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?

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    $\begingroup$ It's a pitty that the ask-johnson-tag does not exist anymore. $\endgroup$ Feb 3, 2020 at 17:16

1 Answer 1


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

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    $\begingroup$ @Jochen Wengenroth: As you can see, the ask-johnson-tag is no longer needed on MO. Good riddance! $\endgroup$ Feb 5, 2020 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. $\endgroup$ Feb 5, 2020 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. $\endgroup$ Feb 5, 2020 at 17:46
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    $\begingroup$ We clearly very much still need the ask-johnson-tag! $\endgroup$ Feb 5, 2020 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. $\endgroup$ Feb 8, 2020 at 23:52

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