Added the Kaplansky Density Theorem.

src/op/notation.tex

@@ -19,6 +19,7 @@
     $P(A)$ & 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} \\ 
 
     $M_n(\complex)$ & 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/index.tex

@@ -2,4 +2,5 @@
 \label{chap:von-neumann-algebras}
 
 \input{./topologies.tex}
+\input{./cayley.tex}
 \input{./vn.tex}

src/op/vn/vn.tex

@@ -176,3 +176,76 @@
     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{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}
+        \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}
+
+