Let $X$ and $Y$ be Banach spaces, and let $\{T_n\}\subset L(X,Y)$, where $L(X,Y)$ denotes the space of bounded linear operators from X to Y.If $T_n\to T$ strongly and $x_n\to x$ weakly, must $T_nx_n\to Tx$ weakly?
For this question, we need to show for $\forall g\in Y^*$, $g(T_nx_n)\to g(Tx)$. And I try to break it into two parts: $|g(T_nx_n)- g(Tx)|\leq|g(T_nx_n)-g(T_nx)|+|g(T_nx)-g(Tx)|$. The second piece will goes to zero as $g\in Y^*$ and $T_n\to T$ strongly. But I have difficulty to control the first piece, or any counter example?
The statement is not true. Take $X=Y=H=l^2(\mathbb{N})$ as a Hilbert space. Let $(e_n)_{n=1}^\infty$ be an orthonormal basis where $e_n$ is defined by $ e_n(j) = \delta_n(j). $ Let $$ S(x_1,x_2,x_3,\ldots,x_n\ldots)=(x_2,x_3,x_4,\ldots) $$ be the unilateral shift on $l^2$. Now, note that $$ \|S^nx\|^2=\sum_{j>n}\|x_j\|^2\to 0 $$ as $n\to\infty$ for all $x=(x_1,x_2,\ldots)\in l^2$, and that $e_{n+1}\to 0$ weakly. This says that $T_n =S^n$ and $x_n = e_{n+1}$ satisfies the given assumption, however, $$ S^n e_{n+1}=e_1 $$ does not converge to $0$ weakly. (Note: But if we require $T_n\to T$ in operator norm, then the conclusion is true.)