Symmetry of the derivative (normed/frechet).

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}