Statement: Let $(X,d)$, $(Y,r)$ be homeomorphic compact metric spaces. Then, there exists a bilipschitz homeomorphism between the two.

Problem: As I have no clue whether the statement is true or false (I'm not well acquainted with the theory of lipschitz functions), any proof/counterexample is welcome.

Rephrasing: Define the Lipschitz distance of two metric spaces as $$d_L(X,Y)\doteq \inf\big|\ln \mathrm{dil}(f)+\ln \mathrm{dil}(f^{-1})\big|$$ where $\mathrm{dil}(f)$ is the dilatation of $f$ and the infimum is taken over all bilipschitz homeomorphisms between X and Y. The question is now whether the Lipschitz distance between two homeomorphic compact metric spaces is always finite.

Comments: It is clear to me that not all homeomorphisms between compact metric spaces are bilipschitz: $x^3\colon [-1,1]\rightarrow [-1,1]$ is a counterexample, which, although trivial, suggests that simply scaling the metrics won't do the trick. Also, any homeomorphism between compact metric spaces is uniformly bicontinuous, thus any such spaces are uniformly isomorphic (where, say à la Weil, uniformities are induced by the respective metrics), which is encouraging towards an affirmative answer, yet it seems difficult to prove (disprove) the existence of any bilipschitz homeomorphism given that the spaces are effectively homeomorphic.

As I would like the statement to be true, I was wondering the following: every compact metric space is complete and second countable ($\Rightarrow$ separable), i.e. Polish, so it is embeddable in Tikhonov's cube $T\doteq I^\mathbb N$. Is there any property of $\mathrm{Homeo}(T)$ that could be of help here?


1 Answer 1


This is hopelessly false. Bilipschitz maps preserve a lot more structure than homeomorphisms, Hausdorff dimension for one thing.

Example 1: equip $[0,1]$ with the metric $d(x,y)=|x-y|^{1/2}$. The topology stayed the same, the Hausdorff dimension changed (it is now $2$). Hence, the new space is not biLipschitz equivalent to the original one.

Example 2, more subtle. The planar set $\{(x,\sqrt{|x|}) : -1\le x\le 1\}$ is homeomorphic to interval $[-1,1]$, but not biLipschitz equivalent to it. The reason is the cusp at the origin: biLipschitz maps cannot form such a thing. Indeed, biLipschitz image of a line segment is a chord-arc curve: there is a constant $M$ such that for every two points $a,b$ on the curve the arc between them has length at most $M|a−b|$ (with $ |a−b| $ being the Euclidean distance.) This property fails for the pair $(\pm \epsilon,\sqrt{\epsilon})$.


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.