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Let $X$ be separable Banach space and $\{x_n\}$ be a bounded sequence, relatively weakly compact sequence in $X$. we set $y_n=\frac{1}{n}\sum_{i=1}^{n}{x_i}$, then (together with the Krein and Eberlein-Smulian theorems), we can assume that there exists a subsequence of $\{y_n\}$ converges weakly to some element $y \in X$.

We suppose that every weak limit point of $\{y_n\}$ must equal $y$.

Can we say that $\{y_n\}$ must converge weakly to $y$?

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  • $\begingroup$ What does it mean for a sequence to equal a single element? It sounds like you're asking if the sequence must be constant, but that clearly isn't right. Is there a typo? $\endgroup$ Jun 6, 2020 at 3:55
  • $\begingroup$ Also the title doesn't make grammatical sense. $\endgroup$ Jun 6, 2020 at 3:55
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    $\begingroup$ Are you asking if the sequence $y_n$ must converge (weakly?) to $y$? $\endgroup$ Jun 6, 2020 at 3:57
  • $\begingroup$ @NateEldredge see my edit $\endgroup$
    – Karim KHAN
    Jun 6, 2020 at 4:04

1 Answer 1

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If $y_n$ does not converge weakly to $y$ then there is a weakly open neighborhood $U$ of $y$ and a subsequence $y_{n_k} \notin U$. By weak compactness this subsequence has a weak limit point $z \notin U$. But $z$ is then also a weak limit point of the original sequence $y_n$, contradicting uniqueness.

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  • $\begingroup$ If $ \{y_n \} $ is relatively weakly compact, then $ \{y_ {n_k} \} $ is a convergent sequence? $\endgroup$
    – Karim KHAN
    Jun 6, 2020 at 4:20
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    $\begingroup$ @KarimKHAN: The subsequence $y_{n_k}$ in my answer isn't necessarily convergent, though it would be possible to find a further subsequence of it which is. $\endgroup$ Jun 6, 2020 at 4:22

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