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?
2 Answers
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 $p$.
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 9780120441433..

1$\begingroup$ Johannes: this is the case when the boundary is sufficiently smooth. But what is H^{3/2} when it is Lipschitz? $\endgroup$ 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? $\endgroup$– MarvinAug 17, 2017 at 4:01