Visualise Lipschitz Equivalence

Question

I have a question on Lipschitz equivalence, defined in the following way:

Let $X$ be a set, and let $d$ and $d'$ be metrics on $X$. We say that $d$ and $d'$ are Lipschitz equivalent if there exists real numbers $c,C>0$ such that for all $x,y \in X$,

$$ c d(x,y) \leq d'(x,y) \leq C d(x,y) $$

I would like to know if it is sensible to visualise the equivalence as follows: let A be a space as seen under metric $d$. Then we can obtain the space A as seen under metric $d'$ by stretching and/or compressing the space by a finite amount. I am picturing it as the space being a giant 2-dimensional rubber surface that I can stretch and compress at will.

Does this make sense?

Answer 1

This is sensible but there is one limitation in thinking about stretching rubber, which is that this viewpoint is mostly limited to length spaces, i.e., metric spaces where the distance is expressed as the infimum of lengths of curves joining a pair of points. In other words, this is OK so long as you take rubber with a grain of salt.

For example, the unit circle has two natural metrics: (1) the Euclidean, where the distance between opposite points is $2$; and (2) the length metric, where the distance between opposite points is $\pi$.

Answer 2

If I understand correctly, your picture would rather refer to a situation where $d'(x,y) = Kd(x,y)\ $ for some fixed $K > 0$.

The situation is a bit different in the case of the Lipschitz equivalence. Indeed, suppose for instance that $c = 1$ and $C = 3$. You may perfectly have three pairs of points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ such that $d'(x_1, y_1) = d(x_1, y_1)$, $d'(x_2, y_2) = 2d(x_2, y_2)$ and $d'(x_3, y_3) = 3d(x_3, y_3)$. Thus in some sense, your compression ratio may depend on the pair of points you choose.