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?


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.

  • $\begingroup$ $|A'_s\cap H|\leq A_s\cap H$ 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

A different symmetrization-based proof is given in this review article by Schechtman (pp. 7-8); see the previous page for references.

  • $\begingroup$ 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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