Suppose $X$ is a reflexive space (possibly nonseparable) which is not superreflexive. Then (by definition) there exists a nonreflexive Banach space $Y$ which is nonreflexive but is finitely representable in $X$, meaning that for each $\lambda >1$, every finite dimensional subspace of $Y$ is $\lambda$isomorphic to a subspace of $X$. Can we always find such $Y$ (i.e. nonreflexive) which is separable? In this spirit, what are examples of reflexive but not superreflexive spaces in which neither $\ell_1$ nor $c_0$ is finitely representable?

$\begingroup$ What does the existence of (twice!) nonreflexive $Y$ have to do with any properties of $X$? There are certainly nonreflexive separable spaces … but I suspect that gremlins ate half the text of your question. $\endgroup$– Harald HancheOlsenJan 4, 2013 at 20:37

1$\begingroup$ Right, Harald; I added the missing part. $\endgroup$– Bill JohnsonJan 4, 2013 at 20:41
1 Answer
The first question is easy: Every non reflexive space has a separable non reflexive subspace (e.g. by the EberleinSmulian theorem or by R. C. James' characterization of non reflexivity).
The second question was a longstanding open problem that was solved by James in the 1970s. Pisier and Xu gave another proofyou can find their paper by using MathSciNet. Their approach is more conceptual and uses interpolation theory but is not easy.

$\begingroup$ Dear Prof. Johnson. Thank you. I've been trying to find the papers with no success yet, but I'll try again. Let me ask then whether the answer to the second question is positive or negative. :) $\endgroup$ Jan 4, 2013 at 22:34

3$\begingroup$ @Bojan Kwitek: examples of the kind of space you asked about in your second question do exist; the PisierXu paper is Random series in the real interpolation spaces between the spaces $v_p$, see link.springer.com/chapter/10.1007%2FBFb0078146 . $\endgroup$ Jan 5, 2013 at 4:14

$\begingroup$ OK, thank you. I haven't spotted this paper. By the way, can we deduce from the fact $\ell_1$ is finitely representable in $X$ that $c_0$ is finitely representable in $X^*$? $\endgroup$ Jan 6, 2013 at 22:56

$\begingroup$ No, Bojan$X=c_0$ is a counterexample. You do get that $\ell_1$ is finitely representable in $X^*$. Having $c_0$ finitely representable in $X^*$ is equivalent to $X$ containing uniformly complemented uniform copies of $\ell_1^n$. $\endgroup$ Jan 28, 2013 at 17:36