# Decomposition of a Lipschitz domain as union of strongly star shaped Lipschitz domains

Definition. We say that $$\Omega$$ is a strongly star shaped domain (with respect to $$x_0$$) in $$\mathbb R ^n$$ if: $$\Omega = \left\{x\in \mathbb R ^n : \| x-x_0 \| < g\Big(\frac{x-x_0}{\| x-x_0 \|}\Big)\right\}$$ and $$\partial \Omega = \left\{x\in \mathbb R ^n : \| x-x_0 \| = g\Big(\frac{x-x_0}{ \| x-x_0 \|}\Big)\right\}$$ with $$g$$ is a continuous, positive function on the unit sphere.

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

Can you help me to find Why we have this result?

I would appreciate if you could answer me.

• The idea is that a Lipschitz domain $\Omega$ for which there exists a ball $B\subset \Omega$ such that every point in $\Omega$ can be joined to any point in $B$ by a line is star-shaped (to see that the defining function is Lipschitz you can appeal to the interior cone characterization of Lipchitz functions). This allows you to show that the domain above a Lipschitz graph satisfies your conclusion and for a general (bounded) Lipschitz domain we cover the boundary by open sets where the boundary is given a the graph of a Lipschitz function. – Jose27 May 13 '17 at 20:57
• Bramble doesn't seem to make your statement. At the bottom of page 2 he states, "Clearly if $\cup^M_{j=1}\Omega_j$ and if (1.7) holds for each $\Omega_j, j=1,\dots,M<\infty$, then it holds for $\Omega$. We will use the fact that any Lipschitz domain can be written as the union of strongly star shaped Lipschitz domains." I can't see anywhere where he claims that that the finite decomposition you've presented is true. – postmortes May 13 '17 at 21:02
• @postmortes: After your quote Bramble states "Hence we may assume, without loss, that $\Omega$ is strongly star-shaped". The only way this reduction is valid is if the decomposition that Mokata mentions is true. – Jose27 May 14 '17 at 2:01
• @Jose27 -- ah, I see what you mean! Thank-you :) – postmortes May 14 '17 at 8:43
• Thank you for your replies, @Jose27 Can you explain your idea, I did not understand it.. – Motaka May 14 '17 at 9:32