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Proposition 7.2.8. Let \( \Omega \) be a bounded domain in \( {\mathbb{C}}^{n} \) with a boundary point \( p \in \) \( \partial \Omega \) which admits a local holomorphic peak function for \( \Omega \) . Let \( K \) be a compact subset of \( \Omega \) and let \( q \in \Omega \) . Then, for every open neighborhood \( U ... | Proof. Note first that a local holomorphic peak function \( h \) at \( p \) generates the local plurisubharmonic peak function \( \log \left| h\right| \) at \( p \) .\n\nSince the Kobayashi pseudo-distance \( {d}_{M} : M \times M \rightarrow \mathbb{R} \) is continuous for any complex manifold \( M \), we may select \(... | Yes |
Theorem 7.3.2. There exists an open neighborhood \( U \) of the origin in \( {\mathbb{C}}^{n} \) such that\n\n\[ \mathop{\lim }\limits_{{v \rightarrow \infty }}\mathop{\sup }\limits_{{\xi \in {\mathbb{C}}^{n},\left| \xi \right| = 1}}\left| {\frac{2 - {S}_{{\Omega }_{v} \cap U}\left( {{p}_{v};\xi }\right) }{2 - {S}_{{\O... | The conclusion of this statement implies: as soon as \( \mathop{\lim }\limits_{{v \rightarrow \infty }}{S}_{{\Omega }_{v} \cap U}\left( {{p}_{v};\xi }\right) \) exists, it will coincide with \( \mathop{\lim }\limits_{{v \rightarrow \infty }}{S}_{{\Omega }_{v}}\left( {{p}_{v};\xi }\right) \) .\n\nWe now demonstrate how ... | No |
Proposition 1.3. Suppose \( X \) is a topological space. Then:\n\n(a) If \( A \subset X \), then \( {A}^{ - } \) is the set of limits of nets from \( A \) . | Proof. (a1) Suppose \( \left\langle {x}_{\alpha }\right\rangle \) is a net in \( A \), and \( {x}_{\alpha } \rightarrow x \) . If \( U \) is any open neighborhood of \( x \), then there exists \( \alpha \) s.t. \( \beta \succ \alpha \Rightarrow {x}_{\beta } \in U \) . In particular, \( {x}_{\alpha } \in U \cap A \), so... | Yes |
Proposition 1.4. Suppose \( X \) is a topological space, \( \left\langle {{x}_{\alpha } : \alpha \in D}\right\rangle \) is a net in \( X \) , and \( {D}^{\prime } \) is cofinal in \( D \) . Then \( {D}^{\prime } \) is directed, and \( \mathop{\lim }\limits_{D}{x}_{\alpha } = x \Rightarrow \mathop{\lim }\limits_{{D}^{\p... | Proof. \( {D}^{\prime } \) is directed: If \( \alpha ,\beta \in {D}^{\prime } \), then \( \alpha ,\beta \in D \), so there exists \( \gamma \in D \) with \( \gamma \succ \alpha \) and \( \gamma \succ \beta .{D}^{\prime } \) is cofinal, so there exists \( \delta \in {D}^{\prime } \) with \( \delta \succ \gamma \), whenc... | Yes |
Proposition 1.5. Suppose \( X \) is a compact topological space, and \( \left\langle {{x}_{\alpha } : \alpha \in D}\right\rangle \) is a net in \( X \) . Then \( \left\langle {x}_{\alpha }\right\rangle \) has a cluster point in \( X \) . | Proof. For all \( \alpha \in D \), set\n\n\[ \n{A}_{\alpha } = \left\{ {{x}_{\beta } : \beta \succ \alpha }\right\} \text{and} \n\]\n\n\[ \n{C}_{\alpha } = {A}_{\alpha }^{ - }\text{.} \n\]\n\nObserve that if \( \alpha ,\beta \in D \), then there exists \( \gamma \in D \) for which \( \gamma \succ \alpha \) and \( \gamm... | Yes |
Proposition 1.6. Suppose \( X \) is a Hausdorff space. Then the following are equivalent:\n\n(i) Each point of \( X \) has a neighborhood base consisting of compact sets.\n\n(ii) Each point of \( X \) has an open neighborhood with compact closure.\n\n(iii) Each point of \( X \) has a compact neighborhood. | Proof. Clearly (i) \( \Rightarrow \) (iii) and (ii) \( \Rightarrow \) (iii), whether \( X \) is Hausdorff or not, while (iii) \( \Rightarrow \) (ii) for Hausdorff spaces by taking a compact (hence closed) neighborhood \( K \) of a point \( x : x \in \operatorname{int}\left( K\right) \), while \( \operatorname{int}{\lef... | Yes |
Proposition 1.7. Suppose \( X \) is a topological space. Then \( X \) is regular if, and only if, \( X \) is locally closed. | Proof. First, suppose \( X \) is regular, \( x \in X \), and \( U \) is open, with \( x \in U \) . Then \( A = X - U \) is closed and does not contain \( x \), so there exists disjoint open \( V \) and \( W \) with \( x \in V \) and \( A \subset W \) . But now \( x \in V \subset {V}^{ - } \subset X - W \subset X - A = ... | Yes |
Proposition 1.8. Suppose \( G \) is a group, and \( {\mathcal{B}}_{e} \) is a nonempty collection of subsets of \( G \) ; each containing the identity, \( e \) ; satisfying:\n\n(i) For all \( {B}_{1} \in {\mathcal{B}}_{e} \) there exists \( {B}_{2} \in {\mathcal{B}}_{e} \) s.t. \( {B}_{2}^{-1} \subset {B}_{1} \) .\n\n(... | Proof. 1. \( \mathcal{T} \) is a topology: Suppose \( \left( {{U}_{\alpha } : \alpha \in \mathcal{O}}\right\} \) is a collection of members of \( \mathcal{T} \) . If \( x \) belongs to their union, \( U \), then \( x \in {U}_{\alpha } \) for some \( \alpha \), so \( {xB} \subset {U}_{\alpha } \subset U \) for some \( B... | Yes |
Proposition 1.9. Suppose \( G \) is a topological group, and \( A \subset G \) . Suppose \( {\mathcal{B}}_{e} \) is a neighborhood base at the identity, e. Then\n\n\[{A}^{ - } = \mathop{\bigcap }\limits_{{B \in {\mathcal{B}}_{e}}}{AB}\] | Proof. We show that if \( x \in {A}^{ - } \), then \( x \in {AB} \) for all \( B \in {\mathcal{B}}_{e} \) ; while if \( x \notin {A}^{ - } \) , then \( x \notin {AB} \) for some \( B \in {\mathcal{B}}_{e} \) .\n\nFirst, suppose \( x \in {A}^{ - } \) . If \( B \in {\mathcal{B}}_{e} \), then \( x{\left( \operatorname{int... | Yes |
Corollary 1.10. If \( G \) is a topological group, and \( {\mathcal{B}}_{e} \) is a neighborhood base at the identity \( e \), then \( \left\{ {{B}^{ - } : B \in {\mathcal{B}}_{e}}\right\} \) is also a neighborhood base at \( e \) . In particular, \( G \) is regular. | Proof. If \( B \in {\mathcal{B}}_{e} \), then \( e \in \operatorname{int}\left( B\right) \subset \operatorname{int}\left( {B}^{ - }\right) \) since \( B \subset {B}^{ - } \) . It remains to show that any open set \( U \), with \( e \in U \), there is a \( B \in {\mathcal{B}}_{e} \) with \( {B}^{ - } \subset U \) . But ... | Yes |
Corollary 1.11. Suppose \( G \) is a topological group, with identity \( e \), and \( {\mathcal{B}}_{e} \) is a neighborhood base at \( e \) . Then the following are equivalent\n\n(i) \( G \) is \( {T}_{0} \) .\n\n(ii) \( G \) is \( {T}_{1} \) .\n\n(iii) \( G \) is \( {T}_{2} \) .\n\n(iv) \( G \) is \( {T}_{3} \) .\n\n... | Proof. We know that \n\nfor general topological reasons. Corollary 1.10 says that (ii) \( \Rightarrow \) (iv), since regular \( + {T}_{1} = {T}_{3} \) . Also, \( \left( \mathrm{v}\right) \Rightarrow \left( \mathrm{{vi}... | Yes |
Proposition 1.12. Suppose \( G \) is a topological group, with identity \( e \) . Then there is a neighborhood base \( {\mathcal{B}}_{e} \) at e such that for all \( B \in {\mathcal{B}}_{e} : B = {B}^{-1} \) . Furthermore, the members of \( {\mathcal{B}}_{e} \) may also be assumed to all be open, or they may be assumed... | Proof. Start with any neighborhood base \( {\mathcal{B}}_{e}^{\prime } \) at \( e \) ; even \( \{ U : U \) open, \( e \in U\} \) will do. Set\n\n\[ \n{\mathcal{B}}_{e} = \left\{ {B \cap {B}^{-1} : B \in {\mathcal{B}}_{e}^{\prime }}\right\} \n\]\n\nSince inversion is a homeomorphism, \( {\mathcal{B}}_{e} \) will consist... | Yes |
Theorem 1.13. Suppose \( G \) is a first countable topological group, and \( U \) is an open set containing the identity, e. Then there exists a neighborhood base \( {\mathcal{B}}_{e} = \) \( \left\{ {{B}_{1},{B}_{2},{B}_{3},\ldots }\right\} \) at e such that:\n\n(a) \( {B}_{1} \subset U \), and\n\n(b) For all \( j : {... | Proof. Start with a countable neighborhood base \( {\mathcal{B}}_{e}^{\prime } \) at \( e \), and manufacture \( {\mathcal{B}}_{e}^{\prime \prime } = \) \( \left\{ {B \cap {B}^{-1} : B \in {\mathcal{B}}_{e}}\right\} ;{\mathcal{B}}_{e}^{\prime \prime } \) is also countable. It is a neighborhood base by the proof of Prop... | Yes |
Theorem 1.14. Suppose \( G \) is a topological group, and suppose \( {\mathcal{B}}_{e} \) is a neighborhood base at the identity, e. Suppose \( A \) is a closed subset of \( G \) and \( K \) is a compact subset of \( G \), with \( A \cap K = \varnothing \) . Then there exists \( B \in {\mathcal{B}}_{e} \) such that \( ... | Proof. The proof comes in two steps.\n\nStep 1. There exists \( B \in {\mathcal{B}}_{e} \) for which \( {\left( AB\right) }^{ - } \cap K = \varnothing \) . Suppose not. Set \( {C}_{B} = {\left( AB\right) }^{ - } \cap K \) and \( \mathcal{C} = \left\{ {{C}_{B} : B \in {\mathcal{B}}_{e}}\right\} .\mathcal{C} \) is a coll... | Yes |
Corollary 1.15. Suppose \( G \) is a topological group, and suppose \( A \) is a closed subset, and \( K \) is a compact subset. Then \( {AK} \) and \( {KA} \) are closed subsets of \( G \) . | Proof. \( {AK} \) is closed: It suffices to show that \( x \notin {AK} \Rightarrow x \notin {\left( AK\right) }^{ - } \) . Suppose \( x \notin {AK} \) . Then \( {A}^{-1}x \cap K = \varnothing \) by property (vi) of set products. \( A \) is closed, so \( {A}^{-1} \) is closed, as is \( {A}^{-1}x \) . By Theorem 1.14, th... | Yes |
Proposition 1.16. Suppose \( A \) and \( B \) are subsets of a topological group \( G \) . Then \( \left( {A}^{ - }\right) \left( {B}^{ - }\right) \subset {\left( AB\right) }^{ - } \) . | Proof. If \( a \in A \), set \( {L}_{a}\left( x\right) = {ax}.{L}_{a} \) is continuous, so \( {L}_{a}^{-1}\left( {\left( AB\right) }^{ - }\right) \) is closed. \( {L}_{a}\left( B\right) = {aB} \subset {AB} \subset {\left( AB\right) }^{ - } \), so \( B \subset {L}_{a}^{-1}\left( {\left( AB\right) }^{ - }\right) \) : thu... | Yes |
Lemma 1.17. Suppose \( X \) and \( Y \) are topological spaces, \( f : X \rightarrow Y \) is a function, and \( \mathcal{B} \) is a subbase for the topology of \( Y \) . If \( {f}^{-1}\left( B\right) \) is open in \( X \) for all \( B \in \mathcal{B} \) , then \( f \) is continuous. | Proof. Suppose \( {f}^{-1}\left( B\right) \) is open in \( X \) for all \( B \in \mathcal{B} \) . Let \( {\mathcal{B}}_{0} \) denote the set of all finite intersections of members of \( \mathcal{B} \) . Since \( {f}^{-1}\left( {{B}_{1} \cap \cdots \cap {B}_{n}}\right) = {f}^{-1}\left( {B}_{1}\right) \cap \cdots \cap \)... | Yes |
Theorem 1.18. Suppose \( \left\langle {{X}_{i} : i \in \mathcal{I}}\right\rangle \) is a family of topological spaces. Then the product topology on \( \prod {X}_{i} \) is the unique topology with the following property: Whenever \( Y \) is a topological space, and\n\n\[ \mathbf{f} = \left\langle {f}_{i}\right\rangle : ... | Proof. Let \( {\pi }_{{i}_{0}} : \prod {X}_{i} \rightarrow {X}_{{i}_{0}} \) be a projection. If \( \mathbf{f} \) is continuous, then each \( {f}_{{i}_{0}} = \) \( {\pi }_{{i}_{0}} \circ \mathbf{f} \) is continuous. On the other hand, if each \( {f}_{i} \) is continuous, then since\n\n\[ {\mathbf{f}}^{-1}\left( {\mathop... | Yes |
Corollary 1.19. Suppose \( \left\langle {{X}_{ij},\left( {i, j}\right) \in \mathcal{I} \times \mathcal{J}}\right\rangle \) is a family of topological spaces parametrized by a set product \( \mathcal{I} \times \mathcal{J} \) . Then the set bijections\n\n\[ \mathop{\prod }\limits_{{i \in \mathcal{I}}}\mathop{\prod }\limi... | Proof. The underlying idea is pretty simple: Show that the product-of-product topology on \( \prod \prod {X}_{i, j} \) has the property specified in Theorem 1.18. Since the situation is symmetric, we use the first bijection. Suppose \( Y \) is a topological space, and \( \mathbf{f} = \left\langle {f}_{i, j}\right\rangl... | Yes |
Corollary 1.20. Suppose \( \left\langle {{G}_{i}, i \in \mathcal{I}}\right\rangle \) is a family of topological groups. Then with the product topology, \( \prod {G}_{i} \) is a topological group. | Proof. 1. Inversion is continuous: Since \( {\left\langle {x}_{i}\right\rangle }^{-1} = \left\langle {x}_{i}^{-1}\right\rangle \), the \( {i}_{0} \) th coordinate function of \( g \mapsto {g}^{-1} \) is \( \left\langle {x}_{i}\right\rangle \mapsto {x}_{{i}_{0}}^{-1} \), that is it is the composite \( g \mapsto {\pi }_{... | Yes |
Proposition 1.21. Suppose \( \left( {X,{\mathcal{T}}_{X}}\right) \) is a topological space, and \( A \) is a subspace, with subspace topology \( {\mathcal{T}}_{A} \) . Then \( \mathcal{T} = {\mathcal{T}}_{A} \) is the unique topology on \( A \) with the following property: Whenever \( Y \) is a topological space, and \... | Proof. If \( f : Y \rightarrow A \) is a function, then for all \( U \subset X : {f}^{-1}\left( U\right) = {f}^{-1}\left( {U \cap A}\right) \) . But \( {f}^{-1}\left( U\right) \) is a typical inverse image of a member of \( {\mathcal{T}}_{X} \), while \( {f}^{-1}\left( {U \cap A}\right) \) is a typical inverse image of... | Yes |
Corollary 1.22. Suppose \( G \) is a topological group, and \( H \) is a subgroup. Then with the induced topology, \( H \) is a topological group. | Proof.\n\n\[ H \times H \hookrightarrow G \times G\xrightarrow[]{\text{ multiplication }}G \]\n\nand\n\n\[ H \hookrightarrow G\xrightarrow[]{\text{ inversion }}G \]\n\nare continuous and take values in \( H \) . Apply Proposition 1.21. | Yes |
Theorem 1.23. Suppose \( G \) is a topological group, and \( H \) is a subgroup. Then:\n\n(a) \( {\mathcal{T}}_{G/H} \) is a topology on \( G/H \) . | Proof. These nine properties are not proved in the order given, but (a) definitely comes first. Let \( \pi : G \rightarrow G/H \) be the natural map.\n\n(a) \( \varnothing = \varnothing H/H \in {\mathcal{T}}_{G/H} \) and \( G/H = {GH}/H \in {\mathcal{T}}_{G/H}. \cup {U}_{i}H/H = \cup \pi \left( {U}_{i}\right) = \) \( \... | Yes |
Proposition 1.24. Suppose \( G \) is a topological group, and \( H \) is a subgroup. Then \( {H}^{ - } \) is also a subgroup. If \( H \) is normal, then so is \( {H}^{ - } \) . | Proof. \( {H}^{ - } \) is closed under multiplication, since \( \left( {H}^{ - }\right) \cdot \left( {H}^{ - }\right) \subset {\left( H \cdot H\right) }^{ - } = \) \( {H}^{ - } \) by Proposition 1.16. \( {H}^{ - } \) is closed under the inversion map \( I \) since \( I \) is a homeomorphism and so preserves closures:\n... | Yes |
Proposition 1.25. Suppose \( G \) is a topological group with identity \( e, H \) is a subgroup of \( G \), and \( C \) is a closed subset of \( G \) with \( e \in \operatorname{int}\left( C\right) \) . Finally, suppose \( H \cap C \) is closed. Then \( H \) is closed. | Proof. The open neighborhoods of \( e \) form a neighborhood base at \( e \), so there exist open neighborhoods \( {W}_{1} \) and \( {W}_{2} \) of \( e \) with \( {W}_{1}{W}_{2} \subset \operatorname{int}\left( C\right) \) . Suppose \( p \in {H}^{ - } \) . Then \( p{W}_{1}^{-1} \cap H \neq \varnothing \), so choose \( ... | Yes |
Proposition 1.26. Suppose \( G \) and \( \widetilde{G} \) are topological groups with identity elements e and \( \widetilde{e} \), respectively. Suppose \( {\mathcal{B}}_{e} \) is a neighborhood base at \( e \), and \( {\mathcal{B}}_{\widetilde{e}} \) is a neighborhood base at \( \widetilde{e} \) . Finally, suppose \( ... | Proof. (a) If \( f \) is continuous, then \( {f}^{-1}\left( {\operatorname{int}\left( \widetilde{B}\right) }\right) \) is open, and \( e \in {f}^{-1}\left( \widetilde{e}\right) \subset \) \( {f}^{-1}\left( {\operatorname{int}\left( \widetilde{B}\right) }\right) \subset {f}^{-1}\left( \widetilde{B}\right) \), so \( e \i... | Yes |
Proposition 1.28. Suppose \( G \) is a complete Hausdorff topological group. Then each right Cauchy net converges. | Proof. Suppose \( \left\langle {x}_{\alpha }\right\rangle \) is a right Cauchy net defined on a directed set \( D \) . Set \( {y}_{\alpha } = {x}_{\alpha }^{-1} \) . Then for each neighborhood \( B \) of the identity \( e \) of \( G \), there exists \( {\alpha }_{0} \in D \) such that for all \( \beta ,\gamma \in D,\be... | Yes |
Proposition 1.29. In any Hausdorff topological group, a convergent net is both left Cauchy and right Cauchy. | Proof. Suppose \( \left\langle {x}_{\alpha }\right\rangle \) is a convergent net defined on a directed set \( D \), with \( {x}_{\alpha } \rightarrow x \) . Suppose \( {B}_{1} \) is an open neighborhood of the identity \( e \) of \( G \) . Choose open neighborhoods \( {B}_{2} \) and \( {B}_{3} \) of \( e \) for which \... | Yes |
Proposition 1.30. Suppose \( G \) is a Hausdorff topological group, and suppose \( A \) is a complete subset of \( G \) . Then \( A \) is closed in \( G \), and any closed subset of \( A \) is also complete. | Proof. \( A \) is closed: Suppose \( x \in {A}^{ - } \) . Then by Proposition 1.3(a), there is a net \( \left\langle {x}_{\alpha }\right\rangle \) defined on a directed set \( D \) such that \( {x}_{\alpha } \in A \) and \( {x}_{\alpha } \rightarrow x \) . But now \( \left\langle {x}_{\alpha }\right\rangle \) is left C... | No |
Proposition 1.31. Suppose \( G \) is a Hausdorff topological group, and suppose \( \left\langle {x}_{\alpha }\right\rangle \) is a left Cauchy net with a cluster point \( x \) . Then \( {x}_{\alpha } \rightarrow x \) . | Proof. Suppose \( \left\langle {x}_{\alpha }\right\rangle \) is a left Cauchy net defined on a directed set \( D \), with a cluster point \( x \) . Suppose \( U \) is open, with \( x \in U \) . Choose an open neighborhood \( {B}_{1} \) of the identity \( e \) in \( G \) for which \( x{B}_{1} \subset U \) . Choose open ... | Yes |
Corollary 1.32. Suppose \( G \) is a Hausdorff topological group, and \( K \) is a compact subset of \( G \) . Then \( K \) is complete. | Proof. Suppose \( \left\langle {x}_{\alpha }\right\rangle \) is a left Cauchy net in \( K \) . Then \( \left\langle {x}_{\alpha }\right\rangle \) has a cluster point \( x \in K \) by Proposition 1.5, and \( \lim {x}_{\alpha } = x \) by Proposition 1.31. | Yes |
Corollary 1.33. Suppose \( G \) is a Hausdorff topological group and suppose \( \left\langle {x}_{\alpha }\right. \) : \( \alpha \in D\rangle \) is a left Cauchy net in \( G \) . Suppose \( {D}^{\prime } \) is cofinal in \( D \), and suppose \( \mathop{\lim }\limits_{{D}^{\prime }}{x}_{\alpha } = x \) . Then \( \mathop... | Proof. In view of Proposition 1.31, it suffices to show that \( x \) is a cluster point of \( \left\langle {{x}_{\alpha } : \alpha \in D}\right\rangle \) . But if \( U \) is an open neighborhood of \( x \), then there exists \( \beta \in {D}^{\prime } \) such that \( \gamma \succ \beta \) and \( \gamma \in {D}^{\prime ... | Yes |
Theorem 1.34. Suppose \( G \) is a first countable Hausdorff topological group, and suppose \( G \) is sequentially complete. Then \( G \) is complete. | Proof. Suppose \( G \) is a first countable Hausdorff topological group in which each left Cauchy sequence converges. Suppose \( \left\langle {x}_{\alpha }\right\rangle \) is a left Cauchy net defined on a directed set \( D \) . Choose any neighborhood base at the identity \( e \) in \( G \) which is in accord with The... | Yes |
Theorem 1.35. Suppose \( G \) is a first countable Hausdorff topological group. Suppose \( {\mathcal{B}}_{e} = \left\{ {{B}_{1},{B}_{2},\ldots }\right\} \) is a neighborhood base at the identity \( e \) in \( G \) for which \( {B}_{j} = {B}_{j}^{-1} \supset {B}_{j + 1}^{2} \) for all \( G \) . Then \( G \) is complete ... | Proof. The \ | No |
Corollary 1.36. Suppose \( G \) is a first countable Hausdorff topological group, and \( H \) is a closed subgroup. Then: If \( G \) is complete, then so are \( H \) and \( G/H \) . | Proof. \( H \) is complete by Proposition 1.30. As for \( G/H \), suppose \( {\mathcal{B}}_{e} = \) \( \left\{ {{B}_{1},{B}_{2},{B}_{3},\ldots }\right\} \) is a neighborhood base at the identity \( e \) of \( G \), with \( {B}_{j} = {B}_{j}^{-1} \supset \) \( {B}_{j + 1}^{2} \) for all \( j \) . Then \( \left\{ {{B}_{j... | Yes |
Example 3. \( {L}^{p} \) -spaces, \( 0 < p < 1 \) . Let \( \left( {X,\mathcal{B},\mu }\right) \) denote a measure space and, as usual, declare two measurable functions to be equivalent when they are equal a.e. Letting \( \left\lbrack f\right\rbrack \) denote the function class of \( f \), set \[ {L}^{p}\left( \mu \righ... | The straightforward inequality \( {\left( s + t\right) }^{p} \leq {s}^{p} + {t}^{p} \) when \( s, t \geq 0 \) : \[ {\left( s + t\right) }^{p} - {s}^{p} = {\int }_{s}^{s + t}p{x}^{p - 1}{dx} = {\int }_{0}^{t}p{\left( x + s\right) }^{p - 1}{dx} \] \[ \leq {\int }_{0}^{t}p{x}^{p - 1}{dx} = {t}^{p} \] yields both the fact ... | Yes |
Theorem 2.1. Suppose \( \\left\\langle {{X}_{i}, i \\in \\mathcal{I}}\\right\\rangle \) is a family of topological vector spaces. Then with the product topology, \( \\prod {X}_{i} \) is a topological vector space. | Proof. \( \\prod {X}_{i} \) is a topological group by Corollary 1.20; the only issue is the joint continuity of scalar multiplication. Letting \( \\mathbb{F} \) denote the base field, note that the \ | No |
Theorem 2.2. Suppose \( X \) is a topological vector space, and \( Y \) is a vector subspace. Then with the induced topology, \( Y \) is a topological vector space. Also, with the quotient topology, \( X/Y \) is a topological vector space that is Hausdorff if, and only if, \( Y \) is closed. | Proof. Again, the only issue is scalar multiplication: \( Y \) is a topological group by Corollary 1.22, while \( X/Y \) is a topological group by Theorem 1.23(h); Theorem 1.23(g) addresses the Hausdorff condition. Letting \( \mathbb{F} \) denote the base field,\n\n\[ \mathbb{F} \times Y \hookrightarrow \mathbb{F} \tim... | Yes |
Proposition 2.3. Suppose \( X \) is a topological vector space over \( \mathbb{F} \), and \( {v}_{1},\ldots ,{v}_{n} \in X \) . Define \( T : {\mathbb{F}}^{n} \rightarrow X \) by \( T\left( {{c}_{1},\ldots ,{c}_{n}}\right) = \sum {c}_{j}{v}_{j} \) . Then \( T \) is continuous. | Proof. In fact,\n\n\[{\mathbb{F}}^{n} \times {X}^{n} \approx \mathop{\prod }\limits_{{i = 1}}^{n}\left( {\mathbb{F} \times X}\right) \xrightarrow[\text{ each factor }]{\text{ mult. in }}\mathop{\prod }\limits_{{i = 1}}^{n}X\xrightarrow[]{\text{ sum }}X\]\n\nis jointly continuous on \( {\mathbb{F}}^{n} \times {X}^{n} \)... | No |
Proposition 2.5. Suppose \( X \) is a topological vector space.\n\n(a) If \( B \) is balanced, then so is \( {B}^{ - } \) . | Proof. (a) If \( 0 \leq \left| c\right| \leq 1 \), and \( x \in {B}^{ - } \), take a net \( \left\langle {x}_{\alpha }\right\rangle ,{x}_{\alpha } \in B \), with \( {x}_{\alpha } \rightarrow x \) . Multiplication by \( c \) is continuous, so \( c{x}_{\alpha } \rightarrow {cx} \) . Since \( c{x}_{\alpha } \in B,{cx} \in... | Yes |
Proposition 2.7. Suppose \( X \) is a topological vector space.\n\n(a) If \( B \) is bounded, and \( V \) is a neighborhood of 0, then there is a real scalar \( {t}_{0} \geq 0 \) such that \( B \subset {cV} \) whenever \( \left| c\right| \geq {t}_{0} \) . | Proof. (a) Choose a balanced neighborhood \( W \) of 0 such that \( W \subset V \) . Then \( B \subset \) \( {c}_{0}W \) for some \( {c}_{0} \) . Set \( {t}_{0} = \left| {c}_{0}\right| \) . If \( \left| c\right| \geq {t}_{0} \), then \( \left| {{c}_{0}/c}\right| \leq 1 \), so \( \left( {{c}_{0}/c}\right) W \subset W \)... | Yes |
Proposition 2.8. Suppose \( B \) is a bounded neighborhood of 0 in a topological vector space \( X \) . Then \( \left\{ {{2}^{-n}B : n = 1,2,\ldots }\right\} \) is a neighborhood base at 0 . In particular, \( X \) is first countable. | Proof. If \( V \) is a neighborhood of 0, then \( \exists {t}_{0} \geq 0 \) such that \( B \subset {cV} \) when \( \left| c\right| \geq {t}_{0} \) . Choose \( n \in \mathbb{N} \) with \( {2}^{n} \geq {t}_{0} \) . Then \( B \subset {2}^{n}V \), so \( {2}^{-n}B \subset V \) . | Yes |
Proposition 2.9. Suppose \( \mathbb{F} = \mathbb{R} \) or \( \mathbb{C} \) . Then the product topology on \( {\mathbb{F}}^{n} \) is the only Hausdorff topological vector space topology on \( {\mathbb{F}}^{n} \) . | Proof. Let \( {\mathcal{T}}_{p} \) denote the product topology, and \( {\mathcal{T}}_{0} \) some other topology making \( {\mathbb{F}}^{n} \) into a Hausdorff topological vector space. Then by Proposition 2.3, \( \left( {{\mathbb{F}}^{n},{\mathcal{T}}_{p}}\right) \rightarrow \) \( \left( {{\mathbb{F}}^{n},{\mathcal{T}}... | Yes |
Corollary 2.10. In any Hausdorff topological vector space, finite-dimensional subspaces are closed. | Proof. It follows from Proposition 2.9 that there is exactly one way to topologize a finite-dimensional vector space over \( \mathbb{R} \) or \( \mathbb{C} \) and make it into a Hausdorff topological vector space, since any topology can be transported to \( {\mathbb{F}}^{n} \) using a basis. (The map is in Proposition ... | Yes |
Corollary 2.11. A locally compact Hausdorff topological vector space is finite-dimensional. | Proof. Suppose \( X \) is a locally compact Hausdorff topological vector space. Let \( V \) be an open neighborhood of 0 for which \( {V}^{ - } \) is compact (Proposition 1.6). Note that \( {V}^{ - } \) is bounded, so \( V \) is bounded, so \( \left\{ {{2}^{-n}V : n = 1,2,\ldots }\right\} \) is a neighborhood base at 0... | Yes |
Proposition 2.13. Suppose \( X \) is a topological vector space, and \( C \) is a convex subset of \( X \) . Then \( {C}^{ - } \) and \( \operatorname{int}\left( C\right) \) are also convex. | Proof. \( {C}^{ - } \) is convex: Suppose \( x \in C \) and \( y \in {C}^{ - } \), and \( 0 \leq t \leq 1 \) . If \( t = 0 \) or 1, then \( {tx} + \left( {1 - t}\right) y \in {C}^{ - } \) trivially, so assume \( 0 < t < 1 \) . In accordance with Proposition 1.3(a), \( y \) is a limit of a net \( \left\langle {y}_{\alph... | Yes |
Theorem 2.15. Suppose \( X \) is a topological vector space, and suppose \( B \) is a convex, balanced subset of \( X \) . Then the following are equivalent:\n\n(i) \( \operatorname{int}\left( B\right) \neq \varnothing \) .\n\n(ii) \( 0 \in \operatorname{int}\left( B\right) \) .\n\n(iii) \( \operatorname{int}\left( B\r... | Proof. Since \( B \neq \varnothing \Rightarrow 0 \in 0 \cdot B \subset B \) : (iii) \( \Rightarrow \) (ii); while (ii) \( \Rightarrow \) (i) trivially. To show (i) \( \Rightarrow \) (iii), assume \( x \in B, y \in \operatorname{int}\left( B\right) \), and \( 0 \leq t < 1 \) . Then \( - y \in B \) since \( B \) is balan... | Yes |
Proposition 2.16. Suppose \( \mu \) is Lebesgue measure on \( \left\lbrack {0,1}\right\rbrack \), and \( 0 < p < 1 \) . Suppose \( C \) is a convex subset of \( {L}^{p}\left( \mu \right) \) . Then \( \operatorname{int}\left( C\right) \neq \varnothing \Rightarrow C = {L}^{p}\left( \mu \right) \) . | Proof. If \( \left\lbrack h\right\rbrack \in \operatorname{int}\left( C\right) \), then \( \left\lbrack 0\right\rbrack \in \operatorname{int}\left( C\right) - \left\lbrack h\right\rbrack = \operatorname{int}\left( {C - \left\lbrack h\right\rbrack }\right) \), the latter equality holding since translation is a homeomorp... | Yes |
Proposition 2.17. Suppose \( X \) and \( Y \) are vector spaces over \( \mathbb{R} \) or \( \mathbb{C} \), and suppose \( T \) is a linear transformation from \( X \) to \( Y \) .\n\n(a) If \( B \) is convex in \( X \), then \( T\left( B\right) \) is convex in \( Y \) . | (a) If \( T\left( x\right), T\left( y\right) \in T\left( B\right) ;x, y \in B \) ; and \( 0 \leq t \leq 1 \) ; then \( {tT}\left( x\right) + \left( {1 - t}\right) T\left( y\right) = \) \( T\left( {{tx} + \left( {1 - t}\right) y}\right) \in T\left( B\right) \) . | Yes |
Proposition 2.18. Suppose \( X \) and \( Y \) are topological vector spaces, and \( T \) is a continuous linear transformation from \( X \) to \( Y \) . If \( B \) is a bounded subset of \( X \), then \( T\left( B\right) \) is a bounded subset of \( Y \) . | Proof. If \( V \) is a neighborhood of 0 in \( Y \), then \( {T}^{-1}\left( V\right) \) is a neighborhood of 0 in \( X \) , so \( B \subset c{T}^{-1}\left( V\right) = {T}^{-1}\left( {cV}\right) \) for some \( c \), whence \( T\left( B\right) \subset {cV} \) . | Yes |
Proposition 2.19. Suppose \( X \) is a vector space over \( \mathbb{C} \), and suppose \( B \) is a nonempty convex subset of \( X \) that is \( \mathbb{R} \)-balanced, that is \( {tB} \subset B \) for \( - 1 \leq t \leq 1 \), and suppose \( \mathcal{F} \) is a family of subsets of \( X \) such that for all \( A \subse... | Proof. Note that \( \{ 0\} = 0 \cdot B \subset B \), so \( 0 \in C \). It is also immediate that \( C \) is convex (since it is an intersection of convex sets).\n\n\( C \) is balanced: If \( x \in C \), and \( \left| c\right| \leq 1 \), write \( c = r{e}^{i\phi }, r \geq 0 \). Then for \( 0 \leq \theta < {2\pi }\)\n\n\... | Yes |
Let \( m \) denote the Lebesgue measure on \( \left\lbrack {0,1}\right\rbrack \) . Recall that if \( p < q \) , then \( {L}^{p}\left( m\right) \supset {L}^{q}\left( m\right) \) in this case, with \( \parallel f{\parallel }_{p} \leq \parallel f{\parallel }_{q} \) for \( f \in {L}^{q}\left( m\right) \). | writing \( 1 = 1/\left( {q/p}\right) + 1/r \), and noting that \( {\left| f\right| }^{p} \in {L}^{q/p}\left( m\right) \), Hölder’s inequality says that\n\n\[ \parallel f{\parallel }_{p}^{p} = {\int }_{0}^{1}{\left| f\right| }^{p} \cdot {1dm} \leq {\begin{Vmatrix}{\left| f\right| }^{p}\end{Vmatrix}}_{q/p} \cdot \paralle... | Yes |
Proposition 3.1. Suppose \( X \) is a locally convex space. Then \( X \) has a neighborhood base at 0 consisting of convex, balanced sets that can all be taken to be open or all closed. | Proof. Suppose \( U \) is open and \( 0 \in U \) . There is a convex set \( C \) with \( 0 \in \operatorname{int}\left( C\right) \) and \( C \subset U \), since \( X \) is locally convex. There is a balanced, open set \( W \) with \( 0 \in W \subset \) int \( \left( C\right) \) by Proposition 2.5(e). Let \( V \) denote... | Yes |
Theorem 3.2. Suppose \( X \) is a vector space over \( \mathbb{R} \) or \( \mathbb{C} \), and suppose \( {\mathcal{B}}_{0} \) is a nonempty family of convex, balanced, absorbent sets satisfying the following two conditions:\n\n( \( \alpha \) ) If \( B \in {\mathcal{B}}_{0} \), then \( \frac{1}{2}B \in {\mathcal{B}}_{0}... | Proof. First note that we do get a topological group via Proposition 1.8: running through conditions (i)-(iv) there:\n\n(i) \( \checkmark - B = B \) when \( B \in {\mathcal{B}}_{0} \) (balanced condition);\n\n(ii) \( \checkmark \frac{1}{2}B + \frac{1}{2}B \subset B \) when \( B \in {\mathcal{B}}_{0} \) ( \( B \) is con... | Yes |
Proposition 3.3. Suppose \( X \) is a vector space over \( \mathbb{R} \), and \( C \) is a convex subset of \( X \) with \( 0 \in C \) . Then:\n\n(a) If \( 0 < s < t \), then \( {sC} \subset {tC} \) . | Proof. (a) If \( x \in C \), then\n\n\[ \n{sx} = t \cdot \left( {\frac{s}{t}x + \left( {1 - \frac{s}{t}}\right) \cdot 0}\right) \in {tC}.\n\] | Yes |
Corollary 3.4. Suppose \( X \) is a vector space over \( \mathbb{R} \), and \( C \) is a convex subset of \( X \) with \( 0 \in C \) . Suppose 0 is an internal point of \( C \) . Then, in the notation of Proposition 3.3:\n\n(a) If \( x \in X \), and \( s > 0 \), then \( {I}_{sx} = s{I}_{x} \) .\n\n(b) If \( x, y \in X ... | Proof. (a) \( t \in {I}_{x} \Leftrightarrow x \in {tC} \Leftrightarrow {sx} \in {stC} \Leftrightarrow {st} \in {I}_{sx} \) .\n\n(b) If \( s \in {I}_{x} \) and \( t \in {I}_{y} \), then \( x \in {sC} \) and \( y \in {tC} \), so \( x + y \in {sC} + {tC} = \left( {s + t}\right) C \) , so \( s + t \in {I}_{x + y} \) . | Yes |
Proposition 3.6. Suppose \( X \) is a topological vector space over \( \mathbb{R} \), and \( p : X \rightarrow \mathbb{R} \) is a gauge. The following are equivalent:\n\n(i) \( p \) is continuous.\n\n(ii) \( p \) is continuous at 0 .\n\n(iii) 0 is interior to \( \{ x \in X : p\left( x\right) \leq 1\} \) . | Proof. (i) \( \Rightarrow \) (ii) is trivial, and (ii) \( \Rightarrow \) (iii) is as well, since the real number 0 is interior to \( ( - \infty ,1\rbrack \) .\n\n(iii) \( \Rightarrow \) (i): Given any \( {x}_{0}, x \in X \) :\n\n\[ p\left( x\right) = p\left( {x - {x}_{0} + {x}_{0}}\right) \leq p\left( {x - {x}_{0}}\rig... | Yes |
Theorem 3.7. Suppose \( X \) is a vector space over \( \mathbb{R} \) or \( \mathbb{C} \), and \( C \) is a convex subset of \( X \) with \( 0 \in C \) . Suppose 0 is an internal point of \( C \) . Then the Minkowski functional \( {p}_{C} \) is a gauge, and\n\n\[ \left\{ {x \in X : {p}_{C}\left( x\right) < 1}\right\} \s... | Proof. The fact that \( {p}_{C} \) is a gauge was established earlier. As for the containments:\n\n\[ {p}_{C}\left( x\right) < 1 \Rightarrow 1 \in {I}_{x} \Rightarrow {p}_{C}\left( x\right) \leq 1 \]\n\n\( \updownarrow \)\n\n\[ x \in C \]\n\nContinuity of \( {p}_{C} \Rightarrow \left\{ {x \in X : {p}_{C}\left( x\right)... | Yes |
Proposition 3.8. Suppose \( X \) is a vector space over \( \mathbb{R} \), and \( C \) is a convex subset of \( X \) with \( 0 \in C \) . Suppose 0 is an internal point of \( C \), and \( {p}_{C} \) is the associated Minkowski functional. Finally, suppose \( f : X \rightarrow \mathbb{R} \) is a linear functional. Then \... | Proof. In the notation of Proposition 3.3, \( {p}_{C}\left( x\right) \) is the left endpoint of the semi-infinite interval \( {I}_{x} \), so \( {p}_{C}\left( x\right) \) is the greatest lower bound for \( {I}_{x} \) . This is exactly what we need.\n\nFirst, suppose \( f\left( x\right) \leq {p}_{C}\left( x\right) \) for... | Yes |
Lemma 3.10. Suppose \( X \) is a locally convex space, and suppose \( C \) is a closed, convex set, with \( 0 \in \operatorname{int}\left( C\right) \) . Suppose \( {x}_{0} \notin C \) . Then there is a continuous linear functional \( F : X \rightarrow \mathbb{R} \) for which \( F\left( C\right) \subset ( - \infty ,1\rb... | Proof. Let \( {p}_{C} \) denote the Minkowski functional for \( C;{p}_{C} \) is a continuous gauge by Theorem 3.7. Since \( {p}_{C}\left( {x}_{0}\right) < 1 \Rightarrow {x}_{0} \in C \) (Theorem 3.7): \( {p}_{C}\left( {x}_{0}\right) \geq 1 \) . If \( {p}_{C}\left( {x}_{0}\right) = 1 \), then \( {p}_{C}\left( {\left( {1... | Yes |
Corollary 3.11. Suppose \( X \) is a locally convex space, and suppose \( C \) is a closed, convex set, with \( 0 \in C \) . Suppose \( {x}_{0} \notin C \) . Then there is a continuous linear functional \( F : X \rightarrow \mathbb{R} \) for which \( F\left( C\right) \subset ( - \infty ,1\rbrack \) and \( F\left( {x}_{... | Proof. Let \( {\mathcal{B}}_{0} \) be a base for the topology at zero consisting of open, convex, balanced sets. Since \( C \) is closed,\n\n\[ \n{x}_{0} \notin C = {C}^{ - } = \mathop{\bigcap }\limits_{{W \in {\mathcal{B}}_{0}}}C + W \n\]\n\nso \( {x}_{0} \notin C + W \) for some \( W \in {\mathcal{B}}_{0} \) . Set\n\... | Yes |
Proposition 3.12. Suppose \( X \) is a locally convex space, and \( {C}_{1} \) and \( {C}_{2} \) are two disjoint nonempty convex sets, with \( {C}_{1} \) closed and \( {C}_{2} \) compact. Then there is a real number \( {r}_{0} \) and a continuous linear functional \( F : X \rightarrow \mathbb{R} \) for which \( F\left... | Proof. Pick any \( {y}_{0} \in {C}_{1} - {C}_{2} \), a closed (Corollary 1.15) convex (Proposition 2.14 applied to \( C = 2{C}_{1}, D = - 2{C}_{2} \), and \( I = \left\lbrack {\frac{1}{2},\frac{1}{2}}\right\rbrack \) ) set. Then \( 0 \in \left( {{C}_{1} - {C}_{2} - {y}_{0}}\right) \) . Also, \( {x}_{0} = - {y}_{0} \not... | Yes |
Proposition 3.14. Suppose \( X \) is a locally convex topological vector space over \( \mathbb{C} \), and \( f : X \rightarrow \mathbb{R} \) is an \( \mathbb{R} \) -linear functional. Then there is a unique \( \mathbb{C} \) -linear functional \( F : X \rightarrow \mathbb{C} \) for which \( f\left( x\right) = \operatorn... | Proof. The formula starts things. First of all, if \( f\left( x\right) = \operatorname{Re}\left( {F\left( x\right) }\right) \) with \( F \) being \( \mathbb{C} \) -linear, then \( f\left( {ix}\right) = \operatorname{Re}\left( {F\left( {ix}\right) }\right) = \operatorname{Re}\left( {{iF}\left( x\right) }\right) = - \ope... | Yes |
Corollary 3.15. Suppose \( X \) is a locally convex topological vector space over \( \mathbb{R} \) or \( \mathbb{C} \), and suppose \( B \) is a nonempty closed, convex, balanced subset of \( X \) . If \( {x}_{0} \notin B \) , then \( \exists F \in {X}^{ * } \) for which \( \left| {F\left( x\right) }\right| \leq 1 \) w... | Proof. Since \( B \) is not empty, \( \{ 0\} = 0 \cdot B \subset B \), that is \( 0 \in B \), so Corollary 3.11 applies to produce a continuous, \( \mathbb{R} \) -linear functional \( f : X \rightarrow \mathbb{R} \) for which \( f\left( x\right) \leq 1 \) when \( x \in B \), but \( f\left( {x}_{0}\right) > 1 \) . If th... | Yes |
Proposition 3.16. Suppose \( X \) is a locally convex topological vector space, and \( Y \) is a subspace. Then any \( f \in {Y}^{ * } \) extends to some \( F \in {X}^{ * } \) . That is, the restriction map \( F \mapsto {\left. F\right| }_{Y} \) from \( {X}^{ * } \) to \( {Y}^{ * } \) is onto. | Proof. First case: Base field \( = \mathbb{R}.\{ x \in Y : f\left( x\right) < 1\} \) is a neighborhood of 0 in \( Y \) , and \( Y \) has the induced topology, so there is a convex, balanced neighborhood \( C \) of 0 in \( X \) such that \( C \cap Y \subset \{ x \in Y : f\left( x\right) < 1\} \) . Thus, letting \( {p}_{... | Yes |
Corollary 3.17. Suppose \( X \) is a Hausdorff locally convex topological vector space over \( \mathbb{R} \) or \( \mathbb{C} \) . Then \( {X}^{ * } \) separates points. That is, if \( x \neq y \), then \( \exists f \in {X}^{ * } \) for which \( f\left( x\right) \neq f\left( y\right) \) . | Proof. Choose any such \( f \in {Y}^{ * }, Y = \operatorname{span}\{ x, y\} \), and continuously extend it to \( X \) . | No |
Proposition 3.19. Suppose \( X \) is a locally convex topological vector space over \( \mathbb{R} \) or \( \mathbb{C} \), and suppose \( A, B \subset X \) and \( D, E \subset {X}^{ * } \) . Then:\n\n(a1) \( A \subset {\left( {A}^{ \circ }\right) }_{ \circ } \) . | Proof. (a1) and (a2) come directly from the definition. | No |
Theorem 3.20 (Bipolar Theorem). Suppose \( X \) is a locally convex topological vector space over \( \mathbb{R} \) or \( \mathbb{C};A, B \subset X \) ; and \( D \subset {X}^{ * } \) . Assume that \( B \) is closed, convex, balanced, and nonempty. Then:\n\n(a) \( {\left( {B}^{ \circ }\right) }_{ \circ } = B \) . | Proof. (a) Follows directly from Corollary 3.15 and Proposition 3.19(a1): \( B \subset \) \( {\left( {B}^{ \circ }\right) }_{ \circ } \), but if \( {x}_{0} \notin B \), then \( \exists f \in {X}^{ * } \) for which \( \left| {f\left( x\right) }\right| \leq 1 \) for all \( x \in B \) but \( \left| {f\left( {x}_{0}\right)... | Yes |
Proposition 3.24. The weak-* topology on \( {X}^{ * } \), and the weak-’ topology on \( {X}^{\prime } \), are their subspace topologies of the product topology on \( \mathop{\prod }\limits_{{x \in X}}\mathbb{F} \) . | Proof. We need only work with \( {X}^{\prime } \), since the weak- \( {}^{\prime } \) topology on \( {X}^{\prime } \) induces the weak-* topology on \( {X}^{ * } \) .\n\nEvery product neighborhood of zero contains a weak- \( {}^{\prime } \) neighborhood of zero:\n\nSuppose \( f \in {X}^{\prime } \), and \( \left| {f\le... | Yes |
Proposition 3.25. \( {X}^{\prime } \) is a closed subspace of \( \mathop{\prod }\limits_{{x \in X}}\mathbb{F} \) . | Proof. \( {X}^{\prime } \) is the intersection of the kernels of all the following continuous linear functionals on \( \mathop{\prod }\limits_{{x \in X}}\mathbb{F} : \) For \( f \in \mathop{\prod }\limits_{{x \in X}}\mathbb{F} \) , \n\n\[ \forall x, y \in X : f \mapsto f\left( x\right) + f\left( y\right) - f\left( {x +... | Yes |
Theorem 3.26 (Banach-Alaoglu). Suppose \( X \) is a locally convex space, and \( U \) is a neighborhood of zero in \( X \) . Then \( {U}^{ \circ } \) is weak-* compact. | Proof. Let \( C \) be a closed, convex, balanced neighborhood of 0 contained in \( U \), and let \( {p}_{C} \) denote its support function. \( {p}_{C} \) is a continuous seminorm by Theorem 3.7. If \( f \in {X}^{\prime } \) and \( \left| {f\left( x\right) }\right| \leq {p}_{C}\left( x\right) \) for all \( x \), then \(... | Yes |
Proposition 3.27. If \( X \) is a locally convex space, then any \( f \in {X}^{\prime } \) which is continuous when \( X \) is equipped with the Mackey topology, belongs to \( {X}^{ * } \) . | Proof. Suppose \( f \in {X}^{\prime } \), and \( f \) is Mackey-topology-continuous. Then \( \{ x \in X \) : \( \left| {f\left( x\right) }\right| \leq 1\} \) contains a Mackey neighborhood of zero, \( {D}_{ \circ } \), with \( D = {\left( {D}_{ \circ }\right) }^{ \circ } \), and \( D \) being weak-* compact, convex, an... | Yes |
Theorem 3.29. Suppose \( X \) is a locally convex space, and \( C \) is an originally closed, convex subset of \( X \) . Then \( C \) is weakly closed. | Proof. Proposition 3.12 (along with Proposition 3.14 if the base field is \( \mathbb{C} \) ) provides a means of separating \( C \) from any \( x \notin C \), by setting \( {C}_{1} = C,{C}_{2} = \{ x\} : \{ y \in X \) : \( \left. {F\left( y\right) > {r}_{0}}\right\} \) is a weakly open neighborhood of \( x \) in \( X -... | No |
Corollary 3.31. Suppose \( X \) is a locally convex space, and \( A \subset X \) . The following are equivalent:\n\n(i) \( A \) is originally bounded, that is \( A \) is bounded when \( X \) is equipped with the original topology.\n\n(ii) \( A \) is weakly bounded, that is \( A \) is bounded when \( X \) is equipped wi... | Proof. (i) \( \Rightarrow \) (ii) If \( A \) is absorbed by all original neighborhoods of zero, then \( A \) is absorbed by weak neighborhoods of zero, since weak neighborhoods of zero are original neighborhoods of zero.\n\n(ii) \( \Rightarrow \) (iv) If \( A \) is absorbed by all weak neighborhoods of zero, then \( \f... | Yes |
Proposition 3.33. Suppose \( X \) is a locally convex space over \( \mathbb{R} \) or \( \mathbb{C} \) . Let \( {\mathcal{B}}_{1} \) be a base at 0 for the topology consisting of convex, balanced sets. Set\n\n\[ \mathcal{F} = \left\{ {{p}_{V} : V \in {\mathcal{B}}_{1}}\right\} \]\n\nwhere \( {p}_{V} \) is the Minkowski ... | Proof. Each \( {p}_{V} \) is a seminorm (Theorem 3.7), and if \( U, V \in {\mathcal{B}}_{1},\exists W \in {\mathcal{B}}_{1} \) with \( W \subset U \cap V \) . But, for example, \( W \subset U \Rightarrow \left( {x \in {tW} \Rightarrow x \in {tU}}\right) \), so that \( \forall x \in X \), \n\n\[ \{ t \geq 0 : x \in {tW}... | Yes |
Corollary 3.34. Suppose \( X \) is a locally convex space over \( \mathbb{R} \) or \( \mathbb{C} \). Then the topology of \( X \) is given by a directed family of seminorms. This family can be chosen to be countable if \( X \) is first countable. | Proof. \( {\mathcal{B}}_{1} \) exists by Proposition 3.1. | No |
Theorem 3.35. Suppose \( X \) is a Hausdorff locally convex space. Then the following are equivalent:\n\n(i) \( X \) is first countable.\n\n(ii) \( X \) is metrizable.\n\n(iii) The topology of \( X \) is given by a translation invariant metric.\n\n(iv) The topology of \( X \) is given by a countable family of seminorms... | Proof. The earlier discussion gives (iv) \( \Rightarrow \) (iii). The implications (iii) \( \Rightarrow \) (ii) and (ii) \( \Rightarrow \) (i) are direct, while (i) \( \Rightarrow \) (iv) comes from Corollary 3.34. | Yes |
Corollary 3.36. Suppose \( X \) is a Hausdorff locally convex space. Then the following are equivalent:\n\n(i) \( X \) is first countable and complete.\n\n(ii) \( X \) is metrizable and complete.\n\n(iii) The topology of \( X \) is given by a complete, translation invariant metric.\n\n(iv) \( X \) is complete, and the ... | Proof. Thanks to Theorem 3.35, the only issue is the variation in \ | No |
Proposition 3.40. Suppose \( X = \bigcup {X}_{n} \) is an LF-space, and \( U \) is a convex subset of \( X \) . Then \( U \) is open in \( X \) if, and only if, \( U \cap {X}_{n} \) is open in \( {X}_{n} \) for all \( n \) . | Proof. If \( U \) is open in \( X \), then \( U \cap {X}_{n} \) is open in \( {X}_{n} \) for all \( n \), by Proposition 3.39(a). On the other hand, suppose \( U \) is convex in \( X \), and \( U \cap {X}_{n} \) is open in \( {X}_{n} \) for all \( n \) . Suppose \( x \in U \) ; then \( x \in {X}_{k} \) for some \( k \)... | Yes |
Corollary 3.41. Suppose \( X = \bigcup {X}_{n} \) is an LF-space, \( Y \) is a locally convex space, and \( T : X \rightarrow Y \) is a linear transformation. Then \( T \) is continuous if, and only if, \( {\left. T\right| }_{{X}_{n}} \) is continuous on \( {X}_{n} \) for each \( n \) . | Proof. If \( T \) is continuous on \( X \), then \( {\left. T\right| }_{{X}_{n}} \) is continuous on \( {X}_{n} \) by definition of the induced topology, so suppose \( {\left. T\right| }_{{X}_{n}} \) is continuous on \( {X}_{n} \) for all \( n \) . If \( V \) is an open convex neighborhood of 0 in \( Y \), then \( {\le... | Yes |
Lemma 3.42. Suppose \( X = \bigcup {X}_{n} \) is an LF-space. Concerning the chain construction, starting at \( k = 1 \) :\n\n(a) If \( U \) is an open, convex, balanced neighborhood of 0 in \( X \), then setting \( {U}_{n} = \) \( U \cap {X}_{n} \) and applying the chain construction produces (in the earlier notation)... | Proof. (a) \( {U}_{n} = {V}_{n} \) by induction on \( n.n = 1 : {U}_{1} = {V}_{1} \) by definition. If \( {U}_{n} = \) \( {V}_{n} \), then since \( {U}_{n} = U \cap {X}_{n} = U \cap {X}_{n + 1} \cap {X}_{n} = {U}_{n + 1} \cap {X}_{n},{V}_{n + 1} = \) \( L\left( {{V}_{n},{U}_{n + 1}}\right) = \operatorname{con}\left( {{... | Yes |
Theorem 3.43. LF-spaces are complete. | Proof. Suppose \( \left\langle {x}_{\alpha }\right\rangle \) is a Cauchy net in an LF-space \( X = \bigcup {X}_{n} \), defined on a directed set \( D \) . Let \( {D}^{\prime } \) be a neighborhood base at 0 in \( X \) consisting of open, convex, balanced sets. If \( U, V \in {D}^{\prime } \), declare \( U \succ V \) wh... | Yes |
Corollary 3.44. Suppose \( X = \bigcup {X}_{n} \) is an LF-space, and \( A \) is a bounded set in \( X \) . Then \( \exists n \) such that \( A \subset {X}_{n} \) . | Proof. Replace \( A \) with \( {\left( {A}^{ \circ }\right) }_{ \circ } \), a nonempty, closed, convex, balanced, bounded set (Proposition 3.19 and Theorem 3.20). This new \( A \) is now complete by Theorem 3.43 and Proposition 1.30, so \( \left( {{X}_{A},{p}_{A}}\right) \) is a Banach space by Proposition 3.30. But \(... | Yes |
Proposition 3.46. The strong dual of an LB-space is a Fréchet space. | Proof. Suppose \( X = \bigcup {X}_{n} \) is an LB-space. Assume we have one norm \( \parallel \bullet \parallel \) defining the various Banach space structures on the spaces \( {X}_{n} \) (Proposition 3.45). Let \( {B}_{n} \) denote the unit ball in \( {X}_{n} \) . Set\n\n\[ \mathcal{B} = \left\{ {{2}^{-k}{\left( {B}_{... | Yes |
Proposition 4.2. Suppose \( X \) is a Hausdorff locally convex space, and \( Y \) is a closed subspace.\n\n(a) If \( X \) is barreled, then \( X/Y \) is barreled.\n\n(b) If \( X \) is infrabarreled, then \( X/Y \) is infrabarreled.\n\n(c) If \( X \) is bornological, then \( X/Y \) is bornological. | Proof. As usual, \( \pi : X \rightarrow X/Y \) is the quotient map. Suppose \( B \) is a convex, balanced subset of \( X/Y \) ; then \( {\pi }^{-1}\left( B\right) \) is also convex and balanced. If \( c > 0 \) and \( x + Y \subset {cB} \), then \( x \in {\pi }^{-1}\left( {cB}\right) = c{\pi }^{-1}\left( B\right) \) sin... | Yes |
Proposition 4.3. Suppose \( X \) and \( Y \) are Hausdorff locally convex spaces.\n\n(a) If \( X \) and \( Y \) are barreled, then \( X \times Y \) is barreled. | Proof. First of all, if \( A \) is bounded in \( X \), then any neighborhood of 0 in \( X \times Y \) of the form \( U \times V \) absorbs \( A \times \{ 0\} \) simply because \( U \) absorbs \( A \), so \( A \times \{ 0\} \) is bounded. Suppose \( B \) is a convex, balanced subset of \( X \times Y \) . The slice \( B ... | Yes |
Proposition 4.4. Suppose \( X \) is a Hausdorff locally convex space over \( \mathbb{C} \). Let \( {\left. X\right| }_{\mathbb{R}} \) denote \( X \) considered as a locally convex space over \( \mathbb{R} \). (a) If \( X \) is barreled, then \( {\left. X\right| }_{\mathbb{R}} \) is barreled. | Proof. The fundamentals here appear in Proposition 2.19. Suppose \( B \) is a convex, \( \mathbb{R} \)-balanced subset of \( X \), that is, a convex, balanced subset of \( {\left. X\right| }_{\mathbb{R}} \). Form \[ C = \mathop{\bigcap }\limits_{{0 \leq \theta < {2\pi }}}{e}^{i\theta }B \] This \( C \) is convex and \(... | Yes |
Theorem 4.5. Suppose \( X \) is a Hausdorff locally convex space of the second category. Then \( X \) is barreled. In particular, Fréchet spaces are barreled. | Proof. Suppose that \( X \) is a Hausdorff locally convex space of the second category, and \( B \) is a barrel in \( X \) . Since \( 0 < r < s \Rightarrow {rB} = s \cdot \left( {r/s}\right) B \subset {sB} \) ( \( B \) is balanced),\n\n\[ \nX = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{nB} \n\]\n\nsince \( B \) is a... | Yes |
Corollary 4.6. LF-spaces are barreled. | Proof. Suppose that \( X = \bigcup {X}_{n} \) is an LF-space, and \( B \) is a barrel in \( X \) . Then for all \( n, B \cap {X}_{n} \) is a barrel in \( {X}_{n} \) : It is closed by Proposition 3.39(a), and all else is trivial. By the above, \( B \cap {X}_{n} \) is a neighborhood of 0 in \( {X}_{n} \), so \( B \) is, ... | No |
Corollary 4.7 (Absorption Principle). Suppose \( X \) is a Hausdorff locally convex space, and suppose \( A \) and \( B \) are two closed, convex, balanced subsets of \( X \) . Assume that \( A \) is bounded and sequentially complete. Then, if \( B \) absorbs every point in \( A \) , then \( B \) absorbs \( A \), that ... | Proof. If \( A = \varnothing \), then \( A \subset B \) and we are done, so suppose \( A \) is nonempty. As in Proposition 3.30, form the space \( \left( {{X}_{A},{p}_{A}}\right) \) . Then \( \left( {{X}_{A},{p}_{A}}\right) \) is a Banach space since \( A \) is sequentially complete, and \( B \cap {X}_{A} \) is a barre... | Yes |
Corollary 4.8. Suppose \( X \) is a Hausdorff locally convex space. If \( X \) is infrabar-reled and sequentially complete, then \( X \) is barreled. | Proof. Suppose \( X \) is an infrabarreled Hausdorff locally convex space that is also sequentially complete, and suppose \( B \) is a barrel in \( X \) and \( A \) is a bounded subset of \( X \) . Then \( {\left( {A}^{ \circ }\right) }_{ \circ } \) is closed, convex, balanced, and bounded; it is also sequentially comp... | Yes |
Corollary 4.9. Infrabarreled spaces are Mackey spaces. | Proof. Suppose \( X \) is an infrabarreled Hausdorff locally convex space. If \( D = \) \( {\left( {D}_{ \circ }\right) }^{ \circ } \) is weak-* compact in \( {X}^{ * } \), then as a subset of \( {X}^{ * } \) with the weak-* topology: \( D \) is nonempty, bounded, closed, convex, balanced, and complete (Corollary 1.32)... | Yes |
Proposition 4.10. Suppose \( X \) is a Hausdorff locally convex space. If \( X \) is first countable, then \( X \) is bornological. | Proof. Suppose \( X \) is a first countable Hausdorff locally convex space. Choose a countable neighborhood base \( {V}_{1},{V}_{2},\ldots \) at 0 such that \( {V}_{1} \supset {V}_{2} \supset {V}_{3} \supset \cdots \) (Theorem 1.13 provides more, but this is all we need.) Suppose \( B \) is a convex, balanced subset of... | Yes |
Corollary 4.11. Normed spaces, Fréchet spaces, and LF-spaces are bornological. | Proof. Normed spaces and Fréchet spaces are first countable. As for LF-spaces, suppose that \( X = \bigcup {X}_{n} \) is an LF-space, and \( B \) is a balanced, convex subset of \( X \) that absorbs all bounded sets. If \( A \) is bounded in \( {X}_{n} \), then each continuous linear functional on \( X \) restricts to ... | Yes |
Theorem 4.12. Suppose \( X \) and \( Y \) are Hausdorff locally convex spaces, and \( T \) : \( X \rightarrow Y \) is a linear transformation. Consider the following three statements:\n\n(i) \( T \) is continuous.\n\n(ii) If \( {x}_{n} \rightarrow 0 \) in \( X \), then \( T\left( {x}_{n}\right) \rightarrow 0 \) in \( Y... | Proof. (i) \( \Rightarrow \) (ii), since continuity \( \Rightarrow \) sequential continuity.\n\nAssume (ii): If \( A \) is bounded but \( T\left( A\right) \) is not bounded in \( Y \), choose a neighborhood \( V \) of 0 in \( Y \) that does not absorb \( T\left( A\right) \) . choose \( T\left( {x}_{n}\right) \in T\left... | Yes |
Lemma 4.13. Suppose \( X \) and \( Y \) are locally convex spaces, and \( c \) is a nonzero scalar. Then for all \( A \subset X \) and \( U \subset Y \) :\n\n\[ \n{cN}\left( {A, U}\right) = N\left( {A,{cU}}\right) = N\left( {{c}^{-1}A, U}\right) .\n\] | Proof. \( T \in N\left( {A,{cU}}\right) \; \Leftrightarrow \;T\left( A\right) \subset {cU} \Leftrightarrow {c}^{-1}T\left( A\right) \subset U \) . But\n\n\[ \n{c}^{-1}T\left( A\right) \subset U\; \Leftrightarrow \;T\left( {{c}^{-1}A}\right) \subset U \Leftrightarrow T \in N\left( {{c}^{-1}A, U}\right) \n\]\nand\n\n\[ \... | Yes |
Proposition 4.14. Suppose \( X \) and \( Y \) are locally convex spaces, and \( \mathcal{F} \) \( {\mathcal{L}}_{c}\left( {X, Y}\right) \) . (a) \( \mathcal{F} \) is bounded in the topology of pointwise convergence if, and only if, for all \( x \in X \) the set \( \{ T\left( x\right) : T \in \mathcal{F}\} \) is bounded... | Proof. The underlying idea is the same for both parts: \[ \mathcal{F} \subset {cN}\left( {A, U}\right) \Leftrightarrow \mathcal{F} \subset N\left( {A,{cU}}\right) \] \[ \Leftrightarrow \forall T \in \mathcal{F} : T\left( A\right) \subset {cU} \Leftrightarrow \mathop{\bigcup }\limits_{{T \in \mathcal{F}}}T\left( A\right... | Yes |
Theorem 4.16 (Banach-Steinhaus/Uniform Boundedness Theorem). Suppose \( X \) and \( Y \) are Hausdorff locally convex spaces, and \( \mathcal{F} \subset {\mathcal{L}}_{c}\left( {X, Y}\right) \) . Consider the following three conditions on \( \mathcal{F} \) :\n\n(i) \( \mathcal{F} \) is equicontinuous.\n\n(ii) \( \mathc... | Proof. (a) (ii) \( \Rightarrow \) (iii) is trivial, since the topology of bounded convergence is finer than the topology of pointwise convergence. To prove that (i) \( \Rightarrow \) (ii), suppose \( \mathcal{F} \) is equicontinuous, \( A \) is bounded in \( X \), and \( U \) is an open neighborhood of 0 in \( Y \) . C... | Yes |
Proposition 4.17. Suppose \( X, Y \), and \( Z \) are Hausdorff locally convex spaces, and \( f : X \times Y \rightarrow Z \) is a separately continuous bilinear map. Suppose \( A \) is bounded in \( X \), and \( Y \) is barreled. Then \( f \) is jointly continuous on \( A \times Y \) . | Proof. Set \( \mathcal{F} = \{ f\left( {a,?}\right) : a \in A\} \subset {\mathcal{L}}_{c}\left( {Y, Z}\right) \) . If \( y \in Y \), then \( f\left( {?, y}\right) \) is continuous from \( X \) to \( Z \), so it sends the bounded set \( A \) to a bounded subset of \( Z \). That is, \( \{ f\left( {a, y}\right) : a \in A\... | Yes |
Corollary 4.18. Suppose \( X, Y \), and \( Z \) are Hausdorff locally convex spaces, and \( f : X \times Y \rightarrow Z \) is a separately continuous bilinear map. Suppose \( X \) is first countable and \( Y \) is barreled. Then \( f \) is jointly continuous provided either \( X \) is normed or \( Y \) is a Fréchet sp... | Proof. Case 1: \( \;X \) is normed. Let \( {A}_{n} \) be the open ball in \( X \) of radius \( n \) . Then \( f \) is continuous on \( {A}_{n} \times Y \) . Thus if \( W \) is open in \( Z \), then \( {f}^{-1}\left( W\right) \cap \left( {{W}_{n} \times Y}\right) \) is relatively open in \( {A}_{n} \times Y \) since \( ... | Yes |
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