Bi-Lipschitz equivalence and fixed points

Two metric spaces {X,Y} are bi-Lipschitz equivalent if there is a bijection {f:X\rightarrow Y} and a constant {L} such that
{L^{-1}d_X(a,b) \le d_Y(f(a),f(b)) \le L\,d_X(a,b)} for all {a,b\in X}. In other words, {f} must be Lipschitz with a Lipschitz inverse; this is asking more than just a homeomorphism (continuous with continuous inverse).

For example, the graph {y=\sqrt{|x|}} is not bi-Lipschitz equivalent to a line. The cusp is an obstruction:

Curve with a cusp

Indeed, the points {A,B = (\pm \epsilon,\sqrt{\epsilon})} are separated by {C=(0,0)} and lie at distance {\sim \sqrt{\epsilon}} from it. So, the images {f(A), f(B)} would lie on opposite sides of {f(C)}, at distance {\sim \sqrt{\epsilon}} from it. But then the distance between {f(A)} and {f(B)} would be of order {\sqrt{\epsilon}}, contradicting {d(A,B)\sim \epsilon}.

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 {B} and the unit sphere {S} 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 {R(x) = -x }, while the ball has no such thing due to Brouwer’s fixed point theorem. But in the infinite-dimensional case {B} and {S} are homeomorphic. Moreover, there exists a Lipschitz map {F\colon B\rightarrow B} such that the displacement function {\|F(x)-x\|} 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 {f\colon B\rightarrow S} exists, then the composition {f^{-1}\circ R\circ f} is a Lipschitz involution of {B} 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.

4 thoughts on “Bi-Lipschitz equivalence and fixed points”

  1. 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?

    1. 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

  2. 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.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s