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We say that $\Omega$ is a strongly star shaped domain (with respect to $0$ for example) in $\mathbb R ^n$ if:

$$\Omega = \{x\in \mathbb R ^n : \left \| x \right \| < g(\frac{x}{\left \| x\right \|})\} $$ and $$\partial \Omega = \{x\in \mathbb R ^n : \left \| x \right \| = g(\frac{x}{\left \| x \right \|})\} $$ with $g$ is a continuous, positive function on the unit sphere.

In this paper, Bramble uses the fact that : Any Lipschitz domain can be written as the union of strongly star shaped Lipschitz domains: $\Omega=\cup_{i=1}^{M}\Omega_i$

Can you help me to find why do we have this result? Do you have any references in which I can find this proposition?


PS: Sorry to ask this question again, but this did not get answered (as I had desired) on M.SE.

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  • $\begingroup$ You can always use a partition of unity such that locally each $\Omega_i$ is star-shaped. $\endgroup$ – T. Amdeberhan May 15 '17 at 6:47
  • $\begingroup$ @ T. Amdeberhan : I do not know, if I understood what you want to say, but if you saw the paper of Bramble he used this proposition (and via the partion of unity) to show an inequality.. $\endgroup$ – Motaka May 15 '17 at 6:54
  • $\begingroup$ Here is link to the Math.SE question. $\endgroup$ – Martin Sleziak May 15 '17 at 23:44
  • $\begingroup$ The property is just a trivial consequence of the definition of Lipschitz domain. $\endgroup$ – Pietro Majer May 16 '17 at 6:01
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A Lipschitz domain $\Omega$ is an open set and any open set is a union of balls, which are strongly star-shaped. So I assume you meant $\overline\Omega$. By definition, any point of $\partial\Omega$ has a nbd in $\overline\Omega$ which is isometric to a sub-graph of a positive $k$-Lipschitz function $f:B(0,r)\to(0,+\infty)$, $$\{(x,t)\, :\, |x|<r, \;0<t<f(x)\}.$$ We can take $r<{f(0)\over 2k+1}$, which makes the latter set strongly star-shaped w.r.to the point $(0,r)$ as it is easy to check.

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  • $\begingroup$ @ Pietro Majer:I totally agree with what you have written Sir, But I can not see how to construct the function $g$ of the definition above^^. Can you explain a little? $\endgroup$ – Motaka May 16 '17 at 12:08
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    $\begingroup$ If $U\subset\mathbb{R}^n$ is the above set, and $p:=(0,r)$, define, for any $x\in\mathbb{R}^n$ of unit norm, $g(x)$ to be the unique $t>0$ such that $p+tx\in\partial U$. You need to prove that this $t$ is actually unique, and that depends continuously from $x$. $\endgroup$ – Pietro Majer May 16 '17 at 19:37
  • $\begingroup$ @ Pieotro Majer: Thanks for your reply, I didn't get the idea, if $\forall x\in \partial \Omega$, there is a function $f$ and a nbd (Let's say $B(0,r)$) such that: $\partial \Omega\cap B(0,r) =\{X=(x,t):\left \| X \right \|< r, 0<t<f(x)\}$. So what is $p=(0,r)$ for you (r?)? can you explain more I will try to complete the proof, Thanks ! $\endgroup$ – Motaka May 17 '17 at 9:50
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    $\begingroup$ $p:=(0,r)\in B(0,r) \times \mathbb{R}\subset \mathbb{R}^{n+1}$ refers to the above answer, where $B(0,r)\subset \mathbb{R}^n$ is the ball of radius $r>0$ centered at the origin $0\in\mathbb{R}^n$ (so also $\Omega$ is an open set of $\mathbb{R}^{n+1}$, in these notations ) Note that the sub-graph of $f$ is an open set in $B(0,r)\times \mathbb{R}\subset \mathbb{R}^{n+1}$, and $(0,r) \in \mathbb{R}^{n+1}$ is a point in this sub-graph. For more details I think you may post a suitable question in StackExchange. $\endgroup$ – Pietro Majer May 17 '17 at 11:39
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    $\begingroup$ k-Lipschitz ensures that the line {p+tx, t>0} has unique intersection with the boundary of the sub-graph set defined in the answer. The continuity of g follows by standard compactness argument. It's an easy computation. Try asking Stack Exchange. $\endgroup$ – Pietro Majer May 17 '17 at 14:52
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You may like to check such results, in particular, Proposition 2.5.4 of the below monograph. Hope you have access to it.

Proposition 2.5.4. Let $\Omega \in \mathcal A_0$ have Lipschitz boundary. Then there exists a finite open covering $\{\Omega_j\}_{j\in\{1,\dots,m\}}$ of $\overline\Omega$ such that, for every $j\in\{1,\dots,m\}$, $\Omega_j \cap \Omega$ is strongly star shaped with Lipschitz boundary.

Carbone L. and De Arcangelis R., Unbounded functionals in the calculus of variations: Representation, relaxation, and homogenization, Chapman & Hall/CRC Monographs and Surveys in Pure and Applied Mathematics, vol. 125, Chapman & Hall/CRC, Boca Raton, FL, 2002

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