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Suppose $X$ is a compact space, $H$ is a Hilbert space, and $f:X \rightarrow B(H)$ is continuous when $B(H)$ is given the strong topology. Does this imply that $f$ is continuous when $B(H)$ is given the operator norm topology?

My guess is that the answer is no, although I can't think of an easy counterexample.

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up vote 2 down vote accepted

Let $H=\ell_2 (\mathbb{R} ) ,$ and let $T_n : \ell_2 \to \ell_2 , $ be defined as follows: $$T_n \left(\sum_{j=1}^{\infty} x_j e_j \right) =x_n e_n ,$$ $$T\equiv 0 .$$ Let $$X=\{T\}\cup\bigcup_{n=1}^{\infty} \{T_n \}$$ and let $\tau_1 $ denote the topology on $X$ induced from $(B(H), \tau ) $ where $\tau$ denotes the strong topology on $B(H) .$ Let $\sigma $ denotes the operator norm topology on $B(H) ,$ and let $f:X\to B(H), $ be defined by $$f(A)=A .$$ Then $(X ,\tau_1 )$ is compact space and $f :(X, \tau_1 ) \to (B(H), \tau )$ is continuous but $f :(X, \tau_1 ) \to (B(H), \sigma )$ is not continuous.

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It may help to notice that $X$ is homeomorphic to $\{1/n : n \ge 1 \} \cup \{0\}$, or to the one-point compactification $\mathbb{N} \cup \{\infty\}$. – Nate Eldredge Jun 15 '14 at 20:51

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