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Corollary 12.45. Suppose \( {\mathbb{F}}_{q} \) is a finite field, \( q = {p}^{r} \), in which multiplication is determined by a monic irreducible polynomial of degree \( r \) (that is, \( {\mathbb{F}}_{q} \) is presented as \( {\mathbb{F}}_{p}\left\lbrack x\right\rbrack /f\left( x\right) \cdot {\mathbb{F}}_{p}\left\lb... | Exercise 12.10. Prove Theorem 12.44 and Corollary 12.45. | No |
Proposition 1.1.4. Let \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) be a Schauder basis for a Banach space \( X \) and \( {\left( {S}_{n}\right) }_{n = 1}^{\infty } \) the natural projections associated with it. Then\n\n\[ \mathop{\sup }\limits_{n}\begin{Vmatrix}{S}_{n}\end{Vmatrix} < \infty . \]\n | Proof. For a Schauder basis the operators \( {\left( {S}_{n}\right) }_{n = 1}^{\infty } \) are bounded a priori. Since \( {S}_{n}\left( x\right) \rightarrow x \) for every \( x \in X \), we have \( \mathop{\sup }\limits_{n}\begin{Vmatrix}{{S}_{n}\left( x\right) }\end{Vmatrix} < \infty \) for each \( x \in X \) . Then t... | Yes |
Proposition 1.1.7. Suppose \( {S}_{n} : X \rightarrow X, n \in \mathbb{N} \), is a sequence of bounded linear projections on a Banach space \( X \) such that\n\n(i) \( \dim {S}_{n}\left( X\right) = n \) for each \( n \) ;\n\n(ii) \( {S}_{n}{S}_{m} = {S}_{m}{S}_{n} = {S}_{\min \{ m, n\} } \), for all integers \( m \) an... | Proof. Let \( 0 \neq {e}_{1} \in {S}_{1}\left( X\right) \) and define \( {e}_{1}^{ * } : X \rightarrow \mathbb{R} \) by \( {e}_{1}^{ * }\left( x\right) {e}_{1} = {S}_{1}\left( x\right) \) . Next we pick \( 0 \neq {e}_{2} \in {S}_{2}\left( X\right) \cap {S}_{1}^{-1}\left( 0\right) \) and define the functional \( {e}_{2}... | Yes |
Proposition 1.1.9. A sequence \( {\left( {e}_{k}\right) }_{k = 1}^{\infty } \) of nonzero elements of a Banach space \( X \) is basic if and only if there is a positive constant \( K \) such that\n\n\[ \begin{Vmatrix}{\mathop{\sum }\limits_{{k = 1}}^{m}{a}_{k}{e}_{k}}\end{Vmatrix} \leq K\begin{Vmatrix}{\mathop{\sum }\l... | Proof. Assume \( {\left( {e}_{k}\right) }_{k = 1}^{\infty } \) is basic, and let \( {S}_{m} : \left\lbrack {e}_{k}\right\rbrack \rightarrow \left\lbrack {e}_{k}\right\rbrack, m = 1,2,\ldots \), be its partial sum projections. If \( m \leq n \) we have\n\n\[ \begin{Vmatrix}{\mathop{\sum }\limits_{{k = 1}}^{m}{a}_{k}{e}_... | Yes |
Theorem 1.3.2. Two bases (or basic sequences) \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) and \( {\left( {y}_{n}\right) }_{n = 1}^{\infty } \) are equivalent if and only if there is an isomorphism \( T : \left\lbrack {x}_{n}\right\rbrack \rightarrow \left\lbrack {y}_{n}\right\rbrack \) such that \( T{x}_{n} = {y}_... | Proof. Let \( X = \left\lbrack {x}_{n}\right\rbrack \) and \( Y = \left\lbrack {y}_{n}\right\rbrack \) . It is obvious that \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) and \( {\left( {y}_{n}\right) }_{n = 1}^{\infty } \) are equivalent if there is an isomorphism \( T \) from \( X \) onto \( Y \) such that \( T{x}_... | Yes |
Lemma 1.3.5. Suppose \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) is a basis for the Banach space \( X \) with basis constant \( {\mathrm{K}}_{\mathrm{b}} \) . Let \( {\left( {u}_{k}\right) }_{k = 1}^{\infty } \) be a block basic sequence of \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) . Then \( {\left( {u}_{k}... | Proof. Suppose that \( {u}_{k} = \mathop{\sum }\limits_{{j = {p}_{k - 1} + 1}}^{{p}_{k}}{a}_{j}{e}_{j}, k \in \mathbb{N} \), is a block basic sequence of \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) . Then, for any scalars \( \left( {b}_{k}\right) \) and integers \( m, n \) with \( m \leq n \) we have\n\n\[ \begin{... | Yes |
Theorem 1.3.9 (Principle of small perturbations). Let \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) be a basic sequence in a Banach space \( X \) with basis constant \( {\mathrm{K}}_{\mathrm{b}} \) . If \( {\left( {y}_{n}\right) }_{n = 1}^{\infty } \) is a sequence in \( X \) such that\n\n\[ 2{\mathrm{\;K}}_{\mathrm... | Proof. For \( n \geq 2 \) and \( x \in \left\lbrack {x}_{n}\right\rbrack \) we have\n\n\[ {x}_{n}^{ * }\left( x\right) {x}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}^{ * }\left( x\right) {x}_{k} - \mathop{\sum }\limits_{{k = 1}}^{{n - 1}}{x}_{k}^{ * }\left( x\right) {x}_{k}, \]\n\nwhere \( {\left( {x}_{n}^{ * }\r... | Yes |
Proposition 1.3.10 (The Bessaga-Pelczyński Selection Principle). Let \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) be a basis for a Banach space \( X \) with basis constant \( {\mathrm{K}}_{\mathrm{b}} \) and dual functionals \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) . Suppose \( {\left( {x}_{n}\right) ... | Proof. Let \( \alpha = \mathop{\inf }\limits_{n}\begin{Vmatrix}{x}_{n}\end{Vmatrix} > 0 \) and suppose \( 0 < v < \frac{1}{4} \) . Pick \( {n}_{1} = 1,{r}_{0} = 0 \) . There exists \( {r}_{1} \in \mathbb{N} \) such that\n\n\[ \begin{Vmatrix}{{x}_{{n}_{1}} - {S}_{{r}_{1}}{x}_{{n}_{1}}}\end{Vmatrix} < \frac{v\alpha }{2{\... | Yes |
Lemma 1.4.1. Let \( X \) be a Banach space.\n\n(i) If \( X \) is separable, then the closed unit ball \( {B}_{{X}^{ * }} \) of \( {X}^{ * } \) is (compact and) metrizable for the weak* topology. | Proof. The proofs of both \( \left( i\right) \) and \( \left( {ii}\right) \) rely on the following simple observation. If \( K \) is a compact set for some topology \( \tau \), and \( {\tau }^{\prime } \) is any Hausdorff topology on \( K \) that is weaker than \( \tau \), then \( \tau \) and \( {\tau }^{\prime } \) co... | Yes |
Every separable infinite-dimensional Banach space contains a basic sequence (i.e., a closed infinite-dimensional subspace with a basis). Furthermore, if \( \epsilon > 0 \), we may find a basic sequence with basis constant at most \( 1 + \epsilon \) . | By the Banach-Mazur theorem (Theorem 1.4.4) we can consider the case in which the separable Banach space \( X \) is a closed subspace of \( \mathcal{C}\left\lbrack {0,1}\right\rbrack \) . Let \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) be a monotone basis for \( \mathcal{C}\left\lbrack {0,1}\right\rbrack \) with b... | Yes |
Lemma 1.5.1. Suppose that \( S \) is a subset of \( {X}^{ * } \) such that \( 0 \in {\bar{S}}^{{\text{weak* }}^{ * }} \) but \( 0 \notin {\bar{S}}^{\parallel \cdot \parallel } \) . Let \( E \) be a finite-dimensional subspace of \( {X}^{ * } \) . Then given \( \epsilon > 0 \) there exists \( {x}^{ * } \in S \) such tha... | Proof. Let us notice that such a set \( S \) exists because the weak* topology and the norm topology of an infinite-dimensional Banach space do not coincide. The fact that \( 0 \notin {\bar{S}}^{\parallel \cdot \parallel } \) implies \( \alpha \leq \begin{Vmatrix}{x}^{ * }\end{Vmatrix} \) for all \( {x}^{ * } \in S \),... | Yes |
Theorem 1.5.2. Suppose that \( S \) is a subset of \( {X}^{ * } \) such that \( 0 \in {\bar{S}}^{{\text{weak* }}^{ * }} \) but \( 0 \notin {\bar{S}}^{\parallel \cdot \parallel } \) . Then for every \( \epsilon > 0, S \) contains a basic sequence with basis constant \( \leq 1 + \epsilon \) . | Proof. Fix a decreasing sequence of positive numbers \( {\left( {\epsilon }_{n}\right) }_{n = 1}^{\infty } \) with \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{\epsilon }_{n} < \infty \) and such that \( \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 - {\epsilon }_{n}}\right) > {\left( 1 + \epsilon \right) }^{-1} ... | Yes |
Proposition 1.5.4. If \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) is a weakly null sequence in an infinite-dimensional Banach space \( X \) such that \( \mathop{\inf }\limits_{n}\begin{Vmatrix}{x}_{n}\end{Vmatrix} > 0 \), then for every \( \epsilon > 0,{\left( {x}_{n}\right) }_{n = 1}^{\infty } \) contains a basic... | Proof. Consider \( S = \left\{ {{x}_{n} : n \in \mathbb{N}}\right\} \) . Since \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) is weakly convergent, the set \( S \) is norm bounded. Furthermore, \( 0 \in {\bar{S}}^{\text{weak }} \) ; hence by Theorem 1.5.2, \( S \) contains a basic sequence with basis constant at most... | Yes |
Lemma 1.5.5. Let \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) be a basic sequence in \( X \) . Suppose that there exists a linear functional \( {x}^{ * } \in {X}^{ * } \) such that \( {x}^{ * }\left( {x}_{n}\right) = 1 \) for all \( n \in \mathbb{N} \) . If \( u \notin \left\lbrack {x}_{n}\right\rbrack \), then the... | Proof. Since \( u \notin \left\lbrack {x}_{n}\right\rbrack \), without loss of generality we can assume \( {x}^{ * }\left( u\right) = 0 \) . Let \( T : X \rightarrow X \) be the operator given by \( T\left( x\right) = {x}^{ * }\left( x\right) u \) . Then \( {I}_{X} + T \) is invertible with inverse \( {I}_{X} - T \) . ... | Yes |
Theorem 1.5.6. Let \( S \) be a bounded subset of a Banach space \( X \) such that \( 0 \notin {\bar{S}}^{\parallel \cdot \parallel } \) . Then the following are equivalent:\n\n(i) S fails to contain a basic sequence,\n\n(ii) \( {\bar{S}}^{\text{weak }} \) is weakly compact and fails to contain 0 . | Proof. (ii) \( \Rightarrow \) (i) Suppose \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \subset S \) is a basic sequence. Since \( {\bar{S}}^{\text{weak }} \) is weakly compact, \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) has a weak cluster point, say \( x \), in \( {\bar{S}}^{\text{weak }} \) . By Mazur’s theorem... | Yes |
Lemma 1.6.1. If \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) is a basic sequence in a Banach space and \( x \) is a weak cluster point of \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \), then \( x = 0 \) . | Proof. Since \( x \) is in the weak closure of the convex set \( \left\langle {{x}_{n} : n \in \mathbb{N}}\right\rangle \) (the linear span of the sequence \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) ), Mazur’s theorem yields that \( x \) belongs to \( \left\lbrack {x}_{n}\right\rbrack \), the norm-closed linear s... | Yes |
Lemma 1.6.2. Let \( A \) be a relatively weakly countably compact subset of a Banach space \( X \) . Suppose that \( x \in X \) is the only weak cluster point of a sequence \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) contained in A. Then \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) converges weakly to \( x \) ... | Proof. Assume that \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) does not converge weakly to \( x \) . Then \( {\left( {x}^{ * }\left( {x}_{n}\right) \right) }_{n = 1}^{\infty } \) fails to converge to \( {x}^{ * }\left( x\right) \) for some \( {x}^{ * } \in {X}^{ * } \) . Hence we may pick a subsequence \( {\left( ... | Yes |
Theorem 1.6.3 (The Eberlein-Šmulian Theorem). Let \( A \) be a subset of a Banach space \( X \) . The following are equivalent:\n\n(i) \( A \) is [relatively] weakly compact,\n\n(ii) \( A \) is [relatively] weakly sequentially compact,\n\n(iii) \( A \) is [relatively] weakly countably compact. | Proof. Since (i) and (ii) both imply (iii), we need only show that (iii) implies both (ii) and (i). We will prove the relativized versions; minor modifications can be made to prove the nonrelativized versions. Note that each of the statements of the theorem implies that \( A \) is bounded.\n\nLet us first do the case (... | Yes |
Proposition 2.1.3. Suppose \( 1 \leq p < \infty \) . Let \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) be a normalized sequence in \( {\ell }_{p} \) [respectively, \( {c}_{0} \) ] such that for each \( j \in \mathbb{N} \) we have \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}\left( j\right) = 0 \) (for ex... | Proof. The gliding hump technique (see Proposition 1.3.10) yields a subsequence \( {\left( {x}_{{n}_{k}}\right) }_{k = 1}^{\infty } \) and a block basic sequence \( {\left( {y}_{k}\right) }_{k = 1}^{\infty } \) of \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) such that \( {\left( {x}_{{n}_{k}}\right) }_{k = 1}^{\inf... | Yes |
Theorem 2.1.9. If \( p \neq r \), every bounded operator \( T : {\ell }_{p} \rightarrow {\ell }_{r} \) is strictly singular. | Proof. This is immediate from Corollary 2.1.6. | No |
Proposition 2.2.1. Every infinite-dimensional closed subspace \( Y \) of \( {\ell }_{p}(1 \leq p < \) \( \infty ) \) [respectively, \( {c}_{0} \) ] contains a closed subspace \( Z \) such that \( Z \) is isomorphic to \( {\ell }_{p} \) [respectively, \( {c}_{0} \) ] and complemented in \( {\ell }_{p} \) [respectively, ... | Proof. Since \( Y \) is infinite-dimensional, for every \( n \) there is \( {x}_{n} \in Y \) with \( \begin{Vmatrix}{x}_{n}\end{Vmatrix} = 1 \) such that \( {e}_{k}^{ * }\left( {x}_{n}\right) = 0 \) for \( 1 \leq k \leq n \) . If not, for some \( N \in \mathbb{N} \) the projection \( {S}_{N}\left( {\mathop{\sum }\limit... | Yes |
Theorem 2.2.3 (The Pelczyński decomposition technique [241]). Let \( X \) and \( Y \) be Banach spaces such that \( X \) is isomorphic to a complemented subspace of \( Y \), and \( Y \) is isomorphic to a complemented subspace of \( X \) . Suppose further that either\n\n(a) \( X \approx {X}^{2} = X \oplus X \) and \( Y... | Proof. Let us put \( X \approx Y \oplus E \) and \( Y \approx X \oplus F \) . If (a) holds, then we have\n\n\[ X \approx Y \oplus Y \oplus E \approx Y \oplus X \]\n\nand by a symmetric argument \( Y \approx X \oplus Y \) . Hence \( Y \approx X \) .\n\nIf \( X \) satisfies \( \left( b\right) \) in particular, we have \(... | Yes |
Theorem 2.2.4. Suppose \( Y \) is a complemented infinite-dimensional subspace of \( {\ell }_{p} \) where \( 1 \leq p < \infty \) [respectively, \( {c}_{0} \) ]. Then \( Y \) is isomorphic to \( {\ell }_{p} \) [respectively, \( {c}_{0} \) ]. | Proof. Proposition 2.2.1 gives an infinite-dimensional subspace \( Z \) of \( Y \) such that \( Z \) is isomorphic to \( {\ell }_{p} \) [respectively, \( {c}_{0} \) ] and \( Z \) is complemented in \( {\ell }_{p} \) [respectively, \( \left. {c}_{0}\right\rbrack \) . Obviously \( Z \) is also complemented in \( Y \) ; t... | Yes |
Theorem 2.3.1. If \( X \) is a separable Banach space, then there exists a continuous operator \( Q : {\ell }_{1} \rightarrow X \) from \( {\ell }_{1} \) onto \( X \) . | Proof. It suffices to show that \( X \) admits of a continuous operator \( Q : {\ell }_{1} \rightarrow X \) such that \( Q\left( \left\{ {\xi \in {\ell }_{1} : \parallel \xi {\parallel }_{1} < 1}\right\} \right) = \{ x \in X : \parallel x\parallel < 1\} \) . Let \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) be a den... | Yes |
Corollary 2.3.2. If \( X \) is a separable Banach space, then \( X \) is isometrically isomorphic to a quotient of \( {\ell }_{1} \) . | Proof. Let \( Q : {\ell }_{1} \rightarrow X \) be the quotient map in the proof of Theorem 2.3.1. Then it follows that \( {\ell }_{1}/\ker Q \) is isometrically isomorphic to \( X \) . | Yes |
Corollary 2.3.3. The space \( {\ell }_{1} \) has an uncomplemented closed subspace. | Proof. Take \( X \) a separable Banach space that is not isomorphic to \( {\ell }_{1} \) . Theorem 2.3.1 yields an operator \( Q \) from \( {\ell }_{1} \) onto \( X \) whose kernel is a closed subspace of \( {\ell }_{1} \) . If \( \ker Q \) were complemented in \( {\ell }_{1} \), then we would have \( {\ell }_{1} = \ke... | Yes |
Theorem 2.3.7. Let \( X \) be a Banach space with the Schur property. Then a subset \( W \) of \( X \) is weakly compact if and only if \( W \) is norm-compact. | Proof. Suppose \( W \) is weakly compact and consider a sequence \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) in \( W \) . By the Eberlein-Šmulian theorem \( W \) is weakly sequentially compact, so \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) has a subsequence \( {\left( {x}_{{n}_{k}}\right) }_{k = 1}^{\infty }... | No |
Corollary 2.3.8. If \( X \) is a reflexive Banach space with the Schur property, then \( X \) is finite-dimensional. | Proof. If a reflexive Banach space \( X \) has the Schur property, then its unit ball is norm-compact by Theorem 2.3.7, and so \( X \) is finite-dimensional. | Yes |
Lemma 2.4.2. Given a series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) in a Banach space \( X \), the following are equivalent:\n\n(a) \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) is unconditionally convergent.\n\n(b) The series \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{x}_{{n}_{k}} \) converge... | Proof. We will establish only \( \left( a\right) \Rightarrow \left( d\right) \) and leave the other easier implications to the reader. Suppose that \( \left( d\right) \) fails. Then there exists \( \epsilon > 0 \) such that for every \( n \) we can find a finite subset \( {F}_{n} \) of \( \{ n + 1,\ldots \} \) with\n\n... | No |
Proposition 2.4.4. Suppose the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) converges unconditionally to some \( x \) in a Banach space \( X \) . Then\n\n(i) \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{\pi \left( n\right) } = x \) for every permutation \( \pi \) .\n\n(ii) \( \mathop{\sum }\limits_... | Proof. Parts \( \left( i\right) \) and \( \left( {ii}\right) \) are immediate. For (iii), given \( {x}^{ * } \in {X}^{ * } \), the scalar series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}^{ * }\left( {x}_{\pi \left( n\right) }\right) \) converges for every permutation \( \pi \) . It is a classical theorem of Riema... | Yes |
Lemma 2.4.6. Let \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) be a formal series in a Banach space \( X \) . The following are equivalent:\n\n(i) \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) is WUC.\n\n(ii) There exists \( C > 0 \) such that for all \( \left( {\xi \left( n\right) }\right) \in {c}_{0... | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) . Put\n\n\[ S = \left\{ {\mathop{\sum }\limits_{{n = 1}}^{\infty }\xi \left( n\right) {x}_{n} \in X : \xi = \left( {\xi \left( n\right) }\right) \in {c}_{00},\parallel \xi {\parallel }_{\infty } \leq 1}\right\} .\n\nThe WUC property implies that \( S \) is wea... | Yes |
Proposition 2.4.7. Let \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) be a series in a Banach space \( X \) . Then \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) is WUC if and only if there is a bounded operator \( T : {c}_{0} \rightarrow X \) with \( T{e}_{n} = {x}_{n} \) . | Proof. If \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) is WUC, then the operator \( T : {c}_{00} \rightarrow X \) defined by \( {T\xi } = \) \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\xi \left( n\right) {x}_{n} \) is bounded for the \( {c}_{0} \) -norm by Lemma 2.4.6. By density, \( T \) extends to a bounde... | Yes |
Proposition 2.4.8. Let \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) be a WUC series in a Banach space \( X \) . Then \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) converges unconditionally in \( X \) if and only if the operator \( T : {c}_{0} \rightarrow X \) such that \( {\mathrm{{Te}}}_{n} = {x}_{n... | Proof. Suppose that \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) is unconditionally convergent. We will show that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\begin{Vmatrix}{T - T{S}_{n}}\end{Vmatrix} = 0 \), where \( {\left( {S}_{n}\right) }_{n = 1}^{\infty } \) are the partial sum projections associat... | Yes |
Theorem 2.4.11. In order that every WUC series in a Banach space \( X \) be unconditionally convergent, it is necessary and sufficient that \( X \) contain no copy of \( {c}_{0} \) . | Proof. Suppose that \( X \) contains no copy of \( {c}_{0} \) and that \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) is a WUC series in \( X \) . By Proposition 2.4.7 there exists a bounded operator \( T : {c}_{0} \rightarrow X \) such that \( T{e}_{n} = {x}_{n} \) for all \( n \) . The operator \( T \) must b... | Yes |
Lemma 2.4.13. Let \( {m}_{0} \) be the set of all sequences of scalars assuming only finitely many different values. Then \( {m}_{0} \) is dense in \( {\ell }_{\infty } \) . | Proof. Let \( a = {\left( {a}_{n}\right) }_{n = 1}^{\infty } \) be a sequence of scalars with \( \parallel a{\parallel }_{\infty } \leq 1 \) . For every \( \epsilon > 0 \) pick \( N \in \mathbb{N} \) such that \( \frac{1}{N} < \epsilon \) . Then the sequence \( b = {\left( {b}_{n}\right) }_{n = 1}^{\infty } \in {m}_{0}... | Yes |
Theorem 2.4.14 (The Orlicz-Pettis Theorem). Suppose \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) is a series in a Banach space \( X \) for which every subseries \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{x}_{{n}_{k}} \) converges weakly. Then \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) converges ... | Proof. The hypothesis easily yields that \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) is a WUC series, so by Proposition 2.4.7, there exists a bounded operator \( T : {c}_{0} \rightarrow X \) with \( T{e}_{n} = {x}_{n} \) for all \( n \) . We will show that \( T \) is compact.\n\nLet us look at \( {T}^{* * } ... | Yes |
Corollary 2.4.15. If a Banach space \( X \) is weakly sequentially complete, then every WUC series in \( X \) is unconditionally convergent. | Proof. If \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \) is WUC, then \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}^{ * }\left( {x}_{n}\right) \) is absolutely convergent for every \( {x}^{ * } \in {X}^{ * } \), which is equivalent to saying that \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{x}^{ * }\left( {x}... | Yes |
Proposition 2.5.2. The space \( {\ell }_{\infty } \) is an isometrically injective space. Hence, if a Banach space \( X \) has a subspace \( E \) isomorphic to \( {\ell }_{\infty } \), then \( E \) is necessarily complemented in \( X \) . | Proof. Suppose \( E \) is a subspace of \( X \) and \( T : E \rightarrow {\ell }_{\infty } \) is bounded. Then \( {Te} = {\left( {e}_{n}^{ * }\left( e\right) \right) }_{n = 1}^{\infty } \) for some sequence \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) in \( {E}^{ * } \) ; clearly \( \parallel T\parallel = \ma... | Yes |
Lemma 2.5.3. Every countably infinite set \( \mathbb{S} \) has an uncountable family of infinite subsets \( {\left\{ {\mathbb{A}}_{i}\right\} }_{i \in \mathcal{I}} \) such that any two members of the family have finite intersection. | Proof. The proof is very simple but rather difficult to spot! Without loss of generality we can identify \( \mathbb{S} \) with the set of the rational numbers \( \mathbb{Q} \) . For each irrational number \( \theta \), take a sequence of rational numbers \( {\left( {q}_{n}\right) }_{n = 1}^{\infty } \) converging to \(... | No |
Theorem 2.5.4. Let \( T : {\ell }_{\infty } \rightarrow {\ell }_{\infty } \) be a bounded operator such that \( {T\xi } = 0 \) for all \( \xi \in {c}_{0} \) . Then there is an infinite subset \( \mathbb{A} \) of \( \mathbb{N} \) such that \( {T\xi } = 0 \) for every \( \xi \in {\ell }_{\infty }\left( \mathbb{A}\right) ... | Proof. We use the family \( {\left( {\mathbb{A}}_{i}\right) }_{i \in \mathcal{I}} \) of infinite subsets of \( \mathbb{N} \) given by Lemma 2.5.3. Suppose that for every such set we can find \( {\xi }_{i} \in {\ell }_{\infty }\left( {\mathbb{A}}_{i}\right) \) with \( T{\xi }_{i} \neq 0 \) . We can assume by normalizati... | Yes |
Theorem 2.5.5 (Phillips and Sobczyk [252, 286]). There is no bounded projection from \( {\ell }_{\infty } \) onto \( {c}_{0} \) . | Proof. If \( P \) is such a projection, we can apply Theorem 2.5.4 to \( T = I - P \), with \( I \) the identity operator on \( {\ell }_{\infty } \), and then it is clear that \( {P\xi } = \xi \) for all \( \xi \in {\ell }_{\infty }\left( \mathbb{A}\right) \) for some infinite set \( \mathbb{A} \), which gives a contra... | No |
Corollary 2.5.6. \( {c}_{0} \) is not isomorphic to a dual space. | Proof. If \( {c}_{0} \) were isomorphic to a dual space, then by the comments that follow the proof of Proposition 2.5.2, \( {c}_{0} \) should be complemented in \( {c}_{0}^{* * } \), which would lead to a contradiction with Theorem 2.5.5. | Yes |
Theorem 2.5.7. Every separable Banach space embeds isometrically in \( {\ell }_{\infty } \) . | Proof. Let \( {\left( {x}_{n}\right) }_{n = 1}^{\infty } \) be a dense sequence in \( X \) . For each integer \( n \) pick \( {x}_{n}^{ * } \in {X}^{ * } \) such that \( \begin{Vmatrix}{x}_{n}^{ * }\end{Vmatrix} = 1 \) and \( {x}_{n}^{ * }\left( {x}_{n}\right) = \begin{Vmatrix}{x}_{n}\end{Vmatrix} \) . The sequence \( ... | Yes |
Corollary 2.5.9. If \( E \) is a closed subspace of a separable Banach space \( X \) and \( E \) is isomorphic to \( {c}_{0} \), then there is a projection \( P \) from \( X \) onto \( E \) . | Proof. Suppose that \( T : E \rightarrow {c}_{0} \) is an isomorphism and let \( \widetilde{T} : X \rightarrow {c}_{0} \) be the extension of \( T \) given by the preceding theorem. Then \( P = {T}^{-1}\widetilde{T} \) is a projection from \( X \) onto \( E \) . (Note that since \( \parallel \widetilde{T}\parallel \leq... | Yes |
To see that \( {\left( {f}_{n}\right) }_{n = 1}^{\infty } \) is a basis for \( {c}_{0} \) we prove that for each \( \xi = {\left( \xi \left( n\right) \right) }_{n = 1}^{\infty } \in {c}_{0} \) we have \( \xi = \mathop{\sum }\limits_{{n = 1}}^{\infty }{f}_{n}^{ * }\left( \xi \right) {f}_{n} \), where \( {f}_{n}^{ * } = ... | Given \( N \in \mathbb{N} \), \n\n\[ \mathop{\sum }\limits_{{n = 1}}^{N}{f}_{n}^{ * }\left( \xi \right) {f}_{n} = \mathop{\sum }\limits_{{n = 1}}^{N}\left( {{e}_{n}^{ * }\left( \xi \right) - {e}_{n + 1}^{ * }\left( \xi \right) }\right) {f}_{n} \]\n\n\[ = \mathop{\sum }\limits_{{n = 1}}^{N}\left( {\xi \left( n\right) - ... | Yes |
Proposition 3.1.3. A basis \( {\left( {u}_{n}\right) }_{n = 1}^{\infty } \) of a Banach space \( X \) is unconditional if and only if there is a constant \( K \geq 1 \) such that for all \( N \in \mathbb{N} \) , \[ \begin{Vmatrix}{\mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n}{u}_{n}}\end{Vmatrix} \leq K\begin{Vmatrix}{\ma... | Proof. Assume \( {\left( {u}_{n}\right) }_{n = 1}^{\infty } \) is unconditional. If \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{u}_{n} \) converges, then so does \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{t}_{n}{a}_{n}{u}_{n} \) for all \( {\left( {t}_{n}\right) }_{n = 1}^{\infty } \in {\ell }_{\infty } \) by... | Yes |
Proposition 3.1.5. Let \( {\left( {u}_{n}\right) }_{n = 1}^{\infty } \) be a basis of a Banach space \( X \) . The following are equivalent:\n\n(i) The basis \( {\left( {u}_{n}\right) }_{n = 1}^{\infty } \) is unconditional.\n\n(ii) The map \( {P}_{A} \) is well defined for every \( A \subseteq \mathbb{N} \) .\n\n(iii)... | Proof. The implication \( \left( i\right) \Rightarrow \left( {ii}\right) \) is a consequence of Proposition 3.1.3.\n\n\( \left( {ii}\right) \Rightarrow \left( {iii}\right) \) follows readily from the uniform boundedness principle.\n\n(iii) \( \Rightarrow \left( {iv}\right) \) and \( \left( {iii}\right) \Rightarrow \lef... | Yes |
Proposition 3.2.1. Let \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) be a basis for a Banach space \( X \) with basis constant \( {\mathrm{K}}_{\mathrm{b}} \) and biorthogonal functionals \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) . Then \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) is a basis f... | Proof. The functionals \( {\left( {e}_{n}^{* * }\right) }_{n = 1}^{\infty } \) satisfy \( {e}_{n}^{* * }\left( {e}_{k}^{ * }\right) = 1 \) if \( n = k \) and 0 otherwise. Hence if suffices to show that the operators \( {T}_{N} : Z \rightarrow Z \) defined by\n\n\[ {T}_{N}\left( {x}^{ * }\right) = \mathop{\sum }\limits_... | Yes |
Proposition 3.2.3. Let \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) be a basis for a Banach space \( X \) with basis constant \( {\mathrm{K}}_{\mathrm{b}} \) and let \( Z = \left\lbrack {e}_{n}^{ * }\right\rbrack \) . The norm on \( X \) defined by\n\n\[ \parallel x{\parallel }_{Z} = \sup \{ \left| {h\left( x\right... | Proof. Let \( x \in X \) . Since \( Z \subseteq {X}^{ * } \),\n\n\[ \parallel x{\parallel }_{Z} \leq \sup \left\{ {\left| {{x}^{ * }\left( x\right) }\right| : {x}^{ * } \in {X}^{ * },\begin{Vmatrix}{x}^{ * }\end{Vmatrix} \leq 1}\right\} = \parallel x\parallel .\n\nFor the reverse inequality pick \( {x}^{ * } \in {S}_{{... | Yes |
Proposition 3.2.5. Let \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) be a basis for a Banach space \( X \) with biorthogonal functionals \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \). Then for every \( {x}^{ * } \in {X}^{ * } \) there is a unique sequence of scalars \( {\left( {a}_{n}\right) }_{n = 1}^{\inf... | Proof. For every \( x \in X \) , \[ \left| {\left( {{x}^{ * } - {S}_{N}^{ * }\left( {x}^{ * }\right) }\right) \left( x\right) }\right| = \left| \left( {{x}^{ * }\left( {x - {S}_{N}\left( x\right) }\right) \mid \leq \begin{Vmatrix}{x}^{ * }\end{Vmatrix}\begin{Vmatrix}{x - {S}_{N}\left( x\right) }\end{Vmatrix}\overset{N ... | Yes |
Proposition 3.2.6. If \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) is a basis for a reflexive Banach space \( X \), then \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) is a basis for \( {X}^{ * } \) . | Proof. Proposition 3.2.5 yields that the linear span of \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) is weak* dense in \( {X}^{ * } \) . Since \( X \) is reflexive, the weak and the weak* topologies of \( {X}^{ * } \) coincide, so that the linear span of \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) ... | Yes |
Proposition 3.2.8. Suppose \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) is a basis for a Banach space \( X \) . The coordinate functionals \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) are a basis for \( {X}^{ * } \) if and only if\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty }}{\begin{Vmatrix}{x}^{ ... | Proof of Proposition 3.2.8. Suppose \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) is a basis for \( {X}^{ * } \) . Every \( {x}^{ * } \in {X}^{ * } \) can be decomposed as \( \left( {{x}^{ * } - {S}_{N}^{ * }{x}^{ * }}\right) + {S}_{N}^{ * }{x}^{ * } \) for each \( N \) . Then\n\n\[ {\begin{Vmatrix}{x}^{ * }\e... | Yes |
The unit vector basis \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) of \( {c}_{0} \) is shrinking, since \( {c}_{0}^{ * } = {\ell }_{1} \) and given \( {x}^{ * } = {\left( {a}_{n}\right) }_{n = 1}^{\infty } \in {\ell }_{1} \), we have | \[ {\begin{Vmatrix}{x}^{ * }\end{Vmatrix}}_{N} = \mathop{\sum }\limits_{{n = N + 1}}^{\infty }\left| {a}_{n}\right| \overset{N \rightarrow \infty }{ \rightarrow }0. \] | No |
Lemma 3.2.14. Suppose \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) is a boundedly complete basis for a Banach space \( X \) with biorthogonal functionals \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) . Then for each \( {x}^{* * } \in {X}^{* * } \) we have\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty ... | Proof. Note that for each \( N \in \mathbb{N} \), \n\n\[ \mathop{\sum }\limits_{{n = 1}}^{N}{x}^{* * }\left( {e}_{n}^{ * }\right) {e}_{n} = {S}_{N}^{* * }\left( {x}^{* * }\right), \]\n\nwhere \( {S}_{N}^{* * } \) is the double adjoint of \( {S}_{N} \). Hence\n\n\[ \begin{Vmatrix}{\mathop{\sum }\limits_{{n = 1}}^{N}{x}^... | Yes |
Theorem 3.2.15. Let \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) be a basis for a Banach space \( X \) with biorthogonal functionals \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \). The following are equivalent:\n\n(i) \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) is a boundedly complete basis for \( X \)... | Proof. \( \left( i\right) \Rightarrow \) (iii) Using Proposition 3.2.3, we need only show that the map is onto. Given \( {h}^{ * } \in {Z}^{ * } \), there exists \( {x}^{* * } \in {X}^{* * } \) such that \( {\left. {x}^{* * }\right| }_{Z} = {h}^{ * } \). By Lemma 3.2.14 the series \( \mathop{\sum }\limits_{{n = 1}}^{\i... | Yes |
Corollary 3.2.16. The space \( {c}_{0} \) has no boundedly complete basis. | Proof. The result follows from Theorem 3.2.15, taking into account that \( {c}_{0} \) is not isomorphic to a dual space (Corollary 2.5.6). | No |
Theorem 3.2.17. Let \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) be a basis for a Banach space \( X \) with biorthogonal functionals \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \). The following are equivalent:\n\n(i) \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) is a shrinking basis for \( X \).\n\n(ii)... | Proof. Just apply Theorem 3.2.15 to the basis \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) of \( Z \) and take into account Corollary 3.2.4. | No |
Theorem 3.2.19 (James [126]). Let \( X \) be a Banach space. If \( X \) has a basis \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) then \( X \) is reflexive if and only if \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) is both boundedly complete and shrinking. | Proof. Assume that \( X \) is reflexive and that \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) is a basis for \( X \) . Then \( {X}^{ * } = Z \) . If not, using the Hahn-Banach theorem, one could find \( 0 \neq {x}^{* * } \in {X}^{* * } \) such that \( {x}^{* * }\left( h\right) = 0 \) for all \( h \in Z \) . By refl... | Yes |
Theorem 3.3.4. Suppose that \( X \) is a Banach space with an unconditional basis. If \( X \) is not reflexive, then either \( {c}_{0} \) is complemented in \( X \), or \( {\ell }_{1} \) is complemented in \( X \) (or both). In either case, \( {X}^{* * } \) is nonseparable. | Proof. The first statement of the theorem follows immediately from Theorem 3.2.19, Theorem 3.3.1, and Theorem 3.3.2. Now, for the latter statement, if \( {c}_{0} \) were complemented in \( X \), then \( {X}^{* * } \) would contain a (complemented) copy \( {\ell }_{\infty } \) . If \( {\ell }_{1} \) were complemented in... | Yes |
Proposition 3.4.2. The sequence \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) of standard unit vectors is a monotone basis for \( \mathcal{J} \) in both norms \( \parallel \cdot {\parallel }_{\mathcal{J}} \) and \( \parallel \cdot {\parallel }_{0} \) . | Proof. We will leave for the reader the verification that \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) is a monotone basic sequence in both norms. To prove that it is a basis, we need only consider the norm \( \parallel \cdot {\parallel }_{0} \) .\n\nSuppose \( \xi \in \mathcal{J} \) . For each \( N \) let\n\n\[{\x... | No |
Proposition 3.4.3. Let \( {\left( {\eta }_{k}\right) }_{k = 1}^{\infty } \) be a normalized block basic sequence with respect to \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) in \( \left( {\mathcal{J},\parallel \cdot {\parallel }_{0}}\right) \). Then for every sequence of scalars \( {\left( {\lambda }_{k}\right) }_{... | Proof. For each \( k \) let \[ {\eta }_{k} = \mathop{\sum }\limits_{{j = {q}_{k - 1} + 1}}^{{q}_{k}}{\eta }_{k}\left( j\right) {e}_{j} \] where \( 0 = {q}_{0} < {q}_{1} < \cdots \), and put \[ \xi = \mathop{\sum }\limits_{{k = 1}}^{n}{\lambda }_{k}{\eta }_{k} \] Suppose \( 1 \leq {p}_{0} < {p}_{1} < \cdots < {p}_{m} \)... | Yes |
Theorem 3.4.6. The space \( \mathcal{J} \) is a subspace of codimension 1 in \( {\mathcal{J}}^{* * } \), and \( {\mathcal{J}}^{* * } \) is isometric to \( \mathcal{J} \) . | Proof. Clearly, \( \mathcal{J} = \left\{ {\xi \in {\mathcal{J}}^{* * } : \mathop{\lim }\limits_{{n \rightarrow \infty }}\xi \left( n\right) = 0}\right\} \) has codimension one in its bidual. To prove the fact that it is isometric to its bidual we observe that\n\n\[ \parallel \xi {\parallel }_{{\mathcal{J}}^{* * }} = \p... | Yes |
Corollary 3.4.7. \( \mathcal{J} \) does not have an unconditional basis. | Proof. The result follows immediately from the separability of \( {\mathcal{J}}^{* * } \), Theorem 3.3.4 and Theorem 3.4.6. | Yes |
Proposition 3.5.2. If a Banach space \( X \) has property (u), then every closed subspace \( Y \) of \( X \) has property (u). | Proof. Let \( {\left( {y}_{s}\right) }_{s = 1}^{\infty } \) be a weakly Cauchy sequence in a closed subspace \( Y \) of \( X \) . Since \( X \) has property (u), there is a WUC series \( \mathop{\sum }\limits_{{i = 1}}^{\infty }{u}_{i} \) in \( X \) such that the sequence \( {\left( {y}_{s} - \mathop{\sum }\limits_{{i ... | Yes |
Proposition 3.5.4. (i) The James space \( \mathcal{J} \) does not have property (u), and so \( \mathcal{J} \) cannot be embedded in any Banach space with an unconditional basis. | Proof. (i) Assume that \( \mathcal{J} \) has property (u). Since the sequence defined for each \( n \) by \( {s}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}{e}_{k} \) is weakly Cauchy in \( \mathcal{J} \), there exists a WUC series in \( \mathcal{J} \) , \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{u}_{k} \), such that t... | Yes |
Theorem 4.1.3. If \( K \) is compact Hausdorff, then the space \( \mathcal{C}\left( K\right) \) is separable if and only if \( K \) is metrizable. | Proof. There is a natural embedding \( s \mapsto {\delta }_{s} \) (the point mass at \( s \) ) of \( K \) into \( \mathcal{M}\left( K\right) \) . This is a homeomorphism for the weak* topology of \( \mathcal{M}\left( K\right) \) . By Lemma 1.4.1 \( \left( i\right) \) , this shows that \( K \) is metrizable if \( \mathc... | Yes |
Proposition 4.1.4. If \( K \) is a totally disconnected compact Hausdorff space, then the collection of simple continuous functions (i.e., functions \( f \) of the form \( f = \) \( \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{j}{\chi }_{{U}_{j}} \), where \( {U}_{1},\ldots ,{U}_{n} \) are disjoint clopen sets) is dense in ... | Proof. This is an easy deduction from the Stone-Weierstrass theorem since the simple functions form a subalgebra of \( \mathcal{C}\left( K\right) \) . | No |
Lemma 4.2.3. Let \( \mathcal{A} \) be a commutative real Banach algebra with an identity \( e \) of norm \( \parallel e\parallel = 1 \) . The following conditions are equivalent:\n\n(i) \( \parallel {2ab}\parallel \leq \begin{Vmatrix}{{a}^{2} + {b}^{2}}\end{Vmatrix}\;\forall a, b \in \mathcal{A} \) .\n\n(ii) \( \parall... | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) If we put \( y = e \) in \( \left( i\right) \), we get\n\n\[ 2\parallel x\parallel \leq \begin{Vmatrix}{{x}^{2} + e}\end{Vmatrix} \leq \begin{Vmatrix}{x}^{2}\end{Vmatrix} + 1 \]\n\nThus for all \( x \) in \( \mathcal{A} \) of norm \( \parallel x\parallel = 1 \... | Yes |
Proposition 4.2.6. A commutative real Banach algebra \( \mathcal{A} \) with an identity \( e \) of norm 1 has the following properties:\n\n(i) If \( x \in \mathcal{A} \) is such that \( \parallel x\parallel \leq 1 \), then \( e + x \in {\mathcal{A}}_{ + } \) .\n\n(ii) \( \mathcal{A} = {\mathcal{A}}_{ + } - {\mathcal{A}... | Proof. Let \( x \) in \( \mathcal{A} \) have \( \parallel x\parallel < 1 \) . By writing \( {\left( 1 + t\right) }^{1/2} \) in its binomial series, valid for scalars \( t \) with \( \left| t\right| < 1 \), we see that the series \( \mathop{\sum }\limits_{{n = 0}}^{\infty }\left( \begin{matrix} 1/2 \\ n \end{matrix}\rig... | Yes |
Lemma 4.2.7. Suppose that \( \mathcal{A} \) satisfies (4.4). Then \( \varphi \left( x\right) \geq 0 \) whenever \( \varphi \in \mathcal{S} \) and \( x \in {\mathcal{A}}_{ + } \) . | Proof. Take \( x \in {\mathcal{A}}_{ + } \) with \( \parallel x\parallel = 1 \) . By Proposition 4.2.6, \( e - x \in {\mathcal{A}}_{ + } \), and by (4.4), \[ \parallel e - x\parallel \leq \parallel \left( {e - x}\right) + x\parallel = 1. \] Hence for \( \varphi \in \mathcal{S} \) we have \[ 1 = \parallel \varphi \paral... | Yes |
Lemma 4.2.8. Suppose that \( \mathcal{A} \) satisfies (4.4). Let \( K \) be the set of all multiplicative states of \( \mathcal{A} \), i.e., | Proof. It is trivial to show that \( K \) is a closed subset of the closed unit ball of \( {\mathcal{A}}^{ * } \) . This ensures that \( K \) is compact for the weak* topology.\n\nSince \( \mathcal{S} \) is convex and compact in the weak* topology of \( {\mathcal{A}}^{ * } \), the Krein-Milman theorem guarantees that \... | Yes |
Lemma 4.3.3. Let \( X \) be a Banach space and \( F \) a linear subspace of \( X \) . Suppose \( V : X \rightarrow \mathcal{C}\left( K\right) \) and \( W : F \rightarrow \mathcal{C}\left( K\right) \) are sublinear maps such that \( W\left( y\right) + V\left( {-y}\right) \geq 0 \) for all \( y \in F \) . If \( \mathcal{... | Proof. For each fixed \( x \in X \) we have\n\n\[ V\left( {x - y}\right) + W\left( y\right) \geq V\left( {-y}\right) - V\left( {-x}\right) + W\left( y\right) \geq - V\left( {-x}\right) \]\n\nfor all \( y \in F \) . That is, \( - V\left( {-x}\right) \) is a lower bound of the set \( \{ V\left( {x - y}\right) + W\left( y... | Yes |
Lemma 4.3.4. Let \( V : X \rightarrow \mathcal{C}\left( K\right) \) be a sublinear map. If \( \mathcal{C}\left( K\right) \) is order-complete, then there is a minimal sublinear map \( W : X \rightarrow \mathcal{C}\left( K\right) \) with \( W\left( x\right) \leq V\left( x\right) \) for all \( x \in X \) . | Proof. Put\n\n\[ \mathcal{S} = \{ U : X \rightarrow \mathcal{C}\left( K\right) : U\text{ is sublinear and }U\left( x\right) \leq V\left( x\right) \text{ for all }x \in X\} .\n\]\n\nThe set \( \mathcal{S} \) is nonempty \( \left( {V \in \mathcal{S}}\right) \) and partially ordered. Let \( \Psi = {\left( {U}_{i}\right) }... | Yes |
Lemma 4.3.5. Suppose that \( \mathcal{C}\left( K\right) \) is order-complete and let \( V : X \rightarrow \mathcal{C}\left( K\right) \) be a sublinear map. If \( V \) is minimal, then \( V \) is linear. | Proof. Given an element \( x \in X \), let us call \( F \) its linear span, \( F = \langle x\rangle \) . Then, \( W\left( {\lambda x}\right) = - {\lambda V}\left( {-x}\right) \) defines a linear map from \( F \) to \( \mathcal{C}\left( K\right) \) . Clearly, \( W\left( {\lambda x}\right) \geq \) \( - V\left( {-{\lambda... | Yes |
Theorem 4.3.7 (Kelley [169]). A Banach space \( X \) is isometrically injective if and only if it is isometrically isomorphic to an order-complete \( \mathcal{C}\left( K\right) \) -space. | Proof. We need only show the forward implication. For that, we are going to identify \( X \) (via an isometric isomorphism) with a suitable \( \mathcal{C}\left( K\right) \) -space which, by the isometric injectivity of \( X \), will, by an appeal to Theorem 4.3.6, be order-complete.\n\nThe trick is to \ | No |
Theorem 4.3.10. \( {L}_{\infty }\left\lbrack {0,1}\right\rbrack \) is isomorphic to \( {\ell }_{\infty } \) . | Proof. First, observe that \( {\ell }_{\infty } \) embeds isometrically into \( {L}_{\infty }\left\lbrack {0,1}\right\rbrack \) via the map\n\n\[ \n{\left( \xi \left( n\right) \right) }_{n = 1}^{\infty } \mapsto \mathop{\sum }\limits_{{n = 1}}^{\infty }\xi \left( n\right) {\chi }_{{A}_{n}}\left( t\right)\n\]\n\nwhere \... | Yes |
Proposition 4.3.11. For every infinite compact Hausdorff space \( K,\mathcal{C}\left( K\right) \) contains a subspace isometric to \( {c}_{0} \) . If \( K \) is metrizable, this subspace is complemented. | Proof. Let \( \left( {U}_{n}\right) \) be a sequence of nonempty, disjoint open subsets of \( K \) . Such a sequence can be found by induction: simply pick \( {U}_{1} \) such that \( {K}_{1} = K \smallsetminus \overline{{U}_{1}} \) is infinite, and then take \( {U}_{2} \subset {K}_{1} \) such that \( {K}_{2} = {K}_{1} ... | Yes |
Proposition 4.3.12. If \( \mathcal{C}\left( K\right) \) is order-complete and \( K \) is metrizable, then \( K \) is finite. | Proof. If \( K \) is infinite, \( \mathcal{C}\left( K\right) \) contains a complemented copy of \( {c}_{0} \) by Proposition 4.3.11. But if, moreover, \( \mathcal{C}\left( K\right) \) is isometrically injective, this would make \( {c}_{0} \) injective, which is false, because \( {c}_{0} \) is uncomplemented in \( {\ell... | Yes |
Corollary 4.3.13. The only isometrically injective separable Banach spaces are finite-dimensional and isometric to \( {\ell }_{\infty }^{n} \) for some \( n \in \mathbb{N} \) . | Proof. If \( X \) is an isometrically injective Banach space, by Theorem 4.3.7, \( X \) can be identified with an order-complete \( \mathcal{C}\left( K\right) \) -space for some compact Hausdorff \( K \) . Since \( X \) is separable, Theorem 4.1.3 yields that \( K \) is metrizable, and by Proposition 4.3.12, \( K \) mu... | Yes |
Proposition 4.4.1. If \( K \) is an infinite compact metric space, then \( \mathcal{C}\left( K\right) \approx \mathcal{C}\left( K\right) \oplus \mathbb{R} \) . Hence \( \mathcal{C}\left( K\right) \) is isomorphic to its hyperplanes. | Proof. By Proposition 4.3.11, \( \mathcal{C}\left( K\right) \approx E \oplus {c}_{0} \approx E \oplus {c}_{0} \oplus \mathbb{R} \) for some subspace \( E \) . Hence \( \mathcal{C}\left( K\right) \approx \mathcal{C}\left( K\right) \oplus \mathbb{R} \) .\n\nThe latter statement of the proposition follows from the fact th... | Yes |
Proposition 4.4.3. (i) If \( K \) is a compact metric space, then \( K \) is homeomorphic to a closed subset of the Hilbert cube \( {\left\lbrack 0,1\right\rbrack }^{\mathbb{N}} \) . | Proof. We have already shown \( \left( i\right) \) in the proof of Theorem 1.4.4. Just take \( {\left( {f}_{n}\right) }_{n = 1}^{\infty } \) a dense sequence in \( \left\{ {f \in \mathcal{C}\left( K\right) : 0 \leq f \leq 1}\right\} \) and define the map \( \sigma : K \rightarrow {\left\lbrack 0,1\right\rbrack }^{\math... | Yes |
Proposition 4.4.5. \( \mathcal{C}\left( \Delta \right) \approx {c}_{0}\left( {\mathcal{C}\left( \Delta \right) }\right) \) . | Proof. Since \( \mathcal{C}\left( \Delta \right) \) is isomorphic to its hyperplanes (Proposition 4.4.1), it is isomorphic to the subspace \( Z = \{ f \in \mathcal{C}\left( \Delta \right) : f\left( {0,0,\ldots }\right) = 0\} \) . For each \( n \in \mathbb{N} \) let \( {\Delta }_{n} = \left\{ {{\left( {s}_{k}\right) }_{... | Yes |
Example 4.5.1. It is easy to construct examples of spaces \( K \) without finite Cantor-Bendixson index. | Let us note, first, that if \( E \) is any closed subset of \( K \) then \( {E}^{\prime } \subset \) \( {K}^{\prime } \), and therefore \( \sigma \left( E\right) \leq \sigma \left( K\right) \) . If \( K \) is a countable compact metric space, then \( {K}_{1} = K \times \gamma \mathbb{N} \) has the property that \( {\le... | Yes |
Lemma 5.1.1. Let \( {\left( {f}_{n}\right) }_{n = 1}^{\infty } \) be a sequence of norm-one, disjointly supported functions in \( {L}_{1}\left( \mu \right) \) . Then \( {\left( {f}_{n}\right) }_{n = 1}^{\infty } \) is a norm-one complemented basic sequence, isometrically equivalent to the canonical basis of \( {\ell }_... | Proof. For any scalars \( {\left( {\alpha }_{i}\right) }_{i = 1}^{n} \) and any \( n \in \mathbb{N} \) , \[ {\begin{Vmatrix}\mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{f}_{i}\end{Vmatrix}}_{1} = {\int }_{\Omega }\left| {\mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{f}_{i}}\right| {d\mu } \] \[ = {\int }_{\Omega... | Yes |
Example 5.2.3. (ii) The closed unit ball of \( {L}_{2}\left( \mu \right) \) is an equi-integrable subset of \( {L}_{1}\left( \mu \right) \) . | Indeed, for every \( f \in {B}_{{L}_{2}\left( \mu \right) } \) and measurable set \( E \), by the Cauchy-Schwarz inequality,\n\n\[ \n{\int }_{E}\left| f\right| {d\mu } \leq {\left( {\int }_{E}1d\mu \right) }^{1/2}{\left( {\int }_{E}{\left| f\right| }^{2}d\mu \right) }^{1/2} \leq {\left( \mu \left( E\right) \right) }^{1... | Yes |
Lemma 5.2.5. Suppose \( \mathcal{F} \) is a bounded subset of \( {L}_{1}\left( \mu \right) \) . Then the following are equivalent:\n\n(i) \( \mathcal{F} \) is equi-integrable;\n\n(ii) \( \mathop{\lim }\limits_{{M \rightarrow \infty }}\mathop{\sup }\limits_{{f \in \mathcal{F}}}{\int }_{\{ \left| f\right| > M\} }\left| f... | Proof. (i) \( \Rightarrow \) (ii) Since \( \mathcal{F} \) is bounded, there is a constant \( A > 0 \) such that \( \mathop{\sup }\limits_{{f \in \mathcal{F}}}\parallel f{\parallel }_{1} \leq A \) . Given \( f \in \mathcal{F} \), by Chebyshev’s inequality,\n\n\[ \mu \left( {\{ \left| f\right| > M\} }\right) \leq \frac{\... | Yes |
Lemma 5.2.6. Suppose \( {\left( {f}_{n}\right) }_{n = 1}^{\infty } \) is an equi-integrable sequence in \( {L}_{1}\left( \mu \right) \) that converges a.e. to some \( g \in {L}_{1}\left( \mu \right) \). Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{\Omega }{f}_{n}{d\mu } = {\int }_{\Omega }{gd\mu }... | Proof. For each \( M > 0 \) let us consider the truncations\n\n\[ {f}_{n}^{\left( M\right) } = \left\{ {\begin{array}{ll} M & \text{ if }{f}_{n} > M, \\ {f}_{n} & \text{ if }\left| {f}_{n}\right| \leq M, \\ - M & \text{ if }{f}_{n} < - M, \end{array}\;{g}^{\left( M\right) } = \left\{ \begin{array}{ll} M & \text{ if }g ... | Yes |
Theorem 5.2.9. The space \( {L}_{1}\left( \mu \right) \) is weakly sequentially complete. | Proof. Let \( {\left( {f}_{n}\right) }_{n = 1}^{\infty } \subset {L}_{1}\left( \mu \right) \) be a weakly Cauchy sequence. Then, no subsequence of \( {\left( {f}_{n}\right) }_{n = 1}^{\infty } \) can be equivalent to the canonical \( {\ell }_{1} \) -basis, which is not weakly Cauchy. Hence the set \( {\left\{ {f}_{n}\r... | Yes |
Corollary 5.2.10. The space \( {c}_{0} \) is not isomorphic to a subspace of \( {L}_{1}\left( \mu \right) \) . | Proof. Since \( {L}_{1}\left( \mu \right) \) is weakly sequentially complete, by Corollary 2.4.15 every WUC series in \( {L}_{1}\left( \mu \right) \) is unconditionally convergent, so by Theorem 2.4.11, \( {L}_{1}\left( \mu \right) \) does not contain a copy of \( {c}_{0} \) . | Yes |
Theorem 5.3.2. Let \( \mathcal{A} \) be a bounded subset of \( \mathcal{M}\left( K\right) \) . The following are equivalent:\n\n(i) \( \mathcal{A} \) is relatively weakly compact;\n\n(ii) \( \mathcal{A} \) is uniformly regular;\n\n(iii) for every sequence of disjoint Borel sets \( {\left( {B}_{n}\right) }_{n = 1}^{\inf... | Proof. (iii) \( \Rightarrow \) (iv) This is immediate, because an open set is a Borel set and\n\n\[ 0 \leq \left| {{\mu }_{n}\left( {U}_{n}\right) }\right| \leq \left| {\mu }_{n}\right| \left( {U}_{n}\right) \overset{n \rightarrow \infty }{ \rightarrow }0. \]\n\n\( \left( {iv}\right) \Rightarrow {\left( iv\right) }^{\p... | Yes |
Theorem 5.4.8. Let \( T : {L}_{1}\left( \mu \right) \rightarrow {L}_{1}\left( \mu \right) \) or \( T : \mathcal{C}\left( K\right) \rightarrow \mathcal{C}\left( K\right) \) be a weakly compact operator. Then \( {T}^{2} \) is compact. | Proof. This is immediate. For example, in the first case, \( T\left( {B}_{{L}_{1}\left( \mu \right) }\right) \) is relatively weakly compact; hence \( {T}^{2}\left( {B}_{{L}_{1}\left( \mu \right) }\right) \) is relatively norm-compact. | Yes |
A Banach space \( X \) has the Radon-Nikodym property if and only if every Lipschitz map from the unit interval \( \left\lbrack {0,1}\right\rbrack \) into \( X \) is differentiable almost everywhere. | Proof. Assume every Lipschitz map on \( \left\lbrack {0,1}\right\rbrack \) is differentiable almost everywhere. Given a bounded linear operator \( T : {L}_{1}\left\lbrack {0,1}\right\rbrack \rightarrow X \), consider\n\n\[ f : \left\lbrack {0,1}\right\rbrack \rightarrow X,\;t \mapsto f\left( t\right) = T\left( {\chi }_... | Yes |
Neither of the spaces \( {L}_{1} \) nor \( {c}_{0} \) has the Radon-Nikodym property. | Indeed, in \( {L}_{1} \) the Lipschitz map\n\n\[ f\left( t\right) = {\chi }_{\left( 0, t\right) },\;0 \leq t \leq 1, \]\n\nis nowhere differentiable. In \( {c}_{0} \) we can consider the Lipschitz map\n\n\[ g\left( t\right) = {\left( \frac{1}{n}\sin \left( nt\right) \right) }_{n = 1}^{\infty },\;0 \leq t \leq 1, \]\n\n... | Yes |
Theorem 5.6.1. Let \( K \) be a compact Hausdorff space. If \( T : \mathcal{C}\left( K\right) \rightarrow X \) is weakly compact, then \( T \) is strictly singular. | Proof. Let \( Y \) be a subspace of \( \mathcal{C}\left( K\right) \) such that \( {\left. T\right| }_{Y} \) is an isomorphism onto its image. Since \( T \) is weakly compact, \( T\left( {B}_{Y}\right) \) is relatively weakly compact, which implies that \( {B}_{Y} \) is weakly compact. But \( T\left( {B}_{Y}\right) \) i... | Yes |
Theorem 5.6.5. Suppose \( X \) is an injective Banach space and \( T : X \rightarrow Y \) is a bounded linear operator. If \( T \) fails to be weakly compact, then there is a closed subspace \( F \) of \( X \) such that \( F \) is isomorphic to \( {\ell }_{\infty } \) and \( {\left. T\right| }_{F} \) is an isomorphism. | Proof. We start by embedding \( X \) isometrically into an \( {\ell }_{\infty }\left( \Gamma \right) \) -space; this can be done by taking \( \Gamma = {B}_{{X}^{ * }} \) and using the embedding \( x \mapsto \widehat{x} \), where \( \widehat{x}\left( {x}^{ * }\right) = {x}^{ * }\left( x\right) \) . Since \( X \) is inje... | Yes |
Proposition 5.7.2. If \( X \) is a nonreflexive subspace of \( {L}_{1}\left( \mu \right) \), then \( X \) contains a subspace isomorphic to \( {\ell }_{1} \) and complemented in \( {L}_{1}\left( \mu \right) \) . | Proof. If \( X \) is nonreflexive, its closed unit ball \( {B}_{X} \) is not weakly compact; therefore, \( {B}_{X} \) is not an equi-integrable set in \( {L}_{1}\left( \mu \right) \) . The proposition then follows from Theorem 5.2.8. | No |
Proposition 5.7.4. Let \( K \) be a compact metric space. If \( X \) is an infinite-dimensional complemented subspace of \( \mathcal{C}\left( K\right) \), then \( X \) contains a complemented subspace isomorphic to \( {c}_{0} \) . | Proof. Again by Proposition 5.7.1, \( X \) is nonreflexive, and hence every projection \( P \) onto it fails to be weakly compact. By Theorem 5.6.3, \( X \) must contain a subspace isomorphic to \( {c}_{0} \), and this subspace must be complemented, because (since \( K \) is metrizable) \( X \) is separable (by Sobczyk... | No |
Theorem 5.7.5. The space \( {\ell }_{\infty } \) is prime. | Proof. Let \( X \) be an infinite-dimensional complemented subspace of \( {\ell }_{\infty } \) . We have already seen that \( X \) cannot be reflexive (Proposition 5.7.1), and hence a projection \( P \) onto \( X \) cannot be weakly compact. In this case we can use Theorem 5.6.5 to deduce that \( X \) contains a copy o... | Yes |
Corollary 5.7.6. There are no infinite-dimensional separable injective Banach spaces. | Proof. Suppose that \( X \) is a separable injective space. Then \( X \) embeds isometrically into \( {\ell }_{\infty } \) by Theorem 2.5.7. Since \( X \) is injective, it embeds complementably into \( {\ell }_{\infty } \), which is a prime space. That forces \( X \) to be isomorphic to \( {\ell }_{\infty } \), a contr... | Yes |
Lemma 6.1.2. Let \( \left( {\Omega ,\sum ,\mu }\right) \) be a probability measure space and suppose \( {\sum }^{\prime } \) is a sub- \( \sigma \) -algebra of \( \sum \) . Then \( \mathbb{E}\left( {\cdot \mid {\sum }^{\prime }}\right) \) is a norm-one linear projection from \( {L}_{p}\left( {\Omega ,\sum ,\mu }\right)... | Proof. Fix \( 1 \leq p \leq \infty \) . It is immediate to check that \( \mathbb{E}{\left( \cdot \mid {\sum }^{\prime }\right) }^{2} = \mathbb{E}\left( {\cdot \mid {\sum }^{\prime }}\right) \) . If \( f \in {L}_{p}\left( \mu \right) \), using Hölder’s inequality in \( {L}_{p}\left( {\Omega ,{\sum }^{\prime },\mu }\righ... | No |
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