Banach spaces $X^*=Y^*$ implies $X=Y$?

Question

Let $X$ and $Y$ be two Banach space. If $X^*$ is isometry to $Y^*$, then $X$ is isometry to $Y$? Here $^*$ means the dual space.

---

If one of $X$ and $Y$ is reflective,the answer is affirmative. Here is my proof. (“=” means isometry)Suppose $X$ is reflective, then $Y^*=X^*$ which is reflective, so $Y^{**}$ is reflective. The classical embedding $Y \mapsto Y^{**}$ shows $Y$ is (isometry to) a closed subspace of $Y^{**}$. Hence $Y$ is reflective by the result that any closed subspace of reflexive space is reflective. So $Y=X^{**}=Y^{**}=Y$.

For non-reflective case, I try to prove but failed. And I guess this claim is false. But It seems to be difficult to give counter examples. Can someone give me some hint or references? Thanks!

Answer 1

A simple counterexample is $$ X = c = \{ \text{convergent sequences}\} $$ and $$ Y = c_0 = \{ \text{null sequences}\}. $$ Then, one can check that $$ X^* = \ell^1 = Y^* $$ (isometric isomorphisms).

The spaces $X$ and $Y$ are isomorphic but not isometrically isomorphic (look at the extremal points of the unit ball).

Answer 2

There exists a Banach space $X$ such that $X^*$ is isometrically isomorphic to $\ell¹$, but $X$ is not even isomorphic to $c_0$. In fact these spaces are actively studied in a subfield of Functional Analysis called "pre-duals of $\ell^1$". See, for example:

[1] Bessaga, Czeslaw; Pełczyński, Aleksander, Spaces of continuous functions. IV: On isomorphical classification of spaces of continuous functions, Stud. Math. 19, 53-62 (1960). ZBL0094.30303.

[2] Benyamini, Y.; Lindenstrauss, J., A predual of (l_1) which is not isomorphic to a C(K) space, Isr. J. Math. 13, 246-254 (1972). ZBL0253.46044.