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I have a question about domains of $\mathbb{R}^d$.

I am reading A First Course in Sobolev Spaces by Giovanni Leoni. This book introduces the following concept to describe the regularity of the boundary of a domain.

(Definition 12.9 in Leoni's book) The boundary $\partial \Omega$ of an open set $\Omega \subset \mathbb{R}^N$ is $locally $ $Lipschitz$ if for each $x_0 \in \partial \Omega$ there exist a neighborhood $A$ of $x_0$, local coordinates $y=(y',y_N) \in \mathbb{R}^{N-1} \times \mathbb{R}$, with $y=0$ at $x=x_0$, a Lipschitz function $f:\mathbb{R}^{N-1} \to \mathbb{R}$, and $r>0$, such that \begin{equation*} \Omega \cap A=\{(y',y_N) \in \Omega \cap A: y' \in Q(0,r), y_N>f(y') \}, \end{equation*} where $Q(0,r)$ is the open ball centered at the origin with radius $r>0$.

My question

In the above definition, $f$ and $r>0$ depend on $x_0$. Therefore, we should write $f=f_{x_0}$, $r=r_{x_0}$.

Can we show the following?

Let $K \subset \partial \Omega$ be a compact subset. Then, $\inf_{x_0 \in K}r_{x_0}>0$ and there exists a positive number $M$ such that \begin{equation*} \sup_{x_0 \in K}\text{Lip}(f_{x_0})\le M, \end{equation*} where $\text{Lip}(f)$ is the Lipschitz constant of $f$.

I think this holds. But I couldn't prove... If you have a nice idea, please let me know.

Thanks in advance.

ADD

That is, I want to prove the following:

There exist positive numbers $r_K$ and $L_{K}$ such that for each $x_0 \in K$ there exist a neighborhood $A$ of $x_0$, local coordinates $y=(y',y_N) \in \mathbb{R}^{N-1} \times \mathbb{R}$, with $y=0$ at $x=x_0$, a Lipschitz function $f:\mathbb{R}^{N-1} \to \mathbb{R}$ with $\text{Lip}(f) \le L_{K}$ and \begin{equation*} \Omega \cap A=\{(y',y_N) \in \Omega \cap A: y' \in Q(0,r_K), y_N>f(y') \}, \end{equation*} where $Q(0,r_K)$ is the open ball centered at the origin with radius $r_K>0$.

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  • $\begingroup$ Who is to say that r is necessarily a continuous function of $x_0$? $\endgroup$ – Ian Jul 12 '17 at 9:35
  • $\begingroup$ You mean we need additional assumptions to show $\inf_{x_{0} \in K}r_{x_0}>0$? $\endgroup$ – sharpe Jul 12 '17 at 10:00
  • $\begingroup$ Just saying that $K$ is compact does not imply that every function on it attains its infimum. The function needs to have some amount of regularity for that. $\endgroup$ – Ian Jul 12 '17 at 10:31
  • $\begingroup$ Do you have an example of locally Lipschitz domain which does not satisfy $\inf_{x_0}r_{x_0}>0$? $\endgroup$ – sharpe Jul 12 '17 at 10:36
  • $\begingroup$ I think domains which have smooth boundary satisfy the conditions $\inf_{x \in K}r_x$ and $\sup_{x \in K} \text{Lip}(f_{x_0})$ \le M. $\endgroup$ – sharpe Jul 12 '17 at 10:53
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For every point $x_0\in K$ you have an open neighborhood $A_{x_0}$ that contains $x_0$. Hence $K\subset\cup_{x_0\in K}A_{x_0}$ and so by compactness you can find $n_K$ points $x_1, \ldots, x_{n_K}\in K$ such that $K\subset\cup_{i=1}^{n_K} A_{x_i}$. So now take $L_K=\max_{i=1,\ldots,{n_K}} L_{x_i}$ and $r_K=\min_{i=1,\ldots,{n_K}}$. If $x\in K$ then there exists $i\in\{1,\ldots,{n_K}\}$ such that $x\in A_{x_i}$.

EDIT Instead of working with cubes use balls (it does not make any difference). For each $x_0\in K$ find a ball $A_{x_0}=B(x_0,r_{x_0})$. Hence $K\subset\cup_{x_0\in K}B(x_0,\frac{r_{x_0}}2)$ and so by compactness you can find $n_K$ points $x_1, \ldots, x_{n_K}\in K$ such that $K\subset\cup_{i=1}^{n_K} B(x_i,\frac{r_{x_i}}2)$. So now take $L_K=\max_{i=1,\ldots,{n_K}} L_{x_i}$ and $r_K=\min_{i=1,\ldots,{n_K}}\frac{r_{x_i}}2$. If $x_0\in K$ then there exists $i\in\{1,\ldots,{n_K}\}$ such that $x_0\in B(x_i,\frac{r_{x_i}}2)$. Then $B(x_0,r_{K}) \subset B(x_i,r_{x_i})$. You can take as function $f$ the function $f_{x_i}$ but you will need to make a translation to change the origin into $x_0$. You can use the same idea with cubes, it is just more cumbersome to write down.

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  • $\begingroup$ How about $\inf_{x \in K}r_x>0$? $\endgroup$ – sharpe Jul 15 '17 at 8:35
  • $\begingroup$ You can do the same. Just take $r_K=\min_{i=1,\dots,n}r_{x_i}>0$. $\endgroup$ – Gio67 Jul 15 '17 at 11:03
  • $\begingroup$ I cannot understand that it is proof. Please give me a proof. $\endgroup$ – sharpe Jul 15 '17 at 11:17
  • $\begingroup$ I just edited it. Since you can cover $K$ with only finitely many $A_{x_i}$ for each $x_0\in K$ you must be in one of these finitely many $A_{x_i}$ . $\endgroup$ – Gio67 Jul 15 '17 at 11:24
  • $\begingroup$ Let $x_0 \in K$. From your argument, there exists a $i=1,\ldots,n$ such that $x_0 \in A_{x_i}$. Therefore, there exists a local coordinates $y=(y',y^N)$ with $y=0$ at $x=x_i$, a Lipschitz function $f$ such that $\Omega \cap A_{x_i}=\{(y',y^N) \in \Omega \cap A_{x_i}: y'\in Q(0,r_{x_i}), y^N >f(y')\}$. $\endgroup$ – sharpe Jul 15 '17 at 11:30

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