If $T_n\to T$ uniformly and $x_n\to x$ weakly, then $T_nx_n\to Tx$ weakly

Question

We say a sequence $(T_k)$ with $T_k\colon H\to H$ is uniformly convergent to $T\colon H\to H$ if $\|T_k-T\|\to 0$. I am interested in the following claim:

For any weakly convergent sequence $(x_k)$ to $x$, the sequence $(T_kx_k)$ is weakly convergent to $Tx$.

This follows from weak sequential continuity of $T$ and uniform convergence of $T_k\to T$, but can these conditions be relaxed?

Answer 1

We have $$T_nx_n-Tx=(T_n-T)x_n+T(x_n-x)$$ Any weakly convergent sequence is bounded therefore $$\|(T_n-T)x_n\|\le \|T_n-T\|\sup_k\|x_k\|\to 0$$ Furthermore $T(x_n-x)$ tends to $0$ weakly. Summarizing $T_nx_n$ tends to $Tx$ weakly.