Slight adjustments.

src/fa/lp/duality.tex

@@ -167,7 +167,7 @@ After the duality of $L^p$ and $L^q$ is established for Hölder conjugate expone
     \norm{g}_{L^\infty(X; H)} \le \sup_{n \in \natp}\norm{g_n}_{L^\infty(X; H)} \le \norm{\phi_g}_{L^1(X; H)^*}
     \]
 
-    The above argument shows that the truncation argument was technically not required. By applying the truncated case again, $\norm{g}_{L^q(X; F)} = \norm{\phi_g}_{L^p(X; E)^*}$.
+    A posteriori, the truncation argument was not required. By applying the truncated case again, $\norm{g}_{L^q(X; F)} = \norm{\phi_g}_{L^p(X; E)^*}$.
 \end{proof}
 
 
@@ -178,7 +178,7 @@ The typical argument for $L^p$ duality requires using the Radon-Nikodym theorem
     Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $H$ be a Hilbert space over $K$, $p, q \in [1, \infty]$ be Hölder conjugates such that one of the following holds:
     \begin{enumerate}[label=(\alph*)]
         \item $p \in (1, \infty)$ and $q \in (1, \infty)$.  
-        \item $p = 1$, $q = \infty$, and $\mu$ is $\sigma$-finite\footnote{This should become localisable. }. 
+        \item $p = 1$, $q = \infty$, and $\mu$ is $\sigma$-finite\footnote{This should become localisable, under the additional hypothesis that $H$ is separable. }. 
     \end{enumerate}
 
     For each $g \in L^q(X, \cm, \mu; H)$, let

src/fa/norm/separable.tex

@@ -12,7 +12,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1), (2): Let $D \subset E$ be a countable dense subset. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $S$ is embedded as a subspace of $\real^D$. By \autoref{theorem:uniform-metrisable}, $\real^D$ is metrisable. By \autoref{proposition:separable-product}, $\real^D$ is separable. Thus $S$ is also metrisable and separable by \autoref{proposition:separable-metric-space}. 
+    (1), (2): Let $D \subset E$ be a countable dense subset. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $S$ is embedded as a subspace of $K^D$. By \autoref{theorem:uniform-metrisable}, $\real^D$ is metrisable. By \autoref{proposition:separable-product}, $K^D$ is separable. Thus $S$ is also metrisable and separable by \autoref{proposition:separable-metric-space}. 
 
     (3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By \autoref{proposition:separable-metric-space}, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable. 
 \end{proof}