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.)



Compactness of the unit ball of a Banach space for topologies finer than the weak* topology
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.


