Timeline for Center of convex figure
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 20, 2022 at 1:40 | comment | added | Saúl RM | Cool, I will read the proof now. I was already finishing one myself so I also left it posted | |
| Oct 20, 2022 at 1:35 | comment | added | fedja | @SaúlRM OK, I found "something interesting" and even posted it as an answer, but the problem about the least Lipschitz constant is still far from the full resolution :-) | |
| Oct 20, 2022 at 0:16 | comment | added | Saúl RM | I don't understand that comment very well, but if you find something interesting with that construction (like a Lipschitz constant less than $\sqrt{2}$) I would be interested :) I tried but I couldn't deduce anything relevant about that center | |
| Oct 20, 2022 at 0:10 | comment | added | fedja | @SaúlRM The idea I'm trying to pursue now is to figure out all $p$ we can use on the line with $1+\delta$ (it seems that you cannot deviate too much from some fixed convex combination of endpoints though the combination may depend on the line direction) and then to place various intervals around $K$ to run into trouble. | |
| Oct 20, 2022 at 0:10 | comment | added | Saúl RM | Cool. I was pretty sure that was going to work but I wasn't able to prove it, and indeed that center (and I think every other construction that has been mentioned until now) would be invariant by isometries so as I explain in my answer it cannot work | |
| Oct 20, 2022 at 0:06 | comment | added | fedja | @SaúlRM It is exactly the same construction as mine only I was too lazy to rotate the axes in my comment, but the thought of considering the average over axis rotations crossed my mind too though I decided to spend some time on trying to prove that you cannot achieve $1$ first, which I'm doing now :-) | |
| Oct 19, 2022 at 23:59 | comment | added | Saúl RM | Other way to obtain Lipschitz constant $\sqrt{2}$ is to fix any unitary vector $v$ and take as center $p_{F,v}$ of $F$ the center of the minimum rectangle parallel to $v$ that contains $F$. Maybe if we define a center $p_F$ to be the average over $v\in\mathbb{S}^1$ of $p_{F,v}$ the resulting center has Lipschitz constant $<\sqrt{2}$? Minimizing the Lipschitz constant seems an interesting question | |
| Oct 19, 2022 at 23:52 | comment | added | fedja | Some Lipschitz constant is trivial. We have such point on the line (the center of the interval). Now just project to two coordinate axis and use the centers of projections for the coordinates of $p$, which gives you Lipschitz constant $\sqrt 2$ immediately. | |
| Oct 19, 2022 at 22:18 | comment | added | Anton Petrunin | I would check baryceter with of vertices of polygon with weights proportional to their external angles. | |
| Oct 19, 2022 at 22:06 | comment | added | Anton Petrunin | No, the map is only $C^{\frac12}$-continuous. | |
| Oct 19, 2022 at 21:29 | comment | added | Christophe Leuridan | Another candidate for $p_F$: the barycenter of the (convex) subset of all points which maximize the distance to $F^c$? | |
| Oct 19, 2022 at 21:11 | comment | added | Christophe Leuridan | Nice counter-examples. So the last argument is false, I suppose that there is still a Lipschitz constant, greater than 1. It would have been nice to give these negative answers in the question, to avoid wrong reflexions. Do you have other natural candidates which fail? | |
| Oct 19, 2022 at 20:13 | comment | added | Anton Petrunin | Please see the example I added to the question. | |
| Oct 19, 2022 at 19:38 | history | answered | Christophe Leuridan | CC BY-SA 4.0 |