$X$ - Banach space and $T_n$ bounded linear operators that is strongly convergent to $T$ as well as $x_n$ is strongly convergent to $x$.
How to prove that $T_n x_n$ is strongly convergent to $Tx$?
I kinda understand that we can prove the case where $T_n x$ is strongly convergent to $Tx$
it is like this $\left\|T x-T_{n} x\right\| \leq\left\|T-T_{n}\right\|\|x\|<\epsilon\|x\|$ ...
But how to proof statement in my case I am not sure. But intuitively it is seems to me obvious thing.
Hint:
$$ \begin{align} | T_n x_n - Tx | &= |T_n x_n - T_n x + T_n x - T x| \\ &\leq |T_n x_n - T_n x| + |T_n x - T x| \\ &= |T_n (x_n - x)| + |T_n x - T x| \end{align} $$
Do you know how to control $T_n x_n \to T_n x$ and $T_n x \to T_x$ separately? It will be helpful to remember that $T_n$ is bounded, and thus continuous.
Once you do this, you'll notice it doesn't quite work, since our bound for $|T_n x_n - T_n x|$ depends on $n$ (do you see why?). We can fix this, though, by using the fact that for large $n$, $T_n \approx T$. Can you modify the above proof, say by using the inequality (which holds for all vectors $y$)
$$|T_n y| \leq \lVert T_n \rVert \lVert y \rVert$$
and then controlling $\lVert T_n \rVert$ using (Edit: Thanks to Nate Eldredge for the correction) the uniform boundedness principle? If you happen to know that a sequence of operators which converge pointwise are bounded in norm, you can also use that directly.
I hope this helps ^_^