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Is it possible to show the existence of an infinite dimensional closed subspace of $\ell_p$ ($1<p<\infty$, $p\neq 2$), not isomorphic to $\ell_p$, in an elementary way?

For $1<p<q<2$, I think we can find such an example isomorphic to $\ell_p(\ell_q^n)$, but the proof I have in mind uses the fact that $\ell_q$ embeds in $L_p(0,1)$, which is not elementary.

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    $\begingroup$ Isomorphic = isometrically isomorphic? $\endgroup$ – LSpice Dec 7 '19 at 20:12
  • $\begingroup$ I mean just (linearly) isomorphic. Maybe $(1+\varepsilon)$-isomorphic for each $\varepsilon>0$. $\endgroup$ – M.González Dec 7 '19 at 20:24
  • $\begingroup$ AFAIK, the proof you outline is the simplest proof for $p<2$. For $p>2$ the simplest proof I know is what Sari mentioned; namely, the result proved by Davie and Figiel independently (using Enflo's basic idea) that in this range $\ell_p$ has a subspace that fails the approximation property. $\endgroup$ – Bill Johnson Dec 9 '19 at 23:35
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Not sure what is considered elementary but I don't think there is an easy and quick argument for it. And for good reasons. $\ell_p$ is block homogeneous; any sequence of disjointly supported (with respect to the unit vector basis) vectors is equivalent to the basis, and most subspaces you can write down by hand will end up being isomorphic to $\ell_p$. So any `elementary' existence proof has to involve subspaces spanned by vectors with mixed supports and some nontrivial invariant. In fact, the first proof of the fact in the question was via Davie's theorem that $\ell_p$ has subspaces without the approximation property which is highly non-trivial. You can also deduce it from Komorowski-Tomczak theorem on the existence of subspaces without local unconditional structure, which is also highly on-trivial. So I think it is safe to say that the answer to your question is negative.

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    $\begingroup$ That $\ell_p$, $2<p$, has a subspace that fails lust was known long before the Komorowski-Tomczak theorem, but also that proof is not elementary. You take random large subspaces $E_n$ of $\ell_p^n$ and consider the $\ell_p$-sum of the $E_n$. These kinds of spaces even fail the GL property. $\endgroup$ – Bill Johnson Dec 9 '19 at 23:39
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One can try to look at relatively simple subspaces constructed by Sobczyk in 1941 [Projections in Minkowski and Banach spaces. Duke Math. J. 8 (1941), 78–106]. These subspaces are uncomplemented subspaces in $l_p$ with $p\ne 2$. Possibly you will be able to show that they are not isomorphic to $l_p$.

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    $\begingroup$ The problem, of course, is that there are uncomplemented subspaces of $\ell_p$ that ARE isomorphic to $\ell_p$. (This fact is however less elementary than that there are subspaces of $\ell_p$ that are not isomorphic to $\ell_p$.) $\endgroup$ – Bill Johnson Dec 9 '19 at 23:30

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