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This must be surely known but I couldn't locate this problem in the literature. It popped out in a priori unrelated approximation problem but if true, would help me greatly.

Let $p\in (1,\infty)$.

Informal version:

Do the spaces $\ell_2^k$ sit well-complemented in all sufficiently large finite-dimenional subspaces of $\ell_p$?

Formally:

Does there exist $C>0$ and two sequences of integers $(d_n)_{n=1}^\infty$ and $(k_n)_{n=1}^\infty$ both increasing to $\infty$ such that each $d_n$-dimensional subspace $X$ of $\ell_p$ contains a $k_n$-dimensional subspace that is $C$-isomorphic to $\ell_2^{k_n}$ and $C$-complemented in $\ell_p$?

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2 Answers 2

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If I understand you correctly, much stronger statement holds for any space with nontrivial type, see Theorem 15.10 in the book of Milman and Schechtman on Asymptotic Theory.

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  • $\begingroup$ That's great, thank you! It precisely the result I have been looking for. $\endgroup$
    – Tomasz Kania
    Aug 11, 2015 at 22:00
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This can be directly calculated. All you need is upper and lower function $L_p$ estimates in a sequence of length $2^n$. Then the Rademacher functions spans well complemented copy of $\ell_2^n$. I think this is explicitly written in Tzafriri's paper `on Banach spaces with an unconditional basis' from late 70's in Israel J. but I cannot access it at the moment.

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  • $\begingroup$ The OP wants a well complemented $\ell_2^k$ in an ARBITRARY high dimensional subspace of $\ell_p$. $\endgroup$
    – Bill Johnson
    Aug 14, 2015 at 20:28
  • $\begingroup$ I see, I missed that. Will this also follow from your paper with Tzafriri on Complemented copies of $\ell_p^n$'s in subspaces of Banach lattices? There should be a local version of that. $\endgroup$
    – Bunyamin Sari
    Aug 14, 2015 at 21:45
  • $\begingroup$ You can do an ultraproduct argument to reduce the problem to infinite dimensional subspaces of super reflexive lattices, but that looks like a silly way of proving it and gives no estimates. $\endgroup$
    – Bill Johnson
    Aug 16, 2015 at 0:58

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