Fixed typo.

preamble.sty

@@ -232,3 +232,10 @@
 \newcommand{\sotlim}{\operatorname*{\text{\small SOT}\text{-}\!\lim}}
 \newcommand{\wotlim}{\operatorname*{\text{\small WOT}\text{-}\!\lim}}
 
+% VNA Types
+\newcommand{\vnI}{\mathrm{I}}
+\newcommand{\vnII}{\mathrm{II}}
+\newcommand{\vnIIo}{\mathrm{II}_{1}}
+\newcommand{\vnIIi}{\mathrm{II}_{\infty}}
+\newcommand{\vnIII}{\mathrm{III}}
+

refs.bib

@@ -268,4 +268,28 @@
   year      = {2002},
   isbn      = {978-1-85233-437-6},
   doi       = {10.1007/978-1-4471-3903-4}
+}
+
+@article{ArensBilinear,
+ ISSN = {00029939, 10886826},
+ URL = {http://www.jstor.org/stable/2031695},
+ author = {Richard Arens},
+ journal = {Proceedings of the American Mathematical Society},
+ number = {6},
+ pages = {839--848},
+ publisher = {American Mathematical Society},
+ title = {The Adjoint of a Bilinear Operation},
+ urldate = {2026-08-13},
+ volume = {2},
+ year = {1951}
+}
+
+
+@MISC {TownesType1,
+    title = {Classification of Type 1 factors},
+    author = {leslie townes},
+    howpublished = {Mathematics Stack Exchange},
+    note = {URL:https://math.stackexchange.com/q/150258 (version: 2012-05-27)},
+    eprint = {https://math.stackexchange.com/q/150258},
+    url = {https://math.stackexchange.com/q/150258}
 }

src/cat/cat/cat-func.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Category]
 \label{definition:category}
-    A \textbf{category} $\catc$ is a collection of objects $\obj{\catc}$, such that for any $A, B, C \in \obj{\catc}$, there exists sets $\mor{A, B}$, $\mor{B, C}$, and a composition law
+    A \textbf{category} $\catc$ is a collection of objects $\obj{\catc}$, such that for any $A, B, C \in \obj{\catc}$, there exist sets $\mor{A, B}$, $\mor{B, C}$, and a composition law
     \[
     \mor{A, B} \times \mor{B, C} \to \mor{A, C}
     \]

src/fa/lc/beqtensor.tex

@@ -58,5 +58,44 @@
     is \textit{the} \textbf{injective cross norm} on $E \otimes_\eps F$. 
 \end{definition}
 
+\begin{definition}[Integral Bilinear Form]
+\label{definition:integral-bilinear-form}
+    Let $E, F$ be locally convex spaces over $K \in \RC$ and $\lambda \in L^2(E, F; K)$ be a bilinear form, then $\lambda$ is \textbf{integral} if there exists equicontinuous subsets $S \subset E^*$ and $T \subset F^*$, and a Radon measure $\mu \in M_R(S \times T; K)$ such that
+    \[
+    \lambda(x, y) = \int_{S \times T} \dpn{x, \phi}{E} \dpn{y, \psi}{F} \mu(d\phi, d\psi)
+    \]
+
+    for all $x, y \in E$. 
+
+    The set $I(E, F)$ is the \textbf{space of integral bilinear forms} on $E$ and $F$. 
+\end{definition}
+
+
+\begin{theorem}
+\label{theorem:injective-dual-bilinear-form}
+    Let $E, F$ be locally convex spaces over $K \in \RC$, then $(E \wh \otimes_\eps F)^* = I(E, F)$. 
+\end{theorem}
+\begin{proof}
+    Let $\lambda \in (E \wh \otimes_\eps F)^*$, then there exists equicontinuous subsets $S \subset E^*$ and $T \subset F^*$ such that for each $x \in E$ and $y \in F$, 
+    \[
+    |\lambda(x, y)| \le \sup_{\phi \in S}\sup_{\psi \in T} |\dpn{x, \phi}{E} \dpn{y, \psi}{F}|
+    \]
+
+    For any $(x, y) \in E \times F$ and $(\phi, \psi) \in S \times T$, let $f_{xy}(\phi, \psi) = \dpn{x, \phi}{E} \dpn{y, \psi}{F}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\Lambda \in C(S \times T; K)^*$ such that the following diagram commutes:
+    \[
+    \xymatrix{
+    & C(S \times T; K) \ar@{->}[rd]^{\Lambda} &  \\
+    E \times F \ar@{->}[ru]^{{(x, y) \mapsto f_{xy}}} \ar@{->}[rr]_{\lambda} &  & K
+    }
+    \]
+
+    Now, since $S$ and $T$ are equicontinuous, using the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, assume without loss of generality that $S$ and $T$ are weak*-compact. In which case, by the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, there exists $\mu \in M_R(S \times T; K)$ such that $\Lambda(f) = \int_{S \times T} f d\mu$ for all $f \in C(S \times T; K)$. Therefore
+    \[
+    \lambda(x, y) = \Lambda(f_{xy}) = \int_{S \times T}f_{xy} d\mu = \int_{S \times T} \dpn{x, \phi}{E}\dpn{y, \psi}{F} d\mu
+    \]
+
+    for all $(x, y) \in E \times F$, and $\lambda \in I(E; F)$. 
+\end{proof}
+
 
 

src/fa/lc/compact.tex

@@ -33,7 +33,6 @@
 \end{definition}
 
 
-
 \begin{lemma}
 \label{lemma:extremal-face}
     Let $E$ be a locally convex space over $\real$, $K \subset E$ be non-empty and compact, and $\phi \in E^*$. Let $\alpha = \sup\bracs{\dpn{x, \phi}{E}|x \in K}$, then $A = \bracs{\phi = \alpha} \cap K$ is a non-empty extreme subset of $K$. 

src/fa/lp/definition.tex

@@ -1,6 +1,12 @@
 \section{Basic Properties}
 \label{section:lp-basic}
 
+\begin{definition}[$B^\infty$ Spaces]
+\label{definition:bounded-borel-function}
+    Let $(X, \cm, \mu)$ be a measure space and $E$ be a normed vector space, then the set $B^\infty(X; E)$ is the \textbf{space of bounded $E$-valued strongly measurable functions} on $X$, and the set $B^\infty(X)$ is the space of bounded complex-valued Borel measurable functions on $X$. 
+\end{definition}
+
+
 \begin{definition}[$\mathcal{L}^p$ Spaces]
 \label{definition:lp-unequivalence}
     Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $f: X \to E$ be strongly measurable, and $p \in [1, \infty)$, then $f$ is \textbf{$p$-integrable} if
@@ -8,17 +14,19 @@
     \norm{f}_{L^p(X; E)} = \norm{f}_{L^p(\mu; E)} = \norm{f}_{L^p(X, \cm, \mu; E)} = \braks{\int \norm{f}_E^p d\mu}^{1/p} < \infty
     \]
 
-    The set $\mathcal{L}^p(X; E) = \mathcal{L}^p(\mu; E) = \mathcal{L}^p(X, \cm, \mu; E)$ is the space of all $p$-integrable functions on $X$.
+    The set $\mathcal{L}^p(X; E) = \mathcal{L}^p(\mu; E) = \mathcal{L}^p(X, \cm, \mu; E)$ is the space of all $E$-valued $p$-integrable functions on $X$.
 \end{definition}
 
 \begin{definition}[Essential Supremum]
 \label{definition:esssup}
     Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be strongly measurable, then $f$ is \textbf{essentially bounded} if
     \[
-    \norm{f}_{L^\infty(X; E)} = \norm{f}_{L^\infty(\mu; E)} = \norm{f}_{L^\infty(X, \cm, \mu; E)} = \inf\bracs{\alpha \ge 0|\mu(\bracs{f > \alpha}) = 0} < \infty
+    \norm{f}_{\mathcal{L}^\infty(X; E)} = \norm{f}_{\mathcal{L}^\infty(\mu; E)} = \norm{f}_{\mathcal{L}^\infty(X, \cm, \mu; E)} = \inf\bracs{\alpha \ge 0|\mu(\bracs{f > \alpha}) = 0} < \infty
     \]
 
-    In which case, $\norm{f}_{L^\infty(X; E)}$ is the \textbf{essential supremum} of $f$.
+    In which case, $\norm{f}_{\mathcal{L}^\infty(X; E)}$ is the \textbf{essential supremum} of $f$. 
+    
+    The set $\mathcal{L}^\infty(X; E) = \mathcal{L}^\infty(\mu; E) = \mathcal{L}^\infty(X, \cm, \mu; E)$ is the space of all $E$-valued essentially bounded functions on $X$. 
 \end{definition}
 
 \begin{definition}[Hölder conjugates]

src/fa/lp/duality.tex

@@ -166,8 +166,8 @@ 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 $p = 1$, $q = \infty$, $H$ 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}

src/fa/norm/absolute.tex

@@ -33,7 +33,7 @@
 
 \begin{theorem}[Riemann's Rearrangement Theorem]
 \label{theorem:riemann-rearrangement}
-    Let $\seq{x_n} \subset \real$ and $N = P \sqcup N$ such that $x_n \ge 0$ for all $n \in P$ and $x_n \le 0$ for all $n \in N$, then
+    Let $\seq{x_n} \subset \real$ and $\natp = P \sqcup N$ be a partition such that $x_n \ge 0$ for all $n \in P$ and $x_n \le 0$ for all $n \in N$, then
     \begin{enumerate}
         \item If $\sum_{n \in P}x_n = \infty$ and $\sum_{n \in N}x_n = -\infty$, then there exists bijections $\sigma, \tau: \natp \to \natp$ such that
         

src/fa/norm/arens.tex

@@ -0,0 +1,117 @@
+\section{The Arens Product}
+\label{section:arens-product}
+
+\begin{definition}[Arens Extension]
+\label{definition:arens-product}
+    Let $E, F, G$ be normed vector spaces over $K \in \RC$ and $\lambda \in L^2(E, F; G)$ be a continuous bilinear mapping, then there exists a unique bilinear mapping $\Lambda_1: E^{**} \times F^{**} \to G^{**}$ such that:
+    \begin{enumerate}
+        \item For each $(x, y) \in E \times F$, $\Lambda_1(x, y) = \lambda(x, y)$. 
+        \item For each $x \in E$, $\Lambda_1(x, \cdot)$ is weak*-continuous. 
+        \item For each $y \in F^{**}$, $\Lambda_1(\cdot, y)$ is weak*-continuous. 
+    \end{enumerate}
+
+    Similarly, there exists a unique bilinear mapping $\Lambda_2: E^{**} \times F^{**} \to G^{**}$ such that:
+    \begin{enumerate}
+        \item For each $(x, y) \in E \times F$, $\Lambda_2(x, y) = \lambda(x, y)$. 
+        \item[(2')] For each $x \in E^{**}$, $\Lambda_2(x, \cdot)$ is weak*-continuous. 
+        \item[(3')] For each $y \in F$, $\Lambda_2(\cdot, y)$ is weak*-continuous. 
+    \end{enumerate}
+
+    The mappings $\Lambda_1, \Lambda_2: E^{**} \times F^{**} \to G^{**}$ are the \textbf{first} and \textbf{second} \textbf{Arens extensions} of $\lambda$, respectively. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Section 1, Theorem 3.2]{ArensBilinear}}}. ]
+    For each $x \in E$, the mapping $\lambda(x, \cdot) \in L(F; G)$ admits an adjoint $\lambda^*(x, \cdot) \in L(G^*; F^*)$, which induces an adjoint of the bilinear map as follows
+    \[
+    \lambda^*: E \times G^* \to F^* \quad \dpn{y, \lambda^*(x, \phi)}{F} = \dpn{\lambda(x, y), \phi}{G}
+    \]
+
+    Applying the above operation again yields a second adjoint
+    \[
+    \lambda^{**}: F^{**} \times G^* \to E^* \quad \dpn{x, \lambda^{**}(y, \phi)}{E} = \dpn{\lambda^*(x, \phi), y}{F^*}
+    \]
+
+    and finally, applying the adjoint operation a third time gives
+    \[
+    \Lambda_1 = \lambda^{***}: E^{**} \times F^{**} \to G^{**} \quad \dpn{\phi, \lambda^{***}(x, y)}{G^*} = \dpn{\lambda^{**}(y, \phi), x}{E^*}
+    \]
+
+    (1): Let $(x, y) \in E \times F$, then for each $\phi \in G^*$, 
+    \[
+        \dpn{\phi, \Lambda_1(x, y)}{G^*} = \dpn{\lambda^{**}(y, \phi), x}{E^*} = \dpn{\lambda^*(x, \phi), y}{F^*} = \dpn{\lambda(x, y), \phi}{G}
+    \]
+
+    By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $\Lambda_1$ is an extension of $\lambda$. 
+
+    (2): Fix $x \in E$ and $\phi \in G^*$, then for each $y \in F^{**}$, 
+    \begin{align*}
+        \dpn{\phi, \Lambda_1(x, y)}{G^*} &= \dpn{\lambda^{**}(y, \phi), x}{E^*} = \dpn{x, \lambda^{**}(y, \phi)}{E} \\
+        &= \dpn{\lambda^*(x, \phi), y}{F^*}
+    \end{align*}
+
+    Since $\lambda^*(x, \phi) \in F^*$, $\Lambda_1(x, \cdot)$ is weak*-continuous. 
+
+    (3): Fix $y \in F^{**}$ and $\phi \in G^*$, then for each $x \in E^{**}$, $\dpn{\phi, \Lambda_1(x, y)}{G^*} = \dpn{\lambda^{**}(y, \phi), x}{E^*}$. Since $\lambda^{**}(y, \phi) \in E^*$, $\Lambda_1(\cdot, y)$ is weak*-continuous. 
+
+    (Uniqueness): By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $E$ is weak*-dense in $E^{**}$, and $F$ is weak*-dense in $F^{**}$, so the extension is uniquely determined.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:arens-extension-continuous}
+    Let $E, F, G$ be normed vector spaces over $K \in \RC$, $\lambda \in L^2(E, F; G)$ be a continuous bilinear mapping, then $\Lambda_1, \Lambda_2 \in L^2(E^{**}, F^{**}; G^{**})$, where
+    \[
+    \norm{\lambda}_{L^2(E, F; G)} = \norm{\Lambda_1}_{L^2(E^{**}, F^{**}; G^{**})} = \norm{\Lambda_2}_{L^2(E^{**}, F^{**}; G^{**})}
+    \]
+\end{proposition}
+\begin{proof}
+    Assume without loss of generality that $\norm{\lambda}_{L^2(E, F; G)} = 1$. Fix $x \in \ol{B_{E}(0, 1)}$, then
+    \[
+    \Lambda_1(x, \ol{B_F(0, 1)}) = \lambda(x, \ol{B_F(0, 1)}) \subset \ol{B_{G^{**}}(0, 1)}
+    \]
+    
+    
+    Since for any $z \in G^{**}$, $\norm{z}_{G^{**}} = \sup_{\phi \in G^*, \norm{\phi}_{G^*} \le 1}\dpn{z, \phi}{G^*}$, the norm on $G^{**}$ is weak*-lower semicontinuous, and $\ol{B_{G^{**}}(0, 1)}$ is weak*-closed. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $\ol{B_{F}(0, 1)}$ is weak*-dense in $\ol{B_{F^{**}}(0, 1)}$. As such, weak*-continuity of $\Lambda_1(x, \cdot)$ and \autoref{proposition:closure-of-image} implies that $\Lambda_1(x, \ol{B_{F^{**}}(0, 1)}) \subset \ol{B_{G^{**}}(0, 1)}$ as well. 
+
+    Now, fix $y \in \ol{B_{F^{**}}(0, 1)}$, then $\Lambda_1(\ol{B_E(0, 1)}, y) \subset \ol{B_{G^{**}}(0, 1)}$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $\ol{B_{E}(0, 1)}$ is weak*-dense in $\ol{B_{E^{**}}(0, 1)}$. Thus the weak*-continuity of $\Lambda_1(\cdot, y)$ and \autoref{proposition:closure-of-image} implies that $\Lambda_1(\ol{B_{E^{**}}(0, 1)}, y) \subset \ol{B_{G^{**}}(0, 1)}$. Therefore
+    \[
+    \Lambda_1(\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}) \subset \ol{B_{G^{**}}(0, 1)}
+    \]
+
+    and $\norm{\lambda}_{L^2(E, F; G)} = \norm{\Lambda_1}_{L^2(E^{**}, F^{**}; G^{**})}$. 
+\end{proof}
+
+
+\begin{definition}[Arens Regularity]
+\label{definition:arens}
+    Let $E, F, G$ be normed vector spaces over $K \in \RC$, $\lambda \in L^2(E, F; G)$ be a continuous bilinear mapping, and $\Lambda_1, \Lambda_2: E^{**} \times F^{**} \to G^{**}$ be its first and second Arens extensions, respectively, then the following are equivalent:  
+    \begin{enumerate}
+        \item $\Lambda_1 = \Lambda_2$. 
+        \item There exists an extension $\Lambda: E^{**} \times F^{**} \to G^{**}$ of $\lambda$ that is separately weak*-continuous. 
+        \item There exists an extension $\Lambda: E^{**} \times F^{**} \to G^{**}$ of $\lambda$ that is separately weak*-continuous when restricted to $\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}$.
+    \end{enumerate}
+    
+    If the above holds, then $\lambda$ is an \textbf{Arens regular} bilinear map, and $\Lambda = \Lambda_1 = \Lambda_2$ is \textit{the} \textbf{Arens extension} of $\lambda$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem 3.3]{ArensBilinear}}}. ]
+    (3) $\Rightarrow$ (1): By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $\ol{B_{E}(0, 1)}$ is weak*-dense in $\ol{B_{E^{**}}(0, 1)}$, and $\ol{B_{F}(0, 1)}$ is weak*-dense in $\ol{B_{F^{**}}(0, 1)}$. Thus the restrictions of $\Lambda_1$ and $\Lambda_2$ to $\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}$ are uniquely determined by the value of $\lambda$ on $\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}$, in the following sense:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\Lambda_1|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}$ is the unique extension of $\lambda|_{\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}}$ such that
+        \begin{enumerate}[label=(\alph*)]
+            \item For each $x \in \ol{B_E(0, 1)}$, $\Lambda_1(x, \cdot)$ is weak*-continuous. 
+            \item For each $y \in \ol{B_{F^{**}}(0, 1)}$, $\Lambda_1(\cdot, y)$ is weak*-continuous. 
+        \end{enumerate}
+        \item $\Lambda_2|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}$ is the unique extension of $\lambda|_{\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}}$ such that
+        \begin{enumerate}[label=(\alph*)]
+            \item For each $x \in \ol{B_{E^{**}}(0, 1)}$, $\Lambda_2(x, \cdot)$ is weak*-continuous. 
+            \item For each $y \in \ol{B_{F}(0, 1)}$, $\Lambda_2(\cdot, y)$ is weak*-continuous. 
+        \end{enumerate}
+    \end{enumerate}
+
+    Since the given extension $\Lambda$ satisfies (i.a), (i.b), (ii.a), and (ii.b), 
+    \[
+    \Lambda_1|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}} = \Lambda = \Lambda_2|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}
+    \]
+
+    As $\Lambda_1, \Lambda_2$ are bilinear, the above implies that $\Lambda_1 = \Lambda_2$. 
+\end{proof}
+
+

src/fa/norm/index.tex

@@ -1,11 +1,13 @@
 \chapter{Normed Vector Spaces}
 \label{chap:normed-spaces}
 
+
 \input{./normed.tex}
 \input{./absolute.tex}
 \input{./linear.tex}
 \input{./separable.tex}
 \input{./multilinear.tex}
+\input{./arens.tex}
 \input{./hilbert.tex}
 \input{./compact.tex}
 \input{./ap.tex}

src/fa/norm/separable.tex

@@ -12,21 +12,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $\seq{x_n} \subset E$ be a dense subset. For each $N \in \natp$, let
-    \[
-    T_N: S \to \real^N \quad y \mapsto (\dpn{x_1, y}{\lambda}, \cdots, \dpn{x_N, y}{\lambda})
-    \]
-
-    Since $\real^N$ is separable, $T_N(S)$ is separable by \autoref{proposition:separable-metric-space}. Thus there exists $\bracs{y_{N, k}}_{k = 1}^\infty \subset S$ such that $\bracs{T_Ny_{N, k}}_{k = 1}^\infty$ is dense in $T_N(S)$.
-
-    Let $y \in S$, then for each $N \in \natp$, there exists $k_N \in \natp$ such that for each $1 \le n \le N$, 
-    \[
-    |\dpn{x_n, y_{N, k_N}}{\lambda} - \dpn{x_n, y}{\lambda}| \le \frac{1}{N}
-    \] 
-
-    Thus for each $N \in \natp$, $\dpn{x_n, y_{N, k_N}}{\lambda} \to \dpn{x_n, y}{\lambda}$ as $N \to \infty$. Since $y_{N, k_N} \to y$ pointwise on a dense subset of $E$ and $\bracsn{y_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $y_{N, k_N} \to y$ in the $\sigma(F, E)$-topology by \autoref{proposition:strong-operator-dense}. 
-
-    (2): Let $\seq{x_n} \subset E$ be a dense subset, then by \autoref{proposition:strong-operator-dense}, the $\sigma(F, E)$-topology on $S$ is induced by $\seq{x_n}$, and hence metrisable by \autoref{theorem:uniform-metrisable}. 
+    (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}

src/fa/notation.tex

@@ -29,6 +29,7 @@
     $E \,\wh{\otimes}_\pi F$ & Projective completion of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
     $p \otimes q$ & Cross seminorm of $p$ and $q$. & \autoref{definition:cross-seminorm} \\
     $N(E; F)$ & Nuclear mappings from $E$ to $F$. & \autoref{definition:nuclear-operator-normed} \\ 
+    $I(E, F)$ & Integral bilinear forms on $E \times F$. & \autoref{definition:integral-bilinear-form} \\ 
     % ---- Order Structures ----
     $x \vee y$, $x \wedge y$ & $\sup$ and $\inf$ in vector lattice. & \autoref{definition:vector-lattice} \\
     $|x|$ & Absolute value $x \vee (-x)$ in a vector lattice. & \autoref{definition:order-absolute-value} \\

src/fa/tvs/bilinear.tex

@@ -94,13 +94,13 @@
 
     Now, let $S \in \sigma$, then $\bracsn{\lambda(x, \cdot)|x \in S}$ is equicontinuous by the $\sigma$-hypocontinuity of $\lambda$. For any $U \in \cn_0(G)$, there exists $V \in \cn_0(F)$ such that $\bigcup_{x \in S}\lambda(x, V \cap F_0) \subset U$. By \autoref{proposition:closure-of-image}, $\bigcup_{x \in S}\Lambda(x, \ol V) \subset \ol U$. Thus \autoref{proposition:tvs-good-neighbourhood-base} implies that $\bracsn{\Lambda(x, \cdot)|x \in S}$ is equicontinuous as well. 
     
-    For each $x_0 \in \ol S$ and $y_0 \in F$, there exists $T \in \tau$ with $y_0 \in \ol T$. As $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$, $\Lambda(x_0, \cdot)$ is a pointwise limit of elements of $\bracsn{\lambda(x, \cdot)|x \in S}$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, 
-    \begin{enumerate}
-        \item $\bracsn{\Lambda(x, \cdot)|x \in \ol S} \subset \ol{\bracsn{\Lambda(x, \cdot)|x \in S}}^{L_s(F; G)}$. 
-        \item $\bracsn{\Lambda(x, \cdot)|x \in \ol S}$ is also equicontinuous. 
-    \end{enumerate}
-
-    so $\Lambda$ is $\ol \sigma$-hypocontinuous. Therefore $\Lambda$ is $(\ol \sigma, \ol \tau)$-hypocontinuous by symmetry. 
+    For each $x_0 \in \ol S$ and $y_0 \in F$, there exists $T \in \tau$ with $y_0 \in \ol T$. As $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$, $\Lambda(x_0, \cdot)$ is a pointwise limit of elements of $\bracsn{\lambda(x, \cdot)|x \in S}$. Thus
+    \[
+    \bracsn{\Lambda(x, \cdot)|x \in \ol S} \subset \ol{\bracsn{\Lambda(x, \cdot)|x \in S}}^{L_s(F; G)}
+    \]
+    
+    
+    By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\bracsn{\Lambda(x, \cdot)|x \in \ol S}$ is also equicontinuous, so $\Lambda$ is $\ol \sigma$-hypocontinuous. Therefore $\Lambda$ is $(\ol \sigma, \ol \tau)$-hypocontinuous by symmetry. 
 \end{proof}
 
 

src/fa/tvs/vector-function.tex

@@ -27,7 +27,7 @@
 
     (T) $\Rightarrow$ (B): Let $f \in \cf$, $S \in \sigma$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\sigma$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
 
-    (T) $\Rightarrow$ (B): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
+    (B) $\Rightarrow$ (T): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
     \begin{align*}
     \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
     &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))

src/measure/bochner-integral/strongly.tex

@@ -1,6 +1,7 @@
 \section{Strongly Measurable Functions}
 \label{section:strongly-measurable}
 
+
 \begin{definition}[Strongly Measurable Function]
 \label{definition:strongly-measurable}
     Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $f: X \to E$, then the following are equivalent:

src/measure/measurable-maps/local-in-measure.tex

@@ -37,7 +37,7 @@
 \label{proposition:convergence-in-measure}
     Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $(Y, d)$ be a separable metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is Cauchy in measure if and only if:
     \begin{enumerate}
-        \item[(L)] $\fF$ is \hyperref[definition:locally-in-measure]{definition:locally-in-measure}. 
+        \item[(L)] $\fF$ is Cauchy \hyperref[locally in measure]{definition:locally-in-measure}. 
         \item[(T)] For each $\eps, \delta > 0$, there exists $F \in \fF$ and $A \in \cf$ such that
         \[
         \sup_{f, g \in F}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps

src/measure/measure/localisable.tex

@@ -198,43 +198,3 @@
         \item[(U)] For all $A \in \cf$, $f|_A = g|_A$ almost everywhere. Since $\cf$ is a scaffold for $\mu$, $f = g$ almost everywhere. 
     \end{enumerate}
 \end{proof}
-
-\begin{corollary}
-\label{corollary:l-infty-dedekind-complete}
-    Let $(X, \cm, \mu)$ be a localisable measure space, then $L^\infty(X; \real)$ is order complete. 
-\end{corollary}
-\begin{proof}
-    Let $\seqi{f} \subset L^\infty(X; \real)$ and $M \in \real$ such that $f_i \le M$ almost everywhere for all $i \in I$. 
-
-    Fix $A \in \cm$ with $\mu(A) < \infty$, and let
-    \[
-    \mathcal{S}_A = \bracs{g \in L^\infty(A; \real)| f_i|_A \le g \text{ almost everywhere }\forall i \in I}
-    \]
-
-    then since $f_i \le M$ almost everywhere for all $i \in I$, $\mathcal{S}_A \ne \emptyset$, and $m_A = \inf_{g \in \mathcal{S}_A}\int g d\mu \in \real$.
-    
-    Let $\seq{g_{A, n}} \subset \mathcal{S}_A$ such that $\seq{g_{A, n}}$ is decreasing pointwise and $\limv{n}\int_A g_{A, n} d\mu \downto m_A$. Take $g_A = \limv{n}g_{A, n}$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $\int g_A d\mu = m_A$. 
-
-    For each $i \in I$, since $g_{A, n} \ge f_i|_A$ almost everywhere for all $n \in \natp$, $g_A \ge f_i|_A$ almost everywhere as well. Thus $g_A \in \mathcal{S}_A$. For any $h \in \mathcal{S}_A$, $g_A \wedge h \in \mathcal{S}_A$ with
-    \[
-    m_A \le \int_A g_A \wedge h d\mu \le \int_A g_A d\mu = m_A 
-    \]
-
-    Thus $g_A \wedge h = g_A$ almost everywhere, so $g_A \le h$ almost everywhere, and $g_A$ is an essential supremum of $\bracsn{f_i|_A}_{i \in I}$. 
-
-    Now, let $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$, then $\one_{A \cap B} g_B + \one_{A \setminus B}M \in \mathcal{S}_A$, and
-    \[
-    m_A \le \int_A g_A \wedge (\one_{A \cap B} g_B + \one_{A \setminus B}M)d\mu \le \int_A g_A d\mu = m_A
-    \]
-
-    Thus $g_A \wedge (\one_{A \cap B} g_B + \one_{A \setminus B}M) = g_A$ almost everywhere, so $g_A|_{A \cap B} \le g_B|_{A \cap B}$ almost everywhere. As the argument is symmetric, $g_A|_{A \cap B} = g_B|_{A \cap B}$ almost everywhere. 
-
-    By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists a measurable function $g:X \to \real$ such that $g|_A = g_A$ for all $A \in \cm$ with $\mu(A) < \infty$. 
-
-    Let $h \in L^\infty(X; \real)$ with $h \ge f_i$ almost everywhere for all $i \in I$, then for any $A \in \cm$ with $\mu(A) < \infty$, 
-    \[
-    \mu(\bracs{h < g} \cap A) \le \mu(\bracs{h|_A < g_A} \cup \bracs{g|_A \ne g_A}) = 0
-    \]
-    
-    As $\mu$ is semifinite, $\mu(\bracs{h < g}) = 0$. Finally, since $g_A \le M$ almost everywhere for all $A \in \cm$ with $\mu(A) < \infty$, $g \le M$ almost everywhere. Therefore $g \in L^\infty(X; \real)$ is indeed the essential supremum of $\seqi{f}$. 
-\end{proof}

src/measure/vector/fin.tex

@@ -117,6 +117,7 @@ Despite not covering the full dual space, the bounded Borel functions still form
     (2) $\Rightarrow$ (1): By the \hyperref[Dominated Convergence Theorem]{theorem:dct-bochner-vector}. 
 \end{proof}
 
+
 \begin{proposition}
 \label{proposition:space-of-measures-extreme-points}
     Let $X$ be an LCH space and $\cm \subset \overline{B_{M_R(X; \complex)}(0, 1)}$ be a compact convex set such that:

src/op/banach/definitions.tex

@@ -18,6 +18,17 @@
     For each $x \in A$, let $L_x \in L(A; A)$ be defined by $y \mapsto xy$, and let $\norm{x}_1 = \norm{L_x}_{L(A; A)}$, then $\norm{x}_1 \le \norm{x}_A$ and $\norm{1}_1 = 1$. On the other hand, $\frac{\norm{x}_A}{\norm{1}_A} \le \norm{x}_1$, so $\norm{\cdot}_1$ is equivalent to $\norm{\cdot}_A$.
 \end{proof}
 
+\begin{definition}[Centre]
+\label{definition:banach-algebra-centre}
+    Let $A$ be a Banach algebra, then
+    \[
+    Z(A) = \bracsn{x \in A|xy = yx \forall y \in A}
+    \]
+
+    is the \textbf{centre} of $A$. 
+\end{definition}
+
+
 \begin{definition}[Homomorphism]
 \label{definition:banach-algebra-homomorphism}
     Let $A, B$ be Banach algebras and $\phi: A \to B$, then $\phi$ is a \textbf{homomorphism} if:

src/op/banach/invertible.tex

@@ -9,7 +9,7 @@
 
 \begin{lemma}[Neumann Series]
 \label{lemma:neumann-series}
-    Let $A$ be a unital banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
+    Let $A$ be a unital Banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
     \[
     x^{-1} = \sum_{n = 0}^\infty (1 - x)^n
     \]

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/c-star/gns.tex

@@ -127,7 +127,7 @@
 
         is a representation of $A$, which is injective if for every $x \in A$, there exists $\phi \in \mathcal{S}$ with $\dpn{x^*x, \phi}{A} \ne 0$. 
 
-        In particular, $A$ is isomorphic to a closed subalgebra of $B([l^2(P(A)); H_\phi])$. 
+        In particular, $A$ is isomorphic to a closed subalgebra of $B([l^2(PS(A)); H_\phi])$. 
     \end{enumerate}
 \end{theorem}
 \begin{proof}
@@ -168,7 +168,7 @@
 
     so $\pi_\phi(x) \ne 0$, and $\pi_{\mathcal{S}}(x) \ne 0$ as well. 
 
-    By \autoref{corollary:cstar-positive-weakstar-dense}, for each $x \in A$, there exists $\phi \in P(A)$ with $\dpn{x^*x, \phi}{A} \ne 0$, so $\pi_{P(A)}$ is injective. By \autoref{theorem:continuity-of-homomorphism-c-star}, $\pi_{P(A)}(A)$ is closed in $B([l^2(P(A)); H_\phi])$. 
+    By \autoref{corollary:cstar-positive-weakstar-dense}, for each $x \in A$, there exists $\phi \in PS(A)$ with $\dpn{x^*x, \phi}{A} \ne 0$, so $\pi_{PS(A)}$ is injective. By \autoref{theorem:continuity-of-homomorphism-c-star}, $\pi_{PS(A)}(A)$ is closed in $B([l^2(PS(A)); H_\phi])$. 
     
 \end{proof}
 

src/op/c-star/state.tex

@@ -31,12 +31,12 @@
 
 \begin{definition}[Pure State]
 \label{definition:pure-state}
-    Let $A$ be a unital $C^*$-algebra and $\phi \in S(A)$, then $\phi$ is a \textbf{pure state} if $\phi$ is an extreme point of $S(A)$. The set $P(A)$ is the collection of all pure states of $A$. 
+    Let $A$ be a unital $C^*$-algebra and $\phi \in S(A)$, then $\phi$ is a \textbf{pure state} if $\phi$ is an extreme point of $S(A)$. The set $PS(A)$ is the collection of all pure states of $A$. 
 \end{definition}
 
 \begin{proposition}
 \label{proposition:state-space-compact-convex}
-    Let $A$ be a unital $C^*$-algebra, then $S(A)$ is a compact convex set, and $S(A)$ is the weak*-closed convex hull of $P(A)$. 
+    Let $A$ be a unital $C^*$-algebra, then $S(A)$ is a compact convex set, and $S(A)$ is the weak*-closed convex hull of $PS(A)$. 
 \end{proposition}
 \begin{proof}
     Since the evaluation map is weak* continuous and
@@ -48,15 +48,15 @@
 
     By \autoref{theorem:cstar-positive-algebraic}, $S(A) \subset \ol{B_{A^*}(0, 1)}$, which is weak* compact by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Therefore $S(A)$ is compact by \autoref{proposition:compact-extensions}. 
 
-    By the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, $S(A)$ is the weak*-closed convex hull of $P(A)$. 
+    By the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, $S(A)$ is the weak*-closed convex hull of $PS(A)$. 
 \end{proof}
 
 \begin{proposition}
 \label{proposition:multiplicative-pure-state}
     Let $A$ be a unital $C^*$-algebra, then:
     \begin{enumerate}
-        \item $\Omega(A) \subset P(A)$. 
-        \item If $A$ is commutative, then $\Omega(A) = P(A)$. 
+        \item $\Omega(A) \subset PS(A)$. 
+        \item If $A$ is commutative, then $\Omega(A) = PS(A)$. 
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -80,7 +80,7 @@
     Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a $C^*$-subalgebra with $1_A \in B$, and $\phi \in S(B)$, then
     \begin{enumerate}
         \item There exists $\Phi \in S(A)$ such that $\Phi|_B = \phi$.
-        \item If $\phi \in P(B)$, then there exists $\Phi \in P(A)$ such that $\Phi|_B = \phi$. 
+        \item If $\phi \in PS(B)$, then there exists $\Phi \in PS(A)$ such that $\Phi|_B = \phi$. 
     \end{enumerate}
 \end{theorem}
 \begin{proof}
@@ -88,7 +88,7 @@
 
     (2): Let $E(\phi) = \bracs{\Phi \in S(A)|\Phi|_B = \phi}$ be the collection of all extensions of $\phi$, then $E(\phi)$ is a weak*-closed convex subset of $S(A)$. By (1), $E(\phi)$ is non-empty, and as such admits an extreme point $\Phi$ by the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}. 
 
-    Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\Phi = (1 - t)\psi + t\rho$. In which case, $\phi = (1 - t)\psi|_B + t\rho|_B$. Since $\phi \in P(B)$, $\phi = \psi|_B = \rho|_B$, so $\psi, \rho \in E(\phi)$. As $\Phi$ is an extreme point of $E(\phi)$, $\Phi = \psi = \rho$. Therefore $\Phi \in P(A)$. 
+    Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\Phi = (1 - t)\psi + t\rho$. In which case, $\phi = (1 - t)\psi|_B + t\rho|_B$. Since $\phi \in PS(B)$, $\phi = \psi|_B = \rho|_B$, so $\psi, \rho \in E(\phi)$. As $\Phi$ is an extreme point of $E(\phi)$, $\Phi = \psi = \rho$. Therefore $\Phi \in PS(A)$. 
 \end{proof}
 
 
@@ -96,15 +96,15 @@
 \label{corollary:cstar-positive-property-probe}
     Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then\footnote{The crude bound seems kind of tragic, but it wouldn't be true otherwise. }
     \begin{align*}
-    \sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in P(A)} \\
+    \sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in PS(A)} \\
     &\subset \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} = \ol{\text{Conv}}(\sigma_A(x))
     \end{align*}
     
 
-    In particular, there exists $\phi \in P(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$. 
+    In particular, there exists $\phi \in PS(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$. 
 \end{corollary}
 \begin{proof}
-    Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in P(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$. 
+    Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in PS(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in PS(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in PS(A)}$. 
 
     Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_A(x)$. In which case,
     \[
@@ -113,7 +113,7 @@
 
     Finally, since $S(A)$ is compact and convex by \autoref{proposition:state-space-compact-convex}, 
     \begin{align*}
-        \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in P(A)}) \\
+        \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in PS(A)}) \\
         &\subset \ol{\text{Conv}}(\sigma_A(x))
     \end{align*}
 
@@ -139,35 +139,35 @@
 \label{corollary:cstar-positive-weakstar-dense}
     Let $A$ be a unital $C^*$-algebra, then:
     \begin{enumerate}
-        \item For each $x \in A$, $x = 0$ if and only if $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. 
-        \item The linear span of $P(A)$ is weak*-dense in $A^*$. 
+        \item For each $x \in A$, $x = 0$ if and only if $\dpn{x, \phi}{A} = 0$ for all $\phi \in PS(A)$. 
+        \item The linear span of $PS(A)$ is weak*-dense in $A^*$. 
     \end{enumerate}
 
     Moreover, for any $x \in A$, 
     \begin{enumerate}[start=2]
-        \item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in P(A)$. 
-        \item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$. 
+        \item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in PS(A)$. 
+        \item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in PS(A)$. 
     \end{enumerate}
     
 \end{corollary}
 \begin{proof}[Proof, {{\cite[Theorem 13.9]{Zhu}}}. ]
-    (1): Let $x \in A$ such that $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. First suppose that $x$ is self-adjoint. By \autoref{theorem:cstar-state-existence}, $\sigma_A(x) = \bracs{0}$, and $\norm{x}_A = [x]_{sp} = 0$ by \autoref{theorem:c-star-normal-spectral-radius}. 
+    (1): Let $x \in A$ such that $\dpn{x, \phi}{A} = 0$ for all $\phi \in PS(A)$. First suppose that $x$ is self-adjoint. By \autoref{theorem:cstar-state-existence}, $\sigma_A(x) = \bracs{0}$, and $\norm{x}_A = [x]_{sp} = 0$ by \autoref{theorem:c-star-normal-spectral-radius}. 
 
-    Now suppose that $x$ is arbitrary. In this case, for each $\phi \in P(A)$, 
+    Now suppose that $x$ is arbitrary. In this case, for each $\phi \in PS(A)$, 
     \[
     0 = \text{Re}(\dpn{x, \phi}{A}) = \dpn{\text{Re}(x), \phi}{A}
     \]
 
     because $\phi$ is Hermitian. Similarly, $\dpn{\text{Im}(x), \phi}{A} = 0$ as well. Thus $\text{Re}(x) = \text{Im}(x) = 0$, and $x = 0$ as well. 
 
-    (2): Since the linear span of $P(A)$ separates points in $A$, it is weak*-dense in $A^*$ by \autoref{lemma:duality-dense}. 
+    (2): Since the linear span of $PS(A)$ separates points in $A$, it is weak*-dense in $A^*$ by \autoref{lemma:duality-dense}. 
 
-    (3): Let $\phi \in P(A)$, then $\phi$ is Hermitian. If $x$ is self-adjoint, then $\dpn{x, \phi}{A} \in \real$. 
+    (3): Let $\phi \in PS(A)$, then $\phi$ is Hermitian. If $x$ is self-adjoint, then $\dpn{x, \phi}{A} \in \real$. 
 
-    On the other hand, if $\dpn{x, \phi}{A} \in \real$, then $\dpn{x, \phi}{A} = \dpn{x^*, \phi}{A}$, and $\dpn{x - x^*, \phi}{A} =0 $. If this holds for all $\phi \in P(A)$, then $x - x^* = 0$ by (1), and $x$ is self-adjoint. 
+    On the other hand, if $\dpn{x, \phi}{A} \in \real$, then $\dpn{x, \phi}{A} = \dpn{x^*, \phi}{A}$, and $\dpn{x - x^*, \phi}{A} =0 $. If this holds for all $\phi \in PS(A)$, then $x - x^* = 0$ by (1), and $x$ is self-adjoint. 
 
-    (4): Let $\phi \in P(A)$, then $\phi$ is positive. Thus if $x$ is positive, $\dpn{x, \phi}{A} \ge 0$. 
+    (4): Let $\phi \in PS(A)$, then $\phi$ is positive. Thus if $x$ is positive, $\dpn{x, \phi}{A} \ge 0$. 
 
-    On the other hand, if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$, then $x$ is self-adjoint by (3). By \autoref{corollary:cstar-positive-property-probe}, $\sigma_A(x) \subset [0, \infty)$. As such, $x$ is positive by \autoref{corollary:spectrum-characterisation-iff}. 
+    On the other hand, if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in PS(A)$, then $x$ is self-adjoint by (3). By \autoref{corollary:cstar-positive-property-probe}, $\sigma_A(x) \subset [0, \infty)$. As such, $x$ is positive by \autoref{corollary:spectrum-characterisation-iff}. 
 \end{proof}
 

src/op/example/bb.tex

@@ -0,0 +1,153 @@
+\section{$L^\infty$}
+\label{section:l-infty-algebra}
+
+
+\begin{proposition}
+\label{proposition:measures-dual-algebra}
+    Let $X$ be a compact Hausdorff space and $\mathscr{M} \subset M_R(X; \complex)$ be a closed subspace such that:
+    \begin{enumerate}
+        \item[(A)] For each $\mu \in \mathscr{M}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{M}$.
+    \end{enumerate}
+
+    and
+    \[
+    J: C(X; \complex) \to \mathscr{M}^* \quad \dpn{\mu, J(f)}{\mathscr{M}} = \int f d\mu
+    \]
+
+    then
+    \begin{enumerate}
+        \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
+        \[
+            \dpn{\mu, \phi^*}{\mathscr{M}^*} = \ol{\dpn{\mu, \phi}{\mathscr{M}^*}}
+        \]
+        \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}
+\begin{proof}
+    Let $\seqi{\mu}$ be a maximal mutually singular family of Radon measures on $X$. Using \autoref{theorem:hilbert-measures-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, identify
+    \[
+    \mathscr{M} = [l^1(I); L^1(\mu_i; \complex)] \quad \mathscr{M}^* = [l^\infty(I); L^\infty(\mu_i; \complex)]
+    \]
+
+    (1): Under the above, $C(X; \complex)$ may be identified as the diagonal
+    \[
+    \bracsn{f \in C(X; \complex)^I|f_i = f_j \forall i, j \in I} \subset [l^\infty(I); L^\infty(\mu_i; \complex)]
+    \]
+
+    By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$. As a result, $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^*$. 
+
+    (2): For each $g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$, let $g^* = \ol g$. For any $\mu \in [l^1(I); L^1(\mu_i; \complex)]$, 
+    \[
+    \dpn{\mu, g^*}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{\mu, \ol g}{[l^1(I); L^1(\mu_i; \complex)]} = \ol{\dpn{\mu, g}{[l^1(I); L^1(\mu_i; \complex)]}}
+    \]
+
+    so the conjugation map is weak*-continuous. 
+
+    (3): Let $f, g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$ and $\mu \in [l^1(I); L^1(\mu_i; \complex)]$, 
+    \[
+        \dpn{\mu, fg}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{f\mu, g}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{g\mu, f}{[l^1(I); L^1(\mu_i; \complex)]}
+    \]
+
+    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}
+
+\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}
+
+

src/op/example/bounded.tex

@@ -18,7 +18,7 @@
 
 \begin{proposition}
 \label{proposition:partial-isometry-characterisation}
-    Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^*T$ is a projection. 
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^*T$ is a projection. In which case, $T^*T$ is a projection onto $\ker(T)^\perp$.  
 \end{proposition}
 \begin{proof}
     ($\Rightarrow$): Suppose that $T$ is a partial isometry. Let $x \in \ker(T)^\perp$, then $\dpn{Tx, Tx}{H} = \norm{x}_H^2$ and $\dpn{T^*Tx, x}{H} = \norm{x}_H^2$. By \hyperref[polarisation]{proposition:polarisation-complex}, for each $x, y \in \ker(T)^\perp$, 
@@ -27,10 +27,18 @@
         &= \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{x + i^ky, x + i^ky}{H} = \dpn{x, y}{H} 
     \end{align*}
 
-    Therefore $T^*T$ is idempotent. As $T^*T$ is self-adjoint, it is a projection. 
+    Therefore $T^*T$ is idempotent. As $T^*T$ is self-adjoint, it is a projection onto $\ker(T)^\perp$. 
 
     ($\Leftarrow$): Suppose that $T^*T$ is a projection, then for each $x \in \ker(T)^\perp$, $\dpn{Tx, Tx}{H} = \dpn{T^*Tx, x}{H} = \norm{x}_H^2$. 
 \end{proof}
+\begin{corollary}
+\label{corollary:partial-isometry-adjoint}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^*$ is a partial isometry. 
+\end{corollary}
+\begin{proof}[Proof, {{\cite[Corollary 12.7]{Zhu}}}. ]
+    Suppose that $T$ is a partial isometry, then $P = T^*T$ is a projection onto $\ker(T)^\perp$ by \autoref{proposition:partial-isometry-characterisation}. In which case, $T(T^*T) = T$ and $(TT^*)^2 = T(T^*T)T^* = TT^*$, so $TT^*$ is a projection, and $T^*$ is a partial isometry by \autoref{proposition:partial-isometry-characterisation}. 
+\end{proof}
+
 
 \begin{theorem}[Polar Decomposition]
 \label{theorem:hilbert-polar-decomposition}
@@ -42,21 +50,34 @@
         \item $\ker P = \ker V$.
     \end{enumerate}
 
-    The pair $(P, V)$ is the \textbf{polar decomposition} of $T$. 
+    The pair $(P, V)$ is the \textbf{polar decomposition} of $T$, and
+    \begin{enumerate}[start=4]
+        \item $V$ is a partial isometry from $\ker(T)^\perp$ to $\ol{T(H)}$.
+        \item $P$ and $V$ are contained in the von Neumann algebra generated by $T$. 
+    \end{enumerate}
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem 12.8]{Zhu}}}. ]
-    Let $P = |T| = \sqrt{T^*T}$, then $P$ is positive (1). For each $x \in H$, 
+\begin{proof}[Proof, {{\cite[Theorem 12.8, Theorem 18.9]{Zhu}}}. ]
+    (1): Let $P = |T| = \sqrt{T^*T}$, then $P$ is positive (1). 
+    
+    (2): For each $x \in H$, 
     \[
         \norm{Px}_H^2 = \dpn{Px, Px}{H} = \dpn{P^*Px, x}{H} = \dpn{T^*Tx, x}{H} = \norm{Tx}_H^2
     \]
 
-    Let $V_0: P(H) \to H$ be defined by $V(Px) = Tx$, then $V_0$ extends to a well-defined isometry $\ol{P(H)} \to H$. Further extend $V_0$ to $V$ by setting its value to $0$ on $P(H)^\perp$, then $V$ is a partial isometry (2). Moreover, for any $x \in H$, $Tx = V_0Px = VPx$ (3).
-
-    Finally, since the initial space of $V$ is $\ol{P(H)}$, $\ker(V) = P(H)^\perp = \ker(P)$ (4). 
+    Let $V_0: P(H) \to H$ be defined by $V(Px) = Tx$, then $V_0$ extends to a well-defined isometry $\ol{P(H)} \to H$. Further extend $V_0$ to $V$ by setting its value to $0$ on $P(H)^\perp$, then $V$ is a partial isometry. 
     
-    It remains to show uniqueness. Let $T = WQ$ be a polar decomposition of $T$ satisfying (1)-(4). By \autoref{proposition:partial-isometry-characterisation}, $W^*W$ is a projection onto $\ker(W)^\perp = \ker(Q)^\perp = \ol{Q(H)}$. Thus $P^2 = T^*T = QW^*WQ = Q^2$, and $P = Q$ by uniqueness of the positive square root. 
+    (3): For any $x \in H$, $Tx = V_0Px = VPx$. 
+    
+    
+    (4), (5): Since the initial space of $V$ is $\ol{P(H)} = \ker(T)^\perp$, $\ker(V) = P(H)^\perp = \ker(P)$. 
+    
+    (Uniqueness): Let $T = WQ$ be a polar decomposition of $T$ satisfying (1)-(4). By \autoref{proposition:partial-isometry-characterisation}, $W^*W$ is a projection onto $\ker(W)^\perp = \ker(Q)^\perp = \ol{Q(H)}$. Thus $P^2 = T^*T = QW^*WQ = Q^2$, and $P = Q$ by uniqueness of the positive square root. 
 
-    Now, since $VP = WP$ and $\ker(V) = \ker(W) = P(H)^\perp$, $V = W$ on $H$, and the polar decomposition is unique. 
+    Since $VP = WP$ and $\ker(V) = \ker(W) = P(H)^\perp$, $V = W$ on $H$, and the polar decomposition is unique. 
+
+    (6): Let $A$ be the von Neumann algebra generated by $T$. By \autoref{theorem:existence-of-projections-vna}, $A$ is a unital $C^*$-algebra, so $P = \sqrt{T^*T} \in A$. To see that $V \in A$, it is sufficient to apply the \hyperref[Bicommutant Theorem]{theorem:bicommutant}. 
+
+    To this end, let $S \in A'$, then $TS = ST = SVP$ and $VSP = VPS = TS$, so $SV$ and $VS$ agree on $\ol{P(H)}$. Since $\ker(V) = \ker(P) = \ol{P(H)}^\perp$, $SV|_{\ker(P)} = 0$. On the other hand, as $SP = PS$, $S(\ker(P)) \subset \ker(P) = \ker(V)$, so $VS|_{\ker(P)} = 0$ as well. Therefore $V \in A'' = A$. 
 \end{proof}
 
 

src/op/example/index.tex

@@ -8,3 +8,4 @@
 \input{./disk.tex}
 \input{./convolution.tex}
 \input{./bc.tex}
+\input{./bb.tex}

src/op/notation.tex

@@ -5,6 +5,7 @@
     \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
     \hline
     $1$ & Identity element of a unital algebra. & \autoref{definition:unital-banach-algebra} \\
+    $Z(A)$ & Centre of a Banach algebra. & \autoref{definition:banach-algebra-centre} \\ 
     $G(A)$ & Invertible group of a unital algebra. & \autoref{definition:banach-algebra-invertible} \\
     $G_0(A)$ & The identity component of $G(A)$. & \autoref{definition:identity-component} \\
     $I(A)$ & The index group of $A$. & \autoref{definition:index-group} \\
@@ -16,9 +17,15 @@
     $\Gamma = \Gamma_A$ & The Gelfand transform on $A$. & \autoref{definition:gelfand-transform} \\ 
     $A[S]$ & $C^*$-subalgebra of $A$ generated by $S \subset A$. & \autoref{definition:generated-subalgebra} \\ 
     $S(A)$ & State space of a $C^*$-algebra $A$. & \autoref{definition:cstar-state} \\ 
-    $P(A)$ & Pure state space of a $C^*$-algebra $A$. & \autoref{definition:pure-state} \\ 
+    $PS(A)$ & Pure state space of a $C^*$-algebra $A$. & \autoref{definition:pure-state} \\ 
     $\dpn{x, y}{\phi}$ & Defined as $\dpn{y^*x, \phi}{A}$, the pseudo inner product associated to a positive linear functional. & \autoref{definition:cstar-state-pseudo-inner-product} \\ 
     $(H_\phi, \pi_\phi, \xi_\phi)$ & GNS triple associated with $\phi \in S(A)$. & \autoref{definition:gns-triple} \\ 
+    $U(T)$ & Cayley transform of $T$. & \autoref{definition:cayley-transform-bounded} \\ 
+    $E_{x, y}$ & $E_{x, y}(B) = \dpn{E(B)x, y}{H}$. & \autoref{definition:spectral-measure} \\ 
+    $\text{Proj}(A)$ & Projections in $A$. & \autoref{definition:vn-projection-lattice} \\
+    $Z(P)$ & Central support of $P \in \text{Proj}(A)$. & \autoref{definition:central-support-vna} \\ 
+    $P \sim Q$ & $P, Q \in \text{Proj}(A)$ are Murray-von Neumann equivalent & \autoref{definition:murray-von-neumann-equivalent} \\ 
+    $P \preceq Q$ & $P$ is Murray-von Neumann subequivalent to $Q$. & \autoref{definition:murray-von-neumann-subequivalent} \\ 
 
     $M_n(\complex)$ & Algebra of $n \times n$ matrices over $\complex$.  & \autoref{definition:matrix-algebra} \\
     $B(H)$ & Algebra of bounded operators on a Hilbert space. & \autoref{definition:hilbert-endomorphism} \\

src/op/vn/cayley.tex

@@ -0,0 +1,53 @@
+\section{The Cayley Transform}
+\label{section:cayley-transform}
+
+\begin{definition}[Cayley Transform]
+\label{definition:cayley-transform-bounded}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $-i \not\in \sigma_A(H)$, then $U(T) = (T - i)(T + i)^{-1}$ is the \textbf{Cayley transform} of $T$. 
+\end{definition}
+
+\begin{theorem}
+\label{theorem:cayley-sa-uni}
+    Let $H$ be a complex Hilbert space and $U_1$ be the set of unitary operators on $H$ with $1$ not in their spectrum, then the Cayley transform $T \mapsto (T - i)(T + i)^{-1}$ is a strong-operator continuous bijection between $B(H)_{sa}$ and $U_1$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 19.3]{Zhu}}}. ]
+    Let $T \in B(H)$ be self-adjoint. By \autoref{proposition:self-adjoint-spectrum}, $\sigma_{B(H)}(T) \subset \real$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, 
+    \[
+    \sigma_{B(H)}[(T-i)(T+i)^{-1}] \subset \bracsn{(t - i)/(t + i)|t \in \real} \subset \partial B_\complex(0, 1) \setminus \bracsn{1}
+    \]
+    
+    Hence $(T - i)(T+i)^{-1}$ is a well-defined unitary element of $B(H)$ whose spectrum does not contain $1$. 
+
+    Since the mapping $t \mapsto -i(t + 1)/(t - 1)$ is the inverse of $t \mapsto (t - i)/(t + i)$ on $\partial B_\complex(0, 1)$, the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous} implies that $T \mapsto -i(T + I)(T - I)^{-1}$ is the inverse of the Cayley transform on $U_1$. 
+
+    For any self-adjoint elements $S, T \in B(H)$,
+    \begin{align*}
+        U(S) - U(T) &= (S + i)^{-1}(S - i) - (T - i)(T+i)^{-1} \\
+        &= (S + i)^{-1}[(S - i)(T + i)  - (S + i)(T - i)](T + i)^{-1} \\
+        &= 2i(S + i)^{-1}(S - T)(T + i)^{-1}
+    \end{align*}
+
+    so for any $x \in H$, 
+    \[
+    \normn{[U(S) - U(T)]x}_H \le 2\normn{(S + i)^{-1}}_{B(H)} \cdot \normn{(S - T)(T+i)^{-1}x}_H
+    \]
+
+    By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\normn{(S+i)^{-1}}_{B(H)} \le 1$. Therefore the Cayley transform is strong-operator continuous. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:functional-calculus-c0-self-adjoint}
+    Let $H$ be a complex Hilbert space and $f \in C_0(\real; \complex)$, then the mapping $T \mapsto f(T)$ is strong-operator continuous on $B(H)_{sa}$. 
+\end{corollary}
+\begin{proof}
+    Let
+    \[
+    g: \partial B_\complex(0, 1) \to \complex \quad z \mapsto \begin{cases}
+        f(-i(z+1)/(z-1)) &z \ne 1 \\
+        0 &z = 1
+    \end{cases}
+    \]
+
+    then since $f \in C_0(\real; \complex)$, $g \in C(\partial B_\complex(0, 1); \complex)$. For each $T \in B(H)_{sa}$, $f(T) = g(U(T))$. By \autoref{proposition:bh-adjoint-strong-continuous}, the mapping $U \mapsto g(U)$ is strong-operator continuous on the set of unitary operators on $H$. By \autoref{theorem:cayley-sa-uni}, $T \mapsto f(T)$ is the composition of two strong-operator continuous mappings. 
+\end{proof}
+

src/op/vn/commutative.tex

@@ -0,0 +1,56 @@
+\section{Commutative von Neumann Algebras}
+\label{section:vn-commutative}
+
+\begin{definition}[Separating Vector]
+\label{definition:separating-vector}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$, and $x \in H$, then $x$ is a \textbf{separating vector} for $A$ if the mapping $A \to H$ defined by $T \mapsto Tx$ is injective.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:maximal-commutative-vn}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a maximal commutative von Neumann algebra if and only if $A = A'$.
+\end{proposition}
+\begin{proof}
+    ($\Leftarrow$): Let $B \supset A$ be a commutative von Neumann algebra, then $B \subset A' = A$. 
+
+    ($\Rightarrow$): For each $T \in (A')_{sa}$, the von Neumann algebra generated by $A$ and $T$ is commutative. As such, $T \in A$. As this holds for all $T \in (A')_{sa}$, $A' = (A')_{sa} + i(A')_{sa} \subset A$.  
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:cyclic-separating-commutant}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a $C^*$-subalgebra with $I \in A$, and $x \in H$, then $x$ is cyclic for $A$ if and only if $x$ is separating for $A'$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 22.1]{Zhu}}}. ]
+    ($\Rightarrow$): Let $T \in A'$ with $Tx = 0$, then $TSx = STx = 0$ for all $S \in A$. In which case, $T(H) \subset \ol{T(Ax)} = \bracs{0}$ by \autoref{proposition:closure-of-image}.
+
+    ($\Leftarrow$): Let $P \in B(H)$ be the orthogonal projection from $H$ onto $\ol{Ax}$, then as $I \in A$, $x \in \ol{Ax}$. Since $\ol{Ax}$ is a reducing subspace for $A$, $P \in A'$. Thus $I, P \in A'$ and $(I - P)x = 0$. Given that $x$ is separating for $A'$, $I = P$, so $\ol{Ax} = H$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:cyclic-is-separating-commutative}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, and $x \in H$ be a cyclic vector for $A$, then $x$ is also a separating vector for $A$. 
+\end{corollary}
+\begin{proof}
+    Since $A$ is commutative, $A \subset A'$. As $x$ is separating for $A'$ by \autoref{proposition:cyclic-separating-commutant}, it is also separating for $A$. 
+\end{proof}
+
+
+\begin{theorem}
+\label{theorem:commutative-has-separating}
+    Let $H$ be a separable Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then $A$ admits a separating vector. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 22.3]{Zhu}}}. ]
+    Let $\seqj{x} \subset H$ be a maximal collection of non-zero vectors such that the spaces $\bracsn{Ax_j|j \in I}$ are mutually orthogonal. Such a collection exists by Zorn's lemma, and must be at most countable given that $H$ is separable. 
+
+    Let $\seq{x_n}$ be an enumeration of such a set, padding by zeroes if necessary, and $x = \sum_{n \in \natp}x_n/2^n$. For any $T \in A$, if $Tx = 0$, then as $\bracsn{Ax_n|n \in \natp}$ are mutually orthogonal, $Tx_n = 0$ for all $n \in \natp$. Since $A$ is commutative, $Ax_n \subset \ker(T)$ for all $n \in \natp$. By maximality of $\seq{x_n}$, $H = [l^2(\natp); \ol{Ax_n}]$, $H \subset \ker(T)$, and $T = 0$. Therefore $x$ is a separating vector. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:maximal-abelian}
+    Let $H$ be a separable Hilbert space and $A \subset B(H)$ be a maximal abelian von Neumann algebra, then $A$ admits a cyclic vector. 
+\end{corollary}
+\begin{proof}[Proof, {{\cite[Corollary 22.4]{Zhu}}}. ]
+    By \autoref{proposition:maximal-commutative-vn}, $A = A'$. By \autoref{theorem:commutative-has-separating}, $A$ admits a separating vector. By \autoref{proposition:cyclic-separating-commutant}, this separating vector for $A$ is a cyclic vector for $A' = A$. 
+\end{proof}
+

src/op/vn/fc.tex

@@ -0,0 +1,80 @@
+\section{The $L^\infty$ Functional Calculus}
+\label{section:linfty-functional-calculus}
+
+
+
+\begin{definition}[$L^\infty$ Functional Calculus]
+\label{definition:linfty-functional-calculus}
+    Let $H$ be a complex Hilbert space, $T \in B(H)$ be normal, and $A \subset B(H)$ be the smallest von Neumann algebra acting on $H$ containing $T$ and $I$, then
+    \begin{enumerate}
+        \item There exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)} \to A$ such that
+        \[
+        T = \int_{\sigma_{B(H)}(T)}\lambda E(d\lambda) \quad T^* = \int_{\sigma_{B(H)}(T)}\ol \lambda E(d\lambda)
+        \]
+        \item Let $\mathscr{E} \subset M_R(\sigma_{B(H)}(T); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $A \subset B(H)$ be the von Neumann algebra generated by $I$ and $T$, then
+        \[
+        I_E: \mathscr{E}^* \to A \quad \phi \mapsto \phi(T) := \int_{\sigma_{B(H)}(T)}\phi dE
+        \]
+
+        is a *-isomorphism. 
+        \item $I_E: \mathscr{E}^* \to A$ is the unique weak* to weak-operator continuous unital *-homomorphism such that $I_E(\text{Id}) = T$. 
+    \end{enumerate}
+    
+    The spectral measure $E$ is the \textbf{resolution of the identity} for $T$, and the mapping $f \mapsto f(T)$ on $\mathscr{E}^*$ is the \textbf{$L^\infty$-functional calculus} of $T$. 
+\end{definition}
+\begin{proof}
+    (1), (2): By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1} applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping
+    \[
+    I_E: \mathscr{E}^{*} \to A \quad \phi \mapsto \int_{\sigma_{B(H)}(T)} \phi dE
+    \]
+
+    is a *-isomorphism that extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$. 
+    
+    For each $\phi \in \mathscr{E}^{*}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $A$ by \autoref{definition:spectral-measure-integral}.
+
+    (3): By uniqueness of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, and (2), the mapping $\phi \mapsto \phi(T)$ is unique. 
+\end{proof}
+
+
+\begin{theorem}
+\label{theorem:vn-projection-norm-dense}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then:
+    \begin{enumerate}
+        \item For each normal operator $T \in A$ and $f \in B^\infty(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$, $f(T) \in A$. 
+        \item The linear span of projections in $A$ is norm dense in $A$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $B \subset \sigma_{B(H)}(T) \setminus \bracs{0}$ be a Borel set. First suppose that $0 \not\in \ol{B}$. By \hyperref[Urysohn's Lemma]{lemma:urysohn}, there exists $f \in C(\sigma_{B(H)}(T); [0, 1])$ with $f(0) = 0$ and $f|_{\ol B} = 1$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, there exists a net $\angles{g_\gamma}_{\gamma \in C} \subset C(\sigma_{B(H)}(T); \complex)$ such that $g_\gamma \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$. As $f\one_B = \one_B$, $fg_\gamma \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as well. 
+
+    By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $h(T) \in A$ for all $h \in C(\sigma_{B(H)}(T); \complex)$ with $h(0) = 0$. In particular, $fg_\gamma(T) \in A$ for all $\gamma \in C$. Thus the \hyperref[$L^\infty$ functional calculus]{definition:linfty-functional-calculus} implies that $\one_B(T) \in A$ as well. 
+
+    If $B$ is arbitrary, then $\one_{B \setminus B_\complex(0, r)} \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as $r \downto 0$. As $\one_{B \setminus B_{\complex}(0, r)}(T) \in A$ for all $r > 0$, $\one_B(T) \in A$ as well. 
+
+    By linearity, $g(T) \in A$ for all $g \in \Sigma(\sigma_{B(H)}(T); \complex)$ with $g(0) = 0$. By \autoref{lemma:separable-metric-space-approx-identity}, $\bracsn{g \in \Sigma(\sigma_{B(H)}(T); \complex)|g(0) = 0}$ is uniformly dense in $\bracsn{f \in B^\infty(\sigma_{B(H)(T)}; \complex)|f(0) = 0}$. Therefore $f(T) \in A$ for all $f \in B^\infty(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:von-neumann-group-connected}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then:
+    \begin{enumerate}
+        \item $G(A)$ is path-connected in the norm topology.
+        \item The unitary group of $A$ is path-connected in the norm topology. 
+        \item $I(A)$ is trivial. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    Using \autoref{theorem:existence-of-projections-vna} and after possibly shrinking $H$, assume without loss of generality that $I \in A$. 
+
+    After choosing and extending a branch of the complex logarithm, let $\phi: \complex \to \complex$ be a Borel measurable function such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $e^{\phi(z)} = z$ for all $z \in \complex \setminus \bracs{0}$. 
+        \item For each $0 < r < R$, $\phi$ is bounded on the annulus $\ol{B(0, R)} \setminus B(0, r)$. 
+    \end{enumerate}
+
+    (1): Let $T \in G(A)$, then there exists $0 < r < R$ such that $\sigma_A(T) \subset \ol{B(0, R)} \setminus B(0, r)$. In which case, $\phi$ is a bounded Borel measurable function on $\sigma_A(T)$. By the \hyperref[Borel functional calculus]{definition:linfty-functional-calculus}, $\phi(T) \in A$ with $T = e^{\phi(T)}$. In which case, the path $t \mapsto e^{t\phi(T)}$ is a norm-continuous path in $G(A)$ from $I$ to $T$.
+    
+    (2): In particular, as $e^{t\phi}(\partial B(0, 1)) \subset \partial B(0, 1)$, the spectrum of $e^{t\phi}$ as an element in the domain of the $L^\infty$ functional calculus, is contained in $\partial B(0, 1)$. Thus if $T$ is unitary, then the path $t \mapsto e^{t\phi(T)}$ lies in the unitary group of $A$ by \autoref{corollary:spectrum-characterisation-iff}.
+\end{proof}
+
+

src/op/vn/index.tex

@@ -2,4 +2,10 @@
 \label{chap:von-neumann-algebras}
 
 \input{./topologies.tex}
-\input{./vn.tex}
+\input{./cayley.tex}
+\input{./vn.tex}
+\input{./commutative.tex}
+\input{./spec.tex}
+\input{./fc.tex}
+\input{./projection.tex}
+\input{./type-decomp.tex}

src/op/vn/projection.tex

@@ -0,0 +1,296 @@
+\section{The Projection Lattice}
+\label{section:vn-projection-lattice}
+
+\begin{definition}[Projection Lattice]
+\label{definition:vn-projection-lattice}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\text{Proj}(A)$ be the set of all projections in $A$, then:
+    \begin{enumerate}
+        \item For any $S \subset \text{Proj}(A)$, let $P$ be the orthogonal projection onto the closed subspace generated by ${\bigcup_{Q \in S}Q(H)}$, then $P = \sup(S) \in A$. 
+        \item For any $S \subset \text{Proj}(A)$, let $P$ be the orthogonal projection onto $\bigcap_{Q \in S}Q(H)$, then $P = \inf(S) \in A$. 
+        \item $\text{Proj}(A)$ is order complete. 
+    \end{enumerate}
+
+    The set $\text{Proj}(A)$ is the \textbf{projection lattice} of $A$. 
+\end{definition}
+\begin{proof}
+    (1): For each $T \in A'$ and $Q \in S$, $TQ = QT$, so $Q(H)$ is a reducing subspace for $T$. As this holds for all $Q \in S$, the closed subspace generated by $\bigcup_{Q \in S}Q(H)$ is a reducing subspace for $T$. Therefore $PT = TP$, and $P \in A$ by the \hyperref[Bicommutant Theorem]{theorem:bicommutant}. 
+
+    (2): For each $T \in A'$ and $Q \in S$, $TQ = QT$, so $Q(H)$ is a reducing subspace for $T$. As this holds for all $Q \in S$, $\bigcap_{Q \in S}Q(H)$ is a reducing subspace for $T$. Therefore $PT = TP$, and $P \in A$ by the \hyperref[Bicommutant Theorem]{theorem:bicommutant}. 
+\end{proof}
+
+\subsection{Central Support of Projections}
+\label{subsection:projection-central-support}
+
+
+
+\begin{definition}[Central Support]
+\label{definition:central-support-vna}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $Z(P) = \inf_{Q \in \text{Proj}(Z(A)), Q \ge P}Q$ is the \textbf{central support} of $P$. 
+\end{definition}
+
+\begin{definition}[Centrally Orthogonal]
+\label{definition:centrally-orthogonal}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\seqi{P} \subset \text{Proj}(A)$, then $\seqi{P}$ is \textbf{centrally orthogonal} if $\bracsn{Z(P_i)}_{i \in I}$ is mutually orthogonal. 
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:central-support-vna}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$. For each $T \in A$, let $R(TP)$ be the orthogonal projection onto $\ol{TP(H)}$, then
+    \[
+    Z(P) = \sup_{T \in A}R(TP)
+    \]
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 24.6]{Zhu}}}. ]
+    Since $Z(P) \in Z(A)$, $Z(P)(H)$ is a reducing subspace of every operator in $A$. As $P \le Z(P)$, $TP(H) \subset T(Z(P)(H)) \subset Z(P)(H)$, so $Z(P) \ge R(TP)$ for all $T \in A$, and $Z(P) \ge \sup_{T \in A}R(TP)$. 
+    
+    On the other hand, for each $S, T \in A$, $S(TP(H)) \subset \bigcup_{R \in A}RP(H)$. As $A$ is a von Neumann algebra, the range of $\sup_{T \in A}R(TP)$ is a reducing subspace for every operator in $A$. Therefore $\sup_{T \in A}R(TP) \in Z(A)$, and $Z(P) \le \sup_{T \in A}R(TP)$.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:central-support-mvn}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $Z(P)Z(Q) \ne 0$. 
+        \item $PAQ \ne \bracsn{0}$. 
+        \item There exists non-zero projections $P_0 \le P$ and $Q_0 \le Q$ such that $P_0 \sim Q_0$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 24.7]{Zhu}}}. ]
+    (1) $\Rightarrow$ (2): For each $T \in A$, let $R(T)$ be the orthogonal projection onto $\ol{T(H)}$. By \autoref{proposition:central-support-vna}, 
+    \[
+    Z(P) = \sup_{T \in A}R(TP) \quad Z(Q) = \sup_{T \in A}R(TQ)
+    \]
+
+    Given that $Z(P)Z(Q) \ne 0$, $Z(P)(H) \not\perp Z(Q)(H)$. Since $Z(P)(H) = \ol{\bigcup_{S \in A}SP(H)}$ and $Z(Q)(H) = \ol{\bigcup_{T \in A}TQ(H)}$, there exists $S, T \in A$ such that $SP(H) \not\perp TQ(H)$. As such, there exists $x, y \in H$ with
+    \[
+    0 \ne \dpn{TQx, SPy}{H} = \dpn{PS^*TQx, y}{H}
+    \]
+
+    so $PAQ \ne \bracsn{0}$. 
+
+    (2) $\Rightarrow$ (3): Let $T \in A$ with $PTQ \ne 0$. Let $P_0 = R(PTQ)$ and $Q_0 = R(QT^*P)$, then $0 \ne P_0 \le P$, $0 \ne Q_0 \le Q$, and $P_0 \sim Q_0$ by \autoref{lemma:mvn-equivalent-adjoint}. 
+
+    (3) $\Rightarrow$ (1): By \autoref{lemma:central-support-mvn-eq}, $Z(P_0) = Z(Q_0)$, so
+    \[
+    Z(P)Z(Q) = Z(P) \wedge Z(Q) \ge Z(P_0) \vee Z(Q_0) \ne 0
+    \] 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:central-support-mvn-eq}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$ with $P \sim Q$, then:
+    \begin{enumerate}
+        \item $Z(P) = Z(Q)$. 
+        \item For any central projection $R \in \text{Proj}(A)$, $PR \sim QR$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    Let $V \in A$ with $P = V^*V$ and $Q = VV^*$, then $V$ is a partial isometry with initial space $P(H)$ and final space $Q(H)$. 
+    
+    (1): Since $Z(P) \ge P$ and $Z(P) \in Z(A)$, 
+    \[
+    Z(P)Q = Z(P)VV^* = VZ(P)V^* = VV^* = Q
+    \]
+
+    and $Z(P) \ge Q$, and $Z(P) \ge Z(Q)$. By symmetry, $Z(Q) \ge Z(P)$, so $Z(P) = Z(Q)$. 
+
+    (2): Let $R$ be a central projection, then
+    \[
+    PR = V^*VR = V^*RV = (VR)^*(VR) \sim (VR)(VR)^* = VRV^* = VV^*R = QR
+    \]
+
+
+\end{proof}
+
+\subsection{Murray-von Neumann Equivalence}
+\label{subsection:mvn-equivalence}
+
+
+\begin{lemma}
+\label{lemma:projection-mental-gymnastics}
+    Let $H$ be a complex Hilbert space and $P, Q \in B(H)$ be projections, then:
+    \begin{enumerate}
+        \item $\ker(PQ) = \ker(Q) + \ker(P) \cap Q(H)$. 
+        \item If $PQ = QP$, then $PQ$ is a projection with $PQ(H) = P(H) \cap Q(H)$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $x \in \ker(PQ)$, then $Q(x) \in \ker(P)$, so $x = Q(x) + (1 - Q)(x) \in  \ker(Q) + \ker(P) \cap Q(H)$.
+    
+    (2): Since $PQ = QP$, $(PQ)^2 = P^2Q^2 = PQ$ and $(PQ)^* = Q^*P^* = QP = PQ$, $PQ$ is a projection. As $PQ(H) = P(Q(H)) \subset P(H)$ and $PQ(H) = Q(P(H)) \subset Q(H)$, $PQ(H) \subset P(H) \cap Q(H)$. On the other hand, $PQ$ is the identity on $P(H) \cap Q(H)$, so $PQ(H) = P(H) \cap Q(H)$. 
+\end{proof}
+
+
+
+
+
+\begin{definition}[Murray-von Neumann Equivalent]
+\label{definition:murray-von-neumann-equivalent}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists $V \in A$ such that $P = V^*V$ and $Q = VV^*$. 
+        \item There exists a partial isometry $V \in A$ from $P(H)$ to $Q(H)$. 
+    \end{enumerate}
+    
+    If the above holds, then $P$ and $Q$ are \textbf{Murray-von Neumann equivalent}, denoted $P \sim Q$. The relation $\sim$ is an equivalence relation on $\text{Proj}(A)$. 
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $V \in A$ with $P = V^*V$ and $Q = VV^*$. By \autoref{proposition:partial-isometry-characterisation}, $V$ is a partial isometry with initial space $\ker(P)^\perp$, and $V^*$ is a partial isometry with initial space $\ker(Q)^\perp$. Therefore $V$ is a partial isometry from $P(H)$ to $Q(H)$.  
+
+    (2) $\Rightarrow$ (1): By \autoref{proposition:partial-isometry-characterisation}, $P = V^*V$ is a projection onto $\ker(V)^\perp$, and $Q = VV^*$ is a projection onto $\ker(V^*)^\perp = V(H)$.
+\end{proof}
+
+\begin{definition}[Murray-von Neumann Subequivalent]
+\label{definition:murray-von-neumann-subequivalent}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then $P$ is \textbf{Murray-von Neumann subequivalent} to $Q$, denoted $P \preceq Q$, if there exists $R \in \text{Proj}(A)$ such that $P \sim R$ and $R \le Q$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:mvn-equivalent-direct-sum}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\seqi{P}, \seqi{Q} \subset \text{Proj}(A)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $\seqi{P}$ is mutually orthogonal.
+        \item $\seqi{Q}$ is mutually orthogonal. 
+        \item For each $i \in I$, $P_i \sim Q_i$. 
+    \end{enumerate}
+
+    then $\sum_{i \in I}P_i \sim \sum_{i \in I}Q_i$. 
+\end{lemma}
+\begin{proof}
+    For each $i \in I$, let $V_i \in A$ such that $P_i = V_i^*V_i$ and $Q_i = V_iV_i^*$, then $V_i$ is a partial isometry with initial space $P_i(H)$ and final space $Q_i(H)$. As $\seqi{P}$ is mutually orthogonal and $\seqi{Q}$ is mutually orthogonal, the sum $\sum_{i \in I}V_i$ converges in strong operator topology to an operator $V$, where
+    \[
+    V^*V = \sum_{i, j \in I}V_i^*V_j = \sum_{i \in I}V_i^*V_i = \sum_{i \in I}P_i
+    \]
+
+    and
+    \[
+    VV^* = \sum_{i, j \in I}V_iV_j^* = \sum_{i \in I}V_iV_i^* = \sum_{i \in I}Q_i
+    \]
+\end{proof}
+
+\begin{lemma}
+\label{lemma:mvn-equivalent-adjoint}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, $T \in A$, and $P, Q \in \text{Proj}(A)$ be orthogonal projections onto $\ol{T(H)}$ and $\ol{T^*(H)}$, respectively, then $P \sim Q$. 
+\end{lemma}
+\begin{proof}
+    Let $T = V|T|$ be the \hyperref[polar decomposition]{theorem:hilbert-polar-decomposition} of $T$, then $V$ is a partial isometry from $\ol{T^*(H)}$ to $\ol{T(H)}$. Since $V \in A$, $P \sim Q$. 
+\end{proof}
+
+\begin{theorem}[Kaplansky's Formula]
+\label{theorem:kaplansky-formula}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then
+    \[
+    [(P \vee Q) - Q] \sim [(P - P \wedge Q)]
+    \]
+\end{theorem}
+\begin{proof}
+    Using \autoref{theorem:existence-of-projections-vna}, assume without loss of generality that $I \in A$. In which case, by (1) of \autoref{lemma:projection-mental-gymnastics},
+    \[
+    
+        [(I - Q)P](H)^\perp = \ker(P(I - Q)) = \ker(Q)^\perp + [\ker(Q) \cap \ker(P)]
+    \]
+
+    and since $P \vee Q$ and $Q$ commute, 
+    \begin{align*}
+        [(I - Q)P](H) &= [\ker(Q)^\perp + [\ker(Q) \cap \ker(P)]]^\perp \\
+        &= \ker(Q) \cap [\ker(Q) \cap \ker(P)]^\perp \\
+        &= \ker(Q) \cap [\ker(Q)^\perp + \ker(P)^\perp]\\
+        &= (I - Q)(H) \cap [Q(H) + P(H)]\\
+        &= (I - Q)(H) \cap [(Q \vee P)(H)] \\
+        &= (P \vee Q)(I - Q)(H) = [(P \vee Q) - Q](H)
+    \end{align*}
+
+    by (2) of \autoref{lemma:projection-mental-gymnastics}. Similarly, 
+    \[
+    [P(I - Q)](H)^\perp = \ker((I - Q)P) = \ker(P) + [\ker(Q)^\perp \cap \ker(P)^\perp]
+    \]
+
+    so
+    \begin{align*}
+        [P(I - Q)](H) &= [\ker(P) + [\ker(Q)^\perp \cap \ker(P)^\perp]]^\perp \\
+        &= \ker(P)^\perp \cap  [\ker(Q) + \ker(P)] \\
+        &= P(H) \cap [(I - Q)(H) + (I - P)(H)] \\
+        &= P(H) \cap [(I - Q) \vee (I - P)](H) \\
+        &= P(H) \cap [I - (P \wedge Q)](H) \\
+        &= [P - (P \wedge Q)](H)
+    \end{align*}
+
+    Therefore
+    \[
+    [(P \vee Q) - Q](H) = [(I - Q)P](H)  \sim [P(I - Q)](H) = [P - (P \wedge Q)](H)
+    \]
+    
+    by \autoref{lemma:mvn-equivalent-adjoint}. 
+\end{proof}
+
+
+\begin{theorem}["Cantor-Bernstein"]
+\label{theorem:murray-von-neumann-subequivalent-partial-order}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$. If $P \preceq Q$ and $Q \preceq P$, then $P \sim Q$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Lemma 25.1]{Zhu}}}. ]
+    Let $U, V \in A$ be partial isometries such that $P = U^*U$, $UU^* \le Q$, $Q = V^*V$, and $VV^* \le P$. Denote $Q_0 = Q$ and $P_0 = P$. For each $n \in \natz$, inductively define $P_{n+1} = VQ_nV^*$ and $Q_{n+1} = UP_nU^*$, then:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $n \in \natz$, $P_n, Q_n \in \text{Proj}(A)$.
+        \item For each $n \in \natz$, $P_n \le P$ and $Q_n \le Q$.
+        \item For each $n \in \natz$, $P_{n+1} \le P_n$ and $Q_{n+1} \le Q_n$. 
+    \end{enumerate}
+
+    As $\seq{P_n}, \seq{Q_n} \subset \text{Proj}(A)$ are non-increasing sequences, by \autoref{theorem:existence-of-projections-vna}, there exists $P_\infty, Q_\infty \in \text{Proj}(A)$ such that $P_n \to P_\infty$ and $Q_n \to Q_\infty$ in the strong operator topology as $n \to \infty$. 
+
+    For each $n \in \natz$, $U(P_n - P_{n+1})U^* = Q_{n+1} - Q_{n+2}$, so
+    \begin{align*}
+    [U(P_n - P_{n+1})]^*[U(P_n - P_{n+1})] &= (P_n - P_{n+1})P(P_n - P_{n+1}) = P_n - P_{n+1} \\
+    [U(P_n - P_{n+1})][U(P_n - P_{n+1})]^* &= U(P_n - P_{n+1})^2U^* = Q_{n+1} - Q_{n+2}
+    \end{align*}
+
+    and $P_n - P_{n+1} \sim Q_{n+1} - Q_{n+2}$. Similarly, $Q_n - Q_{n+1} \sim P_{n+1} - P_{n+2}$. As $P_{n+1} = VQ_nV^*$ for all $n \in \natz$, $P_\infty \sim Q_\infty$ after passing through a strong-operator limit. 
+    
+    For each $N \in \natz$, $\sum_{n = 0}^N (P_n - P_{n+1}) = P - P_{N+1}$, so $P = P_\infty + \sum_{n = 0}^\infty (P_n - P_{n+1})$. Similarly, $Q = Q_\infty + \sum_{n = 0}^\infty (Q_n - Q_{n+1})$. Therefore
+    \begin{align*}
+        P &= P_\infty + \sum_{n = 0}^\infty (P_{2n} - P_{2n+1}) + \sum_{n = 0}^\infty (P_{2n + 1} - P_{2n+2}) \\
+        &\sim Q_\infty + \sum_{n = 0}^\infty (Q_{2n + 1} - Q_{2n+2})  + \sum_{n = 0}^\infty (Q_{2n} - Q_{2n+1}) = Q
+    \end{align*}
+
+    by \autoref{lemma:mvn-equivalent-direct-sum}. 
+\end{proof}
+
+\begin{theorem}[The Comparability Theorem]
+\label{theorem:vna-comparability}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then there exists a central projection $R$ such that $RP \preceq RQ$ and $(I - R)Q \preceq (I - R)P$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 25.4]{Zhu}}}. ]
+    By Zorn's lemma, there exists maximal families $\seqi{P}, \seqi{Q} \subset \text{Proj}(A)$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\seqi{P}$ is mutually orthogonal.
+        \item $\seqi{Q}$ is mutually orthogonal. 
+        \item For each $i \in I$, $P_i \sim Q_i$. 
+        \item For each $i \in I$, $P_i \le P$ and $Q_i \le Q$. 
+    \end{enumerate}
+
+    Let $P_0 = \sum_{i \in I}P_i$ and $Q_0 = \sum_{i \in I}Q_i$, then $P_0 \sim Q_0$ by \autoref{lemma:mvn-equivalent-direct-sum}. By maximality, there exists no non-zero $P', Q' \in \text{Proj}(A)$ such that $P' \le P - P_0$, $Q' \le Q - Q_0$, and $P' \sim Q'$. By \autoref{proposition:central-support-mvn}, $Z(P - P_0) Z(Q - Q_0) = 0$. 
+
+    Let $R = Z(Q - Q_0)$, then $Q - Q_0 \le R$ and $P - P_0 \le (I - R)$, so $(P - P_0)R = 0$ and $(Q - Q_0)R = Q - Q_0$. By \autoref{lemma:central-support-mvn-eq},
+    \[
+    PR = P_0R \sim Q_0R \le QR
+    \]
+
+    and
+    \[
+    Q(I - R) = Q_0(I - R) \sim P_0(I - R) \le P(I - R)
+    \]
+    
+\end{proof}
+
+\begin{corollary}
+\label{corollary:vna-factor-comparability}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a factor, then for any $P, Q \in \text{Proj}(A)$, either $P \prec Q$, $P \sim Q$, or $Q \prec P$. 
+\end{corollary}
+\begin{proof}
+    By the \hyperref[comparability theorem]{theorem:vna-comparability}, there exists a central projection $R$ such that $PR \preceq QR$ and $Q(I - R) \preceq P(I - R)$. As $A$ is a factor, either $R = 0$ or $R = I$. In which case, $P \preceq Q$ or $Q \preceq P$. 
+\end{proof}
+
+
+

src/op/vn/spec.tex

@@ -0,0 +1,336 @@
+\section{The Spectral Theorem}
+\label{section:spectral-theorem}
+
+\begin{definition}[Spectral Measure]
+\label{definition:spectral-measure}
+    Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, and $E: \cb_X \to B(H)$, then $E$ is a \textbf{spectral measure relative to $H$} if:
+    \begin{enumerate}
+        \item For each $B \in \cb_X$, $E(B)$ is an orthogonal projection. 
+        \item $E(\emptyset) = 0$, $E(X) = I_{B(H)}$. 
+        \item For each $B, C \in \cb_X$, $E(B \cap C) = E(B)E(C)$. 
+        \item For each $x, y \in H$, the mapping
+        \[
+        E_{x, y}: \cb_X \to \complex \quad B \mapsto \dpn{E(B)x, y}{H}
+        \]
+
+        is a complex Radon measure on $X$. 
+    \end{enumerate}
+    
+\end{definition}
+
+\begin{lemma}
+\label{lemma:spectral-measure-properties}
+    Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, and $E: \cb_X \to B(H)$ be a spectral measure relative to $H$, then:
+    \begin{enumerate}
+        \item For each $x, y \in H$, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$. 
+        \item For each $x \in H$, $E_{x, x}$ is positive.
+    \end{enumerate}
+
+    Let $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then
+    \begin{enumerate}[start=2]
+        \item For any $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{E}$ as well. 
+        \item Let 
+        \[
+        J: B^\infty(X; \complex) \to \mathscr{E}^* \quad \dpn{\mu, J(f)}{\mathscr{E}} = \int_X f d\mu
+        \]
+
+        then $\mathscr{E}^*$ admits a unique weak*-continuous involution and a unique separately weak*-continuous product, such that $\mathscr{E}^*$ is a commutative unital $C^*$-algebra, and $J$ is a unital *-homomorphism. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $x, y \in H$, $\seqf{B_j} \subset \cb_X$ be disjoint Borel sets, and $B = \bigsqcup_{j = 1}^n B_j$, then for each $1 \le i < j \le n$, $E(B_i)(H) \perp E(B_j)(H)$, so by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz} and the \hyperref[Pythagorean Theorem]{theorem:pythagoras},
+    \begin{align*}
+        \sum_{j = 1}^n |\dpn{E(B_j)x, y}{H}| &= \sum_{j = 1}^n |\dpn{E(B_j)x, E(B_j)y}{H}| \\
+        &\le \sum_{j = 1}^n \norm{E(B_j)x}_H \norm{E(B_j)y}_H \\
+        &\le \braks{\sum_{j = 1}^n \norm{E(B_j)x}_H^2}^{1/2} \cdot \braks{\sum_{j = 1}^n \norm{E(B_j)y}_H^2}^{1/2} \\
+        &= \norm{E(B)x}_H \cdot \norm{E(B)y}_H \le \norm{x}_H \cdot \norm{y}_H
+    \end{align*}
+
+    As the above holds for all finite sequences of disjoint Borel sets, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$.
+
+    (2): For each $B \in \cb_X$, $E(B)$ is a projection, so $E_{x, x}(B) = \dpn{E(B)x, x}{H} \ge 0$.
+
+    (3): For each $x, y \in H$ and $B, C \in \cb_X$, 
+    \[
+    \int_C \one_B dE_{x, y} = \dpn{E(C \cap B)x, y}{H} = \dpn{E(C)E(B)x, y}{H} = E_{E(B)x, y}(C)
+    \]
+
+    By linearity, $fdE_{x, y} \in \mathscr{E}$ for all $f \in \Sigma(X; \complex)$. For each $f \in \Sigma(X; \complex)$, the mapping $\mu \mapsto f d\mu$ is continuous in the total variation norm, so $fd\mu \in \mathscr{E}$ for all $\mu \in \mathscr{E}$ and $f \in \Sigma(X; \complex)$. By \autoref{proposition:lp-simple-dense}, $\Sigma(X; \complex)$ is dense in $L^1(\mu; \complex)$ for all $\mu \in \mathscr{E}$. Therefore $fd\mu \in \mathscr{E}$ for all $f \in L^1(\mu; \complex)$ and $\mu \in \mathscr{E}$. 
+    
+    Finally, let $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, then by the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(\mu; \complex)$ such that $d\nu = f d\mu \in \mathscr{E}$. 
+
+    (4): By (3), for any $\mu \in \mathscr{E}$ and $f \in L^1(\mu; \complex)$, $fd\mu \in \mathscr{E}$ as well. By \autoref{proposition:measures-dual-algebra}, there exists a unique weak*-continuous involution and separately weak*-continuous product on $\mathscr{E}^*$ making $\mathscr{E}^*$ a commutative unital $C^*$-algebra, and $J|_{C(X; \complex)}$ a unital *-homomorphism. Since
+    \begin{enumerate}[label=(\roman*)]
+        \item $J$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-$\sigma(\mathscr{E}^*, \mathscr{E})$ continuous. 
+        \item Conjugation on $B^\infty(X; \complex)$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
+        \item Multiplication on $B^\infty(X; \complex)$ is separately $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
+    \end{enumerate}
+    
+    the mapping $J$ is a unital *-homomorphism.
+\end{proof}
+
+
+\begin{definition}[Integration Against Spectral Measure]
+\label{definition:spectral-measure-integral}
+    Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, $E: \cb_X \to B(H)$ be a spectral measure relative to $H$, $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and
+    \[
+    J: B^\infty(X; \complex) \to \mathscr{E}^* \quad \dpn{\mu, J(f)}{\mathscr{E}} = \int_X f d\mu
+    \]
+
+    Then, $\mathscr{E}^*$ admits a unique weak*-continuous involution and a unique separately weak*-continuous product, such that $\mathscr{E}^*$ is a commutative unital $C^*$-algebra, and $J$ is a unital *-homomorphism. 
+    
+    For each $\phi \in \mathscr{E}^*$, let $I_E(\phi) \in B(H)$ be the operator defined by
+    \[
+    \dpn{I_E(\phi) \cdot x, y}{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}} \quad \forall x, y \in H
+    \]
+
+    then
+
+    \begin{enumerate}
+        \item $I_E$ is a contraction from $\mathscr{E}^*$ to $B(H)$. 
+        \item $I_E$ is continuous from the weak*-topology on $\mathscr{E}^*$ to the weak operator topology on $B(H)$. 
+        \item $I_E$ is an injective unital *-homomorphism. 
+    \end{enumerate}
+    
+    For any $\phi \in \mathscr{E}^*$, $I_E(\phi) = \int_X \phi dE$ is the \textbf{integral} of $\phi$ with respect to $E$. 
+\end{definition}
+\begin{proof}
+    (1): Let $\phi \in \mathscr{E}^*$ and $x, y \in H$, then by \autoref{lemma:spectral-measure-properties}, 
+    \begin{align*}
+    |\dpn{I_E(\phi) \cdot x, y}{H}| &= |\dpn{E_{x, y}, \phi}{\mathscr{E}}| \le \norm{E_{x, y}}_{\mathscr{E}} \cdot \norm{\phi}_{\mathscr{E}^{*}} \\
+    &\le \norm{\phi}_{\mathscr{E}^{*}} \cdot \norm{x}_H \cdot \norm{y}_H
+    \end{align*}
+
+    Since the above holds for all $x, y \in H$, $I_E(\phi) \in B(H)$ with $\norm{I_E(\phi)}_{B(H)} \le \norm{\phi}_{\mathscr{E}^{*}}$. 
+
+    (2): For each $x, y \in H$, $E_{x, y} \in \mathscr{E}$. Since $\angles{\int \phi dE \cdot x, y}_{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}}$ for every $\phi \in \mathscr{E}^{*}$, $I_E$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $B(H)$. 
+
+    (3): By \autoref{lemma:separable-metric-space-approx-identity}, the simple functions $\Sigma(X; \complex)$ are uniformly dense in the bounded Borel functions $B^\infty(X; \complex)$. Since
+    \begin{enumerate}[label=(\roman*)]
+        \item $I_E$ restricted to $J(\Sigma(X; \complex))$ is a *-homomorphism. 
+        \item Multiplication and conjugation are continuous in the uniform norm on $B^\infty(X; \complex)$
+        \item Composition and adjunction are continuous in the operator norm on $B(H)$
+    \end{enumerate}
+
+    the map $I_E$ restricted to $J(B^\infty(X; \complex))$ is a *-homomorphism by continuity. 
+    
+    By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex) \subset B^\infty(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$, so $J(C(X; \complex))$ is weak*-dense in $\mathscr{E}^*$. As
+    \begin{enumerate}[label=(\roman*)]
+        \item $I_E$ restricted to $J(B^\infty(X; \complex))$ is a *-homomorphism. 
+        \item The involution $\phi \mapsto \ol \phi$ is weak*-continuous on $\mathscr{E}^{*}$. 
+        \item The adjunction $T \mapsto T^*$ is weak-operator continuous on $B(H)$. 
+        \item The product $(\phi, \psi) \mapsto \phi \psi$ is separately weak*-continuous on $\mathscr{E}^{*}$. 
+        \item The composition $(S, T) \mapsto ST$ is separately weak-operator continuous on $B(H)$.
+    \end{enumerate}
+
+    the map $I_E$ is a *-homomorphism by the weak* to weak-operator continuity established in (2). Since $E(X) = I_{B(H)}$, $I_E$ is a unital *-homomorphism. 
+
+    Finally, let $\phi \in \mathscr{E}^*$ with $I_E(\phi) = 0$, then $\dpn{I_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}} = 0$ for all $x, y \in H$. As $\mathscr{E}$ is the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, $\phi = 0$. Therefore $I_E$ is an injective unital *-homomorphism. 
+\end{proof}
+
+\begin{theorem}[Spectral Theorem I]
+\label{theorem:spectral-theorem-vn-1}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then:
+    \begin{enumerate}
+        \item There exists a unique spectral measure $E: \cb_{\Omega(A)} \to B(H)$ such that\footnote{Omitting the natural map $C(\Omega(A); \complex) \to \mathscr{E}^*$. }
+        \[
+        T = \int_{\Omega(A)} \Gamma_A T dE \quad \forall T \in A
+        \]
+        \item Let $B \subset B(H)$ be the strong-operator closure of $A$ in $B(H)$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then
+        \[
+        I_E: \mathscr{E}^* \to B \quad \phi \mapsto \int_{\Omega(A)}\phi dE
+        \]
+
+        is a *-isomorphism. 
+    \end{enumerate}
+    
+    The measure $E$ is the \textbf{spectral measure associated with $A$}, and the homomorphism $I_E$ is the \textbf{extended inverse Gelfand transform} of $A$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 20.2]{Zhu}}}. ]
+    (1): By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, $\Gamma_A: A \to C(\Omega(A); \complex)$ is a *-isomorphism. For each $x, y \in H$, $\Gamma_A^{-1}$ induces a mapping
+    \[
+    E_{x, y}: C(\Omega(A); \complex) \to \complex \quad \dpn{f, E_{x, y}}{C(\Omega(A); \complex)} = \dpn{\Gamma_A^{-1}f \cdot x, y}{H}
+    \]
+
+    which, by the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, takes the form of a complex Radon measure on $\Omega(A)$. Thus by the uniqueness part of the Riesz Representation Theorem, such a spectral measure must be unique if it exists. 
+
+    Since $\norm{E_{x, y}}_{C(\Omega(A); \complex)^*} \le \norm{x}_H\norm{y}_H$ for all $x, y \in H$, $\bracsn{E_{x, y}|x, y \in H}$ induces a bounded linear map
+    \[
+    J_E: B^\infty(\Omega(A); \complex) \to B(H) \quad \dpn{J_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*}
+    \]
+
+    with $J_E(f) = \Gamma_A^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$. 
+
+    For any $C \in \cb_{\Omega(A)}$, $\one_C$ is a projection in $B^\infty(\Omega(A); \complex)$. So to see that
+    \[
+    E: \cb_{\Omega(A)} \to B(H) \quad \dpn{E(C)x, y}{H} = E_{x, y}(C)
+    \]
+
+    defines a spectral measure, it is sufficient to show that $J_E$ is a *-homomorphism. 
+
+    Let $x, y \in H$, then as $\Gamma_A$ is a *-isomorphism, for any $f, g \in C(\Omega(A); \complex)$, 
+    \begin{align*}
+        \dpn{fg, E_{x, y}}{C(\Omega(A); \complex)} &= \dpn{\Gamma_A^{-1}f \cdot \Gamma_A^{-1}g \cdot x, y}{H} \\
+        &= \dpn{\Gamma_A^{-1}g \cdot x, (\Gamma_A^{-1}f)^* y}{H} = \dpn{g, E_{x, J_E(f)^*y}}{C(\Omega(A); \complex)}
+    \end{align*}
+
+    As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y} = E_{x, J_E(f)^*y}$. Now, fix $\phi \in B^\infty(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$, 
+    \begin{align*}
+        \dpn{E_{x, y}, \phi f}{C(\Omega(A); \complex)^*} &= \dpn{E_{x, J_E(f)^*y}, \phi}{C(\Omega(A); \complex)^*} = \dpn{J_E(\phi)x, J_E(f)^*y}{H} \\
+        &= \dpn{J_E(f)J_E(\phi)x, y}{H} = \dpn{f, E_{J_E(\phi)x, y}}{C(\Omega(A); \complex)}
+    \end{align*}
+
+    so $\phi E_{x, y} = E_{J_E(\phi)x, y}$ for all $\phi \in B^\infty(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^\infty(\Omega(A); \complex)$, 
+    \begin{align*}
+    \dpn{J_E(\phi \psi)x, y}{H} &= \dpn{E_{x, y}, \phi \psi}{C(\Omega(A); \complex)^*} = \dpn{E_{J_E(\psi) x, y}, \phi}{C(\Omega(A); \complex)^*} \\
+    &= \dpn{J_E(\phi)J_E(\psi)x, y}{H}
+    \end{align*}
+
+    and $J_E$ is a homomorphism. 
+
+    Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_A$ is a *-isomorphism, $J_E(f) = \Gamma_A^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)} = \dpn{J_E(f)x, x}{H} \in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^\infty(\Omega(A); \real)$ and $x \in H$, $\dpn{J_E(\phi)x, x}{H} = \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*} \in \real$ as well. Therefore $J_E(\phi)$ is self-adjoint, and $J_E$ is a *-homomorphism. 
+
+    (2): By \autoref{definition:spectral-measure-integral}, $I_E$ is an injective unital *-homomorphism, so it is sufficient to show that $I_E(\mathscr{E}^*) = B$.
+    
+    Let $J: C(\Omega(A); \complex) \to \mathscr{E}^*$ be defined by $\dpn{\mu, J(f)}{\mathscr{E}} = \int_{\Omega(A)}f d\mu$ for each $\mu \in \mathscr{E}$ and $f \in C(\Omega(A); \complex)$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$, so $J(C(\Omega(A); \complex))$ is weak*-dense in $\mathscr{E}^*$. Since $I_E$ is continuous from the weak* topology on $\mathscr{E}^*$ to the weak operator topology on $B(H)$, $I_E(\mathscr{E}^*) \subset B$ by \autoref{proposition:closure-of-image}. 
+    
+    On the other hand, by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{\mathscr{E}^*}(0, 1)}$ is weak*-compact, so $I_E(\ol{B_{\mathscr{E}^*}(0, 1)})$ is weak-operator compact. As $\Gamma_A: A \to C(\Omega(A); \complex)$ is an isometric isomorphism, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_A(0, 1)}$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_B(0, 1)}$, and $I_E(\mathscr{E}^*) = B$.
+\end{proof}
+
+\begin{theorem}[Spectral Theorem II]
+\label{theorem:spectral-theorem-vn-2}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $Id \in A$, $B \subset B(H)$ be the von Neumann algebra generated by $A$, $E: \cb_{\Omega(A)} \to B(H)$ be the spectral measure associated with $A$, $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $\seqi{\xi} \subset H$ be a maximal family such that the subspaces $\bracsn{A\xi_i|i \in I}$ are mutually orthogonal, then:
+    \begin{enumerate}
+        \item For each $i \in I$, there exists a finite positive Radon measure $\mu_i \in \mathscr{E}$ on $\Omega(A)$ such that for every Borel set $C \in \cb_{\Omega(A)}$, $\mu_i(C) = 0$ if and only if $E_{x, y}(C) = 0$ for all $x, y \in \ol{A\xi_i}$. 
+        \item For each $i \in I$, let $P_i: H \to \ol{A\xi_i}$ be the orthogonal projection onto $\ol{A\xi_i}$, then for any $x, y \in H$, $E_{x, y} = \sum_{i \in I}E_{P_ix, P_iy}$. 
+        \item The natural map $C(\Omega(A); \complex) \to [l^\infty(I); L^\infty(\mu_i; \complex)]$ is injective. Equivalently, $\ol{\bigcup_{i \in I}\supp{\mu_i}} = \Omega(A)$. 
+        \item The space $\mathscr{E}$ is a quotient of $[l^1(I); L^1(\mu_i; \complex)]$ under the mapping
+        \[
+        \mathscr{M}: [l^1(I); L^1(\mu_i; \complex)] \to \mathscr{E} \quad f \mapsto \sum_{i \in I}f_id\mu_i
+        \]
+
+        and $\mathscr{E}^*$ may be identified as a closed subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$ through $\mathscr{M}^*$. 
+        \item There exists a unitary equivalence $U: H \to [l^2(I); L^2(\mu_i; \complex)]$ between $\mathscr{E}^*$ acting on $[l^2(I); L^2(\mu_i; \complex)]$ and $B$ acting on $H$, such that for each $i \in I$, $U|_{\ol{A\xi_i}}$ is an isometry onto the $i$-th factor of $[l^2(I); L^2(\mu_i; \complex)]$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.47]{FollandHarmonic}}}. ]
+    (1): Fix $ i \in I$ and let $\mu_i = E_{\xi_i, \xi_i}$, then for any $C \in \cb_{\Omega(A)}$ with $E_{x, y}(C) = 0$ for all $x, y \in \ol{A\xi_i}$, $\mu_i(C) = 0$. By (1) and (2) of \autoref{lemma:spectral-measure-properties}, $\mu_i$ is a finite positive Radon measure. 
+    
+    By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, for each $S, T \in A$ and $C \in \cb_{\Omega(A)}$, 
+    \[
+    \dpn{E(C)S\xi_i, T\xi_i}{H} = \int_C \Gamma_AS \cdot \ol{\Gamma_AT} dE_{\xi_i, \xi_i}
+    \]
+
+    so $\Gamma_AS \cdot \ol{\Gamma_AT}dE_{\xi_i, \xi_i} = dE_{S\xi_i, T\xi_i} \ll \mu_i$. By (1) of \autoref{lemma:spectral-measure-properties} and completeness of $L^1(\mu_i; \complex)$, $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}}$ is absolutely continuous with respect to $\mu_i$. Therefore for any $C \in \cb_{\Omega(A)}$, $\mu_i(C) = 0$ if and only if $E_{x, y}(C) = 0$ for all $x, y \in \ol{A \xi_i}$. 
+
+    (2): Let $i, j \in I$ with $i \ne j$, $x \in \ol{A\xi_i}$, $y \in \ol{A\xi_j}$, and $f \in C(\Omega(A); \complex)$, then since $\ol{A\xi_i} \perp \ol{A\xi_j}$,
+    \[
+        \int_{\Omega(A)} f dE_{x, y} = \dpn{\Gamma_A^{-1}(f)x, y}{H} = 0
+    \]
+
+    As the above holds for all $f \in C(\Omega(A); \complex)$, $E_{x, y} = 0$. 
+    
+    Given that $\seqi{\xi}$ is maximal, $x = \sum_{i \in I}P_ix$ for all $x \in H$. Thus for any $x, y \in H$, 
+    \[
+        E_{x, y} = \sum_{i, j \in I}E_{P_ix, P_jy} =  \sum_{i \in I}E_{P_ix, P_iy} \in \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])
+    \]
+
+    (3): Let $T \in A$ with $\Gamma_A T = 0$ $\mu_i$-almost everywhere for all $i \in I$. By (1), $E_{P_ix, P_iy} \ll \mu_i$ for all $i \in I$. Thus for any $x, y \in H$, 
+    \[
+    \dpn{Tx, y}{H} = \int_{\Omega(A)}\Gamma_A T dE_{x, y} = \sum_{i \in I}\int_{\Omega(A)}\Gamma_A TdE_{P_ix, P_iy} = 0
+    \]
+
+    Therefore $C(\Omega(A); \complex)$ may be identified as a subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$. 
+
+    (4): By (2), for each $x, y \in H$, $E_{x, y} = \sum_{i \in I}E_{P_ix, P_iy}$. By (1), $E_{P_ix, P_iy} \ll \mu_i$ for all $i \in I$, so $\mathscr{E} \subset \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])$. 
+
+    On the other hand, for each $i \in I$, since $\mu_i$ is a Radon measure, $C(\Omega(A); \complex)$ is dense in $L^1(\mu_i; \complex)$ by \autoref{proposition:radon-cc-dense}. As
+    \begin{align*}
+    \mathscr{E} &\supset \bracsn{E_{x, y}|x, y \in \ol{A\xi_i}} \supset \bracsn{fdE_{\xi_i, \xi_i}|f \in C(\Omega(A); \complex)}  \\
+    &= \bracsn{fd\mu_i|f \in C(\Omega(A); \complex)} 
+    \end{align*}
+    
+
+    and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ is closed, $\mathscr{E} \supset \bracsn{f d\mu_i|f \in L^1(\mu_i; \complex)}$. 
+    
+    Finally, given that the above holds for all $i \in I$, $\mathscr{E} = \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])$. By \autoref{theorem:lp-sum-dual} and \autoref{theorem:lp-duality}, $[l^\infty(I); L^\infty(\mu_i; \complex)] = [l^1(I); L^1(\mu_i; \complex)]^*$, so $\mathscr{E}^*$ may be identified with its image under $\mathscr{M}^*$. 
+    
+    (5): Fix $i \in I$, then for any $S, T \in A$ with $S\xi_i = T\xi_i$,
+    \[
+    \Gamma_AS dE_{\xi_i, \xi_i} = E_{S\xi_i, \xi_i} = E_{T\xi_i, \xi_i} = \Gamma_A T dE_{\xi_i, \xi_i}
+    \]
+
+    so $\Gamma_A S = \Gamma_A T$ $\mu_i$-almost everywhere. Thus the mapping
+    \[
+    U_i: \ol{A\xi_i} \to L^2(\mu_i; \complex) \quad T\xi_i \mapsto \Gamma_AT
+    \]
+
+    is well-defined. Moreover, for any $S, T \in A$, 
+    \[
+    \dpn{S\xi_i, T\xi_i}{H} = \int \Gamma_AS \cdot \ol{\Gamma_A T} dE_{\xi_i, \xi_i} = \dpn{\Gamma_A S, \Gamma_A T}{L^2(\mu_i; \complex)}
+    \]
+
+    so $U_i$ extends into an isometry between $\ol{A\xi_i}$ and $L^2(\mu_i; \complex)$. Thus the mapping
+    \[
+    U: H \to [l^2(I); L^2(\mu_i; \complex)] \quad (Ux)_i = U_i(P_ix)
+    \]
+
+    is an isometry between $H$ and $[l^2(I); L^2(\mu_i; \complex)]$ such that $U(Tx) = \Gamma_AT \cdot Ux$ for all $x \in H$ and $T \in A$. 
+    
+    Finally, given that
+    \begin{enumerate}[label=(\roman*)]
+        \item By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $\mathscr{E}^*$. 
+        \item The weak* topology on $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is equal to the weak operator topology of $[l^\infty(I); L^\infty(\mu_i; \complex)]$ acting on $[l^2(I); L^2(\mu_i; \complex)]$. 
+        \item $A$ is weak-operator dense in $B$. 
+        \item By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, the isomorphism $\phi \mapsto \int \phi dE$ is continuous from the weak* topology on $\mathscr{E}^*$ to the weak operator topology on $B$.
+    \end{enumerate}
+    
+    the mapping $U$ is a unitary equivalence between $\mathscr{E}^*$ acting on $[l^2(I); L^2(\mu_i; \complex)]$ and $B$ acting on $H$. 
+\end{proof}
+
+\begin{corollary}[Representation of Commutative von Neumann Algebras]
+\label{corollary:commutative-von-neumann-linfty}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative von Neumann algebra with $I \in A$, then:
+    \begin{enumerate}
+        \item There exists a LCH space $\Omega$ and a decomposable Radon measure $\mu$ on $\Omega$ such that $A$ is *-isomorphic to $L^\infty(\mu; \complex)$. 
+        \item If $A$ admits a cyclic vector, then $\Omega$ may be taken to be compact. 
+        \item If $H$ is separable, then $\Omega$ may be taken to be compact. 
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    Let $E$ be the spectral measure on $\Omega(A)$ associated with $A$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$. By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, $A$ is *-isomorphic to $\mathscr{E}^*$.
+
+    (1): By (3) of \autoref{lemma:spectral-measure-properties} and \autoref{theorem:hilbert-measures-dual}, $A$ is *-isomorphic to $[l^\infty(I); L^\infty(\mu_i; \complex)]$, where $\seqi{\mu} \subset \mathscr{E}$ is a maximal mutually singular family. Let $\Omega = \bigsqcup_{i \in I}\Omega(A)$, then $\Omega$ is a LCH space. For each $i \in I$, let $\Omega_i$ denote the $i$-th copy of $\Omega(A)$, then
+    \[
+    \mu: \cb_\Omega \to [0, \infty] \quad B \mapsto \sum_{i \in I}\mu_i(B \cap \Omega_i)
+    \]
+
+    is the desired decomposable Radon measure. 
+
+    (2): By \hyperref[Spectral Theorem II]{theorem:spectral-theorem-vn-2}, there exists a single positive Radon measure $\mu \in \mathscr{E}$ on $\Omega(A)$ such that $\mathscr{E}$ is absolutely continuous with respect to it. Therefore the index set in (1) can be taken to be a singleton. 
+
+    (3): If $H$ is separable, then so is $\mathscr{E}$. As such, there exists a single positive Radon measure $\mu \in \mathscr{E}$ on $\Omega(A)$ such that $\mathscr{E}$ is absolutely continuous with respect to it. Therefore the index set in (1) can be taken to be a singleton. 
+\end{proof}
+
+
+\begin{remark}
+\label{remark:spectral-theorem-vn-2}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, $B$ be the von Neumann algebra generated by $A$, $E: \cb_{\Omega(A)} \to B(H)$ be the spectral measure associated with $A$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$.
+
+    By \hyperref[Spectral Theorem II]{theorem:spectral-theorem-vn-2}, there exists a decomposable measure space $\Omega$, corresponding to a number of copies of $\Omega(A)$, such that $\mathscr{E}$ is a quotient of its $L^1$ space, $\mathscr{E}^*$ is a subspace of its $L^\infty$ space, and $H$ is isomorphic to its $L^2$ space. The preceding isomorphisms are all linked by a unitary equivalence between $B$ acting on $H$, and $\mathscr{E}^*$ acting on the $l^2$ direct sum. 
+
+    The complexity of $\Omega$, that is, the number of copies of $\Omega(A)$ that it contains, depends on two factors: 
+    \begin{enumerate}
+        \item The complexity of the von Neumann algebra $B$: If $B$ is sufficiently complex, then $\mathscr{E}$ cannot be expressed as the $L^1$ space of a single measure on $\Omega(A)$. Instead, multiple copies of $\Omega(A)$ are needed to handle mutually singular measures with overlapping supports. For more details on this phenomenon, see \autoref{theorem:hilbert-measures-dual}.
+        \item The size of the Hilbert space $H$ relative to $B$: If $H$ is extremely large, then a large number of vectors are required for $B$ to cover it. As such, many copies of $\Omega(A)$ are required to handle the complexity of $H$. 
+    \end{enumerate}
+
+    More concretely, (1) manifests as the size of the space $\mathscr{E}$, and (2) manifests as the size of the kernel of the mapping $L^1(\Omega) \to \mathscr{E}$. 
+
+    By limiting these two sources of complexity, it is possible to remove the need of multiple copies of $\Omega(A)$. In particular, 
+    \begin{enumerate}
+        \item If $B$ admits a cyclic vector, then only one copy of $\Omega(A)$ is required for the construction in the Spectral Theorem \cite[Theorem 23.1]{Zhu}.
+        \item If $H$ is separable, then at most countably many copies of $\Omega(A)$ are required for the construction in the Spectral Theorem. In which case, the measures can be summed such that $B$ is isomorphic to an $L^\infty$ space on $\Omega(A)$ \cite[Page 24]{FollandHarmonic} \cite[Theorem 23.2]{Zhu}. 
+    \end{enumerate}
+\end{remark}
+

src/op/vn/topologies.tex

@@ -16,6 +16,19 @@ Depending on the topology placed on $H \otimes H$, and the corresponding complet
     By \autoref{proposition:projective-tensor-product-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}. 
 \end{proof}
 
+A natural topology consistent with the ultraweak topology would be the ultrastrong topology. 
+
+\begin{definition}[Ultrastrong Topology]
+\label{definition:bh-ultrastrong-topology}
+    Let $H$ be a complex Hilbert space. For each $x = \seq{x_n} \in L^2(\natp; H)$, let
+    \[
+    \Phi_x: B(H) \to L^2(\natp; H) \quad (\Phi_xT)_n = Tx_n
+    \]
+
+    then the \textbf{ultrastrong}/\textbf{$\sigma$-strong} topology on $B(H)$ is the topology generated by the maps $\bracsn{\Phi_x|x \in L^2(\natp; H)}$. 
+\end{definition}
+
+
 Seeing that $B(H)$ is a dual Banach space, the following fact is immediate:
 
 \begin{proposition}
@@ -36,7 +49,6 @@ Now, a few facts about the more familiar operator topologies:
         \item Every bounded subset of $B(H)$ is relatively compact in the weak operator topology. 
         \item The composition map $(S, T) \mapsto ST$ is separately continuous in the strong and weak operator topologies. 
         \item The composition map $(S, T) \mapsto ST$ is left-hypocontinuous with respect to the strong operator topology and strong-operator bounded subsets of $B(H)$. 
-        \item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology. 
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -45,5 +57,66 @@ Now, a few facts about the more familiar operator topologies:
     (4): By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, every strong-operator bounded subset of $B(H)$ is equicontinuous. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:bh-adjoint-strong-continuous}
+    Let $H$ be a complex Hilbert space, then:
+    \begin{enumerate}
+        \item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology. 
+        \item $T \mapsto T^*$ restricted to the normal operators is continuous in the strong operator topology. 
+        \item For any $f \in C(\complex; \complex)$, the mapping $T \mapsto f(T)$ restricted to any bounded set of normal operators is continuous in the strong operator topology. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Section 19.1]{Zhu}}}. ]
+    (2): Let $S, T \in B(H)$, then
+    \begin{align*}
+        \normn{(S^* - T^*)x}_H^2 &= \normn{S^*x}_H^2 + \normn{T^*x}_H^2 - \dpn{x, ST^*x}{H} - \dpn{ST^*x, x}{H} \\
+        &\le \normn{S^*x}_H^2 + \normn{T^*x}_H^2 - \dpn{x, TT^*x}{H} - \dpn{TT^*x, x}{H} \\
+        &+ |\dpn{x, (T - S)T^*x}{H}| + |\dpn{(T - S)T^*x, x}{H}| \\
+        &\le |\normn{S^*x}_H^2 - \normn{T^*x}_H^2| + 2\norm{x}_H\normn{(T - S)T^*x}_H
+    \end{align*}
+
+    Now, if $S$ and $T$ are normal, then $\normn{S^*x}_H = \norm{Sx}_H$ and $\norm{T^*x}_H = \norm{Tx}_H$, so
+    \begin{align*}
+    |\norm{S^*x}_H^2 - \norm{T^*x}_H^2| &= |\norm{Sx}_H^2 - \norm{Tx}_H^2| \\
+    &\le \norm{(S - T)x}_H (\norm{Sx}_H + \norm{Tx}_H) \\
+    &\le \norm{(S - T)x}_H (\norm{(S - T)x}_H + 2\norm{Tx}_H)
+    \end{align*}
+
+    Therefore
+    \begin{align*}
+    \normn{(S^* - T^*)x}_H^2 &\le \norm{(S - T)x}_H (\norm{(S - T)x}_H + 2\norm{Tx}_H) \\
+    &+ 2\norm{x}_H\normn{(T - S)T^*x}_H
+    \end{align*}
+    
+    and the adjoint map restricted to normal operators is continuous in the strong operator topology. 
+
+    (3): Let $S, T \in B_{B(H)}(0, 1)$ and $x \in H$ and $n \in \natp$, then
+    \begin{align*}
+    \normn{(S^n - T^n)x}_H &\le \sum_{k = 0}^{n-1}\normn{S^{n-1-k}(S - T)T^kx}_{H} \\
+    &\le \sum_{k = 0}^{n - 1}\normn{(S - T)T^kx}_H
+    \end{align*}
+    
+
+    so the mapping $T \mapsto T^n$ on $B_{B(H)}(0, 1)$ is continuous in the strong operator topology. By (2), the mapping $T \mapsto p(T, T^*)$ is strong-operator continuous for all $p \in \complex[z, \ol z]$. 
+
+    By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exist polynomials $p_n \in \complex[z, \ol z]$ such that $p_n \to f$ uniformly on $\ol{B_\complex(0, 1)}$. For any $T \in B_{B(H)}(0, 1)$, $x \in H$, and $n \in \natp$, 
+    \begin{align*}
+    \norm{[f(T) - p_n(T)]x}_H &\le \norm{f(T) - p_n(T)}_{B(H)} \cdot \norm{x}_H \\
+    &\le \norm{x}_H  \cdot \sup_{z \in \ol{B_\complex(0, 1)}}|f(z) - p_n(z)|
+    \end{align*}
+    
+
+    by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}. Thus $f$ is a uniform limit of strong-operator continuous functions on $B_{B(H)}(0, 1)$, and as such also strong-operator continuous by \autoref{proposition:uniform-limit-continuous}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:spatial-isomorphism-sot-continuous}
+    Let $A$ be a $C^*$-algebra, $H_1, H_2$ be a complex Hilbert spaces, $\pi_1: A \to B(H_1)$ and $\pi_2: A \to B(H_2)$ be injective representations of $A$, and $U: H_1 \to H_2$ be an unitary equivalence, then the mapping
+    \[
+    \pi_1(A) \to \pi_2(A) \quad T \mapsto UTU^{-1}
+    \]
+
+    is strong-operator and weak-operator continuous. 
+\end{proposition}
 
 

src/op/vn/type-decomp.tex

@@ -0,0 +1,303 @@
+\section{Type Decomposition}
+\label{section:vna-type-decomposition}
+
+
+\begin{definition}[Finite Projection]
+\label{definition:finite-projection}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $P$ is \textbf{finite} if for any $Q \in \text{Proj}(A)$ with $P \sim Q$ and $Q \le P$, $P = Q$. For any $P \in \text{Proj}(A)$, $P$ is \textbf{infinite} if it is not finite. 
+\end{definition}
+
+\begin{definition}[Abelian Projection]
+\label{definition:abelian-projection}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $P$ is \textbf{abelian} if $PAP$ is abelian. 
+\end{definition}
+
+\begin{definition}[Minimal Projection]
+\label{definition:minimal-projection}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $PAP = \complex P$. 
+        \item There exists no $Q \in \text{Proj}(A)$ with $0 < Q < P$. 
+    \end{enumerate}
+    
+    If the above holds, then $P$ is \textbf{minimal}.
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $Q \in \text{Proj}(A)$ with $Q \le P$, then $Q = PQP$. As $PAP = \complex P$, either $PQP = 0$ or $PQP = P$. 
+
+    (2) $\Rightarrow$ (1): Given that there exists no projections strictly between $0$ and $P$, the only non-zero projection in $PAP$ is $P$ itself. By \autoref{theorem:vn-projection-norm-dense}, the linear span of projections in $PAP$ is norm-dense in $PAP$. Therefore $PAP = \complex P$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:projection-types-gymnastics}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$. 
+    \begin{enumerate}
+        \item If $P$ is minimal, then $P$ is abelian. 
+        \item If $P$ is abelian, then $P$ is finite. 
+        \item If $P$ is finite and $P \sim Q$, then $Q$ is finite. 
+        \item If $P$ is finite and $Q \le P$, then $Q$ is finite. 
+        \item If $P$ is minimal and $P \sim Q$, then $Q$ is minimal. 
+        \item If $P, Q$ are minimal with $P \sim Q$, then for any $U, V \in A$ with $P = U^*U = V^*V$ and $Q = UU^* = VV^*$, there exists $\lambda \in \partial B_{\complex}(0, 1)$ such that $V = \lambda U$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Section 26.1]{Zhu}}}. ]
+    (1): $PAP = \complex P$ is abelian. 
+
+    (2): Let $R \in \text{Proj}(A)$ with $P \sim R \le P$, then there exists $V \in A$ such that $R = V^*V$ and $P = VV^*$. Since $V$ has initial space $R(H) \subset P(H)$ and final space $P(H)$, $V = PVP$ and $V^* = PV^*P$. As $PAP$ is abelian, 
+    \[
+    R = V^*V = PV^*PPVP = PVPPV^*P = VV^* = P
+    \]
+
+    (3): Let $R \in \text{Proj}(A)$ with $Q \sim R \le Q$. Let $V \in A$ with $Q = V^*V$ and $P = VV^*$, then $V$ is a partial isometry with initial space $Q(H)$ and final space $P(H)$. In which case, $P = VQV^*$, and $VRV^* \le VQV^* = P$. Let $U = (VRV^*)V$, then
+    \begin{align*}
+    U^*U &= (VRV^*V)^*(VRV^*V) = V^*VRV^* \cdot VRV^*V \\
+    &= V^*VRV^*V = QRQ = R
+    \end{align*}
+    
+
+    and as $Q = V^*PV$, 
+    \begin{align*}
+        UU^* &= (VRV^*V)(VRV^*V)^* = VRV^*V \cdot V^*VRV^* \\
+        &= VRV^*PVRV^* = VRQRV^* = VRV^*
+    \end{align*}
+    
+    so $VRV^* \sim R \sim Q \sim P$. Given that $P$ is finite, $VRV^* = P$. Therefore
+    \[
+    R = QRQ = V^*VRV^*V = V^*PV = Q
+    \]
+
+    (4): Let $R \in \text{Proj}(A)$ with $Q \sim R \le Q \le P$, then $P \sim (P - Q) + R \le P$, so $P - Q + R = P$, and $Q = R$. 
+
+    (5): Since $P \sim Q$, there exists $V \in A$ with $P = V^*V$ and $Q = VV^*$. Let $R \in \text{Proj}(A)$ with $0 < R \le Q$, then $0 \le V^*RV \le V^*QV = P$. By minimality of $P$, $V^*RV = P$, so
+    \[
+    R = QRQ = VV^*RVV^* = VPV^* = Q
+    \]
+
+    (6): Let $R = U^*V$, then since $U$ and $V$ are partial isometries with initial space $P(H)$ and final space $Q(H)$, 
+    \[
+    PRP = U^*U \cdot U^*V \cdot V^*V = U^*QV = U^*V
+    \]
+
+    so $PRP \in PAP = \complex P$. Thus there exists $\lambda \in \complex$ such that $R = \lambda P$. In which case,
+    \[
+    \lambda U = U \cdot \lambda P = UU^*V = QV = V
+    \]
+
+    and
+    \[
+    P = V^*V = \lambda \ol{\lambda} U^*U = |\lambda|^2 P
+    \]
+
+    so $|\lambda| = 1$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:centrally-orthogonal-sum-properties}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, $\seqi{P} \subset \text{Proj}(A)$ be centrally orthogonal, and $P = \sum_{i \in I}P_i$, then
+    \begin{enumerate}
+        \item For each $T \in A$, $PTP = \sum_{i \in I}P_iTP_i$. 
+        \item If $\seqi{P}$ are abelian, then $P$ is also abelian.
+        \item If $\seqi{P}$ are finite, then $P$ is also finite. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 26.2]{Zhu}}}. ]
+    (1): For each $i \in I$, $Z(P_i) \ge P_i$, so $Z(P_i)P_i = P_i$. For any $i, j \in I$ with $i \ne j$, $Z(P_i)$ and $Z(P_j)$ are orthogonal, so $Z(P_i)P_j = Z(P_i)Z(P_j)P_j = 0$. 
+
+    Let $T \in A$, then by \autoref{proposition:central-support-vna}, $Z(P_i)(H) \supset TP_i(H)$ for all $i \in I$. Therefore
+    \begin{align*}
+    PTP &= \sum_{i, j \in I}P_iTP_j = \sum_{i, j \in I}Z(P_i)P_i \cdot T \cdot Z(P_j)P_j \\
+    &=  \sum_{i, j \in I}Z(P_i)P_i \cdot Z(P_j) \cdot T \cdot Z(P_j)P_j \\
+    &= \sum_{i \in I}Z(P_i)P_i \cdot T \cdot Z(P_i)P_i = \sum_{i \in I}P_i TP_i 
+    \end{align*}
+
+    (2): Let $S, T \in A$, then by (1),
+    \begin{align*}
+        PSP \cdot PTP &= \sum_{i, j \in I}P_iSP_i \cdot P_jTP_j = \sum_{i \in I}P_iSP_i \cdot P_iTP_i \\
+        &= \sum_{i \in I}P_iTP_i \cdot P_iSP_i = PTP \cdot PSP
+    \end{align*}
+
+    (3): Let $R \in \text{Proj}(A)$ with $P \sim R \le P$, and $V \in A$ with $R = V^*V$ and $P = VV^*$, then for each $i \in I$, $Z(P_i)R \sim Z(P_i)P = P_i$, and $Z(P_i)R \le Z(P_i)P = P_i$. As $\seqi{P}$ are finite, $Z(P_i)R = P_i$ for all $i \in I$. Therefore
+    \[
+    R = RP = R\sum_{i \in I}Z(P_i)P_i = \sum_{i \in I}Z(P_i)RP_i = \sum_{i \in I}P_i = P
+    \]
+\end{proof}
+
+\begin{definition}[Type $\vnI$]
+\label{definition:vna-t1}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is of \textbf{type $\vnI$} if for every non-zero central projection $P \in \text{Proj}(Z(A))$, there exists a non-zero abelian projection $Q \in \text{Proj}(A)$ with $P \ge Q$. 
+\end{definition}
+% Todo: add a few equivalent characterisations. 
+
+\begin{definition}[Type $\vnII$]
+\label{definition:vna-t2}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is of \textbf{type $\vnII$} if:
+    \begin{enumerate}
+        \item $A$ has no non-zero abelian projections. 
+        \item For every non-zero central projection $P \in \text{Proj}(Z(A))$, there exists a non-zero finite projection $Q \in \text{Proj}(A)$ with $P \ge Q$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Type $\vnII_1$]
+\label{definition:vna-t21}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a type $\vnII$ von Neumann algebra, then $A$ is of \textbf{type $\vnII_1$} if $I$ is a finite projection. 
+\end{definition}
+
+\begin{definition}[Type $\vnII_\infty$]
+\label{definition:vna-t2inf}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a type $\vnII$ von Neumann algebra, then $A$ is of \textbf{type $\vnII_\infty$} if $A$ has no non-zero finite central projections. 
+\end{definition}
+
+\begin{definition}[Type $\vnIII$]
+\label{definition:vna-t3}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is of \textbf{type $\vnIII$} if $A$ has no non-zero finite projections. 
+\end{definition}
+
+\begin{theorem}[Type Decomposition]
+\label{theorem:vna-type-decomposition}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then there exist unique von Neumann algebras $A_{\vnI}, A_{\vnII_1}, A_{\vnII_{\infty}}, A_{\vnIII} \subset A$\footnote{Not all four types are guaranteed to be present.} of type $\vnI$, $\vnII_1$, $\vnII_\infty$, and $\vnIII$, respectively, such that
+    \[
+    A = A_{\vnI} \oplus A_{\vnII_1} \oplus A_{\vnII_{\infty}} \oplus A_{\vnIII}
+    \]
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 26.3]{Zhu}}}. ]
+    ($\vnI$): By Zorn's lemma, there exists a maximal family $\seqi{P} \subset \text{Proj}(A)$ of centrally orthogonal abelian projections. Let $P = \sum_{i \in I}P_i$, then $P$ is abelian by (2) of \autoref{lemma:centrally-orthogonal-sum-properties}. 
+
+    Let $P_{\vnI} = Z(P)$, then $A_{\vnI} := P_{\vnI}AP_{\vnI}$ is a von Neumann algebra with identity $P_{\vnI}$. Let $R \in \text{Proj}(Z(A_{\vnI})) \setminus \bracs{0}$, then since $0 < R \le P_{\vnI}$, $RP \le R$ is a non-zero abelian projection. Therefore $A_{\vnI}$ is of type $\vnI$. 
+
+    ($\vnII$): Assume without loss of generality that $I \in A$. Since $\seqi{P}$ is maximal and $(I - P_{\vnI}) \in Z(A)$, $(I - P_{\vnI})A(I - P_{\vnI})$ has no non-zero abelian projections. 
+
+    By Zorn's lemma, there exists a maximal family $\seqj{Q} \subset \text{Proj}((I - P_{\vnI})A(I - P_{\vnI}))$ of centrally orthogonal finite projections. Let $Q = \sum_{j \in J}Q_j$, then $Q$ is finite by (3) of \autoref{lemma:centrally-orthogonal-sum-properties}. 
+
+    Let $P_{\vnII} = Z(Q)$ and $A_{\vnII} = P_{\vnII}AP_{\vnII}$, then $A_{\vnII}$ is a von Neumann algebra with identity $P_{\vnII}$. Let $R \in \text{Proj}(Z(A_{\vnII})) \setminus \bracs{0}$, then since $0 < R \le P_{\vnII}$, $RQ \le R$ is a non-zero finite projection by (4) of \autoref{lemma:projection-types-gymnastics}. Thus $A_{\vnII}$ is of type $\vnII$. 
+
+    ($\vnIII$): Let $P_{\vnIII} = I - P_{\vnI} - P_{\vnII}$ and $A_{\vnIII} = P_{\vnIII}AP_{\vnIII}$. Since $P_{\vnIII} \in Z(A)$ and $\seqi{P}$, $\seqj{Q}$ are maximal, $A_{\vnIII}$ has no non-zero finite projections. Therefore $A_{\vnIII}$ is of type $\vnIII$, and $A = A_{\vnI} \oplus A_{\vnII} \oplus A_{\vnIII}$. 
+
+    ($\vnII_1$): By Zorn's lemma, there exists a maximal family $\bracsn{R_k}_{k \in K} \subset \text{Proj}(A_{\vnII})$ of orthogonal central finite projections. Let $P_{\vnII_1} = \sum_{k \in K}R_k$, then $P_{\vnII_1}$ is a central finite projection by (3) of \autoref{lemma:centrally-orthogonal-sum-properties}. Hence $A_{\vnII_1} = P_{\vnII_1}AP_{\vnII_1}$ is of type $\vnII_1$. 
+
+    ($\vnII_\infty$): Let $P_{\vnII_\infty} = P_{\vnII} - P_{\vnII_1}$ and $A_{\vnII_\infty} = P_{\vnII_\infty}AP_{\vnII_\infty}$, then $P_{\vnII_\infty}$ is a central projection. By maximality of $\bracsn{R_k}_{k \in K}$, $A_{\vnII_\infty}$ admits no non-zero finite central projections. Therefore $A_{\vnII_\infty}$ is of type $\vnII_\infty$, $A_{\vnII} = A_{\vnII_1} \oplus A_{\vnII_\infty}$, and
+    \[
+    A = A_{\vnI} \oplus A_{\vnII_1} \oplus A_{\vnII_{\infty}} \oplus A_{\vnIII}
+    \]
+    
+    (Uniqueness): Let $A = A_{\vnI}' \oplus A_{\vnII_1}' \oplus A_{\vnII_{\infty}}' \oplus A_{\vnIII}'$ be a decomposition of $A$ into von Neumann algebras of type $\vnI$, $\vnII_1$, $\vnII_\infty$, and $\vnIII$, respectively. 
+
+    Let $P_{\vnI}'$, $P_{\vnII_1}'$, $P_{\vnII_\infty}'$, and $P_{\vnIII}'$ be the identity elements of $A_{\vnI}'$, $A_{\vnII_1}'$, $A_{\vnII_{\infty}}'$, and $A_{\vnIII}'$, respectively, then
+    \[
+    I = P_{\vnI}' \oplus P_{\vnII_1}' \oplus P_{\vnII_\infty}' \oplus P_{\vnIII}'
+    \]
+
+    is an orthogonal direct sum, and
+    \begin{enumerate}
+        \item[($\vnI$)] Let $P_1 = P'_{\vnI}(I - P_{\vnI})$, then by construction of $P_{\vnI}$, there exists no non-zero abelian projection $R \in \text{Proj}(A)$ with $R \le P_1$. As both $P_{\vnI}'$ and $(I - P_{\vnI})$ are central, $P_1 \in A_{\vnI}'$, so $P_1 = 0$ because $A_{\vnI}'$ is of type $\vnI$. Thus $P_{\vnI}' \le P_{\vnI}$. By symmetry, $P_{\vnI} = P_{\vnI}'$ and $A_{\vnI} = A_{\vnI}'$.
+        \item[($\vnII$, $\vnIII$)] Let $P'_{\vnII} = P_{\vnII_1}' \oplus P_{\vnII_\infty}'$, $A_{\vnII}' = A_{\vnII_1}' \oplus A_{\vnII_\infty}'$, and $P_2 = P'_{\vnII}(I - P_{\vnI} - P_{\vnII})$. By construction of $P_{\vnII}$, there exists no non-zero finite projection $R \in \text{Proj}(A)$ with $R \le P_2$. Since $P_2 \in A_{\vnII}'$ and $A_{\vnII}'$ is of type $\vnII$, $P_2 = 0$ and $P_{\vnII}' \le P_{\vnII}$. By symmetry, $P_{\vnII} = P_{\vnII}'$. Thus $P_{\vnIII} = P_{\vnIII}'$, $A_{\vnII} = A_{\vnII}'$, and $A_{\vnIII} = A_{\vnIII}'$.
+        \item[($\vnII_1$, $\vnII_\infty$)] Let $Q_2 = P'_{\vnII_1}(P_{\vnII} - P_{\vnII_1})$, then there exists no non-zero finite central projection $R \in \text{Proj}(A)$ with $R \le Q_2$. However, since $A_{\vnII_1}'$ is of type $\vnII_1$, $P'_{\vnII_1}$ is itself a finite projection, and every subprojection of $P'_{\vnII_1}$ is finite by (4) of \autoref{lemma:projection-types-gymnastics}. Thus $Q_2 = 0$ and $P_{\vnII_1}' \le P_{\vnII_1}$. By symmetry, $P_{\vnII_1}' = P_{\vnII_1}$. Therefore $P_{\vnII_\infty}' = P_{\vnII_\infty}$, $A_{\vnII_1}' = A_{\vnII_1}$, and $A_{\vnII_\infty}' = A_{\vnII_\infty}$.
+    \end{enumerate}
+    
+\end{proof}
+
+\begin{lemma}
+\label{lemma:abelian-minimal-factor}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a factor, and $P \in \text{Proj}(A)$ be abelian, then $P$ is minimal. 
+\end{lemma}
+\begin{proof}
+    Let $Q \in \text{Proj}(A)$ with $0 \le Q \le P$, then by the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $Q \preceq P - Q$ or $P - Q \preceq Q$. Assume without loss of generality that $Q \preceq P - Q$. 
+    
+    Let $V \in A$ such that $Q = V^*V$ and $VV^* \le P - Q$, then $V$ is a partial isometry with initial and final spaces contained in $P(H)$. Since $P$ is abelian, $V \in PAP$, and $Q = V^*V = VV^* \le P - Q$. Therefore $Q = 0$, and $P$ is minimal. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:type1-bh-matrix}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a factor, $T \in A$, $\seqi{P} \subset \text{Proj}(A)$ be non-zero minimal projections such that $I = \sum_{i \in I}P_i$, and $\bracsn{V_{i, j}}_{i, j \in I} \subset A$ such that $P_i = V_{i, j}^*V_{i, j}$ and $P_j = V_{i, j}V_{i, j}^*$ for all $i, j \in I$, then there exists $\bracsn{\mu_{i, j}}_{i, j \in I} \subset \complex$ such that $T = \sum_{i, j \in I}\mu_{i, j}V_{i, j}$. 
+\end{lemma}
+\begin{proof}
+    Let $i, j \in I$, then since $P_i$ and $P_j$ are minimal, 
+    \begin{align*}
+        (P_iTP_j)^*(P_iTP_j) &= P_jT^*P_iTP_j \in \complex P_j \\
+        (P_iTP_j)(P_iTP_j)^* &= P_iTP_jT^*P_i \in \complex P_i
+    \end{align*}
+    
+
+    If $P_iTP_j \ne 0$, then $P_jT^*P_iTP_j$ and $P_iTP_jT^*P_i$ are positive, and there exists $\lambda > 0$ such that $\lambda P_iTP_j$ is a partial isometry with initial space $P_j(H)$ and final space $P_i(H)$. By \autoref{lemma:projection-types-gymnastics}, there exists $\mu_{i, j} \in \complex$ such that $P_iTP_j = \mu_{i, j} V_{j, i}$. Therefore
+    \[
+    T = \sum_{i, j \in I}P_iTP_j = \sum_{i, j \in I}\mu_{i, j}V_{j, i}
+    \]
+\end{proof}
+
+
+\begin{theorem}[Classification of Type $\vnI$ Factors]
+\label{theorem:type1-bh}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a factor, then the following are equivalent:
+    \begin{enumerate}
+        \item $A$ is of type $\vnI$. 
+        \item There exists a minimal projection $P \in \text{Proj}(A)$. 
+        \item For every non-zero $P \in \text{Proj}(A)$, there exists a non-zero minimal projection $Q \in \text{Proj}(A)$ such that $P \ge Q$. 
+        \item There exists a complex Hilbert space $K$ and a *-isomorphism $\pi: A \to B(K)$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite{TownesType1}}}. ]
+    (1) $\Rightarrow$ (2): Since $A$ is of type $\vnI$, $A$ admits a non-zero abelian projection $Q \in \text{Proj}(A)$. By \autoref{lemma:abelian-minimal-factor}, $Q$ is minimal. 
+
+    (2) $\Rightarrow$ (3): Let $P \in \text{Proj}(A)$ and $Q \in \text{Proj}(A)$ be a minimal projection. By the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $P \preceq Q$ or $Q \preceq P$. 
+
+    If $P \preceq Q$, then there exists $R \in \text{Proj}(A)$ with $P \sim R \le Q$. In which case, $R$ is minimal, and $P$ is also minimal by (5) of \autoref{lemma:projection-types-gymnastics}. 
+
+    If $Q \preceq P$, then there exists $R \in \text{Proj}(A)$ with $Q \sim R \le P$. By (5) of \autoref{lemma:projection-types-gymnastics}, $R$ is minimal with $R \le P$. 
+
+    (3) $\Rightarrow$ (4): By Zorn's lemma, there exists a maximal orthogonal family $\seqi{P} \subset \text{Proj}(A)$ of minimal projections. Since $\seqi{P}$ is maximal and every non-zero projection admits a non-zero minimal subprojection, $I = \sum_{i \in I}P_i$, and $H = \bigoplus_{i \in I}P_iH$.
+
+    For each $i, j \in I$, by the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $P_i \preceq P_j$ or $P_j \preceq P_i$. In both cases, since both projections are minimal, $P_i \sim P_j$. Thus there exists a partial isometry $V_{i, j} \in A$ with initial space $P_i(H)$ and final space $P_j(H)$ such that $P_i = V_{i, j}^*V_{i, j}$ and $P_j = V_{i, j}V_{i, j}^*$. By fixing a particular family\footnote{The partial isometries need not be unique. }, assume without loss of generality that $V_{i, j}^* = V_{j, i}$ for all $i, j \in I$. 
+
+    Fix $i_0 \in I$, let $H_0 = P_{i_0}H$, and define
+    \[
+    U: H \to l^2(I; H_0) \quad (Ux)_i = V_{i, i_0}P_ix
+    \]
+
+    then since $H = \bigoplus_{i \in I}P_iH$ and each $V_{i, i_0}$ is a partial isometry, $U$ is an isometry with inverse
+    \[
+    U^{-1}: l^2(I; H_0) \to H \quad U^{-1}x = \sum_{i \in I}V_{i_0, i}x_i
+    \]
+
+    For each $i \in I$, denote $e_i = \one_{\bracs{i}} \in l^2(I; \complex)$, then for every $x \in l^2(I; H_0)$, 
+    \[
+    UP_iU^{-1}x = U P_i\sum_{j \in I} V_{i_0, j}x_j = e_{i}V_{i, i_0}P_iV_{i_0, i}x_i = e_iP_{i_0}x_i = e_ix_i
+    \]
+    
+    
+    so $UP_iU^{-1}$ is the projection onto the $i$-th component of $l^2(I; H_0)$. 
+    
+    Let $T \in A$, then by \autoref{lemma:type1-bh-matrix}, there exists $\bracsn{\mu_{i, j}}_{i, j \in I} \subset \complex$ such that $T = \sum_{i, j \in I}\mu_{i,j}V_{i, j}$. In which case, for any $x \in l^2(I; \complex)$ and $v \in H_0$, 
+    \[
+        UTU^{-1}(xv) = UT \sum_{i \in I}V_{i_0, i}x_i v = U\sum_{i, j, k \in I}\mu_{i, j}x_k \cdot V_{i, j}V_{i_0, k} \cdot v
+    \]
+
+    For each $i, j \in I$, there exists $\lambda_{i, j} \in \partial B_\complex(0, 1)$ such that $V_{i, j}V_{i_0, i} = \lambda_{i, j}V_{i_0, j}$ by (6) of \autoref{lemma:projection-types-gymnastics}. For every $i \in I$, let $e_i = \one_{\bracs{i}} \in l^2(I; \complex)$, then
+    \begin{align*}
+        UTU^{-1}(xv) &=U\sum_{i, j, k \in I}\mu_{i, j}x_k \cdot V_{i, j}V_{i_0, k} \cdot v =  U\sum_{i, j \in I}\mu_{i, j}x_i \cdot V_{i, j}V_{i_0, i} \cdot v \\
+        &= U\sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot V_{i_0, j} \cdot v = \sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot e_j \cdot V_{j, i_0}P_jV_{i_0, j} \cdot v \\
+        &= \sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot e_j \cdot v \in l^2(I; \complex v)
+    \end{align*}
+
+    Moreover, if $v \ne 0$, then $UTU^{-1}(xv) = 0$ for all $x \in l^2(I; \complex)$ implies that $\mu_{i, j} =0 $ for all $i, j \in I$, and $T = 0$. Thus for any $v \in H_0 \setminus \bracs{0}$, the mapping
+    \[
+    \pi: A \to B(l^2(I; \complex v)) \quad T \mapsto UTU^{-1}|_{l^2(I; \complex v)}
+    \]
+
+    is an injective $*$-homomorphism. 
+
+    Finally, since $\ol{B_A(0, 1)}$ is weak-operator compact and $U$ is an isometry, $\pi(\ol{B_A(0, 1)})$ is also weak-operator compact, and $\pi(A) \subset B(l^2(I; \complex v))$ is a von Neumann algebra. Let $i, j \in I$, then for each $x \in l^2(I; \complex)$, 
+    \begin{align*}
+    \pi(V_{i, j})(xv) &= x_i \cdot e_j \cdot V_{j, i_0}V_{i, j}V_{i_0, i} \cdot v = \lambda_{i, j}x_i \cdot e_j \cdot V_{j, i_0}V_{i_0, j} \cdot v \\
+    &= \lambda_{i, j}x_i \cdot e_j \cdot v
+    \end{align*}
+    
+
+    so $\pi(V_{i, j}) = \lambda_{i, j} e_jv \otimes e_iv$. As $\bracsn{e_iv}_{i \in I}$ is an orthonormal basis for $l^2(I; \complex v)$, $\pi(A) = B(l^2(I; \complex v))$. 
+
+    (4) $\Rightarrow$ (1): Identify $A = \pi(A) = B(K)$, and let $T \in Z(A)$. For each $v \in K$ with $\norm{v}_K = 1$, $T(v \otimes v) = (v \otimes v)T$, so $Tv = T(v \otimes v)v = (v \otimes v)Tv$, and there exists $\lambda \in \complex$ such that $Tv = \lambda v$. 
+    
+    For any $w \in K$ linearly independent from $v$, there exists $\mu \in \complex$ with $Tw = \mu w$, and $\rho \in \complex$ with $T(v + w) = \rho(v + w)$. In which case, $\rho v + \rho w = \lambda v + \mu w$, so $\lambda = \rho = \mu$, and $T = \lambda I$. Therefore $A$ is a factor. 
+
+    For any $v \in K$ with $\norm{v}_K = 1$, $v \otimes v$ is a minimal, and hence abelian projection by (1) of \autoref{lemma:projection-types-gymnastics}. As $v \otimes v \le I$, $A = B(K)$ is of type $\vnI$. 
+\end{proof}
+
+

src/op/vn/vn.tex

@@ -6,7 +6,7 @@
     Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a strong-operator closed $C^*$-subalgebra, then:
     \begin{enumerate}
         \item For any bounded directed family $\cf \subset A_{sa}$, $\sup(\cf) = \sotlim_{T \in \cf}T \in A_{sa}$. 
-        \item For any family of projections $\mathcal{P} \subset A_{sa}$, $\sup(\mathcal{P}) \in A$ is the projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$. 
+        \item For any directed family of projections $\mathcal{P} \subset A_{sa}$, $\sup(\mathcal{P}) \in A$ is the projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$. 
     \end{enumerate}
 
     and
@@ -46,7 +46,7 @@
 
     for all $S \in \cf$ with $S \ge T$. As such a $T$ exists for all $\eps > 0$, $R = \sotlim_{T \in \cf}T$. 
 
-    (2): Assume without loss of generality that $\mathcal{P}$ is directed. By (1), $\sup(\mathcal{P})$ exists in $A$. Since the set of projections in $B(H)$ is strong-operator closed, $\sup(\mathcal{P}) = \sotlim_{P \in \mathcal{P}}P$ is also a projection. For each $x \in \bigcup_{P \in \mathcal{P}}P(H)$, there exists $P \in \mathcal{P}$ with $Px = x$. In which case, 
+    (2): By (1), $\sup(\mathcal{P})$ exists in $A$. Since the set of projections in $B(H)$ is strong-operator closed, $\sup(\mathcal{P}) = \sotlim_{P \in \mathcal{P}}P$ is also a projection. For each $x \in \bigcup_{P \in \mathcal{P}}P(H)$, there exists $P \in \mathcal{P}$ with $Px = x$. In which case, 
     \[
     \dpn{\sup(\mathcal{P})x, x}{H} \ge \dpn{Px, x}{H} = \dpn{x, x}{H} = \norm{x}_H^2
     \]
@@ -89,7 +89,7 @@
 
     By (3) applied to $TT^*/\norm{TT^*}_{B(H)}$, the orthogonal projection onto $\ol{T(H)}$ is in $A$. 
 
-    (5): Let $\mathcal{P}$ be the set of all projections in $A$, then $\mathcal{P} \subset A_{sa}$ is bounded and directed. By (2), $P = \sup_{Q \in \mathcal{P}}Q \in A$, which is the maximum projection in $A$. 
+    (5): Let $\mathcal{P}$ be the set of all projections in $A$, and $\cf \subset 2^{\mathcal{P}}$ be the collection of all finite subsets of $\mathcal{P}$. For each $F \in \cf$, let $P_F$ be the projection onto $\braks{\sum_{P \in F}P}(H)$, then $P_F \ge P$ for all $P \in F$ and $P_F \in A$ by (4). Since $\bracsn{P_F}_{F \in \cf}$ is a bounded and directed family of projections, $\sup_{F \in \cf}P_F \in A$ by (1). As $\sup_{F \in \cf}P_F \in \mathcal{P}$, it is the maximum projection in $A$. 
 
     Let $T \in A$, then by (4), $P$ is greater than the projection onto $\ol{T(H)}$, so $PT = T$. On the other hand, since $PT^* = T^*$, $TP = T$ as well. Therefore $P$ is the multiplicative identity in $A$. 
 \end{proof}
@@ -141,6 +141,20 @@
     Now, since $T \in A''$, $TP = PT$ as well, so $M$ is a reducing subspace for $T$. As $A$ is unital, $x \in M$, so $Tx \in M = \ol{\bracsn{Sx|S \in A}}$. 
 \end{proof}
 
+\begin{lemma}[Amplification]
+\label{lemma:bh-amplification}
+    Let $H$ be a complex Hilbert space, $n \in \natp$, and
+    \[
+    \pi: B(H) \to B(H^n) \quad [\pi(T)(x)]_n = Tx_n
+    \]
+
+    then for each $\seqf{x_j} \subset H$, 
+    \[
+    \max_{1 \le j \le n}\norm{Tx_j}_H \le \norm{\pi(T)(x)}_{H^n} \le n \max_{1 \le j \le n}\norm{Tx_j}_H
+    \]
+\end{lemma}
+
+
 \begin{theorem}[Von Neumann's Bicommutant Theorem]
 \label{theorem:bicommutant}
     Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a unital, self-adjoint subalgebra, then $A''$ is the strong-operator closure of $A$. 
@@ -162,3 +176,82 @@
     so $T \in \ol{A}^{\text{\small SOT}}$. 
 \end{proof}
 
+\begin{definition}[Von Neumann Algebra]
+\label{definition:von-neumann-algebra}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a \textbf{von Neumann algebra acting on $H$} if $A$ is closed in the strong operator topology. 
+\end{definition}
+
+\begin{definition}[Factor]
+\label{definition:vna-factor}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is a \textbf{factor} if $Z(A) = \complex I$. 
+\end{definition}
+
+
+\begin{theorem}[Kaplansky Density Theorem]
+\label{theorem:kaplansky-density}
+    Let $H$ be a Hilbert space, $A \subset B(H)$ be a $C^*$-subalgebra, and $B$ be the strong-operator closure of $A$, then:
+    \begin{enumerate}
+        \item $\ol{B_{A_{sa}}(0, 1)}$ is strong-operator dense in $\ol{B_{B_{sa}}(0, 1)}$.
+        \item $\bracsn{T \in \ol{B_{A}(0, 1)}|T \ge 0}$ is strong-operator dense in $\bracsn{T \in \ol{B_{B}(0, 1)}|T \ge 0}$.
+        \item $\ol{B_{A}(0, 1)}$ is strong-operator dense in $\ol{B_{B}(0, 1)}$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 19.5]{Zhu}}}. ]
+    (1): Let $T \in \ol{B_{B_{sa}}(0, 1)}$ and $\angles{T_\gamma}_{\gamma \in C} \subset A$ be a net such that $T_\gamma \to T$ in the weak operator topology. For each $\gamma \in C$, let $T_\gamma' = (T_\gamma + T_\gamma^*)/2$, then $T_\gamma' \to T$ in the weak operator topology by continuity of the adjoint map in the weak operator topology. 
+
+    As $A_{sa}$ is a subspace of $B(H)$, its strong and weak-operator closures coincide. Thus there exists a net $\angles{S_\gamma}_{\gamma \in C} \subset A_{sa}$ such that $S_\gamma \to T$ in the strong operator topology. In which case, let
+    \[
+    f: \real \to \real \quad t \mapsto \begin{cases}
+        t &t \in [-1, 1] \\
+        1/t &t \in \real \setminus [-1, 1]
+    \end{cases}
+    \]
+
+    then $f \in C_0(\real; \real)$. By \autoref{corollary:functional-calculus-c0-self-adjoint}, the mapping $S \mapsto f(S)$ is strong-operator continuous. As $\norm{T}_{B(H)} \le 1$, $\sigma_{B}(T) \subset [-1, 1]$. Thus $f(T) = T$, and $f(S_\gamma) \to T$ in the strong operator topology. By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $f(S_\gamma)$ is in the closed unit ball of $A_{sa}$ for all $\gamma \in C$. Therefore the closed unit ball of $A_{sa}$ is strong-operator dense in the closed unit ball of $B_{sa}$. 
+
+    (2): Let $T \in \ol{B_{B}(0, 1)}$ with $T \ge 0$ and $\angles{T_\gamma}_{\gamma \in C} \subset A_{sa}$ be a net such that $T_\gamma \to T$ in the strong operator topology. Define
+    \[
+    f: \real \to \real \quad t \mapsto \begin{cases}
+        0 &t \le 0 \\
+        t &t \in [0, 1] \\
+        1/t &t \ge 1
+    \end{cases}
+    \]
+
+    then $f \in C_0(\real; [0, \infty))$. Since $f \ge 0$, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} then implies that $\norm{f(T_\gamma)}_{B(H)} \le 1$ and $f(T_\gamma) \ge 0$ for all $\gamma \in C$. By \autoref{corollary:functional-calculus-c0-self-adjoint}, the mapping $S \mapsto f(S)$ is strong-operator continuous. As $\norm{T}_{B(H)} \le 1$ and $T \ge 0$, $\sigma_{B}(T) \subset [0, 1]$ by \autoref{proposition:positive-spectrum}. Thus $f(T) = T$, and $f(T_\gamma) \to T$ in the strong operator topology. 
+
+    (3): For each $\mathcal{T} \subset B(H)$, let
+    \[
+    M_2(\mathcal{T}) = \bracs{\begin{bmatrix} Q & R \\ S & T \end{bmatrix} \bigg | Q, R, S, T \in \mathcal{T}}
+    \]
+
+    then $M_2(B)$ is the strong-operator closure of $M_2(A)$ in $B(H^2)$. For each $T \in \ol{B_{B}(0, 1)}$, let
+    \[
+    T' = \begin{bmatrix} 0 & T \\ T^* & 0 \end{bmatrix}
+    \]
+
+    then $T' \in \ol{B_{M_2(B)_{sa}}(0, 1)}$. By (1), there exists a net $\angles{(R_\gamma, S_\gamma, T_\gamma)}_{\gamma \in C} \subset A^3$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $\gamma \in C$, 
+        \[
+        \norm{\begin{bmatrix} R_\gamma & T_\gamma \\ T^*_\gamma & S_\gamma \end{bmatrix}}_{B(H^2)} \le 1
+        \]
+
+        In particular, $\norm{T_\gamma}_{B(H)} \le 1$.
+        \item With respect to the strong operator topology on $B(H^2)$,
+        \[
+        \begin{bmatrix} R_\gamma & T_\gamma \\ T^*_\gamma & S_\gamma \end{bmatrix} \to T'
+        \]
+    \end{enumerate}
+
+    Therefore $\angles{T_\gamma} \subset \ol{B_A(0, 1)}$ is a net that converges to $T$ in the strong-operator topology. 
+    
+\end{proof}
+
+
+\begin{remark}
+\label{remark:kaplansky-unitary}
+    The Kaplansky Density Theorem should also apply to the unitary case. Unfortunately, it seems like that the Borel functional calculus is required for an easier proof, so it will be postponed for now. 
+\end{remark}
+
+

src/topology/main/baire.tex

@@ -16,7 +16,7 @@
 \begin{proof}
     $(1) \Rightarrow (2)$: Let $\seq{A_n} \subset 2^X$ be closed with empty interior, then $\seq{A_n}$ are nowhere dense. Hence $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
 
-    $(2) \Rightarrow (3)$: For each $n \in \natp$, let $A_n = U_n^c$, then $A_n$ is closed. For any $\emptyset U \subset A_n$ open, $U \cap U_n \ne \emptyset$ by density of $U_n$, so $A_n$ has empty interior.
+    $(2) \Rightarrow (3)$: For each $n \in \natp$, let $A_n = U_n^c$, then $A_n$ is closed. For any $\emptyset \ne U \subset A_n$ open, $U \cap U_n \ne \emptyset$ by density of $U_n$, so $A_n$ has empty interior.
 
     Suppose that $\bigcap_{n \in \natp}U_n$ is not dense, then there exists $\emptyset \ne V \subset X$ open such that $\bigcup_{n \in \natp}A_n \supset V$, which contradicts the fact that $\bigcup_{n \in \natp}A_n$ has non-empty interior.
 

src/topology/main/cube.tex

@@ -69,7 +69,7 @@
     \mathcal{S} = \bracsn{(E, F) \in \mathcal{B}^2 | \ol{E} \subset F}
     \]
 
-    By \hyperref[Urysohn's Lemma]{lemma:urysohn}, for each $(E, F) \in \mathcal{S}$, there exists $f_{EF} \in C(X; [0, 1])$ such that $f|_E = 1$ and $f|_{F^c} = 0$. For any $x \in X$ and $U \in \cn^o_X(x)$, there exists $E, F \in \mathcal{B}$ such that $x \in E \subset \ol{E} \subset F \subset U$. Thus $f_{EF}(x) = 1$ and $f_{EF}|_{U^c} = 0$. Therefore
+    By \hyperref[Urysohn's Lemma]{lemma:urysohn}, for each $(E, F) \in \mathcal{S}$, there exists $f_{EF} \in C(X; [0, 1])$ such that $f_{EF}|_E = 1$ and $f_{EF}|_{F^c} = 0$. For any $x \in X$ and $U \in \cn^o_X(x)$, there exists $E, F \in \mathcal{B}$ such that $x \in E \subset \ol{E} \subset F \subset U$. Thus $f_{EF}(x) = 1$ and $f_{EF}|_{U^c} = 0$. Therefore
     \[
     \cf = \bracsn{f_{EF}|(E, F) \in \mathcal{S}} \subset C(X; [0, 1])
     \]

src/topology/notation.tex

@@ -26,6 +26,14 @@
     $f \prec U$ & $f \in C_c(X; [0,1])$ with $\mathrm{supp}(f) \subset U$. & \autoref{definition:compactly-supported-01} \\
     $C_0(X; E)$ & Continuous functions vanishing at infinity. & \autoref{definition:vanish-at-infinity} \\
     $BC(X; E)$ & Bounded continuous functions $X \to E$. & \autoref{definition:bounded-continuous-function-space} \\
+    % ---- $L^p$ Spaces ----
+    $B^\infty(X; E)$ & Bounded $E$-valued strongly measurable functions on $X$. & \autoref{definition:bounded-borel-function} \\
+    $B^\infty(X)$ & Bounded $\complex$-valued Borel measurable functions on $X$. & \autoref{definition:bounded-borel-function} \\
+    $\mathcal{L}^p(X; E)$, $\mathcal{L}^p(\mu; E)$, $\mathcal{L}^p(X, \cm, \mu; E)$ & $E$-valued $p$-integrable functions on $X$. & \autoref{definition:lp-unequivalence} \\
+    $\norm{f}_{L^p(X; E)}$ & $L^p$ norm of $f$: $\braks{\int \norm{f}_E^p d\mu}^{1/p}$. & \autoref{definition:lp-unequivalence} \\
+    $\mathcal{L}^\infty(X; E)$, $\mathcal{L}^\infty(\mu; E)$, $\mathcal{L}^\infty(X, \cm, \mu; E)$ & $E$-valued essentially bounded functions on $X$. & \autoref{definition:esssup} \\
+    $\norm{f}_{\mathcal{L}^\infty(X; E)}$ & Essential supremum of $f$. & \autoref{definition:esssup} \\
+    $L^p(X, \cm, \mu; E)$ & $E$-valued $L^p$ space on $(X,\cm,\mu)$; quotient of $\mathcal{L}^p$ by a.e.-equality. & \autoref{definition:lp} \\
 
     % DST 
     $\mathscr{N}$ & The Baire space. & \autoref{definition:the-baire-space} \\