldots,\,n-4,\,n-2,\,n\}.
$$
We shall prove that $\sigma(A_{n+1})=\sigma(B_{n+1})$. Consequently (as the Kac matrix has distinct eigenvalues) the two matrices are similar to each other.
Here is our proof. First, $B_{n+1}^2$ is similar to $A_{n+1}^2$. In fact, if $D$ denotes the $(n+1)\times(n+1)$ diagonal matrix such that $d_{kk}=(-1)^{\lfloor(k-1)/2\rfloor}$ (i.e. $D=\operatorname{diag}(1,1,-1,-1,1,1,-1,-1,\ldots)$), a straightforward computation will show that $DB_{n+1}^2D=A_{n+1}^2$.
Thus $\sigma(B_{n+1}^2)=\sigma(A_{n+1}^2)$. Yet, as pointed out in the answer by Jean Marie, $B_{n+1}$ is similar to $-B_{n+1}$ via the reversal matrix (the mirror image of the identity matrix from left to right). Therefore, nonzero eigenvalues of $B_{n+1}$ must occur in pairs of the form $\pm\lambda$. Hence we have $\sigma(B_{n+1})=\sigma(A_{n+1})$, because the multiplicity of each nonzero eigenvalue of $A_{n+1}^2$ is exactly $2$. | 53,217 |
TITLE: Let $M$ be a smooth manifold and let $N$ be a manifold with boundary. If $dF_{p}$ is an isomorphism, then $F\left(p\right)\in\mbox{int}M $.
QUESTION [3 upvotes]: Let $M$ be a smooth manifold and let $N$ be a manifold with boundary, both with the same dimension $n$. If $dF_{p}$ is an isomorphism, then $F\left(p\right)\in\mbox{int}N $.
I am trying to prove this theorem to prove a result about smooth embeddings. Here is how I am thinking the problem could be solved. Assume thet $F(p)$ is a boundary point of $N$. Then there exists a chart $(V,\psi)$ at $F(p)$ such that $\psi(V) $ is an open subset of the upper half space $\mathbb{H}^{n}$. I guess we have to use some fact about $M$ having no boundary and the the fact that $dF_p$ is an isomorphism to show that there is a contradiction, but I cannot figure that out.
REPLY [5 votes]: By passing to charts near $p$ and $F(p)$, it suffices to prove this in the case $M = \mathbb R^n,N = H^n$. Since $dF_p$ is an isomorphism, by the inverse function theorem (for $\mathbb R^n$, thinking of $H^n$ as a subset of $\mathbb R^n$) we can find an open set $U \ni p$ such that $F(U)$ is open (in $\mathbb R^n$!) and $F : U \to F(U)$ is a diffeomorphism. Since $F(p)\in F(U)$, this rules out $F(p) \in \partial H^n$, because no $\mathbb R^n$-neighbourhood of a boundary point is contained in $H^n$. | 22,597 |
SCSI: What is an Expander?.
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ParalanStore Sells converters and bridges for SCSI and iSCSI, Seagate drives, controllers, host bus adapters, and SCSI cables. | 137,359 |
\begin{appendix}
\subsection{Sketch of Proof of Proposition~\ref{proposition:meanUAVtxPower} \newline(see \cite{azari2019uav} for more detailed derivations)} \label{proof:meanUAVtxPower}
The mean UAV transmit power can be written as
\begin{align} \label{eqn:meanPowerProof}
\mathbb{E}[\pu] = \sum_{\nu \in \{\mathrm{L},\mathrm{N}\}}\int_0^\rMuu f_{\Ru}^\nu(\ru) \mathbb{E}\left[\pu^\nu|\Ru = \ru\right] \mathrm{d}\ru,
\end{align}
where $f_{\Ru}^\nu(\ru) = f_{\Ru}(\ru) \cdot \pruu^\nu(\ru)$ and where the integral in (\ref{eqn:meanPowerProof}) can be written as
\begin{equation} \label{eqn:meanPowerIntegrals}
\begin{aligned}
&\int_0^\rMuu f_{\Ru}^\nu(\ru) \mathbb{E}\left[\pu^\nu|\Ru = \ru\right] \mathrm{d}\ru \\
&\!=\!\int_0^\mathrm{r_m^\nu} \!f_{\Ru}^\nu(\ru) \mathbb{E}[\rho_\u \zeta_{\u\u}^{\eu}] \, \mathrm{d}\ru \!+\! \int_\mathrm{r_m^\nu}^\rMuu \!f_{\Ru}^\nu(\ru) \mathbb{E}[\pumax] \mathrm{d}\ru. \!
\end{aligned}
\end{equation}
The first integral on the right-hand side of (\ref{eqn:meanPowerIntegrals}) is equal to
\begin{equation}
\begin{aligned}
&\int_0^\mathrm{r_m^\nu} \!f_{\Ru}^\nu(\ru) \mathbb{E}[\rho_\u \zeta_{\u\u}^{\eu}] \, \mathrm{d}\ru \\
&=\sum_{i=1}^{j} c_i \int_{\r_i}^\mathrm{r_{i+1}} \ru^{1+\auu^\nu \eu } \cdot e^{-\r_\u^2/(2\sigma_\mathrm{u}^2)} \mathrm{d}\ru
\end{aligned}
\end{equation}
where
\begin{align}
c_i &= \frac{\pur\left(\hat{\tau}_{\mathrm{uu}}^\nu/\guu\right)^{\eu}}{\sigma_\u^2[1-\e^{-r_\mathrm{M}^2/(2\sigma_\mathrm{u}^2)}]} \cdot \pruu^\nu(r_i).
\end{align}
With the change of variable $y = \r_\u^2/2\sigma_\mathrm{u}^2$, we can write
\begin{equation}
\begin{aligned}
&c_i^\nu \int_{\r_i}^\mathrm{r_{i+1}} \ru^{1+\auu^\nu \eu } \cdot e^{-\r_\u^2/(2\sigma_\mathrm{u}^2)} \mathrm{d}\ru \\
& \!= C_i^\nu \Big[ \gamma(1\!+\!\auu^\nu \eu /2,y_{i+1}) \!-\! \gamma(1\!+\!\auu^\nu \eu /2,y_{i}) \Big] \label{eqn:meanPowerFirstIntegral}
\end{aligned}
\end{equation}
where $y_i = \frac{r_i^2}{2\sigma_u^2}$ and
\begin{align}
C_i^\nu &= \frac{(2\sigma_\u^2)^{{\auu^\nu\eu/2}}\pur\left(\hat{\tau}_{\mathrm{uu}}^\nu/\guu\right)^{\eu}}{1-\e^{-r_\mathrm{M}^2/(2\sigma_\mathrm{u}^2)}} \cdot \pruu^\nu(r_i);~i>0,
\end{align}
thus obtaining
\begin{equation} \label{eqn:meanPowerFirstIntegralResult}
\begin{aligned}
&\int_0^\mathrm{r_m^\nu} f_{\Ru}^\nu(\ru) \mathbb{E}[\rho_\u \zeta_{\u\u}^{\eu}] \, \mathrm{d}\ru \\
& = \sum_{i=1}^{j} [C_i^\nu-C_{i+1}^\nu] \, \gamma(1+\auu^\nu \eu k/2,y_{i+1})
\end{aligned}
\end{equation}
where $C_{j+1}^\nu = 0$.
Similarly, the second integral on the right-hand side of (\ref{eqn:meanPowerIntegrals}) is equal to
\begin{equation} \label{eqn:meanPowerSecondIntegral}
\int_\mathrm{r_m^\nu}^\rMuu f_{\Ru}^\nu(\ru) \mathbb{E}[\pumax] \mathrm{d}\ru = \sum_{i=j+1}^{k+1} [B_i^\nu-B_{i-1}^\nu] \, e^{-r_i^2/(2\sigma_\mathrm{u}^2)}
\end{equation}
where $B_j^\nu = 0$, $B_{k+1}^\nu = 0$, and
\begin{align}
B_i^\nu &= \frac{\pumax \, \pruu^\nu(r_i) }{1-\e^{-r_\mathrm{M}^2/(2\sigma_\mathrm{u}^2)}};~i>j.
\end{align}
Proposition~\ref{proposition:meanUAVtxPower} then follows from substituting (\ref{eqn:meanPowerFirstIntegralResult}) and (\ref{eqn:meanPowerSecondIntegral}) into (\ref{eqn:meanPowerIntegrals}) and then into (\ref{eqn:meanPowerProof}).
\subsection{Sketch of Proof of Theorem~\ref{proposition:U2Ucoverage} \newline(see \cite{azari2019uav} for more detailed derivations)} \label{proof:U2Ucoverage}
From Approximation~2, we have $\mathcal{C}_{\mathrm{u}|\Ru}^\N(\ru) = 0$, thus
\begin{equation} \label{eq:U2ULinkCoverage}
\begin{aligned}
\pcovuav &= \sum_{\nu \in \{\mathrm{L},\mathrm{N}\}}\int_0^{\rMuu} \mathcal{C}_{\mathrm{u}|\Ru}^\nu(\ru)\,f_{\Ru}^\nu(\ru) \,\mathrm{d}\ru \\
&= \int_0^\rMuu f_{\Ru}^\L(\ru) \mathcal{C}_{\mathrm{u}|\Ru}^\L(\ru) \mathrm{d}\ru,
\end{aligned}
\end{equation}
where by using Approximation 1 we can write
\begin{equation} \label{CuRu}
\begin{aligned}
\mathcal{C}_{\mathrm{u}|\Ru}^\L(\ru) &\!=\! \mathbb{E}_{I_\mathrm{u}} \!\left\{\mathbb{P}\left[\suu^\L > \frac{\t}{\pu^\L \zuu^\L(\ru)^{-1}}(\mathrm{N_0} \!+\! I_\mathrm{u}) \right] \right\} \\
&\!\!= \sum_{i=1}^{\mathrm{m_{uu}^\L}} \binom{\mathrm{m_{uu}^\L}}{i}(-1)^{i+1} e^{-z_{\u,i}^\L \mathrm{N_0}} \cdot \lapiu^\L(z_{\u,i}^\L).
\end{aligned}
\end{equation}
Under Approximation~2, we can neglect the interference generated by NLoS links and obtain
\begin{equation} \label{eqn:LaplacianProofTheoremOne}
\lapiu^\L(z_{\u,i}^\L) = e^{ -2 \pi (\hat{\lambda}_\u \mathcal{I}_\mathrm{uu}^\L + \lamb \mathcal{I}_\mathrm{gu}^\L)}.
\end{equation}
Theorem~\ref{proposition:U2Ucoverage} then follows by deriving $\mathcal{I}_\mathrm{uu}^\L$ and $\mathcal{I}_\mathrm{gu}^\L$ (with an approach similar to the one in \cite{AzaGerGar19}), by replacing $\pu$ with its mean (from Proposition~\ref{proposition:meanUAVtxPower}), and by substituting $\mathcal{I}_\mathrm{uu}^\L$ and $\mathcal{I}_\mathrm{gu}^\L$ into (\ref{eqn:LaplacianProofTheoremOne}), (\ref{CuRu}), and (\ref{eq:U2ULinkCoverage}).
\end{appendix} | 33,446 |
Nigeria: “There is nothing left in Banki”
Médecins Sans Frontières provided emergency medical assistance and delivered food in Banki, Nigeria near the Cameroon border. It has an estimated population of 15,000 to 20,000, most of whom are displaced people. The situation there is critical with people blocked in the town for security reasons. They critically lack access to food, water and adequate healthcare. Below are two patient testimonies highlighting the critical situation in Nigeria.
Maka
In late July, Médecins Sans Frontières teams referred 55 year-old Nigerian grandmother Maka for urgent medical care to Mora Hospital in Cameroon from Banki, a town 30 kilometres from the Nigeria-Cameroon border. Maka is suffering from severe acute malnutrition, a condition rare in adults and proof of the acute food crisis the population of Banki has been experiencing for the last four months. She was accompanied by her five year-old grandson, as well as her niece who brought her own 11 month-old son. The two children were also suffering from very advanced malnutrition and needed to be admitted to hospital urgently for life-saving treatment.
Maka had lost her smile, she spoke very little and her emaciated face was devoid of any expression. Nine days after she was taken into care by Médecins Sans Frontières teams, Maka was getting better. “I am originally from a village outside of Banki. Violent attacks forced us to leave our village, but four months ago my family and I became trapped in Banki, unable to leave the town. I lost track of some members of my family. For four months, we were unable to leave the town, we couldn't do anything.
Life became very difficult. We received barely two kilograms of food per week, and most of the time it was rice or corn. Everybody got the same amount, whether you were on your own or if you had 10 children. I had to ration the food, because if I had cooked enough for everybody to feel full, we wouldn’t have had anything left to last us until the next food delivery.. I hope that my children who are still in Banki are still receiving at least the small amount of food they are entitled to. That’s what we were promised. But what worries me most of all is the lack of security.”
“We received barely 2 kg of food per week... Everybody got the same amount, whether you were alone or if you had 10 children.”
Today Maka's face shows the hint of a timid smile, reflecting her sincere gratitude. However, she cannot help but worry about the future that awaits her and her family. “I’m happy that my children and I can recover here in Mora. Once we are better, if we can get by and work a little so we can live, I will go back to my husband in Banki. But we can’t do anything. My wish is for my children and family who are still in Banki to join me here in Cameroon. We have some cooking equipment, a little food and we are fighting to survive in safety. We would also like to stay near the hospital. Even though Nigeria is our home, we are not free there and we are too scared to return.”
Dayo
Dayo, 31, was referred to Mora Hospital in Cameroon in late July by Médecins Sans Frontières teams in Banki, Nigeria. She accompanied her sick four year-old son, Barine. The child urgently needed to be admitted to hospital as he was suffering from severe acute malnutrition. Before arriving in Mora, Dayo says her hunger had been so severe that sometimes she felt she was losing her mind. “When somebody spoke to me, I couldn’t even tell if it was a man or woman.” She had refused to take the rare medication that the medical teams in the area prescribed after a consultation. On an empty stomach, the tablets cause unbearable side effects. Nine days after Barine was hospitalised, his health has improved significantly, even though he is not yet able to swallow the doses of therapeutic food required to treat his malnutrition. Unfortunately, two of the five children who Médecins Sans Frontières referred at the same time as Barine have since died. Despite being admitted to hospital, their condition was too severe.
Like Barine and his mother, over 15,000 displaced Nigerians have been living in catastrophic conditions for nearly five months in Banki. Devoid all activity and impossible to leave, Banki now resembles a ghost town. “I come from a village 15 kilometres from Banki. One day, armed men arrived in our village and forbade us from working or travelling. They were violent and terrorised us. My husband, children and I fled into the bush, armed with only machetes and sticks. That’s when hunger set in. We cooked what dried millet and beans we were able to obtain. We could only cook during the day as at night, the fire would have attracted the attention of the people we were trying to hide from. Then our village was burnt down. I lost my mother, father and mother-in-law in the violence.
“There is nothing left in Banki”
We arrived in Banki with nothing, not even a plate or a pot, and I had only the clothes on my back. We couldn’t leave the town and there was nothing to do apart from wait for the delivery of supplies, upon which we were entirely dependent. Luckily, the authorities are distributing some food to the population, but it really is not enough. We receive barely two kilograms of rice or corn per week, and sometimes it has to last two weeks. If we need fuel, we take wood from sheds to burn, and you can find various objects and utensils in abandoned houses. In all the time I have been in Banki, I have yet to see any soap. In addition, we have to be really careful with water, given that the little we receive each day has to be used for drinking, hygiene and washing our clothes.
Although Banki is my home, we are too scared to go back. I heard that in one night alone, three children and two women were kidnapped, along with all their food. I am so worried for my children there. I know that my younger brother is looking after them, but my other son is ill. Each time I get a meal in hospital, I think about the people who are still there.I want all my family to join me here. I would be happy to live with them underneath a tree, as long as it were here. I don’t want to return to Nigeria. There is nothing left in Banki.” | 17,767 |
TITLE: Probability distribution of linear combination two rvs
QUESTION [0 upvotes]: Reading a book a find this results:
- Assume we have a logistic random variable $X$ (with parameters $(\lambda, \theta)$) with pdf
\begin{eqnarray*}
f (x)
&=&
\displaystyle
\frac {\displaystyle \lambda \exp \left\{ -\lambda \left( x - \theta \right) \right\}}
{\displaystyle \left[ 1 + \exp \left\{ -\lambda \left( x - \theta \right) \right\} \right]^2}\,.
\end{eqnarray*}
The last pdf can be re-expressed as a mixture of Laplace distribution
\begin{eqnarray}
f(x)
&=&
\displaystyle
\sum_{k = 0}^\infty \frac {2}{k + 1} {-2 \choose k} \frac {\lambda (k + 1)}{2}
\exp \left\{ -\lambda (k + 1) \mid x - \theta \mid \right\}.
\label{ffx}
\end{eqnarray}
Now, suppose $X$ and $Y$ are independent logistic random variables
with parameters $(\lambda, \theta)$ and $(\mu, \phi)$, respectively.
The cdf of $Z = \alpha X + \beta Y$ can be expressed as:
\begin{eqnarray}
\Pr \left( \alpha X + \beta Y \leq z \right)
&=&
\displaystyle
\sum_{k = 0}^\infty \sum_{l = 0}^\infty
\frac {\displaystyle 4}{\displaystyle (k + 1) (l + 1)} {-2 \choose k} {-2 \choose l}
\Pr \left( \alpha X_k + \beta Y_l \leq z \right),
\end{eqnarray}
where $X_k$ and $Y_l$ are independent Laplace random variable
with parameters $(\lambda (k + 1), \theta)$ and $(\mu (l + 1), \phi)$, respectively.
Please help me to understand how the last, double infinite sum, representation is obtained?
REPLY [2 votes]: This is just swapping integration with summation. For any region $D$ in the plane we have
$$
P((X,Y)\in D)=
\iint_D f_X(x)f_Y(y)dx\,dy.
$$
Plug in the representation of $f_X$ as a mixture of $f_{X_k}$:
$$
f_X(x)=\sum_{k=0}^\infty\frac2{k+1}{-2\choose k}f_{X_k}(x)
$$
(note the $\frac{\lambda(k+1)}2$ is missing because it's part of the Laplace density), and similarly:
$$
f_Y(x)=\sum_{l=0}^\infty\frac2{l+1}{-2\choose l}f_{Y_l}(y)
$$
and obtain
$$
\begin{align}
P((X,Y)\in D)&=
\iint_D \sum_{k=0}^\infty\frac2{k+1}{-2\choose k}f_{X_k}(x)\sum_{l=0}^\infty\frac2{l+1}{-2\choose l}f_{Y_l}(y)dx\,dy\\
&=\sum_{k=0}^\infty\frac2{k+1}{-2\choose k}\sum_{l=0}^\infty\frac2{l+1}{-2\choose l}\iint_D f_{X_k}(x) f_{Y_l}(y)dx\,dy\\
&=\sum_{k=0}^\infty\sum_{l=0}^\infty\frac2{k+1}{-2\choose k}\frac2{l+1}{-2\choose l}P((X_k,Y_l)\in D).
\end{align}
$$ | 91,969 |
\begin{document}
\title{\bf KAM for quasi-linear forced hamiltonian NLS}
\author
{\bf Roberto Feola
\vspace{2mm}
\\ \small
\\ \small
E-mail: rfeola@sissa.it}
\date{}
\maketitle
\begin{abstract}
In this paper we prove the existence of quasi-periodic, small-amplitude, solutions for quasi-linear Hamiltonian perturbations
of the non-linear Sch\"odinger equation on the torus in presence of a quasi-periodic forcing. In particular we prove that such solutions are linearly stable.
The proof is based on a Nash-Moser implicit function theorem. We prove the invertibility of the linearized operator
using a reducibility argument
in which we exploit the pseudo-differential structure of the linearized operator. Due to the multiplicity of the eigenvalues we obtain a block-diagonalization.
\end{abstract}
\tableofcontents
\zerarcounters
\setcounter{equation}{0}
\section{Introduction and Main result}
\label{sec:1}
In the theory of Hamiltonian partial differential equation an important matter is about the existence of quasi-periodic solutions.
This topic has been widely studied in literature using different approach. The classical results on semi-linear PDE's (where the non-linearity does not contains derivatives), have been obtained using KAM theory, see for instance \cite{W,K1,KP}, ora {\em via} Nash-Moser theory \cite{CW}.
In this paper we study the existence of \emph{reducible quasi-periodic solutions} for the hamiltonian NLS equation with unbounded perturbations:
\begin{equation}\label{mega}
iu_{t}=u_{xx}+mu+ \e \ff(\oo t,x,u,u_{x},u_{xx}),
\quad x\in\TTT:=\RRR/2\pi\ZZZ,
\end{equation}
where $\e>0$ is a small parameter, $m>0$ and the nonlinearity is quasi-periodic in time with
diophantine frequency vector $\oo\in\RRR^{d}$ and $\ff(\f,x,z)$, with $\f\in\TTT^{d}$, $z=(z_{0},z_{1},z_{2})\in\CCC^{3}$
is in $C^{q}(\TTT^{d+1}\times\CCC^{3};\CCC)$ in the real sense (i.e. as function of ${\rm Re}(z)$ and ${\rm Im}(z)$).
Note that our case is \emph{quasi-linear}, i.e. our non-linearity contains space derivatives of order $\de=n$, where $n$ is
the order of the highest derivative appearing in the linear constant coefficients term.
The Hamiltonian NLS is on of the most studied model in the literature.
KAM theory for PDE's was infact first developed for the semi-linear NLS with Dirichelet boundary conditions. However already
extending this results to the circle is not completely trivial due to the presence of multiple eigenvalues.
Moreover in our case we have to deal also with the difficulties arising from
unbounded non-linearities. It turns out that dealing with
these two difficulties at the same time requires subtle analysis, already in the case of equation \eqref{mega} with only one derivative.
In order to clarify this point we first discuss
the main ideas needed in order to deal with unbounded non-linearities.
The first result in the case of unbounded perturbation is due to Kuksin in \cite{Ku2} for a class of KdV-type equations, where $\de<n-1$ (\emph{non-critical} unbounded perturbations) and one has simple eigenvalues.
Concerning the NLS equation we mention the results in \cite{ZGY} (reversible case) and in \cite{LY} (hamiltonian case).
These two works are about the NLS in presence of one derivative in the non-linearity, i.e. $\de=n-1$ and with Dirichelet boundary conditions.
In order to deal with this problem (\emph{critical} unbounded perturbations) the authors uses
an appropriate generalization of the ideas developed in \cite{Ku2}. The main point is that one has to deal with time-depending
scalar homological equation, whose solvability is the content of the so called \emph{Kuksin's Lemma}.
In the case of the circle (double eigenvalues) one would get a time-dependent matricial homological equation.
We also mention \cite{BBiP1}-\cite{BBiP2} where a KAM theory is developed to study the case
of a ''weaker'' dispersion law in the derivative Klein-Gordon equation. Again there is a problem of multiplicity of eigenvalues.
It is well known that the ideas used to deal with the case $\de\leq n-1$ do not apply if $\de=n$.
For fully non-linear cases the first results are on the existence of periodic solutions,
see \cite{IPT} on water waves, and \cite{Ba1} and \cite{Ba2} for Kirkhoff and Benjamin-Ono equations.
These results have been obtained by using a Nash-Moser iterative scheme combined
with tecniques of pseudo-differential calculus.
The main point is that the linearized operator has the form $\del_{t}+\DD$ where $\DD$ is a differential operator of order $n$ with non-constant coefficients. The breakthrough idea in \cite{IPT} is conjugate $\DD$ to an operator of the form $\gotD+\RR$ where
$\gotD$ has constant coefficients and $\RR$ is a regularizing pseudo-differential operator of order $k$ sufficiently large (i.e. $\del_{x}^{k}\circ\RR$ is bounded).
For periodic solutions this is enough to invert the linearized operator by Neumann series since $\gotD^{-1} \RR$ is bounded. In the case of quasi-periodic solutions this is not true and substantial new ideas are required.
A very efficient strategy has been developed in a series of papers by Baldi, Berti and Montalto (see cite) mostly on the KdV equation
which were recently extended to the NLS in \cite{FP}. In all the previous papers one strongly uses that the eigenvalues are simple.
In the present paper we deal with the Hamiltonian case
which, due to the presence of double eigenvalues,
has a number of difficulties not only of a technical nature.
We very briefly describe the general strategy and we refer to \cite{FP} for more details.
Here we concentrate in underlining the different problems we have to deal with in order to obtain the result.
\smallskip
\indent {\bf Nash-Moser scheme.}
The first ingredient is a generalized implicit function theorem with parameters (in our case the frequency $\oo$).
This is a well-established iterative scheme which allows to find zeros of a functional provided that one can prove invertibility of its linearization in a neighborhood of the origin.
This is fairly standard material and is based on a formal definition of {\em good parameters} where the algorithm runs through. We restate it in
Proposition \ref{teo4} in order to adapt to our notation.
\smallskip
\indent{\bf Inversion of the linearized operator.} An efficient way to prove bounds on the inverse of a linear operator is to diagonalize it: the so called reducibility. The key point is to control the differences of eigenvalues. In our case we have double eigenvalues and hence we obtain a $2\times 2$ block-diagonal reduction, this is the content of Proposition \ref{teo2ham}. The proof is divided in two steps:
\begin{enumerate}
\item Since we are dealing with unbounded non-linearities, before performing diagonalization, we need to apply some changes of variables in order
to reduce the operator to a constant coefficients unbounded operator plus a bounded remainder. This is a common feature of the above-mentioned literature and the
procedure can be iterated obtaining a constant coefficients unbounded operator plus a remainder which is
regularizing of degree $k$.
In our case we set $k=1$ and we only need to overcome some minor difficulties related to preserving the Hamiltonian structure, the results are detailed in Lemmata \ref{lem:3.88} and \ref{lem:3.9ham}.
\item The previous step gives a precise understanding of the eigenvalues of the matrix which we are diagonalizing. Then by imposing the II Melnikov conditions (quantitative bounds on the difference of eigenvalues)
one diagonalizes by a linear KAM scheme. In our case, since we have double eigenvalues we obtain a block diagonal reduction to a $2\times2$ block diagonal time independent matrix. This is an important difference w.r.t. \cite{} and is due to the fact that we have performed step 1. The II Melnikov conditions which we require are explicitly stated in Proposition \ref{teo2ham}, note that, due to the multiplicity of the eigenvalues we require a weaker condition (see the definition of $\calO_{\infty}^{2\g}$ in \eqref{martina10ham}). This is needed in order to perform the measure estimates.
\end{enumerate}
Once we have diagonalized the bounds on the inverse follow from bounds on the eigenvalues, see Proposition \ref{inverseofl}.
\noindent In the Nash-Moser scheme we need to invert the operator linearized at each approximate solution, namely we perform the diagonalization procedure infinitely many times.
\smallskip
{ \bf Measure estimates.} Now we collect all the Melnikov conditions that we have imposed in the previous steps. In order to conclude the proof we need to show that these conditions are fulfilled for a positive measure set of parameters. The first basic requirement is to prove that we may impose each {\em single} non-resonance condition by only removing a small set of parameters. In our case this is a non trivial problem which we overcome by imposing a non-degeneracy condition (see Hypothesis \ref{hyp3ham}
) and by considering vectors $\oo$ as in \eqref{dio}.
Then we need to show that the union of the resonant sets is still small, this requires proving a ''summability'' condition. This is the most delicate part of the paper where substantial new ideas are needed, see Section 6 for a more
detailed comparison with the case of single eigenvalues \cite{FP}
We consider the equation \eqref{mega}
with
diophantine frequency vector
\begin{equation}\label{dio}
\begin{aligned}
&\oo
\in\Lambda:=\left[\frac{1}{2},\frac{3}{2}\right]^{d}\subset\RRR^{d},
\;\; |{\oo}\cdot\ell|\geq \frac{\g_{0}}{|\ell|^{\tau_0 }}, \; \forall\; \ell\in\ZZZ^{d}\backslash\{0\}.
\end{aligned}
\end{equation}
For instance one can fix $\tau_0=d+1$.
We assume that $\ff(\f,x,z)$, with $\f\in\TTT^{d}$, $z=(z_{0},z_{1},z_{2})\in\CCC^{3}$
is such that
\begin{equation*}
\ff(\f,x,u,u_{x},u_{xx})=f_{1}(\f,x,\x,\h,\x_{x},\h_x,\x_{xx},\h_{xx})+
if_{2}(\f,x,\x,\h,\x_{x},\h_x,\x_{xx},\h_{xx}),
\end{equation*}
where we set
$u=\x+i\h$, with $\x(\f,x),\h(\f,x)\in H^{s}(\TTT^{d+1};\RRR)$ for some $s\geq0$,
and where
\begin{equation}\label{CONTROLLA}
f_{i}(\f ,x, \x_{0},\h_{0},\x_{1},\h_{1},\x_{2},\h_{2}) : \TTT^{d+1}\times\RRR^{6}\to\RRR, \quad i=1,2.
\end{equation}
for some $q\in\NNN$ large enough.
We are interested in the existence of quasi-periodic solution of (\ref{mega}) in $H^{s}$, for some $s$,
for a positive measure sets of $\oo$ that is a function $u(\oo t,x)$ where
$$
u(\f,x) : \TTT^{d}\times\TTT\to\CCC.
$$
In other words we look for non-trivial
$(2\pi)^{d+1}-$periodic solutions $u(\f,x)$ of
\begin{equation}\label{mega2ham}
i\oo\cdot\del_{\f}u=u_{xx}+ \mathtt{m} u+\e \ff(\f,x,u,u_{x},u_{xx})
\end{equation}
in the Sobolev space $H^{s}:=H^{s}(\TTT^{d}\times\TTT; \CCC):=$
\begin{equation}\label{1.1}
\{u(\f,x)=\!\!\!\sum_{(\ell,k)\in\ZZZ^{d}\times\ZZZ}\!\!\!u_{\ell,k}e^{i(\ell\cdot\f+k\cdot x)} :
||u||^{2}_{s}:=\sum_{i\in\ZZZ^{d+1}}|u_{i}|^{2}\langle i\rangle^{2s}<+\infty
\}.
\end{equation}
where $s>\gots_{0}:=(d+2)/2>(d+1)/2$,
$i=(\ell,k)$ and $\langle i\rangle:=\max(|\ell|, |k|,1)$, $|\ell|:=\max\{|\ell_{1}|,\ldots,
|\ell_{d}|\}$, For $s\geq\gots_{0}$ $H^{s}$ is a Banach Algebra and
$H^{s}(\TTT^{d+1})\hookrightarrow C(\TTT^{d+1})$ continuously. We are moreover interested in
studying the linear stability of the possible solutions.
In this paper we assume the following:
\begin{hypo}\label{hyp2ham}
Assume that $\ff$ is
such that
\begin{equation}\label{hamham}
\ff(\oo t,x, u, u_{x},u_{xx})=\del_{\bar{z}_{0}}G(\oo t,x, u,u_{x})-\frac{d}{d x}[\del_{\bar{z}_{1}}G(\oo t,x,u,u_{x})]
\end{equation}
with $\del_{\bar{z}_{i}}=\del_{\x_{i}}+i\del_{\h_{i}}$, $i=0,1$, and
\begin{equation}
G(\oo t, x,u, u_{x}):=F(\oo t, x, \x,\h,\x_{x},\h_{x}) : \TTT^{d+1}\times\RRR^{4} \to \RRR,
\end{equation}
of class $C^{q+1}$.
\end{hypo}
\begin{hypo}\label{hyp3ham}
Assume that $\ff$ is
such that
\begin{equation}\label{nondeg}
\frac{1}{(2\pi)^{d+1}}\int_{\TTT^{d+1}}(\del_{\bar{z}_{1}}\ff)(\f,x,0,0,0)d x d\f=\gote\neq0.
\end{equation}
\end{hypo}
\noindent
Hypothesis \ref{hyp3ham} si quite technical and we will see in the following where we need it. On the contrary
Hypothesis \ref{hyp2ham} is quite natural and it implies that
the equation (\ref{mega}) can be rewritten as an Hamiltonian PDE
\begin{equation}\label{mega3}
u_{t}=i \del_{\bar{u}}\calH(u), \quad \calH(u)=\int_{\TTT}|u_{x}|^{2}+m|u|^{2}+
\e G(\oo t,x,u,u_{x})
\end{equation}
with respect to the non-degenerate symplectic form
\begin{equation}\label{simplectic}
\Omega(u,v):={\rm Re}\int_{\TTT} i u\bar{v}d x, \quad u,v\in H^{s}(\TTT^{d+1}; \CCC),
\end{equation}
where $\del_{\bar{u}}$ is the $L^{2}-$gradient with respect the complex scalar product. The main result of the paper is the following.
\begin{theorem}\label{teo1}
There exists $s:=s(d,\tau_{0})>0$, $q=q(d)\in\NNN$ such that
for every nonlinearity $\ff\in C^{q}(\TTT^{d+1}\times\RRR^{6};\CCC)$ that satisfies
Hypotheses \ref{hyp2ham} and \ref{hyp3ham} if $\e\leq \e_{0}(s,d)$ small enough, then there exists a Lipschitz map
$$
u(\e,\la) : [0,\e_{0}]\times \Lambda\to H^{s}(\TTT^{d+1};\CCC)
$$
such that, if $\la\in \CCCC_{\e}\subset \Lambda$, $u(\e,\la)$ is a solution of (\ref{mega2ham}).
Moreover,
the set $\CCCC_{\e}\subset\Lambda$ is a Cantor set of asymptotically full Lebesgue measure, i.e.
\begin{equation}\label{asy}
|\CCCC_{\e}|\to 1 \quad as \quad \e\to0,
\end{equation}
and
$||u(\e,\lambda)||_{s}\to0$ as $\e\to0$. In addiction, $u(\e,\lambda)$ is {\rm linearly stable}.
\end{theorem}
\section{Functional Setting and scheme of the proof}\label{sec2}
\subsection{Scale of Sobolev spaces}
Due to the complex nature of the NLS we need to work on product spaces.
We will usually denote
\begin{equation}\label{spaces}
\begin{aligned}
{\rm H}^{s}&:={\rm H}^{s}(\TTT^{d+1};\RRR)=H^{s}(\TTT^{d+1};\RRR)\times H^{s}(\TTT^{d+1};\RRR),\\
{\bf H}^{s}&:={\bf H}^{s}(\TTT^{d+1};\CCC)=H^{s}(\TTT^{d+1};\CCC)\times H^{s}(\TTT^{d+1};\CCC)\cap \calU,
\end{aligned}
\end{equation}
where
$$
\calU=\{(h^{+},h^{-})\in H^{s}(\TTT^{d+1};\CCC)\times H^{s}(\TTT^{d+1};\CCC) \; : \; h^{+}=\ol{h^{-}}\}.
$$
There is a one-to-one correspondence between these two spaces given by
${\rm H}^{s}\ni v=(v^{(1)},v^{(2)})\mapsto w=(u,\bar{u})\in{\bf H}^{s}$ with $u=v^{(1)}+i v^{(2)}$.
To simplify the notation, in the paper we should use the same symbol $v$ to indicate a function $v\in {\rm H}^{s}$
or $v\in{\bf H}^{s}$. We will use different symbols in some cases only to avoid confusion.
We also write ${\rm H}^{s}_{x}$ and ${\bf H}^{s}_{x}$ to denote the phase space of functions
in ${\rm H}^{s}(\TTT ;\RRR)=H^{s}(\TTT^{1};\RRR)\times H^{s}(\TTT^{1};\RRR)$ and
${\bf H}^{s}(\TTT ;\CCC)=H^{s}(\TTT^{1};\CCC)\times H^{s}(\TTT^{1};\CCC)\cap \calU,
$
On the product spaces ${\rm H}^{s}$ and ${\bf H}^{s}$ we define, with abuse of notation, the norms
\begin{equation}\label{spaces1}
\begin{aligned}
||z||_{{\rm H}^{s}}&:=\max\{||z^{(i)}||_{s}\}_{i=1,2}, \quad z=(z^{(1)},z^{(2)})\in{\rm H}^{s},\\
||w||_{{\bf H}^{s}}&:=||z||_{H^{s}(\TTT^{d+1};\CCC)}=||z||_{s}, \quad w=(z,\bar{z})\in{\bf H}^{s}, \quad
z=z^{(1)}+iz^{(2)}.
\end{aligned}
\end{equation}
\noindent
For a function $f : \Lambda\to E$ where $\Lambda\subset \RRR^{n}$ and $(E,\|\cdot\|_{E})$ is a Banach space
we define
\begin{eqnarray}\label{lnorm2}
{\it sup \phantom{g}norm:} \; ||f||_{E}^{sup}&\!\!\!\!\!\!:=\!\!\!\!\!\!&||f||^{sup}_{E,\Lambda}:=\sup_{\la\in\Lambda}||f(\oo)||_{E},\\
{\it Lipschitz \phantom{g} semi\!-\!norm:} \;
||f||_{E}^{lip}&\!\!\!\!\!\!:=\!\!\!\!\!\!&||f||_{E,\Lambda}^{lip}:=
\sup_{\substack{\oo_{1},\oo_{2}\in\Lambda \\ \oo_{1}\neq\oo_{2}}}
\frac{||f(\oo_{1})-f(\oo_{2})||_{E}}{|\la_{1}-\la_{2}|}\nonumber
\end{eqnarray}
and for $\g>0$ the weighted Lipschitz norm
\begin{equation}\label{lnorm}
||f||_{E,\g}:=||f||_{E,\Lambda,\g}:=||f||^{sup}_{E}+\g||f||_{E}^{lip}.
\end{equation}
In the paper we will work with parameter families of functions in $\HH_s$,
where $\calH_{s}={\rm H}^{s}, {\bf H}^{s}$, more precisely we consider
$u=u(\oo)\in {\rm Lip}(\Lambda,\HH_s)$
where $\Lambda\subset \RRR^{d}$. In this case we simply write $\|f\|_{\HH_{s},\g}:=\|f\|_{s,\g}$.
Along the paper we shall write also
\begin{equation*}
a\leq_{s} b \;\;\; \Leftrightarrow \;\;\; a\leq C(s) b \;\;\; {\rm for \; some \; constant}\;\; C(s)>0.
\end{equation*}
Moreover to indicate unbounded or regularizing spatial differential operator we shall write $O(\del_{x}^{p})$ for some
$p\in \ZZZ$. More precisely we say that an operator $A$ is $O(\del_{x}^{p})$ if
\begin{equation}\label{pseudo}
A : H_{x}^{s}\to H_{x}^{s-p}, \quad \forall s\geq0.
\end{equation}
Clearly if $p<0$ the operator is regularizing.
\noindent
Now we define the subspaces of trigonometric polynomials
\begin{equation}\label{trig}
H_{{n}}=H_{N_n}:=\big\{u\in L^{2}(\TTT^{d+1}) : u(\f,x):=\sum_{|(\ell,j)|\leq N_{n}}u_{j}(\ell)e^{i(\ell\cdot\f+jx)}\big\}
\end{equation}
where $N_{n}:=N_{0}^{(\frac{3}{2})^{n}}$, and the orthogonal projection
$$
\Pi_{n}:=\Pi_{N_{n}}: L^{2}(\TTT^{d+1})\to H_{n}, \quad \Pi^{\perp}_{n}:=\uno-\Pi_{n}.
$$
This definitions can be extended to the product spaces in \eqref{spaces} in the obvious way. We have the following classical result.
\begin{lemma}
Fo any $s\geq0$ and $\n\geq0$ there exists a constant $C:=C(s,\n)$ such that
\begin{equation}\label{smoothing}
\begin{aligned}
&\|\Pi_{n}u\|_{s+\n,\g}\leq C N_{n}^{\nu}\|u\|_{s,\g}, \;\; \forall u\in H^{s}, \\
& \|\Pi^{\perp}_{n}u\|_{s}\leq C N_{n}^{-\nu}\|u\|_{s+\nu}, \;\; \forall u\in H^{s+\nu}.
\end{aligned}
\end{equation}
\end{lemma}
\noindent
We omit the proof of the Lemma since bounds \eqref{smoothing} are classical estimates for truncated Fourier series
which hold also for the norm in \eqref{lnorm}.
\subsection{Hamiltonian structure}\label{odioham}
Given a function
$u\in {\rm H}^{s}$
if we write
$u=\x+i\h$
one has that the equation
(\ref{mega2ham}) reads
\begin{equation}\label{realsyst}
\left\{
\begin{aligned}
\oo\cdot\del_{\f}\x &=\h_{xx}+\mathtt{m}\h+\e f_{2}(\f,x,\x,\h,\x_{x},\h_{x},\x_{xx},\h_{xx}),\\
-\oo\cdot\del_{\f}\h &=\x_{xx}+\mathtt{m}\x+\e f_{1}(\f,x,\x,\h,\x_{x},\h_{x},\x_{xx},\h_{xx}),
\end{aligned}\right.
\end{equation}
where $f_{i}$ for $i=1,2$ are defined in (\ref{CONTROLLA}).
Equation \eqref{realsyst} is nothing but equation \eqref{mega2ham} written in an explicit way. Now we analyze its Hamiltonian structure.
Thanks to Hypotesis \ref{hyp2ham} we can write
\begin{equation}\label{realsyst2}
\dot{w}=\chi_{H}(w):=J\nabla H(w), \quad w=(\x,\h) \in {\rm H}^{s},\quad J=\left(\begin{matrix}0&1\\ -1& 0 \end{matrix}\right),
\end{equation}
If we consider the space ${\rm H}^{s}$ endowed with
the symplectic form
\begin{equation}\label{symform}
\tilde{\Omega}(w,v):=\int_{\TTT} w\cdot J v d x=(w,Jv)_{L^{2}\times L^{2}}, \quad \forall\; w,v\in
{\rm H}^{s}
\end{equation}
where $\cdot$ is the usual $\RRR^{2}$ scalar product, then
$\chi_{H}$ is the Hamiltonian vector field generator by the hamiltonian function
\begin{equation}\label{realHam}
\begin{aligned}
&H : {\rm H}^{s}\to \RRR ,\qquad
H(w)=\frac{1}{2}\int_{\TTT}|w_{x}|^{2}+\mathtt{m}|w|^{2}+\e F(\oo t,x, w,w_{x}).
\end{aligned}
\end{equation}
Indeed, for any $w,v\in{\rm H}^{s}$
one has
$$
d H(w)[h]=(\nabla H(w), h )_{L^{2}(\TTT)\times L^{2}(\TTT)}=\tilde{\Omega}(\chi_{H}(u),h),
$$
With this notation one has
\begin{equation}\label{eq12ham}
\begin{aligned}
f_{1}
:=
-\del_{\x}F+\del_{\x \x_{x}}F \x_{x}+\del_{\h \x_{x}}F\h_{x}+\del_{\x_{x}\x_{x}}F\x_{xx}+
\del_{\x_{x}\h_{x}}F\h_{xx},\\
f_{2}
:=
-\del_{\h}F+\del_{\x \h_{x}}F \x_{x}+\del_{\h \h_{x}}F\h_{x}+\del_{\x_{x}\h_{x}}F\x_{xx}+
\del_{\h_{x}\h_{x}}F\h_{xx},
\end{aligned}
\end{equation}
where all the functions are evaluated in $(\f,x,\x,\h,\x_{x},\h_{x},\x_{xx},\h_{xx})$.
One can check that the \eqref{realsyst} is equivalent to \eqref{mega2ham}. It is sufficient
to multiply by the constant $i$ the first equation and to add or subtract the second one,
one obtains
\begin{equation}\label{14}
\begin{aligned}
i\oo\cdot\del_{\f}u&=i\oo\cdot\del_{\f}\x-\oo\cdot\del_{\f}\h=
u_{xx}+\mathtt{m}u+\e\ff,\\
i\oo\cdot\del_{\f}\bar{u}&=i\oo\cdot\del_{\f}\x+\oo\cdot\del_{\f}\h
=-\bar{u}_{xx}-\mathtt{m}\bar{u}-\e\ol{\ff}
\end{aligned}
\end{equation}
The classical approach is to consider the ``double'' the NLS in the product space
$H^{s}(\TTT^{d+1};\CCC)\times H^{s}(\TTT^{d+1};\CCC)$ in the
complex independent variables $(u^{+},u^{-})$. One recovers the equation (\ref{mega2ham})
by
studying the system in the subspace $\calU=\{u^{+}=\ol{u^{-}}\}$ (see the (\ref{14})).
On the contrary we prefer to use the real coordinates, because we are working in a differentiable
structure. To define a differentiable structure on complex variables
is more
less natural.
Anyway, one can see in \cite{FP}
how to deal with this problem. There, the authors find an extension of the vector fields
on the complex plane that is merely differentiable. The advantage of that approach, is to deal with
a diagonal linear operator. How we will see in the following of this paper, it is not
necessary to apply the abstract Nash-Moser Theorem proved in
\cite{FP}.
The phase space for the NLS is ${\rm H}^{1}:=H^{1}(\TTT;\RRR)\times H^{1}(\TTT;\RRR)$.
In general we have the following definitions:
\begin{defi}\label{hamilt}
We say that a time dependent linear vector field $\chi(t) : {\rm H}^{s}\to {\rm H}^{s}$ is \emph{Hamiltonian} if $\chi(t)=J\calA(t)$, where $J$ is defined in (\ref{realsyst2}) and $\calA(t)$
is a real linear operator that is self-adjoint with respect the real scalar product on $L^{2}\times L^{2}$.
The corresponding Hamiltonian has the form
$$
H(u):=\frac{1}{2}(\calA(t)u,u)_{L^{2}\times L^{2}}=\int_{\TTT}\calA(t)u\cdot u d x
$$
Moreover, if $\calA(t)=\calA(\oo t)$ is quasi-periodic in time, then the associated operator
$\oo\cdot\del_{\f}\uno-J\calA(\f)$ is called Hamiltonian.
\end{defi}
\begin{defi}\label{hamilt2}
We say that a map $A : {\rm H}^{1}\to{\rm H}^{1}$ is \emph{symplectic} if the symplectic form
$\tilde{\Omega}$ in (\ref{symform}) is preserved, i.e.
\begin{equation}\label{hamilt3}
\tilde{\Omega}(Au,Av)=\tilde{\Omega}(u,v), \quad \forall\;\; u,v\in {\rm H}^{1}.
\end{equation}
If one has a family of symplectic maps $A(\f)$, $\forall\;\f\in\TTT^{d}$ then we say that
the corresponding operator acting on quasi-periodic functions $u(\f,x)$
$$
(Au)(\f,x):=A(\f)u(\f,x),
$$
is symplectic.
\end{defi}
\begin{rmk}\label{hamilt4}
Note that in complex coordinates the phase space is ${\bf H}^{1}:=H^{1}(\TTT;\CCC)\times H^{1}(\TTT;\CCC)$.
The definitions above are the same by using the symplectic form defined in (\ref{simplectic})
and the complex scalar product on $L^{2}$.
\end{rmk}
\noindent
Let $w:=(\x,\h)\in {\rm H}^{s}$. We define the functional
\begin{equation}\label{totale}
\calF(\oo t,x, w):=D_{\oo}w+\e g(\oo t,x, w),\;\;\;\; D_{\oo}=\left(\begin{matrix} \oo\cdot\del_{\f} & -\del_{xx}-m \\
\del_{xx}+m & \oo\cdot\del_{\f}\end{matrix}\right),
\end{equation}
where
\begin{equation}\label{13}
g(\oo t,x , w):=\left(\begin{matrix} -f_{2}(\f,x,\x,\h,\x_{x},\h_{x},\x_{xx},\h_{xx}) \\
f_{1}(\f,x,\x,\h,\x_{x},\h_{x},\x_{xx},\h_{xx})
\end{matrix}\right).
\end{equation}
Now it is more convenient to pass to the complex coordinates. In other words we identify an element
$V:=(v^{(1)},v^{(2)})\in {\rm H}^{s}$ with a function
$v:=v^{(1)}+iv^{(2)}\in H^{s}(\TTT^{d+1};\CCC)$.
Consider the linear operator ${d}_{z}\calF(\oo t ,x,z)$ linearized in some function $z$, and consider the system
\begin{equation}\label{linsyst}
\begin{aligned}
D_{\oo}V+\e { d}_{z}g(\oo t,x , z)V=0, \quad V\in {\rm H}^{s}.
\end{aligned}
\end{equation}
We introduce an invertible linear change of coordinate of the form
\begin{equation}\label{changelinear}
\begin{aligned}
T &: {\rm H}^{s} \to {\rm H}^{s},\\
TV:=&\left(\begin{matrix}\frac{i}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{i}{\sqrt{2}}\end{matrix}\right)\left(\begin{matrix} v^{(1)} \\ v^{(2)} \end{matrix} \right)=\left(\begin{matrix}\frac{i}{\sqrt{2}} v \\
\frac{1}{\sqrt{2}} \bar{v} \end{matrix} \right), \qquad T^{-1}:=\left(\begin{matrix}-\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}}\end{matrix}\right).
\end{aligned}
\end{equation}
\noindent
We postponed the proof of the following Lemma in the Appendix:
\begin{lemma}\label{lemmaccio}
The transformation of coordinates $T$ defined in (\ref{changelinear}) is symplectic.
Moreover, a function $V:=(v^{(1)},v^{(2)})\in {\rm H}^{s}$ is a solution of the system
\begin{equation}\label{lemmaccio1}
d_{z}\calF(\oo t,x,z)V=0,
\end{equation}
if and only if the function
\begin{equation}\label{lemmaccio2}
\left(\begin{matrix}v \\ \bar{v}\end{matrix}\right):=T_{1}^{-1}T V, \quad v\in H^{s}(\TTT^{d+1};\CCC), \quad T_{1}^{-1}:=\left(\begin{matrix}-i\sqrt{2}&0 \\ 0& \sqrt{2} \end{matrix}\right)
\end{equation}
solves the system
\begin{equation}\label{lemmaccio3}
\calL(z) \left(\begin{matrix}v \\ \bar{v}\end{matrix}\right):= T_{1}^{-1}T d_{z}\calF(\oo t,x,z)T^{-1}T_{1} \left(\begin{matrix}v \\ \bar{v}\end{matrix}\right)=0
\end{equation}
In particular the operator $\calL(z) : H^{s}(\TTT^{d+1};\CCC)\times H^{s}(\TTT^{d+1};\CCC)\to
H^{s}(\TTT^{d+1};\CCC)\times H^{s}(\TTT^{d+1};\CCC)$ has the form
\begin{equation}\label{lemmaccio4}
\begin{aligned}
{\calL}(z)
&= \omega\cdot \partial_\f \uno +i (E+A_2) \del_{xx} + i A_1\del_x +i(mE+ A_0)\,,
\end{aligned}
\end{equation}
where
\begin{equation}\label{lemmaccio44}
E=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \quad A_{i}=A_{i}(\f,x, z):=
\left(\begin{matrix}a_{i} & b_{i} \\ -\bar{b}_{i} &-\bar{a}_{i}
\end{matrix}\right)
\end{equation}
with for $i=0,1,2$, and $\forall z
\in H^{s}(\TTT^{d+1};\CCC)$,
\begin{equation}\label{5005}
\begin{aligned}
2a_{i}(\f,x)&:=\e(\del_{z_{i}}\ff)(\f,x,z(\f,x),z_{x}(\f,x),z_{xx}(\f,x)),\\
2b_{i}(\f,x)&:=\e(\del_{\bar{z}_{i}}\ff)(\f,x,z(\f,x),z_{x}(\f,x),z_{xx}(\f,x)),
\end{aligned}
\end{equation}
where we denoted $\del_{z_{i}}:=\del_{z^{(1)}_{i}}-i\del_{z^{(2)}_{i}}$ and $\del_{\bar{z}_{i}}:=\del_{z^{(1)}_{i}}+i\del_{z^{(2)}_{i}}$ for $i=0,1,2$.
\end{lemma}
The operator $\calL$ has further property. It is clearly Hamiltonian with respect to the symplectic form in (\ref{simplectic})
and the corresponding quadratic Hamiltonian
has the form
\begin{equation}\label{linham}
\begin{aligned}
H(u,\bar{u})&=\int_{\TTT}(1+a_{2})|u_{x}|^{2}+\frac{1}{2}\left[b_{2}\bar{u}_{x}^{2}+\bar{b}_{2}u_{x}^{2}
\right]-\frac{i}{2} {\rm Im}(a_{1})(u_{x}\bar{u}-u\bar{u}_{x}) {\rm d}x\\
&+\int_{\TTT}-m|u|^{2}- {\rm Re}(a_{0})|u|^{2}-\frac{1}{2}(
b_{0}\bar{u}^{2}+\bar{b}_{0}u^{2}){\rm d}x.
\end{aligned}
\end{equation}
Note that the symplectic form $\Omega$ in (\ref{simplectic}) is equivalent to the $2-$form
$\tilde{\Omega}$ in (\ref{symform}), i.e. given $u=u^{(1)}+iu^{(2)},v=v^{(1)}+iv^{(2)}\in H^{s}(\TTT^{d+1};\CCC)$,
one has
\begin{equation}\label{equival}
\Omega(u,w)={\rm Re}\int_{\TTT}iu\bar{v}d x=\int_{\TTT}(u^{(1)}v^{(2)}-v^{(1)}u^{(2)})d x=
\tilde{\Omega}(U,V),
\end{equation}
where we set $U=(u^{(1)},u^{(2)}), V=(v^{(1)},v^{(2)})\in H^{s}(\TTT^{d+1};\RRR)\times H^{s}(\TTT^{d+1};\RRR)$.
The (\ref{linham}) is the general form of a linear Hamiltonian operator as $\calL$, and,
the coefficients $a_{i}$ in \eqref{lemmaccio44} have the form
\begin{equation}\label{eq:1111}
\begin{aligned}
&a_{2}(\f,x)\in\RRR, \qquad\qquad\qquad
a_{1}(\f,x)=\frac{\rm d}{{\rm d}x}a_{2}(\f,x)+i {\rm Im}(a_{1})(\f,x),\\
&b_{1}(\f,x)=\frac{\rm d}{{\rm d}x}b_{2}(\f,x),\qquad
a_{0}(\f,x)={\rm Re}(a_{0})(\f,x)+\frac{i}{2}\frac{\rm d}{{\rm d}x}{\rm Im}(a_1)(\f,x)
\end{aligned}
\end{equation}
\subsection{Scheme of the proof}
For better understanding, we divide the prof of Theorem \ref{teo1} in several propositions.
The strategy is essentially the same followed in \cite{BBM} e \cite{FP}. It is based on a Nash-Moser iteration. We consider the operator $\calF$ in (\ref{lemmaccio4}), our aim is to show that there exists
a sequence of functions that converges, in some Sobolev space, to a \emph{solution} of (\ref{mega2ham}).
\begin{defi}[{\bf Good Parameters}]\label{invertibility}
Given $\mu>0,$ $N>1$ let $\ka_{2}:=11\mu+25\nu$,
for any Lipschitz family ${ u}(\la)\in H_{N}\times H_{N}$ with
$||{ u}||_{\gots_{0}+\mu,\g}\leq1$,
we define the set of good parameters
$\oo\in\Lambda$ as:
\begin{subequations}\label{eq104}
\begin{align}
\calG_N({u }):= &\left\{\oo\in \Lambda \; : \;\; ||\calL^{-1}({u }){h }||_{\gots_{0},\g}\leq C(\gots_{0})\g^{-1}||{h }||_{\gots_{0}+\mu,\g},\right. \label{eq104b}\\
& \; ||\calL^{-1}({u }){h }||_{s,\g}\leq C(s)\g^{-1}\left(||{h }||_{s+\mu,\g}+
||{u }||_{s+\mu,\g}||{h }||_{\gots_{0},\g}\right), \label{eq104a}\\
&\forall \gots_{0}\leq s\leq \gots_{0}+\ka_{2}-\mu,\;\;\left.\text{ for all Lipschitz maps ${h }(\la)$}\right\}.\nonumber
\end{align}
\end{subequations}
where $\calL$ is the linearized operator defined in \eqref{lemmaccio4}.
\end{defi}
\noindent
Clearly, Definition \ref{invertibility} depends on $\mu$ and $ N$. For a better understanding of Definition
\ref{invertibility} we refer the reader to Proposition 1.6 in \cite{FP}. Roughly speaking a set $\calG_{N}$
are nothing but set of parameters $\la$ for which some tame estimates hold for the inverse of the linearized operator. The constant $\mu$ represents the loss of regularity due to the presence of the small divisors.
It is important to underline the dependence on $\mu$ of Definition \ref{invertibility}. Indeed, as we will see,
it is not true that the sets $\calG_{N}$ have ``large'' measure for any $\mu$.
\begin{proposition}\label{teo4}
Fix $\g\leq\g_{0}, \mu>\tau> d$. There exist $q\in\NNN$, depending only on $\tau,d,\mu$, such that for any nonlinearity
$\ff\in C^{q}$ satisfying Hypotheses \ref{hyp2ham} and \ref{hyp3ham} the following holds. Let $\calF({ u})$ be defined in Definition
\ref{totale}, then there exists a small constant $\epsilon_0>0$ such that
for any $\e$ with
$0<\e\g^{-1}<\epsilon_0$, there exist constants $C_{\star}, N_0\in \NNN$, a sequence of functions ${u}_{n}$ and a sequence of sets $\calG_{n}(\g,\tau,\mu)\equiv\calG_{n}\subseteq\Lambda$ defined inductively as
$\calG_{0}:=\Lambda$ and $\calG_{n+1}:=\calG_{n}\cap \calG_{N_{n}}(u_{n})$
such that ${ u}_{n} : \calG_{n}\to{ H}^{0}$, $||{ u}_{n}||_{\gots_{0}+\mu,\g}\leq1$ and
\begin{equation}\label{teo41}
||{ u}_{n}-{ u}_{n-1}||_{\gots_{0}+\mu,\g}\leq C_{\star}\e \g^{-1} N_{n}^{-\ka}, \;\; \ka:=18+2\mu,
\end{equation}
with $N_{n}:=N_{0}^{(\frac{3}{2})^{n}}$ defined in \eqref{trig}.
Moreover the sequence converges in norm $||\cdot||_{\gots_{0}+\mu,\g}$ to a function $u_{\infty}$
such that
\begin{equation}\label{teo42}
\calF({ u}_{\infty})=0, \quad \forall\; \oo\in\calG_{\infty}:=\cap_{n\geq0}\calG_{n}.
\end{equation}
\end{proposition}
In the Nash-Moser scheme the main point is to invert, with appropriate bounds, $\calF$ linearized at any ${ u}_n$.
Following the classical Newton scheme we define
$$
{u}_{n+1}= { u}_n -\Pi_{N_{n+1}}d_{z}\calF^{-1}({ u}_n)\Pi_{N_{n+1}}\calF({ u}_{n}).
$$
In principle we do not know wether this definition is well posed since $d_{z}\calF({ u})$
may not be invertible.
To study the invertibility of the linearized operator is a problem substantially different for each equation.
In \cite{FP} the authors work in a reversible contest. Essentially the reversibility condition
introduced there, guarantees that the linearized operator has simple eigenvalues. Hence
it is natural to try to diagonalized $d_{u}\calF$ in order to invert it.
Here, the situation is different. We have that the eigenvalues are multiple, then the diagonalization procedure is more difficult.
For the proof of Proposition \ref{teo4} we refer the reader to the general result
proved in \cite{FP}. In that work is proved an abstract existence result based on a Nash-Moser scheme
on a scale of Banach spaces. Such result applies \emph{tame} functionals. In our case the functional $\calF$ satisfies such properties by Lemma \eqref{lemA2}.
The main step of our approach is to prove the invertibility of the linearized operator
$d_{z}\calF(\oo t,x,z )$, at any $x\in{\rm H}^{s}$.
To do this, we will first prove the following diagonalization result on the operator $\calL$ defined in
(\ref{lemmaccio4}):
\begin{proposition}[{\bf Reducibility}]\label{teo2ham}
Fix $\g\leq\g_{0}$ and $\tau>d$ and consider any $\ff\in C^{q}$ that satisfies Hypotheses \ref{hyp2ham} and \ref{hyp3ham}.
Then there exist
$\h,q\in\NNN$, depending only on $d$, such that
for $0\leq \e\leq \e_{0}$ with $\e_{0}$ small enough the following holds. Consider any subset $\Lambda_{o}\subseteq\Lambda\subseteq \RRR^{d}$ and any Lipschitz families $u(\oo) :\Lambda_{o} \to {\bf H}^{0}$ with $||u||_{\gots_{0}+\h,\g}\leq1$. Consider the linear operator
$\calL : {\bf H}^{s}\to {\bf H}^{s}$ in (\ref{lemmaccio4}) computed at $u$.
then for all $\s=\pm 2,j\in \NNN$
there exist Lipschitz map
$\Omega_{\s,\und{j}} : \Lambda \to {\rm Mat}(2\times 2,\CCC)$
of the form
\begin{equation}\label{1.2.2bis}
\Omega_{\s,\und{j}}= -i \s ({m}_{2}j^{2} + {m}_{0}) \begin{pmatrix}1&0\\ 0 &1 \end{pmatrix} -i\s|{m}_{1}|j \begin{pmatrix}1&0\\ 0 &-1 \end{pmatrix} +i\s R_{\s,\und{j}},
\end{equation}
where $R_{\s,\und{j}}$ is a self-adjoint matrix and
\begin{equation}\label{1.2.2tris}
\begin{aligned}
&|{m}_{2}-1|_{\g}+|{m}_{0}-m|_{\g}\leq \e C,\quad
|R_{j}^{k}|_{\g}\leq \frac{\e C}{\langle j\rangle}, \quad k=\pm j, \; j\in\ZZZ,\\
&\e c\leq |m_{1}|^{sup}\leq \e C, \quad |m_{1}|^{lip}\leq \e^{2}\g^{-1}C.
\end{aligned}
\end{equation}
for any $\s\in\CC$, $j\in \NNN\cup\{0\}$, here and in the following $\CC:=\left\{+1,-1\right\}$.
\noindent
Set
\begin{equation}\label{1.2.2quatuor}
\Omega_{\s,\underline{j}}:=\left(\begin{matrix}\Omega_{\s,j}^{\phantom{g}j} & \Omega_{\s,j}^{-j} \\
\Omega_{\s,-j}^{j} & \Omega_{\s,-j}^{-j} \end{matrix}\right),
\end{equation}
Define $\mu_{\s,j}$ and $\mu_{\s,-j}$ to be the eigenvalues
of $\Omega_{\s,\und{j}}$
Define $\Lambda_{\infty}^{2\g}(u):=\SSSS_{\infty}^{2\g}(u)\cap\calO_{\infty}^{2\g}(u)$ with
\begin{equation}\label{martina10ham}
\begin{aligned}
\SSSS^{2\g}_{\infty}({ u})&:=\left\{\begin{array}{ll}
\oo\in\Lambda_{o} : &|\oo\cdot\ell\!+\!\mu_{\s,{j}}(\oo)-\!\mu_{\s',{j'}}(\oo)|\geq
\frac{2\g|\s j^{2}-\s' j'^{2}|}{\langle\ell\rangle^{\tau}}, \\
&\;\ell\in\ZZZ^{d}, \s,\s'\in\CC, j,j'\in\ZZZ
\end{array}
\right\},\\
\calO_{\infty}^{2\g}(u)&:=\left\{\begin{array}{ll}
\oo\in\Lambda_{o} : &|{\oo}\cdot\ell+\mu_{\s,j}-
\mu_{\s,k}|\geq\frac{2\g}{\langle\ell\rangle^{\tau}\langle j\rangle},\\
&\ell\in\ZZZ^{d}\backslash\{0\}, j\in\ZZZ, k=\pm j, \s\in\CC
\end{array}
\right\},
\end{aligned}
\end{equation}
then we have:
\noindent
(i) for any $s\in(\gots_{0},q-\h)$, if $||z||_{\gots_{0}+\h}<+\infty$ there exist linear
bounded operators $W_{1},W_{2} : {\bf H}^{s}(\TTT^{d+1})\to{\bf H}^{s}(\TTT^{d+1})$ with
bounded inverse, such that $\calL({ u})$ satisfies
\begin{equation}\label{1.2.2ham}
\calL({\bf u})=W_{1}\calL_{\infty}W_{2}^{-1}, \; \calL_{\infty}=\oo\cdot\del_{\f}\uno+\DD_{\infty}, \;
\DD_{\infty}=diag_{(\s,j)\in\CC\times\ZZZ }\{\Omega_{\s,\underline{j}} \},
\end{equation}
\noindent
(ii) for any $\f\in\TTT^{d}$ one has
\begin{equation}\label{1.2.4ham}
W_{i}(\f), W_{i}^{-1}(\f) : {\bf H}^{s}_{x}\to {\bf H}^{s}_{x}, \quad i=1,2.
\end{equation}
with ${\bf H}^{s}_{x}:=H^{s}(\TTT;\CCC)\times H^{s}(\TTT;\CCC)\cap \calU$ and such that
\begin{equation}\label{eq:4.9madonna}
||(W_{i}^{\pm1}(\f)-\uno){ h}||_{{\bf H}^{s}_{x}}\leq
\e\g^{-1} C(s)(||{ h}||_{{\bf H}^{s}_{x}}+||{ u}||_{s+\h+\gots_{0}}
||{ h}||_{{\bf H}^{1}_{x}}).
\end{equation}
\end{proposition}
\begin{rmk}\label{ciccio}
Note that function $ h(t)\in {\bf H}_{x}^{s}$ is a solution of the forced NLS
\begin{equation}\label{1.2.5ham}
\calL(z) h=0
\end{equation}
if and only if the function $v(t):=(v_{1},v_{-1}):=W_{2}^{-1}(\oo t)[h(t)]\in {\bf H}^{s}_{x}$
solves the constant coefficients dynamical system
\begin{equation}\label{1.2.6ham}
\left(\begin{matrix} \del_{t}v_{1} \\ \del_{t} v_{-1}\end{matrix}\right)
+\DD_{\infty}\left(\begin{matrix} v_{1} \\ v_{2} \end{matrix}\right)=\left(\begin{matrix} 0\\ 0 \end{matrix}\right), \quad
\dot{v}_{\s,\underline{j}}=-\Omega_{\s,\underline{j}}v_{\s,\underline{j}},\quad
(\s,j)\in\CC \times\ZZZ,
\end{equation}
where all the eigenvalues of the matrices $\Omega_{\s,\underline{j}}$ are purely imaginary.
Moreover, since $\ol{\Omega_{\s,{j}}^{j}}=-\Omega_{\s,{j}}^{j}$ and
$\ol{\Omega_{\s,-j}^{\phantom{g} j}}=-\Omega_{\s,j}^{\phantom{g} j}$
then one has
$$
\frac{d}{dt}(|v_{1,j}(t)|^{2}+|v_{1,-j}(t)|^{2})=0, \qquad |v_{\s,0}(t)|^{2}=constant
$$
and hence
\begin{equation}\label{1.2.7ham}
\begin{aligned}
||v_{1}(t)||^{2}_{H_{x}^{s}}&=\sum_{j\in\ZZZ}|v_{1,j}(t)|^{2}\langle j\rangle^{2s}\\
&=|v_{1,0}(t)|^{2}+
\sum_{j\in\NNN}(|v_{1,j}(t)|^{2}+|v_{1,-j}(t)|^{2})\langle j\rangle^{2s}\\
&=|v_{1,0}(0)|^{2}+\sum_{j\in\NNN}
(|v_{1,j}(0)|^{2}+|v_{1,-j}(0)|^{2}) \langle j\rangle^{2s}=||v_{1}(0)||^{2}_{H_{x}^{s}}.
\end{aligned}
\end{equation}
Eq. \eqref{1.2.7ham} means that the Sobolev norm in the space of functions depending on $x$, is constant in time.
\end{rmk}
Proposition \ref{teo2ham} is fundamental in order to prove Theorem \ref{teo1}. Of course one can try
to invert the linearized operator without diagonalize it. In addiction to this we are not able to completely diagonalize it due to the multiplicity of the eigenvalues. This is one of the main difference with respect to the
reversible case.
Anyway the result in Proposition \ref{teo2ham}
is enough to prove the stability of the possible solution.
What we obtain is a block-diagonal operator with constant coefficients while in \cite{LY} the authors obtain
a normal form depending on time. Here most of the problems appear because we want to obtain a constant coefficient linear operator.
Another important difference between the case of single eigenvalues and double eigenvalues
stands in the set $\calO_{\infty}^{2\g}$ in \eqref{martina10ham}.
Indeed, as one can see in \eqref{martina10ham}, due to the multiplicity of the eigenvalues, we must impose a very weak non degeneracy condition on the eigenvalues.
Moreover, as we will see in Section \ref{sec6ham}, the measure estimates in the Hamiltonian case are more difficult with respect to the reversible one, and most of the problems
appear due to the presence of the set $\calO_{\infty}^{2\g}$. In order to overcame such problems we will use the additional Hypotheses \ref{hyp3ham}.
In Section \ref{sec:3ham} we will conjugate $\calL$ to a differential linear operator
with constant coefficients plus a \emph{bounded} remainder,
then, in Section \ref{sec:4ham}
we complete block-diagonalize the operator.
Using the reducibility results of Proposition \ref{teo2ham} we are able to prove (see Section \ref{sec:5ham}) the following result:
\begin{lemma}\label{inverseofl}({\bf Right inverse of $\calL$})
Under the hypotheses of Proposition \ref{teo2ham}, set
\begin{equation}\label{eq:4.4.18}
\z:=4\tau+\h+8.
\end{equation}
where $\h$ is fixed in Proposition \ref{teo2ham}.
Consider a Lipschitz family $ u(\oo)$ with $\oo\in\Lambda_{o}\subseteq\Lambda$ such that
\begin{equation}\label{eq:4.4.19}
||u||_{\gots_{0}+\z,\g} \leq1.
\end{equation}
Define the set
\begin{equation}\label{primedimham}
\calP^{2\g}_{\infty}({ u}):=\left\{\begin{array}{ll}
\oo\in\Lambda_{o} : &|\oo\cdot\ell\!+\!\mu_{\s,{j}}(\oo)|\geq
\frac{2\g\langle j\rangle^{2}}{\langle\ell\rangle^{\tau}}, \\
&\;\ell\in\ZZZ^{d}, \s,\in\CC, j\in\ZZZ
\end{array}
\right\}.
\end{equation}
There exists $\epsilon_{0}$, depending only on the data of the problem, such that
if $\e\g^{-1}<\epsilon_{0}$
then, for any $\oo\in\Lambda^{2\g}_{\infty}({u})\cap\calP^{2\g}_{\infty}({u})$
(see (\ref{martina10ham})), and for any Lipschitz family
$g(\oo)\in {\bf H}^{s}$,
the equation $\calL { h}:=\calL(\oo,{ u}(\oo)) h= { g}$,
admits a solution $h\in{\bf H}^{s}$
such that for $\gots_{0}\leq s \leq q-\mu$
\begin{equation}\label{eq:4.4.23}
||{ h}||_{s,\g}\leq C(s)\g^{-1}\left(||{ g}||_{s+2\tau+5,\g}+
||{ u}||_{s+\z,\g}||{ g}||_{\gots_{0},\g}
\right).
\end{equation}
\end{lemma}
Proposition \ref{inverseofl}, combined with Lemma \ref{lemmaccio}, guarantees
the invertibility (with suitable estimates) of the linearized of $\calF$. Of course $d_{z}\calF$ can be inverted only in a
suitable set of parameters depending on the function $z$ on which we linearize.
In principle it can be empty and moreover it is not sufficient to prove that it has positive measure.
Indeed we need to invert the linearized operator
in any approximate solution $u_{n}$ (see Proposition \ref{teo2ham}) so that
$\cap_{n\geq0}\left(\Lambda_{\infty}^{2\g}(u_{n})\cap\calP^{2\g}_{\infty}({u}_{n})\right)$ can have zero measure.
The last part of the paper is
devoted to give some measure estimates of such set.
In the Nash-Moser proposition \ref{teo4} we defined in an implicit way the sets $\calG_n$ in order to ensure bounds on the inverse of $\calL({u}_{n})$. The following Proposition is the main result of Section \ref{sec6ham}.
\begin{proposition}[{\bf Measure estimates}]\label{measurebruttebrutte}
Set $\g_n:=(1+2^{-n})\g$ and consider the set $\calG_{\infty}$ of Proposition \ref{teo4}
with $\mu=\zeta$ defined in Lemma \ref{inverseofl} and fix $\g:=\e^{a}$ for some $a\in(0,1)$. We have
\begin{subequations}\label{eq136tot}
\begin{align}
&\cap_{n\geq0}\calP^{2\g_n}_{\infty}({u}_n)\cap \Lambda^{2\g_n}_{\infty}({ u}_n)\subseteq \calG_{\infty}, \label{eq136b}\\
&|\Lambda\backslash\calG_{\infty}|\to 0, \;\; {\rm as} \;\; \e\to0. \label{eq136}
\end{align}
\end{subequations}
\end{proposition}
Formula (\ref{eq136b}) is essentially trivial. One just need to look at Definition \ref{invertibility}
which fix the sets $\calG_{n}$. It is important because gives us the connection between $\calG_{\infty}$ and the sets we have constructed at each step of the iteration. The (\ref{eq136}) is more delicate. The first point is that we reduce to computing the measure of the left hand side of (\ref{eq136b}).
The strategy described above is similar to that followed in \cite{BBM} and \cite{FP}. It is quite general and can be applied to various case.
The main differences are in the proof
of Proposition \ref{teo2ham}.
Clearly it depends on the unperturbed eigenvalues and on the symmetries one ask for on the system.
\subsection{Proof of Theorem \ref{teo1}}
\noindent
Theorem \ref{teo1} essentially follows by Propositions \ref{teo4} and \ref{measurebruttebrutte}. The measure estimates performed in the last section guarantee that the ``good'' sets defined in Prop. \ref{teo4} are not empty,
but on the contrary have ``full'' measure.
In particular one uses the result of Proposition \ref{teo4} in order to prove Lemma \ref{inverseofl}. Indeed one one has diagonalized the linearized operator it is trivial to get estimate \eqref{eq:4.4.23}.
From formula \eqref{eq:4.4.23} essentially follows \eqref{eq136b}.
Concerning the proof of Proposition \ref{teo4}, we have omitted since it is the same of Proposition $1.6$ in \cite{FP}. The only differences is that in \cite{FP} the authors deal with
a functional that is diagonal plus a non linear perturbation. In this case the situation is slightly different.
However the next Lemma guarantees that the subspaces $H_{n}$ in \eqref{trig} are preserved by the linear part of our functional $\calF$ in \eqref{totale},
\begin{lemma}\label{lemma2}
One has that
\begin{equation}
D_{\oo} : H_{n} \to H_{n}.
\end{equation}
\end{lemma}
\begin{proof}
Let us consider $u=(u^{(1)},u^{2})\in H_{n}$, then
\begin{equation*}
\begin{aligned}
D_{\oo}u&=D_{\oo}\sum_{|(\ell,j)|\leq N_{n}}u^{(i)}_{j}(\ell)e^{i\ell\cdot\f+ijx}\\
&=
\left(\begin{matrix}
\sum_{|(\ell,j)|\leq N_{n}}(i\oo\cdot\ell)u^{1}(\ell)_{j}-[(ij)^{2}+m]u^{(2)}_{j}e^{i\ell\cdot\f+ijx}\\
\sum_{|(\ell,j)|\leq N_{n}}(i\oo\cdot\ell)u^{2}(\ell)_{j}+[(ij)^{2}+m]u^{(1)}_{j}e^{i\ell\cdot\f+ijx}
\end{matrix}\right)\in H_{n}.
\end{aligned}
\end{equation*}
\end{proof}
We fix $\g:=\e^{a}$, $a\in(0,1)$. Then the smallness condition $\e\g^{-1}=\e^{1-a}<\epsilon_{0}$ of Proposition
\ref{teo4} is satisfied. Then we can apply it with $\mu=\zeta$ in (\ref{eq:4.4.18}) (see
Proposition \ref{measurebruttebrutte}).
Hence by (\ref{teo42}) we have that the function
${u}_{\infty}$ in ${\bf H}^{\gots_{0}+\zeta}$ is a solution of the perturbed NLS with frequency
$\oo$. Moreover, one has
\begin{equation}\label{eq:6.1}
|\Lambda\backslash\calG_{\infty}|\stackrel{(\ref{eq136})}{\to}0,
\end{equation}
as $\e$
tends to zero. To complete the proof of the theorem, it remains to prove
the linear stability of the solution.
Since the eigenvalues $\mu_{\s,j}$ are purely imaginary, we know that
the Sobolev norm of the solution ${ v}(t)$ of (\ref{1.2.6ham}) is constant in time. We just need to show
that the Sobolev norm of ${ h}(t)=W_{2}{ v}(t)$, solution of $\calL h=0$
does not grow on time.
Again to do this one can follow the same strategy used in \cite{FP}. In particular one uses the results of Lemma
\ref{lem:3.9ham} in Section \ref{sec:3ham} and estimates \eqref{eq:4.9ham} in Proposition \ref{KAMalgorithmham} in order to get the estimates
\begin{subequations}\label{eq:6.14}
\begin{align}
||{ h}(t)||_{H_{x}^{s}}&\leq K||{ h}(0)||_{H_{x}^{s}},\label{eq:6.14a}\\
\!\!\!\!||{ h}(0)||_{H_{x}^{s}}-\e^{b}K||{ h}(0)||_{H_{x}^{s+1}}\leq&||{h}(t)||_{H_{x}^{s}}\leq
||{ h}(0)||_{H_{x}^{s}}+\e^{b}K||{h}(0)||_{H_{x}^{s+1}}, \label{eq:6.14b}
\end{align}
\end{subequations}
for $b\in(0,1)$.
Clearly the (\ref{eq:6.14}) imply the linear stability of the solution, so we concluded the proof of Theorem \ref{teo1}.
The rest of the paper is devoted to the proof of Propositions \ref{teo2ham},\ref{measurebruttebrutte} and Lemma \ref{inverseofl}.
\section{ Regularization of the linearized operator}\label{sec:3ham}
In this section and in Section \ref{sec:4ham} we apply a reducibility scheme in order to
conjugate the linearized operator to a linear, constant coefficients differential operator.
Here we consider the linearized operator $\calL$ in (\ref{lemmaccio4})
and we construct two operators $\VV_{1}$ and $\VV_{2}$ in order to semi-conjugate
$\calL$ to an operator $\calL_{c}$ of the second order with constant coefficients
plus a remainder of order $O(\del_{x}^{-1})$.
We look for such transformations because, in order to apply a KAM-type algorithm to diagonalize $\calL$, we need first a precise control of the asymptotics of the eigenvalues, and also
some estimates of the transformations
$\VV_{i}$ with $i=1,2$ and their inverse.
The principal result we prove is the following.
\begin{lemma}\label{lem:3.88}
Let $\ff\in C^{q}$ satisfy the Hypotheses of Proposition \ref{teo4} and assume $q>\h_{1}+\gots_{0}$
where
\begin{equation}\label{eq:3.2.0ham}
\h_{1}:=d+2\gots_{0}+10.
\end{equation}
There exists $\epsilon_0>0$ such that, if $\e\g_{0}^{-1}\leq \epsilon_0$ (see (\ref{dio} for the definition of $\g_0$) then, for any $\g\le \g_0$ and for all
${u} \in{\bf H}^{0}$ depending in a Lipschitz way on $\oo\in \Lambda$,
if
\begin{equation}\label{eq:3.2.1bham}
||{ u}||_{\gots_{0}+\h_{1},\g}\leq\e\g^{-1},
\end{equation}
then, for $\gots_{0}\leq s\leq q-\h_{1}$,
the following holds.
\noindent (i)
There exist invertible maps $\VV_1, \VV_2: {\bf H}^{0}\to{\bf H}^{0}$ such that
$\calL_{7}:=\VV_{1}^{-1}\calL\VV_{2}=$
\begin{equation}\label{eq:3.5.9ham}
\oo\cdot\del_{\f}\uno+
i\left(\begin{matrix}m_{2} \!\!\!& 0 \\ 0 & \!\!\!-m_{2}\end{matrix}\right)\del_{xx}+
i\left(\begin{matrix}m_{1} \!\!\!& 0 \\ 0 & \!\!\!-\bar{m}_{1}\end{matrix}\right)\del_{x}
+i \left(\begin{matrix}m_{0} \!\!\!& q_{0}(\f,x) \\ -\bar{q}_{0}(\f,x) & \!\!\!-m_{0}\end{matrix}\right)+
\RR
\end{equation}
with $m_{2},m_{0}\in \RRR$, $m_{1}\in i\RRR$ and $\RR$ is a pseudo-differential operator of order
$O(\del_{x}^{-1})$ (see \eqref{pseudo}).
The $\VV_{i}$ are symplectic maps and moreover for all ${ h}\in {\bf H}^{0}$
\begin{equation}\label{eq:3.2.3ham}
||\VV_{i}{ h}||_{s,\g}+||\VV_{i}^{-1}{ h}||_{s,\g}\leq
C(s)(|| { h}||_{s+2,\g}+
||{ u}||_{s+\h_{1},\g}||{ h}||_{\gots_{0}+2,\g}), \quad i=1,2.
\end{equation}
\noindent
(ii) The coefficient $m_{i}:=m_{i}({ u})$ for $i=0,1,2$ of $\calL_{7}$ satisfies
\begin{equation}\label{eq:3.2.44}
\begin{aligned}
|m_{2}({ u})-1|_\g,|m_{0}({ u})-\mathtt{m}|_\g&\leq \e C,\;\;
|d_{{ u} }m_{i}({ u})[{ h}]|\leq \e C||{ h}||_{\h_{1}},
\;\; i=0,2, \\
|m_{1}({ u})|&\leq \e C,
|d_{{ u} }m_{1}({ u})[{ h}]|\leq \e C||{ h}||_{\h_{1}},
\end{aligned}
\end{equation}
and moreover the constant $m_{1}:=m_{1}(\oo,u(\oo))$ satisfies
\begin{subequations}\label{mammamia}
\begin{align}
&\e c \leq |m_{1}({ u})|, \label{mammamiaa}\\
& \sup_{\oo_{1}\neq\oo_{2}}\frac{|m_{1}(\oo_{1},u(\oo))-m_{1}(\oo_{2},u(\oo))|}{|\la_{1}-\la_{2}|} \leq \e^{2} C\g^{-1}
\label{mammamiab}
\end{align}
\end{subequations}
for some $C>0$.
\noindent
(iii) The operator $\RR:=\RR(u)$ is such that
\begin{eqnarray}\label{eq:3.2.7ham}
\|\RR({ u}){h}\|_{s,\g}&\leq& \e C(s) (\|{ h}\|_{s,\g}+\|{ u}\|_{s+\h_{1},\g}\|{ h}\|_{\gots_{0}}), \label{eq:B15aham}\\
\|d_{{ u}}\RR({ u})[{ h}]{ g}\|_{s}&\leq& \e C(s)\big(\|{ g}\|_{s+1}\|{ h}\|_{\gots_{0}+\h_{1}}+
\|{ g}\|_{2}\|{ h}\|_{s+\h_{1}}\nonumber\\
&+&\|{ u}\|_{s+\h_{1}}\|{ g}\|_{2}\|{ h}\|_{\gots_{0}}\big),\label{eq:B15bham}
\end{eqnarray}
and moreover
\begin{subequations}\label{eq:3.2.7bisham}
\begin{align}
\|q_{0}\|_{s,\g}&\leq \e C(s)(1+\|{ u}\|_{s+\h_{1},\g}),\label{eq:3.2.6aham}\\
\|d_{{ u} }q_{0}({u})[{ h}]\|_{s}&\leq \e C(s)(\|{ h}\|_{s+\h_{1}}+\|{ u}\|_{s+\h_{1}}+
\|{h}\|_{\gots_{0}+\h_{1}}),\label{eq:3.2.6bham}
\end{align}
\end{subequations}
Finally
$\calL_{7}$ is Hamiltonian.
\end{lemma}
\begin{rmk}\label{trenonapoli}
The estimate in \eqref{mammamia} is different from that in \eqref{eq:3.2.44}. As we will see, it is very important
to estimate the Lipschitz norm of the constant $m_{1}$ in order to get the measure estimates in Section 6.
The constant $m_{1}$ depends
in $\oo$ in two way: the first is trough the dependence on $\oo$ of the function $u$; secondly it presents also an explicit dependence on the external parameters. Clearly by \eqref{eq:3.2.44} we can get a bound only on the variation
$
|m_{1}(\oo,u(\oo_{1}))-m_{1}(\oo,u(\oo_{2}))|.
$
To estimate the $|\cdot|^{lip}$ seminorm we need also the \eqref{mammamia}.
\end{rmk}
The proof of Lemma \ref{lem:3.88} is based on the following strategy.
At each step we construct a transformation $\TT_{i}$ that
conjugates $\calL_{i}$ to $\calL_{i+1}$. We fix $\calL_{0}=\calL$. Moreover
the $\TT_{i}$ are symplectic, hence $\calL_{i}$ is Hamiltonian and has the form
\begin{equation}\label{elleham}
\begin{aligned}
\calL_{i}:=
\oo\cdot\del_{\f}\uno+
i(E+A_{2}^{(i)})\del_{xx}+iA_{1}^{(i)}\del_{x}+i(mE+A_{0}^{(i)})+\RR_{i},
\end{aligned}
\end{equation}
with $E$ defined in \eqref{lemmaccio44},
\begin{equation}\label{elle2ham}
A_{j}^{(i)}=A_{j}^{(i)}(\f,x):=
\left(\begin{matrix}a^{(i)}_{j} \!\! &\!\! b^{(i)}_{j} \\ \!\! -\bar{b}^{(i)}_{j} &\!\! -\bar{a}^{(i)}_{j} \end{matrix}\right),\;\;\;\;
j=0,1,2
\end{equation}
and $\RR_{i}$ is a pseudo-differential operator of order $\del_{x}^{-1}$.
Essentially we need to prove bounds like
\begin{eqnarray}\label{eq:B15aaaa}
\|(\TT_{i}^{\pm1}({u})-\uno){ h}\|_{s,\g}
&\!\!\!\leq\!\!\!& \e C(s) (\|{ h}\|_{s,\g}+\|{ u}\|_{s+\ka_{i},\g}\|{ h}\|_{\gots_{0}}), \label{eq:B15cham}\\
\|d_{{u}}(\TT^{\pm1}_{i})({u})[{ h}]{ g}\|_{s}
&\!\!\!\leq\!\!\!& \e C(s)\big(\|{ g}\|_{s+1}\|{ h}\|_{\gots_{0}+\ka_{i}}+
\|{g}\|_{2}\|{ h}\|_{s+\ka_{i}}\nonumber\\
&\!\!\!+\!\!\!&\|{u}\|_{s+\ka_{i}}\|{ g}\|_{2}\|{h}\|_{\gots_{0}}
\big),\label{eq:B15dham}
\end{eqnarray}
for suitable $\ka_{i}$ and on the
coefficients in (\ref{elleham}) we need
\begin{subequations}\label{eq:3.10ham}
\begin{align}
\|a_{j}^{(i)}({u})\|_{s,\g}, \|b_{j}^{(i)}({ u})\|_{s,\g}&\leq \e C(s)(1+\|{ u}\|_{s+\ka_{i},\g}),\label{eq:3.10aham}\\
\!\!\!\!\!\|d_{u}a_{j}^{(i)}({ u})[h]\|_{s}, \|d_{u}b_{j}^{(i)}({ u})[h]\|_{s}&\leq \!\e C(s)(\|{ h}\|_{s+\ka_{i}}\!+\!\|{u}\|_{s+\ka_{i}}\!+\!
\|{ h}\|_{\gots_{0}+\ka_{i}}),\label{eq:3.10bham}
\end{align}
\end{subequations}
for $j=0,1,2$ and $i=1,\ldots,7$ and on $\RR_{i}$ bounds like \eqref{eq:3.2.7ham} with $\ka_{i}$ instead of $\h_{1}$.
The bounds are based on repeated use of classical tame bounds and interpolation estimates of the Sobolev norms. The proof of such properties of the norm can be found in \cite{BBM} in Appendix A.
To conclude one combine the bounds of each transformation to obtain estimates on the compositions. It turn out that
the constant $\h_{1}$ contains all the loos of regularity of each step. We present only the construction
of the transformation that, in the Hamiltonian case, are more involved.
Moreover the difference between Lemma \ref{lem:3.88} and
the result contained in Section 3 of \cite{FP}
is also in equation \eqref{mammamia}. Indeed, in this case we need to prove that non degeneracy hypothesis \ref{hyp3ham}
persists during the steps in order to obtain the same lower bound (possibly with a worse constant) for the constant $m_{1}$ in \eqref{eq:3.2.44}. This fact will be used in Section 6 in order to perform measure estimates.
\paragraph{Step 1. Diagonalization of the second order coefficient}
In this section we want to diagonalize the second order term $(E+A_{2})$ in \eqref{lemmaccio4}.
By a direct calculation one can see that the matrix $(E+A_{2})$ has eigenvalues
$\la_{1,2}:=\sqrt{(1+a_{2})^{2}-|b_{2}|^{2}}$. if we set $a_{2}^{(1)}:=\la_{1}-1$ we have
that $a_{2}^{(1)}\in \RRR$ since $a_{2}\in \RRR$ for any $(\f,x)\in\TTT^{d+1}$ and $a_{i}, b_{i}$ are small.
We define the transformation $\TT^{-1}_{1} : {\bf H}^{0}\to{\bf H}^{0}$ as the matrix
$\TT_{1}^{-1}=\big((\TT_{1}^{-1})_{\s}^{\s'}\big)_{\s,\s'=\pm1}$ with
\begin{equation}\label{11}
\TT^{-1}_{1}:=
\left(\begin{matrix} (2+a_{2}+a_{2}^{(1)})(i\la_{0})^{-1} & b_{2}(i\la_{0})^{-1} \\
-\bar{b}_{2}(i\la_{0})^{-1} & -(2+a_{2}+a_{2}^{(1)})(i\la_{0})^{-1}\end{matrix}
\right),
\end{equation}
where $\la_{0}:=i\sqrt{2\la_{1}(1+a_{2}+\la_{1})}$. Note that ${\rm det}\TT^{-1}_{1}=1$.
One has that
\begin{equation}\label{fiorellinoham}
\TT^{-1}_{1}(E +A_{2})\TT_{1}=\left(\begin{matrix}1+a^{(1)}_{2}(\f,x) & 0\\ 0 & -1-a_{2}^{(1)}(\f,x)\end{matrix}\right).
\end{equation}
Moreover, we have that the transformation is symplectic. We can think that $\TT_{1}$ act on the function of $H^{s}(\TTT^{d+1};\CCC)$ is the following way.
Set $U=(u,\bar{u}),V=(v,\bar{v})\in {\rm H}^{s}$ and let $(MZ)_{\s}$ for $\s\in\{+1,-1\}$ be the first or
the second (respectively) component.
Given a function $u\in H^{s}(\TTT^{d+1};\CCC) $ we define, with abuse of notation, $\TT^{-1}_{1}u:=(\TT^{-1}_{1}U)_{+1}:=
((\TT^{-1}_{1})_{1}^{1})u+((\TT^{-1}_{1})_{1}^{-1})\bar{u}$.
With this notation one has that
\begin{equation*}
\begin{aligned}
\Omega\left(\TT^{-1}_{1}{ u},\TT^{-1}_{1}{ v} \right):={\rm Re}\int_{\TTT}
&i\left((\TT^{-1}_{1})_{1}^{1}(\TT^{-1}_{1})_{-1}^{1}uv+(\TT^{-1}_{1})_{1}^{-1}(\TT^{-1}_{1})_{-1}^{-1}\bar{u}\bar{v}\right)\\
&+i\left((\TT^{-1}_{1})_{1}^{1}(\TT^{-1}_{1})_{-1}^{-1}u\bar{v}+(\TT^{-1}_{1})_{1}^{-1}(\TT^{-1}_{1})_{-1}^{1}\bar{u}v\right){\rm d}x\\
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! ={\rm Re}\int_{\TTT} i{\rm Re}((\TT^{-1}_{1})_{1}^{1}(\TT^{-1}_{1})_{-1}^{1}uv)\\
&+i\left((\TT^{-1}_{1})_{1}^{-1}(\TT^{-1}_{1})_{-1}^{1}(u\bar{v}+\bar{u}v)\right)\\
&+i\left((\TT^{-1}_{1})_{1}^{1}(\TT^{-1}_{1})_{-1}^{-1}-(\TT^{-1}_{1})_{1}^{-1}(\TT^{-1}_{1})_{-1}^{1}\right)u\bar{v}{\rm d} x\\
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! ={\rm Re}\int_{\TTT} iu\bar{v}{\rm d}x=:\Omega(u,v).
\end{aligned}
\end{equation*}
which implies that $\TT_{1}^{-1}$ is symplectic.
Now we can conjugate the operator $\calL$ to an operator $\calL_{1}$ with
a diagonal coefficient of the second order spatial differential operator. Indeed, one has
\begin{equation}\label{eq:3.122}
\begin{aligned}
\calL_{1}&:=\TT_{1}^{-1}\calL \TT_{1}=\oo\cdot\del_{\f}\uno+i \TT_{1}^{-1}(E+A_{2})\TT_{1}\del_{xx}\\
&+
i(2\TT_{1}^{-1}(E+A_{2})\del_{x}\TT_{1}+\TT_{1}^{-1}A_{1}\TT_{1})\del_{x}\\
&+i\left[-i\TT_{1}^{-1}(\oo\cdot\del_{\f}\TT_{1})+\TT_{1}^{-1}(E+A_{2})\del_{xx}\TT_{1}\right.\\
&\left.+
\TT_{1}^{-1}A_{1}\del_{x}\TT_{1}+\TT_{1}^{-1}(mE+A_{0})\TT_{1}
\right];
\end{aligned}
\end{equation}
the (\ref{eq:3.122}) has the form \eqref{elleham}. This identify uniquely the coefficients
$a_{j}^{(1)},b_{j}^{(1)}$ for $j=0,1,2$ and $\RR_{1}$. In particular we have that
$b_{2}^{(1)}\equiv0$ and $\RR_{1}\equiv0$.
Moreover, since the transformation is symplectic, then the new operator $\calL_{1}$ is
Hamiltonian, with an Hamiltonian function
\begin{equation}\label{ham1}
\begin{aligned}
\!\!\!\!\!\!H_{1}(u,\bar{u})&=\int_{\TTT}(1+a_{2}^{(1)})|u_{x}|^{2}-\frac{i}{2}{\rm Im}(a_{1}^{(1)})(u_{x}\bar{u}-u\bar{u}_{x})\!-\!
{\rm Re}(a_{0}^{(1)})|u|^{2}{\rm d}x\\
&+\int_{\TTT}\!\!-m|u|^{2}\!-\!\frac{1}{2}(b_{0}^{(1)}\bar{u}^{2}+\bar{b}_{0}^{(1)}u^{2}){\rm d}x
:=\int_{\TTT}f_{1}(\f,x,u,\bar{u},u_{x},\bar{u}_{x}){\rm d}x,
\end{aligned}
\end{equation}
hence, since $f_{1}$ depends only linearly on $\bar{u}_{x}$, thanks to the Hamiltonian structure,
one has
\begin{equation}
b_{1}^{(1)}(\f,x)=\frac{\rm d}{{\rm d}x}\left(\del_{\bar{z}_{1}\bar{z_{1}}}f_{1}\right)\equiv0.
\end{equation}
This means that we have diagonalized also the matrix of the first order spatial differential operator.
\begin{rmk}\label{rmk23}
It is important to note that $a_{1}^{(1)}(\f,x)$ as the form
$$
a_{1}^{(1)}(\f,x)=\frac{\rm d}{{\rm d}x}a_{2}^{(1)}(\f,x)+\del_{z_{0}\bar{z}_{1}}f_{1}-\del_{z_{1}\bar{z}_{0}}f_{1}
$$
so that the real part of $a_{1}^{(1)}$ depends only on the spatial derivative of $a_{2}^{(1)}$.
\end{rmk}
\paragraph{Step 2. Change of the space variable}
We consider a $\f-$dependent family of diffeomorphisms of the $1-$dimensional torus $\TTT$
of the form
\begin{equation}\label{20ham}
y=x+\x(\f,x),
\end{equation}
where $\x$ is as small real-valued funtion, $2\pi$ periodic in all its arguments. We define the change of variables
on the space of functions as
\begin{equation}\label{21ham}
\begin{aligned}
(\TT_{2} h)(\f,x)&:=\sqrt{1+\x_{x}(\f,x)}h(\f,x+\x(\f,x)), \; {\rm with\;\; inverse} \;\;\;\;\\
(\TT_{2}^{-1}v)(\f,y)&:=\sqrt{1+\widehat{\x}_{x}(\f,y)}v(\f,y+\widehat{\x}(\f,y))
\end{aligned}
\end{equation}
where
\begin{equation}\label{20bisham}
x=y+\widehat{\x}(\f,y),
\end{equation}
is the inverse diffeomorphism of $(\ref{20ham})$.
With a slight abuse of notation we extend the operator to ${\bf H}^s$:
\begin{equation}\label{eq:3.14ham}
{{\TT_{2}}} : {\bf H}^{s}\to {\bf H}^{s}, \quad {\TT_{2}}\left(\begin{matrix} h \\ \bar{h}\end{matrix}\right)=\left(
\begin{matrix} (\TT_{2} h)(\f,x) \\ (\TT_{2} \bar{h})(\f,x)\end{matrix}\right).
\end{equation}
Now we have to calculate the conjugate ${\TT_{2}}^{-1}\calL_{1}{\TT_{2}}$
of the operator $\calL_{1}$ in
$(\ref{eq:3.122})$.
\noindent
The conjugate $\TT_{2}^{-1} a\TT_{2}$ of any multiplication operator
$a : h(\f,x) \to a(\f,x)h(\f,x)$ is the multiplication operator
\begin{equation}\label{con}
v(\f,y)\mapsto(\TT_{2}^{-1}a\sqrt{1+\x_{x}})(\f,y)v(\f,y)=a(\f,y+\widehat{\x}(\f,y))v(\f,y).
\end{equation}
In \eqref{con} we have used the relation
\begin{equation}\label{con2}
0\equiv \x_{x}(\f,x)+\widehat{\x}_{y}(\f,y)+\x_{x}(\f,x)\widehat{\x}_{y}(\f,y),
\end{equation}
that follow by \eqref{20ham} and \eqref{20bisham}.
The conjugate of the differential operators will be
\begin{equation}\label{22ham}
\begin{aligned}
\TT_{2}^{-1}\oo\cdot\del_{\f}\TT_{2}&=\oo\cdot\del_{\f}+[\TT_{2}^{-1}(\oo\cdot\del_{\f}\x)]\del_{y}
-\TT_{2}^{-1}\left(\frac{\oo\cdot\del_{\f}\x_{x}}{2(1+\x_{x})}\right)
,\\
\TT_{2}^{-1}\del_{x}\TT_{2}&=[\TT_{2}^{-1}(1+\x_{x})]\del_{y}-\TT_{2}^{-1}\left(
\frac{\x_{xx}}{2(1+\x_{x})}\right),\\
\TT_{2}^{-1}\del_{xx}\TT_{2}&=[\TT_{2}^{-1}(1+\x_{x})^{2}]\del_{yy}-\TT_{2}^{-1}\left(\frac{2\x_{xxx}+\x_{xx}^{2}}{4(1+\x_{x})^{2}} \right),
\end{aligned}
\end{equation}
where all the coefficients
are periodic functions of $(\f,x)$.
\noindent
Thus, by conjugation, we have that $\calL_{2}=\TT_{2}^{-1}\calL_{1}\TT_{2}$ has the form (\ref{elleham})
with
\begin{equation}\label{24ham}
\begin{aligned}
&\!\!\!\!\!\!1+a_{2}^{(2)}(\f,y)=\TT_{2}^{-1}[(1+a_{2}^{(1)})(1+\x_{x})^{2}],
\\
&\!\!\!\!\!\! a_{1}^{(2)}(\f,y)=
\TT_{2}^{-1}(a_{1}^{(1)}(1+\x_{x}))-i \TT_{2}^{-1}(\oo\cdot\del_{\f}\x),
\\
&\!\!\!\!\!\! a_{0}^{(2)}(\f,y)=i\TT_{2}^{-1}\!\!\left(\frac{\oo\cdot\del_{\f}\x_{x}}{2(1+\x_{x})}\right)
\!-\!\TT_{2}^{-1}\!\left(\frac{\x_{xx}}{2(1+\x_{x})}\right)
\!-\!\TT_{2}^{-1}\!\left(\frac{2\x_{xxx}+\x_{xx}^{2}}{4(1+\x_{x})^{2}} \right)\!,
\\
&\!\!\!\!\!\! b^{(2)}_{0}(\f,y)=\TT_{2}^{-1}(b^{(1)}_{0}),
\end{aligned}
\end{equation}
and $b_{2}^{(2)}=b_{1}^{(2)}=0$.
We are looking for $\x(\f,x)$ such that the coefficient $a_{2}^{(2)}(\f,y)$ does not depend on
$y$, namely
\begin{equation}\label{255}
1+a_{2}^{(2)}(\f,y)=\TT_{2}^{-1}[(1+a_{2}^{(1)})(1+\x_{x})^{2}]=1+a_{2}^{(2)}(\f),
\end{equation}
for some function $a_{2}^{(2)}(\f)$. Since $\TT_{2}$ operates only on the space variables, the $(\ref{255})$
is equivalent to
\begin{equation}\label{266}
(1+a_{2}^{(1)}(\f,x))(1+\x_{x}(\f,x))^{2}=1+a_{2}^{(2)}(\f).
\end{equation}
Hence we have to set
\begin{equation}\label{277}
\x_{x}(\f,x)=\rho_{0}, \qquad \rho_{0}(\f,x):=(1+a_{2}^{(2)}(\f))^{\frac{1}{2}}(\f)(1+a_{2}^{(1)}(\f,x))^{-\frac{1}{2}}-1,
\end{equation}
that has solution $\g$ periodic in $x$ if and only if $\int_{\TTT}\rho_{0} dy=0$. This condition
implies
\begin{equation}\label{288}
a_{2}^{(2)}(\f)=\left(\frac{1}{2\pi}\int_{\TTT}(1+a_{2}^{(1)}(\f,x))^{-\frac{1}{2}}\right)^{-2}-1.
\end{equation}
Then we have the solution (with zero average) of $(\ref{277})$
\begin{equation}\label{299}
\x(\f,x):=(\del_{x}^{-1}\rho_{0})(\f,x),
\end{equation}
where $\del_{x}^{-1}$ is defined by linearity as
\begin{equation}\label{300}
\del_{x}^{-1}e^{ikx}:=\frac{e^{ikx}}{ik}, \quad \forall \; k\in\ZZZ\backslash\{0\}, \quad \del_{x}^{-1}=0.
\end{equation}
In other word $\del_{x}^{-1}h$ is the primitive of $h$ with zero average in $x$.
Moreover, the map $\TT_{2}$ is canonical with respect to the $NLS-$symplectic form, indeed, for any $u,v\in H^{s}(\TTT^{d+1};\CCC)$,
\begin{equation*}
\begin{aligned}
\Omega(\TT_{2} u, \TT_{2} v)=&
{\rm Re}\int_{\TTT}(i\sqrt{1+\x_{x}}u(\f,x
+\f(\f,x)))\sqrt{1+\x_{x}}\bar{v}(\f,x
+\f(\f,x)){\rm d}x\\
&={\rm Re}\int_{\TTT}(1+\x_{x}(\f,x))(iu(\f,x+\x(\f,x)))\bar{v}(\f,x+\x(\f,x)){\rm d}x\\
&={\rm Re}\int_{\TTT}(iu(\f,y))\bar{v}(\f,y){\rm d}y=:\Omega(u,v).
\end{aligned}
\end{equation*}
Thus, conjugating $\calL_{1}$ through the operator $\TT_{2}$ in (\ref{eq:3.14ham})
we obtain the Hamiltonian operator
$\calL_{2}= \TT_{2}^{-1}\calL_{1} \TT_{2}$
with Hamiltonian function given by
\begin{equation}\label{linham2}
\begin{aligned}
\!\!\!\!\!\!\!H_{2}(u,\bar{u})&\!=\!\int_{\TTT}(1+a_{2}^{(2)}(\f))|u_{x}|^{2}\!-\!\frac{i}{2}{\rm Im}(a_{1}^{(2)})(u_{x}\bar{u}\!-\!u\bar{u}_{x})
\!-\!
{\rm Re}(a_{0}^{(2)})|u|^{2}{\rm d}x\\
&\!+\!\int_{\TTT}\!-\!m|u|^{2}\!-\!\frac{1}{2}(b_{0}^{(2)}\bar{u}^{2}+\bar{b}^{(2)}_{0}u^{2}){\rm d}x
\!:=\!\int_{\TTT}f_{2}(\f,x,u,\bar{u},u_{x},\bar{u}_{x}){\rm d}x,
\end{aligned}
\end{equation}
\begin{rmk}\label{rmk24}
As in Remark \ref{rmk23}, the real part of coefficients $a_{1}^{(2)}$ depends on the spatial derivatives
of $a_{2}^{(2)}$, then in this case, again thanks the Hamiltonian structure of the problem,
one has that
$a_{1}^{(2)}(\f,y)=i{\rm Im}(a_{1}^{(2)})(\f,y)$, i.e. it is purely imaginary.
Moreover $b_{2}^{(2)}=b_{1}^{(2)}\equiv 0$ and $\RR_{2}\equiv0$.
\end{rmk}
\paragraph{Step 3: Time reparametrization}
In this section we want to make constant the coefficient of the highest order spatial derivative
operator $\del_{yy}$ of $\calL_{2}$, by a quasi-periodic reparametrization of time.
We consider a diffeomorphism of the torus $\TTT^{d}$ of the form
\begin{equation}\label{32ham}
\theta=\f+\oo\al(\f), \quad \f\in\TTT^{d}, \quad \al(\f)\in\RRR,
\end{equation}
where $\al$ is a small real valued function, $2\pi-$periodic in all its arguments. The induced
linear operator on the space of functions is
\begin{equation}\label{332}
(\TT_{3} h)(\f,y):=h(\f+\oo\al(\f),y),
\end{equation}
whose inverse is
\begin{equation}\label{333}
(\TT_{3}^{-1}v)(\theta,y)=v(\theta+\oo\tilde{\al}(\theta),y),
\end{equation}
where $\f=\theta+\oo\tilde{\al}(\theta)$ is the inverse diffeomorphism of
$\theta=\f+\oo\al(\f)$. We extend the operator to ${\bf H}^{s}$:
\begin{equation}\label{eq:3.15ham}
\TT_{3} : {\bf H}^{s}\to {\bf H}^{s}, \quad \TT_{3}\left(\begin{matrix} h \\ \bar{h}\end{matrix}\right)=\left(
\begin{matrix} (\TT h)(\f,x) \\ (\TT_{3} \bar{h})(\f,x)\end{matrix}
\right).
\end{equation}
\noindent
By conjugation, we have that the differential operator become
\begin{equation}\label{333}
\TT_{3}^{-1}\oo\cdot\del_{\f}\TT_{3}=\rho(\theta)\oo\cdot\del_{\theta},\;\;
\TT_{3}^{-1}\del_{y}\TT_{3}=\del_{y}, \;\;\rho(\theta):=\TT_{3}^{-1}(1+\oo\del_{\f}\al).
\end{equation}
Hence we have $\TT_{3}^{-1}\calL_{2}\TT_{3}=\rho \calL_{3}$ where $\calL_{3}$ has the form \eqref{elleham}
and
\begin{equation}\label{eq:3.166}
\begin{aligned}
1+a_{i}^{(3)}(\theta)&:=\frac{(\TT_{3}^{-1}(1+a_{i}^{(2)}))(\theta)}{\rho(\theta)},
\quad a_{i}^{(3)}(\theta):=\frac{(\TT_{3}^{-1}a_{i}^{(2)})(\theta)}{\rho(\theta)}, \;\; i=0,1, \\
& b_{0}^{(3)}(\theta,y):=\frac{(\TT_{3}^{-1}b_{0}^{(2)})(\theta,y)}{\rho(\theta)},
\end{aligned}
\end{equation}
We look for solution $\al$ such that the coefficient $a_{2}^{(3)}$ is constant in time, namely
\begin{equation}\label{355}
(\TT_{3}^{-1}(1+a_{2}^{(2)}))(\theta)=m_{2}\rho(\theta)=m_{2}\TT_{3}^{-1}(1+\oo\cdot\del_{\f}\al)
\end{equation}
for some constant $m_{2}$, that is equivalent to require that
\begin{equation}\label{366}
1+a_{2}^{(2)}(\f)=m_{2}(1+\oo\cdot\del_{\f}\al(\f)),
\end{equation}
By setting
\begin{equation}\label{377}
m_{2}=\frac{1}{(2\pi)^{d}}\int_{\TTT^{d}}(1+a_{2}^{2}(\f))d\f,
\end{equation}
we can find the (unique) solution of $(\ref{366})$ with zero average
\begin{equation}\label{388}
\al(\f):=\frac{1}{m_{2}}(\oo\cdot\del_{\f})^{-1}(1+a_{2}^{(2)}-m_{2})(\f),
\end{equation}
where $(\oo\cdot\del_{\f})^{-1}$ is defined by linearity
$$
(\oo\cdot\del_{\f})^{-1}e^{i\ell\cdot\f}:=\frac{e^{i\ell\cdot\f}}{i\oo\cdot\ell}, \; \ell\neq0, \quad
(\oo\cdot\del_{\f})^{-1}1=0.
$$
Moreover, the operator $\TT_{3}$ acts only on the time variables, then it is clearly symplectic, since
$$
\Omega(\TT_{3} u, \TT_{3} v)=\Omega(u,v).
$$
Then the operator $\calL_{3}$ is Hamiltonian with hamiltonian function $H_{3}$
\begin{equation}\label{linham3}
\begin{aligned}
H_{3}(u,\bar{u})&=\int_{\TTT}m_{2}|u_{x}|^{2}\!-\!\frac{i}{2}{\rm Im}(a_{1}^{(3)})(u_{x}\bar{u}-u\bar{u}_{x})\!-\!
{\rm Re}(a_{0}^{(3)})|u|^{2}{\rm d}x\\
&+\int_{\TTT}-\frac{1}{2}(b_{0}^{(3)}\bar{u}^{2}+\bar{b}^{(3)}_{0}u^{2}){\rm d}x
:=\int_{\TTT}f_{3}(\f,x,u,\bar{u},u_{x},\bar{u}_{x}){\rm d}x,
\end{aligned}
\end{equation}
\begin{rmk}\label{rmk25}
Also in this case, thanks to the hamiltonian structure of the operator, we have that
the coefficient $a_{1}^{(3)}\in i\RRR$, $b_{2}^{(3)}=b_{1}^{(3)}\equiv 0$ and $\RR_{3}\equiv0$.
\end{rmk}
\paragraph{Step 4. Change of space variable (translation)}
The goal of this section, is to conjugate
$\calL_{3}$ in (\ref{elleham}) with coefficients in (\ref{eq:3.166})
to an operator in which the coefficients of the first order spatial derivative
operator, has zero average in $y$.
Consider the change of the space variable
\begin{equation}\label{eq:3.4.2ham}
z=y+\be(\theta)
\end{equation}
which induces the operators on functions
\begin{equation}\label{eq:3.4.3ham}
\TT_{4} h(\theta,y):=h(\theta,y+\be(\theta)), \quad \TT_{4}^{-1}v(\theta,z-\be(\theta)).
\end{equation}
We extend the operator $ \TT_{4}$ to ${\bf H}^{s}$ as
\begin{equation}\label{eq:3.4.4ham}
\TT_{4}\left(\begin{matrix} h \\ \bar h\end{matrix}\right)=\left(\begin{matrix}(\TT_{4} h)(\theta,y) \\ (\TT_{4}\bar{h})(\theta,y) \end{matrix}\right).
\end{equation}
By conjugation, the differential operators become
\begin{equation}\label{eq:3.4.5ham}
\TT_{4}^{-1}\oo\cdot\del_{\theta}\TT_{4}=\oo\cdot\del_{\theta}+(\oo\cdot\del_{\theta}\be(\theta))\del_{z}, \qquad
\TT_{4}^{-1}\del_{y}\TT_{4}=\del_{z}.
\end{equation}
Hence one has that $\calL_{4}:=\TT_{4}^{-1}\calL_{3}\TT_{4}$ has the form \eqref{elleham}
where
\begin{equation}\label{eq:3.4.7ham}
\begin{aligned}
a_{1}^{(4)}(\theta,z)&:=-i\oo\cdot\del_{\theta}\be(\theta)+(\TT_{4}^{-1}a_{1}^{(3)})(\theta,z), \\
a_{0}^{(4)}(\theta,z):=(&\TT_{4}^{-1}a_{0}^{(3)})(\theta,z), \quad
b_{0}^{(4)}(\theta,z):=(\TT_{4}^{-1}b_{0}^{(4)})(\theta,z).
\end{aligned}
\end{equation}
The aim is to find a function $\be(\theta) $ such that
\begin{equation}\label{eq:3.4.8ham}
\frac{1}{2\pi}\int_{\TTT}a_{1}^{(4)}(\theta,z){\rm d}z=m_{1}, \quad \forall \; \theta\in\TTT^{d},
\end{equation}
for some constant $m_{1}\in\CCC$, independent on $\theta$. By using the (\ref{eq:3.4.7ham}) we have that
the (\ref{eq:3.4.8ham}) become
\begin{equation}\label{eq:3.4.9ham}
-i\oo\cdot\del_{\theta}\be(\theta)=m_{1}-\int_{\TTT}a_{1}^{(3)}(\theta,y){\rm d}y=:V(\theta).
\end{equation}
This equation has a solution periodic in $\theta$ if and only if $V(\theta)$
has zero average in $\theta$. So that
we have to define
\begin{equation}\label{eq:3.4.10ham}
m_{1}:=\frac{1}{(2\pi)^{d+1}}\int_{\TTT^{d+1}}a_{1}^{(3)}(\theta,y){\rm d}\theta{\rm d}y.
\end{equation}
Note also that $m_{1}\in i\RRR$ (see Remark \ref{rmk25}). Then the function $V$
is purely imaginary. Now we can set
\begin{equation}\label{eq:3.4.11ham}
\be(\theta):=i(\oo\cdot\del_{\theta})^{-1}V(\theta),
\end{equation}
to obtain a real diffeomorphism of the torus $y+\be(\theta)$.
Morover one has, for any $u,v\in H^{s}(\TTT^{d+1};\CCC)$
\begin{equation}\label{eq:3.4.13ham}
\Omega(\TT_{4} u,\TT_{4} v)={\rm Re}\int_{\TTT}iu(\f,x+\be(\f))\bar{v}(\f,x+\be(\f))=\Omega(u,v),
\end{equation}
hence $\TT_{4}$ is symplectic. This implies that $\calL_{4}$ is Hamiltonian with
hamiltonian function of the form
\begin{equation}\label{linham4}
\begin{aligned}
\!\!\!\!\!\!H_{4}(u,\bar{u})&=\int_{\TTT}m_{2}|u_{x}|^{2}\!-\!\frac{i}{2}{\rm Im}(a_{1}^{(4)})(u_{x}\bar{u}-u\bar{u}_{x})\!-\!
{\rm Re}(a_{0}^{(4)})|u|^{2}-m|u|^{2}{\rm d}x\\
&+\int_{\TTT}-\frac{1}{2}(b_{0}^{(4)}\bar{u}^{2}+\bar{b}_{0}^{(4)}u^{2}){\rm d}x
:=\int_{\TTT}f_{4}(\f,x,u,\bar{u},u_{x},\bar{u}_{x}){\rm d}x,
\end{aligned}
\end{equation}
\begin{rmk}\label{rmk35ham}
Again one has $b_{2}^{(4)}=b_{1}^{(4)}\equiv 0$ and $\RR_{4}\equiv0$.
\end{rmk}
For simplicity we rename the variables $z=x$ and $\theta=\f$.
\paragraph{Step 5. Descent Method: conjugation by multiplication operator}
In this section we want to eliminate the dependance on $\f$ and $x$ on the coefficient
$c_{9}$ of the operator $\calL_{4}$.
To do this, we consider an operator
of the form
\begin{equation}\label{eq:3.5.1ham}
\TT_{5}:=\left(\begin{matrix}1+z(\f,x) & 0 \\ 0 & 1+\bar{z}(\f,x) \end{matrix}\right),
\end{equation}
where $z : \TTT^{d+1}\to \CCC$. By a direct calculation we have that
\begin{equation}\label{eq:3.5.2ham}
\begin{aligned}
\calL_{4}\TT_{5}-\TT_{5}&\left[\oo\cdot\del_{\f}\uno+
i\left(\begin{matrix}m_{2} \!\!\! & 0 \\ 0 &\!\!\! -m_{2} \end{matrix}\right)\del_{xx}+i
\left(\begin{matrix}m_{1} \!\!\! & 0 \\ 0 &\!\!\! -\ol{m}_{1} \end{matrix}\right)\del_{x}
\right]=\\
&=i\left(\begin{matrix}r_{1}(\f,x) \!\!\! & 0 \\ 0 &\!\!\! -\ol{r}_{1}(\f,x) \end{matrix}\right)\del_{x}
+i\left(\begin{matrix}m+c(\f,x) \!\!\! & d(\f,x) \\ -\ol{d}(\f,x) &\!\!\!-m -
\ol{c}(\f,x) \end{matrix}\right)
\end{aligned}
\end{equation}
where
\begin{equation}\label{eq:3.5.3ham}
\begin{aligned}
r_{1}(\f,x)&:=2m z_{x}(\f,x)+(a_{1}^{(4)}(\f,x)-m_{1})(1+z(\f,x)),\\
c(\f,x)&:=-i(\oo\cdot\del_{\f}z)(\f,x)+a_{0}^{(4)}(\f,x)(1+z(\f,x)),\\
d(\f,x)&:=b_{0}^{(4)}(\f,x)(1+\bar{z}(\f,x)).
\end{aligned}
\end{equation}
We look for $z(\f,x)$ such that $r_{1}\equiv0$. If we look for solutions of the form
$1+z(\f,x)=\exp(s(\f,x))$ we have that $r_{1}=0$ become
\begin{equation}\label{eq:3.5.7ham}
2m_{2} s_{x}+a_{1}^{(4)}-m_{1}=0,
\end{equation}
that has solution
\begin{equation}\label{eq:3.5.8ham}
s(\f,x):=\frac{1}{2m}\del_{x}^{-1}(a_{1}^{(4)}-m_{1})(\f,x)
\end{equation}
where $\del_{x}^{-1}$ is defined in (\ref{300}). Moreover, since
$a_{1}^{(4)}\in i\RRR$, one has that $s(\f,x)\in i\RRR$. Clearly the operator $\TT_{5}$ is invertible for
$\e$ small, then we obtain $\calL_{5}:=\TT_{5}^{-1}\calL_{4}\TT_{5}$ with
\begin{equation}\label{eq:3.5.9bisham}
\calL_{5}:=
\oo\cdot\del_{\f}\uno+i\left(\begin{matrix}m_{2}\!\!\! & 0 \\ 0 & \!\!\! -m_{2}\end{matrix}\right)\del_{xx}+
i\left(\begin{matrix}m_{1}\!\!\! & 0 \\ 0 & \!\!\! -\bar{m}_{1}\end{matrix}\right)\del_{x}+imE
+iA_{0}^{(5)}
\end{equation}
that has the form \eqref{elleham} with
$m_{2}$ and $m_{1}$ are defined respectively in (\ref{377}) and (\ref{eq:3.4.10ham}), while the coefficients of $A_{5}^{(i)}$ are
\begin{equation}\label{eq:3.5.10ham}
\begin{aligned}
a_{0}^{(5)}(\f,x)&:=(1+z(\f,x))^{-1}c(\f,x),\\
b_{0}^{(5)}(\f,x)&:=(1+z(\f,x))^{-1}d(\f,x).
\end{aligned}
\end{equation}
It remains to check that the transformation $\exp(s(\f,x))=1+z$ is symplectic.
One has
\begin{equation}\label{eq:3.5.11ham}
\Omega(e^{s}u,e^{s}v)={\rm Re}\int_{\TTT}ie^{s(\f,x)}u(\f,x)e^{-s(\f,x)}\bar{v}(\f,x){\rm d}x=
\Omega(u,v),
\end{equation}
where we used that $\bar{s}=-s$, that follows by $s\in i\RRR$.
Hence the operator $\calL_{5}$ is Hamiltonian, with corresponding hamiltonian function
\begin{equation}\label{linham5}
\begin{aligned}
\!\!\!\!\!\!\!H_{5}(u,\bar{u})&=\int_{\TTT}m_{2}|u_{x}|^{2}-\frac{i}{2}{\rm Im}(m_{1})(u_{x}\bar{u}-u\bar{u}_{x})-
{\rm Re}(a_{0}^{(5)})|u|^{2}{\rm d}x\\
&\!+\!\int_{\TTT}\!-\!m|u|^{2}\!-\!\frac{1}{2}(b_{0}^{(5)}\bar{u}^{2}+\bar{b}_{0}^{(5)}u^{2}){\rm d}x
:=\int_{\TTT}f_{5}(\f,x,u,\bar{u},u_{x},\bar{u}_{x}){\rm d}x.
\end{aligned}
\end{equation}
Again using the Hamiltonian structure, see (\ref{eq:1111}), we can conclude that
\begin{equation}\label{eq:3.5.12ham}
{\rm Im}(a_{0}^{(5)})(\f,x)=\frac{\rm d}{{\rm d}x}{\rm Im}(m_{1})\equiv0,
\end{equation}
that implies $a_{0}^{(5)}\in \RRR$.
\begin{rmk}\label{rmk45}
We have $b_{2}^{(5)}=b_{1}^{(5)}\equiv 0$ and $\RR_{5}\equiv0$.
\end{rmk}
\paragraph{Step 6. Descent Method: conjugation by pseudo-differential operator}
In this section we want to conjugate $\calL_{5}$ in (\ref{eq:3.5.9bisham}) to an operator of the
form $\oo\cdot\del_{\f}+iM\del_{xx}+iM_{1}\del_{x}+\RR$ where
\begin{equation}\label{eq:3.6.1ham}
M=\left(\begin{matrix}m_{2} & 0 \\ 0 & -m_{2}\end{matrix}\right), \quad M_{1}=
\left(\begin{matrix}m_{1} & 0 \\ 0 & -\bar{m}_{1}\end{matrix}\right),
\end{equation}
and $\RR$ is a pseudo differential operator of order $0$.
We consider an operator of the form
\begin{equation}\label{eq:3.6.2ham}
\tilde{\SSSS}:=\left(\begin{matrix}1+w\Upsilon & 0 \\ 0 & 1+\bar{w}\Upsilon \end{matrix}\right),
\end{equation}
where $w: \TTT^{d+1}\to \RRR$ and $\Upsilon=(1-\del_{xx})\frac{1}{i}\del_{x}$
is defined by linearity as
$$
\Upsilon e^{ijx}=\frac{1}{1+j^{2}}je^{ijx}
$$
We have that the difference
\begin{equation*}\label{eq:3.6.3ham}
\begin{aligned}
\calL_{5}\tilde{\SSSS}-\tilde{\SSSS}&\left[\oo\cdot\del_{\f}\uno\!+\!i
\left(\begin{matrix}m_{2} \!\!\!& 0 \\ 0 & \!\!\!-m_{2}\end{matrix}\right)\del_{xx}\!+\!
i\left(\begin{matrix}m_{1} \!\!\!& 0 \\ 0 & \!\!\!-\bar{m}_{1}\end{matrix}\right)\del_{x}
\!+\!i \left(\begin{matrix}m\!+\!\hat{a}_{0}^{(5)} \!\!\!& b_{0}^{(5)} \\ -\bar{b}_{0}^{(5)} & \!\!\!-\!m\!-\!\hat{a}_{0}^{(5)}\end{matrix}\right)\right]=\\
&=i\left(\begin{matrix}r_{0} & 0 \\ 0 & -\bar{r}_{0} \end{matrix}\right)+
\RR
\end{aligned}
\end{equation*}
where $b_{0}^{(5)}$ is defined in (\ref{eq:3.5.10ham}) and
\begin{equation}\label{eq:3.6.4ham}
\begin{aligned}
r_{0}(\f,x)&:=2m_{2}w_{x}\Lambda\del_{x}+(a_{0}^{(5)}(\f,x)-\hat{a}_{0}^{(5)}(\f)),
\quad \RR
=i\left(\begin{matrix} \tilde{p}_{0} & \tilde{q}_{0} \\ -\bar{\tilde{q}}_{0} & -\bar{\tilde{p}}_{0}
\end{matrix}\right)
\\
\tilde{p}_{0}(\f,x)&:=-i(\oo\cdot\del_{\f}w) \Upsilon +m_{2}w_{xx}\Upsilon +m_{1}w_{x}\Upsilon +
(a_{0}^{(5)}-\hat{a}_{0}^{(5)})w\Upsilon,\\
\tilde{q}_{0}(\f,x)&:=b_{0}^{(5)} \bar{w}\Upsilon-w\Upsilon b_{0}^{(5)}.
\end{aligned}
\end{equation}
We are looking for $w$ such that $r_{0}\equiv0$ or at least $r_{0}$ is ``small'' in some sense. The operator $\RR$ is a pseudo-differential operator of order $-1$.
We can also note that
$$
\Upsilon\del_{x}u=i u-i(1-\del_{xx})^{-1}u
$$
Since the second term is of order $-2$, we want to solve the equation
$$
2imw_{x} +(a_{0}^{(5)}-\hat{a}_{0}^{(5)})u\equiv0.
$$
This equation has solution if and only if
we define
\begin{equation}\label{eq:3.6.5ham}
\hat{a}_{0}^{(5)}(\f):=\frac{1}{2\pi}\int_{\TTT}a_{0}^{(5)}(\f,x){\rm d}x,
\end{equation}
and it is real thanks to (\ref{eq:3.5.12ham}). Now, we define
\begin{equation}\label{eq:3.6.6ham}
w(\f,x):=i\frac{1}{2m} \del_{x}^{-1}(a_{0}^{(5)}-\hat{a}_{0}^{(5)})(\f,x),
\end{equation}
that is a purely imaginary function.
In this way we can conjugate the operator $\calL_{5}$ to an operator of the form
$\oo\cdot\del_{\f}+iM\del_{xx}+iM_{1}\del_{x}+iM_{0}+O(\del_{x}^{-1})$ with the diagonal part of $M_{0}$ constant in the space variable.
Unfortunately, this transformation in not symplectic.
We reason as follow. Let $w=i(w+\bar{w}):=i a$ and consider the Hamiltonian function
$$
H(u,\bar{u})=\frac{1}{2}\int_{\TTT}-(a\Upsilon+\Upsilon a)u\cdot \bar{u}d x.
$$
Since the function $a$ is real, and the operator $\Upsilon : L^{2}(\TTT;\CCC)\to L^{2}(\TTT,\CCC)$
is self-adjoint, then the operator $a\Upsilon+\Upsilon a$ is self-adjont. As consequence the
hamiltonian $H$ is real-valued on $L^{2}$. The corresponding (linear) vector field is
$$
\chi_{H}(u,\bar{u})=-i\left(\begin{matrix}\del_{\bar{u}}H\\ \del_{u}H\end{matrix}\right)
=\left(\begin{matrix}\frac{i}{2}(a\Upsilon+\Upsilon a)u \\
-\frac{i}{2}(a\Upsilon+\Upsilon a)\bar{u}\end{matrix}
\right).
$$
Then, the 1-flow of $\chi_{H}$ generates a symplectic transformation of coordinates, given by
\begin{equation}\label{megaop}
\begin{aligned}
\TT_{6}&:=\exp(\chi_{H}(u,\bar{u})):=\left(\begin{matrix}e^{i\chi} & 0 \\ 0& e^{-i\chi} \end{matrix}\right)\left(\begin{matrix}u\\ \bar{u}\end{matrix}
\right),\\
& e^{i\chi}u:=\left(\sum_{m=0}^{\infty}\frac{1}{m!}\Big(\frac{1}{2}(a\Upsilon+\Upsilon a)\Big)^{m}\right)u.
\end{aligned}
\end{equation}
We can easily check that, the operators in (\ref{megaop}) and (\ref{eq:3.6.2ham}) differs only for an operator
of order $O(\del_{x}^{-2})$. Indeed one has
\begin{equation}\label{megaop2}
\begin{aligned}
e^{i\chi}u&=u+\frac{i}{2}(a\Upsilon+\Upsilon a)u+O(\del_{x}^{-2})=u+\frac{i}{2}a\Upsilon u+
\frac{i}{2}\Upsilon(au)+O(\del_{x}^{-2})\\
&=u+\frac{i}{2}a\Upsilon u+\frac{i}{2}a \Upsilon u
+\frac{i}{2}\frac{1}{i}\del_{x}\Big((1-\del_{xx})^{-1}a_{xx}(1-\del_{xx})^{-1}u\\
&+2(1-\del_{xx})^{-1}a_{x}(1-\del_{xx})^{-1}\del_{x}u\Big)+O(\del_{x}^{-2})\\
&=(\uno+ia\Upsilon)u+O(\del_{x}^{-2}).
\end{aligned}
\end{equation}
In (\ref{megaop2}) we essentially studied the commutator of the pseudo-differential operator
$(1-\del_{xx})^{-1}$ with the operator of multiplication by the function $a$. Since the transformation
$\TT_{6}$ is symplectic we obtain the hamiltonian operator
\begin{eqnarray}\label{pippopippoham}
\calL_{6}&=&\TT_{6}^{-1}\calL_{5}\TT_{6}=\tilde{\calL}_{5}+\tilde\RR,\\
\tilde{\calL}_{5}&:=&\oo\cdot\del_{\f}\uno+im_{2}E\del_{xx}+
i\left(\begin{matrix}m_{1} \!\!\!& 0 \\ 0 & \!\!\!-\bar{m}_{1}\end{matrix}\right)\del_{x}
+imE+i \left(\begin{matrix}\hat{a}_{0}^{(5)}(\f) & b_{0}^{(5)} \\ -\bar{b}_{0}^{(5)} & -\hat{a}_{0}^{(5)}(\f)\end{matrix}\right),\nonumber\\
\tilde\RR&:=&\TT_{6}^{-1}\left[\calL_{5}\TT_{6}-\TT_{6}\tilde{\calL}_{5}
\right],\nonumber
\end{eqnarray}
where $\RR$ is hamiltonian and of order $O(\del_{x}^{-1})$.
\begin{rmk}\label{rmk55}
Here we have that $\calL_{6}$ has the form \eqref{elleham} where
$b_{2}^{(6)}=b_{1}^{(6)}\equiv 0$, $a_{0}^{(6)}:=\hat{a}_{0}^{(5)}$, $b_{0}^{(6)}:=b_{0}^{(5)}$ and $\RR_{6}:=\tilde\RR$.
\end{rmk}
\paragraph{Step 7. Descent Method: conjugation by multiplication operator II}
In this section we want to eliminate the dependance on the time variable of the coefficients $\hat{a}_{0}^{(5)}(\f)$ in (\ref{eq:3.6.5ham}).
Consider the operator
\begin{equation}\label{eq:3.7.1ham}
\TT_{7}:=\left(\begin{matrix} 1+k(\f) & 0 \\ 0 & 1+\bar{k}(\f) \end{matrix}\right),
\end{equation}
with $k : \TTT^{d}\to \CCC$.
By direct calculation we have that
\begin{equation}\label{eq:3.7.2ham}
\begin{aligned}
\!\!\!\!\!\!\!\calL_{6}\TT_{7}-\TT_{7}&\left[\oo\cdot\del_{\f}\uno+i m_{2}E+
i\left(\begin{matrix}m_{1} \!\!\!& 0 \\ 0 & \!\!\!-\bar{m}_{1}\end{matrix}\right)\del_{x}
+i \left(\begin{matrix}m_{0}\!\!\! & 0 \\ 0 &\!\!\! -m_{0} \end{matrix}\right)
\right]\\
&=i\left(\begin{matrix}r_{1} & 0 \\ 0 & -\bar{r}_{1} \end{matrix}\right)+
\left[i\left(\begin{matrix}0 & b_{0}^{(6)} \\ -\bar{b}_{0}^{(6)} & 0 \end{matrix}\right)+
\RR_{6}
\right]\TT_{7},
\end{aligned}
\end{equation}
where
\begin{equation}\label{eq:3.7.3ham}
r_{1}(\f)=\oo\cdot\del_{\f}k(\f)+i({a}_{0}^{(6)}(\f)-m_{0})(1+k(\f)),
\end{equation}
We are looking for $\Gamma$ such that $r_{1}\equiv0$. As done in step 5,
we write $1+k(\f)=\exp(\Gamma(\f))$, then equation $r_{1}\equiv0$ reads
\begin{equation}\label{eq:3.7.4ham}
\oo\cdot\del_{\f}\Gamma(\f)+i({a}_{0}^{(6)}(\f)+m-m_{0})=0,
\end{equation}
that has a unique solution if and only if we define
\begin{equation}\label{eq:3.7.5ham}
m_{0}:=m+\frac{1}{(2\pi)^{d}}\int_{\TTT^{d}}{a}_{0}^{(6)}(\f){\rm d}\f.
\end{equation}
Hence we can set
\begin{equation}\label{eq:3.7.6ham}
\Gamma(\f):=-i(\oo\cdot\del_{\f})^{-1}({a}_{0}^{(6)}+m-m_{0})(\f).
\end{equation}
It turns out that the trasformation $\TT_{7}$ is
invertible, then, by conjugation, we obtain $\calL_{7}:=\TT_{7}^{-1}\calL_{6}\TT_{7}$
with
\begin{equation}\label{eq:3.7.7ham}
\calL_{7}:=
\oo\cdot\del_{\f}\uno+
i\left(\begin{matrix}m \!\!\!& 0 \\ 0 & \!\!\!-m\end{matrix}\right)\del_{xx}+
i\left(\begin{matrix}m_{1} \!\!\!& 0 \\ 0 & \!\!\!-\bar{m}_{1}\end{matrix}\right)\del_{x}
+i \left(\begin{matrix}m_{0} \!\!\!& b_{0}^{(7)} \\ -\bar{b}_{0}^{(7)} & \!\!\!-m_{0}\end{matrix}\right)+
\RR_{7}
\end{equation}
where
we have defined
\begin{equation}\label{eq:3.7.8ham}
\begin{aligned}
\left(\begin{matrix}0 \!\!\!& b_{0}^{(7)} \\ -\bar{b}_{0}^{(7)} & \!\!\! 0\end{matrix}\right)&:=
\TT_{7}^{-1} \left(\begin{matrix}0 & b_{0}^{(6)} \\ -\bar{b}_{0}^{(6)} & 0 \end{matrix}\right)\TT_{7},\quad
{\RR}_{7}:= \TT_{7}^{-1}\RR_{6}\TT_{7}.
\end{aligned}
\end{equation}
Moreover, since by (\ref{eq:3.7.6ham}) the function $\Gamma$ is purely imaginary, then
the transformation is symplectic. Indeed
\begin{equation}\label{eq:3.7.9}
\Omega(e^{\Gamma}u, e^{\Gamma}v):={\rm Re}\int_{\TTT}ie^{\Gamma}ue^{-\Gamma}\bar{v}{\rm d}x=
\Omega(u,v),
\end{equation}
hence the linearized operator $\calL_{7}$ is Hamiltonian.
\subsection{Non-degeneracy Condition}
Here we give the proof of formula \eqref{mammamia}
Let us study the properties of the average of the coefficients of the first order differential operator.
In particular we are interested in how these quantities depends explicitly on $\oo$, see Remark \ref{trenonapoli}.
Consider $a_{1}(\f,x)=a_{1}(\f,x,u)$ where $u$ satisfies \eqref{eq:3.2.1bham} and $a_{i}$ is defined in \eqref{5005}.
One has
\begin{equation}\label{topotopo}
\left|\int_{\TTT^{d+1}}a_{1}(\f,x)\right|\geq \e \gote-C\e\|u\|_{\gots_{0}+\h_{1}}\geq \frac{\gote}{2}\e,
\end{equation}
if $\e\g_{0}^{-1}$ is small enough. Essentially, by using \eqref{eq:B15cham}, \eqref{eq:B15dham}, \eqref{eq:3.10aham} and \eqref{eq:3.10bham}, one can repeat the reasoning followed in \eqref{topotopo} for the average of $a_{1}^{(i)}$ for $i=1,2,3,4$ and prove the \eqref{mammamiaa} with a constant $c< \gote/16$. Let us check \eqref{mammamiab}.
At the starting point there is no explicit dependence on the parameters $\oo$ in $a_{1}$, hence
we get also for $\oo_{1}\neq\oo_{2}$
\begin{equation}\label{topotopo1}
0=\left|\int_{\TTT^{d+1}}a_{1}(\f,x,\oo_{1},u(\oo))-a_{1}(\f,x,\oo_{2},u(\oo))\right|\leq \e^{2}C|\oo_{1}-\oo_{2}|.
\end{equation}
Now, by \eqref{eq:3.122} one has that
$$
a_{1}^{(1)}(\f,x,\oo,u(\oo)):=a_{1}^{(1)}(\f,x,u(\oo)):=i(2\TT_{1}^{-1}(E+A_{2})\del_{x}\TT_{1}+\TT_{1}^{-1}A_{1}\TT_{1})_{1}^{1},
$$
and again we do not have explicit dependence on $\oo$ since the matrix $\TT_{1}$ depends on the external parameters only trough the function $u$. Hence bound \eqref{topotopo1} holds.
\noindent
Now consider the coefficients $a_{1}^{(2)}$ in \eqref{24ham}. There is explicit dependence on $\oo$ only in the term
\begin{equation}\label{topotopo2}
\TT_{2}^{-1}(\oo\cdot\del_{\f}\x)=\sqrt{1+\hat{\x_{y}}(\f,y)}{\oo}\cdot\del_{\f}\x(\f,y+\hat{\x}(\f,y)).
\end{equation}
Recall that the functions $\x$ in \eqref{299} and $\hat{\x}$ depends on $\oo$ only through $u$. Hence one has
\begin{equation}\label{topotopo3}
\begin{aligned}
&\left|\int_{\TTT^{d+1}}\sqrt{1+\hat{\x}_{y}}(\oo_{1}-\oo_{2})\cdot\del_{\f}\x(\f,y+\hat{\x}(\f,y)){\rm d}\f {\rm d}y
\right|=\\
&=\left|\int_{\TTT_{d+1}}\frac{(\oo_{1}-\oo_{2})\cdot\del_{\f}\x(\f,x)}{\sqrt{1+\hat{\x}_{y}(\f,x+\x(\f,x)) }}{\rm d}\f{\rm d}x
\right|\\
&\leq|\oo_{1}-\oo_{2}|\left|\int_{\TTT^{d+1}}\del_{\f}\x(\f,x){\rm d}\f {\rm d}x
\right|\\
&+|\oo_{1}-\oo_{2}|\left|\int_{\TTT_{d+1}}\left(\frac{1}{\sqrt{1+\hat{\x}_{y}}}-1\right)\del_{\f}\x(\f,x)
{\rm d}\f{\rm d}x
\right|
\end{aligned}
\end{equation}
By defining $|u|_{s}^{\infty}:=||u||_{W^{s,\infty}}$ and using the standard estimates of the Sobolev embedding
on the function $\x$ in \eqref{299}
we get
\begin{subequations}
\begin{align}
|\x|^{\infty}_{s}\leq C(s)||\x||_{s+\gots_{0}}&\leq C(s)||\rho_{0}||_{s+\gots_{0}}\leq
\e C(s)(1+||{ u}||_{s+\gots_{0}+2}),\label{eq:3.50aham}.
\end{align}
\end{subequations}
The function $\hat{\x}$ satisfies the same bounds by Lemma \ref{change}. Hence, since the first integral in \eqref{topotopo3} is zero, using the interpolation estimates in Lemma \ref{A}, we get
\begin{equation}\label{topotopo4}
\left|\int_{\TTT^{d+1}}a_{1}^{(2)}(\f,x){\rm d}\f{\rm d}x\right|^{lip}\leq C \e^{2}.
\end{equation}
Let us study the coefficients $a_{1}^{(3)}$ defined in \eqref{eq:3.166}.
In particular one need to control the difference $a_{1}^{(3)}(\oo_1)-a_{1}^{(3)}(\oo_2)$. To do this one can uses standard formul\ae \;
of propagation of errors for Lipschitz functions.
In order to perform the quantitative estimates one can check that
the function $\al({\f})$ defined in (\ref{388}) satisfies the tame estimates (see also Lemma $3.20$ in \cite{FP} ):
\begin{subequations}\label{eq:B24ham}
\begin{align}
|\al|^{\infty}_{s }&\leq\e \g_{0}^{-1}C(s)(1+||{ u}||_{s+d+\gots_{0}+2}),\label{eq:B24aham}\\
|d_{{ u} }\al({ u})[{ h}]|^{\infty}_{s }&\leq \e \g_{0}^{-1}C(s)
(||{ h}||_{s+d+\gots_{0}+2}+
||{ u}||_{s+d+\gots_{0}+2}||{ h}||_{d+\gots_{0}+2}),\label{eq:B24bham}\\
|\al|^{\infty}_{s,\g} &\leq \e\g_{0}^{-1}C(s)
(1+||{ u}||_{s+d+\gots_{0}+2,\g}),\label{eq:B24cham}
\end{align}
\end{subequations}
\noindent
while by the (\ref{333}) one has $\rho=1+\TT_{3}^{-1}(\oo\cdot\del_{\f}\al)$. By using Lemma \ref{change}
and the bounds (\ref{eq:B24ham}) on $\al$ and (\ref{eq:3.2.1bham}) one can prove
\begin{subequations}\label{eq:B34ham}
\begin{align}
|\rho-1|^{\infty}_{s,\g}&\leq
\e\g_{0}^{-1}C(s)(1+||{ u}||_{s+d+\gots_{0}+4,\g}) \label{eq:B34bham}\\
|d_{{ u} }\rho({ u})[{ h}]|^{\infty}_{s}&\leq
\e\g_{0}^{-1}C(s)(||{ h}||_{s+d+\gots_{0}+3}+
||{u}||_{s+d+\gots_{0}+4}||{ h}||_{d+\gots_{0}+3}).\label{eq:B34cham}
\end{align}
\end{subequations}
The bounds above follows by classical tame estimates in Sobolev spaces, anyway the proof can be found
in Section 3 of \cite{FP}. Now by taking the integral of \eqref{topotopo4} and by using \eqref{eq:B24aham}-\eqref{eq:B34cham}, the tame estimates in Lemma \ref{change}
and the \eqref{eq:3.10aham}, \eqref{eq:3.10bham} one obtain the result on the . For the last step one can reason in the same way. Indeed the most important fact is to prove \eqref{topotopo4}. At the starting point we have no explicit dependence on $\la$ in the average of $a_{1}$, but, once that dependence appear, then we have the estimates \eqref{topotopo4} that is quadratic in $\e$.
One has also the following result.
\begin{lemma}\label{lem:3.9ham}
Under the Hypotheses of Lemma \ref{lem:3.88} possibly with smaller $\epsilon_{0}$,
if
(\ref{eq:3.2.1bham}) holds, one has that the $\TT_i$, $i\neq3 $ identify operators $\TT_{i}(\f)$,
of the phase space ${\bf H}^{s}_{x}:={\bf H}^{s}(\TTT)$. Moreover they are invertible
and the following estimates hold for $\gots_{0}\leq s\leq q-\h_{1}$ and i=1,2,4,5,6,7:
\begin{subequations}
\begin{align}
||(\TT_{i}^{\pm1}(\f)-\uno){ h}||_{{\bf H}^{s}_{x}}&\leq \e
C(s)(||{h}||_{{\bf H}^{s}_{x}}+||{ u}||_{s+d+2\gots_{0}+4}
||{ h}||_{{\bf H}^{1}_{x}}),\label{eq:3.93f}
\end{align}
\end{subequations}
\end{lemma}
The Lemma is essentially a consequence if the discussion above. We omit the details because the proof follows basically the same arguments used in
Lemma $3.25$ in \cite{FP}.
\section{Reduction to constant coefficients}\label{sec:4ham}
In this Section we conclude the proof of Proposition \ref{teo2ham} through a
reducibility algorithm. First we need to fix some notations.
Let $b\in \NNN$, we consider the exponential basis $\{ e_{i} : i\in \ZZZ^{b}\}$
of $L^{2}(\TTT^{b})$. In this way we have that $L^{2}(\TTT^{2})$ is the space
$\{u=\sum u_{i}e_{i} : \sum|u_{i}|^{2}<\infty\}$. A linear operator $A : L^{2}(\TTT^{b})\to L^{2}(\TTT^{b})$
can be written as an infinite dimensional matrix
$$
A=(A_{i}^{j})_{i,j\in\ZZZ^{b}}, \quad A_{i}^{j}=(Ae_{j},e_{i})_{L^{2}(\TTT^{b})}, \quad Au=\sum_{i,j}A_{i}^{j}u_{j}e_{i}.
$$
\noindent
where $(\cdot,\cdot)_{L^{2}(\TTT^{d+1})}$ is the usual scalar product on $L^{2}$.
In the following we also use the decay norm
\begin{equation}\label{decayham}
\begin{aligned}
|A|^{2}_{s}:=\sup_{\s,\s'\in \CC}|A_{\s}^{\s'}|^{2}_{s}&:=
\sup_{\s,\s'\in\CC}\sum_{h\in\ZZZ\times\ZZZ^{d}}\langle h\rangle^{2s}
\sup_{k-k'=h}|A_{\s,k}^{\s',k'}|^{2}.
\end{aligned}
\end{equation}
If one has that $A:=A(\oo) $ depends on parameters $\oo\in\Lambda\subset\RRR$ in a Lipschitz way, we define
\begin{equation*}\label{2.1}
\begin{aligned}
|A|_{s}^{sup}&:=\sup_{\la\in\Lambda}|A(\la)|_{s}, \;\;\;
|A|^{lip}_{s}:=\sup_{\la_{1}\neq\la_{2}}\frac{|A(\la_{1})-A(\la_{2})|_{s}}{|\la_{1}-\la_{2}|}, \;\\
|A|_{s,\g}&:=|A|_{s}^{sup}+\g|A|^{lip}_{s}.
\end{aligned}
\end{equation*}
The decay norm we have introduced in (\ref{decayham}) is suitable for the problem we are studying.
Note that
\begin{equation*}
\;\;\;
\forall\; s\leq s' \;\; \Rightarrow \;\; |A_{\s}^{\s'}|_{s}\leq|A_{\s}^{\s'}|_{s'}.
\end{equation*}
Moreover norm (\ref{decayham}) gives information on the polynomial off-diagonal decay of the matrices,
indeed $\forall\; k,k'\in\ZZZ_+\times\ZZZ^{d}$
\begin{equation}\label{decay2}
\begin{aligned}
&|A_{\s,k}^{\s,k'}|\leq \frac{|A_{\s}^{\s'}|_{s}}{\langle k-k' \rangle^{s}},\quad
|A_{i}^{i}|\leq |A|_{0}, \quad |A_{i}^{i}|^{lip}\leq |A|_{0}^{lip}.
\end{aligned}
\end{equation}
In order to prove Prposition \ref{teo2ham} we first prove the following result. We see that
the operator $\calL_{7}$ cannot be diagonalized, but anyway can be block-diagonalized where
the blocks on the diagonal have fixed size. This is sufficient for our analysis.
\begin{theorem}\label{KAMalgorithmham}
Let $\ff\in C^{q}$ satisfy the Hypotheses of Proposition \ref{teo2ham} with $q>\h_{1}+\be+\gots_{0}$
where $\h_{1}$ defined in (\ref{eq:3.2.0ham})
and $\be=7\tau+5$ for some $\tau>d$.
Let $\g\in(0,\g_{0})$,
$\gots_{0}\leq s\leq q-\h_{1}-\be$ and ${ u}(\la)\in{\bf H}^{0}$ be a family of functions depending
on a Lipschitz way on a parameter
$\oo\in\Lambda_{o}\subset\Lambda:[1/2,3/2]$.
Assume that
\begin{equation}\label{eq:4.2ham}
||{u}||_{\gots_{0}+\h_{1}+\be, \Lambda_{o},\g}\leq1.
\end{equation}
Then there exist constants $\epsilon_{0}$, $C$, depending only on the data of the problem,
such that,
if $\e\g^{-1}\leq\epsilon_{0}$,
then there exists a sequence of purely imaginary numbers as in Proposition \ref{teo2ham},
namely
$\Omega_{\s,j}^{\phantom{g} j}, \Omega_{\s,j}^{-j} : \Lambda \to \CCC$
of the form
\begin{equation}\label{1.2.2bisham}
\begin{aligned}
\Omega_{\s,j}^{\phantom{g} j}&:=-i \s m_{2}j^{2}-i\s|m_{1}|j+i\s m_{0}+i\s r_{j}^{j},\\
\Omega_{\s,j}^{-j}&:=i\s r_{j}^{-j},
\end{aligned}
\end{equation}
where
\begin{equation}\label{1.2.2trisham}
m_{2},m_{0}\in\RRR, \;\; m_{1}\in i\RRR, \quad \ol{r_{j}^{k}}=r_{k}^{j},
\;\; k=\pm j
\end{equation}
for any $\s\in\CC$, $j\in \NNN$, moreover
\begin{equation}\label{eq:4.5ham}
|r_{\s,j}^{\phantom{g} k}|_{\g}\leq \frac{\e C}{\langle j\rangle}, \quad \forall\; \s\in\CC,\; j\in\ZZZ, \;\; k=\pm j,
\end{equation}
and
such that, for any $\oo\in \Lambda_{\infty}^{2\g}({ u})$, defined in (\ref{martina10ham}),
there exists a bounded, invertible linear operator
$\Phi_{\infty}(\oo) : {\bf H}^{s}\to {\bf H}^{s}$, with bounded inverse $\Phi_{\infty}^{-1}(\oo)$,
such that
\begin{equation}\label{eq:4.6ham}
\begin{aligned}
\calL_{\infty}(\oo):=\Phi_{\infty}^{-1}(\oo)\circ\calL_{7}\circ\Phi_{\infty}(\oo)&=
{\oo}\cdot\del_{\f}\uno+i \DD_{\infty},\\
{\rm where}\;\;\;\;\;\;\;\;\;
\DD_{\infty}:={\rm diag_{h=(\s,j)\in\CC\times\NNN}}\{\Omega_{\s, \und{j}}(\oo)\},& \quad
\end{aligned}
\end{equation}
with $\calL_7$ defined in \eqref{eq:3.5.9ham} and where
\begin{equation}\label{bambaham}
\Omega_{\s,\underline{j}}:=\left(\begin{matrix}\Omega_{\s,j}^{\phantom{g}j} & \Omega_{\s,j}^{-j} \\
\Omega_{\s,-j}^{\phantom{g}j} & \Omega_{\s,-j}^{-j} \end{matrix}\right)
\end{equation}
Moreover, the transformations $\Phi_{\infty}(\la)$, $\Phi_{\infty}^{-1}$ are symplectic and satisfy
\begin{equation}\label{eq:4.8ham}
|\Phi_{\infty}(\la)-\uno|_{s,\Lambda_{\infty}^{2\g},\g}+
|\Phi_{\infty}^{-1}(\la)-\uno|_{s,\Lambda_{\infty}^{2\g},\g}\leq
\e \g^{-1} C(s)(1+||{ u}||_{s+\h_{1}+\be,\Lambda_{o},\g}).
\end{equation}
In addition to this, for any $\f\in\TTT^{d}$, for any $\gots_{0}\leq s\leq q-\h_{1}-\be$
the operator
$\Phi_{\infty}(\f) : {\bf H}^{s}_{x}\to{\bf H}^{s}_{x}$ is an invertible operator
of the phase space ${\bf H}_{x}^{s}:={\bf H}^{s}(\TTT)$ with inverse
$(\Phi_{\infty}(\f))^{-1}:=\Phi_{\infty}^{-1}(\f)$ and
\begin{equation}\label{eq:4.9ham}
||(\Phi_{\infty}^{\pm1}(\f)-\uno){ h}||_{{\bf H}^{s}_{x}}\leq
\e\g^{-1} C(s)(||{ h}||_{{\bf H}^{s}_{x}}+||{ u}||_{s+\h_{1}+\be+\gots_{0}}
||{ h}||_{{\bf H}^{1}_{x}}).
\end{equation}
\end{theorem}
\begin{rmk}
Note that since the $\Phi_{\infty}$ is symplectic then the operator $\calL_{\infty}$ is hamiltonian.
\end{rmk}
The main point of the Theorem \ref{KAMalgorithmham} is that
the bound on the low norm of $u$ in (\ref{eq:4.2ham}) guarantees the bound
on \emph{higher} norms (\ref{eq:4.8ham}) for the transformations $\Phi_{\infty}^{\pm1}$. This is
fundamental in order to get the estimates on the inverse of $\calL$ in high norms.
Moreover, the definition (\ref{martina10ham}) of the set where the second Melnikov conditions
hold, depends only on the final eigenvalues. Usually in KAM theorems, the non-resonance conditions
have to be checked, inductively, at each step of the algorithm. This
formulation, on the contrary, allow us to discuss the measure estimates only once.
Indeed, the functions $\m_{h}(\oo)$ are well-defined even if
$\Lambda_{\infty}=\emptyset$, so that, we will perform the measure estimates as the last step of the proof of Theorem \ref{teo1}.
\subsection{Functional setting and notations}
\subsubsection{The off-diagonal decay norm }
Here we want to show some important properties of the norm $|\cdot|_{s}$. Clearly the same results hold for the
norm $|\cdot|_{{\bf H}^{s}}:=|\cdot|_{H^{s}\times H^{s}}$. Moreover we will introduce
some characterization of the operators we have to deal with during the diagonalization procedure.
First of all we have following classical results.
\begin{lemma}\label{bubbole}{\bf Interpolation.} For all $s\geq s_{0}>(d+1)/2$
there are $C(s)\geq C(s_{0})\geq1$
such that
if $A=A(\oo)$ and $B=B(\oo)$ depend on the parameter
$\la\in\Lambda\subset\RRR$
in a Lipschitz way, then
\begin{subequations}
\begin{align}
|AB|_{s,\g}&\leq C(s)|A|_{s_0,\g}|B|_{s,\g}
+C(s_{0})|A|_{s,\g}|B|_{s_0,\g},\label{eq:2.11a}\\
|AB|_{s,\g}&\leq C(s)|A|_{s,\g}|B|_{s,\g}.\label{eq:2.11b}\\
\|Ah\|_{s,\g}&\leq C(s)(|A|_{s_0,\g}\|h\|_{s,\g}
+|A|_{s,\g}\|h\|_{s_0,\g}),\label{eq:2.13b}\end{align}
\end{subequations}
\end{lemma}
Lemma \ref{bubbole} implies that for any $n\geq0$ and $s\geq \gots_{0}$ one has
\begin{equation}\label{eq:2.12}
|A^{n}|_{s_{0},\g}\leq [C(s_{0})]^{n-1}|A|^{n}_{s_{0},\g}, \quad
|A^{n}|_{s,\g}\leq n[C(s_{0})|A|_{\gots_{0},\g}]^{n-1}C(s)|A|_{s,\g},
\end{equation}
\begin{equation}\label{eq:2.12bis}
|[A,B]^{n}|_{s,\g}\leq nC(\gots_{0})^{n-1}|A|_{\gots_{0},\g}^{n-1}|B|^{n-1}_{\gots_{0},\g}
\left(|A|_{s,\g}|B|_{\gots_{0},\g}+|A|_{\gots_{0},\g}|B|_{s,\g}\right),
\end{equation}
The following Lemma shows how to invert linear operators which
are ''\emph{near}'' to the identity in norm $|\cdot|_{s}$.
\begin{lemma}\label{inverse} Let $C(s_0)$ be as in Lemma \ref{bubbole}. Consider an operator of the form $\Phi=\uno+\Psi$
where $\Psi=\Psi(\la)$ depends in a Lipschitz way on $\la\in\Lambda\subset\RRR$.
Assume that $C(s_{0})|\Psi|_{s_{0},\g}\leq1/2$. Then $\Phi$ is invertible and,
for all $s\geq s_{0}\geq (d+1)/2$,
\begin{equation}\label{eq:2.14}
\quad |\Phi^{-1}|_{s_{0},\g}\leq 2,\quad
|\Phi^{-1}-\uno|_{s,\g}\leq C(s)|\Psi|_{s,\g}
\end{equation}
Moreover, if one has $\Phi_{i}=\uno+\Psi_{i}$, $i=1,2$ such that $C(s_{0})|\Psi_{i}|_{s_0,\g}\leq1/2$, then
\begin{equation}\label{eq:2.15}
\!\!\!|\Phi^{-1}_{2}-\Phi^{-1}_{1}|_{s,\g}\leq\! C(s)\left(|\Psi_{2}-\Psi_{1}|_{s,\g}\!+\!(|\Psi_{1}|_{s,\g}+|\Psi_{2}|_{s,\g})
|\Psi_{2}-\Psi_{1}|_{s_0,\g}\right).
\end{equation}
\end{lemma}
\begin{proof} See \cite{FP}.
\end{proof}
\subsubsection{T\"opliz-in-time matrices}
We introduce now a special class of operators, the so-called
{\em T\"opliz in time} matrices, i.e.
\begin{equation}\label{eq:2.16}
A_{i}^{i'}=A_{(\s,j,p)}^{(\s',j',p')}:=A_{\s,j}^{\s'j'}(p-p'),
\quad {\rm for} \quad i,i'\in \CC\times\ZZZ\times\ZZZ^{d}.
\end{equation}
To simplify the notation in this case, we shall write
$A_{i}^{i'}=A_{k}^{k'}(\ell)$,
$i=(k,p)=(\s,j,p)\in \CC\times\ZZZ\times \ZZZ^{d}$,
$i'=(k',p')=(\s',j',p')\in \CC\times \ZZZ\times\ZZZ^{d}$,
with $k,k'\in \CC\times\ZZZ$.
They are relevant because
one can identify the matrix $A$ with a one-parameter family of operators, acting on the space
${\bf H}^{s}_{x}$,
which depend on the time, namely
$$
A(\f):=(A_{\s,j}^{\s',j'}(\f))_{\substack{\s,\s'\in \CC \\ j,j'\in\ZZZ}}, \quad
A_{\s,j}^{\s',j'}(\f):=\sum_{\ell\in\ZZZ^{d}}A_{\s,j}^{\s',j'}(\ell)e^{i\ell\cdot\f}.
$$
To obtain the stability result on the solutions we will strongly use this property.
\begin{lemma}\label{1.4} If $A$ is a T\"opliz in time matrix as in (\ref{eq:2.16}), and
$\gots_{0}:=(d+2)/2$, then one has
\begin{equation}\label{aaaaa}
|A(\f)|_{s}\leq C(\gots_{0})|A|_{s+\gots_{0}}, \quad \forall \; \f\in\TTT^{d}.
\end{equation}
\end{lemma}
\begin{proof} See \cite{FP} or \cite{BBM}.
\end{proof}
\begin{defi}\label{def:smooth} {\bf (Smoothing operator)}
Given $N\in\NNN$, we the define the \emph{smoothing operator} $\Pi_{N}$ as
\begin{equation}\label{smoothop}
(\Pi_{N}A)_{\s,j,\ell }^{\s',j',\ell'}=\left\{
\begin{aligned}
& A_{\s,j,\ell}^{\s',j',\ell} \,,\quad |\ell-\ell'|\leq N,\\
& 0 \quad {\rm otherwise}
\end{aligned}\right.
\end{equation}
\end{defi}
\begin{lemma}
Let $\Pi_{N}^{\perp}:=\uno-\Pi_{N}$,
if $A=A(\la)$ is a Lipschitz family $\la\in\Lambda$, then
\begin{equation}\label{eq:2.22}
|\Pi_{N}^{\perp}A|_{s,\g}\leq N^{-\be}|A|_{s+\be,\g}, \quad \be\geq0.
\end{equation}
\end{lemma}
\begin{proof}
See \cite{FP} or \cite{BBM}.
\end{proof}
\begin{lemma}\label{multiop}
Consider $a=\sum_{i}a_{i}e_{i}\in H^{s}(\TTT^{b})$. Then the multiplication operator by the function $a $, i.e.
$h\mapsto a h$ is represented by the matrix $A$ defined as $A_{i}^{i'}=a_{i-i'}$. One has
\begin{equation}\label{multiop1}
|A|_{s}=||a||_{s}.
\end{equation}
Moreover, if $a=a(\la)$ is a Lipschitz family of functions, then
\begin{equation}\label{multiop2}
|A|_{s,\g}=||a||_{s,\g}.
\end{equation}
\end{lemma}
We need some technical lemmata on finite dimensional matrices.
\begin{lemma}\label{finitema}
Given a matrix $M\in\MM_{n}(\CCC)$, where $\MM_{n}(\CCC)$ is the space of the $n\times n$ matrix with coefficients in $\CCC$,
we define the norm $\|M\|_{\infty}:=\max_{i,j=1,\ldots,n} \{A_{i}^{j}\}$. One has
\begin{equation}\label{finitem1}
\|M\|_{\infty}\leq \| M\|_{2}\leq n\|M\|_{\infty},
\end{equation}
where $\|\cdot\|_{2}$ is the $L^{2}-$operatorial norm.
\end{lemma}
\begin{proof}
It follow straightforward by the definitions.
\end{proof}
\begin{lemma}\label{finitema3}
Take two self adjoint matrices $A,B\in\MM_{n}(\CCC)$. Let us define the operator $M : \MM_{n}(\CCC)\to \MM_{n}(\CCC)$
\begin{equation}\label{finitema4}
M : C \mapsto MC:= AC-CB.
\end{equation}
Let $\la_{j}$ and $\be_{j}$ for $j=1,\ldots,n$ be the eigenvalues respectively of $A$ and $B$. Then,
for any $R\in \MM_{n}(\CCC)$
one has that the equation $MC=R$ has a solution with
\begin{equation}\label{finitema5}
\|C\|_{\infty}\leq K \left(\min_{i,j=1,\ldots,n}\{\la_{j}-\be_{i}\}\right)^{-1}\|R\|_{\infty},
\end{equation}
where the constant $K$ depends only on $n$.
\end{lemma}
\begin{proof}
Define the operator $\TT : \MM_{n}(\CCC) \to \CCC^{n^{2}}$ that associate to a matrix the vector of its components.
Then the equation $MC=R$ can be rewritten as
$$
(A\otimes \uno-\uno\otimes B^{T})\TT(C)=\TT(R),
$$
where $\uno$ is the $n\times n$ identity. Then, by using Lemma \ref{finitema}, one has
\begin{equation}\label{finitema6}
\begin{aligned}
\|C\|_{\infty}&=\max_{i=1,\ldots,n^{2}}\|[\TT(C)]_{i}\|_{\infty}\leq n \|(A\otimes \uno-\uno\otimes B^{T})^{-1}\|_{\infty}\max_{i=1,\ldots,n^{2}}|[\TT(R)]_{i}|\\
&\leq n^{2}\|(A\otimes \uno-\uno\otimes B^{T})^{-1}\|_{2}\max_{i=1,\ldots,n^{2}}|[\TT(R)]_{i}|\leq n^{2} c \left(\min_{i,j=1,\ldots,n}\{\la_{j}-\be_{i}\}\right)^{-1}\|R\|_{\infty},
\end{aligned}
\end{equation}
that is the \eqref{finitema5}.
\end{proof}
\subsubsection{Hamiltonian operators}
Here we give a characterization, in terms of the Fourier coefficients, of hamiltonian linear operators.
This is important since we want to show that our algorithm is closed
for such class of operators.
\begin{lemma}\label{lemFouham} Consider a linear operator $B:=(i\s R_{\s}^{\s'}) :{\bf H}^{s}\to {\bf H}^{s}$. Then,
$B$ is hamiltonian with respect to the symplectic form (\ref{simplectic}) if and only if
\begin{equation}\label{eq:2.25ham}
R_{\s,h}^{\s',h'}=R_{-\s',h'}^{-\s,h}, \qquad \ol{R_{\s,h}^{\s',h'}}=R_{\s',-h}^{\s,-h'}
\end{equation}
\end{lemma}
\begin{proof}
In coordinates, an Hamiltonian function for such operator, is a quadratic form real and symmetric,
$$
H=\sum_{\substack{\s,\s'\in\CC \\ h,h'\in\ZZZ^{d+1}}}Q_{\s,h}^{\s',h'}z^{\s}_{h}z^{\s'}_{h'},
$$
where we denote $\ol{z^{\s}_{h}}=z^{-\s}_{-h}$ and $h=(j,p), h'=(j',p')$. This means that, $Q$ satisfies
\begin{equation}\label{antonio}
\ol{Q_{\s,h}^{\s',h'}}=Q_{-\s,-h}^{-\s',-h'}, \quad {Q_{\s,h}^{\s',h'}}=Q_{\s',h'}^{\s,h}
\end{equation}
Now, since the hamiltonian vector field associated to the Hamiltonian $H$ is given by
$B=iJ Q$, then
writing
$$
B=i\left(\begin{matrix}1&0 \\ 0&-1\end{matrix}\right)\left(\begin{matrix}1&0 \\ 0&-1\end{matrix}\right)JQ
$$
we set
$R_{\s}^{\s'}=Q_{-\s}^{\s'}$ follow the (\ref{eq:2.25ham}).
\end{proof}
Since the operator $\calL_{\infty}$ in Theorem \ref{KAMalgorithmham} is hamiltonian,
thanks to the characterization in Lemma \ref{lemFouham}
we can note that the blocks $\Omega_{\s,\und{j}}$ defined in
\eqref{bambaham} as purely imaginary eigenvalues.
\subsection{Reduction algorithm}
We prove Theorem \ref{KAMalgorithmham} by means of the following
Iterative Lemma on the class of linear operators
\begin{defi}\label{classeham}
\begin{equation}\label{eq:4.11ham}
\oo\cdot\del_{\f}\uno+\DD+\RR : \; {\bf H}^{0}\to {\bf H}^{0},
\end{equation}
where $\oo$ is as in \eqref{dio}, and
\begin{equation}\label{eq:4.12ham}
\begin{aligned}
\DD=diag_{(\s,j)\in\CC\times\ZZZ }\{\Omega_{\s,\underline{j}} \}:&=diag_{(\s,j)\in\CC\times \ZZZ}
\left\{\left(\begin{matrix}\Omega_{\s,j}^{\phantom{g}j} & \Omega_{\s,j}^{-j} \\
\Omega_{\s,-j}^{j} & \Omega_{\s,-j}^{-j} \end{matrix}\right)\right\},
\end{aligned}
\end{equation}
where
\begin{equation}\label{eq:4.12bisham}
\begin{aligned}
\Omega_{\s,j}^{\phantom{g} j}&:=-i \s m_{2}j^{2}-i\s|m_{1}|j+i\s m_{0}+i\s r_{j}^{j},\\
\Omega_{\s,j}^{-j}&:=i\s r_{j}^{-j},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eq:4.12trisham}
\begin{aligned}
m_{2},m_{0}\in\RRR, \;\; m_{1}\in i\RRR, \quad \ol{r_{j}^{k}}=r_{k}^{j}, \quad r_{j}^{k}= O(\frac{\e}{\langle j\rangle})
\;\; k=- j, \;\; r_{j}^{k}= O(\frac{\e}{\langle j\rangle}),\; k=j
\end{aligned}
\end{equation}
for any $(\s,j)\in\CC\times\NNN$,
with $\RR$ is a T\"opliz in time Hamiltonian operator such that
$\RR_{\s}^{\s}=O(\e\del_{x}^{-1})$ and $\RR_{\s}^{-\s}=O(\e)$ for $\s=\pm1$.
Moreover we set $\mu_{\s,j}$ for $\s\in\CC$ the eigenvalues of $\Omega_{\s,\und{j}}$.
\end{defi}
Note that the operator
$\calL_{7}$ has the form (\ref{eq:4.11ham}) and satisfies the (\ref{eq:4.12ham}) and (\ref{eq:4.12bisham}) as well as the estimates
(\ref{eq:B15aham}) and (\ref{eq:B15bham}).
Note moreover that for $\calL_{7}$ the matrix $\DD$ is completely diagonal.
This fact is not necessary for our analysis, and it cannot be preserved during the algorithm.
Define
\begin{equation}\label{eq:4.13ham}
N_{-1}:=1, \quad N_{\nu}:=N_{\nu-1}^{\chi}=N_{0}^{\chi^{\n}}, \;\;\forall\;\n\geq0, \;\;\chi=\frac{3}{2}.
\end{equation}
and
\begin{equation}\label{eq:4.14ham}
\al=7\tau+3, \qquad \h_{3}:=\h_{1}+\be,
\end{equation}
where $\h_{1}$ is defined in (\ref{eq:3.2.0ham}) and $\be=7\tau+5$.
Consider
$ \calL_7=\calL_0$.
Note that $\calL_{7}$ belongs to the class of Definition \ref{classeham}. Indeed in this case we have
that
\begin{equation*}
\RR_{0}:=\left(\begin{matrix}0 & q_{0}(\f,x) \\ -\bar{q}_{0}(\f,x) & 0 \end{matrix}\right) + \RR,
\end{equation*}
(see \eqref{eq:3.5.9ham}) and $\RR$ is a pseudo differential operator of order $O(\del_{x}^{-1})$.
We have the following lemma:
\begin{lemma}\label{stimeRham}
The operator $\RR$ defined in Lemma \ref{lem:3.88} satisfies the bounds
\begin{subequations}\label{stimeR1ham}
\begin{align}
|\RR({ u})|_{s,\g}&\leq \e C(s) (1+||{ u}||_{s+\h_{1},\g}), \label{eq:B15aaham}\\
|d_{{ u}}\RR({ u})[{ h}]|_{s}&\leq \e C(s)\left(||{ h}||_{\gots_{0}+\h_{1}}+
||{ h}||_{s+\h_{1}}+||{ u}||_{s+\h_{1}}||{ h}||_{\gots_{0}}
\right),\label{eq:B15bbham}
\end{align}
\end{subequations}
where $\h_{1}$ is defined in Lemma \ref{lem:3.88}.
\end{lemma}
\begin{proof} By the proof of Lemma \ref{lem:3.88} we have that in the operator $\calL_{5}$
in \eqref{eq:3.5.9bisham} the remainder is just a multiplication operator by the functions $a_{0}^{(5)}, b_{0}^{(5)} $. Hence by Remark \ref{multiop} one has that
the decay norm of the operator is finite. We need to check that the transformation $\TT_{6}$ has a finite decay norm.
First of all we have that the function $w$ in \eqref{eq:3.6.6ham} satisfies the following estimates:
\begin{equation}\label{stimeR2ham}
\begin{aligned}
||w||_{s,\g}&\leq_{s}\e(1+||u||_{s+\tau_{1},\g}),\\
||\del_{u}w(u)[h]||_{s}&\leq_{s} \e(||h||_{s+\tau_{1}}+||u||_{s+\tau_{1}}||h||_{\tau_{1}}),
\end{aligned}
\end{equation}
with $\tau_{1}$ a constant depending only on the data of the problem and much small than $\h_{1}$.\footnote{to prove Lemma \ref{lem:3.88} one prove bounds like \eqref{eq:3.2.7ham} and \eqref{eq:3.2.7bisham} on the coefficients of each $\calL_{i}$ with loss of regularity $\tau_{i}$ at each step. The constant $\h_{1}$ of the Lemma is obtained by collecting together the loss of regularity of each step.}
The operator $\tilde{S}=\uno+w\Upsilon $ defined in \eqref{eq:3.6.2ham} satisfies the following estimates in norm $|\cdot|_{s}$ defined in \eqref{decayham}:
\begin{equation}\label{stimeR3ham}
\begin{aligned}
|\tilde{S}-\uno|_{s,\g}&\leq_{s}\e(1+\|u\|_{s+\tau_{1},\g}),\\
|\del_{u}\tilde{S}(u)[h]|_{s}&\leq_{s} \e(\|h\|_{s+\tau_{1}}+\|u\|_{s+\tau_{1}}\|h\|_{\tau_{1}}),
\end{aligned}
\end{equation}
The \eqref{stimeR3ham} follow by the \eqref{stimeR2ham} and the fact that $|\Upsilon|_{s}\leq1$ using Lemma \ref{multiop}.
Clearly also the transformation $\TT_{6}$ defined in \eqref{megaop} satisfies the same estimates as in
\eqref{stimeR3ham}. Hence using Lemma \ref{bubbole} one has that the remainder $\tilde{R}$ of the operator $\calL_{6}$
in \eqref{pippopippoham} satisfies bounds like \eqref{stimeR1ham} with a different constant $\tau_{2}$ (possibly greater than $\tau_{1}$) instead of $\h_{1}$. Now the last transformation $\TT_{7}$ is a multiplication operator, then, by using
again Lemmata \ref{bubbole} and \ref{multiop} one obtain the \eqref{stimeR1ham} on the remainder of the operator
$\calL_{7}$ in \eqref{eq:3.7.7ham}.
\end{proof}
\begin{lemma}\label{teo:KAMham}
Let $q>\h_{1}+\gots_{0}+\be$. There exist constant $C_{0}>0$, $N_{0}\in\NNN$ large, such that
if
\begin{equation}\label{eq:4.15ham}
N_{0}^{C_{0}}\g^{-1}|\RR_{0}|_{\gots_{0}+\be} \leq1,
\end{equation}
then, for any $\nu\geq0$:
\noindent
$({\bf S1})_{\nu}$ There exists operators
\begin{equation}\label{eq:4.16ham}
\calL_{\nu}:=\oo\cdot\del_{\f} \uno +\DD_{\nu}+\RR_{\nu},\;\;
\DD_{\nu}={\rm diag}_{h\in\CC\times\ZZZ}\{\Omega_{\s,\underline{j}}^{\nu}\},
\end{equation}
\noindent
where
\begin{equation}\label{eq:4.16pippoham}
\Omega_{\s,\underline{j}}^{\nu}(\oo)=
\left(\begin{matrix}\Omega_{\s,j}^{\nu,j} & \Omega_{\s,j}^{\nu,-j} \\
\Omega_{\s,-j}^{\nu,j} & \Omega_{\s,-j}^{,\nu-j} \end{matrix}\right),
\end{equation}
\noindent
and
\begin{equation*}
\begin{aligned}
\Omega_{\s,j}^{\nu, j}&:=-i \s m_{2}j^{2}-i\s|m_{1}|j+i\s m_{0}+i\s r_{j}^{\nu,j}=:\Omega_{\s,j}^{0,j}+i\s r_{j}^{\nu,j},\\
\Omega_{\s,j}^{\nu,-j}&:=i\s r_{j}^{\nu,-j}=:\Omega_{\s,j}^{0,-j}+i\s r_{j}^{\nu,-j},
\end{aligned}
\end{equation*}
with $(\s,j)\in \CC\times\ZZZ$, and defined for $\la\in\Lambda_{\nu}^{\g}:=\Lambda_{\nu}^{\g}$,
with
$\Lambda^{\g}_{0}:=\Lambda_{o}$ and for $\nu\geq1$,
\begin{equation}\label{eq:419ham}
\begin{aligned}
\Lambda_{\nu}^{\g}&:=\calP_{\nu}^{\g}(u)\cap\calO_{\nu}^{\g},\\
\SSSS_{\nu}^{\g}(u)&:=
\left\{\oo\in\Lambda_{\nu-1}^{\g} : \begin{array}{ll}
&|i\oo\cdot\ell\!+\! \mu_{h}^{\nu-1}(\oo)\!-\! \mu_{h'}^{\nu-1}(\oo)|\geq
\frac{\g|\s j^{2}-\s'j'^{2}|}{\langle\ell\rangle^{\tau}},\\
&\forall |\ell|\leq N_{\nu-1}, \! h,h'\in\CC\times\ZZZ
\end{array}
\right\},\\
\calO_{\nu}^{\g}(u)&:=\left\{\oo\in\Lambda_{\nu-1}^{\g} : \begin{array}{ll}
&|i\oo\cdot\ell+\mu_{\s,j}^{\nu-1}-\mu_{\s,k}^{\nu-1}|\geq\frac{\g}{\langle\ell\rangle^{\tau}\langle j\rangle}, \\
&\; \ell\in\ZZZ^{d}\backslash\{0\}, j\in\ZZZ
, k=\pm j, \s\in\CC \end{array}
\right\},
\end{aligned}
\end{equation}
where
\begin{equation}\label{automondoschifo2}
\begin{aligned}
\mu^{\nu}_{\s,j}&:=i\s\left(-{m}_{2}j^{2}+{m_{0}}+{r}_{j}^{\nu,j}+{r}_{-j}^{\nu,-j}+\frac{1}{2}a_{j}{b}_{j}\right),\; \\
{b}^{\nu}_{j}&:=\sqrt{\left(-2|{m}_{1}|+\frac{{r}_{j}^{\nu,j}-{r}_{-j}^{\nu,-j}}{a_j}\right)^{2}+4\frac{|{r}_{j}^{\nu,-j}|^{2}}{(a_j)^{2}}}, \\
&a_{j}=j, \; {\rm if} \; j\neq0, \;\; a_{j}=1, \; {\rm if} \; j=0,
\end{aligned}
\end{equation}
are the eigenvalues of the matrix $\Omega_{\s,\und{j}}^{\nu}$.
For $\nu\geq0$ one has $\ol{r_{j}^{\nu,k}}=r_{k}^{\nu,j}$, for $k=\pm j$ and
\begin{equation}\label{eq:4.20ham}
|r_{j}^{\nu,k}|_{\g}:=|r_{j}^{\nu,k}|_{\Lambda_{\nu}^{\g},\g}\leq\frac{\e C}{\langle j\rangle},\qquad
|r_{j}^{\nu,-j}|_{\g}\leq \frac{\e C}{\langle j\rangle}, \quad
|{b}^{\nu}_{j}|_{\g}\leq {\e C}.
\end{equation}
The remainder $\RR_{\nu}$ satisfies $\forall\;s\in[\gots_{0},q-\h_{1}-\be]$ ($\al$ is defined in \eqref{eq:4.14ham})
\begin{equation}\label{eq:4.21ham}
\begin{aligned}
|\RR_{\nu}|_{s}&\leq |\RR_{0}|_{s+\be}N_{\nu-1}^{-\al},\\
|\RR_{\nu}|_{s+\be}&\leq |\RR_{0}|_{s+\be}N_{\nu-1},
\end{aligned}
\end{equation}
\begin{equation}\label{eq:4.21bisham}
|(\RR_{\nu})_{\s}^{-\s}|_{s}+|D(\RR_{\nu})_{\s}^{\s}|_{s}\lessdot|\RR_{\nu}|_{s},\; \s\in\CC, \quad {\rm where} \quad
D:={\rm diag}_{j\in\ZZZ}\{j\}.
\end{equation}
Moreover
there exists a map $\Phi_{\n-1}$ of the form $\Phi_{\nu-1}:=\exp{(\Psi_{\nu-1})}: {\bf H}^{s}\to{\bf H}^{s}$, where $\Psi_{\nu-1}$ is T\"oplitz in time, $\Psi_{\nu-1}:=\Psi_{\nu-1}(\f)$ (see (\ref{eq:2.16})), such that
\begin{equation}\label{eq:4.16hamham}
\calL_{\nu}:=\Phi_{\nu-1}^{-1}\calL_{\nu-1}\Phi_{\nu-1}
\end{equation}
and for $\nu\geq1$ one has:
\begin{equation}\label{eq:4.22ham}
|\Psi_{\nu-1}|_{s,\g}\leq
|\RR_{0}|_{s+\be}^{0} N_{\nu-1}^{2\tau+1}N_{\nu-2}^{-\al}.
\end{equation}
One has that the operators $\Phi_{\nu-1}^{\pm1}$ are symplectic and
the operator $\RR_{\nu}$ is hamiltonian.
Finally the eigenvalues $\mu_{\s,j}^{\nu}$ are purely imaginary.
\noindent
$({\bf S2})_{\nu}$ For all $j\in\ZZZ$
there exists Lipschitz extensions $\tilde{\Omega}_{\s,j}^{\nu,k}: \Lambda \to i \RRR$ of
$\Omega_{\s,j}^{\nu,k}: \Lambda_{\nu}^{\g} \to i \RRR$, for $k=\pm j$, and
$\tilde{\mu}_{h}^{\nu}(\cdot) : \Lambda \to i\RRR$ of $\mu_{h}^{\nu}(\cdot):\Lambda_{\nu}^{\g}\to i\RRR$,
such that for $\nu\geq1$,
\begin{equation}\label{eq:4.23ham}
\begin{aligned}
&|\tilde{\Omega}_{\s,j}^{\nu,k}-\tilde{\Omega}_{\s,j}^{\nu-1,k}|_{\g}\leq |(\RR_{\nu-1})_{\s}^{\s}|_{\gots_{0}}, \quad \s\in\CC, j\in\ZZZ, k= \pm j,\\
&|\tilde{\mu}_{\s,j}^{\nu}-\tilde{\mu}_{\s,j}^{\nu-1}|^{{\rm sup }}\leq|(\RR_{\nu-1})_{\s}^{\s}|_{\gots_{0}}, \quad \s\in\CC, j\in\ZZZ .
\end{aligned}
\end{equation}
\noindent
$({\bf S3})_{\nu}$ Let ${ u}_{1}(\la)$, ${ u}_{2}(\la)$ be Lipschitz families of Sobolev functions, defined for $\la\in\Lambda_{o}$ such that (\ref{eq:4.2ham}), (\ref{eq:4.15ham}) hold with
$\RR_{0}=\RR_{0}({u}_{i})$ with $i=1,2$.
Then for $\nu\geq0$, for any $\la\in\Lambda_{\nu}^{\g_{1}}\cap\Lambda_{\nu}^{\g_{2}}$,
with $\g_{1},\g_{2}\in[\g/2,2\g]$, one has
\begin{subequations}\label{eq:4.24ham}
\begin{align}
|\RR_{\nu}({ u}_{1})-\RR_{\nu}({ u}_{2})|_{\gots_{0}}&\leq
\e N_{\nu-1}^{-\al}||{ u}_{1}-{ u}_{2}||_{\gots_{0}+\h_{3}},
\label{eq:4.24aham}\\
|\RR_{\nu}({ u}_{1})-\RR_{\nu}({ u}_{2})|_{\gots_{0}+\be}&\leq
\e N_{\nu-1}||{ u}_{1}-{ u}_{2}||_{\gots_{0}+\h_{3}},\label{eq:4.24bham}
\end{align}
\end{subequations}
and moreover, for $\nu\geq1$, for any $s\in[\gots_{0},\gots_{0}+\be]$, for any $(\s,j)\in\CC\times\ZZZ$ and $k=\pm j$,
\begin{eqnarray}\label{eq:4.24bisham}
|(r_{\s,j}^{\nu,k}({ u}_{2})-r_{\s,j}^{\nu,k}({ u}_{1}))-(r_{s,j}^{\nu-1,k}({ u}_{2})-r_{\s,j}^{\nu-1,k}({ u}_{1}))|
&\!\!\!\leq\!\!\!&|\RR_{\nu-1}({ u}_{1})-\RR_{\nu-1}({ u}_{2})|_{\gots_{0}},\nonumber\\
|(r_{\s,j}^{\nu,k}({ u}_{2})-r_{\s,j}^{\nu,k}({ u}_{1}))|& \!\!\!\leq\!\!\!& \e C||{ u}_{1}-{ u}_{2}||_{\gots_{0}+\h_{3}}, \label{aaaham}\\
|b^{\nu}_{j}(u_{1})-b^{\nu}_{j}(u_{2})|&\!\!\!\leq\!\!\!&\e C||{ u}_{1}-{ u}_{2}||_{\gots_{0}+\h_{3}}.\label{aaa2ham}
\end{eqnarray}
\noindent
$({\bf S4})_{\nu}$ Let $u_{1},u_{2}$ be as in $({\bf S3})_{\nu}$ and $0<\rho<\g/2$.
For any $\nu\geq0$ one has that, if
\begin{equation}\label{eq:4.25ham}
\begin{aligned}
CN_{\nu-1}^{\tau}&||{ u}_{1}-{ u}_{2}||_{\gots_{0}+\h_{3}}^{{\rm sup}}\leq \rho \e \quad \Rightarrow \\
&P_{\nu}^{\g}({ u}_{1})\subseteq P_{\nu}^{\g-\rho}({ u}_{2}), \quad \calO_{\nu}^{\g}({ u}_{1})\subseteq \calO_{\nu}^{\g-\rho}({ u}_{2}).
\end{aligned}
\end{equation}
\end{lemma}
\proof
We start by proving that ${\bf (Si)_{0}}$ hold for $i=0,\ldots,4$.
${\bf (S1)_{0}}$. Clearly the properties (\ref{eq:4.20ham})-(\ref{eq:4.21ham}) hold by (\ref{eq:4.11ham}), (\ref{eq:4.12ham})
and the form of $\mu_{k}^{0}$ in (\ref{automondoschifo2}), recall that $r_{k}^{0}=0$ .
Moreover, $m_{2}, |m_{1}|$ and $m_{0}$ real imply that $\mu_{k}^{0}$ are imaginary. In addition to this, our hypotheses guarantee that $\RR_{0}$ and
$\calL_{0}$ are hamiltonian operators.
${\bf (S2)_{0}}$. We have to extend the eigenvalues $\mu_{k}^{0}$ from
the set $\Lambda_{0}^{\g}$ to the entire $\Lambda $. Namely we extend
the functions $m_{2}(\la), m_{1}(\la)$ and $m_{0}(\la)$ to a $\tilde{m}_{i}(\la)$ for $i=0,1,2$ which are Lipschitz in $\Lambda$, with the same sup norm
and Lipschitz semi-norm,
by Kirszbraum theorem.
${\bf (S3)_{0}}$. It holds by (\ref{eq:B15aham}) and \eqref{eq:B15bham} for $\gots_{0}$, $\gots_{0}+\be$ using
(\ref{eq:4.2ham}) and (\ref{eq:4.14ham}).
${\bf (S4)_{0}}$. By definition one has
$\Lambda_{0}^{\g}({ u}_{1})=\Lambda_{o}=\Lambda_{0}^{\g-\rho}({ u}_{2})$, then the
(\ref{eq:4.25ham}) follows trivially.
\subsubsection{Kam step}
In this Section we show in detail one step of the KAM iteration. In other words we will
show how to define the transformation $\Phi_{\nu}$ and $\Psi_{\n}$ that trasform the operator
$\calL_{\n}$ in the operator $\calL_{\nu+1}$. For simplicity we shall avoid to write the index,
but we will only write $+$ instead of $\n+1$.
We consider a transformation of the form $\Phi=\exp{(\Psi)}$, with $\Psi:=(\Psi_{\s}^{\s'})_{\s,\s'=\pm1}$, acting on the operator
$$\calL= \oo\cdot\del_{\f} \uno+\DD+\RR
$$
with $\DD$ and $\RR$ as in \eqref{eq:4.16ham},
We define the operator
$$
e^{ad(\Psi)} L:=\sum_{m=0}^{\infty}\frac{1}{m!}[\Psi, L]^{m}, \quad {\rm with} \quad [\Psi, L]^{m}=[\Psi, [\Psi,L]^{m-1}], \quad [\Psi,L]=\Psi L-L\Psi
$$
acting on the matrices $L$. One has that
\begin{equation}\label{pippoham}
e^{ad(\Psi)}L= e^{-\Psi}Le^{\Psi}.
\end{equation}
Clearly the \eqref{pippoham} hold since $\Psi$ is a linear operator.
Then, $\forall\; { h}\in {\bf H}^{s}$, by conjugation one has
\begin{equation}\label{eq:4.1.22ham}
\begin{aligned}
\Phi^{-1}\calL\Phi&=e^{ad(\Psi)}(\oo\cdot\del_{\f}\uno+\DD)+e^{ad(\Psi)}\RR\\
&=\oo\cdot\del_{\f}+\DD+[\Psi,\oo\cdot\del_{\f}\uno+\DD]+\Pi_{N}\RR\\
&+\sum_{m\geq2}\frac{1}{m!}[\Psi,\oo\cdot\del_{\f}\uno+\DD]^{m}+\Pi_{N}^{\perp}\RR+\sum_{m\geq1}\frac{1}{m!}[\Psi,\RR]^{m}
\end{aligned}
\end{equation}
where $\Pi_{N}$ is defined in (\ref{smoothop}).
The smoothing operator $\Pi_{N}$ is necessary for technical reasons: it will be used in order to obtain
suitable estimates on the high norms of the transformation $\Phi$.
In the following Lemma we will show how to solve the \emph{homological} equation
\begin{equation}\label{homeqham}
\begin{aligned}
&[\Psi, \oo\cdot\del_{\f}\uno+\DD]+\Pi_{N}\RR=[\RR], \quad {\rm where}\\
&[\RR]_{\s,j}^{\s',j'}(\ell):=
\left\{
\begin{aligned}
& (\RR)_{\s,j}^{\s',k}(0), \quad \s=\s',\; k=j,-j,\; \ell=0\\
& 0 \quad \quad {\rm otherwise},
\end{aligned}\right.
\end{aligned}
\end{equation}
for $k,k'\in \CC\times\NNN\times\ZZZ^{d}$.
\begin{lemma}[{\bf Homological equation}]\label{lemhomham}
For any $\la\in\Lambda^{\g}_{\nu+1}$ there exists a unique solution $\Psi=\Psi(\f)$ of the homological equation
(\ref{homeqham}), such that
\begin{equation}\label{eq:4.1.33ham}
|\Psi|_{s,\g}\leq C N^{2\tau+1}\g^{-1}
|\RR|_{s,\g}
\end{equation}
Moreover, for $\g/2\leq \g_{1},\g_{2}\leq 2\g$, and if $u_{1}(\la), u_{2}(\la)$ are Lipschitz functions,
then $\forall \; s\in[\gots_{0},\gots_{0}+\be]$,
$\la\in \Lambda^{\g_{1}}_{+}(u_{1})\cap\Lambda^{\g_{2}}_{+}(u_{2})$, one has
\begin{equation}\label{eq:4.1.44ham}
|\Delta_{12}\Psi|_{s}\leq C N^{2\tau+1}\g^{-1}\left(|\RR({u}_{2})|_{s}
||{ u}_{1}-{u}_{2}||_{\gots_{0}+\h_{2}}+
|\Delta_{12}\RR|_{s}\right),
\end{equation}
where we define $\Delta_{12}\Psi=\Psi({u}_{1})-\Psi({u}_{2})$.\\
Finally, one has $\Phi : {\bf H}^{s}\to {\bf H}^{s}$ is symplectic.
\end{lemma}
\begin{proof}
We rewrite the equation \eqref{homeqham} on each component $k=(\s,j,p),k'=(\s',j',p')$ and we get the following
matricial equation
\begin{equation}\label{homeq2ham}
i\oo\cdot(p-p')\Psi_{\s,\und{j},p}^{\s',\und{j'},p'}+\Omega_{\s,\und{j}}\Psi_{\s,\und{j},p}^{\s',\und{j'},p'}-
\Psi_{\s,\und{j},p}^{\s',\und{j'},p'}\Omega_{\s',\und{j'}}=-\RR_{\s,\und{j}}^{\s',\und{j'}}(p-p')
\end{equation}
where $\Omega_{\s,\und{j}}$ is defined in \eqref{eq:4.16ham}
and where we have set
\begin{equation}\label{homeqhomeq}
\Psi_{\s,\und{j},p}^{\s',\und{j'},p'}:=\left(\begin{matrix} \Psi_{\s,j,p}^{\s',j',p'} & \Psi_{\s,j,p}^{\s',-j',p'}\\
\Psi_{\s,-j,p}^{\s',j',p'} & \Psi_{\s,-j,p}^{\s',-j',p'}\end{matrix}
\right)
\end{equation}
the matrix block indexed by $(j,j')$. To solve equation \eqref{homeq2ham} we can use Lemma \ref{finitema3}
with $A:=i\oo\cdot p\uno+\Omega_{\s,\und{j}}$ and $B=i\oo\cdot p'\uno +\Omega_{\s',\und{j'}}$.
Hence if we write $\mu_{\s,h}$ and $\mu_{\s',h'}$ with $h=j,-j$ and $h'=j',-j'$ the eigenvalues respectively of
$\Omega_{\s,\und{j}}$ and $\Omega_{\s',\und{j'}}$,
\begin{equation}\label{homeq3ham}
\begin{aligned}
\|\Psi_{\s,\und{j},p}^{\s',\und{j'},p'}\|_{\infty}&\stackrel{(\ref{eq:419ham})}{\leq}
C \frac{\langle\ell\rangle^{\tau}\g^{-1}}{|\s j^{2}-\s' j'^{2}|} \max_{\substack{h=j,-j, h'=j',-j'}}|\RR_{\s,h}^{\s',h'}(\ell)|, \quad\\
&\qquad \s=\s', j\neq j',\quad{\rm or} \;\s\neq\s', \; \forall j,j'\\
\|\Psi_{\s,\und{j},p}^{\s',\und{j},p'}\|_{\infty}&\stackrel{(\ref{eq:419ham})}{\leq} C\langle\ell\rangle^{\tau}|j|\g^{-1} \max_{\substack{h=j,-j}}|\RR_{\s,h}^{\s',h}(\ell)|, \quad \s=\s',\; j=j',
\end{aligned}
\end{equation}
where we fixed $p-p'=\ell$. Clearly the solution $\Psi$ is T\"opliz in time.
Unfortunately bounds \eqref{homeq3ham}
are not sufficient in order to estimate the decay norm of the matrix $\Psi_{\s}^{\s'}$.
Roughly speaking one needs to prove, for any $\ell$, that
$\Psi_{\s,j}^{\s',j'}(\ell)\approx o(1/\langle j-j'\rangle^{s})$, and $\Psi_{\s,j}^{\s',-j'}\approx o(1/\langle j+j'\rangle^{s})$.
Actually we are able to prove the following.
Assume that either $|j|\leq \frac{C}{\gote}$ or $|j'|\leq \frac{C}{\gote}$ for some large $C>0$ and $\gote$ defined in \eqref{nondeg}.
Assume also that
\begin{equation}\label{quiquo3}
\max_{\substack{h=j,-j, h'=j',-j'}}|\RR_{\s,h}^{\s',h'}(\ell)|=|\RR_{\s,j}^{\s',j'}(\ell)|.
\end{equation}
By \eqref{homeq3ham} we have that
\begin{equation}\label{quiquo}
\begin{aligned}
( |\Psi_{\s,j}^{\s',j'}|^{2}+ |\Psi_{\s,-j}^{\s',-j'}|^{2})\langle j-j'\rangle^{2s}&+
( |\Psi_{\s,j}^{\s',-j'}|^{2}+ |\Psi_{\s,-j}^{\s',j'}|^{2})\langle j+j'\rangle^{2s}\\
& \leq C \frac{\langle\ell\rangle^{2\tau}\g^{-2}}{|\s j^{2}-\s' j'^{2}|^{2}} |\RR_{\s,j}^{\s',j'}(\ell)|^{2}\left(
\langle j-j'\rangle^{2s}+\langle j+j'\rangle^{2s}
\right)\\
&\leq \tilde{C} \frac{\langle\ell\rangle^{2\tau}\g^{-2}}{|\s j^{2}-\s' j'^{2}|^{2}} |\RR_{\s,j}^{\s',j'}(\ell)|^{2}
\langle j-j'\rangle^{2s}
\end{aligned}
\end{equation}
where we used the fact that, for a finite number of $j$ (or finite $j'$), one has
$$
\langle j+j'\rangle\leq K \langle j-j'\rangle,
$$
for some large $K=K(\gote)>0$.
Note also that the smaller is $\gote$ the larger is the constant $K$.
If the \eqref{quiquo3} does not hold one can treat the other cases by reasoning
as done in \eqref{quiquo}.
Assume now that
\begin{equation}\label{quiquo5}
|j|,|j'|\geq \frac{C}{\gote}
\end{equation}
holds. Here the situation is more delicate.
Consider the matrices $\Omega_{\s,\und{j}}, \Omega_{\s',\und{j'}}$ in equation \eqref{homeq2ham}
which have, by \eqref{automondoschifo2}, eigenvalues $\mu_{\s,j}$, $\mu_{\s,-j}$ and $\mu_{\s',j'}$, $\mu_{\s',-j'}$ respectively.
First of all one can note that by \eqref{quiquo5}
\begin{equation}\label{quiquo4}
|\mu_{\s,j}-\mu_{\s,-j}|,\;\geq |m_{1}|\langle j\rangle\geq c\e \gote\langle j\rangle, \quad |\mu_{\s',j'}-\mu_{\s',-j'}|\geq |m_{1}|\langle j' \rangle
\end{equation}
by the \eqref{nondeg}. Hence we can define the invertible matrices
\begin{equation}\label{quiquo10}
U_{\s,\und{j}}:=\left(
\begin{matrix}
\frac{\Omega_{\s,-j}^{-j}-\mu_{\s,j}}{\mu_{\s,j}-\mu_{\s,-j}} & \frac{-\Omega_{\s,j}^{-j}}{\mu_{\s,j}-\mu_{\s,-j}} \\
\frac{-\Omega_{\s,-j}^{j}}{\mu_{\s,j}-\mu_{\s,-j}} &
\frac{\Omega_{\s,j}^{j}-\mu_{\s,-j}}{\mu_{\s,j}-\mu_{\s,-j}}
\end{matrix}
\right),
\end{equation}
and moreover one can check that
\begin{equation}\label{quiquo7}
U_{\s,\und{j}}^{-1}\Omega_{\s,\und{j}}U_{\s,\und{j}}=D_{\s,\und{j}}=\left(\begin{matrix} \mu_{\s,j} & 0 \\
0 &\mu_{\s,-j}\end{matrix}
\right),
\end{equation}
In order to simplify the notation we set
\begin{equation}
f_{\s,j}^{(1)}:=\frac{\Omega_{\s,-j}^{-j}-\mu_{\s,j}}{\mu_{\s,j}-\mu_{\s,-j}}, \qquad f^{(2)}_{\s,j}:= \frac{\Omega_{\s,j}^{j}-\mu_{\s,-j}}{\mu_{\s,j}-\mu_{\s,-j}},
\qquad c_{\s,j}:=\frac{-\Omega_{\s,j}^{-j}}{\mu_{\s,j}-\mu_{\s,-j}}.
\end{equation}
First of all, by using \eqref{quiquo10}, \eqref{quiquo4} and \eqref{eq:4.20ham} one has
\begin{equation}\label{quiquo11}
|f_{\s,j}^{(1)}|+|f_{\s,j}^{(2)}|
\leq 4\frac{C}{c\gote}, \qquad
|c_{\s,j}|\leq \frac{1}{c\e\gote}|r_{j}^{-j}|.
\end{equation}
Hence one has
\begin{equation}\label{quiquo23}
U_{\s}:={\rm diag}_{|j|\geq C/\gote, j\in \NNN}U_{\s,\und{j}}, \qquad |U_{\s}|_{s,\g}\leq \frac{C}{|m_{1}|}|\RR_{\s}^{\s'}|_{s,\g},
\end{equation}
and moreover $U_{\s}$ diagonalizes the matrix $\Omega_{\s}={\rm diag}_{|j|\geq C/\gote}\Omega_{\s,\und{j}}$.
Setting $U_{\s}^{-1}\Psi_{\s}^{\s'}U_{\s}=Y_{\s}^{\s'}$, equation \eqref{homeq2ham}, for $\s,\s'=\pm1$, reads
\begin{equation}\label{quiquo20}
\ii \oo\cdot\del_{\f}Y_{\s}^{\s'}+D_{\s}Y_{\s}^{\s'}-\s' Y_{\s}^{\s'}D_{\s}=
U_{\s}^{-1}R_{\s}^{\s'}U_{\s}.
\end{equation}
For $|\ell|\leq N $ we set
\begin{equation}\label{quiquo21}
Y_{\s,j}^{\s',j'}(\ell)=\frac{(U_{\s}^{-1}\RR_{\s}^{\s'}U_{\s})_{\s,j}^{\s',j'}(\ell)}{\ii\oo\cdot\ell+\mu_{\s,j}-\mu_{\s',j'}}
\end{equation}
and hence
we get the bound
\begin{equation}\label{quiquo22}
|Y_{\s}^{\s'}|_{s}\leq \g^{-1}N^{\tau}|U_{\s}^{-1}\RR_{\s}^{\s'}U_{\s}|_{s},
\end{equation}
where we used the estimates \eqref{eq:419ham} on the small divisors.
By the definition, the estimate \eqref{quiquo23} and the interpolation properties in Lemma \ref{bubbole}
we can bound the decay norm of $\Psi$ as
\begin{equation}\label{homeq4ham}
|\Psi|_{s}\leq C(s)\g^{-1} N^{\tau}|\RR|_{s},
\end{equation}
using that $|\RR|_{s}/|m_{1}|\leq C$ for some constant $C>0$.
Moreover the following hold:
\begin{lemma}\label{regulham}
Define the operator $A$ as
\begin{equation}\label{eqrmk1ham}
A_{k}^{k'}=A_{\s,j}^{\s',j'}(\ell):=\left\{
\begin{aligned}
&\Psi_{\s,j}^{\s,j'}(\ell), \quad \s=\s'\in\CC,\;\; j=\pm j'\in \ZZZ
\;\; \ell\in\ZZZ^{d},\\
&0, \quad {\rm otherwise},
\end{aligned}\right.
\end{equation}
then the operator $\Psi-A$ is regularizing and hold
\begin{equation}\label{regul1ham}
|D(\Psi-A)|_{s}\leq \g^{-1}N^{\tau}|\RR|_{s},
\end{equation}
where $D:=diag\{j\}_{j\in\ZZZ}$.
\end{lemma}
This Lemma will be used in the study of the remainder of the conjugate operator.
In particular we will use it to prove that the reminder is still in the class of operators described in $(\ref{eq:4.12ham})$.
Now we need a bound on the Lipschitz semi-norm of the transformation. Then, given $\oo_{1},\oo_{2}\in \Lambda_{\nu+1}^{\g}$, one has, for $k=(\s,j,p),k'=(\s',j',p')\in\CC\times\ZZZ\times\ZZZ^{d}$, and $\ell:=p-p'$,
\begin{equation}\label{homeq5ham}
\begin{aligned}
{\oo_1}\cdot\ell \Big[\Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{1})&-\Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{2})\Big]+
\Omega_{\s,\und{j}}(\oo_{1})\left[\Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{1})-
\Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{2})\right]+\\
&-\left[\Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{1})-\Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{2})\right]\Omega_{\s',\und{j'}}(\oo_{1})\\
&+(\oo_{1}-\oo_{2})\cdot\ell \Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{2})+\\
&+\left[\Omega_{\s,\und{j}}(\oo_{1})-\Omega_{\s,\und{j}}(\oo_{2})\right]\Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{2})\\
&+
\Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{2})\left[\Omega_{\s',\und{j'}}(\oo_{1})-\Omega_{\s',\und{j'}}(\oo_{2})\right]=\\
&=\RR_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{1})-\RR_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{1}).
\end{aligned}
\end{equation}
First we can note that
\begin{equation}\label{homeq6ham}
\begin{aligned}
|\Omega_{\s,j}^{\phantom{g}j}(\oo_1)-\Omega_{\s',j'}^{\phantom{g}j'}(\oo_2)|&\leq|m_{2}(\oo_{1})-m_{2}(\oo_{2})||\s j^{2}-\s'j'^{2}|+\e\g^{-1}\\
&+
|m_{1}(\oo_{1})-m_{1}(\oo_{2})||\s j-\s'j' |+|m_{0}(\oo_{1})-m_{0}(\oo_{2})|\\
&\leq C |\oo_{1}-\oo_{2}|(\e\g^{-1}|\s j^{2}-\s' j'^{2}|+\e\g^{-1}+\e\g^{-1})
\end{aligned}
\end{equation}
where we used the (\ref{eq:4.16ham}), (\ref{eq:4.20ham}) and (\ref{eq:3.2.44}) to estimate the Lipschitz semi-norm
of the constants $m_{i}$. Following the same reasoning, one can estimate the sup norm of the matrix
$\Omega_{\s,\und{j}}(\oo_{1})-\Omega_{\s,\und{j}}(\oo_{2})$.
Therefore by triangular inequality one has
\begin{equation}\label{homeq7ham}
\begin{aligned}
&\|\Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{1})-\Psi_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{2})\|_{\infty}\lessdot |\RR_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{1})-\RR_{\s,\und{j}}^{\s',\und{j'}}(\ell,\oo_{1})|_{max} N^{\tau}\g^{-1}
+\\
&+|\oo_{1}-\oo_{2}|\left(|\ell|+\e \g^{-1}|\s j^{2}-\s' j^{j'}|\right)\\
&+|\oo_{1}-\oo_{2}|\left(\e\g^{-1}|\s j-\s' j'|\e\g^{-1}\right)\|\RR_{\s,h}^{\s',h'}(\ell,\oo_{2})\|_{\infty}\frac{N^{2\tau+1}\g^{-2}}{|\s j^{2}-\s' j'^{2}|},
\end{aligned}
\end{equation}
for $|\ell|\leq N$, $j\neq j' $ and $\e\g^{-1}\leq1$.
As done for the estimate \eqref{homeq4ham} for a finite number of $j$ of a finite number of $j'$ the bound \eqref{homeq7ham}
is sufficient to get,
for $\oo\in\Lambda_{\nu+1}^{\g}$ and using also the bound \eqref{eq:419ham} with $j=j'$, the estimate
\begin{equation}\label{homeq8ham}
|\Psi|_{s,\g}:=|\Psi|_{s}^{\rm sup}+\g\sup_{\oo_{1}\neq\oo_{2}}\frac{|\Psi(\oo_{1})-\Psi(\oo_{2})|}{|\oo_{1}-\oo_{2}|}
\leq C\g^{-1}N^{2\tau+1}|\RR|_{s,\g},
\end{equation}
that is the \eqref{eq:4.1.33ham}.
On the other hand, in the case of \eqref{quiquo5}, we can reason as follows. Consider the diagonalizing matrix $U_{\s,\und{j}}$ defined in \eqref{quiquo7}
and recall that by \eqref{quiquo23} also the lipschitz semi-norm of $U_{\s}$ is bounded by the lipschitz semi-norm of $\RR_{\s}^{\s'}$.
Hence by \eqref{quiquo20}, \eqref{quiquo21}, using again the interpolation properties of the decay norm in Lemma \eqref{bubbole}
one get the Lipschitz bound in \eqref{homeq8ham}.
Note also that the Lemma \ref{regulham} holds with $|\cdot|_{s,\g}$ and $N^{2\tau+1}$ instead
of $|\cdot|_{s}$ and $N^{\tau}$.
The proof of the bound \eqref{eq:4.1.44ham} is based on the same strategy used to proof \eqref{homeq8ham}. We refer to
the proof of Lemma $4.39$ of \cite{FP}.
Finally we show that $\Psi$ is an hamiltonian vector field, and hence the transformation $\Phi$ is symplectic.
By hypothesis $\RR$ is hamiltonian, hence by Lemma \ref{lemFouham} we have
\begin{equation}\label{homeq11ham}
\left(\ol{\RR_{\s}^{\s}}\right)^{T}=-\RR_{\s}^{\s}, \quad \ol{\RR_{\s}^{-\s}}=\RR_{-\s}^{\s}, \quad
\ol{\RR_{\s}^{\s'}}=\RR_{-\s}^{-\s'},
\quad \forall \; \s,\s'\in\CC.
\end{equation}
Moreover, by inductive hypothesis $({\bf S1})_{\nu}$ one can note that
\begin{equation}\label{homeq12ham}
\left(\ol{\Omega_{\s}}\right)^{T}=-\Omega_{\s}=\Omega_{-\s}.
\end{equation}
By \eqref{homeq11ham}, \eqref{homeq12ham} one can easily note that the solution of the equation
$$
\oo\cdot\del_{\f}\Psi_{\s}^{\s'}+\Omega_{\s}\Psi_{\s}^{\s'}-\Psi_{\s}^{\s'}\Omega_{\s'}=\RR_{\s}^{\s'},
$$
satisfies conditions in \eqref{homeq11ham}, hence, again by Lemma \ref{lemFouham}, $\Psi$ is hamiltonian. This concludes the proof of Lemma \ref{lemhomham}.
\end{proof}
Next Lemma concludes one step of our KAM iteration.
\begin{lemma}[{\bf The new operator $\calL_{+}$}]\label{newopham}
Consider the operator $\Phi=\exp(\Psi)$ defined in Lemma \ref{lemhomham}. Then the operator
$\calL_{+}:=\Phi^{-1}\calL\Phi$ has the form
\begin{equation}\label{newop1ham}
\calL_{+}:=\oo\cdot\del_{\f}\uno+\DD_{+}+\RR_{+},
\end{equation}
where the diagonal part is
\begin{equation}\label{newop2ham}
\begin{aligned}
\DD_{+}&={\rm diag}_{(\s,j)\in\CC\times\ZZZ}\{\Omega_{\s,\underline{j}}^{+}\},
\quad
\Omega_{\s,\underline{j}}^{+}(\la)=
\left(\begin{matrix}\Omega_{\s,j}^{+,j} & \Omega_{\s,j}^{+,-j} \\
\Omega_{\s,-j}^{+,j} & \Omega_{\s,-j}^{+,-j} \end{matrix}\right), \\
&
\begin{aligned}
\Omega_{\s,j}^{+, j}&:=-i \s m_{2}j^{2}-i\s|m_{1}|j+i\s m_{0}+i\s r_{j}^{+,j},\\
\Omega_{\s,j}^{+,-j}&:=i\s r_{j}^{+,-j},\\
r_{j}^{+,h}&:=r_{j}^{h}+\RR_{\s,j}^{\s,h}(0), \quad h=\pm j.
\end{aligned}
\end{aligned}
\end{equation}
with $(\s,j)\in \CC\times\ZZZ, \la\in\Lambda$. The eigenvalues $\mu_{\s,h}^{+}$, with $h=j,-j$, of $\Omega_{\s,\underline{j}}$ satisfy
\begin{equation}\label{newop3ham}
\begin{aligned}
&|r_{j}^{+,h}-r_{j}^{h}|^{lip}\leq |(\RR)_{\s}^{\s}|^{lip}_{\gots_{0}},\\
&|\mu_{\s,h}^{+}-\mu_{\s,h}|^{{sup}}\leq |(\RR)_{\s}^{\s}|_{\gots_{0},\g}, \quad h=j,-j.
\end{aligned}
\end{equation}
The remainder $\RR_{+}$ is such that
\begin{equation}\label{newop4ham}
\begin{aligned}
|\RR_{+}|_{s}&\leq_{s} N^{-\be}|\RR|_{s+\be,\g}+N^{2\tau+1}\g^{-1}|\RR|_{s,\g}|\RR|_{\gots_{0},\g},\\
|\RR_{+}|_{s+\be}&\leq_{s+\be}|\RR|_{s+\be,\g}+N^{2\tau+1}\g^{-1}|\RR|_{s+\be,\g}|\RR|_{\gots_{0},\g},
\end{aligned}
\end{equation}
and $(\RR_{+})_{\s}^{\s}=O(\e\del_{x}^{-1})$ while $(\RR_{+})_{\s}^{-\s}=O(\e)$ for $\s=\pm1$. More precisely,
\begin{equation}\label{eq:4.21trisham}
|(\RR_{+})_{\s}^{-\s}|_{s}+|D(\RR_{+})_{\s}^{\s}|_{s}\lessdot|\RR_{+}|_{s},\; \s\in\CC, \quad {\rm where} \quad
D:={\rm diag}_{j\in\ZZZ}\{j\}.
\end{equation}
Finally, for $\g/2\leq \g_{1},\g_{2}\leq 2\g$, and for $u_{1}(\la),u_{2}(\la)$ Lipschitz functions, then for any
$s\in[\gots_{0},\gots_{0}+\be]$ and $\la\in \Lambda_{+}^{\g_{1}}(u_{1})\cap\Lambda_{+}^{\g_{2}}(u_{2})$
one has
\begin{eqnarray}\label{newop5ham}
|\Delta_{12}\RR_{+}|_{s}&\!\!\!\!\!\!\leq\!\!\!\!\!\!& |\Pi_{N}^{\perp}\Delta_{12}\RR|_{s}+
+N^{2\tau+1}\g^{-1}\Big(|\RR(u_{1})|_{s}+|\RR(u_{2})|_{s} \Big)|\Delta_{12}\RR|_{\gots_{0}}\nonumber\\
&\!\!\!\!\!\!+\!\!\!\!\!\!&N^{2\tau+1}\g^{-1}\Big(\!|\RR(u_{1})|_{s}\!+\!|\RR(u_{2})|_{s}\! \Big)\Big(|\RR(u_{1})|_{\gots_{0}}\!+\!|\RR(u_{2})|_{\gots_{0}} \Big)||u_{1}\!-\!u_{2}||_{\gots_{0}+\h_{3}}\nonumber\\
&\!\!\!\!\!\!+\!\!\!\!\!\! &N^{2\tau+1}\g^{-1}\Big(|\RR(u_{1})|_{\gots_{0}}+|\RR(u_{2})|_{\gots_{0}} \Big)|\Delta_{12}\RR|_{s}
\end{eqnarray}
\end{lemma}
\begin{proof} The \eqref{newop2ham} follow by the \eqref{homeqham}. Note that the term $\RR_{\s,j}^{\s,k}(0)=\RR_{-\s,j}^{-\s,k}$
for $k=j,-j$ and hence the new correction $r_{j}^{+,h}$ does not depend on $\s$. Moreover, by \eqref{decay2} one has
\begin{equation}\label{newop6ham}
|\Omega_{\s,j}^{+,k}-\Omega_{\s,j}^{\phantom{g}k}|^{lip}\leq |(\RR)_{\s}^{\s}|_{\gots_{0}}^{lip}, \quad k=j,-j.
\end{equation}
Moreover, one has
\begin{equation}\label{newop6bisham}
\begin{aligned}
|\mu_{\s,j}^{+}-\mu_{\s,j}|&\leq 2\sup_{h=\pm j}|r_{h}^{+,h}-r_{h}^{h}|+|j||b_{j}^{+}-b_{j}|\\
&\leq 2\sup_{h=\pm j}|r_{h}^{+,h}-r_{h}^{h}|+\frac{|j|}{|j|}\sup_{h=\pm j}|r_{j}^{+,h}-r_{j}^{h}|
\stackrel{(\ref{newop6ham})}{\lessdot}|(\RR)_{\s}^{\s}|,
\end{aligned}
\end{equation}
then the \eqref{newop3ham} follows. Now, by \eqref{eq:4.1.22ham}
one has that
\begin{equation}\label{newop7ham}
\RR_{+}:=\Pi_{N}^{\perp}\RR+\sum_{n\geq2}\frac{1}{n!}[\Psi,\oo\cdot\del_{\f}\uno+\DD]^{n}+
\sum_{n\geq1}\frac{1}{n!}[\Psi,\RR]^{n}:=\Pi_{N}^{\perp}\RR+\BB.
\end{equation}
Here we used the simple fact that $[A,B]^{n}=[A,[A,B]]^{n-1}$ for any $n\geq1$.
Hence we can estimate
\begin{equation*}\label{newop8ham}
\begin{aligned}
|\RR_{+}|_{s,\g}&\leq_{s}|\Pi_{N}^{\perp}\RR|_{s,\g}+\sum_{k\geq2}\frac{1}{k!}|[\Psi,\Pi_{N}\RR]^{k-1}|_{s,\g}+
\sum_{n\geq1}\frac{1}{n!}|[\Psi,\RR]^{n}|_{s,\g}\\
&\leq_{s}
|\Pi_{N}^{\perp}\RR|_{s,\g}+\sum_{n\geq1}\frac{1}{n!}|[\Psi,\RR]^{n}|_{s,\g}
\leq_{s}\!|\Pi_{n}^{\perp}\RR|_{s}\!\\
&+\!\sum_{n\geq1}\frac{(nC(\gots_{0}))^{n-1}}{n!}|\Psi|_{\gots_{0},\g}^{n-1}|\RR|^{n-1}_{\gots_{0},\g}\!\left(|\Psi|_{s,\g}|\RR|_{\gots_{0},\g}\!+\!|\Psi|_{\gots_{0},\g}|\RR|_{s,\g}\right)\\
&\stackrel{(\ref{eq:2.22}),(\ref{eq:4.1.33ham})}{\leq}N^{-\be}|\RR|_{s+\be,\g}+N^{2\tau+1}\g^{-1}|\RR|_{s,\g}|\RR|_{\gots_{0},\g},
\end{aligned}
\end{equation*}
where we assumed that
\begin{equation}\label{newop9ham}
\sum_{n\geq1}\frac{n^{n-1}}{n!}C(\gots_{0})^{n-1}|\Psi|_{\gots_{0},\g}^{n-1}|\RR|^{n-1}_{\gots_{0},\g}<1.
\end{equation}
Now we have to estimate $\Delta_{12}\RR_{+}$ defined for $\la\in\La^{\g_{1}}(u_{1})\cup\Lambda^{\g_{2}}(u_2)$.
We write $\RR_{i}:=\RR(u_{i})$ for $i=1,2$. We first need a technical Lemma used to study the variation
with respect to the function $u$, of the commutator between two operators.
\begin{lemma}\label{newop10ham}
Given operators $A(u), B(u)$ one has that the following identities hold for any $n\geq1$:
\begin{equation}\label{newop11ham}
[A_{1},B_{1}]^{n}=[A_{1},\Delta_{12}B]^{n}+[A_{1},B_{2}]^{n};
\end{equation}
\begin{equation}\label{newop12ham}
[A_{1},B_{2}]^{n}=\Big[A_{1},[A_{2},B_{2}]\Big]^{n-1}+\Big[A_{1},[\Delta_{12}A,B_{2}]\Big]^{n-1};
\end{equation}
\begin{equation}\label{newop13ham}
\begin{aligned}
&\Big[A_{1},[A_{2},B_{2}]\Big]^{n-1}-[A_{2},B_{2}]^{n}=(n-2)\Big[A_{1},\big[\Delta_{12}A,[A_{2},B_{2}]
\big]\Big]^{n-2}\\
&+\Big[\Delta_{12}A,[A_{2},B_{2}]^{n-1}
\Big].
\end{aligned}
\end{equation}
\end{lemma}
\prova We prove the identities by induction. Let us start from the \eqref{newop11ham}. For $n=1$ it clearly holds.
We prove it for $n+1$ assuming that \eqref{newop11ham} holds for $n$. One has
\begin{equation}\label{newop11bisham}
\begin{aligned}
&\left[A_{1},\Delta_{12}B\right]^{n+1}+
[A_{1},B_{2}]^{n+1}=\Big[A_{1},[A_{1},\Delta_{12}B]^{n}\Big]+\Big[A_{1},[A_{1},B_{2}]^{n}\Big]\\
&\stackrel{(\ref{newop11ham})}{=}\Big[A_{1},[A_{1},B_{1}]^{n}\Big]=:[A_{1},B_{1}]^{n+1}.
\end{aligned}
\end{equation}
The remaining formul{\ae} can be proved in the same way.
\EP
By using Lemma \ref{newop10ham}, one can rewrite the term $\BB$ in \eqref{newop7ham}. Then
setting $A_{s}:=|\RR_{1}|_{s}+|\RR_{2}|_{s}$ for any $s\geq0$,
and using \eqref{eq:2.12} and \eqref{newop9ham}, one obtains
\begin{equation*}\label{newop15ham}
\begin{aligned}
|\Delta_{12}\BB|_{s}&
\stackrel{(\ref{eq:4.1.33ham}),(\ref{eq:4.1.44ham})}{\leq_{s}}
N^{2\tau+1}\g^{-1}A_{s}|\Delta_{12}\RR|_{\gots_{0}}+N^{2\tau+1}\g^{-1}A_{\gots_{0}}|\Delta_{12}\RR|_{s}
\\
&+2N^{4\tau+2}\g^{-1}A_{s}A_{\gots_{0}}^{2}||u_{1}-u_{2}||_{\gots_{0}+\h_{2}}\\
&+
2N^{4\tau+2}\g^{-2}A_{s}A_{\gots_{0}}|\Delta_{12}\RR|_{\gots_{0}}
+ N^{4\tau+2}\g^{-2}A_{s}A^{2}_{\gots_{0}}||u_{1}-u_{2}||_{\gots_{0}+\h_{2}}\\
&+
N^{4\tau+2}\g^{-2}A_{\gots_{0}}^{2}|\Delta_{12}\RR|_{s},
\end{aligned}
\end{equation*}
where we used the (\ref{eq:4.1.33ham}) and (\ref{eq:4.1.44ham}). If we assume that
\begin{equation}\label{newop16ham}
N^{2\tau+1}\g^{-1}A_{\gots_{0}}\leq1,
\end{equation}
then, using also \eqref{eq:2.22} we obtain the \eqref{newop5ham}.
Finally by using Lemma \ref{regulham} one can note that $[\Psi,\RR]_{\s}^{\s}=O(\e\del_{x}^{-1})$ while $[\Psi,\RR]_{\s}^{-\s}=O(\e)$ for $\s=\pm1$, this implies that the new remainder $\RR_{+}$ has the same properties.
\end{proof}
Clearly we proved Lemma \ref{newopham} by assuming the \eqref{newop9ham} and \eqref{newop16ham}. These hypotheses
have to be verified inductively at each step. In the next Section we prove that the procedure described
above, can be iterated infinitely many times.
\subsection{Conclusions and Proof of Theorem \ref{KAMalgorithmham}}
To complete the proof of Lemma \ref{teo:KAMham} we proceed by induction. The proof of the iteration is essentially standard and based on the estimates of the previous Section.
We omit the proof of properties $({\bf S1})_{\nu+1},({\bf S2})_{\nu+1}$ and $({\bf S3})_{\nu+1}$ since one can repeat
almost word by word the proof of Lemma \ref{teo:KAM} in Section \ref{sec:4}.
The $(S4)_{\nu+1}$ is fundamental different. The difference depends on the multiplicity of the eigenvalues. Moreover the
result is weaker. This is why, in this case, the set of good parameters is smaller. We will see this fact in Section 6.
\noindent
${\bf (S4)}_{\nu+1}$ Let $\oo\in\Lambda_{\nu+1}^{\g}$, then
by (\ref{eq:419ham}) and the inductive hypothesis ${\bf (S4)_{\nu}}$ one has
that $\Lambda_{\nu+1}^{\g}({ u}_{1})\subseteq\Lambda_{\nu}^{\g}({ u}_{1})\subseteq\Lambda_{\nu}^{\g-\rho}({u}_{2})\subseteq\Lambda_{\nu}^{\g/2}({ u}_{2})$.
Hence the eigenvalues $\mu_{h}^{\nu}(\oo,{u}_{2}(\oo))$ are well defined by the ${\bf (S1)_{\nu}}$.
Now, since $\la\in\Lambda_{\nu}^{\g}({u}_{1})\cap\Lambda_{\nu}^{\g/2}({u}_{2})$,
we have for $h=(\s,j) \in\CC\times\ZZZ$
and setting $h'=(\s',j')\in\CC\times\ZZZ$
\begin{equation}\label{eq:4.2.22ham}
\begin{aligned}
|(\mu_{h}^{\nu}&-\mu_{h'}^{\nu})(\oo,{ u}_{2}(\oo))-(\mu_{h}^{\nu}-\mu_{h'}^{\nu})(\oo,{ u}_{1}(\oo))|\leq
|\s j^{2}-\s'j'^{2}| |m_{2}(u_1)-m_{2}(u_2)|\\
&+|m_{0}(u_1)-m_{0}(u_2)||\s-\s'|
+\max_{\substack{j}}|r_{j}^{\nu,j}(\oo,{ u}_{2}(\oo))-r_{j}^{\nu,j}(\oo,{ u}_{1}(\oo))|\\
&+|j||b^{\nu}_{j}(u_{1})-b^{\nu}_{j}(u_{2})|+|j'||b^{\nu}_{j'}(u_{1})-b^{\nu}_{j'}(u_{2})|\\
&\stackrel{(\ref{eq:3.2.44}),(\ref{eq:4.24bisham}),(\ref{aaa2ham})}{\leq}
\e C\left(|\s j^{2}-\s'j'^{2}|+||j|+|j'||\right) ||{u}_{2}-{ u}_{1}||_{\gots_{0}+\h_{2}},
\end{aligned}
\end{equation}
The (\ref{eq:4.2.22ham}) implies that for any $|\ell|\leq N_{\nu}$ and $j\neq \pm j'$,
\begin{equation}\label{eq:4.2.23ham}
\begin{aligned}
|i\oo\cdot\ell+\mu_{\s,j}^{\nu}({ u}_{2})-\mu_{\s',j'}^{\nu}({u}_{2})|
&\stackrel{(\ref{eq:419ham}),(\ref{eq:4.2.22ham})}{\geq}
\g|\s j^{2}-\s' j'^{2}|\langle\ell\rangle^{-\tau}\\
&-C|\s j^{2}-\s'j'^{2}| ||{ u}_{2}-{ u}_{1}||_{\gots_{0}+\h_{2}}\\
&\stackrel{{\bf (S4)_{\nu}}}{\geq}(\g-\rho)|\s j^{2}-\s' j'^{2}|\langle\ell\rangle^{-\tau},
\end{aligned}
\end{equation}
where we used that, for any $\la\in\Lambda_{0}$, one has $C\e N^{\tau}_{\nu}||{ u}_{1}-{ u}_{2}||_{\gots_{0}+\h_{2}}\leq \rho$ (note that this condition is weaker with respect to the hypothesis in $({\bf S4})_{\nu}$.
Now, the (\ref{eq:4.2.23ham}),
imply
that if $\la\in \calP_{\nu+1}^{\g}({u}_{1})$ then
$\la\in \calP_{\nu+1}^{\g-\rho}({ u}_{2})$.
Now assume that $\la\in \calO_{\nu+1}^{\g}(u_{1})$. We have two cases: if $|j|\geq 4|\oo| |\ell|/\e \gote$, then
we have no small divisors. Indeed one has
\begin{equation*}\label{nanonano}
\begin{aligned}
b_{j}^{\nu}(u)^{2}&=\left(-2|m_{1}|+\frac{r_{j}^{\nu,j}-r_{-j}^{\nu,-j}}{j}\right)^{2}+4\frac{|r_{j}^{\nu,-j}|^{2}}{|j|^{2}}\geq\left(2|m_{1}|-\frac{\e C}{|j|}\right)^{2}\\
&\stackrel{(\ref{eq:3.2.44})}{\geq}|m_{1}|^{2}\left(2-\frac{\e C}{|j|\e \gote}\right)^{2}\geq
|m_{1}|^{2}\left(2-\frac{\e \gote}{4|\oo||\ell|}\right)^{2}\geq \frac{|m_{1}|^{2}}{4}\geq \frac{(\e \gote)^{2}}{4},
\end{aligned}
\end{equation*}
for any $u$.
Hence it is obvious that
\begin{equation}\label{eq:4.2.23bisham}
\begin{aligned}
|i\oo\cdot\ell+\mu_{\s,j}^{\nu}({ u}_{2})-\mu_{\s,-j}^{\nu}({u}_{2})|&\geq \frac{4|\oo||\ell|}{\e \gote}|b_{j}^{\nu}(u_{2})|-|\oo\cdot\ell|\\
&{\geq} |\oo||\ell| \geq \frac{\g-\rho}{\langle \ell\rangle^{\tau}\langle j\rangle}.
\end{aligned}
\end{equation}
Let us consider the case $|j|\leq 4|\oo| |\ell|/\e \gote$: one has
\begin{equation*}\label{eq:4.2.23trisham}
\begin{aligned}
|i\oo\cdot\ell+\mu_{\s,j}^{\nu}({ u}_{2})-\mu_{\s,-j}^{\nu}({u}_{2})|
&\stackrel{(\ref{eq:419ham}),(\ref{eq:4.2.22ham})}{\geq}
\g\langle\ell\rangle^{-\tau}\langle j\rangle^{-1}-\e C|j| ||{ u}_{2}-{ u}_{1}||_{\gots_{0}+\h_{2}}\\
&{\geq}\frac{1}{\langle\ell\rangle^{\tau}\langle j\rangle}\left(\g-\e|j|^{2}{CN_{\nu}^{-\al+\tau+2}}\right)\geq\frac{\g-\rho}{\langle\ell\rangle^{\tau}\langle j\rangle}
\end{aligned}
\end{equation*}
that is the ${\bf (S4)}_{\nu+1}$.
\noindent
\emph{Proof of Theorem \ref{KAMalgorithmham}}
We want apply Lemma \ref{teo:KAMham} to the linear operator
$\calL_{0}=\calL_{7}$ defined in (\ref{eq:3.5.9ham})
where
$$
\RR_{0}:=\left(\begin{matrix}0 & q_{0}(\f,x) \\ -\bar{q}_{0}(\f,x)& 0 \end{matrix}\right)+\RR_{7},
$$
with $\RR_{7}$ defined in \eqref{eq:3.7.8ham}. One has that $\RR_{0}$ satisfies the $(iii)$ of Lemma \ref{lem:3.88}.
Then
\begin{equation}\label{eq:4.1.2ham}
\begin{aligned}
|\RR_{0}|_{\gots_{0}+\be}&\stackrel{(\ref{eq:3.2.7ham})}{\leq}\e C(\gots_{0}+\be)(1+||{\bf u}||_{\be+\gots_{0}+\h_{1},\g})
\stackrel{(\ref{eq:4.2})}{\leq}2\e C(\gots_{0}+\be), \qquad \Rightarrow \\
&N_{0}^{C_0}|\RR_{0}|_{\gots_{0}+\be}^{0}\g^{-1}\leq 1,
\end{aligned}
\end{equation}
if $\e\g^{-1}\leq \epsilon_{0}$ is small enough, that is the (\ref{eq:4.15ham}). Then we have to prove that
in the set $\cap_{\nu\geq0}\Lambda_{\nu}^{\g}$
there exists
a final transformation
\begin{equation}\label{eq:4.1.3ham}
\Phi_{\infty}=\lim_{\nu\to\infty}\tilde{\Phi}_{\nu}=\lim_{\nu\to\infty}\Phi_{0}\circ\Phi_{1}\circ\ldots\circ\Phi_{\nu}.
\end{equation}
and the normal form
\begin{equation}\label{eq:4.1.1ham}
\Omega_{\s,\und{j}}^{\infty}:=\Omega_{\s,\und{j}}^{\infty}(\la)=\lim_{\nu\to+\infty}\tilde{\Omega}^{\nu}_{\s,\und{j}}(\la)
=\tilde{\Omega}^{0}_{\s,\und{j}}(\la)+
\lim_{\nu\to+\infty}\left(\begin{matrix} i\s \tilde{r}_{j}^{\n,j} & i\s \tilde{r}_{j}^{\nu,-j} \\ i\s \tilde{r}_{-j}^{\nu,j} &
i\s \tilde{r}_{-j}^{\nu,-j}\end{matrix}\right).
\end{equation}
The proof that limits in \eqref{eq:4.1.3ham} and \eqref{eq:4.1.1ham} exist uses the bounds of Lemma \ref{teo:KAMham}. We refer the reader
to \cite{FP}
for more details.
The following Lemma gives us a connection between the Cantor sets defined in Lemma
\ref{teo:KAMham} and Theorem \ref{KAMalgorithmham}.
Again the proof is omitted since it is essentially the same of Theorem $4.27$ in Section 4 of \cite{FP}.
\begin{lemma}\label{lem:4.4ham}
One has that
\begin{equation}\label{eq:4.1.12ham}
\Lambda^{2\g}_{\infty}\subset\cap_{\nu\geq0}\Lambda_{\nu}^{\g}.
\end{equation}
\end{lemma}
Since one prove that in $\Lambda^{2\g}_{\infty}$ the limit in \eqref{eq:4.1.3ham} exists in norm $|\cdot|_{s,\g}$
one has
\begin{equation}\label{101ham}
\begin{aligned}
\calL_{\n}&\stackrel{(\ref{eq:4.16ham})}{=}\oo\cdot\del_{\f}\uno+\DD_{\n}+\RR_{\n}\stackrel{|\cdot|_{s,\g}}{\to}
\oo\cdot\del_{\f}\uno+\DD_{\infty}=:\calL_{\infty}, \\
& \DD_{\infty}:=diag_{(\s,j)\in C\times\ZZZ}\Omega_{\s,\und{j}}^{\infty}.
\end{aligned}
\end{equation}
and moreover
\begin{equation}\label{102ham}
\calL_{\infty}=\Phi_{\infty}^{-1}\circ\calL_{0}\circ\Phi_{\infty},
\end{equation}
that is the (\ref{eq:4.6ham}),
while the (\ref{eq:4.5ham}) follows by the smallness in \eqref{eq:4.20ham} and the convergence
Finally, Lemma \ref{bubbole},
Lemma \ref{1.4} and (\ref{eq:4.8ham}) implies the bounds (\ref{eq:4.9ham}).
This concludes the proof.
\EP
\section{Inversion of the linearized operator}\label{sec:5ham}
In this Section we prove the invertibility of $\calL(u)$, and consequently of $d_{u}F(u)$ (see \ref{lemmaccio}),
by showing the appropriate tame estimates on the inverse.
The following Lemma resume the results obtained in the previous Sections.
We have the following result
\begin{lemma}\label{lemma5.8ham}
Let $\calL=W_{1}\calL_{\infty}W_{2}^{-1}$ where
\begin{equation}\label{eq:4.4.1ham}
W_{i}=\VV_{i}\Phi_{\infty}, \quad
\VV_{1}:=\TT_{1}\TT_{2}\TT_{3}\rho\TT_{4}\TT_{5}\TT_{6}\TT_{7}, \quad
\VV_{2}=\TT_{1}\TT_{2}\TT_{3} \TT_{4}\TT_{5}\TT_{6}\TT_{7}.
\end{equation}
where $\VV_{i}$ and $\Phi_{\infty}$ are defined in Lemmata \ref{lem:3.88} and \ref{KAMalgorithmham}.
Let $\gots_{0}\leq s\leq q-\be-\h_{1}-2$, with $\h_{1}$ define in (\ref{eq:3.2.0ham}) and $\be$ in Theorem (\ref{KAMalgorithmham}). Then, for $\e\g^{-1}$ small enough, and
\begin{equation}\label{eq:4.4.2ham}
||{ u}||_{\gots_{0}+\be+\h_{1}+2,\g}\leq1,
\end{equation}
one has for any $\la\in\Lambda^{2\g}_{\infty}$,
\begin{equation}\label{eq:4.4.3ham}
\begin{aligned}
||W_{i}{h}||_{s,\g}+||W_{i}^{-1}{h}||_{s,\g}&
\leq C(s)\left(||{ h}||_{s+2,\g}+
||{u}||_{s+\be+\h_{1}+4,\g}||{ h}||_{\gots_{0},\g}\right),
\end{aligned}
\end{equation}
for $i=0,1$. Moreover, $W_{i}$ and $W_{i}^{-1}$ symplectic.
\end{lemma}
\begin{proof}
Each $W_i$ is composition of two operators, the $\VV_i$ satisfy the (\ref{eq:3.2.3ham}) while $\Phi_\infty$ satisfies (\ref{eq:4.8ham}). We use
Lemma \ref{bubbole}
in order to pass to the operatorial norm. Then Lemma \ref{lem5}
implies the bounds (\ref{eq:4.4.3ham}). Moreover the transformations $W_{i}$ and $W_{i}^{-1}$ symplectic because they are composition of symplectic transformations $\VV_{i}$,$\VV_{i}^{-1}$ and
$\Phi_{\infty}$, $\Phi_{\infty}^{-1}$ .
\end{proof}
Thanks to Lemma \ref{lemma5.8ham} the proof of Proposition \ref{teo2ham} is almost concluded.
We fix the constants $\h=\h_{1}+\be+2$ (the constant $\h$ has to be chosen) and $q>\gots_{0}+\h$.
Let $\Omega_{\s,j}^{\phantom{g},j}$ and $\Omega_{\s,j}^{-j}$ be the functions defined in \eqref{eq:4.1.1ham}, and consequently $\mu_{\s,j}$ the eigenvalues of the matrices $\Omega_{\s,\und{j}}$. Therefore
by Lemmata \ref{KAMalgorithmham} and \ref{lemma5.8ham} item $(i)$ in Proposition \ref{teo2ham} hold. The item $(ii)$
follows simply by applying a dynamical point of view to our operator. Indeed
by Lemma \ref{lem:3.9ham} and \eqref{eq:4.9ham} in Lemma \ref{KAMalgorithmham} one has that each transformation in
\eqref{eq:4.4.1ham} is a quasi-periodic in time transformation of the phase space ${\bf H}^{s}_{x}$ plus a reparametrization of the time with $\TT_{3}$.
Under a transformation of the form ${\bf u}=A(\oo t){\bf v}$, one has that
\begin{equation}\label{eq:6.4ham}
\begin{aligned}
&\del_{t}{\bf u}=L(\oo t){\bf u}\;\;\; \leftrightarrow\;\;\;\;
\del_{t}{\bf v}=L_{+}(\oo t){\bf v}, \\
&L_{+}(\oo t)=A(\oo t)^{-1} L(\oo t)A(\oo t)-A(\oo t)^{-1}\del_{t}A(\oo t)
\end{aligned}
\end{equation}
by conjugation. Moreover the transformation $A(\oo t)$ acts on the functions ${\bf u}(\f,x)$ as
\begin{equation}\label{eq:6.5ham}
\begin{aligned}
( {A}{\bf u})(\f,x)&:=(A(\f){\bf u}(\f,\cdot))(x):=A(\f){\bf u}(\f,x),\\
&({A}^{-1}{\bf u})(\f,x)=A^{-1}(\f){\bf u}(\f,x).
\end{aligned}
\end{equation}
Then on the space of quasi-periodic functions one has that the operator
\begin{equation}\label{eq:6.6ham}
\calL:=\oo\cdot\del_{\f}-L(\f),
\end{equation}
associated to the system (\ref{eq:6.4ham}), is transformed by $ {A}$ into
\begin{equation}\label{eq:6.7ham}
{A}^{-1}\calL {A}=\oo\cdot\del_{\f}-L_{+}(\f),
\end{equation}
that represent the system in (\ref{eq:6.4ham}) acting on quasi-periodic functions. The same consideration hold for a transformation of the type $\TT_{3}$ as explained in Section
\ref{sec:5}.
Now we prove the following Lemma that is the equivalent result of Lemma \ref{inverseofl} in the Hamiltonian case.
\begin{lemma}[{\bf Right inverse of $\calL$}]\label{inverseoflham}
Under the hypotheses of Proposition \ref{teo2ham}, let us set
\begin{equation}\label{eq:4.4.18ham}
\z:=4\tau+\h+8
\end{equation}
where $\h$ is fixed in Proposition \ref{teo2ham}.
Consider a Lipschitz family ${\bf u}(\oo)$ with $\oo\in\Lambda_{o}\subseteq\Lambda\subseteq\RRR^{d}$
such that
\begin{equation}\label{eq:4.4.19ham}
||{\bf u}||_{\gots_{0}+\z,\g} \leq1.
\end{equation}
Define the set
\begin{equation}\label{primedim}
\calP^{2\g}_{\infty}({ u}):=\left\{\begin{array}{ll}
\oo\in\Lambda_{o} : &|\oo\cdot\ell\!+\!\mu_{\s,{j}}(\oo)|\geq
\frac{2\g\langle j\rangle^{2}}{\langle\ell\rangle^{\tau}}, \\
&\;\ell\in\ZZZ^{d}, \s,\in\CC, j\in\ZZZ
\end{array}
\right\}.
\end{equation}
There exists $\epsilon_{0}$, depending only on the data of the problem, such that
if $\e\g^{-1}<\epsilon_{0}$
then,
for any $\oo\in\Lambda^{2\g}_{\infty}({u})\cap\calP^{2\g}_{\infty}({u})$
(see (\ref{martina10ham})), and for any Lipschitz family
${ g}(\oo)\in {\bf H}^{s}$,
the equation $\calL { h}:=\calL(\oo,{ u}(\oo)){ h}= { g}$,
where $\calL$ is the linearized operator $\calL$ in \eqref{lemmaccio4}, admits a solution
\begin{equation}\label{eq:4.4.22ham}
{ h}:=\calL^{-1}{ g}:=W_{2}\calL_{\infty}^{-1}W_{1}^{-1}{ g}
\end{equation}
such that
\begin{equation}\label{eq:4.4.23ham}
||{ h}||_{s,\g}\leq C(s)\g^{-1}\left(||{ g}||_{s+2\tau+5,\g}+
||{ u}||_{s+\z,\g}||{g}||_{\gots_{0},\g}
\right), \quad \gots_{0}\leq s \leq q-\z.
\end{equation}
\end{lemma}
\begin{proof}
A direct consequence of Lemma \ref{lemma5.8ham} is that, once one has conjugated the operator
$\calL$ in \eqref{lemmaccio4} to a block-diagonal operator $\calL_{\infty}$ in \eqref{eq:4.6ham} is essentially
trivial to invert it:
\begin{lemma}\label{inverselinftyham}
For ${ g}\in{\bf H}^{s}$, consider the equation
\begin{equation}\label{eq:4.4.7ham}
\calL_{\infty}({u}){h}={ g}.
\end{equation}
If $\oo\in \Lambda^{2\g}_{\infty}({\bf u})\cap \calP^{2\g}_{\infty}(u)$ (defined in \eqref{martina10ham} and \eqref{primedim}), then there exists a unique solution
$\calL_{\infty}^{-1}{g}:={h}\in{\bf H}^{s}$. Moreover, for all Lipschitz family ${ g}:=
{g}(\oo)\in {\bf H}^{s}$ one has
\begin{equation}\label{eq:4.4.8ham}
||\calL_{\infty}^{-1}{ g}||_{s,\g}\leq C \g^{-1}||{ g}||_{s+2\tau+1,\g}.
\end{equation}
\end{lemma}
\begin{proof}
One can follows the same strategy used for Lemma
$5.44$ in \cite{FP}
and conclude using Lemma \ref{finitema}.
\end{proof}
In order to conclude the proof of Lemma \ref{inverseofl} it is sufficient
to collect the results of Lemmata \ref{lemma5.8ham} and \ref{inverselinftyham}. In particular one uses \eqref{eq:4.4.3ham} and \eqref{eq:4.4.8ham}
to obtain the estimate
\begin{equation}\label{eq:4.4.25ham}
\begin{aligned}
||{h}||_{s,\g}&=\|W_{2}\calL_{\infty}^{-1}W_{1}^{-1}{ g}\|_{s,\g}\\
&\leq C(s)\g^{-1} \left( ||{ g}||_{s+2\tau+5,\g}+
||{ u}||_{s+4\tau+\be+10+\h_{1},\g}||{ g}||_{\gots_{0},\g}
\right),
\end{aligned}
\end{equation}
\end{proof}
Note that by Lemma \ref{lemmaccio} the estimates \eqref{eq:4.4.25ham} holds also for the linearized operator
$d_{u}\calF(u)$.
\section{Measure estimates}\label{sec6ham}
In Section \ref{sec:3ham}, \ref{sec:4ham} and \ref{sec:5ham} we prove that in the set
$\Lambda^{2\g}_{\infty}({\bf u}_{n})\cap \calP^{2\g}_{\infty}(u)$ we have good bounds (see \eqref{eq:4.4.23ham}) on the inverse of $\calL(u_{n})$
. We also give a precise characterization of this set in terms of the eigenvalues of $\calL$.
Now in the Nash-Moser proposition \ref{teo4} we defined in an implicit way the sets $\calG_n$ in order to ensure bounds on the inverse of $\calL({ u}_{n})$.
In this section we prove Proposition \ref{measurebruttebrutte} which is the analogous
analysis performed in Section $6$ of \cite{FP}.
\begin{proposition}[{\bf Measure estimates}]\label{measurebruttebrutte}
Set $\g_n:=(1+2^{-n})\g$ and consider the set $\calG_{\infty}$ of Proposition \ref{teo4}
with $\mu=\zeta$ defined in Lemma \ref{inverseoflham} and fix $\g:=\e^{a}$ for some $a\in(0,1)$. We have
\begin{subequations}\label{eq136totham}
\begin{align}
&\cap_{n\geq0}\calP^{2\g_n}_{\infty}({u}_n)\cap \Lambda^{2\g_n}_{\infty}({ u}_n)\subseteq \calG_{\infty}, \label{eq136bham}\\
&|\Lambda\backslash\calG_{\infty}|\to 0, \;\; {\rm as} \;\; \e\to0. \label{eq136ham}
\end{align}
\end{subequations}
\end{proposition}
{\bf Proof of Proposition \ref{measurebruttebrutte}}.
Let $({ u}_{n})_{\geq0}$ be the sequence of approximate solutions introduced in Proposition
\ref{teo4} which is well defined in $\calG_{n}$ and satisfies the hypothesis
of Proposition \ref{teo2ham}. $\calG_{n}$ in turn is defined in
Definition \ref{invertibility}.
We now define inductively a sequence of nested sets $G_{n}\cap H_{n}$ for $n\geq0$.
Set $G_{0}\cap H_{0}=\Lambda$ and
\begin{equation}\label{eq142bis}
\begin{aligned}
G_{n+1}&:=
\left\{\begin{aligned}
\oo\in G_{n} \; :\; &|i\oo\cdot\ell+\mu_{\s,j}({ u}_{n})-\mu_{\s',j'}({ u}_{n})|
\geq\frac{2\g_{n}|\s j^{2}-\s' j'^{2}|}{\langle\ell\rangle^{\tau}}, \\
&\;\forall\ell\in\ZZZ^{n},\;\; \s,\s'\in\CC, \;\; j,j'\in\ZZZ
\end{aligned}\right\},\\
H_{n+1}&:=
\left\{\begin{aligned}
\oo\in H_{n} \; :\; &|i\oo\cdot\ell+\mu_{\s,j}({ u}_{n})-\mu_{\s,-j}({ u}_{n})|
\geq\frac{2\g_{n}}{\langle\ell\rangle^{\tau}\langle j\rangle}, \\
&\;\forall\ell\in\ZZZ^{n}\backslash\{0\},\;\; \s\in\CC, \;\; j\in\ZZZ
\end{aligned}\right\},\\
P_{n+1}&:=
\left\{\begin{aligned}
\oo\in P_{n} \; :\; &|i\oo\cdot\ell+\mu_{\s,j}({ u}_{n})|
\geq\frac{2\g_{n}\langle j\rangle^{2}}{\langle\ell\rangle^{\tau}}, \\
&\;\forall\ell\in\ZZZ^{n},\;\; \s\in\CC, \;\; j\in\ZZZ
\end{aligned}\right\},
\end{aligned}
\end{equation}
Recall that $\mu_{\s,j}(u_{n})$ and $\mu_{\s,-j}(u_{n})$ are the eigenvalues of the matrices $\Omega_{\s,\und{j}}$
defined in Proposition \ref{teo2ham} in \eqref{1.2.2bis}.
The following Lemma implies \eqref{eq136bham}.
\begin{lemma}\label{megalemma} Under the Hypotheses of Proposition \ref{measurebruttebrutte},
for any $n\geq0$, one has
\begin{equation}\label{eq137}
P_{n+1}\cap G_{n+1}\cap H_{n+1}\subseteq \calG_{n+1}.
\end{equation}
\end{lemma}
\begin{proof}
For any $n\geq0$
and if $\la\in G_{n+1}$, one has
by Lemmata
\ref{inverselinftyham} and \ref{inverseofl}, (recalling that $\g\leq \g_{n}\leq 2\g$ and $2\tau+5<\zeta$)
\begin{equation}\label{eq139}
\begin{aligned}
||\calL^{-1}({u}_{n}){h }||_{s,\g}&\leq C(s)\g^{-1}\left(
||{ h}||_{s+\zeta,\g}+||{u}_{n}||_{s+\zeta,\g}||{h}||_{\gots_{0},\g}\right),\\
||\calL^{-1}({u}_{n})||_{\gots_{0},\g}&\leq C(\gots_{0})\g^{-1}
N_{n}^{\zeta}||{ h}||_{\gots_{0},\g},
\end{aligned}
\end{equation}
for $\gots_{0}\leq s\leq q-\mu$, for any ${h}(\la)$ Lipschitz family.
The (\ref{eq139}) are nothing but the (\ref{eq104}) in Definition \ref{invertibility} with $\mu=\zeta$ .
It represents the loss of regularity that you have when you perform the
regularization procedure in Section \ref{sec:3ham} and during the diagonalization algorithm in
Section \ref{sec:4ham}. This justifies our choice of $\mu$ in Proposition \ref{measurebruttebrutte}.
\end{proof}
\noindent
Now we prove formula \eqref{eq136ham} that is the most delicate point.
It turns out, by an explicit computation, that we can write for $j\neq0$,
\begin{equation}\label{luna2}
\mu_{\s,j}-\mu_{\s,-j}{:=}i\s\sqrt{
(-2|m_{1}|j+r_{j}^{j}-r_{-j}^{-j})^{2}+4|r_{j}^{-j}|}:=j b_{j}=jb_{j}(u_{n}),
\end{equation}
where $r_{j}^{k}$, for $j,k\in \NNN$ are the coefficients of the matrix $R_{\s,\und{j}}$ in \eqref{1.2.2bisham},
and we define
\begin{equation}\label{luna3}
\psi(\oo,u_{n}):={\oo}\cdot\ell+j b_{j}(u_{n}).
\end{equation}
Now we write
for any $\ell\in \ZZZ^{d}\backslash\{0\}$ and $j\in \ZZZ$,
\begin{equation}\label{luna}
H_{n}:=\bigcap_{\substack{\s\in\CC, \\(\ell,j)\in\ZZZ^{d+1}}}\!\!\! A^{\s}_{\ell,j}(u_{n})
:=\bigcap_{\substack{\s\in\CC,\\ (\ell,j)\in\ZZZ^{d+1}}}\left\{\oo\in H_{n-1} : |i\oo\cdot\ell+j b_{j}(u_{n})|\geq\frac{\g_{n}}{\langle j\rangle\langle\ell\rangle^{\tau}}
\right\}.
\end{equation}
Clearly one need to estimate the measure of $\bigcap_{n\geq0}H_{n}$. The strategy to get such estimate is quite standard and it is the following:
\begin{itemize}
\item[{\bf a}.] First one give an estimate of the resonant set for fixed $(\s,j,\ell)\in \CC\times\ZZZ\times\ZZZ^{d}$ (namely $(A^{\s}_{\ell,j})^{c}$).
This point require a lower bound
on the Lipschitz sub-norm of the function $\psi$ in \eqref{luna3}. In this way we can give an estimate of the measure of the bad set using the standard arguments to estimate the measure of sub-levels of Lipschitz functions. This is in general non trivial but
in the case of the sets $G_{n}$ and $P_{n}$ there is a well established strategy to follow
that uses that $\mu_{\s,j}\sim O(j^{2})$. In the case of the sets $H_{n}$ the problem
is more difficult since $\mu_{\s,j}\sim O(\e j)$, hence, even if $j$ is large, it could happen that $\mu_{\s,j}\sim \oo\cdot\ell$.
However we prove such lower bound (see \eqref{luna10}) using result of Lemma \ref{nani2} and
non-degeneracy condition on $m_{1}$ (see \eqref{mammamia}). Moreveor we use deeply the fact that we have $d$ parameters $\oo_{i}$ for $i=1,\ldots,d$ to move. On the contrary in Section $6$ of \cite{FP}
the authors
performed the estimates by choosing a diophantine direction
$\bar{\oo}$ and using as frequency the vector $\oo=\la \bar{\oo}$ with $\la\in[1/2,3/2]$, hence using just one parameter.
In this case this is not possible.
\item[{\bf b}.] Item ${\bf a.}$ provides and estimate like $|(A^{\s}_{j,\ell})^{c}|\sim \g/(j|\ell|^{\tau})$.
The second point is to have some summability of the series in $j$ since one need to control
$\bigcup_{j,\ell}(A^{\s}_{\ell,j})^{c}$. One can sum over $\ell$ by choosing $\tau$ large enough. In principle on can think
to weaker the Melnikov conditions and ask for a lower bound of the type
\begin{equation}\label{albero}
|\psi|\geq\g/|j|^{2}|\ell|^{\tau}.
\end{equation}
This can cause two problems. If one ask \eqref{albero} it may be very difficult to prove the lower bound on the Lipschitz norm.
Secondly in the reduction algorithm one must have a remainder $\RR$ that support
the loss of $2$ derivatives in the space. Our strategy is different: we use results in Lemmata \ref{luna4} and \ref{luna6} to prove
that the number of $j$ for which $(A^{\s}_{\ell,j})^{c}\neq \emptyset$ is controlled by $|\ell|$.
\item[{\bf c}.] Finally one has to prove some ``relation'' between the sets $H_{n}$ and $H_{k}$ for $k\neq n$. Indeed
the first two points imply only that the set $H_{n}$ has large measure as $\e\to 0$. But in principle as $n$ varies this sets can be unrelated, so that the intersection can be empty. Roughly speaking in Lemma \ref{luna11} we prove that
lots of resonances at the step $n$ have been already removed at the step $n-1$. In other words we prove that, if $|\ell|$ is sufficiently small, if $\psi(u_{n-1})$ satisfies the Melnikov conditions, then also $\psi(u_{n})$ automatically has the good bounds.
Again this point is different from the case studied in Section $6$ of \cite{FP}.
Indeed with double eigenvalues one is able to prove the previous claim
only for $n$ large enough and not for any $n$. This is the reason in this case the set of good parameters is small, but in any case of full measure.
\end{itemize}
\noindent
In the following Lemma we resume the key result one need to prove Proposition \ref{measurebruttebrutte}.
\begin{lemma}\label{cavallo2}
For any $n\geq0$ one has
\begin{equation}\label{frodo5}
|P_{n}\backslash P_{n+1}|,|G_{n}\backslash G_{n+1}|,|H_{n}\backslash H_{n+1}|\leq C\sqrt{\g}.
\end{equation}
Moreover, if $n\geq\bar{n}(\e)$ (where $\bar{n}(\e)$ is defined in Lemma \ref{luna11}), then one has
\begin{equation}\label{frodo6}
|P_{n}\backslash P_{n+1}|,|G_{n}\backslash G_{n+1}|,|H_{n}\backslash H_{n+1}|\leq C\sqrt{\g} N_{n}^{-1}.
\end{equation}
In particular $\bar{n}(\e)$ has the form
\begin{equation}\label{frodo10}
\bar{n}(\e):=a {\rm log}{ {\rm log}\left[b\frac{1}{c \g \e}\right]},
\end{equation}
with $a,b,c>0$ independent on $\e$.
\end{lemma}
By Lemma \ref{cavallo2} follows the \eqref{eq136}. Indeed on one hand we have
\begin{equation}\label{frodo11}
\begin{aligned}
|\Lambda\backslash \cap_{n\geq0} H_{n}|&\leq
\sum_{n=0}^{\bar{n}(\e)}|H_{n}\backslash H_{n+1}|+\sum_{n> \bar{n}(\e)}|H_{n}\backslash H_{n+1}|\leq C \g\bar{n}(\e).
\end{aligned}
\end{equation}
The same bounds holds for $|\Lambda\backslash \cap_{n\geq0}G_{n}|,|\Lambda\backslash \cap_{n\geq0}P_{n}|.$
Now, fixing $\g:=\g(\e)=\e^{a}$ with $a\in (0,1)$, one has that
$$
|\Lambda\backslash \calG_{\infty}|\leq C\sqrt{\g(\e)}(1+\bar{n}(\e))\to0, \quad {\rm as} \quad \e\to0.
$$
This concludes the proof of Proposition \ref{measurebruttebrutte}. It remains to check Lemma \ref{cavallo2} following the strategy in three point explained above.
We will give the complete proof only for the sets $H_{n}$ that is more difficult.
The inductive estimates on $G_{n}$ and $P_{n}$ is very similar, anyway one can follows essentially word by word the proof of Proposition $1.10$ in Section $6$ of \cite{FP}. Similar measure estimates can be also found in \cite{BBM}.
\begin{lemma}\label{luna4}
If $|b_{j}||j|\geq 2|{\oo}\cdot\ell|$ or $|b_{j}||j|\leq |{\oo}\cdot\ell|/2$ then
$(A^{\s}_{\ell,j}(u_{n}))^{c}=\emptyset$.
\end{lemma}
\begin{proof} Lemma follows by the fact that ${\oo}$ is diophantine with constant $\tau_{0}$ and $\tau>\tau_{0}$
and from the smallness of $|m_{1}|$.
\end{proof}
\noindent
Thanks Lemma \ref{luna4} in the following we will consider only the $j\in \SSSS_{\ell,n}\subseteq\ZZZ$ where
\begin{equation}\label{luna5}
\SSSS_{\ell,n}:=\left\{j\in\ZZZ \; \frac{|{\oo}\cdot\ell|}{2}\leq |j|b_{j}(u_n)\leq2|{\oo}\cdot\ell|
\right\}
\end{equation}
for some constant $C>0$. In order to estimate the measure of $(A^{\s}_{\ell,j}(u_{n}))^{c}$ we need the following technical Lemma.
\begin{lemma}\label{luna6}
If $j\in \SSSS_{\ell,n}\cap (A_{\ell,n})^{c}$, where
$$A_{\ell}:=\{j\in\ZZZ : |j|\leq4|\ell| C/\gote\},$$ then
one has that $|b_{j}(u_{n})|\geq |m_{1}(u_{n})|/2$.
\end{lemma}
\prova
It follow by
\begin{equation}\label{nano}
\begin{aligned}
b_{j}^{2}&=\left(-2|m_{1}|+\frac{r_{j}^{j}-r_{-j}^{-j}}{j}\right)^{2}+4\frac{|r_{j}^{-j}|^{2}}{|j|^{2}}\geq\left(2|m_{1}|-\frac{\e C}{|j|}\right)^{2}\\
&\stackrel{(\ref{eq:3.2.44})}{\geq}|m_{1}|^{2}\left(2-\frac{\e C}{|j|\e \gote}\right)^{2}\geq
|m_{1}|^{2}\left(2-\frac{1}{4|\ell|}\right)^{2}\geq \frac{|m_{1}|^{2}}{4}.
\end{aligned}
\end{equation}
\EP
\noindent
An consequence of Lemmata \ref{luna4} and \ref{luna6} is that we need to study the sets $A^{\s}_{\ell,j}$ only for
\begin{equation}\label{padre}
|j|\leq \frac{C|\ell|}{\e \gote}.
\end{equation}
It is essentially what explained in item ${\bf b}.$ Note the here we used the non-degeracy of the constant $m_{1}$.
\begin{lemma}\label{nani2}
For any $n\geq0$ and $j\in \SSSS_{\ell,n}$ one has
\begin{equation}\label{nani3}
|b_{j}(u_n)|^{lip}\leq K \frac{1}{|j|}\left[|m_{1}|^{lip}|j|+\e C\right],
\end{equation}
for some $K>0$.
\end{lemma}
\prova
One can note that,
\begin{equation}\label{nani4}
\begin{aligned}
|b_{j}(\oo_1)-&b_{j}(\oo_{2})|=\left|\frac{ b^2_{j}(\oo_1)-b^{2}_{j}(\oo_{2})}{b_{j}(\oo_1)+b_{j}(\oo_{2})}\right|\leq\\
&\leq|\oo_{1}-\oo_{1}|\left[|m_{1}|^{lip}+\frac{1}{|j|}(|r_{j}^{j}|^{lip}+|r_{-j}^{-j}|^{lip}+|r_{j}^{-j}|^{lip})\right],
\end{aligned}
\end{equation}
using that
\begin{equation}\label{nani5}
\frac{|(-2|m_{1}(\oo_{1})|+(r_{j}^{j}-r_{-j}^{-j})(\oo_{1})/j)|+|(-2|m_{1}(\oo_{1})|+(r_{j}^{j}-r_{-j}^{-j})(\oo_{1})/j)|}{b_{j}(\oo_1)+b_{j}(\oo_{2})}\leq2,
\end{equation}
and that the same bound holds also for $|(r_{j}^{-j})(\oo_1)|/|j|(b_{j}(\oo_1)+b_{j}(\oo_{2}))$.
\EP
\noindent
An immediate consequence of \eqref{nani3} is that
\begin{equation}\label{nani6}
|j||b_{j}|^{lip}\stackrel{(\ref{eq:3.2.44})}{\leq}4|\ell|\frac{C}{\gote}2K\e C, \qquad j\in \SSSS_{\ell,n}\cap A_{\ell}
\end{equation}
\begin{equation}\label{nani7}
|j||b_{j}|^{lip}\stackrel{(\ref{mammamia})}{\leq}K|j|\frac{1}{|j|}\left[\e|m_{1}(0)|C\frac{|\ell|}{\e \gote}+\e C\right]
\leq \tilde{K}\e |\ell|, \qquad j\in\SSSS_{\ell,n} \cap(A_{\ell})^{c}
\end{equation}
\noindent
By Lemmata \ref{luna6} and \ref{nani2} we deduce the following fundamental estimates on the function $\psi$ defined in \eqref{luna3}. First we note that,
since there exists $i\in\{1,\ldots,d\}$ such that $|\ell_{i}|\geq |\ell|/2d$, one has
$$
|\del_{\oo_i}\oo\cdot\ell|\geq\frac{|\ell|}{2d}.
$$
Hence one has
\begin{equation}\label{luna10}
\begin{aligned}
|\psi|^{lip}&\geq \left(\frac{|\ell|}{2d}-|j||b_{j}|^{lip}
\right)\stackrel{(\ref{nani6})}{\geq}\frac{|\ell|}{4d},
\end{aligned}
\end{equation}
for $\e$ small enough for any $j\in \SSSS_{\ell,n}$. The \eqref{luna10} is fundamental in order to estimate
the measure of a single resonant set and this is what we claimed in item ${\bf a}$. The following Lemma
is the part ${\bf c}.$ of the strategy,
\begin{lemma}\label{luna11}
For $|\ell|\leq N_{n}$ one has that for any $\e>0$ there exists $\bar{n}:=\bar{n}(\e)$ such that
if $n\geq\bar{n}(\e)$ then
\begin{equation}\label{luna122}
(A^{\s}_{\ell,j}(u_{n}))^{c}\subseteq (A^{\s}_{\ell,j}(u_{n-1}))^{c}.
\end{equation}
\end{lemma}
\begin{proof}
We first have to estimate
\begin{equation}\label{speriamobene}
\begin{aligned}
|j||b_{j}(u_{n})-b_{j}(u_{n-1})|&\leq
4\max_{h=\pm j}\{|r_{j}^{-h}(u_{n})-r_{j}^{ -h}(u_{n-1})|\}\\
&+2|m_{1}(u_{n})-m_{1}(u_{n-1})| |j|.
\end{aligned}
\end{equation}
By Lemma \ref{teo:KAMham}, using the $({\bf S4})_{n+1}$ with
$\g=\g_{n-1}$ and $\g-\rho=\g_{n}$, and with ${ u}_{1}={u}_{n-1}$, ${ u}_{2}={u}_{n}$,
we have
\begin{equation}\label{eq162}
\Lambda_{n+1}^{\g_{n-1}}({ u}_{n-1})\subseteq\Lambda_{n+1}^{\g_{n}}({ u}_{n}),
\end{equation}
since, for $\e\g^{-1}$ small enough, and $n\geq \bar{n}(\e)$ defined as
\begin{equation}
\bar{n}(\e):=\frac{1}{\log(3/2)}\log\left[\frac{1}{(\ka-\tau-3)\log N_{0}}\log\left(\frac{1}{C\g \e}\right)\right]
\end{equation}
\begin{equation}\label{eq163}
C N_{n}^{\tau}\sup_{\la\in G_{n}}||{ u}_{n}-{ u}_{n-1}||_{\gots_{0}+\mu}
\leq \e (\g_{n-1}-\g_{n})=:\e \rho=\e \g2^{-n}.
\end{equation}
where $\ka$ is defined in (\ref{teo41}) with
$\nu=2$, $\mu=\zeta$ defined in (\ref{eq:4.4.18}) with $\h=\h_{1}+\be$, $\mu>\tau$
(see Lemmata \ref{measurebruttebrutte}, \ref{megalemma} and (\ref{eq:4.14ham}), (\ref{eq:3.2.0ham})).
We also note that,
\begin{equation}\label{eq164}
G_{n}\cap H_{n}\stackrel{(\ref{eq142bis}), (\ref{martina10})}{\subseteq}\Lambda_{\infty}^{2\g_{n-1}}({u}_{n-1})
\stackrel{(\ref{eq:4.1.12ham})}{\subseteq}\Lambda_{n+1}^{\g_{n-1}}({ u}_{n-1})
\stackrel{(\ref{eq162})}{\subseteq}
\Lambda_{n+1}^{\g_{n}}({ u}_{n}).
\end{equation}
This means that
$\la\in H_{n}\cap G_{n}\subset \Lambda_{n+1}^{\g_{n-1}}({u}_{n-1})\cap\Lambda_{n+1}^{\g_{n}}({ u}_{n})$,
and hence, we can apply the ${\bf (S3)}_{\nu}$, with $\nu=n+1$, in Lemma \ref{teo:KAMham} to get
for any $h,k=\pm j$,
\begin{equation*}\label{eq165}
\begin{aligned}
\!\!\!\!\!\!|r_{h}^{k}({ u}_{n})-&r_{h}^{k}({ u}_{n-1})|\leq
|r_{h}^{n+1,k}({ u}_{n})-r_{h}^{n+1,k}({ u}_{n-1})|\\
&+|r_{h}^{k}({ u}_{n})-r_{h}^{n+1,k}({ u}_{n})|
+
|r_{h}^{k}({ u}_{n-1})-r_{h}^{n+1,k}({ u}_{n-1})|\\
\stackrel{(\ref{eq:3.2.6aham}),(\ref{eq:4.24bisham}),(\ref{teo41})}\leq& C\e^{2}\g^{-1} N_{n}^{-\ka}
+\e\left(1+||{u}_{n-1}||_{\gots_{0}+\h_{1}+\be}+||{ u}_{n}||_{\gots_{0}+\h_{1}+\be}\right)N_{n}^{-\al}.
\end{aligned}
\end{equation*}
Now, first of all $\ka>\al$ by (\ref{teo41}), (\ref{eq:4.14ham}), moreover
$\h_{1}+\be<\h_{5}$ then by ${\bf (S1)}_{n}$, $({\bf S1})_{n-1}$, one has
$||{u}_{n-1}||_{\gots_{0}+\h_{5}}+||{u}_{n}||_{\gots_{0}+\h_{5}}\leq 2$, we obtain
\begin{equation}\label{eq166}
|r_{h}^{k}({u}_{n})-r_{h}^{k}({ u}_{n-1})|\stackrel{(\ref{eq165})}{\leq}
\e N_{n}^{-\al}.
\end{equation}
Then, by (\ref{speriamobene}), (\ref{eq:3.2.44}) and (\ref{eq166}) one has that
\begin{equation}\label{eq167}
|(\mu_{\s,j}-\mu_{\s,-j})(u_{n})-(\mu_{\s,j}-\mu_{\s,-j})(u_{n-1})|\leq C\e |j| N_{n}^{-\al},
\end{equation}
hence for $|\ell|\leq N_{n}$, and $\la\in G_{n}\cap H_{n}$, we have
\begin{equation}\label{eq168}
\begin{aligned}
|i\oo\cdot\ell+\mu_{\s,j}({ u}_{n})-\mu_{\s,j}({ u}_{n})|
&\stackrel{(\ref{eq167})}{\geq}\frac{2\g_{n-1}}{\langle\ell\rangle^{\tau}\langle j\rangle}
-C\e |j|N_{n}^{-\al}
\geq\frac{2\g_{n}}{\langle\ell\rangle^{\tau}\langle j\rangle},
\end{aligned}
\end{equation}
since $j\in\SSSS_{\ell,n}$, hence $|j|\leq 4|\oo||\ell|/\e\gote$, and $n$ is such that $N_{n}^{\tau-\al+2}\lessdot\g2^{-n}\e$.
The \eqref{eq168} implies the \eqref{luna122}.
\end{proof}
\noindent
An immediate consequence of Lemma \ref{luna11} is the following.
\begin{proof}{\emph{Proof of Lemma \ref{cavallo2}.}}
First of all, write
\begin{equation}\label{luna14}
\begin{aligned}
&H_{n}\backslash H_{n+1}:=\bigcup_{\substack{\s\in\CC, j\in\ZZZ \\ \ell\in\ZZZ^{d}}}
(A_{\ell, j}^{\s}({ u}_{n}))^{c}.\\
\end{aligned}
\end{equation}
By using Lemma \ref{luna11} and equation \eqref{luna5},
we obtain
\begin{equation}\label{luna15}
\begin{aligned}
&H_{n}\backslash H_{n+1}\subseteq H_{n}^{(1)}\cup H_{n}^{(2)}\cup H_{n}^{(3)}\cup H_{n}^{(4)}\\
&H_{n}^{(1)}:=\Big(\bigcup_{\substack{\s\in\CC,\\
\;
j\in \SSSS_{\ell}\cap A_{\ell} \\ |\ell|\leq N_{n} }}(A_{\ell, j}^{\s}({ u}_{n}))^{c}\Big),
\quad H_{n}^{(2)}:=\Big(\bigcup_{\substack{\s\in\CC,\\
\;
j\in \SSSS_{\ell}\cap A_{\ell} \\ |\ell|> N_{n} }}
(A_{\ell, j}^{\s}({ u}_{n}))^{c}\Big), \\
&H_{n}^{(3)}:=\Big(\bigcup_{\substack{\s\in\CC,\\
\;
j\in \SSSS_{\ell}\cap (A_{\ell})^{c} \\ |\ell|\leq N_{n} }}
(A_{\ell, j}^{\s}({ u}_{n}))^{c}\Big), \quad H_{n}^{(4)}:=\Big(\bigcup_{\substack{\s\in\CC,\\
\;
j\in \SSSS_{\ell}\cap (A_{\ell})^{c} \\ |\ell|> N_{n} }}
(A_{\ell, j}^{\s}({ u}_{n}))^{c}\Big).
\end{aligned}
\end{equation}
One has that the cardinality if the set $\SSSS_{\ell,n}\cap A_{\ell}$ is less than $4|\ell|C/\gote$. This implies that
\begin{equation}\label{stime1}
|H^{(2)}|\leq\sum_{|\ell|> N_{n}}\frac{4|\ell|C\g_{n}}{\gote\langle j\rangle\langle\ell\rangle^{\tau}}\frac{4d}{|\ell|}\lessdot
C\g N_{n}^{-1}.
\end{equation}
Let us estimate the measure of the sets $H^{(i)}$ for $i=3,4$.
The cardinality of $\SSSS_{\ell,n}\cap(A_{\ell})^{c}$ is less than $K|\ell|/\e \gote$, hence we have to study the case $j\in\SSSS_{\ell,n}\cap(A_{\ell})^{c}$ more carefully. We introduce the sets
\begin{equation}\label{nani7}
B_{\ell,j}^{\s}:=\left\{\oo\in H_{n-1} : |i\oo\cdot\ell+j b_{j}(u_{n})|\geq\frac{\g'_{n}\al_n}{ \langle\ell\rangle^{\tau_{1}}}
\right\},
\end{equation}
for $\ell\in\ZZZ^{d}\backslash\{0\}$, $j\in\SSSS_{\ell,n}\cap (A_{\ell})^{c}$,
where $\al_{n}:={\rm inf}_{j}|b_{j}(u_{n})|$, $\g'_{n}=(1+2^{-n})\g'$, $\g'\leq\g_0$ and $\tau_1>0$.
We have the following result.
\begin{lemma}\label{nani10}
Given $\g'$ and $\tau_1$, there exist $\g$ and $\tau$ such that
if $\la\in B_{\ell,j}^{c}$ then $\la\in A_{\ell,j}^{\s}$ for $j\in\SSSS_{\ell,n}\cap (A_{\ell})^c$.
\end{lemma}
\begin{proof}
First of all
\begin{equation*}
j\in\SSSS_{\ell}, \;\; \Rightarrow b_{j}\geq\frac{|\oo\cdot\ell|}{2|j|}, \;\; \Rightarrow \al_{n}\geq \frac{\g_{0}}{2\langle\ell\rangle^{\tau_0}\langle j\rangle},
\end{equation*}
hence
\begin{equation*}
|\oo\cdot\ell+jb_{j}|\geq \frac{\g'_{n}\al_{n}}{\langle\ell\rangle^{\tau_1}}\geq \frac{\g'_n\g_{0}}{\langle j\rangle\langle\ell\rangle^{\tau_{1}+\tau_0}2}\geq \frac{\g_{n}}{\langle j\rangle\langle \ell\rangle^{\tau}},
\end{equation*}
if $\g'\g_{0}\geq 2\g$ and $\tau\geq \tau_{1}+\tau_0$.
\end{proof}
\noindent
By Lemma \ref{nani10} follows that
\begin{equation}\label{frodo3}
|H_{n}^{(4)}|\leq \sum_{|\ell|>N_{n}}\sum_{j\in\SSSS_{\ell,n}\cap (A_{\ell})^{c}}|B_{\ell j}^{\s}|\leq
\sum_{|\ell|>N_{n}}\frac{4|\ell|K\g'_{n}\al_{n}}{\e\gote \langle\ell\rangle^{\tau_{1}}}\frac{4d}{|\ell|}
\lessdot C\g' N_{n}^{-1}
\end{equation}
Unfortunately, for the sets $H_{n}^{(1)}$ and $H_{n}^{(3)}$ we cannot provide an estimate like \eqref{frodo3}; by the summability of the series in $\ell$ we can only conclude
\begin{equation}\label{frodo4}
|H_{n}^{(1)}|, |H^{(3)}_{n}|\leq C\g'.
\end{equation}
This implies the \eqref{frodo5} for any $n\geq0$. Moreover by Lemma \ref{luna11} we have that
if $n\geq \bar{n}(\e)$ then $H_{n}^{(1)}=H_{n}^{(3)}=\emptyset$, hence the \eqref{frodo6} follows by \eqref{stime1} and \eqref{frodo3}.
Lemma \ref{cavallo2} implies
(\ref{eq136ham}) by choosing, for instance, $\g:=(\g')^{2}\leq\g_0\leq1$.
\end{proof}
\appendix
\section{Techical Lemmata}
The following are results on the properties of algebra, tame product of the norms introduced above and some classical lemmata
on composition operators an change of variables.
\begin{lemma}\label{A} Let $s_{0}>d/2$.
For $s\geq s_{0}$, and by setting $|u|^{\infty}_{s }:=\sum_{|\al|\leq s}||D^{\al}u||_{L^{\infty}}$
the norm in $W^{s,\infty}$, one has
\begin{equation}\label{A5}
||uv||_{s}\leq C(s_{0})||u||_{s}||v||_{s_{0}}+C(s)||u||_{s_{0}}||v||_{s}, \quad \forall\; u,v\in H^{s}(\TTT^{s}).
\end{equation}
and for $s\geq0$, $s\in\NNN$ one has
\begin{equation}\label{A7}
||uv||_{s}\leq\frac{3}{2}||u||_{L^{\infty}}||v||_{s}+C(s)|u|^{\infty}_{s }||v||_{0},
\forall\; u\in W^{s,\infty}, v\in H^{s}.
\end{equation}
If $u:=u(\la)$ and $v:=v(\la)$ depend in a lipschitz way on $\la\in\Lambda\subset\RRR^{d}$,
all the previous statements hold also for the norms
$|\cdot|^{\infty}_{s }$, $||\cdot||_{s,\g}$ and $|\cdot|^{\infty}_{s,\g }$.
\end{lemma}
Now we recall classical tame estimates for composition of functions.
\begin{lemma}\label{lemA2}{\bf Composition of functions} Let $f : \TTT^{d}\times B_{1}\to\CCC$, where
$B_{1}:=\left\{y\in\RRR^{m} : |y|<1\right\}$. it induces the composition operator on $H^{s}$
\begin{equation}\label{A11}
\tilde{f}(u)(x):=f(x,u(x),Du(x),\ldots,D^{p}u(x))
\end{equation}
where $D^{k}$ denotes the partial derivatives $\del_{x}^{\al}u(x)$ of order $|\al|=k$.
Assume $f\in C^{r}(\TTT^{d}\times B_{1})$. Then
\noindent
$(i)$ For all $u\in H^{r+p}$ such that $|u|_{p,\infty}<1$, the composition operator $(\ref{A11})$
is well defined and
\begin{equation}\label{A12}
||\tilde{f}(u)||_{r}\leq C||f||_{C^{r}}(||u||_{r+p}+1),
\end{equation}
where the constant $C$ depends on $r,p,d$. If $f\in C^{r+2}$, then, for all $|u|^{\infty}_{s },
|h|^{\infty}_{p }<1/2$, one has
\begin{equation}\label{A13}
\begin{aligned}
||\tilde{f}(u+h)-\tilde{f}(u)||_{r}&\leq C||f||_{C^{r+1}}(||h||_{r+p}+|h|^{\infty}_{p }||u||_{r+p}),\\
||\tilde{f}(u+h)-\tilde{f}(u)-\tilde{f}'(u)[h]||_{r}&\leq
C||f||_{C^{r+2}}|h|^{\infty}_{p}(||h||_{r+p}+|h|^{\infty}_{p }||u||_{r+p}).
\end{aligned}
\end{equation}
\noindent
$(ii)$ the previous statement also hold replacing $||\cdot||_{r}$ with the norm $|\cdot|_{\infty}$.
\end{lemma}
\begin{proof} For the proof see \cite{Ba2} and \cite{Moser-Pisa-66}.
\end{proof}
\begin{lemma}\label{lemA3}{\bf Lipschitz estimate on parameters} Let $d\in\NNN$,
$d/2<s_{0}\leq s$, $p\geq0$, $\g>0$. Let $F : \Lambda\times H^{s}\to\CCC$, for
$\Lambda\subset\RRR^{d}$, be a $C^{1}-$map in $u$
satisfying the tame esitimates: $\forall\; ||u||_{s_{0}+p}\leq1$, $h\in H^{s+p}$,
\begin{subequations}\label{A14}
\begin{align}
||F(\la_{1},u)-F(\la_{2},u)||_{s}&\leq C(s) |\la_{1}-\la_{2}|(1+||u||_{s+p}), \quad
\la_{1},\la_{2}\in\Lambda\label{A14aa}\\
\sup_{\la\in\Lambda}||F(\la,u)||_{s}&\leq C(s)(1+||u||_{s+p}),\label{A14a}\\
\sup_{\la\in\Lambda}||\del_{u}F(\la,u)[h]||_{s}&\leq C(s)
(||h||_{s+p}+||u||_{s+p}||h||_{s_{0}+p})\label{A14b}.
\end{align}
\end{subequations}
Let $u(\la)$ be a Lipschitz family of functions with
$||u||_{s_{0}+p,\g} \leq1$. Then one has
\begin{equation}\label{A15}
||F(\cdot,u)||_{s,\g}\leq C(s)(1+||u||_{s+p,\g}).
\end{equation}
The same statement holds when the norms $||\cdot||_{s}$ are replaced by $|\cdot|^{\infty}_{s }$.
\end{lemma}
\begin{proof} See Appendix A in \cite{FP}.
\end{proof}
\begin{lemma}\label{change}{\bf (Change of variable)} Let $p : \RRR^{d}\to \RRR^{d}$
be a $2\pi-$periodic function in $W^{s,\infty}$, $s\geq1$, with $|p|^{\infty}_{1 }\leq1/2$.
Let $f(x)=x+p(x)$. Then one has
$(i)$ $f$ is invertible, its inverse is $f^{-1}(y)=g(y)=y+q(y)$ where $q$ is $2\pi-$periodic,
$q\in W^{s,\infty}(\TTT^{d};\RRR^{d})$ and $|q|^{\infty}_{s }\leq C|p|^{\infty}_{s }$. More precisely,
\begin{equation}\label{A18}
|q|_{L^{\infty}}=|p|_{L^{\infty}}, \; |Dq|_{L^{\infty}}\leq 2|Dp|_{L^{\infty}},
\; |Dq|^{\infty}_{s-1 }\leq C|Dp|^{\infty}_{s-1 },
\end{equation}
where the constant $C$ depends on $d,s$.
Moreover, assume that $p=p_{\la}$ depends in a Lipschitz way by a parameter
$\la\in\Lambda\subset\RRR^{d}$, an suppose, as above, that $|D_{x}p_{\la}|_{L^{\infty}}\leq1/2$
for all $\la$. Then $q=q_{\la}$ is also Lipschitz in $\la$, and
\begin{equation}\label{A19}
|q|^{\infty}_{s,\g}\leq C\left(|p|^{\infty}_{s,\g}+
\left[\sup_{\la\in\Lambda}|p_{\la}|^{\infty}_{s+1 }\right]
|p|_{L^{\infty},\g}
\right)\leq C|p|^{\infty}_{s+1,\g},
\end{equation}
the constant $C$ depends on $d,s$ (it is independent on $\g$).
$(ii)$ If $u\in H^{s}(\TTT^{d};\CCC)$, then $u\circ f(x)=u(x+p(x))\in H^{s}$, and, with the same $C$ as
in $(i)$ one has
\begin{subequations}\label{A20}
\begin{align}
||u\circ f-u||_{s}&\leq C(|p|_{L^{\infty}}||u||_{s+1}+|p|^{\infty}_{s }||u||_{2}),\label{A20b}\\
||u\circ f||_{s,\g}&\leq
C(|u|_{s+1,\g}+|p|^{\infty}_{s,\g}||u||_{2,\g})\label{A20c}.
\end{align}
\end{subequations}
The
(\ref{A20b}) and (\ref{A20c}) hold also for $u\circ g$ and if one replace norms $||\cdot||_{s}$, $||\cdot||_{s,\g}$ with $|\cdot|_{s}^{\infty}$, $|\cdot|_{s,\g}^{\infty}$.
\end{lemma}
\begin{lemma} \label{lem5} {\bf (Composition).} Assume that
for any $||u||_{s_{0}+\mu_{i},\g}\leq1$ the operator $\calQ_{i}(u)$ satisfies
\begin{equation}\label{A35}
||\calQ_{i}h||_{s,\g}\leq C(s)(||h||_{s+\tau_{i},\g}+||u||_{s+\mu_{i},\g}||h||_{s_{0}+\tau_{i}\g}),
\quad i=1,2.
\end{equation}
Let $\tau:=\max\{\tau_{1},\tau_{2}\}$, and $\mu:=\max\{\mu_{1},\mu_{2}\}$. Then,
for any
\begin{equation}\label{A36}
||u||_{s_{0}+\tau+\mu,\g}\leq1,
\end{equation}
one has that the composition operator $\calQ:=\calQ_{1}\circ \calQ_{2}$ satisfies
\begin{equation}\label{A37}
||\calQ h||_{s,\g}\leq C(s)(||h||_{s+\tau_{1}+\tau_{2},\g}+||u||_{s+\tau+\mu,\g}||h||_{s_{0}+\tau_{1}+\tau_{2},\g}).
\end{equation}
\end{lemma}
\begin{proof} It is sufficient to apply the estimates (\ref{A35}) to $\calQ_{1}$ first, then to $\calQ_{2}$ and
using the condition (\ref{A36}).
\end{proof}
\smallskip
\noindent
{{\bf Proof of Lemma \ref{lemmaccio}.}} We first show that $T$ is symplectic. Consider $W=(w^{(1)},w^{(2)}),V=(v^{(1)},v^{(2)})\in H^{s}(\TTT^{d+1};\RRR)\times H^{s}(\TTT^{d+1};\RRR) $ and set $w=w^{(1)}+iw^{(2)}$, $v=v^{(1)}+iv^{(2)}$,
then one has
\begin{equation}\label{lemmaccio5}
\begin{aligned}
\tilde{\Omega}(TW,TV)&:=\int_{\TTT}\left(\begin{matrix}\frac{i}{\sqrt{2}} w \\ \frac{1}{\sqrt{2}} \bar{w}\end{matrix}\right)\cdot J\left(\begin{matrix}\frac{i}{\sqrt{2}} v \\ \frac{1}{\sqrt{2}} \bar{v}\end{matrix}\right)d x=
\int_{\TTT}W JVd x=:\tilde{\Omega}(W,V).
\end{aligned}
\end{equation}
To show the (\ref{lemmaccio3}) is sufficient to apply the definition of $T_{1}$.
First of all consider the linearized operator in some $z=(z^{(1)},z^{(2)})$
\begin{equation}\label{linop}
D_{z}\calF(\oo t, x, z)=D_{\oo}+\e D_{z}g(\oo t,x, z)=D_{\oo}+\e \del_{z_{0}}g+\e\del_{z_{1}}g \del_{x}+\e\del_{z_{2}}g\del_{xx}
\end{equation}
where $D_{\oo}$ and $g$ are defined in (\ref{totale}) and (\ref{13}) and
\begin{equation}\label{linop2}
\del_{z_{i}}g:=(a^{(i)}_{jk})_{j,k=1,2}:=(\del_{z^{(j)}_{i}}g_{k})_{j,k=1,2}.
\end{equation}
All the coefficients $a^{(i)}_{jk}$ are evaluated in $(z^{(1)},z^{(2)},z^{(1)}_{x},z^{(2)}_{x},z^{(1)}_{xx},z^{(2)}_{xx})$.
By using the definitions (\ref{linop}), (\ref{linop2}) and recalling that $g=(g_{1},g_{2})=(-f_{1},f_{2})$ and
$\ff=f_{1}+if_{2}$, one can check with an explicit computation that
$$
\calL(z)=T_{1}^{-1}T d_{z}\calF(\oo t,x,z)T^{-1}T_{1}
$$
has the desired form.
\EP
\thanks{This research was supported by the European Research Council under
FP7 ``Hamiltonian PDEs and small divisor problems: a dynamical systems approach'' grant n. 306414-HamPDEs} | 199,561 |
TITLE: Choosing $5$ elements from first $14$ natural numbers so that at least two of the five numbers are consecutive
QUESTION [1 upvotes]: Let $n$ be the number of five element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find $n$.
My work I have made a block of two consecutive numbers (like $(1,2), (2,3), (13,14)$ etc.). Now we can choose this block in $13$ ways. Now we have to choose $3$ numbers from the rest $12$ numbers. We can do it in $12 \choose 3$ ways. So, by multiplication principle we come to know that there are $13 \times {12\choose 3}$ ways .
Am I right? Please tell where I had made the mistake?
All the $5$ elements are distinct. I didn't ask to arrange the group. I ask the number of sets only.
REPLY [0 votes]: Finding the number of five element sets with the property that there are no consecutive numbers in it comes to finding the number of sums $$n_1+n_2+n_3+n_4+n_5+n_6=9$$ where $n_1$ and $n_6$ are nonnegative integers and $n_2,n_3,n_4,n_5$ are positive integers.
If we are working in set $\{1,\dots,14\}$ then e.g. solution $(0,2,3,1,2,1)$ represents the subset $\{1,4,8,10,13\}$.
This comes to the same as finding the number of sums $$m_1+m_2+m_3+m_4+m_5+m_6=5$$ where $m_1,m_2,m_3,m_4,m_5,m_6$ are nonnegative integers.
Here solution $(0,1,2,0,1,1)$ represents the subset $\{1,4,8,10,13\}$.
With stars and bars we find that there are: $$\binom{10}{5}$$ possibilities.
In total there are $\binom{14}5$ five element subsets of $\{1,\dots,14\}$ so:$$\binom{14}5-\binom{10}5$$of them will have a least one pair of consecutive numbers. | 141,762 |
\begin{document}
\begin{abstract}
Given a function $f: (a,b) \rightarrow \mathbb{R},$
L\"owner's theorem states $f$ is monotone when extended to self-adjoint matrices via the functional calculus, if and only if $f$ extends to a self-map of the complex upper half plane. In recent years, several generalizations of L\"owner's theorem have been proven in several variables.
We use the relaxed
Agler, \McCarthyc and Young theorem on locally matrix monotone functions in several commuting variables to generalize results in the noncommutative case. Specifically, we show that a real free function defined over an operator system must analytically continue to a noncommutative upper half plane as map into another noncommutative upper half plane.
\end{abstract}
\maketitle
\section{Introduction}
Let $f: (a,b) \rightarrow \mathbb{R}.$
L\"owner answered the question of when such a function
is monotone when $f$ is extended to self-adjoint matrices (or even operators in general) via the functional calculus, which has found various applications.
Specifically, we say $f$ is \dfn{matrix monotone} on $(a,b) \subset \R$ if
$$A \leq B \Rightarrow f(A) \leq f(B)$$
whenever $A$ and $B$ are self-adjoint matrices of the same size with spectrum in $(a, b),$ $f$ is being applied in the sense of the functional calculus on self-adjoint operators, and $\leq$ is interpreted to mean that the difference is positive semidefinite.
Let $\Pi$ be the upper half plane in $\mathbb{C}.$ L\"owner's theorem states the following:
\begin{theorem}[L\"owner \cite{lo34}]\label{LownerOne}
Let $f: (a,b) \to \R$ be a bounded Borel function. The function $f$ is matrix monotone if and only if $f$ is real analytic and analytically continues to the upper half plane
as a function from $\Pi \cup (a,b)$ into $\overline{\Pi}$.
\end{theorem}
For a modern treatment of L\"owner's theorem, see e.g. \cite{don74,bha97,bha07}. For applications, see e. g. \cite{Boche2004e, Jorswieck2007f, wigner, wigner2} .
Now, one could ask what kind of functions preserve inequalities of, say,
two block matrices of size $2$ by $2.$
That is, given a function $f(X_{11},X_{12},X_{21},X_{22})$ and an inequality
$$\bpm
A_{11} & A_{12} \\
A_{21} & A_{22}
\epm \leq \bpm
B_{11} & B_{12} \\
B_{21} & B_{22}
\epm,$$
when can we say
$$f(A_{11},A_{12},A_{21},A_{22}) \leq
f(B_{11},B_{12},B_{21},B_{22})?$$
For example, is the function given by the formula
$X_{11} - X_{12}X_{22}^{-1}X_{21},$ the Schur complement, monotone in the above sense on positive block $2$ by $2$ matrices? It turns out the Schur complement is indeed monotone, which has certainly been known for some time\cite{liu99}, and can be shown via elementary arguments-- for example, its inverse appears in the formula for block inversion of a matrix as a diagonal entry. However, we are interested in an effective systematic way of classifying such functions as in L\"owner's theorem, and that is what we will establish in the noncommutative context.
\section{The noncommutative context}
We now describe the noncommutative context in which we desire to prove a generalization of L\"owner's theorem. First, we must give the appropriate generalization of the functional calculus (see \cite{vvw12} for a more thorough introduction). We also note various noncommutative generalizations to the free functional calculus of L\"owner's theorem were considered by the current author and Tully-Doyle \cite{pastd14}, and by Palfia \cite{palfia} previously, and to other functional calculi by Hansen \cite{han03} and Agler, \McCarthy, and Young \cite{amyloew}. Moreover, this work fits into a greater effort to systematize the theory of matrix inequalities \cite{heltonPositive, heltmc12, helmconvex04, helmc04, HKN}.
Let $R$ be a real topological vector space.
We define the \dfn{matrix universe over $R$,} denoted $\mathcal{M}(R)$, to be
$$\mathcal{M}(R) = \bigcup_{n\in\mathbb{N}} M_n(\mathbb{C})\otimes_{\mathbb{R}} R,$$
where $M_n(\mathbb{C})$ denotes the space of $n$ by $n$ matrices. We endow $\mathcal{M}(R)$ with the disjoint union topology.
Given $\mathcal{U} \subset \mathcal{M}(R),$
we use $\mathcal{U}_n$ to denote
$\mathcal{U} \cap M_n(\mathbb{C})\otimes R$.
We define the \dfn{Hermitian matrix universe over $R$,} denoted $\mathcal{S}(R)$, to be
$$\mathcal{S}(R) = \bigcup_{n\in\mathbb{N}} S_n(\mathbb{C})\otimes_{\mathbb{R}} R,$$
where $S_n(\mathbb{C})$ denotes the space of $n$ by $n$ Hermitian matrices.
For a concrete example, if we take $R = S_2(\mathbb{C}),$ the $2$ by $2$ Hermitian matrices over $\mathbb{C},$ $\mathcal{M}(S_2(\mathbb{C}))$ consists of all block $2$ by $2$ matrices and $\mathcal{S}(S_2(\mathbb{C}))$ consists of all block $2$ by $2$ Hermitian matrices. For another example, taking $R = \mathbb{R}^2,$ the set $\mathcal{M}(\mathbb{R}^2)$ consists of all pairs of same-sized matrices and $\mathcal{S}(\mathbb{R}^2)$ consists of all pairs of same-sized Hermitian matrices.
We define a {\bf domain} $D \subset \mathcal{M}(R)$
to satisfy the following two axioms:
\begin{enumerate}
\item $X \oplus Y \in D \Leftrightarrow X, Y \in D$
\item $X \in D_n \Rightarrow U^*XU \in D$ for all $n$ by $n$ unitaries
$U$ over $\mathbb{C}.$
\end{enumerate}
Let $D \subset \mathcal{M}(R_1)$ be a free domain.
We say a function $f:D \rightarrow \mathcal{M}(R_2)$ is a {\bf free function} if
\begin{enumerate}
\item $f|_{D_n}$ maps into $\mathcal{M}(R_2)_n$
\item $f(X\oplus Y) = f(X) \oplus f(Y),$
\item $S^{-1}f(X)S = f(S^{-1}XS)$ for all $n$ by $n$ invertible matrices $S$
over $\mathbb{C}$ such that $X, S^{-1} XS \in D_n.$
\end{enumerate}
We note any noncommutative rational expression gives a free function on its domain of definition. For example, the Schur complement, $X_{11} - X_{12}X_{22}^{-1}X_{21},$ gives a free function on the subset $D \subseteq \mathcal{M}(S_2(\mathbb{C}))$ where $X_{22}^{-1}$ is defined. For another example, the matrix geometric mean,
$X_1^{1/2}(X_1^{-1/2}X_2X_1^{-1/2})^{1/2} X_1^{1/2}$
defines a free function on the subset $D \subseteq \mathcal{S}(\mathbb{R}^2)$ of all pairs of positive definite matrices.
If $R$ is a real operator system, that is, $R$ is a real subspace containing $1$ in a $C^*$ algebra of self-adjoint elements, for each $n$ there is a natural ordering on $S_n(\mathbb{C})\otimes R,$ since matrices over $R$ are themselves elements of a larger $C^*$-algebra. That is, given $A, B \in S_n(\mathbb{C})\otimes R,$ we say $A \leq B$ if $B - A$ is positive semidefinite as an element of $S_n(\mathbb{C})\otimes R.$
Accordingly, given $R_1$ and $R_2$ real operator systems and a domain $D \subseteq \mathcal{S}(R_1)$, we say a free function $f: D \rightarrow \mathcal{S}(R_2)$ is \dfn{matrix monotone} if
$A \leq B \Rightarrow f(A) \leq f(B)$
whenever $A$ and $B$ have the same size.
Define
$\Pi(R) = \{\text{Im } X > 0\}$
where $A > B$ if the difference is strictly positive definite, that is, it is self-adjoint and its spectrum is a subset of $(0,\infty)$, and $\text{Im } X = (X - X^*)/2i$.
We show the following theorem.
\begin{theorem}[Noncommutative L\"owner theorem over operator systems]
Let $R_1$ and $R_2$ be closed real operator systems.
Let $D \subseteq \mathcal{S}(R_1)$ be a free domain.
Suppose each $D_n$ is convex and open as a subset of $S_n(\mathbb{C})\otimes R$.
A function $f: D \rightarrow \mathcal{S}(R_2)$ is matrix monotone
if and only if
$f$ extends to a continuous free function $F: \Pi(R_1) \cup D \rightarrow \overline{\Pi(R_2)}.$
\end{theorem}
We note that such a function must be analytic on each $\Pi(R_1)_n$ due to the draconian nature of free functions. See \cite{vvw12}. We also point out that the case where $R_1 = \mathbb{R}^d$ as a diagonal algebra and $R_2 = \mathbb{R}$ was explored in \cite{pastd14, palfia}, and that the current work simplifies the proof of the main result of those works if we are willing to use the commutative L\"owner theorem from \cite{amyloew} as a black box. Moreover, if we are given a rational expression, such as the Schur complement, on a nice finite dimensional operator system, such as a matrix algebra, one can apply the algorithms in \cite{HKN} which make the rational convex Positivstellensatz
\cite{pascoePosSS} effective to check that a function is matrix monotone in our sense.
Finally, we should comment that the setting of operator systems is equivalent to defining an Archimedian matrix ordering
on $\mathcal{S}(R)$, where $R$ is an abstract real vector space, by the Choi-Effros Theorem \cite{effrosChoi}.
That is, we might have alternatively defined an ordering on $\mathcal{S}(R)$ using any
proper closed Archimedian matrix convex cone, but the result is the same.
Before we arrive at the proof of our Theorem, we should revisit our Schur complement. Our domain $D \subset \mathcal{S}(S_2(\mathbb{C}))$ is the set of positive definite block $2$ by $2$ matrices upon which our function, defined by the formula
$$f\bpm
X_{11} & X_{12} \\
X_{21} & X_{22}
\epm = X_{11} - X_{12}X_{22}^{-1}X_{21},$$
is a free function $f: D \rightarrow \mathcal{S}(\mathbb{R}).$
According to our Theorem, $f$ will be matrix monotone if and only if $f$ extends to a continuous free function from
$D \cup \Pi(S_2(\mathbb{C}))$ to $\overline{\Pi(\mathbb{R})}.$
It is clear that extension of $f$ to the new domain must still be given by the same formula as before.
Either using the algorithms in \cite{HKN, pascoePosSS} or by brute force, one can see that
$$\textrm{Im } f =
\bpm 1 \\ (X_{22}^*)^{-1} X_{12}^* \epm^*
\left[ \textrm{Im } \bpm
X_{11} & X_{12} \\
X_{21} & X_{22}
\epm\right] \bpm 1 \\ (X_{22}^*)^{-1}X_{12}^* \epm$$
which is manifestly positive definite whenever $\textrm{Im } \bpm
X_{11} & X_{12} \\
X_{21} & X_{22}
\epm$ is positive definite-- that is $f$ maps $\Pi(S_2(\mathbb{C}))$ to $\Pi(\mathbb{R}).$ That is, our Theorem now implies that the Schur complement is matrix monotone.
Another example of a matrix monotone function, is the matrix geometric mean and various generalizations, see \cite{Lawson2011, Bhatia2012}. In the two parameter case it is not immediately clear to the author how to show the
function
$X_1^{1/2}(X_1^{-1/2}X_2X_1^{-1/2})^{1/2} X_1^{1/2}$
continues to a map from $\Pi(\mathbb{R}^2)$ to
$\overline{\Pi(\mathbb{R})}$ without going through the generalization of L\"owner's theorem.
\section{The proof of the main result}
$(\Rightarrow)$
The proof will go by viewing, for each $n,$ $f|_{D_n}$ as a matrix monotone function in several commuting variables in the sense of Agler, \McCarthy, and Young.
Agler, \McCarthy, and Young extended L\"owner's theorem to several commuting
variables for the class of locally matrix monotone functions \cite{amyloew}.
Subsequently, it was generalized to remove some technical assumptions by the author in \cite{pascoemollifier}.
Let $E$ be an open subset of $\mathbb{R}^d.$
Let $CSAM^d_n(E)$ denote the $d$-tuples of commuting self-adjoint matrices of size $n$ with joint spectrum contained in $E$.
We say that a function $f: E \rightarrow \mathbb{R}$ is \dfn{locally matrix monotone} if for any $C^1$
path $\gamma: (-1,1) \rightarrow CSAM^d_n(E)$ such that $\gamma'(0)_i > 0$, there exists
an $\epsilon >0$ such that for all $-\epsilon < t_1 < t_2 < \epsilon,$
$g(\gamma(t_1)) \leq g(\gamma(t_2)).$
We recall the following theorem.
\begin{theorem}[Agler, \McCarthy, and Young \cite{amyloew}, Pascoe \cite{pascoemollifier}] \label{amyold}
Let $E$ be an open subset of $\mathbb{R}^d.$
Let $g: E \rightarrow \mathbb{R}$ be a locally matrix monotone function. Then $g$ is analytic, and $g$ extends to a (unique) continuous function on $\Pi^d \cup E$ which maps into $\overline{\Pi}$ which is analytic on $\Pi^d.$
\end{theorem}
We note that the original formulation of Agler, \McCarthy, and Young applied only to $C^1$ functions $g$, and via an argument using mollifiers it was shown that the theorem holds for arbitrary functions.
We note that is sufficient to show that on each nonempty $D_n$, our function $f$ analytically continues to $\Pi(R_1)_n$ taking values in $\overline{\Pi(R_2)_n}$. It is an elementary, but perhaps somewhat involved, exercise to show that the induced extension of $f$ will be a free function on $\Pi(R_1).$ Namely, the edge-of-the-wedge theorem will ensure that the extension of $f$ actually analytically continues through each $D_n$ as a function on $\Pi(R_1)_n \cup D_n$ and the rest of the properties will follow by analytic continuation.
Now, we note that it is sufficient to show that for every (completely) positive unital linear functional $l: M(R_2)_n \rightarrow \mathbb{C}$ that $f_{n,l} = l \circ f|_{D_n}$
extends analytically to $D_n \cup \Pi(R_1)_n$ as a map taking values with positive imaginary part.
This is obvious when $R_2$ is finite dimensional, and an exercise in functional analysis otherwise.
Fix $P \in D_n.$
Let $K_1, \ldots K_m > 0$ be positive elements of $\mathcal{S}(R_1)_n.$
Let $C$ be the cone generated by $K_1, \ldots, K_m$
and let $S$ be the span of $K_1, \ldots, K_m$.
We will show that $f_{l,n}$ uniquely analytically continues to
$P + S + iC.$ Taking larger and larger sets of $K_i$ will give an analytic continuation $f_{n,l}$ to the whole of $\Pi(R_1)_n.$ That is, the sets $P +S+ iC$ exhaust $\Pi(R_1)_n.$
Define the function $g(h) = f_l(P + \sum_i h_i K_i))$.
Now $g(h)$ is a locally matrix monotone function in the sense of Theorem \ref{amyold} as a function on $\mathbb{R}^m$ which induces the unique analytic continuation of $f_{n,l}$ to
the desired space taking values in $\overline{\Pi}.$ So, we are done.
$(\Leftarrow)$ The converse direction is easy and follows from a computation of the derivative for directions pointing into the upper half plane. See \cite[Lemma 4.8]{pastd14} where the details are essentially the same.
\bibliography{references}
\bibliographystyle{plain}
\end{document} | 32,242 |
It’s August and you know what that means – German Football!
All the talk seems to center on who is destined to capture the trophy of Bundesliga come Spring. But the favorites tend to be favorites for a reason. I’d rather look at and breakdown the “sleeper” teams who can impact the league and create drama just like us fans want to have. So without further ado, let’s being…
Leverkusen
It has been a mixed transfer window for the side; they have lost key players like Gonzalo Castro and Son Heung-Min, but have brought in quality replacement in Charles Aranguiz and Javier Hernandez. The loss of Castro to Dortmund was a tough blow for the side, since the midfielder was their main tempo controller and enabled the attackers to play at the sheer pace that Leverkusen have become famous for.
They had a successful season last time out, securing fourth place; and were a joy to watch. Midfielder Hakan Calhanoglu and forwards Karim Bellarabi and Stefan Kiessling still remain and they have decent quality at the back; if the new signings bed in well then Bayer should have a strong season ahead.
Borussia Monchengladbach
Gladbach had a great season last time out, claiming third spot on the league table. Over the summer they lost one of their stars however; as striker Max Kruse departed for Wolfsburg. The loan spell of Christoph Kramer also ended as he returned to parent club Leverkusen. They have purchased attackers Josip Drmic and Thorgan Hazard to compensate, but the team chemistry that Kruse had is tough, if not impossible, to replace.
They are a side that relies on patient passing and build up play, which is why Kruse was so essential; he participated in the build-up and ensured that the opportunities the team created were taken. Unless they manage to fit their new players into the system right away, it is tough to see them achieving the successes of last season, and they could end up being a possession controlling side with no bite.
These teams will be the likely sleepers in this season’s Bundesliga, but below them are several others that could pose a credible threat as per the odds reflected at online bookmaker bet365. Frankfurt are a team who are always worth watching; their games tend to have massive amounts of goals both for and against and makes for good viewing. Similarly, Hoffenheim provide entertaining games and play an ‘all-in’ sort of strategy that is great for viewers. Augsburg, of course, is worthy of keeping an eye on although it seems extremely unlikely that they will manage to repeat the previous season’s achievements.
Overall, it seems like the Bundesliga is set for another great season. With the foreign money pouring in as clubs from other leagues buy German talent, the German clubs are focusing more on the youth academies, an approach markedly different from the other major leagues. It certainly makes for exciting viewing and also results in a great number of talents being unearthed; the German national team is testament enough to that. | 131,588 |
practical engagement
for leaders
In this 4-hour session, participants will discuss:
The definitions of engagement and consultation – are they the same or different?
Opportunities of good quality engagement for both the organisation and the community
Identifying stakeholders and understanding their different roles and influence in the engagement
Ensuring that engagement practice is considered during decision-making processes
Frameworks, policies and tools kits
Understand the resources needed for successful engagement
The relationship between internal and external engagement practice
Different engagement methods – why so many?
Reporting engagement practice
Something you can do tomorrow to improve your engagement practice.
NOTE: If working in local, state or federal government organisations, we include details of legislative requirements for consultation (and/or engagement).
This session will be delivered at a level, recognising that participants are unlikely to be the hands-on operatives of engagement practice, rather the leaders and decision-makers, who also determine both whether to engagement and what resources will be made available for the engagement.
Who should attend? This session has been designed for Executives, Senior Managers and Decision Makers (including Elected Members) of organisations that engage.
Pre-requisites: none
Recommended Experience: Engagement – none, Leadership – minimum of 12 months
Delivery: This session can be delivered at your own location or in a public training facility.
We recognise that Executives, Senior Managers and Decision Makers (including Elected Members) makers are busy people. We designed the practical engagement for leaders module to facilitate leaders, whilst exercising their leadership and decision-making duties, to recognise the practice of engagement in their roles.
| 42,336 |
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Highlights from Whitman College’s 131st Commencement included a keynote speech by the social activist Wanjiru Kamau-Rutenberg ’01, the awarding of an honorary doctorate to federal judge James L. Robert ’69 and inspirational words from President Kathleen Murray to 374 seniors to “model empathy, grace and respect.” Read more about the ceremony.
Sure, college seniors often toss their mortarboards to end commencement. But how many decorate them beforehand? Check out this photo gallery of Whittie mortarboard artistry.
*See other file. Photo gallery could run there instead.
Parents of graduating seniors abounded with pride in their children and in Whitman. Attendees celebrated what the graduates learned, how they grew, what they majored in, how they excelled, plus praised Whitman professors, clubs and services, among many other things. Read more about parental reflections.
A Whitman education lasts a lifetime. Several lifetimes, in many cases. Alumni from a few years ago to many decades ago returned to Walla Walla to watch the next generation graduate. In fact, twin sisters
Maggie Gose '17, a biology major from Bellevue, Washington, and Chloe Gose '17, a sociology major, not only received hugs from grandpa Jerry Gose ’50, who attended, but also are the great-great-granddaughters of Christopher Columbus Gose, a member of Whitman’s first graduating class in 1886. Read about Whitman legacy families at Commencement.
Which students held leadership roles at Commencement? Read biographies of Arie Knops ’17, class speaker; Blake Ladenburg ’17, senior fund chair; Emily Volpert ’17, class banner designer; and Gabriel Merrill-Steskal ’18 and Grace Pyles ’18, marshals. | 206,636 |
A decade ago–long before the current controversies over what big companies are doing with our data–a lot of people were already irate about ad networks that followed their activity across sites in order ever more precisely to target marketing messages. A feature called Do Not Track arose as a simple, comprehensible way for browser users to take back their privacy. To opt out of being tracked, you’d check a box in your browser’s settings.
Notably, this didn’t opt out of advertising–just the technology used to target ads. With Do Not Track checked, no web server or embedded code would associate your behavior at a given site with actions elsewhere on the web. It was a great idea.
And now it’s dead.
Oh, for all practical purposes, DNT died years ago. But Apple’s removal of the Do Not Track preference from Safari for Macs and iOS in an update in early February officially signaled the end of what might have been a workable understanding between consumers and the advertisers that rely on ad-tech networks to target them.
Apple’s move follows the dissolution of a World Wide Web Consortium (W3C) project, the Tracking Protection Working Group, which shut down after eight years on January 17. The release notes for Safari 12.1called Do Not Track an “expired” standard, which is sadly accurate.
In October 2018, on its public mailing list, the W3C group discussed how to describe Do Not Track’s failure in a preface to its final piece of work. After some back and forth, the group agreed on the language that appears:
…there has not been sufficient deployment of these extensions (as defined) to justify further advancement, nor have there been indications of planned support among user agents, third parties, and the ecosystem at large.
It’s an artful self-own by the group’s participants, which included representatives from ad industry trade groups, large advertisers, and ad delivery platforms, as well as ones from privacy groups, governments, and browser makers. After a flurry of work from 2011 to 2013, the group hadn’t met face to face since 2013, according to its notes.
The working group’s existence used to imply that the ad industry was actively moving towards a consensus on self-regulation when it came to online privacy. But DNT turned out to be a useful fig leaf, not a solution. “The best way to sabotage a process is by wholeheartedly participating in it,” says Alan Toner, a privacy and data protection special adviser at the Electronic Frontier Foundation (EFF), who represents his organization at the W3C.
In the ultimate irony, Apple told me via a spokesperson that it removed Do Not Track after the W3C group shuttered because, if enabled, it could help ad networks “fingerprint” a browser, a technique used by tracking systems to defeat ad blockers by identifying unique characteristics in a user’s browser configuration.
It could have all been so different.
ONE-CLICK PRIVACY
Do Not Track bubbled up from the seeming success of the Federal Trade Commission’s National Do Not Call registry, which went into effect in 2003. It allowed consumers to register their phone numbers as being unwelcoming to commercial solicitations. Companies making calls to people other than customers have to purge these numbers from calling databases. (Do Not Call was ultimately a failure, because it only prevented scrupulous parties from calling, not those who blithely ignored the law or were engaged in outright scams.)
Initially, Do Not Track was going to be a similar kind of central registry. But in 2009, privacy advocates Chris Soghoian and Sid Stamm implemented the idea as a simple Firefox plug-in. The plug-in would add a Do Not Track header to the metadata a browser sends to a server on initiating a connection. If a user had enabled Do Not Track, the value of the header would be “1”; otherwise, “0.” It was that simple. It didn’t matter from a technical perspective that no server knew how to interpret that header at the time and therefore ignored it; the policy details could be worked out later.
This straightforward idea caught fire, and within a couple of years, all the major browser had added an option to express a preference. Stamm, now an associate professor at the Rose-Hulman Institute of Technology, says the header was “a way to shout, ‘Hey, I don’t like this!'” He developed the plug-in with Soghoian because “people were really unaware how much data was collected about them.”
Stamm and Soghoian, who is now a privacy and cybersecurity adviser to U.S. Senator Ron Wyden (D-OR), were part of a group of privacy advocates and security engineers who advocated for DNT. By 2011, the FTC appeared poised to recommend that DNT evolve from a nascent browser feature into a regulatory requirement. The W3C opened a working group to study how to turn DNT into a fully recognized standard that would define how it could be implemented.
Arvind Narayanan, now an associate professor at Princeton and part of that early DNT-formulating group, said via an email statement that the prospect of federal legislation brought ad players to the table. But when that legislation didn’t materialize, “the prolonged negotiations in fact proved useful to the industry to create the illusion of a voluntary self-regulatory process, seemingly preempting the need for regulation.”
The moment passed. Those involved in the ad industry, whether social networks or ad-tech firms, had little interest in pursuing DNT if they could avoid it. Publishers didn’t demand the technology as a way to protect visitors to their sites; advertisers didn’t act as though it affected them directly.
One of the wrenches in the works was the issue of whether Do Not Track was really a binary deviation. As a two-position switch, it was either off or on. But if a user hadn’t considered the matter–or didn’t even know DNT existed–a third state existed: not yet decided. If a user hadn’t chosen to turn on DNT, browsers either left it turned off–or didn’t send DNT info one way or the other to websites.
Microsoft broke the model. In 2012, the company opted to preset Internet Explorer’s Do Not Track to the “on” state without requiring a user to pick or confirm that choice. Though the move defaulted to the most privacy-friendly option, it also put a crimp in Google’s ad hegemony, which Microsoft would not have seen as a bad thing.
Companies that were part of the ad economy already had reason to be wary of DNT; a DNT that stopped users from being tracked without them explicitly opting in looked like an existential threat. “Do Not Track started with one leg cut off the moment Microsoft used it as a marketing tool, by turning it on by default,” says Dan Jaye, a veteran online-ad veteran, most recently the founder of aqfer.
Sam Tingleff, the chief technology officer of the Interactive Advertising Bureau’s Tech Lab, provides another reason why the ad-business players took issue with DNT: From their perspective, it was too simple. A user could only turn it on or off for the browser as a whole, without per-site whitelist or blacklist options, something that the W3C group was working to elaborate on.
The lack of legislative or regulatory action, Microsoft’s DNT misstep (which the company reversed–too late–in 2015), and the W3C’s stalled movement forward left the DNT checkbox in place but without any power. Narayanan says that it was clear to him Do Not Track had failed by 2013. The corpse only stopped kicking recently. Do Not Track died before consumers had a chance to gain a taste for being tracked.
In the absence of consumers’ ability to express a preference and without U.S. regulation governing tracking, what did the ad industry expect would happen? It’s not clear.
Even as DNT made its way into browsers, tracking and advertising bloat became ridiculous. Some mainstream content sites such as those affiliated with newspapers might have 70 to 100 individual pieces of remotely loaded JavaScript or tracking images, inflating a relatively simple page of text into multiple megabytes while also bringing browsers to a crawl as they executed all that code. | 115,231 |
TITLE: Discrete sets and linear maps
QUESTION [1 upvotes]: Let $A\subset \mathbb{R}^n$ be a discrete set and $f\colon \mathbb{R}^n \longrightarrow \mathbb{R}^k$, with $k<n$, a surjective linear map. Can you ensure that $f(A)$ is discrete?
I think this is true but I haven't been able to prove it. This has been my attemp so far:
Let $y=f(x)$ for some $x\in A$. Since $A$ is discrete, there is some open neighbourhood $U$ of $x$ such that $U\cap A=\{x\}$. Now, since $f$ is an open map, we have that $f(U)$ is an open neighbourhood of y. But now I am not able to prove that $f(U)\cap f(A)=\{y\}$.
REPLY [2 votes]: Let $\{q_k\}_{k=1}^\infty$ be the set of rational numbers indexed by $\mathbb{N}$—i.e., each rational number $q_k$ has been given a natural-numbered "label", $k$, determined by an arbitrary bijection $\mathbb{N} \to \mathbb{Q}$. Working in $\mathbb{R}^2$, consider $A = \{(q_k, k) \ | \ k \in \mathbb{N} \}$, which is discrete as any two distinct points are separated from each other vertically by at least $1$ unit of distance. Now let $f: \mathbb{R}^2 \to \mathbb{R}$ be the projection map $(x,y) \mapsto x$. You can check that $f$ is both linear and surjective, and notice that $f(A) = \mathbb{Q}$, which is no longer discrete: for instance, any open ball centered at $0 \in \mathbb{Q}$ contains (infinitely many) rational numbers $1/n$ for all $n$ sufficiently large. | 190,917 |
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What is going on with Transcend?
I am about to leave also and I never thought I'd see myself type that!
Over the past few months they have had post after post on their own company board regarding the bad sound quality and all the problems the VR software has been having. They rolled out this platform WAY before it was ready and expect the MLSs to just deal with the problems while they "work" on them. These same problems have been going on for months. So yes, while we agreed to terms on PTO, we only earn so much PTO per pay period, which is not even close to the amount of time lost due to the technical problems. It is one thing to accept employment and realize that the company doesn't pay for downtime, but any reasonable person would assume that the downtime would be short-lived. Not months and months as this has been.
There is one account that the voice is uploaded from another system and it's only the voice files dictated on 2 specific ports there, not all of their dictation. The company is working to get the problem resolved with the client presently. I'm holding out because overall I don't think the company is bad.
Transcendite MT
essential for transcribing. If the big bosses had to listen to the scratches and static for hours on end, day after day, maybe this wouldn't happen, but most of them don't even know what a foot pedal and headset are. I find that the sound quality from the internet files are far worse than anything I ever encountered on DVI, C-phone, or Lanier systems telephone systems.
Only good thing I have heard lately is that a friend of mine, also an MT recently filed for disability based on hearing loss and/or inability to hear the files that she was required to do and won the case for total disability. She does have some hearing loss, but when she took her computer to the hearing and dialed in and had the adjudicator listen, there was no question that she would never be able to continue in this line of work. Maybe a few more workman comp cases would wake them up. | 216,220 |
The officials of the two countries signed the agreements in the fields of anti-terrorism, defence, security, trade and science and technology.
Prime Minister Nawaz Sharif and Sri Lankan President Maithripala Sirisena were present at the signing ceremony.
Speaking on the occasion, Nawaz Sharif said there are vast opportunities for bilateral trade adding that the quantum of trade will be taken to one billion dollars per annum. He said Pakistan is ready to give Sri Lanka the most favoured nation (MFN) status in trade.
He said Pakistan is cooperating with Sri Lanka in the defence sector and the two countries will work together to choke financial assistance to terrorists.
President Sirisena addressing on the occasion said his country enjoys close relationship with Pakistan in all sectors. He said Sri Lankan people acknowledge the role of Nawaz Sharif in promotion of bilateral relations. He said the leadership of Prime Minister Nawaz Sharif has created political stability in Pakistan.
Earlier, Sri Lankan President Maithripala Sirisena warmly received Pakistani.
| 297,042 |
Film notes on Dirty Wars
Dirty Wars is a 2013 American documentary film based on the book Dirty Wars: The World Is a Battlefield by Jeremy Scahill. The film is directed by Richard Rowley based on a screenplay written by Scahill and David Riker.
Release
Dirty Wars premiered at the 2013 Sundance Film Festivalon January 18, 2013. The film competed in the U.S. documentary section,[4] and it won the Cinematography award.[5]
The film was released in four theaters in New York City, Los Angeles, and Washington, DC on June 7, 2013. Over the opening weekend, it grossed an estimated $66,000, a theater average of $16,500.[4]
Reception
Critical reception
Dirty Wars received critical acclaim. Review aggregation website Rotten Tomatoes certified the film as “fresh” with a score of 84% based on 61 reviews.[6] Metacritic rated the film 76 based on 18 reviews, indicating “generally favorable reviews”.[7] Trevor Johnston found the film to be.“[8]However, Douglas Valentine found “…the film is so devoid of historical context, and so contrived, as to render it a work of art, rather than political commentary. And as art, it is pure self-indulgence.“[9][10]
Accolades
Dirty Wars was nominated for a 2013 Academy Award for Best Documentary Feature.[11] The film won the 2013 Golden Reel Award for Best Sound Editing: Sound Effects, Foley, Dialogue, ADR and Music in a Feature Documentary, given by the Motion Picture Sound Editors society.[12]
Further reading
An excerpt from the review at Free Thought Blog:
.”
An excerpt from the review at CineVue:
“It’s explosive subject matter that points fingers all the way up to POTUS himself in attempting to expose the secret actions of military branch JSOC (the Joint Special Operations Command), which Scahill characterises as President Obama’s own private paramilitary unit. Extending far beyond the geographical theatre of conflict, Dirty Wars follows the reporter from the employment of militia in Somalia, to missile strikes in Yemen. It’s this attack which strikes the most fear into Scahill’s heart; he already knows the astronomical scale of the hit list, but he soon discovers that American citizens are on it. “The means of defence against foreign danger have been always the instruments of tyranny at home,” said James Madison over 200 years ago – it’s chilling how keenly they echo through history.”
One thought on “Rick Rowley Thursday – Watch: ‘Dirty Wars’ (2013)” | 559 |
Sjögren Syndrome (SS)
(Sjögren's Syndrome)).—see GI mucosal or submucosal atrophy and diffuse infiltration by plasma cells and lymphocytes may cause symptoms (eg, dysphagia).
Symptoms and Signs
Glandular manifestations
SS.
SS severe disease (eg, severe vasculitis or visceral involvement), treat with corticosteroids, cyclophosphamide, or rituximab.
Resources In This Article
Drugs Mentioned In This Article
- Drug NameSelect Trade
- cevimelineEVOXAC
- PilocarpineISOPTO CARPINE, PILOPINE HS, SALAGEN
- HydroxychloroquinePLAQUENIL
- cyclosporineNEORAL, SANDIMMUNE
- prednisoneRAYOS
- rituximabRITUXAN
- cyclophosphamideCYTOXAN (LYOPHIL | 68,675 |
Recent bookmarks
Nature's Changing Course by Durrant
Fandoms: Harry Potter - J. K. Rowling
12 Dec 2013
Summary
Snape wakes up expecting to celebrate his seventeenth birthday. Instead he finds out he's missed the last twenty years and the wizarding world is celebrating the death of the Dark Lord.
Cast adrift in this strange world, where he is seen as both hero and villain for things he has no recollection of, he struggles to come to terms with suddenly being famous and the unlikely attentions of one Harry Potter.
We Are A Hurricane by MoMoMomma
Fandoms: Teen Wolf (TV)
15 Dec 2013
Summary
Peter isn't sure exactly what part of "Being A Responsible Uncle" includes acquiescing to your nephew's pleas and fucking him through a heat, but, then again, Peter hasn't been what you'd call a 'responsible uncle' for a very long time. Apparently, longer than even he originally thought. | 146,358 |
“Moon Monster”
OT3M x 30 (10 rounds)
10 Burpees
10 Toes to Bar
3 Snatches or Cleans (climbing)
Virtual Access:
WOD Guidance
Build on the Bar each round. Modify reps to complete burpees and toes to bar in less than 90 seconds. Focus on your barbell today!
Question of the Day
If you could see one movie for the first time again what would it be?!
Moon Monster
OT3M x 30:
10 burpees
10 K2C
3 hang power snatch 35-45#
Good one!
OT2M x 10
8 burpees
8 single leg v-ups
8 db snatches (35# db)
last round 10:10:10
Finished each round around 0:53-0:56
Moon Monster @ 8:30 am with Coach James- thank you!
6 burpees to dumbbells
10 knee ups
power cleans up to 65#
Moon Monster
OT3M X 30
10 burpees
10 Toes2Bar 5/5
3MuscleSnatch 65#/75# for2sets/65#
Qod: Napoleon Dynamite | 360,987 |
It turns out someone does like Windows Vista, along with Office and the other stuff Microsoft sells.
– reports the New York Times, quoting Microsoft’s Kevin Johnson:
Customer demand for Windows Vista this quarter continued to build with double-digit growth in multi-year agreements by businesses and with the vast majority of consumers purchasing premium editions.
(emphasis mine)
…lot of folks spend so much time bitching about Vista and Office that they overlook one key point: Folks are buying this stuff.
– says Between the Lines. Donna Bogatin goes on in her “the-blogosphere-is-all-wrong-except-me” style:
The disconnect between tech blogosphere negative Microsoft hype and positive Microsoft reality continues to astound. Yesterday, Microsoft reported 27% revenue growth, fastest first quarter since 1999…
Typical Vista gloom and doom blogger headlines: “No one is lining up for Windows Vista in San Francisco,” “The top five things about Windows Vista that still suck,” “Is Windows XP too good for Microsoft’s own good?”…
If Vista were truly the nightmare it is made out to be in the blogosphere, wouldn’t there be a massive consumer Microsoft revolt?
Time for a reality check. Product quality, customer satisfaction and market success have very little to do with each other when you have a monopoly.
The Vista problems are real, they are not fantasies created by bloggers. But how exactly are consumers supposed to revolt? They still need computers, and despite Apple’s respectable growth, they still represent a fraction of the consumer PC market. Try to buy a PC today, it’s hard to NOT end up with Vista (even I got one)
Customer demand for Vista? No, it’s customer demand for computers, in a market with no choice. I’m not “making this up”, Donna. It’s all in Microsoft’s 10-Q:
…
The increased “demand” for premium versions comes from another well-documented fact, i.e. Microsoft’s new segmentation, castrating Vista Home Basic and essentially making Home Premium the equivalent of XP Home – a hidden price increase, by any measure.
A true measure of “demand” for Vista would be corporate licenses and retail sales, and both are behind. But not for long: eventually, after the release of SP1 corporate IT will give in, too – who wants to be “left behind”, after all.
This isn’t liking Vista at all – it’s assimilation by the Borg.
Related posts: Between the Lines, Insider Chatter, Seeking Alpha, All about Microsoft, Tom Foremski: IMHO, Silicon Valley Watcher, Mark Evans, Computerworld, Gaffney3.com, Seeking Alpha Software stocks, Todd Bishop’s Microsoft Blog, Alice Hill’s Real Tech News, Paul Mooney, Between the Lines , TechCrunch, All about Microsoft and Parislemon (who, like me, did not overdose of $Kool-aid$)
Update (1/11/08): A UK Government report advises school to avoid upgrading to Vista, or deploying Office 2007.
See further update here.
Been using vista for over a year now… no problems.
The problem imo isn’t MS’ monopoly, but the blogsphere bias towards the company.
Here’s how this goes down – Microsoft releases a new OS, everybody lines up to bash it because all the “cool kids” do it. Apple sells a service pack and everybody is cheering, bringing flowers and worshiping it.
Alex, I personally spent hours fixing Vista crap, if you search this blog, you’ll find some evidence, and of course even more by Googling it.
The fact that PC Manufacturers were pushed to allow customers roll back to XP is not urban legend, it’s reality.
But this will all be history in two years, we’ll be in a Vista world.
The problem isn’t the monopoly directly; it’s buggy software with undocumented problems/feautures. It’s the fact that MS does not really HAVE TO release user-friendly and working software anymore, since their market is almost “guaranteed”. So, indirectly, it’s all about the monopoly.
HI
I am another early vista adopter. I did have some NVidia problems with my dual core AMD four monitor vista machine but everything has been fine now for months. Love it in fact. Find going back to XP a clunky experience. I am a windows developer and have mostly MS software on the machine. I do use quite a bit of none MS software and it seems to work pretty well.
Jim
Zoli, I believe that the pc manufacturers, mainly dell, simply gave into the cries of the blogosphere, which in its turn made some consumers worry and ask for XP instead of Vista when they wouldn’t care otherwise.
It’s perfectly normal for a brand new OS to have initial driver issues. That was the case with 95, 98, XP and same in Vista. It’s impossible to roll out a new major version of kernel and have no compatibility issues. Try to do so results in a huge mess (for example, IE team has attempted to just that).
To me windows just works. Because the worse case scenario is a linux desktop, as much progress that it made in recent years, it’s still light years behind on drivers and usability. OSX in this case counts for nothing because it runs on a completely cosed system and it’s very easy to perfect stability and drivers when you have only a couple of hundred, maybe thousands hardware combination as opposed to millions which windows and linux must support.
Sure, Vista has flaws and bugs, just like any other operating system. Some people are unfortunate that their hardware combination causes them to have more issues. While I haven’t had a single crash or problem in over a year on Vista, trying to install Ubuntu 7.04 on the very same box resulted in absolutely nothing – live cd didn’t event boot.
Vista has been very solid in my experience and i very much enjoy the new office.
Windows monopoly is a double sided coin. On one side you have all the bad things that come with corporate power. On the other side you have a single OS and a single standard which gives developers the ability to develop once and reach 90% of the market. Without one solid platform, the IT world would easily be in chaos that linux is now.
The only reason why linux is seeing a rise and development now is because of the major players backing select distributions. Now you can develop linux software and test it in 3-6 major distros instead of 30-60.
Some things need to be centralized and have standards behind them. More often than not both are achieved by corporate backing and there’s just no getting away from it.
I can’t agree with Zoli more. Without knowing, I designed a great benchmark for Vista: I bought a new Thinkpad X61, identical to a work machine I had, loaded with XP. I was happy to get the new Thinkpad with Vista on it- as many, I was curious and wanted to play with it. As I have written in my blog, MSFT messed up with the most basic parameter, speed. Everything takes so much longer on the Vista laptop.
True, Windows was slower than DOS when it first came out but it was dramatically better in UI and ease of use. What is dramatically better about Vista???
I provide consumer PC support services for a living. As far as Vista is concerned, I am now receiving at least 3 calls a weeks from people who want Vista removed from their new PC/laptop purchase and replaced with XP. Even after upgrading their systems to 2GB memory and disabling Vista’s security feature, they still don’t think it’s fast enough for the work they do. And if it means having to buy a new webcam when drivers are unavailable for the embedded webcam, so be it. I hear about people whose new Vista PCs sit mostly unused, with their owners doing most of their work on their older Windows machines. Even with Microsoft’s extension of XP sales, retailers still do not want to go to the trouble of getting you a PC with XP because of their very low profit margins. I have machines with both XP and Vista (both of which are tweaked for performance), and after six months I hardly use the Vista PC anymore. Too many issues come up with Vista.
Sure Vista is the best thing MS has done for Linux! My User database has grown 1000% and I no Longer worry about complaints. like their systems incompatablities the only hard thing is setting up shockwave with wine and installing java into debian the only 2 hard things to setup, but then agin without viruses please leave me some work to do:) In any case most of these clients after about 2 months initail setteling in then I wont hear from them unless they buy new hardware they want hookedup. Since switching from Dev/support/tech only with MS to Linux Dev/support/tech I have actually increased my Income at least 200% | 257,841 |
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ClickFree MD700 120GB Offers Simple Plug ‘n’ Play Driver Free Backup
For those of you who really know that they ought to back up their valuable data but somehow never quite get around to it (alarmingly at least one hand appears to be raised here at TFTS Towers) ClickFree’s new 120GB MD700 could be the solution on account of the fact that it offers driver free plug ‘n’ play auto-backup capabilities straight out of the box.
Based on a single 2.5” 120GB SATA 5400 RPM hard disk drive, the ClickFree MD700 sports a 8MB onboard buffer cache and works by simply connecting the drive to your PC via a free USB port – and which point you’re free to let the drive take care of things completely automatically, without the need to install drivers or run through laborious configuration screens. Backing up your data really couldn’t be any simpler.
ClickFree MD700 120GB Specifications:
- Hard Drive: 2.5 inch SATA
- Capacity: 120GB
- Drive RPM: 5400
- Cache Buffer: 8MB
- Interface: USB 2.0 and 1.1 compliant
- Dimensions: 4.52″ x 0.67″ x 2.99″ in (115 mm x 17 mm x 76 mm)
- Weight: 6.35 oz (180 g)
- PC Requirements: Computer running Microsoft Windows Vista (Home Basic, Home Premium, Ultimate or Business), Windows XP (Home, Professional or Media Center Edition) or Windows 2000 (with Service Pack 4))
- Available USB ports (1.1 or 2.0)
The ClickFree MD700 retails for around $170 at the time or writing.?
Rovio Adds Angry Birds Space To The Former Android Market, Offers Ad-Free & HD Versions | 65,458 |
Oh Lizzo. herbs; taking (iffy) supplements; and of course, posing for before-and-after photos. On some level, Lizzo broadcast her dieting routine to her more than 9 million followers for promotional reasons: JJ Smith, the creator of the detox plan she used, is a Black woman who has built a large platform on Instagram and elsewhere teaching people how to “clear out” their bodies, specifically their livers. Like other wellness “gurus,” Smith also promotes unfounded dieting tips that purport to help participants lose up to 50 pounds in five months and shrink their waistlines. She’s even appeared on Dr. Oz, a syndicated show on which Donald Trump’s favorite quack peddles disinformation about wellness to millions of viewers.
On another level, Lizzo wanted to show the world that fat people, especially fat Black women, can do with their bodies what thinner people do without succumbing to diet culture. Yes, this is as odd as it sounds, but Lizzo still attempted to make the argument in good faith: “I would normally be so afraid and ashamed to post things like this online because I feel as a fat girl, people just expect if you are doing some sort of thing for health, that you’re doing it for some dramatic weight loss,” she said in her Instagram Story after she began receiving backlash. “That is not the case. In reality, November stressed me the fuck out. I drank a lot. I ate a lot of spicy things that fucked my stomach up and I wanted to reverse it and get my health back to where I was.” But whatever her reasons, Lizzo’s disclosure immediately drummed up controversy: Some fans were disappointed that their patron saint of body positivity broadcast her detox online. Others were adamant that Lizzo should be allowed to do with her body as she pleases—after all, hadn’t people praised Adele’s reveal of her own post–weight-loss body earlier this year? Still others couldn’t understand why a celebrity attempting a detox was worthy of discussion to begin with.
I’m in this latter group. As I watched Lizzo’s Instagram Story—and then watched her defend the detox before scrubbing her feed clean of the whole fiasco—a simple question crossed my mind: Why did Lizzo share such a personal decision with 9 million followers? She’s well aware that she’s been anointed a paragon of body positivity whether or not she wants to be, most likely because any fat woman who dares to live is categorized as such. In a world that still sees thinness as the ideal and larger bodies as transgressing the social order, our sheer unwillingness to apologize for being fat is considered an act of bravery. Like many pop artists, Lizzo sings openly about the ups and downs; but unlike most of them, her honesty and self-love has been deemed a form of body activism without either her input or her consent. “It bothered me for a long time that all people could talk about or think about was my size,” Lizzo told David Letterman in a recent episode of his Netflix series My Next Guest Needs No Introduction. “I didn’t like when people condemned me for it, and it also kind of rubbed me the wrong way when I was praised.” Lizzo, alongside other fat women in the internet age, including Adele, face an unwinnable conundrum: She’s unapologetically herself, which might inspire others to reject diet culture—but she doesn’t want to be idolized for simply living in the only way she knows how.
Perhaps that’s why Lizzo shared her detox story: If she can’t avoid the inevitable public focus on her body, she can at least use her visibility to debunk the idea that fat women are only invested in wellness as a means of weight loss. After all, she has previously posted her workout routine as a way to counter social media’s insistence that fat people can’t be healthy. If she’s going to be positioned as a figurehead of a movement, she might as well leverage that status to challenge widely held stereotypes about the limitations and habits of fat bodies. But even that logic has proved to be a double-edged sword for Lizzo, and for other fat women online and off: We’re all capable of succumbing to the allure of diet culture because we’re all indoctrinated in it and must actively fight to unlearn it. In a social-media ecosystem that encourages users to overshare as a way to build followings, brands, and in some cases entire careers, unlearning in public can be both useful and profitable—but it can also backfire. For social-media influencers or Instagram users who supplement their income by selling advertisements and creating branded content based on their own personal lives, oversharing is good business. For those who want to use their own experience to educate others, it can seem like a necessary evil.
Either way, the lines between sharing as a form of public good, sharing as a way to accrue clout, and sharing in a way that causes unintentional collateral damage are likely to get blurry. In this case, Lizzo did the latter—sharing her detox to dispel the myth that all people diet to lose weight, while also falling into the trap of sharing a before-and-after photo and, very likely, encouraging disordered eating. It’s unlikely that she did so with bad intent, but the damage is done nevertheless. Detoxing is a form of dieting, regardless of how it’s marketed, and dieting is nearly always a gateway to disordered eating. Not a single person, including someone who has been declared a body-positive icon (and seemingly embraced the label), is immune to the trappings of fatphobic diet culture. The idea that fat bodies—or anybody perceived as fat—is a work in progress that can be perfected through restrictive dieting (which can sometimes take the form of detoxing), is so pervasive that there’s a billion-dollar industry built around it. Unlearning the idea that a fat body is incomplete, that it can be inhabited and taken care of without being worked on, is a lifelong process that isn’t linear.
Asking Lizzo to speed up the pace of her own unlearning because she’s a famous woman in a fat body is both unrealistic and unfair—especially given the sheer number of thinner celebrities who are paid to promote controversial dieting teas on Instagram. In 2018, the Guardian published an eye-opening piece about Flat Tummy Tea becoming an Instagram phenomenon by tapping influencers, including the Kardashians, to promote their products. Hiring high-profile brand ambassadors to hawk Flat Tummy Tea’s products (including appetite suppressants) bought the company both goodwill and trust, despite the reality that the company, and others that rushed to imitate it, promote disordered eating—via laxatives, no less—in ways that have a demonstrable negative impact. “Dietary supplements sold for detox or weight loss are snake oil, plain and simple,” Dr. S Bryn Austin, a Harvard Medical School professor who specializes in eating- disorder risks, told the Guardian. “The liver and kidneys already do the so-called detox, and adding junk products into the diet only makes their job harder. And weight loss claims for these products are either [an] outright sham or a result of adulteration of the products with potentially dangerous stimulants, laxatives, or diuretics.” In other words, bodies detox themselves, and no amount of supplements, smoothies, or apple cider vinegar do more than our own bodies naturally do to regulate us.
Dieting can never be touted as a body-positive choice because it literally relies on starving the body to achieve a specific aim.
However, influencers and celebrities, including Beyoncé—who has promoted a detox on national television—have for years normalized the idea that “cleaning out our bodies” is not a diet, but simply a step toward “wellness.” And attempting to dress up detoxes—or the keto diet, or so-called gut health—in order to maintain membership in the body-positive community merely reinforces the very diet culture fat liberation is attempting to dismantle. The reality that our bodies, most of the time, will clean and restore themselves without the intervention of teas and juice fasts and supplements is a difficult sell when detoxes are treated as a “healthy” form of restricting. After all, if Lizzo endorses detoxing—or the Kardashians or any other number of celebrities—then how dangerous can it really be? No one should police Lizzo’s body or what she chooses to do with it, but that has not been a reality for her. She’s been subjected to intense scrutiny by mere virtue of being a fat public figure. She can’t even get dressed and go to a basketball game without making headlines, so it’s understandable that she, or any other fat celebrity, would want to reclaim her body for herself, not subjected to any other person’s whims about who she should be or how she should be in the world.
However, sharing a detox in public doesn’t remove the bulls-eye; it only increases its size, making her body—once again—a target for the judgment and disdain that has always surrounded fatness in a consumer culture. Everyone who embarks on a fat-liberation journey must unlearn at their own pace throughout their lives. However, there is also zero need for her to share this kind of routine online—or attempt to declare detoxing a body-positive action. “I’m so proud of myself. I’m proud of my results,” Lizzo said in her Instagram Story. “My sleep has improved. My hydration, my inner peace, my mental stability. My fucking body, my fucking skin, my eyes, like I feel and look like a bad bitch. And I think like that’s it. I’m a big girl who did a smoothie detox.”
It’s not that Lizzo is a fat woman doing a smoothie detox; it’s seeing a smoothie detox as essential to rebalancing her life that raises concern. Dieting can never be touted as a body-positive choice because it literally relies on starving the body to achieve a specific aim. No matter how it’s spun, no matter how individual the choice, it feeds into an existing, shaming, and often physically harmful culture of fatphobia. “Self-love” does not make a movement. Instead, fat liberation seeks to shift the very paradigm of diet culture—to expose its shallowness, its dangerousness, and its willingness to profit from the shame and pain of fat people. Though Lizzo has benefited from fat liberation, she might not be interested in being its face; and that’s okay. But at the bare minimum, can she—and all of us—keep our detoxes off the internet? For the good of ourselves, and the actual good of our bodies. | 39,194 |
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- Research article
- Open Access
- Open Peer Review
Undergraduate student drinking and related harms at an Australian university: web-based survey of a large random sample
© Hallett et al; licensee BioMed Central Ltd. 2011
- Received: 13 September 2011
- Accepted: 16 January 2012
- Published: 16 January 2012
Abstract ≥.
Trial registration
Keywords
- Sexual Assault
- Hazardous Drinking
- Standard Drink
- Hazardous Alcohol
- Student Drinking
Background
A high prevalence of hazardous drinking by university students has been reported in many countries [1–3] with this population group often drinking more than their non-university/college student peers [4–7]. In large-scale national surveys in the United States, 37-44% of students report binge drinking (more than five standard drinks per occasion; each containing 12 g ethanol) in the previous two weeks [8–10] with men drinking more than women, although this difference has narrowed over time [10–12]. Among New Zealand (NZ) university students, 37% have reported one or more binge episodes in the previous week [13].
Factors within the university environment contribute to these high levels of consumption leading to a range of negative consequences [5, 7, 14]. These include: social, physical and psychological harms to the student e.g. academic impairment, blackouts, injury, suicide, unintended sexual activity and sexual coercion; harm to other people including interpersonal and sexual violence; and costs to the institution such as property damage and student attrition [13, 15–21]. The secondhand effects of people's drinking on others are also assuming greater importance for advocacy in alcohol control policy, both for the victims experiencing assaults, sexual violence and impacts on studying [18, 22] and for the wider community experiencing litter, noise and vandalism [23].
Australian studies report between 70-96% of university students regularly consume alcohol [24–29] with 50% drinking to intoxication at least weekly [30, 31]. However, previous studies have relied on convenience samples [24–39] and most are at least a decade old [24–33, 38]. The one Australian study that used a random sample [40] surveyed only international students and therefore is not generalizable to all university students. This study found, that 66% consumed alcohol and 2% drank five standard drinks or more per occasion once or more a week.
There is significant support for the use of the Internet to collect epidemiological data particularly among university populations [41–45]. Online surveys permit fast application and wide accessibility [46, 47]. With their capacity for interactivity, automaticity, respondent anonymity and cost effectiveness [48–50], ability to facilitate more honest and thoughtful responses [42, 51] and good validity and reliability [43, 52–57], a carefully conducted online survey may help overcome many of the barriers associated with collecting epidemiological data [58–60]. Unlike the proportionate cost to attain large sample sizes using traditional modes of survey implementation, marginal costs are low and therefore they are advantageous for large sample sizes [59, 61]. In addition, unique features such as complex logic and branching [62] and real-time error checking and automated data entry [63] allow statistical processing to occur in real time [64]. This enables web-survey technology to deliver concurrent feedback interventions [65]. It may be ethically obligatory when surveys identify harmful behaviours among respondents to provide feedback. This may be an efficient option given that provision of immediate feedback in this context has been shown to change behaviour [65, 66].
As part of a larger efficacy trial of a web-based alcohol screening and brief intervention [65, 67], this study estimated the frequency and quantity of alcohol consumption, and prevalence of hazardous drinking and secondhand effects among a large sample of undergraduate students attending a university in Perth, Western Australia.
Methods
Participants
A random sample of 13,000 undergraduate students aged 17-25 years, enrolled full-time and studying on campus at a Western Australian university, were invited to complete a web survey on alcohol consumption, secondhand effects, attitudes toward nutrition/ingredient labeling [68] and tobacco use [69]. Women made up 52.4% of the sample and 20.6% were non-residents. The term 'non-resident' refers to students enrolled at the university that are not permanent residents of Australia or New Zealand and includes those on student visas and humanitarian visas.
Procedure
We adopted a survey recruitment procedure described in detail elsewhere [59, 65, 67, 70]. Four weeks after the start of the first semester of 2007 the University Surveys Office accessed the enrolment database to identify a random sample of 13,000 full-time undergraduates aged 17-25 years. A personally addressed letter from the research team was sent to each student, inviting them to participate in the survey. The letter explained that they would soon receive a hyperlink to the questionnaire in an email message, that responses would be confidential and that the research team was independent of the university administration. Students were offered the opportunity to win one of 40 AU$100 gift vouchers for participating. After 1 week, a reminder email was sent encouraging completion of the questionnaire to those who had not yet responded. A second reminder was sent 10 days later.
Measures
The questionnaire included items on: past alcohol use [71]; current alcohol use [Alcohol Use Disorders Identification Test (AUDIT) [72]]; peak consumption in the previous 4 weeks [73]; height and weight (in order to estimate Blood Alcohol Concentration); secondhand effects of drinking [22]; attitudes toward nutrition/ingredient labelling on alcohol packaging [68]; and tobacco use [74]. The use of standardised instruments for measuring personal use [Alcohol Use Disorders Identification Test (AUDIT) [72]] and secondhand effects [22] make it comparable to studies carried out in other countries. The complete wording and layout of all items can be seen at:.
Data analysis plan
Descriptive statistics were computed for the following: demographic data [age (17-19, 20-25 year olds), sex, and citizenship (Australian and NZ residents, non-residents)] of respondents and the sample; early or late response to the survey; the quantity and frequency of alcohol use; AUDIT scores; and the number of secondhand effects. Three AUDIT subscale scores were calculated to measure alcohol consumption (AUDIT items 1-3), dependence (items 4-6) and problems (or adverse consequences) (items 7-10) [72, 75]. Total AUDIT scores were divided into four ordinal categories: moderate (0-7), hazardous (8-15), harmful (16-19), and dependent (20-40) [72]; and binary categories of hazardous (≥ 8) and non-hazardous (< 8).
The representativeness of responders to the random sample was assessed using chi-squared tests. The association between participant demographics and being either early or late responder, or to having an AUDIT score ≥ 8, was assessed using chi-squared tests. T-tests were used to compare the mean AUDIT measure for the three subscales (alcohol consumption, dependence and problems) against participant age, sex, and citizenship and to compare total AUDIT score between early and late responders.
The association of secondhand effects experience to participant demographics and frequency of consuming six or more drinks on one occasion (item three in the AUDIT [72]) was also assessed using chi-squared tests. The association between frequencies of consuming six or more drinks (60 g ethanol) on secondhand effects was assessed using multivariable logistic regression after adjusting for gender, age and citizenship. Results are presented as odds-ratio and associated 99% confidence intervals.
The associations between age, sex and citizenship and hazardous drinking were analysed using binary logistic regression. To protect against small effects being considered as being statistically significant due to the large sample size in the study, p-values of < 0.01 will be considered as statistically significant. The assumptions behind the statistical models fitted were assessed to ensure validity of results.
This study received ethical approval from Curtin University [HR 189/2005] and is registered with the Australian and New Zealand Clinical Trial Register [ACTRN12608000104358].
Results
Demographic comparison of responders vs. study sample and early vs. late responders
Consumption characteristics
Ninety percent of respondents had consumed alcohol in the last 12 months, with a mean volume per typical occasion of 5.1 (SD = 5.0) standard drinks for women and 8.7 (SD = 8.6) for men. The National Health and Medical Research Council (Australia) thresholds for acute harm (40 g/60 g ethanol for women/men) [76] were exceeded at least once in the last 4 weeks by 48% of respondents.
AUDIT subscale scores and hazardous drinking (AUDIT Score ≥ 8) by demographic characteristics
Men had higher odds of drinking at hazardous levels compared to women (OR: 2.0; 95% CI: 1.8-2.2). Australian/NZ residents had higher odds compared to non-residents (OR: 5.1; 95% CI: 4.3-6.0) and the association with age was non-significant (p = 0.113).
Significantly higher mean AUDIT scores (p < 0.001) were observed for men and Australian/NZ residents compared to women and non-residents in all AUDIT subscales (shown in Table 2). There were significant differences in relation to age in the AUDIT Consumption subscale with higher mean scores for 17-19 year olds compared to 20-24 year olds, but not the other subscales.
Secondhand effects
Secondhand effects experience by demographic characteristics
Men were more likely than women to experience being 'pushed, hit or otherwise assaulted' (8.7% vs. 4.8%; p < 0.001) and to have been a victim of another crime off campus (2.8% vs. 1.8%; p = 0.007) while women were more likely to experience an unwanted sexual advance (13.8% vs. 7.1%; p < 0.001) and to have had to 'baby-sit' or take care of another student who had too much to drink (28.8% vs. 25.1%; p = 0.001). Those aged 17-19 years were more likely than 20-25 year olds to have had a serious argument (13.7% vs. 11.1%; p = 0.001); been assaulted (7.2% vs. 5.6%; p = 0.005); had to 'baby-sit' another student (31.9% vs. 21.6%; p < 0.001); had their studying or sleep interrupted (22.1% vs. 19.4%; p = 0.004) or to have experienced unwanted sexual advances (12.1% vs. 9.5%; p = 0.001).
Effects* of frequency of consuming six or more drinks (60 g ethanol) on secondhand effects
Discussion
This study is the first known prevalence study of student drinking completed in Australia with undergraduate students. The vast majority of students were current drinkers (90%) and there was a high prevalence of hazardous drinking (48%), with a higher prevalence among men compared with women, and in Australian/NZ residents compared with non-residents. A relatively large proportion of students' experienced secondhand effects from other people's drinking.
The survey had a response rate of 56%, which is higher than large college surveys in the early 2000s in the United States (52%) [10], but lower than online surveys in New Zealand using similar procedures (63-82%) [13, 59]. Higher response rates for online surveys have been linked to pre-notification, personalised contacts and follow up reminders [77]. Both this and the New Zealand studies incorporated pre-notification, personalised emails and follow-up notices. However, the earlier New Zealand study used up to nine follow-up contacts (compared to five in this study) including a telephone reminder, which may explain some of the difference. Follow-up notices are likely to increase response rates though larger numbers of notices may not appreciably affect response if the contact develops resistance to participation [78]. It is also possible that the novelty factor of online surveys may have reduced in the years since the New Zealand studies and factors such as proliferation of junk mail, bombardment with online questionnaires and demands on student time may also have impacted on response rates [47].
The level of alcohol consumption reported in this study is less than that reported in New Zealand, for both men and women [13]. Although gender convergence in drinking has been reported elsewhere [10–12, 79] and a similar trend appears to be occurring in Australia [80], this study shows a large discrepancy between men and women. However, there are no older prevalence studies from which to assess attenuation trends.
Large numbers of people were affected by other students' drinking. Of particular note was the 0.9% (n = 36) of women and 1.1% (n = 35) of men who reported being a victim of sexual assault in the previous 4 weeks. This is slightly higher than that found in a New Zealand sample [81] though with overlapping confidence intervals. While the New Zealand sample was limited to those who had consumed alcohol in the previous 4 weeks, our sample included non-heavy drinkers and may highlight the impact that hazardous alcohol consumption can have on all students. Extrapolated to the entire student population this may mean approximately 140 students at this university experience sexual assault in this context each month.
A limitation of this study was the imprecision in the specificity of crimes listed in the secondhand effects questions and the reliance on respondents to attribute responsibility for the effect. As only yes or no responses were available, multiple experiences of the same effect were not captured and therefore the prevalence of these effects may be underestimated. Given the high prevalence of some of these effects further research in this area is warranted. Our estimates may be biased by selective non-response but conversely computerised questionnaires are known to increase reporting of high-risk behaviour [42, 51].
Universities with large on-campus resident populations may have higher levels of drinking than commuter universities due to students' greater proximity to peers [82]. As this study is based on a single commuter university and has a high proportion of students on temporary visas, the findings may be limited in their generalisability.
This study highlights the need for university programs to target drinking in this population. With half of male, and over a third of female, respondents drinking at hazardous levels, population approaches are needed. The literature suggests that programs should also address environmental factors, particularly the availability and promotion of alcohol on and around campus [22, 83].
Conclusions
Hazardous alcohol use among undergraduate students remains an issue of concern although there is a lack of prevalence data on this population's alcohol consumption in Australia. Some alcohol related harms such as sexual assault are only detected with large population samples. Web-based surveys are a cost-effective approach for measuring health behaviours in student populations, with a relatively high response rate. It is suggested that this research is replicated in other Australian universities, particularly residential campuses. Such surveys are required to develop trend data which will facilitate intervention program development.
Declarations
Acknowledgements
This work was supported by the Western Australian Health Promotion Foundation (Healthway) [15166]. The authors gratefully acknowledge Deputy Vice Chancellor Jane den Hollander, Alice Tsang and Nerissa Wood, other university administration staff and the study participants for their support of the research. The Centre for Behavioural Research in Cancer Control is supported by Cancer Council WA.
Authors’ Affiliations
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The Law Office of Stefani K. Nolan is located in Jacksonville, Florida, where I assist clients mainly in the area.
I offer a free 30-minute consultation to meet with me, the attorney, to answer your questions and provide you insight of our office and how we handle our cases. We want you, the client, to feel comfortable with our office so that you can turn over your legal issues to us and let us do the work for you. You will find that we are service-oriented and passionate about the outcome of your case in regards to the impact it will have on your life. It is important that you feel comfortable in hiring an attorney that will provide competent, skillful representation on your behalf, especially protecting the two most important assets in your life: your family and your finances.
Our office serves Duval, Clay, Nassau and St. Johns counties and is centrally located at 1616 Jork Road, Suite 203, Jacksonville, Florida, which is off Atlantic Boulevard between the Hart Bridge/Emerson Expressway and University Boulevard.
If you need representation in the following areas, give us a call; we can help: | 56,162 |
Information for "IEEE Winnipeg Section History" Basic information Display titleIEEE Winnipeg Section History Default sort keyWinnipeg Page length (in bytes)503 Page ID546716:55, 21 November 2008 Latest editorAdministrator1 (talk | contribs) Date of latest edit15:40, 12 "" | 74,345 |
All of us have procrastinated at one point or the other, and chances are that you have done the same. individual.
- Loss of opportunities: As a result of procrastination, you are bound to lose out on several opportunities that come across your way. While at that point it may have seemed like a great idea to shelve the opportunity for later, the problem is that the ‘later’ never comes and you are bound to blow out a precious opportunity, one that could have helped to set your career on a new trajectory altogether.
- Loss of time: Often, you are bound to become aware that you are procrastinating and as a result, you realize that you have lost valuable time that you could have better utilized for something else altogether. While it is a good thing to become aware of when you are procrastinating, the fact remains that you cannot go back in time and any time you have lost as a result of your procrastination, is lost forever.
- Shoddy time management: If you happen to procrastinate regularly, then chances are that you would suck at time management or any activity that requires you to finish tasks in a given period.
- Goal setting: As a result of your tendency to procrastinate often, you would not be able to accomplish most of your goals. As a result of this action on your part, you would not be able to achieve your complete potential and would have to make do with what you can, for the moment.
- Career: If you continue to procrastinate regularly, then chances are that your career will take a hit. NO company likes to retain employees who cannot manage the most simple of tasks or schedules, and are not motivated enough to improve themselves or their career. A proven ability to meet deadlines on time or even monthly targets can result in your termination and soon it will only be a matter of time before you find it hard to land a job anywhere.
- Lowered self-esteem: Often when you procrastinate, you are bound to ask yourself why you feel that you cannot do a certain task and as a result, both desperation and lowered self-esteem sets in. And if you are a habitual procrastinator, then this can result in chronic depression and total lack of self-esteem.
These are some of the negative effects of procrastination; the fact remains that no one is immune to this and everyone would have procrastinated at one point of time or the other. You need to do all you can to combat the insidious effects of procrastination and take effective measures at the earliest. | 132,488 |
Ash and his friends enter a small town where all the girls are lined up to watch Aaron, one of the Elite Four, conduct a public training session. Aaron knows his Bug-type Pokémon so well, he can even understand what Bug-type Pokémon say. Aaron plans to challenge Cynthia for the top Trainer spot in Sinnoh, and the crowds are certainly impressed by his Vespiquen, Drapion, and Skorupi. When Aaron learns that Dawn and Ash know Cynthia, he takes them to his private training center in the woods. Team Rocket follows them and puts syrup on a tree to lure Aaron's Pokémon, but it only ends up attracting Beedrill!
Dawn notices a photo of Aaron and a Wurmple. This Wurmple was Aaron's greatest friend, but Aaron doesn't seem to have a Wurmple now—where did it go? Before Aaron can explain, Team Rocket's balloon descends from the sky and vacuums up his Pokémon along with Pikachu and Piplup. Team Rocket escapes, but Aaron is determined to find and rescue his Pokémon. When he was young, he was a poor loser who blamed Wurmple for his defeat, driving Wurmple away. Ever since that mistake, he's worked hard to understand Bug-types and vowed not to lose a Pokémon ever again!
Team Rocket doesn't get far—their kidnapped Pokémon decide to fight back, and the balloon crashes in the forest. As Seviper and Carnivine try to stop Pikachu, Piplup, and Aaron's Pokémon from escaping, Aaron and our heroes arrive to save the day. Team Rocket makes a second attempt to vacuum up the Pokémon, but a Beautifly stops them with a powerful SolarBeam attack. Team Rocket crashed near the tree where Aaron met Wurmple, and this Beautifly is the Evolution of Aaron's Wurmple! All this time, it's been training to become strong enough to help Aaron. Now it's certainly strong enough to help defeat Team Rocket, and Team Rocket is sent packing by the combined attacks of our heroes' Pokémon. As Aaron happily trains for his big challenge, Ash heads to Hearthome City for his own big challenge: a Gym Battle against Fantina! | 313,130 |
I made an PV surface simulation with the Ladybug tools including the Ladybug_sunpath Shading component.
In the first simulation I used simple surfaces which cover most of the roof.
In the second simulation I splitted these surfaces into smaller parts so I could get a better view of how much energy on which spot is generated. Unfortunately the results of both simulations differ quite a lot. I used for both simulations the same parameters. (the energy per year is written in white)
undevided surface
devided surface.
What could be the issue here?
(I used the standard template: 020_Photovoltaics shaded analysis) | 210,588 |
not just let them choose. So I added a simple update to use the jquery UI date picker on the review dialog. Also, the grid default sort was never set. Using the aaSorting initialization parameter, you can specify which column should have the default sort in a grid. I added this to both grid instances.
Style
None of what I built had much style. It really doesn’t look like it is part of a site for that matter as it was simply a POC without any real use cases. The patterns have sort of an Admin feel which might not map to a great consumer Beer application. That aside, here is where the MVC approach is nice. Without touching server code, we (or better yet a designer) can improve the markup, add some CSS classes and define some style rules and quickly have us looking much better.
Data Access Layer
I referred to use of my datacontext as a data access layer. There is certainly a lot of room for opinion on approaches for better abstracting that away. One approach would be to take all the domain classes like Beer and Brewery and move them to a Clarity.BeerApp.Data.Model assembly, leaving the context in the Data assembly. My entities really don’t need to know anything about EF and can be separated. Then I can build mock contexts and support better testing. I may do this in the future and post an update.
Routing Improvments
Similar to the DataTables parameter binding improvements I suggested in the posts, there may be other ways to take better advantage of MVC features. I didn’t explore this, but there may be some benefit to registering routes with the app to better handle some of the Review edits which would simplify the url creation. Or maybe the Review views should have their own controller. That is all open for debate.
In summary, this sample and set of posts was a starting point for me to look into combining Entity Framework 4.2, ASP.NET MVC 3 and the jquery DataTables into a web app. I built off of some other’s ideas referenced in the initial post and hope this might help as a reference for some else learning about these technologies.38 comments , permalink
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Shoreline Master Plans: Local Discretion
or
Department of Ecology Mandate?
September 17 - Before the Community Development Committee
October 2 - Before the County Council
County Council Contacts:
Pierce County is in the process of reviewing and ratifying its Shoreline Master Plan update. Part of the update process has included responding to the Department of Ecology's required changes. In a letter dated June 27, Pierce County wrote of the "...apparent inability to exercise local discretion in certain areas of shoreline management."
Specifically, Pierce County writes:
"Pierce County would prefer to prohibit the dumping of dredged material within the State's designated Nisqually Reach Aquatic Reserve due to various factors including the existence of alternative dredged material disposal sites within the County."Further:
"Pierce County would also prefer to prohibit certain aquaculture activities in proximity to estuaries and within select bays and inlets that are developed with high density residential densities."
"Obligating the County to accept uses in inappropriate areas that will result in unavoidable impacts and user conflicts is unfortunate."Not allowed by Ecology:
18. Aquaculture is prohibited in Estuaries within 300 feet of the mouth of freshwater streams (as measured at extreme low tide).
19. Aquaculture is prohibited adjacent to residential neighborhoods in Horsehead Bay, Wollochet Bay, Lay Inlet and adjacent to Raft Island due to water quality and visual impacts.
d. Disposal of dredged material within the Nisqually Reach Aquatic Reserve
17. With the exception of Olympia Oyster propagation which is a conditional use, new commercial shellfish aquaculture operations are prohibited within the Nisqually Reach Aquatic Reserve.
For complete document showing what was struck by DOE, added by DOE, and why, see this link:
For DOE's "recommended changes" see this document: | 5,271 |
\begin{document}
\begin{frontmatter}
\title{Existence and uniqueness of global weak solutions to strain-limiting viscoelasticity with Dirichlet boundary data\tnoteref{t1,t2}}
\tnotetext[t1]{M.~Bul\'{\i}\v{c}ek's work is supported by the project 20-11027X financed by GA\v{C}R. M.~Bul\'{\i}\v{c}ek is a member of the Ne\v{c}as Center for Mathematical Modeling.
V.~Patel is supported by the UK Engineering and Physical Sciences Research Council [EP/L015811/1].}
\author[Bulicek-address]{Miroslav Bul\'{i}\v{c}ek}
\ead{mbul8060@karlin.mff.cuni.cz}
\author[Ox-address]{Victoria Patel}
\ead{victoria.patel@maths.ox.ac.uk}
\author[Sengul-address]{Yasemin \c Seng\"{u}l\corref{cor1}}
\cortext[cor1]{Corresponding author. Tel: +90 216 483 9541}
\ead{yasemin.sengul@sabanciuniv.edu}
\author[Ox-address]{Endre S\"{u}li}
\ead{endre.suli@maths.ox.ac.uk}
\address[Bulicek-address]{Charles University, Faculty of Mathematics and Physics, Sokolovsk\'{a}~83, 186~75 Praha~8, Czech Republic.}
\address[Ox-address]{Mathematical Institute, University of Oxford, Andrew Wiles Building, Woodstock Road, Oxford OX2 6GG, UK.}
\address[Sengul-address]{Faculty of Engineering and Natural Sciences, Sabanci University, Tuzla 34956, Istanbul, Turkey.}
\begin{abstract}
We consider a system of evolutionary equations that is capable of describing certain viscoelastic effects in linearized yet nonlinear models of solid mechanics. The essence of the paper is that the constitutive relation, involving the Cauchy stress, the small strain tensor and the symmetric velocity gradient, is given in an implicit form. For a large class of implicit constitutive relations we establish the existence and uniqueness of a global-in-time large-data weak solution. We then focus on the class of so-called limiting strain models, i.e., models for which the magnitude of the strain tensor is known to remain small a~priori, regardless of the magnitude of the Cauchy stress tensor. For this class of models, a new technical difficulty arises, which is that the Cauchy stress is only an integrable function
over its domain of definition, resulting in the underlying function spaces being nonreflexive and thus the weak compactness of bounded sequences of elements of these spaces is lost. Nevertheless, even for problems of this type we are able to provide a satisfactory existence theory, as long as the initial data have finite elastic energy and the boundary data fulfill natural compatibility conditions.
\end{abstract}
\begin{keyword}
nonlinear viscoelasticity\sep strain-limiting theory \sep evolutionary problem \sep global existence \sep weak solution \sep regularity \MSC{35M13\sep 35K99 \sep 74D10 \sep 74H20}
\end{keyword}
\end{frontmatter}
\setcounter{equation}{0}
\numberwithin{equation}{section}
\section{Introduction}
This paper is devoted to the study of the following nonlinear system of partial differential equations (PDEs). We assume that $\Omega \subset \mathbb{R}^{d}$ is a given bounded open domain and we denote the parabolic cylinder by $Q:=(0,T)\times \Omega$ and its spatial boundary by $\Gamma:=(0,T)\times \partial \Omega$, where $T>0$ is the length of the time interval of interest. For given data $\bG:\R^{d\times d}_{sym}\to \R^{d\times d}_{sym}$, $\bef:Q\to \R^d$, $\bu_I:\Omega\to \R^d$, $\bv_0:\Omega\to \R^d$, $\bu_{\Gamma}: \Gamma \to \R^d$ and \(\alpha\), $ \beta>0$, we seek a couple $(\bu, \bT):Q \to \R^d \times \R^{d\times d}_{sym}$ satisfying
\begin{subequations}\label{model}
\begin{alignat}{2}
\partial^2_{tt}\bu - \diver \bT & = \bef &&\quad \textrm{in $Q$},\label{linear-moment}\\
\label{cons-law}
\alpha \beps(\bu) + \beta \beps(\partial_t\bu) &= \bG(\bT)&&\quad \textrm{in $Q$},\\
\bu(0) = \bu_{I},\quad \partial_t\bu(0) & = \bv_{0} &&\quad \textrm{in $\Omega$},\label{init-data}\\
\bu & = \bu_{\Gamma} &&\quad \text{on $\Gamma$}.\label{bound-data}
\end{alignat}
\end{subequations}
Here, \eqref{linear-moment} represents an approximation\footnote{In fact, the density $\varrho$ of the solid should also appear in \eqref{linear-moment}. In principle $\varrho$ could be a function of space and time and should satisfy the balance of mass equation. Since we are dealing with small strains here, i.e. the case when the deformation of the solid is small, under the assumption that the solid is homogeneous at initial time $t=0$, we can consider the density to be equal to a constant for all times $t\in(0,T)$. We shall therefore scale the density to be identically equal to one for simplicity; see also the discussion in~\cite{BuMaRa12}. We note however that under suitable assumptions it is not too difficult to extend the results presented herein to the case of variable density.} of the balance of linear momentum, where $\bef$ is the density of the external body forces, $\bu$ is the displacement, $\bT$ denotes the Cauchy stress tensor and the operator $\diver$ denotes the standard divergence operator with respect to the spatial variables $x_1,\ldots, x_d$.
The Cauchy stress tensor $\bT$ is implicitly related to the small strain tensor $\beps(\bu):=\frac12 (\nabla \bu + (\nabla \bu)^{\rm T})$ and to the symmetric velocity gradient $\beps(\partial_t\bu):= \partial_t(\beps(\bu))$ via \eqref{cons-law}. The initial displacement and the initial velocity are given by \eqref{init-data} and the Dirichlet boundary condition for the displacement is represented by \eqref{bound-data}. A more detailed discussion concerning the relevance of \eqref{model} to problems in viscoelasticity is contained in Section~\ref{physics}.
It remains to specify the form of the implicit constitutive law \eqref{cons-law}. The minimal assumptions imposed on the mapping $\bG$ throughout the paper are the following. We assume that the function $\bG:\Rsym \to \Rsym$ is a continuous mapping such that, for some $p\in [1,\infty)$, some positive constants $C_1$ and $C_2$, and for all \(\bT\), $\bW \in \Rsym$, the following inequalities hold:
\begin{align}
\tag{A1} \big(\bG(\bT) - \bG(\bW)\big) \cdot (\bT - \bW) &\ge 0,\label{A1}\\
\tag{A2} \bG(\bT) \cdot \bT &\geq C_{1} |\bT|^{p} - C_{2},\label{A2}\\
\tag{A3} |\bG(\bT)| &\leq C_{2} (1 + |\bT|)^{p-1},\label{A3}
\end{align}
where $| \cdot |$ stands for the usual Frobenius matrix norm.
Assumptions \eqref{A1}--\eqref{A3} are sufficient for the existence and uniqueness of a weak solution provided that $p\in (1,\infty)$. For $p =1$, however, we must impose a more restrictive assumption due to the lack of compactness experienced when working in \( L^1\). Namely, we will assume that there exists a strictly convex function $\phi \in \mathcal{C}^2(\R_+; \R_+)$ such that $\phi(0)=\phi'(0)=0$, $|\phi''(s)|\le C(1+s)^{-1}$ for every \(s\in \mathbb{R}_+\), and for all $\bT\in \Rsym$ there holds
\begin{align}\tag{A4}\label{A4}
\bG(\bT)=\frac{\phi'(|\bT|) \bT}{|\bT|}.
\end{align}
In order to simplify the exposition and avoid nonessential technical details concerning the choice of appropriate function spaces that admit suitable trace theorems, we shall assume that there exists a function $\bu_0:Q \to \R^d$ fulfilling, in an appropriate sense, the initial and boundary conditions
\[
\begin{aligned}
\bu_0(0)&=\bu_I &&\text{in }\Omega,\\
\partial_t \bu_0(0)&=\bv_0 &&\text{in } \Omega,\\
\bu_0&=\bu_{\Gamma} &&\text{on } \Gamma.
\end{aligned}
\]
We shall henceforth formulate all assumptions on the initial and boundary data in terms of $\bu_0$, rather than $\bu_I$, $\bv_0$ and $\bu_\Gamma$. We note that since the function spaces for $\bu_0$ stated below are the same as those for the weak solution $\bu$, it is in fact necessary that such a $\bu_0$ exists. Otherwise our construction of a weak solution would not be possible.
\subsection{Statement of the main results}\label{mathematics}
First, we formulate our result for the case when $p>1$.
\begin{theorem}\label{T1}
Let $p'>2d/(d+2)$, let $\bG$ satisfy \eqref{A1}, \eqref{A2} and \eqref{A3}, and let \(\alpha\), $ \beta > 0$ be arbitrary. Assume that the data satisfy the following hypotheses:
\begin{equation}
\begin{split}
\bu_0&\in W^{1,p'}(0,T; W^{1,p'}(\Omega;\R^d))\cap W^{2,p}(0,T; (W^{1,p'}_0(\Omega;\R^d))^*)\cap \mathcal{C}^1([0,T];L^2(\Omega;\R^d)),\\
\bef&\in L^{p}(0,T; (W^{1,p'}_0(\Omega;\R^d))^*).
\end{split}\label{data-as}
\end{equation}
Then, there exists a couple $(\bu, \bT)$ fulfilling
\begin{align}
\bu &\in \mathcal{C}^{1}([0, T]; L^{2}(\Omega;\R^d))\cap W^{1, p'}(0, T; W^{1, p'}(\Omega;\R^d))\cap W^{2,{p}}(0, T; (W_{0}^{1,p' }(\Omega;\R^d))^{*}), \label{FSu}\\
\bT &\in L^{p}(0, T; L^{p}(\Omega;\Rsym))\label{FST}
\end{align}
and solving \eqref{model} in the following sense:
\begin{align}\label{WF}
\langle \partial_{tt}\bu, \bw \rangle + \int_{\Omega} \bT \cdot \nabla \bw &= \langle \bef, \bw \rangle && \forall\, \bw \in W_{0}^{1,p'}(\Omega;\R^d),\quad
\textrm{for a.e. }\,t \in (0, T),
\\
\label{T-const}
\alpha \beps(\bu) + \beta \partial_{t} \beps(\bu)&=\bG(\bT) &&\textrm{a.e. in }Q,
\end{align}
and
\begin{equation}
\bu-\bu_0 =\b0 \quad \textrm{ a.e. on }\Gamma\qquad \textrm{and}\qquad \bu(0)-\bu_0(0)=\partial_t \bu(0)-\partial_t\bu_0(0)=\b0\quad \textrm{ a.e. in }\Omega. \label{bcint}
\end{equation}
Furthermore, the function \( \bu \) is unique. If, additionally, the mapping \( \bG \) is strictly monotonic, then \( \bT \) is also unique.
\end{theorem}
Before proceeding, we will first comment on the assertions of Theorem~\ref{T1}. The proof of Theorem~\ref{T1} is based on the relevant a~priori estimates, for which the assumption \eqref{data-as} seems to be both optimal and minimal. The function spaces considered in \eqref{FSu}, \eqref{FST} correspond to the structural assumptions imposed on $\bG$, namely the coercivity assumption \eqref{A2} and the growth condition \eqref{A3}. Since $p>1$, we have a ``standard" function space setting, so the nonlinearity in \eqref{T-const} can be identified by using a modification of Minty's method. Theorem~\ref{T1} can also be understood as an extension of the results established in ~\cite{BuMaRa12}; similarly as here, the authors of~\cite{BuMaRa12} treated a viscoelastic solid model of generalized Kelvin--Voigt type, but they considered a constitutive relation for the Cauchy stress of the following explicit form:
\begin{align*}
\bT&=\bT_{el}(\beps(\bu)) + \bT_{vis}(\partial_t\beps(\bu)) \qquad \textrm{a.e. in }Q.
\end{align*}
The regularity results for such models are available in~\cite{BuKaSt13}. It is remarkable that while \eqref{T-const} can be fully justified from the physical point of view via implicit constitutive theory, see \cite{Raj-03}, the above explicit form $\bT=\bT_{el} + \bT_{vis}$ can be justified for particular choices of $\bT_{el}$ and $\bT_{vis}$ only.
In contrast with the case of $p>1$, almost none of what was said above applies in the case $p=1$, or for the limit, as $p\to 1_+$, of the sequence of solutions constructed in Theorem~\ref{T1}. Indeed, for similar models in the purely steady elastic setting, it was demonstrated in \cite{BeBuMaSu17} that $\bT$ is, in general, a Radon measure and therefore one can hardly consider \eqref{T-const} pointwise in $Q$. Nevertheless, it was shown there that under some structural assumptions on $\bG$ (corresponding to \eqref{A4}), one may hope for $\bT$ to be integrable. A similar situation was also studied in~\cite{BeBuGm20} but with $p\to \infty$, which, in general, leads to solutions $\bu$ in $BV$ spaces. However, under a structural assumption related to~\eqref{A4}, one can again overcome such difficulties and show the existence of a solution that belongs to a Sobolev space.
A similar situation can be expected in our setting when $p=1$. Therefore, in order to avoid difficulties associated with the interpretation of $\partial_{tt}\bu$ and the interpretation of the sense in which the initial data are attained, we assume here, for simplicity, that the right-hand side $\bef\in L^2(Q;\mathbb{R}^d)$.
Thus, inspired by \cite{BeBuMaSu17}, if \( p =1\) we assume in addition to \eqref{A1}--\eqref{A3} that we have \eqref{A4}.
It then follows from the structural assumptions that for all $s\in \R_+$ we have
\begin{equation*}
\begin{split}
\frac{C_1 s}{2} - C_2 &\le \phi(s)\le C_2 s,\\
0&\le \phi'(s) \le C_2.
\end{split}
\end{equation*}
Since $\phi$ is convex, we deduce that there exists an $L>0$ such that
\begin{equation}\label{dfL}
L:=\lim_{s\to \infty} \phi'(s) \ge \phi'(t) \qquad \forall \,t\in \R.
\end{equation}
The number $L$ plays an essential role in the subsequent analysis, in particular in the assumptions on the initial and boundary data. Indeed, thanks to \eqref{A4}, we see that
\begin{equation}\label{dfL2}
L=\lim_{|\bW|\to \infty} |\bG(\bW)|\ge |\bG(\bT)| \qquad \forall \,\bT\in \Rsym.
\end{equation}
Hence, if \eqref{cons-law} is satisfied then we must necessarily have
\begin{equation}\label{nec}
|\alpha \beps(\bu) + \beta \partial_t \beps(\bu)|\le L \qquad \textrm{ a.e. in }Q.
\end{equation}
Consequently, if such a $\bu$ should exist then it is natural to assume the same requirement as \eqref{nec} also for the initial and boundary data, that is, we must have
\begin{equation}\label{nec0}
|\alpha \beps(\bu_0) + \beta \partial_t \beps(\bu_0)|\le L \qquad \textrm{ a.e. in }Q.
\end{equation}
In fact, we require in the existence analysis that \eqref{nec0} is satisfied with a strict inequality sign; such a condition is called the {\it safety strain condition}.
\tikzstyle{dot}=[draw,fill=white,circle,inner sep=0pt,minimum size=4pt]
\def\deltazero{
\begin{tikzpicture}[scale=2]
\footnotesize
\draw[ultra thin] (-0.2, 0.0) -- (3.2, 0.0) node[below]{$s$};
\draw[ultra thin] ( 0.0, -0.2) -- (0.0, 3.2) node[left ]{$\phi(s)$};
\draw[ultra thin] ( -0.05, 1) -- (0.05, 1);
\draw[variable=\t,domain=0:3.2,thick,samples=400]
plot (\t ,{\t - ln(\t +1)}) ;
\draw[variable=\t,domain=0:3.2,thick,dashed,samples=400]
plot ({\t} ,{(1+(\t)^(2))^(1/2)-1} ) ;
\end{tikzpicture}
}
\def\deltaone{
\begin{tikzpicture}[scale=1.5]
\footnotesize
\draw[ultra thin] (-0.2, 0.0) -- (2.2, 0.0) node[below]{$\abs{\bG(\bT)}$};
\draw[ultra thin] ( 0.0, -0.2) -- (0.0, 2.2) node[left ]{$|\bT|$};
\draw[variable=\t,domain=0:2.2,dotted,samples=400]
plot ({\t} ,{1}) (-0.1,1) node{1} ;
\draw[variable=\t,domain=0:2,thick,dashdotted,samples=400]
plot ({\t},{\t/(pow(1+pow(\t,10),1/10))}) ;
\draw[variable=\t,domain=0:2,thick,samples=400]
plot ({\t} ,{(\t)/(1+\t)} ) ;
\draw[variable=\t,domain=0:2,thick,dashed,samples=400]
plot ({\t} ,{(\t)/((1+(\t)^(2))^(1/2))} ) ;
\end{tikzpicture}
}
\begin{theorem}\label{T4.1}
For some strictly convex $\phi\in \mathcal{C}^2(\R_+; \R_+)$, let $\bG$ satisfy \eqref{A1}--\eqref{A4} with $p=1$. Assume that the data satisfy the following hypotheses:
\begin{equation}
\begin{split}
\bu_0&\in W^{1,\infty}(0,T; W^{1,2}(\Omega;\R^d))\cap W^{2,1}(0,T;L^2(\Omega;\R^d)),\\
\bef&\in L^{2}(0,T; L^2(\Omega;\R^d)),
\end{split}\label{data-as2}
\end{equation}
with the safety strain condition
\begin{equation}\label{compt}
\|\alpha \beps(\bu_0) + \beta\partial_t \beps(\bu_0)\|_{L^{\infty}(Q; \Rsym)}< L
\end{equation}
and the following bound holds for every \( \delta>0 \):
\begin{equation}\label{compt2}
\esssup_{(t,x)\in (\delta,T)\times \Omega} |\partial_{tt} \beps (\bu_0(t,x))|< \infty.
\end{equation}
Then, there exists a unique couple $(\bu,\bT)$ fulfilling
\begin{align}
\bu &\in W^{1,\infty}(0, T; W^{1, 2}(\Omega;\R^d))\cap \mathcal{C}^1([0,T]; L^2(\Omega;\mathbb{R}^d))\cap W^{2,{2}}_{loc}(0, T; L^2(\Omega;\R^d) ), \label{FSu2}\\
\beps(\bu) &\in L^{\infty}(Q;\Rsym), \label{FSe2}\\
\partial_t\beps(\bu) &\in L^{\infty}(Q;\Rsym), \label{FSe22}\\
\bT &\in L^{1}(0, T; L^{1}(\Omega;\Rsym))\label{FST2}
\end{align}
and satisfying
\begin{align}\label{WF2}
\int_{\Omega}\partial_{tt}\bu \cdot \bw + \bT \cdot \nabla \bw\dx &= \int_{\Omega}\bef \cdot \bw \dx &&\forall \, \bw \in W_{0}^{1,\infty}(\Omega;\R^d),\quad \text{for a.e. }\,t \in (0, T),\\
\label{T-const2}
\alpha \beps(\bu) + \beta \partial_{t} \beps(\bu)&=\bG(\bT) &&\textrm{a.e. in }Q,
\end{align}
and
\begin{equation}
\bu-\bu_0 =\b0 \quad \textrm{ a.e. on }\Gamma\qquad \textrm{and}\qquad \bu(0)-\bu_0(0)=\partial_t \bu(0)-\partial_t\bu_0(0)=\b0 \quad \textrm{ a.e. in }\Omega. \label{bcint2}
\end{equation}
\end{theorem}
This theorem answers the question of existence of weak solutions to the problem under the assumptions \eqref{A1}--\eqref{A3} when $p\to 1_+$ and therefore provides an existence result for limiting strain models for which the symmetric displacement gradient and symmetric velocity gradient remain bounded; see Section~\ref{physics} for the physical background and the importance of the model. We note that a very similar existence result was established recently in~\cite{BuPaSeSu20}; there are however certain essential differences, which make the results of the present paper much stronger. First, in~\cite{BuPaSeSu20} the authors only consider the prototypical model
\begin{align}\label{A4p}
\bG(\bT):=\frac{\bT}{(1+|\bT|^q)^{\frac{1}{q}}},
\end{align}
while we are able to cover here a more general class of models under hypothesis \eqref{A4}. The corresponding potential $\phi$ (whose existence is assumed in \eqref{A4}) is, for the model \eqref{A4p}, given by
$$
\phi(s):=\int_0^s \frac{t}{(1+t^q)^{\frac{1}{q}}}\dt,\qquad s \in \mathbb{R}_+.
$$
The role of the parameter $q$ in \eqref{A4p} is indicated in Fig.~\ref{Fig1}.
\begin{figure}[h]
\centering
\deltaone{}
\caption{Dependence of $|\bG|$ on $|\bT|$ for the prototype model~\eqref{A4p}. The three curves correspond to $q=1$ (solid curve), $q=2$ (dashed curve) and $q=10$ (dash-dotted curve). Clearly, $|\bT|$ tends to $1$ more rapidly with increasing~$q$.}
\label{Fig1}
\end{figure}
Second, the paper~\cite{BuPaSeSu20} is concerned with the spatially periodic setting, which simplifies the analysis in an essential way, most notably with regard to the derivation of the relevant a~priori estimates. Finally, in~\cite{BuPaSeSu20} the initial data are assumed to be quite regular (they are supposed to belong to the Sobolev space $W^{k,2}(\Omega;\mathbb{R}^d)$ with $k>\frac{d}{2}$), which is related to the choice of the method used therein to prove the existence of a weak solution. In this paper we do not require such strong regularity of the initial data. Nevertheless, since in our context here it is difficult to describe the correct space-time trace spaces, because we are dealing with $L^{\infty}$-type spaces and symmetric gradients, and since we want to state the result in its full generality (so as to be able to admit time-dependent boundary data), we do assume a certain compatibility condition via an a~priori prescribed space-time function $\bu_0$ that we use in order to impose the initial and boundary conditions. Indeed, the existence of $\bu_0$ satisfying the safety strain condition \eqref{compt} is necessary for the existence of a solution. Next, the assumption \eqref{data-as2}$_1$, requiring temporal regularity of $\bu_0$, is required in order to ensure that $\bu_0$ and $\partial_t \bu_0$ have meaningful traces at time $t=0$. Finally, the assumption~\eqref{compt2} prescribes the required temporal smoothness of the boundary data, but since it only involves $t \in (\delta, T)$ with $\delta>0$, it has nothing to do with either the regularity of the initial condition or its compatibility with the boundary data. We give several examples for simplified settings in the following remark.
\begin{remark}
We discuss two cases of boundary and initial data from \eqref{init-data}--\eqref{bound-data} for which it is easy to construct a function $\bu_0$ that satisfies the assumptions \eqref{data-as2}--\eqref{compt2}.
\bigskip
\noindent \textsf{Boundary data independent of time.} Suppose that $\bu_{\Gamma}$ is independent of time and $\bu_I\in W^{1,2}(\Omega;\R^d)$ satisfies the compatibility condition $\bu_I|_{\partial \Omega}=\bu_{\Gamma}$. Because the
boundary data are independent of time, it is natural to assume that $\bv_0 \in W^{1,2}_0(\Omega; \R^d)$, where
\begin{equation}\label{case1}
\|\alpha \beps(\bu_I) +\beta \beps(\bv_0)\|_{L^\infty(\Omega;\mathbb{R}^{d\times d}_{sym})} <L.
\end{equation}
Then we can set
$$
\bu_0(t,x):=\mathrm{e}^{-\frac{\alpha t}{\beta}}\bu_I(x) +\frac{\alpha \bu_I(x) + \beta \bv_0(x)}{\alpha}(1-\mathrm{e}^{-\frac{\alpha t}{\beta}}).
$$
Indeed, a direct computation yields that
$$
\partial_t \bu_0(t,x)= \bv_0(x)\,\mathrm{e}^{-\frac{\alpha t}{\beta}},
$$
and thus, $\bu_0(0,x)=\bu_I(x)$, $\partial_t \bu_0(0,x)=\bv_0(x)$ for $x \in \Omega$ and $\bu_0|_{\Gamma} = \bu_{\Gamma}$. Moreover,
$$
\alpha \beps(\bu_0) + \beta \partial_t \beps(\bu_0)= \alpha \beps(\bu_I) + \beta \beps(\bv_0).
$$
Consequently, $\bu_0$ satisfies \eqref{compt} provided \eqref{case1} holds. The validity of \eqref{compt2} is obvious.
\bigskip
\noindent \textsf{Time-dependent boundary data.} In this setting, we a~priori assume the existence of some $\tilde{\bu}$ such that $\tilde{\bu}(0,x)=\bu_I(x)$ for $x \in \Omega$ and $\tilde{\bu}|_{\Gamma}=\bu_{\Gamma}$. In addition, it is natural to assume the compatibility condition $\bv_0(\cdot)=\partial_t \bu_{\Gamma}(0,\cdot)$ on $\partial \Omega$. We adopt the following assumption on $\tilde{\bu}$ and $\bv_0$:
\begin{equation}\label{case2}
\|\alpha \beps(\tilde{\bu})+ \beta(\partial_t\beps(\tilde{\bu})-\partial_t \beps(\tilde{\bu}(0,\cdot)) +\beps(\bv_0(\cdot)))\|_{L^\infty(Q;\mathbb{R}^{d \times d}_{sym})} <L.
\end{equation}
We define
$$
\bu_0(t,x):= \tilde{\bu}(t,x)+\frac{\beta (\bv_0(x)-\partial_t \tilde{\bu}(0,x))}{\alpha}(1-\mathrm{e}^{-\frac{\alpha t}{\beta}}).
$$
Clearly, $\bu_0(0,x)=\tilde{\bu}(0,x)=\bu_I(x)$ for $x \in \Omega$ and $\bu_0=\bu_{\Gamma}$ on $\Gamma$. The time derivative of $\bu_0$ is
$$
\partial_t \bu_0(t,x)= \partial_t\tilde{\bu}(t,x)+(\bv_0(x)-\partial_t \tilde{\bu}(0,x))\,\mathrm{e}^{-\frac{\alpha t}{\beta}}
$$
and thus $\partial_t \bu_0(0,x)= \bv_0(x)$ for $x \in \Omega$. In addition, since
$$
\alpha \beps(\bu_0) + \beta \partial_t \beps (\bu_0)=\alpha \beps(\bu_I)+ \beta(\partial_t\beps(\bu_I)-\partial_t \beps(\bu_I(0)) +\beps(\bv_0)),
$$
we see that \eqref{compt} is equivalent to \eqref{case2}. The assumption \eqref{compt2} is then only related to our extension of the boundary data inside of $\Omega$ and the temporal regularity of the boundary data.
\end{remark}
\subsection{Relevance to the modelling of viscoelastic solids}\label{physics}
With these results in mind, we will now discuss the importance of such problems.
We often encounter materials exhibiting viscoelastic response. By definition, viscoelasticity involves the material response of both elastic solids and viscous fluids, which can be modelled linearly or nonlinearly (see \cite{Sengul-visco-review} for an extensive overview). On the other hand, it is well-known that implicit constitutive theories allow for a more general structure in modelling than explicit ones (cf. \cite{Raj-03}, \cite{Raj-06}), where the strain could be given as a function of the stress. Indeed, this is the case in our constitutive relation \eqref{cons-law} in system \eqref{model}. Rajagopal's main contribution \cite{Raj-10} to the theory was to show that a nonlinear relationship between the stress and the strain can be obtained after linearizing the strain. The relation \eqref{cons-law} was first obtained by Erbay and \c{S}eng\"{u}l in \cite{Erbay-Sengul} as a result of the linearization of the relation between the stress and the strain tensors under the assumption that the magnitude of the strain is small. For models of this type it is possible that once the magnitude of the strain has reached a certain limiting value (as is the case in Theorem \ref{T4.1}), any further increase of the magnitude of the stress will cause no changes in the strain. These models are called \textit{strain-limiting (strain-locking) models} and such behaviour has been observed in numerous experiments (see \cite{Sengul-strain-lim-review} and references therein). For a further discussion of such models in the purely elastic setting or in the setting of the generalized Kelvin--Voigt model we refer to \cite{BuMaRa12}, and in the viscoelastic setting to \cite{Erbay-Sengul, Sengul-strain-lim-review}.
Now we introduce some basic kinematics in order to discuss these limiting strain models from a mathematical perspective.
We denote by $\bu(\mathbf{X}, t):=\bx(\mathbf{X},t)-\mathbf{X}$ the displacement of a given body at a space-time point $(\mathbf{X},t)$, where $\mathbf{X}$ is the position vector in the reference configuration and
$\bx(\mathbf{X},t)$ is the position vector in the current configuration. We denote the deformation of the body, which is assumed to be stress-free initially, by $\boldsymbol{\chi}(\mathbf{X}, t)$. The deformation gradient
is defined as $\mathbf{F} = \partial \boldsymbol{\chi}/ \partial \mathbf{X} $. By the polar decomposition theorem, we can ensure the existence of positive definite, symmetric tensors $\bU$, $\bV$, and a rotation $\bR$ such that
\[\bF= \bR \bU = \bV \bR,\]
where $\bU$ and $\bV$ are the \textit{right} and \textit{left Cauchy--Green stretch tensor}, respectively.
Moreover, we know that each of these decompositions is unique and
\[\bC = {\bU}^{2} = {\bF}^{\rm T} \bF, \quad \bB = {\bV}^{2} = \bF {\bF}^{\rm T},\]
where $\bB$, $\bC$ are called the \textit{right} and \textit{left Cauchy--Green deformation tensor}, respectively. We define the velocity as $\bv = \partial \boldsymbol{\chi}/ \partial t$ and denote by $\bD$ the symmetric part of the gradient of the velocity field $\bL = \partial \bv / \partial \bx $.
Under the assumption that
\begin{equation}\label{smallness}
\| \nabla \bu \|_{L^\infty(Q;\mathbb{R}^{d\times d})} = O(\delta), \qquad 0<\delta \ll 1,
\end{equation}
one can obtain the linearized strain, mentioned previously, as
\begin{equation}\label{lin-strain}
\beps (\bu)= \frac{1}{2} \left[\nabla \bu + (\nabla \bu)^{\rm T}\right].
\end{equation}
As is explained in \cite{Sengul-strain-lim-review}, in the purely elastic setting, starting from the following constitutive relation between the stress and the strain
\begin{equation}\label{implicit}
\bG(\bT, \bB) = \mathbf{0},
\end{equation}
for frame-indifferent and isotropic bodies, one can obtain the representation
\begin{equation}\label{representation}
\begin{split}
\bG(\bT, \bB) & = \chi_{0} \bI + \chi_{1} \bT + \chi_{2} \bT + \chi_{3} {\bT}^{2} + \chi_{4} \bB^{2} + \chi_{5} (\bT \bB + \bB \bT) \\
& \quad + \chi_{6} (\bT^{2} \bB + \bB \bT^{2}) + \chi_{7} (\bB^{2} \bT + \bT \bB^{2}) + \chi_{8} (\bT^{2} \bB^{2} + \bB^{2} \bT^{2}),
\end{split}
\end{equation}
where the functions $\chi_i$, $i=0,\dots,8$, depend only on the scalar invariants of $\bT$ and $\bB$, which can be expressed in terms of
\[\mathrm{tr}\,{\bT}, \mathrm{tr}\,{\bB}, \mathrm{tr}\,{\bT^{2}}, \mathrm{tr}\,{\bB^{2}}, \mathrm{tr}\,{\bT^{3}}, \mathrm{tr}\,{\bB^{3}}, \mathrm{tr}\,{\bT \bB}, \mathrm{tr}\,{\bT^{2}} \bB, \mathrm{tr}\,{\bT \bB^{2}}, \mathrm{tr}\,{\bT^{2} \bB^{2}}.\]
Under the smallness assumption (\ref{smallness}), we have that \( |\bB - (\bI + \beps)|= O(\delta^2) \), with $\beps=\beps(\bu)$. Thus, at the end of the linearization process, \eqref{representation} gives a nonlinear relationship between \( \bT \) and \( \beps\).
In many studies a simpler subclass of constitutive relations than \eqref{representation} is considered, namely
\begin{equation}\label{B-T}
\bB = \tilde{\chi}_{0} \bI + \tilde{\chi}_{1} \bT + \tilde{\chi}_{2} \bT^{2}.
\end{equation}
Under the assumption \eqref{smallness}, the equality \eqref{B-T} becomes
\begin{equation}\label{quad-T}
\beps = \bar{\chi}_{0} \bI+ \bar{\chi}_{1} \bT + \bar{\chi}_{2} \bT^{2},
\end{equation}
with some invariant-dependent coefficients $\bar{\chi}_i$, $i=0,1,2$. The analysis of a limiting strain problem with a constitutive relation of the form \( \beps = \bG(\bT) \), which is a more general version of \eqref{quad-T}, with a bounded mapping \( \bG \), as those considered here, was also studied in \citep{BuMaRaSu}, \cite{BeBuMaSu17}, where the authors highlight the analytical difficulties associated with such models, most notably the lack of weak compactness of approximations to the stress tensor in \( L^1(\Omega;\mathbb{R}^{d \times d}_{sym})\). We rely on methods developed in \cite{BeBuMaSu17} in order to show that (\ref{WF2}) holds for our proposed solution of the problem. The additional time-dependence here presents further difficulties in the analysis. In particular, we must develop suitable space-time estimates.
Next we focus on the derivation of the constitutive relation of interest in this paper.
In the viscoelastic setting, as is explained in \cite{Erbay-Sengul}, instead of \eqref{implicit} one would start with a general implicit constitutive relation of the form
\begin{equation}\label{implicit-visco}
\bG(\bT, \bB, \bD) = \mathbf{0}.
\end{equation}
For simplicity and in view of (\ref{B-T}), we study the following subclass of such models:
\begin{equation}\label{visco-cons}
\alpha \bB + \beta \bD = \gamma_{0} \bI + \gamma_{1} \bT + \gamma_{2} \bT^{2},
\end{equation}
where $\gamma_{i} = \gamma_{i}(I_{1}, I_{2}, I_{3})$, $i = 0, 1, 2,$ $I_{1} = \text{tr} \bT, I_{2} = \frac{1}{2} \text{tr} \bT^{2}, I_{3} = \frac{1}{3} \text{tr} \bT^{3},$ and $\alpha$, $\beta$ are nonnegative constants. We note that under assumption (\ref{smallness}) we may interchange derivatives with respect to \( \bx \) and \( \mathbf{X}\). In particular, the linearized counterpart of \( \mathbf{D}\) can be identified with \( \beps_t = \beps(\bu_t) \). Therefore, assuming \eqref{smallness} and writing the right-hand side of \eqref{visco-cons} more generally as a nonlinear function of $\bT$, one obtains \eqref{cons-law} as required.
Models of the type (\ref{visco-cons}) were considered in \cite{RaSa14} in order to describe viscoelastic solid bodies. The model is a generalization of the classical (linear) Kelvin--Voigt model, which in one space dimension involves the constitutive relation
\begin{equation}\label{K-V-linear}
\sigma = E\epsilon + \eta\epsilon_t,
\end{equation}
where \(\sigma\) denotes the scalar stress, \( \epsilon\) the scalar strain, and \( E\), \( \eta\) are constants signifying the modulus of elasticity and the viscosity, respectively. As mentioned before, it is worth noting that similar models have been considered in \cite{BuMaRa12,BuKaSt13}, where the authors assumed that the stress \( \bT \) was a sum of the elastic \( \bT_{el} \) and viscous \( \bT_{vis}\) parts. Considering implicit relations for each component separately, they obtained \( \bT_{el} = \tens{H}(\beps)\), \( \bT_{vis} = \tens{G}(\beps_t)\) for nonlinear mappings \( \bH\), \( \bG\). However, the assumptions that were made there on \( \tens{H}\) and \( \tens{G}\) result in a problem that is no longer of strain-limiting type. This, together with the additive decomposition of the stress considered there, led to an analysis that is very different from the one performed here.
Some analysis (albeit limited) of the problem (\ref{model}) is available in the literature, which we now discuss. In one space dimension the authors of \cite{Erbay-Sengul} derived the equation
\begin{equation}\label{strain-rate}
\sigma_{xx} + \beta \sigma_{xxt} = g(\sigma)_{tt},
\end{equation}
using the equation of motion \eqref{linear-moment} together with the constitutive relation \eqref{cons-law} and setting $\alpha = 1$, where, as in (\ref{K-V-linear}), \( \sigma\) refers to the scalar stress. In (\ref{strain-rate}), the nonlinearity $g$ corresponds to $\bG$ in the current case.
The authors investigated conditions on the function $g$ under which travelling wave solutions exist. Furthermore, in \cite{ErErSe} the authors proved the local-in-time existence of solutions for equation \eqref{strain-rate}.
In this work, we use the same set of equations without deriving a single equation on account of the fact that we are working in a higher-dimensional setting. In particular, the symmetric gradient does not reduce to a classical gradient operator as in the one-dimensional case, a property that is exploited in \cite{Erbay-Sengul} and \cite{ErErSe}.
A related problem is studied in \cite{ErSe20} where the authors looked at the stress-rate case instead of the strain-rate case. In the one-dimensional setting, this resulted in the equation
\begin{equation}\label{stress-rate}
\sigma_{xx} + \gamma \sigma_{ttt} = h(\sigma)_{tt}.
\end{equation}
The constitutive law for that study was $\epsilon + \gamma \sigma_{t} = h(\sigma)$ instead of \eqref{cons-law}. The authors pointed out that travelling wave solutions of equations \eqref{strain-rate} and \eqref{stress-rate} will coincide. However, we do not attempt to explore the stress-rate problem here.
We close this section with a thermodynamical justification of the model (\ref{model}).
We will show in particular that the total energy of the system is constant and the sum of the kinetic energy and the elastic energy is a decreasing function of time.
We suppose that the constitutive relation can be written as
\begin{align*}
\beps + \beta\partial_t\beps = \frac{\partial \varphi}{\partial\bT}(\bT) =: \bG(\bT)
\end{align*}
where \( \varphi\) is a function from \( \mathbb{R}^{d\times d}\) to \( \mathbb{R}_+\) defined by \( \varphi(\bT) = \phi(|\bT|)\) and $\beps=\beps(\bu)$.
We shall suppose that \( \phi({0}) = \phi'(0) = 0\) and assume that \( \phi \in \mathcal{C}^2(\mathbb{R}_+;\mathbb{R}_+) \) is strictly convex.
Clearly this is the case if (\ref{A4}) holds.
Under these assumptions, \( \varphi\) is also strictly convex, noting that \( \phi \) is strictly increasing on \( [0, \infty ) \). Furthermore \( \bG\) is monotone. Next, we define the convex conjugate \( \varphi^*\) by
\[
\varphi^*(\beps) = \sup_{\bT \in \mathbb{R}^{d\times d}_{sym}} \big( \, \beps\cdot \bT - \varphi(\bT)\big).
\]
We note that \( \varphi^*\) is also convex and, for any \( \bT \in \mathbb{R}^{d\times d}_{sym}\), the following identity holds:
\[
\varphi^*(\bG(\bT)) + \varphi(\bT) = \bG(\bT) \cdot \bT.
\]
Thus, the function \( \bG^{-1} = \frac{\partial\varphi^*}{\partial\bT}\) is also monotone. With these facts in mind, formally testing (\ref{linear-moment}) against \( \partial_t \bu \) and assuming the absence of all body forces, we obtain
\begin{equation}\label{energy-1}
\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int_{\Omega} |\partial_t \bu|^2 \,\mathrm{d}x + \int_\Omega \bT \cdot \partial_t \beps(\bu) \,\mathrm{d}x = 0.
\end{equation}
However, the integrand in the second term on the right-hand side can be rewritten as
\begin{align*}
\bT \cdot \partial_t \beps &= \frac{\partial\varphi^*}{\partial\bT}\cdot \partial_t \beps + \left( \bT - \frac{\partial\varphi^*}{\partial\bT}(\beps) \right)\cdot \partial_t \beps
\\
&= \partial_t (\varphi^*(\beps)) + \frac{1}{\beta} \left( \bT - \frac{\partial\varphi^*}{\partial\bT}(\beps) \right)\cdot ( \bG(\bT) - \beps)
\\
&= \partial_t (\varphi^*(\beps)) + \frac{1}{\beta} \left( \bT - \bG^{-1}(\beps) \right)\cdot ( \bG(\bT) - \beps).
\end{align*}
Substituting this back into (\ref{energy-1}) and defining \( \bT_0 := \bG^{-1}(\beps)\), we see that
\begin{equation}\label{energy-2}
\frac{\mathrm{d}}{\mathrm{d}t }\left( \int_\Omega \frac{1}{2}|\partial_t \bu|^2 + \varphi^*(\beps) \,
\mathrm{d}x\right) + \frac{1}{\beta}\int_\Omega\left( \bT - \bT_0\right)\cdot ( \bG(\bT) - \bG(\bT_0))\,\mathrm{d}x =0.
\end{equation}
Recalling that \( \bG \) is monotone, we deduce that
\begin{align*}
\sup_{t\in (0,T)} \left(\int_\Omega \frac{1}{2}|\partial_t \bu|^2 + \varphi^*(\beps) \,
\mathrm{d}x\right) \leq \int_\Omega \frac{1}{2}|\bv_0|^2 + \varphi^*(\beps(\bu_I))\,\mathrm{d}x.
\end{align*}
In particular the sum of the kinetic energy and elastic energy is decreasing. The extra term that appears in (\ref{energy-2}) corresponds to the dissipation; thus energy is conserved in accordance with the laws of thermodynamics.
The structure of the remainder of the paper is as follows. In Section \ref{Thm1} we prove Theorem \ref{T1}. We structure the proof in the following way. First, in Section \ref{galerkin} we use a Galerkin method and find a weak solution to an approximate problem. In Section \ref{s:uni-est}, we obtain uniform bounds on the sequence of Galerkin solutions, and use these in Section \ref{s:limit} in order to take the limit as $n \to \infty$. Finally, we show that the limit is the correct one in Section \ref{s:id-non}. Uniqueness is then proved in Section \ref{s:un}. In Section \ref{regularity} we obtain further temporal and spatial regularity estimates for these solutions. Finally, in Section \ref{limiting} we look at the case when $p=1$ and give the proof of Theorem~\ref{T4.1}.
\section{Proof of Theorem~\ref{T1}}\label{Thm1}
To prove the existence of a weak solution, we use a compactness argument based on a sequence of Galerkin approximations. However, since $\bG$ is not invertible in general, we introduce the following regularization:
$$
\bG_n(\bT):= \bG(\bT) + n^{-1} |\bT|^{p-2}\,\bT.
$$
Note that for all $n\in \mathbb{N}$, the regularized mapping still satisfies \eqref{A1}--\eqref{A3} (with $C_2$ replaced by $(C_2+1)$) and in addition the inequality \eqref{A1} is strict whenever $\bT\neq \bW$. Therefore, it directly follows from the theory of monotone operators that there exists a continuous inverse $\bG_n^{-1}:\Rsym \to \Rsym$.
\subsection{Galerkin approximation}\label{galerkin}
Let $\{\bom_{j}\}_{j=1}^{\infty}$ be a basis of $W_{0}^{2d, 2}(\Omega;\R^d)$, which is orthonormal in $L^{2}(\Omega; \R^d)$.\footnote{Such a basis can be found by looking for eigenfunctions $\bom_j \in W_{0}^{2d, 2}(\Omega;\R^d)$ of the problem
$$
-\Delta^{2d} \bom_j = \lambda_j \bom_j\qquad \mbox{on $\Omega$.}$$
} We denote by $P^n$ the projection of $W^{2d,2}_0(\Omega;\R^d)$ onto the linear hull of $\{\bom_{j}\}_{j=1}^{n}$, which is continuous.
We look for $\bu^{n}$ of the form
\[\bu^{n}(t, x) = \bu_0(t,x) + \sum_{i=1}^{n} C_{i}^{n}(t) \bom_{i}(x),\]
by solving, for all $j = 1, 2, \ldots, n$ and almost all $t\in (0,T)$, the following problem:
\begin{subequations}\label{Gn}
\begin{align}
\int_{\Omega} \partial^2_{tt} \bu^{n} \cdot \omega_{j} +\bG_n^{-1}\left(\alpha \beps(\bu^{n}) + \beta \partial_{t} \beps(\bu^{n})\right) \cdot \nabla \bom_{j} \dx &= \langle \bef, \bom_j\rangle, \label{G1} \\
\bu^{n}(0) &= \bu_{0}(0),\label{G2}\\
\partial_{t} \bu^{n}(0) &= \partial_t \bu_0(0). \label{G3}
\end{align}
\end{subequations}
We note that \eqref{G2} and \eqref{G3} are equivalent to $\vec{C}^n(0)=\mathbf{0}$ and $\partial_t \vec{C}^n(0)=\mathbf{0}$, respectively.
Since $\bG_n^{-1}$ is continuous and the basis functions $\{\bom_{j}\}_{j=1}^\infty $ are orthonormal in $L^2(\Omega;\mathbb{R}^d)$, the equation \eqref{G1} reduces to
\[\partial_{tt} C^{n}_{i}(t) = F_i(t,\vec{C}^{n}(t), \partial_{t} \vec{C}^{n}(t) ),\]
where $F_i$ are Carath\'{e}odory mappings. Hence, using standard Carath\'{e}odory theory for a system of ordinary differential equations, we deduce that there exists a solution on some maximal time interval $(0, T^{*})$. Furthermore, either we must have $|\vec{C}^{n}(t)| + |\partial_{t}\vec{C}^{n}(t)| \to \infty$ as $t \to T^{*}_{-}$ or we can extend the solution to the whole interval $(0, T)$. We shall next show that the latter is true by establishing uniform bounds on the sequence of Galerkin approximations.
\subsection{Uniform bounds}\label{s:uni-est}
First, let us define
\[
\bT^n:=\bG_n^{-1}\left(\alpha \beps(\bu^{n}) + \beta \partial_{t} \beps(\bu^{n})\right),
\]
which is clearly equivalent to
\begin{equation}\label{Tn}
\alpha \beps(\bu^{n}) + \beta \partial_{t} \beps(\bu^{n})=\bG(\bT^n) + n^{-1} |\bT^n|^{p-2}\bT^n.
\end{equation}
Then, we multiply \eqref{G1} by $C^n_j$ and also by $\partial_t C^n_j$ and sum the resulting identities with respect to $j=1,\ldots, n$ to obtain
\begin{equation}\label{test1}
\begin{split}
\int_{\Omega} \partial_{tt} \bu^n \cdot \partial_t (\bu^n - \bu_0)+ \bT^n \cdot \partial_t \beps(\bu^n-\bu_0)\dx &= \langle \bef, \partial_t (\bu^n-\bu_0)\rangle, \\
\int_{\Omega} \partial_{tt}\bu^n \cdot (\bu^n-\bu_0)+ \bT^n \cdot \beps(\bu^n-\bu_0)\dx &= \langle \bef, (\bu^n - \bu_0)\rangle.
\end{split}
\end{equation}
Next, it follows from \eqref{Tn} that
\[\bT^n \cdot \partial_t \beps(\bu^n) = \frac{1}{\beta} \left(\bG(\bT^n) \cdot \bT^n + n^{-1}|\bT^n|^p - \alpha \, \bT^n \cdot \beps(\bu^n)\right).\]
Also, we can write
\[ \int_{\Omega} \partial_{tt}(\bu^n-\bu_0) \cdot (\bu^n-\bu_0) \dx =\ddt \int_{\Omega} \partial_{t}(\bu^n-\bu_0) \cdot (\bu^n-\bu_0)\dx - \int_{\Omega} | \partial_{t} (\bu^n-\bu_0)|^{2}\dx.\]
Using these two identities in \eqref{test1}, we obtain
\begin{equation}\label{test2}
\begin{split}
&\frac{1}{2} \ddt\int_{\Omega}|\partial_{t} (\bu^n-\bu_0)|^2 \dx + \frac{1}{\beta} \int_{\Omega} \bG(\bT^n)\cdot \bT^n + n^{-1}|\bT^n|^p \dx \\
&= \frac{\alpha}{\beta} \int_{\Omega} \bT^n \cdot \beps(\bu^n)\dx + \langle \bef, \partial_t (\bu^n-\bu_0)\rangle +\int_{\Omega} \bT^n\cdot \partial_t \beps(\bu_0) - \partial_{tt} \bu_0 \cdot \partial_t(\bu^n-\bu_0)\dx,
\end{split}
\end{equation}
and
\begin{equation}\label{test2b}
\begin{split}
&\ddt\frac{\alpha}{\beta} \int_{\Omega} \partial_{t}(\bu^n-\bu_0)\cdot (\bu^n-\bu_0)\dx + \frac{\alpha}{\beta} \int_{\Omega} \bT^n \cdot \beps(\bu^n)\dx \\
&= \frac{\alpha}{\beta} \int_{\Omega} |\partial_{t} (\bu^n-\bu_0)|^{2}-\partial_{tt}\bu_0 \cdot (\bu^n-\bu_0) +\bT^n \cdot \beps(\bu_0)\dx +\frac{\alpha}{\beta}\langle \bef, (\bu^n - \bu_0)\rangle.
\end{split}
\end{equation}
By summing these equalities we find that one term cancels and we deduce that
\begin{equation}\label{test3}
\begin{split}
&\frac{1}{2} \ddt\int_{\Omega}|\partial_{t} (\bu^n-\bu_0)|^2 +\frac{2\alpha}{\beta} \partial_{t}(\bu^n-\bu_0)\cdot (\bu^n-\bu_0)\dx+ \frac{1}{\beta} \int_{\Omega} \bG(\bT^n)\cdot \bT^n + n^{-1}|\bT^n|^p \dx \\
&=\langle\bef, \partial_t (\bu^n-\bu_0)\rangle +\int_{\Omega} \bT^n\cdot \partial_t \beps(\bu_0) - \partial_{tt} \bu_0 \cdot \partial_t(\bu^n-\bu_0)\dx\\
&\quad +\frac{\alpha}{\beta} \int_{\Omega} |\partial_{t} (\bu^n-\bu_0)|^{2}-\partial_{tt}\bu_0 \cdot (\bu^n-\bu_0) +\bT^n \cdot \beps(\bu_0)\dx +\langle \bef, (\bu^n - \bu_0)\rangle.
\end{split}
\end{equation}
Next, we define on \( [0, T]\) the function
$$
Y^n:=\frac14\int_{\Omega}|\partial_{t}(\bu^n-\bu_0)|^{2} + |\bu^n-\bu_0|^{2} + \left|\partial_{t} (\bu^n-\bu_0) + \frac{2 \alpha}{\beta} (\bu^n-\bu_0)\right|^2\dx.
$$
Using this, we can rewrite the first term on the left-hand side of \eqref{test3} as
$$
\frac{1}{2} \ddt\int_{\Omega}|\partial_{t} (\bu^n-\bu_0)|^2 +\frac{2\alpha}{\beta} \partial_{t}(\bu^n-\bu_0)\cdot (\bu^n-\bu_0)\dx=Y^n - \left(\frac{\alpha^{2}}{\beta^{2}} + \frac{1}{4} \right)\ddt \int_{\Omega} |\bu^n-\bu_0|^2 \dx.
$$
Consequently, using this identity in \eqref{test3}, applying \eqref{A2} to the second term on the left-hand side, and the H\"{o}lder inequality to the terms on the right-hand side together with the Poincar\'e and Korn inequalities, it follows that
\begin{equation}
\begin{split}
&\ddt Y^n + \frac{C_1}{\beta} \int_{\Omega} |\bT^n|^p \dx \le C\left(1+ Y^n\right)+C(\|\beps(\bu_0)\|_{p'}+\|\partial_t \beps(\bu_0)\|_{p'})\|\bT^n\|_p\\
&+ C(\|\beps(\bu^n)\|_{p'}+\|\partial_t \beps(\bu^n)\|_{p'}+\|\beps(\bu_0)\|_{p'}+\|\partial_t \beps(\bu_0)\|_{p'})(\|\bef\|_{(W^{1,p'}_0)^*} +\|\partial_{tt} \bu_0\|_{(W^{1,p'}_0)^*}),
\end{split}\label{tt1}
\end{equation}
where $C$ is a generic constant that is independent of \( n \).
To bound the right-hand side, we use \eqref{Tn} to observe that
\[\partial_{t} \left(\mathrm{e}^{\frac{\alpha}{\beta} t} \beps(\bu^n)\right) = \frac{\mathrm{e}^{\frac{\alpha}{\beta} t}}{\beta} (\bG(\bT^n)+n^{-1}|\bT^n|^{p-2}\bT^n),\]
which, after integration with respect to time, gives
\[\beps(\bu^n(t))= \mathrm{e}^{-\frac{\alpha}{\beta} t} \beps(\bu_{0}(0)) + \mathrm{e}^{-\frac{\alpha}{\beta} t} \int_{0}^{t} \frac{\mathrm{e}^{\frac{\alpha}{\beta}\tau}}{\beta} (\bG(\bT^n(\tau) + n^{-1}|\bT^n(\tau)|^{p-2}\bT^n(\tau))\dtau.\]
Using properties of the Bochner integral, it follows that
\begin{equation}
\begin{split}
\|\beps(\bu^n(t))\|_{p'}^{p'} &\leq C \left(\int_{0}^{t} \| \bG(\bT^n) + n^{-1}|\bT^n|^{p-2}\bT^n\|_{p'}^{p'}\dtau + \| \bu_{0}(0) \|_{1,p'}^{p'} \right)\\
&\leq C \left(\int_{0}^{t} \|\bT^n\|_p^p\dtau + \| \bu_{0}(0) \|_{1,p'}^{p'}+1 \right), \label{tt3}
\end{split}
\end{equation}
where for the second inequality we have used \eqref{A3}. Consequently, using \eqref{tt3} and \eqref{Tn}, we have also the following bound on the time derivative:
\begin{equation}
\begin{split}
\|\partial_t \beps(\bu^n(t))\|_{p'}^{p'} &\leq C \left(1+ \| \bu_{0}(0) \|_{1,p'}^{p'}+ \|\bT^n(t)\|_p^p +\int_{0}^{t} \|\bT^n\|_p^p\dtau \right). \label{tt4}
\end{split}
\end{equation}
Hence, using \eqref{tt3} and \eqref{tt4} for the terms appearing on the right-hand side of \eqref{tt1}, and applying Young's inequality to the resulting right-hand side, we see that
\begin{equation}
\begin{split}
&\ddt \left(Y^n + \frac{C_1}{4\beta} \int_0^t \|\bT^n\|_p^p \dtau\right) + \frac{C_1}{4\beta} \|\bT^n\|_p^p \le C\left(Y^n + \frac{C_1}{4\beta} \int_0^t \|\bT^n\|_p^p \dtau\right)\\
&\qquad +C\sup_{t\in [0,T]}\|\bu_0(t)\|_{1,p'}^{p'} + C\left(\|\partial_t \beps(\bu_0)\|^{p'}_{p'}+\|\bef\|^p_{(W^{1,p'}_0)^*} +\|\partial_{tt} \bu_0\|^p_{(W^{1,p'}_0)^*}\right).
\end{split}\label{tfinal}
\end{equation}
Thus, using Gr\"{o}nwall's lemma and the assumptions on the data, we have that
\begin{equation}
\begin{split}
\sup_{t\in (0,T)} Y^n(t) + \int_0^T\|\bT^n\|_p^p \dtau\le C(\bu_0,\bef)+ Y^n(0)=C(\bu_0,\bef).
\end{split}\label{tfinali}
\end{equation}
Finally, from the definition of $Y^n$, the bounds \eqref{tt3}, \eqref{tt4}, and Korn's inequality, we deduce that
\begin{equation}
\sup_{t\in (0,T)} \left(\|\partial_t \bu^n\|_2^2+\|\bu^n\|_2^2 + \|\bu^n\|_{1,p'}^{p'}\right)+\int_0^T \|\bT^n\|_p^{p} + \|\partial_t \bu^n\|_{1,p'}^{p'} \dt \le C(\bu_0,\bef).\label{AE-n}
\end{equation}
It remains to provide a bound on $\partial_{tt}\bu^n$. We define the set $\mathcal{V}:=\{\bw \in W^{2d, 2}_{0}(\Omega;\R^d), \| \bw\|=1\}$. Using the orthonormality of the basis and the continuity of $P^n$, we deduce from \eqref{G1} that
\begin{equation*}
\begin{split}
\| \partial_{tt} \bu^n(t) \|_{(W^{2d, 2}_{0}(\Omega;\R^d))^{*}} &= \sup_{\bw \in \mathcal{V}} \int_{\Omega} \partial_{tt} \bu^{n}(t)\cdot \bw \dx \\
& = \sup_{\bw \in \mathcal{V}} \int_{\Omega} \partial_{tt} \bu^{n}(t) \cdot P^{n} \bw \dx \\
& = \sup_{\bw \in \mathcal{V}}\left(\langle \bef, \bw\rangle - \int_{\Omega} \bT^{n}(t) \cdot \nabla (P^{n} \bw) \dx\right)\\
& \leq \sup_{\bw \in \mathcal{V}} (\|\bef(t)\|_{(W^{1,p'}_0(\Omega;\R^d))^*}+\|\bT^{n} (t)\|_{p}) \| P^{n} \bw \|_{1,p'} \\
& \leq C\sup_{\bw \in \mathcal{V}}(\|\bef(t)\|_{(W^{1,p'}_0(\Omega;\R^d))^*}+\|\bT^{n} (t)\|_{p}) \|P^n\bw\|_{2d,2} \\
&\leq C (\|\bef(t)\|_{(W^{1,p'}_0(\Omega;\R^d))^*}+\|\bT^{n} (t)\|_{p}),
\end{split}
\end{equation*}
where we have used the fact that $W^{2d,2}(\Omega;\R^d)$ is continuously embedded into $W^{1,p'}(\Omega;\R^d)$.
Therefore, it follows from \eqref{AE-n} that
\begin{equation}\label{AE-2-n}
\int_{0}^{T} \| \partial_{tt} \bu^{n} \|^{p}_{(W^{2d, 2}_{0}(\Omega;\R^d))^{*}} \dt\le C\int_0^T \|\bef\|^p_{(W^{1,p'}_0(\Omega;\R^d))^*}+\|\bT^{n} \|_{p}^p \dt \le C(\bu_0,\bef).
\end{equation}
\subsection{Limit $n\to \infty$}\label{s:limit}
Using the bounds from Section \ref{s:uni-est} in conjunction with the reflexivity and separability of the underlying spaces, we can find a subsequence, that we do not relabel, such that
\begin{equation}\label{C1}
\begin{aligned}
\bG(\bT^{n}) & \rightharpoonup \bar{\bG} &&\text{weakly in }L^{p'}(0, T; L^{p'}(\Omega;\Rsym)), \\
\bu^{n} & \overset{*}{\rightharpoonup} \bu &&\text{weakly$^*$ in } W^{1, \infty}(0, T; L^{2}(\Omega;\R^d)), \\
\bu^{n} & \rightharpoonup \bu &&\text{weakly in } W^{1, p'}(0, T; W^{1, p'}(\Omega;\R^d)), \\
\bT^{n} & \rightharpoonup \bT &&\text{weakly in } L^{p}(0, T; L^{p}(\Omega;\Rsym)), \\
\partial_{tt} \bu^{n} & \rightharpoonup \partial_{tt} \bu &&\text{weakly in } L^{p}(0, T; (W_{0}^{2d, 2}(\Omega;\R^d))^{*}).
\end{aligned}
\end{equation}
Hence, we see that $\bT$ fulfills \eqref{FST} and $\bu$ belongs to the first two spaces indicated in \eqref{FSu}. In addition, thanks to the fact that $W^{1,p'}(\Omega;\R^d)$ is compactly embedded into $L^2(\Omega;\R^d)$, using the Aubin--Lions lemma (for a further subsequence, not indicated,) we even have that
\begin{equation}\label{C2}
\begin{aligned}
\bu^{n} & \to \bu &&\text{strongly in }\mathcal{C}([0,T];L^2(\Omega;\R^d), \\
\partial_{t} \bu^{n} & \to \partial_{t} \bu &&\text{strongly in }L^{2}(0, T; L^{2}(\Omega;\R^d))\cap \mathcal{C}([0,T]; (W^{2d,2}_0(\Omega; \R^d))^* ).
\end{aligned}
\end{equation}
Thus, it follows directly from the fact that $\bu^n(0)=\bu_0(0)$ and $\partial_t \bu^n(0)=\partial_t \bu_0(0)$ and the above convergence result (\ref{C2}) that
$$
\bu(0)=\bu_0 \quad \textrm{and} \quad \partial_t \bu(0)=\partial_t \bu_0(0).
$$
Next, we let $n\to \infty$ in \eqref{G1}. Let $\phi \in \mathcal{C}^{\infty}([0, T])$ be arbitrary. We multiply \eqref{G1} by $\phi$ and integrate the result over $(0, T)$ to get
\[\int_{0}^{T} \langle \partial_{tt} \bu^{n}, \bom_{j} \phi \rangle \dt + \int_{0}^{T} \int_{\Omega} \bT^{n} \cdot \nabla(\phi \bom_{j})\dx \dt = \int_0^T \langle \bef, \bom_j \rangle \dt ,\]
for every \( j\in \{ 1,\ldots, n \}\).
Thus, for a fixed $j$, we can let $n \to \infty$ and using the weak convergence result \eqref{C1} we deduce that
\[\int_{0}^{T} \langle \partial_{tt} \bu, \bom_{j} \phi \rangle \dt + \int_{0}^{T} \int_{\Omega} \bT \cdot \nabla(\phi \bom_{j})\dx \dt = \int_0^T \langle \bef, \bom_j \rangle \dt.\]
Since $j$ and $\phi$ were arbitrary and recalling that $\{\bom_{j}\}_{j=1}^{\infty}$ forms a basis of $W^{2d,2}_0(\Omega; \R^d)$, it follows that
\begin{equation}\label{WFe}
\langle \partial_{tt}\bu, \bw \rangle + \int_{\Omega} \bT \cdot \nabla \bw \dx = \langle \bef, \bw \rangle \qquad \forall\, \bw \in W_{0}^{2d,2}(\Omega;\R^d),\quad \text{for a.e. }\,t \in (0, T).
\end{equation}
Consequently, thanks to the density of $W^{2d,2}_0(\Omega; \R^d)$ in $W^{1,p'}_0(\Omega;\R^d)$, we see that for almost all $t\in (0,T)$ we have $\partial_{tt}\bu \in (W^{1,p'}_0(\Omega;\R^d))^*$. Furthermore, we have
$$
\|\partial_{tt}\bu^n(t)\|_{(W^{1,p'}_0(\Omega;\R^d))^*}=\sup_{\bw\in W^{1,p'}_0(\Omega;\R^d); \, \|\bw\|=1}\left[ -\int_{\Omega}\bT^n(t)\cdot \nabla \bw\dx +\langle \bef (t), \bw \rangle\right].
$$
Thus, using \eqref{AE-n} and \eqref{C1}, it follows that
\begin{equation}
\int_0^T \|\partial_{tt}\bu^n\|_{(W^{1,p'}_0(\Omega;\R^d))^*}^p \dt \le C\int_0^T \|\bT^n\|_p^p + \|\bef\|_{(W^{1,p'}_0(\Omega;\R^d))^*}^p \dt\le C(\bu_0,\bef).\label{time-u}
\end{equation}
Hence, \eqref{WFe} can be strengthened so that \eqref{WF} holds. In addition, by standard parabolic interpolation and the fact that $\partial_t \bu_0 \in \mathcal{C}([0,T]; L^2(\Omega; \R^d))$, we see that $\bu$ satisfies \eqref{FSu}.
Finally, letting $n\to \infty$ in \eqref{Tn} and using \eqref{C1}, we see that
\begin{equation}\label{T-constw}
\alpha \beps(\bu) + \beta \partial_{t} \beps(\bu)=\overline{\bG} \quad \textrm{a.e. in }Q.
\end{equation}
Hence, in order to show \eqref{T-const} and deduce the existence of a weak solution, it remains to show that $\overline{\bG}=\bG(\bT)$ a.e. in $Q$.
\subsection{Identification of the nonlinearity}\label{s:id-non}
In order to identify the nonlinearity, we will use monotone operator theory. Let $\phi\in \mathcal{C}^1_0([0,T])$ be an arbitrary nonnegative function. We multiply both identities in \eqref{test1} by $\phi$ and integrate the result over $(0,T)$. With the help of integration by parts and the fact that $\bu^n(0)=\bu_0(0)$ and $\phi(T)=0$, we observe that
\begin{equation}\label{id1}
\begin{split}
&\int_0^T\int_{\Omega} \bT^n \cdot \partial_t \beps(\bu^n) \phi\dx \dt=\int_0^T\int_{\Omega}\frac{|\partial_{t} (\bu^n-\bu_0)|^2 \phi'}{2}+ \bT^n\cdot \partial_t \beps(\bu_0)\phi\dx\dt \\
&\qquad+\int_0^T \langle \bef, \partial_t (\bu^n-\bu_0)\rangle\phi - \langle\partial_{tt} \bu_0, \partial_t(\bu^n-\bu_0)\rangle\phi \dt
\end{split}
\end{equation}
and
\begin{equation}\label{id2}
\begin{split}
\int_0^T\int_{\Omega} \bT^n \cdot \beps(\bu^n)\phi\dx \dt&=\int_0^T\int_{\Omega} \partial_{t}(\bu^n-\bu_0)\cdot (\bu^n-\bu_0)\phi'\dx\dt \\
&\quad+\int_0^T\int_{\Omega} |\partial_{t} (\bu^n-\bu_0)|^{2}\phi+\bT^n \cdot \beps(\bu_0)\phi\dx \dt \\
&\quad+\int_0^T\langle \bef, (\bu^n - \bu_0)\rangle\phi -\langle\partial_{tt}\bu_0, (\bu^n-\bu_0)\rangle \phi \dt.
\end{split}
\end{equation}
Next, we use the weak convergence results \eqref{C1} and the strong convergence results \eqref{C2} to identify the limits on the right-hand sides of \eqref{id1} and \eqref{id2}. In particular, we obtain
\begin{equation}\label{id1l}
\begin{split}
\lim_{n\to \infty}&\int_0^T\int_{\Omega} \bT^n \cdot \partial_t \beps(\bu^n) \phi\dx \dt=\int_0^T\int_{\Omega}\frac{|\partial_{t} (\bu-\bu_0)|^2 \phi'}{2}+\bT\cdot \partial_t \beps(\bu_0)\phi\dx\dt \\
& \qquad +\int_0^T \langle \bef, \partial_t (\bu-\bu_0)\rangle\phi - \langle\partial_{tt} \bu_0, \partial_t(\bu-\bu_0)\rangle\phi \dt
\end{split}
\end{equation}
and
\begin{equation}\label{id2l}
\begin{split}
\lim_{n\to \infty}\int_0^T\int_{\Omega} \bT^n \cdot \beps(\bu^n)\phi\dx \dt&=\int_0^T\int_{\Omega} \partial_{t}(\bu-\bu_0)\cdot (\bu-\bu_0)\phi'\dx\dt \\
&\quad+\int_0^T\int_{\Omega} |\partial_{t} (\bu-\bu_0)|^{2}\phi+\bT \cdot \beps(\bu_0)\phi\dx \dt \\
&\quad+\int_0^T\langle \bef, (\bu - \bu_0)\rangle\phi -\langle\partial_{tt}\bu_0, (\bu-\bu_0)\rangle \phi \dt.
\end{split}
\end{equation}
Next, we use \eqref{WF} to evaluate the terms on the right-hand sides of \eqref{id1l}, \eqref{id2l}.
We note that, thanks to the regularity of $\bu$, both
$\bu-\bu_0$ and $\partial_{t} (\bu-\bu_0)$
are admissible test functions in \eqref{WF}. Using these two choices as the test function $\bw$, multiplying each of the resulting equalities by $\phi$ and integrating over $(0, T)$, we may apply integration by parts in order to obtain the following identities:
\begin{equation}\label{id1lf}
\begin{split}
&\int_0^T\int_{\Omega} \bT \cdot \partial_t \beps(\bu) \phi\dx \dt=\int_0^T\int_{\Omega}\frac{|\partial_{t} (\bu-\bu_0)|^2 \phi'}{2}+\bT\cdot \partial_t \beps(\bu_0)\phi\dx\dt \\
& \qquad +\int_0^T \langle \bef, \partial_t (\bu-\bu_0)\rangle\phi - \langle\partial_{tt} \bu_0, \partial_t(\bu-\bu_0)\rangle\phi \dt
\end{split}
\end{equation}
and
\begin{equation}\label{id2lf}
\begin{split}
\int_0^T\int_{\Omega} \bT \cdot \beps(\bu)\phi\dx \dt&=\int_0^T\int_{\Omega} \partial_{t}(\bu-\bu_0)\cdot (\bu-\bu_0)\phi'\dx\dt \\
&\quad+\int_0^T\int_{\Omega} |\partial_{t} (\bu-\bu_0)|^{2}\phi+\bT \cdot \beps(\bu_0)\phi\dx \dt \\
&\quad+\int_0^T\langle \bef, (\bu - \bu_0)\rangle\phi -\langle\partial_{tt}\bu_0, (\bu-\bu_0)\rangle \phi \dt.
\end{split}
\end{equation}
Comparing \eqref{id1l} with \eqref{id1lf} and \eqref{id2l} with \eqref{id2lf}, we see that
\begin{equation}
\label{jk}
\limsup_{n\to \infty} \int_{Q} \phi\bT^{n} \cdot (\alpha \beps(\bu^{n}) + \beta \partial_t \beps(\bu^{n}))\dx \dt \leq \int_{Q} \phi\bT \cdot (\alpha \beps(\bu) + \beta \partial_t \beps(\bu))\dx \dt.
\end{equation}
Therefore, using the nonnegativity of $\phi$, we observe that
\begin{equation}\label{MN}
\begin{split}
\limsup_{n\to \infty}\int_{Q} \phi\bG(\bT^{n})\cdot \bT^{n}\dx \dt &\le\limsup_{n\to \infty}\int_{Q} \phi(\bG(\bT^{n})+n^{-1}|\bT^n|^{p-2}\bT^n)\cdot \bT^{n}\dx \dt \\
&\overset{\eqref{Tn}}=\limsup_{n\to \infty}\int_{Q} \phi\bT^n \cdot (\alpha \beps(\bu^{n}) + \beta \partial_t \beps(\bu^{n})) \dx \dt \\
&\overset{\eqref{jk}}\le \int_{Q} \phi\bT \cdot (\alpha \beps(\bu) + \beta \partial_t \beps(\bu)) \dx \dt\\
&\overset{\eqref{T-constw}}=\int_{Q} \phi\bT \cdot \overline{\bG} \dx \dt.
\end{split}
\end{equation}
The inequality (\ref{MN}) is the key to identifying the nonlinearity.
Let $\bW \in L^{p}(Q,\Rsym)$ be arbitrary. Using the monotonicity assumption \eqref{A1}, the weak convergence results \eqref{C1}, the bound \eqref{MN} and the nonnegativity of $\phi$, we obtain
\[0 \leq \limsup_{n\to \infty} \int_{Q} \phi \left(\bG(\bT^{n}) - \bG(\bW)\right)\cdot\left(\bT^{n} - \bW\right)\dx \dt \leq \int_{Q} \phi \left(\overline{\bG} - \bG(\bW)\right)\cdot (\bT - \bW)\dx \dt.\]
Setting $\bW = \bT - \kappa \bB$ for an arbitrary $\bB \in L^{p'}(Q;\Rsym)$ and $\kappa>0$, we divide through by $\kappa$ to deduce that
$$
0\le \int_{Q} \phi \left(\overline{\bG} - \bG(\bT - \kappa \bB)\right)\cdot \bB\dx \dt.
$$
Hence, since $\bG$ is continuous, we may let $\kappa \to 0_+$ to deduce that
$$
0\le \int_{Q} \phi \left(\overline{\bG} - \bG(\bT)\right)\cdot \bB\dx \dt.
$$
As $\bB$ and $\phi$ are arbitrary, we conclude that
\[\overline{\bG} = \bG(\bT) \quad \text{a.e. in }Q.\]
Thus we have proved the existence of a weak solution.
\subsection{Uniqueness of solution}\label{s:un}
To complete the proof of Theorem \ref{T1} it remains to show uniqueness of the weak solution.
To this end, let $(\bu_{1}, \bT_{1})$ and $(\bu_{2}, \bT_{2})$ be two weak solutions of (\ref{model}) emanating from the same data. We denote $\bu:=\bu_1-\bu_2$. Then, using \eqref{WF}, we see that
\[\langle \partial_{tt} \bu, \bw \rangle + \int_{\Omega} (\bT_{1} - \bT_{2})\cdot \beps(\bw)\dx = 0 \quad \forall\, \bw \in W_{0}^{1, p'}(\Omega;\R^d) \textrm{ and a.e. }t\in (0,T).\]
Since $\bu$ and $\partial_t \bu$ belong to $W^{1,p'}_0(\Omega;\R^d)$ for almost all $t\in (0,T)$, we can set $\bw=\bu$ and $\bw= \partial_{t} \bu$ in the above to deduce that, for almost all $t$, the following holds:
\begin{equation*}
\begin{split}
&\frac{1}{2} \ddt \| \partial_{t} \bu\|_{2}^{2} + \int_{\Omega} (\bT_{1} - \bT_{2}) \cdot \partial_t \beps(\bu) \dx = 0, \\
& \ddt \int_{\Omega} \partial_{t} \bu \cdot \bu \dx + \int_{\Omega} (\bT_{1} - \bT_{2}) \cdot \beps(\bu)\dx = \int_{\Omega}|\partial_{t} \bu |^2 \dx.
\end{split}
\end{equation*}
Therefore,
\[\ddt\left(\int_{\Omega} \frac{\beta}{2} | \partial_{t} \bu |^{2} + \alpha \partial_{t} \bu \cdot \bu \dx \right) + \int_{\Omega} (\bT_{1} - \bT_{2}) \cdot \left(\beta \partial_t\beps(\bu) + \alpha \beps(\bu)\right)\dx = \int_{\Omega} \alpha|\partial_{t} \bu|^2 \dx.\]
Using the same procedure as in the a~priori estimates and also the constitutive relation \eqref{T-const}, we obtain
\begin{equation*}
\begin{split}
& \frac{1}{4} \ddt \int_{\Omega} \beta |\partial_{t} \bu|^{2} + \beta |\bu|^{2} + \beta \left|\partial_{t} \bu + \frac{2 \alpha}{\beta} \bu \right|^{2}\dx + \int_{\Omega} (\bG(\bT_1)-\bG(\bT_2))\cdot (\bT_{1} - \bT_{2}) \,\mathrm{d}x\\
& =\int_{\Omega} \alpha|\partial_{t} \bu|^2 + \left(\beta+\frac{\alpha^2}{\beta}\right)|\bu|^2\dx
\\&
\le C(\alpha,\beta) \int_{\Omega} \beta |\partial_{t} \bu|^{2} + \beta |\bu|^{2} + \beta \left|\partial_{t} \bu + \frac{2 \alpha}{\beta} \bu \right|^{2}\dx.
\end{split}
\end{equation*}
The second term on the left-hand side is nonnegative thanks to \eqref{A1} so we may apply Gr\"{o}nwall's inequality. Since $\bu(0)=\partial_t \bu(0)=\mathbf{0}$, we deduce that $\bu = \mathbf{0}$ a.e. in $Q$. In addition, by monotonicity, we also obtain that $\big(\bG(\bT_{1}) - \bG(\bT_{2})\big)\cdot (\bT_{1} - \bT_{2}) = 0$ a.e. in~$Q$. This proves that $\bu_{1} = \bu_{2}$ a.e. in $Q$, and if $\bG$ is strictly monotone then also $\bT_{1} = \bT_{2}$.
\section{Regularity estimates}\label{regularity}
\def\mA{\mathcal{A}}
In this section we prove the higher regularity estimates for the solution from Theorem \ref{T1}. We note that this is an essential part in the proof of the existence of a solution for the limiting strain model, i.e., the case~$p=1$, as it
involves passing to the limit $p \rightarrow 1_+$.
As the focus turns to the limiting strain model, in this part we will now assume that there exists a strictly convex $\mathcal{C}^2$-function $F:\Rsym\to \R^d$ such that, for all $\bT\in \Rsym$,
\begin{equation}\label{basic-e}
\frac{\partial F(\bT)}{\partial \bT} = \bG(\bT).
\end{equation}
In this case, $\bG$ is strictly monotone. To simplify the subsequent notation, for an arbitrary $\bT\in \Rsym$, we denote
$$
\mathcal{A}(\bT):=\frac{\partial^2 F(\bT)}{\partial \bT \partial \bT} = \frac{\partial \bG(\bT)}{\partial \bT}, \qquad \mA^{ij}_{kl}(\bT):=\frac{\partial \bG_{ij}(\bT)}{\partial \bT_{kl}}.
$$
We also define a new scalar product on $\Rsym$ by
\begin{equation}\label{scalar}
(\bV,\bW)_{\mA}:= \mA(\bT)\bV \cdot \bW = \sum_{i,j,k,l=1}^d \frac{\partial \bG_{ij}(\bT)}{\partial \bT_{kl}} \bV_{ij} \bW_{kl}.
\end{equation}
The fact that (\ref{scalar}) does indeed define a scalar product follows from the fact that~$\bG$ has a potential~$F$. In particular, we know that for all $\bT\in \Rsym$ there holds $\frac{\partial \bG_{ij}(\bT)}{\partial \bT_{kl}} =\frac{\partial \bG_{kl}(\bT)}{\partial \bT_{ij}}$, i.e., symmetry, and also $\mA$ is positive definite as a result of the convexity assumption.
In what follows, we will split the regularity estimates. First, we focus on time regularity and then we consider regularity with respect to the spatial variable. Here, we provide only a formal proof. Nevertheless, the time regularity proof is in fact fully rigorous since it can be deduced already at the level of Galerkin approximations. The spatial regularity proof is only formal, but can be justified by using a standard difference quotient technique. We emphasise that we do not impose any coercivity and growth assumptions on $\mA$ here because, in the case $p=1$, we lose such information.
We note that when $p\in (1,\infty)$ one can usually assume that
\begin{equation}
\begin{aligned}
|(\bV,\bW)_{\mA}|\le C_3(1+|\bT|)^{p-2}\, |\bV|\, |\bW|,
\qquad (\bW,\bW)_{\mA} \ge C_4(1+|\bT|)^{p-2}\,
|\bW|^2.
\end{aligned}\label{simplas}
\end{equation}
Under assumption \eqref{simplas}, the regularity estimates can be deduced in an easier way. However, they are not included here as the more challenging case of $p=1$ is our primary interest.
Also, it is worth observing that our prototype models \eqref{A4p} do not satisfy \eqref{simplas}$_2$ and in general, the assumption \eqref{simplas}$_2$ cannot be satisfied when $p=1$.
Defining the convex conjugate \(F^*\) of \( F\) as in Section~\ref{physics}, we recall that, from the definition of~\( \bG\), we have that
\begin{equation}
F(\bT) + F^*(\bG(\bT)) = \bG(\bT) \cdot \bT.\label{basic-e2}
\end{equation}
\subsection{Time regularity}
Here, we improve the bound on the time derivative. This bound will be used for the limiting strain model in order to pass to the limit in the term $\partial_{tt}\bu$ in the weak formulation. We formulate the following lemma locally in time in order to keep the initial data as general as possible.
\begin{lemma}\label{time-r}
Let $p\in (1,\infty)$ and suppose that \eqref{basic-e} holds with $\bG$ fulfilling \eqref{A1}--\eqref{A3}. Assume that $\bef\in L^2(0,T; L^2(\Omega;\R^d))$ and $\bu_0 \in W^{2,p'}_{loc}(0,T; W^{1,p'}(\Omega;\R^d))$. Then for any weak solution to~\eqref{model} and for every~$\delta>0$, the following bound holds:
\begin{equation}\label{WFfinallocal}
\begin{split}
&\sup_{t\in (\delta,T)}\int_{\Omega} F^*(\bG(\bT))\dx+\int_{\delta}^T\|\partial_{tt}\bu\|_2^2\dt\\
& \le C(\alpha,\beta) \left(\int_{\frac{\delta}{2}}^T\int_{\Omega} |\bef|_2^2 + |\partial_t \bu|_2^2 + |\partial_{tt} \bu_0|_2^2 + |\partial_t \bu_0|_2^2+|\bT \cdot \partial_t(\beta\partial_t \beps(\bu_0)+\alpha \beps(\bu_0))|\dx \dt\right)\\
&\quad +\frac{C(\alpha,\beta)}{\delta}\int_0^{\delta}\int_{\Omega}F^*(\alpha \beps(\bu(\tau)) + \beta \partial_t \beps(\bu(\tau)))+|\partial_t \bu(\tau)|^2\dx \dtau.
\end{split}
\end{equation}
If additionally $\bu_0 \in W^{2,p'}(0,T; W^{1,p'}(\Omega;\R^d))$, then we have the following global-in-time bound:
\begin{equation}\label{WFfinal}
\begin{split}
&\sup_{t\in (0,T)}\int_{\Omega} F^*(\bG(\bT))\dx+\int_0^T\|\partial_{tt}\bu\|_2^2\dt\\
& \le C(\alpha,\beta) \left(\int_Q |\bef|_2^2 + |\partial_t \bu|_2^2 + |\partial_{tt} \bu_0|_2^2 + |\partial_t \bu_0|_2^2+|\bT \cdot \partial_t(\beta\partial_t \beps(\bu_0)+\alpha \beps(\bu_0))|\dx \dt\right)\\
&\quad +C(\alpha,\beta)\int_{\Omega}F^*(\alpha \beps(\bu_0(0)) + \beta \partial_t \beps(\bu_0(0))) + |\partial_t \bu_0(0)|^2\dx.
\end{split}
\end{equation}
\end{lemma}
\begin{proof}
Recalling that $\bef\in L^2(0,T;L^2(\Omega,\R^d))$, we set $\bw:=\partial_{tt}(\bu-\bu_0)$ in \eqref{WF} to observe that, for almost all $t\in (0,T)$,
\begin{equation*}
\int_{\Omega} |\partial_{tt}\bu|^2+ \bT \cdot \partial_{tt}\beps(\bu)\dx = \int_{\Omega} \bef\cdot \partial_{tt}(\bu-\bu_0) + \partial_{tt}\bu\cdot \partial_{tt}\bu_0+\bT \cdot \partial_{tt}\beps(\bu_0)\dx.
\end{equation*}
This identity can be rewritten as
\begin{equation}
\begin{split}\label{wtf}
&\beta\|\partial_{tt}\bu\|_2^2+ \int_{\Omega}\bT \cdot (\beta \partial_{tt}\beps(\bu)+ \alpha\partial_t\beps(\bu))\dx \\
&\quad = \beta\int_{\Omega} \bef\cdot \partial_{tt}(\bu-\bu_0) + \partial_{tt}\bu\cdot \partial_{tt}\bu_0+\bT \cdot \partial_{tt}\beps(\bu_0)+\frac{\alpha}{\beta}\bT \cdot \partial_t\beps(\bu)\dx.
\end{split}
\end{equation}
First, we evaluate the last term on the right-hand side. Setting $\bw:=\partial_t(\bu-\bu_0)$ in \eqref{WF}, we see that
\begin{equation*}
\int_{\Omega}\bT \cdot \partial_{t}\beps(\bu)\dx =-\frac12 \ddt \|\partial_t \bu\|_2^2 + \int_{\Omega} \bef\cdot \partial_{t}(\bu-\bu_0) + \partial_{tt}\bu\cdot \partial_{t}\bu_0+\bT \cdot \partial_{t}\beps(\bu_0)\dx.
\end{equation*}
From the second term on the left-hand side of (\ref{wtf}), using \eqref{cons-law}, we see that
$$
\begin{aligned}
\int_{\Omega}\bT \cdot (\beta \partial_{tt}\beps(\bu)+ \alpha\partial_t\beps(\bu))\dx&=\int_{\Omega} \bT\cdot \partial_t \bG(\bT)\,\mathrm{d}x=\int_{\Omega}\partial_t(\bT\cdot \bG(\bT)) - \partial_t \bT \cdot \bG(\bT)\dx \\
&=\int_{\Omega}\partial_t(\bT\cdot \bG(\bT) - F(\bT))\dx=\ddt \int_{\Omega} F^*(\bG(\bT))\dx.
\end{aligned}
$$
Thus, using these two identities in \eqref{wtf} and applying Young's inequality, we obtain the following:
\begin{equation}\label{WFt2}
\begin{split}
&\ddt \left(\int_{\Omega} F^*(\bG(\bT))+\frac{\alpha |\partial_t \bu|^2}{2\beta}\dx \right)+\frac{\beta}{2}\|\partial_{tt}\bu\|_2^2\\
&\quad \le C(\alpha,\beta) (\|\bef\|_2^2 + \|\partial_t \bu\|_2^2 + \|\partial_{tt} \bu_0\|_2^2 + \|\partial_t \bu_0\|_2^2) + \frac{1}{\beta}\int_{\Omega} \bT \cdot \partial_t(\beta\partial_t \beps(\bu_0)+\alpha \beps(\bu_0)).
\end{split}
\end{equation}
Integrating (\ref{WFt2}) over \( (0, T) \) and using the fact that
$$
F^*(\bG(\bT(0)))=F^*(\alpha \beps(\bu_0) + \beta \partial_t \beps(\bu_0)),
$$
we deduce \eqref{WFfinal}. Similarly, integrating \eqref{WFt2} over $(\tau,t)$ where $\delta/2\le \tau\le \delta\le t\le T$ are arbitrary, we deduce that
\begin{equation}\label{WFt221}
\begin{split}
&\sup_{t\in (\delta,T)}\left(\int_{\Omega} F^*(\bG(\bT))+\frac{\alpha |\partial_t \bu|^2}{2\beta}\dx \right)+\int_{\delta}^T\|\partial_{tt}\bu\|_2^2\,\mathrm{d}t\\
&\quad \le C(\alpha,\beta) {\int_{\frac{\delta}{2}}^T \int_{\Omega}} |\bef|^2 + |\partial_t \bu|^2 + |\partial_{tt} \bu_0|^2 + |\partial_t \bu_0|^2 +|\bT \cdot \partial_t(\beta\partial_t \beps(\bu_0)+\alpha \beps(\bu_0))|\dx \dt \\
&\qquad + C(\alpha, \beta)\int_{\Omega}F^*(\alpha \beps(\bu(\tau)) + \beta \partial_t \beps(\bu(\tau))) + |\partial_t \bu(\tau)|^2\dx.
\end{split}
\end{equation}
Integrating with respect to $\tau \in (\delta/2,\delta)$ and dividing by $\delta$, we directly obtain \eqref{WFfinallocal}.
\end{proof}
\subsection{Spatial regularity}
Here, we will improve the spatial regularity of a weak solution. In particular,
we prove a weighted bound on $\nabla \bT$, which is a key tool for obtaining the existence of a weak solution for the limiting strain model, i.e., in the case $p=1$.
\begin{lemma}
\label{reg-s}
Let all of the assumptions of Lemma~\ref{time-r} be satisfied. In addition, assume that $\partial_t \bu_0(0)\in W^{1,2}(\Omega;\R^d)$ and
$$
\int_0^T\int_{\Omega} |\mA(\bT)||\bT|^2 +|\mA(\bT)||\bef|^2\dx\dt< \infty.
$$
Then, for an arbitrary open set $\Omega' \subset \overline{\Omega'} \subset \Omega$ and any $\delta>0$, we have the following bound:
\begin{equation}\label{sp-r-l}
\begin{split}
&\sup_{t\in (\delta,T)} \|\partial_t \nabla \bu\|_{L^2(\Omega')} + \sum_{k=1}^d\int_{\delta}^T \int_{\Omega'}(\partial_k \bT, \partial_k \bT)_{\mA(\bT)} \dx \dt \\
&\quad \le C(\Omega',\delta)\int_0^T\int_{\Omega} |\bT| |\bG(\bT)| + |\mA(\bT)||\bT|^2 + |\bef|^2 + |\nabla \bu|^2 +|\partial_t \nabla \bu|^2 +|\mA(\bT)||\bef|^2\dx\dt.
\end{split}
\end{equation}
If, additionally, $\bu_0 \in \mathcal{C}^1([0,T]; W^{1,2}(\Omega;\R^d))$, then we also have
\begin{equation}\label{sp-r}
\begin{split}
&\sup_{t\in (0,T)} \|\partial_t \nabla \bu\|_{L^2(\Omega')} + \sum_{k=1}^d\int_0^T \int_{\Omega'}(\partial_k \bT, \partial_k \bT)_{\mA(\bT)} \dx \dt \\
&\quad \le C(\Omega')\int_0^T\int_{\Omega} |\bT| |\bG(\bT)| + |\mA(\bT)||\bT|^2 + |\bef|^2 + |\nabla \bu|^2 +|\partial_t \nabla \bu|^2 +|\mA(\bT)||\bef|^2\dx\dt\\
&\qquad + C\|\partial_t \nabla \bu_0(0)\|_2^2.
\end{split}
\end{equation}
\end{lemma}
\begin{proof}
Fix an arbitrary nonnegative smooth compactly supported $\varphi\in \mathcal{C}^{\infty}_0(\Omega)$. Then, we can choose $\bw:= -\diver (\varphi^2 \nabla \partial_t \bu)$ in \eqref{WF} and integrate by parts to deduce the following identity:
\begin{equation}
\label{startr}
\begin{split}
\frac{\beta}{2}\ddt \int_{\Omega}|\partial_t \nabla \bu \varphi|^2\dx + \int_{\Omega}\sum_{i,j,k=1}^d \partial_{k}\bT_{ij} \partial_j(\varphi^2 \beta \partial_t \partial_k \bu_i) \dx = -\beta\int_{\Omega} \bef\cdot \diver (\varphi^2 \nabla \partial_t \bu)\dx.
\end{split}
\end{equation}
Similarly, setting $\bw:= -\diver (\varphi^2 \nabla \bu)$ in \eqref{WF} leads to
\begin{equation}
\label{startr2}
\begin{split}
&\alpha \ddt \int_{\Omega} \partial_{t}\nabla \bu \cdot \nabla \bu \varphi^2\dx + \int_{\Omega}\sum_{i,j,k=1}^d \partial_{k}\bT_{ij} \partial_j(\varphi^2 \alpha \partial_k \bu_i) \dx \\
&\qquad = -\alpha\int_{\Omega} \bef\cdot \diver (\varphi^2 \nabla \bu)\dx+\alpha\int_{\Omega} |\partial_{t}\nabla\bu \varphi|^2\dx.
\end{split}
\end{equation}
Summing these two identities, we deduce that
\begin{equation}
\label{startr3}
\begin{split}
\frac{\beta}{4}\ddt \int_{\Omega}|\partial_t \nabla \bu \varphi|^2 +\left|\partial_t \nabla \bu \varphi+\frac{2\alpha}{\beta}\nabla \bu\varphi\right|^2\dx + \int_{\Omega}\sum_{i,j,k=1}^d \partial_{k}\bT_{ij} \partial_j(\varphi^2 (\alpha \partial_k \bu_i+\beta \partial_t \partial_k \bu_i)) \dx \\
= -\int_{\Omega} \bef\cdot \diver (\varphi^2 (\beta\nabla \partial_t \bu+\alpha\nabla \bu))\dx+\frac{2\alpha^2}{\beta} \int_{\Omega}\partial_{t}\nabla \bu\cdot \nabla \bu \varphi^2\dx+\alpha\int_{\Omega} |\partial_{t}\nabla\bu \varphi|^2\dx.
\end{split}
\end{equation}
Now we show that the second integral on the left-hand side is the key source of information. We use \eqref{cons-law}, integration by parts and the symmetry of $\bT$ in order to observe that
\begin{equation}\label{sttt}
\begin{aligned}
&\int_{\Omega}\sum_{i,j,k=1}^d \partial_{k}\bT_{ij} \partial_j(\varphi^2 (\alpha \partial_k \bu_i+\beta \partial_t \partial_k \bu_i)) \dx\\
&=\sum_{i,j,k=1}^d \int_{\Omega}\partial_{k}\bT_{ij} (\varphi^2 (\alpha \partial_k \partial_j\bu_i+\beta \partial_t \partial_k \partial_j\bu_i)) +2\partial_{k}\bT_{ij} \varphi\partial_j\varphi (\alpha \partial_k \bu_i+\beta \partial_t \partial_k \bu_i) \dx\\
&=\sum_{i,j,k=1}^d \int_{\Omega}\partial_{k}\bT_{ij} \varphi^2 \partial_k(\alpha \beps_{ij}(\bu)+\beta \partial_t \beps_{ij}(\bu)) +4\partial_{k}\bT_{ij} \varphi\partial_j\varphi (\alpha \beps_{ik}(\bu)+\beta \partial_t \beps_{ik}(\bu)) \dx\\
&\quad -2\sum_{i,j,k=1}^d \int_{\Omega}\partial_{k}\bT_{ij} \varphi\partial_j\varphi (\alpha \partial_i \bu_k+\beta \partial_t \partial_i \bu_k) \dx\\
&=\sum_{i,j,k=1}^d \int_{\Omega}\partial_{k}\bT_{ij} \varphi^2 \partial_k\bG_{ij}(\bT) -4\bT_{ij} \partial_{k}(\varphi\partial_j\varphi) \bG_{ik}(\bT)-4\bT_{ij} \varphi\partial_j\varphi \partial_{k}\bG_{ik}(\bT) \dx\\
&\quad +\sum_{i,j,k=1}^d \int_{\Omega}\bT_{ij} \partial_{kj}(\varphi^2 )\partial_i(\alpha \bu_k+\beta \partial_t \bu_k) \dx+2\sum_{i,j,k=1}^d \int_{\Omega}\bT_{ij} \varphi\partial_j\varphi \partial_i(\alpha \partial_{k}\bu_k+\beta \partial_t \partial_{k}\bu_k)\dx\\
&=\int_{\Omega}\sum_{k=1}^d (\partial_k \bT \varphi, \partial_k \bT \varphi)_{\mA(\bT)} -4\sum_{i,j,k=1}^d \bT_{ij} \partial_{k}(\varphi\partial_j\varphi) \bG_{ik}(\bT)-4\sum_{i,j,k=1}^d \bT_{ij} \varphi\partial_j\varphi \partial_{k}\bG_{ik}(\bT) \dx\\
&\quad -\sum_{i,j,k=1}^d \int_{\Omega}\partial_j\bT_{ij} \partial_{k}(\varphi^2 )\partial_i(\alpha \bu_k+\beta \partial_t \bu_k) \dx-\sum_{i,j,k=1}^d \int_{\Omega}\bT_{ij} \partial_{k}(\varphi^2 )\partial_{ij}(\alpha \bu_k+\beta \partial_t \bu_k) \dx\\
&\quad +2\sum_{i,j,k=1}^d \int_{\Omega}\bT_{ij} \varphi\partial_j\varphi \partial_i \bG_{kk}(\bT)\dx
\\&=: \sum_{m=1}^6 I_m.
\end{aligned}
\end{equation}
We need to determine what bounds can be deduced from \eqref{sttt}. In particular, we show that the terms $I_2, \ldots, I_6$ can be bounded in terms of $I_1$ and the data. The simplest bound is for $I_2$. In particular, it directly follows that
$$
|I_2| \le C(\varphi) \int_{\Omega} |\bT|\, |\bG(\bT)|\dx.
$$
Letting $\delta_{nk}$ denote the Kronecker delta, in order to bound \( I_3\) we first rewrite it in the following way:
$$
\begin{aligned}
\sum_{i,j,k=1}^d \bT_{ij} \varphi\partial_j\varphi \partial_{k}\bG_{ik}(\bT)&=\sum_{i,j,k,l,m,n=1}^d \delta_{nk}\bT_{ij} \varphi\partial_j\varphi \mA^{ik}_{lm}(\bT)\partial_{n}\bT_{lm}\\
&=\sum_{j,n=1}^d \left(\sum_{i,k,l,m=1}^d \mA^{ik}_{lm}(\bT)\partial_{n}\bT_{lm} \delta_{nk}\bT_{ij} \varphi\partial_j\varphi\right).
\end{aligned}
$$
Using the Cauchy--Schwarz inequality and the fact that $\mA$ generates a scalar product, by applying Young's inequality we find that
$$
\begin{aligned}
|I_3| &\le C \int_{\Omega}\left|\sum_{j,n=1}^d \left(\sum_{i,k,l,m=1}^d \mA^{ik}_{lm}(\bT)\partial_{n}\bT_{lm} \delta_{nk}\bT_{ij} \varphi\partial_j\varphi\right) \right|\dx \\
&\le C \int_{\Omega}\left|\sum_{j,n=1}^d \left(\sum_{i,k,l,m=1}^d \mA^{ik}_{lm}(\bT)\partial_{n}\bT_{lm} \varphi \partial_{n}\bT_{ik} \varphi \right)^{\frac12}\left(\sum_{i,k,l,m=1}^d \mA^{ik}_{lm}(\bT) \delta_{nm}\bT_{lj} \partial_j \varphi \delta_{nk}\bT_{ij} \partial_j\varphi\right)^{\frac12} \right|\dx\\
&\le \frac{I_1}{8} +C(\varphi) \int_{\Omega}|\mA(\bT)| |\bT|^2\dx.
\end{aligned}
$$
The term $I_6$ can be bounded in a very similar way. In particular, we have
$$
|I_6|\le \frac{I_1}{8} +C(\varphi) \int_{\Omega}|\mA(\bT)| |\bT|^2\dx.
$$
For $I_4$, we use the equation \eqref{linear-moment} and Young's inequality to obtain
$$
\begin{aligned}
|I_4|&=\left|\sum_{i,k=1}^d \int_{\Omega}(\bef_i -\partial_{tt} \bu_i) \partial_{k}(\varphi^2 )\partial_i(\alpha \bu_k+\beta \partial_t \bu_k) \dx\right|\\
&\le C(\varphi) \int_{\Omega} |\bef|^2 + |\partial_{tt}\bu|^2 + |\partial_t \nabla \bu \varphi|^2 + |\nabla \bu \varphi|^2 \dx.
\end{aligned}
$$
Finally, to evaluate $I_5$, we first recall the following identity
\begin{equation}
\begin{aligned}
&\partial_{ij}(\alpha \bu_k+\beta \partial_t \bu_k)\\
&= \partial_i(\alpha\beps_{jk}(\bu)+\beta \partial_t\beps_{jk}(\bu)) +\partial_j(\alpha\beps_{ik}(\bu)+\beta \partial_t\beps_{ik}(\bu))-\partial_k(\alpha\beps_{ij}(\bu)+\beta \partial_t\beps_{ij}(\bu)).
\end{aligned}\label{ID12}
\end{equation}
Then, we can rewrite $I_5$ with the help of \eqref{cons-law} to find that
$$
I_5=-\sum_{i,j,k=1}^d \int_{\Omega}\bT_{ij} \partial_{k}(\varphi^2 )\left(\partial_i \bG_{jk}(\bT) +\partial_j \bG_{ik}(\bT)-\partial_k\bG_{ij}(\bT)\right) \dx.
$$
Hence, we see that we are in the same situation as with the term $I_3$ and we can deduce that
$$
|I_5|\le \frac{I_1}{8} +C(\varphi) \int_{\Omega}|\mA(\bT)| |\bT|^2\dx.
$$
Thus we have suitable bounds on the left-hand side of (\ref{startr3}).
Next we rewrite the first term on the right-hand side of \eqref{startr3} in the following way:
\begin{equation*}
\begin{split}
&\int_{\Omega} \bef\cdot \diver (\varphi^2 (\alpha \nabla \bu +\beta \partial_t \nabla \bu))\dx\\
&=\sum_{i,j=1}^d \int_{\Omega} \bef_i (\partial_j (\varphi^2) (\alpha \partial_j \bu_i +\beta \partial_t \partial_j\bu_i)+\varphi^2 (\alpha \partial_{jj}\bu_i +\beta \partial_t \partial_{jj} \bu_i))\dx\\
&=\sum_{i,j=1}^d \int_{\Omega} \bef_i (\partial_j (\varphi^2) (\alpha \partial_j \bu_i +\beta \partial_t \partial_j\bu_i)+\varphi^2(2\partial_j \bG_{ij}(\bT)-\partial_i \bG_{jj}(\bT))\dx.
\end{split}
\end{equation*}
Hence, using Young's inequality in the first term and a procedure similar to the one used for $I_3$ in the second, we get
\begin{equation}
\label{ops}
\begin{split}
&\left|\int_{\Omega} \bef\cdot \diver (\varphi^2 (\alpha \nabla \bu +\beta \partial_t \nabla \bu))\dx\right|\\
&\quad \le \frac{I_1}{8}+C(\varphi)\int_{\Omega} |\bef|^2 + |\nabla \bu|^2 +|\partial_t \nabla \bu|^2 +|\mA(\bT)||\bef|^2\dx.
\end{split}
\end{equation}
Substituting the above bounds into \eqref{startr3} and using a similar procedure to the one used in the proof of Lemma~\ref{time-r}, we deduce \eqref{sp-r} and \eqref{sp-r-l}.
\end{proof}
\section{Limiting strain - Proof of Theorem~\ref{T4.1}}\label{limiting}
As in the proof of Theorem \ref{T1}, in order to prove Theorem \ref{T4.1} we first introduce an approximate problem. However, we are able to make use of the knowledge obtained from Theorem \ref{T1}. Indeed, we define a function on \( \Rsym\) by
\begin{equation}\label{Gndf}
\bG^n(\bT):= \bG(\bT) + n^{-1} \bT.
\end{equation}
Since $\bG$ satisfies \eqref{A1}--\eqref{A3} with $p=1$, it is evident that $\bG^n$ satisfies \eqref{A1}--\eqref{A3} with $p=2$. Therefore, thanks to Theorem~\ref{T1}, we know that there exists a couple $(\bu^n,\bT^n)$, fulfilling\footnote{We assume a slightly different restriction on $\bu_0$ than in~Theorem~\ref{T1}. However, the proof of Theorem~\ref{T1} can be easily adapted to this case.}
\begin{align}
\bu^n &\in \mathcal{C}^{1}([0, T]; L^{2}(\Omega;\R^d))\cap W^{1, 2}(0, T; W^{1,2}(\Omega;\R^d))\cap W^{2,2}(0, T; (W_{0}^{1,2 }(\Omega;\R^d))^{*}), \label{FSun}\\
\bT^n &\in L^{2}(0, T; L^{2}(\Omega;\Rsym))\label{FSTn}
\end{align}
and satisfying
\begin{equation}\label{WFn}
\langle \partial_{tt}\bu^n, \bw \rangle + \int_{\Omega} \bT^n \cdot \nabla \bw \dx = \int_{\Omega} \bef \cdot \bw\dx \qquad \forall\, \bw \in W_{0}^{1,2}(\Omega;\R^d)\quad \text{for a.e. }\,t \in (0, T),
\end{equation}
and
\begin{equation}\label{T-constn}
\alpha \beps(\bu^n) + \beta \partial_{t} \beps(\bu^n)=\bG^n(\bT^n)= \bG(\bT^n) + n^{-1}\bT^n \quad \textrm{a.e. in }Q.
\end{equation}
We note that we can replace the duality pairing by the integral over $\Omega$ in the term containing $\bef$ thanks to the assumed regularity of $\bef$.
Moreover, we know that\footnote{In case that $\Omega$ is not a Lipschitz domain, the identity below is not understood in the sense of traces but in the sense that $\bu-\bu_0 \in W^{1,1}_0(\Omega;\R^d)$ for almost all $t\in (0,T)$, where $W^{1,1}_0(\Omega;\R^d)$ defined as the closure of $C^\infty_0(\Omega;\R^d)$ in the norm of $W^{1,1}(\Omega;\R^d)$.}
$$
\bu^n=\bu_0 \quad \textrm{ on } \Gamma \cup (\{0\}\times \Omega), \qquad \partial_t\bu^n=\partial_t\bu_0 \quad \textrm{ on } \{0\}\times \Omega.
$$
We want to consider the limit as $n\to \infty$ in order to prove the existence of a solution to the limiting strain problem in the sense of Theorem~\ref{T4.1}.
\subsection{A~priori $n$-independent bounds}
We start with bounds that are independent of the order of approximation. For this purpose, we use and mimic many steps from preceding sections. We start with the first uniform bound. Setting $\bw:=\beta \partial_t(\bu^n-\bu_0)+\alpha(\bu-\bu_0)$ in \eqref{WFn}, after exactly the same algebraic manipulations as those used for \eqref{test3} we deduce that
\begin{equation}\label{test3n}
\begin{split}
&\frac{1}{4} \ddt\int_{\Omega}\beta|\partial_{t} (\bu^n-\bu_0)|^2 +\beta\left|\partial_{t} (\bu^n-\bu_0)+\frac{2\alpha}{\beta}(\bu^n-\bu_0)\right|^2\dx+ \int_{\Omega} \bG^n(\bT^n)\cdot \bT^n \dx \\
&=\int_{\Omega} \bT^n\cdot (\alpha \beps(\bu_0)+\beta\partial_t \beps(\bu_0)) \dx +\alpha \int_{\Omega} |\partial_{t} (\bu^n-\bu_0)|^{2}\,\mathrm{d}x\\
&\quad +\int_{\Omega}(\bef-\partial_{tt} \bu_0) \cdot (\alpha(\bu^n-\bu_0)+ \beta\partial_t (\bu^n-\bu_0))\dx+ \frac{2\alpha^2}{\beta}\int_{\Omega}\partial_t (\bu^n-\bu_0) \cdot (\bu^n-\bu_0)\dx.
\end{split}
\end{equation}
In order to obtain the required a~priori estimate, we need to use the safety strain condition. In particular, it follows from \eqref{compt} that there exists a $\delta>0$ such that
\begin{equation}
|\alpha \beps(\bu_0)+\beta\partial_t \beps(\bu_0)|\le L-2\delta \qquad \textrm{a.e. in }Q, \label{compt-A}
\end{equation}
where $L$ is as defined in \eqref{dfL2}. In addition, if we define \( F(\bT) := \phi(|\bT|) \), it follows from the convexity of $\phi$ that, for any $\tilde{\delta}>0$, there exists a $C_{\tilde{\delta}}$ such that, for all $\bT\in \R^{d\times d}_{sym}$,
\begin{equation}
F(\bT)\ge (L-\tilde{\delta})|\bT| -C_{\tilde{\delta}}.\label{lowerbo}
\end{equation}
We choose \( \tilde{\delta} = \delta\) as in (\ref{compt-A}) and let \( C_\delta\) be the corresponding constant from (\ref{lowerbo}).
Since $C_{\delta}$ depends in principle on $\bu_0$ and $F$, and \( \delta\) is now given, we do not trace the dependence of $C$ on $\delta$ in what follows.
Consequently, for the second term on the left-hand side of \eqref{test3n}, we can use \eqref{basic-e2} and \eqref{T-constn} to deduce that
$$
\bG^n(\bT^n)\cdot \bT^n = n^{-1}|\bT^n|^2 + F(\bT^n) + F^*(\bG(\bT^n))\ge (L-\delta)|\bT^n|+ n^{-1}|\bT^n|^2 -C(\delta).
$$
Furthermore, the first term on the right-hand side of \eqref{test3n} can be bounded by using \eqref{compt-A} in the following way:
$$
\int_{\Omega} \bT^n\cdot (\alpha \beps(\bu_0)+\beta\partial_t \beps(\bu_0)) \dx\le (L-2\delta)\|\bT^n\|_1.
$$
Therefore, it follows from \eqref{test3n}, the above bounds and H\"{o}lder's inequality that
\begin{equation}\label{test3n2}
\begin{split}
&\frac{1}{4} \ddt\int_{\Omega}\beta|\partial_{t} (\bu^n-\bu_0)|^2 +\beta\left|\partial_{t} (\bu^n-\bu_0)+\frac{2\alpha}{\beta}(\bu^n-\bu_0)\right|^2\dx+ \delta \|\bT^n\|_1+ n^{-1}\|\bT^n\|_2^2\\
&\le C \left( \int_{\Omega}\beta|\partial_{t} (\bu^n-\bu_0)|^2 +\beta\left|\partial_{t} (\bu^n-\bu_0)+\frac{2\alpha}{\beta}(\bu^n-\bu_0)\right|^2\dx+\|\bef\|_2^2 + \|\partial_{tt}\bu_0\|_2^2 +1\right).
\end{split}
\end{equation}
The application of Gr\"{o}nwall's lemma then leads to
\begin{equation}\label{aen1}
\sup_{t\in (0,T)} \left(\|\partial_t \bu(t)\|_2^2 + \|\bu(t)\|_2^2\right) + \int_0^T \|\bT^n\|_1 + n^{-1}\|\bT^n\|_2^2 \dt \le C(\bef,\bu_0),
\end{equation}
where we have used the assumption \eqref{data-as2} on the data. It also follows from \eqref{T-const} and the above bound that
$$
\int_{Q}|\alpha \beps(\bu)+\beta\partial_t \beps(\bu)|^2 \dx \dt \le \int_{Q} (L+n^{-1}|\bT^n|)^2 \dx \dt \le C(\bef,\bu_0).
$$
Consequently, since $\beps(\bu(0))\in L^{\infty}(\Omega,\R^{d\times d})$ and $\bG^n(\bT^n)\cdot \bT^n$ is nonnegative, arguing similarly as in the bound \eqref{tt3}, we deduce with the help of Korn's inequality and \eqref{test3n} that
\begin{equation}
\int_Q |\bG^n(\bT^n)\cdot \bT^n|\dx\dt +\int_0^T \|\partial_t\bu^n\|_{1,2}^2 + \|\bu^n\|_{1,2}^2\dt\le C(\bef,\bu_0).\label{aen2}
\end{equation}
\subsection{Regularity via $n$-independent bounds}
The bounds \eqref{aen1}, \eqref{aen2} are not sufficient to pass to the limit $n\to \infty$, since we only have a~priori control on $\bT^n$ in a nonreflexive space $L^1(Q;\R^{d \times d})$. In particular, at best we have that the weak star limit of $\bT^n$ is a measure. Therefore, the pointwise relation \eqref{T-const2} is neither meaningful nor likely to be valid in this case. Instead, we improve our information by using the regularity technique introduced in Section~\ref{regularity}. Namely, we use Lemma~\ref{time-r} and Lemma~\ref{reg-s}. First, we define an approximatinon $F_n$ of the potential $F$ by
$$
F_n(\bT):= F(\bT) + \frac{|\bT|^2}{2n}.
$$
Then we have that
$$
\frac{\partial F^n(\bT)}{\partial \bT}=\bG_n(\bT) = \bG(\bT) + n^{-1}\bT.
$$
We may now apply the results from Section~\ref{regularity} with $p=2$, replacing $(\bu, F, \bG)$ with the triple $(\bu^n, F_n, \bG_n)$. We note that using the definition of $\bG_n$ we may define~\( \mathcal{A}_n \) in an analogous way to~\( \mathcal{A}\). In particular, we write
$$
\begin{aligned}
(\mA_n(\bT^n))_{ijkl}&:=\frac{\partial}{\partial\bT^n_{kl}}\left(\frac{\phi'(|\bT^n|)}{|\bT^n|}\bT^n_{ij} + n^{-1} \bT^n_{ij}\right)\\
&=\delta_{ik}\delta_{jl} \left(n^{-1}+\frac{\phi'(|\bT^n|)}{|\bT^n|}\right)+ \left(\frac{\phi''(|\bT^n|)|\bT^n| - \phi'(|\bT^n|)}{|\bT^n|}\right)\frac{\bT^n_{ij}\bT^n_{kl}}{|\bT^n|^2}.
\end{aligned}
$$
Consequently, using the fact that $\phi'(0)=0$ and $\phi''(s)\le C(1+s)^{-1}$, we see that
\begin{equation}\label{Abbound}
|\mA_n(\bT^n)| \le Cn^{-1} +\frac{C}{1+|\bT^n|}.
\end{equation}
With this in mind, let us first discuss regularity with respect to time. We see that all assumptions of Lemma~\ref{time-r} are satisfied. Therefore we have, for every~$\delta>0$, that the following inequality holds:
\begin{equation}\label{WFfinallocaln}
\begin{split}
&\sup_{t\in (\delta,T)}\int_{\Omega} F_n^*(\bG_n(\bT^n))\dx+\int_{\delta}^T\|\partial_{tt}\bu^n\|_2^2\dt\\
& \le C(\alpha,\beta) \left(\int_{\frac{\delta}{2}}^T\int_{\Omega} |\bef|_2^2 + |\partial_t \bu^n|_2^2 + |\partial_{tt} \bu_0|_2^2 + |\partial_t \bu_0|_2^2+|\bT^n \cdot \partial_t(\beta\partial_t \beps(\bu_0)+\alpha \beps(\bu_0))|\dx \dt\right)\\
&\quad +\frac{C(\alpha,\beta)}{\delta}\int_0^{\delta}\int_{\Omega}F_n^*(\alpha \beps(\bu^n(\tau)) + \beta \partial_t \beps(\bu^n(\tau)))+|\partial_t \bu^n(\tau)|^2\dx \dtau.
\end{split}
\end{equation}
We focus on the bound of the right-hand side. For the second integral on the right-hand side, it follows from the properties of the convex conjugate function and the uniform bounds \eqref{aen1}, \eqref{aen2} that
$$
\begin{aligned}
&\int_0^{\delta}\int_{\Omega}F_n^*(\alpha \beps(\bu^n) + \beta \partial_t \beps(\bu^n))+|\partial_t \bu^n|^2\dx \dtau
=\int_0^{\delta}\int_{\Omega}F_n^*(\bG_n(\bT^n))+|\partial_t \bu^n|^2\dx \dtau
\\&\quad
\le \int_{Q}\bG_n(\bT^n)\cdot \bT^n+|\partial_t \bu^n|^2\dx \dt \le C(\bu_0,\bef).
\end{aligned}
$$
For the first term on the right-hand side of \eqref{WFfinallocaln}, we use H\"{o}lder's inequality, the assumptions on the data \eqref{data-as2}, \eqref{compt}, \eqref{compt2} and the uniform bound \eqref{aen1} in order to deduce that
$$
\begin{aligned}
&\int_{\frac{\delta}{2}}^T\int_{\Omega} |\bef|_2^2 + |\partial_t \bu^n|_2^2 + |\partial_{tt} \bu_0|_2^2 + |\partial_t \bu_0|_2^2+|\bT^n \cdot \partial_t(\beta\partial_t \beps(\bu_0)+\alpha \beps(\bu_0))|\dx \dt\\
&\quad \le C(\bu_0,\bef) + \|| \partial_{tt}\beps(\bu_0)|+|\partial_t\beps(\bu_0)|\|_{L^{\infty}((\frac{\delta}{2},T)\times \Omega)}\int_{0}^T\int_{\Omega} |\bT^n|\dx \dt
\\&\quad
\le C(\bu_0, \bef).
\end{aligned}
$$
It follows from the above bounds and \eqref{WFfinallocaln} that, for every $\delta>0$, we have
\begin{equation}\label{WFfinallocalnn}
\begin{split}
&\sup_{t\in (\delta,T)}\int_{\Omega} F_n^*(\bG_n(\bT^n))\dx+\int_{\delta}^T\|\partial_{tt}\bu^n\|_2^2\dt\le C(\bef,\bu_0).
\end{split}
\end{equation}
Similarly, in case that \eqref{compt2} holds even for $\delta=0$, we can use \eqref{WFfinal}. By a very similar computation to the one above we deduce that
\begin{equation}
\begin{split}
&\sup_{t\in (0,T)}\int_{\Omega} F_n^*(\bG_n(\bT^n))\dx+\int_0^T\|\partial_{tt}\bu^n\|_2^2\dt\\
&\quad \le C(\bef,\bu_0) + C\int_{\Omega}F_n^*(\alpha \beps(\bu_0(0)) + \beta \partial_t \beps(\bu_0(0)))\dx \\
&\quad \le C(\bef,\bu_0) + C\int_{\Omega}F^*(\alpha \beps(\bu_0(0)) + \beta \partial_t \beps(\bu_0(0)))\dx \\
&\quad \le C(\bef,\bu_0),
\end{split}\label{WFfinalnnn}
\end{equation}
using the fact that $F_n^* \le F^*$ and assumptions \eqref{compt}, \eqref{compt2} with $\delta=0$.
Next, we consider the spatial regularity estimates. For an arbitrary open set $\Omega' \subset \overline{\Omega'} \subset \Omega$ and for any $\delta>0$, it follows from \eqref{sp-r-l} that
\begin{equation}\label{sp-r-ln}
\begin{split}
&\sup_{t\in (\delta,T)} \|\partial_t \nabla \bu^n\|_{L^2(\Omega')} + \sum_{k=1}^d\int_{\delta}^T \int_{\Omega'}(\partial_k \bT^n, \partial_k \bT^n)_{\mA_n(\bT^n)} \dx \dt \\
&\quad \le C(\Omega',\delta)\int_Q |\bT^n| |\bG_n(\bT^n)| + |\mA_n(\bT^n)||\bT^n|^2 + |\bef|^2 + |\nabla \bu^n|^2 +|\partial_t \nabla \bu^n|^2 +|\mA_n(\bT^n)||\bef|^2\dx\dt.
\end{split}
\end{equation}
Since $|\bT^n| |\bG_n(\bT^n)|=|\bT^n\cdot \bG_n(\bT^n)|$, we can use \eqref{aen1}, \eqref{aen2} to deduce that
$$
\int_Q |\bT^n| |\bG_n(\bT^n)| + |\bef|^2 + |\nabla \bu^n|^2 +|\partial_t \nabla \bu^n|^2\dx\dt\le C(\bu_0,\bef).
$$
Thus, it only remains to bound the terms involving $\mA_n$ on the right-hand side of \eqref{sp-r-ln}.
To this end, we note that
$$
\int_Q |\mA_n(\bT^n)||\bT^n|^2 +|\mA_n(\bT^n)||\bef|^2\dx\dt \le C\int_{Q}n^{-1}|\bT^n|^2 + |\bT^n| + |\bef|^2\le C(\bu_0,\bef),
$$
where the last inequality follows from \eqref{aen1} and the assumptions on $\bef$. Using these inequalities for the terms appearing on the right-hand side of \eqref{sp-r-ln}, we immediately deduce that
\begin{equation}\label{sp-r-lnn}
\begin{split}
&\sup_{t\in (\delta,T)} \|\partial_t \nabla \bu^n\|_{L^2(\Omega')} + \sum_{k=1}^d\int_{\delta}^T \int_{\Omega'}(\partial_k \bT^n, \partial_k \bT^n)_{\mA_n(\bT^n)} \dx \dt \le C(\bu_0,\bef,\Omega').
\end{split}
\end{equation}
Similarly, if $\bu_0 \in \mathcal{C}^1([0,T]; W^{1,2}(\Omega;\R^d))$ we can use \eqref{sp-r} and perform similar computations to find that
\begin{equation}\label{sp-rn}
\begin{split}
&\sup_{t\in (0,T)} \|\partial_t \nabla \bu^n\|_{L^2(\Omega')} + \sum_{k=1}^d\int_0^T \int_{\Omega'}(\partial_k \bT^n, \partial_k \bT^n)_{\mA_n(\bT^n)} \dx \dt \le C(\Omega',\bu_0,\bef).
\end{split}
\end{equation}
Next, we focus on the bounds on the second derivatives of $\partial_t \bu^n$ and $\bu^n$. It follows from \eqref{T-constn} and the Cauchy--Schwarz inequality that
$$
\begin{aligned}
&|\partial_k (\alpha \beps(\bu^n) + \beta\partial_t \beps(\bu^n))|^2=(\partial_k (\alpha \beps(\bu^n) + \beta\partial_t \beps(\bu^n)))\cdot \partial_k \bG_n(\bT^n) \\
&\quad = (\partial_k (\alpha \beps(\bu^n) + \beta\partial_t \beps(\bu^n)), \partial_k \bT^n)_{\mA_n(\bT^n)}\\
&\quad \le (\partial_k (\alpha \beps(\bu^n) + \beta\partial_t \beps(\bu^n)), \partial_k (\alpha \beps(\bu^n) + \beta\partial_t \beps(\bu^n)))_{\mA_n(\bT^n)}^{\frac12} (\partial_k \bT^n, \partial_k \bT^n)_{\mA_n(\bT^n)}^{\frac12}\\
&\quad \le C|\partial_k (\alpha \beps(\bu^n) + \beta\partial_t \beps(\bu^n))|(\partial_k \bT^n, \partial_k \bT^n)_{\mA_n(\bT^n)}^{\frac12}.
\end{aligned}
$$
Therefore,
$$
\begin{aligned}
&|\partial_k (\alpha \beps(\bu^n) + \beta\partial_t \beps(\bu^n))|^2\le C(\partial_k \bT^n, \partial_k \bT^n)_{\mA_n(\bT^n)}.
\end{aligned}
$$
Using this and \eqref{sp-r-lnn}, simple algebraic manipulations imply that
\begin{equation}\label{sp-r-lnn1}
\begin{split}
\int_{\delta}^T \int_{\Omega'} |\nabla(\alpha \beps(\bu^n) + \beta\partial_t \beps(\bu^n))|^2\dx \dt \le C(\bu_0,\bef,\Omega').
\end{split}
\end{equation}
\subsection{Convergence results as $n\to \infty$ based on uniform bounds}
From the uniform bounds \eqref{aen1}, \eqref{aen2}, we see that we can find a subsequence, that we do not relabel, such that
\begin{align}
\bu^n &\rightharpoonup \bu &&\textrm{weakly in }W^{1,2}(0,T; W^{1,2}(\Omega;\R^d)),\label{Kn1}\\
\bu^n &\rightharpoonup^* \bu &&\textrm{weakly$^*$ in } W^{1,\infty}(0,T; L^2(\Omega;\R^d)),\label{Kn2}\\
n^{-1}\bT^n &\to \b0 &&\textrm{strongly in }L^2(0,T; L^2(\Omega; \R^{d\times d})).\label{Kn3}
\end{align}
In addition, using the regularity estimates \eqref{WFfinallocalnn}, \eqref{sp-r-lnn1}, as well as the Aubin--Lions lemma, we deduce that
\begin{align}
\bu^n &\rightharpoonup \bu &&\textrm{weakly in }W^{2,2}_{loc}(0,T; L^{2}(\Omega;\R^d)),\label{Kn4}\\
\bu^n &\rightharpoonup \bu &&\textrm{weakly in } W^{1,2}_{loc}(0,T; W^{2,2}_{loc}(\Omega;\R^d)),\label{Kn5}\\
\bu^n &\to \bu &&\textrm{strongly in }W^{1,2}_{loc}(0,T; W^{1,2}_{loc}(\Omega; \R^{d})).\label{Kn6}
\end{align}
Next, we focus on the limiting passage in \eqref{T-constn}. Since the mapping $\bG$ is bounded, we know that
\begin{align}
\bG(\bT^n) &\rightharpoonup^* \overline{\bG} \qquad \textrm{weakly$^*$ in }L^{\infty}(Q;\R^{d\times d}). \label{Kn7}
\end{align}
Our goal is to identify $\overline{\bG}$. We first note that from \eqref{T-constn}, \eqref{Kn2} and \eqref{Kn3} we must have
\begin{equation}
\overline{\bG}=\alpha \beps(\bu) + \beta \partial_t \beps(\bu) \qquad \textrm{a.e. in }Q. \label{aennn}
\end{equation}
Next, we want to show that there exists a $\tilde{\bT}$ such that $\overline{\bG}=\bG(\tilde{\bT})$. To do so, we appeal to Chacon's biting lemma to deduce from \eqref{aen1} that there exists a $\bT\in L^1(Q; \R^{d\times d})$ and a nondecreasing sequence of sets $Q_1\subset Q_2\subset \cdots$, with $|Q\setminus Q_i|\to 0$ as $i\to \infty$, such that for each $i\in \mathbb{N}$ there holds
\begin{align}
\bT^n &\rightharpoonup \bT \qquad\textrm{weakly in }L^{1}(Q_i;\R^{d\times d}).\label{Kn8}
\end{align}
However, thanks to \eqref{Kn6}, \eqref{aennn} and Egoroff's theorem, we also know that for every $\varepsilon>0$ and every $i\in \mathbb{N}$ there exists a $Q_{i,\varepsilon}\subset Q_i$, with $|Q_i\setminus Q_{i,\varepsilon}|\le \varepsilon$, such that
$$
\alpha \beps(\bu^n) + \beta \partial_t\beps(\bu^n)\to \overline{\bG}\qquad \textrm{strongly in } L^{\infty}(Q_{i,\varepsilon};\mathbb{R}^{d\times d}).
$$
Therefore, using the monotonicity of $\bG$ and the above convergence result, we deduce that, for an arbitrary $\bW\in L^1(Q;\R^{d\times d})$, that
$$
\begin{aligned}
0&\le \lim_{n\to \infty}\int_{Q_{i,\varepsilon}}(\bG(\bT^n)-\bG(\bW))\cdot (\bT^n-\bW)\dx \dt\\
&= \int_{Q_{i,\varepsilon}}\bG(\bW)\cdot (\bW-\bT)-\overline{\bG}\cdot \bW\dx \dt+ \lim_{n\to \infty}\int_{Q_{i,\varepsilon}}\bG(\bT^n)\cdot \bT^n\dx \dt\\
&\le \int_{Q_{i,\varepsilon}}\bG(\bW)\cdot (\bW-\bT)-\overline{\bG}\cdot \bW\dx \dt+ \lim_{n\to \infty}\int_{Q_{i,\varepsilon}}\bG_n(\bT^n)\cdot \bT^n\dx \dt\\
&= \int_{Q_{i,\varepsilon}}\bG(\bW)\cdot (\bW-\bT)-\overline{\bG}\cdot \bW\dx \dt+ \lim_{n\to \infty}\int_{Q_{i,\varepsilon}}(\alpha \beps(\bu^n) + \beta \partial_t\beps(\bu^n))\cdot \bT^n\dx \dt\\
&=\int_{Q_{i,\varepsilon}}(\overline{\bG}-\bG(\bW))\cdot (\bT-\bW)\dx \dt.
\end{aligned}
$$
Since $\bG$ is a monotone mapping and $\bW$ is arbitrary, we can use Minty's method to deduce that
$$
\overline{\bG}=\bG(\bT) \qquad \textrm{ a.e. in }Q_{i,\varepsilon}.
$$
Recalling that $\varepsilon>0$ and $i\in \mathbb{N}$ are arbitrary, we see that \eqref{T-const2} follows from \eqref{aennn} and the above identity. Additionally, setting $\bW:=\bT$ in the above and using the fact that $\overline{\bG}=\bG(\bT)$, we see that
$$
\begin{aligned}
\lim_{n\to \infty}\int_{Q_{i,\varepsilon}}|(\bG(\bT^n)-\bG(\bT))\cdot (\bT^n-\bT)|\dx \dt= \lim_{n\to \infty}\int_{Q_{i,\varepsilon}}(\bG(\bT^n)-\bG(\bT))\cdot (\bT^n-\bT)\dx \dt=0.
\end{aligned}
$$
Consequently, we must have that
$$
\bT^n \to \bT \qquad \textrm{ a.e. in } Q_{i,\varepsilon},
$$
as a result of the strict monotonicity of $\bG$. However, as before, since $\varepsilon>0$ and $i\in \mathbb{N}$ are arbitrary, we deduce that
\begin{align}
\bT^n \to \bT \qquad \textrm{ a.e. in } Q. \label{Kn9}
\end{align}
Using \eqref{aen1}, \eqref{Kn9} and Fatou's lemma, it follows that
\begin{equation}
\int_{Q}|\bT|\dx \dt \le C(\bu_0,\bef).\label{aest1nl}
\end{equation}
Next, we focus on the boundary and initial conditions for $\bu$. It is evident from the convergence result \eqref{Kn1}, combined with the fact that $\bu^n=\bu_0$ on $\Gamma$ and $\bu^n(0)=\bu_0$ on $\Omega$, that we must have $\bu=\bu_0$ on $\Gamma$ as well. Furthermore, it follows that
$$
\|\bu(t)-\bu_0(0)\|_{1,2} \to 0 \qquad \textrm{ as } t\to 0_+.
$$
Concerning the attainment of the initial condition for $\partial_t\bu(0)$ we need to proceed slightly differently since we have control on $\partial_{tt}\bu$ locally in $(0,T)$. We integrate \eqref{test3n} over a time interval $(0,t)$, where $0<t<T$, and since we know that for each $n$ the initial datum is attained we deduce that
\begin{equation}\label{test3naa}
\begin{split}
&\frac{1}{4} \int_{\Omega}\beta|\partial_{t} (\bu^n-\bu_0)(t)|^2 +\beta\left|\partial_{t} (\bu^n-\bu_0)(t)+\frac{2\alpha}{\beta}(\bu^n-\bu_0)(t)\right|^2\dx \\
&=\int_0^t\int_{\Omega} \bT^n\cdot ((\alpha \beps(\bu_0)+\beta\partial_t \beps(\bu_0))-\bG_n(\bT^n)) +\alpha |\partial_{t} (\bu^n-\bu_0)|^{2}\dx \dtau\\
& \quad+\int_0^t\int_{\Omega}(\bef-\partial_{tt} \bu_0) \cdot (\alpha(\bu^n-\bu_0)+ \beta\partial_t (\bu^n-\bu_0))+ \frac{2\alpha^2}{\beta}\partial_t (\bu^n-\bu_0) \cdot (\bu^n-\bu_0)\dx\dtau.
\end{split}
\end{equation}
Our goal is to let $n\to \infty$. Since $t>0$, we can use the ``local" convergence result \eqref{Kn4} to let $n\to \infty$ in the left-hand side of \eqref{test3naa}. To bound also the right-hand side, we first use the safety strain condition \eqref{compt}, which implies that there exists a $\bT_0\in L^1(Q;\R^{d\times d})$ such that
$$
\alpha \beps(\bu_0)+\beta\partial_t \beps(\bu_0)=\bG(\bT_0)\qquad \textrm{ a.e. in }Q.
$$
Thus, using the monotonicity of $\bG$, we see that
$$
\begin{aligned}
\bT^n\cdot ((\alpha \beps(\bu_0)+\beta\partial_t \beps(\bu_0))-\bG_n(\bT^n))&\le \bT^n\cdot (\bG(\bT_0)-\bG(\bT^n))\le \bT_0\cdot (\bG(\bT_0)-\bG(\bT^n)).
\end{aligned}
$$
Using the convergence results \eqref{Kn1}--\eqref{aennn} applied to all terms in \eqref{test3naa} with the above inequality yields the following:
\begin{equation}\label{test3naab}
\begin{split}
&\frac{1}{4} \int_{\Omega}\beta|\partial_{t} (\bu-\bu_0)(t)|^2 +\beta\left|\partial_{t} (\bu-\bu_0)(t)+\frac{2\alpha}{\beta}(\bu-\bu_0)(t)\right|^2\dx \\
&\le\int_0^t\int_{\Omega} \bT_0\cdot ((\alpha \beps(\bu_0)+\beta\partial_t \beps(\bu_0))-\bG(\bT)) +\alpha |\partial_{t} (\bu-\bu_0)|^{2}\dx\dtau\\
& \quad +\int_0^t\int_{\Omega}(\bef-\partial_{tt} \bu_0) \cdot (\alpha(\bu-\bu_0)+ \beta\partial_t (\bu-\bu_0))+ \frac{2\alpha^2}{\beta}\partial_t (\bu-\bu_0) \cdot (\bu-\bu_0)\dx\dtau\\
&\le C\int_0^t \|\bT_0\|_1 +\|\bef\|_2 +\|\partial_{tt}\bu_0\|_2 +1 \dtau.
\end{split}
\end{equation}
Letting $t\to 0_+$, we see that
$$
\lim_{t\to 0_+}(\|\bu(t)-\bu_0(0)\|_2^2 +\|\partial_t\bu(t)-\partial_t\bu_0(0)\|^2_2)=0.
$$
In addition, it also follows from \eqref{Kn4} that $\bu\in \mathcal{C}^{1}_{loc}(0,T; L^2(\Omega;\R^d))$, which combined with the above result gives that $\bu\in \mathcal{C}^{1}([0,T]; L^2(\Omega;\R^d))$.
\subsection{Validity of the equation in the limit}
To summarize the results so far, we have found a couple $(\bu,\bT)$ that satisfies \eqref{FSu}--\eqref{FST2} and \eqref{T-const2}, \eqref{bcint2}. It remains to show \eqref{WF2}. To do so, we use the method developed in \cite{BeBuMaSu17}. Let $g$ be a smooth nonnegative nonincreasing function satisfying
$$
g(s)=\left\{\begin{aligned}
&1, &&\textrm{for }s\in [0,1],\\
&0, &&\textrm{for }s>2.
\end{aligned}
\right.
$$
For each \( k \in \mathbb{N}\), let us define
$$
g_k(s):=g(s/k).
$$
It is clear that $g_k\nearrow 1$. Next let $\bv\in \mathcal{C}^{\infty}_0(Q; \R^d)$ be arbitrary but fixed. Thanks to \eqref{Kn4} and \eqref{Kn9}, all terms in \eqref{WF2} are well-defined for almost all $t\in (0,T)$ and we just need to check that the equality holds.
Using the properties of $g_k$, we have
\begin{equation}\label{last1}
\begin{split}
I&:=\int_Q \partial_{tt}\bu \cdot \bv+ \bT \cdot \nabla \bv - \bef \cdot \bv \dx\dt \\
&= \lim_{k\to \infty}\int_Q \partial_{tt}\bu \cdot \bv g_k(|\bT|)+ \bT \cdot \nabla \bv g_k(|\bT|) - \bef \cdot \bv g_k(|\bT|) \dx\dt.
\end{split}
\end{equation}
Using \eqref{Kn4}, \eqref{Kn8}, the fact that $\bT^n\in L^2_{loc}(0,T; W^{1,2}_{loc}(\Omega; \R^{d\times d}))$, which follows from \eqref{sp-r-lnn}, and the fact that $g_k(|\bT^n|)$ is supported only in the set where $|\bT^n|\le k$, we can rewrite the right-hand side of \eqref{last1} in the following way:
\begin{equation}\label{last2}
\begin{split}
I&= \lim_{k\to \infty}\lim_{n\to \infty}\int_Q \partial_{tt}\bu^n \cdot \bv g_k(|\bT^n|)+ \bT^n \cdot \nabla \bv g_k(|\bT^n|) - \bef \cdot \bv g_k(|\bT^n|) \dx\dt\\
&= \lim_{k\to \infty}\lim_{n\to \infty}\int_Q \partial_{tt}\bu^n \cdot \bv g_k(|\bT^n|)+ \bT^n \cdot \nabla (\bv g_k(|\bT^n|)) - \bef \cdot \bv g_k(|\bT^n|) \dx\dt\\
&\quad -\lim_{k\to \infty}\lim_{n\to \infty}\int_Q \bT^n \cdot (\nabla g_k(|\bT^n|)\otimes \bv )\dx \dt\\
&=-\lim_{k\to \infty}\lim_{n\to \infty}\int_Q \bT^n \cdot (\nabla g_k(|\bT^n|)\otimes \bv )\dx \dt,
\end{split}
\end{equation}
where for the last equality we have used \eqref{WFn} with $\bw:=\bv g_k(|\bT^n|)$. It remains to show that the right-hand side of \eqref{last2} vanishes. We define
$$
M_{k,n}(s):= \int_0^s \frac{g'_k(t)}{\frac{\phi'(t)}{t}+n^{-1}}\dt \le \int_0^s \frac{t g'_k(t)}{\phi'(t)}\dt=:M_k(s).
$$
Then, using that $|g_k'(s)|\le Cs^{-1} \chi_{\{s\in (k,2k)\}}$, we see that
\begin{equation}\label{prym}
M_k(s)\left\{\begin{aligned}&\le C\min\{s,k\} &&\textrm{for all } s\ge 0,\\
&=0 &&\textrm{for }s\le k.\end{aligned}\right.
\end{equation}
Next we can use the structural assumption \eqref{A4} to rewrite the term under the limit in \eqref{last2} as
\begin{equation}\label{smus}
\begin{aligned}
-\int_Q& \bT^n \cdot (\nabla g_k(|\bT^n|)\otimes \bv )\dx \dt \\&= -\int_Q \bG_n(\bT^n) \cdot (\nabla |\bT^n|\otimes \bv )\frac{g'_k(|\bT^n|)}{\frac{\phi'(|\bT^n|)}{|\bT^n|}+n^{-1}}\dx \dt\\
&= -\int_Q \bG_n(\bT^n) \cdot (\nabla M_{k,n}(|\bT^n|)\otimes \bv )\dx \dt\\
&= \int_Q \diver \bG_n(\bT^n) \cdot \bv M_{k,n}(|\bT^n|) \dx \dt+\int_Q \bG_n(\bT^n) \cdot \nabla \bv M_{k,n}(|\bT^n|)\dx \dt.
\end{aligned}
\end{equation}
For the first term on the right-hand side of (\ref{smus}), we use the definition of $\mA_n$ alongside the Cauchy--Schwarz inequality to obtain
$$
\begin{aligned}
&| \diver \bG_n(\bT^n) \cdot \bv M_{k,n}(|\bT^n|) |=\left|\sum_{i,j,a,b=1}^d (\mA_n(\bT^n))^{ij}_{ab} \partial_j \bT_{ab}^n \bv_i M_{k,n}(|\bT^n|) \right|\\
&=\left|\sum_{m=1}^d \sum_{i,j,a,b=1}^d (\mA_n(\bT^n))^{ij}_{ab} \partial_m \bT_{ab}^n \delta_{mj}\bv_i M_{k,n}(|\bT^n|) \right|\\
&\le \left|\sum_{m=1}^d \left(\partial_m \bT^n, \partial_m \bT^n\right)^{\frac12}_{\mA_n(\bT^n)}\left(\sum_{i,j,a,b=1}^d (\mA_n(\bT^n))^{ij}_{ab} \delta_{mj}\bv_i\delta_{ma}\bv_b M^2_{k,n}(|\bT^n|)\right)^{\frac12} \right|\\
&\le \left|\sum_{m=1}^d \left(\partial_m \bT^n, \partial_m \bT^n\right)^{\frac12}_{\mA_n(\bT^n)}\left((n^{-1}+\frac{C}{1+|\bT^n|})|\bv|^2 M^2_{k,n}(|\bT^n|)\right)^{\frac12} \right|.
\end{aligned}
$$
Using this bound in \eqref{smus} and then in \eqref{last2}, recalling the fact that $\bv$ is compactly supported, we deduce with the help of H\"{o}lder's inequality and the uniform bound \eqref{sp-r-ln} that
\begin{equation}\label{last12}
\begin{split}
|I|&\le \lim_{k\to \infty}\lim_{n\to \infty} \int_Q \left|\sum_{m=1}^d \left(\partial_m \bT^n, \partial_m \bT^n\right)^{\frac12}_{\mA_n(\bT^n)}\left(\left( n^{-1}+\frac{C}{1+|\bT^n|}\right) |\bv|^2 M^2_{k,n}(|\bT^n|)\right)^{\frac12} \right| \dx\dt\\
&\le C(\bv)\lim_{k\to \infty}\lim_{n\to \infty} \left(\int_Q \left(n^{-1}+\frac{C}{1+|\bT^n|} \right) M^2_{k,n}(|\bT^n|) \dx\dt\right)^{\frac12}\\
&=C(\bv)\lim_{k\to \infty} \left(\int_Q \frac{ M^2_{k}(|\bT|)}{|\bT|} \dx\dt\right)^{\frac12},
\end{split}
\end{equation}
where for the last equality we used \eqref{Kn9} and the boundedness of $M_k$. Consequently, using that $\bT\in L^1(Q;\R^{d\times d})$ and the structure of $M_k$ \eqref{prym}, we deduce that
\begin{equation*}
\begin{split}
|I|&\le C(\bv)\lim_{k\to \infty} \left(\int_Q \frac{ M^2_{k}(|\bT|)}{|\bT|} \dx\dt\right)^{\frac12}\le C(\bv)\lim_{k\to \infty} \left(\int_{Q\cap \{|\bT|>k\}} |\bT| \dx\dt\right)^{\frac12}=0.
\end{split}
\end{equation*}
Since $\bv$ is arbitrary, we see that \eqref{WF2} holds for almost all $t\in (0,T)$ and all smooth compactly supported $\bw$. Finally, using a weak$^*$ density argument based on \cite[Lemma A.3]{BeBuMaSu17} we deduce that \eqref{WF2} holds for an arbitrary $\bw\in W^{1,2}_0(\Omega,\R^d)$ fulfilling $\beps(\bw)\in L^{\infty}(Q;\R^{d\times d})$. This concludes the proof of the existence of a solution as asserted in Theorem~\ref{T4.1}.
\subsection{Uniqueness of the solution}
It remains to prove the uniqueness of such weak solutions.
Let $(\bu_1,\bT_1)$ and $(\bu_2,\bT_2)$ be two solutions emanating from the same data and denote $\bv:=\bu_1-\bu_2$. Then it follows from \eqref{WF2} that, for almost all $t\in (0,T)$ and for every $\bw\in W^{1,\infty}_0(\Omega;\mathbb{R}^d)$,
\begin{equation}\label{uniq1}
\int_{\Omega} \partial_{tt} \bv \cdot \bw + (\bT_1-\bT_2)\cdot \beps(\bw)\dx =0.
\end{equation}
Since $\partial_t \beps(\bv)$ and $\beps(\bv)$ belong to $L^{\infty}(\Omega;\R^{d\times d})$ for almost all $t\in (0,T)$, we can again use the weak$^*$ density argument as in the previous section to deduce that \eqref{uniq1} holds with $\bw:=\alpha \bv + \beta \partial_t \bv$. Consequently, since we have
$$
\alpha \bv + \beta \partial_t \bv=\bG(\bT_1)-\bG(\bT_2),
$$
we can use the monotonicity of $\bG$ and integration over $(t_0, t)$, with $0<t_0<t < T$, to deduce from \eqref{uniq1} that
$$
\begin{aligned}
0&\ge 2\int_{t_0}^t \int_{\Omega}\partial_{tt} \bv \cdot (\alpha \bv + \beta \partial_t \bv)\dx \dtau \\
&\quad= \beta \int_{\Omega}|\partial_t \bv(t)|^2 - |\partial_t \bv(t_0)|^2 +2\alpha \partial_t \bv(t)\cdot \bv(t) - 2\alpha \bv(t_0)\cdot \bv(t_0)\dx -2\alpha \int_{t_0}^t\int_{\Omega} |\partial_t \bv|^2\dx \dtau.
\end{aligned}
$$
We note that this procedure is rigorous for every such $t_0>0$ thanks to the regularity of $\bu_1$ and $\bu_2$ asserted in \eqref{FSu2}.
Since $\bv\in \mathcal{C}^1([0,T]; L^2(\Omega;\R^d))$ as a result of \eqref{FSu2}, we can use \eqref{bcint2} and let $t_0\to 0_+$ in the above inequality in order to deduce that
$$
\begin{aligned}
0&\ge \beta \int_{\Omega}|\partial_t \bv(t)|^2 +2\alpha \partial_t \bv(t)\cdot \bv(t) \dx -2\alpha \int_{0}^t\int_{\Omega} |\partial_t \bv|^2\dx \dtau\\
&=\beta \int_{\Omega}|\partial_t \bv(t)|^2 +2\alpha \partial_t \bv(t)\cdot\left( \int_0^t\partial_t \bv(\tau)\dtau\right) \dx -2\alpha \int_{0}^t\int_{\Omega} |\partial_t \bv|^2\dx \dtau\\
&\ge \frac{\beta}{2} \left(\|\partial_t \bv (t)\|_2^2 - C(\alpha,\beta,T)\int_0^t \|\partial_t \bv(\tau)\|_2^2 \dtau\right)
\\&= \mathrm{e}^{-tC(\alpha,\beta,T)}\ddt \left( \mathrm{e}^{-tC(\alpha,\beta,T)}\int_0^t \|\partial_t \bv(\tau)\|_2^2 \dtau\right).
\end{aligned}
$$
Simple integration with respect to $t$ then gives that $\partial_t \bv \equiv 0$ almost everywhere in $Q$ and consequently $\bu_1=\bu_2$. By strict monotonicity, we necessarily also have that $\bT_1=\bT_2$ almost everywhere in $Q$ and, hence, uniqueness follows.
\bibliographystyle{plainnat}
\bibliography{bibliography}
\end{document} | 213,200 |
TITLE: Show that $f=0$ given $\int_{a}^{b} x^n\,f(x)\,dx=0$
QUESTION [4 upvotes]: Let $f:[a,b]\rightarrow \mathbb {R} $ a continious function such that $$ \int_{a}^{b} x^n\,f(x)\,dx=0$$ for all $n \in \mathbb N$.
Show that $f$ is identically $0$.
I notice that $f$ is bounded and it touches its bounds. But I don't know how to continue.
REPLY [1 votes]: Assuming you mean for all $ n \geq 0 $ rather than some fixed $ n $, this is true. One possible proof is to first note that since this result holds for all $ n \geq 0 $, for any polynomial function $ p: [a,b] \to \mathbb R $, $ \int_a^b p(x) f(x) dx = 0 $. Using Stone-Weierstrass, we then have that for any continuous function $ g: [a, b] \to \mathbb R $, $ \int_a^b g(x) f(x) dx = 0 $. If $ f $ is itself continuous, then simply take $ g = f $ to conclude. If f is not continuous, then you can only conclude that $ f = 0 $ a.e., so I only know how do to this with Lebesgue integration.
In this case, we use the fact that for a function $ f \in L^1([-\pi, \pi]) $, if every Fourier coefficient $ a_n = \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) e^{-inx} dx = 0 $ then $ f = 0 $ a.e. For a reference, this is Theorem 3.1 in chapter 4 of Stein and Shakarchi's Real Analysis. As $ \cos(nx) $ and $ \sin(nx) $ are continuous for all $ n \in \mathbb Z $, and as $ e^{inx} = \cos(nx) + i \sin(nx) $, we have that every Fourier coefficient of $ f $ is 0, so $ f = 0 $ a.e. by the theorem cited. | 75,467 |
Saturday, August 20, 2011
KARACHI: AN UPDATE
B.RAMAN.
2. The victims in the four days of fresh violence were mainly Mohajirs supporting the Muttahida Qaumi Movement (MQM) of Altaf Hussain mainly representing Mohajir migrants from Uttar Pradesh and Gujarat, the Mohajir Qaumi Movement (Haqiqi) mainly representing migrants from Bihar, Pashtun supporters of the Awami National Party (ANP), Balochs without political affiliation and Barelvi Sunnis of the Sunni Tehreek.
3.The number of victims in the Sindhi community, which supports the ruling Pakistan People’s Party (PPP) of President Asif Ali Zardari and the various Sindhi nationalist parties, and in the Punjabi community, which supports the Pakistan Muslim League (PML) of Nawaz Sharif, has reportedly been low, but exact figures are not available.
4.The deterioration in the situation has been partly the outcome of the alleged action of the Government in releasing the leaders and cadres of the anti-Altaf Hussain MQM (H) who had been arrested and jailed by Pervez Musharraf when he was the President as part of a secret deal with Altaf under which the MQM observed restraint in Karachi in return for the jailing of Altaf’s opponents in the Mohajir community.
5. The MQM of Altaf sees the release of Altaf’s Mohajir opponents by the PPP-led Government as a revival of Benazir Bhutto’s policy (1988-90 and 1993-95) of pitting the MQM (H) against the MQM in the streets of Karachi .
6. The present spell of violence, which started as business and smuggling related clashes between the Barelvi Mohajirs of the MQM and the Deobandi Pashtuns of the ANP, has since assumed a wider dimension with Mohajirs killing Mohajirs. The ethnic and sectarian strife, which one saw at the beginning of the present spell of violence, has been aggravated by gang warfare between rival Mohajir mafia gangs.
7. It is pure and simple criminal violence not motivated by any political ideology or religious goal. The violence is about who controls the mafia economy of Karachi. There have been increasing demands for Army intervention since none of the groups involved in the violence has any confidence in the police, which is controlled by Rehman Malik, the Interior Minister belonging to the PPP.
8. The demand for Army intervention has come from the Mohajirs of the MQM, who allege that there has been Taliban infiltration into Karachi under the cover of the ANP, the Pashtuns of the ANP, who look upon the violence as the result of the Mohajir mafia warfare, the Balochs, who find themselves caught in the violence between the Mohajirs and the Pashtuns, the Barelvi organisations and all major business organisations.
9. The only organisations not in favour of an Army intervention are the PPP, the Sindhi nationalist parties, the MQM (H) and the PML (N). The Army, while expressing its concern over the continuing violence, has said that it is for the civilian Government to deal with the situation.
10. The Army is unlikely to intervene unless there are targeted attacks on military, Air Force and naval personnel in uniform performing duty or on military, Air Force and naval establishments or the Karachi port.
11. The latest round of violence has targeted the Police . A bus carrying police officers in mufti was attacked killing four of them. The death of an Air Force employee has also been reported, but he was reportedly on a private visit to Karachi. The Army, the Air Force and the Navy have not so far been targeted. The attack on PNS Mehran, the headquarters of the naval air wing in Karachi in May, was not related to the ongoing ethnic and sectarian violence. It was a pure and simple terrorist attack in which the Tehrik-e-Taliban Pakistan (TTP) was suspected.
12. Sections of the Pakistani media have carried highly pessimistic accounts of the situation in Karachi---with the “Dawn” of Karachi even saying that Pakistan is unravelling.
13. Pakistan is not unravelling. The cycle of violence in Karachi---sometimes up, sometimes down—will continue, but none of the contending parties is likely to force a strategic confrontation, which could lead to the destabilisation of Pakistan. Periodic tactical confrontations will continue till the policing of Karachi improves and the criminal-politician and criminal-police nexus is broken. That is not for tomorrow. ( 21:52 PM 2 comments:
SOMEWHERE ONLY WE KNOW
"Somewhere Only We Know"
( "Somewhere Only We Know" composed and played by English alternative rock band Keane.It became one of the greatest hits of 2004 worldwide.)
So if you have a minute why don't we go
Talk about it somewhere only we know?
This could be the end of everything
So why don't we go
So why don't we go
This could be the end of everything
So why don't we go
Somewhere only we know?
Posted by B.RAMAN at 1:14 AM 1 comment:
| 292,713 |
TITLE: consecutive states in Markov chain
QUESTION [0 upvotes]: Let $S = \{1,2\}$ denote a state space and consider a Markov process on $S$ with a positive transition matrix (ie, all entries are strictly positive). Let $S_\infty = \times_{t=1}^\infty S$ denote the underlying probability space and $\mathbb{P}$ denote the probability measure on $S_\infty$ induced by the Markov transition matrix.
For each $n$, let $F_n := \{ s \in S_\infty : \text{there are at most $n$ consecutive occurrences of $1$ or $2$} \}$. Is it true that $\lim_{n \to \infty}
\mathbb{P}(F_n) = 1$?
REPLY [0 votes]: Of course not. Let $A_i$ be the event that $i$ occurs infinitely often. Then $P(A_1\cap A_2)=1$ since $P(A_1\cap A_2)=P(A_1)+P(A_2)-P(A_1\cup A_2)=2-P(A_1\cup A_2)\geq 1$, since the probability of $1$ or $2$ occuring infinitely often is 1 by ergodicity. It's a fact that any finite irreducible Markov chain with no absorbing states is ergodic in that it visits each state infinitely often with probability 1. Furthermore it will visit each set of positive measure at least once, meaning that it will definitely visit the event that there are more than $n$ consecutive 1's or 2's. | 84,426 |
Am I the only one who used to make batches of cookie dough and freeze it?? I never actually baked the cookies and while a delicious college treat, not super ideal with eggs and all. Introducing my high protein edible cookie dough (aka chickpea energy bites!).
A lot of days, it still takes some finagling for me to figure out how to eat enough protein to maintain muscle. Largely because I don’t particularly crave protein or large meals and simply eating by calories alone, my protein intake wasn’t nearly high enough for my goals or to help my body go in to major repair mode.
What I am always able to eat enough of is chocolate.
Which means this is a dessert that’s helping me hit all my goals, WOHOOOOO.
Why not make my chocolate even better by recreating that cookie dough taste using BEANS!!!
- No dairy
- No eggs
- No gluten
I swear if you haven’t ever tried this, you’ll be floored. Why it tastes like cookie dough I can’t even explain, I’m a runner, not a chef remember. It’s heaven.
I have zero control around heaven.
Thus I decided to turn that edible cookie dough in to bites, so I would automatically be portion controlled.
Edible Cookie Dough Chickpea Energy Bites
I made a massive batch and to be honest, I’d recommend smaller, ha!
I thought I was saving time, so I’m sharing here a half recipe of what I made.
These freeze ok, but not as good as other things we’ve made. Enjoy Life Dark chocolate chips (to keep it dairy free)
Directions
While I used the Vitamix for these chickpea energy bites, it would again have been easier in my food processor…I was being lazy.
- Add almonds to food processor and chop away (can skip this and use salted almond butter)
- Remove almonds and pour in the chickpeas
- You don’t want to create a soup, just getting the chickpeas blended
- In a large bowl, combine all ingredients and stir
- Place in fridge for 20-30 minutes, you know enough time to go read a few blogs
- Using a cookie scoop, scoop them out and on to wax paper
- They aren’t perfectly round, but it’s easy and really…well I think you’ve seen the trend here
- You can layer them in a Tupperware container with wax in between
- Keep in the fridge
Nutrition Info
Makes roughly 30 balls using the scoop
60 calories
3.7 grams fat
9.8 grams carbs
3.6 grams protein
Looking for more high protein ideas?
- High protein 3 ingredient pancakes
- High protein post workout smoothies
- High protein cookie dough (no beans!)
Have you ever made a dessert with chickpeas?
Are you a lazy cook? Just me?
Other ways to connect with Amanda
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Cyndi
These sound so good!! I love chickpeas and have been wanting to make some protein bites.
Emily Swanson
I can’t remember, but I think I’ve used them to make some sort of bars? I’m glad you got such a good trainer who worked out a formula for you that has worked so well; it’s amazing how our bodies need that protein to repair!
Jennifer
I also had a surgery and for a fitness freak like me it was so tough, but I went a different way. Someone told me about strong black seed oil. I call it a miracle worker, not only did it help me maintain my weight, but also worked as an anti-inflammatory and pain relief agent. I usually make its tea and now thinking of combining this cookie doe and my tea for a very healthy snack. Looking forward to it…thanks for the recipe :) | 84,003 |
Olinguito, Bassaricyon neblina
The olinguito had been mistakenly identified for more than 100 years and is also the first carnivore species to be discovered in the American continents in 35 years. (Mark Gurney/Smithsonian/Getty Images)
Zookeepers tried — unsuccessfully — for years to get a little creature named Ringerl to breed with their olingos, thought to be her brethren. But it turned out Ringerl is another species altogether: the olinguito, a “little, adorable olingo,” the first new mammalian carnivore discovered in the Western Hemisphere in 35 years. Weighing about two pounds, the olinguito is different than its cousin the olingo in a variety of aesthetic ways, including having a rounder face, a shorter tail, smaller ears and darker fur. And though Ringerl spent her life in zoos, her true brethren are living successfully in the wild, with a population estimated in the tens of thousands potentially distributed through Central and South America.
MORE FROM WEATHER.COM: Amazing Hybrid Animals
Eclypse is a zorse. Her mother is a Chapman's zebra and her father is a horse. (Yannik Willing/AFP/Getty Images)
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TITLE: Proof for value of sum of sine and cosine
QUESTION [1 upvotes]: I have come across those sums of sine and cosine, while trying to show that windmills dont move without external force. Although it is clear that should be the case, i'm stuck in proofing it. I wonder if there is some way to proof
$$\sum_{i=0}^{n-1} \cos\left(\frac{2\pi i}{n}\right) = 0 \tag{1} \label{1} $$
In case of sine the proof is easy because of simple symmetry
$$\sum_{i=0}^{n-1} \sin\left(\frac{2\pi i}{n}\right) = 0 \tag{2}$$
Also if n is even $\sum_{i=0}^{n-1} \cos\left(\frac{2\pi i}{n}\right) = 0$ is easy to show.
So my question is: How do you prove or at least argue that $\eqref{1}$ holds for all n?
REPLY [1 votes]: Let
$$
\zeta=e^{2\pi i/n}=\cos\left(\frac{2\pi}n\right)+i\sin\left(\frac{2\pi}n\right).
$$
then for all $k=0,...,n-1$,
$$
\zeta^k=e^{2\pi ik/n}=\cos\left(\frac{2k\pi}n\right)+i\sin\left(\frac{2k\pi}n\right).
$$
Since $\zeta^n=1$ but $\zeta\neq1$ we must have
$$
1+\zeta+\zeta^2+\cdots+\zeta^{n-1}=0.
$$
Take real and imaginary parts. | 161,157 |
By Gina Perille, Globe Correspondent | September 29, 2004
Few plays are forever relevant, and attempts to make them so can amount to untold dramatic indignities. Fortunately, Richard McElvain chose a sturdy play to adapt and made bold writing choices in his re-imagining of Sophocles's "Antigone" at Boston Playwrights' Theatre.
McElvain sets the action in a modern-day, war-torn society, complete with press conferences, political coverups, and street-wise vernacular. We see Creon, ruler of Thebes, blustering through a photo-op and a citywide decree of victory over a civil war insurgence. Creon's chorus is made up of Brooks Brothers-clad staff members who swarm about him, affirming his every move.
Nearby is Antigone -- daughter of Oedipus and sister of two fallen soldiers from opposite sides of the conflict. She learns of Creon's executive order prohibiting consecration and burial of one of her brothers. When Antigone defies the order and later refuses to repent, Creon refuses to pardon her, thereby setting off a chain of tragic events, including two blood-soaked suicides.
Despite the title, "Antigone" proves to be as much about Creon and his unilateral decisions as it does about the woman he punishes for not changing her mind. McElvain's adaptation deftly presents two compelling, if unreasonable, characters. Marianna Bassham infuses Antigone with a passionate recklessness that is not hysterical, while McElvain, who also performs in the production, pounds out Creon's moral clarity with a resolve that would be admirable if it were not so frightening.
It's impossible to look through McElvain's forcibly modern lens and not see the parallels with today's newspaper headlines, but his adaptation gives equal time to the universal dangers of inflexibility and insulation.
This Nora Theatre Company production is not without missteps, however -- the most glaring of which is the wandering format for the delivery of the chorus sections. Greek choruses have long been the recipients of creative license in terms of structure, cadence, and rhythm. But there's free form and then there's free fall. The quasi-sung, quasi-choreographed sections here are misguided at best. One limp segment, combined with the corporate costumes, drifts closer to a tableau from "How to Succeed in Business Without Really Trying" than anyone could have intended.
The members of the chorus also play a range of individual roles and more than redeem themselves in the other portrayals. Donna Sorbello is masterful as Eurydice, wife of Creon, generating tremendous emotion out of the simplest movement. Eric Mello conjures eerie tenderness as the unburied brother Polynices, and Jim Spencer captures the conflicted essence of Haemon, son of Creon and lover of Antigone.
Also, Sylvia Ann Soares is effectively disturbing as the prophecy-spewing Tiresias. Rounding out the cast is Ed Peed, who plays the Guard with occasionally too much flair, and Jessica Burke in the essentially thankless role of Ismene, the sister of Antigone who embodies what obedient women "should" do in society.
McElvain's adaptation is raw and urgent, providing a fresh context for the message Sophocles intended so many centuries ago: It takes more courage to admit error than it does to show resolve. McElvain's update brings that message forth in a most graphic way.
AntigonePlay by Sophocles; original adaptation by Richard McElvain. Directed by: Daniel Gidron. Set, Brynna Bloomfield. Lights, Kathy Peter. Costumes, Jacqueline Dalley. Original Music and Sound, Dewey Dellay. Lights, Kathy Peter. Presented by the Nora Theatre Company.At: Boston Playwrights Theatre through Oct. 3; 617-491-2026. | 95,529 |
Ski Stats
Green Runs:13%
Blue Runs:40%
Red Runs:37%
Black Runs:10%
Artificial:800 guns
Cross Country:45km
Self Catered
£525
8 April 2017 (7 nights)
Scheduled flights
Accommodation
Transfer
7 nights | Guide prices per person
Residence.
One of the most distinctive ski resort in the Alps, Avoriaz is perched high up on a snow shelf looking down over a cliff edge towards Morzine and beyond.
There are no cars, making for a calm, safe atmosphere. Lifts are very close to the accommodation, which varies greatly depending on the age of the building. There are about 25 restaurants (some good, but pricey) and reasonable nightlife. The ski school is excellent. Children are well catered for, including a very good kids' village and Geneva is only two hours away.
Overall, Avoriaz ski holidays are blessed with a number of qualities to establish the resort's excellent reputation in the region. It's good for its snow, its challenging skiing and the fact that you can always ski to your door.
Green Runs:13%
Blue Runs:40%
Red Runs:37%
Black Runs:10%
Artificial:800 guns
Cross Country:45km
Avoriaz is arguably one of the most enjoyable resorts in the region with Portes du Soleil on the doorstep this is a vast area of skiing to enjoy. Sitting at 1,800m, it can guarantee good snow, unlike most other resorts. With a large area of varied runs there is enough skiing to satisfy all levels. For more experience skiers there are some great black pistes, the most notable of which is 'The Wall', which plunges you down into Switzerland and should only be tried by those with plenty of tough miles under their belts. But don't despair beginners and intermediates - there are countless motorway pistes across the whole network to keep you busy.
A great spot for freestyle, there is a specialist park, a half-pipe, a slalom course and extensive off-piste. Additionally, there is a giant "ecological park" called the Stash with loads of natural modules set off-piste between the trees. Every season they manage to add new features so that you will never have a dull moment finding new areas to discover.
A large variety of restaurants allows you to try as many culinary variations as possible. Table du Marche, which is a sister restaurant to similarly cool places in St Tropez and Marrakech, is probably the best restaurant in town thanks to the excellent team of dynamic and experienced chefs. The cosy wood-panelled restaurant of Salle a Manger feels relaxed and welcoming and is the perfect setting for the rustic French food that is served. Found right at the foot of the slopes, Trappeurs serves delicious and hearty Savoyard dishes on its sunny terrace. If you like your alpine meals with a side serving of Savoyard cheese then Les Fontaines place is for you. The fondue and raclette here are excellent, just like the welcoming staff.
Avoriaz has vibrant night life with plenty of great bars and night clubs to occupy the evenings. Chapka is a great spot for a relaxed evening with friends, with a lively atmosphere and a trendy bar. Fantastique is a friendly and welcoming bar with great drinks and a snack food at the end of the day. With a varied cocktail menu and live local bands and DJs on selected nights it promises a good time. If you want to stay out until late, Le Yak is open until 4am and offers the best clubbing in the area. For great live music go to The Place or the Wild Horse Saloon and for those looking for something different theres even a bowling alley that's open until 2am.
The new Aquariaz, a bowling alley and the Altiform Fitness Centre along with the shops and restaurants in Avoriaz with keep those who arent skiing busy during the day. Easy access to Morzine, Les Gets and other resorts in the Portes du Soliel area means that you can explore the surrounding area easily. Trips to Annecy and Evian are also possible. | 400,483 |
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July 12, 2008
Political Observations
A few things have come to mind this week, all to do with Inconsistency.
- All the people/newspapers who a few years ago were outraged at the imprisonment of Tony Martin are now backing imprisoning everyone who carries a knife. So apparently you shouldn’t go to jail if you shoot someone with a gun, but if you just carry a knife you should.
- On last week’s Question Time the audience appeared to clap everything people said. This meant that they would clap a point by one of the commentators, and then applaud an exactly opposing argument by another panelist. This suggests both that the panelists were making strong arguments, and that the audience was full of idiots.
- It was commonly noted that during the G8 meeting politicians undertook an 8 course meal, whilst simultaneously talking about Global food shortages, and in Gordon Brown’s case telling people not to waste food. The latter isn’t hypocritical, assuming Mr. Brown finished his meal.
June 13, 2008
David Davis
David Davis’ resignation appears to have caused shockwaves throughout British Politics. He could have stayed in the safe and comfortable position of Shadow Home Secretary, and pushed for his issues when the Conservatives get elected in two years time. Instead he has chosen to gamble.
Their Opinion
People’s reaction to David Davis’ actions seem to be strongly split among those who agree or disagree with his argument.
Jackie Smith, Home Secretary, and Hazel Blears, Communities secretary, when interviewed ignored the issues and just talked about Conservative ‘disarray’, and David Blunkett called it ‘political theatre’. Conservatives who made their opinion know publically seemed to back his decision, this includes David Cameron and Dominic Grieve, the new shadow home secretary.
Civil Liberties campaigning groups, such as NO2ID and Liberty have backed his stand. Newspapers position is consistent with their editorial stand on the issues, for example The Independant is describing him as a ‘Freedom Fighter’, whilst the Sun claims he has “gone stark raving mad”, is “a quitter” and describes it as “Treachery”.
A real Motive?
A lot of people have claimed that this isn’t the real motive for his resignation, but its quite clear that Cameron & Co are backing his move, albeit cautiously. His replacement, Dominic Grieve, takes the same stance as he does on the issues so there will be no change of Conservative Policy, which means there’s no rejection of his views in central office. Whilst I’m sure many Conservatives are pissed off that he taken an action that won’t improve their electoral prospects, I doubt he was being muffled by party policy.
Another assertion, made by both The Sun and The Daily Express, is that this is a politically motived act of treachery towards David Cameron. Its certainly not Cameron’s style of politics. Its hard to see what this gets David Davis politically. He has lost his position as Shadow Home Secretary, and may loose his seat as an MP. The level of risk to his seat is, I believe, quite low – he has a strong majority and liberal democrats won’t be opposing him, but certainly the move from his current position to a back bencher is a long step down. The Sun reconcile the lack of motivation with their cynical claims by asserting that he has gone mad, an unsound argument if ever I heard one.
Opposition
The liberal democrats won’t be standing against him, because they agree with him on the issues. If I were they, I would campaign for him, the only way you can get political advantage out of this is to show that you have a non-partisan principled stand on the issue.
The Labour party haven’t made their decision clear. I would be very surprised if they put a candidate forward, their narrative of these events is one that tries to undermine David Davis at every turn (‘disarray’, ‘theatre’) the most sensible way to continue this approach is to not stand a candidate against him. Furthermore, they are unlikely to win, (at the last general election their candidate placed 3rd with 6,000 votes to David Davis’ 22,000). It would also cost the party campaigning funds that it can ill-afford.
The BNP won’t be running a candidate against him, since they agree with his position, and UKIP don’t know what they are doing. Unsurprising for a party who published its manifesto for the 2005 General Election with a typo in the headline of its first page. Even worse than one of my blog entries!
Kelvin Mackenzie, former editor of the Sun, has said he will stand. This is the first time that Rupert Murdoch has directly pushed his own candidate, instead of backing existing parties. Quote of the week: “The Sun has always been very up for 42 days and perhaps even 420 days.” Since he will have financial support from Mr. Murdoch its highly likely that he will have a funding advantage. This entirely suits Mr. Davis, since it will result in a debate, it also suits the Labour party since they have someone to fight their battles. I doubt Mr. Mackenzie will win, since I find it impossible to believe that anyone could like an editor of a tabloid newspaper.
Observations
Don’t believe the ‘cost to the tax payer’ argument: the total cost of a by-election according to the BBC is £2000. The cost to political parties is far greater.
Whilst there are clear negative effects on the Conservative party, people are completely ignoring the positive effects: no news about nannygate! This is a far more interesting story than the alleged corruption of the Conservative Party Chairman, whilst the corruption story is potentially more damaging. The real concern people have here is the risk factor.
For years people have called on politicians to take actions that are nonpartisan, based on principle, and creative. This is certainly creative, due to its unprecedented nature, as already established this is based on principle and its clearly not a partisan act – the man is operating off his own bat. If there is a rejection of this action by the British public I can only conclude one thing: that they are more hypocritical and two faced than the politicians.
May 23, 2008
Oil!
Recently Oil prices reached a high of $135/barrel, this appears to have caused consternation amongst many. Gordon Brown has noted this as a problem when reflecting on his loss in Crewe and Nantwich, whilst it cropped up in this week’s Question Time. , that also contained the ever-annoying Hazel Blears.
There are many proposed solutions here – hybrid cars, more fuel efficient cars, better public transport. These all reduce the amount of fuel consumed by the economy. The ever-annoying Hazel Blears argued that we should seek to increase supply – newer fuel reserves. Whilst it is obvious that rising fuel prices, will make utilizing more expensive fields profitable, and thus actioned, as a government strategy this isn’t viable.
Firstly because fossil fuels are an inherently finite resource – they will run out, so this strategy is only ever going to work in the short term. On a more practical note, many scientists seem skeptical about the viability of further north sea exploration. Apparently we gave most of the oil to the Norwegians when drawing the borders.
Unfortunately many ignorant talking heads appear to have taken it upon themselves to propose a tax decrease. This, again, is a short term solution – tax decreases, so the cost of fuel decreases, demand increases, so the price increases. This would encourage exploration – due to the long term price rise, but either way it encourages increased consumption – so the fuel reserves run out sooner. After the immediate gain, we loose out in the long run.
There is another alternative for improvement. It involves a fantastic piece of technology. Its very cheap. By consuming less fuel it also reduces environmental problems. Additionally its healthy for you (though not as much as people think). My suggestion is …
BUY A BIKE
Overwhelming people’s transport options can be fulfilled by cycling – very few people in the UK travel long distance as part of their daily journeys, many of those who do can get their fuel costs subsidized by their business (e.g. sales Reps, road haulage), most of our journeys consist of urban driving. These needs can nearly always be fulfilled by bicycling.
I don’t know how to ride a bike (I never learnt whilst a child) and I’m almost sorry that this isn’t more of a stigma. Does this make me a hypocrit? Of course not … I walk to work.
May 10, 2008
Clinton Out!
After Obama destroyed Clinton’s last chances of taking the popular vote or pledged delegates in Indiana and North Carolina – the knives were out in the American press. It had gone beyond any sort of reasonable analysis and had entered the realm of pop-culture references. For example Politico described her campaign strategy as the Death Star, whilst Huff Post went the more personal “EVERYONE BUT HILLARY KNOWS IT”, and CBS implies her campaign should be euthenized.
Clearly She can see the light at the end of the tunnel – but doesn’t know its the train thats coming to run her over. Anyhow, rumour mill suggests that Obama is going to claim victory on May 20th – so it’ll be interested to see if she intend to push it beyond then, or call it quits.
April 21, 2008
Council Elections 2008
Coventry
It seems a bit unfair of me to have already mentioned the London Mayoral elections, given their massive media coverage, without having talked about our local vote on 1st May. Its council election time!
Practical Points
The election is on May 1st. It is now too late to apply for a postal vote, Your polling card will specify what what ward you live in, and where you need to vote. If you can’t be bothered to move, you can find out here . I live in Whoberley, which has a wikipedia page that really needs improving, and if you live on campus you are likely to be in the Wainbody ward. Each ward has three councillors, for a total of 54 councillors in Coventry City Council. One will be elected this election. You can find out who is running here If you live in whoberley and are confused by the map on the polling card then I’ve created a Google Map .
Background
Historically Coventry has been a Labour stronghold. They held control of the council every year from 1973 to 2002 with the exception of 1978. The conservatives took over in 2006, with a period of no overall control in between. Both Wainbody and Whoberley have currently three conservative councillors. In wainbody in 2007 the conservatives took 56% of the vote, approximately 2.5 times the number of votes that the second place candidate came in. A student (Emma Biermann) stood for election as the Green Party candidate, coming in 4th place with nearly 9% of the vote.
Whats worth noting is that there are approximately 4,000 students living on campus at University of Warwick, the university registers everyone to vote, and the total turnout in wainbody at the last council elections is also about 4,000 people. In other words a highly motivated student populace could not only swing a close vote – they could vote in whomever they wanted. (Assuming of course that not too high a percentage of campus dwellers are ineligible to vote.)
In Whoberley the conservative majority in 2007 was only 127 votes. Furthermore, the incumbent Conservative Councillor (Joan Griffin) has been booted from their ticket due to health scares, but is running as an independant. I suspect this will split the vote, and offer somewhat of a spoiler for the conservative party.
Whoberley
I’d really like to cover the liberal democrats at these elections, but they don’t really seem to be making much of an effort to get my vote. The Labour Party has put out a full, front, page ad in the Coventry Times, whilst the Conservatives have sent out quite good looking flyers. I have received no such information from Brian Rees Lewis (who also ran as the Lib Dem candidate last time). I can’t even find his name listed on their website. Their website is also incredibly uninformative. They have 9 days to shove something under my door if they want me to care.
Joan Griffin seems to be emphasizing school discipline and minimising the number of drinking licenses handed out. Both of these policies put me off. The conservatives and their candidate Roger Bailey have made a series of commitments for the next few years:
“Continue the ‘Contact and Connect’ service for our elderly residents” & “Keeping council tax rises below pensions” – I think is probably a good thing overall, and that policies should be sympathetic to the needs of those on fixed incomes, but its not really a key issue for me.
“Keeping Pool Meadow open” & “Keeping weekly refuse collections” – I don’t believe any party has the balls to close it, so this seems like rabble rousing to me. Also neither of these are new policies, so making pledges on them is a pretty weak campaigning effort.
Expanding recycling – thats a great pledge, but the council also wanted to allow building on Hearsall Common before a residents petition (organised by Lib Dems in the apparently vastly more active neighbouring region) stopped it . There’s more to environmental issues that just recycling and its a very vague pledge – they don’t specify what they will actually do to improve recycling.
The Labour candidate for Whoberley, Bally Singh, seems to be making a strong effort – lots of emphasis on consultation and a blog. He also also emphasises working with other people – quite important since I expect that Coventry will remain a Conservative controlled council even if he is elected, so he will need to force change from a minority position – a hard task to accomplish. He provides 3 policy pledges:
“Protect our local environment” – he specifies problems here well (Hearsall Common, Watchmakers Building, litter) though not solutions. He also a more broad ranged view of environment than the Conservative Pledges.
“Challenge Anti Social behaviour” – I have no sympathy for ‘anti-anti social behaviour policy’. He somewhat redeems himself by discussing how they can work with the local MP to try and get more activities for people to do.
“Improve our council services” – He doesn’t actually say what he will do, but he does talk about consulting with the community, so this pledge is somewhat worthless, but not entirely.
Your Vote
Please do vote in your local council elections. I’ve presented what I think is a useful summary of information, so you have no excuse.
April 11, 2008
Tibet
Recently a lot of campaigners have argued the case that China shouldn’t be allowed to hold the olympics due to its ongoing occupation of tibet. It has been proposed that a boycott would reduce the positive publicity that the Chinese would be getting from their hosting activities. Gordon Brown announced the other day that he wouldn’t be present at the opening ceremonies. Media fetishism suddenly turned this into some kind of important stand again Chinese policies on tibet. This was denounced by Downing Street in an effort not to upset diplomatic relations.
But even if he hadn’t turned up – how is that progress? I can’t imagine Wen Jiabao trembling in his boots. Do we think that not having to shake some fat Scottish bloke’s hand is the biggest stand we can make on foreign policy? Is that what being a world power means these days? We fight wars against 3rd world countries actively and via proxy and yet here we are facing the another major country and all we can manage is a minor diplomatic snub? Will the historians of the 22nd Century ever enscribe the sentence “Chinese human rights abuses were cleaned up thanks to a second rate act of gesture politics by an unpopular 1/2 term prime minister?” I sincerely doubt it.
The key issue here is that whilst people have been caught up in the idea of denying China the publicity and support that the Olympics brings – we politcally, legally and by implication morally acquiesce to their acts. We’ve somehow managed to completely the obvious fact that Britain recognises Chinese rule over Tibet. We signed a treaty with the Qing empire in 1904 and ratified it in 1906. We officially considered it part of the Republic of China when dealing with Chiang Kai-shek in the second world war. When the west decided to be less hostile to the PRC the Tibet issue was ignored.
Bottom line: if you really want a stand from Britain on tibet – don’t boycott the olympics. Call for an Act of Parliament changing Britain’s recognise Chinese borders to exclude tibet. Now that really would be a stand on the matter.
March 03, 2008
(Little) Uber Tuesday II
Follow-up to Uber Tuesday from True Contradictions
I haven’t blogged about the US primaries in nearly a month, so its time to get back on the prediction box again. In the last month the big day turned into a score draw, followed by 11 straight primary defeats for Clinton. (The Obama camp count 12, since the results for democrats abroad, who voted on uber tuesday, weren’t in until afterwards.) So surely the race is over now? Obama’s got the delegate lead, after 11 wins he must have the momentum.
Well not quite. Tomorrow Texas, Ohio, Rhode Island and Vermont all vote, with Texas and Ohio holding the lions share of the votes. Both of these states are traditionally expected to be Clinton strongholds. Both Bill and that clever guy with no hair, Carville have both said that these are must win states for Clinton. Her campaign has focussed on them heavily in the last month, since a defeat in either would really signal its end. Polls initially going her way have turned against her in the last week or so.
The real question at this point is whether Obama’s mail to the Canadians saying he really loves NAFTA and he’s just using it as a political football will hurt him. I suspect not his late in the day. Clinton has made a resurgence in the polls – she was down an average of 7 pts in Texas from polling data 5 days ago, its back to a level playing field again. Its entirely possible her message sits better with people in Texas and Ohio than Obama’s – I can’t really see why she is gaining otherwise. And the numbers frequently lie. Especially given the inherent pro-Obama bias of the texas polling system – I wouldn’t be surprised for him to take more delegates, whilst Hillary wins the popular vote. Hillary will probably win Ohio by 5-10% or so.
More interesting than this is speculation on the final race for the presidency. McCain has been doing very well in nationwide polling recently – of the last 5 nationwide polls I’ve seen Obama has only won one of them. Still the difference is narrow enough at the moment to not really be important. McCain’s campaign is in a poor state of organisation, and highly susceptible to a heavy tv spend over the summer months – as Clinton did in 1996 against Bob Dole. Such a strategy could only really work if Obama were to take the nomination early, months of uncertainty would help McCain a lot more.
Another issue is fundraising – In Janury Obama hauled 31 million, and early february this seemed to accelerate even more for both Democratic candidates. The question is how sustainable this is. McCain has huge fundraising potential, since his campaign hasn’t reached out to many republicans, who would certainly be more interested in seeing him in the White House than Clinton (I will continue to ignore Ann Coulter since I belive she is one of the few people whose endorsement would do more harm than good) and probably Obama as well. Consequently I wouldn’t be too surprised if McCain is competitive in fundraising by the time the conventions roll around.
I’ve already mentioned a few possible running mate choices for the democrats, but lets lengthen that list a bit and increase the chance of me being correct!
McCain
Condi
She has excellent brand name recognition, good experience, she’s female, african American, clever and would help get conservatives on board with McCain. She’s a fairly long shot though, and would provide Democratic amunition to tie McCain in with an unpopular administration.
The Midwest
States in this region are looking competitive, and are necessary Republican holds, so two term governors from this region such as Tim Pawlenty of Minnesota are possibles. Pawlenty is McCain national campaign co-chair.
Charlie Christ
Popular florida governor, who would help secure that state for the republicans, and has sound conservative credentials. Unfortunately he’s not married, and rumoured to be gay – if there’s anything that could put off Republicans more, or the ‘Reagan Democrats’ that McCain needs the votes of then he’s running mate being ‘outed’ as gay would be it. Rush Limbaugh and Ann Coulter would have a field day. To quote a republican strategist – ‘The republican party is the party of family values – and where’s his family?’ Still he’s probably one of the strongest options for McCain.
Obama
Bloomberg
Obama should really look for a middle of the road – cross partisan democrat who has experience and popularity. Good thing Michael Bloomberg has recently announced he won’t be running! Bloomberg doesn’t have a long term Washington History, but his business experience is invaluable, he also has unmatchable fundraising ability. Whether Bloomberg would be willing to understudy is another matter.
Tim Kaine
Virginia governor who jumped on the Obama bandwagon early. He’s also in a key swing state, and has a lot of experience at the statewide level.
Bill Richardson
Even though I mentioned him in my previous running mate rundown – its worth mentioning that he hasn’t thrown his support behind Hillary as one would expect him to. He’s also being heavily courted by the Obama camp. Even if he isn’t named as VP – I read a suggestion to name him as secretary of state at the same time – in order to bolster Obama’s lack of experience.
Janet Napolitano
Governor of Arizona and thus could help Obama in the south or west of the country. Being a female helps balance the ticket, and the geography works too. Also a former Attourney General of Arizona, so comes with strong experience. Not being a washington insider helps Obama’s message of change. For the same basic reasons Kathleen Sebelius .
Clinton
I’m not going to bother suggesting vice presidential nominees, since I think Clinton would have a massive uphill struggle getting elected if she were democratic nominee.
February 05, 2008
Uber Tuesday
Follow-up to Florida from True Contradictions
When I originally started bloggin the republican race was looking vastly more interesting than the democratic one, several series candidates, from different wings of the party all competing with each other. Originally strong candidates were looking incredibly weak (Guiliani, McCain at the time) and Huckabee was riding a strong wave with practically no money or support. Since then the tables have turned. The republican race has settled down, with McCain a clear leader in both support and pledged delegates, whilst the democratic field has narrows with the loss of Edwards, in many ways Obama’s rise has made it less predictable.
Democrats
Whats worth noting is that Obama is coming good at the right time – 2 weeks ago he was nationally behind Clinton by 12%, one week ago that had dropped to 6%, on Sunday he was listed as 3% behind, and today CNN released poll figures showing that Obama was ahead by 3%. This is the first poll to show Obama ahead of Clinton nationally, and also the only poll showing that! Super Tuesday could really be the end of the line for the Clinton campaign – that is to say if she doesn’t win here its only downhill. Even a close 2nd for Obama will be perceived as a victory for him, and might possibly give enough momentum for the nomination.
Due to the sheer size of uber tuesday I have decided to present the results in a different manner to previously. Firstly I have prepared a spreadsheet that approximates the total number of votes I believe each candidate shall receive. I believe I will be massively more inaccurate today than previously, since as well the sheer size of the event Edward’s dropout has left massive numbers of undecided voters in the polling numbers.
Now the first thing to bear in mind about my figures are that Obama trails Clinton – this isn’t because he is behind nationally, but because many of Clinton’s strongest states (California and New York for example) vote on uber tuesday. If Obama can survive this, then he is likely to take major delegate hauls on, for example, Feb 12th (Maryland, Virginia and DC). The totals, for those who can’t be bothered to read are 871 – 816.
North East
I expect Clinton to take her home state of New York, where she is senator by a significant margin, whilst Obama will take votes upstate, I believethe city will be highly pro-Clinton. New Jersey seems a lot closer, but still goes for Clinton. Its a similar story in Massachusetts. Conneticut I believe will go Obama, due to a recent flood of new voter registration, especially amongster younger voters.
North
Minnesota appears to be favouring Obama, in light of backing from a pair of house representatives, the same with North Dakota.
South
In the south Obama leads in Georgia, and I would expect him to take the state on the back of African American support. i essentially believe white southerners will vote for Clinton and against any sort of change at all. So Alabama is a tie, and Tennessee (with its lower black populace) goes to Clinton. I also believe she will take Arkansas by a high margin, as Bill was a popular govenor there, additionally she has strong backing by the states’ current leading democrats. Oklahoma neighbours Arkansas and seems to be favouring Clinton. Arizona used to be a Hillary lead, but Obama has been endorsed by the governer and a house representative recently, so I expect it shall be close.
South West
I’ve listed California as a Clinton win, though large margin has been wiped out by Obama’s strong campaigning. His strength amongst black minorities and college educated white voters will be cancelled out by women and the elderly. Unfortunately, about 1/2 of democrats in California tend to vote by absentee ballots, and a lot early, this gives Hillary the edge, because of her past lead here. I believe New Mexico will go to Hillary, governor Richardson sat heside her at the super bowl, and her strength with hispanic voters will do her well here.
Limitations
I have no idea what the hell will happen in Alaska, American Samoa, Kansas, and Democrats abroad. I believe Alaska to go Obama but its a large state with no polling data, a small populace and a republican bent. Basically I’m pissing in the wind and guessing that they hate Hillary. The last poll in Idaho was conducted in mid december and shows al gore in the lead – I’m going for Obama, since Hillary seems to admit that he is in the lead there. I’ve given Obama a small lead in Kansas due to the strong support by State Senate members.
Republicans
I can’t be bothered doing a proper Republican rundown, since McCain will win and the democratic race has become far more exciting. I might write something on it tomorrow.
February 04, 2008
Drama
With Lost returning to the television screens, and its mass fans worldwide, its easy to ignore the fact that its an inconsistent program. Sometimes its plotting drags, and whilst most characters are well developed now, both Hurley and Charlie started out as joke figures in my mind. With this in mind I decided to list a few programs that I feel ought to be viewed by anyone interested in serious drama. This is by no means an exhaustive list – since I generally don’t watch much tv drama (too much crap around).
The Sopranos
Now complete family crime drama, the show juxtaposes the family and business interests of its main character, Tony Soprano. This allows for in detailed characterisation, and the opportunity to bring a variety of characters together, from his daughter’s university friends to hardened criminals. Tony is a fascinating study in and of himself – my personal favourite is the manner in which he eats, it has a very special idiom to it. He plays with his food obsessively, then eventually his takes his fork, stabs one element of it, and eats it – swallowing quickly. Often it seems as though his eating habbits are a compulsion, a comfort for a man who has difficulty expressing his emotions. The strong supporting cast and character driven nature of the show are also highlights.
Dexter
The show is set from the point of view of its serial killing title character. Fascinating insite is given to the relationship between people, through Dexter’s description of how he ‘fakes’ human emotions. The show is driven forward by his discovery of how Dexter became the way he is. Another point of interest is they way his moral code guides the way he acts and the judgements implicit and explicit within its development.
Oz
This nineties prison drama, set primarily in the experimental wing of the Oswald State penitentiary. The fast moving plot frequently focusses on the infighting between different ethnic groups within the prison. Overall narration is done in the style of classical greek theatre – with interludes where the key theme of each episode is discussed. The show deliberately chooses to offer ambiguity in response to key criminal issues – such as the debate between reforming and incarcerating prisoners.
House of Cards
Miniseries from 1990 that charts the rise of its lead character – Francis Urquhart – in a post Thatcherite conservative party. The series had the good fortune to be aired at the same time as the internal struggle surrounding Thatcher’s departure, however, it still stands the test of time. Urquhart is portrayed as a Machavellian political machine, only interested in his own political rise. The show depicts the corresponding fall of those manipulated by Urquhart’s underhand dealings. There are frequent parallels with Shakespearean work, notably Richard III. My favourite aspects are the short solliloques, and frequent glances to camera given by Urquhart that reveal his true intent and the brief shots of rats that increasingly appear at times throughout the show.
January 31, 2008
Electoral Update
So Florida went as expected, but big things have happened since! Firstly Guiliani dropped out. Thats fairly sensible on his behalf – he had no chance of winning from this point. Secondly McCain having won Florida, took the backing of both Rudy and Also Arnie, which lead to some obvious and hilarious headlines . Schwarzenegger for president is all I have to say. Which gives Mitt the Mormon (whilst writing this article I actually mistyped Moron) a problem, possible unsurmountable, of how to make a comeback. He has less backing, less momentum, and lower national polling numbers.
Finally Edwards has now also dropped out – leaving a straight Clinton vs Obama showdown. In some ways I’m surprised Edwards left it this long, since his wife’s cancer problems had resurfaced a while ago. I suspect that his campaign didn’t have much money left to run on at this point as well. One of the interesting things is that Edwards hasn’t yet endorsed anyone. If he leaves it till after super-tuesday its unlikely to have any significant impact. The only real motivation I can see here is that it leaves the door open for him being a vice presidential nomination for either candidate.
Which brings me onto another interesting question that a lot of people keep on asking – who could be a vice presidential nominee. I won’t bother covering Edwards, since I’m sure everyone reading this blog was alive in 2004, but I’ll offer a few of options either way.
Bill Richardson
Popular new mexico governor, with hispanic background. Has strong experience as energy minister and foreign affairs minion under Bill Clinton. Also has plenty of hair. Would offer more to Obama than Clinton by adding experience to his idealism and strong backing amongst the hispanic community thats crucial in swing states like new mexico, and especially florida.
Evan Bayh
Former governor of Indiana, and current senator from Indiana, Bayh nearly made a presidential run of his own this year – forming a committee and raising $10 million but decided not to enter the race due to strong competition from other candidates. Considered to be a fiscally conservative democrat Bayh has served as chair of the Democratic Leadership Council (the same position Bill Clinton held before his Presidential run). He also serves on the committee for Armed Forces, which might offer something to counter McCain. Additionally has been considered as a running mate for both Al Gore and John Kerry, and Bill Clinton has suggested that he would make a good President. In my opinion more likely for Hillary Clinton than Barack Obama.
As to republicans, Guiliani could make a viable running mate for possible candidate, as would popular former florida governer Jeb Bush. Bush would have strong fundraising potential, and also help the swing state of Florida remain republican.
If Michael Bloomber decides that he won’t run, he would make an ideal running mate for anyone out of McCain, Clinton or Obama. Currently New York Major, practically limitless fundraising abilities (he’s worth $11 billion), fiscally conservative but socially pseudo-liberal. Success in business is often seen as a good administrative indicator in politicians. Been a member of both parties in the past, and is often seen as advocating better cooperation and less partisan politics. I wouldn’t be too surprised if there was a tussel for him as a running mate. | 186,579 |
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Join Lisa Bock for an in-depth discussion in this video Modifying browser settings, part of IT Security Foundations: Core Concepts.
- A web browser is the graphical user interface to the Internet. A web browser supports obtaining information in a wide variety of formats from viewing content on a webpage to emailing, watching video, chatting, and playing games. It is the portal to the outside world and steps should be taken to ensure safe browsing. Internet Explorer is installed by default on a Windows operating system, and like most browsers can be configured to provide another layer of security such as deciding if you want to allow cookies, how and when sites can use your location information, and blocking unwanted popups.
Hypertext transfer protocol or HTTP is the protocol used to request and retrieve webpages. It's a stateless protocol in that the server that houses the webpage doesn't keep any information about what a client requested in the past such as items in a shopping cart and browsing activity for generating recommendations or obtaining marketing or advertising information. Cookies used by many major websites are small text files put on end devices to store state information to make browsing easier by preserving information about preferences and sign-in information for cookie creation.
For example, I visit an eCommerce site from my computer for the first time. When the first HTTP request comes to the site, a site creates a cookie that has at least a name-value pair such as unique ID 678. However, in most cases, there is a lot more information in the cookie. Some cookies can be dangerous. Settings will allow you to enable, disable, delete, or block cookies. Browser settings can be done on any browser.
Internet Explorer is built into the Windows operating system so we're going to take a look at the browser settings in Internet Explorer. If I go to the upper right-hand side and I select tools, and down below internet options. I'm going to go to the privacy tab. Under the privacy tab, you can see that it is set at medium. And what does that mean? Medium, well, that will block third party cookies that do not have a compact privacy policy. Third party cookies are used for tracking, and a compact privacy policy which is generally on a first party cookie tells explicit information about how your information is going to be used.
We're going to drop this down to accept all cookies, which is something I wouldn't recommend. If we except all cookies, this is going to save cookies from any website. However, if we totally restrict it and block all cookies, this is going to block all cookies from all websites, and cookies that are already on this computer cannot be read by websites. Well, this is very restrictive and it may alter the way your browsing sessions go because it won't remember anything about you and might make browsing a little more difficult.
Down below, looking at location. Location services lets sites ask for where you are, for example, weather or mapping services. Popup blocker. This limits or blocks popups on sites that you visit. In private browsing, deletes information such as passwords, search history, and page history once you close the tab used so that specific websites do not have access to user browsing patterns. We can see that, if I go back to tools, and I select safety.
Do not track request are set on by default. If I want to turn them off, I would have to go in and manually say turn off. Here we can learn more about do not track.: Modifying browser settings | 262,309 |
The Falling Man: photo by Richard Drew for Associated Press, Sept. 11, 2001.
1 comment:
Tom, everything your doing here is so marvelous--the image-text back-n-forth is right on, keeping tight pace between the present and internal, the always moving motion of the eye and the ear on the light of what reigns as experience. | 281,316 |
TITLE: Find $\lim \limits_{x\to 0} \frac{\sin({\pi \sqrt{\cos x})}}{x} $
QUESTION [3 upvotes]: So I am having trouble finding this limit:
$$\lim \limits_{x\to 0} \frac{\sin({\pi \sqrt{\cos x})}}{x}$$
The problem is I can't use the derivative of the composition of two functions nor can I use other techniques like l'Hôpital's theorem.
I tried numerous techniques to calculate this limit but in vain so if you have any simple idea that is in the scope of my knowledge ( I am a pre-calculus student ), please do let me know without actually answering the question.
REPLY [2 votes]: In this answer I will use the fact that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to derive the limit
\begin{align}
\lim_{x \to 0} \frac{\sin \left( \pi\sqrt{\cos x} \right)}{x}&=\lim_{x \to 0} \frac{\sin \left(\pi - \pi \sqrt{\cos x} \right)}{x}\\
&= \lim_{x \to 0} \frac{\sin \left(\pi\left(1 - \sqrt{\cos x}\right) \right)}{\pi\left(1 - \sqrt{\cos x}\right) }\frac{\pi\left(1 - \sqrt{\cos x}\right) }x\\
&=\lim_{x \to 0}\frac{\pi\left(1 - \sqrt{\cos x}\right) }x\\
&= \lim_{x \to 0} \frac{\pi\left ( 1 - \cos x \right) }{x \left (1 + \sqrt{\cos x}\right)}\\
&= \frac{\pi}2 \lim_{x \to 0} \frac{1 - \cos x }x\\
&= \frac{\pi}2 \lim_{x \to 0} \frac{1 - \cos^2 x }{x\left ( 1 + \cos x\right) }\\
&= \frac{\pi}4 \lim_{x \to 0} \frac{sin^2 x}{x}\\
&= \frac{\pi}4 \lim_{x \to 0} \sin x \\
&= 0
\end{align}
REPLY [0 votes]: $$sin(\alpha) = sin(\pi - \alpha)$$
$$sin(\pi\sqrt{cos{x}}) = sin(\pi -\pi\sqrt{cos{x}})$$
$$ sin(\pi -\pi\sqrt{cos{x}}) \sim_{x\to 0} \pi -\pi\sqrt{cos{x}}$$
$$\lim_{x\to 0} \frac{sin(\pi\sqrt{cos{x}})}{x} = \lim_{x\to 0} \frac{\pi -\pi\sqrt{cos{x}}}{x} = \pi \cdot \lim_{x\to 0} \frac{1 -\sqrt{cos{x}}}{x} = \pi \cdot \lim_{x\to 0} \frac{(1 -\sqrt{cos{x}})(1 +\sqrt{cos{x}})}{x(1 +\sqrt{cos{x}})} = \pi \cdot \lim_{x\to 0} \frac{1 - cos{x}}{x(1 +\sqrt{cos{x}})} = \pi \cdot \lim_{x\to 0} \frac{0.5x^2}{x(1 +\sqrt{cos{x}})} = \pi \cdot \lim_{x\to 0} \frac{0.5x}{(1 +\sqrt{cos{x}})}=0$$ | 26,369 |
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TITLE: Set of diameter $\le 1$ contained in set of constant width $1$
QUESTION [9 upvotes]: I'm reading the paper Minimal universal covers in $E^n$ by H.G. Eggleston and they state that every set $A\subseteq{\bf R}^2$ of diameter at most $1$ (the diameter of $A$ is defined as $\sup_{x,y\in A}|x-y|$) is contained in a set which has width $1$ along any direction. I can't see how the proof of this should work, though.
EDIT: Maybe the proof is something along the lines of constructing a thing that has width $1$ along any rational direction, then it should has width $1$ along all directions. For this, it should suffice to show that for any $A$ of diameter $\le 1$ you can add a translate $I$ of the unit interval (embedded into ${\bf R}^2$) such that $A\cup I$ has diameter $1$. Then you are able to add an infinite amount of intervals at different angles without the diameter exceeding $1$.
REPLY [0 votes]: WLOG we can assume that $X$ is convex. $p_1,\
p_2\in X$ s.t. $|p_1-p_2|=1$. Then there is a circular sector $V_1$ at $p_1$ whose angle is $\theta_1$.
And $V_1$ contains arc $p_2p_3$. Here $\theta_1$ is the largest when $V_1-X$ is empty.
Now we do the same thing at $p_2$. Hence we have $\sum_{i=1}^{2m+1}
\ \theta_i =\pi,\ \theta_i>0$ and $\bigcup_{i=1}^{2m+1} \ V_i$ contains $X$. | 62,645 |
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Over the last century and a half, Pease Awning continues to evolve and change with the times. Decade after decade Pease Awning has the reputation of a high quality custom manufacturer that can create, build and install any type of Shade Product available on the market. We look forward to passing our knowledge to future generations and servicing your grandchildren as we serviced your grandparents. We also look forward to future inventions and will continue to strive to bring our customer’s the most current proven best quality products to fit your lifestyle.
Ted Franklin, President | 357,324 |
Harry S. Truman once said, “We must have strong minds, ready to accept facts as they are.” Show Harry your strong mind and wrap your head around these.
403
Total number of male registered members on LawDate.com, which bills itself as the “first full-service social networking site designed with lawyers in mind and the non-lawyers who want to meet them.”
613
Total number of female registered members on LawDate.com.
50
Number of years retired Supreme Court Justice David Souter has put a restriction on public access to his papers, “barring anyone—researchers, historians, friends, journalists—from viewing the material,” according to The Legal Times.
100
Distance in yards a judge ordered R&B singer Chris Brown to keep from Rihanna, according to The New York Times.
10
Distance in yards a judge ordered R&B singer Chris Brown to keep from Rihanna when attending “industry events,” according to The New York Times.
15
Percentage of corporate legal bills that are audited by a third-party before being paid, according to a quote in a CFO.com article attributed to Michael Rynowecer, president of legal-industry research firm BTI Consulting Group.
7
Percentage drop in spending by Fortune 1000 companies in the first three months of 2009 on outside legal services, according to CFO.com.
70,422
Number of followers “Rex Gradeless,” Saint Louis University School of Law student and founder of the blog Social Media Law Student has on Twitter, which is said to be the most of any legal Twitter user.
126
Average number of followers per Twitter user, according to guardian.co.uk.
97,281
Number of women lawyers in the country of Italy, according to The Council of Bars and Law Societies of Europe. With a female population of 30,699,543, approximately one in every 315 women is a lawyer.
71,079
Number of women lawyers in the United Kingdom, according to The Council of Bars and Law Societies of Europe. With a female population of 30,488,622, approximately one in every 430 women is a lawyer.
14
Number of women lawyers in the country of Liechtenstein, according to The Council of Bars and Law Societies of Europe. With a female population of 17,450, approximately one in every 1,250 women is a lawyer.
3:1
Student-faculty ratio at the new University of California at Riverside Law School, according to The Los Angeles Times.
400,000
Approximate number of members of the American Bar Association.
2.5
Estimated percentage of member dues the ABA uses for lobbying efforts.
45
Number of days a small, lucky number of law students will have to consider an offer from Skadden, Arps, according to a Bloomberg story.
52
Number of years a Wake Forest University law school graduate was missing his class ring after leaving it in an airport bathroom before being reunited with it this week, according to NBC Chicago.
$10-$50 million
Rough estimate of assets and liabilities for Masry & Vititoe. Best known as the real-life “Erin Brockovich” firm, Masry & Vititoe has filed for chapter 11 bankruptcy protection, according to The Ventura County Star.
$228,552
Base salary for attorney Ed Masry (played by Albert Finney in the movie Erin Brockovich) at the time of his death in 2005, according to The Ventura County Star.
$256,271,286
Approximated amount grossed worldwide by the film Erin Brockovich, according to BoxOfficeMojo.com.
9
Ranking of the movie Erin Brockovich, according to the ABA Journal’s list of “The 25 Greatest Legal Movies.”
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Guano Dubango
August 28, 2009 at 2:53 am
300,000: Number of Female Lawyers in the USA; compare: 3: Number of BEAUTIFUL female lawyers in New York City. Will someone please recruit some BEAUTIES to become lawyers in New York City? Thank you!
BL1Y
August 28, 2009 at 3:17 am
Why would you want more female lawyers? A woman making 5 figures will be happy with a lawyer making 6, but once they become lawyers making 6 figures, they’ll only look for men making 7, which rules out most male lawyers. Not to mention the affect working in a stressful, unrewarding industry packed with people with personality disorders will have on her. And of course there’s the seamlessweb ass.
Guano Dubango
August 28, 2009 at 3:24 am
That is why I only ask for BEAUTIES, who will, presumably, not have the seamless web ass, which I also presume is something like a fat ass? If the woman is truly a BEAUTY, she will have to have a nice ass, as well as a nice body, and a nice face. The failure to have any of the foregoing will result in automatic disqualification by the New York Bar Association. I do not think we will be able to increase the number of female bar members in any appreciable manner, but I figured I’d do what I can for the males of New York City through this idea.
BL1Y
August 28, 2009 at 6:00 am
Now we know Guano doesn’t work at a law firm, or else he would have witnessed seamlessweb ass in action.
Anony
August 28, 2009 at 7:01 am
Who knew there was a law date.com?
Doug
August 28, 2009 at 7:12 am
I’m on lawdate.com. Two dates. Two disasters.
Anonymous
August 28, 2009 at 7:45 am
bl1y, you should try lawdate and report back to us.
BL1Y
August 28, 2009 at 8:23 am
I’d rather (and will likely) date my hand.
Anonymous
August 28, 2009 at 8:24 am
did Masry get a cut of the movie money?
Dannie
August 28, 2009 at 10:57 am
Doug, you need to tell us more about your lawdate.com experiences. total disasters?
Doug
August 28, 2009 at 11:21 am
Put it this way: They were both uglier than you’ve imagined Alma to be. YIKES!
Dannie
August 28, 2009 at 11:58 am
man, bummer. sounds awful.
Doug
August 28, 2009 at 12:20 pm
Took a long, long shower after each. Felt so dirty just being in the same place as them.
BL1Y
August 28, 2009 at 12:34 pm
Doug: Did they not have pictures online?
Doug
August 28, 2009 at 1:11 pm
Pictures, yes. Were they accurate, no.
Lady lawyer
August 28, 2009 at 7:35 pm
Doug, what would the ladies say about you?
BL1Y
August 29, 2009 at 12:35 am
Doug: Outdated or “the angles?” Either way, you should feel completely justified in just walking away from the date from the beginning if the woman has lied about her appearance.
Alma Federer
August 29, 2009 at 9:33 am
You men are pigs. Do you really think you are all that? You probably make Drew Carey look like George Clooney. In other words, we are tired hearing about how ugly the women are. You men are even BIGGER losers! | 296,137 |
PLEASANTVILLE, N.Y. -- The Pace University Athletics Marketing and Promotions Department has announced the promotional schedule for the upcoming 2014 football season.
Due to renovations, all home games will be played at Anne M. Dorner Middle School in Ossining with kick-off for all contests scheduled for noon.
The Setters open the 2014 season against Alderson Broaddus University on Saturday, Sept. 6. The first 500 fans who take the shuttle bus from campus will receive Pace Wayfarer sunglasses. Donations will be accepted throughout the game for the National ALS Association. The following weekend, Pace hosts Merrimack College for Military Appreciation Day. The first 500 fans who take the bus for the Saturday, Sept. 13 game will receive Pace Athletics dog tags. The next home game will be on Saturday, Sept. 20 as the Setters take on Southern Connecticut State University. The first 500 fans who take the bus to the game will receive a free Pace Athletics draw string bag. The Setters return home on Saturday, Oct. 4 as Pace hosts LIU Post. The game will feature Take-A-Kid to the game. Setters squad game day T-shirts will be sold at the bus kiosks for $10 with $3 going to the P4K Dance Marathon.
Pace football will face Stonehill College on Oct. 18, which features the annual "Pace Goes Pink" game. The first 500 fans will receive "Pace Goes Pink" wristbands and the 2014 edition of the "Pace Goes Pink" T-shirt will be sold during the game for $5. All proceeds will be donated to the Northern Westchester Hospital Breast Institute. Pace football ends its home schedule on Saturday, Nov. 1 with a game against University of New Haven. The graduating football seniors will be honored prior to the game and the first 500 fans to take the bus to the game will receive Pace Athletics phone wallets. To view the complete 2014 Pace football schedule, click here. | 130,717 |
When I have been faced with grief in the past, I cook. It takes my mind off of things and keeps me busy. I don't like to cry. The rhythmic chopping, thoughts of what needs to go into the saute pan and grabbing things from the spice drawer fill my head rather than the sadness I really feel. Later that night, it did hit me.
So our Christmas plans were turned a bit upside down. That night, I cooked some of the food intended for the dinner party for my family and some dear friends who helped us get through that rough experience. The next day, I was looking at a pound of cod ling, trying to figure out what to do with it.
My husband has always had a soft side for creamy chowder. And it's one of those things my kids really like. We often fight for the last spoonfuls. This seemed like a good use for the fish and I had most of the ingredients in the house. Fast forward about 45 minutes later (after I went out for clam juice) and we had delicious fish chowder. It made me happy to hear the yummy noises c0ming from everyone's mouths and it certainly kept my mind off of things for a few more hours. Before we finished, my husband snapped a few good photos. For whatever reason, I feel compelled to show you these creations rather than only write about it. I thought these really captured the day.
As with all of my recipes, I look at this as simple, tasty, and fairly healthy (I mean, I could have used heavy cream, right?!). I might have to search for some other options for clam juice besides Snow's (only because I'm not really sure about the quality) but in a pinch, it turned out just fine. Thanks for listening...
Amy
Amy's Fish Chowder
Serves 4 (or in our case 3 adults and 2 little ones)
2 tsp Olive oil
2 Tbl unsalted butter, divided
1/2 medium onion, chopped
Ground pepper
A few saffron threads
2 celery stalks, chopped
2 russet potatoes, peeled and cubed into bite size pieces
1 Tbl flour
1 bottle of clam juice (about 8 oz)
2 cups whole milk
about 3/4 lb of firm white fish, cleaned and cut into cubes
salt and pepper to taste
1. Heat a medium sized pot. It's important to make sure there is enough depth so that the fish will be able to cook in the soup. Over medium heat, add olive oil, 1 Tbl butter and onion. Add a few grinds of pepper and the saffron. After a few minutes, add the celery, then the potato.
Tip: While the onion cooks, chop the celery. While the celery cooks, peel and cube the potatoes.
2. Add the remaining tablespoon of butter, allow it to melt, then add the flour. Mix and cook for 2-4 minutes. Lower the heat a little and then add the clam juice (it's better if it's a room temperature) and stir to combine. Add the milk and stir. Bring it up to a boil and lower to a simmer.
Tip: While the soup is beginning to come to a boil, cut the fish.
3. When the soup is simmering, carefully add the fish into the soup and stir. Keep the soup at a simmer in order to cook the fish, about 5 minutes. Add salt and pepper to taste. Cook the soup until it reaches the desired thickness. Remove a piece of fish and cut with a fork to make sure it's cooked. Check the potatoes for doneness as well. They should be soft but not mushy.
4. Serve with some crusty sourdough bread and a nice glass of white wine, like a chardonnay.
Tip: Before you add the butter and flour (roux), you can add white wine or sherry for more flavor (about 2 Tbl). Make sure it has evaporated before you add the roux.
You can also add other fish like crab, clams or shrimp. If you are adding cooked fish or seafood, for example crab, add at the very end.
Bacon is another good flavor in a chowder. Add chopped bacon to the onions and celery and cook until it is crisp. Reduce butter to only 1 Tbl (add with the flour only).
If you don't have saffron, just eliminate from the recipe. It's nice but not necessary. | 270,015 |
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Amid social media discussion about the shocking video that emerged of a United Airlines employee brawling with an ex-NFL player at Newark International Airport, Alec Baldwin decided he had an important perspective to share on the employee’s right to a safe workplace environment.
Unfortunately for Baldwin, many begged to differ with the actor’s view that the employee was the victim. In response to an Instagram post on Baldwin’s comment, thousands suggested the compulsively outspoken actor should “sit this one out” given his own recent history regarding workplace safety. Specifically, they slammed Baldwin for lecturing anyone on workplace safety when he faces a criminal investigation and is being sued over his role in the fatal shooting of cinematographer Halyna Hutchins on the set of “Rust,” a Western film he starred in and was producing.
Baldwin’s latest online controversy emerged after he decided to comment on the viral video of last Thursday’s airport brawl between a passenger and an United Airlines employee, according to The Shaderoom celebrity news account, which shared a screenshot Tuesday morning of Baldwin’s comment. Within two hours, The Shaderoom Post on Baldwin’s comment had been viewed more than 2 million times.
“Alec, have a seat sir,” wrote actor Laura Govan. Her response garnered more than 18,000 likes. Govan also wrote: “FYI You should still be grieving over the lady you killed.”
“Sir, you shot and killed someone at YOUR workplace. Please sit this one out,” added Wendy Osefo, a public affairs professor, political commentator and co-star of “The Real Housewives of Potomac.” Her comment garnered more than 13,000 likes within several hours.
The fight video showed the unidentified airline employee shoving ex-Denver Broncos cornerback Brendan Langley last Thursday. It’s not clear what prompted the shoving, but the video showed that Langley reacted by unleashing a flurry of blows that sent the employee reeling. Rather than stay down, the bloodied employee got back up and tried to fight Langley some more.
In choosing to see the employee as the victim, Baldwin expressed support for his right to be safe in his workplace. He said, “The guy working at the airport is the victim. He came to work to do a job. The other guy, with his big mouth, is guilty of workplace abuse, where people come to work with an expectation of safety, even civility. This (expletive) who hit this guy should be put on a no fly list.”
While Langley was arrested and charged with simple assault, as TMZ reported, United Airlines also revealed to TMZ and to the New York Post that it had issues with the employee’s actions and fired him.
Meanwhile, Govan mostly received praise for telling Baldwin to “have a seat.” Many expressed disbelief that Baldwin would voice support for the employee, saying the video showed him striking Langley first, but they also expressed anger over Baldwin’s arrogance and “privilege” in thinking he should have a voice on this incident.
“The nerve of him,” one person wrote, while another said that Baldwin seems to think that people have forgotten that he killed someone.
“He needs to be quiet,” one person wrote. “You killed someone at your work place sir so sit down and don’t comment.”
Some people came to Baldwin’s defense Tuesday, saying that it was “low” for anyone to bring up Hutchins’ death. They argued that Hutchins’ death wasn’t Baldwin’s fault. Rather it was, as Baldwin has said, “a tragic accident.”
Hutchins was killed Oct. 21 on the Santa Fe, New Mexico set of the low-budget Western film. Baldwin was handling a Colt .45 revolver while rehearsing a scene with Hutchins and director Joel Souza. For reasons yet to be determine, the gun was loaded with a live round, when it only should have been loaded with dummy rounds.
The gun also fired, with the round striking both Hutchins and Souza. Hutchins was killed and Souza was wounded. Baldwin has denied pulling the trigger and said that he accepted the assistant director’s word that the gun was safe to use.
The criminal investigation by New Mexico authorities is ongoing. For his role in the shooting and as a producer on the film, Baldwin has been named as a defendant in lawsuits filed by Hutchins’ family and by the the film’s script supervisor, Mamie Mitchell.
The family’s lawsuit, filed in February on behalf of Hutchins’ husband, Matthew Hutchins, and her 9-year-old son, Andros, alleges that Baldwin “recklessly” shot and killed the cinematographer. The suit also alleges that Baldwin, the film’s other producers and other defendants failed to “perform industry standard safety checks and follow basic gun safety rules while using real guns to produce the movie ‘Rust,’ with fatal consequences.”
The New Mexico Environment Department’s Occupational Health and Safety Bureau in April issued a hefty fine of $139,793 against Rust Movie Productions, the production company behind “Rust,” for a number of workplace safety violations. The bureau said the company failed across the board to maintain and enforce industry standard protocols for using guns on the set — failures that led to Hutchins’ death. | 174,585 |
TITLE: Affine-local property that is not stalk-local
QUESTION [3 upvotes]: From Vakil's foundations of algebraic geometry, page 158, 5.3.2"
We say such a property is affine-local: a property is affine-local if we can check
it on any affine cover. (For example, reducedness is an affine-local property. Do
you understand why?) Note that if U is an open subscheme of X, then U inherits
any affine-local property of X. Note also that any property that is stalk-local (a
scheme has property P if and only if all its stalks have property Q) is necessarily
affine-local (a scheme has property P if and only if all of its affine open sets have
property R, where an affine scheme has property R if and only if all its stalks have
property Q). But it is sometimes not so obvious what the right definition of Q is;
see for example the discussion of normality in the next section.
He says stalk-local properties are necessarily affine local. I feel like the converse is not necessarily true but I can't come up with a counterexample. What is an affine local property that is not stalk-local?
REPLY [3 votes]: I think locally noetherian is such a property. A scheme is locally noetherian if and only if all of its affine opens are noetherian. But there exists a non-noetherian ring $A$ such that $A_{\mathfrak{p}}$ is noetherian for all prime ideals $\mathfrak{p}$. An example is $\prod_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$. | 126,920 |
MANCHESTER, UK, April 10, 2019– Freewheel Holidays, the UK specialist in European self-guided cycling holidays, offers itineraries in four former Soviet-bloc countries: Romania, The Czech Republic, Hungary and Poland.
Freewheel Holidays’ itineraries focus on colorful histories and today’s newly energized and creative populations that emerged in the early 1990s. These multi-day, bespoke bicycle vacations involve leisure pedaling and immersive, on-own sightseeing. All include accommodations, breakfasts, bike rentals, briefings, route planning and sightseeing suggestions, applicable bus/train and luggage transfers and more.
Romania
Bucharest, Romania, is gateway to Transylvania's Heritage that moves into the countryside, steeped in legend with castles, picturesque villages and Saxon fortifications overlaying Roman remains. In Transylvania, the legendary home of Dracula, cyclists may share routes with a horse and cart on rough paths where the map says “major thoroughfare.” This itinerary explores, among others,
The per person double rate is from £789. See
Czech Republic
The Czech Republic (Czechia) is a young country as of 1993 when a bloodless revolution divided the former Communist Czechoslovakia into The Czech Republic and Slovakia. Elbe Cycle Path: Prague to Dresden begins in the UNESCO World Heritage city of Prague. Icons here include:
Visitors may visit the Jewish Museum in Pinkas Synagogue with walls covered with heartbreaking drawings made by children of the Terezin (Theresienstadt) concentration camp from 1942-1944.
Cycling along the Moldau, guests arrive in Mělník where an imposing castle marks the wine-growing Central Bohemian region. Next up are Theresienstadtand the Paleolithic roots of 10th century Litoměřice with colourful Gothic, Renaissance and Baroque facades. Cycling continues along River Elbe with fairy tale castles and the fortress "Königsstein" enroute to magical Dresden. The per person double rate is from £619. See
Prague is the gateway for Bike and Barge: Prague and The Elbe, new for 2019. A barge accommodates guests who hop on and off adjacent bicycle paths to explore, among others,
The per person double rate is from £759. See
Hungary
Freewheel Holiday’s Hungarian Nature and Spas, new for 2019, blends nature with a handful of the more than 400 hot springs and spas of thermal waters rich in minerals and trace elements. The gateway city is Budapest before transferring into the country for cycling around Lake Tisza (named for one of central Europe’s main rivers). Guests discover active thermal waters and a bird sanctuary before visiting a stud farm in a region known for Nonius, a breed favored by the Hungarian royal court. Guests may tour in a “Puszta-Taxi” (covered horse carriage) to view cattle, spiral-horned sheep and typical draw well (buckets on ropes lowered into the well). Guests ride into the largest continuous natural grassland of Central and Western Europe before reaching a community favored by hunters and by fans of open-air-spas and thermal baths. Before returning to Budapest is a visit to Hungary’s largest open-air thermal spa with 17 pools, swimming- wave- spring- and thermal pools. The per person double rate is from £619. See
Poland
Krakow and the Danajec Path utilizes cycle paths paralleling foaming rivers flanked by mountains. One ride is along River Dunajec, from the foot of the High Tatras to one of Poland’s most beautiful towns, Stary Sacz, and the neighbouring Nowy Sacz, which still evokes the Habsburg Empire. Krakow, the gateway city, is a UNESCO World Heritage site whose historic architecture was unscathed in World War II. Among its architectural treasures are
In the country cyclists ascend and descend past farm carts laden with hay. They’re introduced to old-world traditions of Gorale (locals) and sample œoscypek, a cheese of ewes. A cable car assists a cycle to a village that is an open-air museum of timbered architecture. Crossing into Slovakia guests imbibe the beauty of Dunajec River Gorge followed by Slivovitz (plum Bračdy brandy). The per person double rate is from £619.. | 259,025 |
Secondary shaping together nicely in the spring April 16, 2014 Kyle Griffin is among the team's leading returning tacklers, and he spoke after practice Tuesday to discuss how the secondary is looking in spring ball. [read more]
Women's Tennis beat Rams for first time in DI era to close regular season April 15, 2014 A 5-2 regular season finale win over Colorado State for the first time since joining Division I cemented the third consecutive season of finishing at or above .500 for Northern Colorado Women's Tennis. [read more]
Cheer team to host open tryouts April 15, 2014 The Northern Colorado cheer team will be holding open tryouts on May 3 at the Butler-Hancock Sports Pavilion. [read more]
Baseball loses 14-3 at No. 15 University of Washington April 14, 2014 Northern Colorado ends road trip to Pacific Northwest at 0-4 and returns home this weekend to continue WAC play. [read more]
/sports/fball/2013-14/photos/0001/griffin_0831.jpg?max_width=988&max_height=298 /sports/wten/2013-14/photos/Walters-West.jpg?max_width=988&max_height=298 /sports/mgolf/2013-14/photos/Cheer_Tryouts.jpg?max_width=988&max_height=298 /sports/bsb/2013-14/photos/Moseley_Landon_2_0329.jpg?max_width=988&max_height=298
April 14, 2014 | Women's Soccer Twenty-one Bears named to Dean's Citation List
April 14, 2014 | Softball Duffy takes Big Sky Player of the Week honor
April 13, 2014 | Men's Tennis Bobcats sneak by Men's Tennis in conference season finale
April 13, 2014 | Women's Tennis Women's Tennis finishes Big Sky regular season with loss at Montana State
April 13, 2014 | Baseball Bears' bats silenced in series finale at Seattle
April 12, 2014 | Women's Volleyball Volleyball goes undefeated Saturday in home spring tournament
April 12, 2014 | Men's Tennis Bears Men's Tennis still in line for Big Sky Championship despite loss to Grizzlies
April 12, 2014 | Baseball Baseball drops second game of series in Seattle
April 12, 2014 | Women's Tennis Grizzlies take down Women's Tennis on road
April 12, 2014 | Softball Softball takes Portland State series with big Game 3
April 12, 2014 Offensive line working with fresh faces
April 11, 2014 | Softball Offensive bursts contribute to Friday split for Softball
April 11, 2014 | Track & Field Bears have nine qualifying times at Nebraska Quadrangular
April 11, 2014 | Baseball Bears drop series-opener to Seattle University
April 11, 2014 Trio of Bears enjoying clean bills of health in spring ball
April 10, 2014 Football - Spring Game Promo
April 9, 2014 | Women's Swimming & Diving 2013-2014 Winter academic all-WAC announced
April 9, 2014 | Baseball Top of order leads Bears in loss to Air Force
April 8, 2014 | Men's Golf Men's Golf finishes 17th in Redhawk Invitational
April 8, 2014 | Women's Golf Women's Golf finishes Cowgirl Classic in 19th place
April 8, 2014 | Softball Softball - Portland State Series Preview - 4/8/2014
April 8, 2014 Anticipation building for football spring game
April 7, 2014 | Women's Golf Women's Golf in 19th after first day of Cowgirl Classic
April 6, 2014 | Baseball Baseball loses Sunday heartbreaker
April 6, 2014 | Women's Soccer Bears Soccer leaves Regis Spring Tournament with two wins
April 5, 2014 | Men's Tennis Men's Tennis falls just short at home against Sacramento State
April 5, 2014 | Baseball Baseball starts off as pitchers duel
April 5, 2014 | Softball Softball turns tide on Sacramento State in series finale
April 4, 2014 | Track & Field Track and field qualifies five for Big Sky Championship at Tom Benich Invite
April 4, 2014 | Baseball Baseball doesn't hold on to early lead
April 4, 2014 | Women's Tennis Senior Day successful for Women's Tennis against UND
April 4, 2014 | Softball Sacramento State upends UNC Softball in doubleheader
April 4, 2014 Join Bears for final luncheon of year
April 2, 2014 | Softball Early lead not enough as softball falls to Colorado State in extra innings, 6-5
April 2, 2014 Football's offense taking shape
April 1, 2014 | Men's Golf Men's golf shines at Whiting-Turner
April 1, 2014 | Baseball Comeback rally falls short at Air Force
April 1, 2014 | Women's Tennis Three seniors to play final home match for Women's Tennis
April 1, 2014 | Softball College Sports Madness names Hudson Big Sky Player of the Week
March 31, 2014 | Men's Golf Men's golf sits in third place after day one of Towson Invitational
March 31, 2014 | Softball Holliday named Big Sky Pitcher of the Week
March 31, 2014 | Men's Basketball 2014 Student Athlete Advisory Committee (SAAC) Video
March 30, 2014 | Baseball First career home run not enough for baseball
March 29, 2014 | Women's Soccer Bears Soccer beat alumni in annual spring match
March 29, 2014 | Track & Field Eight Bears qualify for Big Sky Championships Saturday
March 29, 2014 | Baseball Baseball outhits UND in both games
UPCOMING
4/17/14 1:00 PM
Men's Tennis
IPFW
at Northern Colorado
4/17/14 All Day
Track and Field
vs. Cal State LA Twilight
@ Los Angeles, Calif
4/18/14 3:00 PM
Baseball
Grand Canyon
at Northern ColoradoLive stats
4/18/14 All Day
Track and Field
vs. Mt. Sac Relays
@ Walnut, Calif.
4/19/14 12:00 PM
Baseball
Grand Canyon
at Northern ColoradoLive stats | Video | Audio
4/19/14 2:00 PM
Softball
Northern Colorado
at Idaho StateLive stats | Video
4/19/14 4:00 PM
Softball
Northern Colorado
at Idaho StateLive stats | Video
4/19/14 All Day
Track and Field
vs. Long Beach Invitational
@ Long Beach, Calif.
4/20/14 12:00 PM
Baseball
Grand Canyon
at Northern ColoradoLive stats | Video | Audio
4/20/14 2:00 PM
Softball
Northern Colorado
at Idaho StateLive stats | Video
RESULTSBox Score | Recap | Live stats | Video
4/12/14 Final - 5 innings
Softball
2Portland State
19at Northern Colorado | 398,193 |
How To Work as an Online Test Scorer and Make up to $18 an Hour
And since you are only guaranteed employment until the papers run out, you are in a race against all your phantom coworkers to score as many papers as you can, as fast as possible. Scorers do not see any identifying characteristics of the students.
First, no one scorer will score a student's entire test.
But the visit was a look inside a process that is often mysterious to teachers and parents who wonder what happens when their students' answers go into the corporate black box of Pearson.
For example, test-scoring jobs never have a guaranteed end date. You sometimes have to look at words a little phonetically.
- Pearson recruited them through its own website, personal referrals, job fairs, Internet job search engines, local newspaper classified ads and even Craigslist and Facebook.
- Monthly Review | The Loneliness of the Long-Distance Test Scorer
- You may work on a month-long project without ever speaking to another human being, never mind seeing the children who actually wrote the papers.
- The papers are also a testament to the persistence of racism, describing teenagers kicked out of stores or denied service or jobs because of the color of their skin.
- Top 10 best cryptocurrency to invest in 2019
Scorer Mark Farmer said he knows that if he scores 1, different portions of questions in a day, he is likely to make a mistake. But this operation in a suburban office complex outside Columbus is a very focused assembly line operation: Rodriguez-Rabin has worked for the testing company Pearson on various work from home grading standardized tests since Signs taped to posts inside the scoring room remind scorers of their mission: Supervisors will "backread" about five percent of scored questions to see if a grader scored them properly.
Score SAT Essays | SAT Suite of Assessments – The College Board. Proving their reliability so they will continue to get more contracts.
The first grade will be used, unless the scores are not "adjacent" - off by a single point up or down. The Human Resources people who interview and hire you are temps, as are most of the supervisors.
ACT Writing Test: Readers
But the results are not necessarily judged by teachers. Grading the Common Core: Test scoring is a huge business, dominated by a few multinational corporations, which arrange the work in order to extract maximum profit.
Scorers who repeatedly fail to match these so-called validity papers are let go. As Valerie Gomm read several paragraphs from a fifth-grade essay on a laptop screen, she consulted a heavily bitcoin trading signals twitter sheaf of papers that prescribed the criteria for evaluating reading comprehension, written expression and conventions like spelling and punctuation.
But rather than addressing the problems outlined above, it seems more likely that this move will only transfer the absurdities in current state tests to a national level, with the danger that they will take on an even greater legitimacy.
Become an SAT Essay Scorer
Just as test scorers are never allowed to know the effects of our scores on students, we never get a chance to meet them, to see how they have developed as writers, thinkers, or human beings, or to know what life in their communities or families is like. The papers are also a testament to the persistence of racism, describing etf tipping point trading system kicked out of stores or denied service or jobs because of the color of their skin.
Take your career to the next level.
Then, Pearson has some other measures in place that double check about 15 percent of the scored answers. Odell noted that grading 55 to 80 an hour seems high, but many answers are easy to grade.
If answers are supposed to be really simple, the pace can be even faster. Equally bad, the fanaticism surrounding accountability via testing, which claims it will result in higher-quality teachers, is doing nothing of the sort. This is the reality of test scoring. In the test-scoring centers in which I have worked, located in downtown St.
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But, given that this is the process through which so many students are learning to write and to think, one would hope for more. Since starting as a full-time grader April 13, she has scored a succession of five third grade math questions. This cannot be good for quality, but as long as the statistics match up and the project finishes on time, the companies are happy.
When I was a student, I envisioned my work at home disney store being graded by qualified teachers in another part of the country, who taught the grade level work from home grading standardized tests subject corresponding to the tests. The chasm between rich and poor is at times felt in the writing itself, as some students come from unimaginable privilege, while many more endure heartbreaking experiences in foster homes.
OT- Information About Scoring/Grading Standardized Tests From Pearson Education - art-martem.com Forum
They work with just one rn work from home sc at a time, grading that single question a few hundred times a day. Read a passage from a novel written in the first person, and a poem written in the third person, and describe how the poem might change if it were written in the first person.
No points are given for creativity on these tests, although some scorers have told me that, until recently, a number of states did factor creativity into their scores. Pearson sends all its communications to home scorers via e-mail, now supplemented by automated phone calls telling you to check your inbox.
Error-free communications are rare. Nor, to my knowledge, have they begun to outsource this work to India. Community Rules apply to all content you upload or otherwise submit to this site. Some scores forex view double-checked Pearson and the scorers acknowledge that even after training and even with the guides, scoring errors can still happen.
Opening range trading system never hear how the students work from home grading standardized tests. Graders score questions, not whole tests Rather than having scorers work through full exams one by one, Pearson takes advantage of exams being given forex broker start bonus online this spring.
So if 40 percent of papers received 3s the previous year on a scale of 1 to 6then a similar percentage should receive 3s this year.
Or that this money was subsidizing temporary jobs with no health care and no hope for transitioning into long-term employment—jobs which, in a better world, would not exist. With each question broken out and assigned to scorers randomly, a student will likely have a different scorer for each question.
You may work on a month-long project without ever speaking to another human being, never mind seeing the children who actually wrote the papers.
Scoring Job Opportunities
Though the odds might seem slim, our collective goal, as students, teachers, parents—and even test scorers—should work at home disney store to liberate education from this farcical numbers game. Unfortunately, the joke is on us, as the Obama administration pushes for even more high-stakes standardized testing. Log in as many hours as you can and score as much as possible!
The material on this site may not be reproduced, distributed, transmitted, cached or otherwise used, except with the prior written permission of Advance Ohio.
Work from Home
Parcc, formally work from home grading standardized tests as the Partnership for Assessment of Readiness for College and Careersand the Smarter Balanced Assessment Consortiumanother test development group, along with contractors like Pearson, worked with current classroom teachers and state education officials to develop the questions and set detailed criteria for grading student responses.
Then I would print it out and read it the next day while I was working at the scoring center. Why would people in their right minds want to leave etf tipping point trading system assessment in the hands of poorly trained, overworked, low-paid temps, working for companies interested only in cranking out acceptable numbers and improving their bottom line? | 247,734 |
Self Glow
Self Glow% | 150,567 |
When you run the setup program for either of these, make a note of the
location where the C++ compiler is installed. This information will be useful to
you later.
Tabs:
Set the Tab Size and Indent Size to 5.
Visual Studio and Visual C++\Examples directory, and
its name is Project_Sample.
Do the following steps, in order:
Tip: If the Solution Explorer window is not
visible, select Solution Explorer from the View menu.
Also, if you do not see main.asm in the Solution Explorer
window, look at the tabs along the bottom of the window. Click the
Solution Explorer tab.:
In the output window. 12 through 16 require the building of 16-bit applications.
Except for a few exceptions, which are noted in the book, your 16-bit
applications will run under Windows XP and Windows Vista..
9.0\Common7\IDE folder. If you are using Windows Vista, it will display
a verification window before copying the file.:
There are two different versions of the ZIP file:
Step 2: Extract the ZIP file into the c:\Irvine\Examples directory on
your computer.
Step 3: Do the following:
This file assembles, links, and debugs a single assembly language
source file. Before using it, install Visual Studio 2008 in the following
directory:
C:\Program Files\Microsoft Visual Studio 9.0
Next, install the Irvine 5th edition link libraires and include
files in the following directory: C:\Irvine
Finally, copy this batch file to a location on your system path.
We recommend the following directory:
C:\Program Files\Microsoft Visual Studio errors in the. | 337,301 |
Idea No.
11456
Dressy Dress Sweet 16 with DJ
Award
Date
July 2005
From
Leslie in Walnut Creek California USA
Honorable Mention
Sweet 16 Party! This can be a tough one, because kids this age don't want anything corny, or parents hovering around. So my husband and I rented a local women's club so we could host a dance party. By renting the club, you don't have to worry about the noise, and there is plenty of parking.
We sent out two sets of invitations: one to the girls, which were pink and feminine and told the girls to wear "dressy dresses." Girls this age don't get many chances to dress up for the boys, so they loved it. The boys' invitations were less "girly"--just a flyer with a cartoon squirrel dancing at the top (my daughter is a dancer), with the basic information and a request not to bring a gift. This is big for boys--they never know what to get, and this way, it lets them off the hook. We did tell them to wear "collared shirts" in deference to the girls, who would be very dressy.
We decorated the clubhouse with different shades of pink balloons, just blowing them up with helium, letting them bump up to the ceiling with the long ribbon streamers hanging down. This lends a little atmosphere to the empty hall. We also did a banner with her name and "Sweet 16", and put round tables around the edge of the dance floor, with folding chairs. We used white tablecloths, and little round mirrors with votive candles and chocolate candy hearts as the centerpieces. We also had a big birthday poster to sign, with an 8 X 10 photo of my daughter and lots of pink hearts all around. Everyone signed it, and it's one of her most prized possessions.
The main reason the party was a success was the DJ, one of my daughter's classmates who runs a DJ business with some friends. He knew exactly what songs the kids liked, even though the lyrics would burn your ears. The choice of DJ is the MOST critical factor to a successful teenage party. If the kids think the DJ is lame, they won't stay. My daughter and her friends go to Catholic school, and they aren't allowed to dance inappropriately or listen to dirty lyrics, so having a night without the deans checking their every move was huge.
We hosted about 150 kids from her school, and that means we needed helpers. We had dads at the door with a list of invited guests, and no one else got in unless the birthday girl said OK. These same dads also moved kids out of the parking lot and into the party. (Next year we would definitely include wristbands with the invitations. No wristband, no entry.) I also hired a couple of college kids at $50 each (one boy, one girl) to help with clean-up, serve snacks, and to advise girls dancing in a lewd way that they were making fools of themselves. (Funny how girls will listen to a 20-year-old stranger, but not their parents!)
They also took some great pictures. This way, mom and dad aren't hovering and creating a chilling effect, but you do have "eyes and ears" out on the dance floor.
Food was easy. We started the party at 7 pm, so kids would have already had dinner. I sent up a three-tiered silver tray in the food area, and piled it with Krispy Kremes. That, plus cheese, crackers and salami was all we needed. We also had big ice chests filled with bottled waters outside the doors, in the kitchen, and some waiting in the fridge. The kids went through water like crazy! At about 11 pm, my mom friends starting making simple peanut butter and jelly sandwiches on white bread, and cut them into quarters. We piled a serving tray, and sent our helpers out into the crowd. They snapped those up in a hurry, and no mess. I started making grilled cheese sandwiches, too, and did the same thing. So easy, and the kids loved them.
We only had one case of a kid showing up drunk, and my husband delt with him by sitting him down and making him drink water until his parents showed up. He actually came back to the kitchen to apologize. It could have been a nightmare, but because we planned for every eventuality and had lots of parent help, it was the best night ever for my daughter and her friends. I would do it again, and we probably | 297,757 |
TITLE: Does $\sum\limits_{k=1}^n\frac{a_i-a_k}{a_i+a_k}\cdot\frac{a_j-a_k}{a_j+a_k}=0$ for all $i\neq j$ imply $a_1=a_2=\cdots=a_n$?
QUESTION [8 upvotes]: $$\forall i,\forall j\neq i,\quad\sum_{k=1}^n\frac{a_i-a_k}{a_i+a_k}\cdot\frac{a_j-a_k}{a_j+a_k}=0.$$
We can't have two different $a_i=0$ because of the denominators; but we can allow one $a_i=0$, if the terms $k=i$ and $k=j$ are excluded from the sum.
For $n=3$, these equations are easy to solve:
\begin{align*}
(1,2):\quad&\frac{a_1-a_3}{a_1+a_3}\cdot\frac{a_2-a_3}{a_2+a_3}=0\\
(1,3):\quad&\frac{a_1-a_2}{a_1+a_2}\cdot\frac{a_3-a_2}{a_3+a_2}=0\\
(2,3):\quad&\frac{a_2-a_1}{a_2+a_1}\cdot\frac{a_3-a_1}{a_3+a_1}=0.
\end{align*}
Indeed we just get $a_1=a_2=a_3$.
For $n=4$, the first of $6$ equations is
$$(1,2):\quad\frac{a_1-a_3}{a_1+a_3}\cdot\frac{a_2-a_3}{a_2+a_3}+\frac{a_1-a_4}{a_1+a_4}\cdot\frac{a_2-a_4}{a_2+a_4}=0.$$
(For the other $5$, just permute the indices.) I multiplied to clear the denominators, then added equations $(1,2)$ and $(3,4)$ to get
$$4(a_1a_2-a_3a_4)^2=0$$
and thus
$$a_1a_2=a_3a_4,\quad a_1a_3=a_2a_4,\quad a_1a_4=a_2a_3.$$
These imply that $a_1^2=a_2^2=a_3^2=a_4^2$; and we can't have $a_i=-a_j$, again because of the denominators. So $a_1=a_2=a_3=a_4$.
Does this continue for $n\geq5$?
If the variables are non-negative real numbers, then we can arrange them in order, $a_1\geq a_2\geq a_3\geq\cdots\geq a_n\geq0$; equation $(1,2)$ is then a sum of non-negative terms, so each term must vanish, which gives $a_2=a_3=\cdots=a_n$. Then equation $(2,3)$ has only its first term remaining, which gives $a_1=a_2$.
What if some of the variables are negative, or complex numbers?
We might define $b_{ij}=\dfrac{a_i-a_j}{a_i+a_j}$ to simplify the equations to $\sum_kb_{ik}b_{jk}=0$. Collecting these into an antisymmetric matrix $B$, we see that the system of equations is just saying that
$$BB^T=-B^2=B^TB=D$$
is some diagonal matrix. But I don't think this tells us enough about $B$ itself.
The defining equation for $b_{ij}$ can be rearranged to
$$a_j=\frac{1-b_{ij}}{1+b_{ij}}a_i$$
so in particular
$$a_3=\frac{1-b_{13}}{1+b_{13}}a_1=\frac{1-b_{23}}{1+b_{23}}a_2=\frac{1-b_{23}}{1+b_{23}}\cdot\frac{1-b_{12}}{1+b_{12}}a_1;$$
cancelling $a_1$,
$$(1+b_{31})(1+b_{23})(1+b_{12})=(1-b_{31})(1-b_{23})(1-b_{12});$$
expanding,
$$2b_{12}b_{23}b_{31}+2b_{12}+2b_{23}+2b_{31}=0.$$
In this process I divided by some things that might be $0$, but this resulting cubic equation is valid nonetheless.
I think we can dispense with $a_i$ now. In summary, we need to solve the system of equations
\begin{align*}
\forall i,\forall j,\quad&b_{ij}+b_{ji}=0\\
\forall i,\forall j,\forall k,\quad&b_{ij}b_{jk}b_{ki}+b_{ij}+b_{jk}+b_{ki}=0\\
\forall i,\forall j\neq i,\quad&\sum_kb_{ik}b_{jk}=0.
\end{align*}
Is the only solution $b_{ij}=0$?
REPLY [6 votes]: $\def\C{\mathbb{C}}$This answer solves the system of equations\begin{gather*}
\sum_{k = 1}^n \frac{a_i - a_k}{a_i + a_k} · \frac{a_j - a_k}{a_j + a_k} = 0 \quad (\forall i ≠ j) \tag{$*$}
\end{gather*}
in $\C$ and the italic letter $i$ is not the imaginary unit $\mathrm{i}$.
On the one hand, suppose $(a_1, \cdots, a_n) \in \C^n$ is a solution to ($*$). For any $i, j, k$,\begin{gather*}
\frac{a_i - a_k}{a_i + a_k} - \frac{a_j - a_k}{a_j + a_k} = \frac{2(a_i - a_j) a_k}{(a_i + a_k) (a_j + a_k)}\\
= \frac{a_i - a_j}{a_i + a_j} · \frac{2(a_i + a_j) a_k}{(a_i + a_k) (a_j + a_k)} = \frac{a_i - a_j}{a_i + a_j} \left( 1 - \frac{a_i - a_k}{a_i + a_k} · \frac{a_j - a_k}{a_j + a_k} \right),
\end{gather*}
thus\begin{gather*}
\sum_{k = 1}^n \left( \frac{a_i - a_k}{a_i + a_k} - \frac{a_j - a_k}{a_j + a_k} \right) = \sum_{k = 1}^n \frac{a_i - a_j}{a_i + a_j} \left( 1 - \frac{a_i - a_k}{a_i + a_k} · \frac{a_j - a_k}{a_j + a_k} \right)\\
= n · \frac{a_i - a_j}{a_i + a_j} - \frac{a_i - a_j}{a_i + a_j} \sum_{k = 1}^n \frac{a_i - a_k}{a_i + a_k} · \frac{a_j - a_k}{a_j + a_k} \stackrel{(*)}{=} n · \frac{a_i - a_j}{a_i + a_j}. \tag{1}
\end{gather*}
Define $c_i = \dfrac{1}{n} \sum\limits_{k = 1}^n \dfrac{a_i - a_k}{a_i + a_k}$ for all $i$, then (1) implies that $\dfrac{a_i - a_j}{a_i + a_j} = c_i - c_j$, i.e.\begin{gather*}
(1 - c_i + c_j) a_i = (1 - c_j + c_i) a_j \quad (\forall 1\leqslant i, j \leqslant n). \tag{2}
\end{gather*}
Note that for any $1 \leqslant i < j < k \leqslant n$ with $a_i, a_j, a_k ≠ 0$,$$
\begin{cases}
(1 - c_i + c_j) a_i = (1 - c_j + c_i) a_j\\
(1 - c_j + c_k) a_j = (1 - c_k + c_j) a_k\\
(1 - c_k + c_i) a_k = (1 - c_i + c_k) a_i
\end{cases}
$$
imply that$$
(1 - c_i + c_j)(1 - c_j + c_k)(1 - c_k + c_i) = (1 - c_j + c_i)(1 - c_k + c_j)(1 - c_i + c_k),
$$
which is simplified to be\begin{gather*}
(c_i - c_j)(c_j - c_k)(c_k - c_i) = 0. \tag{3}
\end{gather*}
Case 1: If $a_{i_0} = 0$ for some $i_0$, then $a_i ≠ 0$ for all $i ≠ i_0$ because of non-zero denominators in ($*$), and (2) implies that $c_i = c_{i_0} + 1$ for all $i ≠ i_0$. Thus for any $i, j ≠ i_0$, (2) implies that $a_i = a_j$.
Case 2: If $a_i ≠ 0$ for any $i$, then (3) implies that among any $c_i, c_j, c_k$, there are at least two equal to each other. Thus all $c_i$'s assume at most two values, and whenever $c_i = c_j$ for some $i$ and $j$, (2) implies that $a_i = a_j$.
To summarize, all possible $(a_1, \cdots, a_n)$'s (up to permutation) are of the form$$
(\underbrace{a, \cdots, a}_{m \text{ copies of } a}, \underbrace{b, \cdots, b}_{n - m \text{ copies of } b})
$$
where $2 \leqslant m \leqslant n$ (since $n \geqslant 3$), $a, b \in \C$ and $a ≠ b$. Now without loss of generality assume that $a_1 = a_2 = a$, then$$
0 \stackrel{(*)}{=} \sum_{k = 1}^n \frac{a_1 - a_k}{a_1 + a_k} · \frac{a_2 - a_k}{a_2 + a_k} = \sum_{k = 1}^n \left( \frac{a - a_k}{a + a_k} \right)^2 = (n - m) \left( \frac{a - b}{a + b} \right)^2,
$$
which implies that $m = n$. Therefore all $a_i$'s are equal.
On the other hand, it is easy to verify that $(a_1, \cdots, a_n) = (a, \cdots, a)$ $(a \in \C^*)$ are indeed solutions to ($*$). Therefore they are all the solutions. | 79,812 |
THE SCOTTISH SHETLAND SHEEPDOG CLUB
WELCOME TO THE SCOTTISH SHETLAND
SHEEPDOG CLUB
In Shetland
through the ages, livestock has evolved on a diminutive
scale in proportion to the landscape. Similarly the type of
dogs that herded the livestock also evolved as a diminutive
working dog, and this became known as the
Sheltie. The Sheltie was probably a generic collie, and
it was almost certainly the local shepherd's dog.
There was
no breed standard, the shepherd just needed a tough,
courageous, intelligent little dog. An
engraving of 1840 in Lerwick shows a Sheltie in the foreground. Then in Lerwick,
1908, the 'Shetland Collie Club' was formed in an attempt to
preserve the breed.
(Founded 1909)
Christmas time in 1914 the
Kennel Club recognised the breed. The preferred breed name
'Shetland Collie' was refused, and the breed was
named the
'Shetland Sheepdog'.
Therefore the club represented by this website can
be justifiably proud of being the originator of the
Sheltie as a pedigree dog!
The Shetland Sheepdog, or Sheltie is a
particularly well proportioned and handsome dog.
The Sheltie is
a dog of great intelligence and trainability. It
still has many of the characteristics of a working
sheepdog, and makes a fine guard dog, for it is
loyal and affectionate to its owners, but wary of
strangers.
The club
requested Kennel Club recognition in 1909, but this was refused;
later the same year The Scottish Shetland Sheepdog Club
was formed.
As the Kennel Club refused to recognise the breed,
the Scottish club produced its own stud book, effectively providing
a standard for the breed. Eventually at
The November 2018 Open show schedule are ready to download - Schedule
Breed Judge: Mr M Vincent (Sylbecq)
The June Open show results are now on here
The Scottish Shetland Sheepdog Club 2018 Calendar is now for sale.
And The Scottish Shetland Sheepdog Club 2019 Calendar will soon be on sale.
Also Norman is looking for New Models for the 2020 Calender. Photos to be of Shelties only, no human friends and please no show poses either. Thanks.
Please contact Mr Norman Ritchie, e-mail norman.ritchie@btinternet.com
Closing date the 31st of July.
Please contact:- Mr Norman Ritchie
The newsletter editor
would be would be most grateful if anyone has any articles
they would wish to submit for publication. Please forward
any article to
Mrs Helen Moir, The Oaks, Cabrich, Kirkhill, Inverness IV5 7PH
or E-mail h.moir@uwclub.net
by the 31st of August please.
Visitors since
23rd January, 2004.
The Web Site was Last Updated
17 January 2012
Webmaster: webmaster@scottish-sheltie.org.uk
© The Scottish Shetland Sheepdog Club
2004 - 2012 | 180,393 |
TITLE: Notation question in calculus of variations -- QFT
QUESTION [2 upvotes]: these two integrals below are equal, but I am not understanding where the $x'$ variable comes from.
\begin{align}
I_0&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] -\big[\varphi(x)\big]^4 +J(x)\varphi(x)\right\}}\\
&=e^{ -i\int d^4x'\big[ \frac{\delta}{\delta J(x')} \big]^4 } e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}}
\end{align}
The gist of what I have been doing is to write
\begin{align}
I_0&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}}e^{ -i\int d^4x \big[\varphi(x)\big]^4} \\
&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}} \left[ 1+\left( -i\int d^4x \big[ \varphi(x) \big]^4 \right) +...\right]\\
&=\left[ 1+\left( -i\int d^4x \left[\dfrac{d}{dJ} \right]^4\right)+... \right]e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}}
\end{align}
The point is to pull the $\varphi^4$ term out of the integral by writing $\varphi$ as $d/dJ$ except in this case I have to use the variation $\delta$ instead and I don't see why. I used the $d/dJ$ trick in simpler examples without an integral in the exponent, and I am not seeing the connection to the variational notation $\delta$ which appears in the second of the first two equations above. Obviously, I can recombine the prefactor sum in my final equation into $\text{exp}$, but I do not see the point of the $x'$ variable. I hope it is clear that if I had $\delta/\delta J(x')$ in my last equation instead of $d/dJ$ then I would get the form of the second equation which is the correct form. Why can't I just write it as $d/dJ$ like I did before? Please give me a tip, thanks.
REPLY [1 votes]: The issue is really what it means to to compute one functional derivative $\delta/\delta f$. Once we get that part, we can raise it to the nth power and get the result.
How to compute one variation?
\begin{align}
\dfrac{\delta}{\delta J(x')} K&=\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}} \\
&=\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x J(x)\varphi(x)}e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] \right\}} \\
&=A\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x J(x)\varphi(x)} \\
\end{align}
Let $F=e^{ i\int d^4x J(x)\varphi(x)} $. The next step is what follows. Instead of using $\Delta$, you need to use $\varepsilon$ times a Dirac delta. This is part of the definition of the functional derivative I guess, If anyone want to say a little more about that would be nice, but I think it's just part of the definition. However, wikipedia calls the Dirac delta a "test function" so maybe it is an ansatz of some type. Anyway, the answer follows:
\begin{align}
\dfrac{\delta}{\delta J(x')}F&=\lim\limits_{\varepsilon\to0}\dfrac{F[J(x)+\varepsilon\delta(x-x')]-F[J(x)]}{\varepsilon}\\
&=\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\int d^4x [J(x)+\varepsilon\delta(x-x')]\varphi(x)}-e^{ i\int d^4x J(x)\varphi(x)} }{ \varepsilon }\\
&=\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\int d^4x [J(x)+\varepsilon\delta(x-x')]\varphi(x)}-e^{ i\int d^4x J(x)\varphi(x)} }{ \varepsilon }\\
&=F\,\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\int d^4x \varepsilon\delta(x-x')\varphi(x)}-1 }{ \varepsilon }\\
&=F\,\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\varepsilon\varphi(x')}-1 }{ \varepsilon }
\end{align}
use l'Hopital's rule
\begin{align}
\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\varepsilon\varphi(x')}-1 }{ \varepsilon }\stackrel{*}{=}\lim\limits_{\varepsilon\to0}\dfrac{i\varphi(x') e^{ i\varepsilon\varphi(x')}}{ 1 }=i\varphi(x')
\end{align}
therefore
\begin{align}
\left[\dfrac{\delta}{\delta J}\right]^4F=\left[\varphi(y)\right]^4 F
\end{align}
then
\begin{align}
AF+\left(-i\int d^4x \left[\dfrac{d}{dJ}\right]^4 \right)AF+...&=AF+\left(-i\int d^4x' \left[\varphi(x')\right]^4\right)AF+...\\
&=e^{ -i\int d^4x' \big[\varphi(x')\big]^4}AF \\
&=e^{ -i\int d^4x \big[\varphi(x)\big]^4} AF
\end{align}
Plugging in F and A, we get the expected result. | 182,812 |
TITLE: Under what conditions $f$ and $g$ are open maps?
QUESTION [0 upvotes]: Let the composition of two arbitrary function $f$ and $g$ be an open function. Under what conditions $f$ and $g$ are open maps?
REPLY [0 votes]: Let $f:X\to Y$ and $g:Y\to Z$ be functions between topological spaces and let $h$ denote the composition of $f$ and $g$. Assume that $h$ is open. Then you have
If $g$ is injective and continuous, then for an open subset $U$ of $X$
we have $f(U)=g^{-1}(h(U)),$ so $f$ is open.
If $f$ is surjective and continuous, then for an open subset $V$ of $Y$
we have $g(V)=h(f^{-1}(V)),$ so $g$ is open.
In general, we have $h(f^{-1}(V))=g(V\cap f(X))$ for a set $V\subseteq Y.$ | 17,473 |
Bharath Ane Nenu is set for a grand release on April 20th. Mahesh Babu has given exclusive interview on the occasion.
About holiday trip>
Tour has been rejuvenating. Everytime we used to go after film’s release. This is the first time we have gone on vacation before release. I can say its best pre-release phase in my career as Bharath AneNenu is surrounded by positive vibe. Pre release buzz and talk is extraordinary. Entire team is confident
About politics?
I am not a great political enthusiast. I am excited when heard the script one year back. I like the way Koratala siva garu designed my character. At the same time Im feared of doing CM role as its of great responsiblities. I don’t know much about politics but in the process of doing this film I got awareness. I am honored to do this film.
Any preparation for the role?
Full credit to Koratala Siva for writing interesting dialogues with detailing and logics as its a political drama. We need to be cautious and careful in doing such responsible genre. I have never mouthed such lengthy dialogues before. With Siva garu’s help I pulled it off. I didn’t do any home work. I have gone through few speeches of Bava Galla Jayadev in parliament.
Inspiration of any CM?
We didn’t want to imitate anybody. My bodylanguage as CM will be fresh. Even my costumes look formal and youthful.
Masterpiece?
Upon listening to the story itself I was confident that its going to be masterpiece. When Devi Sri Prasad gave us theme song. Assembly set erected by art director. We shot those episodes like real assembly sessions. With all these grandeur, we had positive vibes.
Satires on political scenario?
No, our film will not target any political party or leader and it will not have any satires on them. We are sure every leader will watch and appreciate the film. Its honest political drama.
Political interest?
After doing the film, did you gain interest on politics. In future do you have any political aspirations. No, I dedicate my entire life to cinema.
Responsibility?
We have made honest film. However, its ultimately entertainment purpose. But it gives a message that everybody in society are entailed with responsibility.
Politics not a joke?
Politics is responsible job, its not a joke. I play CM’s role who takes stringent steps to safeguard people’s interest. I hope it motivates real life leaders.
Release day, best in career?
Im happy BAN got preponed. April 20th is our mother’s birthday. I can’t ask for more. Her blessings are with me. Interestingly, Chandrababu Naidu’s birthday is also on sameday. Its a good day indeed.
About Kiara Advani?
We were looking for fresh face as CM’s girl friend. After watching MS Dhoni-The untold story, we thought she’s apt for the role.
Spectacular sets?
Art director Suresh has done terrific job. Be it Assembly set or Vachadayyo Sami song set, it’s spectacular.
Mahesh’s next film is with Vamsi Paidipally, the film goes on floors from June 7th in USA. Pooja Hegde is pairing superstar.
| 324,135 |
Best – […]
Tag: wishlist
I was at Barnes and Noble today and saw Dilbert 2.0, a 20th anniversary collection of Dilbert comics from Andrews McMeel. The collection is fashioned in the very same way as previous all-in-one collections like The Complete Calvin and Hobbes and The Complete Far Side and includes over 4000 comic strips from Dilbert over the […] | 283,599 |
Quotations About / On: FREEDOM
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- 31.
Every farewell combines loss and new freedom.(Mason Cooley (b. 1927), U.S. aphorist. City Aphorisms, Ninth Selection, New York (1992).)
- 32.
The history of the world is none other than the progress of the consciousness of freedom.(Georg Wilhelm Friedrich Hegel (1770-1831), German philosopher. The Philosophy of History, "Introduction," sct. 3 (1837).)
- 33.
Freedom of conscience entails more dangers than authority and despotism.(Michel Foucault (1926-1984), French philosopher. Madness and Civilization, ch. 7 (1965).)More quotations from: Michel Foucault, freedom
- 34.
- 35.
The spirit of truth and the spirit of freedomthese are the pillars of society.(Henrik Ibsen (1828-1906), Norwegian dramatist. Lona Hessel, in Pillars of Society, act 4.)More quotations from: Henrik Ibsen, freedom, truth
- 36.
There's something contagious about demanding freedom.(Robin Morgan (b. 1941), U.S. feminist author, poet. Sisterhood Is Powerful, introduction (1970).)More quotations from: Robin Morgan, freedom
- 37.
Freedom is the most contagious virus known to man.(Hubert H. Humphrey (1911-1978), U.S. Democratic politician, vice president. speech, Oct. 29, 1964, New York City.)More quotations from: Hubert H Humphrey, freedom
- 38.
- 39.
- 40.
Freedom is a possession of inestimable value.(Marcus Tullius Cicero (106-43 B.C.), Roman orator, philosopher, statesman. Paulus.)More quotations from: Marcus Tullius Cicero, | 236,213 |
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Date of Award
1999
Document Type
Dissertation
Degree Name
Doctor of Philosophy
School
School of Education
Program
Counseling Psychology, Ph.D.
First Advisor
Jimmy Kijai
Second Advisor
Nancy J. Carbonell
Third Advisor
Frederick A. Kosinski, Jr.
Abstract
Problem. Eating disorders, in the form of anorexia nervosa and bulimia nervosa, have been recognized as significant mental-health issues for the last three decades, and the incidence is rising as we approach the new millennium. Currently, many women who come into a mental-health setting due to depression, anxiety, low self-esteem, relationship issues, sexual issues, etc., are also struggling with eating-disordered behaviors, thoughts, and feelings. These behaviors, thoughts, and feelings may remain well hidden from the counselor throughout the course of therapy or until they become severe and more difficult to treat. Because eating disorders are very complex involving psychological, physical, and mental functioning, and because the symptoms become progressively more severe, early detection and intervention are essential for optimal outcome. Numerous assessment instruments exist but are not employed until obvious signs of eating disorders are exhibited. This study was designed to develop a subscale from the MMPI-2 items which will screen for eating disorders. Since the MMPI-2 is widely used early in the process of psychological evaluation, it was deemed the desirable instrument to use.
Method. The methodology for this study involved scale development and included four phases. Phase 1 was the initial study in which 354 MMPI answer sheets from eating-disordered individuals were compared to 238 MMPI answer sheets from non-eating-disordered individuals in order to determine MMPI items which differentiate between the two groups. Phase 2 utilized expert judges to evaluate the pertinence of each item on an eating-disorder questionnaire and to assign directionality. Phase 3 involved administering the items remaining after the first two phases to a new research sample comprised of eating-disordered and non-eating-disordered subjects. Phase 4 entailed eliminating the items which did not meet the total correlation criterion, and computing internal consistency using Cronbach’s alpha coefficient for the remaining items.
Results. This research resulted in the development of a 68-item proposed MMPI-2 subscale to screen for anorexia nervosa and bulimia nervosa. Each item met the differential criterion at the .01 level and the correlation coefficient at .33. The proposed MMPI-2 subscale has a reliability of .971 and its composition is unifactorial.
Conclusions. This research establishes the efficacy of utilizing the MMPI-2 to screen for eating disorders. Additional administration of the instrument is needed before it should move from research into practice.
Subject Area
Eating disorders--Diagnosis, Minnesota Multiphasic Personality Inventory.
Recommended Citation
Woodka, Roseann M., "Screening for Eating Disorders Utilizing the Minnesota Multiphasic Personality Inventory" (1999). Dissertations. 1559.
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