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In this article and in the book of Ladyzhenskaya et al (1968) - Linear and Quasilinear Elliptic Equations we have the following definition of what is a domain of type (A):

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There is no example of a large class that fits into this definition. By $|B_\rho|$ we mean the Lebesgue $N$-dimensional measure of $B_\rho$

My question is the following: A uniform Lipschitz bounded domain (connected, open) domain is a domain of type (A)?

P.S. In the book of Mariano Giaquinta - An introduction to the regularity theory for ellitic systems, harmonic maps and minimal graphs (2012) we have the following information:

enter image description here

Clearly it's not the same of (A) domains but maybe it helps...

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    $\begingroup$ The second condition is actually kind of a complement of the first. (A) effectively says that the density at the boundary is bounded away from 1 and the other that it is bounded away from 0. If you replace $\Omega$ with its complement, you turn one into the other. And both hold rather obviously for Lipschitz domains, as can be shown by looking at a local graph. $\endgroup$
    – mlk
    Commented Aug 22 at 9:57
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    $\begingroup$ @Bogdan See also this question $\endgroup$
    – Hannes
    Commented Aug 22 at 17:14
  • $\begingroup$ Thank you very much! What a nice result! $\endgroup$
    – Bogdan
    Commented Aug 22 at 17:30

2 Answers 2

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$\newcommand\Om\Omega\newcommand\th\theta\newcommand\p\partial$After the previous answer was given, the OP changed "Lipschitz connected" to "uniform Lipschitz". If the latter is understood in the sense of the Wikipedia definition with $r$, $h$, and the Lipschitz constant (say $K$) the same for all points $p\in\p\Om$, then the answer becomes yes.

Indeed, then for each $p\in\p\Om$ there is a unit vector $u$ such that for all $s\in(0,r]$ $$|B_p(s)\cap\Om|\le|B_p(s)\cap C_{p,u,K}|=s^d|B_p(1)\cap C_{p,u,K}| =s^d(1-\th)|B_p(1)|=(1-\th)|B_p(s)|, $$ where $d$ is the dimension of the ambient Euclidean space (say $E$), $|\cdot|$ is the Lebesgue measure, $B_p(s)$ is the open ball in $E$ centered at $p$ of radius $s$, $$ C_{p,u,K}:=\{x\in E\colon u\cdot(x-p)\le K\|\pi_{u^\perp}(x-p)\|\},$$ $\cdot$ is the inner product, $\pi_{u^\perp}$ is the orthoprojector onto the orthogonal complement to $u$, $\|\cdot\|$ is the Euclidean norm (so that $C_{p,u,K}$ is a cone, with vertex $p$), and $\th:=1-|B_p(1)\cap C_{p,u,K}|/|B_p(1)|$, so that $\th$ is a strictly positive real number, which depends only on $K$ (but not on $p$ or $u$).

Thus, we do have $$|B_p(s)\cap\Om|\le(1-\th)|B_p(s)|$$ for all $p\in\p\Om$ and all $s\in(0,r]$.

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  • $\begingroup$ So, it seems that the number of connected components of $B_p(s)\cap\Omega$ has no importance. Then why the author gives the definition in this more complicated way? $\endgroup$
    – Bogdan
    Commented Aug 23 at 5:12
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    $\begingroup$ @Bogdan : I guess the author wanted to also cover the cases such as two open balls that are disjoint but with a common boundary point. $\endgroup$
    – Iosif Pinelis
    Commented Aug 23 at 11:49
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A counterexample: $n=2$, $$\Omega=B^-_{(0,0)}(2)\cup B^+_{(1,0)}(1)\cup B^+_{(-1,0)}(1),$$ where $B^\pm_C(r):=B_C(r)\cap\Pi^\pm$, $B_C(r)$ is the open ball of radius $r$ centered at $C$, and $\Pi^\pm:=\{(x,y)\in\Bbb R^2\colon\pm y\ge0\}$.

Here is a picture of $\Omega$, with the bad point $(0,0)$ on its boundary:

enter image description here

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    $\begingroup$ That is not a Lipschitz-domain though, as no matter how you rotate it, the boundary is not the graph of a Lipschitz-function in any neighborhood of the origin. $\endgroup$
    – mlk
    Commented Aug 22 at 11:48
  • $\begingroup$ @mlk : The domain was said to be, not Lipschitz, but Lipschitz connected, which seems to mean that any two points of it can be connected by a Lipschitz path $\gamma\colon[0,1]\to\Omega$ with the same Lipschitz constant for all such paths. Also, I think the boundary is the graph of a Lipschitz function: parametrize the boundary by the arclength. $\endgroup$
    – Iosif Pinelis
    Commented Aug 22 at 12:03
  • $\begingroup$ It is the image of a Lipschitz function, but not the graph; in general one uses Lipschitz-domains precisely to avoid the type of cusp you have in your example, as they tend to cause all kind of problems in PDE. But you are right, it is not clear if the question meant (Lipschitz connected) or Lipschitz, connected domains. However I have never seen the former anywhere, while the latter is standard and also the one used in the second example. $\endgroup$
    – mlk
    Commented Aug 22 at 12:10
  • $\begingroup$ @mlk : Sorry, I did not read your previous comment carefully enough. On the other hand, I can hardly read "Lipschitz connected, open and bounded" (with a comma already there) as "Lipschitz, connected, open and bounded". Also, what do you mean by "the second example"? $\endgroup$
    – Iosif Pinelis
    Commented Aug 22 at 12:15
  • $\begingroup$ I mean the second quote in the question, which was talking about domains of "class $C^1$ or Lipschitz". And you are right that without the comma the sentence can only be read in one way, but that could have been a typo, so maybe we have to wait for clarification. $\endgroup$
    – mlk
    Commented Aug 22 at 12:28

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