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Suppose $X$ and $Y$ are Banach spaces whose dual spaces are isometrically isomorphic. It is certainly true that $X$ and $Y$ need not be isometrically isomorphic, but must it be true that there is a continuous (not necessarily isometric) isomorphism of $X$ onto $Y$?

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up vote 13 down vote accepted

Indeed, $\ell_1$ provides a strong counterexample. As noted by Matt, the spaces C(X), where X is countable and compact, provide nonisomorphic Banach spaces whose duals are isomorphic to $\ell_1$. If X is countable and compact (and Hausdorff, of course!) then X is homeomorphic to a closed ordinal interval [0, a] (equipped with its natural order topology) for some countable ordinal a; this result is known as the Mazurkiewicz-Sierpinski theorem. About 50 years ago Bessaga and Pelczynski showed that if a and b are infinite countable ordinals and a < b, then C([0, a]) is isomorphic to C([0, b]) if and only if b < a^*w*, where w denotes the first infinite ordinal. Thus C([0, w]) is in fact isomorphic to C([0, w^2]), contrary to Gerald's assertion above. Moreover, the least infinite ordinal b such that C([0, b]) is not isomorphic to C([0, w]) is b = w^w. Combining the Mazurkiewicz-Sierpinski result with the Bessaga=Pelczynski result, one has that each space C(X), where X is countable and compact, is isomorphic to C([0, w^(w^*a*)]) for a unique countable ordinal a. In particular, there are uncountably many nonisomorphic Banach spaces whose duals are isometrically isomorphic to $\ell_1$, namely the spaces C([0, w^(w^*a*)]), where a varies over the set of countable ordinals.

$\ell_1$ has other preduals too. For example, a 1972 paper from the Israel Journal of Mathematics by Yoav Benyamini and Joram Lindenstrauss exhibits a Banach space E whose dual is isometrically isomorphic to $\ell_1$, but E is not isomorphic to any space of the form C(X).

Although Gerald's assertion that the Cantor-Bendixson rank distinguishes between spaces of continuous functions on countable compact spaces is not correct, there is a rank/index that does distinguish between the isomorphism classes of these spaces, namely the Szlenk index. In fact the Szlenk index may be used to prove the 'only if' direction of the Bessaga-Pelczysnki result.

I strongly recommend Chapter 2 of the book Biorthogonal Systems in Banach Spaces by Hajek et al if you want to read about the Szlenk index and a proof of the Bessaga-Pelczynski result. That book also contains a sketch proof of the Mazurkiewicz-Sierpinski theorem; for a full account of that theorem I recommend Section 8 of Semadeni's classic book Banach spaces of continuous functions.

If you want a different example, I think the James-tree space JT would do. JT is separable but its dual is (at least I think so - it is worth checking!) isometrically isomorphic to the dual of the direct sum $*JT \oplus H*$, where H is Hilbert space of dimension equal to the cardinality of the continuum. In particular, $*JT \oplus H*$ cannot be isomorphic to JT because they have different density characters, but their duals are isometrically isomorphic. If you want to check the details of this example consult Chapter 13 of the Albiac and Kalton book recommended above by Matthew.

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If X is any countable, compact topological space, then consider the Banach space C(X). The dual space is the finite measures on X, but as any measure is countably additive, and X is countable, taking some ordering we get an isometric isomorphism between C(X)* and $\ell_1$. But not all such C(X) are isomorphic: you can find such X by taking a countable ordinal w and considering the closed interval [0,w]. This is all very nicely explained in the book "Topics in Banach Space Theory" by Albiac and Kalton.

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One may also remark that $c_{0}$ $\times$ (ℓ∞$/c_{0}$) is a predual of (ℓ∞)$^*$.

However, $c_{0}$ $\times$ (ℓ∞$/c_{0}$) is not isomorphic to ℓ∞.

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Nice example Ady, that one has never crossed my mind. –  Philip Brooker Jan 23 '10 at 0:49
More striking than this is that the separable space C[0,1] has dual which is isometrically isomorphic to the dual of the non separable space QC[0,1] (the uniform closure of the step functions on [0,1], so it is also a C(K) space). As Philip mentioned, JT provides a similar example, but of course is not isomorphic to a C(K) space. –  Bill Johnson Jan 23 '10 at 16:29
IMO, to be "more striking" is only a question of taste. Since, in my example, the two preduals are homeomorphic, due to an easy application of the Michael Selection Theorem. –  Ady Feb 1 '10 at 20:58
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There are non-isomorphic preduals of $\ell_1$, but I'm afraid I don't know a reference for this. Recently, Argyros and Haydon solved a long-standing open problem by constructing a predual of $\ell_1$ with extraordinary properties; $c_0$ would not have done the job ...

Actually, I now realize I'm not quite sure about the isometric version -- that is, what happens if you assume the duals are isometric and not just isomorphic. But I would be very surprised if the answer to your question was yes.

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If K, L are countable compact metric spaces, then the isomorphism of C(K) and C(L) depends on the Cantor-Bendixson rank ... If the first empty derived set of K is K^(n) and the first empty derived set of L is L^(m) and n \ne m, then C(K) and C(L) are not isomorphic. So, for example, C([0,w]) and C([0,w^2]) are not isomporphic. But (as noted previously) their duals are both isometric to l^1.

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Thanks to Philip for correcting this. –  Gerald Edgar Jan 23 '10 at 0:56
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