1
$\begingroup$

This question already has an answer here:

I would like to prove that if $T:X\rightarrow Y$ is a compact operator, then for every weak convergent sequence $(x_n)_{n\in\mathbb N}$ with $x_n\rightharpoonup x$ for some $x\in X$ it follows that $Tx_n\rightarrow Tx$ with respect to the norm on $Y$.

I started the proof with,

Since $(x_n)_{n\in\mathbb N}$ is weak convergent there exists a positive constan $C$ sucht that $||x_n||\leq C$ for all $n$. Hence by compactness it follows that $Tx_n$ contains a convergent subsequence i.e $Tx_{n_k}\rightarrow y$ for some $y\in Y$. Now it should be somehow possible to deduce that $y=Tx$. But I am not able to move on.

Could someone give me hint how to proceed? or even how to find a better start for my proof?

Thanks in advance!

$\endgroup$

marked as duplicate by David Mitra, user63181, Namaste, user147263, Claude Leibovici May 27 '14 at 14:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

Denote by $T^*$ the adjoint of $T$. For each integer $k$ and each $u\in Y^*$, $$\langle Tx_{n_k},u\rangle_{Y,Y^*}=\langle x_{n_k},T^*u\rangle_{X,X^*}$$ hence taking the limit $k\to \infty$, $$\langle y,u\rangle_{Y,Y^*}=\langle x,T^*u\rangle_{X,X^*}=\langle Tx,u\rangle_{Y,Y^*}.$$ We conclude using Hahn-Banach theorem.

$\endgroup$
  • $\begingroup$ Just to put everything together, First I show that every subsequence of $Tx_n$ has convergent subsequnce with limit $Tx$. Then since the image of bounded sets has a compact closure I can conclude that $Tx_n\rightarrow Tx$? $\endgroup$ – Thorben May 27 '14 at 13:20
  • $\begingroup$ Once you showed that each subsequence converges to $Tx$ you are done. $\endgroup$ – Davide Giraudo May 27 '14 at 20:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.