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Let $H$ be a complex Hilbert space, and let $\mathcal{B}(H)$ denote the algebra of bounded operators on $H$. It is known that the strong operator topology and the norm topology on $\mathcal{B}(H)$ coincide if and only if $H$ is finite dimensional. But the strong operator topology cannot be characterized using sequences, so we have to use nets. My question is: Can the strong and norm topologies coincide sequentially- id est, every strong convergent sequence is norm convergent - in a certain subalgebra $A \subseteq \mathcal{B}(H)$ with infinite dimension?

Thank you very much!

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    $\begingroup$ What do you mean by asking for two topologies to "coincide sequentially"? Are you asking if every sequence that converges in SOT automatically converges in norm? Because if so, a simple counterexample is given by taking $P_n\in {\mathcal B}(\ell_2)$ to be the orthogonal projection onto the span of the $n$th standard basis vector; we have $P_n\to 0$ in SOT while $\Vert P_n\Vert=1$ for all $n$ $\endgroup$
    – Yemon Choi
    Commented Sep 24, 2019 at 13:46
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    $\begingroup$ Yes, that what I mean. It is easy to see that it cannot happen in the whole algebra, but are there any infinite-dimensional subalgebras with this property? $\endgroup$
    – javi1996
    Commented Sep 24, 2019 at 14:07
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    $\begingroup$ So just to try and pin down a precise question: you are asking if there is an infinite-dimensional $*$-subalgebra of ${\mathcal B}(\ell_2)$ in which every SOT-convergent sequence is automatically norm-convergent? $\endgroup$
    – Yemon Choi
    Commented Sep 24, 2019 at 14:13
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    $\begingroup$ Yes, but I'm not including the hypothesis of being *-subalgebras, just only a subalgebra. $\endgroup$
    – javi1996
    Commented Sep 24, 2019 at 14:16
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    $\begingroup$ It might be worth pointing out that if $H$ is separable then the SOT-topology is metrizable when you restrict to norm bounded subsets. $\endgroup$
    – Jesse Peterson
    Commented Sep 26, 2019 at 17:52

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