# A generalisation of a Hardy inequality

I found the following result in Brezis' book.

Let $\Omega$ be a bounded open set of class $C^1$. Let $d(x):=\operatorname{dist}(x,\partial\Omega)$. Then there exists $C>0$ such that $$\|\frac{u}{d} \|_{L^2(\Omega)}\le C\|\nabla u\|_{L^2(\Omega)}\qquad\forall\ u\in H^1_0(\Omega).$$ Viceversa, if $u\in H^1(\Omega)$ is such that $\frac{u}{d}\in L^2(\Omega)$, then $u\in H^1_0(\Omega)$.

I'm wondering if the first implication of the statement still holds with the following hypotheses.

Let $\Omega$ be a Lipschitz domain and let $\Gamma\subsetneq\partial\Omega$ be "good enough". $d(x):=\operatorname{dist}(x,\Gamma)$. Then there exists $C>0$ such that $$\|\frac{u}{d} \|_{L^2(\Omega)}\le C\|\nabla u\|_{L^2(\Omega)}\qquad\forall\ u\in H^1_{0,\Gamma}(\Omega).$$

I guess I found the answer in this paper: https://link.springer.com/article/10.1007/s11118-015-9463-8

• What is your question? The statement as it is written is incorrect. Consider e.g. the case where $\Omega$ is a ball and $\Gamma$ is a single point on the boundary. – Hans Engler Aug 31 '18 at 12:15
• Right. I mean $\Gamma$ should have at least positive $(d-1)$-Lebesgue measure and should be a "good set" in some sense (that I do not know). – avati91 Aug 31 '18 at 12:18