Added some linfty.

src/fa/lp/duality.tex

@@ -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, under the additional hypothesis that $H$ is separable. }. 
+        \item $p = 1$, $q = \infty$, $H$ is separable, and $\mu$ is localisable. 
     \end{enumerate}
 
     For each $g \in L^q(X, \cm, \mu; H)$, let
@@ -203,25 +203,27 @@ The typical argument for $L^p$ duality requires using the Radon-Nikodym theorem
 
     By \autoref{theorem:lp-dual-function}, $g \in L^q(X; H)$. 
 
-    (Arbitrary): In the case of (a), by \autoref{lemma:lp-functional-support}, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^p(X; H)$, $\dpn{f, \phi}{L^p(X; H)} = \dpn{\one_A \cdot f, \phi}{L^p(X; H)}$. In the case of (b), $A = X$ is a $\sigma$-finite set satisfying the same restriction condition. 
+    (Arbitrary): In the case of (a), by \autoref{lemma:lp-functional-support}, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^p(X; H)$, $\dpn{f, \phi}{L^p(X; H)} = \dpn{\one_A \cdot f, \phi}{L^p(X; H)}$. In the case of (b), $A = X$ is a localisable set satisfying the same restriction condition. 
 
-    Let $\seq{A_n} \subset \cm$ such that $\mu(A_n) < \infty$ for all $n \in \natp$, and $A = \bigsqcup_{n \in \natp}A_n$. By the finite case, there exists $\seq{g_n} \subset L^q(X; H)$ such that for each $n \in \natp$ and $f \in L^p(X; H)$, 
+    Let $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$. By the finite case, there exists $g_F \in L^q(F; H)$ such that for every $f \in L^p(X; H)$, 
     \[
-    \int \dpn{f, g_n}{H} d\mu = \dpn{\one_{A_n} \cdot f, \phi}{L^p(X; H)}
-    \] 
-
-    Let $g = \sum_{n = 1}^\infty g_n$. If $q < \infty$, then $g \in L^q(X; H)$ by the \hyperref[Monotone Convergence Theorem]{theorem:mct}. Otherwise,
-    \[
-    \norm{g}_{L^\infty(X; H)} \le \sup_{n \in \natp}\norm{g_n}_{L^\infty(X; H)} \le \norm{\phi}_{L^1(X; H)^*}
+    \int_F \dpn{f, g_F}{H} d\mu = \dpn{\one_{F} \cdot f, \phi}{L^p(X; H)}
     \]
+
+    In the case of (a), there exists a countable exhaustion $\seq{F_n} \subset \cm$ of $A$ with sets of finite measure. For each $n \in \natp$, a representative of $g_{F_n}$ may be taken to have separable range. In the case of (b), such a representative may be chosen for every $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$. Thus by the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists a measurable function $g: X \to H$ such that $g|_{F} = g_F$ almost everywhere for all $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$. 
+
+    If $q < \infty$, then $g \in L^q(X; H)$ by the \hyperref[Monotone Convergence Theorem]{theorem:mct}. Otherwise,
+    \[
+    \norm{g}_{L^\infty(X; H)} \le \sup_{\substack{F \in \cm \\ F \subset A \\ \mu(F) < \infty}}\norm{g_F}_{L^\infty(F; H)} \le \norm{\phi}_{L^1(X; H)^*}
+    \]
+
+    Hence $g \in L^q(X; H)$ with $\norm{g}_{L^q(X; H)} \le \norm{\phi}_{L^1(X; H)^*}$. 
     
-    For every $f \in L^p(X; H)$, 
+    Finally, let $f \in L^p(X; H)$, then there exists $\seq{F_n} \subset \cm$ such that $F_n \upto \bracsn{f \ne 0} \cap A$ and $\mu(F_n) < \infty$ for all $n \in \natp$. In which case, by the \hyperref[Dominated Convergence Theorem]{theorem:dct},
     \begin{align*}
-    \int \dpn{f, g}{H} d\mu &= \sum_{n = 1}^\infty \int \dpn{f, g_n}{H} d\mu = \sum_{n = 1}^\infty \dpn{\one_{A_n} \cdot f, \phi}{L^p(X; H)} \\
-    &= \dpn{f, \phi}{L^p(X; H)}
+        \int \dpn{f, g}{H} d\mu &= \limv{n}\int_{F_n} \dpn{f, g}{H} d\mu = \limv{n}\int_{F_n} \dpn{f, g_{F_n}}{H} d\mu \\
+        &= \limv{n}\dpn{\one_{F_n} \cdot f, \phi}{L^p(X; H)} = \limv{n}\dpn{f, \phi}{L^p(X; H)}
     \end{align*}
-    
-    by the \hyperref[Dominated Convergence Theorem]{theorem:dct}.
 
     Therefore the mapping is surjective, and hence an isomorphism. 
 \end{proof}