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Sep 11, 2017 at 16:20 comment added Bunyamin Sari 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.
Sep 11, 2017 at 11:54 vote accept user512365
Sep 11, 2017 at 11:27 history answered Bill Johnson CC BY-SA 3.0