Added the Bochner integral.

src/fa/tvs/continuous.tex

@@ -51,3 +51,30 @@
     \end{enumerate}
     The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
 \end{definition}
+
+\begin{theorem}[Linear Extension Theorem (TVS)]
+\label{theorem:linear-extension-theorem-tvs}
+    Let $E$ be a TVS over $K \in \RC$, $F$ be a complete Hausdorff TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
+    \begin{enumerate}
+        \item There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_D = T$.
+        \item[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
+    \end{enumerate}
+
+\end{theorem}
+\begin{proof}
+    By (3) of \autoref{definition:continuous-linear}, $T \in UC(D; F)$. By \autoref{theorem:uniform-continuous-extension}, there exists a unique $\ol T \in C(E; F)$ such that $\ol T|_D = T$. 
+
+    It remains to show that $\ol T$ is linear. Since $\ol T$ is continuous, the maps
+    \[
+    A: E \times E \to E \quad (x, y) \mapsto \ol Tx + \ol Ty - \ol T(x + y)
+    \]
+
+    and
+    \[
+    M: K \times E \to E \quad (\lambda x) \mapsto \lambda \ol Tx - Tx
+    \]
+
+    are continuous. Thus $\bracs{A = 0} \supset D \times D$ and $\bracs{M = 0} \supset K \times D$ are both closed. By density of $D$, $\bracs{A = 0} = E \times E$ and $\bracs{M = 0} = K \times E$. Therefore $T$ is linear. 
+\end{proof}
+
+

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

@@ -153,14 +153,34 @@
 
 \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 $L(E; F)$ is the \textbf{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 $L(E; F_w)$ is the \textbf{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}