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*}

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

The proof is basically correct now, with a couple of flaws that can be fixed.

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.

  • $\begingroup$ 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? $\endgroup$ – Nirav Feb 18 '14 at 2:01
  • $\begingroup$ By the way thanks for the feedback. I did forget to consider the case that the cone $C$ lies in the ball. $\endgroup$ – Nirav Feb 18 '14 at 2:06
  • 1
    $\begingroup$ @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.) $\endgroup$ – user129131 Feb 18 '14 at 4:56

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.