Timeline for Lipschitz parametrization of a symmetric convex curve
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 19, 2013 at 19:54 | comment | added | Sergei Ivanov | @djoke: the map is onto, but indeed it fails to be injective if the target has corners. I overlooked that. It seems that it can be approximated by a homeomorphism with the same Lipschitz constant if the target is peicewise smooth. I am not sure about the case of infinitely many corners. | |
| Jun 19, 2013 at 16:19 | comment | added | djoke | $Lip(f)=1/4 \sqrt{2} \pi$ | |
| Jun 19, 2013 at 16:18 | comment | added | djoke | The square in not a counterexample. Namely, assume as we may that the square $|\gamma|=2\pi$. Then by using arc-length parametrization $f: S^1\to \gamma$ we obtain that $Lip(f)=\sqrt{2}{4}\pi$. | |
| Jun 19, 2013 at 14:50 | comment | added | Anton Petrunin | @djoke, it is not hard to perturb the map into a homeomorphism by making the Lipschitz constant little worse. On the other hand if you want to keep the constant then you can not do this --- square is an example. | |
| Jun 19, 2013 at 10:40 | comment | added | djoke | @Sergei I meant that the map should be a homeomorphism, otherwise the question is trivial. Your projection maybe is not onto? | |
| Jun 19, 2013 at 9:04 | history | answered | Sergei Ivanov | CC BY-SA 3.0 |