Adjusted organisation in the TVS chapter.

src/fa/tvs/space-of-linear.tex

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