1
$\begingroup$

Let $\Omega$ be a domain in $\mathbb{R}^n$. Then I can find a bounded subdomain $\Omega_0$ in $\Omega$ with $r=\mathrm{dist}(\Omega_0,\partial\Omega)>0$. Here domain means an open and connected subset of $\mathbb{R}^n$.

I want to find a bounded Lipschitz domain $\Omega_1$ in $\mathbb{R}^n$ which contains $\Omega_0$ and contained in $\Omega$.

I have a rough idea about the proof, but I'm not satisfying this strategy because I cannot make a rigorous proof based on this idea.

Cover $\Omega_0 \cup \partial \Omega_0$ by a family of open cubes whose diameter is less than $r$. Then by compactness, there exists a finite family $\mathcal{F}$ of cubes which covers $\Omega_0 \cup \partial \Omega_0$. Enlarge each $Q$ in $\mathcal{F}$ so that for any $Q\in \mathcal{F}$, there exists $Q'\in\mathcal{F}$ with $Q\neq Q^\prime$ such that $Q\cap Q' \neq \varnothing$.

Define $\Omega_1$ be the union of the above cubes. Then $\Omega_1$ is clearly open and connected. So it is a domain in $\mathbb{R}^n$.

Is union of two Lipschitz domains (with nonemtpy intersection) is Lipschitz? If so, then since cubes are Lipschitz domain and $\Omega_1$ is a finite union of Lipschitz domains, so I can prove the claim.

It seems obvious in 2D by picture but I cannot make any progress. One of my trial was changing parameter when two domain meets.

I can find such a domain with $C^1$-boundary using Sard's lemma. So the existence is clear. But I want to find different method based on elementary arguments.

Thanks for help.

$\endgroup$
1
$\begingroup$

It is not true. Take two squares in $\mathbb{R^2}$ that touch only at one corner, say $Q_1=(0,1)\times (0,1)$ and $Q_2=(1,2)\times (1,2)$. Then each cube is Lipschitz but the union is not, since at the point $(1,1)$ you cannot write the boundary as the graph of a function.

$\endgroup$
  • $\begingroup$ Thanks you for your answer. But that case is not counterexample of my claim since $Q_1 \cap Q_2 =\varnothing$. I want to make two domain into one domain by overlapping. $\endgroup$ – Will Kwon Jul 4 '17 at 2:40
  • $\begingroup$ but it is easy to modify this example to make the two domains overlap. You just take an handle that connects $Q_1$ with $Q_2$ away from $(1,1)$. $\endgroup$ – Gio67 Jul 4 '17 at 2:46
  • $\begingroup$ Thanks. I didn't think that. $\endgroup$ – Will Kwon Jul 4 '17 at 2:49
  • 1
    $\begingroup$ but if the cubes meet along faces (instead of corners), then I think that their union would be Lipschitz, so your construction could work. It's tricky though. Maybe working with dyadic cubes might help. en.wikipedia.org/wiki/Dyadic_cubes $\endgroup$ – Gio67 Jul 4 '17 at 2:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.