Compactness in the weak* topology

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.

Solution 1:

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.

Solution 2:

(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.