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Math 401 Paper 1, Side note 3: Levy's concentration theorem
Basic definitions
Lipschitz function
$\eta$-Lipschitz function
Let (X,\operatorname{dist}_X) and (Y,\operatorname{dist}_Y) be two metric spaces. A function f:X\to Y is said to be $\eta$-Lipschitz if there exists a constant L\in \mathbb{R} such that
\operatorname{dist}_Y(f(x),f(y))\leq L\operatorname{dist}_X(x,y)
for all x,y\in X. And \eta=\|f\|_{\operatorname{Lip}}=\inf_{L\in \mathbb{R}}L.
That basically means that the function f should not change the distance between any two pairs of points in X by more than a factor of L.
Levy's concentration theorem in High-dimensional probability by Roman Vershynin
Levy's concentration theorem (Vershynin's version)
This theorem is exactly the 5.1.4 on the High-dimensional probability by Roman Vershynin.
Isoperimetric inequality on \mathbb{R}^n
Among all subsets A\subset \mathbb{R}^n with a given volume, the Euclidean ball has the minimal area.
That is, for any \epsilon>0, Euclidean balls minimize the volume of the $\epsilon$-neighborhood of A.
Where the volume of the $\epsilon$-neighborhood of A is defined as
A_\epsilon(A)\coloneqq \{x\in \mathbb{R}^n: \exists y\in A, \|x-y\|_2\leq \epsilon\}=A+\epsilon B_2^n
Here the \|\cdot\|_2 is the Euclidean norm. (The theorem holds for both geodesic metric on sphere and Euclidean metric on \mathbb{R}^n.)
Isoperimetric inequality on the sphere
Let \sigma_n(A) denotes the normalized area of A on n dimensional sphere S^n. That is \sigma_n(A)\coloneqq\frac{\operatorname{Area}(A)}{\operatorname{Area}(S^n)}.
Let \epsilon>0. Then for any subset A\subset S^n, given the area \sigma_n(A), the spherical caps minimize the volume of the $\epsilon$-neighborhood of A.
The above two inequalities is not proved in the Book High-dimensional probability.
To continue prove the theorem, we use sub-Gaussian concentration (Chapter 3 of High-dimensional probability by Roman Vershynin) of sphere \sqrt{n}S^n.
This will leads to some constant C>0 such that the following lemma holds:
The "Blow-up" lemma
Let A be a subset of sphere \sqrt{n}S^n, and \sigma denotes the normalized area of A. Then if \sigma\geq \frac{1}{2}, then for every t\geq 0,
\sigma(A_t)\geq 1-2\exp(-ct^2)
where A_t=\{x\in S^n: \operatorname{dist}(x,A)\leq t\} and c is some positive constant.
Proof of the Levy's concentration theorem
Proof:
Without loss of generality, we can assume that \eta=1. Let M denotes the median of f(X).
So \operatorname{Pr}[|f(X)\leq M|]\geq \frac{1}{2}, and \operatorname{Pr}[|f(X)\geq M|]\geq \frac{1}{2}.
Consider the sub-level set A\coloneqq \{x\in \sqrt{n}S^n: |f(x)|\leq M\}.
Since \operatorname{Pr}[X\in A]\geq \frac{1}{2}, by the blow-up lemma, we have
\operatorname{Pr}[X\in A_t]\geq 1-2\exp(-ct^2)
And since
\operatorname{Pr}[X\in A_t]\leq \operatorname{Pr}[f(X)\leq M+t]
Combining the above two inequalities, we have
\operatorname{Pr}[f(X)\leq M+t]\geq 1-2\exp(-ct^2)
Levy's concentration theorem in Metric Structures for Riemannian and Non-Riemannian Spaces by M. Gromov
Levy's concentration theorem (Gromov's version)
The Levy's lemma can also be found in Metric Structures for Riemannian and Non-Riemannian Spaces by M. Gromov.
3\frac{1}{2}.19The Levy concentration theory.
Theorem 3\frac{1}{2}.19 Levy concentration theorem:
An arbitrary 1-Lipschitz function f:S^n\to \mathbb{R} concentrates near a single value a_0\in \mathbb{R} as strongly as the distance function does.
That is
\mu\{x\in S^n: |f(x)-a_0|\geq\epsilon\} < \kappa_n(\epsilon)\leq 2\exp(-\frac{(n-1)\epsilon^2}{2})
where
\kappa_n(\epsilon)=\frac{\int_\epsilon^{\frac{\pi}{2}}\cos^{n-1}(t)dt}{\int_0^{\frac{\pi}{2}}\cos^{n-1}(t)dt}
a_0 is the Levy mean of function f, that is the level set of f^{-1}:\mathbb{R}\to S^n divides the sphere into equal halves, characterized by the following equality:
\mu(f^{-1}(-\infty,a_0])\geq \frac{1}{2} \text{ and } \mu(f^{-1}[a_0,\infty))\geq \frac{1}{2}
Hardcore computing may generates the bound but M. Gromov did not make the detailed explanation here.
Detail proof by Takashi Shioya.
The central idea is to draw the connection between the given three topological spaces,
S^{2n+1},CP^nand\mathbb{R}.
First, we need to introduce the following distribution and lemmas/theorems:
OBSERVATION
consider the orthogonal projection from \mathbb{R}^{n+1}, the space where S^n is embedded, to \mathbb{R}^k, we denote the restriction of the projection as \pi_{n,k}:S^n(\sqrt{n})\to \mathbb{R}^k. Note that \pi_{n,k} is a 1-Lipschitz function (projection will never increase the distance between two points).
We denote the normalized Riemannian volume measure on S^n(\sqrt{n}) as \sigma^n(\cdot), and \sigma^n(S^n(\sqrt{n}))=1.
Definition of Gaussian measure on \mathbb{R}^k
We denote the Gaussian measure on \mathbb{R}^k as \gamma^k.
d\gamma^k(x)\coloneqq\frac{1}{\sqrt{2\pi}^k}\exp(-\frac{1}{2}\|x\|^2)dx
x\in \mathbb{R}^k, \|x\|^2=\sum_{i=1}^k x_i^2 is the Euclidean norm, and dx is the Lebesgue measure on \mathbb{R}^k.
Basically, you can consider the Gaussian measure as the normalized Lebesgue measure on \mathbb{R}^k with standard deviation 1.
Maxwell-Boltzmann distribution law
It is such a wonderful fact for me, that the projection of
n+1dimensional sphere with radius\sqrt{n}to\mathbb{R}^kis a Gaussian distribution asn\to \infty.
For any natural number k,
\frac{d(\pi_{n,k})_*\sigma^n(x)}{dx}\to \frac{d\gamma^k(x)}{dx}
where (\pi_{n,k})_*\sigma^n is the push-forward measure of \sigma^n by \pi_{n,k}.
In other words,
(\pi_{n,k})_*\sigma^n\to \gamma^k\text{ weakly as }n\to \infty
Proof
We denote the n dimensional volume measure on \mathbb{R}^k as \operatorname{vol}_k.
Observe that \pi_{n,k}^{-1}(x),x\in \mathbb{R}^k is isometric to S^{n-k}(\sqrt{n-\|x\|^2}), that is, for any x\in \mathbb{R}^k, \pi_{n,k}^{-1}(x) is a sphere with radius \sqrt{n-\|x\|^2} (by the definition of \pi_{n,k}).
So,
\begin{aligned}
\frac{d(\pi_{n,k})_*\sigma^n(x)}{dx}&=\frac{\operatorname{vol}_{n-k}(\pi_{n,k}^{-1}(x))}{\operatorname{vol}_k(S^n(\sqrt{n}))}\\
&=\frac{(n-\|x\|^2)^{\frac{n-k}{2}}}{\int_{\|x\|\leq \sqrt{n}}(n-\|x\|^2)^{\frac{n-k}{2}}dx}\\
\end{aligned}
as n\to \infty.
(n-\|x\|^2)^{\frac{n-k}{2}}=\left(n(1-\frac{\|x\|^2}{n})\right)^{\frac{n-k}{2}}\to n^{\frac{n-k}{2}}\exp(-\frac{\|x\|^2}{2})