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 $\ell_1$?

  • 1
    $\begingroup$ Did you mean "Must $X$ contain a complemented copy of $\ell_1$?" $\endgroup$ – user31373 Jun 19 '12 at 16:10
  • $\begingroup$ @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 $\ell_1$? $\endgroup$ – Philip Brooker Jun 21 '12 at 2:12
  • $\begingroup$ 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. $\endgroup$ – Philip Brooker Jun 21 '12 at 2:19
  • $\begingroup$ @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.) $\endgroup$ – user31373 Jun 21 '12 at 2:33
  • $\begingroup$ @LeonidKovalev: good point about the title; I actually didn't notice that myself! $\endgroup$ – 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:

  1. Norm separability.
  2. 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.

  • $\begingroup$ An exposition of the James-Lindenstrauss construction is given in the last chapter of Albiac-Kalton (specifically, Theorem 13.1.6 on page 313). $\endgroup$ – t.b. Jun 21 '12 at 14:43
  • $\begingroup$ @t.b.: thanks for adding that reference; Albiac-Kalton is a great resource. $\endgroup$ – Philip Brooker Jun 21 '12 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.