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Karim KHAN
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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 equalconverge weakly to $y$?

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 equal $y$?

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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Karim KHAN
  • 199
  • 1
  • 5

Existence of weak limit for bouded sequence $\{y_n\}$ such that for every weak limit point $\{y_n\}$ must equal $y$

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 equal $y$?