\section{Spaces of Linear Maps}
\label{section:space-linear-map-new}

\begin{definition}[Space of Bounded Linear Maps]
\label{definition:bounded-linear-map-space}
    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $k \in \nat$. The space $B_{\sigma}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in \mathfrak{B}(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform topology.

    Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\sigma}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$. 
\end{definition}

\begin{proposition}[{{\cite[III.3.3]{SchaeferWolff}}}]
\label{proposition:bounded-linear-map-space-bounded}
    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $A \subset B_\sigma(E; F)$, then the following are equivalent:
    \begin{enumerate}
        \item $A \subset B_\sigma(E; F)$ is bounded with respect to the $\sigma$-uniform topology. 
        \item For each $V \in \cn_F(0)$, $\bigcap_{T \in A}T^{-1}(V)$ absorbs every $S \in \sigma$.
        \item For every $S \in \sigma$, $\bigcup_{T \in A}T(A)$ is bounded in $F$. 
    \end{enumerate}
    
\end{proposition}
% Proof omitted because it is obvious. 

\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}.

    The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
\end{definition}


\begin{proposition}
\label{proposition:strong-operator-dense}
    Let $E, F$ be TVSs over $K \in \RC$ and $\net{T} \subset L(E; F)$ and $T \in L_s(E; F)$. If
    \begin{enumerate}
        \item[(a)] There exists a dense subset $S \subset E$ such that $T_\alpha x \to Tx$ strongly for all $x \in S$.
        \item[(b)] $\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous. 
    \end{enumerate}

    
    then $T_\alpha \to T$ in $L_s(E; F)$.
\end{proposition}
\begin{proof}
    Let $x \in E$, $U \in \cn_F(Tx)$, and $V \in \cn_F(Tx)$ be balanced such that $V + V + V \subset U$. By (b), there exists a balanced neighbourhood $W \in \cn_E(0)$ such that $T(W) \cup \bigcup_{\alpha \in A}T_\alpha(W) \subset V$. By (a), there exists $y \in S \cap (x + W)$ and $\alpha_0 \in A$ such that for all $\alpha \ge \alpha_0$, $T_\alpha y - Ty \in V$. In which case, for any $\alpha \ge \alpha_0$,
    \[
    T_\alpha x - Tx = \underbrace{T_\alpha x - T_\alpha y}_{\in V} + \underbrace{T_\alpha y - Ty}_{\in V} + \underbrace{Ty - Tx}_{\in V} \in U
    \]
\end{proof}



\begin{definition}[Weak Operator Topology]
\label{definition:weak-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 $F_w^E$ is the \textbf{weak operator topology}.

    The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology.
\end{definition}

\begin{definition}[Bounded Convergence Topology]
\label{definition:bounded-convergence-topology}
    Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}, or the \textbf{uniform topology}. 

    The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
\end{definition}

\begin{definition}[Topology of Precompact Convergence]
\label{definition:compact-operator-topology}
    Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{K} \subset 2^E$ be the collection of precompact subsets of $E$, then the $\mathfrak{K}$-uniform topology on $L(E; F)$ is the \textbf{topology of precompact convergence}. 

    The space $L_c(E; F)$ denotes $L(E; F)$ equipped with the topology of precompact convergence.
\end{definition}


\begin{proposition}
\label{proposition:operator-space-completeness}
    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
    \begin{enumerate}
        \item $\hom(E; F)$ is a closed subspace of $F^E$ with respect to the product topology.
        \item $B(E; F)$ is a closed subspace of $F^E$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): For each $x, y \in E$ and $\lambda \in K$, the mappings
    \[
    \phi_{x, y}: F^E \to F \quad T \mapsto Tx + Ty - T(x + y)
    \]

    and
    \[
    \psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - \lambda Tx
    \]

    are continuous with respect to the product topology. Since
    \[
    \hom(E; F) = \bigcap_{x, y \in E}\bracsn{\phi_{x, y} = 0} \cap \bigcap_{\substack{x \in E \\ \lambda \in K}}\bracsn{\psi_{x, \lambda} = 0}
    \]

    and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^E$. 

    (2): By \autoref{definition:bounded-function-space} and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^E$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.

\end{proof}