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The Maurey-Pisier theorem states that if $p_X$ is the supremum of those $p$ such that the Banach space $X$ has Rademacher type $p$, then $\ell_{p_X}$ is finitely representable in $X$.

For $1\leq p<\infty$, let us say the Schauder basis $(e_i)_{i=1}^\infty$ has \emph{block type} $p$ if there exists a constant $C$ such that for any natural number $n$, any $0=k_0<\ldots <k_n$, any scalars $(a_i)_{i=1}^{k_n}$, if $y_i=\sum_{j=k_{i-1}+1}^{k_i} a_je_j$, \begin{equation}\bigl(\int_0^1 \|\sum_{i=1}^n r_i(t)y_i\|^pdt\bigr)^{1/p} \leq C \bigl(\sum_{i=1}^n \|y_i\|^p\bigr)^{1/p}.\end{equation} Here, $(r_i)_{i=1}^\infty$ is the sequence of Rademacher functions on $[0,1]$.

This definition was given on page 24 of the article "Infinite dimensional geometric moduli and type-cotype theory" by V.D. Milman and A Perelson, in the book "Geometric Aspects of Banach Spaces: Essays in Honour of Antonio Plans."

My question, which doesn't seem to be directly stated in that article, is does the block version of the Maurey-Pisier theorem hold:

$Q:$ Is it true that, if $p_X$ is the supremum of those $p\in [1,\infty)$ such that the basis $(e_i)$ has block type $p$, $\ell_{p_X}$ is block finitely representable in $(e_i)$?

The article of Milman and Perelson references "the variant of Maurey-Pisier's theorem for blocks of a given sequence as it was done, for example, in [MSch2]." However, in the bibliography of that article, there is no [MSch2] entry. I have looked through some of the bibliography items listed in that article, but I am unable to find anywhere a definitive answer to the question $Q$.

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  • $\begingroup$ [MSch2] is Springer Lecture Notes 1200 by Milman and Schechtman. Or you can deduce the result from Rosenthal's JFA 28, 197-225 (1978) paper. $\endgroup$ Commented Nov 9, 2016 at 19:39
  • $\begingroup$ In my copy of the Milman Perelson article (I'm not sure if there are multiple copies), the Milman Schechtman book is [MSch], and there is no [MSch2]. I've been through the proof of the Maurey-Pisier theorem in the Milman Schechtman book, and I don't recall a block version of Maurey-Pisier. The closest I've seen was Propositions 12.5 and 12.6, but these require unconditionality. For the non-block version, this is obtained in a claim on page 88, and the block analogue of this breaks down (it requires a type/cotype estimate on disjointly (not just successively) supported vectors. $\endgroup$
    – user
    Commented Nov 9, 2016 at 20:41
  • $\begingroup$ I've looked through the paper of Rosenthal, and I couldn't readily see how to deduce an answer to question Q from the results there. The proofs of Krivine's theorem I know first pass to a 1 subsymmetric basis bfr in the original basis, and then find a common approximate eigenvalue for the doubling and tripling operators. It is not until the second step where you know the right p so that l_p is block fr. At this moment, I don't see how to perform the first step of passing to a 1 subsymmetric basis while guaranteeing the new basis doesn't have better block type than the original. $\endgroup$
    – user
    Commented Nov 10, 2016 at 12:29
  • $\begingroup$ IIRC you use a Brunel-Sucheston finite Ramsey argument to get a subset of your finite length but long sequence that is sub symmetric. At the start you need all subsets of your sequence to push against the limit. But I am saying this from memory and could well be wrong. Email Schechtman and/or Milman. $\endgroup$ Commented Nov 10, 2016 at 20:04

2 Answers 2

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I believe that the answer is NO. I do not know a counterexample, however, in positive direction one can do the following.

If the basis is unconditional you can define type/cotype on disjointly supported vectors. Then the corresponding result is true. This is written here (with slightly different language), see Theorem 5.6.

The reference that you are looking for must be the paper by Milman and Sharir. A warning though, what they call block version is actually disjointly supported version (not successive blocks) i mentioned above.

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  • $\begingroup$ It is clear that the summing basis does not have any block type better than 2, but is it clear that it does not have any block type better than 1? Otherwise the non-block finite representability of $\ell_1^n$ in the summing basis isn't a counterexample. Is there an easy counterexample for the same question for block cotype? $\endgroup$
    – user
    Commented Nov 14, 2016 at 22:19
  • $\begingroup$ I agree that one has to compute the block type of summing basis. So i revised my answer. I even believed there are unconditional counterexamples but i do not know any now. $\endgroup$ Commented Nov 14, 2016 at 23:04
  • $\begingroup$ You may also want to check if Example 6.4 of the paper by Knaust, Odell and Schlumprecht is relevant. $\endgroup$ Commented Nov 14, 2016 at 23:45
  • $\begingroup$ I didn't compute exactly the best block type of the summing basis, but by comparing it to the Rademacher functions, you can deduce it must be somewhere in [1,2], and none of these $\ell_p$ spaces is block finitely representable in the summing basis. This does give a counterexample without knowing the exact block type. $\endgroup$
    – user
    Commented Nov 15, 2016 at 18:25
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It seems as though the answer is negative, at least for type, although I don't believe this is remarked anywhere in the literature. If $(s_i)$ is the summing basis of $c_0$, for any scalars $(a_i)_{i=1}^n$, $\|\sum_{i=1}^n a_i s_i\|\geqslant |\sum_{i=1}^n a_i|$. This means that $$\int_0^1 \|\sum_{i=1}^n r_i(t) s_i\|_{c_0} dt \geqslant \int_0^1 |\sum_{i=1}^n r_i(t)|dt \geqslant n^{1/2}/A_1,$$ where $A_1$ is the constant from Khintchine's inequality. From this it follows that $(s_i)$ couldn't have block type greater than $2$. But no $\ell_p$, $1\leqslant p<\infty$, is block finitely representable in the summing basis.

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