## Poincare inequality from different approach

$\mathbf{Problem:}$ Let $\Omega$ be convex and $u\in C^1(\Omega)$. Suppose $S\subset\Omega$ with $\displaystyle \frac{|S|}{|\Omega|}>0$,

$\displaystyle |u(x)-u_S|\leq \frac{d^n}{n|S|}\int_{\Omega}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy$

where $d$ is the diameter of $\Omega$ and $\displaystyle u_S=\frac{1}{|S|}\int_S u(y)dy$.

$\mathbf{Proof\,1: }$

$\displaystyle u(y)-u(x)=\int_0^1\frac{d}{dt}\left(u(ty+(1-t)x)\right)dt=\int_0^1\nabla u(ty+(1-t)x)\cdot(x-y)dt$

$\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S|u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^1|\nabla u(ty+(1-t)x)||x-y|dtdy$

Set $z=ty+(1-t)x$, then $t|x-y|=|z-x|$. Since $|x-y|\leq d$, $\displaystyle t\geq \frac{|z-x|}{d}$. We also have $t^ndy=dz$

Then

$\displaystyle |u(x)-u_S|\leq \frac{d}{|S|}\int_\Omega\int_\frac{|z-x|}{d}^\infty \frac{|\nabla u(z)|}{t^n}\,dtdz=\frac{d^n}{(n-1)|S|}\int_\Omega\frac{|\nabla u(z)|}{|z-x|^{n-1}}dz$.

$\mathbf{Proof\,2: }$ Using

$\displaystyle u(x)-u(y)=\int_{0}^{|x-y|}\frac{d}{dt}u(x+tw)dt=\int_0^{|x-y|}\nabla u(x+tw)\cdot wdt$

where $\displaystyle w=\frac{x-y}{|x-y|}$. Then

$\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S |u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^{|x-y|}|\nabla u(x+tw)|dtdy$

$\displaystyle \leq \frac{1}{|S|}\int_\Omega\int_0^{\infty}|\nabla u(x+tw)|\chi_{\{t<|x-y|\}}dtdy$

$\displaystyle \leq \frac{1}{|S|} \int_0^\infty\int_{|w|=1}\int_0^d|\nabla u(x+tw)|\chi_{\{t<|x-rw|\}}r^{n-1}dwdrdt$

$\displaystyle \leq \frac{d^n}{n|S|}\int_0^\infty\int_{|w|=1}|\nabla u(x+tw)|\chi_{\{t<|x-\xi(w)|\}}dwdt$

$\displaystyle =\frac{d^n}{n|S|}\int_\Omega \frac{|\nabla u(z)|}{|z-x|^{n-1}}dz$

where $\xi(w)$ is the intersection of ray $x+tw$ with $\partial \Omega$ (it is unique because $\Omega$ is convex).

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$