# Is a “SSS domains” 's function Lipschitzian?

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??

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: $$\mathcal D\cap U=\{x=(\underbrace{x_1,...,x_{d-1}}_{x^{'}},x_d)\in U: 0<x_d<\phi(x^{'})\}.$$

• Did $g$ change its name to $h$? (Also, but less importantly, I wonder why the definition specifies $\partial\Omega$ the way it does. I would have thought this follows from the form of $\Omega$, given the stated requirements on $g$ of course.) – Harald Hanche-Olsen Jun 8 '17 at 7:53
• Yeah it's "$g$". The definition is like that , you can find it in the book of Galdi .. I think it's "obvious" that $g$ is Lipschitz, because it modilises $\Omega$ which is Lipschitz, but I don't know how to prove it.. – Motaka Jun 8 '17 at 8:09

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 $\Omega$.
• Yeah, your $g$ is not Lipschitz, but why $\Omega$ Ω is still Lipschitz? Can you explain me a little.? – Motaka Jun 9 '17 at 14:15
• I have added a picture. I do not see any reason why $\Omega$ is not Lipschitz. – gerw Jun 10 '17 at 20:25
• Can you give me the program's name that you used to draw the graph of $\Omega$? Thank's a lot – Motaka Jun 16 '17 at 10:53