\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}
\label{proposition:equicontinuous-bounded}
    Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, then for any ideal $\sigma \subset \mathfrak{B}(E)$, $\alg$ is a bounded subset of $B_\sigma(E; F)$. 
\end{proposition}
\begin{proof}
    Let $S \in \sigma$ and $U \in \cn_F(0)$, then there exists $V \in \cn_E(0)$ such that $\bigcup_{T \in \alg}T(V) \subset U$. Since $S$ is bounded, there exists $\lambda \in K$ such that $S \subset \lambda V$. Therefore $\bigcup_{T \in \alg}T(S) \subset \lambda U$, and $\alg$ is bounded in $B_\sigma(E; F)$. 
\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')] $E$ is barrelled and $F$ is locally convex.
    \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 topology and the compact-open topology 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}

\begin{lemma}
\label{lemma:equicontinuous-bilinear}
    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg \subset L^2(E, F; G)$ be continuous bilinear maps, then the following are equivalent:
    \begin{enumerate}
        \item $\alg$ is equicontinuous.
        \item $\alg$ is equicontinuous at $0$. 
    \end{enumerate}
\end{lemma}
\begin{proof}
    (2) $\Rightarrow$ (1): For each $(x_0, y_0), (x, y) \in E \times F$ and $\lambda \in \alg$, 
    \[
    \lambda(x, y) - \lambda(x_0, y_0) = \lambda(x - x_0, y - y_0) + \lambda(x - x_0, y_0) + \lambda(x_0, y - y_0)
    \]

    For each $U \in \cn_G(0)$ circled, there exists circled neighbourhoods $V \in \cn_E(0)$ and $W \in \cn_F(0)$ such that $\lambda(V \times W) \subset U$ for all $\lambda \in \alg$. In which case, there exists $\mu > 0$ such that $y_0 \in \mu W$ and $x_0 \in \mu V$. Thus if $(x, y) - (x_0, y_0) \in \mu^{-1}(V \times W)$, then for every $\lambda \in \alg$,
    \[
    \lambda(x - x_0, y_0) \in \lambda(\mu^{-1} V \times \mu W) = \lambda(V \times W) \subset U
    \]

    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}
    Let $E$ be a locally convex space over $K \in \RC$ and $\alg \subset E^*$ be equicontinuous, then $\alg$ is precompact with respect to $\sigma(E^*, E)$.
\end{theorem}
\begin{proof}
    For each $x \in E$, $\alg(x) = \bracsn{\dpn{x, \phi}{E}|\phi \in \alg}$ is precompact by \autoref{proposition:equicontinuous-bounded}. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, 
    \begin{enumerate}
        \item[(C2)] The $\sigma(E^*, E)$-closure of $\alg$ in $\prod_{x \in E}K$ is equicontinuous. 
        \item[(C3)] The $\sigma(E^*, E)$-closure of $\alg$ in $\prod_{x \in E}K$ is compact. 
    \end{enumerate}

    By \autoref{proposition:operator-space-completeness}, the $\sigma(E^*, E)$-closure of $\alg$ in $\prod_{x \in E}\ol{\alg(x)}$ is a subset of $\hom(E; K)$. Hence the $\sigma(E^*, E)$-closure of $\alg$ in $E^*$ is compact. 
\end{proof}