trace embedding For simplicity, let $\Omega$ is a smooth bounded domain in $\mathbb{R}^N$. I know that for $N \ge 3,$ the trace map $W^{1,2}(\Omega) \hookrightarrow L^{q}(\partial \Omega)$ is compact for $q \in [1,\frac{2(N-1)}{N-2})$. For example, see Theorem 6.2, Chapter 2 of the book Direct Methods
in the Theory of Elliptic Equations written by J. Necas.
I'd like to know that for $N=2,$ is the fact that $W^{1,2}(\Omega) \hookrightarrow L^{q}(\partial \Omega)$ is compact for $q\in [1,\infty)$ true?
It seems to be true, but I cannot find the reference that states it explicitly. Is it true or not?
 I would be grateful for any comment about it.
 A: It would be helpful to sketch a proof for the case $N \ge 3$ first. 


*

*trace map $W^{1,2}(\Omega) \to W^{1/2,2}(\partial \Omega)$ is bounded,

*inclusion $W^{1/2,2}(\partial \Omega) \to L^1(\partial \Omega)$ is compact, 

*inclusion $W^{1/2,2}(\partial \Omega) \to L^{q^*}(\partial \Omega)$ with $q^* = \frac{2(N-1)}{N-2}$ is bounded. 


By interpolation, the inclusion $W^{1/2,2}(\partial \Omega) \to L^{q^*}(\partial \Omega)$ is compact for all $q \in [1,q^*)$, and the claim follows. 
The only modification needed in the case $N=2$ is that this time $W^{1/2,2}(\partial \Omega) \to L^{q}(\partial \Omega)$ is bounded for all $q<q^*=\infty$, but not for $q=q^*$. This is because the Sobolev embedding doesn't work if $\textrm{order of derivatives} \times \textrm{exponent} = \textrm{dimension}$. The claim follows in the same fashion. 
If you need a reference, you can find these embeddings in Hitchhiker's guide to the fractional Sobolev spaces. 
A: I found a reference for my question which is the result for more general domain, see http://www.ams.org/journals/proc/2009-137-12/S0002-9939-09-10045-X/S0002-9939-09-10045-X.pdf
