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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 $\left<A_n,D\right>=\operatorname{tr}(D^*A_n)\to 0$ 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 $$\sum_{i=1}^\infty \left<A_ne_i,De_i\right>\to 0~~~\forall D\in B_2(H)$$ If we choose $D$ such that $De_i=x$ and $De=0$ otherwise, then we have $$\left<A_ne_i,x\right>\to 0$$ 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 $\left<A,B\right>=\operatorname{tr}(B^*A)$. )

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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.

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