Let $\Omega$ to be an open of $ \mathbb R^d$
- We say that $\Omega$ is a strongly star-shaped domain (with respect to the origin $0$) if :
$\Omega = \{x\in \mathbb R ^n : \left \| x \right \| < g(\frac{x}{\left \| x \right \|})\}\cup\{0\} $ 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.
I'm looking for the proof of :
If $\Omega$ is a Lipschitz domain, is a $g$ Lipschtiz function??
I appreciate your answers and your help.
PS: A domain of $\mathbb{R}^d$ with Lipschitz boundary is an open subset $\mathcal D \subset \mathbb{R}^d$, wich is locally the subgraph of a Lipschitz function with respect to some choise of orthogonal coordinates. In other words, for any $x_0\in \partial \mathcal D$, up to an orthogonal change of coordinates, there is an open set $U\subset \mathbb R^d$, and a Lipschitz function $\phi:\mathbb R^{d-1}\rightarrow \mathbb R^{+}$ such that: $$\mathcal D\cap U=\{x=(\underbrace{x_1,...,x_{d-1}}_{x^{'}},x_d)\in U: 0<x_d<\phi(x^{'})\}.$$
No, I don't think so. Take a function $g$, such that $$g( \cos(\phi), \sin(\phi)) = 1 + \operatorname{sign}(\phi) \, \sqrt{|\phi|}$$ for small values of $\phi$. Then, $g$ is not Lipschitz. However, the singularity of the derivative of $g$ turns into an horizontal tangent for $\partial\Omega$ at $(1,0)$, and $\Omega$ is still Lipschitz (it could be even $C^1$?).
Edit: Here is a picture of $\Omega$.