Symmetry of the derivative (normed/frechet).

src/fa/norm/index.tex

@@ -2,3 +2,5 @@
 \label{chap:normed-spaces}
 
 \input{./src/fa/norm/normed.tex}
+\input{./src/fa/norm/linear.tex}
+\input{./src/fa/norm/multilinear.tex}

src/fa/norm/linear.tex

@@ -0,0 +1,13 @@
+\section{Linear Maps}
+\label{section:normed-linear-maps}
+
+\begin{proposition}
+\label{proposition:normed-linear-map-space}
+    Let $E, F$ be normed vector spaces, then the topology on $L_b(E; F)$ is induced by the \textbf{operator norm}
+    \[
+    \norm{\cdot}_{L(E; F)}: L(E; F) \to [0, \infty) \quad T \mapsto \sup_{\substack{x \in E \\ \norm{x}_E = 1}}Tx
+    \]
+\end{proposition}
+\begin{proof}
+    By \ref{proposition:lc-spaces-linear-map}.
+\end{proof}

src/fa/norm/multilinear.tex

@@ -0,0 +1,20 @@
+\section{Multilinear Maps}
+\label{section:normed-multilinear}
+
+\begin{proposition}
+\label{proposition:bilinear-separate}
+    Let $E, F, G$ be normed spaces and $T: E \times F \to G$ be a bilinear map. If:
+    \begin{enumerate}
+        \item For each $x \in E$, $y \mapsto T(x, y)$ is a continuous linear map from $F$ to $G$.
+        \item For each $y \in F$, $x \mapsto T(x, y)$ is a continuous linear map from $E$ to $G$.
+        \item $E$ is a Banach space.
+    \end{enumerate}
+    then $T \in L^2(E, F; G)$.
+\end{proposition}
+\begin{proof}
+    For each $y \in F$, let $T_y \in L(E; G)$ be defined by $x \mapsto T(x, y)$. Let $x \in X$, then
+    \[
+    \sup_{y \in B_F(0, 1)}\norm{T_yx}_G = \sup_{y \in B_F(0, 1)}\norm{T(x, y)}_G < \infty
+    \]
+    by continuity of $y \mapsto T(x, y)$. By the Uniform Boundedness Principle (\ref{theorem:uniform-boundedness}), $M = \sup_{y \in B_F(0, 1)}\norm{T_y}_{L(E; G)} < \infty$. Thus for any $x \in E$ and $y \in F$, $\norm{T(x, y)}_G \le M\norm{x}_E\norm{y}_F$.
+\end{proof}

src/fa/norm/normed.tex

@@ -61,3 +61,22 @@
     \]
     so $\sum_{n = 1}^\infty Tx_n = y$.
 \end{proof}
+
+\begin{theorem}[Uniform Boundedness Principle]
+\label{theorem:uniform-boundedness}
+    Let $E, F$ be normed spaces and $\mathcal{T} \subset L(E; F)$. If
+    \begin{enumerate}
+        \item For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
+        \item $E$ is a Banach space.
+    \end{enumerate}
+    then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} < \infty$.
+\end{theorem}
+\begin{proof}
+    For each $n \in \natp$, let $A_n = \bracs{x \in X|\norm{Tx}_F \le n \forall T \in \mathcal{T}}$, then each $A_n$ is closed with $\bigcup_{n \in \natp}A_n = E$. By the Baire Category Theorem (\ref{theorem:baire}), there exists $n \in \natp$ and $U \subset E$ open such that $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
+
+    Let $x \in U$ and $r > 0$ such that $\overline{B(x, r)} \subset U$, then for any $y \in E$ with $\norm{y}_E \le r$ and $T \in \mathcal{T}$,
+    \[
+    \norm{Ty} = \norm{Ty + Tx - Tx}_E = \normn{T\underbrace{(x + y)}_{\in U}}_E + \norm{Tx}_E \le 2n
+    \]
+    so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
+\end{proof}