Timeline for Center of convex figure
Current License: CC BY-SA 4.0
28 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2022 at 20:29 | history | bounty ended | Anton Petrunin | ||
| Oct 20, 2022 at 21:59 | comment | added | Anton Petrunin | @fedja I was thinking about metric spaces $X$ that admit a short retraction $\mathrm{Haus}\,X\to X$; here $\mathrm{Haus}\,X$ denotes the set of all nonempty compact sets of $X$ equipped with Hausdorff metric. Evidently injective spaces and descrete spaces are among the examples. I am not sure, maybe all length spaces with this property are injective. | |
| Oct 20, 2022 at 21:58 | comment | added | Saúl RM | @AntonPetrunin maybe you are referring to the center from the answer, and not the one we mentioned in the comments? I would have referenced the comment but I don't know if that is possible | |
| Oct 20, 2022 at 21:55 | comment | added | Anton Petrunin | @SaúlRM no, this center is only $C^{\frac12}$ continuous. | |
| Oct 20, 2022 at 21:53 | comment | added | Anton Petrunin | @YoavKallus thanks (evidently I did not read this answer to the end). | |
| Oct 20, 2022 at 21:00 | comment | added | Saúl RM | Also, @fedja it seems like this topic has been well studied, I don't think there is no need to contact me anymore, but thanks anyways :) (I will delete this message later if I remember) | |
| Oct 20, 2022 at 20:37 | comment | added | Saúl RM | In Oliver Klein's thesis it claims that for centers invariant by translations, the constant $\frac{4}{\pi}$ is the minimal one, and it is satisfied by the Steiner center. And taking into account the comment above by Fedja it shoud indeed be the same constant for centers that are not necessarily invariant by translation. Btw I think the Steiner point is the one we mentioned in the comments to this answer | |
| Oct 20, 2022 at 18:22 | comment | added | Yoav Kallus | In an answer to a post linked in the question (mathoverflow.net/questions/120240), Günter Rote seems to claim "The best Lipschitz constant (for the Hausdorff distance) is achieved by the so-called Steiner point". Is that a different constant than what is discussed here? | |
| Oct 20, 2022 at 16:51 | comment | added | Saúl RM | @fedja I agree, it would make sense finding the constant or some good upper/lower bounds before sending anything | |
| Oct 20, 2022 at 16:48 | comment | added | fedja | @AntonPetrunin Now, once the problem is solved (at least in the originally posted formulation) the question "What led you to asking it?" becomes legitimate (especially if you suggest that SaulRM makes a formal publication). So what's the underlying story here (if there is one)? | |
| Oct 20, 2022 at 16:38 | comment | added | fedja | @SaúlRM It could be a nice project and it may make a good Monthly article, but then (IMHO) it would make sense to try to figure out the optimal Lipschitz constant before sending anything to a journal. Journal publishing process is easy, BTW: you type the article in TeX, proofread it 5 times, go to the journal webpage and follow the instructions for uploading files. With some of them (Monthly included) you are also strongly encouraged to use their TeX header (downloadable from the journal webpage) to have the journal format from the start. No rocket science. I'll send you an e-mail later. :-). | |
| Oct 20, 2022 at 14:35 | comment | added | Saúl RM | Thanks for the information! I have left my email in my profile in case @fedja is interested in collaborating. | |
| Oct 20, 2022 at 14:28 | comment | added | Anton Petrunin | (I would write it as elementary as possible and send it to Math Monthly or Math Intelligencer) | |
| Oct 20, 2022 at 14:23 | comment | added | Anton Petrunin | I think that the central idea is yours + you may try to cooperate with Fedja (he knows how to publish). | |
| Oct 20, 2022 at 14:00 | comment | added | Saúl RM | @AntonPetrunin I'm not sure how publishing works, I am a first year PhD student right now. Also, fedja published a complete proof a few minutes before me | |
| Oct 20, 2022 at 13:55 | comment | added | Saúl RM | @fedja Wow, that seems like it's going to work. I will think about it later in detail (and your proof too, I am just studying ultrafilters so it will be a good exercise) | |
| Oct 20, 2022 at 13:35 | comment | added | fedja | More precisely, since the isometry invariant version must assign to an equilateral triangle its center, we can place two such triangles in the "hourglass" configuration and get the lower bound $L\ge 2/\sqrt 3$ immediately. | |
| Oct 20, 2022 at 13:07 | comment | added | fedja | Actually your original argument works too. If we have an $L$-Lipschitz $p$, then we can average $p(v+K)-v$ over big disks and take a limit to get a version commuting with translations and then $R^{-1}p(RK)$ over rotations $R$ to get it commuting with rotations too, without changing $L$. This observation should allow to get fairly good bounds, especially from below. I hope I didn't miss anything. | |
| Oct 20, 2022 at 13:02 | vote | accept | Anton Petrunin | ||
| Oct 20, 2022 at 11:29 | comment | added | Anton Petrunin | You should publish it. | |
| Oct 20, 2022 at 11:27 | vote | accept | Anton Petrunin | ||
| Oct 20, 2022 at 11:28 | |||||
| Oct 20, 2022 at 11:21 | vote | accept | Anton Petrunin | ||
| Oct 20, 2022 at 11:27 | |||||
| Oct 20, 2022 at 3:15 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Oct 20, 2022 at 1:51 | comment | added | fedja | Yep, you proof is simpler. The underlying idea is essentially the same but you use one direction only, which makes both the argument and its quantification fairly simple. Congratulations! Now let's try to tighten the bounds :-) | |
| Oct 20, 2022 at 1:39 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Oct 20, 2022 at 0:08 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Oct 19, 2022 at 23:37 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Oct 19, 2022 at 23:31 | history | answered | Saúl RM | CC BY-SA 4.0 |