We say, following this definition, that a domain $\Omega\subset \mathbb{R}^{n}$ is weakly Lipschitz if it can locally be flattened by a Lipschitz homeomomorphism $\phi$ (i.e., a Lipschitz continuous map $\phi$ such that its inverse $\phi^{-1}$ is still Lipschitz).

My question is the following: given an hyperplane $H\subset \mathbb{R}^n$, we consider the projection $\pi_H:\mathbb{R}^n\to H$. Then, is it true that $\pi(\Omega)$ is still a weakly Lipschitz domain?

I've found here that the answer is negative for strongly Lipschitz domains, and I don't have enough intuition to see what can possibly go wrong with weakly Lipschitz domain (although I'm afraid there can be something weird)

Any reference or help is welcomed. Thank you!


In the question that you linked to I described a simple counter-example, a "curved croissant". I think the same example works here.

To be specific: let $\phi = \arg(x + i y) \in (-\pi, \pi)$, and consider $$\Omega = \biggl\{(x,y,z) \in \mathbb{R}^3: \Bigl(\sqrt{x^2 + y^2} - 1\Bigr)^2 + (z - \phi)^2 + \frac{\phi^2}{\pi^2} < 1\biggr\} .$$ This is clearly Lipschitz, but its projection onto the $xy$ plane is not weakly Lipschitz at $(-2, 0)$.

Here is $\Omega$:

A croissant

And here is its projection:

A flat croissant

  • $\begingroup$ Wow, beautiful! $\endgroup$ – Gil Sanders Oct 18 at 10:34
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    $\begingroup$ @Mateusz: I was wondering whether the class of weakly lipschitz domain is at least preserved by lipschitz homeomorphisms. Is this statement true or do you know any counterexample? $\endgroup$ – guido giuliani Oct 19 at 22:26
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    $\begingroup$ Bi-Lipschitz homeomorphisms of $\mathbb{R}^n$ obviously preserve this class. Bi-Lipschitz homeomorphisms from $\Omega$ onto $\Omega'$ need not: the $\Omega$ defined above can be mapped onto $$\Omega'=\biggl\{(x,y,z) \in \mathbb{R}^3: \Bigl(\sqrt{x^2 + y^2} - 1\Bigr)^2 + z^2 + \frac{\phi^2}{\pi^2} < 1\biggr\}$$ by the map $(x,y,z) \mapsto (x,y,z-\phi)$. However, $\Omega'$ is not weakly Lipshitz. (At least if I got the definition of a weakly Lipschitz domain right.) $\endgroup$ – Mateusz Kwaśnicki Oct 19 at 22:35
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    $\begingroup$ Yes sorry: I also forgot to mention that I was speaking of bi-Lipschitz homeomorphisms defined on some open set containing $\Omega$. This should be ok, right? $\endgroup$ – guido giuliani Oct 20 at 15:06
  • $\begingroup$ @guidogiuliani: Yes, this should work with no problems. $\endgroup$ – Mateusz Kwaśnicki Oct 20 at 18:21

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