Interior cone condition preserved on a small perturbation of the domain. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:08:24Z http://mathoverflow.net/feeds/question/54184 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54184/interior-cone-condition-preserved-on-a-small-perturbation-of-the-domain Interior cone condition preserved on a small perturbation of the domain. alext87 2011-02-03T11:24:37Z 2011-02-06T15:01:46Z <p>I'm looking for a proof in the literature or just a proof of: </p> <p>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 </p> <p>$\Omega_\delta =${$x\in\Omega : dist(x,\partial\Omega)>\delta$}</p> <p>There is a $\delta_0$ sufficiently small such that $\Omega_\delta$ for $0&lt;\delta&lt;\delta_0$ satisfies the interior cone condition with parameters $r/2$ and $\theta/2$.</p> <p>Note: I'm also only interested when $\Omega$ lies on one side of its boundary.</p> http://mathoverflow.net/questions/54184/interior-cone-condition-preserved-on-a-small-perturbation-of-the-domain/54190#54190 Answer by Tapio Rajala for Interior cone condition preserved on a small perturbation of the domain. Tapio Rajala 2011-02-03T12:47:43Z 2011-02-03T12:47:43Z <p>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 $\Omega$.</p> <p><strong>The claim is not true.</strong> Consider the following domain:</p> <p><a href="http://users.jyu.fi/~tamaraja/temp/domain.jpg" rel="nofollow"> <img src="http://users.jyu.fi/~tamaraja/temp/domain.jpg" width="409"></a></p> http://mathoverflow.net/questions/54184/interior-cone-condition-preserved-on-a-small-perturbation-of-the-domain/54523#54523 Answer by fedja for Interior cone condition preserved on a small perturbation of the domain. fedja 2011-02-06T15:01:46Z 2011-02-06T15:01:46Z <p>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.</p> <p>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.</p>