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.

Is there any infinite dimensional Banach space where the unit ball and unit sphere are definitively known or known not to be bi-Lipschitz equivalent?

According to Benyamini’s 2002 article/book chapter “Introduction to the Uniform Classification of Banach Spaces”, the only space for which the answer is known is the Gowers-Maurey example of a separable reflexive Banach space that is not isomorphic to any proper subspace. For it the answer is negative, because a bi-Lipschitz equivalence B->S would have a point of Gateaux differentiability, and the derivative would yield an isomorphism of the space onto proper subspace. See page 10 in https://books.google.com/books?id=_j_0ojpk_RIC

In l2(R), map an element of B, x = (x1, x2, x3, …) to (sqr(1 – norm(x)), x1, x2, x3, …) in S. norm(f(x) – f(x)) = K + norm(x – y) <= norm(x – y) where 0 <= K = 2 – norm(x) – norm(y) -2 * sqr(1 + norm(x) * norm(y) – norm(x) – norm(y)) S.

This is not a bijection, though. The image is a hemisphere, not a sphere.