If $\{T_n\},\{S_n\}$ are two sequences of bounded operators on a Banach space $X$, such that $\{T_n\}$ converges weakly to $T$, and $\{S_n\}$ converges strongly to $S$, does it follow that $T_nS_n\to TS$ strongly?
I've shown the conclusion is true if both sequence converge strongly, but would appreciate any hints to help me answer the question in case only the second sequence in the product converges strongly (and the first weakly).
($T_n\to T$ weakly means that $T_n x\to Tx$ in the weak topology of $X$. $S_n\to S$ strongly means that $S_n x\to Sx$ in the norm topology of $X$ for any $x\in X$.)
I've reduced this question to showing that $$ ||S(T_n-T)x||\to 0 $$ for any $x\in X$.
This might seem redundant but also seems correct to me.
If we take the sequence $(S_n)$ to be the constant sequence of $\textbf{Identity operators}$ then this sequence converges strongly to the Identity operator. Since $(T_n)$ converges to $T$ weakly and in this case the sequence $T_nS_n$ is nothing but the sequence $(T_n)$, which clearly fails to converge strongly.