Timeline for Sobolev extension problems of $W^1_\infty(\Omega)$
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 29 at 0:39 | vote | accept | Javier | ||
| Jun 27 at 23:27 | answer | added | TaQ | timeline score: 1 | |
| Jun 22 at 17:01 | answer | added | TaQ | timeline score: 2 | |
| Jun 22 at 5:51 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing
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| Jun 22 at 3:53 | history | edited | Javier | CC BY-SA 4.0 |
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| Jun 22 at 3:52 | history | edited | Javier | CC BY-SA 4.0 |
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| Jun 21 at 17:46 | comment | added | TaQ | Whatever reasonable meaning you give to "$C(\Omega)$", for the two-dimensional $\Omega$ in my first comment above, the claim $W^1_\infty(\Omega)\subset W^1_\infty(\mathbb{R}^n)|_\Omega$ does not hold. Then only $(*)$ remains, and the truth of this depends on the definition of "$C(\Omega)$". | |
| Jun 21 at 17:19 | comment | added | TaQ | It would be "natural" to take $C(\Omega)$ the space of all continuous functions $\Omega\to\mathbb R$ with the topology of uniform convergence on compact subsets, but then the resulting "Sobolev space" would only be Fréchet but not Banach, and this is contrary to the usual purpose of using Sobolev spaces. | |
| Jun 21 at 17:12 | comment | added | TaQ | You do not define what you mean by "$C(\Omega)$", but if the given definition of Sobolev space is intended to give the usual Banach space, and hence $C(\Omega)$ is taken to be some Banach space of bounded functions, then the claim $f\in W^1_\infty(\Omega)$ need not hold for $\Omega$ arbitrary bounded connected open set For example, in dimension two consider the set of points $(t,s)$ with $0<t<1$ and $\sin(t^{-1})<s<\sin(t^{-1})+1\,$. | |
| Jun 21 at 16:05 | history | edited | Javier | CC BY-SA 4.0 |
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| S Jun 21 at 16:01 | review | First questions | |||
| Jun 21 at 18:22 | |||||
| S Jun 21 at 16:01 | history | asked | Javier | CC BY-SA 4.0 |