Extension of Sobolev functions on non-connected set

Question

Let $D = D_1 \cup D_2 \cup...\cup D_m$ be a subset of $\mathbb{R}^d$, where each $D_j$ is a (connected and) bounded domain with Lipschitz boundary, and $\min_{i,j} dist(D_i,D_j) \geq c > 0$ for some $c$. Let a function $f$ defined on $D$. Is there a Sobolev extension theorem for $f$ from $D$ to $\mathbb{R}^d$?

Answer 1

We can indeed extend Sobolev functions suitably on such a domain. I assume you know already that we can do this for connected, bounded Lipschitz domains and want to extend this to such domains you described.

In this case, extend for every $i$ the functions $f_{|D_i}$ to $\mathbb{R}^d$, and denote this extension by $E_i(f)$. Now, choose smooth cutoff functions $\phi_i\in C_c^\infty(B_{c/3}(D_i))$ such that $\phi_i\equiv 1$ on $D_i$. ($B_{c/3}(D_i)$ is all the points having distance at most $c/3$ away from $D_i$). Then, the extension by zero of each of the $\phi_i E(f_i)$ is a Sobolev function on the full space.

By assumption on the separation of the subdomains, $B_{c/3}(D_i)$ and $B_{c/3}(D_j)$ are still well-separated for $i\neq j$. Then we set $$ E(f)=\begin{cases}{\sum_{i=1}^m E_i(f)(x)\phi_i}(x) & \text{if }x\in\bigcup_i B_{c/3}(D_i) \\ 0\quad &\text{else}\end{cases}. $$ This should give you the result (note that for estimating the norm of this extension, any norm of the $\phi_i$ can be absorbed into the constant, and each of the extensions on the smaller domain depends continuously on the norm of $f_{|D_i}$ and therefore on $f$).