Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In many references, Poincaré inequality is presented in the following way :

Let $\Omega\subset \mathbb R^d$ an open bounded set. We can find a constant $C$ which depend of $\Omega$ such that for all $u\in H^1_0(\Omega)$, we have \begin{equation} \lVert u\rVert_{L^2}\leq C\lVert \nabla u\rVert_{(L^2(\Omega))^d}. \end{equation}

In fact it works if $\Omega$ is bounded in one direction. An other sufficient condition is that we can find $v\neq 0$ such that Lebesgue measure of $\{\lambda\in\mathbb R,\lambda v\in \Omega\}$ is finite).

My question, maybe a little vague, is the following: is there a "nice" necessary and sufficient condition on $\Omega$ to have Poincaré's inequality?

share|cite|improve this question
You know Ziemers Book 'Weakly Differentiable Functions'? Chapter 4 is dedicated to Poincaré type inequalities. – user20266 Jan 7 '12 at 11:19
Yes, but when I looked at it I didn't think about this question. And some pages are missing in Gooble book (which is normal). Anyway, this book is at the library of my university, so I will have a look at it Monday. – Davide Giraudo Jan 7 '12 at 11:29
One generalization I know from one of my teachers can be found here:… This needs $\Omega$ to have Lipschitz boundary, and increases the space of admissible functions $H_0^1(\Omega)$ to a closed subspace of $H^1(\Omega)$ which does not contain the non-zero constant functions. – Beni Bogosel Jan 7 '12 at 17:38
Beni Bogosel: Thanks, I didn't know this result. @Thomas I look at this book, but I didn't find the answer. Maybe should I ask it at MathOverfow. – Davide Giraudo Jan 11 '12 at 10:40
@DavideGiraudo: I suppose such condition can be that $\Omega$ is regular enough such that the Rellich Kondrachov theorem holds. – Beni Bogosel Apr 30 '12 at 8:02
up vote 4 down vote accepted

Both historically and statistically, the one and only correct name for the inequality in question $$ \int\limits_{\Omega}\!|u(x)|^2dx\leqslant C\!\int\limits_{\Omega}\!|\nabla u(x)|^2dx \quad \forall\,u\in H_0^1(\Omega)\tag{$\ast$} $$ is to be the Friedrichs inequality. Whenever the Sobolev space $H_0^1(\Omega)$ is defined as a closure of the subspace $C_0^{\infty}(\Omega)$ in $H^1(\Omega)$, the Friedrichs inequality $(\ast)$ stays valid for any open set $\Omega\subset\mathbb{R}^d$ of finite thickness, e.g., bounded in at least one direction. Otherwise, a nonsmooth boundary $\partial\Omega$ requires some correct definition of a zero trace on $\partial\Omega$, in which case the validity of inequality $(\ast)$ depends wholly on the nonsmooth domain geometry, while for certain simple generalizations of the zero trace concept, the necessary and sufficient conditions for $(\ast\ast)$ to be valid have already been found. But this is not the case for the true Poincaré inequality that can be written in the form $$ \int\limits_{\Omega}\!|u(x)|^2dx\leqslant C\Bigl(\Bigl|\int\limits_{\Omega}\!u(x)dx\Bigr|^2+ \int\limits_{\Omega}\!|\nabla u(x)|^2dx\Bigr) \quad \forall\,u\in H^1(\Omega)\tag{$\ast\ast$}, $$ or in some other equivalent form. Inequality $(\ast\ast)$ is valid for a bounded domain satisfying, e.g., the cone condition, though the cone condition is not necessary for $(\ast\ast)$ to be valid. Alternatively, there is a bounded domain $\Omega$ with just a single singular point $a\in\partial\Omega$ such that $\partial\Omega\backslash\{a\}\in C^1$ while the inequality $(\ast\ast)$ is not valid. But still no condition on the geometry of the nonsmooth bounded domain $\Omega$ necessary and sufficient for the validity of $(\ast\ast)$ has yet been found. And so far, domains for which the inequality $(\ast\ast)$ is valid remain tagged as the Nikodim domains (see p. 330 in R.E. Edwards "Functional Analysis. Theory and Applications". Dover Publ., N.Y., 1995).

share|cite|improve this answer
Thank you very much for your answer. – Davide Giraudo Mar 23 '14 at 15:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.