4
$\begingroup$

I am interested in the following question about James' quasi-reflexive Banach space $\mathcal{J}$:

Does there exists a non-Hilbertian subspace $X$ of $\mathcal{J}$ such that $X$ isomorphically embeds into every non-Hilbertian subspace of itself?

Here, by "subspace" I mean "closed, infinite-dimensional vector subspace", and by "Hilbertian" I mean "isomorphic to $\ell_2$".

I vaguely recall having found, one year ago, a paper proving that the answer to this question was no, or at least giving a similar/partial result suggesting that the answer should be no. Problem is, I don't manage to find this paper again, I don't even remember who were the authors and what was the exact result they proved. Do some of you recall having seen something like that?

$\endgroup$
1
  • 6
    $\begingroup$ It's a pitty that the ask-johnson-tag does not exist anymore. $\endgroup$ – Jochen Wengenroth Feb 3 '20 at 17:16
2
$\begingroup$

A weaker `block version' is true for the conditional spreading basis (the summing basis) of $\mathcal{J}$: Every seminormalized block basis of the spreading basis has a subsequence either equivalent to an unconditional basis ($\ell_2$) or a convex block sequence equivalent to the basis itself. The result holds in general in spaces with a convex block homogeneous conditional spreading basis. See the section 5 of the following paper

A study of conditional spreading sequences Spiros A. Argyros, Pavlos Motakis, Bünyamin Sari

$\endgroup$
10
  • 1
    $\begingroup$ @Jochen Wengenroth: As you can see, the ask-johnson-tag is no longer needed on MO. Good riddance! $\endgroup$ – Bill Johnson Feb 5 '20 at 0:37
  • $\begingroup$ @Bunyamin Sari: In this paper: jstor.org/stable/2041285?seq=1#metadata_info_tab_contents, the authors prove that James' space has uncountably many pairwise nonequivalent unconditional basic sequences. This contradicts your claim that the only one is $\ell_2$. Actually, $\mathcal{J}$ itself cannot have the property I'm asking about: the same paper shows it has non-Hilbertian reflexive subspaces, hence it cannot isomorphically embed into these subspaces. $\endgroup$ – N. de Rancourt Feb 5 '20 at 12:49
  • $\begingroup$ I agree that the result I quoted doesn't answer your question, sorry I was too quick. I edited the answer. This doesn't contradict the results you quoted either because it involves block subspaces of the conditional basis. $\endgroup$ – Bunyamin Sari Feb 5 '20 at 17:46
  • $\begingroup$ We clearly very much still need the ask-johnson-tag! $\endgroup$ – Bunyamin Sari Feb 5 '20 at 17:47
  • $\begingroup$ Let $X:=\left(\sum_n \oplus E_n \right)_2$ with $d_n(X):=\sup d(E,\ell_2^n)\to\infty$. Is there $Y:=\left(\sum_n \oplus F_n \right)_2\subset X$ so that $d_n(Y)\to \infty$ in much slower pace (so that $X$ doesn't embed into $Y$). This is surely true, and would answer the OP's question in negative. $\endgroup$ – Bunyamin Sari Feb 8 '20 at 23:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.