Timeline for Image of trace operator on $W^{2,1}(\mathbb{R}^2)$
Current License: CC BY-SA 4.0
7 events
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| Apr 15 at 1:28 | comment | added | Liding Yao | @vmist The first google result gives me "Trace Operators in Besov and Triebel-Lizorkin Spaces" by Cornelia Schneider in 2010. It mentions the $p>1$ in the beginning. My guess of $\operatorname{Tr}W^{2,1}$ is $B_{11}^1$. Note that this doesn't imply by any Besov-Triebel-Lizorkin cases since $W^{2,1}\not\subset F_{1\infty}2$ so it needs a careful proof. Certainly Prof. Hajlasz is more expert on this than me. | |
| Apr 14 at 23:57 | comment | added | vmist | @LidingYao do you mean the $s=\frac{1}{p}$ endpoint? i.e. that the trace is surjective $T: W^{\frac{1}{p},p}(\Omega) \to L^p(\partial\Omega)$ for $p>1$? And if so I would love a reference. | |
| Apr 14 at 23:40 | comment | added | vmist | The surjectivity in the W^{1,1} case is a theorem by Gagliardo, and was asked previously mathoverflow.net/questions/145393/… which includes references. Having worked on it more I think the image of the trace on $W^{s,1}$ is the Besov space $B^{s-1}_{1,1}$ for all $s>1$ as you suggested, though a full proof still eludes me. | |
| Apr 14 at 23:03 | comment | added | Piotr Hajlasz | I also believe it is not surjective. I could try to write a counterexample, but I am overwhelmed now with the end of the semester. | |
| Apr 14 at 17:53 | comment | added | Liding Yao | I believe that it's not surjection. For p>1 and s>1/p the image given by Besov space, which collapse to L^p at the endpoint. I believe the similar reverse result via Besov spaces should holds for p=1. Do you have a quick reference for the first paragraph. | |
| Apr 14 at 16:42 | history | edited | vmist | CC BY-SA 4.0 |
added 292 characters in body
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| Apr 14 at 11:03 | history | asked | vmist | CC BY-SA 4.0 |