Do dual Banach spaces admitting $c_0$ as a quotient contain complemented copies of $\ell_1$? The following question is bothering me. Suppose that we have Banach space $X$ such that $X^*$ has a quotient isomorphic to $c_0$. Must $X^*$ contain a complemented copy of $\ell_1$?
 A: Since every Banach space with a subspace isomorphic to $\ell_1$ has nonseparable dual, to construct a counterexample to the question it suffices to find a Banach space $X$ such that $X^{\ast\ast}$ is norm separable and $X^\ast$ has a quotient isomorphic to $c_0$. To this end we use the James-Lindenstrauss construction, which yields (amongst other things) the following result: Let $Y$ be a separable Banach space. Then there exists a separable Banach space $Z$ such that $Z^{\ast\ast}/Z$ is isomorphic to $Y$. This result was proved by Joram Lindenstrauss in his paper On James's paper "Separable conjugate spaces", Israel J. Math 9(3) (1971), pp.279-284. (Robert James earlier obtained this result for the case where $Y$ is finite dimensional.
To give the claimed counterexample, we also mention the notion of a three-space property for Banach spaces. In particular, a property of a Banach space is a three-space property if whenever $E$ is a Banach space, $F \subseteq E$ is a closed linear subspace and two of the spaces $E$, $F$ and $E/F$ have the property, then all three of the spaces $E$, $F$ and $E/F$ necessarily have the property; a classical example of a three-space property of Banach spaces is reflexivity. We shall call upon the fact that the following properties are both three-space properties:


*

*Norm separability.

*Norm separability of the dual.


Let $Z$ be a separable Banach space such that $Z^{\ast\ast}/Z$ is isomorphic to $c_0$. We claim that $Z^{\ast\ast\ast}$ is norm separable; once this is established, taking $X=Z^{\ast}$ gives the desired counterexample. To this end, let us first notice that since norm separability is a three-space property, $Z^{\ast\ast}$ is norm separable (i.e., take $E=Z^{\ast\ast}$ and $F=Z$ above). Moreover, this implies that $Z^\ast$ is norm separable. In particular, $Z$ and $Z^{\ast\ast}/Z$ both have separable dual, hence taking again $E=Z^{\ast\ast}$ and $F=Z$ above and applying the fact that norm separability of the dual is a three-space property, we conclude that $Z^{\ast\ast\ast}$ is norm separable, as claimed.
