TITLE: Weak convergence implies strong convergence
QUESTION [6 upvotes]: Assume that the sequence $(A_n)$ of bounded linear operators on a Hilbert space $H$ converges weakly to an operator $A$. Assume also that $\|A_nx\|\to \|Ax\|$ for all $x\in H$.
        Prove that $(A_n)$ converges strongly to $A$, i.e. $A_n x\to Ax$ for all $x\in H$.
To be honest, I don't have too much of an idea on this one. We have
$$<A_n,x> \to <A,x> \leq ||A_nx-Ax||xx||$$
I am thinking it is likely going to be a trick with the inequality (and I'm pretty sure the above isn't correct). It seems to me obvious but yet it is eluding me. I think my struggle is figuring out what that inner product looks like for the weak convergence of an operator.

REPLY [2 votes]: $||A_nx-Ax||^2=||A_nx||^2-<A_nx,Ax>-<Ax,A_nx>+||Ax||^2 \to ||Ax||^2-||Ax||^2-||Ax||^2+||Ax||^2=0$
in second and third term, I used $A_n$ convergence weakly $A$.