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I know that it is a standard fact that multiplication is not sequentially jointly WOT continuous. However, can we find a sequence $A_n$ of self-adjoint bounded operators on a Hilbert space converging WOT to $A$, but $$A_n^2 \not\rightarrow A^2$$ in WOT?

From my experience, counterexamples to the WOT-continuity of multiplication are not given by the squaring function, but instead by two different sequences $A_n, B_n$. Thank you for any help, self-studying the weak operator topology is proving unintuitive to say the least.

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Let $A_n$ be a sequence of self-adjoint operators such that $A_n$ (resp. $A_n^2$) converge in WOT to $A$ (resp. $A^2$).

Then for any $x$, $y_n=A_n(x)$ converges weakly to $A(x)$. Besides, $\|A_n(x)\|^2=\langle A_n(x),\,A_n(x) \rangle = \langle A_n^2(x),\,x\rangle \rightarrow \langle A^2(x),\,x\rangle = \|A(x)\|^2$.

So $\|y_n-A(x)\|^2=\|y_n\|^2+\|A(x)\|^2-2\langle y_n,\,A(x)\rangle \rightarrow 2\|A(x)\|^2-2\langle A(x),\,A(x)\rangle =0$.

So $A_n$ converges strongly.

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  • $\begingroup$ Very nice, so this implies you may take any non-norm converging sequence that converges WOT as a counterexample. $\endgroup$ – Rick Sanchez Aug 3 at 22:36
  • $\begingroup$ Your argument doesn't show that $A_n$ converges in norm to $A$, only that $A_n$ converges strongly to $A$ $\endgroup$ – Aweygan Aug 5 at 14:45
  • $\begingroup$ Right, I corrected the mistake. $\endgroup$ – Mindlack Aug 5 at 15:16

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