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It is a well known result that for any compact, convex set $\Omega \subset \mathbb{R}^n$ and for any $\varepsilon > 0$ there are convex sets $\Omega_1 \subset \Omega \subset \Omega_2$ with C^2 boundaries $\partial \Omega_1$ and $\partial \Omega_2$ such that $d(\Omega, \partial \Omega_1) < \varepsilon$ and $d(\Omega, \partial \Omega_2) < \varepsilon$ where $d$ denotes the Hausdorff distance between sets (see for instance section 4.3 from Eggleston's Convexity).

It is also generally implied that convexity is a critical requirement and the above won't hold for non-convex domains. I'm trying to understand what fails in the non-convex case since (my) intuition seems to imply this should hold also for non-convex domains: a draw is far from a proof, but it looks like I could draw a regular (say $C^2$) set as close as desired to the non-convex set.

EDIT:

The set motivating me for asking this question is that made of a circle without a sector: circle without a sector

Aitor B
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1 Answers1

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Let $\Omega = [-1,1]^2\setminus ((0,1]\times {0})$.

$\Omega$ is a closed square from whom we removed a line going to its center. You cannot construct a $C^2$ curve outside $\Omega$ going closer than $1$ to $0$ since the only way to achieve this is to use the line as you path. However, such curve cannot be differentiable everywhere because it must have a cusp.

Edit : please note that $\Omega$ is not compact in this example. So it does not fully answer the question. I am thinking about it.

nicomezi
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  • This is a nice example of a case I wasn't considering. Indeed, this shows a limitation on approximating some non-convex sets. However, the example I was having in mind is a little bit different. I'll edit my question. – Aitor B Oct 22 '19 at 10:28
  • See my edit please – nicomezi Oct 22 '19 at 10:29
  • In the annulus case one could always approximate it by smaller / bigger annulus, which have smooth boundary right? – Aitor B Oct 22 '19 at 10:41
  • Why do I have to approximate all the boundary with just one curve? Isn't it possible to have two unconnected components of the boundary, as in the annulus itself? Aditionally, do you have any hint on what might fail on the example I added to the question? – Aitor B Oct 22 '19 at 10:57
  • You are right, the annulus is not a good example. Nothing fails in your example if you get rid of the convex hypothesis on $\Omega$ and $\Omega_i$. – nicomezi Oct 22 '19 at 11:10