[Math] $T$ compact operator implies that every weakly convergent sequence $x_n$ implies strong convergence of $Tx_n$

Question

I'm working on a problem regarding compact operators and weakly convergent sequences. We know that an operator $T$ on a Hilbert space $H$ is compact iff for every bounded sequence $(x_n)_n \subset H$ its image $(Tx_n)_n \subset H$ admits a convergent subsequence (one can use this as a definition of a compact operator. I wanna show the following:

Let $T$ be a compact operator on $H$. Then for every sequence $(x_n)_n \subset H$ which converges weakly to $0$, the sequence $(Tx_n)_n)$ converges to $0$ in norm.

Can someone help me?

Answer 1

Suppose otherwise. Then there is a subsequence $(x_{n_{k}})$ such that $ ||T(x_{n_{k}})||>\epsilon$ for all $k$.

This subsequence $(x_{n_{k}})$ still weakly converges to 0. So its bounded and hence our sequence $||T(x_{n_{k}})||$ has a convergent subsequence. Note that since $T$ is weak to weak continous, we have $T(x_{n_{k}})$ converges weakly to 0. In particular our convergent subsequence must have limit 0. But this contradicts our choice of subsequence.