Fixed typo.

preamble.sty

@@ -170,10 +170,6 @@
 \newcommand{\calr}{\mathcal{R}}
 \newcommand{\scp}{\mathscr{P}}
 
-% Jokes
-\newcommand{\lol}{\boxed{\text{LOL.}}}
-\newcommand{\ez}{\boxed{\mathbb{EZ}}}
-
 % Colours
 \newcommand{\pblue}[1]{\textcolor[rgb]{0, 0.44, 0.75}{#1}}
 \newcommand{\poran}[1]{\textcolor{orange}{#1}}
@@ -230,3 +226,16 @@
 \newcommand{\conv}{\text{Conv}}
 \newcommand{\aconv}{\text{AbsConv}}
 
+% Limits
+\newcommand{\slim}{\operatorname*{s\text{-}\!\lim}}
+\newcommand{\wlim}{\operatorname*{w\text{-}\!\lim}}
+\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

@@ -257,4 +257,39 @@
   address   = {Upper Saddle River, NJ},
   year      = {2000},
   isbn      = {0-13-181629-2}
+}
+
+@book{RyanTensor,
+  author    = {Ryan, Raymond A.},
+  title     = {Introduction to Tensor Products of Banach Spaces},
+  series    = {Springer Monographs in Mathematics},
+  publisher = {Springer},
+  address   = {London},
+  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}
 }

spec.toml

@@ -10,7 +10,7 @@ indirectReferences = true
 [website]
 font = "roboto"
 fontSize = 16
-lineHeight = 1.3
+lineHeight = 1.5
 textAlign = "left"
 lineWidth = 45
 primaryColour = "violet"

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/duality/adjoint.tex

@@ -0,0 +1,75 @@
+\section{Adjoint Maps}
+\label{section:adjoint-maps}
+
+\begin{definition}[Adjoint Map]
+\label{definition:adjoint-map}
+    Let $E, F$ be vector spaces over a field $K$, and $T \in \hom(E; F)$ be a linear map, then the mapping
+    \[
+    T^*: F^* \to E^* \quad \dpn{x, T^*\phi}{E} = \dpn{Tx, \phi}{F}
+    \]
+
+    is the \textbf{algebraic adjoint} of $T$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:adjoint-weak-continuous}
+    Let $\dpn{E, F}{\lambda}$ and $\dpn{G, H}{\mu}$ be dualities over $K \in \RC$ and $T \in \hom(E; G)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $T$ is $\sigma(E, F)$-$\sigma(G, H)$ continuous. 
+        \item $T^*(H) \subset F$. 
+    \end{enumerate}
+
+    If the above holds, then
+    \begin{enumerate}[start=2]
+        \item $T^*|_{H}$ is $\sigma(H, G)$-$\sigma(F, E)$ continuous. 
+        \item $T^{**} = T$. 
+    \end{enumerate}
+
+    and the restriction of $T^*$ to $H$ is the \textbf{adjoint} of $T$ with respect to $\dpn{E, F}{\lambda}$ and $\dpn{G, H}{\mu}$. 
+\end{proposition}
+% Proof omitted due to obviousness. 
+
+\begin{proposition}
+\label{proposition:adjoint-polar-gymnastics}
+    Let $\dpn{E, F}{\lambda}$ and $\dpn{G, H}{\mu}$ be dualities over $K \in \RC$, $T: E \to G$ be a $\sigma(E, F)$-$\sigma(G, H)$ continuous linear map, $A \subset E$, and $B \subset G$, then:
+    \begin{enumerate}
+        \item $T(A)^\circ = (T^{*})^{-1}(A^\circ)$.
+        \item If $T(A) \subset B$, then $T^{*}(B^\circ) \subset A^\circ$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): 
+    \begin{align*}
+        T(A)^\circ &= \bracsn{\phi \in H| \text{Re}\dpn{Tx, \phi}{\mu} \le 1 \forall x \in A} \\
+        &= \bracsn{\phi \in H| \text{Re}\dpn{x, T^*\phi}{\lambda} \le 1 \forall x \in A} = (T^{*})^{-1}(A^\circ)
+    \end{align*}
+
+    (2): 
+    \begin{align*}
+        T^*(B^\circ) &= T^*(\bracs{\phi \in H| \text{Re}\dpn{y, \phi}{\mu} \le 1 \forall y \in B}) \\
+        &\subset T^*(\bracs{\phi \in H| \text{Re}\dpn{y, \phi}{\mu} \le 1 \forall y \in T(A)}) \\
+        &= T^*(\bracs{\phi \in H| \text{Re}\dpn{Tx, \phi}{\mu} \le 1 \forall x \in A})\\
+        &= T^*(\bracs{\phi \in H| \text{Re}\dpn{x, T^*\phi}{\lambda} \le 1 \forall x \in A}) \subset A^\circ
+    \end{align*}
+\end{proof}
+
+\begin{corollary}
+\label{corollary:adjoint-kernel-gymnastics}
+    Let $\dpn{E, F}{\lambda}$ and $\dpn{G, H}{\mu}$ be dualities over $K \in \RC$ and $T: E \to G$ be a $\sigma(E, F)$-$\sigma(G, H)$ continuous linear map, then:
+    \begin{enumerate}
+        \item $\ker(T^*) = T(E)^\perp = \bracs{\phi \in H| \dpn{y, \phi}{\mu} = 0 \forall y \in T(E)}$. 
+        \item $T^*$ is injective if and only if $T(E)$ is $\sigma(G, H)$-dense in $G$. 
+    \end{enumerate}
+\end{corollary}
+
+\begin{proposition}
+\label{proposition:adjoint-continuity}
+    Let $\dpn{E, F}{\lambda}$ and $\dpn{G, H}{\mu}$ be dualities over $K \in \RC$, $T: E \to G$ be a $\sigma(E, F)$-$\sigma(G, H)$ continuous linear map, $\sigma \subset 2^E$ be a saturated ideal of $\sigma(E, F)$-bounded sets, $\tau \subset 2^G$ be a saturated ideal of $\sigma(G, H)$-bounded sets, then the following are equivalent:
+    \begin{enumerate}
+        \item $T^*$ is continuous with respect to the $\tau$-uniform topology on $H$ and the $\sigma$-uniform topology on $F$. 
+        \item $T(\sigma) \subset \tau$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $A \in \sigma$, then there exists $B \in \tau$ such that $T^*\phi(A) \subset \ol{B_K(0, 1)}$ for all $\phi \in H$ with $\phi(B) \subset \ol{B_K(0, 1)}$. In which case, $T^*(B^\circ) \subset A^\circ$.  Assume without loss of generality that $A$ and $B$ are convex, circled, and closed. By \autoref{proposition:adjoint-polar-gymnastics} applied to $T^*$ and the \hyperref[Bipolar theorem]{theorem:bipolar}, $T(A) \subset B$. Therefore $T(\sigma) \subset \tau$. 
+\end{proof}

src/fa/duality/definitions.tex

@@ -78,4 +78,38 @@
     (3) $\Rightarrow$ (1): Let $x \in E$ such that $\dpn{x, y}{\lambda} = 0$ for all $y \in F_0$, then since $F_0$ is $\sigma(F, E)$-dense in $F$, $\dpn{x, y}{\lambda} = 0$ for all $y \in F$. Hence $x = 0$. 
 \end{proof}
 
+\begin{theorem}[Goldstine]
+\label{theorem:goldstine-weak}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, $A \subset E$ be non-empty, convex, circled, and $\sigma(E, F)$-compact, and $B$ be the closed unit ball of $E_A^*$, then $B \cap F$ is $\sigma(E_A^*, E_A)$-dense in $B$. 
+\end{theorem}
+\begin{proof}
+    For any $S \subset E$ or $S \subset F$, denote $S^\circ$ as the polar of $S$ with respect to $\dpn{E, F}{\lambda}$. For any $S \subset E_A$ or $S \subset E_A^*$, denote $S^\bullet$ as the polar of $S$ with respect to $\dpn{E_A, E_A^*}{E_A}$. 
+
+
+    Since $B_{E_A}(0, 1)$ is circled and $A = \ol{B_{E_A}(0, 1)}^{E_A}$ is compact in $E$, 
+    \[
+    B \cap F = \bracsn{\phi \in F| \text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall x \in A} = A^\circ
+    \]
+
+    is the polar of $A$ with respect to $\dpn{E, F}{\lambda}$. Now, as $A$ is convex, circled, and compact, the \hyperref[Bipolar Theorem]{theorem:bipolar} implies that
+    \begin{align*}
+    A^{\circ\bullet} &= \bracsn{x \in E_A|\text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall \phi \in A^\circ} \\
+    &= E_A \cap \bracsn{x \in E|\text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall \phi \in A^\circ} = E_A \cap A^{\circ\circ} = A
+    \end{align*}
+
+    Given that $B \cap F$ is a convex and circled subset of $E_A^*$, 
+    \[
+    \ol{B \cap F}^{\sigma(E_A^*, E_A)} = (B \cap F)^{\bullet\bullet} = A^{\circ\bullet\bullet} = A^\bullet = B
+    \]
+\end{proof}
+
+\begin{corollary}
+\label{corollary:weak-dense-unit-ball}
+    Let $E$ be a normed space over $K \in \RC$, then $E \cap \ol{B_{E^{**}}(0, 1)}$ is $\sigma(E^{**}, E^*)$-dense in $\ol{B_{E^{**}}(0, 1)}$. 
+\end{corollary}
+\begin{proof}
+    By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{E^*}(0, 1)}$ is convex, circled, and $\sigma(E^*, E)$-compact. By \hyperref[Goldstine's Theorem]{theorem:goldstine-weak}, $E \cap \ol{B_{E^{**}}(0, 1)}$ is $\sigma(E^{**}, E^*)$-dense in $\ol{B_{E^{**}}(0, 1)}$. 
+\end{proof}
+
+
 

src/fa/duality/index.tex

@@ -4,5 +4,6 @@
 \input{./definitions.tex}
 \input{./polar.tex}
 \input{./mackey.tex}
+\input{./adjoint.tex}
 
 

src/fa/duality/mackey.tex

@@ -66,6 +66,22 @@
     On the other hand, let $\mathcal{T} \subset 2^E$ be a locally convex topology consistent with $\dpn{E, F}{\lambda}$. By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, every $\mathcal{T}$-equicontinuous set is relatively $\sigma(F, E)$-compact. Therefore $\mathcal{T}$ is coarser than the topology of uniform convergence on relatively $\sigma(E, F)$-compact, convex, and circled sets. 
 \end{proof}
 
+\begin{corollary}
+\label{corollary:mackey-bounded}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, $\topo$ be a topology on $E$ consistent with $\dpn{E, F}{\lambda}$, and $B \subset E$, then $B$ is bounded with respect to $\topo$ if and only if $B$ is bounded with respect to $\sigma(E, F)$. 
+\end{corollary}
+\begin{proof}
+    Assume without loss of generality that $\topo = \tau(E, F)$. Suppose that $B$ is $\sigma(E, F)$-bounded. Let $A \subset F$ be convex, circled, and $\sigma(F, E)$-compact. By continuity of dual pairing, 
+    \[
+    A = \bracsn{\phi \in F|\ |\dpn{x, \phi}{\lambda}| \le 1 \forall x \in B}
+    \]
+
+    is a convex, circled, and closed subset. Given that $B$ is $\sigma(E, F)$-bounded, $\sup_{x \in B}|\dpn{x, \phi}{\lambda}| < \infty$ for all $x \in E$. Thus $A$ is absorbing and hence a barrel. 
+
+    Since $B$ is compact, the auxiliary space $E_B$ is a Banach space. In particular, $E_B$ is barrelled by \autoref{proposition:baire-barrel}. By continuity of the inclusion map, $A \cap E_B$ is a barrel in $E_B$. Therefore there exists $\lambda > 0$ such that $B \subset \lambda A \cap E_B \subset \lambda A$.
+\end{proof}
+
+
 \begin{definition}[Mackey Space]
 \label{definition:mackey-space}
     Let $E$ be a separated locally convex space over $K \in \RC$, then $E$ is a \textbf{Mackey space} if $E$ is equipped with the Mackey topology of $\dpn{E, E^*}{E}$. 
@@ -73,10 +89,10 @@
 
 \begin{proposition}
 \label{proposition:barreled-mackey}
-    Let $E$ be a separated barreled space over $K \in \RC$, then $E$ is a Mackey space. 
+    Let $E$ be a separated barrelled space over $K \in \RC$, then $E$ is a Mackey space. 
 \end{proposition}
 \begin{proof}
-    Let $\cf \subset E^*$ be a $\sigma(E^*, E)$-compact set and $U \in \cn_{K}(0)$ be a barrel, then $V = \bigcap_{\phi \in \cf}\phi^{-1}(U)$ is convex, circled, and closed. For each $x \in E$, $\cf(x) = \bracs{\dpn{x, \phi}{E}|\phi \in \cf}$ is bounded. Thus $V$ is absorbing and hence a barrel. Since $E$ is barreled, $V \in \cn_E(0)$. Therefore the Mackey topology is contained in the topology of $E$, and $E$ is a Mackey space. 
+    Let $\cf \subset E^*$ be a $\sigma(E^*, E)$-compact set and $U \in \cn_{K}(0)$ be a barrel, then $V = \bigcap_{\phi \in \cf}\phi^{-1}(U)$ is convex, circled, and closed. For each $x \in E$, $\cf(x) = \bracs{\dpn{x, \phi}{E}|\phi \in \cf}$ is bounded. Thus $V$ is absorbing and hence a barrel. Since $E$ is barrelled, $V \in \cn_E(0)$. Therefore the Mackey topology is contained in the topology of $E$, and $E$ is a Mackey space. 
 \end{proof}
 
 

src/fa/lc/barrel.tex

@@ -1,4 +1,4 @@
-\section{Barreled Spaces}
+\section{Barrelled Spaces}
 \label{section:barrel}
 
 \begin{definition}[Barrel]
@@ -6,7 +6,7 @@
     Let $E$ be a TVS over $K \in \RC$ and $D \subset E$, then $D$ is a \textbf{barrel} if it is convex, circled, radial, and closed. 
 \end{definition}
 
-\begin{definition}[Barreled Space]
+\begin{definition}[Barrelled Space]
 \label{definition:barreled-space}
     Let $E$ be a locally convex space over $K \in \RC$, then the following are equivalent:
     \begin{enumerate}

src/fa/lc/beqtensor.tex

@@ -0,0 +1,101 @@
+\section{The Injective Tensor Product}
+\label{section:beq-tensor-product}
+
+\begin{lemma}
+\label{lemma:tensor-product-dual-injection}
+    Let $E, F$ be locally convex spaces over $K \in \RC$, then the canonical map
+    \[
+    E \otimes F \to L^2(E^*, F^*; K) \quad (x \otimes y)(\phi, \psi) = \dpn{x, \phi}{E}\dpn{y, \psi}{F}
+    \]
+
+    is injective. 
+\end{lemma}
+\begin{proof}
+    Let $\lambda = \sum_{j = 1}^n x_j \otimes y_j \in E \otimes F$ such that $\lambda(\phi, \psi) = 0$ for all $\phi \in E^*$ and $\psi \in F^*$. Assume without loss of generality that $\bracsn{x_j}_1^n \subset E$ is a linearly independent set. Fix $\phi \in E^*$, then for every $\psi \in F^*$, 
+    \[
+        0 = \lambda(\phi, \psi) = \sum_{j = 1}^n \dpn{x_j, \phi}{E} \dpn{y_j, \psi}{F} = \angles{\sum_{j = 1}^n x_j\dpn{y_j, \psi}{F}, \phi}_E
+    \]
+    
+    By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $\sum_{j = 1}^n x_j\dpn{y_j, \psi}{F} = 0$. Since $\bracs{x_j}_1^n \subset E$ is linearly independent, $\dpn{y_j, \psi}{F} = 0$ for each $1 \le j \le n$. 
+
+    As the above holds for all $\psi \in F^*$, the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility} implies that $y_j = 0$ for each $1 \le j \le n$. 
+\end{proof}
+
+\begin{definition}[Bi-Equicontinuous Convergence]
+\label{definition:beq}
+    Let $E, F$ be locally convex spaces over $K \in \RC$ and
+    \[
+    \sigma = \bracsn{S \times T| S \subset E^* \text{ equipcontinuous}, T \subset F^* \text{ equicontinuous}}
+    \]
+
+    be the product of all equicontinuous subsets of $E^*$ and $F^*$, then the $\sigma$-topology on $L^2(E, F; K)$ is the \textbf{topology of bi-equicontinuous convergence} on $L^2(E, F; K)$. Under this topology, $L^2(E, F; K)$ is a locally convex space. 
+\end{definition}
+\begin{proof}
+    By \autoref{proposition:lc-spaces-linear-map}, the $\sigma$-topology is a vector space topology. 
+\end{proof}
+
+
+
+\begin{definition}[Injective Tensor Product]
+\label{definition:beq-tensor-product}
+    Let $E, F$ be locally convex spaces over $K \in \RC$, and identify $E \otimes F$ as a subspace of $L^2(E, F; K)$, then $E \otimes F$ equipped with the topology of bi-equicontinuous convergence is the \textbf{injective tensor product} of $E$ and $F$, denoted $E \otimes_\eps F$. 
+    
+    The Hausdorff completion $E \wh{\otimes}_\eps F$ of $E \otimes_\eps F$ is the \textbf{injective completion} of $E$ and $F$. 
+\end{definition}
+
+\begin{definition}[Injective Cross Seminorm]
+\label{definition:beq-cross-norm}
+    Let $E, F$ be locally convex spaces over $K \in \RC$, $S \subset E^*$ and $T \subset F^*$ be equicontinuous, and $\lambda = \sum_{j = 1}^n x_j \otimes y_j \in E \otimes F$, then
+    \[
+    [\lambda]_{S, T} = \braks{\sum_{j = 1}^n x_j \otimes y_j}_{S, T} = \sup_{\phi \in S, \psi \in T} \abs{\sum_{j = 1}^n \dpn{x_j, \phi}{E}\dpn{y_j, \psi}{F}}
+    \]
+
+    is the \textbf{injective cross seminorm} of $\lambda$ with respect to $S$ and $T$. The family of all such norms induces the topology on $E \otimes_\eps F$. In particular, if $E, F$ are normed vector spaces, then
+    \[
+    \norm{\lambda}_{E \otimes_\eps F} = \norm{\sum_{j = 1}^n x_j \otimes y_j} = \sup_{\substack{\phi \in E^* \\ \norm{\phi}_{E^*} \le 1}}\sup_{\substack{\psi \in E \\ \norm{\psi}_{F^*} \le 1}}\abs{\sum_{j = 1}^n \dpn{x_j, \phi}{E}\dpn{y_j, \psi}{F}}
+    \]
+
+    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

@@ -1,6 +1,27 @@
 \section{Compact Convex Sets}
 \label{section:compact-convex}
 
+
+\begin{theorem}[Mazur]
+\label{theorem:convex-hull-complete}
+    Let $E$ be a locally convex space over $K \in \RC$ and $A \subset E$ be compact, then
+    \begin{enumerate}
+        \item $\conv(A)$ is totally bounded in $E$. 
+        \item If $E$ is complete, then $\conv(A)$ is relatively compact. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $U \in \cn_E(0)$ be convex and circled, then since $A$ is compact, there exists $B \subset A$ finite such that $A \subset B + U$. As such, $\conv(A) \subset \conv(B + U) = \conv(B) + U$. Since $B$ is finite, $\conv(B)$ is compact. Thus there exists $C \subset \conv(B)$ finite such that
+    \[
+    \conv(A) \subset \conv(B + U) = \conv(B) + U \subset C + U
+    \]
+
+    which yields a finite covering of $\conv(A)$ using $U$. 
+
+    (2): By \autoref{proposition:compact-uniform}. 
+\end{proof}
+
+
 \begin{definition}[Extreme Point]
 \label{definition:extreme-point}
     Let $E$ be a vector space over $\real$, $K \subset E$, and $x \in K$, then $x$ is \textbf{extremal} if there exists no $y, z \in K$ such that $x \in (y, z) \subset K$. 
@@ -12,7 +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 $K$. 
@@ -83,3 +103,43 @@
     
 \end{proof}
 
+\begin{lemma}
+\label{lemma:compact-null-auxiliary}
+    Let $E$ be a Banach space over $K \in \RC$ and $A \subset E$ be compact, then:
+    \begin{enumerate}
+        \item There exists a null sequence $\seq{x_n} \subset E$ such that $A \subset \ol{\conv}(\seq{x_n})$.
+        \item There exists a compact, convex, and circled set $B \subset E$ such that $A$ is compact in $E_B$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma III.9.1]{SchaeferWolff}}}. ]
+    (1): Assume without loss of generality that $A \ne \emptyset$. Let $\bracsn{\lambda_n}_0^\infty \subset (0, \infty)$ such that $\sum_{n \in \natz}\lambda_n = 1$. For each $n \in \natz$, let $r_n = \lambda_{n+1}^2$ and $A_n \subset A$ be finite such that $A \subset \bigcup_{x \in A_n}B_E(x, r_n)$. 
+
+    Define $B_0 = \lambda_0^{-1}A_0$. For each $n \in \natp$, write $A_n = \bracsn{x_j}_1^k$. By definition of $A_{n-1}$, there exists $\bracsn{y_j}_1^k \subset A_{n-1}$ such that $d(x_j, y_j) < r_{n-1}$ for all $1 \le j \le k$. For every $1 \le j \le k$, let $z_j = (x_j - y_j)/\lambda_n$, and define $B_n = \bracsn{z_j}_1^k$. 
+
+    By the above construction, $A_n \subset \sum_{j = 0}^n \lambda_j B_j$ for all $n \in \natz$. As $\sum_{n \in \natz}\lambda_n = 1$, 
+    \[
+    A \subset \ol{\conv}\braks{\bigcup_{n \in \natz}A_n} \subset \ol{\conv}\braks{\bracs{0} \cup \bigcup_{n \in \natz}B_n}
+    \]
+
+    Finally, for each $n \in \natp$, $B_n$ is finite with $\norm{z}_E \le r_{n-1}/\lambda_{n} = \lambda_n$ for all $z \in B_n$. As $B_0$ is finite as well, any enumeration of $\bracs{0} \cup \bigcup_{n \in \natz}B_n$ yields a null sequence. 
+
+    (2): Using (1) and \hyperref[Mazur's Theorem]{theorem:convex-hull-complete}, assume without loss of generality that there exists a null sequence $\seq{x_n} \subset E$ such that $A$ is the closed convex hull of $\seq{x_n}$. 
+
+    Since $\seq{x_n} \subset E$ is a null sequence, there exists $\seq{\lambda_n} \subset [1, \infty)$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\lambda_n \to \infty$ as $n \to \infty$. 
+        \item $\lambda_n x_n \to 0$ as $n \to \infty$. 
+    \end{enumerate}
+
+    Let $B = \ol{\aconv}(\seq{\lambda_n x_n})$, then by \hyperref[Mazur's Theorem]{theorem:convex-hull-complete}, $B$ is a compact, convex, and circled subset of $E$ with $\seq{x_n} \subset B$. In addition, $\seq{x_n} \subset E_B$ with $\norm{x_n}_{E_B} \le \lambda_n^{-1}$ for all $n \in \natp$. Thus $\seq{x_n}$ is a null sequence in $E_B$ as well. 
+
+    Now, let $A'$ be the closed convex hull of $\seq{x_n}$ with respect to $E_B$. Since the inclusion $E_B \to E$ is continuous, $A'$ is a compact convex set in $E$ by \autoref{proposition:compact-extensions}. As such, $A' = A$ by \autoref{proposition:closure-of-image} and \autoref{proposition:compact-closed}. Therefore $A$ is a compact subset of $E_B$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:auxiliary-weak-dense}
+    Let $E$ be a separated locally convex space over $K \in \RC$, $A \subset E$ be compact, convex, and circled, and $
+\end{lemma}
+
+
+

src/fa/lc/continuous.tex

@@ -20,6 +20,7 @@
     $(4) \Rightarrow (3)$: Let $U \in \cn_F(0)$ be convex, circled, and radial, then its gauge $[\cdot]_U$ is a continuous seminorm on $F$ by \autoref{definition:locally-convex}. Thus there exists a continuous seminorm $[\cdot]_E$ such that $[Tx]_U \le [x]_E$. In which case, $V = \bracs{x \in E| [x]_E < 1} \in \cn_E(0)$ with $T(V) \subset U$. Therefore $T$ is continuous at $0$, and continuous by \autoref{definition:continuous-linear}.
 \end{proof}
 
+
 \begin{proposition}
 \label{proposition:tvs-convex-multilinear}
     Let $\seqf{E_j}$ and $F$ be locally convex spaces, and $T: \prod_{j = 1}^n E_j \to F$ be $n$-linear map, then the following are equivalent:

src/fa/lc/convex.tex

@@ -135,7 +135,7 @@
 \label{definition:seminorm-topology}
     Let $E$ be a vector space over $K \in \RC$ and $\seqi{[\cdot]}$ be seminorms, then:
     \begin{enumerate}
-        \item For each $i \in I$, $d_i: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_i$ is a pseudo-metric.
+        \item For each $i \in I$, $d_i: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_i$ is a pseudometric.
         \item The topology induced by $\seqi{d}$ makes $E$ a topological vector space.
         \item For each $i \in I$, $[\cdot]_i: E \to [0, \infty)$ is continuous.
     \end{enumerate}

src/fa/lc/index.tex

@@ -13,5 +13,6 @@
 \input{./hahn-banach.tex}
 \input{./spaces-of-linear.tex}
 \input{./tensor.tex}
+\input{./beqtensor.tex}
 \input{./nuclear.tex}
 \input{./nuclear-space.tex}

src/fa/lc/inductive.tex

@@ -17,7 +17,7 @@
         is a fundamental system of neighbourhoods for $E$ at $0$.
         \item If $E$ is spanned by $\bigcup_{i \in I}T_i(E_i)$, then
         \[
-        \fB = \bracs{\Gamma\paren{\bigcup_{i \in I}T_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
+        \fB = \bracs{\aconv\paren{\bigcup_{i \in I}T_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
         \]
 
         is a fundamental system of neighbourhoods for $E$ at $0$. 
@@ -42,7 +42,7 @@
 
     (6): If $E$ is spanned by $\bigcup_{i \in I}T_i(E_i)$, then each set in $\fB$ is radial. Hence $\fB$ is a family of neighbourhoods of $E$ at $0$. 
 
-    Let $U \in \cn_E(0)$ be convex, circled, and radial, then for each $i \in I$, $T_i^{-1}(U) \in \cn_{E_i}(0)$, so $U \supset \bigcup_{i \in I}T_i[T_i^{-1}(U)]$. Since $U$ is convex and circled, $U \supset \Gamma\paren{\bigcup_{i \in I}T_i[T_i^{-1}(U)]} \in \fB$. Therefore $\fB$ forms a fundamental system of neighbourhoods for $E$ at $0$.
+    Let $U \in \cn_E(0)$ be convex, circled, and radial, then for each $i \in I$, $T_i^{-1}(U) \in \cn_{E_i}(0)$, so $U \supset \bigcup_{i \in I}T_i[T_i^{-1}(U)]$. Since $U$ is convex and circled, $U \supset \aconv\paren{\bigcup_{i \in I}T_i[T_i^{-1}(U)]} \in \fB$. Therefore $\fB$ forms a fundamental system of neighbourhoods for $E$ at $0$.
 \end{proof}
 
 \begin{definition}[Locally Convex Direct Sum]
@@ -61,7 +61,7 @@
         
         \item The family
         \[
-        \fB = \bracs{\Gamma\paren{\bigcup_{i \in I}\iota_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
+        \fB = \bracs{\aconv\paren{\bigcup_{i \in I}\iota_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
         \]
 
         is a fundamental system of neighbourhoods for $E$ at $0$. 

src/fa/lc/nuclear-space.tex

@@ -54,7 +54,7 @@
 
 \begin{theorem}
 \label{theorem:nuclear-lp}
-    Let $E$ be a nuclear space over $K \in \RC$, $U \in \cn_E(0)$, and $p \in [1, \infty]$, then there exists $V \in \cn_E(0)$ with $V \subset U$ such that $\wh E_V$ is isomorphic to a subspace of $l^p(\natp; K)$ with equal norms.  
+    Let $E$ be a nuclear space over $K \in \RC$, $U \in \cn_E(0)$, and $p \in [1, \infty]$, then there exists $V \in \cn_E(0)$ with $V \subset U$ such that $\wh E_V$ is isometrically isomorphic to a subspace of $l^p(\natp; K)$.
 \end{theorem}
 \begin{proof}[Proof, {{\cite[III.7.3]{SchaeferWolff}}}. ]
     Assume without loss of generality that $U$ is convex and circled, and the canonical projection $\pi_U: E \to \wh E_U$ is nuclear. In which case, there exists an equicontinuous sequence $\seq{\phi_n} \subset E^*$, $\seq{y_n} \subset B_{\wh E_U}(0, 1)$, and $\seq{\lambda_n} \subset K$ such that
@@ -81,5 +81,161 @@
     Finally, let $V = T^{-1}(B_{l^p(\natp; K)})$, then $V \subset U$, and $\wh E_V$ is isomorphic to $\ol{T(E)}$, with equal norms. 
 \end{proof}
 
+\begin{corollary}
+\label{corollary:complete-nuclear-projective-limit}
+    Let $E$ be a complete nuclear space over $K \in \RC$, then $E$ is a projective limit of Hilbert spaces over $K$. For any Fréchet space $F$, $F$ is nuclear if and only if it is the projective limit of a sequence $\seq{H_n}$ of Hilbert spaces such that the mapping $H_m \to H_n$ is nuclear for all $1 \le m < n < \infty$. 
+\end{corollary}
 
 
+\begin{summary}
+\label{summary:nuclear-extension}
+    Every subspace and separated qoutient space of a nuclear space is nuclear. The product of nuclear spaces is nuclear. The locally convex direct sum of countably many nuclear spaces is nuclear. 
+\end{summary}
+\begin{proof}
+    See \autoref{proposition:nuclear-quotient}, \autoref{proposition:nuclear-subspace}, \autoref{proposition:nuclear-direct-sum}, and \autoref{proposition:nuclear-product}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:nuclear-subspace}
+    Let $E$ be a nuclear space over $K \in \RC$ and $F \subset E$ be a subspace, then $F$ is also nuclear.
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem III.7.4]{SchaeferWolff}}}. ]
+    Firstly, a setup about auxiliary spaces and subspaces is required. Let $U \in \cn_E(0)$ be convex and circled, then the composition of the inclusion map $\iota: F \to E$ and the canonical projection $\pi_U: E \to E_U$ factors through $F_{U \cap F}$ as follows:  
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_U} & E_U \\
+    F \ar@{->}[u]^{\iota} \ar@{->}[r]_{\pi_{U \cap F}} & F_{U \cap F} \ar@{->}[u]_{\widehat \pi_U}
+    }
+    \]
+
+    where $\widehat \pi_U$ is an isometric embedding. As a result, the factored map $\widehat \pi_U: F_{U \cap F} \to E_U$ extends to an isometric embedding on the completions: 
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_U} & E_U \ar@{->}[r] & \widehat E_{U} \\
+    F \ar@{->}[u]^{\iota} \ar@{->}[r]_{\pi_{U \cap F}} & F_{U \cap F} \ar@{->}[u]_{\widehat \pi_U} \ar@{->}[r] & \widehat F_{U \cap F} \ar@{->}[u]_{\widehat \pi_U}
+    }
+    \]
+
+    which enables identifying $\widehat F_{U \cap F}$ as a closed subspace of $\widehat E_{U}$.
+
+    To start the proof, let $U \in \cn_E(0)$ be a given convex and circled neighbourhood. Since $E$ is nuclear, there exists a convex and circled neighbourhood $V \in \cn_E(0)$ with $V \subset U$ such that the induced map $\widehat \pi_U: \widehat E_V \to \widehat E_U$ is nuclear. By prior discussion, the following diagram commutes:
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_V} & \widehat E_V \ar@{->}[r]^{\widehat \pi_{U}} & \widehat E_U \\
+    F \ar@{->}[u] \ar@{->}[r] & \widehat F_{V \cap F} \ar@{->}[u] \ar@{->}[r]_{\widehat \pi_{U \cap F}} & \widehat F_{U \cap F} \ar@{->}[u]
+    }
+    \]
+
+    Thus the induced map $\widehat \pi_{U \cap F}: \widehat F_{V \cap F} \to \widehat F_{U \cap F}$ corresponds to the restriction of $\widehat \pi_U$ to $\widehat F_{V \cap F}$. Since $\widehat \pi_U$ is nuclear, there exists $\seq{\phi_n} \subset E_V^*$ and $\seq{y_n} \subset \widehat E_U$ such that
+    \[
+    \widehat \pi_U x = \sum_{n = 1}^\infty y_n\dpn{x, \phi_n}{\widehat E_V} \quad \forall x \in \widehat E_V
+    \]
+
+    and $\sum_{n \in \natp} \norm{y_n}_{\widehat E_U}\norm{\phi_n}_{E_V^*} < \infty$. 
+
+    Now, using \autoref{theorem:nuclear-lp}, further assume without loss of generality that $\widehat E_{U}$ is a Hilbert space. Let $P: \widehat E_{U} \to \widehat F_{U \cap F}$ be the orthogonal projection of $\widehat E_U$ onto $\widehat F_{U \cap F}$, then
+    \[
+    \widehat \pi_{U \cap F}x = \sum_{n = 1}^\infty Py_n \dpn{x, \phi_n}{\widehat F_{V \cap F}} \quad \forall x \in \widehat F_{V \cap F}
+    \]
+
+    with
+    \[
+    \normn{\widehat \pi_{U \cap F}}_{N(\widehat F_{V \cap F}; \widehat F_{U \cap F})}
+    \le \sum_{n \in \natp} \norm{Py_n}_{\widehat F_{U \cap F}} \norm{\phi_n}_{F_{V \cap F}^*} \le \sum_{n \in \natp} \norm{y_n}_{\widehat E_U}\norm{\phi_n}_{E_V^*} < \infty
+    \]
+
+    Therefore the induced map $\widehat \pi_{U \cap F}$ is nuclear, and $F$ is a nuclear space.
+\end{proof}
+
+
+
+
+\begin{proposition}
+\label{proposition:nuclear-quotient}
+    Let $E$ be a nuclear space over $K \in \RC$, and $F$ be a closed subspace of $E$, then $E/F$ is also nuclear. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem III.7.4]{SchaeferWolff}}}. ]
+    Firstly, a setup about auxiliary spaces and quotients is required. Let $p: E \to E/F$ be the canonical projection and $U \in \cn_E(0)$ be a convex and circled neighbourhood, then the composition of maps $E \to E/F \to (E/F)_{p(U)}$ factors through $E_{U}$ as follows: 
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_U} \ar@{->}[d]_{p} & E_U \ar@{->}[d] \\
+    E/F \ar@{->}[r]_{\pi_{p(U)}} & (E/F)_{p(U)}
+    }
+    \]
+
+    This extends through the completion
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_U} \ar@{->}[d]_{p} & E_U \ar@{->}[d] \ar@{->}[r] & \widehat E_U \ar@{->}[d] \\
+    E/F \ar@{->}[r]_{\pi_{p(U)}} & (E/F)_{p(U)} \ar@{->}[r] & \widehat{(E/F)}_{p(U)}
+    }
+    \]
+
+    and yields that $\widehat{(E/F)}_{p(U)}$ is a quotient space of $\widehat E_{U}$. 
+
+    To begin the proof, let $U \in \cn_E(0)$ be a convex and circled neighbourhood. Since $E$ is nuclear, there exists a convex and circled neighbourhood $V \in \cn_E(0)$ with $V \subset U$ such that the induced map $\widehat \pi_{U}: \widehat E_V \to \widehat E_{U}$ is nuclear. The composition of maps $\wh E_V \to \wh E_U \to \wh{(E/F)}_{p(U)}$ then factors through $\widehat{(E/F)}_{p(V)}$ as $\widehat \pi_{p(U)}$:
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_V} \ar@{->}[d]_{p} & \widehat E_V \ar@{->}[d] \ar@{->}[r]^{\widehat \pi_U} & \widehat E_U \ar@{->}[d]^{\widehat p} \\
+    E/F \ar@{->}[r]_{\pi_{p(V)}} & \widehat{(E/F)}_{p(V)} \ar@{->}[r]_{\widehat \pi_{p(U)}} & \widehat{(E/F)}_{p(U)}
+    }
+    \]
+
+    Since $\wh \pi_U: \wh E_V \to \wh E_U$ is nuclear, there exists $\seq{\phi_n} \subset E_V^*$ and $\seq{y_n} \subset \wh E_U$ such that
+    \[
+    \wh \pi_U x = \sum_{n = 1}^\infty y_n \dpn{x, \phi_n}{\wh E_V} \quad \forall x \in \wh E_V
+    \]
+
+    and $\sum_{n \in \natp}\norm{y_n}_{\wh E_U}\norm{\phi_n}_{E_V^*} < \infty$. 
+
+    Now, using \autoref{theorem:nuclear-lp}, further assume without loss of generality that $\wh E_V$ is a Hilbert space. Identify $(\widehat{E/F})_{p(V)}$ as a closed subspace of $\widehat E_V$, and let $P: \widehat E_V \to (\widehat{E/F})_{p(V)}$ be the orthogonal projection of $\widehat E_V$ onto $(\widehat{E/F})_{p(V)}$. This allows rewriting
+    \[
+        \widehat \pi_{p(U)}x = \sum_{n = 1}^\infty \widehat p(y_n) \dpn{Px, \phi_n}{\wh E_V} = \sum_{n = 1}^\infty \widehat p(y_n) \dpn{x, P\phi_n}{(\widehat{E/F})_{p(V)}}
+    \]
+
+    where
+    \begin{align*}
+    \normn{\widehat \pi_{p(U)}}_{N((\widehat{E/F})_{p(V)}; (\widehat{E/F})_{p(U)})} &\le \sum_{n \in \natp}\normn{\widehat p(y_n)}_{(\widehat{E/F})_{p(U)}}\norm{P\phi_n}_{(\widehat{E/F})_{p(V)}} \\
+    &\le \sum_{n \in \natp}\norm{y_n}_{\wh E_U}\norm{\phi_n}_{E_V^*} < \infty
+    \end{align*}
+
+    Therefore $\widehat \pi_{p(U)}$ is nuclear, and $E/F$ is a nuclear space. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:nuclear-direct-sum}
+    Let $\seq{E_n}$ be nuclear spaces over $K \in \RC$, then $\bigoplus_{n = 1}^\infty E_n$ is also nuclear. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem III.7.4]{SchaeferWolff}}}. ]
+    For each $n \in \natp$, identify $E_n$ as a subspace of $\bigoplus_{n = 1}^\infty E_n$. Let $F$ be a Banach space and $T \in L(\bigoplus_{n = 1}^\infty E_n; F)$. For each $n \in \natp$, $E_n$ is a nuclear space, so $T|_{E_n}: E_n \to F$ is a nuclear operator, and there exists $\bracsn{\phi_{n, k}}_{k = 1}^\infty \subset E_n^*$ equicontinuous, $\bracsn{y_{n, k}}_{k = 1}^\infty \subset B_F(0, 1)$, and $\bracsn{\lambda_{n, k}}_{k = 1}^\infty \subset K$ such that $\sum_{k \in \natp}|\lambda_{n, k}| \le 2^{-n}$ and
+    \[
+    Tx = \sum_{k = 1}^\infty \lambda_{n, k}y_{n, k} \dpn{x, \phi_{n, k}}{E_n}
+    \]
+
+    for all $x \in E_n$. Thus for any $x \in \bigoplus_{n = 1}^\infty E_n$, 
+    \begin{align*}
+    Tx &= \sum_{n = 1}^\infty \sum_{k = 1}^\infty \lambda_{n, k}y_{n, k}\dpn{x_n, \phi_{n, k}}{E_n} \\
+    &= \sum_{n = 1}^\infty \sum_{k = 1}^\infty \lambda_{n, k}y_{n, k}\dpn{x, \phi_{n, k} \circ \pi_n}{\bigoplus_{n = 1}^\infty E_n}
+    \end{align*}
+    
+    where $\sum_{n \in \natp}\sum_{k \in \natp}|\lambda_{n, k}| \le \sum_{n \in \natp}2^{-n} < \infty$ and $\bracsn{y_{n, k}|n, k \in \natp} \subset B_F(0, 1)$. 
+
+    Finally, for each $n \in \natp$, let $U_n = \bigcap_{k \in \natp}\phi_{n, k}^{-1}(B_K(0, 1))$, then $U_n \in \cn_{E_n}(0)$ by equicontinuity of $\bracsn{\phi_{n, k}}_{k = 1}^\infty \subset E_n^*$. Let $U = \aconv(\bigcup_{n \in \natp}U_n)$, then $U \in \cn_{\bigoplus_{n = 1}^\infty E_n}(0)$ and $U \subset \bigcap_{n \in \natp}\bigcap_{k \in\natp}(\phi_{n, k} \circ \pi_n)^{-1}(B_K(0, 1))$. Hence $\bracsn{\phi_{n, k} \circ \pi_n|n, k \in \natp}$ is equicontinuous, and $T$ is a nuclear operator. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:nuclear-product}
+    Let $\seqi{E}$ be nuclear spaces over $K \in \RC$, then $\prod_{i \in I}E_i$ is nuclear. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem III.7.4]{SchaeferWolff}}}. ]
+    Let $F$ be a Banach space and $T \in L(\prod_{i \in I}E_i; F)$, then there exists $J \subset I$ finite and $\wh T \in L(\prod_{j \in J}E_j; F)$ such that the following diagram commutes: 
+    \[
+    \xymatrix{
+    \prod_{i \in I} E_i \ar@{->}[r]^{T} \ar@{->}[d]_{\pi_J} & F \\
+    \prod_{j \in J}E_j \ar@{->}[ru]_{\widehat T} & 
+    }
+    \]
+
+    By \autoref{proposition:finite-lc-product}, $\prod_{j \in J}E_j = \bigoplus_{j \in J}E_j$. By \autoref{proposition:nuclear-direct-sum}, $\bigoplus_{j \in J}E_j$ is a nuclear space, so $\widehat T: \bigoplus_{j \in J}E_j \to F$ is a nuclear operator. As the composition of a continuous operator and a nuclear operator, $T$ is nuclear by \autoref{proposition:nuclear-gymnastics}. Therefore $\prod_{i \in I}E_i$ is a nuclear space. 
+\end{proof}
+

src/fa/lc/projective.tex

@@ -23,7 +23,7 @@
     By (U) of \autoref{definition:tvs-projective-limit} and \autoref{definition:tvs-initial}, $E$ is equipped with the projective topology generated by the projection maps $E \to E_i$. By \autoref{proposition:lc-projective-topology}, $E$ is locally convex.
 \end{proof}
 
-\begin{proposition}[{{\cite[II.5.4]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:complete-lc-projective-limit}
     Let $E$ be a separated complete locally convex space over $K \in \RC$, $\mathcal{B} \subset \cn_E(0)$ be a fundamental system of neighbourhoods consisting of convex, circled, and radial sets, directed under inclusion.
 
@@ -44,7 +44,7 @@
 
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[II.5.4]{SchaeferWolff}}}]
     (1): Since $V \supset U$, $[\cdot]_V \ge [\cdot]_U$, so $M_V \supset M_U$. Thus $\ker(\pi_V) \supset M_U$. By (U) of the \hyperref[quotient]{definition:tvs-quotient}, $\pi_V$ factors through $E_U$ as $\pi^U_V$, so $\pi^U_V \in L(E_U; E_V)$.
 
     (2): Since $\mathcal{B}$ is a fundamental system of neighbourhoods, it is downward-directed under inclusion. For any $U, V, W \in \mathcal{B}$ with $U \subset V \subset W$, $M_U \supset M_V \supset M_W$. Thus $\pi^U_W = \pi^V_W \circ \pi^U_V$.

src/fa/lc/tensor.tex

@@ -26,9 +26,9 @@
     \end{enumerate}
 
     
-    The space $E \otimes_\pi F$ is the \textbf{projective tensor product} of $E$ and $F$, and the mapping $\iota \in L^2(E, F; E \otimes_\pi F)$ is the \textbf{canonical embedding}.
+    The space $E \otimes_\pi F$ is the \textbf{projective tensor product} of $E$ and $F$, and the mapping $\iota \in L^2(E, F; E \otimes_\pi F)$ is the canonical embedding.
 
-    The space $E \widetilde{\otimes}_\pi F$ denotes the Hausdorff completion of $E \otimes_\pi F$.
+    The space $E \wh{\otimes}_\pi F$ denotes the Hausdorff completion of $E \otimes_\pi F$.
 \end{definition}
 \begin{proof}
     Let $E \otimes_\pi F = E \otimes F$ be the \hyperref[tensor product]{definition:tensor-product} of $E$ and $F$ as vector spaces. Let $\mathscr{T} \subset 2^{2^X}$ be the collection of all locally convex topologies satisfying (1) and (2), and let $\mathcal{S}$ be the projective topology on $E \otimes_\pi F$ generated by $\mathscr{T}$. 
@@ -53,6 +53,32 @@
     In constructing the \hyperref[projective tensor product]{definition:projective-tensor-product}, it may be more natural to obtain its topology as a projective topology using its universal property. However, doing so requires taking a least upper bound across \textit{all continuous linear maps defined on} $E \times F$, a collection too big to be a set. As such, constructing it as a projective topology is logically dubious, or at the very least beyond my abilities. 
 \end{remark}
 
+\begin{proposition}
+\label{proposition:projective-tensor-product-dual}
+    Let $E, F$ be locally convex space over $K \in \RC$, then
+    \[
+    (E \wh \otimes_\pi F)^* = (E \otimes_\pi F)^* \iso L^2(E, F; K) \iso L(E; F^*) \iso L(F; E^*)
+    \]
+
+    where:
+    \begin{enumerate}
+        \item The dual pairing between $E \otimes_\pi F$ and $L^2(E, F; K)$ is given by
+        \[
+        \angles{\sum_{k = 1}^n x_k \otimes y_k, \lambda}_{E \otimes_\pi F} = \sum_{k = 1}^n \lambda(x_k, y_k)
+        \]
+        \item The dual pairing between $E \otimes_\pi F$ and $L(E; F^*)$ is given by
+        \[
+        \angles{\sum_{k = 1}^n x_k \otimes y_k, T}_{E \otimes_\pi F} = \sum_{k = 1}^n \dpn{y_k, Tx_k}{F}
+        \]
+        \item The dual pairing between $E \otimes_\pi F$ and $L(F; E^*)$ is given by
+        \[
+        \angles{\sum_{k = 1}^n x_k \otimes y_k, T}_{E \otimes_\pi F} = \sum_{k = 1}^n \dpn{x_k, Ty_k}{E}
+        \]
+    \end{enumerate}
+\end{proposition}
+% Proof omitted because the hardest part of this result is writing it down. 
+
+
 
 
 \begin{definition}[Cross Seminorm]

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]
@@ -132,7 +140,7 @@
 
 \begin{theorem}[{{\cite[III.6.5]{SchaeferWolff}}}]
 \label{theorem:l1-tensor}
-    Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $K \in \RC$, then the map $L^1(X; K) \td{\otimes}_\mu E \to L^1(X; E)$ defined by extending
+    Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $K \in \RC$, then the map $L^1(X; K) \td{\otimes}_\pi E \to L^1(X; E)$ defined by extending
     \[
     L^1(X; K) \times E \to L^1(X; E) \quad f \otimes x \mapsto x \cdot f
     \]

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. 
+        \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/lp/seq.tex

@@ -1,6 +1,22 @@
-\section{$l^p$ Direct Sums}
+\section{Sequence Spaces}
 \label{section:lp-direct-sum}
 
+\begin{definition}[$c_0$-Direct Sum]
+\label{definition:c0-direct-sum}
+    Let $\seqi{X}$ be normed vector spaces over $K \in \RC$. For any $x \in \prod_{i \in I}X_i$, $x$ \textbf{vanishes at infinity} if for each $\eps > 0$, $\bracs{i \in I| \norm{x_i}_{X_i} \ge \eps}$ is finite. The space
+    \[
+    [c_0(I); X_i] = \bracs{x \in \prod_{i \in I}X_i \bigg | x \text{ vanishes at infinity}}
+    \]
+
+    equipped with the uniform norm
+    \[
+    \norm{x}_{[c_0(I); X_i]} = \sup_{i \in I}\norm{x_i}_{X_i}
+    \]
+
+    is the \textbf{$c_0$-direct sum} of $\seqi{X}$.
+\end{definition}
+
+
 \begin{definition}[$l^p$-Direct Sum]
 \label{definition:lp-direct-sum}
     Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$, then the \textbf{$l^p$-direct sum} of $\seqi{X}$ is the space
@@ -63,6 +79,42 @@
     \]
 \end{proof}
 
+
+\begin{theorem}
+\label{theorem:c0-sum-dual}
+    Let $\seqi{X}$ be normed vector spaces over $K \in \RC$. For each $y \in [l^1(I); X_i^*]$, let
+    \[
+    \phi_y: [c_0(I); X_i] \to K \quad x \mapsto \sum_{i \in I}\dpn{x_i, y_i}{X_i}
+    \]
+
+    then the mapping
+    \[
+    [l^1(I); X_i^*] \to [c_0(I); X_i]^* \quad y \mapsto \phi_y
+    \]
+
+    is an isometric isomorphism. 
+\end{theorem}
+\begin{proof}
+    By \hyperref[Hölder's Inequality]{proposition:lp-direct-sum-gymnastics}, for each $y \in [l^1(I); X_i^*]$, $\norm{\phi_y}_{[c_0(I); X_i]^*} \le \norm{y}_{[l^1(I); X_i^*]}$. 
+
+    Let $\phi \in [c_0(I); X_i]^*$, then there exists $y \in [l^\infty(I); X_i^*]$ such that for each $i \in I$ and $x_i \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[c_0(I); X_i]} = \dpn{x_i, y_i}{X_i}$. 
+
+    Let $J \subset I$ be finite and $\alpha \in (0, 1)$, then there exists $x \in [c_0(I); X_i]$ such that
+    \begin{enumerate}
+        \item $\{i \in I|x_i \ne 0\} \subset J$. 
+        \item $\norm{x}_{[c_0(I); X_i]} \le 1$. 
+        \item For each $j \in J$, $\dpn{x_j, y_j}{X_j} \ge \alpha\norm{y_j}_{X_j^*}$. 
+    \end{enumerate}
+
+    Thus
+    \[
+    \alpha\sum_{j \in J}\norm{y_j}_{X_j^*} \le \sum_{j \in J}\dpn{x_j, y_j}{X_j} = \dpn{x, \phi}{[c_0(I); X_i]} \le \norm{\phi}_{[c_0(I); X_i]^*}
+    \]
+
+    As the above holds for all $\alpha \in (0, 1)$ and $J \subset I$ finite, $y \in [l^1(I); X_i^*]$ with $\norm{y}_{[l^1(I); X_i^*]} \le \norm{\phi}_{[c_0(I); X_i]^*}$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $\dpn{x, \phi_y}{[c_0(I); X_i]} = \dpn{x, \phi}{[c_0(I); X_i]}$ for all $x \in [c_0(I); X_i]$. Hence the map is an isometric isomorphism.  
+\end{proof}
+
+
 \begin{theorem}
 \label{theorem:lp-sum-dual}
     Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$ and $q \in (1, \infty]$ be Hölder conjugates. For each $y \in [l^q(I); X_i^*]$, let
@@ -78,7 +130,7 @@
     is an isometric isomorphism. 
 \end{theorem}
 \begin{proof}
-    Let $\phi \in [l^p(I); X_i]^*$, then there exists $y \in [l^\infty(I); X_i^*]$ such that for each $i \in I$ and $x \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[l^p(I); X_i]} = \dpn{x_i, y_i}{X_i}$. 
+    Let $\phi \in [l^p(I); X_i]^*$, then there exists $y \in [l^\infty(I); X_i^*]$ such that for each $i \in I$ and $x_i \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[l^p(I); X_i]} = \dpn{x_i, y_i}{X_i}$. 
 
     Since the $q = \infty$ case has been ruled out, assume that $q \in (1, \infty)$. For each $\alpha \in (0, 1)$, there exists $x \in [l^\infty(I); X_i]$ with $\norm{x_i}_{X_i} \le 1$ and $\dpn{x_i, y_i}{X_i} \ge \alpha \norm{y_i}_{X_i^*}$. For each $J \subset I$ finite and $i \in I$, let $F_J(i) = \one_{J}(i) \cdot \norm{y_i}_{X_i^*}^{q - 1}$, then by \autoref{lemma:holder-conjugate-gymnastics},
     \[

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/ap.tex

@@ -0,0 +1,246 @@
+\section{The Approximation Property}
+\label{section:approximation-property}
+
+\begin{definition}[Approximation Property]
+\label{definition:approximation-property}
+    Let $E$ be a separated locally convex space over $K \in \RC$, then the following are equivalent:
+    \begin{enumerate}
+        \item The closure of $E^* \otimes E$ in $L_c(E; E)$ contains the identity map.
+        \item $E^* \otimes E$ is dense in $L_c(E; E)$. 
+        \item For each locally convex space $F$ over $K$, $E^* \otimes F$ is dense in $L_c(E; F)$. 
+        \item For each locally convex space $F$ over $K$, $F^* \otimes E$ is dense in $L_c(F; E)$. 
+    \end{enumerate}
+
+    If the above holds, then $E$ has the \textbf{approximation property}. 
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $T \in L_c(E; E)$ and $A \subset E$ be precompact, then $T(A)$ is also precompact by \autoref{proposition:totally-bounded-image}. Let $U \in \cn_E(0)$, then there exists $S \in E^* \otimes E$ such that $Sx - x \in U$ for all $x \in T(A)$. In which case, $STx - Tx \in U$ for all $x \in A$. 
+
+    (1) $\Rightarrow$ (3): Let $T \in L_c(E; F)$ and $A \subset E$ be precompact, and $U \in \cn_F(0)$, then there exists $S \in E^* \otimes E$ such that $Sx - x \in T^{-1}(U)$ for all $x \in A$. In which case, $TS \in E^* \otimes F$ and $TSx - Tx \in U$ for all $x \in A$. 
+
+    (1) $\Rightarrow$ (4): Let $T \in L_c(F; E)$ and $A \subset F$ be precompact, then $T(A)$ is also precompact. Let $U \in \cn_E(0)$, then there exists $S \in E^* \otimes E$ such that $Sx - x \in U$ for all $x \in T(A)$. Thus $STx - Tx \in U$ for all $x \in A$.  
+\end{proof}
+
+\begin{proposition}
+\label{proposition:approximation-property-associated}
+    Let $E$ be a locally convex space over $K \in \RC$. If there exists a fundamental system of convex and circled neighbourhoods $\fB \subset \cn_E(0)$ such that for each $V \in \fB$, $\wh E_V$ has the approximation property, then $E$ has the approximation property. 
+\end{proposition}
+\begin{proof}
+    Let $V \in \fB$, $\pi_V: E \to \wh E_V$ be the canonical projection, and $A \subset E$ be precompact, then $\pi_V(A)$ is precompact as well. Since $\wh E_V$ has the approximation property, there exists $T \in E_V^* \otimes \wh E_V$ such that $Tx - x \in \pi_V(V)$ for all $x \in \pi_V(A)$. As $E_V$ is dense in $\wh E_V$, there exists $S \in E_V^* \otimes E_V$ such that $Sx - Tx \in \pi_V(V)$ for all $x \in \pi_V(A)$. In which case, $Sx - x \in 2\pi_V(V)$ for all $x \in \pi_V(A)$, and $S \circ \pi_V(x) - \pi_V(x) \in 2\pi_V(V)$. 
+    
+    Write $S = \sum_{j = 1}^n \phi_j \otimes y_j$. For each $1 \le j \le n$, choose any representative $x_j \in \pi_V^{-1}(y_j)$, then for any $x \in A$, 
+    \[
+        \pi_V \braks{x - \sum_{j = 1}^n x_j\dpn{x, \phi_j \circ \pi_V}{E}} = S \circ \pi_V(x) - \pi_V(x) \in -2\pi_V(V) = 2\pi_V(V)
+    \]
+
+    Finally, since $\ker(\pi_V) = \bigcap_{\lambda > 0}\lambda V \subset V$, $x - \sum_{j = 1}^n x_j\dpn{x, \phi_j \circ \pi_V}{E} \in -3V = 3V$. Therefore if $R = \sum_{j = 1}^n (\phi_j \circ \pi_V) \otimes x_j \in E^* \otimes E$, then $Rx - x \in 3V$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:approximation-property-hilbert}
+    Every subspace of a product of Hilbert spaces has the approximation property. Every subspace of a projective limit of Hilbert spaces has the approximation property. 
+\end{corollary}
+
+\begin{lemma}
+\label{lemma:compact-operator-topology-banach-dual}
+    Let $E, F$ be Banach spaces over $K \in \RC$ and $\phi \in L_c(E; F)^*$, then there exists $\seq{x_n} \subset E$ and $\seq{\psi_n} \subset F^*$ such that:
+    \begin{enumerate}
+        \item $\limv{n}x_n = 0$. 
+        \item $\sum_{n \in \natp}\norm{\psi_n}_{F^*} < \infty$. 
+        \item For each $T \in L(E; F)$, $\dpn{T, \phi}{L_c(E; F)} = \sum_{n = 1}^\infty \dpn{Tx_n, \psi_n}{F}$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    Since $\phi \in L_c(E; F)^*$, there exists $A \subset E$ compact and $\alpha > 0$ such that $|\dpn{T, \phi}{L_c(E; F)}| \le \alpha\sup_{x \in A}\norm{Tx}_F$ for all $T \in L(E; F)$. After rescaling $A$, assume without loss of generality that $\alpha = 1$, so that $|\dpn{T, \phi}{L_c(E; F)}| \le \sup_{x \in A}\norm{Tx}_F$ for all $T \in L(E; F)$. 
+
+    By \autoref{lemma:compact-null-auxiliary}, there exists $\seq{x_n} \subset E$ with $\limv{n}x_n = 0$ and $A \subset \ol{\conv}(\seq{x_n})$. Since $E$ is complete, \hyperref[Mazur's Theorem]{theorem:convex-hull-complete} implies that $\ol{\conv}(\seq{x_n})$ is compact as well. Thus for each $T \in L(E; F)$, 
+    \[
+    T(A) \subset T(\ol{\conv}(\seq{x_n})) = \ol{\conv}(T(\seq{x_n}))
+    \]
+
+    In particular, 
+    \[
+    |\dpn{T, \phi}{L_c(E; F)}| \le \sup_{x \in A}\norm{Tx}_F \le \sup_{n \in \natp}\norm{Tx_n}_F
+    \]
+
+    As $\seq{x_n}$ is a null sequence in $E$, $\seq{Tx_n} \in c_0(\natp; F)$ for each $T \in L(E; F)$. Let $L = \bracs{\seq{Tx_n}|T \in L(E; F)}$, then $L$ is a subspace of $c_0(\natp; F)$. By the above estimate, $\phi$ factors through $L$ as follows: 
+    \[
+    \xymatrix{
+    L_c(E; F) \ar@{->}[rd]_{\phi} \ar@{->}[r] & L \ar@{->}[d]^{\widehat \phi} \\
+    & K
+    }
+    \]
+
+    The \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach} then yields an extension $\Phi$ of $\wh \phi$ as shown below:
+    \[
+    \xymatrix{
+    L_c(E; F) \ar@{->}[rd]_{\phi} \ar@{->}[r] & L \ar@{->}[d]^{\widehat \phi} \ar@{->}[r] & c_0(\mathbb{N}^+; F) \ar@{->}[ld]^{\Phi} \\
+    & K & 
+    }
+    \]
+
+    By \autoref{theorem:c0-sum-dual}, there exists $\seq{\psi_n} \in l^1(\natp; F^*)$ such that for each $y \in c_0(\natp; F)$, $\dpn{y, \Phi}{c_0(\natp; F)} = \sum_{n = 1}^\infty \dpn{y_n, \psi_n}{F}$. In particular, for each $T \in L(E; F)$, 
+    \begin{align*}
+        \dpn{T, \phi}{L_c(E; F)} &= \dpn{\seq{Tx_n}, \wh \phi}{L} = \dpn{\seq{Tx_n}, \Phi}{c_0(\natp; F)} \\
+        &= \sum_{n = 1}^\infty \dpn{Tx_n, \psi_n}{F}
+    \end{align*}
+\end{proof}
+
+\begin{remark}
+\label{remark:compact-operator-topology-banach-dual}
+    In \autoref{lemma:compact-operator-topology-banach-dual}, I would like to say that the mapping from $E \wh \otimes_\pi F^*$ to $L_c(E; F)^*$ defined by (3) is surjective. However, it does not seem right to me that this mapping is continuous at all. As such, I decided against mentioning the projective completion for this lemma. 
+\end{remark}
+
+
+\begin{theorem}
+\label{theorem:approximation-property-dual}
+    Let $E$ be a Banach space over $K \in \RC$, then the following are equivalent:
+    \begin{enumerate}
+        \item $E$ has the approximation property. 
+        \item For any Banach space $F$, the closure of $F^* \otimes E$ in $L(F; E)$ is $\mathcal{K}(F; E)$.
+        \item For any Banach space $F$, the canonical map $F^* \wh \otimes_\pi E \to L(F; E)$ is injective. 
+        \item The canonical map $E^* \wh \otimes_\pi E \to L(E; E)$ is injective.
+    \end{enumerate}
+
+    and the following are equivalent:
+    \begin{enumerate}[label=(\arabic**)]
+        \item $E^*$ has the approximation property. 
+        \item For any Banach space $F$, the closure of $E^* \otimes F$ in $L(E; F)$ is $\mathcal{K}(E; F)$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem III.9.5]{SchaeferWolff}}} and {{\cite[Proposition 4.6]{RyanTensor}}}. ]
+    (1) $\Rightarrow$ (2): Let $T \in \mathcal{K}(F; E)$, then $T(B_F(0, 1))$ is precompact in $E$. Thus for any $\eps > 0$, there exists $S \in E^* \otimes E$ such that $\norm{Sy - y}_{E} < \eps$ for all $y \in T(B_F(0, 1))$. In which case, $\norm{STx - Tx}_{E} < \eps$ for all $x \in B_F(0, 1)$. Therefore $ST \in F^* \otimes E$ with $\norm{ST - T}_{L(F; E)} \le \eps$. 
+
+    (2) $\Rightarrow$ (1): Let $A \subset E$ be compact and $\eps > 0$. By \autoref{lemma:compact-null-auxiliary}, there exists a convex, circled, and compact set $B \subset E$ such that $A$ is compact as a subset of $E_B$. 
+
+    Since $B$ is compact, the inclusion $E_B \to E$ is compact. By (2) applied to the inclusion map, there exists $T \in E_B^* \otimes E$ such that $\norm{Tx - x}_{E} \le \eps \norm{x}_{E_B}$ for all $x \in E_B$. 
+
+    Write $T = \sum_{j = 1}^n \phi_j \otimes x_j$, then as $A$ is compact in $E_B$, \hyperref[Goldstine's Theorem]{theorem:goldstine-weak} and the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} imply that there exists $\seqf{\psi_j} \subset E^*$ such that $|\dpn{x, \phi_j - \psi_j}{E_B}| \le \eps/\sum_{j = 1}^n \norm{x_j}_E$ for all $x \in A$ and $1 \le j \le n$. In which case, 
+    \begin{align*}
+        \norm{x - \sum_{j = 1}^n x_j\dpn{x, \psi_j}{E}}_E &\le \norm{x - \sum_{j = 1}^n x_j\dpn{x, \phi_j}{E_B}}_E \\
+        &+  \sum_{j = 1}^n \norm{x_j}_E|\dpn{x, \phi_j - \psi_j}{E_B}| \\
+        &\le \eps\norm{x}_{E_B} + \eps \le \eps\braks{1 + \sup_{x \in A}\norm{x}_{E_B}}
+    \end{align*}
+
+    for all $x \in A$. Therefore $S = \sum_{j = 1}^n \psi_j \otimes x_j \in E^* \otimes E$ with $\norm{Sx - x}_{E} \le \eps\braks{1 + \sup_{x \in A}\norm{x}_{E_B}}$ for all $x \in A$, and $E$ has the approximation property. 
+
+    (1) $\Rightarrow$ (3): Let $T \in F^* \wh \otimes_\pi E$ such that $Tx = 0$ for all $x \in F$. By \autoref{theorem:metrisable-tensor-product}, there exists $\seq{\phi_n} \subset F^*$ and $\seq{x_n} \subset E$ such that $T = \sum_{n = 1}^\infty \phi_n \otimes x_n$. $\sum_{n \in \natp}\norm{\phi_n}_{F^*}\norm{x_n}_E < \infty$, $\limv{n}x_n = 0$, and $\sum_{n \in \natp}\norm{\phi_n}_{F^*} < \infty$. 
+
+    Let $A$ be the closure of $\seq{x_n}$, then as $\seq{x_n}$ is a null sequence, $A$ is compact. Let $S \in L(E; F^{**}) = (F^* \wh \otimes_\pi E)^*$ and $\eps > 0$, then there exists $R \in E^{*} \otimes F^{**}$ such that $\norm{Rx - Sx}_{F^{**}} \le \eps$ for all $x \in A$. Write $R = \sum_{k = 1}^m \psi_k \otimes y_k$, then by \autoref{proposition:projective-tensor-product-dual}, 
+    \begin{align*}
+        \dpn{T, R}{F^* \wh \otimes_\pi E} &= \sum_{n = 1}^\infty \dpn{\phi_n, Rx_n}{F^*} = \sum_{n = 1}^\infty \angles{\phi_n, \sum_{k = 1}^m y_k \dpn{x_n, \psi_k}{E}}_{F^*} \\
+        &= \sum_{k = 1}^m \sum_{n = 1}^\infty \dpn{\phi_n, y_k}{F^*} \dpn{x_n, \psi_k}{E}
+    \end{align*}
+
+    Since $\sum_{n \in \natp}\norm{\phi_n}_{F^*} < \infty$, assume without loss of generality that $\bracsn{y_k}_1^m \subset F$ with \hyperref[Goldstine's Theorem]{theorem:goldstine-weak}. This allows rewriting
+    \[
+    \dpn{T, R}{F^* \wh \otimes_\pi E} = \sum_{k = 1}^m \dpn{Ty_k, \psi_k}{E} = 0
+    \]
+
+    so
+    \[
+    |\dpn{T, S}{F^* \wh \otimes_\pi E}| \le |\dpn{T, R}{F^* \wh \otimes_\pi E}| + \eps \sum_{n \in \natp}\norm{\phi_n}_{F^*} = \eps \sum_{n \in \natp}\norm{\phi_n}_{F^*}
+    \]
+
+    As the above holds for all $\eps > 0$, $\dpn{T, S}{F^* \wh \otimes_\pi E} = 0$. Therefore $T = 0$ as an element of $F^* \wh \otimes_\pi E$. 
+
+
+    $\neg$ (1) $\Rightarrow$ $\neg$ (4): Suppose that $E$ suffers from a lack of the approximation property, then $\text{Id}$ is not in the closure of $E^* \otimes E$ in $L_c(E; E)$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in L_c(E; E)^*$ such that $\dpn{\text{Id}, \phi}{L_c(E; E)} = 1$, but $\dpn{T, \phi}{L_c(E; E)} = 0$ for all $T \in E^* \otimes E$. 
+
+    By \autoref{lemma:compact-operator-topology-banach-dual}, there exists a null sequence $\seq{x_n} \subset E$ and $\seq{\psi_n} \in l^1(\natp; E^*)$ such that for each $T \in L(E; E)$, 
+    \[
+    \dpn{T, \phi}{L_c(E; E)} = \sum_{n = 1}^\infty \dpn{Tx_n, \psi_n}{E}
+    \]
+
+    In particular, for any $x \in E$ and $\eta \in E^*$, 
+    \begin{align*}
+        0 &= \dpn{\eta \otimes x, \phi}{L_c(E; E)} = \sum_{n = 1}^\infty \dpn{x_n, \eta}{E} \dpn{x, \psi_n}{E} \\
+        &= \angles{\sum_{n = 1}^\infty x_n\dpn{x, \psi_n}{E}, \eta}_E
+    \end{align*}
+
+    By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $\sum_{n = 1}^\infty x_n \dpn{x, \psi_n}{E} = 0$. Thus $\sum_{n = 1}^\infty x_n \dpn{x, \psi_n}{E} =0 $ for all $x \in E$. 
+
+    As $\seq{x_n}$ is a null sequence and $\seq{\psi_n} \in l^1(\natp; E^*)$, $\sum_{n \in \natp}\norm{x_n}_E\norm{\psi_n}_{E^*} < \infty$. This yields that $\sum_{n = 1}^\infty \psi_n \otimes x_n \in E^* \wh \otimes_\pi E$ with $\braks{\sum_{n = 1}^\infty \psi_n \otimes x_n} x = 0$ for all $x \in E$. 
+
+    However, since $1 = \dpn{\text{Id}, \phi}{L_c(E; E)} = \sum_{n = 1}^\infty \dpn{x_n, \psi_n}{E}$, $ \sum_{n = 1}^\infty \psi_n \otimes x_n \ne 0$ as an element of $E^* \wh \otimes_\pi E$. Therefore the canonical mapping from $E^* \wh \otimes_\pi E$ to $L(E; E)$ is not injective. 
+
+
+    (1*) $\Rightarrow$ (2*): Let $T \in \mathcal{K}(E; F)$ be compact, then $T^* \in \mathcal{K}(F^*; E^*)$ is compact by \hyperref[Schauder's Theorem]{theorem:compact-adjoint}, and $T^*(B_{F^*}(0, 1))$ is relatively compact.
+    
+    Let $\eps > 0$, then since $E^*$ has the approximation property, there exists $S \in E^{**} \otimes E^*$ such that $\norm{S\phi - \phi}_{E^*} \le \eps$ for all $\phi \in T^*(B_{F^*}(0, 1))$. 
+    By \hyperref[Gantmacher's Theorem]{theorem:weakly-compact-biadjoint}, $T^{**}(E^{**}) \subset F$. Thus $T^{**}S^* \in E^{***} \otimes F \subset L(E, F)$. For any $x \in E$ and $\phi \in B_{F^*}(0, 1)$, 
+    \begin{align*}
+        \dpn{T^{**}S^*x - T^{**}x, \phi}{F} &= \dpn{T^{**}S^*x, \phi}{F} - \dpn{Tx, \phi}{F} \\
+        &= \dpn{x, ST^*\phi}{E} - \dpn{x, T^*\phi}{E} \\
+        |\dpn{T^{**}S^*x - Tx, \phi}{F}| &\le \eps \norm{x}_E
+    \end{align*}
+
+    As this holds for all $\phi \in B_{F^*}(0, 1)$, $\norm{T^{**}S^*x - Tx}_F \le \eps \norm{x}_E$ by \autoref{proposition:dual-norm}. Therefore $\norm{T^{**}S^* - T}_{L(E; F)} \le \eps$. 
+    
+    (2*) $\Rightarrow$ (1*): Let $A \subset E^*$ be compact. Using \hyperref[Mazur's Theorem]{theorem:convex-hull-complete}, assume without loss of generality that $A$ is also convex and circled. 
+
+    Since $A$ is compact, it is norm bounded and hence equicontinuous, so the polar $U := A^\circ \in \cn_E(0)$ with respect to $\dpn{E, E^*}{E}$ is a convex and circled neighbourhood of $0$.
+
+    The canonical projection $\pi_U: E \to E_U$ induces an adjoint map $\pi_U^*: (E_U)^* \to E^*$. For each $\phi \in (E_U)^*$ with $\norm{\phi}_{(E_U)^*} \le 1$, $\pi_U^*\phi = \phi \circ \pi_U \in U^\circ$. As $A$ is already compact, convex, and circled, the \hyperref[Bipolar Theorem]{theorem:bipolar} implies that $U^{\circ} = A^{\circ\circ} = A$ and $\phi \in A$. Hence $\pi_U^* \in L((E_U)^*; (E^*)_A)$. On the other hand, for any $\phi \in A$, $U \subset \phi^{-1}(B_K(0, 1))$. As such, $\phi$ factors through $E_U$ as follows: 
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_U} \ar@{->}[rd]_{\phi} & E_U \ar@{->}[d]^{\widehat \phi} \\
+    & K
+    }
+    \]
+
+    where $\normn{\wh \phi}_{(E_U)^*} \le 1$. Thus $\pi_U^*$ is an isomorphism between $(E_U)^*$ and $(E^*)_A$. 
+    
+    Identify $(E_U)^*$ with $(E^*)_A$, then the inclusion $\iota_A: (E^*)_A \to E^*$ corresponds exactly to the adjoint of $\pi_U: E \to E_U$. Since $A$ is compact, $\iota_A: (E^*)_A \to E^*$ is compact, so \hyperref[Schauder's Theorem]{theorem:compact-adjoint} implies that $\pi_U: E \to \wh E_U$ is compact as well. 
+
+    Let $\eps > 0$, then by assumption applied to $\pi_U \in \mathcal{K}(E; \wh E_U)$, there exists $T \in E^* \otimes \wh E_U$ such that $\norm{T - \pi_U}_{L(E; \wh E_U)} \le \eps$. In which case, $T^* \in (E_U)^{**} \otimes E^{*} = (E^*)_A^* \otimes E^*$ with $\norm{T^* - \iota_A}_{L((E^*)_A; E^*)} \le \eps$ as well. 
+    
+    Finally, since $A$ is compact, \hyperref[Goldstine's Theorem]{theorem:goldstine-weak} and the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} allow assuming without loss of generality that $T^*$ takes the form of an element of $E_U \otimes E^*$ on $(E^*)_A$. In which case, $T^*$ indeed corresponds to an element of $E^{**} \otimes E^*$ such that $\norm{T^*\phi - \phi}_{E^*} \le \eps$ for all $\phi \in A$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:approximation-property-dual}
+    Let $E$ be a Banach space over $K \in \RC$. If $E^*$ has the approximation property, then so does $E$. 
+\end{corollary}
+\begin{proof}
+    By (3) of \autoref{theorem:approximation-property-dual}, for any Banach space $F$, the canonical map from $F^{*} \wh \otimes_\pi E^*$ to $L(F; E^*)$ is injective. Since $L(F; E^*)$ is canonically isomorphic to $L(E; F^*)$, the canonical map from $F^* \wh \otimes_\pi E^*$ to $L(E; F^*)$ is then injective. 
+
+    Now, let $F := E^*$, then the above yields an injection from $E^{**} \wh \otimes_\pi E^*$ to $L(E; E^{**})$. Let $T \in E \wh \otimes_\pi E^*$. By \autoref{theorem:metrisable-tensor-product}, there exists $\seq{x_n} \subset E$ and $\seq{\phi_n} \subset E^*$ such that $\sum_{n \in \natp}\norm{x_n}_{E}\norm{\phi_n}_{E^*} < \infty$ and $T = \sum_{n =1}^\infty x_n \otimes \phi_n$. As an operator, for each $x \in E$, 
+    \[
+    Tx = \sum_{n = 1}^\infty x_n \dpn{x, \phi_n}{E} \in E
+    \]
+
+    Therefore the restriction of the canonical map $E^{**} \wh \otimes_\pi E^* \to L(E; E^{**})$ to $E \wh \otimes_\pi E^*$ yields an injection into $L(E; E)$. By (4) of \autoref{theorem:approximation-property-dual}, $E$ has the approximation property. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:approximation-property-nuclear}
+    Let $E$ and $F$ be Banach spaces over $K \in \RC$. If $E^*$ or $F$ has the approximation property, then the canonical map
+    \[
+    E^* \otimes_\pi F \to N(E; F) \quad \braks{\sum_{j = 1}^n \phi_j \otimes y_j}(x) = \sum_{j = 1}^n y_j \dpn{x, \phi_j}{E}
+    \]
+
+    extends to an isometric isomorphism. 
+\end{corollary}
+\begin{proof}
+    If $F$ has the approximation property, then the canonical map $E^* \wh \otimes_\pi F \to N(E; F)$ is injective by (3) of \autoref{theorem:approximation-property-dual}. 
+
+    If $E^*$ has the approximation property, then by (3) of \autoref{theorem:approximation-property-dual}, the canonical map
+    \[
+    F^{**} \wh \otimes_\pi E^{*} \to N(F^*; E^*) \iso N(E; F^{**})
+    \]
+    
+    is injective. Restricting to $F \wh \otimes_\pi E^*$ yields an injection into $N(E; F)$. 
+\end{proof}
+
+\begin{corollary}[Existence of Continuous Trace]
+\label{corollary:trace-existence-approx}
+    Let $E$ a Banach space over $K \in \RC$ with the approximation property, then there exists a unique $\tr \in N(E; E)^*$ such that for each $\phi \in E^*$ and $x \in E$, $\tr(\phi \otimes y) = \dpn{y, \phi}{E}$.
+\end{corollary}
+\begin{proof}
+    By (U) of the \hyperref[projective tensor product]{definition:projective-tensor-product} and the isomorphism $E^* \wh \otimes_\pi E \iso N(E; E)$ from \autoref{corollary:approximation-property-nuclear}. 
+\end{proof}
+
+
+
+

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/compact.tex

@@ -0,0 +1,59 @@
+\section{Compact Operators}
+\label{section:compact-operator}
+
+\begin{definition}[Compact Operator]
+\label{definition:compact-operator}
+    Let $E, F$ be locally convex spaces over $K \in \RC$ and $T \in L(E; F)$, then $T$ is \textbf{compact} if there exists $U \in \cn_E(0)$ such that $T(U)$ is relatively compact in $F$. 
+
+    The set $\mathcal{K}(E; F)$ is the \textbf{space of compact operators} from $E$ to $F$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:compact-normed-complete}
+    Let $E$ be a normed space over $K \in \RC$ and $F$ be a complete Hausdorff topological vector space over $K$, then $\mathcal{K}(E; F)$ is a closed subspace of $L_b(E; F)$. 
+\end{proposition}
+\begin{proof}
+    Let $T \in \ol{\mathcal{K}(E; F)}$ and $U \in \cn_F(0)$ be circled, then there exists $S \in \mathcal{K}(E; F)$ such that $Sx - Tx \in U$ for all $x \in B_E(0, 1)$. Since $S$ is compact, there exists $Y \subset F$ finite with $S(B_E(0, 1)) \subset Y + U$. In which case, $T(B_E(0, 1)) \subset Y + U + U = Y + 2U$. Therefore $T(B_E(0, 1))$ is totally bounded, and as $F$ is complete, relatively compact in $F$. 
+\end{proof}
+
+\begin{theorem}[Schauder]
+\label{theorem:compact-adjoint}
+    Let $E, F$ be normed vector spaces over $K \in \RC$ with $F$ being complete and $T \in L(E; F)$, then $T$ is compact if and only if $T^* \in L(F^*; E^*)$ is compact. 
+\end{theorem}
+\begin{proof}
+    ($\Rightarrow$): Let $\cf \subset F^*$ be bounded, then $\cf$ is equicontinuous and hence relatively compact in the weak* topology by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Since $F$ is complete, $\ol{T(B_E(0, 1))}$ is compact. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\cf$ is relatively compact with respect to the topology of uniform convergence on $\ol{T(B_E(0, 1))}$. Thus $T^*(\cf) = \bracsn{\phi \circ T|\phi \in \cf}$ is relatively compact with respect to the topology of uniform convergence on $B_E(0, 1)$. In other words, $T^*(\cf)$ is relatively compact in $E^*$, and $T^*$ is a compact operator. 
+
+    ($\Leftarrow$): By \autoref{proposition:operator-space-completeness}, $E^*$ is complete. The preceding case then implies that $T^{**} \in L(E^{**}; F^{**})$ is compact. As such, its restriction to $E$, being identified with $T$, is also compact. 
+\end{proof}
+
+\begin{theorem}[Gantmacher]
+\label{theorem:weakly-compact-biadjoint}
+    Let $E, F$ be Banach spaces over $K \in \RC$, and $T \in L(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $T(B_E(0, 1))$ is relatively $\sigma(F, F^*)$-compact. 
+        \item $T^{**}(E^{**}) \subset F \subset F^{**}$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    Let $B_E$ be the closed unit ball of $E$, and $B_{E^{**}}$ be the closed unit ball of $E^{**}$. By \hyperref[Goldstine's Theorem]{theorem:goldstine-weak}, $B_E$ is $\sigma(E^{**}, E^*)$-dense in $B_{E^{**}}$. Since $T^{**}$ is a $\sigma(E^{**}, E^*)$-$\sigma(F^{**}, F^*)$-continuous extension of $T$, 
+    \[
+    T^{**}(B_{E^{**}}) = T^{**}\paren{\ol{B_E}^{\sigma(E^{**}, E^*)}} \subset \ol{T(B_E)}^{\sigma(F^{**}, F^*)}
+    \]
+
+    by \autoref{proposition:closure-of-image}. On the other hand, $B_{E^{**}}$ is $\sigma(E^{**}, E^*)$-compact by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Hence \autoref{proposition:compact-extensions} implies that $T^{**}(B_{E^{**}})$ is $\sigma(F^{**}, F^*)$-closed, so
+    \[
+    T^{**}(B_{E^{**}}) = T^{**}\paren{\ol{B_E}^{\sigma(E^{**}, E^*)}} = \ol{T(B_E)}^{\sigma(F^{**}, F^*)}
+    \]
+
+    The above equality shows that the following five statements are equivalent:
+    \begin{enumerate}[label=(\roman*)]
+        \item $T(B_E)$ is relatively $\sigma(F, F^*)$-compact. 
+        \item $\ol{T(B_E)}^{\sigma(F^{**}, F^*)} = \ol{T(B_E)}^{\sigma(F, F^*)}$. 
+        \item $\ol{T(B_E)}^{\sigma(F^{**}, F^*)} \subset F$. 
+        \item $T^{**}(B_{E^{**}}) \subset F$. 
+        \item $T^{**}(E^{**}) \subset F$. 
+    \end{enumerate}
+
+    where (i) is equivalent to (1), and (v) is equivalent to (2). 
+\end{proof}
+

src/fa/norm/index.tex

@@ -1,9 +1,14 @@
 \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}
+\input{./schauder.tex}

src/fa/norm/linear.tex

@@ -1,4 +1,4 @@
-\section{Linear Maps}
+\section{Linear Maps Between Normed Spaces}
 \label{section:normed-linear-maps}
 
 \begin{proposition}

src/fa/norm/multilinear.tex

@@ -1,4 +1,4 @@
-\section{Multilinear Maps}
+\section{Multilinear Maps Between Normed Spaces}
 \label{section:normed-multilinear}
 
 \begin{proposition}

src/fa/norm/schauder.tex

@@ -0,0 +1,48 @@
+\section{Schauder Bases}
+\label{section:schauder-bases}
+
+\begin{definition}[Schauder Basis]
+\label{definition:schauder-basis}
+    Let $E$ be a separable Banach space over $K \in \RC$ and $\seq{x_n} \subset E$, then $\seq{x_n}$ is a \textbf{Schauder basis} of $E$ if for each $x \in E$, there exists a unique $\seq{\lambda_n(x)} \in K^\natp$ such that
+    \[
+    x = \limv{N}\sum_{n = 1}^N \lambda_n(x) x_n
+    \]
+
+    The sequence of mappings $\seq{\lambda_n} \subset K^{E}$ is the \textbf{coefficient forms} of $\seq{x_n}$, and the Schauder basis $\seq{x_n}$ is \textbf{normalised} if $\norm{x_n}_E = 1$ for all $n \in \natp$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:schauder-basis-functional}
+    Let $E$ be a separable Banach space over $K \in \RC$, $\seq{x_n} \subset E$ be a normalised Schauder basis, and $\seq{\lambda_n} \subset K^{E}$ be its coefficient forms, then:
+    \begin{enumerate}
+        \item $\seq{\lambda_n} \subset E^*$ is equicontinuous. 
+        \item For each $N \in \natp$ and $x \in E$, let $P_Nx = \sum_{n = 1}^N \dpn{x, \lambda_n}{E}x_n$, then $P_N \to \text{Id}$ uniformly on compact sets as $N \to \infty$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[III.9.6]{SchaeferWolff}}}. ]
+    By uniqueness of the basis decomposition, $\seq{\lambda_n} \subset \hom(E; K)$. For each $x \in E$, let
+    \[
+    \norm{x}_{E'} = \sup_{N \in \natp} \norm{\sum_{n = 1}^N \lambda_n(x)x_n}_E
+    \]
+
+    then $\norm{x}_{E} \le \norm{x}_{E'}$, and $E$ is complete with respect to $\norm{\cdot}_{E'}$. By the \hyperref[Open Mapping Theorem]{theorem:open-mapping}, $\norm{\cdot}_E$ is equivalent to $\norm{\cdot}_{E'}$, and there exists $C \ge 0$ such that $\norm{x}_{E'} \le C\norm{x}_{E}$ for all $x \in E$. 
+
+    (1): For each $N \in \natp$ and $x \in E$, since $\seq{x_n}$ is normalised, 
+    \begin{align*}
+        |\lambda_N(x)| &= \norm{\lambda_N(x)x_N}_E = \norm{\sum_{n = 1}^{N}\lambda_n(x)x_n - \sum_{n = 1}^{N-1}\lambda_n(x)x_n }_E \\
+        &\le 2\norm{x}_{E'} \le 2C\norm{x}_{E}
+    \end{align*}
+
+    Hence $\seq{\lambda_n} \subset E^*$ with $\sup_{n \in \natp}\norm{\lambda_n}_{E^*} \le 2C$. 
+    
+    (2): Since $\sup_{N \in \natp}\norm{P_N}_{L(E; E)} \le C$, $\seq{P_N}$ is equicontinuous. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $P_N \to \text{Id}$ uniformly on compact sets as $N \to \infty$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:separable-schauder-ap}
+    Let $E$ be a separable Banach space over $K \in \RC$ with a Schauder basis, then $E$ enjoys the approximation property. 
+\end{corollary}
+\begin{proof}
+    By \autoref{proposition:schauder-basis-functional}, there exists $\seq{P_N} \subset L(E; E)$ such that $P_N \to \text{Id}$ uniformly on compact sets as $N \to \infty$. 
+\end{proof}
+

src/fa/norm/separable.tex

@@ -12,25 +12,66 @@
     \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}
 
+\begin{lemma}
+\label{lemma:compact-embed}
+    Let $E$ be a normed vector space over $K \in \RC$ and $A \subset [0, 1]$ be closed, then $C(A; E)$ embeds isometrically into $C([0, 1]; E)$. 
+\end{lemma}
+\begin{proof}
+    First note that if $0 \not\in A$ or $1 \not\in A$, $C(A; E)$ embeds isometrically into $C(A \cup \bracs{0, 1}; E)$ through extension by $0$. Thus assume without loss of generality that $A$ contains the endpoints $0$ and $1$. 
+
+    Let $U = [0, 1] \setminus A$, then there exists $\seq{(a_n, b_n)} \subset [0, 1]^2$ such that  $U = \bigsqcup_{n \in \natp}(a_n, b_n)$. For each $f \in C(A; E)$, let
+    \[
+    Tf: [0, 1] \to E \quad x \mapsto \begin{cases}
+        f(x) &x \in A \\
+         \frac{b_n - x}{b_n - a_n}f(a_n) + \frac{x - a_n}{b_n - a_n}f(b_n) &x \in (a_n, b_n) \subset [0, 1]
+    \end{cases}
+    \]
+
+    then the mapping $f \mapsto Tf$ is an isometric embedding into $E^{[0, 1]}$ with respect to the uniform norm.
+    
+    Since $U$ is open and $Tf$ is affine on each component of $U$, $Tf$ is continuous on $U$. It remains to show that $Tf$ is continuous on $A$. Let $x \in A$ and $\eps > 0$, then there exists $\delta > 0$ such that $\norm{f(y) - f(x)}_E < \eps$ for all $y \in (x -\delta, x + \delta) \cap A$. Now, a case analysis: 
+    \begin{enumerate}
+        \item If there exists $y \in (x - \delta, x) \cap A$, then for any $z \in U \cap (y, x)$, there exists $n \in \natp$ such that $(a_n, b_n) \subset (y, x)$ and $z \in (a_n, b_n)$. In which case, since $\norm{f(a_n) - f(x)}_E < \eps$ and $\norm{f(b_n) - f(x)}_E < \eps$, $\norm{Tf(z) - Tf(x)}_E < \eps$. Thus $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x)$.
+        \item Otherwise, $x = 0$ or $Tf|_{(x - \delta, x)}$ is an affine function. Either way, there exists $y \in (x - \delta, x)$ such that $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x) \cap [0, 1]$. 
+    \end{enumerate}
+
+    Thus there exists $y \in (x - \delta, x)$ with $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x) \cap [0, 1]$. Similarly, there exists $y' \in (x, x + \delta)$ with $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (x, y') \cap [0, 1]$. Therefore $Tf$ is continuous at $x$. Since this holds for all $x \in U$ and $x \in A$, $Tf \in C([0, 1]; E)$. 
+\end{proof}
+
+\begin{theorem}[Banach-Mazur]
+\label{theorem:banach-mazur}
+    Let $E$ be a separable normed vector space over $K \in \RC$, then there exists an isometric embedding $\iota \in L(E; C([0, 1]; K))$. 
+\end{theorem}
+\begin{proof}
+    Let $B$ be the closed unit ball of $E^*$, equipped with the weak* topology. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, the linear mapping
+    \[
+    E \to C(B; K) \quad x(\phi) = \dpn{x, \phi}{E}
+    \]
+
+    is an isometric embedding. By \autoref{proposition:separable-dual}, $B$ is a compact metric space. The \hyperref[Alexandroff-Hausdorff Theorem]{theorem:cantor-universality} then provides a continuous surjection $f: 2^{\natp} \to B$. Thus the composition map
+    \[
+    C(B; K) \to C(2^{\natp}; K) \quad g \mapsto g \circ f
+    \]
+
+    is a linear isometric embedding. Let $\mathcal{C} \subset [0, 1]$ be the Cantor set, then $\mathcal{C}$ is homeomorphic to $2^{\natp}$ through \autoref{proposition:cantor-space-embedding}. Hence $C(2^{\natp}; K)$ is isometrically isomorphic to $C(\mathcal{C}; K)$. 
+
+    Finally, \autoref{lemma:compact-embed} provides yet another linear isometric embedding $C(\mathcal{C}; K) \to C([0, 1]; K)$. Composing the above maps as follows
+    \[
+    \xymatrix{
+E \ar@{->}[r] & C(B; K) \ar@{->}[r] & C(2^{{\mathbb N}^+}; K) \ar@{->}[r] & C(\mathcal{C}; K) \ar@{->}[r] & C([0, 1]; K)
+}
+    \]
+
+    yields the desired embedding. 
+\end{proof}
+
+
+
 \begin{proposition}
 \label{proposition:separable-banach-borel-sigma-algebra}
     Let $E$ be a separable normed vector space, then the Borel $\sigma$-algebra on $E$ is generated by the following families of sets:
@@ -57,3 +98,4 @@
 \end{proof}
 
 
+

src/fa/notation.tex

@@ -17,6 +17,8 @@
     $L_s(E; F)$ & $L(E; F)$ with strong operator topology. & \autoref{definition:strong-operator-topology} \\
     $L_w(E; F)$ & $L(E; F)$ with weak operator topology. & \autoref{definition:weak-operator-topology} \\
     $L_b(E; F)$ & $L(E; F)$ with topology of bounded convergence. & \autoref{definition:bounded-convergence-topology} \\
+    $L_c(E; F)$ & $L(E; F)$ with topology of precompact convergence. & \autoref{definition:compact-operator-topology} \\
+    $\mathcal{K}(E; F)$ & Space of compact operators from $E$ to $F$. & \autoref{definition:compact-operator} \\ 
     $\widehat{E}$ & Hausdorff completion of TVS $E$. & \autoref{definition:tvs-completion} \\
     % ---- Locally Convex ----
     $\mathrm{Conv}(A)$ & Convex hull of $A$. & \autoref{definition:convex-hull} \\
@@ -24,9 +26,10 @@
     $[\cdot]_A$ & Gauge of a radial set $A$. & \autoref{definition:gauge} \\
     $\rho_M$ & Quotient of seminorm $\rho$ by subspace $M$. & \autoref{definition:quotient-norm} \\
     $E \otimes_\pi F$ & Projective tensor product of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
-    $E \,\widetilde{\otimes}_\pi F$ & Projective completion of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
+    $E \,\wh{\otimes}_\pi F$ & 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/order/order.tex

@@ -54,3 +54,9 @@
 \label{definition:order-vector-complete}
     Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{order complete} if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
 \end{definition}
+
+\begin{definition}[Monotone Complete]
+\label{definition:monotone-complete}
+    Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{monotone complete} if for any order bounded directed set $A \subset E$, $\sup(A)$ exists. 
+\end{definition}
+

src/fa/tvs/bilinear.tex

@@ -0,0 +1,170 @@
+\section{Bilinear Mappings}
+\label{section:bilinear-tvs}
+
+
+
+\begin{theorem}
+\label{theorem:separate-joint-bilinear}
+    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds:
+    \begin{enumerate}
+        \item[(B)] $E$ is Baire.
+        \item[(B')] $E$ is barrelled and $G$ is locally convex. 
+    \end{enumerate}
+
+    and that
+    \begin{enumerate}
+        \item[(M)] $E$ and $F$ are both metrisable. 
+        \item[(E)] For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg} \subset L(F; G)$ is equicontinuous. 
+    \end{enumerate}
+
+    then $\alg$ is equicontinuous. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[III.5.1]{SchaeferWolff}}}. ]
+    Let $\seq{(x_n, y_n)} \subset E \times F$ and $\seq{\lambda_n} \subset \alg$ such that $(x_n, y_n) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and \autoref{proposition:equicontinuous-net}. By (B) or (B') and the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_n(x_n, y_n) \to 0$ as $n \to \infty$ by \autoref{proposition:equicontinuous-net}. By (M) and \autoref{proposition:equicontinuous-net}, $\alg$ is equicontinuous at $0$, and hence equicontinuous by \autoref{lemma:equicontinuous-bilinear}. 
+\end{proof}
+
+\begin{definition}[Hypocontinuity]
+\label{definition:hypocontinuity}
+    Let $E, F, G$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal of bounded sets, and $\lambda: E \times F \to G$ be a separately continuous bilinear map, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $S \in \sigma$ and $V \in \cn_G(0)$, there exists $U \in \cn_F(0)$ such that $\lambda(S \times U) \subset V$. 
+        \item For each $S \in \sigma$, $\bracs{\lambda(x, \cdot)|x \in S} \subset F^*$ is equicontinuous. 
+    \end{enumerate}
+    
+    If the above holds, then $\lambda$ is \textbf{$\sigma$-hypocontinuous}. 
+
+    For any ideal $\tau \subset 2^F$ of bounded sets, $\lambda$ if \textbf{$(\sigma, \tau)$-hypocontinuous} if $\lambda$ is $\sigma$-hypocontinuous and $\tau$-hypocontinuous. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:separate-hypocontinuous}
+    Let $E, F, G$ be TVSs over $K \in \RC$, and $\lambda: E \times F \to G$ be a separately continuous bilinear map. 
+    If one of the following holds:
+    \begin{enumerate}
+        \item[(B)] $E$ is Baire.
+        \item[(B')] $E$ is barrelled and $G$ is locally convex. 
+    \end{enumerate}
+
+    then $\lambda$ is $B(E)$-hypocontinuous. 
+\end{proposition}
+\begin{proof}
+    Since $\lambda$ is separately continuous, the mapping
+    \[
+    E \to L(F; G) \quad x \mapsto \lambda(x, \cdot)
+    \]
+
+    is continuous with respect to the strong operator topology on $L(F; G)$. As such, for each $B \subset E$ bounded, $\bracs{\lambda(x, \cdot)|x \in B}$ is bounded in $L_s(F; G)$. By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracs{\lambda(x, \cdot)|x \in B}$ is equicontinuous. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:hypocontinuous-restriction}
+    Let $E, F, G$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be ideals of bounded sets, and $\lambda: E \times F \to G$ be a bilinear mapping. 
+    \begin{enumerate}
+        \item If $\lambda$ is $\sigma$-hypocontinuous, then $\lambda$ is continuous on $S \times F$ for all $S \in \sigma$. 
+        \item If $\lambda$ is $(\sigma, \tau)$-hypocontinuous, then $\lambda$ is uniformly continuous on $S \times T$ for all $S \in \sigma$ and $T \in \tau$. 
+    \end{enumerate}
+\end{proposition}
+% Omitted for obviousness
+
+\begin{proposition}
+\label{proposition:hypocontinuous-linear-extension}
+    Let $E, F, G$ be TVSs over $K \in \RC$, $E_0 \subset E$ and $F_0 \subset F$ be dense subspaces, and $\sigma \subset B(E_0)$ and $\tau \subset B(F_0)$ be ideals of bounded sets. Denote $\ol{\sigma}$ and $\ol \tau$ as the ideals generated by $\bracsn{\ol{S}|S \in \sigma}$ and $\bracsn{\ol{T}|T \in \tau}$, respectively. If
+    \begin{enumerate}[label=(\alph*)]
+        \item $G$ is a complete Hausdorff TVS. 
+        \item $\ol\sigma$ covers $E$ and $\ol{\tau}$ covers $F$. 
+    \end{enumerate}
+
+    Then, for any $(\sigma, \tau)$-hypocontinuous bilinear map $\lambda: E_0 \times F_0 \to G$, there exists a unique $\Lambda: E \times F \to G$ such that:
+    \begin{enumerate}
+        \item $\Lambda|_{E_0 \times F_0} = \lambda$. 
+        \item $\Lambda$ is bilinear and $(\ol\sigma, \ol\tau)$-hypocontinuous.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    By (2) of \autoref{proposition:hypocontinuous-restriction}, for each $S \in \sigma$ and $T \in \tau$, $\lambda|_{S \times T}$ is uniformly continuous. Thus (a) and \autoref{theorem:uniform-continuous-extension} imply that there exists a unique continuous extension of $\lambda$ to $\ol S \times \ol T$. By (b) and the \hyperref[gluing lemma]{lemma:glue-function}, there exists a unique $\Lambda: E \times F \to G$ such that:
+    \begin{enumerate}
+        \item $\Lambda|_{E_0 \times F_0} = \lambda$. 
+        \item[(2')] For each $S \in \sigma$ and $T \in \tau$, $\Lambda|_{\ol S \times \ol T}$ is continuous. 
+    \end{enumerate}
+
+    so the extension is unique by (1) of \autoref{proposition:hypocontinuous-restriction}. 
+
+    It remains to show that $\Lambda$ is bilinear and $(\ol \sigma, \ol \tau)$-hypocontinuous. To this end, observe that for each $x \in E_0$, $\lambda(x, \cdot) \in L(F_0; G)$, and extends to a unique element of $L(F; G)$ by the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-tvs}. For each $S \in \sigma$ and $T \in \tau$, $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$. As such, $\Lambda(x, \cdot)$ is the unique continuous extension of $\lambda(x, \cdot)$ to an element of $L(F; G)$, so $\Lambda(x, \cdot) \in L(F; G)$ for all $x \in E_0$. By symmetry, $\Lambda(\cdot, y) \in L(E; G)$ for all $y \in F_0$. 
+
+
+    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 $\bracsn{\lambda(x, \cdot)|x \in S}$. 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}
+
+
+
+
+\begin{proposition}
+\label{proposition:multilinear-identify}
+	Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
+	\begin{enumerate}
+        \item The map
+        \[
+        I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
+        \]
+
+        defined by
+        \[
+        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
+        \]
+
+        is an isomorphism.
+        \item The map
+        \[
+        I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
+        \]
+
+        defined by
+        \[
+        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
+        \]
+
+        is an isomorphism.
+    \end{enumerate}
+
+    which allows the identification
+    \[
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
+    \]
+
+    under the map $I$ in (2).
+\end{proposition}
+\begin{proof}
+    (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
+    \[
+    I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
+    \]
+
+    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
+    \[
+    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in \mathfrak{B}(F)
+    \]
+
+    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
+
+    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
+
+    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
+
+    On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
+
+    (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
+    \[
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F)
+    \]
+
+    Thus (2) holds for all $k \in \natp$.
+\end{proof}
+

src/fa/tvs/complete-metric.tex

@@ -133,3 +133,20 @@
 
     Thus by \autoref{proposition:successive-approximation-all}, $B_F(0, t) \subset T(B_E(0, r)) \in \cn_F(0)$ for all $r > r_0$. As $r_0 > 0$ is arbitrary, $T(U) \in \cn_F(0)$ for all $U \in \cn_E(0)$. Therefore $T$ is open by translation-invariance of the topology on $E$.
 \end{proof}
+
+\begin{theorem}[Closed Graph Theorem]
+\label{theorem:closed-graph}
+    Let $E, F$ be complete metric TVSs over $K \in \RC$ and $T \in \hom(E; F)$. If its graph $\Gamma(T) \subset E \times F$ is closed, then $T \in L(E; F)$. 
+\end{theorem}
+\begin{proof}
+    Given that $E$ and $F$ are both complete metric TVSs, $E \times F$ is a complete metric TVS by \autoref{proposition:product-complete}. Since $\Gamma(T) \subset E \times F$ is a closed subspace of $E \times F$, it is also a complete metric TVS over $K$ by \autoref{proposition:complete-closed}. 
+    
+    Let $\pi_1: E \times F \to E$ and $\pi_2: E \times F \to F$ be the projection maps of $E \times F$. As $\Gamma(T)$ is the graph of a function, $\pi_1|_{\Gamma(T)}: \Gamma(T) \to E$ is a continuous bijection. By the \hyperref[Open Mapping Theorem]{theorem:open-mapping}, it is an isomorphism. Therefore $T$ may be expressed as the following composition of continuous linear maps
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_1|_{\Gamma(T)}^{-1}} & \Gamma(T) \ar@{->}[r]^{\pi_2|_{\Gamma(T)}} & F
+    }
+    \]
+
+\end{proof}
+

src/fa/tvs/equicontinuous.tex

@@ -88,28 +88,6 @@
     and $\lambda(x_0, y - y_0) \in U$ as well. Therefore $\alg$ is equicontinuous at $(x_0, y_0)$. 
 \end{proof}
 
-
-
-\begin{theorem}
-\label{theorem:separate-joint-bilinear}
-    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds:
-    \begin{enumerate}
-        \item[(B)] $E$ is Baire.
-        \item[(B')] $E$ is barrelled and $G$ is locally convex. 
-    \end{enumerate}
-
-    and that
-    \begin{enumerate}
-        \item[(M)] $E$ and $F$ are both metrisable. 
-        \item[(E)] For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg} \subset L(F; G)$ is equicontinuous. 
-    \end{enumerate}
-
-    then $\alg$ is equicontinuous. 
-\end{theorem}
-\begin{proof}[Proof, {{\cite[III.5.1]{SchaeferWolff}}}. ]
-    Let $\seq{(x_n, y_n)} \subset E \times F$ and $\seq{\lambda_n} \subset \alg$ such that $(x_n, y_n) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and \autoref{proposition:equicontinuous-net}. By (B) or (B') and the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_n(x_n, y_n) \to 0$ as $n \to \infty$ by \autoref{proposition:equicontinuous-net}. By (M) and \autoref{proposition:equicontinuous-net}, $\alg$ is equicontinuous at $0$, and hence equicontinuous by \autoref{lemma:equicontinuous-bilinear}. 
-\end{proof}
-
 % TODO: Replace this with a more general version involving polars in the future. 
 \begin{theorem}[Banach-Alaoglu]
 \label{theorem:alaoglu}

src/fa/tvs/index.tex

@@ -14,4 +14,5 @@
 \input{./inductive.tex}
 \input{./vector-function.tex}
 \input{./space-of-linear.tex}
-\input{./equicontinuous.tex}
+\input{./equicontinuous.tex}
+\input{./bilinear.tex}

src/fa/tvs/space-of-linear.tex

@@ -20,69 +20,6 @@
 \end{proposition}
 % Proof omitted because it is obvious. 
 
-
-\begin{proposition}
-\label{proposition:multilinear-identify}
-	Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
-	\begin{enumerate}
-        \item The map
-        \[
-        I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
-        \]
-
-        defined by
-        \[
-        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
-        \]
-
-        is an isomorphism.
-        \item The map
-        \[
-        I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
-        \]
-
-        defined by
-        \[
-        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
-        \]
-
-        is an isomorphism.
-    \end{enumerate}
-
-    which allows the identification
-    \[
-    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
-    \]
-
-    under the map $I$ in (2).
-\end{proposition}
-\begin{proof}
-    (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
-    \[
-    I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
-    \]
-
-    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
-    \[
-    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in \mathfrak{B}(F)
-    \]
-
-    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
-
-    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
-
-    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
-
-    On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
-
-    (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
-    \[
-    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F)
-    \]
-
-    Thus (2) holds for all $k \in \natp$.
-\end{proof}
-
 \begin{definition}[Strong Operator Topology]
 \label{definition:strong-operator-topology}
     Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^E$ is the \textbf{strong operator topology}.
@@ -120,11 +57,19 @@
 
 \begin{definition}[Bounded Convergence Topology]
 \label{definition:bounded-convergence-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}.
+    Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}, or the \textbf{uniform topology}. 
 
     The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
 \end{definition}
 
+\begin{definition}[Topology of Precompact Convergence]
+\label{definition:compact-operator-topology}
+    Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{K} \subset 2^E$ be the collection of precompact subsets of $E$, then the $\mathfrak{K}$-uniform topology on $L(E; F)$ is the \textbf{topology of precompact convergence}. 
+
+    The space $L_c(E; F)$ denotes $L(E; F)$ equipped with the topology of precompact convergence.
+\end{definition}
+
+
 \begin{proposition}
 \label{proposition:operator-space-completeness}
     Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:

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/index.tex

@@ -8,5 +8,6 @@
 \input{./measurable-maps/index.tex}
 \input{./lebesgue-integral/index.tex}
 \input{./bochner-integral/index.tex}
+\input{./weak-integral/index.tex}
 \input{./lcg/index.tex}
 \input{./notation.tex}

src/measure/measurable-maps/index.tex

@@ -8,4 +8,4 @@
 \input{./metric.tex}
 \input{./approx.tex}
 \input{./in-measure.tex}
-\input{./locally-in-measure.tex}
+\input{./local-in-measure.tex}

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/radon/c0.tex

@@ -125,7 +125,7 @@
     
     As such a $\phi \in C_0(X; E)$ exists for all $\eps > 0$ and $\seqf{A_j}$, $\norm{I_\mu}_{C_0(X; E)^*} \ge \norm{\mu}_{\text{var}}$. Therefore the map $\mu \mapsto I_\mu$ is isometric. 
 
-    (Surjective): Let $B = \bracsn{\phi \in E^*|\norm{\phi}_{E^*} \le 1}$ and equip it with the weak*-topology and
+    (Surjective): Let $B = \bracsn{\phi \in E^*|\norm{\phi}_{E^*} \le 1}$ and equip it with the weak* topology and
     \[
     T: C_0(X; E) \to C_0(X \times B; K) \quad (Tf)(x, \phi) = \dpn{f(x), \phi}{E}
     \]

src/measure/sets/borel.tex

@@ -8,6 +8,32 @@
 \end{definition}
 
 
+
+\begin{lemma}
+\label{lemma:monotone-borel-characterisation}
+    Let $X$ be a topological space, then $\cb_X$ is the smallest subset of $2^X$ that:
+    \begin{enumerate}
+        \item contains open and closed subsets of $X$. 
+        \item is closed under countable intersections.
+        \item is closed under countable disjoint unions. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 8.2.4]{CohnMeasure}}}. ]
+    Let $\cf \subset 2^X$ be the smallest subset of $2^X$ satisfying the lemma. Since $\cb_X$ satisfies the lemma, $\cf \subset \cb_X$. On the other hand, let
+    \[
+    \cf_0 = \bracs{A \subset X| A \in \cf, A^c \in \cf}
+    \]
+
+    then by definition, $\cf_0$ is closed under complements. Let $\seq{A_n} \subset \cf_0$, then
+    \[
+    \bigcup_{n \in \natp}A_n = \bigsqcup_{n \in \natp}A_n \setminus \bigcup_{k = 1}^{n-1} A_k
+     = \bigsqcup_{n \in \natp}A_n \cap \bigcap_{k = 1}^{n - 1}A_k^c
+    \]
+
+    Since $\cf_0 \subset \cf$ is closed under complements, $\seq{A_n^c} \subset \cf$ as well. By (2) and (3), $\bigcup_{n \in \natp}A_n \in \cf$ and $\bigcap_{n \in \natp}A_n^c \in \cf$. Thus $\bigcup_{n \in \natp}A_n \in \cf$ as well. By (1), $\cf_0$ is a $\sigma$-algebra that contains all open subsets of $X$, so $\cf \supset \cf_0 \supset \cb_X$.  
+\end{proof}
+
+
 \begin{definition}[Borel $\sigma$-Algebra on $\ol{\real}$]
 \label{definition:borel-sigma-algebra-extended}
     The family

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/measure/weak-integral/definition.tex

@@ -0,0 +1,76 @@
+\section{Weak Integrals*}
+\label{section:weak-integral}
+
+\begin{definition}[Weakly Measurable]
+\label{definition:weakly-measurable}
+    Let $(X, \cm)$ be a measurable space, $E$ be a locally convex space over $K \in \RC$, and $f: X \to E$, then $f$ is \textbf{weakly measurable} if for each $\phi \in E^*$, $\phi \circ f: X \to K$ is Borel measurable. 
+\end{definition}
+
+As I know so little about weak integrals, I will Dunning-Kruger myself right now, give an opinion, and laugh about it later. My gripe with seeing the definition comes from the need to test against \textit{every} continuous linear functional. To me, this seems quite inflexible: consider integrating a distribution-valued function. \textit{Surely} it is wiser to only test this function against test functions rather than \textit{the dual of $\mathcal{D}'$ (dual with respect to $\mathcal{D}'$ with the bounded convergence topology)}. As such, it may be more productive to consider a more flexible form of testing, such as using duality. 
+
+
+\begin{definition}[Weak Integrability*]
+\label{definition:weakly-integrable}
+    Let $(X, \cm, \mu)$ be a measure space, $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, and $f: X \to E$, then $f$ is \textbf{Dunford $\lambda$-integrable*} if:
+    \begin{enumerate}[label=(I\arabic*)]
+        \item $f$ is weakly measurable. 
+        \item For each $\phi \in F$, $\phi \circ f \in L^1(X; K)$.
+        \item For each $A \in \cm$, the mapping
+        \[
+        F \to K \quad \phi \mapsto \int_A \dpn{f(x), \phi}{\lambda} d\mu 
+        \]
+
+        is a continuous linear functional on $F$. 
+    \end{enumerate}
+
+    For each $A \in \cm$, the element $\phi \mapsto \int_A \dpn{f(x), \phi}{\lambda} d\mu$ of $F^*$ is the \textbf{Dunford $\lambda$-integral} of $f$ over $A$, denoted $\int_A^* f d\mu$. 
+
+    The function $f$ is \textbf{Pettis $\lambda$-integrable*} if it satisfies (I1), (I2), and
+    \begin{enumerate}
+        \item[(I3+)] For each $A \in \cm$, the mapping
+        \[
+        F \to K \quad \phi \mapsto \int_A \dpn{f(x), \phi}{\lambda} d\mu 
+        \]
+
+        is a $\sigma(F, E)$-continuous linear functional on $F$. 
+    \end{enumerate}
+
+    In which case, for each $A \in \cm$, $\int_A^* f d\mu$ is the \textbf{Pettis $\lambda$-integral} of $f$ over $A$. 
+\end{definition}
+
+It is at this point that I start to understand why the bidual setup is useful: existence. 
+
+
+\begin{proposition}
+\label{proposition:dunford-existence}
+    Let $(X, \cm, \mu)$ be a measure space, $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, and $f: X \to E$. If
+    \begin{enumerate}[label=(I\arabic*)]
+        \item $f$ is weakly measurable. 
+        \item For each $\phi \in F$, $\phi \circ f \in L^1(X; K)$.
+        \item[(C)] $F$ is a Fréchet space. 
+    \end{enumerate}
+
+    then $f$ is \hyperref[Dunford $\lambda$-integrable]{definition:weakly-integrable}. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Section 3.3]{RyanTensor}}}. ]
+    Let $T: F \to L^1(X; K)$ be defined by $T\phi = \phi \circ f$. To see that $T$ is continuous, it is sufficient to apply the \hyperref[Closed Graph Theorem]{theorem:closed-graph}. 
+    
+    Let $\seq{\phi_n} \subset F$, $\phi \in F$, $\seq{g_n} \subset L^1(X; K)$, and $g \in L^1(X; K)$ such that
+    \begin{enumerate}[label=(\roman*)]
+        \item $g_n = \phi_n \circ f$ almost everywhere for all $n \in \natp$. 
+        \item $\phi_n \to \phi$ and $g_n \to g$ as $n \to \infty$.
+    \end{enumerate}
+    
+    
+    By passing through a subsequence using \autoref{theorem:cauchy-in-measure-limit}, assume further without loss of generality that $g_n \to g$ almost everywhere. In which case, $\phi_n \circ f \to g$ almost everywhere as well, and $\phi \circ f = g$ almost everywhere. Thus $T$ is continuous. 
+
+    Now, let $T^*: L^1(X; K)^* \to F^*$ be the adjoint of $T$. For each $A \in \cm$, the mapping $\Phi_A: L^1(X; K) \to K$ defined by $g \mapsto \int_A g d\mu$ is a continuous linear functional on $L^1(X; K)$. As such, for each $\phi \in F$, 
+    \[
+        \int_A \dpn{f, \phi}{\lambda} d\mu = \int_A T\phi d\mu = \dpn{T\phi, \Phi_A}{L^1(X; K)} = \dpn{\phi, T^*\Phi_A}{F}
+    \]
+
+    Therefore $T^*\Phi_A = \int_A^* f d\mu$ is the desired Dunford $\lambda$-integral of $f$ over $A$. 
+\end{proof}
+
+
+

src/measure/weak-integral/index.tex

@@ -0,0 +1,5 @@
+\chapter{Weak Integrals*}
+\label{chap:weak-integral}
+
+\input{./definition.tex}
+

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/index.tex

@@ -11,4 +11,6 @@
 \input{./order.tex}
 \input{./positive.tex}
 \input{./state.tex}
-\input{./gns.tex}
+\input{./gns.tex}
+\input{./non-unital.tex}
+\input{./quotients.tex}

src/op/c-star/involution.tex

@@ -20,9 +20,9 @@
 
 \begin{proposition}
 \label{proposition:c-star-algebra-gymnastics}
-    Let $A$ be a $C^*$ algebra, then:
+    Let $A$ be a $C^*$-algebra, then:
     \begin{enumerate}
-        \item For each $x \in A$, $\norm{x}_A = \normn{x^*}_A\norm{x}_A$. 
+        \item For each $x \in A$, $\norm{x}_A^2 = \normn{x^*}_A\norm{x}_A$. 
     \end{enumerate}
 
     If $A$ is unital, then

src/op/c-star/non-unital.tex

@@ -0,0 +1,151 @@
+\section{Non-Unital $C^*$-Algebras}
+\label{section:non-unital-cstar}
+
+\begin{proposition}
+\label{proposition:c-star-unitisation}
+    Let $A$ be a non-unital $C^*$-algebra and $\td A$ be its unitisation, then there exists a unique norm $\norm{\cdot}_{\td A}: \td A \to [0, \infty)$ such that:
+    \begin{enumerate}
+        \item For each $x \in A$, $\norm{x}_{\td A} = \norm{x}_A$. 
+        \item $(\td A, \norm{\cdot}_{\td A})$ is a unital $C^*$-algebra. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem 15.1]{Zhu}}}. ]
+    For each $x \in A$ and $\lambda \in \complex$, let
+    \[
+    \norm{x + \lambda}_{\td A} = \sup_{\substack{y \in A \\ \norm{y}_A \le 1}} \norm{xy + \lambda y}_A
+    \]
+
+    be the operator seminorm corresponding to $\td A$ acting on $A$. If $\norm{xy + \lambda y}_A = 0$ for all $y \in A$, then $xy = -\lambda y$ for all $y \in A$. Given that $A$ is non-unital, $\lambda = 0$. Since $A$ is a $C^*$-algebra, $\norm{x}_A^2 = \norm{xx^*}_A = 0$, and $x = 0$ as well. Thus $\norm{\cdot}_{\td A}$ is indeed a norm on $\td A$. 
+
+    (1): Let $x \in A \setminus \bracs{0}$, then since $A$ is a Banach algebra, $\norm{x}_{\td A} \le \norm{x}_A$. On the other hand, as $A$ is a $C^*$-algebra, 
+    \[
+    \norm{x}_{\td A} \ge \norm{x \cdot \frac{x^*}{\norm{x}_A}}_A =  \frac{\norm{x}_A^2}{\norm{x}_A} = \norm{x}_A
+    \]
+
+    (2): As $\norm{\cdot}_{\td A}$ is the operator norm corresponding to $\td A$ acting on $A$, $(\td A, \norm{\cdot}_{\td A})$ is a Banach algebra. Moreover, for any $x \in A$ and $\lambda \in \complex$, 
+    \begin{align*}
+        \norm{x + \lambda}_{\td A}^2 &= \sup_{\substack{y \in A \\ \norm{y}_A \le 1}}\norm{(x + \lambda)y}_A^2 
+        = \sup_{\substack{y \in A \\ \norm{y}_A \le 1}}\norm{y^*(x + \lambda)^*(x + \lambda)y}_A  \\
+        &\le \sup_{\substack{y \in A \\ \norm{y}_A \le 1}}\norm{(x + \lambda)^*(x + \lambda)y}_A = \norm{(x + \lambda)^*(x+\lambda)}_{\td A}
+    \end{align*}
+
+    so $(\td A, \norm{\cdot}_{\td A})$ is a $C^*$-algebra. 
+
+    Finally, \autoref{corollary:c-star-unique-norm} implies that there can be at most one norm on $\td A$ making it a unital $C^*$-algebra, so the constructed norm is unique. 
+\end{proof}
+
+
+\begin{definition}[Approximate Identity]
+\label{definition:banach-approximate-identity}
+    Let $A$ be a Banach algebra and $\angles{e_\beta}_{\beta \in B} \subset A$ be a net, then $\angles{e_\beta}_{\beta \in B}$ is an \textbf{approximate identity} of $A$ if:
+    \begin{enumerate}
+        \item For each $\beta \in B$, $\norm{e_\beta}_{A} \le 1$. 
+        \item For every $x \in A$, $e_\beta x \to x$ and $x e_\beta \to x$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Increasing Approximate Identity]
+\label{definition:increasing-approximate-identity}
+    Let $A$ be a $C^*$-algebra and $\angles{e_\beta}_{\beta \in B} \subset A$ be an approximate identity, then $\angles{e_\beta}_{\beta \in B}$ is \textbf{increasing} if:
+    \begin{enumerate}
+        \item For each $\beta \in B$, $e_\beta \ge 0$. 
+        \item For each $\beta, \gamma \in B$ with $\beta \le \gamma$, $e_\beta \le e_\gamma$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{lemma}
+\label{lemma:cstar-approximate-identity-existence}
+    For each $m, n \in \natp$ with $m \le n$, 
+    \begin{enumerate}
+        \item $\sup_{t \in [0, \infty)} \paren{1/n + t}^{-2}t \le n/4$. 
+        \item For every $t \in [0, \infty)$, $m^{-1}(m^{-1} + t)^{-1} \ge n^{-1}(n^{-1} + t)^{-1}$. 
+        \item For every $t \in [0, \infty)$, $(1/n + t)^{-1}t = 1 - (1/n + t)^{-1}/n$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): For each $t \in [0, \infty)$, 
+    \begin{align*}
+        0 \le \paren{t - \frac{1}{n}}^2 &= t^2 - \frac{2t}{n} + \frac{1}{n^2} = t^2 + \frac{2t}{n} + \frac{1}{n^2} - \frac{4t}{n} \\
+        0 &\le \paren{t + \frac{1}{n}}^2 - \frac{4t}{n}
+    \end{align*}
+
+    so $\frac{4t}{n} \le \paren{t + \frac{1}{n}}^2$ and $\frac{n}{4} \ge \paren{\frac{1}{n} + t}^{-2}t$. 
+    
+    (2): For each $t \in [0, \infty)$, 
+    \[
+    \frac{1}{m}\paren{\frac{1}{m} + t}^{-1} = \frac{1}{1 + mt} \ge \frac{1}{1 + nt} = \frac{1}{n}\paren{\frac{1}{n} + t}^{-1}
+    \]
+
+    (3): For every $t \in [0, \infty)$, 
+    \begin{align*}
+        \paren{\frac{1}{n} + t}^{-1}t &= \paren{\frac{1}{n} + t}^{-1}\braks{\paren{\frac{1}{n} + t} - \frac{1}{n}} \\
+        &= \paren{\frac{1}{n} + t}^{-1}\paren{\frac{1}{n} + t} - \frac{1}{n}\paren{\frac{1}{n} + t}^{-1} \\
+        &= 1 - \frac{1}{n}\paren{\frac{1}{n} + t}^{-1}
+    \end{align*}
+    
+\end{proof}
+
+
+
+\begin{theorem}
+\label{theorem:cstar-approximate-identity-existence}
+    Let $A$ be a unital $C^*$-algebra, $I \subset A$ be a left ideal, and $\cf \subset 2^I$ be the collection of all finite subsets of $I$, directed under inclusion. For each $F \in \cf$, let
+    \[
+    p_F = \sum_{x \in F}x^*x \quad e_F = \paren{\frac{1}{|F|} + p_F}^{-1}p_F
+    \]
+    
+    then $\angles{e_F}_{F \in \cf} \subset I \cap \ol{B_A(0, 1)}$ is an increasing net of positive elements such that $xe_F \to x$ for all $x \in I$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 15.2]{Zhu}}}. ]
+    Since $p_F$ is positive, $1/|F| + p_F$ is invertible by \autoref{proposition:positive-spectrum}, and the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} implies that $0 \le e_F \le 1_A$. As $I \subset A$ is a left ideal, $e_F \in I \cap \ol{B_A(0, 1)}$. Now,
+    \begin{align*}
+        &\sum_{x \in F}[x(e_F - 1_A)]^*[x(e_F - 1_A)] = \sum_{x \in F}(e_F - 1_A)x^*x(e_F - 1_A) \\
+        &= (e_F - 1_A)p_F(e_F - 1_A) = e_F^2p_F - 2e_Fp_F + p_F \\
+        &= (1/|F| + p_F)^{-2}p_F \cdot \braks{p_F^2 - 2\paren{\frac{1}{|F|} + p_F}p_F + \paren{\frac{1}{|F|} + p_F}^2}
+    \end{align*}
+
+    where
+    \begin{align*}
+        &p_F^2 - 2\paren{\frac{1}{|F|} + p_F}p_F + \paren{\frac{1}{|F|} + p_F}^2 \\
+        &= - p_F^2 -\frac{2p_F}{|F|} + \frac{1}{|F|^2} + \frac{2p_F}{|F|} + p_F^2 = \frac{1}{|F|^2}
+    \end{align*}
+
+    so
+    \[
+    \sum_{x \in F}[x(e_F - 1_A)]^*[x(e_F - 1_A)] = \frac{1}{|F|^2}(1/|F| + p_F)^{-2}p_F
+    \]
+    
+    By (1) of \autoref{lemma:cstar-approximate-identity-existence}, $\sup_{t \in [0, \infty)} (1/|F| + t)^{-2}t \le |F|/4$, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} shows that
+    \[
+    0 \le \sum_{x \in F}[x(e_F - 1_A)]^*[x(e_F - 1_A)] \le \frac{1}{|F|^2} \cdot \frac{|F|}{4} = \frac{1}{4|F|}
+    \]
+
+    Thus for each $x \in F$, $0 \le [x(e_F - 1_A)]^*[x(e_F - 1_A)] \le 1/(4|F|)$. In particular, $\norm{xe_F - x}_A \le \sqrt{1/4|F|}$. 
+
+    To see that $\angles{e_F}_{F \in \cf}$ is increasing, let $F, G \in \cf$ with $F \subset G$, then $p_F \le p_G$, and $(1/|F| + p_F)^{-1} \ge (1/|F| + p_G)^{-1}$ by \autoref{lemma:cstar-inversion-order-reversing}. By (2) of \autoref{lemma:cstar-approximate-identity-existence}, 
+    \[
+    \frac{1}{|F|}\paren{\frac{1}{|F|} + t}^{-1} \ge \frac{1}{|G|} \paren{\frac{1}{|G|} + t}^{-1}
+    \]
+
+    for all $t \in [0, \infty)$. For each $t \in [0, \infty)$, rewrite
+    \begin{align*}
+    \paren{\frac{1}{|F|} + t}^{-1}t &= 1 - \frac{1}{|F|}\paren{\frac{1}{|F|} + t}^{-1} \\
+    \paren{\frac{1}{|G|} + t}^{-1}t &= 1 - \frac{1}{|G|}\paren{\frac{1}{|G|} + t}^{-1}
+    \end{align*}
+    
+    Thus by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, 
+    \begin{align*}
+        e_F &= 1 - \frac{1}{|F|}\paren{\frac{1}{|F|} + p_F}^{-1} \le 1 - \frac{1}{|F|}\paren{\frac{1}{|F|} + p_G}^{-1} \\
+        &\le 1 - \frac{1}{|G|}\paren{\frac{1}{|G|} + p_G}^{-1} = e_G
+    \end{align*}
+    
+\end{proof}
+
+\begin{corollary}
+\label{corollary:cstar-approximate-identity-existence-actual}
+    Let $A$ be a $C^*$-algebra, then $A$ admits an increasing approximate identity.
+\end{corollary}
+\begin{proof}
+    Identify $A$ as a self-adjoint two-sided ideal of its unitisation $\td A$, which is a $C^*$-algebra by \autoref{proposition:c-star-unitisation}. Applying \autoref{theorem:cstar-approximate-identity-existence} to $A$ and $\bracs{x^*|x \in A}$ as left and right ideals, respectively, yields that the net constructed by the theorem is an increasing approximate identity for $A$. 
+\end{proof}
+

src/op/c-star/order.tex

@@ -82,6 +82,19 @@
     The condition in the sign decomposition that $x^+x^- = x^-x^+ = 0$ is essential. Otherwise I may use silly decompositions like $0 = 1 - 1$. 
 \end{remark}
 
+\begin{lemma}
+\label{lemma:cstar-inversion-order-reversing}
+    Let $A$ be a unital $C^*$-algebra, $x, y \in G(A)$ be positive elements with $x \le y$, then $x^{-1} \ge y^{-1}$.
+\end{lemma}
+\begin{proof}
+    Since $y - x \ge 0$ and $x$ is invertible, $y^{-1/2}(y - x)y^{-1/2} \ge 0$ as well. As such, 
+    \[
+    y^{-1/2}xy^{-1/2} \le y^{-1/2}yy^{-1/2} = 1
+    \]
+
+    Thus $\sigma_A(y^{-1/2}xy^{-1/2}) \subset \ol{B_\complex(0, 1)}$, and $\sigma_A(y^{1/2}x^{-1}y^{1/2}) \subset \complex \setminus B_\complex(0, 1)$. Hence $y^{1/2}x^{-1}y^{1/2} \ge 1$, and $x^{-1} \ge y^{-1}$. 
+\end{proof}
+
 
 
 

src/op/c-star/quotients.tex

@@ -0,0 +1,54 @@
+\section{Quotients of $C^*$-Algebras}
+\label{section:cstar-quotient}
+
+\begin{lemma}
+\label{lemma:two-sided-ideal-self-adjoint}
+    Let $A$ be a unital $C^*$-algebra and $I \subset A$ be a closed two-sided ideal, then for each $x \in I$, $x^* \in I$ as well. 
+\end{lemma}
+\begin{proof}
+    By \autoref{theorem:cstar-approximate-identity-existence}, there exists an increasing net $\angles{e_\beta}_{\beta \in B} \subset I \cap \ol{B_A(0, 1)}$ such that $e_\beta x \to x$ and $xe_\beta \to x$ for all $x \in I$. As $A$ is a $C^*$-algebra, 
+    \[
+    \norm{e_\beta x - x}_A = \norm{x^*e_\beta - x^*}_A \to 0
+    \]
+
+    Given that $I$ is two-sided, $x^*e_\beta \in I$ for all $\beta \in B$. Since $I$ is closed, the above implies that $x^* \in I$ as well. 
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:cstar-ideal-quotient-norm}
+    Let $A$ be a unital $C^*$-algebra, $I \subset A$ be a closed two-sided ideal, and $\angles{e_\beta}_{\beta \in B}$ be an increasing approximate identity for $I$, then for each $x \in A$, 
+    \[
+    \norm{x + I}_{A/I} = \inf\bracsn{\norm{x - y}_A|y \in I} = \lim_{\beta \in B}\norm{xe_\beta - x}_A
+    \]
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 15.6]{Zhu}}}. ]
+    For each $y \in I$, $ye_\beta \to y$. Thus
+    \[
+    \limsup_{\beta \in B}\norm{xe_\beta - x}_A = \limsup_{\beta \in B}\norm{(x - y)(1 - e_\beta)}_A \le \norm{x - y}_A
+    \]
+
+    As the above holds for all $y \in I$, $\limsup_{\beta \in B}\norm{xe_\beta - x}_A \le \norm{x + I}_{A/I}$. 
+
+    On the other hand, since $I$ is a two-sided ideal, $xe_\beta \in I$ for all $\beta \in B$. Thus $\liminf_{\beta \in B}\norm{xe_\beta - x}_A \ge \norm{x + I}_{A/I}$.
+\end{proof}
+
+\begin{theorem}
+\label{theorem:cstar-quotient}
+    Let $A$ be a unital $C^*$-algebra and $I \subset A$ be a closed two-sided ideal, then $A/I$ equipped with the quotient norm is a $C^*$-algebra. 
+\end{theorem}
+\begin{proof}
+    Under the quotient structures, $A/I$ is an involutive Banach algebra. It remains to show that $\norm{x + I}_{A/I}^2 = \norm{x^*x + I}_{A/I}$ for all $x \in A$. 
+
+    By \autoref{lemma:two-sided-ideal-self-adjoint}, $I$ is a $C^*$-algebra. Thus \autoref{theorem:cstar-approximate-identity-existence} implies the existence of an increasing approximate identity $\angles{e_\beta}_{\beta \in B} \subset I$ for $I$. 
+
+    For each $x \in A$ and $y \in I$, 
+    \begin{align*}
+        \norm{x + I}_{A/I}^2 &= \lim_{\beta \in B}\norm{x - xe_\beta }_A^2 = \lim_{\beta \in B}\norm{(x - xe_\beta )^*(x - x e_\beta)}_A \\
+        &= \lim_{\beta \in B}\norm{(1_A - e_\beta)x^*x(1_A - e_\beta)}_A \\ 
+        &= \lim_{\beta \in B}\norm{(1_A - e_\beta)(x^*x + y)(1_A - e_\beta)}_A \le \norm{x^*x + y}_{A}
+    \end{align*}
+
+    As the above holds for all $y \in I$, $\norm{x + I}_{A/I}^2 \le \norm{x^*x + I}_{A/I}$. Thus $A/I$ equipped with the quotient norm is a $C^*$-algebra. 
+\end{proof}
+

src/op/c-star/sa.tex

@@ -20,6 +20,7 @@
     By \autoref{proposition:complex-conjugation-properties}. 
 \end{proof}
 
+
 \begin{definition}[Normal]
 \label{definition:c-star-normal}
     Let $A$ be an involutive algebra over $\complex$ and $x \in A$, then the following are equivalent:

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/index.tex

@@ -3,5 +3,6 @@
 
 \input{./banach/index.tex}
 \input{./c-star/index.tex}
+\input{./vn/index.tex}
 \input{./example/index.tex}
 \input{./notation.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

@@ -0,0 +1,11 @@
+\chapter{Von Neumann Algebras}
+\label{chap:von-neumann-algebras}
+
+\input{./topologies.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

@@ -0,0 +1,122 @@
+\section{Topologies on $B(H)$}
+\label{section:topologies-on-bh}
+
+Let $H$ be a complex Hilbert space. Thanks to its self-duality, there is a natural dual pairing
+\[
+B(H) \times (H \otimes H) \to \complex \quad \dpn{T, \phi \otimes x}{B(H)} =  \dpn{Tx, \phi}{H}
+\]
+
+Depending on the topology placed on $H \otimes H$, and the corresponding completion, a handful of different topologies arise on $B(H)$. In fact, the above duality produces a predual for $B(H)$, being the trace class operators: 
+
+\begin{definition}[Ultraweak Topology]
+\label{definition:bh-ultraweak-topology}
+    Let $H$ be a complex Hilbert space, then the dual of $H \wh \otimes_\pi H$ is $B(H)$, and the $\sigma(B(H), H \wh \otimes_\pi H)$-topology is the \textbf{ultraweak}/\textbf{$\sigma$-weak} topology on $B(H)$. 
+\end{definition}
+\begin{proof}
+    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}
+\label{proposition:bh-ultraweak-bounded}
+    Let $H$ be a complex Hilbert space, then every bounded subset of $B(H)$ is relatively compact with respect to the ultraweak topology. 
+\end{proposition}
+\begin{proof}
+    By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. 
+\end{proof}
+
+Now, a few facts about the more familiar operator topologies:  
+
+\begin{proposition}
+\label{proposition:bh-operator-topologies-facts}
+    Let $H$ be a complex Hilbert space, then:
+    \begin{enumerate}
+        \item The dual of $B(H)$ with respect to its strong and weak operator topologies is $H \otimes H$. 
+        \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 $B(H)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): By \autoref{proposition:bh-ultraweak-bounded}. 
+
+    (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

@@ -0,0 +1,257 @@
+\section{Von Neumann Algebras}
+\label{section:vna}
+
+\begin{theorem}[Existence of Projections]
+\label{theorem:existence-of-projections-vna}
+    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 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
+    \begin{enumerate}[start=2]
+        \item Let $T \in A_{sa}$ with $0 \le T \le I$ and $P \in B(H)$ be the orthogonal projection onto $\ol{T(H)}$, then $P = \sotlim_{n \to \infty}T^{1/n} \in A$. 
+        \item For each $T \in A$, the orthogonal projection onto $\ol{T(H)}$ is in $A$. 
+    \end{enumerate}
+
+    Finally, 
+    \begin{enumerate}[start=4]
+        \item There exists a maximum projection $P \in A$ such that $PT = TP = T$ for all $T \in A$, which is the multiplicative unit of $A$. 
+    \end{enumerate}
+    
+    
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Section 17]{Zhu}}}. ]
+    (1): After rescaling, assume without loss of generality that $-I \le T \le I$ for all $T \in \cf$. Since $\cf \subset A_{sa}$, $\norm{T}_{B(H)} = [T]_{sp} \le 1$ by \autoref{theorem:c-star-normal-spectral-radius}, where the spectral radius is taken with respect to $B(H)$. 
+    
+    Thus $\cf \subset \ol{B_{A}(0, 1)}$, and is relatively compact in the weak operator topology by the \hyperref[Banach-Alaoglu Theorem]{proposition:bh-ultraweak-bounded}. As such, $\bigcap_{T \in \cf}\ol{\bracs{S \in \cf|S \ge T}}^{\text{\small WOT}} \ne \emptyset$. Let $R \in \bigcap_{T \in \cf}\ol{\bracs{S \in \cf|S \ge T}}^{\text{\small WOT}}$. Since $A$ is strong-operator closed, so is $A_{sa}$ by \autoref{proposition:bh-operator-topologies-facts}. Thus for each $T \in A_{sa}$, $\bracs{S \in A_{sa}|S \ge T}$ is closed in the weak operator topology, and $R \in A_{sa}$ with $R \ge T$ for all $T \in \cf$. 
+
+    Let $S \in B(H)$ be self-adjoint such that $S \ge T$ for all $T \in \cf$, then $S \ge T$ for all $T \in \ol{\cf}^{\text{\small WOT}}$. In particular, $S \ge R$, thus $R$ is indeed the supremum of $\cf$. 
+
+    Finally, let $x \in H$ and $\eps > 0$, then there exists $T \in \cf$ such that $\dpn{(R - T)x, x}{H} \le \eps$. For any $S \in \cf$ with $S \ge T$,
+    \[
+    \normn{(R - S)^{1/2}x}_H^2 = \dpn{(R - S)x, x}{H} \le \dpn{(R - T)x, x}{H} \le \eps
+    \] 
+
+    By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} and \autoref{theorem:c-star-normal-spectral-radius},
+    \[
+    \normn{(R - S)^{1/2}}_{B(H)} = \norm{R - S}_{B(H)}^{1/2} \le \sqrt{2}
+    \]
+    
+    so
+    \[
+    \norm{(R - S)x}_H \le \normn{(R - S)^{1/2}}_{B(H)} \cdot \normn{(R - S)^{1/2}x}_H \le \sqrt{2 \eps}
+    \]
+
+    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): 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, 
+    \[
+    \dpn{\sup(\mathcal{P})x, x}{H} \ge \dpn{Px, x}{H} = \dpn{x, x}{H} = \norm{x}_H^2
+    \]
+
+    Thus $x \in \sup(\mathcal{P})(H)$, and $\sup(\mathcal{P})(H) \supset {\ol{\bigcup_{P \in \mathcal{P}}P(H)}}$. 
+
+    On the other hand, let $Q \in B(H)$ be the orthogonal projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$, then $Q$ is also an upper bound of $\mathcal{P}$. Therefore
+    \[
+    \ol{\bigcup_{P \in \mathcal{P}}P(H)} = Q(H) \supset \sup(\mathcal{P})(H)
+    \]
+
+    (3): As $0 \le T \le I$, $\sigma_{B(H)}(T) \subset [0, 1]$ by the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}. For each $n \in \natp$ and $t \in [0, 1]$, let $f_n(t) = t^{1/n}$, then $f_n$ is an increasing sequence of continuous functions on $[0, 1]$. By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\bracsn{f_n(T)}_1^\infty = \bracsn{T^{1/n}}_1^\infty$ is an increasing sequence that lies between $0$ and $I$. 
+
+    Let $n \in \natp$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:stone-weierstrass}, there exist polynomials $\seq{p_{n, k}} \subset \real[x]$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $k \in \natp$, $p_{n, k}(0) = 0$. 
+        \item $p_{n, k} \to f_n$ uniformly on $\sigma_{B(H)}(T)$. 
+    \end{enumerate}
+
+    As $A$ is a uniformly closed subalgebra of $B(H)$, $f_n(T) \in A$ for all $n \in \natp$. 
+    
+    By (1), the supremum $Q = \sup_{n \in \natp}T^{1/n} = \sotlim_{n \to \infty}T^{1/n}$ exists in $A$. Since
+    \[
+    Q^2 = \sotlim_{n \to \infty}T^{2/n} = \sotlim_{n \to \infty}T^{1/n} = Q
+    \]
+
+    and $Q$ is self-adjoint, $Q$ is a projection. 
+    
+    For each $x \in H \setminus \ker(T)$, $\dpn{Qx, x}{H} \ge \dpn{Tx, x}{H} > 0$ because $T$ is positive. As such, $\ker(Q) \subset \ker(T)$. For any $x \in \ker(T)$, $\dpn{Qx, x}{H} = \limv{n}\dpn{T^{1/n}x, x}{H} = 0$, so $\ker(Q) \supset \ker(T)$. Since both operators are self-adjoint, $\ol{Q(H)} = \ol{T(H)} = P(H)$, and $P = Q$. 
+
+    (4): Assume without loss of generality that $T \ne 0$. Let $x \in H$, then
+    \[
+    \norm{T^*x}_H^2 = \dpn{T^*x, T^*x}{H} = \dpn{TT^*x, x}{H}
+    \]
+
+    Thus $\ker(T^*) = \ker(TT^*)$, and
+    \[
+    \ol{T(H)} = \ker(T^*)^\perp = \ker(TT^*)^\perp = \ol{TT^*(H)}
+    \]
+
+    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 $A$, 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}
+
+\begin{lemma}
+\label{lemma:invariant-projection-test}
+    Let $H$ be a complex Hilbert space, $M \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $M$, and $T \in B(H)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $T(M) \subset M$. 
+        \item $PTP = TP$. 
+    \end{enumerate}
+\end{lemma}
+% Proof omitted due to obviousness
+
+
+\begin{definition}[Reducing Subspace]
+\label{definition:reducing-subspace}
+    Let $H$ be a complex Hilbert space, $M \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $P$, and $T \in B(H)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $T(M) \subset M$ and $T^*(M) \subset M$. 
+        \item $TP = PT$. 
+    \end{enumerate}
+
+    If the above holds, then $M$ is a \textbf{reducing subspace} of $T$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Corollary 18.3]{Zhu}}}. ]
+    (1) $\Rightarrow$ (2): By \autoref{lemma:invariant-projection-test}, $PTP = TP$ and $PT^*P = T^*P$. Thus $TP = PTP = PT$. 
+
+    (2) $\Rightarrow$ (1): Since $TP = PT$, $PTP = PT$, and $T(M) \subset M$ by \autoref{lemma:invariant-projection-test}. Similarly, $T^*P = PT^*$ implies that $PT^*P = PT^*$, and $T^*(M) \subset M$ by \autoref{lemma:invariant-projection-test}. 
+\end{proof}
+
+\begin{definition}[Commutant]
+\label{definition:commutant}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$, then
+    \[
+    A' = \bracs{T \in B(H)| TS = ST \forall S \in A}
+    \]
+
+    is the \textbf{commutant} of $A$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:bicommutant-pointwise}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a unital, self-adjoint subalgebra, $T \in A''$, and $x \in H$, then $Tx \in \ol{\bracsn{Sx|S \in A}}$.
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 18.4]{Zhu}}}. ]
+    Let $M = \ol{\bracsn{Sx|S \in A}}$, then since $A$ is self-adjoint, $M$ is a reducing subspace for each element of $A$. Let $P \in B(H)$ be the orthogonal projection onto $M$, then $PS = SP$ for all $S \in A$. As such, $P \in A'$. 
+
+    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 $A$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Section 18.3]{Zhu}}}. ]
+    ($\overline{A}^{\text{\small SOT}} \subset A''$): By separate continuity of composition, $A''$ is a strong-operator closed subset that contains $A$. Hence $A''$ contains the strong-operator closure of $A$. 
+
+    ($\overline{A}^{\text{\small SOT}} \supset A''$): Let $T \in A''$ and $x = \seqf{x_j} \in H^n$. For each $S \in B(H)$, denote $S^{(n)} = (S, \cdots, S)$ ($n$-copies), then
+    \begin{enumerate}[label=(\roman*)]
+        \item $A^{(n)} = \bracsn{S^{(n)}|S \in A}$ is a unital, self-adjoint subalgebra of $B(H^n)$. 
+        \item $T^{(n)} \in (A^{(n)})''$. 
+    \end{enumerate}
+
+    By \autoref{lemma:bicommutant-pointwise},
+    \[
+    (Tx_1, \cdots, Tx_n) \in \ol{\bracsn{(Sx_1, \cdots, Sx_n)|S \in A}}
+    \]
+
+    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/dst/analytic.tex

@@ -0,0 +1,34 @@
+\section{Analytic Sets}
+\label{section:analytic-sets}
+
+\begin{definition}[Analytic Set]
+\label{definition:analytic-set}
+    Let $X$ be a Polish space and $A \subset X$, then $A$ is \textbf{analytic} if there exists a Polish space $Z$ and $f \in C(Z; X)$ such that $f(Z) = A$. 
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:analytic-sets}
+    Let $X$ be a Polish space, then:
+    \begin{enumerate}
+        \item For each family $\seq{A_n} \subset X$ of analytic sets, $\bigcup_{n \in \natp}A_n$ is analytic. 
+        \item For each family $\seq{A_n} \subset X$ of analytic sets, $\bigcap_{n \in \natp}A_n$ is analytic. 
+        \item For any $A \in \cb_X$, $A$ is analytic. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 8.2.1-8.2.3]{CohnMeasure}}}. ]
+    For each $n \in \natp$, let $Z_n$ be a Polish space and $f_n \in C(Z_n; X)$ such that $A_n = f_n(Z_n)$. 
+
+    (1): By \autoref{proposition:polish-space-extension}, $Z = \bigsqcup_{n \in \natp}Z_n$ is a Polish space. Let $f \in C(Z; X)$ be the gluing of $\seq{f_n}$, then $\bigcup_{n \in \natp}A_n = f(Z)$.
+
+    (2): By \autoref{proposition:polish-space-extension}, $Z = \prod_{n \in \natp}Z_n$ is a Polish space. For each $m, n \in \natp$, $\bracs{f_m \circ \pi_m = f_n \circ \pi_n}$ is closed. Thus
+    \[
+    \Delta := \bigcap_{m \in \natp}\bigcap_{n \in \natp}\bracs{f_m \circ \pi_m = f_n \circ \pi_n}
+    \]
+
+    is closed, and Polish by \autoref{proposition:polish-space-extension}. Hence $\bigcap_{n \in \natp}A_n = f_1 \circ \pi_1(\Delta)$ is also analytic.
+
+    (3): By \autoref{proposition:polish-space-extension}, every open and closed subset of $X$ is analytic. Thus (1), (2), and \autoref{lemma:monotone-borel-characterisation} imply that every element of $\cb_X$ is analytic.
+\end{proof}
+
+

src/topology/dst/index.tex

@@ -0,0 +1,6 @@
+\chapter{Polish Spaces and Analytic Sets}
+\label{chap:polish-spaces}
+
+\input{./polish.tex}
+\input{./analytic.tex}
+\input{./zero.tex}

src/topology/dst/polish.tex

@@ -0,0 +1,65 @@
+\section{Polish Spaces}
+\label{section:polish}
+
+
+
+
+\begin{definition}[Polish Space]
+\label{definition:polish-space}
+    Let $X$ be a topological space, then $X$ is \textbf{Polish} if it is completely metrisable and second countable. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:polish-space-extension}
+    The following spaces are Polish:
+    \begin{enumerate}
+        \item Closed subspace of a Polish space.
+        \item Open subspace of a Polish space. 
+        \item Countable products of Polish spaces. 
+        \item Countable disjoint union of Polish spaces. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $X$ be a Polish space with complete metric $d$ and $A \subset X$ be a closed subset, then $A$ is second countable. By \autoref{proposition:complete-closed}, $A$ is complete with respect to $d$. 
+
+    (2): Let $X$ be a Polish space with complete metric $d$ and $U \subset X$ be open. Assume without loss of generality that $U \subsetneq X$ and $d(X \times X) \subset [0, 1]$. Define
+    \[
+    d_U: U \times U \to [0, \infty] \quad (x, y) \mapsto d(x, y) + \abs{\frac{1}{d(x, U^c)} - \frac{1}{d(y, U^c)}}
+    \]
+
+    then $d_U$ is a metric on $U$. Since $d \le d_U$, the topology induced by $d_U$ is finer than the topology induced by $d$. On the other hand, the mapping $x \mapsto d(x, U^c)$ is continuous, so the topology induced by $d_U$ is coarser than the topology induced by $d$. Therefore $d_U$ induces the subspace topology of $U$. 
+
+    Now, let $\seq{x_n} \subset U$ be a Cauchy sequence with respect to $d_U$, then there exists $N \in \natp$ such that $d_U(x_m, x_n) \le 1$ for all $m, n \ge N$. Thus
+    \[
+    \frac{1}{d(x_n, U^c)} \le \frac{1}{d(x_N, U^c)} + d(x_n, x_N) + \abs{\frac{1}{d(x_n, U^c)} - \frac{1}{d(x_N, U^c)}} \le \frac{1}{d(x_N, U^c)} + 1
+    \]
+
+    and $\delta = \inf_{n \in \natp}d(x_n, U^c) > 0$. Since $d \le d_U$, $\seq{x_n}$ is Cauchy with respect to $d$ as well. Thus as $\bracsn{x \in X|d(x, U^c) \ge \delta} \subset U$ is a closed subset of $X$, there exists $x \in U$ such that $x_n \to x$ as $n \to \infty$. Therefore $U$ is complete with respect to $d_U$. 
+
+    (3): By \autoref{proposition:separable-product}, \autoref{proposition:product-complete}, and \autoref{theorem:uniform-metrisable}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:polish-subspace}
+    Let $X$ be a Polish space and $Y \subset X$, then $Y$ is Polish if and only if it is $G_\delta$ in $X$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 8.1.5]{CohnMeasure}}}. ]
+    ($\Rightarrow$): Suppose that $Y$ is Polish. Let $d_X: X^2 \to [0, 1]$ and $d_Y: Y^2 \to [0, 1]$ be complete metrics on $X$ and $Y$, respectively. For each $n \in \natp$, let $\mathcal{U}_n \subset 2^X$ be the collection of subsets of $X$ such that for each $U \in \mathcal{U}_n$,
+    \begin{enumerate}[label=(\roman*)]
+        \item $U$ is a non-empty open subset of $X$. 
+        \item $\sup_{x, y \in U}d_X(x, y) \le 1/n$. 
+        \item $\sup_{x, y \in U \cap Y}d_Y(x, y) \le 1/n$. 
+    \end{enumerate}
+    
+    Let $U_n = \bigcup_{U \in \mathcal{U}_n}U$, then $Y \subset \ol{Y} \cap \bigcap_{n \in \natp}U_n$. On the other hand, let $x \in \ol{Y} \cap \bigcap_{n \in \natp}U_n$. Let $V_1 \in \mathcal{U}_1 \cap \cn_X(x)$ and $x_1 \in V_1$. For each $n \in \natp$ with $n \ge 2$, let $V_n \in \mathcal{U}_n \cap \cn_X(x)$ with $V_n \subset V_{n-1}$ and $x_n \in V_n \cap Y$, then by (ii) and (iii), $\seq{x_n}$ is Cauchy with respect to $d_X$ and $d_Y$. In particular, there exists $y \in Y$ such that $x_n \to y$ with respect to $d_Y$ as $n \to \infty$. Since $d_Y$ induces the subspace topology on $Y$, $x_n \to y$ with respect to $d_X$ as $n \to \infty$ as well. Therefore $x = y$, and $Y \supset \ol{Y} \cap \bigcap_{n \in \natp}U_n$. 
+
+    As every closed subset of $X$ is $G_\delta$, $Y = \ol{Y} \cap \bigcap_{n \in \natp}U_n$ is also $G_\delta$. 
+
+    ($\Leftarrow$): Suppose that $Y$ is $G_\delta$ in $X$. Let $\seq{U_n} \subset 2^X$ be open sets such that $Y = \bigcap_{n \in \natp}U_n$. In which case, $Y$ is homeomorphic to the diagonal
+    \[
+    \Delta = \bracs{x \in \prod_{n \in \natp}U_n \bigg | x_m = x_n \forall m, n \in \natp}
+    \]
+
+    For each $n \in \natp$, $U_n$ is Polish by (2) of \autoref{proposition:polish-space-extension}. As a closed subspace of a product of Polish spaces, $\Delta$ is Polish by (1) and (3) of \autoref{proposition:polish-space-extension}. Therefore $Y$ is also Polish. 
+\end{proof}
+

src/topology/dst/zero.tex

@@ -0,0 +1,124 @@
+\section{Zero Dimensional Spaces}
+\label{section:zero-dimensional}
+
+\begin{definition}[Zero-Dimensional]
+\label{definition:zero-dimentional}
+    Let $X$ be a topological space, then $X$ is \textbf{zero-dimensional} if $X$ admits a base consisting of clopen sets. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:zero-dimensional-extension}
+    Let $\seqi{X}$ and $X$ be zero-dimensional spaces, then:
+    \begin{enumerate}
+        \item For any $A \subset X$, $A$ is zero-dimensional. 
+        \item $\prod_{i \in I}X_i$ is zero-dimensional.
+        \item $\bigsqcup_{i \in I}X_i$ is zero-dimensional. 
+    \end{enumerate}
+\end{proposition}
+% Proof omitted. 
+
+\begin{definition}[Cantor Space]
+\label{definition:cantor-space}
+    Let $2 = \bracs{0, 1}$ be equipped with the discrete topology, then $2^{\natp}$ is the \textbf{Cantor space}. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:cantor-space-embedding}
+    The mapping
+    \[
+    2^{\natp} \to [0, 1] \quad \seq{x_n} \mapsto 2\sum_{n \in \natp} \frac{x_n}{3^n}
+    \]
+
+    is an embedding. 
+\end{proposition}
+
+
+\begin{definition}[Baire Space]
+\label{definition:the-baire-space}
+    Let $\natp$ be equipped with the discrete topology, then $\mathscr{N} = (\natp)^{\natp}$ is the \textbf{Baire space}. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:baire-universality}
+    Let $X$ be a non-empty Polish space, then there exists a surjective mapping $f \in C(\mathscr{N}; X)$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 8.2.7]{CohnMeasure}}}. ]
+    Let $d: X^2 \to [0, \infty)$ be a complete metric on $X$. To construct the desired map, it is sufficient to construct $\bracsn{C(n_1, \cdots, n_N)| \bracsn{n_k}_1^N \subset \natp, N \in \natp} \subset 2^X$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $N \in \natp$, $\bracsn{n_k}_1^N \subset \natp$, $C(n_1, \cdots, n_N) \subset X$ is closed and non-empty. 
+        \item For each $N \in \natp$, $\bracsn{n_k}_1^N \subset \natp$, $\text{diam}(C(n_1, \cdots, n_N)) \le 1/N$. 
+        \item For each $N \in \natp$ and $\bracsn{n_k}_1^{N} \subset \natp$,
+        \[
+        C(n_1, \cdots, n_{N}) = \bigcup_{n_{N+1} \in \natp}C(n_1, \cdots, n_{N+1})
+        \]
+        \item $X = \bigcup_{n_1 \in \natp}C(n_1)$. 
+    \end{enumerate}
+
+    Since $X$ is Polish, there exists a countable dense subset $\seq{x_k} \subset X$. For each $n_1 \in \natp$, let $C(n_1) = \ol{B(x_{n_1}, 1/2)}$, then $\bracsn{C(n_1)|n_1 \in \natp}$ satisfies (i) and (ii) by definition. As $\seq{x_k}$ is dense in $X$, $X = \bigcup_{n_1 \in \natp}C(n_1)$, and (iv) is also satisfied. 
+
+    Let $N \in \natp$ and suppose inductively that $\bracsn{C(n_1, \cdots, n_k)| \bracsn{n_j}_1^k \subset \natp,1 \le k \le N}$ has been constructed to satisfy (i)-(iv) for each $1 \le k \le N$. 
+    
+    Fix $\bracsn{n_k}_1^N \subset \natp$. By \autoref{proposition:separable-metric-space}, there exists a countable dense subset $\seq{y_k} \subset C(n_1, \cdots, n_N)$. For each $n_{N+1} \in \natp$, let $C(n_1, \cdots, n_{N+1}) = \ol{B(y_{n_{N+1}}, 1/[2(N+1)])} \cap C(n_1, \cdots, n_N)$, then $\bracsn{C(n_1, \cdots, n_{N+1})|n_{N+1} \in \natp}$ satisfies (i) and (ii) by definition. Since $\seq{y_k}$ is dense in $C(n_1, \cdots, n_N)$,
+    \[
+     C(n_1, \cdots, n_{N}) = \bigcup_{n_{N+1} \in \natp}C(n_1, \cdots, n_{N+1})
+    \]
+
+    so (iii) is also satisfied. 
+
+    Finally, suppose that $\bracsn{C(n_1, \cdots, n_N)| \bracsn{n_k}_1^N \subset \natp, N \in \natp} \subset 2^X$ has been constructed to satisfy (i)-(iv). Since $X$ is complete, for each $\seq{n_k} \in \mathscr{N}$, there exists a unique $f(\seq{n_k}) \in X$ such that $\bracs{f(\seq{n_k})} = \bigcap_{N \in \natp}C(n_1, \cdots, n_N)$. For each $x \in X$, (iii) and (iv) imply that there exists $\seq{n_k} \in \mathscr{N}$ such that $x = f(\seq{n_k})$. As such, $f: \mathscr{N} \to X$ is a surjective mapping. For any $\seq{m_k}, \seq{n_k} \in \mathscr{N}$ and $N \in \natp$ with $m_k = n_k$ for each $1 \le k \le N$, $d(f(\seq{m_k}), f(\seq{n_k})) \le 1/N$. Therefore $f \in C(\mathscr{N}; X)$. 
+\end{proof}
+
+\begin{theorem}[Alexandroff-Hausdorff]
+\label{theorem:cantor-universality}
+    Let $X$ be a non-empty compact metrisable space, then there exists a surjective mapping $f \in C(2^{\natp}; X)$. 
+\end{theorem}
+\begin{proof}[Proof, adapted from {{\cite[Proposition 8.2.7]{CohnMeasure}}}. ]
+    Let $d: X \times X \to [0, \infty)$ be a metric on $X$. Since $X$ is compact, for each $N \in \natp$, there exists $K_N \in \natp$ and $\seqf{x_{N, k}|1 \le k \le K_N} \subset X$ such that $X = \bigcup_{k = 1}^{K_N}B(x_{N, k}, 1/N)$.
+
+    To construct the desired map, it is sufficient to construct
+    \[
+    \bracs{C(n_1, \cdots, n_N) \bigg | \bracsn{n_k}_1^N \in \prod_{n = 1}^{N}\bracs{1, \cdots, K_n}, N \in \natp} \subset 2^X
+    \]
+    
+    
+    such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $N \in \natp$ and $\bracsn{n_k}_1^N$, $C(n_1, \cdots, n_N) \subset X$ is closed and non-empty. 
+        \item For each $N \in \natp$ and $\bracsn{n_k}_1^N$, $\text{diam}(C(n_1, \cdots, n_N)) \le 4/N$. 
+        \item For each $N \in \natp$ and $\bracsn{n_k}_1^{N}$,
+        \[
+        C(n_1, \cdots, n_{N}) = \bigcup_{n_{N+1} = 1}^{K_{N+1}}C(n_1, \cdots, n_{N+1})
+        \]
+        \item $X = \bigcup_{n_1 = 1}^{K_1}C(n_1)$. 
+    \end{enumerate}
+
+    For each $1 \le n_1 \le K_1$, let $C(n_1) = \ol{B(x_{1, n_1}, 1)}$, then $\bracsn{C(n_1)|1 \le n_1 \le K_1}$ satisfies (i), (ii), and (iv) by definition. 
+
+    Let $N \in \natp$ and suppose inductively that
+    \[
+    \bracs{C(n_1, \cdots, n_K) \bigg | \bracsn{n_k}_1^K \in \prod_{n = 1}^{K}\bracs{1, \cdots, K_n}, 1 \le K \le N} \subset 2^X
+    \]
+    
+    
+    has been constructed to satisfy (i)-(iv) for each $1 \le K \le N$. Fix $\bracsn{n_k}_1^N \in \prod_{n = 1}^{N}\bracs{1, \cdots, K_n}$. Since $X = \bigcup_{k = 1}^{K_{N+1}}B(x_{N+1, k}, 1/(N+1))$, there exists $\bracsn{y_k}_1^{K_{N+1}} \subset C(n_1, \cdots, n_N)$ such that $C(n_1, \cdots, n_N) \subset \bigcup_{k = 1}^{K_{N+1}}B(y_k, 2/(N+1))$. For each $1 \le n_{N+1} \le K_{N+1}$, let $C(n_1, \cdots, n_{N+1}) = C(n_1, \cdots, n_N) \cap \ol{B(y_{n_{N+1}}, 2/(N+1))}$, then $\bracsn{C(n_1, \cdots, n_{N+1})|1 \le n_{N+1} \le K_{N+1}}$ satisfies (i)-(iii). 
+
+    Now, suppose that
+    \[
+    \bracs{C(n_1, \cdots, n_N) \bigg | \bracsn{n_k}_1^N \in \prod_{n = 1}^{N}\bracs{1, \cdots, K_n}, N \in \natp} \subset 2^X
+    \]
+    
+    has been constructed to satisfy (i)-(iv). Since $X$ is complete, for each $\seq{n_k} \in \prod_{n \in \natp}\bracs{1, \cdots, K_n}$, there exists a unique $f(\seq{n_k}) \in X$ such that $\bracs{f(\seq{n_k})} = \bigcap_{N \in \natp}C(n_1, \cdots, n_N)$. For each $x \in X$, (iii) and (iv) imply that there exists $\seq{n_k} \in \prod_{n \in \natp}\bracs{1, \cdots, K_n}$ such that $x = f(\seq{n_k})$. As such, $f: \prod_{n \in \natp}\bracs{1, \cdots, K_n} \to X$ is a surjective mapping. For any $\seq{m_k}, \seq{n_k} \in \prod_{n \in \natp}\bracs{1, \cdots, K_n}$ and $N \in \natp$ with $m_k = n_k$ for each $1 \le k \le N$, $d(f(\seq{m_k}), f(\seq{n_k})) \le 4/N$. Therefore $f \in C(\prod_{n \in \natp}\bracs{1, \cdots, K_n}; X)$. 
+
+    Finally, for each $n \in \natp$, there exists $L_n \in \natp$ and a surjective mapping $g_n: 2^{L_n} \to \bracs{1, \cdots, K_n}$. Let
+    \[
+    g = \prod_{n \in \natp}g_n: \prod_{n \in \natp}2^{L_n} \to \prod_{n \in \natp}\bracs{1, \cdots, K_n}
+    \]
+    
+    be the product of $\seq{g_n}$, then $g \in C(\prod_{n \in \natp}2^{L_n}; \prod_{n \in \natp}\bracs{1, \cdots, K_n})$, and
+    \[
+    f \circ g: 2^{\natp} \iso \prod_{n \in \natp}2^{L_n} \to X
+    \]
+
+    is the desired continuous surjection. 
+\end{proof}
+

src/topology/index.tex

@@ -6,4 +6,5 @@
 \input{./functions/index.tex}
 \input{./metric/index.tex}
 \input{./groups/index.tex}
+\input{./dst/index.tex}
 \input{./notation.tex}

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/c0.tex

@@ -64,4 +64,43 @@ For details regarding the complex-valued cased, in particular its properties as
 
 \end{proof}
 
+\begin{theorem}
+\label{theorem:c0-beq-tensor}
+    Let $X$ be a LCH space and $E$ be a complete locally convex space over $K \in \RC$, then the canonical map
+    \[
+    C_0(X; K) \otimes_\eps E \to C_0(X; E) \quad \sum_{j = 1}^n f_j \otimes y_j \mapsto \sum_{j = 1}^n y_j \cdot f_j
+    \]
+
+    extends into an isomorphism between $C_0(X; K) \wh{\otimes}_\eps E$ and $C_0(X; E)$. Moreover, if $E$ is a Banach space, then the isomorphism is an isometry. 
+\end{theorem}
+\begin{proof}
+    To see that the canonical map is continuous, let $\rho: E \to [0, \infty)$ be a continuous seminorm on $E$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists an equicontinuous family $T \subset E^*$ such that for each $y \in E$, $\rho(y) = \sup_{\phi \in T}|\dpn{y, \phi}{E}|$. 
+
+    Let $\lambda = \sum_{j = 1}^n f_j \otimes y_j \in C_0(X; K) \otimes_\eps E$, then
+    \[
+    \sup_{x \in X}\rho(\lambda(x)) = \sup_{x \in X}\sup_{\phi \in T}|\dpn{\lambda(x), \phi}{E}| = \sup_{x \in X}\sup_{\phi \in T}\abs{\sum_{j = 1}^n f_j(x) \dpn{y_j, \phi}{E}}
+    \]
+
+    Since the evaluation maps $\bracsn{\pi_x: C_0(X; K) \to K|x \in X}$ are equicontinuous, the uniform seminorm on $C_0(X; E)$ with respect to $\rho$ is bounded above by a cross seminorm of the injective tensor product. Thus the inclusion is continuous. 
+
+    On the other hand, let $T \subset E^*$ be equicontinuous, then there exists a continuous seminorm $\rho: E \to [0, \infty)$ such that $|\phi| \le \rho$ for all $\phi \in T$. In which case, for any $\lambda = \sum_{j = 1}^n f_j \otimes y_j \in C_0(X; K) \otimes_\eps E$, $I \in C_0(X; K)^*$, and $\phi \in T$, 
+    \begin{align*}
+        \abs{\sum_{j = 1}^n \dpn{f_j, I}{C_0(X; K)}\dpn{y_j, \phi}{E}} &= \abs{\angles{\sum_{j = 1}^nf_j\dpn{y_j, \phi}{E}, I}_{C_0(X; K)}} \\
+        &\le \norm{I}_{C_0(X; K)} \cdot \sup_{x \in X} \abs{\angles{\sum_{j = 1}^n f_j(x) y_j, \phi}_{E}} \\
+        &\le \norm{I}_{C_0(X; K)} \cdot \sup_{x \in X} \rho(\lambda(x))
+    \end{align*}
+    
+    Hence the cross seminorm corresponding to $B_{C_0(X; K)^*}(0, 1)$ and $T$ is bounded above by the uniform seminorm on $C_0(X; E)$ with respect to $\rho$, so the inclusion is an embedding. 
+
+    Finally, by \autoref{proposition:c0-properties}, $C_0(X; K)$ is complete. By \autoref{proposition:c0-tensor}, $C_0(X; K) \otimes_\eps E$ is dense in $C_0(X; E)$. Therefore the canonical map extends to an isomorphism through the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-tvs}.
+\end{proof}
+
+\begin{corollary}
+\label{corollary:c0-seq-beq-tensor}
+    Let $I$ be a set and $E$ be a complete locally convex space over $K \in \RC$, then $c_0(I; K) \wh \otimes_\eps E \iso c_0(I; E)$. 
+\end{corollary}
+\begin{proof}
+    By \autoref{theorem:c0-beq-tensor}. 
+\end{proof}
+
 

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/metric/metric.tex

@@ -68,11 +68,6 @@
     (2) $\Rightarrow$ (3): Since $\fF$ is Cauchy, there exists $\seq{E_n} \subset \fF$ such that for each $n \in \natp$, $E_n \supset E_{n+1}$ and $\sup_{y, z \in E_n}d(y, z) \le 1/n$. For each $n \in \natp$, let $x_n \in E_n$, then there exists a subsequence $\seq{n_k}$ and $x \in X$ such that $x = \limv{n}x_n$. In which case, $x \in \bigcap_{n \in \natp}\overline{E_n}$. For each $n \in \natp$, $\sup_{y, z\in E_n}d(y, z) \le 1/n$, so $B_X(x, 2/n) \supset E_n$. Therefore $\fF \to x$. 
 \end{proof}
 
-\begin{definition}[Polish Space]
-\label{definition:polish-space}
-    Let $X$ be a topological space, then $X$ is \textbf{Polish} if it is completely metrisable and second countable. 
-\end{definition}
-
 
 
 \begin{theorem}[Banach's Fixed Point Theorem]

src/topology/notation.tex

@@ -26,4 +26,15 @@
     $f \prec U$ & $f \in C_c(X; [0,1])$ 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} \\ 
 \end{tabular}