# From weak and weak star to norm convergence

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.

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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. – 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.
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. – Quickbeam2k1 Feb 21 '13 at 20:58