Let $(\mathcal{X} , \|\cdot \|_\mathcal{X})$ be a Banach space and $\mathcal{X}'$ its topological dual. We denote by $\| \cdot \|_{\mathcal{X}'}$ the dual norm and define also the topological dual $\mathcal{X}''$ of the Banach space $(\mathcal{X}',\|\cdot\|_{\mathcal{X}'})$. The unit ball of $\mathcal{X}'$ is denoted by
$$\mathcal{B} = \{ y \in \mathcal{X}', \ \| y\|_{\mathcal{X}'} \leq 1\}.$$
We consider three topologies on $\mathcal{X}'$, on which we recap basic facts:
- The norm topology, for which $\mathcal{B}$ is not compact as soon as $\mathcal{X}$ is infinite dimensional (Riesz' theorem).
- The weak* topology, which is the coarsest topology such that the linear functionals $y \mapsto y(x)$ are continuous for any $x \in \mathcal{X}$. The Banach-Alaoglu theorem states that $\mathcal{B}$ is compact for the weak*-topology.
- The weak topology, which is the coarsest topology such that the linear functionals $y \mapsto z(y)$ are continuous for $z \in \mathcal{X}''$.
The weak* topology is weaker than the weak topology, which is weaker than the norm topology. Moreover, the unit ball $\mathcal{B}$ is not compact for the weak topology as soon as the space is not reflexive (otherwise, the weak and weak* topologies coincide).
My questions are the following: Are there intermediate topologies between the weak* and the weak topology for which the unit ball $\mathcal{B}$ is compact? Or can we say in some sense that the weak* topology is the finest for which the unit ball is compact?
I am not expecting a unique answer for every non-reflexive infinite dimensional Banach spaces, but possibly characterisations of the spaces for which the weak* is indeed the only topology between the weak* and the weak topology.
If it helps, the same questions can be considered for the specific cases:
- $(\mathcal{X},\mathcal{X}',\mathcal{X}'') = (c_0(\mathbb{Z}), \ell_1(\mathbb{Z}), \ell_\infty(\mathbb{Z}))$ where $c_0(\mathbb{Z})$ is the space of vanishing sequences endowed with the norm $\|\cdot\|_\infty$.
- $(\mathcal{X},\mathcal{X}',\mathcal{X}'') = (\mathcal{C}(\mathbb{T}), \mathcal{M}(\mathbb{T}), \mathcal{M}'(\mathbb{T}))$ where $\mathbb{T}$ is the torus, $\mathcal{C}(\mathbb{T})$ the space of continuous periodic functional endowed with the supremum norm, and $\mathcal{M}(\mathbb{T}) the space of finite Radon measure.
(Motivation: I try to understand what is the largest topology for which $\mathcal{B}$ is compact beyond the weak* topology in order to use the Krein-Millmann theorem ensuring the existence of extreme points for convex compact sets.)
Best Answer
As pointed out by Goulifet, my previous answer was wrong. In fact almost the exact opposite is true: on any dual Banach space there is no locally convex vector space topology strictly stronger than the weak${}^*$ topology that makes the unit ball compact. That's because any stronger topology for which the unit ball is compact would have to agree with the weak${}^*$ topology on the unit ball (if two compact Hausdorff topologies are comparable, they are equal), so by Krein-Smulian they would have the same continuous linear functionals (anything continuous for the new, stronger topology would be continuous for the weak* topology on the unit ball and therefore weak* continuous).
We conclude that the new topology would have to equal the weak* topology using an argument kindly supplied by Jochen Wengenroth in the comments.