Weak convergence of a sequence of elements of a compact set implies strong convergence

Asked Apr 23 '15 at 11:12

Active Feb 07 '22 at 16:35

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Let E be a Banach space and $K \subset E$ a compact subset in the strong topology. Let $(x_n)_{n \geq 1} \subset K$ such that $x_n \rightharpoonup x$ in $\sigma (E, E^*)$.
K is compact, so $(x_n)_n$ has a convergent subsequence $(x_{n_k})_k$ which has the limit $\bar{x} \in K$. Then we have: $$x_{n_k} \to \bar{x} \Rightarrow x_{n_k} \rightharpoonup \bar{x}, \;\; \text{i.e} \;\; <f, x_{n_k}> \; \to \; <f, \bar{x}>, \; \forall f \in E^*.$$ But $$ x_n \rightharpoonup x, \; \text{i.e.} \; <f, x_n> \;\; \to \;\; <f, x>, \;\; \forall f \in E^*. $$ It follows that $$ <f, \bar{x}> = <f, x> \;\; \text{i.e.} \;\; \bar{x} = x. $$ So $$ x_{n_k} \to x. $$ How can I show that $(x_n)$ is convergent?

Thank you!

asked Apr 23 '15 at 11:12

g.pomegranate

2,215

1

See here. – Daniel Fischer Apr 23 '15 at 11:18
As above, we have that $(x_{n_k})_k$ has a subsequence converging at x and it follows that $x_n \to x$. Am I right? Thank you for your help! – g.pomegranate Apr 23 '15 at 11:28
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Since $x_n \rightharpoonup x$, every convergent subsequence of $(x_n)$ must converge to $x$. By (sequential) compactness of $K$, every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent subsequence $\bigl(x_{n_{k_m}}\bigr)$. Criterion $\leadsto x_n \to x$. – Daniel Fischer Apr 23 '15 at 11:33

Weak convergence of a sequence of elements of a compact set implies strong convergence

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