# Weak continuous convergence of operators

Let $T_n$ and $T$ be linear maps from Banach space $X$ to a Banach space $Y$.

Suppose $T_n$ satisfies $T_nx \to Tx$ (convergence in the $Y$ norm) for all $x \in X$.

Let $x_n \rightharpoonup x$ in $X$ (weak convergence). Under what conditions on $T_n$ and $T$ does this imply that $$T_nx_n \rightharpoonup Tx?$$

Let us assume all maps are bounded if necessary.

What if instead of $x_n \rightharpoonup x$ we hypothesised $x_n \to x$?

Let $X=X^*$ be a separable (for simplicity) Hilbert space, take $x_n=e_n$ the canonical basis. $Y=\Bbb R$. Take also $T_n=e^*_n$ - the canonical basis in $X^*$ (by Riesz representation, we can say that $T_n=x_n$ and $T_ny = (e_n,y)_X$).
Then $\forall x\, T_nx\to0$ (so $T=0$), $x_n\rightharpoonup 0$ (so $x=0$), yet $1=T_nx_n \not\to Tx=0$