Compact embedding of $W^{k,p}(\Omega)$ into $W^{k-1,p*}(\Omega)$, where $p^*$ is the Sobolev conjugate

Question

How do I see that the embedding of $$W^{k,p}(\Omega) \hookrightarrow W^{k-1,p^*}(\Omega)$$ for $1 \leq p < \infty$, the Sobolev conjugate $\frac{1}{p^*}=\frac{1}{p} - \frac{1}{n}$, and $\Omega$ open and bounded in $\mathbb{R}^n$ is a compact operator? I think the embedding is just the identity map right? But I do not see how it is a compact operator, i.e. that every bounded sequence in $W^{k,p}(\Omega)$ contains a converging sequence in $W^{k-1,p^*}(\Omega)$. Does anyone know how to prove this?

Answer 1

The embedding is not compact. First off, you would require a Lipschitz boundare. Then we have the following result due to Rellich and Kondrachov.

Let $\Omega \subset \mathbb{R}^n$ be a bounded Lipschitz domain. Further, let $m_1,m_2 \in \mathbb{N}_0$ and $p_1,p_2 \in [1,\infty)$. If $$m_1 - \frac{n}{p_1}>m_2-\frac{n}{p_2} \quad \text{and} \quad m_1>m_2,$$ then $W^{m_1,p_1}(\Omega)$ is continuously, compactly embedded in $W^{m_2,p_2}(\Omega)$ via the identity mapping.

In your case we take $m_1=k, m_2=k-1, p_1=p, p_2=p^*$. Then we have $$k-\frac{n}{p}>k-1-\frac{n}{p*} \Leftrightarrow \frac{1}{n}>\frac{1}{p}-\frac{1}{p^*}.$$ But by assumption you have an equality there and because of that compactness of the embedding it not true (but the continuity still holds). See here and here for an counter-example in the case $k=1,p=2$ that Rellich's theorem is indeed sharp.

Sometimes one also sees the version of Rellich's theorem that $W^{m_1,p_1}$ is compactly embedded in $W^{m_2,p_2}$ if $p_1 \in [1,n)$ and $p_2 \in [1,\frac{np}{n-p}]$. So as before, we notice that the Sobolev conjugate is excluded.