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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
May 7, 2016 at 15:28 comment added Aleksei Lissitsin @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.
May 4, 2016 at 15:27 comment added Bill Johnson 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).
May 4, 2016 at 14:08 comment added Bill Johnson 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).
May 3, 2016 at 9:57 history asked Aleksei Lissitsin CC BY-SA 3.0