Timeline for Lipschitz-regularity of partition of unity
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Aug 30, 2023 at 14:16 | comment | added | Pietro Majer | I'd say it's essentially the same. | |
| Aug 29, 2023 at 15:37 | comment | added | shuhalo | @PietroMajer: Is there a specific reason to choose the squares of the distance functions rather than the distance functions themselves? | |
| S Jun 16, 2022 at 19:56 | history | bounty ended | Carlos_Petterson | ||
| S Jun 16, 2022 at 19:56 | history | notice removed | Carlos_Petterson | ||
| Jun 16, 2022 at 19:56 | vote | accept | Carlos_Petterson | ||
| Jun 16, 2022 at 19:20 | answer | added | Willie Wong | timeline score: 4 | |
| Jun 16, 2022 at 17:43 | comment | added | Willie Wong | Explicit upper bounds certainly would depend on the minutiae of the cover. You can see this in the one dimensional case. Let $K = [0,1]$ and there are no upper bound to the Lipschitz constant of $\psi_U$ over the class of all covers of $K$ by pairs of open intervals. | |
| Jun 16, 2022 at 9:51 | history | edited | Carlos_Petterson | CC BY-SA 4.0 |
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| Jun 15, 2022 at 15:30 | history | edited | Carlos_Petterson | CC BY-SA 4.0 |
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| S Jun 15, 2022 at 15:29 | history | bounty started | Carlos_Petterson | ||
| S Jun 15, 2022 at 15:29 | history | notice added | Carlos_Petterson | Authoritative reference needed | |
| Jun 14, 2022 at 5:34 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Jun 13, 2022 at 22:56 | comment | added | Carlos_Petterson | @LSpice Thanks, its all fixed now; sorry about the typos. | |
| Jun 13, 2022 at 22:56 | history | edited | Carlos_Petterson | CC BY-SA 4.0 |
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| Jun 13, 2022 at 22:41 | history | edited | Carlos_Petterson | CC BY-SA 4.0 |
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| Jun 13, 2022 at 18:28 | comment | added | Carlos_Petterson | @FedorPetrov I clarified the notation; I use K-U to denote the complement of $U$ in $K$. | |
| Jun 13, 2022 at 18:28 | history | edited | Carlos_Petterson | CC BY-SA 4.0 |
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| Jun 13, 2022 at 13:57 | comment | added | Carlos_Petterson | @PietroMajer Ah true, but then can we upper-bound the Lipschitz constant in a similar fashion? | |
| Jun 13, 2022 at 13:49 | comment | added | Pietro Majer | Well, what I wrote is just this: if, for a given $x\in K$, $U$ is the only open set in $\mathcal U$ such that $x\in U$, then $\psi_V(x)=0$ for all $V\neq U$, so $\psi_U(x)=1$; since $\psi_U(y)=0$ on $\partial U$ we need the Lipschitz constant of $\psi_U$ at least 1/diam(U). | |
| Jun 13, 2022 at 11:30 | comment | added | Carlos_Petterson | @PietroMajer This I am having a lot of trouble seeing; I was also suspecting that $L=1/r$ where $r>0$ is the Lebesgue number of $\mathscr{U}$. (Ideally, do you have a reference to this or your claim?) | |
| Jun 13, 2022 at 11:21 | comment | added | Pietro Majer | Note that if an open set $U$ can't be removed from the covering $\mathcal U$, i.e. $K\cap U\setminus \bigcup_{U\neq V\in\mathcal U}V\neq\emptyset$, then the $\psi_U$ supported in $U$ in any partition of unity subordinated to $\mathcal U$ must have Lipschitz constant at least 1/diam(U). | |
| Jun 13, 2022 at 10:52 | history | asked | Carlos_Petterson | CC BY-SA 4.0 |