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Is it possible to have two (separable) Banach spaces, $X$ and $Y$, that are not isometrically isomorphic, and yet their dual spaces $X^*$ and $Y^*$ are isometrically isomorphic?

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    $\begingroup$ $\ell_1$, has $c_0$ and $c$ (the space of bounded sequences that have a limit at infinity with the sup norm) as pre-duals. $c$ and $c_0$ are not isometrically isomorphic. There are many other pre-duals of $\ell_1$. $\endgroup$ – David Mitra Sep 25 '13 at 19:14
  • $\begingroup$ @DavidMitra: That is an answer :) $\endgroup$ – Eric Stucky Sep 25 '13 at 19:19
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Yes. For example, $\ell_1$, has $c_0$ and $c$ (the space of sequences that have a limit at infinity with the sup norm) as pre-duals. $c$ and $c_0$ are not isometrically isomorphic (see here). There are many other pre-duals of $\ell_1$.

See this paper for some results on when a Banach space has a unique isometric predual.

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  • $\begingroup$ Thanks, I didn't realize there was such an easy example $\endgroup$ – Owen Sizemore Sep 25 '13 at 21:27
  • $\begingroup$ After thinking more I am especially embarrassed, being an operator algebraist. Namely given $C^*$-algebras A and B, consider the von Neumann algebra $A^{**}$ and $B^{**}$. Any vN algebra has a unique predual. So if $A^{**}=B^{**}$ then $A^*=B^*$. But we can get this with many different $A$ and $B$. $\endgroup$ – Owen Sizemore Sep 27 '13 at 13:19

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