From 07fe8b35c017dbcc749182c57f5772586bea8e35 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Wed, 6 May 2026 00:27:05 -0400 Subject: [PATCH] Added the Banach-Steinhaus theorem. --- src/fa/tvs/equicontinuous.tex | 62 +++++++++++++++++++++++++ src/fa/tvs/index.tex | 1 + src/topology/uniform/equicontinuous.tex | 32 ++++--------- src/topology/uniform/uc.tex | 4 +- 4 files changed, 74 insertions(+), 25 deletions(-) create mode 100644 src/fa/tvs/equicontinuous.tex diff --git a/src/fa/tvs/equicontinuous.tex b/src/fa/tvs/equicontinuous.tex new file mode 100644 index 0000000..c96de44 --- /dev/null +++ b/src/fa/tvs/equicontinuous.tex @@ -0,0 +1,62 @@ +\section{Equicontinuous Families of Linear Maps} +\label{section:equicontinuous-linear} + +\begin{proposition}[{{\cite[IV.4.2]{SchaeferWolff}}}] +\label{proposition:equicontinuous-linear} + Let $E, F$ be TVSs over $K \in RC$ and $\alg \subset \hom(E; F)$, then the following are equivalent: + \begin{enumerate} + \item $\alg$ is uniformly equicontinuous. + \item $\alg$ is equicontinuous. + \item $\alg$ is equicontinuous at $0$. + \item For each $V \in \cn_F(0)$, there exists $U \in \cn^o(E)$ such that $\bigcup_{T \in \alg}T(U) \subset V$. + \item For each $V \in \cn_F(0)$, $\bigcap_{T \in \alg}T^{-1}(V) \in \cn_E(0)$. + \end{enumerate} +\end{proposition} +\begin{proof} + (5) $\Rightarrow$ (1): Let $V \in \cn_F(0)$, then $U = \bigcap_{T \in \alg}T^{-1}(V) \in \cn_E(0)$. Thus for any $x, y \in E$ with $x - y \in U$, $Tx - Ty \in V$ for all $T \in \alg$. +\end{proof} + +\begin{proposition}[{{\cite[IV.4.3]{SchaeferWolff}}}] +\label{proposition:equicontinuous-linear-closure} + Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, and $\alg'$ be the closure of $\alg$ in $F^E$ with respect to the product topology, then $\alg'$ is equicontinuous and hence $\alg' \subset L(E; F)$. +\end{proposition} +\begin{proof} + By \autoref{proposition:operator-space-completeness}, $\alg' \subset \hom(E; F)$. By \autoref{theorem:arzela-ascoli}, $\alg'$ is equicontinuous. +\end{proof} + +\begin{theorem}[Banach-Steinhaus] +\label{theorem:banach-steinhaus} + Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$. Suppose that one of the following holds: + \begin{enumerate} + \item[(B)] $E$ is a Baire space. + \item[(B')] $E$ and $F$ are locally convex with $E$ being barreled. + \end{enumerate} + + and that + \begin{enumerate} + \item[(E2)] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$. + \end{enumerate} + + then + \begin{enumerate} + \item[(E1)] $\alg$ is equicontinuous. + \item[(C1)] The product uniformity and the compact uniformity on $\cf$ coincide. + \item[(C2)] The closure of $\alg$ in $F^E$ is with respect to the product topology is an equicontinuous subset of $L(E; F)$. + \end{enumerate} +\end{theorem} +\begin{proof}[Proof, {{\cite[IV.4.2]{SchaeferWolff}}}. ] + (B) + (E2) $\Rightarrow$ (E1): Let $V \in \cn_F(0)$ be closed and circled, then $U = \bigcap_{T \in \alg}T^{-1}(V)$ is circled and closed. By (E2), $U$ is absorbing, so $E = \bigcup_{n \in \natp}nU$. Since $E$ is Baire, there exists $n \in \natp$, $W \in \cn_E(0)$, and $x \in E$ such that $x + W \subset nU$. As $U$ is circled, + \[ + W \subset nU - nU = nU + nU = 2nU + \] + + so $U \in \cn_E(0)$, and $\alg$ is equicontinuous by \autoref{proposition:equicontinuous-linear}. + + (B') + (E2) $\Rightarrow$ (E1): Let $V \in \cn_F(0)$ be convex, circled, and closed, then $U = \bigcap_{T \in \alg}T^{-1}(V)$ is convex, circled, and closed. By (E2), $U$ is absorbing, and hence a barrel in $E$. By (B'), $U \in \cn_E(0)$, $\alg$ is equicontinuous by \autoref{proposition:equicontinuous-linear}. + + (E1) $\Rightarrow$ (C1) + (C2): By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} and \autoref{proposition:equicontinuous-linear-closure}. +\end{proof} + + + + diff --git a/src/fa/tvs/index.tex b/src/fa/tvs/index.tex index 91ce811..1a4b5dc 100644 --- a/src/fa/tvs/index.tex +++ b/src/fa/tvs/index.tex @@ -13,3 +13,4 @@ \input{./inductive.tex} \input{./vector-function.tex} \input{./space-of-linear.tex} +\input{./equicontinuous.tex} \ No newline at end of file diff --git a/src/topology/uniform/equicontinuous.tex b/src/topology/uniform/equicontinuous.tex index da54876..13a450c 100644 --- a/src/topology/uniform/equicontinuous.tex +++ b/src/topology/uniform/equicontinuous.tex @@ -13,7 +13,6 @@ Let $(X, \fU)$ and $(Y, \fV)$ be uniform spaces, and $\cf \subset UC(X; Y)$, then $\cf$ is \textbf{uniformly equicontinuous} if for every $V \in \fV$, there exists $U \in \fU$ such that $(f \times f)(V) \subset \fU$ for all $f \in \cf$. \end{definition} - \begin{theorem}[Arzelà-Ascoli] \label{theorem:arzela-ascoli} Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, and $\cf \subset C(X; Y)$. If @@ -24,6 +23,7 @@ then \begin{enumerate}[label=(C\arabic*)] \item The product uniformity and the compact uniformity on $\cf$ coincide. + \item The closure of $\cf$ in $Y^X$ with respect to the product topology is equicontinuous. \end{enumerate} In addition, if $\cf$ satisfies (E1) and @@ -32,11 +32,11 @@ \end{enumerate} then - \begin{enumerate}[label=(C\arabic*), start=1] - \item $\cf$ is a precompact subset of $Y^X$ with respect to the compact uniformity. + \begin{enumerate}[label=(C\arabic*), start=3] + \item $\cf$ is a precompact subset of $C(X; Y)$ with respect to the compact uniformity. \end{enumerate} - Conversely, if $X$ is a LCH space, then (C2) implies (E1) + (E2). + Conversely, if $X$ is a LCH space, then (C3) implies (E1) + (E2). \end{theorem} \begin{proof} (E1) $\Rightarrow$ (C1): By \autoref{proposition:compact-uniform-open}, the compact-open topology coincides with the compact-uniform topology on $C(X; Y)$ and thus $\cf$. @@ -53,32 +53,18 @@ so the product uniformity and the compact uniformity coincide. - (E1) + (E2) $\Rightarrow$ (C2): By \autoref{proposition:totally-bounded-product}, $\prod_{x \in X}\cf(x)$ is totally bounded. Since $\cf \subset \prod_{x \in X}\cf(x)$, $\cf$ is also totally bounded. As the pointwise and compact uniformities coincide, $\cf$ is totally bounded with respect to the compact uniformity. By \autoref{proposition:product-complete} and \autoref{proposition:compact-uniform}, $\prod_{x \in X}\ol{\cf(x)}$ is compact and hence complete. Therefore the closure of $\cf$ with respect to the compact uniformity is compact. + (E1) $\Rightarrow$ (C2): Let $\cf'$ be the closure of $\cf$ in $Y^X$ with respect to the product topology. Let $x \in X$ and entourage $U$ of $Y$. Using \autoref{proposition:goodentourages}, assume without loss of generality that $U$ is closed. Since $\cf$ is equicontinuous, there exists $V \in \cn_X(x)$ such that $(f(x), f(y)) \in U$ for all $f \in \cf$ and $y \in V$. For any element $g \in \cf'$, $(g(x), g(y)) \in \ol U = U$ for all $y \in V$. Therefore $\cf'$ is also equicontinuous. - (C2) $\Rightarrow$ (E1): Assume that $X$ is a LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_X(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j} \subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_j \times g)(V) \subset U$. For each $1 \le j \le n$, $f_j \in C(X; Y)$, so there exists $V_j \in \cn_X(x)$ with $V_j \subset V$ such that for any $y \in V_j$, $(f_j(x), f_j(y)) \in U$. Let $W = \bigcap_{j = 1}^n V_j$, then for any $g \in \cf$ with $(f_j \times g)(V) \subset U$ and $y \in W$, + (E1) + (E2) $\Rightarrow$ (C3): Using (C2), assume without loss of generality that $\cf$ is closed in $Y^X$ with respect to the product topology. In which case, $\cf$ is a closed subset of $\prod_{x \in X}\ol{\cf(x)}$ with respect to the product topology. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff} and \autoref{proposition:compact-extensions}, $\cf$ is compact in the product topology. By (C1), $\cf$ is also compact in the compact uniform topology. + + (C3) $\Rightarrow$ (E1): Assume that $X$ is a LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_X(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j} \subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_j \times g)(V) \subset U$. For each $1 \le j \le n$, $f_j \in C(X; Y)$, so there exists $V_j \in \cn_X(x)$ with $V_j \subset V$ such that for any $y \in V_j$, $(f_j(x), f_j(y)) \in U$. Let $W = \bigcap_{j = 1}^n V_j$, then for any $g \in \cf$ with $(f_j \times g)(V) \subset U$ and $y \in W$, \[ (g(x), f_j(x)), (f_j(x), f_j(y)), (f_j(y), g(y)) \in U \circ U \circ U \] Therefore $g(W) \subset (U \circ U \circ U)(g(x))$, and $\cf$ is equicontinuous. - (C2) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the compact uniformity, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. + (C3) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the compact uniformity, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. \end{proof} -\begin{corollary} -\label{corollary:arzela-locally-compact} - Let $X$ be a LCH space, $Y$ be a uniform space, and $\cf \subset C(X; Y)$ such that: - \begin{enumerate}[label=(E\arabic*)] - \item $\cf$ is equicontinuous. - \item For each $x \in X$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact in $Y$. - \end{enumerate} - - then $\cf$ is a precompact subset of $C(X; Y)$ with respect to the compact uniformity. -\end{corollary} -\begin{proof} - By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\cf$ is a precompact subset of $Y^X$. By \autoref{proposition:lch-compactly-generated}, $C(X; Y)$ is a closed subset of $Y^X$. Therefore $\cf$ is a precompact subset of $C(X; Y)$. -\end{proof} - - - diff --git a/src/topology/uniform/uc.tex b/src/topology/uniform/uc.tex index 2ca2fe9..30a9063 100644 --- a/src/topology/uniform/uc.tex +++ b/src/topology/uniform/uc.tex @@ -51,7 +51,7 @@ \end{enumerate} - known as the \textbf{initial uniformity} on $X$ generated by $\seqi{f}$. + The collection $\fU$ is the \textbf{initial uniformity} on $X$ generated by $\seqi{f}$. \end{definition} \begin{proof} (3): Since the diagonal is mapped to the diagonal and $\fB$ is closed under intersections, it is sufficient to verify (UB3) for $\fB$. Let $J \subset I$ be finite and $\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j) \in \fB$, then there exists $\bracs{V_j}_{j \in J}$ such that $V_j \circ V_j \subset U_j$ for each $j \in J$. In which case, for any $(x, y), (y, z) \in f_j^{-1}(V_j)$, $(f(x), f(y)), (f(y), f(z)) \in V_j$ and $(f(x), f(z)) \in U_j$. Thus $(f_j \times f_j)^{-1}(V_j) \circ (f_j \times f_j)^{-1}(V_j) \subset (f_j \times f_j)^{-1}(U_j)$, and @@ -65,7 +65,7 @@ (U): For any $i \in I$, $\mathfrak{V} \supset (f_i \times f_i)^{-1}(\fU_i)$. By (F2), $\mathfrak{V} \supset \fB$, so $\mathfrak{V} \supset \fU$. - (4): Let $J \subset I$ finite and $\seqj{U_j}$ such that $U_j \in \fU_j$ for each $j \in J$, then + (4): Let $J \subset I$ finite and $\seqj{U}$ such that $U_j \in \fU_j$ for each $j \in J$, then \[ (f \times f)^{-1}\paren{\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j)} = \bigcap_{j \in J}[(f_j \circ f) \times (f_j \circ f)]^{-1}(U_j) \]