Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm looking for a proof in the literature or just a proof of:

Let $\Omega\subset\mathbb{R}^d$ be an open and bounded domain with satisfying the interior cone condition with parameters $r$ and $\theta$. Let $\Omega_\delta$ be the $\delta$-interior of $\Omega$ that is

$ \Omega_\delta = ${$x\in\Omega : dist(x,\partial\Omega)>\delta$}$ $

There is a $\delta_0$ sufficiently small such that $\Omega_\delta$ for $0<\delta<\delta_0$ satisfies the interior cone condition with parameters $r/2$ and $\theta/2$.

Note: I'm also only interested when $\Omega$ lies on one side of its boundary.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

I am assuming that you are using the same definition of interior cone condition which I have heard. That is, from every point $x \in \Omega$ there is some truncated cone from $x$ with an opening angle $\theta$ and radius $r$ inside $\Omega$.

The claim is not true. Consider the following domain:

share|improve this answer

What is true, however, is that if $\Omega$ satisfies the cone condition with $\theta$ and $r$, then each point in $\Omega_\delta$ has a $(\theta/2,r/2)$-cone attached to it and contained in $\Omega_{\delta\theta/20}$. The conclusion is that you still can exhaust your domain with domains with a uniform cone condition whose boundaries are almost equidistant to the boundary of $\Omega$ (just take the unions of those cones). Most likely, that's all you really need.

The proof is next to trivial. If $a\in\Omega_\delta$ and $K$ is the cone for $a$ in $\Omega$, then all points lying in the shrinked cone that are not more than $\delta/2$ away from $a$ are in $\Omega_{\delta/2}$ by the triangle inequality but all farther points are far even from the boundary of $K$ (here is where the aperture comes into the bounds) and, thereby, from $\partial \Omega$ as well.

share|improve this answer

Your Answer

 
discard

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.