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Suppose that $E$ is a Banach space and let $E^*$ denote its dual space with canonical norm $\lVert\bullet\rVert_{E^*}$. Suppose that $\lvert\bullet\rvert_{E^*}$ is an equivalent norm on $E^*$. The Banach–Alaoglu theorem tells us that the unit ball with respect to the norm $\lVert\bullet\rVert_{E^*}$ is weak-* compact. Is it true that the unit ball with respect to the norm $\lvert\bullet\rvert_{E^*}$ is also weak-* compact?

More generally, if $E$ is a locally convex topological vector space, and the polar topology of $E^*$ is induced by some norm $\lvert\bullet\rvert_{E^*}$. Is it true that the unit ball is weak-* compact?

If $E$ is reflexive, then the unit ball of $E^*$ with respect to the norm $\lvert\bullet\rvert_{E^*}$ is strongly closed and convex, hence weakly closed and therefore weak-* closed. I don't know in general, whether it's true.

Any ideas? Thanks!

Tomasz Kania
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Yai0Phah
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  • @G.Sassatelli Thanks for clarification. Yes. – Yai0Phah Mar 01 '16 at 16:46
  • Banach-Alaoglu has nothing to do with norms, really. If $E$ is a Hausdorff locally convex space, and $U$ a neighbourhood of $0$ in $E$, then the polar of $U$ is $\sigma(E^{\ast},E)$-compact. – Daniel Fischer Mar 01 '16 at 16:48

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No. If this were true then the isomorphic Banach spaces $L_\infty$ and $\ell_\infty$ would have isomorphic preduals, but $L_1$ is not isomorphic to $\ell_1$. Indeed, you are asking whether all bounded isomorphisms $X^*\to X^*$ are weakly*-to-weakly* continuous (in which case they would have pre-adoints being isomorphisms on their own). This is not the case.

The drastic case of $\ell_1$ shows that that you can have uncountably many balls (closed, symmetric around the origin, convex sets with non-empty interior) in $\ell_1$ that are not homeomorphic to in the whatever weak* topology.

Community
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Tomasz Kania
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    Unless the nontrivial thing mentioned in the answer, everything seems right to me (wait for reviews from others). Facts used: 1. The dual of $X^$, with respect to weak- topology, is $X$ itself; 2. Suppose $T\colon X\to Y$ is continuous linear between TVSs, then the transpose ${}^tT$ is well-defined and weak-$\to$weak- continuous; 3. A linear map between normed spaces $T\colon E\to F$ is continuous if and only if it's weak-weak continuous. – Yai0Phah Mar 01 '16 at 19:43