# Non-super reflexive space

Suppose $X$ is a reflexive space (possibly non-separable) which is not super-reflexive. Then (by definition) there exists a non-reflexive Banach space $Y$ which is non-reflexive 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. non-reflexive) which is separable? In this spirit, what are examples of reflexive but not super-reflexive spaces in which neither $\ell_1$ nor $c_0$ is finitely representable?

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What does the existence of (twice!) non-reflexive $Y$ have to do with any properties of $X$? There are certainly non-reflexive separable spaces … but I suspect that gremlins ate half the text of your question. –  Harald Hanche-Olsen Jan 4 '13 at 20:37
Right, Harald; I added the missing part. –  Bill Johnson Jan 4 '13 at 20:41

@Bojan Kwitek: examples of the kind of space you asked about in your second question do exist; the Pisier-Xu paper is Random series in the real interpolation spaces between the spaces $v_p$, see link.springer.com/chapter/10.1007%2FBFb0078146 . –  Philip Brooker Jan 5 '13 at 4:14
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 $X^*$? –  Bojan Kwitek Jan 6 '13 at 22:56
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 $\ell_1^n$. –  Bill Johnson Jan 28 '13 at 17:36