Weak convergence + compactness = strong convergence?

Question

Let $X$ be a Banach space and $K$ a compact subset of $X$. If $(x_n)_n$ is a sequence such that $x_n\in K$ for all $n$ and $(x_n)_n$ converges weakly to some $x\in X$, i.e. $x^*(x_n)\to x^*(x)$ for all $x^*\in X^*$ the dual space of $X$.

I know that we have the strong convergence for some subsequence. But, do we have the strong convergence of the whole sequence?

A compact space does not a have a strictly weaker Hausdorff topology. — Jochen, Mar 18 '15 at 14:33

score 0 · Answer 1 · edited Apr 13 '17 at 12:21

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I just realized that the sequence $(x_n)_n$ has the following property:

"Every subsequence has a subsequence which converges to $x$." It follows that the whole sequence must converge to $x$. For more details see the questions:

Question 1

Question 2

Question 3

edited Apr 13 '17 at 12:21

Community

1

answered Mar 19 '15 at 01:52

user165633

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Weak convergence + compactness = strong convergence?

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