On the Banach–Alaoglu theorem: is the unit ball of an equivalent norm also weak-* compact?

Question

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!

Answer 1

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.