When is a Lipschitz homeomorphism of metric spaces bi-Lipschitz?

Solution 1:

Lemma. Suppose that $X, Y$ are smooth compact Riemannian $n$-manifolds, $f: X\to Y$ is a diffeomorphism. Then $f$ is bilipschitz with respect to the Riemannian distance functions.

Proof. It clearly suffices to show that $f$ is Lipschitz (considering the inverse we will get the bilipschitz property).

Edit: I do not know why did I write such a complicated proof at first. Here is a much shorter one. Let $UX$ denote the unit tangent bundle of $X$. Then the map $df: UX\to TY$ is continuous, since we assumed that $f$ is smooth. Therefore, since the Riemannian metric on $Y$ is smooth, the function $\phi(u)=|df(u)|$, $u\in UX$ is continuous too. Let $C$ denote its maximum (which exists since $UX$ is compact). Therefore, for every unit speed piecewise-smooth path $c: [a,b]\to X$ the length of its image $$ L(f\circ c)= \int_{a}^b |(f\circ c)'|dt\le C \int_a^b |c'(t)|dt= C L(c), $$ where $L(c)$ is the length of the path in $X$ and $L(f\circ c)$ is the length in $Y$. Therefore, $f$ increases length of paths at most by the factor $C$. Since the distance functions on $X$ and $Y$ are defined by minimizing lengths of paths between points, it follows that $$ d_Y(f(p), f(q))\le C d_X(p, q), p, q\in X. $$ Hence, $f$ is $C$-Lipschitz. qed