Suppose we have a domain $\Omega\subset \mathbb{R}^n$ which is homeomorrphic to the unit ball $B(0,1)\subset \mathbb{R}^n$ and such that $\partial \Omega$ is of class $C^1$ (technically, this means that for every point in the domain we can give a $C^1$-diffeomorphism to the half-space in dimension $n$).

A map $f\colon (X,d_X)\to (Y,d_Y)$ between metric spaces is said to be bi-Lipschitz if there is a constant $K>0$ such that $$ \frac{1}{K} d_X(x_1,x_2) \leq d_Y(f(x_1), f(x_2)) \leq K d_X(x_1,x_2)$$ for all $x\in X$. The spaces $X$ and $Y$ are said to be Lipschitz equivalent if there is a surjective bi-Lipschitz map between them (any bi-Lipschitz map is necessarily injective).

I wonder if in this case the domain $\Omega\subset \mathbb{R}^n$ is Lipschitz equivalent to the unit ball $B(0,1)\subset \mathbb{R}^n$. If true, is the condition $C^1$ necessary, sufficient?

EDIT: My original question did not assume that the domain was homeomorphic to the ball. However, in this case it is clearly false, so I added this condition.

  • $\begingroup$ You need the domain to be homeomorphic to the ball if you want bi-Lipschitz equivalence, and simply connectedness is not enough if $n>2$. A small neighborhood of the unit sphere is simply connected but not homeomorphic to the ball. $\endgroup$ – Joonas Ilmavirta May 24 '15 at 13:49
  • $\begingroup$ But a small neighbourhood of the sphere would not be open if I understand your example? $\endgroup$ – Mauricio G Tec May 24 '15 at 13:56
  • $\begingroup$ I meant something like the set $A=\{x\in\mathbb R^n;0.9<|x|<1.1\}$, which is open, bounded, smooth and (if $n>2$) simply connected but not homeomorphic to the ball. $\endgroup$ – Joonas Ilmavirta May 24 '15 at 14:02
  • $\begingroup$ Yes I understood it now. So it's clearly false then. $\endgroup$ – Mauricio G Tec May 24 '15 at 14:03
  • $\begingroup$ Yes, but it is a more interesting problem whether a bounded $C^1$ domain which is homeomorphic to the ball is always bi-Lipschitz equivalent to it. $\endgroup$ – Joonas Ilmavirta May 24 '15 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.