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Proposition 12.1.8. The space \( {L}_{p} \) is finitely representable in \( {\ell }_{p} \) for \( 1 \leq p < \infty \) . | Proof. For each \( n \) let \( {E}_{n} = \left\lbrack {{\chi }_{\left( \left( k - 1\right) /{2}^{n}, k/{2}^{n}\right) } : 1 \leq k \leq {2}^{n}}\right\rbrack \) be the subspace spanned in \( {L}_{p} \) by the characteristic functions on dyadic intervals. Since \( {E}_{n} \) is isometrically isomorphic to a subspace of ... | Yes |
Proposition 12.1.10 \( \left( {L}_{q}\right. \) -subspaces of \( \left. {L}_{p}\right) \) . (i) For \( 1 \leq p \leq 2,{L}_{q} \) embeds in \( {L}_{p} \) if and only if \( p \leq q \leq 2 \) . (ii) For \( 2 < p < \infty ,{L}_{q} \) embeds in \( {L}_{p} \) if and only if \( q = 2 \) or \( q = p \) . Moreover, if \( {L}_... | Proof. Let \( 1 \leq p, q < \infty \) and suppose that \( {L}_{q} \) embeds in \( {L}_{p} \) . Then, since \( {\ell }_{q} \) embeds in \( {L}_{q} \), it follows that \( {\ell }_{q} \) embeds in \( {L}_{p} \) . This implies, by Theorem 6.4.18, that either \( q = p \), or \( q = 2 \), or \( 1 \leq p < q < 2 \) . It remai... | Yes |
Lemma 12.1.11. Let \( E \) be a finite-dimensional normed space and suppose \( {\left( {x}_{j}\right) }_{j = 1}^{N} \) is an \( \epsilon \) -net on the surface of the unit ball \( \{ e : \parallel e\parallel = 1\} \), where \( 0 < \epsilon < 1 \) . Suppose \( T : E \rightarrow X \) is a linear map such that \( 1 - \eps... | Proof. First suppose \( \parallel e\parallel = 1 \) . Pick \( j \) such that \( \begin{Vmatrix}{e - {x}_{j}}\end{Vmatrix} \leq \epsilon \) . Then\n\n\[ \parallel {Te}\parallel \leq \begin{Vmatrix}{{Te} - T{x}_{j}}\end{Vmatrix} + \left( {1 + \epsilon }\right) \]\n\nand so\n\n\[ \parallel T\parallel \leq \parallel T\para... | Yes |
Proposition 12.1.12. Let \( X, Y \) be infinite-dimensional Banach spaces.\n\n(i) The ultraproduct \( {X}_{\mathcal{U}} \) is finitely representable in \( X \) . | Proof. (i) Let \( E \) be a finite-dimensional subspace of \( {X}_{\mathcal{U}} \) and suppose \( \epsilon > 0 \) . We can (by selecting representatives for a basis in \( E \) ) suppose \( E \subset {\ell }_{\infty }\left( X\right) \) and that \( \parallel \cdot {\parallel }_{\mathcal{U}} \) is a norm on \( E \) . Choo... | Yes |
Theorem 12.1.15. Let \( X \) be a Banach space. Then\n\n(i) \( X \) fails to have type \( p > 1 \) if and only if \( {\ell }_{1} \) is finitely representable in \( X \) .\n\n(ii) \( X \) fails to have cotype \( q < \infty \) if and only if \( {\ell }_{\infty } \) is finitely representable in \( X \) . | Proof. We will use Proposition 7.2.5. For \( \left( i\right) \) it suffices to note that \( {\alpha }_{N}\left( X\right) = \sqrt{N} \) for every \( N \) . Thus for fixed \( N \) and all \( n \) we can find \( {\left( {x}_{nk}\right) }_{k = 1}^{N} \) such that\n\n\[{\left( \mathop{\sum }\limits_{{k = 1}}^{N}{\begin{Vmat... | No |
Proposition 12.1.18. If \( 1 < p < \infty \), then the space \( {\ell }_{p} \) is superreflexive. | Proof. Fix \( 1 < p < \infty \) . Let \( X \) be a Banach space that is finitely representable in \( {\ell }_{p} \) and let \( Y \) be a separable subspace of \( X \) . By transitivity, \( Y \) is finitely representable in \( {\ell }_{p} \), so that by Theorem 12.1.9, the space \( Y \) is isomorphic to a subspace of \(... | Yes |
Proposition 12.1.19. If a Banach space \( Y \) is crudely finitely representable in a superreflexive Banach space \( X \), then \( Y \) is superreflexive. | Proof. Let \( Z \) be a separable subspace of \( Y \) . By transitivity, \( Z \) is crudely finitely representable in \( X \) . Proposition 12.1.13 yields that \( Z \) is finitely representable in \( X \) equipped with an equivalent norm; hence \( Z \) is reflexive. Using again Lemma 12.1.17, we deduce that \( Y \) is ... | Yes |
Corollary 12.1.21. The space \( {L}_{p} \) is superreflexive for \( 1 < p < \infty \) . | Proof. We know that \( {L}_{p} \) is finitely representable in \( {\ell }_{p} \) and that \( {\ell }_{p} \) is superreflexive. Now just apply Proposition 12.1.19. | Yes |
Proposition 12.2.1. Let \( T : X \rightarrow Y \) be an operator with closed range. Suppose \( y \in \) \( Y \) is such that the equation \( {T}^{* * }{x}^{* * } = y \) has a solution \( {x}^{* * } \in {X}^{* * } \) with \( \begin{Vmatrix}{x}^{* * }\end{Vmatrix} < 1 \) . Then the equation \( {Tx} = y \) has a solution ... | Proof. This is almost immediate. We must show that \( y \in T\left( {U}_{X}\right) \), where \( {U}_{X} \) is the open unit ball of \( X \) .
First suppose \( y \notin T\left( X\right) \) . In this case there exists \( {y}^{ * } \in {Y}^{ * } \) with \( {T}^{ * }\left( {y}^{ * }\right) = 0 \) but \( {y}^{ * }\left( y\... | Yes |
Proposition 12.2.2. Let \( T : X \rightarrow Y \) be an operator with closed range and suppose \( K : X \rightarrow Y \) is a finite-rank operator. Then \( T + K \) also has closed range. | Proof. Suppose \( T + K \) does not have closed range. Then there is a bounded sequence \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) with \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {T + K}\right) \left( {x}_{n}\right) = 0 \) but \( d\left( {{x}_{n},\ker \left( {T + K}\right) }\right) \geq 1 \) for all... | Yes |
Theorem 12.2.3. Let \( X \) be a Banach space, \( A = {\left( {a}_{jk}\right) }_{j, k = 1}^{m, n} \) an \( m \times n \) real matrix, and \( B = {\left( {b}_{jk}\right) }_{j, k = 1}^{p, n} \) a \( p \times n \) real matrix. Let \( {y}_{1},\ldots ,{y}_{m} \in X,{y}_{1}^{ * },\ldots ,{y}_{p}^{ * } \in \) \( {X}^{ * } \),... | Proof. Consider the operator \( {T}_{0} : {\ell }_{\infty }^{n}\left( X\right) \rightarrow {\ell }_{\infty }^{m}\left( X\right) \) defined by\n\n\[ {T}_{0}\left( {{x}_{1},\ldots ,{x}_{n}}\right) = {\left( \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{jk}{x}_{k}\right) }_{j = 1}^{m}. \]\n\nWe claim that \( {T}_{0} \) has clos... | Yes |
Theorem 12.2.4 (The Principle of Local Reflexivity). Let \( X \) be a Banach space. Suppose that \( F \) is a finite-dimensional subspace of \( {X}^{* * } \) and \( G \) is a finite-dimensional subspace of \( {X}^{ * } \) . Then given \( \epsilon > 0 \) there exist a subspace \( E \) of \( X \) containing \( F \cap X \... | Proof. Given \( \epsilon > 0 \) let us take \( v > 0 \) such that \( \left( {1 + v}\right) {\left( 1 - 3v\right) }^{-1} < 1 + \epsilon \) and pick a \( v \) -net \( {\left( {x}_{j}^{* * }\right) }_{j = 1}^{N} \) in \( \left\{ {{x}^{* * } \in F : \begin{Vmatrix}{x}^{* * }\end{Vmatrix} = 1}\right\} \) . Let \( S : {\math... | Yes |
Proposition 12.3.6. Suppose \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) is a nonconstant spreading sequence in a Banach space X.\n\n(i) If \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) fails to be weakly Cauchy, then \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) is a basic sequence equivalent to the canonica... | Proof. (i) If \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) is not weakly Cauchy, then no subsequence can be weakly Cauchy (by the spreading property), and so by Rosenthal's theorem (Theorem 11.2.1), some subsequence is equivalent to the canonical \( {\ell }_{1} \) -basis; but then again, this means that the entire ... | Yes |
Lemma 12.3.10. Suppose \( \mathcal{X} \) is a spreading sequence space. Then there is a spreading sequence space \( \mathcal{Y} \) that is block finitely representable in \( \mathcal{X} \) such that the canonical basis of \( \mathcal{Y} \) is unconditional with unconditional basis constant \( {K}_{u} = 1 \) . | Proof. By the previous remarks we can assume that the canonical basis \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) of \( \mathcal{X} \) is 2-unconditional, and that \( \mathcal{X} \) is not isomorphic to \( {c}_{0} \) . Thus, if we let \( {y}_{n} = \) \( \mathop{\sum }\limits_{{j = 1}}^{n}{\left( -1\right) }^{j}{e}... | Yes |
Lemma 12.3.12. Suppose \( \mathcal{X} \) is a 1-unconditional spreading sequence space. Then\n\n(i) There exists a sequence \( {\left( {\xi }_{n}\right) }_{n = 1}^{\infty } \) in \( \mathcal{X}\left( {\mathbb{Q}}_{0}\right) \) with \( {\begin{Vmatrix}{\xi }_{n}\end{Vmatrix}}_{\mathcal{X}\left( {\mathbb{Q}}_{0}\right) }... | Proof. (i) Let us start by observing that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\begin{Vmatrix}{T}_{2}^{n}\end{Vmatrix}}^{\frac{1}{n}} = \mathop{\inf }\limits_{n}{\begin{Vmatrix}{T}_{2}^{n}\end{Vmatrix}}^{\frac{1}{n}} \]\n\n(12.3)\n\nand so\n\n\[ \begin{Vmatrix}{T}_{2}^{n}\end{Vmatrix} \geq {2}^{n\theta... | Yes |
Theorem 12.3.13 (Dvoretzky’s Theorem). The space \( {\ell }_{2} \) is finitely representable in every infinite-dimensional Banach space. | Proof. An immediate conclusion from Krivine’s theorem is that some \( {\ell }_{p}(1 \leq \) \( p < \infty \) ) or \( {c}_{0} \) is finitely representable in every infinite-dimensional Banach space \( X \) . In the case of \( {c}_{0} \) this implies that \( {\ell }_{\infty } \) is finitely representable in \( X \), and ... | Yes |
Theorem 13.1.1. Suppose \( \left( {X,\parallel \cdot {\parallel }_{X}}\right) \) is an n-dimensional normed space. Then there exists an n-dimensional ellipsoid of maximum volume contained in \( {B}_{X} \) . | Proof. Pick an ellipsoid \( {\mathcal{E}}_{n} \) contained in \( {B}_{X} \) . This ellipsoid corresponds to the image \( S\left( {B}_{{\ell }_{2}^{n}}\right) \) of some linear map \( S : {\ell }_{2}^{n} \rightarrow X \) . Let \( \parallel \cdot {\parallel }_{\mathrm{E}} \) be the Euclidean norm induced on \( X \) by \(... | Yes |
Lemma 13.1.2. Suppose \( \left( {X,\parallel \cdot {\parallel }_{X}}\right) \) is an \( n \) -dimensional normed space. Let \( x \in X \) be such that \( \parallel x{\parallel }_{X} = \parallel x{\parallel }_{{\mathcal{E}}_{X}} = 1 \) . Then \( \parallel x{\parallel }_{{X}^{ * }} = 1 \) . | Proof. We must see that \( \langle x, y\rangle \leq 1 \) whenever \( \parallel y{\parallel }_{X} \leq 1 \) . For \( t > 0 \) we have\n\n\[ \parallel \left( {1 + t}\right) x - {ty}{\parallel }_{{\mathcal{E}}_{X}} \geq \parallel \left( {1 + t}\right) x - {ty}{\parallel }_{X} \geq \left( {1 + t}\right) \parallel x{\parall... | Yes |
Lemma 13.1.3. Suppose that \( \left( {X,\parallel \cdot {\parallel }_{X}}\right) \) is an \( n \) -dimensional normed space and that \( T \in \mathcal{B}\left( X\right) \) . Then there is \( x \in X \) with \( \parallel x{\parallel }_{{\mathcal{E}}_{X}} = \parallel x{\parallel }_{X} = 1 \) such that\n\n\[ \left| {\oper... | Proof. For any \( t > 0 \) pick \( {x}_{t} \) such that \( {\begin{Vmatrix}{x}_{t}\end{Vmatrix}}_{{X}_{\mathcal{E}}} = 1 \) and\n\n\[ {\begin{Vmatrix}{x}_{t} + tT\left( {x}_{t}\right) \end{Vmatrix}}_{X} = \parallel I + {tT}{\parallel }_{{\mathcal{E}}_{X} \rightarrow X} \]\n\n(13.4)\n\nSince \( {\mathcal{E}}_{X} \) is a... | Yes |
Theorem 13.1.4. Suppose \( \left( {X,\parallel \cdot {\parallel }_{X}}\right) \) is an n-dimensional normed space. Then,\n\n\[{\pi }_{2}\left( {I}_{X \rightarrow {\mathcal{E}}_{X}}\right) \leq \sqrt{n}\] | Proof. Suppose \( {x}_{1},\ldots ,{x}_{k} \in X \) . Consider the operator \( T : X \rightarrow X \) given by \( T = \) \( \mathop{\sum }\limits_{{i = 1}}^{k}{x}_{i} \otimes {x}_{i} \), that is,\n\n\[{Tu} = \mathop{\sum }\limits_{{i = 1}}^{k}\left\langle {{x}_{i}, u}\right\rangle {x}_{i}\]\n\nWe note that\n\n\[\operato... | Yes |
Theorem 13.1.5 (John). If \( \left( {X,\parallel \cdot {\parallel }_{X}}\right) \) is an n-dimensional normed space, then \( d\left( {X,{\ell }_{2}^{n}}\right) \leq \sqrt{n} \) | Proof. We have \( \parallel I{\parallel }_{{\mathcal{E}}_{X} \rightarrow X} = 1 \) and \( \parallel I{\parallel }_{X \rightarrow {\mathcal{E}}_{X}} \leq {\pi }_{2}\left( {I}_{X \rightarrow {\mathcal{E}}_{X}}\right) \leq \sqrt{n} \) . Geometrically,\n\n\[ \n{\mathcal{E}}_{X} \subset {B}_{X} \subset \sqrt{n}{\mathcal{E}}... | Yes |
Proposition 13.1.6. If \( X = {\ell }_{\infty }^{n} \) (or \( X = {\ell }_{1}^{n} \) ), then \( d\left( {X,{\ell }_{2}^{n}}\right) = \sqrt{n} \). | Proof. Let \( S : {\ell }_{\infty }^{n} \rightarrow {\ell }_{2}^{n} \) be a linear isomorphism. Let \( D = \parallel S{\parallel }_{{\ell }_{\infty }^{n} \rightarrow {\ell }_{2}^{n}} \) and \( C = \) \( {\begin{Vmatrix}{S}^{-1}\end{Vmatrix}}_{{\ell }_{2}^{n} \rightarrow {\ell }_{\infty }^{n}} \) . We have\n\n\[ \n{C}^{... | Yes |
Theorem 13.1.7 (The Kadets-Snobar Theorem). Let \( X \) be a Banach space of dimension \( n \) . Then for every Banach space \( Y \) containing \( X \) as a subspace there is a projection \( P \) of \( Y \) onto \( X \) with \( \parallel P\parallel \leq \sqrt{n} \) . | Proof. According to Theorem 13.1.4, there is an operator \( S : X \rightarrow {\ell }_{2}^{n} \), where \( n = \) \( \dim X \), such that \( \begin{Vmatrix}{S}^{-1}\end{Vmatrix} = 1 \) and \( {\pi }_{2}\left( S\right) \leq \sqrt{n} \) . Using Theorem 8.2.13, we see that \( S \) extends to a bounded operator \( T : Y \r... | Yes |
Theorem 13.2.2 (The Concentration of Measure Phenomenon). Let \( f \) be a Lipschitz function on \( {\mathcal{S}}^{n - 1} \) with Lipschitz constant 1. Then for \( t > 0 \) , \[ {\sigma }_{n}\left( {\left| {f - \bar{f}}\right| > t}\right) \leq 4{e}^{-n{t}^{2}/{72}{\pi }^{2}}, \] where \[ \bar{f} = {\int }_{{\mathcal{S}... | Proof. We shall assume that \( \bar{f} = 0 \), and so \( \left| {f\left( x\right) }\right| \leq 1 \) for all \( x \in {\mathcal{S}}^{n - 1} \) . Let us first extend \( f \) to \( {\mathbb{R}}^{n} \smallsetminus \{ 0\} \) by putting \[ f\left( x\right) = \parallel x{\parallel }_{2}f\left( {x/\parallel x{\parallel }_{2}}... | Yes |
Lemma 13.3.1. Let \( \left( {{Y}_{0},\parallel \cdot \parallel }\right) \) be an m-dimensional Euclidean space. Suppose \( \epsilon > \) 0 . Then there is an \( \epsilon \) -net \( {\left\{ {x}_{j}\right\} }_{j = 1}^{N} \) for \( \left\{ {x \in {Y}_{0} : \parallel x\parallel = 1}\right\} \) with \( N \leq {\left( 1 + \... | Proof. Pick a maximal subset \( {\left\{ {x}_{j}\right\} }_{j = 1}^{N} \) of \( \left\{ {x \in {Y}_{0} : \parallel x\parallel = 1}\right\} \) with the property that \( \begin{Vmatrix}{{x}_{i} - {x}_{j}}\end{Vmatrix} \geq \epsilon \) whenever \( i \neq j \) . It is clear that this is an \( \epsilon \) -net. The open bal... | Yes |
Theorem 13.3.2. Suppose \( \parallel \cdot {\parallel }_{\mathrm{F}} \) is a norm on \( {\mathbb{R}}^{n} \) with \( \parallel x{\parallel }_{\mathrm{F}} \leq \parallel x{\parallel }_{2} \) for all \( x \in {\mathbb{R}}^{n} \) . (a) Suppose \( 0 < \epsilon < \frac{1}{3} \) . Then there is a \( k \) -dimensional subspace... | Proof. (a) Let us fix some \( k \) -dimensional subspace \( {Y}_{0} \) of \( {\mathbb{R}}^{n} \) and use Lemma 13.3.1 to pick an \( \epsilon /3 \) -net \( {\left\{ {x}_{j}\right\} }_{j = 1}^{N} \) for \( \left\{ {x \in {Y}_{0} : \parallel x{\parallel }_{2} = 1}\right\} \) with \( N \leq {\left( 1 + 6/\epsilon \right) }... | Yes |
Lemma 13.3.3. Let \( \parallel \cdot {\parallel }_{\mathrm{F}} \) be a norm on \( {\mathbb{R}}^{n} \) with \( \parallel x{\parallel }_{\mathrm{F}} \leq \parallel x{\parallel }_{2} \) for all \( x \) . Then:\n\n(a) \( {\theta }_{\mathrm{F}} \geq 1/\parallel I{\parallel }_{\mathrm{F} \rightarrow {\ell }_{2}^{n}} \) .\n\n... | Proof. Part (a) follows from \( 1 \leq \parallel I{\parallel }_{\mathrm{F} \rightarrow {\ell }_{2}^{n}}\parallel \xi {\parallel }_{\mathrm{F}} \) for all \( \xi \in {\mathcal{S}}^{n - 1} \) .\n\nSince \( \parallel G{\parallel }_{2} \) and \( G/\parallel G{\parallel }_{2} \) are independent random variables,\n\n\[ \math... | Yes |
Theorem 13.3.4. Suppose \( 1 \leq p < \infty \) and \( n \in \mathbb{N} \) . Then for \( 0 < \epsilon < 1/3 \), the space \( {\ell }_{p}^{n} \) contains a subspace \( {X}_{0} \) with \( \dim {X}_{0} = k \) and \( {d}_{{X}_{0}} \leq 1 + \epsilon \), provided:\n\n- \( k \leq c{n}^{2/p}{\epsilon }^{2}{\left| \log \epsilon... | Proof. We consider \( {\mathbb{R}}^{n} \) equipped with the norm \( \parallel \cdot {\parallel }_{\mathrm{F}} = \parallel \cdot {\parallel }_{p}\left( {1 \leq p < \infty }\right) \) and denote the corresponding \( {\theta }_{\mathrm{F}} \) simply by \( {\theta }_{p} \) .\n\nIf \( p > 2 \), by Hölder’s inequality we hav... | Yes |
Proposition 13.3.5 (The Dvoretzky-Rogers Lemma). Let \( \left( {X,\parallel \cdot {\parallel }_{X}}\right) \) be an \( n \) - dimensional normed space and suppose that \( \parallel \cdot {\parallel }_{{\mathcal{E}}_{X}} \) is the norm induced on \( X \) by the John ellipsoid. Then there is an orthonormal basis \( {\lef... | Proof. We must recall the definition of the John ellipsoid of \( X \) as the ellipsoid of maximal volume contained in \( {B}_{X} \) . We pick \( {\left( {e}_{j}\right) }_{j = 1}^{n} \) inductively such that \( {\begin{Vmatrix}{e}_{1}\end{Vmatrix}}_{{\mathcal{E}}_{X}} = \) \( {\begin{Vmatrix}{e}_{1}\end{Vmatrix}}_{X} = ... | Yes |
Lemma 13.3.6. There is an absolute constant \( c > 0 \) such that if \( G \) is an m-dimensional Gaussian, then\n\n\[ \mathbb{E}\left( {\parallel G{\parallel }_{\infty }}\right) \geq c{\left( \log m\right) }^{1/2},\;m = 2,3,\ldots \] | Proof. If \( g \) is a scalar Gaussian and \( s > 0 \), then\n\n\[ \mathbb{P}\left( {\left| g\right| > s}\right) = \sqrt{\frac{2}{\pi }}{\int }_{s}^{\infty }{e}^{-\frac{1}{2}{\xi }^{2}}{d\xi } \geq \sqrt{\frac{2}{\pi }}s{e}^{-2{s}^{2}}. \]\n\nThus\n\n\[ \mathbb{P}\left( {\parallel G{\parallel }_{\infty } > s}\right) \g... | Yes |
Theorem 13.3.7 (Dvoretzky’s Theorem). There is an absolute constant \( c > 0 \) with the following property: If \( \left( {X,\parallel \cdot {\parallel }_{X}}\right) \) is an n-dimensional normed space and \( 0 < \epsilon < 1/3 \), then \( X \) has a subspace \( {X}_{0} \) with \( \dim {X}_{0} = k \) and \( {d}_{{X}_{0... | Proof. Let \( \parallel \cdot {\parallel }_{{\mathcal{E}}_{X}} \) be the norm induced on \( X \) by the John ellipsoid. By the Dvoretzky-Rogers lemma, we can pass to a subspace \( Y \) of \( X \) with \( m = \dim Y \geq n/2 \), and with the property that \( \left( {Y,\parallel \cdot {\parallel }_{{\mathcal{E}}_{X}}}\ri... | Yes |
Lemma 13.4.4. Suppose \( E \) is a finite-dimensional subspace of \( X \) and let \( v > 0 \) . Then there is a finite-codimensional subspace \( Y \) of \( E \) such that\n\n\[ \parallel e + y\parallel \geq \left( {1 - v}\right) \parallel e\parallel ,\;e \in E, y \in Y. \] | Proof. Let \( {\left\{ {x}_{i}\right\} }_{i = 1}^{N} \) be a \( v \) -net for \( S = \{ x \in E : \parallel x\parallel = 1\} \) . We use the Hahn-Banach theorem to construct \( {\left\{ {x}_{i}^{ * }\right\} }_{i = 1}^{N} \) in \( {X}^{ * } \) such that \( \begin{Vmatrix}{x}_{i}^{ * }\end{Vmatrix} = 1 \) and \( {x}_{i}... | Yes |
Theorem 13.4.5 (Lindenstrauss and Tzafriri [200]). Suppose \( X \) is an infinite-dimensional Banach space in which every closed subspace is complemented. Then \( X \) is isomorphic to a Hilbert space. | Proof. By Theorem 13.4.3 we must prove that there is \( \lambda \geq 1 \) such that every finite-dimensional subspace of \( X \) is \( \lambda \) -complemented. Suppose that this is not the case and\n\nfor a finite-dimensional subspace \( E \) of \( X \), set,\n\n\[ \lambda \left( E\right) = \inf \{ \parallel P\paralle... | Yes |
Problem 14.1.1. When are two separable Banach spaces homeomorphic as topological spaces? | This question was the subject of intense research in the years after the Second World War and was resolved beautifully by Kadets (announced in 1965) [145]:\n\nTheorem 14.1.2. All sep | No |
Theorem 14.1.3 (Mazur and Ulam [217]). Let \( X \) and \( Y \) be real normed spaces and suppose that \( \Phi \) is an isometry mapping \( X \) onto \( Y \) that carries 0 into 0 . Then \( \Phi \) is a bounded linear operator. In particular, if two real Banach spaces \( X \) and \( Y \) are isometric as Banach spaces, ... | Proof. It suffices to show that \( \Phi \) preserves midpoints of line segments, i.e.,\n\n\[ \Phi \left( \frac{x + y}{2}\right) = \frac{\Phi \left( x\right) + \Phi \left( y\right) }{2} \]\n\n(14.2)\n\nfor all \( x, y \in X \) . Indeed, if this is the case, setting \( y = 0 \) and using the assumption that \( \Phi \left... | Yes |
Lemma 14.1.7. Let \( f : X \rightarrow Y \) be a map between two metric spaces.\n\n(iii) If \( f \) is uniformly continuous and \( X \) is metrically convex, then \( {\omega }_{f}\left( s\right) < \infty \) for all \( s > 0 \) . | Proof. We do (iii) and leave the other statements as an exercise. We need to show that for \( s > 0 \) there is \( {C}_{s} > 0 \) such that \( d\left( {f\left( x\right), f\left( y\right) }\right) \leq {C}_{s} \) whenever \( d\left( {x, y}\right) \leq s \) . From the definition of uniform continuity there exists \( {\de... | No |
Proposition 14.1.8 (Corson and Klee [53]). Let \( f : X \rightarrow Y \) be a uniformly continuous map. If \( X \) is metrically convex, then for every \( \theta > 0 \) there exists a constant \( {K}_{\theta } > 0 \) such that \( d\left( {f\left( x\right), f\left( y\right) }\right) \leq {K}_{\theta }d\left( {x, y}\righ... | Proof. Fix \( \theta > 0 \) . Given \( x, y \) in \( X \) with \( d\left( {x, y}\right) \geq \theta \), let \( m \) be the smallest integer such that \( d\left( {x, y}\right) /m < \theta \) . By the metric convexity of \( X \) we may choose points \( x = {x}_{0},{x}_{1},\ldots ,{x}_{m} = y \) in \( X \) with \( d\left(... | Yes |
Lemma 14.1.14. Let \( X \) be an unbounded metrically convex space. Given \( x \in X \) and \( r > 0 \) there is \( y \in X \) such that \( d\left( {x, y}\right) = r \) . | Proof. Since \( X \) is unbounded, there is \( z \in X \) such that \( d\left( {x, z}\right) = R > r \) . Let \( t = r/R \in \left( {0,1}\right) \) . The metric convexity of \( X \) yields \( y \in X \) such that \( d\left( {x, y}\right) = {td}\left( {x, z}\right) = r. \) | Yes |
Lemma 14.1.15. Let \( f : X \rightarrow Y \) be a map between two unbounded metric spaces.\n\n(i) Suppose that for some constants \( A \geq 1 \) and \( B \geq 0 \) the inequalities\n\n\[ \frac{1}{A}d\left( {x, y}\right) - B \leq d\left( {f\left( x\right), f\left( y\right) }\right) \leq {Ad}\left( {x, y}\right) + B \]\n... | Proof. (i) Suppose \( d\left( {x, y}\right) > \theta \) for a given \( \theta > 0 \) . Then,\n\n\[ d\left( {f\left( x\right), f\left( y\right) }\right) \leq {Ad}\left( {x, y}\right) + B \leq {Ad}\left( {x, y}\right) + \frac{B}{\theta }d\left( {x, y}\right) = \left( {A + \frac{B}{\theta }}\right) d\left( {x, y}\right) .... | Yes |
Lemma 14.1.17. A metric space \( X \) contains an \( \alpha \) -separated \( \alpha \) -net for every \( \alpha > 0 \) . | Proof. For \( \alpha > 0 \), let \( \mathcal{I} = \{ \mathcal{N} \subseteq X : d\left( {{s}_{1},{s}_{2}}\right) \geq \alpha \) for all \( \left( {{s}_{1},{s}_{2}}\right) \in \mathcal{N} \times \mathcal{N} \) , \( \left. {{s}_{1} \neq {s}_{2}}\right\} \) . The set \( \mathcal{I} \) is nonempty, since it contains \( \var... | Yes |
Proposition 14.1.21. Let \( X \) and \( Y \) be normed spaces and suppose \( f : X \rightarrow Y \) is a uniform homeomorphism. Then \( f \) is a (surjective) coarse Lipschitz embedding. | Proof. Since \( f \) is uniformly continuous, Proposition 14.1.8 yields for every \( \theta > 0 \) a constant \( {K}_{\theta } \) such that \( \parallel f\left( x\right) - f\left( y\right) \parallel \leq {K}_{\theta }\parallel x - y\parallel \) for every \( x, y \in X \) with \( \parallel x - y\parallel \geq \theta \) ... | Yes |
Proposition 14.2.2. Let \( X \) be a finite-dimensional Banach space, and let \( f \) be a Lipschitz map from an open set in \( X \) to a (possibly infinite-dimensional) Banach space \( Y \) . Iff is Gâteaux differentiable at a point \( x \), then \( f \) is Fréchet differentiable at \( x \) . | Proof. Given \( \epsilon > 0 \), by compactness of the unit sphere of \( X \) we may choose an \( \epsilon \) -net \( {\left\{ {u}_{i}\right\} }_{i = 1}^{N} \) in \( {S}_{X} \) . Let \( \delta > 0 \) be such that \( \begin{Vmatrix}{f\left( {x + t{u}_{i}}\right) - f\left( x\right) - t{D}_{f}\left( x\right) \left( {u}_{i... | Yes |
Lemma 14.2.4. Let \( f : X \rightarrow Y \) be a Lipschitz map between Banach spaces. Suppose \( E \) is a finite-dimensional subspace of \( X \) . Define \( g : X \rightarrow Y \) by\n\n\[ g\left( x\right) = {\int }_{E}f\left( {x - \xi }\right) \varphi \left( \xi \right) {d\lambda }\left( \xi \right) ,\;x \in X, \]\n\... | Proof. Using the properties of the Bochner integral, we have\n\n\[ \parallel g\left( y\right) - g\left( y\right) \parallel = \begin{Vmatrix}{{\int }_{E}\left( {f\left( {x - \xi }\right) - f\left( {y - \xi }\right) }\right) \varphi \left( \xi \right) {d\lambda }\left( \xi \right) }\end{Vmatrix} \]\n\n\[ \leq {\int }_{E}... | Yes |
Lemma 14.2.5. Let \( f : X \rightarrow Y \) be a Lipschitz map between Banach spaces. Suppose \( E \) is a \( k \) -dimensional subspace of \( X \) . Define maps \( {\left( {g}_{n}\right) }_{n = 1}^{\infty } \) from \( X \) to \( Y \) by\n\n\[ \n{g}_{n}\left( x\right) = {2}^{nk}{\int }_{E}f\left( {x - \xi }\right) \var... | Proof. Write \( f\left( x\right) = {2}^{nk}{\int }_{E}f\left( x\right) \varphi \left( {{2}^{n}\xi }\right) {d\xi } \) . Then \n\n\[ \n\begin{Vmatrix}{{g}_{n}\left( x\right) - f\left( x\right) }\end{Vmatrix} = {2}^{nk}\begin{Vmatrix}{{\int }_{E}\left( {f\left( {x - \xi }\right) - f\left( x\right) }\right) \varphi \left(... | Yes |
Lemma 14.2.9. Let \( h : X \rightarrow \left\lbrack {0,\infty }\right\rbrack \) be a measurable function. Suppose there exists a finite-dimensional subspace \( E \) of \( X \) such that \( h\left( {x + \xi }\right) = 0 \) a.e. \( \xi \in E \) for all \( x \in X \) , i.e., \( \lambda \left( {\{ \xi \in E : h\left( {x + ... | Proof. Pick a probability measure \( \mu \) on \( \left( {X,\mathcal{B}\left( X\right) }\right) \) such that \( \mu \left( {X \smallsetminus E}\right) = 0 \), and \( \mu \left( A\right) = 0 \) if and only if \( A \cap E \) has Lebesgue measure zero. Then we have \( h\left( {x + \xi }\right) = \) \( {0\mu } \) -a.e \( \... | Yes |
Lemma 14.2.11. Let \( h \) be a Haar-null map on a separable infinite-dimensional Banach space \( X \) . For every \( \epsilon > 0 \) there exist a probability measure space \( \left( {\Omega ,\sum ,\mathbb{P}}\right) \) and a random variable \( \eta : \Omega \rightarrow X \) with \( \parallel \eta \parallel < \epsilon... | Proof. We know that there exist a probability space \( \left( {\Omega ,\sum ,\mathbb{P}}\right) \) and a random variable \( \eta : \Omega \rightarrow X \) such that \( \mathbb{E}\left( {h\left( {x + \eta }\right) }\right) = 0 \) for all \( x \in X \) . Since \( X \) is separable, we can decompose \( \Omega \) into coun... | Yes |
Lemma 14.2.12. Let \( X \) be a separable infinite-dimensional Banach space.\n\n(i) If \( {\left( {h}_{n}\right) }_{n = 1}^{\infty } \) is a sequence of Haar-null maps on \( X \), then \( h = \mathop{\sum }\limits_{{n = 1}}^{\infty }{h}_{n} \) is Haar-null.\n\n(ii) If \( {\left( {A}_{n}\right) }_{n = 1}^{\infty } \) is... | Proof. For each \( n \) choose a probability measure space \( \left( {{\Omega }_{n},{\mathbb{P}}_{n}}\right) \) and a random variable \( {\eta }_{n} : {\Omega }_{n} \rightarrow X \) such that \( \mathbb{E}\left( {{h}_{n}\left( {x + {\eta }_{n}}\right) }\right) = 0 \) for all \( x \in X \) and (by Lemma 14.2.11) \( \beg... | Yes |
Theorem 14.2.15 (Heinrich and Mankiewicz [124]). Let \( X \) and \( Y \) be Banach spaces and suppose \( f : X \rightarrow Y \) is a Lipschitz embedding. Assume \( X \) is separable and that \( Y \) has the Radon-Nikodym property. Then \( X \) is linearly isomorphic to a subspace of \( Y \) . More specifically, if \( {... | Proof. By Theorem 14.2.13 there exists at least one point \( {x}_{0} \in X \) where \( f \) is Gâteaux differentiable. Then, by the discussion leading to the estimate (14.11), \( {D}_{f}\left( {x}_{0}\right) \) is an isomorphic embedding of \( X \) into \( Y \) and \( \begin{Vmatrix}{{D}_{f}\left( {x}_{0}\right) }\end{... | Yes |
Theorem 14.2.19 (Infinite-dimensional weak* Rademacher Theorem). Let \( X \) and \( Z \) be separable Banach spaces and let \( f : X \rightarrow {Z}^{ * } \) be a Lipschitz map. Then\n\n(a) \( f \) is weak* differentiable outside a Haar-null set.\n\n(b) \( \begin{Vmatrix}{{D}_{f}^{ * }\left( x\right) }\end{Vmatrix} \le... | Proof. To show part (c) let \( a \) be the inverse of \( \operatorname{Lip}\left( {f}^{-1}\right) \) . Choose an increasing sequence \( {\left( {E}_{n}\right) }_{n = 1}^{\infty } \) of finite-dimensional subspaces of \( X \) whose union is dense. For each \( n \), consider the set \( {D}_{n} \) of all points \( x \in X... | No |
Proposition 14.2.21 (Existence of linear Hahn-Banach extension operators). Let \( Z \) be a Banach space, and let \( V \) be a separable subspace of \( Z \) . There exist a separable space \( W \) with \( V \subseteq W \subseteq Z \) and a linear map \( T : {W}^{ * } \rightarrow {Z}^{ * } \) such that \( \parallel T\pa... | Proof. Certainly, if \( Z \) is separable, then the result is clear, since it suffices to take \( W = Z \) . Having disposed of this case, we may assume that \( Z \) is nonseparable. Then, in particular, \( Z \) is infinite-dimensional and \( V \) can be assumed to be also infinite-dimensional. Pick a sequence \( {\lef... | Yes |
Corollary 14.2.22. Let \( Y \) be a separable subspace of a dual space \( {Z}^{ * } \) . Then there exists a separable Banach space \( W \) such that \( Y \) embeds linearly isometrically into \( {W}^{ * } \) and \( {W}^{ * } \) embeds linearly isometrically into \( {Z}^{ * } \) . | Proof. Choose a separable subspace \( V \subseteq Z \) that is isometrically norming for \( Y \) . By Proposition 14.2.21, there exist \( V \subseteq W \subseteq Z \) and a linear isometric embedding \( T : {W}^{ * } \rightarrow {Z}^{ * } \) such that \( Q \circ T = {I}_{{W}^{ * }} \) . Since \( W \) is also isometrica... | Yes |
Theorem 14.2.23. If a separable Banach space \( X \) Lipschitz embeds into the dual \( {Z}^{ * } \) of a Banach space \( Z \), then \( X \) is isomorphic to a subspace of \( {Z}^{ * } \) . Quantitatively, if \( f : X \rightarrow {Z}^{ * } \) is a Lipschitz embedding, then there is a linear embedding \( T : X \rightarro... | Proof. Assume \( f : X \rightarrow {Z}^{ * } \) is a Lipschitz embedding. Since \( X \) is separable, \( f\left( X\right) \) is contained in a separable subspace \( Y \) of \( {Z}^{ * } \) . By Corollary 14.2.22 we can find a separable Banach space \( W \) and isometric linear embeddings \( {S}_{0} : Y \rightarrow {W}^... | Yes |
Theorem 14.2.27 (Ribe [268]). If there is a coarse Lipschitz embedding of a Banach space \( X \) into a Banach space \( Y \), then \( X \) is crudely finitely representable in \( Y \) . | Proof. Suppose \( X \) coarse Lipschitz embeds into \( Y \) . By Corollary 14.1.26 there is a Lipschitz embedding \( f : X \rightarrow {Y}_{\mathcal{U}} \) for some free ultrafilter \( \mathcal{U} \) on \( \mathbb{N} \) . Since \( {\left\lbrack {Y}_{\mathcal{U}}\right\rbrack }^{* * } \) is finitely representable in \( ... | Yes |
Corollary 14.2.29. Let \( X \) and \( Y \) be Banach spaces such that there exists a coarse Lipschitz embedding from \( X \) into \( Y \) . If \( Y \) has type \( p \) [respectively, cotype \( q \) ], then \( X \) has type p [respectively, cotype q]. | Proof. Use Theorem 14.2.27 and the definitions of type and cotype. | No |
Corollary 14.2.30. If a Banach space \( X \) coarse Lipschitz embeds into a Hilbert space, then it is isomorphic to a Hilbert space. | Proof. This follows directly from Theorem 14.2.27 and Theorem 12.1.6. | Yes |
Let \( q : {\ell }_{\infty } \rightarrow {\ell }_{\infty }/{c}_{0} \) be the canonical quotient map. We start from the existence of a continuum of infinite subsets \( {\left( {\mathbb{A}}_{i}\right) }_{i \in \mathcal{I}} \) of \( \mathbb{N} \) with the property that \( {\mathbb{A}}_{i} \cap {\mathbb{A}}_{j} \) is finit... | If \( x = \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}q\left( {\xi }_{{i}_{n}}\right) \), where \( {a}_{1} \geq {a}_{2} \geq {a}_{3} \geq \cdots \geq 0 \), we let\n\n\[ f\left( x\right) \left( k\right) = \left\{ \begin{array}{ll} {a}_{1} & \text{ if }k \in {\mathbb{A}}_{{i}_{1}}, \\ {a}_{n} & \text{ if }k \in {\mat... | Yes |
Proposition 14.3.5 (Heinrich and Mankiewicz [124]). Let \( f \) be a Lipschitz embedding of a Banach space \( X \) into a Banach space \( Y \) . Assume that \( f \) is Gâteaux differentiable at a point \( {x}_{0} \in X \) . If \( f\left( X\right) \) is a Lipschitz retract of \( Y \) and \( X \) is complemented in \( {X... | Proof. By a simple translation argument, without loss of generality we may assume that \( {x}_{0} = 0 \) and \( f\left( 0\right) = 0 \) . We will write \( {D}_{f} \) instead of \( {D}_{f}\left( {x}_{0}\right) \) . Let \( g : Y \rightarrow X \) be a Lipschitz map such that \( g \circ f = {I}_{X} \) . For all \( {y}_{1},... | Yes |
Theorem 14.3.8. Let \( X \) and \( Y \) be separable Banach spaces such that \( Y \) has the Radon-Nikodym property and \( X \) is complemented in \( {X}^{* * } \) . If \( X \) is Lipschitz isomorphic to a Lipschitz retract of \( Y \), then \( X \) is linearly isomorphic to a complemented subspace of \( Y \) . | Proof. Let \( f : X \rightarrow Y \) and \( g : Y \rightarrow X \) be Lipschitz maps such that \( g \circ f = {I}_{X} \) . Theorem 14.2.13 ensures the existence of points in \( X \) where \( f \) is Gâteaux differentiable. Let \( {D}_{f} : X \rightarrow Y \) be the Gâteaux derivative of \( f \) at one of these points. ... | Yes |
Theorem 14.3.12. If \( X \) is Lipschitz isomorphic to \( {\ell }_{1} \) and is a dual space, then \( X \) is linearly isomorphic to \( {\ell }_{1} \) . | Proof. The extra hypothesis on \( X \) guarantees that it is complemented in its bidual. By Theorem 14.3.8 we obtain that \( X \) is complemented in \( {\ell }_{1} \) . Since \( {\ell }_{1} \) is a prime space, \( X \) must be isomorphic to \( {\ell }_{1} \) . | Yes |
Lemma 14.4.1. A convex function \( f : \mathbb{R} \rightarrow \mathbb{R} \) is differentiable at a point \( x \) if and only if | \[ \mathop{\lim }\limits_{{t \rightarrow {0}^{ + }}}\frac{f\left( {x + t}\right) + f\left( {x - t}\right) - {2f}\left( x\right) }{t} = 0. \] | No |
Lemma 14.4.2. Let \( g : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be a convex function. Then for all \( x \in {\mathbb{R}}^{n} \) and all \( t > 0 \) | Proof. Pick \( x = {\left( {x}_{i}\right) }_{i = 1}^{n} \in {\mathbb{R}}^{n} \) and \( t > 0 \) . Convexity shows that\n\n\[ \mathop{\sup }\limits_{{\parallel h{\parallel }_{1} \leq t}}g\left( {x + h}\right) - g\left( x\right) = \mathop{\max }\limits_{\substack{{1 \leq i \leq n} \\ {\epsilon = \pm 1} }}\left( {g\left( ... | Yes |
A convex function \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is Fréchet differentiable at a point \( x \in {\mathbb{R}}^{n} \) if and only if the \( n \) partial derivatives of \( f \) at \( x \) exist and are finite. In particular, \( f \) is Fréchet differentiable at \( x \) if and only if it is Gâteaux differ... | Suppose the \( n \) partial derivatives of \( f \) exist at \( x \in {\mathbb{R}}^{n} \) . Define the linear map \( T : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) by\n\n\[ T\left( h\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\frac{\partial f}{\partial {x}_{i}}\left( x\right) {h}_{i},\;h = \left( {{h}_{1},\ldots ,{h}_{... | Yes |
Proposition 14.4.5. Suppose \( E \) is a finite-dimensional normed space. If \( f : E \rightarrow \mathbb{R} \) is a convex function, then it is continuous on \( E \) and Fréchet differentiable at every point of a dense subset of \( E \) . | Proof. Without loss of generality we may and do assume that \( E \) is \( {\mathbb{R}}^{n} \) equipped with the norm \( \parallel \cdot {\parallel }_{1} \) . The continuity of a convex function \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) when \( n > 1 \) follows from Lemma 14.4.2 and the 1-dimensional case. For \... | Yes |
Lemma 14.4.7. Suppose \( \left( {E,\parallel \cdot \parallel }\right) \) is a finite-dimensional normed space.\n\n(i) If \( x \in {\Omega }_{\parallel \cdot \parallel } \), then\n\n\[ \begin{Vmatrix}{{D}_{\parallel \cdot \parallel }\left( x\right) }\end{Vmatrix} = {D}_{\parallel \cdot \parallel }\left( x\right) \left( ... | Proof. (i) For \( u \in E \) and \( t > 0 \), by the reverse triangle inequality,\n\n\[ \left| \frac{\parallel x + {tu}\parallel - \parallel x\parallel }{t}\right| \leq \frac{\parallel x + {tu} - x\parallel }{\left| t\right| } = \parallel u\parallel .\n\nMaking \( t \rightarrow {0}^{ + } \) gives \( \left| {{D}_{\paral... | Yes |
Lemma 14.4.9. Let \( \left( {E,\parallel \cdot \parallel }\right) \) be a finite-dimensional normed space. Pick \( x \in E \) with \( \parallel x\parallel = 1 \) a point of differentiability of the norm. Then \( {D}_{\parallel \cdot \parallel }\left( x\right) \) is the only 1-Lipschitz map \( \varphi : E \rightarrow \m... | Proof. By Lemma 14.4.7 (i), \( {D}_{\parallel \cdot \parallel }\left( x\right) : E \rightarrow R \) is 1-Lipschitz and \( {D}_{\parallel \cdot \parallel }\left( x\right) \left( {\alpha x}\right) = \alpha \) for all \( \alpha \in \mathbb{R} \) . Conversely, let \( \varphi : E \rightarrow \mathbb{R} \) be a 1-Lipschitz m... | Yes |
Corollary 14.4.11. Every onto isometry \( \Phi : X \rightarrow Y \) between Banach spaces such that \( \Phi \left( 0\right) = 0 \) is linear. | Proof. Theorem 14.4.10 applied to \( \Phi \) shows that \( \Phi = {T}^{-1} \) is a linear isometry. | No |
Theorem 14.4.12 (Godefroy and Kalton [105]). Let \( X \) be a separable Banach space. If there exists an isometry \( \Phi \) from \( X \) into a Banach space \( Y \), then \( Y \) contains a closed linear subspace that is linearly isometric to \( X \) . | Proof. We may and do assume that \( \Phi \left( 0\right) = 0 \) and that \( \left\lbrack {\Phi \left( X\right) }\right\rbrack = Y \) . By Theorem 14.4.10, there is a norm-one quotient map \( Q : Y \rightarrow X \) such that \( Q \circ \Phi = {I}_{X} \) . We can therefore apply Theorem 14.3.3 with \( g = \Phi \), and th... | Yes |
Theorem 14.5.2 (Enflo-Lindenstrauss). Let \( 1 \leq p < \infty \) . If the spaces \( {L}_{p}\left( {\mu }_{1}\right) \) and \( {L}_{q}\left( {\mu }_{2}\right) \) are uniformly homeomorphic, then either they are of the same finite dimension or \( p = q \) . | Proof. Suppose \( 1 \leq p < q < \infty \) and that both \( {L}_{p}\left( {\mu }_{1}\right) \) and \( {L}_{q}\left( {\mu }_{2}\right) \) are of infinite dimension. If \( {L}_{p}\left( {\mu }_{1}\right) \) and \( {L}_{q}\left( {\mu }_{2}\right) \) are uniformly homeomorphic, by Proposition 14.1.21, \( {L}_{p}\left( {\mu... | Yes |
Theorem 14.5.3 (Bourgain-Enflo-Gorelik). For \( 1 \leq p < \infty \) with \( p \neq 2 \) the spaces \( {\ell }_{p} \) and \( {L}_{p} \) are not uniformly homeomorphic. | Here the case \( p = 1 \) is due to Enflo in the 1970s (unpublished; see a proof in [23, Theorem 10.13]); the case \( 1 < p < 2 \) was established in 1987 by Bourgain [29], and the case \( 2 < p < \infty \) was not settled until 1994 by Gorelik [112]. | No |
Proposition 14.5.5. Let \( X \) and \( Y \) be Banach spaces, and let \( f : X \rightarrow Y \) be a coarse Lipschitz map. If \( {\operatorname{Lip}}_{\infty }\left( f\right) > 0 \), then for every \( \theta > 0,\epsilon > 0 \), and \( 0 < \delta < 1 \), there exist \( x, y \in X \) with \( \parallel x - y\parallel > \... | Proof. Suppose \( \theta ,\delta ,\epsilon \) are given. For \( \eta > 0 \) as small as we wish we pick \( {\theta }^{\prime } > \theta \) such that \( {\operatorname{Lip}}_{{\theta }^{\prime }}\left( f\right) < \left( {1 + \eta }\right) {\operatorname{Lip}}_{\infty }\left( f\right) \) . Then we choose \( x, y \in X \)... | Yes |
Lemma 14.5.6. Suppose \( 1 \leq p < \infty \) . Let \( {\left( {e}_{i}\right) }_{i = 1}^{\infty } \) be the canonical basis of \( {\ell }_{p} \) , and for \( N \in \mathbb{N} \), denote by \( {E}_{N} \) the closed linear span of \( \left\{ {{e}_{i} : i > N}\right\} \) . Let \( x, y \in {\ell }_{p} \) , \( \delta \in \l... | Proof. Given \( 0 < \eta < 1 \), pick \( N \in \mathbb{N} \) such that \( \mathop{\sum }\limits_{{i = 1}}^{N}{\left| {v}_{i}\right| }^{p} \geq \left( {1 - {\eta }^{p}}\right) \parallel v{\parallel }_{p}^{p} \) .\n\n(i) We may clearly assume that \( p > 1 \) . Let now \( z \in {E}_{N} \) be that \( \parallel z{\parallel... | Yes |
Proposition 14.5.7. Let \( 1 \leq p < q < \infty \) and suppose that \( f : {\ell }_{q} \rightarrow {\ell }_{p} \) is a coarse Lipschitz map. Then for every \( t > 0 \) and \( \epsilon > 0 \) there exist \( u \in {\ell }_{q},\tau > t, N \in \mathbb{N} \), and a compact subset \( K \) of \( {\ell }_{p} \) such that\n\n\... | Proof. We assume that \( {\operatorname{Lip}}_{\infty }\left( f\right) > 0 \), since if \( {\operatorname{Lip}}_{\infty }\left( f\right) = 0 \), the conclusion is clear. We choose a small \( \delta > 0 \) (to be specified later). Then we pick \( \theta \) large enough (also to be detailed later) that \( {\operatorname{... | Yes |
Corollary 14.5.8. If \( 1 \leq p < q < \infty ,{\ell }_{q} \) does not coarse Lipschitz embed into \( {\ell }_{p} \) . | Proof. Let \( f : {\ell }_{q} \rightarrow {\ell }_{p} \) be a coarse Lipschitz map. With the notation of the previous proposition, we can find a sequence \( {\left( {u}_{n}\right) }_{n = 1}^{\infty } \) in \( u + \tau {B}_{{E}_{N}} \) such that \( {\begin{Vmatrix}{u}_{n} - {u}_{m}\end{Vmatrix}}_{q} \geq \tau \) for \( ... | Yes |
Corollary 14.5.10. If \( 1 \leq q < p < \infty ,{\ell }_{q} \) does not coarse Lipschitz embed into \( {\ell }_{p} \) . | Proof. Suppose that \( {\ell }_{q} \) coarse Lipschitz embeds into \( {\ell }_{p} \) . Then, using homogeneity, there exist \( f : {\ell }_{q} \rightarrow {\ell }_{p} \) and \( B \geq 1 \) such that\n\n\[ \parallel x - y{\parallel }_{q} \leq \parallel f\left( x\right) - f\left( y\right) {\parallel }_{p} \leq B\parallel... | Yes |
Example 14.6.2. (a) If \( X = {\ell }_{p} \) equipped with its natural norm, then\n\n\[{\bar{\rho }}_{{\ell }_{p}}\left( \tau \right) = {\left( 1 + {\tau }^{p}\right) }^{1/p} - 1,\;\tau > 0,\] | and so \( {\ell }_{p} \) is asymptotically uniformly smooth if \( 1 < p < \infty \) . The natural norm actually has an optimal (i.e., minimal) modulus of asymptotic uniform smoothness among all equivalent norms. | Yes |
Lemma 14.6.4 (The Gorelik Principle). Let \( E \) and \( X \) be Banach spaces. Suppose \( \varphi : E \rightarrow X \) is a homeomorphism whose inverse \( {\varphi }^{-1} \) is Lipschitz. Let \( b \) and \( c \) be positive constants such that \( c > \operatorname{Lip}\left( {\varphi }^{-1}\right) \cdot b \), and let ... | Proof. Put \( a = \operatorname{Lip}\left( {\varphi }^{-1}\right) \cdot b \) . We first construct a compact subset \( \widetilde{K} \) of the ball \( c{B}_{E} \) such that if \( \psi : \widetilde{K} \rightarrow E \) is a continuous map with \( \parallel x - \psi \left( x\right) \parallel \leq a \) for every \( x \in \w... | Yes |
Proposition 15.1.2. There is a norm-one operator \( T : \mathcal{X} \rightarrow X \) defined by \( T{e}_{n} = {x}_{n} \) for \( n \in \mathbb{N} \) . The operator \( T \) is a quotient map. | Proof. It is easy to see that \( \xi \in X \) implies that \( \mathop{\sum }\limits_{{j = 1}}^{\infty }\xi \left( j\right) {x}_{j} \) must converge and that\n\n\[\n\begin{Vmatrix}{\mathop{\sum }\limits_{{j = 1}}^{\infty }\xi \left( j\right) {x}_{j}}\end{Vmatrix} \leq \parallel \xi {\parallel }_{\mathcal{X}}\n\]\n\nThus... | Yes |
Lemma 15.1.3. \( {T}^{ * }\left( {X}^{ * }\right) \cap \mathcal{Y} = \{ 0\} \), and \( {T}^{ * }{X}^{ * } + \mathcal{Y} \) is norm closed. | Proof. It is enough to note that if \( {x}^{ * } \in {X}^{ * } \) and \( {\xi }^{ * } \in \mathcal{Y} \), then\n\n\[ \n{\begin{Vmatrix}{T}^{ * }{x}^{ * }\end{Vmatrix}}_{\mathcal{X}} = \begin{Vmatrix}{x}^{ * }\end{Vmatrix} \leq {\begin{Vmatrix}{T}^{ * }{x}^{ * } + {\xi }^{ * }\end{Vmatrix}}_{{\mathcal{X}}^{ * }}\n\]\n\n... | No |
Theorem 15.1.6. For every separable Banach space \( X \) there is a separable Banach space \( \mathcal{Z} \) such that \( {\mathcal{Z}}^{* * }/\mathcal{Z} \) is isomorphic to \( X \) . Furthermore, \( {\mathcal{Z}}^{ * } \) has a shrinking basis. | Proof. We take \( \mathcal{Z} = \ker T \) in the above construction. We show that \( \mathcal{X} \) can then be identified canonically with \( {\mathcal{Z}}^{* * } \) . More precisely, we show that under the pairing between \( \mathcal{X} \) and \( \mathcal{Y} \) we can identify \( \mathcal{Y} \) with \( {\mathcal{Z}}^... | Yes |
Theorem 15.3.1 (Pelczyński's universal basis space). There is a unique separable Banach space \( U \) with a basis and with the property that every Banach space with a basis is isomorphic to a complemented subspace of \( U \) . | Proof. To prove the existence of \( U \) it suffices to construct a Banach space \( X \) with a basis \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) such that every normalized basic sequence (in any Banach space) is equivalent to a complemented subsequence of \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) . Then th... | No |
Theorem 15.3.2. There is a unique Banach space \( {U}_{1} \) with an unconditional basis \( {\left( {u}_{n}\right) }_{n = 1}^{\infty } \) and with the property that every Banach space with an unconditional basis is isomorphic to a complemented subspace of \( {U}_{1} \) . | Proof. Suppose \( X \) is the space constructed in the preceding proof. Then we can define a norm on \( {c}_{00} \) by\n\n\[ \parallel \xi {\parallel }_{{U}_{1}} = \mathop{\sup }\limits_{{{\epsilon }_{j} = \pm 1}}{\begin{Vmatrix}\mathop{\sum }\limits_{{j = 1}}^{\infty }{\epsilon }_{j}\xi \left( j\right) {e}_{j}\end{Vma... | No |
Proposition 15.4.2. Suppose a basis \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) of a Banach space \( X \) satisfies a lower 2-estimate on blocks. Then,\n\n(i) The formula\n\n\[ \parallel \parallel x\parallel \parallel = \max \left\{ {\parallel x\parallel ,\sup {\left( \mathop{\sum }\limits_{{j = 1}}^{n}{\begin{Vma... | Proof. We leave the verification of \( \left( i\right) \) to the reader. To show (ii), suppose\n\n\[ \mathop{\sup }\limits_{n}\begin{Vmatrix}{\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}{x}_{k}}\end{Vmatrix} < \infty \]\n\nbut the series \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}{x}_{k} \) does not converge. The... | No |
Proposition 15.4.4. The norm \( \parallel \cdot {\parallel }_{\mathcal{X}} \) has the following properties:\n\n(i) For every \( B \in \mathcal{B} \),\n\n\[ \parallel \xi {\parallel }_{B} = \parallel \xi {\parallel }_{\mathcal{X}},\;\xi \in {c}_{00}\left( B\right) .\n\]\n\n(ii) If \( {E}_{1},\ldots ,{E}_{n} \) are disjo... | Proof. (i) follows directly from (15.8).\n\n(ii) Given \( \epsilon > 0 \), pick disjoint segments \( {\left( {S}_{jk}\right) }_{k = 1}^{{m}_{n}} \) for \( j = 1,2,\ldots, n \) such that\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}\mathop{\sum }\limits_{{k = 1}}^{{m}_{n}}{\begin{Vmatrix}{S}_{jk}{E}_{j}\xi \end{Vmatrix}}^{2... | Yes |
Lemma 15.4.6. Suppose \( \xi \in {c}_{00} \) is supported on \( \left\lbrack {1, N}\right\rbrack \) and \( \eta \in {c}_{00} \) is supported on \( \lbrack N + 1,\infty ) \) . Then\n\n\[ \parallel \xi + \eta {\parallel }_{\mathcal{X}} \leq {\left( \parallel \xi {\parallel }_{\mathcal{X}}^{2} + \parallel \eta {\parallel ... | Proof. Let \( \delta = \mathop{\sup }\limits_{{m \geq N + 1}}{\begin{Vmatrix}{T}_{m}\eta \end{Vmatrix}}_{\mathcal{X}} \) . Suppose \( \epsilon > 0 \) and pick disjoint segments \( {\left( {S}_{j}\right) }_{j = 1}^{m} \) such that\n\n\[ \parallel \xi + \eta {\parallel }_{\mathcal{X}}^{2} < \mathop{\sum }\limits_{{j = 1}... | Yes |
Theorem 15.4.8. There is a Banach space \( \mathcal{Y} \) such that \( {\mathcal{Y}}^{ * } \) is separable and \( {\mathcal{Y}}^{* * }/\mathcal{Y} \) is isometric to \( {\ell }_{2}\left( \mathcal{I}\right) \), where \( \mathcal{I} \) has the cardinality of the continuum. | Proof. We use the space \( \mathcal{J} \) but with the basis of Problem 3.11, which is a special case of the construction of Theorem 15.1.6. It is trivial to see that the basis \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) of the space \( \mathcal{X} \) constructed in Section 15.1 has an exact lower 2-estimate on bl... | No |
Theorem 15.4.9. The space \( {\mathcal{Y}}^{ * } = \mathcal{J}\mathcal{T} \) has nonseparable dual, but \( {\ell }_{1} \) does not embed into \( \mathcal{J}\mathcal{T} \) . | Proof. Obviously, \( \mathcal{J}{\mathcal{T}}^{ * } \) is nonseparable. Since \( \mathcal{J}\mathcal{T} \) is a dual space, it is complemented in its bidual, and so \( \mathcal{J}{\mathcal{T}}^{* * } = \mathcal{J}\mathcal{T} \oplus W \), where \( W \) can be identified as the dual of the space \( \mathcal{J}{\mathcal{T... | Yes |
Theorem 15.4.10. Let \( X \) be any separable dual space. Then there is a Banach space \( Z \) such that \( {Z}^{* * }/Z \) is isomorphic to \( {\ell }_{2}{\left( X\right) }_{i \in \mathcal{I}} \) where \( \mathcal{I} \) has the cardinality of the continuum. | Proof. Let \( X = {Y}^{ * } \) and construct \( \mathcal{Z} \) as in Section 15.1 such that \( {\mathcal{Z}}^{* * }/\mathcal{Z} \approx Y \) . Using the canonical basis of \( \mathcal{Z} \) as in Theorem 15.4.8 will give us a space \( Z \) such that \( {Z}^{* * }/Z \) is isomorphic to \( {\ell }_{2}{\left( {Y}^{ * }\ri... | Yes |
Lemma 1.4.1. Setting \( F\left( x\right) \mathrel{\text{:=}} {P}_{x}\left( {\{ \mathbf{0}\} }\right) \) and\n\n\[ k = {i}_{1} + {i}_{2}N + \cdots + {i}_{n}{N}^{n - 1}\;\left( { \in {\mathbb{N}}_{0}}\right) ,\]\n\nwe get the formula\n\n\[ {P}_{x}\left( {\{ k\} }\right) = F\left( {x + k}\right) \]\n\nwhere we have identi... | Proof. To see this, identify functions on \( \left\lbrack {0,1}\right\rbrack \) with 1-periodic functions on \( \mathbb{R} \), and note that the second formula in (1.4.1) yields \( {\tau }_{{i}_{n}}\cdots {\tau }_{{i}_{1}}\left( x\right) = \left( {x + k}\right) /{N}^{n} \) where \( k \) is given by (1.4.2). Hence, if \... | Yes |
Lemma 1.5.1. Let \( N, W, F,{P}_{x} \), and \( h \) be as described above; see (1.5.2)-(1.5.3). Then \( h \) satisfies the following cocycle identity:\n\n\[ h\left( x\right) W\left( x\right) = {P}_{Nx}\left( {N\mathbb{Z}}\right) ,\;x \in \mathbb{R}. \] | Proof. We calculate the left-hand side in (1.5.6), using the earlier equations:\n\n\[ h\left( x\right) W\left( x\right) \underset{\begin{matrix} \text{ by (1.4.3) } \\ \text{ and (1.5.3) } \end{matrix}}{ = }W\left( x\right) \mathop{\sum }\limits_{{j \in \mathbb{Z}}}F\left( {x + j}\right) \]\n\n\[ \underset{\text{by per... | Yes |
Theorem 1.5.2. Let \( N, W, F,{P}_{x} \), and \( h \) be as described above. Let \( x \in \mathbb{R} \), and suppose that \( {P}_{x}\left( {\{ \mathbf{0}\} }\right) > 0 \) . Then the following two conditions are equivalent.\n\n(a) The limit on the right-hand side below exists, and\n\n\[ F\left( x\right) = \mathop{\lim ... | Proof. (a) \( \Rightarrow \) (b). An iteration of the identity (1.5.6) in Lemma 1.5.1 above yields\n\n\[ {P}_{x}\left( {{N}^{k}\mathbb{Z}}\right) = \left( {\mathop{\prod }\limits_{{j = 1}}^{k}W\left( \frac{x}{{N}^{j}}\right) }\right) h\left( \frac{x}{{N}^{k}}\right) .\n\]\n\n(1.5.7)\n\nUsing (1.5.5), and working in \( ... | Yes |
Lemma 2.5.1. (Kolmogorov) Let \( N \geq 2 \) be fixed, and let\n\n\[ \Omega = \{ 0,1,\ldots, N - 1{\} }^{\mathbb{N}} \]\n\nFor \( n = 1,2,\ldots \), let\n\n\[ {P}^{\left( n\right) } : {\mathfrak{A}}_{n} \rightarrow \mathbb{C} \]\n\nbe a sequence of linear functionals such that (i)-(iii) hold:\n\n(i) \( {P}^{\left( n\ri... | Proof of Lemma 2.5.1. The proof of Kolmogorov's extension result may be given several forms, but we note that the argument we used above (in a special case), based on an application of the Stone-Weierstraß theorem, also works in general. | No |
Lemma 2.6.2. There is a unique mapping \( \rho : M\left( \Omega \right) \rightarrow M\left( X\right) \) which satisfies\n\n\[ \rho \left( {fg}\right) = \rho \left( f\right) \rho \left( g\right) \]\n\n(2.6.3)\n\nand\n\n\[ \rho \left( {\chi }_{A\left( {{i}_{1},\ldots ,{i}_{n}}\right) }\right) = {\chi }_{{\tau }_{{i}_{1}}... | Proof. Recalling (2.4.9), we note that\n\n\[ {\chi }_{A\left( {{i}_{1},\ldots ,{i}_{n}}\right) }\left( \omega \right) = {\delta }_{{i}_{1},{\omega }_{1}}\cdots {\delta }_{{i}_{n},{\omega }_{n}},\;\omega \in \Omega . \]\n\nAs a result,\n\n\[ {\chi }_{A\left( {{i}_{1},\ldots ,{i}_{n}}\right) }{\chi }_{A\left( {{j}_{1},\l... | Yes |
Lemma 2.6.3. Let \( X, W \), and \( N \) be as described in the beginning of this chapter, and let \( \left\{ {{P}_{x} \mid x \in X}\right\} \) be the process obtained in the conclusion of Lemma 2.4.1. Then\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{{N - 1}}W\left( {{\tau }_{i}x}\right) {P}_{{\tau }_{i}x}\left\lbrack {f\le... | Proof of Lemma 2.6.3. It follows from (2.4.6) and the arguments in the proof of Lemma 2.4.1 that it is enough to verify (2.6.6) for \( f \in {C}_{\text{fin }}\left( \Omega \right) \), or for \( f \in {\mathfrak{A}}_{n} \) . Let \( f \in {\mathfrak{A}}_{n} \) . Then\n\n\[ \mathop{\sum }\limits_{i}W\left( {{\tau }_{i}x}\... | Yes |
Theorem 2.7.1. Let \( X, W \), and \( N \) be as described in the beginning of this chapter, and suppose in addition that (2.4.1) is satisfied. Let \( R = {R}_{W} \) be the Ruelle operator on \( {L}^{\infty }\left( X\right) \), and let \( \left\{ {{P}_{x} \mid x \in X}\right\} \) be the process on \( \Omega \) from Lem... | Proof. (2.7.1) \( \Rightarrow \) (2.7.2). This follows by Lemma 2.6.3 and the cocycle property. Let \( V \) be a cocycle, and define \( h = {h}_{V} \) by (2.7.1). Then for \( x \in X \),\n\n\[\left( {Rh}\right) \left( x\right) = \mathop{\sum }\limits_{i}W\left( {{\tau }_{i}x}\right) h\left( {{\tau }_{i}x}\right) = \mat... | Yes |
Corollary 2.7.3. The \( 1 - 1 \) correspondence between the harmonic functions \( h \) in Theorem 2.7.1 and the cocycles \( V \) is an order-isomorphism when the ordering of both families of functions is defined in the pointwise sense. | Proof. We noted that when \( V \) is a given cocycle, then the corresponding function \( h \) is defined by (2.7.1); and conversely, \( V \) may be computed from \( h \) via the formula (2.7.6).\n\nSince each \( {P}_{x} \) is a positive measure,(2.7.1) yields the implication\n\n\[ \n{V}_{1} \leq {V}_{2} \Rightarrow {h}... | Yes |
Corollary 2.7.4. Let \( v : \Omega \rightarrow \mathbb{C} \) be a bounded measurable function, and let \( X, W \) , and \( R = {R}_{W} \) be as stated in the theorem. Then\n\n\[ V\left( {x,\omega }\right) \mathrel{\text{:=}} v\left( \omega \right) ,\;\omega \in \Omega ,\]\n\n(2.7.8)\n\nis a cocycle if and only if \( v ... | Proof. Let \( v : \Omega \rightarrow \mathbb{C} \) be given, and suppose \( V \) in (2.7.8) is a cocycle. Then\n\n\[ v\left( \omega \right) = V\left( {x,\omega }\right) = V\left( {{\tau }_{{\omega }_{1}}x,\left( {{\omega }_{2},{\omega }_{3},\ldots }\right) }\right) = v\left( {{\omega }_{2},{\omega }_{3},\ldots }\right)... | Yes |
Corollary 2.7.6. Let \( E \subset \Omega \), and let \( X,\sigma ,{\tau }_{i}, N \), and \( W \) be as described in the beginning of the chapter. Set\n\n\[ \n{h}_{E}\left( x\right) \mathrel{\text{:=}} {P}_{x}\left\lbrack {\chi }_{E}\right\rbrack \]\n\n(2.7.11)\n\nwhere \( {\chi }_{E} \) denotes the indicator function o... | Proof. This follows directly from Corollary 2.7.4 above, if we note that\n\n\[ \n{\chi }_{E} \circ {\sigma }^{\Omega } = {\chi }_{{\left( {\sigma }^{\Omega }\right) }^{-1}E}. \]\n | Yes |
Proposition 2.8.2. Let \( v \) be \( R \) -invariant, and let \( h \) satisfy \( {Rh} = h \) . Then the measure\n\n\[ d{v}_{h} \mathrel{\text{:=}} {hdv} \]\n\n(2.8.3)\n\nis invariant.\n\nConversely, if \( {v}_{1} \) is a \( \mathcal{B} \) -measure on \( X \) which is assumed \( \sigma \) -invariant, and if \( {v}_{1} \... | Proof. (Part one!)\n\n\[ \int f \circ {\sigma d}{v}_{h} = \int f \circ {\sigma hdv} \]\n\n\[ = \int R\left( {f \circ {\sigma h}}\right) {dv} = \int {fRhdv} \]\n\n\[ = \int {fhdv} = \int {fd}{v}_{h} \]\n\nwhich implies that \( {v}_{h} \) is \( \sigma \) -invariant.\n\nTo prove (2.8.4) under the assumptions in the second... | Yes |
Lemma 3.1.1. Let \( W \) and \( f \) be bounded 1-periodic functions on \( \mathbb{R} \). (If the two functions are measurable, they will be assumed only essentially bounded, and a.e. 1-periodic.)\n\nThen the function\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{{N - 1}}W\left( \frac{x + k}{N}\right) f\left( \frac{x + k}{N}\... | Proof. Note that the expression in (3.1.12) is the right-hand side in (3.1.9), or equivalently in (3.1.10), and so it is the function \( {R}_{W}f \). We now give the explicit argument as to why it is 1-periodic. Note that the individual terms in the sum (3.1.12) have period \( N \) and not 1 . Let\n\n\[ g\left( x\right... | Yes |
Lemma 3.2.1. Let \( W : \left\lbrack {0,1}\right\rbrack \rightarrow \left\lbrack {0,1}\right\rbrack \) be a given measurable function and extend \( W \) from \( \left\lbrack {0,1}\right\rbrack \) to \( \mathbb{R} \) by periodicity. Then the densities for the probabilities \( {P}_{x} \) in (1.2.4) are\n\n\[ W\left( {{\t... | Proof. A direct calculation, using (3.2.1) and the periodicity of \( W \), yields\n\n\[ W\left( {{\tau }_{{\omega }_{s}}{\tau }_{{\omega }_{s - 1}}\cdots {\tau }_{{\omega }_{1}}x}\right) = W\left( \frac{x + k}{{N}^{s}}\right) \;\text{ for }1 \leq s \leq n. \]\n\nIt is understood that the integer \( k \in {\mathbb{N}}_{... | Yes |
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