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The following question is related to the previous question Minimality properties of James' space; I post it as a new question since the system does not allow me to add a comment.

Question Consider the following class of non-Hilbertian spaces: $X_{p,2}=(\sum_{n=1}^\infty \oplus\ell^p_n)_2$, $1\le p\le \infty$, $p\neq 2$. Is it true that the only infinite dimensional Banach space that is isomorphically embedded into anyone of them is the Hilbert space?

Notice that all these spaces are subspaces of the space $\mathcal{J}$.

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  • $\begingroup$ For somebody outside this community: $\ell^p_n$ means the $n$-dimensional $\ell^p$? $(\Sigma\cdots)_2$ means the $\ell^2$-sum? Certainly "is the Hilbert space" means "is isomorphic to a Hilbert space". $\endgroup$
    – YCor
    Commented Feb 6, 2020 at 16:13
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    $\begingroup$ @YCor Yes to all. $\endgroup$
    – Bunyamin Sari
    Commented Feb 6, 2020 at 16:17
  • $\begingroup$ How do you know these spaces are subspaces of $\mathcal{J}$? I suppose its sufficient to prove the case $p=\infty$, whence the remaining cases follow due to the fact that all finite-dimensional spaces embed almost isometrically into $\ell_\infty^n$ for sufficiently large $n$. $\endgroup$
    – Ben W
    Commented Feb 8, 2020 at 21:59
  • $\begingroup$ @Ben W The standard basis of $\mathcal J$ is skipped Hilbertian so whenever you have finite dimensional spaces with a gap in between their supports, they add in $\ell_2$ sense. $\endgroup$
    – Bunyamin Sari
    Commented Feb 8, 2020 at 22:17
  • $\begingroup$ @BunyaminSari Right but can you really find $\ell_\infty^n$ uniformly in $\mathcal{J}$ in the first place? $\endgroup$
    – Ben W
    Commented Feb 8, 2020 at 22:35

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I think such a space has type 2 and cotype 2 so by Kwapien's theorem it is isomorphic to a Hilbert space.

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  • $\begingroup$ I would, then, be interested in the answer to the original question when we restrict our attention to $p>2$. $\endgroup$
    – N. de Rancourt
    Commented Feb 6, 2020 at 18:00
  • $\begingroup$ Yes; you only need embedding for one $\infty>p>2$ and one $p<2$ to apply Kwapien's theorem. If you have embedding for all $p>2$ and uniform control on the isomorphism constants, then again you get a positive answer (use the fact that any $n$ dimensional subspace of $L_p$ is $n^{|1/p-1/2|}$ isomorphic to $\ell_p^n$). $\endgroup$
    – Bill Johnson
    Commented Feb 7, 2020 at 14:54
  • $\begingroup$ OTOH, if $X$ embeds into $X_{p,2}$ for all $p<2$, then $X$ need not be isomorphic to a Hilbert space. Take $p_n \uparrow 2$ and $m_n \to \infty$. Then $(\sum_{n=1}^\infty \ell_{p_n}^{m_n})_2$ embeds into $X_{p,2}$ for all $p<2$ but is not isomorphic to a Hilbert space if $m_n \to \infty$ sufficiently quickly. $\endgroup$
    – Bill Johnson
    Commented Feb 7, 2020 at 15:04
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    $\begingroup$ Proposition 21.6 and Remark 2 following In Tomczak-Jaegermann's book "Banach-Mazur Distances..." yield that for for all $2<s<p<\infty$, $\ell_s$ is isomorphic to a subspace of $L_2(\ell_p)$, with uniformly bounded isomorphism constant for $s<(2+p)/2$. It follows that if $s_n\downarrow2$ and $m_n \to \infty$ quickly, then $(\sum_{n=1}^\infty \ell_{s_n}^{m_n})_2$ embeds into $X_{p,2}$ for all $p>2$ but is not isomorphic to a Hilbert space. $\endgroup$
    – Bill Johnson
    Commented Feb 8, 2020 at 17:26
  • $\begingroup$ Did I misunderstand the OP's question? It sounded like he was asking whether every Banach space is a Hilbert space that embeds isomorphically into $(\oplus\ell_p^n)_{\ell_2}$, $p\in[1,2)\cup(2,\infty]$. But the space itself is not isomorphic to a Hilbert space, so the answer is no. $\endgroup$
    – Ben W
    Commented Feb 8, 2020 at 21:57

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