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Let $X$ be a Banach space, and let $X^*$ denote its continuous dual space.

Under the weak* topology, do compactness and sequential compactness coincide?

That is, is a subset of $X^*$ weakly* compact if and only if it is weakly* sequentially compact? Does one imply the other?

Perhaps I should make this next one a separate question, but I'd prefer to keep all of this in one place.

Is the weak* topology on $X^*$ Hausdorff? Is the weak topology on $X$ Hausdorff?

Motivation: I would like to say that if a subset of $X^*$ is weakly* compact, then it is weakly* closed, and that if a subset of $X$ is weakly compact, then it is weakly closed.

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For (ii) it seems that the following could be relevant: Let $\mathcal{F} \subset \mathcal{F'}$ be a weak and strong topology respectively on a space $X$. If $\mathcal{F'}$ is compact and $\mathcal{F}$ is Hausdorff, then $\mathcal{F} = \mathcal{F'}$ – PEV Mar 16 '11 at 17:26
What have you tried in order to show the second part? It is pretty straightforward... – Mariano Suárez-Alvarez Mar 16 '11 at 17:43
up vote 12 down vote accepted

Let me answer your second question first.

The weak$^{\ast}$-topology is Hausdorff (let me treat the real case, the complex case is similar): If $\phi \neq \psi$ are two linear functionals then there is $x \in X$ such that $\phi(x) \lt r \lt \psi(x)$. The sets $U = \{f \in X^{\ast} \,:\,f(x) \lt r\}$ and $V = \{f \in X^{\ast}\,:\,f(x) \gt r\}$ are weak$^{\ast}$-open (since evaluation at $x$ is weak$^{\ast}$-continuous) and disjoint neighborhoods of $\phi$ and $\psi$, respectively.

That the weak topology is Hausdorff is shown similarly, using Hahn-Banach.

Next, if $X$ is separable, then the unit ball in the dual space is metrizable with respect to the weak$^{\ast}$-topology: pick a countable dense set $\{x_{n}\}_{n \in \mathbb{N}}$ of the unit ball of $X$ and verify that \[ d(\phi,\psi) = \sum_{n=1}^{\infty} 2^{-n} \frac{|\phi(x_n) - \psi(x_n)|}{1+|\phi(x_n) - \psi(x_n)|} \] defines a metric compatible with the weak$^{\ast}$-topology. Hence the unit ball is sequentially compact in the weak$^{\ast}$-topology (this can be shown directly using Arzelà-Ascoli, by the way).

Using a standard Baire category argument, one can show that weak$^{\ast}$-compact sets are norm-bounded: Indeed, if $K$ is weak$^{\ast}$-compact, it is a Baire space. Write $B^{\ast}$ for the closed unit ball in $X^{\ast}$. Clearly $K = \bigcup_{n = 1}^{\infty} (K \cap n \cdot B^{\ast})$, so at least one of the closed subsets $K \cap n \cdot B^{\ast}$ of $K$ must have non-empty interior. By compactness finitely many translates of $n\cdot B^{\ast}$ must cover $K$, thus $K$ is bounded in norm and hence $K$ is a closed subset of a large enough ball.

Conclusion: If $X$ is separable then every weak$^{\ast}$-compact subset of $X^{\ast}$ is sequentially compact.

I don't know if the converse is true.

If $X$ is not separable, then weak$^{\ast}$-compactness does not imply weak$^{\ast}$-sequential compactness, the standard example is mentioned in Florian's post.

Since you might be interested in the weak topology as well, there's a rather difficult result due to Eberlein:

Recall that a space is countably compact if every countable open cover has a finite subcover. A sequentially compact space is countably compact.

Theorem (Eberlein) If a subset of a Banach space is weakly countably compact then it is weakly compact and weakly sequentially compact.

and finally:

Theorem (Eberlein-Šmulian) A bounded subset of a Banach space is weakly sequentially compact if and only if it is weakly compact. In particular, if the unit ball is weakly sequentially compact then $X$ is reflexive.

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Why do we need the assumption of boundedness in Eberlein-Smulian? By Eberlein, countable compactness implies both compactness and sequential compactness (in the weak topology). But in any topological space, both compactness and sequential compactness imply countable compactness. So all three concepts coincide, no? – Jesse Madnick Apr 3 '11 at 21:04
@Jesse: You're right, it is superfluous. I think I've written this in this way because I learned it with that hypothesis and it is hard to get rid of old habits... – t.b. Apr 4 '11 at 0:19

(i) No. Consider $\ell^\infty$ (bounded sequences). The unit ball of ${\ell^\infty} ^*$ is compact by Alaoglu's theorem, but not sequencially compact: the sequence of functionals $a\mapsto a(n)$ (picking the nth element of the sequence $a\in \ell^\infty$) is bounded, but does not have a *-weakly convergent subsequence.

(ii) Yes. For the weak* topology this follows directly from the definition; for the weak topology this follows from the Hahn-Banach theorem.

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