# On the embedding of a Sobolev space on a unbounded domain

I have a question about a Sobolev space.

Let $D \subset \mathbb{R}^d$ be a connected open subset with smooth boundary. When $D$ is bounded, Relich's theorem states that $W^{1,2}(D)$ (Sobolev space on $D$ with Neumann boundary condition) is compactly embedded in $L^2(D)$. But I am interested in the case $D$ is unbouned.

Question

When $D=\{(x,y) \in \mathbb{R}^2: y>0 \}$, can we show that $W^{1,2}(D)$ is compactly embedded in $L^2(D)$ or $W^{1,2}_{0}(D)$ is compactly embedded in $L^2(D)$ ? ($W_{0}^{1,2}(D)$ is the Sobolev space on $D$ with Dirichlet boundary condition)

If you know related research, please let me know.

No. Let $\phi\in C^\infty_0(B_1)$ where $B_1$ is the ball of radius 1.
Let $\phi_n(x,y) = \phi(x,y - 10n)$. By construction $(\phi_n)$ is a sequence of functions in $C^\infty_0(D)$ with bounded $W^{1,2}$ norm, but has no converging subsequence in $L^2$.
• Thank you for your reply. I understand $\{\phi_n\}$ is bounded in $W^{1,2}(D)$ (also in $W_{0}^{1,2}(D)$). Therefore, there exists $\psi$ and $\{n_k\} \subset \{n\}$ such that $\phi_{n_{k}} \to \psi$ weakly in $W^{1,2}(D)$. But what is $\psi$? – sharpe Aug 9 '16 at 17:35
• $\psi=0$! the mother of all weak limits – daw Aug 9 '16 at 19:35
• I've almost understood. Because $\{\phi_n\}$ is bounded in $W_{0}^{1,2}(D)$ (also in $L^2(D)$), there exists subsequence $\{n_k\}$ and $\psi \in L^2(D)$ such that $\phi_{n_k} \to \psi$ weakly in $L^2(D)$. Take $\varphi \in C_{0}^{\infty}(D)$ arbitrary. Then, $\int_{D} \phi_{n_{k}} \varphi\,dx=0$ for large ebough $k$. This implies $\int_{D}\psi \varphi\,dx=0$. Therefore $\psi=0$. Is this correct? – sharpe Aug 10 '16 at 6:38