Let $E$ be a Banach space. Is it possible that $E$ is super-reflexive and $\ell_p$ is crudely finitely representable in $E$ for all $p\in (1,2)$?
It seems unlikely but I cannot find an argument off the top of my head.
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No. If $\ell_p$ is finitely crudely representable in a Banach space $X$, then $\ell_p$ is $1+\epsilon$ finitely representable in $X$ for all $\epsilon >0$ by Krivine's theorem. You can find this in the book of Milman and Schechtman.
answered Sep 11, 2017 at 11:27
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1$\begingroup$ Just adding a comment since the answer may not be obvious to everyone. The OP's assumption implies $\ell_1$ is finitely representable too because for any $n$ you can pick $p$ close enough to 1 and $\varepsilon$ small enough so that the basis distance to $\ell_1^n$ and $\ell_p^n$ is small. The key is what Bill said: that you can find $(1+\varepsilon)$ copy of such $\ell_p^n$ by Krivine's theorem. $\endgroup$ Sep 11, 2017 at 16:20