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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.

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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:

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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.

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