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I have a question about a property of Lipschitz domain.

Let $D \subset \mathbb{R}^d$ be a bounded domain (connected open subset ). $D$ is called a bounded Lipschitz domain if there exist positive constants $\delta$, $M$ such that for each $x_0 \in \partial \Omega$ there exist a neighborhood $U_{x_0}$ of $x_0$, local coordinates $y=(y',y_d) \in \mathbb{R}^{d-1} \times \mathbb{R}$, with $y=0$ at $x_0$, and a Lipschitz continous function $f_{x_0}:\mathbb{R}^{d-1} \to \mathbb{R}$, such that \begin{equation*} D \cap U_{x_0}=\{(y',y_N) \in D \cap U_{x_0} \mid y' \in B(0,\delta),\, y_N>f(y') \},\quad \text{Lip}(f) \le M, \end{equation*} where we define $\text{Lip}(f)=\inf \{L \ge 0 \mid |f(x)-f(y)| \le L|x-y|,\, x,y \in B(0,\delta) \}$.

In the above definition, $B(0,\delta)$ denotes the open ball in $\mathbb{R}^d$ centered at the origin with radius $1$.

My question

We denote by $m$ the Lebesgue measure on $D$.

Let $D \subset \mathbb{R}^d$ be a bounded Lipschitz domain. Then, can we show the following?

There exists a positive constant $C \ge 1$ such that $$C^{-1}r^d \le m(\bar{D} \cap B(x,r)) \le Cr^d$$ for any $x \in \bar{D}$, $r \in (0,\text{diam}(D)]$.

It is clear that $m(\bar{D} \cap B(x,r)) \le Cr^d$ for any $x \in \bar{D}$, $r \in (0,\text{diam}(D)]$.

If you know a proof or references, please let me know.

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  • $\begingroup$ You can show (more or less directly from your boxed definition) that $D$ satisfies a uniform interior cone condition, which implies your lower bound in a straightforward way. $\endgroup$ – user101142 Aug 19 '18 at 17:44
  • $\begingroup$ I'm sorry but I do not know the uniform interior cone condition well. Can you give me detailed proof? $\endgroup$ – sharpe Aug 19 '18 at 18:00
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This is probably somewhat over-the-top, but anyway: The nice paper [1] by Hajlasz, Koskela and Tuominen says that your desired inequality is true for Sobolev extension domains, so for domains $D$ for which there exists a continuous linear operator $E \colon W^{1,p}(D) \to W^{1,p}(\mathbb{R}^n)$ such that $(Eu)_{\restriction D} = u$. It is a classical result that a Lipschitz domain is such an extension domain, e.g. Theorem 1.4.3.1 in Grisvard [2].

[1] Hajłasz, Piotr; Koskela, Pekka; Tuominen, Heli, Sobolev embeddings, extensions and measure density condition, J. Funct. Anal. 254, No. 5, 1217-1234 (2008). ZBL1136.46029.

[2] Grisvard, Pierre, Elliptic problems in nonsmooth domains, Classics in Applied Mathematics 69. Philadelphia, PA: Society for Industrial and Applied Mathematics (SIAM) (ISBN 978-1-611972-02-3/pbk). xx, 410 p. (2011). ZBL1231.35002.

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  • $\begingroup$ Thank you for your comment. That became an extremely useful resource for me. $\endgroup$ – sharpe Aug 19 '18 at 19:08

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