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src/fa/lc/continuous.tex

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+\section{Continuous Linear Maps}
+\label{section:tvs-convex-morphism}
+
+
+\begin{proposition}
+\label{proposition:tvs-convex-morphism}
+    Let $E, F$ be locally convex spaces and $T \in \hom(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $T$ is uniformly continuous.
+        \item $T$ is continuous.
+        \item $T$ is continuous at $0$.
+        \item For every continuous seminorm $[\cdot]_F$ on $F$, there exists a continuous seminorm $[\cdot]_E$ on $E$ such that $[Tx]_F \le [x]_E$ for all $x \in E$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    $(1) \Leftrightarrow (2) \Leftrightarrow (3)$: By \ref{definition:continuous-linear}.
+
+    $(2) \Rightarrow (4)$: $x \mapsto [Tx]_F$ is a continuous seminorm on $E$.
+
+    $(4) \Rightarrow (3)$: Let $U \in \cn_F(0)$ be convex, circled, and radial, then its gauge $[\cdot]_U$ is a continuous seminorm on $F$ by \ref{definition:locally-convex}. Thus there exists a continuous seminorm $[\cdot]_E$ such that $[Tx]_U \le [x]_E$. In which case, $V = \bracs{x \in E| [x]_E < 1} \in \cn_E(0)$ with $T(V) \subset U$. Therefore $T$ is continuous at $0$, and continuous by \ref{definition:continuous-linear}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tvs-convex-multilinear}
+    Let $\seqf{E_j}$ and $F$ be locally convex spaces, and $T: \prod_{j = 1}^n E_j \to F$ be $n$-linear map, then the following are equivalent:
+    \begin{enumerate}
+        \item $T$ is continuous.
+        \item For every continuous seminorm $[\cdot]_F$ on $F$, there exists continuous seminorms $\seqf{[\cdot]_j}$ on $\seqf{E_j}$, such that for every $x \in \prod_{j = 1}^n E_j$,
+        \[
+        [Tx]_F \le \prod_{j = 1}^n [x_j]_{E_j}
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    $(1) \Rightarrow (2)$: By continuity of $T$, there exists continuous seminorms $\seqf{[\cdot]_j}$ on $\seqf{E_j}$ such that for any $x \in \prod_{j = 1}^n E_j$, $\max_{1 \le j \le n}[x_j]_{E_j} < 1$ implies that $[Tx]_F < 1$. In which case, the inequality follows from linearity.
+\end{proof}