Ngô Quốc Anh

February 22, 2010

The Poincaré inequality: W^{1,p} vs. W_0^{1,p}

Filed under: Giải Tích 6 (MA5205), Giải tích 8 (MA5206), Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 1:50

In mathematics, the Poincaré inequality is a result in the theory of Sobolev spaces, named after the French mathematician Henri Poincaré. The inequality allows one to obtain bounds on a function using bounds on its derivatives and the geometry of its domain of definition. Such bounds are of great importance in the modern, direct methods of the calculus of variations. A very closely related result is the Friedrichs’ inequality.

This topic will cover two versions of the Poincaré inequality, one is for $W^{1,p}(\Omega)$ spaces and the other is for $W_o^{1,p}(\Omega)$ spaces.

The classical Poincaré inequality for $W^{1,p}(\Omega)$ spaces. Assume that $1\leq p \leq \infty$ and that $\Omega$ is a bounded open subset of the $n$dimensional Euclidean space $\mathbb R^n$ with a Lipschitz boundary (i.e., $\Omega$ is an open, bounded Lipschitz domain). Then there exists a constant $C$, depending only on $\Omega$ and $p$, such that for every function $u$ in the Sobolev space $W^{1,p}(\Omega)$,

$\displaystyle \| u - u_{\Omega} \|_{L^{p} (\Omega)} \leq C \| \nabla u \|_{L^{p} (\Omega)}$,

where

$\displaystyle u_{\Omega} = \frac{1}{|\Omega|} \int_{\Omega} u(y) \, \mathrm{d} y$

is the average value of $u$ over $\Omega$, with $|\Omega|$ standing for the Lebesgue measure of the domain $\Omega$.

Proof. We argue by contradiction. Were the stated estimate false, there would exist for each integer $k = 1,...$ a function $u_k \in W^{1,p}(\Omega)$ satisfying

$\displaystyle \| u_k - (u_k)_{\Omega} \|_{L^{p} (\Omega)} \geq k \| \nabla u_k \|_{L^{p} (\Omega)}$.

We renormalize by defining

$\displaystyle {v_k} = \frac{{{u_k} - {{({u_k})}_\Omega }}}{{{{\left\| {{u_k} - {{({u_k})}_\Omega }} \right\|}_{{L^p}(\Omega )}}}}, \quad k \geqslant 1$.

Then

$\displaystyle {({v_k})_\Omega } = 0, \quad {\left\| {{v_k}} \right\|_{{L^p}(\Omega )}} = 1$

and therefore

$\displaystyle\| \nabla v_k \|_{L^{p} (\Omega)} \leqslant \frac{1}{k}$.

In particular the functions $\{v_k\}_{k\geq 1}$ are bounded in $W^{1,p}(\Omega)$.

By mean of the Rellich-Kondrachov Theorem, there exists a subsequence ${\{ {v_{{k_j}}}\} _{j \geqslant 1}} \subset {\{ {v_k}\} _{k \geqslant 1}}$ and a function $v \in L^p(\Omega)$ such that

$\displaystyle v_{k_j} \to v$ in $L^p(\Omega)$.

Passing to a limit, one easily gets

$\displaystyle v_\Omega = 0, \quad {\left\| {{v}} \right\|_{{L^p}(\Omega )}} = 1$.

On the other hand, for each $i=\overline{1,n}$ and $\varphi \in C_0^\infty(\Omega)$,

$\displaystyle\int_\Omega {v{\varphi _{{x_i}}}dx} = \mathop {\lim }\limits_{{k_j} \to \infty } \int_\Omega {{v_{{k_j}}}{\varphi _{{x_i}}}dx} = - \mathop {\lim }\limits_{{k_j} \to \infty } \int_\Omega {{v_{{k_j},{x_i}}}\varphi dx} = 0$.

Consequently, $v\in W^{1,p}(\Omega)$ with $\nabla v=0$ a.e. Thus $v$ is constant since $\Omega$ is connected. Since $v_\Omega=0$ then $v \equiv 0$. This contradicts to $\|v\|_{L^p(\Omega)}=1$.

The Poincaré inequality for $W_0^{1,2}(\Omega)$ spaces. Assume that $\Omega$ is a bounded open subset of the $n$-dimensional Euclidean space $\mathbb R^n$ with a Lipschitz boundary (i.e., $\Omega$ is an open, bounded Lipschitz domain). Then there exists a constant $C$, depending only on $\Omega$ such that for every function $u$ in the Sobolev space $W_0^{1,2}(\Omega)$,

$\displaystyle \| u \|_{L^2(\Omega)} \leq C \| \nabla u \|_{L^2(\Omega)}$.

Proof. Assume $\Omega$ can be enclosed in a cube

$\displaystyle Q=\{ x \in \mathbb R^n: |x_i| \leqslant a, 1\leqslant i \leqslant n\}$.

Then for any $x \in Q$, we have

$\displaystyle\begin{gathered} {u^2}(x) = {\left( {\int_{ - a}^{{x_1}} {{u_{{x_1}}}(t,{x_2},...,{x_n})dt} } \right)^2} \hfill \\ \qquad\leqslant ({x_1} + a)\int_{ - a}^{{x_1}} {{{({u_{{x_1}}})}^2}dt} \hfill \\ \qquad\leqslant 2a\int_{ - a}^a {{{({u_{{x_1}}})}^2}dt} . \hfill \\ \end{gathered}$.

Thus

$\displaystyle\int_{ - a}^a {{u^2}(x)dx} \leqslant 4{a^2}\int_{ - a}^a {{{({u_{{x_1}}})}^2}dt}$.

Integration over $x_2,...,x_n$ from $-a$ to $a$ gives the result.

The Poincaré inequality for $W_0^{1,p}(\Omega)$ spaces. Assume that $1\leq p and that $\Omega$ is a bounded open subset of the $n$-dimensional Euclidean space $\mathbb R^n$ with a Lipschitz boundary (i.e., $\Omega$ is an open, bounded Lipschitz domain). Then there exists a constant $C$, depending only on $\Omega$ and $p$, such that for every function $u$ in the Sobolev space $W_0^{1,p}(\Omega)$,

$\displaystyle \| u \|_{L^{p^\star} (\Omega)} \leq C \| \nabla u \|_{L^{p} (\Omega)}$,

where $p^\star$ is defined to be $\frac{np}{n-p}$.

Proof. The proof of this version is exactly the same to the proof of $W^{1,p}(\Omega)$ case.

Remark. The point $u =0$ on the boundary of $\Omega$ is important. Otherwise, the constant function will not satisfy the Poincaré inequality. In order to avoid this restriction, a weight has been added like the classical Poincaré inequality for $W^{1,p}(\Omega)$ case. Sometimes, the Poincaré inequality for $W_0^{1,p}(\Omega)$ spaces is called the Sobolev inequality.