Let $X, Y$ be a banach spaces, $A\in \mathcal{L} (X,Y)$ , there is a proposition that affirms that if $X$ is reflexive , and for any $ x_n\rightarrow x$ weakly in x we have that $ Ax_n\rightarrow x$ when $ n\rightarrow \infty$ then $ A$ is compact.
I have proved it using Eberlain-Smulyain theorem, so it trivial this way the proof I think. But I can't find any counterexample that $A$ doesn't have to be comapct if $X$ is not reflexive.
Consider $X=Y=\ell^1(\mathbb N)$, and let $A=I$, the identity operator. We know weak convergence implies strong convergence (see here). Thus for any $(x_n)$ converging weakly to $x$ we know $x_n\to x$ in norm. Thus $Ix_n\to Ix$ clearly. However, the identity operator on a Banach space is compact if and only if the space is finite dimensional.