# 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: 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. Anonymous says:

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. Anonymous says:

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.

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

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