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Theorem 2.5. Let \( f\left( x\right) = x \cdot \sin \frac{1}{x}, x \in \left( {0,\infty }\right) \) . Then\n\n\[ \left| {f\left( y\right) - f\left( x\right) }\right| \leq \sqrt{2\left| {y - x}\right| },\;\forall x, y \in \left( {0,\infty }\right) . | Proof. Define \( {J}_{0} = \left\lbrack {\frac{1}{{\alpha }_{1}},\infty }\right) ,{J}_{n} = \left\lbrack {\frac{1}{{\alpha }_{n + 1}},\frac{1}{{\alpha }_{n}}}\right), n = 1,2,\ldots \) From (2.1) we see that \( f \) is monotone in each interval \( {J}_{k}, k = 0,1,\ldots \) Moreover,\n\n\[ f\left( \frac{1}{{\alpha }_{n... | Yes |
Theorem 3.3. Let us assume we are given a pointwise canonical modulus \( \mathcal{R} \) . Then \( \bar{g} = \widetilde{n} \) . | Proof. We begin by considering a simple special case. Trivially, if \( {\mathcal{P}}_{N} \) is not equivalent to \( \widetilde{b} \) then every countable, Liouville isometry acting simply on a Bernoulli, regular manifold is trivially Milnor-Serre and almost everywhere right-elliptic. It is easy to see that \( {N}_{\lam... | No |
Theorem 3.4. Let us assume \( \psi \) is greater than \( M \) . Then\n\n\[ \n\mathfrak{l}\left( {-1 - \mathcal{N},{J}^{\prime \prime }}\right) \neq {H}^{-8} \]\n\n\[ \n\sim \mathop{\sup }\limits_{{m \rightarrow 0}}{\bar{p}}^{-1}\left( {q}^{9}\right) \cup \log \left( {2 \land \begin{Vmatrix}{M}^{\prime \prime }\end{Vmat... | Proof. See [22]. | No |
Lemma 4.3. Let \( {\mathcal{Y}}^{\prime } \) be a solvable, embedded, Hardy random variable acting finitely on a Noetherian, invariant random variable. Let \( \widehat{S} \supset \pi \) be arbitrary. Then \( {\nu }^{\prime } \neq \bar{\Omega }\left( \overline{\mathbf{m}}\right) \) . | Proof. Suppose the contrary. Let \( \mathcal{I} < \sqrt{2} \) be arbitrary. Note that if \( {n}^{\prime \prime } \) is left-closed then \( {0}^{-1} = \) \( \cosh \left( \frac{1}{e}\right) \) . In contrast, if Cauchy’s criterion applies then \( \widetilde{L} \supset \infty \) . Now \( \psi \) is bounded by \( \overline{... | Yes |
Lemma 5.3. Let \( E = \eta \) be arbitrary. Assume \( \parallel \mathfrak{p}\parallel = {F}_{L} \) . Then there exists an one-to-one, algebraic and freely meager surjective ring. | Proof. We follow [7]. It is easy to see that if Erdős’s criterion applies then \( \begin{Vmatrix}{\mathcal{K}}^{\left( T\right) }\end{Vmatrix} > \widetilde{b} \) . By an approximation argument, \( \mathcal{W} \leq e \) . By existence, \( \bar{s} = \varnothing \) . Hence if the Riemann hypothesis holds then \( \mathbf{s... | Yes |
Proposition 5.4. Let \( h \) be an anti-meager subring. Let \( \widehat{\mathcal{A}} \) be a Cantor isomorphism. Then every countably reversible monodromy is countable and left-Noetherian. | Proof. See [5]. | No |
Lemma 1. Let \( x \) and \( y \) be (not necessary positive) integers and let \( n \) be a positive integer. Given an arbitrary prime \( p \) (in particular, we can have \( p = 2 \) ) such that \( \gcd \left( {n, p}\right) = 1, p \mid x - y \) and neither \( x \), nor \( y \) is divisible by \( p \) (i.e., \( p \nmid x... | Proof. We use the fact that\n\n\[ \n{x}^{n} - {y}^{n} = \left( {x - y}\right) \left( {{x}^{n - 1} + {x}^{n - 2}y + {x}^{n - 3}{y}^{2} + \cdots + {y}^{n - 1}}\right) .\n\]\n\nNow if we show that \( p \nmid {x}^{n - 1} + {x}^{n - 2}y + {x}^{n - 3}{y}^{2} + \cdots + {y}^{n - 1} \), then we are done. In order to show this,... | Yes |
Lemma 2. Let \( x \) and \( y \) be (not necessary positive) integers and let \( n \) be an odd positive integer. Given an arbitrary prime \( p \) (in particular, we can have \( p = 2 \) ) such that \( \gcd \left( {n, p}\right) = 1, p \mid x + y \) and neither \( x \), nor \( y \) is divisible by \( p \), we have\n\n\[... | Proof. Since \( x \) and \( y \) can be negative, using Lemma 1 we obtain\n\n\[ \n{v}_{p}\left( {{x}^{n} - {\left( -y\right) }^{n}}\right) = {v}_{p}\left( {x - \left( {-y}\right) }\right) \Rightarrow {v}_{p}\left( {{x}^{n} + {y}^{n}}\right) = {v}_{p}\left( {x + y}\right) . \n\]\n\nNote that since \( n \) is an odd posi... | Yes |
Theorem 2 (Second Form of LTE). Let \( x, y \) be two integers, \( n \) be an odd positive integer, and \( p \) be an odd prime such that \( p \mid x + y \) and none of \( x \) and \( y \) is divisible by \( p \) . We have\n\n\[ \n{v}_{p}\left( {{x}^{n} + {y}^{n}}\right) = {v}_{p}\left( {x + y}\right) + {v}_{p}\left( n... | Proof. This is obvious using Theorem 1. See the trick we used in proof of Lemma 2. | No |
Theorem 3 (LTE for the case \( p = 2 \) ). Let \( x \) and \( y \) be two odd integers such that \( 4 \mid x - y \) . Then\n\n\[ \n{v}_{2}\left( {{x}^{n} - {y}^{n}}\right) = {v}_{2}\left( {x - y}\right) + {v}_{2}\left( n\right) .\n\] | Proof. We showed that for any prime \( p \) such that \( \gcd \left( {p, n}\right) = 1, p \mid x - y \) and none of \( x \) and \( y \) is divisible by \( p \), we have\n\n\[ \n{v}_{p}\left( {{x}^{n} - {y}^{n}}\right) = {v}_{p}\left( {x - y}\right) \n\]\n\nSo it suffices to show that\n\n\[ \n{v}_{2}\left( {{x}^{{2}^{n}... | Yes |
Theorem 4. Let \( x \) and \( y \) be two odd integers and let \( n \) be an even positive integer. Then\n\n\[ \n{v}_{2}\left( {{x}^{n} - {y}^{n}}\right) = {v}_{2}\left( {x - y}\right) + {v}_{2}\left( {x + y}\right) + {v}_{2}\left( n\right) - 1.\n\] | Proof. We know that the square of an odd integer is of the form \( {4k} + 1 \) . So for odd \( x \) and \( y \) we have \( 4 \mid {x}^{2} - {y}^{2} \) . Now let \( m \) be an odd integer and \( k \) be a positive integer such that \( n = m \cdot {2}^{k} \) . Then\n\n\[ \n{v}_{2}\left( {{x}^{n} - {y}^{n}}\right) = {v}_{... | Yes |
Find all positive integers \( n \) for which there exist positive integers \( x, y \) and \( k \) such that \( \gcd \left( {x, y}\right) = 1, k > 1 \) and \( {3}^{n} = {x}^{k} + {y}^{k} \) | Solution. \( k \) should be an odd integer (otherwise, if \( k \) is even, then \( {x}^{k} \) and \( {y}^{k} \) are perfect squares, and it is well known that for integers \( a, b \) we have \( 3 \mid {a}^{2} + {b}^{2} \) if and only if \( 3 \mid a \) and \( 3 \mid b \), which is in contradiction with \( \gcd \left( {x... | Yes |
Let \( p \) be a prime number and \( m > 1 \) be a positive integer. Show that if for some positive integers \( x > 1, y > 1 \) we have\n\n\[ \frac{{x}^{p} + {y}^{p}}{2} = {\left( \frac{x + y}{2}\right) }^{m} \]\n\nthen \( m = p \) . | Solution. One can prove by induction on \( p \) that \( \frac{{x}^{p} + {y}^{p}}{2} \geq {\left( \frac{x + y}{2}\right) }^{p} \) for all positive integers \( p \) . Now since \( \frac{{x}^{p} + {y}^{p}}{2} = {\left( \frac{x + y}{2}\right) }^{m} \), we should have \( m \geq p \) . Let \( d = \) \( \gcd \left( {x, y}\rig... | Yes |
Find all positive integers \( a, b \) that are greater than 1 and satisfy\n\n\[ {b}^{a} \mid {a}^{b} - 1.\text{.} \] | Solution. Let \( p \) be the least prime divisor of \( b \) . Let \( m \) be the least positive integer for which \( p \mid {a}^{m} - 1 \) . Then \( m \mid b \) and \( m \mid p - 1 \), so any prime divisor of \( m \) divides \( b \) and is less than \( p \) . Thus, not to run into a contradiction, we must have \( m = 1... | Yes |
Problem 4. Find all positive integer solutions of the equation \( {x}^{2009} + {y}^{2009} = {7}^{z} \) | Solution. Factor 2009. We have \( {2009} = {7}^{2} \cdot {41} \) . Since \( x + y \mid {x}^{2009} + {y}^{2009} \) and \( x + y > 1 \), we must have \( 7 \mid x + y \) . Removing the highest possible power of 7 from \( x, y \), we get \( {v}_{7}\left( {{x}^{2009} + {y}^{2009}}\right) = {v}_{7}\left( {x + y}\right) + {v}... | Yes |
Proposition 1.1.1. On a compact \( n \) -dimensional manifold \( M \), a gradient steady or expanding Ricci soliton is necessarily an Einstein metric. | Proof. We shall only prove the steady case and leave the expanding case as an exercise. Our argument here follows that of Hamilton [63].\n\nLet \( {g}_{ij} \) be a complete steady gradient Ricci soliton on a manifold \( M \) so that\n\n\[ \n{R}_{ij} + {\nabla }_{i}{\nabla }_{j}f = 0.\n\]\n\nTaking the trace, we get\n\n... | No |
Proposition 1.3.4 (Hamilton [59]). Let \( {R}_{abcd} = {M}_{\alpha \beta }{\varphi }_{ab}^{\alpha }{\varphi }_{cd}^{\beta } \) . Then under the Ricci flow (1.1.5), \( {M}_{\alpha \beta } \) satisfies the evolution equation | \[ \frac{\partial {M}_{\alpha \beta }}{\partial t} = \Delta {M}_{\alpha \beta } + {M}_{\alpha \beta }^{2} + {M}_{\alpha \beta }^{\# } \] where \( {M}_{\alpha \beta }^{2} = {M}_{\alpha \gamma }{M}_{\beta \gamma } \) is the operator square and \( {M}_{\alpha \beta }^{\# } = \left( {{C}_{\alpha }^{\gamma \eta }{C}_{\beta ... | Yes |
(i) \( \bar{\lambda }\left( {g}_{ij}\right) \) is nondecreasing along the Ricci flow whenever it is nonpositive; moreover, the monotonicity is strict unless we are on a gradient expanding soliton; | Proof. Let \( {f}_{0} \) be a minimizer of \( \lambda \left( {{g}_{ij}\left( t\right) }\right) \) at \( t = {t}_{0} \) and solve the backward heat equation\n\n\[ \frac{\partial f}{\partial t} = - {\Delta f} + {\left| \nabla f\right| }^{2} - R \]\n\nto obtain \( f\left( t\right), t \leq {t}_{0} \), with \( {\int }_{M}{e... | Yes |
Lemma 1.5.7 (Perelman [103]). If \( {v}_{ij} = \delta {g}_{ij}, h = {\delta f} \), and \( \eta = {\delta \tau } \), then\n\n\[ \delta \mathcal{W}\left( {{v}_{ij}, h,\eta }\right) \]\n\n\[ = {\int }_{M} - \tau {v}_{ij}\left( {{R}_{ij} + {\nabla }_{i}{\nabla }_{j}f - \frac{1}{2\tau }{g}_{ij}}\right) {\left( 4\pi \tau \ri... | Proof. Arguing as in the proof of Lemma 1.5.2, the first variation of the functional \( \mathcal{W} \) can be computed as follows,\n\n\[ {\delta W}\left( {{v}_{ij}, h,\eta }\right) \]\n\n\[ = {\int }_{M}\left\lbrack {\eta \left( {R + {\left| \nabla f\right| }^{2}}\right) + \tau \left( {-{\Delta v} + {\nabla }_{i}{\nabl... | Yes |
Proposition 1.5.8. If \( {g}_{ij}\left( t\right), f\left( t\right) \) and \( \tau \left( t\right) \) evolve according to the system\n\n\[ \left\{ \begin{array}{l} \frac{\partial {g}_{ij}}{\partial t} = - 2{R}_{ij}, \\ \frac{\partial f}{\partial t} = - {\Delta f} + {\left| \nabla f\right| }^{2} - R + \frac{n}{2\tau }, \... | Proof. Using Lemma 1.5.7, we have\n\n(1.5.11)\n\n\[ \frac{d}{dt}\mathcal{W}\left( {{g}_{ij}\left( t\right), f\left( t\right) ,\tau \left( t\right) }\right) \]\n\n\[ = {\int }_{M}{2\tau }{R}_{ij}\left( {{R}_{ij} + {\nabla }_{i}{\nabla }_{j}f - \frac{1}{2\tau }{g}_{ij}}\right) {\left( 4\pi \tau \right) }^{-\frac{n}{2}}{e... | Yes |
(i) \( \mu \left( {{g}_{ij}\left( t\right) ,\tau - t}\right) \) is nondecreasing along the Ricci flow; moveover, the monotonicity is strict unless we are on a shrinking gradient soliton; | Proof. Fix any time \( {t}_{0} \), let \( {f}_{0} \) be a minimizer of \( \mu \left( {{g}_{ij}\left( {t}_{0}\right) ,\tau - {t}_{0}}\right) \) . Note that the backward heat equation\n\n\[ \frac{\partial f}{\partial t} = - {\Delta f} + {\left| \nabla f\right| }^{2} - R + \frac{n}{2\tau } \]\n\nis equivalent to the linea... | Yes |
Proposition 2.1.2. If the scalar curvature \( R \) of the solution \( {g}_{ij}\left( t\right) ,0 \leq t \leq T \) , to the Ricci flow is nonnegative at \( t = 0 \), then it remains so on \( 0 \leq t \leq T \) . | Proof. Let \( f \) be the function constructed in Lemma 2.1.1 and recall\n\n\[ \frac{\partial R}{\partial t} = {\Delta R} + 2{\left| \operatorname{Ric}\right| }^{2} \]\n\nFor any small constant \( \varepsilon > 0 \) and large constant \( A > 0 \), we have\n\n\[ \frac{\partial }{\partial t}\left( {R + \varepsilon {e}^{A... | Yes |
Lemma 2.3.3.\n\n\[ \frac{d}{dt}\varphi \left( t\right) \leq \sup \left\{ {\frac{\partial \psi }{\partial t}\left( {y, t}\right) \mid y \in Y\text{ satisfies }\psi \left( {y, t}\right) = \varphi \left( t\right) }\right\} .\n\] | Proof. Choose a sequence of times \( \left\{ {t}_{j}\right\} \) decreasing to \( t \) for which\n\n\[ \mathop{\lim }\limits_{{{t}_{j} \rightarrow t}}\frac{\varphi \left( {t}_{j}\right) - \varphi \left( t\right) }{{t}_{j} - t} = \frac{{d\varphi }\left( t\right) }{dt} \]\n\nSince \( Y \) is compact, we can choose \( {y}_... | Yes |
Given any unit vector \( Y \) at any point \( q \in M \) with \( \tau = \bar{\tau } \) , consider an \( \mathcal{L} \) -shortest geodesic \( \gamma \) connecting \( p \) to \( q \) and extend \( Y \) along \( \gamma \) by solving the ODE (3.2.8). Then the Hessian of the \( \mathcal{L} \) -distance function \( L \) on \... | Proof. As said before, we pretend that the shortest \( \mathcal{L} \) -geodesics between \( p \) and \( q \) are unique so that \( L\left( {q,\bar{\tau }}\right) \) is smooth. Otherwise, the inequality is still valid, by a standard barrier argument, when understood in the sense of distributions (see, for example, [112]... | Yes |
Proposition 5.1.9. Let \( \left( {M,{g}_{ij}\left( t\right) }\right) \) be a solution of the normalized Ricci flow on a compact surface with \( \chi \left( M\right) > 0 \) . Then there exist positive constants \( c \) and \( C \) depending only on the initial metric such that \[ {\left| {M}_{ij}\right| }^{2} \leq C{e}^... | Now we consider a modification of the normalized Ricci flow. Consider the equation \( \left( {5.1.25}\right) \) \[ \frac{\partial }{\partial t}{g}_{ij} = 2{M}_{ij} = \left( {r - R}\right) {g}_{ij} + 2{\nabla }_{i}{\nabla }_{j}\varphi . \] As we saw in Section 1.3, the solution of this modified flow differs from that of... | Yes |
Proposition 5.1.10. On a compact surface there are no shrinking Ricci solitons other than constant curvature. | Proof. By definition, a shrinking Ricci soliton on a compact surface \( M \) is given by\n\n\( \left( {5.1.26}\right) \)\n\n\[ \n{\nabla }_{i}{X}_{j} + {\nabla }_{j}{X}_{i} = \left( {R - r}\right) {g}_{ij} \]\n\nfor some vector field \( X = {X}_{j} \) . By contracting the above equation by \( R{g}^{-1} \), we have\n\n\... | Yes |
Proposition 5.2.5. Suppose that the initial metric of the solution to the Ricci flow on \( {M}^{3} \times \lbrack 0, T) \) has positive Ricci curvature. Then for any \( \varepsilon > 0 \) we can find \( {C}_{\varepsilon } < + \infty \) such that\n\n\[ \left| {{R}_{ij} - \frac{1}{3}R{g}_{ij}}\right| \leq {\varepsilon R}... | Proof. Again we consider the ODE system (5.2.2). Let \( {M}_{\alpha \beta } \) be diagonalized with eigenvalues \( \lambda \geq \mu \geq \nu \) at \( t = 0 \) . We saw in the proof of Lemma 5.2.4 the inequalities \( \lambda \geq \mu \geq \nu \) persist for \( t > 0 \) . We only need to show that there are positive cons... | Yes |
Corollary 7.3.7 (Perelman [104]). For any \( l < \infty \) one can find \( A = A\left( l\right) < \) \( \infty \) and \( \theta = \theta \left( l\right) ,0 < \theta < 1 \), with the following property. Suppose we are in the situation of the lemma above, with \( \delta < \bar{\delta }\left( {A,\theta ,\varepsilon }\righ... | Proof. We know from Proposition 7.3.3 that on the standard solution,\n\n\[{\int }_{0}^{\theta }{Rdt} \geq \text{ const. }{\int }_{0}^{\theta }{\left( 1 - t\right) }^{-1}{dt}\]\n\n\[= - \text{const.} \cdot \log \left( {1 - \theta }\right) \text{.}\]\n\nBy choosing \( \theta = \theta \left( l\right) \) sufficiently close... | Yes |
Lemma 2.3. Suppose \( \psi \in {\Psi }_{1} \) and \( \rho \in \mathcal{Z}\left( \psi \right) \) . Then\n\n\[ \n{\mathcal{C}}^{ * }\left( {\rho ,\psi }\right) \geq 0 \n\] | We introduce the smooth weight\n\n\[ \n\omega \left( s\right) = \frac{\sqrt{\pi }}{{\mathcal{L}}_{2}}\exp \left\{ \frac{{\left( s - {s}_{0}\right) }^{2}}{4{\mathcal{L}}_{2}^{2}}\right\} \;\text{ with }\;{\mathcal{L}}_{2} = {\mathcal{L}}^{400}, \n\]\n\n(2.15)\n\nwhich is positive for \( \sigma = 1/2 \) . Lemma 2.3 impli... | No |
Proposition 2.6. Assume that (A) holds. Then\n\n\[ \n{\Xi }_{3}^{ * } = o\left( {\mathfrak{a}\mathcal{P}}\right) \n\] | Under Assumption (A), a contradiction is immediately derived from (2.18), Proposition 2.4, 2.5 and 2.6, This proves Theorem 1. | No |
Lemma 3.1. Assume (A) holds. Then\n\n\[ \mathop{\sum }\limits_{{{D}^{4} < n \leq {P}^{2}}}\frac{\nu {\left( n\right) }^{2}}{n} \ll {\mathcal{L}}^{-{2011}} \] | Proof. Let\n\n\[ \phi \left( s\right) = \zeta {\left( s\right) }^{-2}L{\left( s,\chi \right) }^{-2}\mathop{\sum }\limits_{n}\frac{\nu {\left( n\right) }^{2}}{{n}^{s}} \]\n\nwhich has the Euler product representation\n\n\[ \phi \left( s\right) = \mathop{\prod }\limits_{p}{\phi }_{p}\left( s\right) \;\left( {\sigma > 1}\... | Yes |
Lemma 3.2. Assume (A) holds. Then we have\n\n\\[ \mathop{\sum }\limits_{{{D}^{4} < n \leq {D}^{8}}}\frac{\nu {\left( n\right) }^{2}{\tau }_{2}{\left( n\right) }^{2}}{n} \ll {\mathcal{L}}^{-{2007}}. \\] | Proof. As the situation is analogous to Lemma 3.1 we give a sketch only. It can be verified that the function\n\n\\[ {\phi }^{ * }\left( s\right) = \zeta {\left( s\right) }^{-8}L{\left( s,\chi \right) }^{-8}\mathop{\sum }\limits_{n}\frac{\nu {\left( n\right) }^{2}{\tau }_{2}{\left( n\right) }^{2}}{{n}^{s}} \\]\nis anal... | Yes |
Lemma 3.3. For any \( s \) and any complex numbers \( c\left( n\right) \) we have\n\n\[ \mathop{\sum }\limits_{{\psi \in \Psi }}{\left| \mathop{\sum }\limits_{{n \leq P}}\frac{c\left( n\right) \psi \left( n\right) }{{n}^{s}}\right| }^{2} \ll \mathcal{P}\mathop{\sum }\limits_{{n \leq P}}\frac{{\left| c\left( n\right) \r... | Proof. The first assertion follows by the orthogonality relation; the second assertion follows by the large sieve inequality. \( ▱ \) | No |
Lemma 3.4. The inequality\n\n\[ \left| {{X}_{1}\left( {{D}^{80},\psi }\right) }\right| + \left| {{X}_{2}\left( {{D}^{80},\psi }\right) }\right| + {\int }_{1}^{{D}^{80}}\frac{\left| {{X}_{1}\left( {x,\psi }\right) }\right| + \left| {{X}_{2}\left( {x,\psi }\right) }\right| }{x}{dx} < {\mathcal{L}}^{1171} \]\n\nholds for ... | Write\n\[ {X}_{3}\left( {x,\psi }\right) = \mathop{\sum }\limits_{{{D}^{4} < n \leq x}}\frac{\nu \left( n\right) \psi \left( n\right) }{{n}^{{s}_{0}}}\;\text{ for }\;x > {D}^{4}. \]\n\nAssume that (A) holds. By Cauchy's inequality, the second assertion of Lemma 3.2 and Lemma 3.1,\n\n\[ \mathop{\sum }\limits_{{\psi \in ... | No |
Lemma 4.1. Let\n\n\\[ \n{\\Omega }_{1} = \\left\\{ {s : \\;1/2 - {\\left( {100}\\mathcal{L}\\right) }^{-1}\\log \\mathcal{L} < \\sigma < 1 + {\\left( {100}\\mathcal{L}\\right) }^{-1}\\log \\mathcal{L},\\;\\left| {t - {2\\pi }{t}_{0}}\\right| < {\\mathcal{L}}_{1} + 5}\\right\\} .\n\\]\n\nIf \\( s \\in {\\Omega }_{1} \\)... | Proof. By the Stieltjes integral we may write\n\n\\[ \nF{\\left( s,\\psi \\right) }^{20} = 1 + {\\int }_{1}^{{D}^{80}}{x}^{{s}_{0} - s}d\\left\\{ {{X}_{1}\\left( {x,\\psi }\\right) }\\right\\} .\n\\]\n\nFor \\( s \\in {\\Omega }_{1} \\) and \\( 1 \\leq x \\leq {D}^{80} \\) we have\n\n\\[ \n\\left| {x}^{{s}_{0} - s}\\ri... | Yes |
Lemma 4.2. If \( s \in {\Omega }_{1} \), then\n\n\[ F\left( {s,\psi }\right) G\left( {s,\psi }\right) = 1 + O\left( {\mathcal{L}}^{-{227}}\right) . \] | Proof. We have\n\n\[ F\left( {s,\psi }\right) G\left( {s,\psi }\right) - 1 = \mathop{\sum }\limits_{{{D}^{4} < n \leq {D}^{8}}}\frac{\varsigma \left( n\right) \psi \left( n\right) }{{n}^{s}} = {\int }_{{D}^{4}}^{{D}^{8}}{x}^{{s}_{0} - s}d\left\{ {{X}_{4}\left( {x,\psi }\right) }\right\} . \]\n\nThus, similar to (4.2), ... | Yes |
If \( s \in {\Omega }_{2} \), then \( \frac{{F}^{\prime }}{F}\left( {s,\psi }\right) = O\left( \mathcal{L}\right) \) | Assume \( \left| w\right| \leq {\left( {200}\mathcal{L}\right) }^{-1}\log \mathcal{L} \), so that \( s + w \in {\Omega }_{1} \) . By Lemma 4.1 and 4.2, \( {\mathcal{L}}^{-{88}} \ll \left| {F\left( {s + w,\psi }\right) }\right| \ll {\mathcal{L}}^{88}. \) Thus the logarithm \( \mathfrak{l}\left( {s, w}\right) \mathrel{\t... | Yes |
Lemma 4.5. If\n\n\\[ \n\\frac{1}{2} + {\\alpha }^{2} < \\sigma < 1,\\;\\left| {t - {2\\pi }{t}_{0}}\\right| < {\\mathcal{L}}_{1} + 2, \n\\]\n\nthen\n\n\\[ \nA\\left( {s,\\psi }\\right) \n\\neq 0 \n\\] | Proof. We discuss in two cases.\n\nCase 1. \\( 1/2 + {\\mathcal{L}}^{-1} \\leq \\sigma < 1 \\) .\n\nBy Lemma 4.2 and trivial estimation,\n\n\\[ \n\\frac{F\\left( {1 - s,\\bar{\\psi }}\\right) }{F\\left( {s,\\psi }}\\right) } \\ll {D}^{c} \n\\]\n\nHence, by (4.5),\n\n\\[ \n\\mathcal{B}\\left( {s,\\psi }}\\right) \\ll {P... | Yes |
Lemma 4.7. Suppose \( \rho = 1/2 + {i\gamma } \) is a zero of \( \mathcal{A}\left( {s,\psi }\right) \) satisfying \( \left| {\gamma - {2\pi }{t}_{0}}\right| < {\mathcal{L}}_{1} + 2 \) . Then the function \( \mathcal{A}\left( {\rho + w,\psi }\right) \) has exactly three zeros inside the circle \( \left| w\right| = \alph... | Proof. In a way similar to the proof of Lemma 4.6, it is direct to verify that\n\n\[ \left| {\mathcal{A}\left( {\rho + w,\psi }\right) - \left( {1 - {P}^{-{2w}}}\right) }\right| < \left| {1 - {P}^{-{2w}}}\right| \]\n\nif \( \left| w\right| = \alpha \left( {1 + {c}^{\prime }\alpha \mathcal{L}}\right) \) . Hence, the fun... | Yes |
Lemma 4.8. Assume that \( \rho \) is a zero of \( L\left( {s,\psi }\right) L\left( {s,{\chi \psi }}\right) \) in \( \Omega \) . Then we have\n\n\[ \widetilde{Z}{\left( \rho ,\psi \right) }^{-1} = - G\left( {\rho ,\psi }\right) F\left( {1 - \rho ,\bar{\psi }}\right) + O\left( {\mathcal{L}}^{-{100}}\right) . \] | Proof. It follows from Lemma 4.4 that\n\n\[ F\left( {\rho ,\psi }\right) + \widetilde{Z}\left( {\rho ,\psi }\right) F\left( {1 - \rho ,\bar{\psi }}\right) \ll {\mathcal{L}}^{-{179}}. \]\n\nThe result follows by multiplying both sides by \( \widetilde{Z}{\left( \rho ,\psi \right) }^{-1}G\left( {\rho ,\psi }\right) \) an... | Yes |
Lemma 5.1. Suppose \( \psi \left( {\;\operatorname{mod}\;p}\right) \in \Psi ,\left| {\sigma - 1/2}\right| \leq \alpha ,\left| {t - {2\pi }{t}_{0}}\right| < {\mathcal{L}}_{1} + 2, u = 0 \) and \( \left| v\right| < {L}^{20} \) . Then \[ \frac{Z\left( {s + w,\psi }\right) - Z\left( {s,\psi }\right) {\left( p{t}_{0}\right)... | Proof. We have \[ \frac{Z\left( {s + w,\psi }\right) }{Z\left( {s,\psi }\right) } = \exp \left\{ {{\int }_{0}^{w}\frac{{Z}^{\prime }}{Z}\left( {s + {w}^{\prime },\psi }\right) d{w}^{\prime }}\right\} . \] Assume \( \left| {w}^{\prime }\right| \leq \left| w\right| \) . By (2.6) and the Stirling formula, \[ \frac{{Z}^{\p... | Yes |
Lemma 5.2. Let \( \\psi \) and \( s \) be as in Lemma 5.1. Then\n\n\[ \n\\frac{Y\\left( {s + {\\beta }_{1},\\psi }\\right) Y\\left( {s + {\\beta }_{2},\\psi }\\right) Y\\left( {s + {\\beta }_{3},\\psi }\\right) }{Y\\left( {s,\\psi }\\right) } = {\\left( p{t}_{0}\\right) }^{{\\beta }_{3}}Z{\\left( s,\\psi \\right) }^{-1... | Proof. The left side is\n\n\[ \nZ{\\left( s,\\psi \\right) }^{-1}\\mathop{\\prod }\\limits_{{1 \\leq j \\leq 3}}\\left( \\frac{Y\\left( {s + {\\beta }_{j},\\psi }\\right) }{Y\\left( {s,\\psi }\\right) }\\right) .\n\]\n\nBy (2.6) and the Stirling formula, for \( \\left| w\\right| < {5\\alpha } \) ,\n\n\[ \n\\frac{{Y}^{\... | Yes |
Lemma 5.3. If \( x \leq {t}_{0}^{1.02} \), then\n\n\[ \Delta \left( x\right) = \omega \left( {1/2 + {2\pi ix}}\right) \left( {1 + O\left( \alpha \right) }\right) + O\left( \varepsilon \right) \]\n\n(5.8)\n\nif \( x > {t}_{0}^{1.02} \), then\n\n\[ \Delta \left( x\right) \ll \exp \left\{ {-{\left( {10}^{-2}{\mathcal{L}}_... | Proof. By the Mellin transform (see [1], Lemma 2) we have\n\n\[ {\Delta }_{1}\left( x\right) = {\int }_{0}^{\infty }\exp \left\{ {\left( {{s}_{0} - 1}\right) \log y - {\mathcal{L}}_{2}^{2}{\log }^{2}y - {2\pi ixy}}\right\} {dy} \]\n\nwhere the logarithm vanishes at \( y = 1 \) . This yields, by substituting \( y = {e}^... | Yes |
Lemma 5.4. (i). If \( 1/2 \leq \sigma \leq 2 \), then\n\n\[ \delta \left( s\right) \ll {\mathcal{L}}^{c}{\left| s\right| }^{-2} \] | Proof. (i). Using partial integration twice we obtain\n\n\[ \delta \left( s\right) = \frac{1}{s\left( {s + 1}\right) }{\int }_{0}^{\infty }{\Delta }^{\prime \prime }\left( x\right) {x}^{s + 1}{ds}. \]\n\nBy (5.10) we have\n\n\[ {\Delta }^{\prime \prime }\left( x\right) = - 4{\pi }^{2}{\int }_{-\infty }^{\infty }{\left(... | No |
Lemma 5.7. We have\n\[ \n{L}^{\prime }\left( {1,\chi }\right) \gg \frac{D}{\varphi \left( D\right) }.\n\] | Proof. The right side of the equality\n\n\[ \n\frac{1}{2\pi i}{\int }_{\left( 1\right) }\zeta \left( {1 + s}\right) L\left( {1 + s,\chi }\right) \frac{{D}^{4s}{\omega }_{1}\left( s\right) }{s}{ds} = \mathop{\sum }\limits_{n}\frac{\nu \left( n\right) }{n}g\left( \frac{{D}^{4}}{n}\right) .\n\]\n\nis\n\[ \n\gg \mathop{\su... | Yes |
Lemma 5.8. If\n\n\\[ \n\\alpha \\leq \\left| {s - 1}\\right| \\leq {10\\alpha } \n\\]\n\nthen\n\n\\[ \nL\\left( {s,\\chi }\\right) = {L}^{\\prime }\\left( {1,\\chi }\\right) \\left( {s - 1}\\right) + O\\left( {\\alpha }_{2}\\right) \n\\]\n\nwhere\n\n\\[ \n{\\alpha }_{2} = {\\mathcal{L}}^{-{15}} \n\\] | Proof. This follows from the relation\n\n\\[ \nL\\left( {s,\\chi }\\right) = L\\left( {1,\\chi }\\right) + {L}^{\\prime }\\left( {1,\\chi }\\right) \\left( {s - 1}\\right) + {\\int }_{1}^{s}\\left( {s - w}\\right) {L}^{\\prime \\prime }\\left( {w,\\chi }\\right) {dw}, \n\\]\n\n(A) and a simple bound for \\( {L}^{\\prim... | Yes |
Lemma 5.9. Suppose \( \psi \in {\Psi }_{1},\left| {\sigma - 1/2}\right| \leq \alpha ,\left| {t - {2\pi }{t}_{0}}\right| \leq {\mathcal{L}}_{1} + {10} \) and \( \left| {s - \rho }\right| \gg \alpha \) for any zero \( \rho \) of \( L\left( {s,\psi }\right) \) . Then\n\n\[ \frac{L\left( {s + {\beta }_{1},\psi }\right) }{L... | Proof. It is known that\n\n\[ \frac{{L}^{\prime }}{L}\left( {{s}^{\prime },\psi }\right) = \mathop{\sum }\limits_{{\left| {\rho - {s}^{\prime }}\right| < 1}}\frac{1}{{s}^{\prime } - \rho } + O\left( {1/\alpha }\right) \] \n\n(5.16)\n\nfor \( \left| {\Re \left\{ {s}^{\prime }\right\} - 1/2}\right| \leq \alpha \) and \( ... | Yes |
Proposition 7.1. Assume \( {\mathbf{a}}_{1} \) and \( {\mathbf{a}}_{2} \) satisfy (7.2). We have\n\n\[ \n{\Theta }_{1}\left( {{\mathbf{a}}_{1},{\mathbf{a}}_{2}}\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{\psi \in {\Psi }_{1}}}\frac{1}{2\pi i}{\int }_{\mathcal{J}\left( 1\right) }\mathcal{C}\left( {s,\psi }\right... | Proof of Proposition 7.1: Initial steps\n\nFor notational simplicity we write \( {\Theta }_{1} \) for \( {\Theta }_{1}\left( {{\mathbf{a}}_{1},{\mathbf{a}}_{2}}\right) \) . Note that for any \( \psi \in \Psi ,\mathcal{C}\left( {s,\psi }\right) \) is analytic if \( \sigma > 1 \) and \( \left| {t - {2\pi }{t}_{0}}\right|... | Yes |
Lemma 8.2. Suppose \( T < x < P \) . Then for \( \mu = 6,7 \) ,\n\n\[ \mathop{\sum }\limits_{{m < x}}\frac{\chi \left( m\right) }{{m}^{1 - {\beta }_{j}}}{\left( \frac{x}{m}\right) }^{{\beta }_{\mu }}\log \frac{x}{m} = {L}^{\prime }\left( {1,\chi }\right) {\mathcal{F}}_{j\mu }\left( x\right) + O\left( {\mathcal{L}}^{-6}... | Proof. The sum is equal to\n\n\[ \frac{1}{2\pi i}{\int }_{\left( 1\right) }L\left( {1 - {\beta }_{j} + s,\chi }\right) \frac{{x}^{s}{ds}}{{\left( s - {\beta }_{\mu }\right) }^{2}}. \]\n\nWe move the contour of integration to the vertical segments\n\n\[ s = \alpha + {it}\;\text{ with }\;\left| t\right| \geq D, \]\n\n\[ ... | Yes |
Lemma 8.4. Suppose \( {dr} < P{T}^{-2} \) and \( T < x < P \) . Then for \( \mu = 6,7 \) ,\n\n\[ \mathop{\sum }\limits_{{n < x}}\frac{\chi \left( n\right) {\xi }_{0j}\left( {n;d, r}\right) }{n}{\left( \frac{x}{n}\right) }^{-{\beta }_{\mu }}\log \frac{x}{n} = {L}^{\prime }\left( {1,\chi }\right) \Pi \left( {d, r}\right)... | Proof. The sum is equal to\n\n\[ \frac{1}{2\pi i}{\int }_{\left( 1\right) }\left( {\mathop{\sum }\limits_{n}\frac{\chi \left( n\right) {\xi }_{0j}\left( {n;d, r}\right) }{{n}^{1 + s}}}\right) \frac{{x}^{s}{ds}}{{\left( s + {\beta }_{\mu }\right) }^{2}}. \]\n\n(8.9)\n\nThe contour of integration is moved in the same way... | Yes |
Lemma 10.1. Write\n\n\[ \n{\mathcal{V}}_{1j}\left( y\right) = \mathop{\sum }\limits_{m}\frac{\chi \left( m\right) \widetilde{f}\left( {\log \left( {ym}\right) /\log P}\right) }{{m}^{1 - {\beta }_{j}}}. \]\n\nIf \( 1 \leq y \leq {P}^{0.5}/T \), then\n\n\[ \n{\mathcal{V}}_{1j}\left( y\right) \ll {T}^{-c} \]\n\n(10.2)\n\n... | Proof. The inequalities (10.2) follow by the Polya-Vinogradov inequality and partial summation. Write\n\n\[ \n{P}_{1}^{\prime } = {P}^{0.504},\;{P}_{2}^{\prime } = {P}^{0.502},\;{P}_{3}^{\prime } = {P}^{0.5}. \]\n\nThen\n\[ \n\widetilde{f}\left( {\log y/\log P}\right) = \frac{500}{\log P} \cdot \frac{1}{2\pi i}{\int }_... | Yes |
Lemma 11.2. Suppose \( \sigma = 1/2 \) and \( \left| {t - {2\pi }{t}_{0}}\right| < {\mathcal{L}}_{1} \) . Then\n\n\[ \n{\widetilde{J}}_{1}\left( {s,\psi }\right) = Z\left( {s,{\chi \psi }}\right) {\widetilde{J}}_{2}\left( {1 - s,\bar{\psi }}\right) + O\left( {{E}_{2}\left( {s,\psi }\right) }\right) \n\] \n\nwhere \n\n\... | Proof. Suppose \( {0.5} \leq z \leq {0.504} \) . In a way similar to the proof of Lemma 6.1, it can be deduced that\n\n\[ \n\mathop{\sum }\limits_{n}\frac{{\chi \psi }\left( n\right) }{{n}^{s}}g\left( \frac{{P}^{z}}{n}\right) = L\left( {s,{\chi \psi }}\right) - Z\left( {s,{\chi \psi }}\right) \mathop{\sum }\limits_{n}\... | Yes |
Lemma 12.1. We have\n\n\[ \mathop{\sum }\limits_{l}\frac{\chi \left( l\right) {\varkappa }_{13}\left( {dl}\right) }{{l}^{1 - {\beta }_{j}}} \ll \left\{ \begin{array}{ll} {T}^{-c} & \text{ if }d \leq {P}_{1}^{\prime \prime }/T, \\ {\alpha }_{1} & \text{ if }{P}_{1}^{\prime \prime }/T < d \leq {P}_{1}^{\prime \prime }, \... | Proof. Note that \( {\varkappa }_{13}\left( n\right) \ll {\alpha }_{1} \) if \( {P}_{1}^{\prime \prime }/T \leq n \leq {P}_{1}^{\prime \prime }T \) . In the case \( d \leq {P}_{1}^{\prime \prime } \) the results follow by the Polya-Vinogradov inequality and partial summation. On the other hand, in the case \( {P}_{1}^{... | Yes |
Lemma 12.3. If \( {P}_{1}^{\prime \prime } < {dr} < {P}_{2} \), then\n\n\[ \mathop{\sum }\limits_{l}\frac{\chi \left( l\right) {\bar{\varkappa }}_{13}\left( {drl}\right) {\xi }_{j}\left( {l;d, r}\right) }{l} = \frac{{L}^{\prime }\left( {1,\chi }\right) \Pi \left( {d, r}\right) }{\log {P}_{1}}\left( {-1 + \left( {-2{\be... | Proof. The left side is equal to\n\n\[ - \frac{1}{\log {P}_{1}}\frac{\partial }{\partial w}\left( {{\left( dr/{P}_{1}^{\prime \prime }\right) }^{{\beta }_{6} - w}\mathop{\sum }\limits_{{l < {P}_{2}^{\prime \prime }/{dr}}}\frac{\chi \left( l\right) {\xi }_{j}\left( {l;d, r}\right) }{{l}^{1 - {\beta }_{6} + w}}}\right) {... | Yes |
Lemma 15.2. If \( \left| {s - 1}\right| < {5\alpha } \), then | \[ {\mathcal{M}}_{1}\left( {1,1;s}\right) = \mathop{\prod }\limits_{{\left( {q, D}\right) = 1}}\frac{1 - \chi \left( q\right) {q}^{-2}}{1 - {q}^{-2}} + O\left( {\alpha }_{1}\right) . \] | No |
For \( \sigma \geq 9/{10} \) the function\n\n\[ \n{\mathcal{U}}_{1j}\left( s\right) \mathrel{\text{:=}} \frac{1}{\zeta {\left( s\right) }^{2}L{\left( s,\chi \right) }^{2}}\mathop{\sum }\limits_{n}\frac{\chi \left( n\right) {\tau }_{2}\left( n\right) {\varpi }_{1j}\left( n\right) }{{n}^{s}} \n\]\n\nis analytic and bound... | By Lemma 15.3, we can move the contour of integration in the same way as in the proof of Lemma 8.4 to obtain\n\n\[ \n\mathop{\sum }\limits_{{n < T}}\frac{\chi \left( n\right) {\tau }_{2}\left( n\right) {\varpi }_{1j}\left( n\right) }{n} = {L}^{\prime }{\left( 1,\chi \right) }^{2}{\mathcal{U}}_{2j}\left( 1\right) + O\le... | No |
Lemma 16.2. The function\n\n\[ \n{\mathcal{U}}_{2j}\left( s\right) = \frac{1}{\zeta {\left( s\right) }^{3}L{\left( s,\chi \right) }^{3}}\mathop{\sum }\limits_{n}\frac{{\varpi }_{2j}\left( n\right) \left( {\nu * \chi }\right) \left( n\right) }{{n}^{s}} \n\]\n\nis analytic and bounded for \( \sigma > 9/{10} \) . Further ... | We have\n\n\[ \n\mathop{\sum }\limits_{{n < T}}\frac{{\varpi }_{2j}\left( n\right) \left( {\nu * \chi }\right) \left( n\right) }{n} = \mathop{\sum }\limits_{n}\frac{{\varpi }_{2j}\left( n\right) \left( {\nu * \chi }\right) \left( n\right) }{n}g\left( \frac{T}{n}\right) + O\left( {1/{\mathcal{L}}^{10}}\right) \n\]\n\n\[... | Yes |
Proposition 1.1 (Diamond). Suppose that \( \pi : {D}_{p} \rightarrow {\mathrm{{GL}}}_{2}\left( A\right) \) is a continuous representation where \( A \) is an Artinian local ring with residue field \( k \), a finite field of characteristic p. Suppose \( \pi \approx \left( \begin{matrix} {\chi }_{1} & \varepsilon \\ 0 & ... | Proof (taken from [Dia, Prop. 6.1]). We may replace \( \pi \) by \( \pi \otimes {\chi }_{2}^{-1} \) and we let \( \varphi = {\chi }_{1}{\chi }_{2}^{-1} \) . Then \( \pi \cong \left( \begin{matrix} \varphi & \varepsilon & t \\ 0 & 1 & \end{matrix}\right) \) determines a cocycle \( t : {D}_{p} \rightarrow M\left( 1\right... | Yes |
Proposition 1.6.\n\n\[ \n\\# {H}_{L}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q}, X}\\right) /\\# {H}_{{L}^{ * }}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q},{X}^{ * }}\\right) = {h}_{\\infty }\\mathop{\\prod }\\limits_{{q \\in \\sum }}{h}_{q} \n\]\n\nwhere\n\[ \n\\begin{cases} {h}_{q} & = \\# {H}^{0}\left( {{\\mat... | Proof. Adapting the exact sequence of Poitou and Tate (cf. [Mi2, Th. 4.20]) we get a seven term exact sequence\n\n\[ \n0 \\rightarrow {H}_{L}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q}, X}\\right) \\rightarrow {H}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q}, X}\\right) \\rightarrow \\mathop{\\prod }\\limits_{{q \\... | Yes |
Proposition 1.7. If \( q \notin \sum \), and \( X \) is an arbitrary finite \( \operatorname{Gal}\left( {{\mathbf{Q}}_{\sum }/\mathbf{Q}}\right) \) - module of p-power order,\n\n\[ \n\# {H}_{{L}^{\prime }}^{1}\left( {{\mathbf{Q}}_{\sum \cup q}/\mathbf{Q}, X}\right) /\# {H}_{L}^{1}\left( {{\mathbf{Q}}_{\sum }/\mathbf{Q}... | Proof. Consider the short exact sequence of inflation-restriction: \n\nThe proposition follows when we note that\n\n\[ \n\# {H}^{0}\left( {{\mathbf{Q}}_{q},{X}^{ * }}\right) = \# {H}^{1}{\left( {\mathbf{Q}}_{q}^{\mathr... | Yes |
Proposition 1.8. If \( q \in \mathcal{M}\;\left( {q \neq p}\right) \) and \( X = {V}_{{\lambda }^{n}} \) then \( {h}_{q} = 1 \) . | Proof. This is a straightforward calculation. For example if \( q \) is of type (A) then we have\n\n\[ \n{L}_{n, q} = \ker \left\{ {{H}^{1}\left( {{\mathbf{Q}}_{q},{V}_{{\lambda }^{n}}}\right) \rightarrow {H}^{1}\left( {{\mathbf{Q}}_{q},{W}_{{\lambda }^{n}}/{W}_{{\lambda }^{n}}^{0}}\right) \oplus {H}^{1}\left( {{\mathb... | Yes |
Proposition 1.9. (i) If \( X = {V}_{{\lambda }^{n}} \) then | \[ {h}_{p}{h}_{\infty } = \# {\left( \mathcal{O}/\lambda \right) }^{3n}\# {H}^{0}\left( {{\mathbf{Q}}_{p},{V}_{{\lambda }^{n}}^{ * }}\right) /\# {H}^{0}\left( {\mathbf{Q},{V}_{{\lambda }^{n}}^{ * }}\right) \] in the unrestricted case.\n\nProof. Case (i) is trivial. | No |
Corollary 1. In case (i), \( {J}_{H}{\widehat{\left( N)(\overline{\mathbf{Q}}\right) }}_{\mathfrak{m}} \simeq {\mathbf{T}}_{\mathfrak{m}}^{2}\; \) and \( \;{\mathrm{{Ta}}}_{\mathfrak{m}}\left( {{J}_{H}\left( N\right) \left( \overline{\mathbf{Q}}\right) }\right) \simeq \) \( {\mathbf{T}}_{\mathrm{m}}^{2} \) . In case (i... | In each case the first isomorphisms of Corollary 1 follow from the theorem together with the rank 2 result alluded to previously. Corollary 2 and the second isomorphisms of corollary 1 then follow on applying duality (2.4). (In the proof and in all applications we will only use the notion of a Gorenstein \( {\mathbf{Z}... | Yes |
Proposition 2.7. Suppose that \( \mathfrak{m} \) is a maximal ideal of \( \mathbf{T} = {\mathbf{T}}_{H}\left( N\right) \) associated to an irreducible representation. Suppose that \( q + {Np} \) . Then\n\n\[ \left( {\Delta }_{q}\right) = {\left( q - 1\right) }^{2}\left( {{T}_{q}^{2}-\langle q\rangle {\left( 1 + q\right... | The proof is a trivial generalization of that of Proposition 2.6. | No |
Proposition 4.6.\n\n\[ \left\langle {{f}_{\varphi },{f}_{\varphi }}\right\rangle = \frac{1}{{16}{\pi }^{3}}{N}^{2}\left\{ {\mathop{\prod }\limits_{\substack{{q \mid N} \\ {q \notin {S}_{\varphi }} }}\left( {1 - \frac{1}{q}}\right) }\right\} {L}_{N}\left( {2,{\varphi }^{2}\overline{\widehat{\chi }}}\right) {L}_{N}\left(... | Proof. One begins with a formula of Petersson that for an eigenform of weight 2 on \( {\Gamma }_{1}\left( N\right) \) says\n\n\[ \langle f, f\rangle = {\left( 4\pi \right) }^{-2}\Gamma \left( 2\right) \left( \frac{1}{3}\right) \pi \left\lbrack {{\mathrm{{SL}}}_{2}\left( \mathbf{Z}\right) : {\Gamma }_{1}\left( N\right) ... | Yes |
Proposition 1. Suppose that \( \mathcal{O} \) is a complete discrete valuation ring and that \( \varphi : S \rightarrow T \) is a surjective local \( \mathcal{O} \) -algebra homomorphism between complete local Noetherian \( \mathcal{O} \) -algebras. Suppose further that \( {\mathfrak{p}}_{T} \) is a prime ideal of \( T... | Proof. First we consider the case where \( u = 0 \) . We may assume that the generators \( {x}_{1},\ldots ,{x}_{r} \) lie in \( {\mathfrak{p}}_{T} \) by subtracting their residues in \( T/{\mathfrak{p}}_{T} \rightarrow \mathcal{O} \) . By (ii) we may also write\n\n\[ S \simeq \mathcal{O}\llbracket {x}_{1},\ldots ,{x}_{... | Yes |
Example2.1.1 求 \( f\left( x\right) = \sin {2x} + \sin x + \cos x \) 的值域. | 由恒等式 \( {\sin }^{2}x + {\cos }^{2}x = 1 \) 自然可以想到,令 \( t = \sin x + \cos x \) 那么就可以得到 \( \sin {2x} = {t}^{2} - 1 \) ,所以原函数可化为\n\n\[ f\left( x\right) = {t}^{2} - 1 + t, t \in \left\lbrack {-\sqrt{2},\sqrt{2}}\right\rbrack \]\n\n所以值域为 \( \left\lbrack {-\frac{5}{4},1 + \sqrt{2}}\right\rbrack \) . | Yes |
对 \( x > 0 \) ,求\n\n\[ f\left( x\right) = \frac{{x}^{3} + x}{\left( {{x}^{2} + 2}\right) \left( {2{x}^{2} + 1}\right) } \]\n\n的值域. | 由结构,上下同除 \( {x}^{2} \) 得到\n\n\[ f\left( x\right) = \frac{x + \frac{1}{x}}{\left( {{x}^{2} + 2}\right) \left( {2 + \frac{1}{{x}^{2}}}\right) } = \frac{x + \frac{1}{x}}{2{x}^{2} + \frac{2}{{x}^{2}} + 5} \]\n\n于是令 \( t = x + \frac{1}{x} \in \lbrack 2, + \infty ) \) ,得到\n\n\[ f\left( x\right) = \frac{t}{2{t}^{2} + 1} = \fr... | Yes |
实数 \( x, y \) 满足 \( 4{x}^{2} - {5xy} + 4{y}^{2} = 5 \) ,记 \( S = {x}^{2} + {y}^{2} \) ,求 \( \frac{1}{{S}_{\min }} + \frac{1}{{S}_{\max }} \) | 由于式子是对称的, 所以和差换元可以很好的化解变量乘法 \( {xy} \) . 直接使用和差换元,令 \( x = a + b, y = a - b \) 则 \( S = {x}^{2} + {y}^{2} = 2\left( {{a}^{2} + {b}^{2}}\right) \).\n\n代入条件 \( 4{x}^{2} - {5xy} + 4{y}^{2} = 5 \) ,得 \( 3{a}^{2} + {13}{b}^{2} = 5 \) . 即 \( {a}^{2} \in \left\lbrack {0,\frac{3}{5}}\right\rbrack \) . 亦即\n\n\[ S = 2\left( {{a}... | Yes |
Example2.2.2(Nesbitt 不等式) 证明: 对 \( a, b, c > 0 \) ,有\n\n\[ \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \geq \frac{3}{2} \] | Proof: 由结构,使用和差换元处理分母,令 \( x = b + c, y = c + a, z = a + b \) . 得到\n\n\[ a = \frac{y + z - x}{2} \]\n\n\[ b = \frac{z + x - y}{2} \]\n\n\[ c = \frac{x + y - z}{2} \]\n\n代回原式,即对 \( x, y, z > 0 \) ,证明\n\n\[ a = \frac{y + z - x}{2x} + \frac{z + x - y}{2y} + c = \frac{x + y - z}{2z} \]\n整理后得到\n\n\[ \frac{1}{2}\left\lbrack ... | No |
已知 \( x, y \) 为正数,且 \( \frac{2x}{{3x} + y} + \frac{y}{x + {2y}} \) 的最大值为 \( a + b\sqrt{2} \) (其中 \( a, b \) 为有理数,求 \( {ab} \) 的值. | 设 \( m = {3x} + y, n = x + {2y} \) ,则\n\n\[ x = \frac{{2m} - n}{5}, y = \frac{{3n} - m}{5} \]\n\n原式 \( = \frac{1}{5}\left( {\frac{{4m} - {2n}}{m} + \frac{{3n} - m}{n}}\right) = \frac{1}{5}\left( {7 - \left( {\frac{2n}{m} + \frac{m}{n}}\right) }\right) \)\n\n运用均值不等式,上式 \( \leq \frac{1}{5}\left( {7 - 2\sqrt{2}}\right) \)... | Yes |
若实数 \( a, b, c \) 满足 \( {a}^{2} + {b}^{2} = {c}^{2} \) ,求 \( \frac{b}{a - {2c}} \) 的取值范围. | 对条件进行变形, 得到\n\n\[ \n{\left( \frac{a}{c}\right) }^{2} + {\left( \frac{b}{c}\right) }^{2} = 1 \n\]\n\n这是典型的三角变换的式子,我们令 \( \frac{b}{c} = \sin t,\frac{a}{c} = \cos t \) 代入式子得到\n\n\[ \n\frac{b}{a - {2c}} = \frac{\sin t}{\cos t - 2} \n\]\n\n利用刚刚提到的万能换元代换得到\n\n\[ \n\frac{\sin t}{\cos t - 2} = \frac{2u}{1 - {u}^{2} - 2\left( {... | Yes |
设实数 \( x, y \) 满足 \( {x}^{2} + {2xy} - 1 = 0 \) ,求 \( {x}^{2} + {y}^{2} \) 的最小值. | 条件式子是齐次的,这意味着通过恒等式 \( {\sin }^{2}x + {\cos }^{2}x = 1 \) 可以对式子化解. 当然我们可以换一种方式论述,即令 \( {x}^{2} + {y}^{2} = {r}^{2} \) ,那么可令 \( x = r\sin t, y = r\cos t \) ,代入条件式子,得到\n\n\[ \n{r}^{2}\left( {{\sin }^{2}t + 2\sin t\cos t}\right) = 1 \n\]\n\n所以就有\n\n\[ \n{r}^{2} = \frac{1}{{\sin }^{2}t + \sin {2t}} \n\]\n\n\[ \n= \frac{2}{1... | Yes |
Example2.3.3 求函数 \( y = \sqrt{1 - {4x}} + \sqrt{2 + x} \) 的值域. | Answer: 本例其实用柯西不等式或是其他什么方法会更简便些, 但本例可以很好的说明三角换元的结构性变换。由两个式子 \( \sqrt{1 - {4x}},\sqrt{2 + x} \) ,我们很容易的就能发现其平方和之间具有某种联系, 具体而言, 令\n\n\[ u = \sqrt{1 - {4x}}, v = \sqrt{2 + x} \]\n\n那么就有\n\n\[ y = u + v,{u}^{2} + 4{v}^{2} = 9 \]\n\n这也就是常见的三角换元, 后面的步骤留做习题. | No |
设 \( x, y, z \) 是正实数,求\n\n\[\n\frac{{4xz} + {yz}}{{x}^{2} + {y}^{2} + {z}^{2}}\n\]\n\n的最大值. | 首先对式子进行齐次化, 即令\n\n\[{x}^{2} + {y}^{2} + {z}^{2} = {k}^{2}\]\n\n引入球面三角换元\n\n\[x = k\sin \alpha \cos \beta\]\n\n\[y = k\sin \alpha \sin \beta\]\n\n\[z = k\cos \alpha\]\n\n由条件中 \( x, y, z \) 都是正实数,得到 \( \alpha ,\beta \in \left( {0,\frac{\pi }{2}}\right) \) ,代入得到\n\n\[\frac{{4xz} + {yz}}{{x}^{2} + {y}^{2} + {z}^{2}} = \fra... | Yes |
已知 \( x, y \) 为正数,且 \( \frac{2x}{{3x} + y} + \frac{y}{x + {2y}} \) 的最大值为 \( a + b\sqrt{2} \) (其中 \( a, b \) 为有理数,求 \( {ab} \) 的值. | 注意到两个分式都是齐次式,直接设 \( t = \frac{y}{x} \) ,则原式变为\n\n\[\n\frac{2}{3 + t} + \frac{t}{1 + {2t}} = \frac{{t}^{2} + {7t} + 2}{{t}^{2} + {7t} + 3}\n\]\n\n则又回到了我们熟悉的简单有理分式求最值上了, 后面步骤留做习题。 | No |
Example2.4.2 若不等式 \( \sqrt{x} + \sqrt{y} \leq k\sqrt{{2x} + y} \) 对于任意正实数 \( x, y \) 恒成立,求 \( k \) 的取值范围. | 显然 \( k > 0 \) . 由 \( {\left( \sqrt{x} + \sqrt{y}\right) }^{2} \leq {k}^{2}\left( {{2x} + y}\right) \) ,得\n\n\[ \left( {2{k}^{2} - 1}\right) x - 2\sqrt{xy} + \left( {{k}^{2} - 1}\right) y \geq 0 \]\n\n对于 \( x, y > 0 \) 恒成立.\n\n令 \( t = \sqrt{x}y > 0 \) ,则得\n\n\[ f\left( t\right) = \left( {2{k}^{2} - 1}\right) {t}^{2} -... | Yes |
Proposition 2.3 (Szemerédi’s theorem, again). Write \( {\nu }_{\text{const }} : {\mathbb{Z}}_{N} \rightarrow {\mathbb{R}}^{ + } \) for the constant function \( {\nu }_{\text{const }} \equiv 1 \) . Let \( 0 < \delta \leq 1 \) and \( k \geq 1 \) be fixed. Let \( N \) be a large integer parameter, and let \( f : {\mathbb{... | A direct proof of Proposition 2.3 can be found in [40]. A formulation of Szemerédi's theorem similar to this one was also used by Furstenberg [10]. Combining this argument with the one in Gowers gives an explicit bound on \( c\left( {k,\delta }\right) \) of the form \( c\left( {k,\delta }\right) \geq \exp \left( {-\exp... | Yes |
Lemma 3.4. Let \( \nu \) be a \( k \) -pseudorandom measure. Then \( {\nu }_{1/2} \mathrel{\text{:=}} \left( {\nu + {\nu }_{\text{const }}}\right) /2 = \) \( \left( {\nu + 1}\right) /2 \) is also a \( k \) -pseudorandom measure (though possibly with slightly different bounds in the \( O\left( \right) \) and \( o\left( ... | Proof. It is clear that \( {\nu }_{1/2} \) is non-negative and has expectation \( 1 + o\left( 1\right) \) . To verify the linear forms condition (3.1), we simply replace \( \nu \) by \( \left( {\nu + 1}\right) /2 \) in the definition and expand as a sum of \( {2}^{m} \) terms, divided by \( {2}^{m} \) . Since each term... | Yes |
Lemma 5.2. Suppose that \( \nu \) is \( k \) -pseudorandom (as defined in Definition 3.3). Then we have\n\n\[ \n{\begin{Vmatrix}\nu - {\nu }_{\text{const }}\end{Vmatrix}}_{{U}^{d}} = \parallel \nu - 1{\parallel }_{{U}^{d}} = o\left( 1\right) \n\]\n\nfor all \( 1 \leq d \leq k - 1 \) . | Proof. By (5.7) it suffices to prove the claim for \( d = k - 1 \) . Raising to the power \( {2}^{k - 1} \) , it suffices from (5.4) to show that\n\n\[ \n\mathbb{E}\left( {\left. {\mathop{\prod }\limits_{{\omega \in \{ 0,1{\} }^{k - 1}}}\left( {\nu \left( {x + \omega \cdot h}\right) - 1}\right) }\right| \;x \in {\mathb... | Yes |
Proposition 5.3 (Generalised von Neumann). Suppose that \( \nu \) is \( k \) -pseudorandom. Let \( {f}_{0},\ldots ,{f}_{k - 1} \in {L}^{1}\left( {\mathbb{Z}}_{N}\right) \) be functions which are pointwise bounded by \( \nu + {\nu }_{\text{const }} \), or in other words\n\n\[ \left| {{f}_{j}\left( x\right) }\right| \leq... | Proof. By replacing \( \nu \) with \( \left( {\nu + 1}\right) /2 \) (and by dividing \( {f}_{j} \) by 2), and using Lemma 3.4, we see that we may in fact assume without loss of generality that we can improve (5.11) to\n\n\[ \left| {{f}_{j}\left( x\right) }\right| \leq \nu \left( x\right) \text{ for all }x \in {\mathbb{... | Yes |
Lemma 5.4 (Cauchy-Schwarz). Let \( \nu : {\mathbb{Z}}_{N} \rightarrow {\mathbb{R}}^{ + } \) be any measure. Let \( {\phi }_{0},{\phi }_{1},\ldots ,{\phi }_{k - 1} \) : \( {\mathbb{Z}}_{N}^{k - 1} \rightarrow {\mathbb{Z}}_{N} \) be functions of \( k - 1 \) variables \( {y}_{1},\ldots ,{y}_{k - 1} \), such that \( {\phi ... | Proof of Lemma 5.4. Consider the quantity \( {J}_{d} \) . Since \( {\phi }_{k - d - 1} \) does not depend on \( {y}_{k - d - 1} \), we may take all quantities depending on \( {\phi }_{k - d - 1} \) outside of the \( {y}_{k - d - 1} \) average. This allows us to write\n\n\[ \n{J}_{d} = \mathbb{E}\left( {G\left( {y,{y}^{... | Yes |
Lemma 5.5 ( \( \nu \) covers its own cubes uniformly). For \( n = 0,2 \), we have\n\n\[ \n\\mathbb{E}\\left( {{\\left| W\\left( x, h\\right) - 1\\right| }^{n}\\mathop{\\prod }\\limits_{{\\omega \\in \\{ 0,1{\\} }^{k - 1}}\\nu \\left( {x + \\omega \\cdot h}\\right) \\mid x \\in {\\mathbb{Z}}_{N}, h \\in {\\mathbb{Z}}_{N... | Proof. Expanding out the square, it then suffices to show that\n\n\[ \n\\mathbb{E}\\left( {W{\\left( x, h\\right) }^{q}\\mathop{\\prod }\\limits_{{\\omega \\in \\{ 0,1{\\} }^{k - 1}}\\nu \\left( {x + \\omega \\cdot h}\\right) \\mid x \\in {\\mathbb{Z}}_{N}, h \\in {\\mathbb{Z}}_{N}^{k - 1}}\\right) = 1 + o\\left( 1\\ri... | Yes |
Lemma 6.1 (Lack of Gowers uniformity implies correlation). Let \( \nu \) be a \( k \) -pseudorandom measure, and let \( F \in {L}^{1}\left( {\mathbb{Z}}_{N}\right) \) be any function. Then we have the identities\n\n\[ \langle F,\mathcal{D}F\rangle = \parallel F{\parallel }_{{U}^{k - 1}}^{{2}^{k - 1}} \]\n\n(6.4)\n\nand... | Proof. The identity (6.4) is clear just by expanding out both sides using (6.3), (5.4). To prove (6.5) we may of course assume \( F \) is not identically zero. By (6.1) and (6.4) it suffices to show that\n\n\[ \left| {\langle f,\mathcal{D}F\rangle }\right| \leq \parallel f{\parallel }_{{U}^{k - 1}}\parallel F{\parallel... | Yes |
Proposition 6.2 (Uniform distribution wrt basic Gowers anti-uniform functions). Suppose that \( \nu \) is \( k \) -pseudorandom. Let \( K \geq 1 \) be a fixed integer, let \( \Phi : {I}^{K} \rightarrow \mathbb{R} \) be a fixed continuous function, let \( \mathcal{D}{F}_{1},\ldots ,\mathcal{D}{F}_{K} \) be basic Gowers ... | Proof. We will prove this result in two stages, first establishing the result for \( \Phi \) polynomial and then using a Weierstrass approximation argument to deduce the general case. Fix \( K \geq 1 \), and let \( {F}_{1},\ldots ,{F}_{K} \in {L}^{1}\left( {\mathbb{Z}}_{N}\right) \) be fixed functions obeying the bound... | No |
Proposition 7.2 (Each function generates a \( \sigma \) -algebra). Let \( \nu \) be a \( k \) -pseudorandom measure, let \( 0 < \varepsilon < 1 \) and \( 0 < \eta < 1/2 \) be parameters, and let \( G \in {L}^{\infty }\left( {\mathbb{Z}}_{N}\right) \) be function taking values in the interval \( I \mathrel{\text{:=}} \l... | Proof. Observe from Fubini's theorem and (2.4) that\n\n\[ \n{\int }_{0}^{1}\mathop{\sum }\limits_{{n \in \mathbb{Z}}}\mathbb{E}\left( {{\mathbf{1}}_{G\left( x\right) \in \left\lbrack {\varepsilon \left( {n - \eta + \alpha }\right) ,\varepsilon \left( {n + \eta + \alpha }\right) }\right\rbrack }\left( {\nu \left( x\righ... | Yes |
Proposition 7.3. Let \( \nu \) be a \( k \) -pseudorandom measure. Let \( K \geq 1 \) be an fixed integer and let \( \mathcal{D}{F}_{1},\ldots ,\mathcal{D}{F}_{K} \in {L}^{\infty }\left( {\mathbb{Z}}_{N}\right) \) be basic Gowers anti-uniform functions. Let \( 0 < \varepsilon < 1 \) and \( 0 < \eta < 1/2 \) be paramete... | Proof. The claim (7.4) follows immediately from (7.1). Now we prove (7.5) and (7.6). Since each of the \( {\mathcal{B}}_{\varepsilon ,\eta }\left( {\mathcal{D}{F}_{j}}\right) \) are generated by \( O\left( {1/\varepsilon }\right) \) atoms, we see that \( \mathcal{B} \) is generated by \( {O}_{K,\varepsilon }\left( 1\ri... | Yes |
Proposition 8.2 (Iterative Step). Let \( \nu \) be a \( k \) -pseudorandom measure, and let \( f \in \) \( {L}^{1}\left( {\mathbb{Z}}_{N}\right) \) be a non-negative function satisfying \( 0 \leq f\left( x\right) \leq \nu \left( x\right) \) for all \( x \in {\mathbb{Z}}_{N} \) . Let \( 0 < \) \( \eta \ll \varepsilon \l... | Then we have the estimates \[ {\begin{Vmatrix}\left( 1 - {\mathbf{1}}_{{\Omega }_{K}}\right) \mathbb{E}\left( f \mid {\mathcal{B}}_{K}\right) \end{Vmatrix}}_{{L}^{\infty }\left( {\mathbb{Z}}_{N}\right) } \leq 1 + {O}_{K,\varepsilon }\left( {\eta }^{1/2}\right) \] \( {}^{17} \) For the specific purpose of \( k \) -term ... | Yes |
Lemma 9.4. \( \nu \left( n\right) \geq 0 \) for all \( n \in {\mathbb{Z}}_{N} \), and furthermore we have \( \nu \left( n\right) \geq {k}^{-1}{2}^{-k - 5}\widetilde{\Lambda }\left( n\right) \) for all \( {\epsilon }_{k}N \leq n \leq 2{\epsilon }_{k}N \) (if \( N \) is sufficiently large depending on \( k \) ). | Proof. The first claim is trivial. The second claim is also trivial unless \( {Wn} + 1 \) is prime. From definition of \( R \), we see that \( W{n + 1} > R \) if \( N \) is sufficiently large. Then the sum over \( d \mid {Wn} + 1, d \leq R \) in (9.2) in fact consists of just the one term \( d = 1 \) . Therefore \( {\L... | Yes |
Proposition 9.5 (Goldston-Yıldırm). Let \( m, t \) be positive integers. For each \( 1 \leq i \leq \) \( m \), let \( {\psi }_{i}\left( \mathbf{x}\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{j = 1}}^{t}{L}_{ij}{x}_{j} + {b}_{i} \), be linear forms with integer coefficients \( {L}_{ij} \) such that \( \left| {L}_... | \[ \mathbb{E}\left( {{\Lambda }_{R}{\left( {\theta }_{1}\left( \mathbf{x}\right) \right) }^{2}\ldots {\Lambda }_{R}{\left( {\theta }_{m}\left( \mathbf{x}\right) \right) }^{2} \mid \mathbf{x} \in B}\right) = \left( {1 + {o}_{m, t}\left( 1\right) }\right) {\left( \frac{W\log R}{\phi \left( W\right) }\right) }^{m}. \] | Yes |
Lemma 9.7. The measure \( \nu \) constructed in Definition 9.3 obeys the estimate \( \mathbb{E}\left( \nu \right) = \) \( 1 + o\left( 1\right) \) . | Proof. Apply Proposition 9.5 with \( m \mathrel{\text{:=}} t \mathrel{\text{:=}} 1,{\psi }_{1}\left( {x}_{1}\right) \mathrel{\text{:=}} {x}_{1} \) and \( B \mathrel{\text{:=}} \left\lbrack {{\epsilon }_{k}N,2{\epsilon }_{k}N}\right\rbrack \) (taking \( N \) sufficiently large depending on \( k \), of course). Comparing... | Yes |
Lemma 9.9. Let \( m \geq 1 \) be a parameter. There is a weight function \( \tau = {\tau }_{m} : \mathbb{Z} \rightarrow {\mathbb{R}}^{ + } \) such that \( \tau \left( n\right) \geq 1 \) for all \( n \neq 0 \), and such that for all distinct \( {h}_{1},\ldots ,{h}_{j} \in \left\lbrack {{\epsilon }_{k}N,2{\epsilon }_{k}N... | Proof. We observe that\n\n\[ \mathop{\prod }\limits_{{p \mid \Delta }}\left( {1 + {O}_{m}\left( {p}^{-1/2}\right) }\right) \leq \mathop{\prod }\limits_{{1 \leq i < j \leq m}}{\left( \mathop{\prod }\limits_{{p \mid {h}_{i} - {h}_{j}}}\left( 1 + {p}^{-1/2}\right) \right) }^{{O}_{m}\left( 1\right) }.\]\n\nBy the arithmeti... | Yes |
Lemma 10.1 (Local factor estimate). If \( p \leq w\left( N\right) \), then \( {\omega }_{X}\left( p\right) = 0 \) for all non-empty \( X \) ; in particular, \( {E}_{p} = 1 \) when \( p \leq w\left( N\right) \) . If instead \( p > w\left( N\right) \), then \( {\omega }_{X}\left( p\right) = {p}^{-1} \) when \( \left| X\r... | Proof. The first statement is clear, since the maps \( {\theta }_{j} : {\mathbb{Z}}_{p}^{t} \rightarrow {\mathbb{Z}}_{p} \) are identically 1 when \( p \leq w\left( N\right) \) . The second statement (when \( p > w\left( N\right) \) and \( \left| X\right| = 1 \) ) is similar since in this case \( {\theta }_{j} \) unifo... | Yes |
Lemma 10.3. The Euler products \( \mathop{\prod }\limits_{p}{E}_{p}^{\left( j\right) } \) for \( j = 1,2 \) are absolutely convergent in the domain \( {\mathcal{D}}_{1/{6m}}^{m} \) . In particular, \( {G}_{1},{\widetilde{G}}_{2} \) can be continued analytically to this domain. Furthermore, we have the estimates | Proof. First consider \( j = 1 \) . From (10.8) and Taylor expansion we have the crude bound \( {E}_{p}^{\left( 1\right) }\left( {z,{z}^{\prime }}\right) = 1 + {O}_{m}\left( {p}^{-2 + 4/{6m}}\right) \) in \( {\mathcal{D}}_{1/{6m}}^{m} \), which gives the desired convergence and also the \( {C}^{m}\left( {\mathcal{D}}_{... | Yes |
Lemma 10.4. [17] Let \( R \) be a positive real number. Let \( G = G\left( {z,{z}^{\prime }}\right) \) be an analytic function of \( {2m} \) complex variables on the domain \( {\mathcal{D}}_{\sigma }^{m} \) for some \( \sigma > 0 \), and suppose that\n\n\[ \parallel G{\parallel }_{{C}^{m}\left( {\mathcal{D}}_{\sigma }^... | Proof. While this lemma is essentially in [17], we shall give a complete proof in the Appendix for sake of completeness. | Yes |
Lemma 10.5. If \( p \leq w\left( N\right) \), then \( {\omega }_{X}\left( p\right) = 0 \) for all non-empty \( X \) ; in particular, \( {E}_{p} = 1 \) when \( p \leq w\left( N\right) \) . If instead \( p > w\left( N\right) \), then \( {\omega }_{X}\left( p\right) = {p}^{-1} \) when \( \left| X\right| = 1 \) and \( {\om... | Proof. When \( p \leq w\left( N\right) \) then \( W\left( {x + {h}_{i}}\right) + 1 \equiv 1\left( {\;\operatorname{mod}\;p}\right) \) and the claim follows. When \( p > w\left( N\right) \) and \( \left| X\right| \geq 1,{\omega }_{X}\left( p\right) \) is equal to \( 1/p \) when the residue classes \( \left\{ {{h}_{i}\le... | Yes |
Lemma 10.6. Let \( 0 < \sigma < 1/{6m} \) . Then the Euler products \( \mathop{\prod }\limits_{p}{E}_{p}^{\left( l\right) } \) for \( l = 0,1,2 \) are absolutely convergent in the domain \( {\mathcal{D}}_{\sigma }^{m} \) . In particular, \( {G}_{0},{G}_{1},{G}_{2} \) can be continued analytically to this domain. Furthe... | Proof. The estimates for \( {G}_{1} \) and \( {G}_{2} \) proceed exactly as in Lemma 10.3 (the additional factors of \( {\lambda }_{p}\left( {z,{z}^{\prime }}\right) \) which appear on both the numerator and denominator of \( {E}_{p}^{\left( 1\right) } \) cancel to first order, and thus do not present any new difficult... | Yes |
Corollary 1. The sum of the angles of a quadrilateral is less than four right angles. |  | No |
Corollary 3. There do not exist lines that are everywhere equidistant. | As Corollary 3 states, lines are never equidistant. Instead the distance between sensed parallels varies from point to point as shown in the following theorem. | No |
Theorem 3.1. Three distinct points \( X\left( {{x}_{1},{x}_{2},1}\right), Y\left( {{y}_{1},{y}_{2},1}\right) \), and \( Z\left( {{z}_{1},{z}_{2},1}\right) \) are collinear iff the determinant\n\n\[ \left| \begin{array}{lll} {x}_{1} & {y}_{1} & {z}_{1} \\ {x}_{2} & {y}_{2} & {z}_{2} \\ 1 & 1 & 1 \end{array}\right| = 0 \... | Proof. \( X, Y, Z \) are collinear iff there is a line \( u\left\lbrack {{u}_{1},{u}_{2},{u}_{3}}\right\rbrack \) such that\n\n\[ {u}_{1}{x}_{1} + {u}_{2}{x}_{2} + {u}_{3} = 0 \]\n\n\[ {u}_{1}{y}_{1} + {u}_{2}{y}_{2} + {u}_{3} = 0 \]\n\n\[ {u}_{1}{z}_{1} + {u}_{2}{z}_{2} + {u}_{3} = 0 \]\n\nor\n\n\[ \left\lbrack {{u}_{... | Yes |
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