Added the existence of projections.

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 +226,9 @@
 \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}}
+

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/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/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/vn/index.tex

@@ -2,3 +2,4 @@
 \label{chap:von-neumann-algebras}
 
 \input{./topologies.tex}
+\input{./vn.tex}

src/op/vn/vn.tex

@@ -0,0 +1,101 @@
+\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 bounded commuting family $\cf \subset A_{sa}$, $\sup(\cf)$ exists in $A_{sa}$. 
+        \item For any family of projections $\mathcal{P} \subset A_{sa}$, $\sup(\mathcal{P}) \in A$ is the projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$. 
+    \end{enumerate}
+
+    and
+    \begin{enumerate}[start=3]
+        \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=5]
+        \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[Theorem 17.1]{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): Assume without loss of generality that $A$ is the smallest strong-operator closed $C^*$-subalgebra of $B(H)$ containing $\cf$. In which case, by separate continuity of composition in the strong operator topology, $A$ is also commutative. Thus $A_{sa}$ is a lattice by the \hyperref[Gelfand-Naimark Theorem]{section:gelfand-naimark}, and $\cf$ may be extended into a directed family. By (1), $\sup(\cf)$ exists in $A$. 
+
+    (3): Assume without loss of generality that $\mathcal{P}$ is directed. By (1), $\sup(\mathcal{P})$ exists in $A$. Since the set of projections in $B(H)$ is strong-operator closed, $\sup(\mathcal{P}) = \sotlim_{P \in \mathcal{P}}P$ is also a projection. For each $x \in \bigcup_{P \in \mathcal{P}}P(H)$, there exists $P \in \mathcal{P}$ with $Px = x$. In which case, 
+    \[
+    \dpn{\sup(\mathcal{P})x, x}{H} \ge \dpn{Px, x}{H} = \dpn{x, x}{H} = \norm{x}_H^2
+    \]
+
+    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
+    \[
+    \ol{\bigcup_{P \in \mathcal{P}}P(H)} = Q(H) \supset \sup(\mathcal{P})(H)
+    \]
+
+    (4): 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 $x \in T(H) \setminus \bracs{0}$, $\dpn{Qx, x}{H} \ge \dpn{Tx, x}{H} > 0$ 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 $P = Q$. 
+
+    (5): 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 (4) applied to $TT^*/\norm{TT^*}_{B(H)}$, the orthogonal projection onto $\ol{T(H)}$ is in $A$. 
+
+    (6): Let $\mathcal{P}$ be the set of all projections in $A$, then $\mathcal{P} \subset A_{sa}$ is bounded and directed. By (3), $P = \sup_{Q \in \mathcal{P}}Q \in A$, which is the maximum projection in $A$. 
+
+    Let $T \in A$, then by (5), $P$ 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}
+
+
+