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Let me recall subj:

If $s>0$, $A$ and $B$ are two subsets of $\mathbb{S}^{n}$, $|A|=|B|$ ($|\cdot|$ stands for the Lebesgue measure on the sphere) and $B$ is a cup $B=\{ (x_1,x_2,\dots,x_n)\in \mathbb{S}^n, x_n\leq t \}$ (for some $t\in [-1,1]$), then $|A_s|\geq |B_s|$, where $A_s$ means $s$-neighborhood of the set.

It leads to measure concentration inequalities for the sphere and so has numerous applications. So I guess that Levy's initial proof was simplified, maybe not once. What is the easiest proof of the inequality and where to read it?

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The shortest and most amazing proof (in my opinion) is by Steiner symmetrization around half of a great circle. Given $A$, and given a half great circle $\gamma$, rotate the sphere so that $\gamma$ is a meridian arc. Then for each latitude sphere $H$, you can replace $A \cap H$ by the spherical cap in $H$ centered at $H \cap \gamma$. Let $A'$ be the result. Then it is not hard to show that $|A'_s| \le |A_s|$ for all $s > 0$; in fact even each $|A'_s \cap H| \le |A_s \cap H|$. And you can show that you can pick a sequence of half great circles such that $A$ converges to $B$ under symmetrization, and that some of the inequalities are strict unless $A$ is congruent to $B$.

Of course this is just an outline, but it is an accurate summary (I hope) of the Steiner symmetrization argument. It also works in Euclidean or hyperbolic space using a line rather than half of a line.

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  • $\begingroup$ $|A'_s\cap H|\leq A_s\cap H$ is induction propose if we use induction on dimension, right? $\endgroup$ – Fedor Petrov Oct 26 '10 at 10:40
  • $\begingroup$ ah, no, $A_s\cap H$ does not depend only on $A\cap H$, but on close lattitudes too... $\endgroup$ – Fedor Petrov Oct 26 '10 at 10:43
  • $\begingroup$ Yes, it's a proof by induction, even though $A_s \cap H$ depends on nearby latitudes. $\endgroup$ – Greg Kuperberg Oct 26 '10 at 10:55
  • $\begingroup$ Ah, for $A'$ it depend only on one latitude (though maybe different from $H$), so here everything is OK. It remains to show that a sequence of symmetrizations converge to a hat, It should be standard. $\endgroup$ – Fedor Petrov Oct 26 '10 at 13:32
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    $\begingroup$ Although it's been a long time, I think I learned it from Figiel, Lindenstrauss, Milman. I'm also not sure whether this proof is essentially different from the ones that you have seen. It may mainly be a matter of abbreviated vs detailed explanation. $\endgroup$ – Greg Kuperberg Mar 6 '17 at 14:54
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A different symmetrization-based proof is given in this review article by Schechtman (pp. 7-8); see the previous page for references.

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  • $\begingroup$ Sorry, is a link OK? It downloads some file of 1.1kb size. $\endgroup$ – Fedor Petrov Oct 26 '10 at 19:01
  • $\begingroup$ It's a PostScript file. If you have a Mac and double-click on the file, it will automatically be converted into a PDF file. For Windows, it's a bit more difficult. $\endgroup$ – Deane Yang Oct 27 '10 at 1:24
  • $\begingroup$ I may read .ps files, but not of 1kb size $\endgroup$ – Fedor Petrov Oct 27 '10 at 7:27
  • $\begingroup$ The .ps file is actually 700kB. The link works for me, on two different computers. $\endgroup$ – Mark Meckes Oct 27 '10 at 11:16

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