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.

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.

• In your reference, I cannot find embeddings you mentioned above. Would you tell me more? – 04170706 Aug 4 '17 at 17:41
• Well, what about Theorems 6.5 and 7.1 (and their extensions)? You're right that the trace theorem is not covered there. It would help much if you sketched the proof you know for the case $N \ge 3$, otherwise I don't know your background knowledge. – Michał Miśkiewicz Aug 5 '17 at 18:50
• Thanks. I will check the proof for the case $N \ge 3.$ I'd like to know if you think the claim for N=2 is true, although I cannot find the reference which states it exactly. – 04170706 Aug 6 '17 at 0:29
• I think it is, and I guess I included the proof in the answer. – Michał Miśkiewicz Aug 7 '17 at 18:41

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