Timeline for A question about whether an operator can be lipschitz or not
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| S Jul 24, 2018 at 7:13 | history | bounty ended | CommunityBot | ||
| S Jul 24, 2018 at 7:13 | history | notice removed | CommunityBot | ||
| Jul 20, 2018 at 10:44 | history | edited | Hheepp | CC BY-SA 4.0 |
added 1 character in body
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| Jul 17, 2018 at 16:04 | comment | added | Hheepp | @ Mateusz Kwasnicki: Yes you are right. You can absorb b(.,.) into a(.,.). Also, you can put any assumtion on c(.,.), to make the integral meaningful. In other words, i want to see with any extra assumptions on coefficients, is it possible to show that the operator is contraction or not? | |
| Jul 16, 2018 at 9:34 | comment | added | Mateusz Kwaśnicki | Apparently some assumptions on the coefficients are required: if $c(\zeta,x)$ is everywhere zero, then $\mathcal{B}$ is not even defined. Also, I fail to see why $b(\zeta, x)$ is needed, as it can be absorbed into $a(\zeta, x)$. | |
| S Jul 16, 2018 at 6:13 | history | bounty started | Hheepp | ||
| S Jul 16, 2018 at 6:13 | history | notice added | Hheepp | Draw attention | |
| Jul 14, 2018 at 14:20 | history | edited | Hheepp | CC BY-SA 4.0 |
I replaced the space $C^{\sigma/2, \sigma}(X)$ with the correct on, $C^{\sigma, \sigma/2}(X)$.
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| Jul 14, 2018 at 14:17 | comment | added | Hheepp | @Andrew : Thanks Andrew, you are right. I have made mistake in typing. The correct Space is $C^{\sigma/2, \sigma}(X)$. The range of $\sigma$, in my problem belongs to $(0,1)$. | |
| Jul 13, 2018 at 18:47 | comment | added | Andrew | What values $\sigma$ can take? And shouldn't it be $C^{\sigma, \sigma/2}(X)$ instead of $C^{\sigma/2, \sigma}(X)$? | |
| Jul 13, 2018 at 15:55 | review | Close votes | |||
| Jul 14, 2018 at 19:23 | |||||
| Jul 13, 2018 at 11:48 | history | asked | Hheepp | CC BY-SA 4.0 |