From c02d873dddf01155b06f38f192ea7d8ba8e25c6e Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sun, 9 Aug 2026 21:18:50 -0400 Subject: [PATCH] Added bilinear forms. --- src/fa/lc/continuous.tex | 1 + src/fa/tvs/bilinear.tex | 170 +++++++++++++++++++++++++++++++++ src/fa/tvs/equicontinuous.tex | 22 ----- src/fa/tvs/index.tex | 3 +- src/fa/tvs/space-of-linear.tex | 63 ------------ 5 files changed, 173 insertions(+), 86 deletions(-) create mode 100644 src/fa/tvs/bilinear.tex diff --git a/src/fa/lc/continuous.tex b/src/fa/lc/continuous.tex index ebedf43..d4f4b17 100644 --- a/src/fa/lc/continuous.tex +++ b/src/fa/lc/continuous.tex @@ -20,6 +20,7 @@ $(4) \Rightarrow (3)$: Let $U \in \cn_F(0)$ be convex, circled, and radial, then its gauge $[\cdot]_U$ is a continuous seminorm on $F$ by \autoref{definition:locally-convex}. Thus there exists a continuous seminorm $[\cdot]_E$ such that $[Tx]_U \le [x]_E$. In which case, $V = \bracs{x \in E| [x]_E < 1} \in \cn_E(0)$ with $T(V) \subset U$. Therefore $T$ is continuous at $0$, and continuous by \autoref{definition:continuous-linear}. \end{proof} + \begin{proposition} \label{proposition:tvs-convex-multilinear} Let $\seqf{E_j}$ and $F$ be locally convex spaces, and $T: \prod_{j = 1}^n E_j \to F$ be $n$-linear map, then the following are equivalent: diff --git a/src/fa/tvs/bilinear.tex b/src/fa/tvs/bilinear.tex new file mode 100644 index 0000000..d418a4e --- /dev/null +++ b/src/fa/tvs/bilinear.tex @@ -0,0 +1,170 @@ +\section{Bilinear Mappings} +\label{section:bilinear-tvs} + + + +\begin{theorem} +\label{theorem:separate-joint-bilinear} + Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds: + \begin{enumerate} + \item[(B)] $E$ is Baire. + \item[(B')] $E$ is barrelled and $G$ is locally convex. + \end{enumerate} + + and that + \begin{enumerate} + \item[(M)] $E$ and $F$ are both metrisable. + \item[(E)] For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg} \subset L(F; G)$ is equicontinuous. + \end{enumerate} + + then $\alg$ is equicontinuous. +\end{theorem} +\begin{proof}[Proof, {{\cite[III.5.1]{SchaeferWolff}}}. ] + Let $\seq{(x_n, y_n)} \subset E \times F$ and $\seq{\lambda_n} \subset \alg$ such that $(x_n, y_n) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and \autoref{proposition:equicontinuous-net}. By (B) or (B') and the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_n(x_n, y_n) \to 0$ as $n \to \infty$ by \autoref{proposition:equicontinuous-net}. By (M) and \autoref{proposition:equicontinuous-net}, $\alg$ is equicontinuous at $0$, and hence equicontinuous by \autoref{lemma:equicontinuous-bilinear}. +\end{proof} + +\begin{definition}[Hypocontinuity] +\label{definition:hypocontinuity} + Let $E, F, G$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal of bounded sets, and $\lambda: E \times F \to G$ be a separately continuous bilinear map, then the following are equivalent: + \begin{enumerate} + \item For each $S \in \sigma$ and $V \in \cn_G(0)$, there exists $U \in \cn_F(0)$ such that $\lambda(S \times U) \subset V$. + \item For each $S \in \sigma$, $\bracs{\lambda(x, \cdot)|x \in S} \subset F^*$ is equicontinuous. + \end{enumerate} + + If the above holds, then $\lambda$ is \textbf{$\sigma$-hypocontinuous}. + + For any ideal $\tau \subset 2^F$ of bounded sets, $\lambda$ if \textbf{$(\sigma, \tau)$-hypocontinuous} if $\lambda$ is $\sigma$-hypocontinuous and $\tau$-hypocontinuous. +\end{definition} + +\begin{proposition} +\label{proposition:separate-hypocontinuous} + Let $E, F, G$ be TVSs over $K \in \RC$, and $\lambda: E \times F \to G$ be a separately continuous bilinear map. + If one of the following holds: + \begin{enumerate} + \item[(B)] $E$ is Baire. + \item[(B')] $E$ is barrelled and $G$ is locally convex. + \end{enumerate} + + then $\lambda$ is $B(E)$-hypocontinuous. +\end{proposition} +\begin{proof} + Since $\lambda$ is separately continuous, the mapping + \[ + E \to L(F; G) \quad x \mapsto \lambda(x, \cdot) + \] + + is continuous with respect to the strong operator topology on $L(F; G)$. As such, for each $B \subset E$ bounded, $\bracs{\lambda(x, \cdot)|x \in B}$ is bounded in $L_s(F; G)$. By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracs{\lambda(x, \cdot)|x \in B}$ is equicontinuous. +\end{proof} + +\begin{proposition} +\label{proposition:hypocontinuous-restriction} + Let $E, F, G$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be ideals of bounded sets, and $\lambda: E \times F \to G$ be a bilinear mapping. + \begin{enumerate} + \item If $\lambda$ is $\sigma$-hypocontinuous, then $\lambda$ is continuous on $S \times F$ for all $S \in \sigma$. + \item If $\lambda$ is $(\sigma, \tau)$-hypocontinuous, then $\lambda$ is uniformly continuous on $S \times T$ for all $S \in \sigma$ and $T \in \tau$. + \end{enumerate} +\end{proposition} +% Omitted for obviousness + +\begin{proposition} +\label{proposition:hypocontinuous-linear-extension} + Let $E, F, G$ be TVSs over $K \in \RC$, $E_0 \subset E$ and $F_0 \subset F$ be dense subspaces, and $\sigma \subset B(E_0)$ and $\tau \subset B(F_0)$ be ideals of bounded sets. Denote $\ol{\sigma}$ and $\ol \tau$ as the ideals generated by $\bracsn{\ol{S}|S \in \sigma}$ and $\bracsn{\ol{T}|T \in \tau}$, respectively. If + \begin{enumerate}[label=(\alph*)] + \item $G$ is a complete Hausdorff TVS. + \item $\ol\sigma$ covers $E$ and $\ol{\tau}$ covers $F$. + \end{enumerate} + + Then, for any $(\sigma, \tau)$-hypocontinuous bilinear map $\lambda: E_0 \times F_0 \to G$, there exists a unique $\Lambda: E \times F \to G$ such that: + \begin{enumerate} + \item $\Lambda|_{E_0 \times F_0} = \lambda$. + \item $\Lambda$ is bilinear and $(\ol\sigma, \ol\tau)$-hypocontinuous. + \end{enumerate} +\end{proposition} +\begin{proof} + By (2) of \autoref{proposition:hypocontinuous-restriction}, for each $S \in \sigma$ and $T \in \tau$, $\lambda|_{S \times T}$ is uniformly continuous. Thus (a) and \autoref{theorem:uniform-continuous-extension} imply that there exists a unique continuous extension of $\lambda$ to $\ol S \times \ol T$. By (b) and the \hyperref[gluing lemma]{lemma:glue-function}, there exists a unique $\Lambda: E \times F \to G$ such that: + \begin{enumerate} + \item $\Lambda|_{E_0 \times F_0} = \lambda$. + \item[(2')] For each $S \in \sigma$ and $T \in \tau$, $\Lambda|_{\ol S \times \ol T}$ is continuous. + \end{enumerate} + + so the extension is unique by (1) of \autoref{proposition:hypocontinuous-restriction}. + + It remains to show that $\Lambda$ is bilinear and $(\ol \sigma, \ol \tau)$-hypocontinuous. To this end, observe that for each $x \in E_0$, $\lambda(x, \cdot) \in L(F_0; G)$, and extends to a unique element of $L(F; G)$ by the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-tvs}. For each $S \in \sigma$ and $T \in \tau$, $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$. As such, $\Lambda(x, \cdot)$ is the unique continuous extension of $\lambda(x, \cdot)$ to an element of $L(F; G)$, so $\Lambda(x, \cdot) \in L(F; G)$ for all $x \in E_0$. By symmetry, $\Lambda(\cdot, y) \in L(E; G)$ for all $y \in F_0$. + + + Now, let $S \in \sigma$, then $\bracsn{\lambda(x, \cdot)|x \in S}$ is equicontinuous by the $\sigma$-hypocontinuity of $\lambda$. For any $U \in \cn_0(G)$, there exists $V \in \cn_0(F)$ such that $\bigcup_{x \in S}\lambda(x, V \cap F_0) \subset U$. By \autoref{proposition:closure-of-image}, $\bigcup_{x \in S}\Lambda(x, \ol V) \subset \ol U$. Thus \autoref{proposition:tvs-good-neighbourhood-base} implies that $\bracsn{\Lambda(x, \cdot)|x \in S}$ is equicontinuous as well. + + For each $x_0 \in \ol S$ and $y_0 \in F$, there exists $T \in \tau$ with $y_0 \in \ol T$. As $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$, $\Lambda(x_0, \cdot)$ is a pointwise limit of elements of $\bracsn{\lambda(x, \cdot)|x \in S}$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, + \begin{enumerate} + \item $\bracsn{\Lambda(x, \cdot)|x \in \ol S} \subset \ol{\bracsn{\Lambda(x, \cdot)|x \in S}}^{L_s(F; G)}$. + \item $\bracsn{\Lambda(x, \cdot)|x \in \ol S}$ is also equicontinuous. + \end{enumerate} + + so $\Lambda$ is $\ol \sigma$-hypocontinuous. Therefore $\Lambda$ is $(\ol \sigma, \ol \tau)$-hypocontinuous by symmetry. +\end{proof} + + + + +\begin{proposition} +\label{proposition:multilinear-identify} + Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then + \begin{enumerate} + \item The map + \[ + I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F) + \] + + defined by + \[ + (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1}) + \] + + is an isomorphism. + \item The map + \[ + I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F) + \] + + defined by + \[ + IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k) + \] + + is an isomorphism. + \end{enumerate} + + which allows the identification + \[ + \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F) + \] + + under the map $I$ in (2). +\end{proposition} +\begin{proof} + (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and + \[ + I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot) + \] + + Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case, + \[ + T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in \mathfrak{B}(F) + \] + + by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$. + + In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$. + + It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous. + + On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well. + + (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then + \[ + \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F) + \] + + Thus (2) holds for all $k \in \natp$. +\end{proof} + diff --git a/src/fa/tvs/equicontinuous.tex b/src/fa/tvs/equicontinuous.tex index 89542e8..96f2518 100644 --- a/src/fa/tvs/equicontinuous.tex +++ b/src/fa/tvs/equicontinuous.tex @@ -88,28 +88,6 @@ and $\lambda(x_0, y - y_0) \in U$ as well. Therefore $\alg$ is equicontinuous at $(x_0, y_0)$. \end{proof} - - -\begin{theorem} -\label{theorem:separate-joint-bilinear} - Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds: - \begin{enumerate} - \item[(B)] $E$ is Baire. - \item[(B')] $E$ is barrelled and $G$ is locally convex. - \end{enumerate} - - and that - \begin{enumerate} - \item[(M)] $E$ and $F$ are both metrisable. - \item[(E)] For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg} \subset L(F; G)$ is equicontinuous. - \end{enumerate} - - then $\alg$ is equicontinuous. -\end{theorem} -\begin{proof}[Proof, {{\cite[III.5.1]{SchaeferWolff}}}. ] - Let $\seq{(x_n, y_n)} \subset E \times F$ and $\seq{\lambda_n} \subset \alg$ such that $(x_n, y_n) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and \autoref{proposition:equicontinuous-net}. By (B) or (B') and the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_n(x_n, y_n) \to 0$ as $n \to \infty$ by \autoref{proposition:equicontinuous-net}. By (M) and \autoref{proposition:equicontinuous-net}, $\alg$ is equicontinuous at $0$, and hence equicontinuous by \autoref{lemma:equicontinuous-bilinear}. -\end{proof} - % TODO: Replace this with a more general version involving polars in the future. \begin{theorem}[Banach-Alaoglu] \label{theorem:alaoglu} diff --git a/src/fa/tvs/index.tex b/src/fa/tvs/index.tex index 22335c1..c2f2936 100644 --- a/src/fa/tvs/index.tex +++ b/src/fa/tvs/index.tex @@ -14,4 +14,5 @@ \input{./inductive.tex} \input{./vector-function.tex} \input{./space-of-linear.tex} -\input{./equicontinuous.tex} \ No newline at end of file +\input{./equicontinuous.tex} +\input{./bilinear.tex} \ No newline at end of file diff --git a/src/fa/tvs/space-of-linear.tex b/src/fa/tvs/space-of-linear.tex index dc56d1f..51ecc53 100644 --- a/src/fa/tvs/space-of-linear.tex +++ b/src/fa/tvs/space-of-linear.tex @@ -20,69 +20,6 @@ \end{proposition} % Proof omitted because it is obvious. - -\begin{proposition} -\label{proposition:multilinear-identify} - Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then - \begin{enumerate} - \item The map - \[ - I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F) - \] - - defined by - \[ - (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1}) - \] - - is an isomorphism. - \item The map - \[ - I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F) - \] - - defined by - \[ - IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k) - \] - - is an isomorphism. - \end{enumerate} - - which allows the identification - \[ - \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F) - \] - - under the map $I$ in (2). -\end{proposition} -\begin{proof} - (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and - \[ - I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot) - \] - - Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case, - \[ - T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in \mathfrak{B}(F) - \] - - by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$. - - In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$. - - It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous. - - On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well. - - (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then - \[ - \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F) - \] - - Thus (2) holds for all $k \in \natp$. -\end{proof} - \begin{definition}[Strong Operator Topology] \label{definition:strong-operator-topology} Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^E$ is the \textbf{strong operator topology}.