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.)

  • $\begingroup$ You can just take any $x'\in X'$, $\|x'\|>1$, and let the weak $*$ open sets together with $\{ x'\}$ generate a stronger topology. This still induces the same topology on $B$ as before. $\endgroup$ Jul 28, 2022 at 14:57
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    $\begingroup$ So perhaps you really want vector space topologies? $\endgroup$ Jul 28, 2022 at 14:57
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    $\begingroup$ Perhaps relevant: it is a well known classical Banach space result that the finest topology (not necessarily lc or linear) on a dual Banach space which agrees with the weak star topology on balls is the Mackey topology (uniform convergence on weak compacta). If one assumes linearity, then one only need look at the unit ball. $\endgroup$
    – ottone
    Jul 28, 2022 at 17:26
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    $\begingroup$ If you look just at the trace of the unit ball, there is not. Any two comparable compact Hausdorff topologies on a set coincide. $\endgroup$ Jul 28, 2022 at 17:27
  • $\begingroup$ @ChristianRemling: you are right, we actually want a vector space topology that makes, for instance, continuous and functional via elements of $\mathcal{X}$ plus possibly one additional functional via an element of $\mathcal{X}'' \backslash \mathcal{X}$$ (as in Nik Weaver approach, see below). $\endgroup$
    – Goulifet
    Jul 29, 2022 at 18:55

1 Answer 1


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.

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    $\begingroup$ For the James space $J$ there is essentially only one continuous linear functional on $J'$ which is not weak$^*$-continuous (all others are multiples since $J$ is co-one-dimensional in $J''$). $\endgroup$ Jul 28, 2022 at 14:03
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    $\begingroup$ Thank you! I was having doubts about that point and trying to figure it out. I'll correct my answer. $\endgroup$
    – Nik Weaver
    Jul 28, 2022 at 14:09
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    $\begingroup$ I have some problem with your construction. Take the following example where $\mathcal{X} = c_0$ is the set of vanishing sequences. Then, $\mathcal{X}' = \ell_1$ and $\mathcal{X}'' = \ell_\infty$. Consider the sequence of sequences $u_n = e_n = (0,\ldots , 0, 1 , 0 ,\ldots) \in \mathcal{B}$ and $f = 1 \in \ell_\infty \backslash c_0$. Clearly, $u_n \rightarrow 0$ for the weak* topology and $\langle u_n , f \rangle = 1$ for any $n$. However, I cannot say that there exists $u \in \ell_1$ such that $u_n \rightarrow u$. $\endgroup$
    – Goulifet
    Jul 29, 2022 at 18:51
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    $\begingroup$ About the last point, we see that the sequence $u \in \ell_1$ should satisfy that $\langle u , e_n \rangle = 0$ for any $n$ and $\langle u, 1 \rangle = 1$, the two requirements being impossible. This seems to prove that it is not possible to extract from $(u_n)$ a subsequence (of sequences) that converges in $\ell_1$. Do you agree? $\endgroup$
    – Goulifet
    Jul 29, 2022 at 18:52
  • $\begingroup$ Oh gosh, you are absolutely right! I'll have to correct my answer. $\endgroup$
    – Nik Weaver
    Jul 29, 2022 at 20:55

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