Added a continuity result in strong operator topology.

src/op/vn/topologies.tex

@@ -16,6 +16,19 @@ Depending on the topology placed on $H \otimes H$, and the corresponding complet
     By \autoref{proposition:projective-tensor-product-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}. 
 \end{proof}
 
+A natural topology consistent with the ultraweak topology would be the ultrastrong topology. 
+
+\begin{definition}[Ultrastrong Topology]
+\label{definition:bh-ultrastrong-topology}
+    Let $H$ be a complex Hilbert space. For each $x = \seq{x_n} \in L^2(\natp; H)$, let
+    \[
+    \Phi_x: B(H) \to L^2(\natp; H) \quad (\Phi_xT)_n = Tx_n
+    \]
+
+    then the \textbf{ultrastrong}/\textbf{$\sigma$-strong} topology on $B(H)$ is the topology generated by the maps $\bracsn{\Phi_x|x \in L^2(\natp; H)}$. 
+\end{definition}
+
+
 Seeing that $B(H)$ is a dual Banach space, the following fact is immediate:
 
 \begin{proposition}
@@ -36,7 +49,6 @@ Now, a few facts about the more familiar operator topologies:
         \item Every bounded subset of $B(H)$ is relatively compact in the weak operator topology. 
         \item The composition map $(S, T) \mapsto ST$ is separately continuous in the strong and weak operator topologies. 
         \item The composition map $(S, T) \mapsto ST$ is left-hypocontinuous with respect to the strong operator topology and strong-operator bounded subsets of $B(H)$. 
-        \item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology. 
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -45,5 +57,55 @@ Now, a few facts about the more familiar operator topologies:
     (4): By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, every strong-operator bounded subset of $B(H)$ is equicontinuous. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:bh-adjoint-strong-continuous}
+    Let $H$ be a complex Hilbert space, then:
+    \begin{enumerate}
+        \item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology. 
+        \item $T \mapsto T^*$ restricted to the normal operators is continuous in the strong operator topology. 
+        \item For any $f \in C(\complex; \complex)$, the mapping $T \mapsto f(T)$ restricted to any bounded set of normal operators is continuous in the strong operator topology. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Section 19.1]{Zhu}}}. ]
+    (2): Let $S, T \in B(H)$, then
+    \begin{align*}
+        \normn{(S^* - T^*)x}_H^2 &= \normn{S^*x}_H^2 + \normn{T^*x}_H^2 - \dpn{x, ST^*x}{H} - \dpn{ST^*x, x}{H} \\
+        &\le \normn{S^*x}_H^2 + \normn{T^*x}_H^2 - \dpn{x, TT^*x}{H} - \dpn{TT^*x, x}{H} \\
+        &+ |\dpn{x, (T - S)T^*x}{H}| + |\dpn{(T - S)T^*x, x}{H}| \\
+        &\le |\normn{S^*x}_H^2 - \normn{T^*x}_H^2| + 2\norm{x}_H\normn{(T - S)T^*x}_H
+    \end{align*}
 
+    Now, if $S$ and $T$ are normal, then $\normn{S^*x}_H = \norm{Sx}_H$ and $\norm{T^*x}_H = \norm{Tx}_H$, so
+    \begin{align*}
+    |\norm{S^*x}_H^2 - \norm{T^*x}_H^2| &= |\norm{Sx}_H^2 - \norm{Tx}_H^2| \\
+    &\le \norm{(S - T)x}_H (\norm{Sx}_H + \norm{Tx}_H) \\
+    &\le \norm{(S - T)x}_H (\norm{(S - T)x}_H + 2\norm{Tx}_H)
+    \end{align*}
+
+    Therefore
+    \begin{align*}
+    \normn{(S^* - T^*)x}_H^2 &\le \norm{(S - T)x}_H (\norm{(S - T)x}_H + 2\norm{Tx}_H) \\
+    &+ 2\norm{x}_H\normn{(T - S)T^*x}_H
+    \end{align*}
+    
+    and the adjoint map restricted to normal operators is continuous in the strong operator topology. 
+
+    (3): Let $S, T \in B_{B(H)}(0, 1)$ and $x \in H$ and $n \in \natp$, then
+    \begin{align*}
+    \normn{(S^n - T^n)x}_H &\le \sum_{k = 0}^{n-1}\normn{S^{n-1-k}(S - T)T^kx}_{H} \\
+    &\le \sum_{k = 0}^{n - 1}\normn{(S - T)T^kx}_H
+    \end{align*}
+    
+
+    so the mapping $T \mapsto T^n$ on $B_{B(H)}(0, 1)$ is continuous in the strong operator topology. By (2), the mapping $T \mapsto p(T, T^*)$ is strong-operator continuous for all $p \in \complex[z, \ol z]$. 
+
+    By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exist polynomials $p_n \in \complex[z, \ol z]$ such that $p_n \to f$ uniformly on $\ol{B_\complex(0, 1)}$. For any $T \in B_{B(H)}(0, 1)$, $x \in H$, and $n \in \natp$, 
+    \begin{align*}
+    \norm{[f(T) - p_n(T)]x}_H &\le \norm{f(T) - p_n(T)}_{B(H)} \cdot \norm{x}_H \\
+    &\le \norm{x}_H  \cdot \sup_{z \in \ol{B_\complex(0, 1)}}|f(z) - p_n(z)|
+    \end{align*}
+    
+
+    by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}. Thus $f$ is a uniform limit of strong-operator continuous functions on $B_{B(H)}(0, 1)$, and as such also strong-operator continuous by \autoref{proposition:uniform-limit-continuous}. 
+\end{proof}
 

src/op/vn/vn.tex

@@ -141,6 +141,20 @@
     Now, since $T \in A''$, $TP = PT$ as well, so $M$ is a reducing subspace for $T$. As $A$ is unital, $x \in M$, so $Tx \in M = \ol{\bracsn{Sx|S \in A}}$. 
 \end{proof}
 
+\begin{lemma}[Amplification]
+\label{lemma:bh-amplification}
+    Let $H$ be a complex Hilbert space, $n \in \natp$, and
+    \[
+    \pi: B(H) \to B(H^n) \quad [\pi(T)(x)]_n = Tx_n
+    \]
+
+    then for each $\seqf{x_j} \subset H$, 
+    \[
+    \max_{1 \le j \le n}\norm{Tx_j}_H \le \norm{\pi(T)(x)}_{H^n} \le n \max_{1 \le j \le n}\norm{Tx_j}_H
+    \]
+\end{lemma}
+
+
 \begin{theorem}[Von Neumann's Bicommutant Theorem]
 \label{theorem:bicommutant}
     Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a unital, self-adjoint subalgebra, then $A''$ is the strong-operator closure of $A$.