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Let $X$ be a Banach space, and let $B$ be the closed unit ball of $X^*$, equipped with the weak-* topology. Alaoglu's theorem says that $B$ is compact. If $X$ is separable, then $B$ is metrizable, and in particular it is sequentially compact.

What about the converse? If $B$ is sequentially compact, must $X$ be separable?

This question was inspired by this one.

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  • $\begingroup$ Wouldn't $B$ be weak* sequentially compact in any reflexive space by Eberlein's theorem? $\endgroup$ Commented Jul 9, 2012 at 17:42
  • $\begingroup$ @DavidMitra: You are quite right. I had forgotten about Eberlein. Want to post this as an answer? $\endgroup$ Commented Jul 9, 2012 at 18:00
  • $\begingroup$ See also math.stackexchange.com/questions/1234722/… $\endgroup$ Commented Apr 22, 2015 at 12:43

3 Answers 3

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No, not necessarily. Any reflexive Banach space has a weakly compact unit ball; so, by the Eberlein-Šmulian Theorem, any reflexive Banach space has a weak* sequentially compact unit ball.

Also, and more generally, it follows from Rosenthal's $\ell_1$ Theorem (a Banach space $X$ does not contain $\ell_1$ isomorphically if and only if every every bounded sequence in $X$ has a weakly Cauchy subsequence) that if $X^*$ does not contain $\ell_1$ isomorphically, then the unit ball of $X^*$ is weak* sequentially compact.

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For an explicit counterexample: let $H$ be a non-separable Hilbert space, and suppose we have a sequence of vectors $x_n$ in the unit ball of $H$. Let $H_0$ be the closed linear span of the $x_n$, so that $H_0$ is a separable Hilbert space. Then the $x_n$ have a subsequence converging weakly in $H_0$. But because of the orthogonal decomposition $H = H_0 \oplus H_0^\perp$, it is easy to see that this subsequence also converges weakly in $H$.

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To characterize Banach spaces for which the dual ball is weak* sequentially compact is an open question! See Chapter XIII of Diestel, Sequences and Series in Banach Spaces for a discussion on that topic.

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