I haven't found this yet and I'm somehow not sure if my idea is correct.

The Problem: Let $X$ be a separable Banach-Space, let $x_k\to x$ weakly and such that for every $\lambda_k \to \lambda$ weakly-* there holds $\lambda_k(x_k)\to \lambda(x)$. Then $x_k\to x$ strongly (in the norm of $X$).

My idea was to give a proof with contradiction. Hence assume there holds for some $\epsilon >0$ and a subsequence of $x_k$ denoted again by $k$: $$ \epsilon <||x_k-x||=|\lambda^*_k(x_k-x)| $$ for a functional provided by Hahn-Banach theorem with norm 1. From that and the separabilty we conclude that there is a further subsequence such that $\lambda_{k_l}^*\to \lambda^*$ weakly-* . Since each subsequence of $x_k$ also converges weakly to $x$ we use the "weakly-*" assumption to receive a contradiction since for the previous subsequence $$|\lambda^*_{k_l}(x_{k_l}-x)|=|\lambda^*_{k_l}(x_{k_l})-\lambda^*_{k_l}(x)|\to 0$$

Somehow this seems to easy and I feel like I'm not using especially the weak convergence in the right way.

  • 1
    $\begingroup$ I made an edit fixing what must surely have been a typo in the post: I inserted an $\epsilon$ in what seems the intended place. $\endgroup$ – Harald Hanche-Olsen Feb 21 '13 at 20:40

I think the reason for your confusion is the condition made on the convergence $x_k \rightarrow x$. The condition that for any $\lambda_k \rightarrow \lambda$ weak* we have $|\lambda_k(x_k) - \lambda(x)| \rightarrow 0$ implies $x_k \rightarrow x$ weakly, by taking $\lambda_k$ to be the constant sequence. You certainly use this condition, so there's nothing to worry about.


I think your proof looks just fine. There is nothing to add and nothing to subtract. Oh, except you need to take a subsequence to get the first inequality. But you must already have intended that, or else why did you say further subsequence? Hahn–Banach and compactness are powerful tools. You should not be surprised that they can yield strong results with little effort.

  • $\begingroup$ Thanks. You're right. I forgot to mention the first subsequence. Somehow I was doubting since this looked somehow like only exploiting $||x_k||\to ||x||$ (I considered $x=0$ first and then this is of course norm conergence :)). But this instead of the weakly-* assumption from above does not lead to strong convergence in general separable spaces. Additionally the weak convergence of $x_k$ is somehow only implicitly used in the weakly-* assumption. $\endgroup$ – Quickbeam2k1 Feb 21 '13 at 20:58

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.