Corollary 37.2.3.label Let $H$ be a complex Hilbert space and $f \in C_{0}(\real; \complex)$, then the mapping $T \mapsto f(T)$ is strong-operator continuous on $B(H)_{sa}$.
Proof. Let
\[g: \partial B_{\complex}(0, 1) \to \complex \quad z \mapsto \begin{cases}f(-i(z+1)/(z-1)) &z \ne 1 \\ 0 &z = 1\end{cases}\]
then since $f \in C_{0}(\real; \complex)$, $g \in C(\partial B_{\complex}(0, 1); \complex)$. For each $T \in B(H)_{sa}$, $f(T) = g(U(T))$. By Proposition 37.1.5, the mapping $U \mapsto g(U)$ is strong-operator continuous on the set of unitary operators on $H$. By Theorem 37.2.2, $T \mapsto f(T)$ is the composition of two strong-operator continuous mappings.$\square$
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