Added the bicommutant theorem.

src/op/vn/topologies.tex

@@ -35,17 +35,14 @@ Now, a few facts about the more familiar operator topologies:
         \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 For any bounded subset $B \subset B(H)$, the composition map $(S, T) \mapsto ST$ restricted to $B \times B(H)$ is continuous in the strong operator topology. 
+        \item The composition map $(S, T) \mapsto ST$ is left-hypocontinuous with respect to the strong operator topology and strong-operator bounded subsets of $B(H)$. 
         \item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology. 
     \end{enumerate}
 \end{proposition}
 \begin{proof}
     (2): By \autoref{proposition:bh-ultraweak-bounded}. 
 
-    (4): Let $\angles{S_\alpha}_{\alpha \in A} \subset B$, $\angles{T_\alpha}_{\alpha \in A} \subset B(H)$, and $(S, T) \in B \times B(H)$ such that $S_\alpha \to S$ and $T_\alpha \to T$ in the strong operator topology. Since $\{S_\alpha| \alpha \in A\}$ is equicontinuous, for any $x \in H$,
+    (4): By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, every strong-operator bounded subset of $B(H)$ is equicontinuous. 
-    \[
-    \lim_{\alpha \in A} S_\alpha T_\alpha x = \lim_{\alpha \in A}S_\alpha Tx = \lim_{\alpha \in A}STx
-    \]
 \end{proof}
 
 

src/op/vn/vn.tex

@@ -6,18 +6,17 @@
     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]
+    \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=5]
+    \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}
     
@@ -47,9 +46,7 @@
 
     for all $S \in \cf$ with $S \ge T$. As such a $T$ exists for all $\eps > 0$, $R = \sotlim_{T \in \cf}T$. 
 
-    (2): Assume without loss of generality that $A$ 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 $A$. 
+    (2): Assume without loss of generality that $\mathcal{P}$ is directed. By (1), $\sup(\mathcal{P})$ exists in $A$. Since the set of projections in $B(H)$ is strong-operator closed, $\sup(\mathcal{P}) = \sotlim_{P \in \mathcal{P}}P$ is also a projection. For each $x \in \bigcup_{P \in \mathcal{P}}P(H)$, there exists $P \in \mathcal{P}$ with $Px = x$. In which case, 
-
-    (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, 
     \[
     \dpn{\sup(\mathcal{P})x, x}{H} \ge \dpn{Px, x}{H} = \dpn{x, x}{H} = \norm{x}_H^2
     \]
@@ -61,7 +58,7 @@
     \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 $I$. 
+    (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*)]
@@ -78,9 +75,9 @@
 
     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 $P = Q$. 
+    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 $P = Q$. 
 
-    (5): Assume without loss of generality that $T \ne 0$. Let $x \in H$, then
+    (4): Assume without loss of generality that $T \ne 0$. Let $x \in H$, then
     \[
     \norm{T^*x}_H^2 = \dpn{T^*x, T^*x}{H} = \dpn{TT^*x, x}{H}
     \]
@@ -90,12 +87,78 @@
     \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 $A$. 
+    By (3) 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 $A$. 
+    (5): Let $\mathcal{P}$ be the set of all projections in $A$, then $\mathcal{P} \subset A_{sa}$ is bounded and directed. By (2), $P = \sup_{Q \in \mathcal{P}}Q \in A$, which is the maximum projection in $A$. 
 
-    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 $A$. 
+    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{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}
+