Timeline for Composition operators on fractional-order (periodic) Sobolev spaces
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Jul 17, 2015 at 14:35 | answer | added | Jean Duchon | timeline score: 0 | |
| Jul 17, 2015 at 12:58 | history | edited | F. H. | CC BY-SA 3.0 |
added 21 characters in body
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| Jul 17, 2015 at 9:22 | history | rollback | F. H. |
Rollback to Revision 4
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| Jul 16, 2015 at 20:50 | history | edited | F. H. | CC BY-SA 3.0 |
Skip other case
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| Jul 16, 2015 at 13:49 | history | edited | F. H. | CC BY-SA 3.0 |
added 19 characters in body
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| Jul 16, 2015 at 12:59 | history | edited | F. H. | CC BY-SA 3.0 |
Corrected flaw and revised question.
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| Jul 14, 2015 at 12:02 | answer | added | Jean Duchon | timeline score: 1 | |
| Jul 14, 2015 at 7:26 | history | edited | F. H. | CC BY-SA 3.0 |
Corrected error in citation.
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| Jul 13, 2015 at 15:47 | comment | added | Joonas Ilmavirta | The claim you wish to conclude is false. Take a compactly supported smooth $u$ so that $u(0)=0$ but $\nabla u(0)\neq0$. If $p\geq1$ is not a natural number, then there is $s>0$ so that $u^p\notin H^s(\mathbb R^n)$ (because classical derivatives fail to exist to high orders) although $u\in H^s(\mathbb R^n)$ for all $s$. | |
| Jul 13, 2015 at 15:14 | history | asked | F. H. | CC BY-SA 3.0 |