Sobolev space on composite domain

Question

Let $\Omega_1$ and $\Omega_2$ be two sufficiently smooth domains in $\mathbb R^2$. Consider the composite domain $\Omega =Ω_1 \cup Ω_2$. Also, consider the sobolev space $H_0^1 (Ω)$. Let $ Y_1 = H_0^1(Ω_1) $ and $ Y_2 = H_0^1(Ω_2)$. View $Y_1$ and $Y_2$ as closed subspaces of $H_0^1(Ω)$ (by extending functions on $Ω$ to be zero.) I am looking for a reference to the fact that $Y_1 + Y_2$ is dense in $H_0^1 (Ω)$.

Thanks for any help.

Answer 1

Here is a proof in the case that $\Omega_1$ and $\Omega_2$ are open: Since $C_c^\infty(\Omega)$ is dense in $H_0^1(\Omega)$, it is sufficient to consider $\varphi \in C_c^\infty(\Omega)$. Further, we define $$ d_i(x) := \operatorname{dist}(x, \partial\Omega_i).$$

Then, $$ \varphi_i := \varphi \frac{d_i}{d_1 + d_2} \in H_0^1(\Omega_i).$$ This follows from:

• $d_1 + d_2$ is uniformly positive on the compact support $K$ of $\varphi$
• thus, $d_i/(d_1+d_2)$ is Lipschitz on $K$
• Lipschitz functions act as multipliers on $H_0^1(\Omega)$

Moreover, $\varphi = \sum_{i = 1}^2 \varphi_i$.

$%this is some vertical spacing$

In the case that $\Omega_1$ or $\Omega_2$ are not open, it might not be possible to extend $H_0^1(\Omega_i)$ functions by $0$ to $H_0^1(\Omega)$ functions: Consider $\Omega_i = [i, i+2]$.