Housekeeping.

src/fa/norm/normed.tex

@@ -73,21 +73,14 @@
 \label{theorem:uniform-boundedness}
     Let $E, F$ be normed vector 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.
+        \item[(B)] $E$ is a Banach space.
+        \item[(E2)] For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
     \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 \hyperref[Baire Category Theorem]{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 $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
+    By the \autoref{theorem:banach-steinhaus} theorem, $\mathcal{T}$ is equicontinuous. Therefore there exists $r > 0$ such that $\bigcup_{T \in \mathcal{T}}T[B_E(0, r)] \subset B_F(0, 1)$. In which case, $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le r^{-1}$. 
 \end{proof}