Compactness of Sobolev spaces

Question

I have a question about Sobolev spaces.

I am reading A First Course in Sobolev Spaces by Giovanni Leoni. In this book, a sufficient and necessary condition is stated (Exercise 11.18):

Let $\Omega \subset \mathbb{R}^d$ be an open subset and let $1 \le p <\infty$. The embedding \begin{equation*} W^{1,p}(\Omega) \to L^{p}(\Omega) \end{equation*} is compact if and only if \begin{equation*} \lim_{n \to \infty} \sup \left\{ \int_{\Omega \setminus \Omega_n}|u|^p\,dx :\|u\|_{W^{1,p}(\Omega)} \le 1 \right\}=0, \end{equation*} where $\Omega_n=\left\{x \in \Omega: \text{dist}(x,\partial \Omega)>1/n, |x|<n \right\}$.

I would like to know the solution of this exercise. If you know references and papers related to this exercise, please let me know.

Thanks in advance.

Answer 1

• Case $1 \le p < N$: let $\{u_n\} \subset B_{W^{1,p}(\Omega)}(0,1)$, we need to show that there exist a subsequence $\{u_{n_k}\}$ and a function $u \in L^p(\Omega)$ such that $u_{n_k} \to u$ in $L^p(\Omega)$. The main difficulty is that we cannot apply directly the Rellich-Kondrachov Theorem since we don't have any information about the regularity of the domain $\Omega$. Here is a way to around this problem:

Fact: (I'll leave this as an exercise -insert diabolic laughter-) Let $B$ be an open set and let $A \subset B$ such that $\text{dist}(A,\partial B) \ge \eta$, for some fiven $\eta > 0$. Then There exists an open set $C$ of class $C^1$ such that $A \subset C \subset B$ and $\text{dist}(C,\partial B) > \eta/2.$

Consider $\Omega_1 \subset \Omega$ as in the statement of the problem and find $U_1$ open and of class $C^1$ as above. Applying the RKT to $\{u_n\}$ restricted to $U_1$ we can find a subsequence $\{u_n^1\} \subset \{u_n\}$ and $u^1 \in L^p(U_1)$ such that $u_n^1 \to u^1$ in $L^p(U_1)$. Now consider $\Omega_2$ and find $U_2$ as above. Then you can apply the RKT in $U_2$ to $\{u_n^1\}$ to get a further subsequence $\{u_n^2\}$ and a function $u^2$ such that $u_n^2 \to u^2$ in $L^p(U_2)$. Notice that by uniqueness of the weak limit, $u^1 = u^2$ a.e. in $U_1 \cap U_2$. By a standard diagonal procedure, $v_n = u_n^n$ converges to a function $u$ in $L^p(u_n)$ for every $n$. $$ \begin{align} \|v_n - u\|_{L^p(\Omega)} = &\ \|v_n - u\|_{L^p(\Omega_n)} + \|v_n - u\|_{L^p(\Omega \setminus \Omega_n)} \\ \le &\ \|v_n - u\|_{L^p(U_n)} + \|v_n\|_{L^p(\Omega \setminus \Omega_n)} + \|u\|_{L^p(\Omega \setminus \Omega_n)} \\ \to &\ 0 \ \text{as } n \to \infty. \end{align} $$

• Case $p > N$: I'll show the result for a regular bounded domain $U$, the conclusion then follows by a diagonal argument exactly as above.

We have the following facts, which again I'll leave for you to check:

1. $W^{1,p}(U) \hookrightarrow C^{0,1-\frac{N}{p}}(\bar{U})$.
2. $C^{0,1-\frac{N}{p}}(\bar{U}) \hookrightarrow \hookrightarrow C^{0,\alpha}(\bar{U})$ for every $0 < \alpha < 1 - \frac{N}{p}$.
3. $C^{0,\alpha}(\bar{U}) \hookrightarrow L^p(U)$.

Since the composition of a compact operator with a continuous operator is compact the result follows by combining the embeddings in 1,2,3.

• Case $p = N$: As in the previous case, let $U$ be a bounded regular domain (the diagonal argument can then be applied). For $q < N$ we have $$W^{1,N}(U) \hookrightarrow W^{1,q}(U) \hookrightarrow\hookrightarrow L^s(U), \quad s < q^*.$$ To conclude the proof it is enough to notice that for $q > \frac{N}{2}$, $q^* > N$.

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Proof of necessity: assume by contradiction that $$\limsup_{n \to \infty} \sup_{u \in B(0,1)}\int_{\Omega \setminus \Omega_n}|u|^p = \eta > 0.$$ Then we can find $\bar{n}$ and $\{u_n\}$ such that if $n \ge \bar{n}$ $$\int_{\Omega \setminus \Omega_n}|u_n|^p \ge \frac{\eta}{2}.$$ By assumption $W^{1,p}(\Omega) \hookrightarrow \hookrightarrow L^p(\Omega)$ then $\{u_n\}$ has subsequence (not relabeled) that converges. To get a contradiction we can show that this subsequence is not Cauchy.

To this end, fix $n$ and consider $u_n \cdot \chi_{\Omega \setminus \Omega_m}$. By Lebesgue's Dominated Convergence Theorem, $u_n \cdot \chi_{\Omega \setminus \Omega_m} \to 0$ in $L^p$ as $m \to \infty$. Then there exists $\bar{m}$ such that if $m \ge \bar{m}$ $$\int_{\Omega \setminus \Omega_m}|u_n|^p \le \frac{\eta}{4}.$$ Then for $m,n$ large we have $$ \begin{align} \|u_n - u_m\|_{L^p(\Omega)} \ge &\ \|u_n - u_m\|_{L^p(\Omega \setminus \Omega_m)} \\ \ge &\ \|u_m\|_{L^p(\Omega \setminus \Omega_m)} - \|u_n\|_{L^p(\Omega \setminus \Omega_m)} \\ \ge &\ \frac{\eta}{2} - \frac{\eta}{4} = \frac{\eta}{4} > 0. \end{align} $$