7
$\begingroup$

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?

$\endgroup$
6
  • 2
    $\begingroup$ You know Ziemers Book 'Weakly Differentiable Functions'? Chapter 4 is dedicated to Poincaré type inequalities. $\endgroup$
    – user20266
    Jan 7, 2012 at 11:19
  • $\begingroup$ 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. $\endgroup$ Jan 7, 2012 at 11:29
  • 2
    $\begingroup$ 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. $\endgroup$ Jan 7, 2012 at 17:38
  • $\begingroup$ 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. $\endgroup$ Jan 11, 2012 at 10:40
  • 1
    $\begingroup$ @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 $\endgroup$ Apr 30, 2012 at 8:02

1 Answer 1

7
$\begingroup$

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).

$\endgroup$
1
  • 1
    $\begingroup$ Thank you very much for your answer. $\endgroup$ Mar 23, 2014 at 15:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .