We say that $\Omega$ is a star-shaped domain (with respect to the origin) of $\mathbb R ^n$ if : 


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


I showed that there is a $\mathcal C^1$ diffeomorphism between $\Omega$ and the unit ball (Euclidean norm $\left \| . \right \|\_{2}$).
$$\begin{array}{ccccc}
\Phi & : & B & \to & \Omega \\
 & & y & \mapsto & y\;h(\frac{y}{\left \| y \right \|}) \\
\end{array}$$
$\Phi$ have some properties: 


• $\Phi$ is well defined.


• $\Phi(\partial B)=\partial \Omega$.


• $\Phi$ is a bijection.


• $\Phi$ is a smooth function.


**Now I would like to show the existence of a Lipschitzian bijection between this domain $\Omega$ and a cube in $\mathbb R ^n$ (norm $\left \| . \right \|\_{\infty}$).**


I appreciate your answers and your help.



A Bi-Lipschitzian application
There might be none. If the boundary of $\Omega$ presents a cusp, then it cannot be flattened even into a corner by a Lipschitz map (in particular, you $\Phi$ must have unbounded first derivative).


**Edit:** here are some details. Observe that a Lipschitz map from a bounded set can be extended with the same Lipschitz constant to the closure of the domain (it maps Cauchy sequences to Cauchy sequences). This extension would send the boundary of the starting domain bijectively into the boundary of the cube. All you have to do to construct a counter-example is ensure this cannot happen; e.g. in dimension 2 take a starting domain with a cusp, i.e. a point where the left-hand part of the boundary meets the right-hand part with a vanishing angle. This cannot be sent by a Lipschitz map to the boundary of a square without folding, and folding would prevent bijectivity.


