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}
+
+