Cleanup

src/fa/norm/normed.tex

@@ -21,12 +21,13 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1) $\Rightarrow$ (2): Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is also circled. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$ and $\abs{\lambda}^{-1}U \subset V$.
+    (1) $\Rightarrow$ (2): Using \autoref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is also circled. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$ and $\abs{\lambda}^{-1}U \subset V$.
 
-    Let $\norm{\cdot}_E: E \to [0, \infty)$ be the gauge (\ref{definition:gauge}) of $U$. For each $r > 0$, let $B_E(0, r) = \bracs{x \in E|\norm{x}_E < r}$, then
+    Let $\norm{\cdot}_E: E \to [0, \infty)$ be the \hyperref[gauge]{definition:gauge} of $U$. For each $r > 0$, let $B_E(0, r) = \bracs{x \in E|\norm{x}_E < r}$, then
     \[
     \bracs{\lambda U|\lambda > 0} = \bracs{B_E(0, r)|r > 0}
     \]
+
     is a fundamental system of neighbourhoods at $0$ for $E$. Therefore $\norm{\cdot}_E$ induces the topology on $E$.
 \end{proof}
 
@@ -51,14 +52,17 @@
     \[
     \norm{y_n}_F \le \gamma^{n - 1}\norm{y_1}_F = \gamma^{n - 1}\norm{y}_F
     \]
+
     Since $\norm{x_n}_E \le C\norm{y_n}_F$,
     \[
     \sum_{k \in \natp}\norm{x_k}_E \le C\norm{y}_F\sum_{k \in \nat_0}\gamma^k = \frac{C\norm{y}_F}{1 - \gamma}
     \]
+
     In addition,
     \[
     \norm{y - \sum_{k = 1}^n Tx_k}_F = \norm{y_{n+1}}_F \le \gamma^n \norm{y}_F
     \]
+
     so $\sum_{n = 1}^\infty Tx_n = y$.
 \end{proof}
 
@@ -72,11 +76,12 @@
     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 $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
+    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 $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
 \end{proof}