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Corollary 21.1.17 If \( X \) is reflexive, then \( X \) is separable if and only if \( {X}^{\prime } \) is separable. | Proof: From the above theorem, if \( {X}^{\prime } \) is separable, then so is \( X \) . Now suppose \( X \) is separable with a dense subset equal to \( D \) . Then since \( X \) is reflexive, \( J\left( D\right) \) is dense in \( {X}^{\prime \prime } \) where \( J \) is the James map satisfying \( {Jx}\left( {x}^{ * ... | Yes |
Proposition 21.2.2 Definition 21.2.1 is well defined, the integral is linear on simple functions and\n\n\[ \begin{Vmatrix}{{\int }_{\Omega }x\left( s\right) {d\mu }}\end{Vmatrix} \leq {\int }_{\Omega }\parallel x\left( s\right) \parallel {d\mu } \]\n\nwhenever \( x \) is a simple function. | Proof: It suffices to verify that if \( \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}{\mathcal{X}}_{{E}_{k}}\left( s\right) = 0 \), then \( \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}\mu \left( {E}_{k}\right) = 0 \) . Let \( f \in {X}^{\prime } \) . Then\n\n\[ f\left( {\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}{\mathcal{X}... | Yes |
The Bochner integral is well defined and if \( x \) is Bochner integrable and \( f \in {X}^{\prime } \), then \( f\left( {{\int }_{\Omega }x\left( s\right) {d\mu }}\right) = {\int }_{\Omega }f\left( {x\left( s\right) }\right) {d\mu } \) and the triangle inequality is valid, \( \begin{Vmatrix}{{\int }_{\Omega }x\left( s... | Proof: Theorem 21.2.4 shows \( {\int }_{\Omega }\parallel x\left( s\right) \parallel {d\mu } < \infty \) and that the definition of the integral is well defined. It remains to verify the triangle inequality on Bochner integral functions and the claim about passing a continuous linear functional inside the integral. Fir... | Yes |
Corollary 21.2.6 Let an \( X \) valued function \( x \) be Bochner integrable and let \( L \in \) \( \mathcal{L}\left( {X, Y}\right) \) where \( Y \) is another Banach space. Then \( {Lx} \) is a \( Y \) valued Bochner integrable function and\n\n\[ L\left( {{\int }_{\Omega }x\left( s\right) {d\mu }}\right) = {\int }_{\... | Proof: From Theorem 21.2.4 there is a sequence of simple functions \( \left\{ {y}_{n}\right\} \) having the properties listed in that theorem. Then consider \( \left\{ {L{y}_{n}}\right\} \) which converges pointwise to \( {Lx} \) . Since \( L \) is continuous and linear,\n\n\[ {\int }_{\Omega }{\begin{Vmatrix}L{y}_{n} ... | Yes |
Corollary 21.2.8 Suppose \( Y \) is a reflexive Banach space and \( X \) is a Banach space such that there exists a continuous one to one mapping, \( g : X \rightarrow Y \) such that \( g\left( X\right) \) is a closed subset of \( Y \) . Then \( X \) is reflexive. | Proof: By the open mapping theorem, \( g\left( X\right) \) and \( X \) are homeomorphic since \( {g}^{-1} \) must also be continuous. Therefore, since \( g\left( X\right) \) is reflexive because it is a closed subspace of a reflexive space, it follows \( X \) is also reflexive. | Yes |
Lemma 21.2.9 Suppose \( V \) is a reflexive Banach space and that \( V \) is a dense subset of \( W \), another Banach space in the topology of \( W \) . Then \( {i}^{ * }{W}^{\prime } \) is a dense subset of \( {V}^{\prime } \) where here \( i \) is the inclusion map of \( V \) into \( W \) . | Proof: First note that \( {i}^{ * } \) is one to one. If \( {i}^{ * }{w}^{ * } = 0 \) for \( {w}^{ * } \in {W}^{\prime } \), then this means that for all \( v \in V \) ,\n\n\[ \n{i}^{ * }w\left( v\right) = {w}^{ * }\left( v\right) = 0 \n\]\n\nand since \( V \) is dense in \( W \), this shows \( {w}^{ * } = 0 \) .\n\nCo... | Yes |
Corollary 21.2.10 Let \( E \) and \( F \) be reflexive Banach spaces and let \( A \) be a closed operator \( A : D\left( A\right) \subseteq E \rightarrow F \) . Suppose also that \( D\left( A\right) \) is dense in \( E \) . Then making \( D\left( A\right) \) into a Banach space by using the above graph norm given in 21... | Proof: First note that \( E \times F \) is a reflexive Banach space and \( \mathcal{G}\left( A\right) \) is a closed subspace of \( E \times F \) so it is also a reflexive Banach space. Now \( D\left( A\right) \) is isometric to \( \mathcal{G}\left( A\right) \) and so it follows \( D\left( A\right) \) is a dense subspa... | Yes |
Theorem 21.2.11 Let \( X, Y \) be separable Banach spaces and let \( A : D\left( A\right) \subseteq X \rightarrow \) \( Y \) be a closed operator where \( D\left( A\right) \) is a dense separable subset of \( X \) with respect to the graph norm on \( D\left( A\right) \) described above \( {}^{1} \). Suppose also that \... | Proof: First of all, consider the assertion that \( x \) is strongly measurable into \( D\left( A\right) \). Letting \( f \in D{\left( A\right) }^{\prime } \) be given, there exists a sequence, \( \left\{ {g}_{n}\right\} \subseteq {i}^{ * }{X}^{\prime } \) such that \( {g}_{n} \rightarrow f \) in \( D{\left( A\right) }... | Yes |
Theorem 21.2.12 Let \( X \) and \( Y \) be separable Banach spaces and let \( A : D\left( A\right) \subseteq \) \( X \rightarrow Y \) be a closed operator. Also let \( \left( {\Omega ,\mathcal{F},\mu }\right) \) be a \( \sigma \) finite measure space and let \( x : \Omega \rightarrow X \) be Bochner integrable such tha... | Proof: Consider the graph of \( A \) , \[ G\left( A\right) \equiv \{ \left( {x,{Ax}}\right) : x \in D\left( A\right) \} \subseteq X \times Y. \] Then since \( A \) is closed, \( G\left( A\right) \) is a closed separable Banach space with the norm \( \parallel \left( {x, y}\right) \parallel \equiv \max \left( {\parallel... | Yes |
Lemma 21.3.1 Let \( x \in X \) and suppose \( A \) is strongly measurable. Then\n\n\[ s \rightarrow A\left( s\right) x \]\n\nis strongly measurable as a map into \( Y \) . | Proof: Since \( A \) is assumed to be strongly measurable, it is the pointwise limit of simple functions of the form\n\n\[ {A}_{n}\left( s\right) \equiv \mathop{\sum }\limits_{{k = 1}}^{{m}_{n}}{A}_{k}^{n}{\mathcal{X}}_{{E}_{k}^{n}}\left( s\right) \]\n\nwhere \( {A}_{k}^{n} \) is in \( \mathcal{L}\left( {X, Y}\right) \... | Yes |
Lemma 21.3.3 The above definition is well defined. Furthermore, if 21.3.12 holds then \( s \rightarrow \parallel A\left( s\right) \parallel \) is measurable and if 21.3.13 holds, then \[ \begin{Vmatrix}{{\int }_{\Omega }A\left( s\right) {d\mu }}\end{Vmatrix} \leq {\int }_{\Omega }\parallel A\left( s\right) \parallel {d... | Proof: It is clear that in case \( s \rightarrow A\left( s\right) x \) is measurable for all \( x \in X \) there exists a unique \( \Psi \in \mathcal{L}\left( {X, Y}\right) \) such that \[ \Psi \left( x\right) = {\int }_{\Omega }A\left( s\right) {xd\mu } \] This is because \( x \rightarrow {\int }_{\Omega }A\left( s\ri... | Yes |
Lemma 21.3.6 Let \( H \) be a Hilbert space and suppose \( A \in \mathcal{L}\left( {H, H}\right) \) is a compact operator. Then\n\n1. \( A \) is a compact operator if and only if whenever if \( {x}_{n} \rightarrow x \) weakly in \( H \), it follows that \( A{x}_{n} \rightarrow {Ax} \) strongly in \( H \) . | Proof: Consider \( \Rightarrow \) of 1. Suppose then that \( {x}_{n} \rightarrow x \) weakly. Since \( \left\{ {x}_{n}\right\} \) is weakly bounded, it follows from the uniform boundedness principle that \( \left\{ \begin{Vmatrix}{x}_{n}\end{Vmatrix}\right\} \) is bounded. Let \( {x}_{n} \in \widehat{B} \) for \( \wide... | Yes |
Lemma 21.3.8 Let \( A \in \mathcal{L}\left( {H, H}\right) \) and suppose it is self adjoint and compact. Let \( B \) denote the closed unit ball in \( H \) . Let \( e \in B \) be such that\n\n\[ \left| \left( {{Ae}, e}\right) \right| = \mathop{\max }\limits_{{x \in B}}\left| \left( {{Ax}, x}\right) \right| . \]\n\nThen... | Proof: From the above observation, \( \left( {{Ax}, x}\right) \) is always real and since \( A \) is compact, \( \left| \left( {{Ax}, x}\right) \right| \) achieves a maximum at \( e \) . It remains to verify \( e \) is an eigenvector. If \( \left| \left( {{Ae}, e}\right) \right| = 0 \) for all \( e \in B \), then \( A ... | Yes |
Theorem 21.3.10 Let \( A\left( s\right) \in \mathcal{L}\left( {H, H}\right) \) be a compact self adjoint operator and \( H \) is a separable Hilbert space such that \( s \rightarrow A\left( s\right) x \) is strongly measurable. Then there exist real numbers \( {\left\{ {\lambda }_{k}\left( s\right) \right\} }_{k = 1}^{... | Proof: It is simply a repeat of the above proof of the Hilbert Schmidt theorem except at every step when the \( {e}_{k} \) and \( {\lambda }_{k} \) are defined, you use the Kuratowski measurable selection theorem, Theorem 21.3.4 on Page 698 to obtain \( {\lambda }_{k}\left( s\right) \) is measurable and that \( s \righ... | Yes |
Theorem 21.5.3 \( {L}^{p}\left( {\Omega ;X}\right) \) is complete. Also every Cauchy sequence has a subsequence which converges pointwise. | Proof: If \( \left\{ {x}_{n}\right\} \) is Cauchy in \( {L}^{p}\left( {\Omega ;X}\right) \), extract a subsequence \( \left\{ {x}_{{n}_{k}}\right\} \) satisfying\n\n\[ \n{\begin{Vmatrix}{x}_{{n}_{k + 1}} - {x}_{{n}_{k}}\end{Vmatrix}}_{p} \leq {2}^{-k} \n\]\n\nand apply Lemma 21.5.2. The pointwise convergence of this su... | Yes |
Theorem 21.5.5 If \( x \) is strongly measurable and \( {x}_{n}\left( s\right) \rightarrow x\left( s\right) \) a.e. (for \( s \) off a set of measure zero) with \[ \begin{Vmatrix}{{x}_{n}\left( s\right) }\end{Vmatrix} \leq g\left( s\right) \text{ a.e. } \] where \( {\int }_{\Omega }{gd\mu } < \infty \), then \( x \) is... | Proof: The measurability of \( x \) follows from Theorem 21.1.10 if convergence happens for each \( s \) . Otherwise, \( x \) is measurable by assumption. Then \( \begin{Vmatrix}{{x}_{n}\left( s\right) - x\left( s\right) }\end{Vmatrix} \leq \) \( {2g}\left( s\right) \) a.e. so, from Fatou’s lemma, \[ {\int }_{\Omega }{... | Yes |
Theorem 21.5.7 Let \( \left( {\Omega ,\mathcal{F},\mu }\right) \) be a finite measure space and let \( X \) be a separable Banach space. Let \( \left\{ {f}_{n}\right\} \subseteq {L}^{1}\left( {\Omega ;X}\right) \) be uniformly integrable and bounded such that \( {f}_{n}\left( \omega \right) \rightarrow f\left( \omega \... | Proof: Let \( \varepsilon > 0 \) be given. Then by uniform integrability there exists \( \delta > 0 \) such that if \( \mu \left( E\right) < \delta \) then \[ {\int }_{E}\begin{Vmatrix}{f}_{n}\end{Vmatrix}{d\mu } < \varepsilon /3 \] By Fatou’s lemma the same inequality holds for \( f \) . Also Fatou’s lemma shows \( f ... | Yes |
Theorem 21.5.8 Let \( 1 \leq p < \infty \) and let \( p < r \leq \infty \) . Then \( {L}^{r}\left( {\left\lbrack {0, T}\right\rbrack, X}\right) \) is a Borel subset of \( {L}^{p}\left( {\left\lbrack {0, T}\right\rbrack ;X}\right) \) . Letting \( C\left( {\left\lbrack {0, T}\right\rbrack ;X}\right) \) denote the functio... | Proof: First consider the claim about \( {L}^{r}\left( {\left\lbrack {0, T}\right\rbrack ;X}\right) \) . Let\n\n\[ \n{B}_{M} \equiv \left\{ {x \in {L}^{p}\left( {\left\lbrack {0, T}\right\rbrack ;X}\right) : \parallel x{\parallel }_{{L}^{r}\left( {\left\lbrack {0, T}\right\rbrack ;X}\right) } \leq M}\right\} .\n\]\n\nT... | Yes |
Lemma 21.5.9 Let \( \left( {\Omega ,\mu }\right) \) be a regular measure space where \( \Omega \) is a locally compact Hausdorff space. Then \( {C}_{c}\left( {\Omega ;X}\right) \) the space of continuous functions having compact support and values in \( X \) is dense in \( {L}^{p}\left( {0, T;X}\right) \) for all \( p ... | Proof: First is it shown the simple functions are dense in \( {L}^{p}\left( {0, T;X}\right) \) . Let \( f \in {L}^{p}\left( {0, T;X}\right) \) and let \( \left\{ {x}_{n}\right\} \) denote a sequence of simple functions which converge to \( f \) pointwise which also have the property that\n\n\[ \begin{Vmatrix}{{x}_{n}\l... | Yes |
Lemma 21.7.4 Suppose \( \nu \) is a complex measure defined on \( \mathcal{S} \) a \( \sigma \) algebra where \( \left( {\Omega ,\mathcal{S}}\right) \) is a measurable space, and let \( \mu \) be a measure on \( \mathcal{S} \) with \( \left| {\nu \left( E\right) }\right| \leq {r\mu }\left( E\right) \) and suppose there... | Proof: Let \( B\left( {p,\delta }\right) \subseteq \mathbb{C} \smallsetminus \overline{B\left( {0, r}\right) } \) and let \( E \equiv {h}^{-1}\left( {B\left( {p,\delta }\right) }\right) \) . If \( \mu \left( E\right) > 0 \) . Then \[ \left| {\frac{1}{\mu \left( E\right) }{\int }_{E}{hd\mu } - p}\right| \leq \frac{1}{\m... | Yes |
Theorem 21.8.3 If \( X \) is a Banach space and \( {X}^{\prime } \) has the Radon Nikodym property, then if \( \left( {\Omega ,\mathcal{S},\mu }\right) \) is a finite measure space,\n\n\[{\left( {L}^{p}\left( \Omega ;X\right) \right) }^{\prime } \cong {L}^{{p}^{\prime }}\left( {\Omega ;{X}^{\prime }}\right)\]\n\nand in... | Proof: Let \( l \in {\left( {L}^{p}\left( \Omega ;X\right) \right) }^{\prime } \) and define \( F\left( E\right) \in {X}^{\prime } \) by\n\n\[F\left( E\right) \left( x\right) \equiv l\left( {{\mathcal{X}}_{E}\left( \cdot \right) x}\right) .\] | Yes |
Theorem 21.8.6 (Riesz representation theorem) Let \( \\left( {\\Omega ,\\mathcal{S},\\mu }\\right) \) be \( \\sigma \) finite and let \( {X}^{\\prime } \) have the Radon Nikodym property. Then for\n\n\[ \n\\Lambda \\in {\\left( {L}^{p}\\left( \\Omega ;X,\\mu \\right) \\right) }^{\\prime }, p \\geq 1 \n\]\n\n there exis... | Proof: The above lemma gives the existence part of the conclusion of the theorem. Uniqueness is done as before. | No |
Corollary 21.8.7 If \( {X}^{\prime } \) is separable, then for \( \left( {\Omega ,\mathcal{S},\mu }\right) \) a \( \sigma \) finite measure space, | \[ {\left( {L}^{p}\left( \Omega ;X\right) \right) }^{\prime } \cong {L}^{{p}^{\prime }}\left( {\Omega ;{X}^{\prime }}\right) . \] | No |
Corollary 21.8.9 If \( X \) is separable and reflexive and \( \left( {\Omega ,\mathcal{S},\mu }\right) \) a \( \sigma \) finite measure space, then if \( p \in \left( {1,\infty }\right) \), then \( {L}^{p}\left( {\Omega ;X}\right) \) is reflexive. | Proof: This is just like the scalar valued case. | No |
Lemma 21.8.10 Let \( B = \overline{B\left( {\mathbf{0}, L}\right) } \) be a closed ball in \( {L}^{\infty }\left( {0, T, H}\right) \) . Then \( B \) is a Polish space with respect to the weak \( * \) topology. The closure is taken with respect to the usual topology. | Proof: Let \( {\left\{ {\mathbf{z}}_{k}\right\} }_{k = 1}^{\infty } = X \) be a dense countable subspace in \( {L}^{1}\left( {0, T, H}\right) \) . You start with a dense countable set and then consider all finite linear combinations having coefficients in \( \mathbb{Q} \) . Then the metric on \( B \) is\n\n\[ d\left( {... | Yes |
Proposition 21.9.1 Let \( E \) be a Banach space and let \( \left\{ {u}_{n}\right\} \) be a sequence in \( {L}^{2}\left( {\Gamma, E}\right) \) and let \( G\left( x\right) \) be a weakly compact set in \( E \), and \( {u}_{n}\left( x\right) \in G\left( x\right) \) a.e. for each \( n \) . Let \( \lim \sup \left\{ {{u}_{n... | Proof: Let \( H = \left\{ {w \in {L}^{2}\left( {\Gamma, E}\right) : w\left( x\right) \in H\left( x\right) }\right. \) a.e. \( \} \) . Then \( H \) is convex. If you have \( {w}_{i} \in H \), then since each \( H\left( x\right) \) is convex, it follows that \( \lambda {w}_{1}\left( x\right) + \) \( \left( {1 - \lambda }... | Yes |
Lemma 21.10.1 Suppose \( V \subseteq W \) and the injection map is compact, hence continuous. Suppose also that \( W \subseteq U \) with continuous injection. Then for any \( \varepsilon > 0 \) there exists \( {C}_{\varepsilon } \) such that for all \( v \in V \) , \[ \parallel v{\parallel }_{W} \leq \varepsilon \paral... | Proof: Suppose not. Then there exists \( \varepsilon > 0 \) for which things don’t work out. Thus there exists \( {v}_{n} \in V \) such that \[ {\begin{Vmatrix}{v}_{n}\end{Vmatrix}}_{W} > \varepsilon {\begin{Vmatrix}{v}_{n}\end{Vmatrix}}_{V} + n{\begin{Vmatrix}{v}_{n}\end{Vmatrix}}_{U} \] Dividing by \( {\begin{Vmatrix... | Yes |
Theorem 21.10.4 Let \( V \subseteq W \subseteq U \) where these are Banach spaces such that the injection map of \( V \) into \( W \) is compact and the injection map of \( W \) into \( U \) is continuous. Let \( \Omega \) be an open set in \( {\mathbb{R}}^{m} \) and let \( \mathcal{A} \) be a bounded subset of \( {L}^... | Proof: Let \( \infty > M \geq \mathop{\sup }\limits_{{u \in {L}^{p}\left( {\Omega ;V}\right) }}\parallel u{\parallel }_{{L}^{p}\left( {\Omega ;V}\right) }^{p} \) . Let \( \left\{ {\psi }_{n}\right\} \) be a mollifier with support in \( B\left( {\mathbf{0},1/n}\right) \) . I need to show that \( \mathcal{A} \) has an \(... | Yes |
Corollary 21.10.7 Let \( E \subseteq W \subseteq X \) where the injection map is continuous from \( W \) to \( X \) and compact from \( E \) to \( W \) . Then if \( \gamma > \alpha \), the embedding of \( {C}^{0,\gamma }\left( {\left\lbrack {0, T}\right\rbrack, E}\right) \) into \( {C}^{0,\alpha }\left( {\left\lbrack {... | Proof: Let \( \phi \in {C}^{0,\gamma }\left( {\left\lbrack {0, T}\right\rbrack, E}\right) \n\n\[ \frac{\parallel \phi \left( t\right) - \phi \left( s\right) {\parallel }_{X}}{{\left| t - s\right| }^{\alpha }} \leq {\left( \frac{\parallel \phi \left( t\right) - \phi \left( s\right) {\parallel }_{W}}{{\left| t - s\right|... | Yes |
Corollary 21.10.9 Let \( E \subseteq W \subseteq X \) where the injection map is continuous from \( W \) to \( X \) and compact from \( E \) to \( W \) . Let \( p \geq 1 \), let \( q > 1 \), and define\n\n\[ S \equiv \left\{ {u \in {L}^{p}\left( {\left\lbrack {a, b}\right\rbrack ;E}\right) : }\right. \text{ for some }C... | Proof: The first part was done earlier. Therefore, we just prove the new stuff which involves a bound on the \( {L}^{1} \) norm of the derivative. It suffices to show \( S \) has an \( \eta \) net in \( {L}^{p}\left( {\left\lbrack {a, b}\right\rbrack ;W}\right) \) for each \( \eta > 0 \) .\n\nIf not, there exists \( \e... | Yes |
Theorem 22.1.3 If \( \mathop{\lim }\limits_{{\mathbf{y} \rightarrow \mathbf{x}}}\mathbf{f}\left( \mathbf{y}\right) = \mathbf{L} \) and \( \mathop{\lim }\limits_{{y \rightarrow x}}\mathbf{f}\left( \mathbf{y}\right) = {\mathbf{L}}_{1} \), then \( \mathbf{L} = {\mathbf{L}}_{1} \) . | Proof: Let \( \varepsilon > 0 \) be given. There exists \( \delta > 0 \) such that if \( 0 < \left| {\mathbf{y} - \mathbf{x}}\right| < \delta \) and \( \mathbf{y} \in D\left( \mathbf{f}\right) \), then\n\n\[ \parallel \mathbf{f}\left( \mathbf{y}\right) - \mathbf{L}\parallel < \varepsilon ,\begin{Vmatrix}{\mathbf{f}\lef... | Yes |
Theorem 22.1.5 Suppose \( \mathbf{f} : D\left( \mathbf{f}\right) \subseteq V \rightarrow {\mathbb{F}}^{m} \) . Then for \( \mathbf{x} \) a limit point of \( D\left( \mathbf{f}\right) \) , \[ \mathop{\lim }\limits_{{y \rightarrow x}}\mathbf{f}\left( \mathbf{y}\right) = \mathbf{L} \] if and only if \[ \mathop{\lim }\limi... | Proof: Suppose 22.1.1. Then letting \( \varepsilon > 0 \) be given there exists \( \delta > 0 \) such that if \( 0 < \parallel y - x\parallel < \delta \), it follows \[ \left| {{f}_{k}\left( y\right) - {L}_{k}}\right| \leq \parallel \mathbf{f}\left( y\right) - \mathbf{L}\parallel < \varepsilon \] which verifies 22.1.2.... | Yes |
Theorem 22.1.6 For \( f : D\left( f\right) \rightarrow W \) and \( x \in D\left( f\right) \) a limit point of \( D\left( f\right), f \) is continuous at \( x \) if and only if\n\n\[ \mathop{\lim }\limits_{{y \rightarrow x}}f\left( y\right) = f\left( x\right) \] | Proof: First suppose \( f \) is continuous at \( x \) a limit point of \( D\left( f\right) \) . Then for every \( \varepsilon > 0 \) there exists \( \delta > 0 \) such that if \( \parallel x - y\parallel < \delta \) and \( y \in D\left( f\right) \), then \( \left| {f\left( x\right) - f\left( y\right) }\right| < \vareps... | Yes |
Find \( \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {3,1}\right) }}\left( {\frac{{x}^{2} - 9}{x - 3}, y}\right) \) . | It is clear that \( \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {3,1}\right) }}\frac{{x}^{2} - 9}{x - 3} = 6 \) and \( \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {3,1}\right) }}y = 1 \) . Therefore, this limit equals \( \left( {6,1}\right) \) . | Yes |
Find \( \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {0,0}\right) }}\frac{xy}{{x}^{2} + {y}^{2}} \) . | First of all, observe the domain of the function is \( {\mathbb{R}}^{2} \smallsetminus \{ \left( {0,0}\right) \} \), every point in \( {\mathbb{R}}^{2} \) except the origin. Therefore, \( \left( {0,0}\right) \) is a limit point of the domain of the function so it might make sense to take a limit. However, just as in th... | Yes |
Theorem 22.2.2 The derivative is well defined. | Proof: First note that for a fixed vector \( \mathbf{v},\mathbf{o}\left( {t\mathbf{v}}\right) = \mathbf{o}\left( t\right) \) . This is because\n\n\[ \mathop{\lim }\limits_{{t \rightarrow 0}}\frac{\mathbf{o}\left( {t\mathbf{v}}\right) }{\left| t\right| } = \mathop{\lim }\limits_{{t \rightarrow 0}}\parallel \mathbf{v}\pa... | Yes |
Lemma 22.2.3 Let \( \mathbf{f} \) be differentiable at \( \mathbf{x} \) . Then \( \mathbf{f} \) is continuous at \( \mathbf{x} \) and in fact, there exists \( K > 0 \) such that whenever \( \parallel \mathbf{v}\parallel \) is small enough,\n\n\[ \left| \right| \mathbf{f}\left( {\mathbf{x} + \mathbf{v}}\right) - \mathbf... | Proof: From the definition of the derivative,\n\n\[ \mathbf{f}\left( {\mathbf{x} + \mathbf{v}}\right) - \mathbf{f}\left( \mathbf{x}\right) = D\mathbf{f}\left( \mathbf{x}\right) \mathbf{v} + \mathbf{o}\left( \mathbf{v}\right) .\n\nLet \( \parallel \mathbf{v}\parallel \) be small enough that \( \frac{\mathbf{o}\left( {\p... | Yes |
Theorem 22.3.1 (The chain rule) Let \( U \) and \( V \) be open sets \( U \subseteq X \) and \( V \subseteq \) \( Y \) . Suppose \( \mathbf{f} : U \rightarrow V \) is differentiable at \( \mathbf{x} \in U \) and suppose \( \mathbf{g} : V \rightarrow {\mathbb{F}}^{q} \) is differentiable at \( \mathbf{f}\left( \mathbf{x... | Proof: This follows from a computation. Let \( B\left( {\mathbf{x}, r}\right) \subseteq U \) and let \( r \) also be small enough that for \( \parallel \mathbf{v}\parallel \leq r \), it follows that \( \mathbf{f}\left( {\mathbf{x} + \mathbf{v}}\right) \in V \) . Such an \( r \) exists because \( \mathbf{f} \) is contin... | Yes |
Theorem 22.4.2 Let \( f : U \subseteq X \rightarrow Y \) where \( f \) takes bounded sets to precompact sets. Then \( {Df}\left( x\right) \) also takes bounded sets in \( X \) to precompact sets in \( Y \) . | Proof: If this is not so, then there exists a bounded set \( B \) in \( X \) and for some \( \varepsilon > \) 0 a sequence of points \( {Df}\left( x\right) {b}_{n} \) such that all these points are further apart than \( \varepsilon \) . Without loss of generality, one can assume \( B = B\left( {0, r}\right) \), a ball.... | Yes |
Theorem 22.5.1 Let \( \mathbf{f} : U \subseteq {\mathbb{F}}^{n} \rightarrow {\mathbb{F}}^{m} \) and suppose \( \mathbf{f} \) is differentiable at \( \mathbf{x} \) . Then all the partial derivatives \( \frac{\partial {f}_{i}\left( \mathbf{x}\right) }{\partial {x}_{j}} \) exist and if \( J\mathbf{f}\left( \mathbf{x}\righ... | \[ \frac{\partial \mathbf{f}\left( \mathbf{x}\right) }{\partial {x}_{i}} \equiv \mathop{\lim }\limits_{{t \rightarrow 0}}\frac{\mathbf{f}\left( {\mathbf{x} + t{\mathbf{e}}_{i}}\right) - \mathbf{f}\left( \mathbf{x}\right) }{t}. \] | No |
Lemma 22.6.1 Let \( Y \) be a normed vector space and suppose \( \mathbf{h} : \left\lbrack {0,1}\right\rbrack \rightarrow Y \) is differentiable and satisfies\n\n\[ \begin{Vmatrix}{{\mathbf{h}}^{\prime }\left( t\right) }\end{Vmatrix} \leq M \]\n\nThen\n\n\[ \parallel \mathbf{h}\left( 1\right) - \mathbf{h}\left( 0\right... | Proof: Let \( \varepsilon > 0 \) be given and let\n\n\[ S \equiv \{ t \in \left\lbrack {0,1}\right\rbrack : \text{ for all }s \in \left\lbrack {0, t}\right\rbrack ,\left| \right| \mathbf{h}\left( s\right) - \mathbf{h}\left( 0\right) \left| \right| \leq \left( {M + \varepsilon }\right) s\} \]\n\nThen \( 0 \in S \) . Let... | Yes |
Theorem 22.6.2 Suppose \( U \) is an open subset of \( X \) and \( \mathbf{f} : U \rightarrow Y \) has the property that \( D\mathbf{f}\left( \mathbf{x}\right) \) exists for all \( \mathbf{x} \) in \( U \) and that, \( \mathbf{x} + t\left( {\mathbf{y} - \mathbf{x}}\right) \in U \) for all \( t \in \left\lbrack {0,1}\ri... | Proof: Let\n\n\[ \mathbf{h}\left( t\right) \equiv \mathbf{f}\left( {\mathbf{x} + t\left( {\mathbf{y} - \mathbf{x}}\right) }\right) . \]\n\nThen by the chain rule,\n\n\[ {\mathbf{h}}^{\prime }\left( t\right) = D\mathbf{f}\left( {\mathbf{x} + t\left( {\mathbf{y} - \mathbf{x}}\right) }\right) \left( {\mathbf{y} - \mathbf{... | Yes |
Theorem 22.6.3 Let \( X \) be a normed vector space having basis \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{n}}\right\} \) and let \( Y \) be another normed vector space having basis \( \left\{ {{\mathbf{w}}_{1},\cdots ,{\mathbf{w}}_{m}}\right\} \) . Let \( U \) be an open set in \( X \) and let \( \mathbf{f} :... | Proof: Let \( \mathbf{v} = \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}{\mathbf{v}}_{k} \) . Then\n\n\[ \n\mathbf{f}\left( {\mathbf{x} + \mathbf{v}}\right) - \mathbf{f}\left( \mathbf{x}\right) = \mathbf{f}\left( {\mathbf{x} + \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}{\mathbf{v}}_{k}}\right) - \mathbf{f}\left( \mathbf{x}... | Yes |
Theorem 22.6.4 Suppose \( \mathbf{f} : U \rightarrow Y \) where \( U \) is an open set in \( X \), a normed linear space. Suppose that \( \mathbf{f} \) is Gateaux differentiable on \( U \) and that the Gateaux derivative is continuous on an open set containing \( \mathbf{x} \) . Then \( \mathbf{f} \) is Frechet differe... | Proof: Denote by \( G\left( \mathbf{x}\right) \in \mathcal{L}\left( {X, Y}\right) \) the Gateaux derivative. Thus\n\n\[ G\left( \mathbf{x}\right) \mathbf{v} \equiv \mathop{\lim }\limits_{{\lambda \rightarrow 0}}\frac{\mathbf{f}\left( {\mathbf{x} + \lambda \mathbf{v}}\right) - \mathbf{f}\left( \mathbf{x}\right) }{\lambd... | Yes |
Lemma 22.6.5 Let \( \parallel x\parallel = \mathop{\sup }\limits_{{{\begin{Vmatrix}{y}^{ * }\end{Vmatrix}}_{{X}^{\prime }} \leq 1}}\left| \left\langle {{y}^{ * }, x}\right\rangle \right| \) . | Proof: Let \( f\left( {kx}\right) = k\parallel x\parallel \) . Then\n\n\[ \mathop{\sup }\limits_{{\parallel {kx}\parallel \leq 1}}\left| {\langle f, x\rangle }\right| = \mathop{\sup }\limits_{{\left| k\right| \leq 1/\parallel x\parallel }}\left| k\right| \parallel x\parallel = 1 \]\n\nThen by Hahn Banach theorem, there... | Yes |
Show \( f \) is differentiable at \( u \in X \) . | Consider the Gateaux differentiability.\n\n\[ \mathop{\lim }\limits_{{t \rightarrow 0}}\frac{f\left( {u + {tv}}\right) - f\left( u\right) }{t} = \mathop{\lim }\limits_{{t \rightarrow 0}}\frac{t{\int }_{\Omega }\nabla u \cdot \nabla {vdx}}{t} + t\frac{1}{2}{\int }_{\Omega }\nabla v \cdot \nabla v \]\n\nso it converges t... | Yes |
Lemma 22.8.1 Suppose \( U \) is an open set in \( X \times Y \) . Then the set, \( {U}_{\mathbf{y}} \) defined by\n\n\[ \n{U}_{\mathbf{y}} \equiv \{ \mathbf{x} \in X : \left( {\mathbf{x},\mathbf{y}}\right) \in U\}\n\]\n\nis an open set in \( X \) . Here \( X \times Y \) is a finite dimensional vector space in which the... | Proof: In finite dimensions it doesn't matter how this norm is defined because all are equivalent. It obviously satisfies most axioms of a norm. The only one which is not obvious is the triangle inequality. I will show this now.\n\n\[ \n\left| \right| \left( {\mathbf{x},\mathbf{y}}\right) + \left( {{\mathbf{x}}_{1},{\m... | Yes |
Corollary 22.8.2 Let \( U \subseteq \mathop{\prod }\limits_{{i = 1}}^{n}{X}_{i} \) be an open set and let\n\n\[ \n{U}_{\left( {\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{i - 1},{\mathbf{x}}_{i + 1},\cdots ,{\mathbf{x}}_{n}\right) } \equiv \left\{ {\mathbf{x} \in {\mathbb{F}}^{{r}_{i}} : \left( {{\mathbf{x}}_{1},\cdots ,{\ma... | Proof: Let \( \mathbf{z} \in {U}_{\left( {\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{i - 1},{\mathbf{x}}_{i + 1},\cdots ,{\mathbf{x}}_{n}\right) } \) . Then \( \left( {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{i - 1},\mathbf{z},{\mathbf{x}}_{i + 1},\cdots ,{\mathbf{x}}_{n}}\right) \equiv \) \( \mathbf{x} \in U \) by definition... | Yes |
Example 22.9.2 Let\n\n\\[ \nf\\left( {x, y}\\right) = \\left\\{ \\begin{array}{l} \\frac{{xy}\\left( {{x}^{2} - {y}^{2}}\\right) }{{x}^{2} + {y}^{2}}\\text{ if }\\left( {x, y}\\right) \\neq \\left( {0,0}\\right) \\\\ 0\\text{ if }\\left( {x, y}\\right) = \\left( {0,0}\\right) \\end{array}\\right.\n\\]\n\nFrom the defin... | Now\n\n\\[ \n{f}_{xy}\\left( {0,0}\\right) \\equiv \\mathop{\\lim }\\limits_{{y \\rightarrow 0}}\\frac{{f}_{x}\\left( {0, y}\\right) - {f}_{x}\\left( {0,0}\\right) }{y} = \\mathop{\\lim }\\limits_{{y \\rightarrow 0}}\\frac{-{y}^{4}}{{\\left( {y}^{2}\\right) }^{2}} = - 1\n\\]\n\nwhile\n\n\\[ \n{f}_{yx}\\left( {0,0}\\rig... | Yes |
Theorem 22.10.2 If \( Y \) is a Banach space, then \( \mathcal{L}\left( {X, Y}\right) \) is also a Banach space. | Proof: Let \( \left\{ {L}_{n}\right\} \) be a Cauchy sequence in \( \mathcal{L}\left( {X, Y}\right) \) and let \( x \in X \). \[ \begin{Vmatrix}{{L}_{n}x - {L}_{m}x}\end{Vmatrix} \leq \parallel x\parallel \begin{Vmatrix}{{L}_{n} - {L}_{m}}\end{Vmatrix}. \] Thus \( \left\{ {{L}_{n}x}\right\} \) is a Cauchy sequence. Let... | Yes |
Lemma 22.10.3 Let \( \left( {X,\parallel \cdot \parallel }\right) \) is a Banach space, and if \( A \in \mathcal{L}\left( {X, X}\right) \) and \( \parallel A\parallel = \) \( r < 1 \), then \[ {\left( I - A\right) }^{-1} = \mathop{\sum }\limits_{{k = 0}}^{\infty }{A}^{k} \in \mathcal{L}\left( {X, X}\right) \] where the... | Proof: First of all, why does the series make sense? \[ \begin{Vmatrix}{\mathop{\sum }\limits_{{k = p}}^{q}{A}^{k}}\end{Vmatrix} \leq \mathop{\sum }\limits_{{k = p}}^{q}\begin{Vmatrix}{A}^{k}\end{Vmatrix} \leq \mathop{\sum }\limits_{{k = p}}^{q}\parallel A{\parallel }^{k} \leq \mathop{\sum }\limits_{{k = p}}^{\infty }{... | Yes |
Let \( X, Y, Z \) be Banach spaces and suppose \( U \) is an open set in \( X \times Y \). Let \( f : U \times \Lambda \rightarrow Z \) satisfy \( f\left( {\cdot ,\cdot ,\lambda }\right) \) is in \( {C}^{1}\left( U\right) \) and suppose for each \( \lambda \), \[ f\left( {{x}_{0},{y}_{0},\lambda }\right) = 0,{D}_{1}f{\... | Proof: It is just a repeat of the above proof except you use the uniform contraction principle, Corollary 7.11.4 to get the fixed point. | No |
Theorem 22.10.9 (inverse function theorem) Let \( {x}_{0} \in U \), an open set in \( X \), and let \( f : U \rightarrow Y \) where \( X, Y \) are finite dimensional normed vector spaces. Suppose\n\n\[ f\text{is}{C}^{1}\left( U\right) \text{, and}{Df}{\left( {x}_{0}\right) }^{-1} \in \mathcal{L}\left( {Y, X}\right) \te... | Proof: Apply the implicit function theorem to the function\n\n\[ F\left( {x, y}\right) \equiv f\left( x\right) - y \]\n\nwhere \( {y}_{0} \equiv f\left( {x}_{0}\right) \) . Thus the function \( y \rightarrow x\left( y\right) \) defined in that theorem is \( {f}^{-1} \) . Now let\n\n\[ W \equiv B\left( {{x}_{0},\delta }... | Yes |
Say \( X = {\mathbb{R}}^{2} \) and \( \Lambda = \mathbb{R} \) . Let \( f\left( {x, y,\lambda }\right) = x + {xy} + {y}^{2} + \lambda \) . Then | \[ {D}_{1}f\left( {0,0,0}\right) = \left( {1,0}\right) \] this \( 1 \times 2 \) matrix mapping \( {\mathbb{R}}^{2} \) to \( \mathbb{R} \) . Thus \( {X}_{2} = {\left( 0,\alpha \right) }^{T} : \alpha \in \mathbb{R} \) and \( {X}_{1} = {\left( \alpha ,0\right) }^{T} \) : \( \alpha \in \mathbb{R} \) . In this case, \( {Y}_... | Yes |
Let \( l \geq 1 \) and let \( {t}_{m} \in \mathbb{C} \) . Then if \( \mathbf{h} \in {X}^{l} \), then whenever \( \left| z\right| \) is small enough, 22.13.35 holds. Also the coefficients satisfy\n\n\[ \n{a}_{{k}_{1}\cdots {k}_{l}}\left( {x,{t}_{l}{h}_{l},\cdots ,{t}_{1}{h}_{1}}\right) = \left( {\mathop{\prod }\limits_{... | Proof: Let \( C \) be small enough that the circles \( {t}_{m}C \) for all \( m = 1,\cdots, l \) and \( C \) have radius less than \( \frac{\delta }{l} \) . First assume \( {t}_{m} \neq 0 \) for all \( m \) . Then\n\n\[ \n{a}_{{k}_{1}\cdots {k}_{l}}\left( {x,{t}_{l}{h}_{l},\cdots ,{t}_{1}{h}_{1}}\right)\n\]\n\n\[ \n= {... | Yes |
Lemma 22.13.3 Suppose\n\n\[ g\left( {x + {zh}}\right) = g\left( x\right) + \mathop{\sum }\limits_{{m = 1}}^{\infty }{b}_{m}\left( {x, h}\right) {z}^{m} \]\n\nfor all \( z \) small enough. Then\n\n\[ {b}_{1}\left( {x,{h}_{1} + {h}_{2}}\right) = {b}_{1}\left( {x,{h}_{1}}\right) + {b}_{1}\left( {x,{h}_{2}}\right) . \] | Proof: Recall that\n\n\[ f\left( {x + \mathop{\sum }\limits_{{m = 1}}^{l}{z}_{m}{h}_{m}}\right) = \mathop{\sum }\limits_{{{k}_{l} = 0}}^{\infty }\cdots \mathop{\sum }\limits_{{{k}_{1} = 0}}^{\infty }{a}_{{k}_{1}\cdots {k}_{l}}\left( {x,{h}_{l},\cdots ,{h}_{1}}\right) {z}_{1}^{{k}_{1}}\cdots {z}_{l}^{{k}_{k}} \]\n\nand ... | Yes |
Lemma 22.13.4 Suppose \( a\left( {x,{h}_{l},\cdots ,{h}_{1}}\right) \) is multilinear, \( \left( {{h}_{i} \rightarrow a\left( {x,{h}_{l},\cdots ,{h}_{1}}\right) }\right. \) is linear), \[ \begin{Vmatrix}{a\left( {x,{h}_{l},\cdots ,{h}_{1}}\right) }\end{Vmatrix} \leq C\mathop{\prod }\limits_{{m = 1}}^{l}\begin{Vmatrix}{... | Proof: If \( l = 1 \), the conclusion is obvious and is nothing more than the definition of the derivative. \[ f\left( {x + h}\right) - f\left( x\right) - a\left( {x, h}\right) = o\left( {\parallel h\parallel }\right) \] and so from the definition of the derivative, \( a\left( {x, h}\right) = {Df}\left( x\right) h \) .... | Yes |
Theorem 22.13.6 Let \( X \) and \( Y \) be two complex Banach spaces and let \( U \) be an open set in \( X \). Then \( f : U \rightarrow Y \) is analytic on \( U \) if and only if \( {Df}\left( x\right) \) exists for each \( x \in U \) and in this case, \( f \in {C}^{\infty }\left( U\right) \), and if \( h \in X \), t... | Proof: We know\n\n\[ f\left( {x + {zh}}\right) = f\left( x\right) + \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}\left( {x, h}\right) {z}^{n}. \]\n\nDifferentiating, we obtain\n\n\[ {D}^{k}f\left( {x + {zh}}\right) {h}^{k} = k!{a}_{k}\left( {x, h}\right) + \mathop{\sum }\limits_{{n = k + 1}}^{\infty }n\left( {n - 1}... | No |
Lemma 22.14.1 The spaces \( X \) and \( Y \) with the given norms are Banach spaces and if \( L : X \rightarrow Y \) is defined as \( {L\phi }\left( t\right) = {\phi }^{\prime }\left( t\right) \) for all \( t \in {B}_{1} \), then \( L \) is one to one, onto and continuous. | Proof: It is clear that \( X \) and \( Y \) are both normed linear spaces. It remains to show they are Banach spaces. Suppose \( \left\{ {\phi }_{n}\right\} \) is a Cauchy sequence in \( X \) . Then \( {\phi }_{n} \rightarrow \phi \) uniformly and \( {\phi }_{n}^{\prime } \rightarrow \psi \) uniformly where \( \psi \) ... | Yes |
Define \( \partial U \) to be those points \( \mathbf{x} \) with the property that for every \( r > 0, B\left( {\mathbf{x}, r}\right) \) contains points of \( U \) and points of \( {U}^{C} \) . Then for \( U \) an open set, \[ \partial U = \bar{U} \smallsetminus U \] | Proof: First consider claim 23.0.1. Let \( \mathbf{x} \in \bar{U} \smallsetminus U \) . If \( B\left( {\mathbf{x}, r}\right) \) contains no points of \( U \), then \( \mathbf{x} \notin \bar{U} \) . If \( B\left( {\mathbf{x}, r}\right) \) contains no points of \( {U}^{C} \), then \( \mathbf{x} \in U \) and so \( \mathbf... | Yes |
Lemma 23.0.4 \( \mathbf{f} \in C\left( {\bar{\Omega } \times \left\lbrack {a, b}\right\rbrack ;{\mathbb{R}}^{p}}\right) \) if and only if \( t \rightarrow \mathbf{f}\left( {\cdot, t}\right) \) is in \( C\left( {\left\lbrack {a, b}\right\rbrack ;C\left( {\bar{\Omega };{\mathbb{R}}^{p}}\right) }\right) \) . | Proof: \( \Rightarrow \) By uniform continuity, if \( \varepsilon > 0 \) there is \( \delta > 0 \) such that if \( \left| {t - s}\right| < \delta \) , then for all \( \mathbf{x} \in \bar{\Omega },\parallel \mathbf{f}\left( {\mathbf{x}, t}\right) - \mathbf{f}\left( {\mathbf{x}, s}\right) \parallel < \frac{\varepsilon }{... | Yes |
Corollary 23.1.1 If \( f \in C\left( {\left\lbrack {a, b}\right\rbrack ;X}\right) \) where \( X \) is a normed linear space, then there exists a sequence of polynomials which converge uniformly to \( f \) on \( \left\lbrack {a, b}\right\rbrack \) . | The polynomials are of the form\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{m}\left( \begin{matrix} m \\ k \end{matrix}\right) {\left( {l}^{-1}\left( t\right) \right) }^{k}{\left( 1 - {l}^{-1}\left( t\right) \right) }^{m - k}f\left( {l\left( \frac{k}{m}\right) }\right) \]\n\nwhere \( l \) is a linear one to one and onto map... | Yes |
Lemma 23.1.4 Let \( \mathbf{g} \in {C}^{\infty }\left( {{\mathbb{R}}^{p};{\mathbb{R}}^{p}}\right) \) and let \( {\left\{ {\mathbf{y}}_{i}\right\} }_{i = 1}^{\infty } \) be points of \( {\mathbb{R}}^{p} \) and let \( \eta > 0 \) . Then there exists \( \mathbf{e} \) with \( \parallel \mathbf{e}\parallel < \eta \) and \( ... | Proof: Let \( S = \left\{ {\mathbf{x} \in {\mathbb{R}}^{p} : \det D\mathbf{g}\left( \mathbf{x}\right) = 0}\right\} \) . By Sard’s lemma, \( \mathbf{g}\left( S\right) \) has measure zero. Let \( N \equiv { \cup }_{i = 1}^{\infty }\left( {\mathbf{g}\left( S\right) - {\mathbf{y}}_{i}}\right) \) . Thus \( N \) has measure ... | Yes |
Lemma 23.1.5 Let \( \mathbf{f} \in C\left( {\bar{\Omega };{\mathbb{R}}^{p}}\right) \) and let \( {\left\{ {\mathbf{y}}_{i}\right\} }_{i = 1}^{\infty } \) be points not in \( \mathbf{f}\left( {\partial \Omega }\right) \) and let \( \delta > 0 \) . Then there exists \( \mathbf{g} \in {C}^{\infty }\left( {\bar{\Omega };{\... | Proof: Pick \( \widetilde{\mathbf{g}} \in {C}^{\infty }\left( {\bar{\Omega };{\mathbb{R}}^{p}}\right) ,\parallel \widetilde{\mathbf{g}} - \mathbf{f}{\parallel }_{\infty ,\bar{\Omega }} < \delta \) . \( \mathbf{g} \equiv \widetilde{\mathbf{g}} - \mathbf{e} \) where \( \mathbf{e} \) is from the above Lemma 23.1.4 and \( ... | Yes |
Lemma 23.1.7 Suppose \( \mathbf{g},\widehat{\mathbf{g}} \) both satisfy the above definition, \[ \operatorname{dist}\left( {\mathbf{f}\left( {\partial \Omega }\right) ,\mathbf{y}}\right) > \delta > \parallel \mathbf{f} - \mathbf{g}{\parallel }_{\infty ,\overline{\Omega }},\operatorname{dist}\left( {\mathbf{f}\left( {\p... | Proof: From the triangle inequality, if \( t \in \left\lbrack {0,1}\right\rbrack \) , \[ \parallel \mathbf{f} - \left( {t\mathbf{g} + \left( {1 - t}\right) \widehat{\mathbf{g}}}\right) {\parallel }_{\infty } \leq t\parallel \mathbf{f} - \mathbf{g}{\parallel }_{\infty } + \left( {1 - t}\right) \parallel \mathbf{f} - \wi... | Yes |
Lemma 23.1.9 Let \( K \) be a compact set and \( C \) a closed set in \( {\mathbb{R}}^{p} \) such that \( K \cap C = \) \( \varnothing \) . Then \[ \operatorname{dist}\left( {K, C}\right) \equiv \inf \{ \parallel \mathbf{k} - \mathbf{c}\parallel : \mathbf{k} \in K,\mathbf{c} \in C\} > 0. \] | Proof: Let \[ d \equiv \inf \{ \parallel \mathbf{k} - \mathbf{c}\parallel : \mathbf{k} \in K,\mathbf{c} \in C\} \] Let \( \left\{ {\mathbf{k}}_{i}\right\} ,\left\{ {\mathbf{c}}_{i}\right\} \) be such that \[ d + \frac{1}{i} > \begin{Vmatrix}{{\mathbf{k}}_{i} - {\mathbf{c}}_{i}}\end{Vmatrix} \] Since \( K \) is compact,... | Yes |
1. (homotopy invariance) If \( \mathbf{h} \in C\left( {\bar{\Omega } \times \left\lbrack {0,1}\right\rbrack ,{\mathbb{R}}^{p}}\right) \) and \( \mathbf{y}\left( t\right) \notin \mathbf{h}\left( {\partial \Omega, t}\right) \) for all \( t \in \left\lbrack {0,1}\right\rbrack \) where \( \mathbf{y} \) is continuous, then\... | Consider 1., the first property about homotopy. This follows from Theorem 23.2.1 applied to \( H\left( {\mathbf{x}, t}\right) \equiv \mathbf{h}\left( {\mathbf{x}, t}\right) - \mathbf{y}\left( t\right) \) . | Yes |
1. If \( \mathbf{y} \notin \mathbf{f}\left( {\bar{\Omega } \smallsetminus {\Omega }_{1}}\right) \) and \( {\Omega }_{1} \) is an open subset of \( \Omega \), then \( d\left( {\mathbf{f},\Omega ,\mathbf{y}}\right) = d\left( {\mathbf{f},{\Omega }_{1},\mathbf{y}}\right) \) . | Proof: Consider 1. You can take \( {\Omega }_{2} = \varnothing \) in 2 of Theorem 23.2 .2 or you can modify the proof of 2 slightly. | No |
Theorem 23.3.3 (Borsuk) Let \( \mathbf{f} \in C\left( {\bar{\Omega };{\mathbb{R}}^{p}}\right) \) be odd and let \( \Omega \) be symmetric with \( \mathbf{0} \notin \mathbf{f}\left( {\partial \Omega }\right) \) . Then \( d\left( {\mathbf{f},\Omega ,\mathbf{0}}\right) \) equals an odd integer. | Proof: Let \( {\psi }_{n} \) be a mollifier which is symmetric, \( \psi \left( {-\mathbf{x}}\right) = \psi \left( \mathbf{x}\right) \) . Also recall that \( \mathbf{f} \) is the restriction to \( \bar{\Omega } \) of a continuous function, still denoted as \( \mathbf{f} \) which is defined on all of \( {\mathbb{R}}^{p} ... | Yes |
Lemma 23.4.2 Let \( \mathrm{g} : \overline{B\left( {\mathbf{0}, r}\right) } \rightarrow {\mathbb{R}}^{p} \) be one to one and continuous where here \( B\left( {\mathbf{0}, r}\right) \) is the ball centered at \( \mathbf{0} \) of radius \( r \) in \( {\mathbb{R}}^{p} \) . Then there exists \( \delta > 0 \) such that\n\n... | Proof: For \( t \in \left\lbrack {0,1}\right\rbrack \), let\n\n\[ \mathbf{h}\left( {\mathbf{x}, t}\right) \equiv \mathbf{g}\left( \frac{\mathbf{x}}{1 + t}\right) - \mathbf{g}\left( \frac{-t\mathbf{x}}{1 + t}\right) \]\n\nThen for \( \mathbf{x} \in \partial B\left( {\mathbf{0}, r}\right) ,\mathbf{h}\left( {\mathbf{x}, t... | Yes |
Theorem 23.4.3 (invariance of domain) Let \( \Omega \) be any open subset of \( {\mathbb{R}}^{p} \) and let \( \mathbf{f} : \Omega \rightarrow {\mathbb{R}}^{p} \) be continuous and locally one to one. Then \( \mathbf{f} \) maps open subsets of \( \Omega \) to open sets in \( {\mathbb{R}}^{p} \) . | Proof: Let \( \overline{B\left( {{\mathbf{x}}_{0}, r}\right) } \subseteq \Omega \) where \( \mathbf{f} \) is one to one on \( \overline{B\left( {{\mathbf{x}}_{0}, r}\right) } \) . Let \( \mathbf{g} \) be defined on \( \overline{B\left( {\mathbf{0}, r}\right) } \) given by\n\n\[ \mathbf{g}\left( \mathbf{x}\right) \equiv... | Yes |
Corollary 23.4.4 If \( p > m \) there does not exist a continuous one to one map from \( {\mathbb{R}}^{p} \) to \( {\mathbb{R}}^{m} \) . | Proof: Suppose not and let \( \mathbf{f} \) be such a continuous map,\n\n\[ \mathbf{f}\left( \mathbf{x}\right) \equiv {\left( {f}_{1}\left( \mathbf{x}\right) ,\cdots ,{f}_{m}\left( \mathbf{x}\right) \right) }^{T}. \]\n\nThen let \( \mathbf{g}\left( \mathbf{x}\right) \equiv {\left( {f}_{1}\left( \mathbf{x}\right) ,\cdot... | Yes |
Corollary 23.4.5 If \( \mathbf{f} \) is locally one to one and continuous, \( \mathbf{f} : {\mathbb{R}}^{p} \rightarrow {\mathbb{R}}^{p} \), and\n\n\[ \mathop{\lim }\limits_{{\left| \mathbf{x}\right| \rightarrow \infty }}\left| {\mathbf{f}\left( \mathbf{x}\right) }\right| = \infty \]\n\nthen \( \mathbf{f} \) maps \( {\... | Proof: By the invariance of domain theorem, \( \mathbf{f}\left( {\mathbb{R}}^{p}\right) \) is an open set. It is also true that \( \mathbf{f}\left( {\mathbb{R}}^{p}\right) \) is a closed set. Here is why. If \( \mathbf{f}\left( {\mathbf{x}}_{k}\right) \rightarrow \mathbf{y} \), the growth condition ensures that \( \lef... | Yes |
Theorem 23.4.6 (Brouwer fixed point) Let \( B = \overline{B\left( {\mathbf{0}, r}\right) } \subseteq {\mathbb{R}}^{p} \) and let \( \mathbf{f} : B \rightarrow B \) be continuous. Then there exists a point \( \mathbf{x} \in B \), such that \( \mathbf{f}\left( \mathbf{x}\right) = \mathbf{x} \) . | Proof: Assume there is no fixed point. Consider \( \mathbf{h}\left( {\mathbf{x}, t}\right) \equiv t\mathbf{f}\left( \mathbf{x}\right) - \mathbf{x} \) for \( t \in \left\lbrack {0,1}\right\rbrack \) . Then for \( \parallel \mathbf{x}\parallel = r \) ,\n\n\[ \mathbf{0} \notin t\mathbf{f}\left( \mathbf{x}\right) - \mathbf... | Yes |
Theorem 23.4.9 There does not exist a retract of \( \overline{B\left( {\mathbf{0}, r}\right) } \) onto its boundary, \( \partial B\left( {\mathbf{0}, r}\right) \) . | Proof: Suppose \( \mathbf{f} \) were such a retract. Then for all \( \mathbf{x} \in \partial B\left( {\mathbf{0}, r}\right) ,\mathbf{f}\left( \mathbf{x}\right) = \mathbf{x} \) and so from the properties of the degree, the one which says if two functions agree on \( \partial \Omega \), then they have the same degree,\n\... | Yes |
Theorem 23.4.10 Let \( \Omega \) be a symmetric open set in \( {\mathbb{R}}^{p} \) such that \( \mathbf{0} \in \Omega \) and let \( \mathbf{f} : \partial \Omega \rightarrow V \) be continuous where \( V \) is an \( m \) dimensional subspace of \( {\mathbb{R}}^{p}, m < p \) . Then \( \mathbf{f}\left( {-\mathbf{x}}\right... | Proof: Suppose not. Using the Tietze extension theorem on components of the function, extend \( \mathbf{f} \) to all of \( {\mathbb{R}}^{p},\mathbf{f}\left( \bar{\Omega }\right) \subseteq V \) . (Here the extended function is also denoted by \( \mathbf{f} \) .) Let \( \mathbf{g}\left( \mathbf{x}\right) = \mathbf{f}\lef... | Yes |
Theorem 23.4.11 Let \( n \) be odd and let \( \Omega \) be an open bounded set in \( {\mathbb{R}}^{p} \) with \( \mathbf{0} \in \Omega \) . Suppose \( \mathbf{f} : \partial \Omega \rightarrow {\mathbb{R}}^{p} \smallsetminus \{ \mathbf{0}\} \) is continuous. Then for some \( \mathbf{x} \in \partial \Omega \) and \( \lam... | Proof: Using the Tietze extension theorem, extend \( \mathbf{f} \) to all of \( {\mathbb{R}}^{p} \) . Also denote the extended function by \( \mathbf{f} \) . Suppose for all \( \mathbf{x} \in \partial \Omega ,\mathbf{f}\left( \mathbf{x}\right) \neq \lambda \mathbf{x} \) for all \( \lambda \in \mathbb{R} \) . Then\n\n\[... | Yes |
Lemma 23.5.1 Let \( {\left\{ {K}_{i}\right\} }_{i = 1}^{\infty } \) be the connected components of \( {\mathbb{R}}^{p} \smallsetminus C \) where \( C \) is a closed set. Then \( \partial {K}_{i} \subseteq C \) . | Proof: Since \( {K}_{i} \) is a connected component of an open set, it is itself open. See Theorem 7.13.10. Thus \( \partial {K}_{i} \) consists of all limit points of \( {K}_{i} \) which are not in \( {K}_{i} \) . Let \( \mathbf{p} \) be such a point. If it is not in \( C \) then it must be in some other \( {K}_{j} \)... | No |
Theorem 23.5.3 (product formula) Let \( {\left\{ {K}_{i}\right\} }_{i = 1}^{\infty } \) be the bounded components of \( {\mathbb{R}}^{p} \smallsetminus \) \( \mathbf{f}\left( {\partial \Omega }\right) \) for \( \mathbf{f} \in C\left( {\bar{\Omega };{\mathbb{R}}^{p}}\right) \), let \( \mathbf{g} \in C\left( {{\mathbb{R}... | Proof: The compact set \( \mathbf{f}\left( \bar{\Omega }\right) \cap {\mathbf{g}}^{-1}\left( \mathbf{y}\right) \) is contained in \( {\mathbb{R}}^{p} \smallsetminus \mathbf{f}\left( {\partial \Omega }\right) \) and so, \( \mathbf{f}\left( \bar{\Omega }\right) \cap {\mathbf{g}}^{-1}\left( \mathbf{y}\right) \) is covered... | Yes |
Proposition 23.5.4 Let \( H \) be a compact set and let \( \mathbf{f} : H \rightarrow {\mathbb{R}}^{p}, p \geq 2 \) be one to one and continuous so that \( H \) and \( \mathbf{f}\left( H\right) \equiv C \) are homeomorphic. Suppose \( {H}^{C} \) has only one connected component so \( {H}^{C} \) is connected. Then \( {C... | Proof: I want to show that \( {C}^{C} \) has no bounded components so suppose it has a bounded component \( K \) . Extend \( \mathbf{f} \) to all of \( {\mathbb{R}}^{p} \) and let \( \mathbf{g} \) be an extension of \( {\mathbf{f}}^{-1} \) to all of \( {\mathbb{R}}^{p} \) . Then, by the above Lemma 23.5.1, \( \partial ... | Yes |
Proposition 23.5.5 Let \( B \) be the ball \( B\left( {\mathbf{0},1}\right) \) with \( {S}^{p - 1} \) its boundary, \( p \geq 2 \) . Suppose \( \mathbf{f} : {S}^{p - 1} \rightarrow C \equiv \mathbf{f}\left( {S}^{p - 1}\right) \subseteq {\mathbb{R}}^{p} \) is a homeomorphism. Then \( {C}^{C} \) also has exactly two comp... | Proof: I need to show there is only one bounded component of \( {C}^{C} \) . From Proposition 23.5.4, there is at least one. Otherwise, \( {S}^{p - 1} \) would have no bounded components which is obviously false.\n\nLet \( \mathbf{f} \) be a continuous extension of \( \mathbf{f} \) off \( {S}^{p - 1} \) and let \( \mat... | Yes |
Theorem 23.5.6 Let \( {S}^{p - 1} \) be the unit sphere in \( {\mathbb{R}}^{p}, p \geq 2 \) . Suppose \( \gamma : {S}^{p - 1} \rightarrow \) \( \Gamma \subseteq {\mathbb{R}}^{p} \) is one to one onto and continuous. Then \( {\mathbb{R}}^{p} \smallsetminus \Gamma \) consists of two components, a bounded component (calle... | Proof: \( {\gamma }^{-1} \) is continuous since \( {S}^{p - 1} \) is compact and \( \gamma \) is one to one. By the Jordan separation theorem, \( {\mathbb{R}}^{p} \smallsetminus \Gamma = {U}_{o} \cup {U}_{i}\; \) where these on the right are the connected components of the set on the left, both open sets. Only one of t... | Yes |
Corollary 23.5.7 Let \( B \) be an open ball and let \( \gamma : \bar{B} \rightarrow {\mathbb{R}}^{p} \) be one to one and continuous. Let \( {U}_{i},{U}_{o} \) be as in the above theorem, the bounded and unbounded components of \( \gamma {\left( \partial B\right) }^{C} \) . Then \( {U}_{i} = \gamma \left( B\right) \) ... | Proof: By connectedness and the observation that \( \gamma \left( B\right) \) contains no points of \( C \equiv \mathbf{\gamma }\left( {\partial B}\right) \), it follows that \( \mathbf{\gamma }\left( B\right) \subseteq {U}_{i} \) or \( {U}_{o} \) . Suppose \( \mathbf{\gamma }\left( B\right) \subseteq {U}_{i} \) . I wa... | Yes |
Lemma 23.6.1 Let \( \Omega \) be a bounded open set in \( {\mathbb{R}}^{p},\mathbf{f} \in C\left( {\bar{\Omega };{\mathbb{R}}^{p}}\right) \), and suppose \( {\left\{ {\Omega }_{i}\right\} }_{i = 1}^{\infty } \) are disjoint open sets contained in \( \Omega \) such that\n\n\[ \mathbf{y} \notin \mathbf{f}\left( {\bar{\Om... | Proof: By assumption, the compact set \( {\mathbf{f}}^{-1}\left( \mathbf{y}\right) \equiv \{ \mathbf{x} \in \bar{\Omega } : \mathbf{f}\left( \mathbf{x}\right) = \mathbf{y}\} \) has empty intersection with\n\n\[ \bar{\Omega } \smallsetminus { \cup }_{j = 1}^{\infty }{\Omega }_{j} \]\n\nand so this compact set is covered... | Yes |
Lemma 23.6.2 If a component \( L \) of \( {\mathbb{R}}^{p} \smallsetminus \mathbf{f}\left( C\right) \) intersects a component \( H \) of \( {\mathbb{R}}^{p} \smallsetminus \) \( \mathbf{f}\left( {\partial K}\right), K \) a component of \( {\mathbb{R}}^{p} \smallsetminus C \), then \( L \subseteq H \) . | Proof: First note that by Lemma 23.5.1 for \( K \in \mathcal{K},\partial K \subseteq C \). Next suppose \( L \in \mathcal{L} \) and \( L \cap H \neq \varnothing \) where \( \mathcal{H} \) is as described above. Suppose \( \mathbf{y} \in L \smallsetminus H \) so \( L \) is not contained in \( H \). Since \( \mathbf{y} \... | Yes |
Proposition 23.6.4 Let \( \Omega \) be an open connected bounded set in \( {\mathbb{R}}^{p}, p \geq 1 \) such that \( {\mathbb{R}}^{p} \smallsetminus \partial \Omega \) consists of two, three if \( n = 1 \), connected components. Let \( \mathbf{f} \in C\left( {\bar{\Omega };{\mathbb{R}}^{p}}\right) \) be continuous and... | Proof: First suppose \( n \geq 2 \) . By the Jordan separation theorem, \( {\mathbb{R}}^{p} \smallsetminus \mathbf{f}\left( {\partial \Omega }\right) \) consists of two components, a bounded component \( B \) and an unbounded component \( U \) . Using the Tietze extention theorem, there exists \( \mathbf{g} \) defined ... | Yes |
Lemma 23.7.1 Let \( \mathbf{h} : {\mathbb{R}}^{p} \rightarrow {\mathbb{R}}^{p} \) be a homeomorphism. Then\n\n\[ 1 = d\left( {\mathbf{h},\Omega ,\mathbf{h}\left( \mathbf{y}\right) }\right) d\left( {{\mathbf{h}}^{-1},\mathbf{h}\left( \Omega \right) ,\mathbf{y}}\right) \]\n\nwhenever \( \mathbf{y} \notin \partial \Omega ... | Proof: It is known that \( \mathbf{y} \notin \partial \Omega \) . Let \( \mathcal{L} \) be the components of \( {\mathbb{R}}^{p} \smallsetminus \mathbf{h}\left( {\partial \Omega }\right) \) . Thus \( \mathbf{h}\left( \mathbf{y}\right) \notin \mathbf{h}\left( {\partial \Omega }\right) \) . The product formula gives\n\n\... | Yes |
Lemma 23.7.3 The following formula holds\n\n\[ d\left( {{\mathbf{h}}_{-\mathbf{y}} \circ \mathbf{g} \circ {\mathbf{h}}_{\mathbf{z}},{\mathbf{h}}_{-\mathbf{z}}\left( \Omega \right) ,{\mathbf{h}}_{-\mathbf{y}}\left( \mathbf{y}\right) }\right) = d\left( {\mathbf{g} \circ {\mathbf{h}}_{\mathbf{z}},{\mathbf{h}}_{-\mathbf{z}... | In other words, \( d\left( {\mathbf{g},\Omega ,\mathbf{y}}\right) = d\left( {\mathbf{g}\left( {\cdot + \mathbf{z}}\right) ,\Omega - \mathbf{z},\mathbf{y}}\right) = d\left( {\mathbf{g}\left( {\cdot + \mathbf{z}}\right) - \mathbf{y},\Omega - \mathbf{z},\mathbf{0}}\right) \) . | Yes |
Theorem 23.7.4 You have a function \( d \) which has integer values \( d\left( {\mathbf{g},\Omega ,\mathbf{y}}\right) \in \mathbb{Z} \) whenever \( \mathbf{y} \notin \mathbf{g}\left( {\partial \Omega }\right) \) for \( \mathbf{g} \in C\left( {\bar{\Omega };{\mathbb{R}}^{p}}\right) \) . Assume it satisfies the following... | Proof: First note that \( \mathbf{h} \rightarrow d\left( {\mathbf{h},\Omega ,\mathbf{y}}\right) \) is continuous on \( C\left( {\bar{\Omega },{\mathbb{R}}^{p}}\right) \) . Say \( \parallel \mathbf{g} - \mathbf{h}{\parallel }_{\infty } < \) \( \operatorname{dist}\left( {\mathbf{h}\left( {\partial \Omega }\right) ,\mathb... | Yes |
Theorem 23.8.1 Let \( \Omega \) be a bounded open set in \( {\mathbb{R}}^{n} \) and let \( \mathbf{f} \in C\left( {\bar{\Omega };{\mathbb{R}}_{m}^{n}}\right) \) where \( {\mathbb{R}}_{m}^{n} = \left\{ {\mathbf{x} \in {\mathbb{R}}^{n} : {x}_{k} = 0\text{for}k > m}\right\} \) . Thus \( \mathbf{x} \) concludes with a colu... | Proof: To save space, let \( \mathbf{g} = \operatorname{id} - \mathbf{f} \) . Then there is no loss of generality in assuming at the outset that \( \mathbf{y} \) is a regular value for \( \mathbf{g} \) . Indeed, everything above was reduced to this case. Then for \( \mathbf{x} \in {\mathbf{g}}^{-1}\left( \mathbf{y}\rig... | Yes |
Theorem 23.8.3 Let \( \Omega \) be an open bounded set in \( V \) a real normed \( n \) dimensional vector space. Then there exists a topological degree \( d\left( {f,\Omega, y}\right) \) for \( f \in C\left( {\bar{\Omega }, V}\right), y \notin \) \( f\left( {\partial \Omega }\right) \) which satisfies all the properti... | Proof: There is an isomorphism \( \theta : {\mathbb{R}}^{n} \rightarrow V \) which also preserves all topological properties. This follows from the properties of finite dimensional vector spaces. In fact, every algebraic isomorphism is automatically a homeomorphism preserving all topological properties. Then it is pret... | Yes |
Theorem 23.8.4 Let \( \Omega \) be a bounded open set in \( V \) an \( n \) dimensional normed linear space and let \( f \in C\left( {\bar{\Omega };{V}_{m}}\right) \) where \( {V}_{m} \) is an \( m \) dimensional subspace. Let \( y \in {V}_{m} \smallsetminus \left( {I - f}\right) \left( {\partial \Omega }\right) \) . T... | Proof: Letting \( \left\{ {{v}_{1},\cdots ,{v}_{m}}\right\} \) be a basis for \( {V}_{m} \), let a basis for \( V \) be\n\n\[ \left\{ {{v}_{1},\cdots ,{v}_{m},{v}_{m + 1},\cdots ,{v}_{n}}\right\} \]\n\nLet \( \theta \) be the isomorphism which satisfies \( \theta {\mathbf{e}}_{i} = {v}_{i} \) where the \( {\mathbf{e}}_... | Yes |
Theorem 23.9.6 Let \( D \) be the Leray Schauder degree just defined and let \( \Omega \) be a bounded open set \( y \notin \left( {I - F}\right) \left( {\partial \Omega }\right) \) where \( F \) is always a compact mapping. Then the following properties hold:\n\n1. \( D\left( {I,\Omega, y}\right) = 1 \)\n\n2. If \( {\... | Proof: The mapping \( x \rightarrow 0 \) is clearly compact. Then an approximating sequence is \( {F}_{k},{F}_{k}x = 0 \) for all \( k \) . Then\n\n\[ D\left( {I,\Omega, y}\right) = \mathop{\lim }\limits_{{k \rightarrow \infty }}d\left( {{\left. I\right| }_{{V}_{k}},\Omega \cap {V}_{k}, y}\right) = 1 \]\n\nFor the seco... | Yes |
Theorem 23.9.7 Let \( B = \overline{B\left( {0, r}\right) } \) and let \( F : B \rightarrow B \) be compact. Then \( F \) has a fixed point. | Proof: Suppose it does not. Then consider \( D\left( {I - {tF}, B\left( {0, r}\right) ,0}\right) \) . If \( t = 1 \) , \( 0 \notin \left( {I - {tF}}\right) \left( {\partial B}\right) \) since otherwise, there would be a fixed point. If \( t < 1 \) there is no point of \( \partial B \) which \( I - {tF} \) sends to 0 be... | Yes |
Theorem 23.9.8 Let \( K \) be a closed bounded convex subset of a Banach space \( X \) and suppose \( F : K \rightarrow K \) is compact. Then \( F \) has a fixed point. | Proof: By Theorem 16.2.5, \( K \) is a retract. Thus there is a continuous function \( R : X \rightarrow K \) which leaves points of \( K \) unchanged. Then you consider \( F \circ R \) . It is still a compact mapping obviously. Let \( B\left( {0, r}\right) \) be so large that it contains \( K \) . Then from the above ... | Yes |
Let \( g : \left\lbrack {0, T}\right\rbrack \times {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) be continuous. Let \( F : C\left( {\left\lbrack {0, T}\right\rbrack ;{\mathbb{R}}^{n}}\right) \rightarrow \) \( C\left( {\left\lbrack {0, T}\right\rbrack ;{\mathbb{R}}^{n}}\right) \) be given by \[ F\left( y\right) \left... | Proof: Let \( {r}_{M} \) be the radial projection in \( {\mathbb{R}}^{n} \) onto \( \overline{B\left( {0, M}\right) } \) . Then \( F \circ {r}_{M} \) is compact because \( \left| {g\left( {s,{r}_{M}y}\right) }\right| \) is bounded. It also maps into a compact subset of \( C\left( {\left\lbrack {0, T}\right\rbrack ;{\ma... | Yes |
Theorem 23.9.10 Let \( f : X \rightarrow X \) be a compact map. Then either\n\n1. There is a fixed point for \( {tf} \) for all \( t \in \left\lbrack {0,1}\right\rbrack \) or\n\n2. For every \( r > 0 \), there exists a solution to \( x = {tf}\left( x\right) \) for \( t \in \left( {0,1}\right) \) such that \( \parallel ... | Proof: Suppose there is \( {t}_{0} \in \left\lbrack {0,1}\right\rbrack \) such that \( {t}_{0}f \) has no fixed point. Then \( {t}_{0} \neq 0.{t}_{0}f \) obviously has a fixed point if \( {t}_{0} = 0 \) . Thus \( {t}_{0} \in (0,1\rbrack \) . Then let \( {r}_{M} \) be the radial retraction onto \( \overline{B\left( {0, ... | Yes |
Theorem 23.9.11 Let \( X \) be an infinite dimensional Banach space and let \( 0 \notin \partial \Omega \) where \( \Omega \) is an open bounded subset of \( X \) . Let \( F : \bar{\Omega } \rightarrow X \) be compact. Suppose that \( {Fx} \neq {\lambda x} \) for all \( x \in \partial \Omega \) and that \( 0 \notin \ov... | Proof: Recall that \( D\left( {I - F,\Omega ,0}\right) \equiv \mathop{\lim }\limits_{{k \rightarrow \infty }}d\left( {I - {F}_{k},\Omega \cap {V}_{k},0}\right) \) where \( {F}_{k} \) has values in a finite dimensional subspace \( {V}_{k} \) ,\n\n\[ \mathop{\sup }\limits_{{x \in \bar{\Omega }}}\begin{Vmatrix}{{F}_{k}\le... | Yes |
Corollary 23.9.12 Let \( X \) be an infinite dimensional Banach space. Let \( 0 \in {\Omega }_{0} \subseteq \) \( \Omega \) be two open sets. Let \( F : \bar{\Omega } \rightarrow X \) be a compact mapping which satisfies\n\n1. \( \parallel {Fx}\parallel \leq \parallel x\parallel \) for \( x \in \partial {\Omega }_{0} \... | Proof: First note that \( \overline{\Omega \smallsetminus {\Omega }_{0}} \) is like an annulus with both edges included. Suppose \( F \) does not have a fixed point in \( \overline{\Omega \smallsetminus {\Omega }_{0}} \) . What if \( t = 1 \) and \( x \in \partial \Omega \) ? Could \( 0 = \left( {I - F}\right) \left( x... | Yes |
Theorem 24.1.2 Let \( I \) be \( {C}^{1}, I \) is non constant, satisfy the Palais Smale condition, and \( {I}^{\prime } \) is Lipschitz continuous on bounded sets. Also suppose that \( c \in \mathbb{R} \) is such that either \( \left\lbrack {I\left( u\right) \in \left\lbrack {c - \delta, c + \delta }\right\rbrack }\ri... | Proof: Suppose \( \left\lbrack {I\left( u\right) \in \left\lbrack {c - \delta, c + \delta }\right\rbrack }\right\rbrack = \varnothing \) for some \( \delta > 0 \) . Then \( \left\lbrack {I\left( u\right) \leq c + \delta /2}\right\rbrack \subseteq \) \( \left\lbrack {I\left( u\right) \leq c - \delta /2}\right\rbrack \) ... | Yes |
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