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Let $O\subset\mathbb{R}^n$ be a open set which is star shaped with respect to the origin. How does one prove that there exists an increasing sequence of star shaped (w.r.t the origin) domains $O_i$ such that :

  • $O_i$ has a smooth boundary
  • $O_i\subset O$
  • $\lambda (O\backslash O_i)\to 0$ when $i\to\infty$, or, even better for my purpose, $O\backslash O_i$ is contained in an $\varepsilon_i$-neighborhood of $\partial O$.

For my purpose, it wouldn't hurt to assume that $O$ is bounded if that helps. This kind of approximation is handy for estimating the distributional laplacian of the distance function on a Riemannian manifold.

That seems like a pretty standard fact but I haven't been able to locate or derive a clean proof of this. I apologize if the question is not up to the standards of MO.

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    $\begingroup$ How about taking domains given by the set of points $p$ such that $dist(p,\partial O)>1/i$? (In case $O$ is bounded,but otherwise just look at the problem on $\mathbb S^n$). I don't know if those domains are star-shaped, but they look like it on the examples I've (quickly) considered. $\endgroup$ – Loïc Teyssier Mar 2 '16 at 12:10
  • $\begingroup$ Oops, I forgot one crucial requirement: I want the $O_i$ to have smooth boundary. $\endgroup$ – Thomas Richard Mar 2 '16 at 12:27
  • $\begingroup$ You could choose a radial diffeomorphism to euclidean space and take preimages of balls . $\endgroup$ – Mohan Ramachandran Mar 2 '16 at 21:06
  • $\begingroup$ I am not sure this works without any additional knowledge on the diffeomorphism. $\endgroup$ – Thomas Richard Mar 4 '16 at 11:33
  • $\begingroup$ @MohanRamachandran Please be so kind and post the links. $\endgroup$ – Sebastian Goette Mar 7 '16 at 9:58
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This is not really an answer, but an equivalent and hopefully easier reformulation of the question that does not fit in one comment.

You can describe the starshaped domain $O$ by a function $\rho\colon S^{n-1}\to(0,\infty]$ such that $$O\setminus\{0\}=\{\,x\in\mathbb R^n\setminus\{0\}\mid |x|<\rho(x/|x|)\,\}\;.$$ Then $O$ is open if and only if $\rho$ is lower smicontinuous. You want to approximate $\rho$ by smooth functions $\rho_i$ from below. This turns out to be impossible (both with respect to Lebesgue measure and with respect to distance) if $\rho$ jumps from $\infty$ down to a finite number.

If $\rho$ is bounded by $C$, the "$\varepsilon_i$" condition is equivalent to the following: all $\rho_i<\rho_{i+1}<\rho$ for all $i$, and for each $\varepsilon>0$, there exists an $i$ such that for each $p\in S^{n-1}$, there is $q\in S^{n-1}$ with $|q-p|<\varepsilon$ such that $\rho_i(p)>\rho(q)-\varepsilon$. The idea is that there exists $r\ge \rho(q)$ such that $|rq-\rho_i(p)\,p|<(1+C)\varepsilon$, and there is a boundary point of $O$ somewhere on the line segment from $\rho_i(p)\,p$ to $rq$. I am sure one can show that smooth approximations exist that satisfy this.

If $\rho$ is unbounded, one has to demand instead that for each $\varepsilon>0$, there exists an $i$ such that for each $p\in S^{n-1}$, there is $q\in S^{n-1}$ with $|q-p|<\varepsilon/\rho_i(p)$ such that $\rho_i(p)>\rho(q)-\varepsilon$. Just adapt the argument above.

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  • $\begingroup$ Thanks. I had come to this reduction at some point, but failed to get a clean proof from there. $\endgroup$ – Thomas Richard Mar 2 '16 at 13:26
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    $\begingroup$ At least I see how to get a sequence of Lipschitz functions from here. If someone knows exactly how good one can smooth those, then you are done. $\endgroup$ – Sebastian Goette Mar 2 '16 at 16:35

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