Gaussian isoperimetric inequality

In mathematics, the Gaussian isoperimetric inequality, proved by Boris Tsirelson and Vladimir Sudakov,^[1] and later independently by Christer Borell,^[2] states that among all sets of given Gaussian measure in the n-dimensional Euclidean space, half-spaces have the minimal Gaussian boundary measure.

Let ${\displaystyle \scriptstyle A}$ be a measurable subset of ${\displaystyle \scriptstyle \mathbf {R} ^{n}}$ endowed with the standard Gaussian measure ${\displaystyle \gamma ^{n}}$ with the density ${\displaystyle {\exp(-\|x\|^{2}/2)}/(2\pi )^{n/2}}$. Denote by

${\displaystyle A_{\varepsilon }=\left\{x\in \mathbf {R} ^{n}\,|\,{\text{dist}}(x,A)\leq \varepsilon \right\}}$

the ε-extension of A. Then the Gaussian isoperimetric inequality states that

${\displaystyle \liminf _{\varepsilon \to +0}\varepsilon ^{-1}\left\{\gamma ^{n}(A_{\varepsilon })-\gamma ^{n}(A)\right\}\geq \varphi (\Phi ^{-1}(\gamma ^{n}(A))),}$

where

${\displaystyle \varphi (t)={\frac {\exp(-t^{2}/2)}{\sqrt {2\pi }}}\quad {\rm {and}}\quad \Phi (t)=\int _{-\infty }^{t}\varphi (s)\,ds.}$

The original proofs by Sudakov, Tsirelson and Borell were based on Paul Lévy's spherical isoperimetric inequality.

Sergey Bobkov proved Bobkov's inequality, a functional generalization of the Gaussian isoperimetric inequality, proved from a certain "two point analytic inequality".^[3] Bakry and Ledoux gave another proof of Bobkov's functional inequality based on the semigroup techniques which works in a much more abstract setting.^[4] Later Barthe and Maurey gave yet another proof using the Brownian motion.^[5]

The Gaussian isoperimetric inequality also follows from Ehrhard's inequality.^[6][7]

• Concentration of measure
• Borell–TIS inequality