8
$\begingroup$

It was proved by W.B. Johnson and H.P. Rosenthal [Studia Math. 43 (1972), 77–92] that every Banach space $X$ with $X^{**}$ separable is hereditarily reflexive: every infinite dimensional closed subspace of $X$ contains an infinite dimensional reflexive subspace.

Suppose that $X$ is separable and $X^{**}/X$ reflexive. Is $X$ hereditarily reflexive?

Of course, we would have a positive answer if each infinite dimensional closed subspace of such a space $X$ contains an infinite dimensional subspace $Y$ with $Y^{**}$ separable.

$\endgroup$
  • 1
    $\begingroup$ Gowers' dichotomy theorem reduces to the case that $𝑋$ is HI. Have you asked Argyros whether there exists a hereditarily non reflexive HI space $X$ with $X^{**}/X$ reflexive? $\endgroup$ – Bill Johnson Jul 14 at 18:11
  • 1
    $\begingroup$ I have looked at Spiros A. Argyros, Alexander D. Arvanitakis, Andreas G. Tolias. Saturated extensions, the attractors method and Hereditarily James Tree Spaces. pp. 1-90 in "Methods in Banach space theory". J.M.F. Castillo and W.B. Johnson eds. Cambridge University Press, 2006. $\endgroup$ – M.González Jul 14 at 19:04
5
$\begingroup$

The question has a negative answer:

Following the idea in Bill Johnson's comment, I looked at the work of Argyros. In this paper (see the reference below), there are several examples of hereditarily indecomposable Banach spaces containing no infinite dimensional reflexive subspaces.

One of these spaces $X$ has the additional property that $X^{**}/X$ is isomorphic to $\ell_2(\Gamma)$ with $\Gamma$ uncountable: see Proposition 5.4 in the paper.

[reference] S.A. Argyros, A.D. Arvanitakis, A.G. Tolias. Saturated extensions, the attractors method and Hereditarily James Tree Spaces. pp. 1-90 in "Methods in Banach space theory". J.M.F. Castillo and W.B. Johnson eds. London Math. Soc. Lecture Note Ser. 337, Cambridge University Press, 2006.

| cite | improve this answer | |
$\endgroup$
  • 5
    $\begingroup$ Just one of the many amazing examples constructed by Argyros et al. $\endgroup$ – Bill Johnson Jul 15 at 13:22

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.