Let $\Omega\subset \mathbb{R}^d$ be an open and bounded domain with Lipschitz smooth boundary. Let $\delta>0$ and $ \Omega_\delta = \{ x\in\Omega : \inf_{y\in\partial\Omega} \left\|x-y\right\|_{L_2(\Omega)}\geq \delta \} $

Is there a $\hat{\delta}>0$ such that $\forall 0<\delta<\hat{\delta}$ the space $\Omega_{\delta}$ has a Lipschitz smooth boundary?

This statement seems like it should be true. I would really appreciate some help.

Thanks in advance.


1 Answer 1


My answer is Yes. Of course, I presume that you assume $\Omega$ is on one side only of its boundary.

The boundary $\partial\Omega$ is compact. By assumptions, it has an atlas with finitely many charts, each one corresponding to a piece $\Gamma_j$ which is the graph of a Lipschitz function: in appropriate coordinates, $\Gamma_j$ is given by $x_d=\phi_j(x_1,\ldots,x_{d-1})$.

Let us choose $\delta$ smaller than one tenth of $$\delta_0:=\min_{x\in\bar\Omega}\\,\max_j\{d(x,\partial\Gamma_j)\}>0$$

Let $\bar x$ be on the boundary of $\Omega_\delta$, not on $\partial\Omega$. Let $j$ be such that $B(\bar x;9\delta)\cap\partial\Omega\subset\Gamma_j$. There exists a point $\bar y\in\partial\Gamma_j$ such that $d(\bar x,\bar y)=\delta$. Wlog, we have $\bar y=0$ and the graph is locally $x_d=\phi_j(x_1,\ldots,x_{d-1})$. Let $M$ be the Lipschitz constant of $\phi_j$. Locally, the boundary of $\Omega_\delta$ is the upper envellop of the balls $B(y;\delta)$ with $y\in \Gamma_j$; as a matter of fact, for $x\in B(\bar x;\delta)$, the closest points of $\partial\Omega$ to $x$ belong to $\Gamma_j$. It is easy to see that it is the graph of a function $\phi_j^\delta$ whose Lipschitz constant is at most $M$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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