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I'm having trouble digesting the proof of Theorem 9.6 in Milman and Schechtman's classic book "Asymptotic theory of finite dimensional normed spaces" (pg. 54). I'm new to functional analysis, so this is surely a simple misunderstanding on my part.

Theorem 9.6

Here, $M_{\|~\|}$ is denoting the average norm for points on $S^{n-1}$. The first line in the computation of $M_{\|\cdot\|}$ is confusing me:

$$M_{\|\cdot\|} \approx \left(\int_{S^{n-1}} \|\sum a_i e_i\|^2 d\mu(\bar{a})\right)^{1/2}$$

We're actually trying to show a lower bound here, so isn't the $\approx$ hiding a $1/\sqrt{n}$ factor? Isn't this bad for the rest of the argument? Earlier in the book, in the proof of Dvoretzky's Theorem, $M_{\|\cdot\|}$ is given as $\int_{S^{n-1}} \|\sum a_i e_i\| d\mu(\bar{a})$, which is what I'd have expected here also.

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Are you confusing the median $M_{\|\ \|}$ with the mean/average $A_{\|\ \|}$? –  user31373 Jul 14 '12 at 19:11

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By Remark 1 at the bottom of page 19, the median $M_{\|\ \|}$ is within an additive constant $C_p$ from the $L^p$ mean $\left(\int_{S^{n-1}}\|x\|^p\right)^{1/p}$, for any $1\le p<\infty$. They use this remark with $p=2$. John's ellipsoid theorem 3.3 is invoked here only to justify the application of 5.1 and the subsequent remark 1. They do not use $\|x\|\ge |x|/\sqrt{n}$ under any of the integral signs; as you said, this would destroy the argument. But the better bound $\|e_i\|\ge 1/4$ holds for half of basis vectors, and this is enough to make the cotype inequality effective.

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I'd indeed confused median with mean, and I didn't remember that remark. Thanks for your help! –  arnab Jul 14 '12 at 21:06

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