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](https://mathoverflow.net/questions/1380/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?
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> 
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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).



Unique preduals up to (nonisometric) isomorphism?
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.


