5
$\begingroup$

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?

$\endgroup$

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

Let $(x_n)_{n\in \mathbb{N}}$ converge weakly to $0$. Note that $(x_n)$ is bounded. Since $T$ is weak-to-weak continuous, it follows that $Tx_n\rightharpoonup 0$. Hence any cluster point of $(Tx_n)$, be it strong or weak, must be $0$. On the other hand, your $T$ is compact. If $(Tx_n)$ would not converge to $0$ in norm, then there would be a subsequence $(Tx_{n_k})$ that converges weakly to $0$ but not strongly. But since $T$ is compact, there would be a further subsequence $(Tx_{n_{k_l}})$ converging strongly to $0$ which is absurd!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .