Timeline for A smooth function such that the second derivative of its absolute value is a distribution of positive order
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 27, 2021 at 18:59 | answer | added | Michael Renardy | timeline score: 4 | |
| May 27, 2021 at 14:03 | comment | added | Christian Remling | @Bazin: Thank you for clarifying. It's clear enough actually, with hindsight, but I somehow reinterpreted "positive order" as "finite order". | |
| May 27, 2021 at 10:27 | comment | added | Bazin | @Christian Remling: I want to find a $C^\infty$ function $f$ such that, with $g=\vert f\vert$, the distribution second derivative of $g$ is not a Radon measure. | |
| May 27, 2021 at 10:24 | history | edited | Bazin | CC BY-SA 4.0 |
added 25 characters in body
|
| May 27, 2021 at 5:16 | answer | added | Jochen Wengenroth | timeline score: 0 | |
| May 26, 2021 at 19:54 | comment | added | Jochen Wengenroth | @bathalf15320 No, as every measure, $2\delta$ has order $0$. | |
| May 26, 2021 at 19:46 | history | edited | Bazin | CC BY-SA 4.0 |
added 50 characters in body
|
| May 26, 2021 at 17:33 | comment | added | bathalf15320 | What's wrong with $g(x)=x$? The second derivative of $|g|$ is then $2\delta$ which I imagine is a distribution of positive order. | |
| May 26, 2021 at 15:38 | comment | added | Bazin | I meant locally Lipschitz-continuous. | |
| May 26, 2021 at 15:38 | comment | added | Nate Eldredge | Do you mean locally Lipschitz? Or was $f$ supposed to be compactly supported or something? | |
| May 26, 2021 at 15:33 | comment | added | Johannes Hahn | It is not "easy to verify", because it's wrong in general. $f=g=x\mapsto x^2$ is not Lipschitz | |
| May 26, 2021 at 15:19 | history | asked | Bazin | CC BY-SA 4.0 |