Added a continuity result in strong operator topology.

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 $A$.