Timeline for Fractional-order Rellich–Kondrashov Theorem
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2017 at 19:17 | answer | added | anonymous | timeline score: 0 | |
| S Jan 15, 2017 at 12:38 | history | suggested | Martin Sleziak | CC BY-SA 3.0 |
added doi link and link to arxiv version
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| Jan 15, 2017 at 12:23 | review | Suggested edits | |||
| S Jan 15, 2017 at 12:38 | |||||
| Jan 15, 2017 at 12:01 | comment | added | Todd Trimble | On the Cyrillic spelling that had been in the title, see the meta discussion here: meta.mathoverflow.net/questions/3104/… | |
| Jan 15, 2017 at 11:53 | history | edited | Todd Trimble | CC BY-SA 3.0 |
changed the title in response to a meta discussion
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| Jan 14, 2017 at 15:56 | vote | accept | anonymous | ||
| Jan 14, 2017 at 15:22 | answer | added | Hannes | timeline score: 5 | |
| Jan 14, 2017 at 14:32 | history | edited | anonymous | CC BY-SA 3.0 |
Clarify what $p^*$ is. Fix constraints on exponents/dimensions.
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| Jan 14, 2017 at 14:26 | comment | added | anonymous | @Hannes You're right. Thanks for being thorough! | |
| Jan 14, 2017 at 14:19 | comment | added | Hannes | Also, adding to my previous comment: I think one should expect $W^{s,p}(\Omega) \hookrightarrow W^{s-\epsilon,q}(\Omega)$ compactly for $q \in [1,p^*(n,\epsilon))$ since you spend only $\epsilon$ of differentiability - assuming that was not what you asked for. | |
| Jan 14, 2017 at 14:15 | comment | added | Hannes | I suggest to add that $p^* = p^*(n,s) = \frac{np}{n-sp}$. I am aware that the paper also just writes $p^*$ but at least in my world $p^*$ is usually the Sobolev conjugate corresponding to $s=1$. | |
| Jan 14, 2017 at 11:39 | history | edited | anonymous | CC BY-SA 3.0 |
The L2-Laplacian representation requires a factor
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| Jan 14, 2017 at 11:00 | history | asked | anonymous | CC BY-SA 3.0 |