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It's well known that there are Banach spaces which has a unique isometric predual-- for example, any von Neumann algebra. As other questions on here (for example, Isomorphisms of Banach Spaces ) have shown, there are also Banach spaces $E$ with non-isomorphic (either isometric, or not) $E_1$ and $E_2$ with $E_1^*$ and $E_2^*$ both (isometrically) isomorphic to $E$.

However, I don't know the answer to the following:

Is there a Banach space $E$ such that, if $F$ is another Banach space with $T:F^*\rightarrow E^*$ a (not necessarily isometric) isomorphism, then necessarily $T$ is weak$^*$-continuous?

This is clearly true if $E$ is finite-dimensional, or more generally, reflexive. But I cannot think of a non-reflexive example. (I sort of think that the James space might be an example, but I can't see how to show this).

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up vote 7 down vote accepted

No. Take $E=F$ and perturb the identity on $E^*$ by a rank one operator that is not weak$^*$ continuous; such an operator existing is equivalent to $E$ being non reflexive.

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Ah, so it was quite easy... – Matthew Daws Apr 13 '11 at 13:13

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