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What is the easiest (preferably without calculations) way to see that the mean value of $\max(x_1,x_2,\dots,x_n)$ on the sphere $\mathbb{S}^{d-1}= \{ (x_1,\dots,x_n):\ x_1^2+\dots+x_n^2=1 \}$ behaves like $\sqrt{\log(n)/n}$, or at least that is is much more then $1/\sqrt{n}$ for large $n$? The same (and less or more a priori equivalent) question concerns the standard Gaussian measure and expectation of $\infty$-norm w.r.t. it.

The proofs I know (for example, the one which V. Milman attributes to Figiel) use too many integrals.

And by the way, how to put {,} in math here? \ { does not work for me

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\lbrace and \rbrace. –  Gerry Myerson Nov 9 '10 at 12:01
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Or \\{ and \\}. –  Andrey Rekalo Nov 9 '10 at 12:02
    
Oh, thanks! $\ \ $ –  Fedor Petrov Nov 9 '10 at 12:50
    
doesn't the fact that on your hypersphere ($d$ should be $n$ in your question), $\|x\|_\infty \ge 1/\|x\|_1 \ge 1/\sqrt{n}$ help? –  Suvrit Nov 10 '10 at 15:04
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1 Answer 1

up vote 15 down vote accepted

For some positive $c$ bounded away from zero, the probability that a standard gaussian variable is larger than $c\sqrt{\log n}$ is $1/n$. It follows that the probability that at least one variable out of $n$ independent standard gaussians is larger than $c\sqrt{\log n}$ is $1-(1-1/n)^n$ which tends to $1-1/e$. From that one gets that the expectation of the maximum of $n$ standard gaussians is at least (basically) $(1-1/e)c\sqrt{\log n}$. As you said the question about variables on the sphere is equivalent to that.

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Thanks, that's clear indeed. –  Fedor Petrov Nov 9 '10 at 12:53
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