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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!

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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

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  • $\begingroup$ 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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