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](http://www.math.tamu.edu/~bramble/psfiles/infsup1018.pdf),
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.*



Decompostition of a Lipschitz domain
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. 


