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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.

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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 $\| f_n - f\|_{L^q(\partial\Omega)} \to 0$.

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