Weak convergence and strong convergence on $B(H)$

Question

Let $\mathcal{A} \subset B(H)$ be a weak closed convex bounded set of self-adjoint operators. If $A_n \rightarrow_{wo} A\in \mathcal{A}$, do we have $A_n \rightarrow A$ strongly?($A_n$ is a sequence in $\mathcal{A}$)

Answer 1

Let $H=\ell^2(\mathbb{N})$ and let $\mathcal{A}$ be the set of all self-adjoint elements in the unit ball of $B(H)$. Define $S_n\colon H\to H, S_n \xi(k)=\xi(k+n)$. Then $\|S_n\|=1$, hence $A_n:=\frac 1 2(S_n+S_n^\ast)\in \mathcal{A}$.

The adjoint of $S_n$ is given by $$ S_n^\ast\xi(k)=\begin{cases}\xi(k-n)&\colon k\geq n\\0&\colon k<n\end{cases} $$ It is easy to see that $S_n\to 0$ strongly and consequently $S_n^\ast\to 0$ weakly. Thus, $A_n\to 0$ weakly.

However, let $\xi=\delta_0=(1,0,\dots)$. Then $A_n\xi=\frac 1 2 \delta_n$ has norm $\frac 1 2$. Hence $A_n$ does not converge to $0$ strongly.