If $T_nx\to Tx$ and $x_n\to x$ then $T_nx_n\to Tx$

Question

Let $X,Y$ be Banach spaces, $T_n:X\to Y$ linear and bounded with $\lim_{n\to\infty}T_nx=Tx$ for every $x\in X$. If $x_n\to x$ in $X$ then we have $T_nx_n\to Tx$ in $Y$.

What I'm trying to do is to check for triangle inequalities like

$$\left\|{T_nx_n-Tx}\right\|\le \left\|{T_nx_n-Tx_n}\right\|+\left\|{Tx_n-Tx}\right\|$$

It is easily proved $T$ is linear and bounded so we have no issue with $\left\|{Tx_n-Tx}\right\|$.

But $\left\|{T_nx_n-Tx_n}\right\|$ is another matter, since we don't have $T_n\to T$ with the operator norm.

Could anyone give me a hint?

Thank you.

Answer 1

Hint: Write $\|T_nx_n-Tx\|$ $\leq \|T_nx_n-T_nx\|+\|T_nx-Tx\|$.

You know that $\lim_nT_n x=Tx$, you have to show $\lim_n\|T_nx_n-T_nx\|=0$.

Since $\lim_nT_nx=Tx$, for every $x\in X$, $\sup_n\|T_n(x)\|<+\infty$. Banach Steinhaus implies that there exists $A>0$ such that $\|T_n\|<A$, thus $\|T_nx_n-T_nx\|\leq A\|x-x_n\|$.

https://en.wikipedia.org/wiki/Uniform_boundedness_principle