Question

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?

Answer 1

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.

Answer 2

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