Added some linfty.

src/op/c-star/gelfand.tex

@@ -65,13 +65,5 @@
     By \autoref{theorem:gelfand-naimark}, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^*$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the \hyperref[Stone-Nakano Theorem]{theorem:stone-nakano-extremely-disconnected} implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.
 \end{proof}
 
-\begin{corollary}
-\label{corollary:linfinity-extremely-disconnected}
-    Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected. 
-\end{corollary}
-\begin{proof}
-    By \autoref{corollary:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremely disconnected. 
-\end{proof}
-
 
 

src/op/example/bb.tex

@@ -19,9 +19,9 @@
         \item $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^*$. 
         \item There exists a unique weak*-continuous involution on $\mathscr{M}^*$ such that $J(f^*) = J(f)^*$ for all $f \in C(X; \complex)$, given by
         \[
-            \dpn{\mu, \phi^*}{C(X; \complex)^*} = \ol{\dpn{\mu, \phi}{C(X; \complex)^*}}
+            \dpn{\mu, \phi^*}{\mathscr{M}^*} = \ol{\dpn{\mu, \phi}{\mathscr{M}^*}}
         \]
-        \item There exists a unique seperately weak*-continuous bilinear map on $\mathscr{M}^*$ such that $J(fg) = J(f)J(g)$ for all $f, g \in C(X; \complex)$.
+        \item There exists a unique separately weak*-continuous bilinear map on $\mathscr{M}^*$ such that $J(fg) = J(f)J(g)$ for all $f, g \in C(X; \complex)$.
         \item $\mathscr{M}^{*}$ equipped with the above involution and product is a commutative unital $C^*$-algebra. 
     \end{enumerate}
 \end{proposition}
@@ -40,7 +40,7 @@
 
     (2): For each $g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$, let $g^* = \ol g$. For any $\mu \in [l^1(I); L^1(\mu_i; \complex)]$, 
     \[
-    \dpn{\mu, g^*}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{f, \ol g}{[l^1(I); L^1(\mu_i; \complex)]} = \ol{\dpn{f, g}{[l^1(I); L^1(\mu_i; \complex)]}}
+    \dpn{\mu, g^*}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{\mu, \ol g}{[l^1(I); L^1(\mu_i; \complex)]} = \ol{\dpn{\mu, g}{[l^1(I); L^1(\mu_i; \complex)]}}
     \]
 
     so the conjugation map is weak*-continuous. 
@@ -53,4 +53,101 @@
     so the composition map is separately weak*-continuous. 
 
     (4): $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is a commutative unital $C^*$-algebra. 
-\end{proof}
+\end{proof}
+
+\begin{proposition}
+\label{proposition:linfty-von-neumann-algebra}
+    Let $(X, \cm, \mu)$ be a localisable measure space, and let $L^\infty(X; \complex)$ act on $L^2(X; \complex)$ by multiplication, then:
+    \begin{enumerate}
+        \item The weak* topology on $L^\infty(X; \complex)$ is equal to the weak operator topology of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$.
+        \item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): Let $A \subset B(L^2(X; \complex))$ be the von Neumann algebra generated by $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$. By \autoref{theorem:lp-duality}, $L^\infty(X; \complex)$ is the dual of $L^1(X; \complex)$, so (1) and the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu} imply that the closed unit ball of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is weak-operator closed. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_{L^\infty(X; \complex)}(0, 1)} = \ol{B_A(0, 1)}$. Therefore $L^\infty(X; \complex) = A$. 
+\end{proof}
+
+\begin{theorem}[Order Structure of $L^\infty$]
+\label{theorem:l-infty-dedekind-complete}
+    Let $(X, \cm, \mu)$ be a localisable measure space, then
+    \begin{enumerate}
+        \item $L^\infty(X; \real)$ is order complete. 
+        \item For each $\phi \in L^1(X; \real)$ with $\phi \ge 0$ and bounded directed subset $S \subset L^\infty(X; \real)$,
+        \[
+        \sup_{f \in S} \dpn{\phi, f}{L^1(X; \real)} = \bigg\langle\phi, \sup_{f \in S}f\bigg\rangle_{L^1(X; \real)}
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    By \autoref{proposition:linfty-von-neumann-algebra}, $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra. 
+
+    (1): Since $L^\infty(X; \real)$ is a lattice, it is order complete by \autoref{theorem:existence-of-projections-vna}.
+
+    (2): By \autoref{theorem:existence-of-projections-vna}, $\sup(S) = \sotlim_{f \in S}f = \wotlim_{f \in S}f$. By (1) of \autoref{proposition:linfty-von-neumann-algebra}, $\sup_{f \in S}\dpn{\phi, f}{L^1(X; \real)} = \dpn{\phi, \sup_{f \in S}f}{L^1(X; \real)}$.
+\end{proof}
+
+
+\begin{corollary}
+\label{corollary:linfinity-extremely-disconnected}
+    Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected. 
+\end{corollary}
+\begin{proof}
+    By \autoref{theorem:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremely disconnected. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:l-infty-order-isotone}
+    Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be localisable measure spaces, and $T: L^\infty(X; \complex)\to L^\infty(Y; \complex)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $T$ is an isometric isomorphism.
+        \item For each $f, g \in L^\infty(X; \complex)$, $f \ge g$ if and only if $Tf \ge Tg$. 
+    \end{enumerate}
+    
+    then:
+    \begin{enumerate}
+        \item $T^*(L^1(Y; \complex)) \subset L^1(X; \complex)$.
+        \item $T$ is weak*-continuous. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    Let $\phi \in L^1(Y; \complex)$ with $\phi \ge 0$, then $T^*\phi \in L^\infty(X; \complex)^*$. For each $\seq{B_n} \subset \cm$ and $B \in \cm$ with $B_n \upto B$, $\one_B = \sup_{n \in \natp}\one_{B_n}$ as an element of $L^\infty(X; \complex)$. Thus (b) and \autoref{theorem:l-infty-dedekind-complete} imply that
+    \begin{align*}
+    \sup_{n \in \natp}\dpn{\one_{B_n}, T^*\phi}{L^\infty(X; \complex)} &= \sup_{n \in \natp} \dpn{\phi, T\one_{B_n}}{L^1(Y; \complex)} = \dpn{\phi, T\one_B}{L^1(Y; \complex)} \\
+    &= \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}
+    \end{align*}
+
+    Hence the mapping $B \mapsto \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}$ is a finite positive measure on $(X, \cm)$, which is absolutely continuous with respect to $\mu$. By the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(X; \complex)$ with $f \ge 0$ such that $\int_B f d\mu = \dpn{\one_{B}, T^*\phi}{L^\infty(X; \complex)}$ for all $B \in \cm$. 
+
+    Let $g \in L^\infty(X; [0, 1])$. By \autoref{lemma:separable-metric-space-approx-identity}, there exists simple functions $\seq{g_n} \subset \Sigma(X; [0, 1])$ such that $g_n \upto g$ pointwise. In which case, $g = \sup_{n \in \natp}g_n$ as an element of $L^\infty(X; \real)$, so (b) and \autoref{theorem:l-infty-dedekind-complete} imply that, 
+    \begin{align*}
+    \dpn{\phi, Tg}{L^1(Y; \complex)} &= \sup_{n \in \natp}\dpn{\phi, Tg_n}{L^1(Y; \complex)} = \sup_{n \in \natp}\dpn{f, g_n}{L^1(X; \complex)} \\
+    &= \dpn{f, g}{L^1(X; \complex)}
+    \end{align*}
+
+    By linearity, $\dpn{\phi, Tg}{L^1(Y; \complex)} = \dpn{f, g}{L^1(X; \complex)}$ for all $g \in L^\infty(X; \complex)$. Therefore $T^*(L^1(Y; \complex)) \subset L^1(X; \complex)$, and $T$ is weak*-continuous. 
+\end{proof}
+
+\begin{theorem}[Uniqueness of $L^\infty$]
+\label{theorem:linfty-uniqueness}
+    Let $X$ be a compact Hausdorff space, $\mu, \nu: \cb_X \to [0, \infty)$ be Radon measures on $X$, and $\Phi: L^\infty(\mu; \complex) \to L^\infty(\nu; \complex)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $\Phi$ is a *-isomorphism. 
+        \item $\Phi|_{C(X; \complex)}$ is the identity.
+    \end{enumerate}
+
+    then:
+    \begin{enumerate}
+        \item $\mu$ and $\nu$ are equivalent. 
+        \item $L^\infty(\mu; \complex) = L^\infty(\nu; \complex)$. 
+        \item $\Phi$ is the identity map. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 21.4]{Zhu}}}. ]
+    As $\Phi$ is a *-isomorphism, $\Phi f \ge \Phi g$ if and only if $f \ge g$ for any $f, g \in L^\infty(\mu; \complex)$. 
+    
+    (1): By \autoref{lemma:l-infty-order-isotone}, there exists $f \in L^1(\mu; \complex)$ such that $\int f g d\mu = \int g d\nu$ for all $g \in C(X; \complex)$. Thus the uniqueness of the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0} implies that $\nu = f d\mu$. As the argument is symmetric, the two measures are equivalent. 
+
+    (3): By \autoref{lemma:l-infty-order-isotone}, $\Phi$ is also weak*-continuous. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex)$ is weak*-dense in $L^\infty(\mu; \complex)$ and $L^\infty(\nu; \complex)$. As $\Phi$ is the identity on $C(X; \complex)$, $\Phi$ is the identity on $L^\infty(\mu; \complex)$. 
+\end{proof}
+
+