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I am reading a paper http://www.math.tamu.edu/~johnson/TF3.4.pdf by Bill Johnson and Andrzej Szankowski and having trouble grasping why $d_n(Z_m) \leq d_n(\ell_{p_{m+1}} ) = n^{|p_{m+1}-2|}$ in the end of page 13 holds.

Here $d_n(X) = \sup d(E, \ell_2^n)$, where $d$ is the Banach-Mazur distance and $\sup$ is taken over all $n$-dimensional subspaces $E$ of $X$. So, in fact, there are two questions:

1) Why $d_n(\ell_p) = n^{|p-2|}$?

2) Why $d_n(Z_m) \leq d_n(\ell_{p_{m+1}})$? Here $Z_m = \sum_{i > m} \oplus_2 \ell_{p_i}^{k_i}$ with $|p_i - 2| \leq |p_{m+1} - 2|$ and $k_i \in \mathbb N$.

Let me note that it is relatively easy to see that $d_n(\ell_p^m) = n^{|1/p - 1/2|}$. See, e.g., [Handbook of Banach space geometry, p. 43] for $d(\ell_p^n, \ell_2^n) = n^{|1/p - 1/2|}$. (Also Banach-Mazur distance estimate finite-dimensional $\ell_p$ spaces or Banach-Mazur distance between $\ell^p$-norms .)

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    $\begingroup$ Aleksei, if you take an $\ell_2$ sum of spaces, $d_2$ of the sum is at most the supremum if $d_2$ of the individual spaces. D. R. Lewis proved that $d_n(L_p)$ is $n^{|1/p -1/2|}$. That fact should be in Tomczak's book and is mentioned on the page you reference. Other things on the page follow from arithmetic unless we made a computational mistake. Certainly we did not claim the equality in your display (1). $\endgroup$ Commented May 4, 2016 at 14:08
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    $\begingroup$ I look again at the passage and see why you are confused. On the penultimate line of the page the first equality should be an inequality The inequality follows from what I wrote in my first comment plus $|1/2 - 1/p| \le |p-2|$ ($p\ge 1$ for us). $\endgroup$ Commented May 4, 2016 at 15:27
  • $\begingroup$ @BillJohnson Thank you for the quick answer! Regarding the $\ell_2$-sum result, now that I knew it was true, it was easy to see. I am still looking for the direct reference to $d_n(\ell_p)$ result. Once I find it, I'll post it here too, for completeness. $\endgroup$ Commented May 7, 2016 at 15:28

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