0
$\begingroup$

I'm attempting to answer the following:

Let X be a reflexive Banach space, and Y be a normed space. Suppose T : X → Y is a linear operator with the property that if a sequence $x_n$ converges weakly to 0 in X, then $T(x_n)$ converges weakly to 0 in Y . Prove that T is bounded.

I feel like I can use the Banach-Alaoglu Theorem to get there, namely that it guarantees that every bounded sequence in X has a weakly convergent subsequence, but I'm not sure how to proceed.

Any help is much appreciated!

$\endgroup$
  • $\begingroup$ I think X must be a reflexive space $\endgroup$ – Guy Fsone Dec 11 '17 at 19:33
0
$\begingroup$

By Contradiction: If T is not bounded then for all n there eixst $x_n\in X$ with $\|x_n\|=1$ such that $$\|T(x_n)\|\ge n$$

the sequence $x_n$ is bounded then by Banach-Alaoglu theorem, we can extract a sub-sequence $x_{n_j}$ which converges weakly in X. And by assumption $(Tx_{n_j})$ converges weakly in $Y$ and Hence $(T(x_{n_j}))_j$ is a bounded sequence in $Y$. This goes in contradiction with the fact that $$\|T(x_{n_j})\|\ge n_j$$

Since every weakly convergence sequence in bounded.

$\endgroup$
  • $\begingroup$ Thank you! It seems that the 'weakly convergent to 0' part of the problem is unnecessary - we just need weakly convergent. Am I right in saying this? $\endgroup$ – OGBerglemir Dec 11 '17 at 21:55
  • $\begingroup$ it is necesary if you take $x_n-x\to 0$ instead of $x_n\to x$ $\endgroup$ – Guy Fsone Dec 11 '17 at 21:57
  • $\begingroup$ OK, so then by Banach-Alaoglu $x_{n_k} \rightarrow x$ weakly, so taking $y_n = x_{n_k} - x$, we have that $y_n \rightarrow 0$ weakly? $\endgroup$ – OGBerglemir Dec 11 '17 at 22:03
  • $\begingroup$ @OGBerglemir Exactly $\endgroup$ – Guy Fsone Dec 11 '17 at 22:08
2
$\begingroup$

Assume $T$ is not bounded, then there is a sequence $(x_n)$ such that $\|x_n\|=1$ and $Tx_n\to\infty$. Then we can extract a subsequence (denoted the same) such that $x_n \rightharpoonup x$, and by the properties of $T$, $Tx_n\rightharpoonup Tx$. Weakly converging sequences are bounded, contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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