Squaring not weak operator topology continuous on self-adjoint operators?

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.

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.
• Your argument doesn't show that $A_n$ converges in norm to $A$, only that $A_n$ converges strongly to $A$ Aug 5, 2019 at 14:45