# Function extension in a Sobolev space

Let $\Omega$ be a domain of $R^n$ and let $H^2(\Omega)$ be the usual Sobolev space.

Let $\emptyset\ne \omega_1\subset\omega_2$ be open subsets of $\Omega$, and let $\theta \in H^2(\omega_1)$.

I am wondering about the existence of a function $\tilde{\theta} \in H^2(\Omega)$ such that :

1) $\tilde{\theta}=\theta$ on $\omega_1$,

2) $\tilde{\theta}$ is constant on $\Omega-\omega_2$.

Thanks!

-
add comment

## 1 Answer

The answer is yes, with the caveat indicated in the comment below. Consider an arbitrary extension $\hat{\theta}\in H^2(\Omega)$. Now choose a compactly supported smooth function $\eta$ such that

$${\rm supp}\;\eta\subset \Omega\setminus \omega_2,\;\;\eta\equiv 1 \;\;\mbox{on \omega_1}.$$

The function $\tilde{\theta}=\eta\cdot \hat{\theta}$ has the properties you asked for.

-
Provided that the function $\theta \in H^2(\omega_1)$ has a Sobolev-extension to begin with. –  Tapio Rajala Jul 13 '12 at 14:07
If there is no extension, then the answer is obviously negative. However extensions do exist under mild regularity assumption. See e.g. Adams' book on Sobolev spaces. –  Liviu Nicolaescu Jul 13 '12 at 14:36
add comment