I asked about the compact trace embedding (trace embedding). In the answer,

For $N \ge3,$

  • 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.

I don't understand the part "By interpolation, the inclusion $W^{1/2,2}(\partial \Omega) \to L^{q}(\partial \Omega)$ is compact for all $q \in [1,q^*)$" .

Would you tell me about it in detail? I would be grateful for any comment about it.


Let $\{f_n\}$ be given, such that $f_n \rightharpoonup f$ in $W^{1/2,2}(\partial\Omega)$. We have to show $\| f_n - f\|_{L^q(\partial\Omega)} \to 0$.

By Hölder's inequality, we have $$ \| f_n - f \|_{L^q(\partial\Omega)} \le \| f_n - f \|_{L^1(\partial\Omega)}^{\lambda}\, \| f_n - f \|_{L^{q^*}(\partial\Omega)}^{1-\lambda}, $$ where $\lambda \in (0,1)$ is determined by $1/q = \lambda/1 + (1-\lambda)/q^*$. This inequality is typically called an 'interpolation inequality, because you can interpolate between the norms in $L^1$ and $L^{q^*}$.

Now, since the embedding to $L^1$ is compact, you have $$ \| f_n - f \|_{L^1(\partial\Omega)} \to 0 $$ and since the embedding to $L^{q^*}$ is bounded, you have $$ \| f_n - f \|_{L^{q^*}(\partial\Omega)} \le C $$ for some $C > 0$.

Putting things together, you have the desired $\| f_n - f\|_{L^q(\partial\Omega)} \to 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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