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I am trying to solve the following exercise: Given $\phi: B(\mathcal{H}) \to \mathbb{C}$ linear functional, prove that $\phi$ is continuous with respect to weak operator topology if and only if $\phi$ is continuous with respect to strong operator topology.

However, I don't know how to approach it, and I don't have clear in mind how weak and strong operator topologies are defined for linear functional - I only studied how these topologies are defined for operators $x \in B(\mathcal{H})$. Can anyone help me understand the exercise and solve it?

Notation: $B(\mathcal{H})$ is the set of bounded, linear operators acting on Hilbert space $\mathcal{H}$. $\mathbb{C}$ is the set of complex numbers.

MBlrd
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  • Does this answer your question? A linear operator between Banach spaces is weakly continuous iff norm continuous? – Anne Bauval Mar 31 '24 at 13:23
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    @AnneBauval Thank you for your link. There I can see that there is equivalence with norm operator topology, whereas in the exercise above the equivalence is with respect to strong operator topology. However, it is helping me better understand their definitions, so still very helpful! – MBlrd Mar 31 '24 at 13:37
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    I didn't pay enough attention. I rétracter my close vote. – Anne Bauval Mar 31 '24 at 13:38
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    Continuous with respect to weak operator topology means $\phi$ is continuous as a function from $B(H)$, equipped with weak operator topology, to $\mathbb{C}$, equipped with the usual topology. Or, to put it another way, given any $T_\lambda \to T$ in $B(H)$, where the convergence is in weak operator topology, then $\phi(T_\lambda) \to \phi(T)$. Continuous with respect to strong operator topology is similar. – David Gao Mar 31 '24 at 15:27
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    For a proof of this result, see Theorem 4.2.6 of Murphy’s “$C^\ast$-algebras and operator theory”. – David Gao Mar 31 '24 at 15:29

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