# A Bi-Lipschitzian application

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 $\left \| . \right \|_{\infty}$).

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).
• @ Benoît Kloeckner: Unfortunately I did not receive your idea well, if I understood what you mean, it is that we can not find a bijection bi-Lipschitz between the cube and the domain $\Omega$, is that right? But from what I have shown before, the ball unit is in "smoooth bijection" with the the $\mathcal C^1$ $\Omega$, and on the other hand there is a diffeomorphism betwin a unit cube, and a unit ball jf.burnol.free.fr/agreg161007CubeBoule.pdf ;So we will have a bijection between the domain and the cube, right? – Motaka Apr 4 '17 at 13:25
• @Mokata: there is not even a Lipschitz bijection from some $\Omega$ to the cube (or, for that matter, the ball). I gather that your map is between the open domain and the open ball (otherwise your proof cannot hold). Then remember that a $C^1$ map on a noncompact set need not be Lipschitz. This discussion confirms me something that I was not sure of, I think your questions are more suited for math.SE than MO. – Benoît Kloeckner Apr 4 '17 at 13:31
• Last comment: bottomline is, $C^1$ property on an open domain do not extend to the closure, while the Lipschitz property does. So if there where a Lipschitz bijection from the open $\Omega$ to the open cube, there would be one from the closure of $\Omega$ to the closed cube. – Benoît Kloeckner Apr 4 '17 at 13:57