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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.

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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.

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  • $\begingroup$ Can you explain more? Ps: Now I am interested in showing the lipschitzian bijection between the cube and the domain $\endgroup$
    – Motaka
    Commented Apr 4, 2017 at 11:26
  • $\begingroup$ @Mokata: the other direction seems more elementary, probably doable by hand. $\endgroup$
    – Benoît Kloeckner
    Commented Apr 4, 2017 at 13:09
  • $\begingroup$ @ 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? $\endgroup$
    – Motaka
    Commented Apr 4, 2017 at 13:25
  • $\begingroup$ @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. $\endgroup$
    – Benoît Kloeckner
    Commented Apr 4, 2017 at 13:31
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    $\begingroup$ 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. $\endgroup$
    – Benoît Kloeckner
    Commented Apr 4, 2017 at 13:57

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