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Let $Y$ be a Banach space isomorphic to $\ell_p$, $1<p<\infty$. Is it true that any finite subset of $\ell_p$ is isometric to some finite subset of $Y$?

It seems to me that it is an interesting question. It can be regarded as a special case of a recent question Under which conditions is it possible to find points with same distances under bi-Lipschitz map (which was closed) and is related to my (unanswered) question Isometric embeddings of finite subsets of $\ell_2$ into infinite-dimensional Banach spaces

In connection with this question it is worth mentioning that there is a theory developed by Krivine in Ann. Math. (2) 104, 1-29 (1976) (with important additions by Maurey and Pisier (Stud. Math. 58, 45-90 (1976)) and further simplifications by other authors, see Chapter 12 in the book by Benyamini-Lindenstrauss on Geometric Nonlinear Functional Analysis or Part II in the book Milman-Schechtman on Asymptotic Theory) which implies that for any $\varepsilon>0$ any finite-dimensional subspace of $\ell_p$ embeds into any Banach space isomorphic to $\ell_p$ with distortion $\le (1+\varepsilon)$.

Added on 4/3/2017: In a recent paper James Kilbane proved that the set of possible counterexamples (if they exist) is small in a certain sense.

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  • $\begingroup$ Is $p$ any number in $[1,\infty]$? Also I understand that $\ell^p$ means $\ell^p(\mathbf{N})$. $\endgroup$
    – YCor
    Commented Dec 27, 2015 at 12:46
  • $\begingroup$ @YCor By $\ell_p$ I meant $\ell_p(\mathbb{N})$, but I meant $p\in(1,\infty)$ only (I corrected this). For $p=1,\infty$ one can easily answer the question using the fact that $\ell_1$ and $\ell_\infty$ are isomorphic to spaces having the uniqueness of geodesics property. Thank you for your question. $\endgroup$
    – Mikhail Ostrovskii
    Commented Dec 27, 2015 at 16:30
  • $\begingroup$ @MichailOstrovskii: For the benefit of the non experts you should add that $\ell_p$ is $1+\epsilon$ finitely representable in every isomorphic of $\ell_p$ for every $\epsilon >0$. $\endgroup$
    – Bill Johnson
    Commented Dec 27, 2015 at 18:59
  • $\begingroup$ @BillJohnson: I added such information, thank you for the comment $\endgroup$
    – Mikhail Ostrovskii
    Commented Dec 27, 2015 at 19:40

1 Answer 1

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We (me and the author posing the question) have answered this question in the negative - for each $p \neq 2$ we have found a space $X$ such that $X$ is isomorphic to $\ell_p$ but there are 5 element subsets of $\ell_p$ that do not embed isometrically into $X$. Our paper is on arxiv here: https://arxiv.org/abs/1708.01570

The question about $\ell_2$ is still open.

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