# 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.

1. Hi, is there any book where I can get the proof (or alternative proof) of $W^{1,p}$?

Comment by Daniel — May 2, 2011 @ 11:17

• You can read it from Evans’s book, page 263.

Comment by Ngô Quốc Anh — May 2, 2011 @ 12:21

2. Hi, why can you apply in the first proof the Rellich-Kondrachov theorem? As stated in Wikipedia, this requires that p is smaller than n.

Comment by Marc — July 4, 2011 @ 3:14

• Thanks Marc for the question. In view of the Morrey theorem, if $p>n$, you can embed $W^{1,p}(\Omega)$ into some Holder space. Thanks to Arzela-Ascoli theorem, you may prove the desired result. For the critical case, $p=n$, note that $p^\star \to +\infty$ as $p \to n$ from the left.

Comment by Ngô Quốc Anh — July 4, 2011 @ 13:59

3. Hi again, I found the general version of the Rellich-Kondrachov theorem, from which indeed it follows that the embedding W^1,p —> L^p is compact, for all p=1. I still wanted to ask if you have a proof of the fact, that for n>=2 (I know the proof of the case n=1), if the domain is connected, then if v in W^1,p satisfies grad v = 0 distributionally (!!!), it follows that v is constant (a.e.). Thanks!

Comment by Marc — July 4, 2011 @ 4:14

• You may read the proof of Corollary 2.1.9 from a book entitled “Weakly Differentiable Functions” by Ziemer.

Comment by Ngô Quốc Anh — July 5, 2011 @ 1:21

4. Hi , Why is it that $(v_k)_\Omega =0$ in the proof of classical inequality. Can you give me a explanation. 🙂 Nice work.

Comment by Ananda — June 15, 2012 @ 22:42

• Hi, my guess is the following: from the choice of $v_k$, by taking the average, it is automatically zero (just look at the formula for $v_k$ in terms of $u_k$).

Comment by Ngô Quốc Anh — June 15, 2012 @ 23:28

5. Hi! How does the constant $C (\Omega)$ in the inequality for $W^{1,p}(\Omega)$ depend on explicitly on the domain? And is there any result about the relationship between $C(\Omega)$ and $C(\tilde\Omega)$ for e.g. $\Omega \subset \tilde\Omega$? Thanks for your answer!

Comment by Paul — May 9, 2013 @ 17:46

• Hi, thanks for your interest in my blog. Concerning to your question, the constant $C(\Omega)$ is known as the best constant in the literature.

It is really hard to find any precise formula the $C(\Omega)$ appearing in the Poincaré inequality. Loosely speaking, an estimate like lower or upper bounds for $C(\Omega)$ could be a very good and deep result, surely worth publication.

As you may know, the best constant for the Sobolev inequality took a very long time and by many great mathematicians to figure it out such as Thierry Aubin and Giorgio Talenti. For the Poincare inequality, we can find the best constant for some special domains such as convex domains.

Comment by Ngô Quốc Anh — May 10, 2013 @ 6:44

6. Hi, about the inequality for $W^{1,p}$, what happens when the domain $\Omega$ is unbounded? Thanks for your answer!

Comment by Dang Son — November 16, 2013 @ 18:57

• Thank Dang Son for raising the question.

At the moment, I am not so sure about the answer but I guess it cannot be true. Well, for the case of $W_0^{1,p}(\mathbb R)$, it is pretty easy to construct counter-examples by enlarging the support of the standard bubble functions. This could work in $W^{1,p}(\mathbb R)$ but I have not calculated yet.

Comment by Ngô Quốc Anh — November 17, 2013 @ 23:26

7. Hi! Thank you for useful information. I have a question. Why this limit is equal to zero?

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

Comment by Liza — November 27, 2013 @ 3:19

• Dear Liza,

Since $\nabla v_{k_j}=(v_{k_j,x_1} \cdots v_{k_j,x_n})$, you can estimate using the Holder inequality as follows

$\begin{array}{lcl} \displaystyle \Big|\int_\Omega {{v_{{k_j},{x_i}}}} \varphi dx \Big| &\leqslant&\displaystyle {\left( {\int_\Omega {|v_{{k_j},{x_1}}^2 + \cdots + v_{{k_j},{x_n}}^2{|^{p/2}}} dx} \right)^{1/p}}{\left( {\int_\Omega {{\varphi ^{p/(p - 1)}}} dx} \right)^{(p - 1)/p}} \hfill \\ &=&\displaystyle {\left( {\int_\Omega {|\nabla {v_{{k_j}}}{|^p}} dx} \right)^{1/p}}{\left( {\int_\Omega {{\varphi ^{p/(p - 1)}}} dx} \right)^{(p - 1)/p}} \hfill \\ &=&\displaystyle \underbrace {{{\left\| {\nabla {v_{{k_j}}}} \right\|}_{{L^p}(\Omega )}}}_{ \to 0}{\left( {\int_\Omega {{\varphi ^{p/(p - 1)}}} dx} \right)^{(p - 1)/p}}.\end{array}$

Comment by Ngô Quốc Anh — November 27, 2013 @ 3:55

8. Hi! I want to say thank you, i get more useful information here. And today I have a question about the Poincare inequality.
Suppose that $u \in H^1_0 \cap H^2$, so the inequality below is true:

$\displaystyle\|\nabla u\|^2_{L^2} \leq C\|\Delta u\|^2_{L^2}.$

Comment by Dang Son — February 19, 2014 @ 22:15

• Hi, thanks for your interest in my blog. Regarding to your question, it should be true. The point is that on the Sobolev space $H^2$, the usual norm is given as the following

$\displaystyle \| u\|_{H^2}^2 = \|\Delta u\|_{L^2}^2 + \|\nabla u\|_{L^2}^2 + \| u\|_{L^2}^2.$

Having the fact that the trace of $u$ vanishes on the boundary, the leading term $\|\Delta u\|_{L^2}$ should dominate the lower order term $\|\nabla u\|_{L^2}$, up to some multiplicative constant.

I hope this is clear for you.

Comment by Ngô Quốc Anh — February 19, 2014 @ 22:22

9. Around Poincare inequality, it is interesting. And i have seen an inequality in some paper. It’s like this

$\|\Delta u\|_{L^2}^2 \leq \lambda_n \|\nabla u\|_{L^2}^2$

where $\lambda_n$ is the eigenvalue of Laplace operator.

Comment by Dang Son — April 11, 2014 @ 9:34

• Hi Dang Son,

Again, thanks for raising the question.

Technically, there is no way (or impossible) to control terms of higher order derivatives in terms of terms of lower ones, such as those you said. However, and I am not sure if you are aware of, one can have some kind of inverse Poincare inequality. When I took a course named Finite Element Method when I was in graduate school, I realized that the following holds

$\displaystyle\int_\Omega |\nabla u|^2 dx \leqslant \frac{C}{h} \int_\Omega u^2 dx$

for any quasi-uniform mesh.

Comment by Ngô Quốc Anh — April 11, 2014 @ 12:26

10. Hi, there is a book where i can get the proof of poincaré inequality on the unit cube? [0,1]^n

Comment by dan — November 15, 2015 @ 22:54

• Hi, I do not know a particular book at the moment, I just googled and found this

Boundary Value Problems and Fourier Expansions by Charles R. MacCluer

Let me know if you are not satisfied with this.

Comment by Ngô Quốc Anh — November 15, 2015 @ 23:07

11. […] am reading POincare inequality from Evans ( can be found here : https://anhngq.wordpress.com/2010/02/22/the-poincare-inequality/ ) and i am having some trouble understanding it . I would be glad if someone could explain the […]

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