1
$\begingroup$

Let $\Omega$ be a Lipschitz domain in $\mathbb{R}^2$, and let $\mathscr{T}:H^2(\Omega) \to L^2(\partial \Omega)$ be the trace operator defined in the usual way. Is there a characterization of its image? It is clearly a subspace of $H^1(\partial\Omega)$; is it necessarily a closed subspace?

$\endgroup$
1
  • $\begingroup$ In this answer mathoverflow.net/a/275418/14551 the book is cited where traces of functions from Sobolev spaces on a Lipschitz boundary are characterized. $\endgroup$
    – Andrew
    Aug 17, 2017 at 10:09

2 Answers 2

2
$\begingroup$

From Elliptic problems in nonsmooth domains by P. Grisvard:

Theorem 1.5.1.3. Let $\Omega$ be a bounded open subset of $\mathbb R^n$ with a Lipschitz boundary $\Gamma$. Then the mapping $u\to \gamma u$ which is defined for $u\in C^{0,1}(\overline{\Omega})$, has a unique continuous extension as an operator from $W^1_p(\Omega)$ onto $W^{1-\frac{1}{p}}_p(\Gamma)$. This operator has a right continuous inverse independent of $p$.

$\endgroup$
1
$\begingroup$

For $1<p<\infty$, the image of the trace operator on $W^{m,p}(\Omega)$ is $W^{m-\frac{1}{p},p}(\partial\Omega)$, see https://en.wikipedia.org/wiki/Sobolev_space#Traces

As a reference they give

Adams, Robert A.; Fournier, John (2003) [1975]. Sobolev Spaces. Pure and Applied Mathematics. 140 (2nd ed.). Boston, MA: Academic Press. ISBN 978-0-12-044143-3..

$\endgroup$
2
  • 1
    $\begingroup$ Johannes: this is the case when the boundary is sufficiently smooth. But what is H^{3/2} when it is Lipschitz? $\endgroup$
    – user113499
    Aug 17, 2017 at 2:32
  • 1
    $\begingroup$ As user113499 says, I don't believe the reference you cite gives an answer. Standard definitions of the space $H^{3/2}\left(\partial\Omega\right)$ generally require that the domain have a boundary which is $C^{1,1}$ not $C^{0,1}$. Or perhaps I am missing something in the definitions in Adams and Fournier? $\endgroup$
    – Marvin
    Aug 17, 2017 at 4:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.