# Weak convergence of Hilbert Schmidt operators

So I am stuck at this situation. Let $$\{A_n\}$$ be a weakly convergent sequence in $$B_2(H)$$ converging to $$0$$ in the weak topology on $$B_2(H)$$. Which means that $$\left=\operatorname{tr}(D^*A_n)\to 0$$ for each $$D\in B_2(H)$$. I want to prove/disprove that $$\|A_n\|_2\to 0$$,i.e $$A_n\to 0$$ . Clearly $$\sum_{i=1}^\infty \left\to 0~~~\forall D\in B_2(H)$$ If we choose $$D$$ such that $$De_i=x$$ and $$De=0$$ otherwise, then we have $$\left\to 0$$ for each $$i$$ and $$x\in H$$. Can we infer from here that $$\|A_n\|_2\to 0$$?

( $$H$$ denotes a Hilbert space with orthonormal basis $$\{e_i\}$$ and $$B_2(H)$$ denotes the Banach algebra of bounded linear operators on $$H$$ with Schatten 2-norm, also know as Hilbert Schmidt operators. $$B_2(H)$$ is a Hilbert space as well with inner product $$\left=\operatorname{tr}(B^*A)$$. )

Well, the space of Hilbert-Schmidt operators is a Hilbert space, so you are asking whether weak convergence to zero implies norm convergence to zero in a Hilbert space. The answer is no. For instance, let $$A_n$$ be the rank 1 projection onto $$e_n$$. This converges weakly but not in norm to zero.