Two metric spaces are bi-Lipschitz equivalent if there is a bijection and a constant such that
for all . In other words, must be Lipschitz with a Lipschitz inverse; this is asking more than just a homeomorphism (continuous with continuous inverse).
For example, the graph is not bi-Lipschitz equivalent to a line. The cusp is an obstruction:
Indeed, the points are separated by and lie at distance from it. So, the images would lie on opposite sides of , at distance from it. But then the distance between and would be of order , contradicting .
There are other pairs of spaces which are homeomorphic but not bi-Lipschitz equivalent, like a circle and Koch snowflake.
What is the simplest example of two spaces that are not known to be bi-Lipschitz equivalent? Probably: the unit ball and the unit sphere in an infinite-dimensional Hilbert space.
In finite dimensions these are not even homeomorphic: e.g., the sphere has a self-homeomorphism without fixed points, namely , while the ball has no such thing due to Brouwer’s fixed point theorem. But in the infinite-dimensional case and are homeomorphic. Moreover, there exists a Lipschitz map such that the displacement function is bounded below by a positive constant: no hope for anything like Brouwer’s fixed point theorem.
It’s hard to see what an obstruction to bi-Lipschitz equivalence could be: there are no cusps, nothing is fractal-like, dimensions sort of agree (both infinite), topologically they are the same… Here is a possible direction of attack (from Geometric Nonlinear Functional Analysis by Benyamini and Lindenstrauss). If a bi-Lipschitz map exists, then the composition is a Lipschitz involution of with displacement bounded from below. So, if such a map could be ruled out, the problem would be answered in the negative. As far as I know, it remains open.