Lemma 40.2.2.label Let $H$ be a complex Hilbert space, then the strong and weak operator topologies on $U(B(H))$ coincide.
Proof. Let $S, T \in U(B(H))$ and $\xi \in H$, then
\begin{align*}\norm{(S - T)\xi}_{H}^{2}&= \dpn{(S - T)\xi, (S - T)\xi}{H}\\&= \norm{S\xi}_{H}^{2} + \norm{T\xi}_{H}^{2} - 2\text{Re}\angles{S\xi, T\xi}_{H} \\&= 2\norm{\xi}_{H}^{2} - 2\text{Re}\angles{S\xi, T\xi}_{H}\end{align*}
which tends to $0$ as $S \to T$ in the weak operator topology.$\square$
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