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 $\left<A,B\right>=\operatorname{tr}(B^\*A)$. )



Weak convergence of Hilbert Schmidt operators
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.


