# Domains of Lipschitz class are domains of type A.

Let $\Omega\subset\mathbb{R}^n$ be an open bounded set. We say that $\Omega$ is of type $A$ if there exists a constant, $A$, such that \begin{equation} |\Omega\cap B_{\rho}(x_0)|\geq A\rho^n \end{equation} for ever ball $B_{\rho}(x_0)$, with $x_0\in\Omega$ and $0<\rho<\text{diam }\Omega$.

I am trying to show that if $\Omega$ is of Lipschitz class then it is of type $A$.

Let $B_{\rho}(x)$ be a ball with $x\in\Omega$ and $0<\rho<\text{diam }\Omega$.

Lipschitz domains observe the cone property. This means that there exists a cone, $C$, such that for all $x\in\Omega$, $C$ is congruent to the cone with vertex at $x$ denoted as $C_x$ and $C_x\subset\Omega$.

Suppose that $\text{diam }C\geq\rho$. Then, the cone, $\tilde{C}_{x}\equiv C_{x}\cap B_{\rho}(x)\subset \Omega(x, \rho)$, is congruent to $C\cap B_{\rho}(x)$ and \begin{equation*} 0<\theta\alpha(n)\rho^n\equiv A_1\rho^n=|\tilde{C}_{x}|< |\Omega(x, \rho)|, \end{equation*} for some fixed $0<\theta<1$. If, on the other hand, $\text{diam }C<\rho$ then $\tilde{C_x}=C_x\subset \Omega (x, \rho)$ and for $A_2=(\text{diam }\Omega)^n|C|^{-1}>\rho^n|C|^{-1}$, we have \begin{equation*} \rho^nA_2^{-1}\leq |C_x|\leq |\Omega(x, \rho^n)| \end{equation*}So if we let $A=\min\{A_1, A_2^{-1}\}$ then we find that for any ball $B_{\rho}(x)$ with $x\in\Omega$ and $0<\rho<\text{diam }\Omega$ we have \begin{equation*} |\Omega(x, \rho)|\geq A\rho^n.\end{equation*}

• Thanks...I'll try to fix my proof or come up with a new one. – Nirav Feb 13 '14 at 8:33
• I have a new proof now. – Nirav Feb 16 '14 at 7:11

Your computation of the volume of $\tilde C_{x_k}$ implicitly assumes that $C_{x_k}$ is big enough so that its intersection with $B_{\rho_k}(x_k)$ has lateral height $\rho_k$. But this is only true if $\rho_k$ is less than or equal to the lateral height of cone $C$. You should also consider the other case, when $C_{x_k}$ is entirely contained in $B_{\rho_k}(x_k)$. Here you actually need the given bound on $\rho$: $$|\Omega \cap B_{\rho_k}(x_k) |\ge |C| \ge A(\operatorname{diam}\Omega)^n\ge A\rho_k^n$$ where $A$ is simply chosen to be $(\operatorname{diam}\Omega)^n/|C|$.
Also, the inequality $\dots <\alpha(n)\frac{(\text{diam }\Omega)^n}{k}$ and its consequence $|\Omega(x_k,\rho_k)|\to 0$ are not needed (or used) in the present version of the proof.
• If $A=(\text{diam }\Omega)^n |C|^{-1}$ then shouldn't your last inequality read \begin{equation}\alpha(n)\frac{\rho_k^n}{k}\geq|\Omega\cap B_{\rho_k}(x_k)|\geq |C|>A^{-1}\rho_k^n\quad\forall\ k\in\mathbb{N}\end{equation} and then conclude that the inequality fails for $k$ large enough? – Nirav Feb 18 '14 at 2:01
• By the way thanks for the feedback. I did forget to consider the case that the cone $C$ lies in the ball. – Nirav Feb 18 '14 at 2:06
• @Nirav Those are details. By the way -- if you re-read your proof, you'll see there is no actual argument by contradiction there: you just give a lower estimate on $|\Omega \cap B_\rho(x)|/\rho^n$ (and I would write the proof in this direct way myself.) – user129131 Feb 18 '14 at 4:56