Theorem 37.2.2.label 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 $U_{1}$.
Proof, [Theorem 19.3, Zhu93]. Let $T \in B(H)$ be self-adjoint. By Proposition 36.4.6, $\sigma_{B(H)}(T) \subset \real$. By the Spectral Mapping Theorem,
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 Spectral Mapping Theorem 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 $S, T \in B(H)$,
so for any $x \in H$,
By the continuous functional calculus, $\normn{(S+i)^{-1}}_{B(H)}\le 1$. Therefore the Cayley transform is strong-operator continuous.$\square$
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