Added the Banach-Steinhaus theorem.

src/fa/tvs/equicontinuous.tex

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+\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}
+
+
+
+