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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.

Evangelopoulos Foivos
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1 Answers1

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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 $\Omega_i = [i, i+2]$.

gerw
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  • Brilliant! Could you provide me with a reference to the fact that Lipschitz functions act as multiplies on $H_0^1$? – Evangelopoulos Foivos May 25 '21 at 13:48
  • This follows from the product rule: $\nabla (u v) = u \nabla v + v \nabla u$, Hölder's inequality and the fact that the Lipschitz functions coincide with $W^{1,\infty}(\Omega)$ (for $\Omega$ nice enough). – gerw May 25 '21 at 13:56
  • I see. However this relies to the fact that $C^{0,1} (Ω) = W^{1,\infty}(Ω)$ for very nice $Ω$ (like quasiconvex). What happens when $Ω$ is not that nice? Do you think the statement is still true? – Evangelopoulos Foivos May 25 '21 at 14:11
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    This should also hold for Lipschitz domains, i.e., it is not very restrictive. Moreover, we only need that Lipschitz functions belong to $W^{1,\infty}(\Omega)$ and this holds without any assumption on $\Omega$. – gerw May 25 '21 at 15:36