Can we classify weakly compact subsets of $l^{\infty}$?

Question

My question is as stated. The motivation behind this is that I want to prove that such subsets are norm-separable.

Answer 1

Edit: my previous answer was plain wrong as the set $B_K$ need not be a ball of any space (for example, take the unit vector basis of $c_0$; $c_0$ does not contain any infinite-dimensional reflexive subspaces). Here is an easier argument.

Suppose that $K\subset \ell_\infty$ is weakly compact. As $\ell_\infty$ is isometric to the dual space of $\ell_1$, it carries the weak* topology, which is coarser than the weak topology of $\ell_\infty$. In particular, $K$ is also compact in the weak* topology and the two relative topologies agree on $K$. However, as $\ell_1$ is separable, weak* compact subsets of $\ell_\infty$ are second-countable. One may use Mazur's lemma (as explained here) to conclude that a set is weakly separable if and only if it is norm-separable.

Suppose that $K\subset \ell_\infty$ is weakly compact. Then, by Mazur's lemma, so is the closed convex hull of $K\cup \{0\}$, call it $B_K$. Consider $\ell_\infty$ as the dual space to $\ell_1$. Since the weak* topology on $\ell_\infty$ is coarser than the weak topology, by compactness, they must be equal on $B_K$. Since $B_K$ is weakly compact, $X_K$ is reflexive. However, the weak*-topology and the weak topologies of $\ell_\infty$ coincide on $X_K$, so $X_K$ is weak*-closed. Then, by reflexivity, $X^*_K$ is isomorphic to $\ell_1 / (X_K)_\perp$, where $(X_K)_\perp$ is the pre-annihilator of $X_K$ so $X_K$ (and hence $K$ itself) is norm-separable.

This can be generalised as follows. If $\Gamma$ is infinite, then each weakly compact subset of $\ell_\infty(\Gamma)$ has density at most $|\Gamma|$. (For example, for regular cardinals, it follows from Theorem 2.6 here.)

Answer 2

Another way to prove this is to use the following result:

Lemma: If $K \subset \ell_{\infty}$ is weakly compact, then it is weakly metrisable.

Proof: For $x,y \in K$ we set,

$$ d(x,y) = 2^{-n} |x_n-y_n|. $$

Since $K$ is norm bounded (by the uniform boundedness principle) the sum is finite and we can readily check this defines a metric on $K.$ Moreover it is a locally convex space defined by the coordinate maps $x \mapsto x_n.$

Since $(K,d)$ is generated by fewer functionals compared to the weak topology, it is coarser and so the identity map,

$$ (K,w) \rightarrow (K,d) $$

is continuous. Since $(K,w)$ is compact and $(K,d)$ is Hausdorff the map is a homeomorphism, so the topologies coincide.

Hence it follows that $K$ is weakly separable, by using total boundedness and taking $\frac1n$-nets.

Now let $\{x_n\}_{n=1}^{\infty}$ be a weakly dense subset of $K$ and set,

$$ C = \left\{ \sum_{i=1}^n t_i x_{k_i} \mid n \in \mathbb N, t_1, \dots, t_n \in [0,1] \cap \mathbb Q, \sum_{i=1}^n t_i = 1, k_1 < \dots < k_n \right\}. $$

That is, the rational convex hull of the $x_n$'s. One can easily show this is weakly dense in $\mathrm{conv}\ K,$ so by Mazur's lemma,

$$ \overline{\mathrm{conv}}^{\lVert\cdot\rVert} C = \overline{\mathrm{conv}}^w C = \overline{\mathrm{conv}}\ K. $$

Since $C$ is countable, $\overline{\mathrm{conv}}\ K$ and hence $K$ is norm separable.