Skip to main content

Timeline for A Bi-Lipschitzian application

Current License: CC BY-SA 3.0

9 events
when toggle format what by license comment
Jan 7, 2019 at 16:47 vote accept Motaka
Apr 4, 2017 at 13:57 comment added Benoît Kloeckner 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.
Apr 4, 2017 at 13:46 comment added Motaka In all that I wrote, it was about, open ball, open cube, and open domain. Perhaps i didn't inderstand your idea, since I do not usually discuss in English, anyway thanks for Your time and your help.
Apr 4, 2017 at 13:31 comment added Benoît Kloeckner @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.
Apr 4, 2017 at 13:25 comment added Motaka @ 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?
Apr 4, 2017 at 13:09 comment added Benoît Kloeckner @Mokata: the other direction seems more elementary, probably doable by hand.
Apr 4, 2017 at 13:09 history edited Benoît Kloeckner CC BY-SA 3.0
Added explanations
Apr 4, 2017 at 11:26 comment added Motaka Can you explain more? Ps: Now I am interested in showing the lipschitzian bijection between the cube and the domain
Apr 4, 2017 at 10:15 history answered Benoît Kloeckner CC BY-SA 3.0