Let $B$ be a Banach space, $B^*$ be the dual of $B$ and $\mathcal B$ be the closed unit ball in $B^*$.

By the Banach-Alaoglu theorem, $\mathcal B$ is closed in the weak-* topology. My question is whether in general the weak-* topology on $B^*$ is the strongest topology such that this assertion holds?

Thank you

  • $\begingroup$ Following the nice answer of @xyzzyz, I would restrict the question to topologies such that $B^*$ is a TVS. $\endgroup$ – the_lar Apr 1 '13 at 17:47
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    $\begingroup$ The following facts may be relevant. The strongest linear topology on $E'$ which agrees with the weak star topology on the ball is locally convex and is, in fact, the topology of uniform convergence on the compacta of $E$. It coincides with the finest topology which agrees with the weak star topology on multiples of the unit ball. This is the essential content of the Banach-Dieudonné theorem. $\endgroup$ – jbc Apr 1 '13 at 17:57

No. The compactness of unit ball depends only on the open sets intersecting the unit ball. You can add lots and lots of new open sets far from the unit ball without making the unit ball noncompact.

For instance, introduce a new topology on $B^*$: a subset $U \subset B^*$ is open if and only if there's some set $V$ open in the weak-* topology such that $U \cap \mathcal{B} = V \cap \mathcal{B}$. This condition ensures that the induced topology on $\mathcal{B}$ is the same as topology induced from weak-* topology. Now, since these topologies are the same, $\mathcal{B}$ is compact in this new topology. The complement of the $\mathcal{B}$ in $B^*$ has a discrete topology, though.

Probably you were thinking of some topology that's related to the Banach space structure?

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    $\begingroup$ I like your answer, and indeed I felt that my question was a bit vague. But indeed, I was thinking of a topology that would at least make $B^*$ into a topological vector space. It would be even nicer if it was some "known" or otherwise nice topology. $\endgroup$ – the_lar Apr 1 '13 at 17:44

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