TITLE: strong convergence of product of operators
QUESTION [4 upvotes]: I'm trying to prove the next:
Let $H$ be a Hilbert space. Consider $\{T_{n}\}$ and $\{S_{n}\}$ sequences of $\mathcal{B}(H)$ such that $S_{n}\xrightarrow{s} S$ and $T_{n}\xrightarrow{s}T;$ here $S_{n}\xrightarrow{s} S$ means that $S_{n}$ converges strongly to $S,$ i.e. for each $x\in H,$ $S_{n}x\rightarrow Sx.$
Then $S_{n}T_{n}\xrightarrow{s}ST.$
So, for $x\in H,$ We have $||(S_{n}T_{n}-ST)x||\leq||(S_{n}-S)T_{n}x||+||S||||(T_{n}-T)x||.$
The second term of right side of the inequality above converges to zero. My doubt comes from the other term. If $y_{n}=T_{n}x$ it would seem such term converges to zero too because of the strong convergence of $S_{n},$ but such $y_{n}$ depends of $n;$ I have doubts about it. 
Any kind of help is thanked in advanced.

REPLY [4 votes]: One uses Uniform Boundedness Principle to conclude that $\sup_{n}\|S_{n}\|<\infty$. With such, one has $$\|S_{n}T_{n}x-S_{n}Tx\|\leq\left(\sup_{n}\|S_{n}\|\right)\cdot\|T_{n}x-Tx\|$$ and we split $$\|S_{n}T_{n}x-STx\|\leq\|S_{n}T_{n}x-S_{n}Tx\|+\|S_{n}Tx-STx\|.$$
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