# Open sets and Poincaré's inequality

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 $$\lVert u\rVert_{L^2}\leq C\lVert \nabla u\rVert_{(L^2(\Omega))^d}.$$

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?

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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: mathproblems123.wordpress.com/2011/10/05/… 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. en.wikipedia.org/wiki/Rellich%E2%80%93Kondrachov_theorem – Beni Bogosel Apr 30 '12 at 8:02

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).

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Thank you very much for your answer. – Davide Giraudo Mar 23 '14 at 15:23