# Dual without quotients isomorphic to $c_0$

The following question is bothering me. Suppose we have a dual Banach space $X^*$ and assume $X^*$ has a quotient isomorphic to $c_0$. Must $X^*$ contain a complemented copy of $\ell_1$?

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Did you mean "Must $X$ contain a complemented copy of $\ell_1$?" – user31373 Jun 19 '12 at 16:10
@Leonid: I suspect that this question is not so much about duality, but more about universality (in particular, universal surjectivity). Perhaps I am reading too much into it, but the way I interpret the question is: Suppose $X^\ast$ has a quotient isomorphic to $c_0$. Is $X^\ast$ surjectively universal for the class of separable Banach spaces? That is, does $X^\ast$ necessarily admit continuous linear surjections onto all separable Banach spaces? Equivalently, does $X^\ast$ contain a complemented subspace isomorphic to $\ell_1$? – Philip Brooker Jun 21 '12 at 2:12
In any case, the answer is no for both the OP's question as given and to your suggested possible correction to the question. In the case suggested by you - i.e., does $X$ contain a complemented copy of $\ell_1$? - I am sure you have no trouble coming up with a counterexample, e.g., take $X=c_0$ or, more generally, $X=C(K)$ for infinite compact Hausdorff $K$. I will post a negative answer to the OP's question later today when I have the time to do so. – Philip Brooker Jun 21 '12 at 2:19
@PhilipBrooker Thank you. To tell the truth, I did not feel inclined to think about the question without being certain that it's stated correctly. (For one thing, the title says "Dual without quotients isomorphic to $c_0$, contrary to the body of the question.) – user31373 Jun 21 '12 at 2:33
@LeonidKovalev: good point about the title; I actually didn't notice that myself! – Philip Brooker Jun 21 '12 at 14:39

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:
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.