Proposition 37.1.5.label Let $H$ be a complex Hilbert space, then:
- (1)
The adjoint map $T \mapsto T^{*}$ is continuous in the weak operator topology and the ultraweak topology.
- (2)
$T \mapsto T^{*}$ restricted to the normal operators is continuous in the strong operator topology.
- (3)
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.
Proof, [Section 19.1, Zhu93]. (2): Let $S, T \in B(H)$, then
Now, if $S$ and $T$ are normal, then $\normn{S^*x}_{H} = \norm{Sx}_{H}$ and $\norm{T^*x}_{H} = \norm{Tx}_{H}$, so
Therefore
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
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 Stone-Weierstrass Theorem, 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 $n \in \natp$,
by the 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 Proposition 7.3.2.$\square$
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