I found that there is three possible definitions of bi-Lipschitz function. I can see to some extent how they are related to each other.
Definition 1: bi-Lipschitz function. Given metric spaces $(X,d_X)$, $(Y,d_Y)$, a function $f:X \to Y$ is called bi-Lipschitz if there exists a constant $K_{1}>0$ such that for all $x_1,x_2 \in X$, we have that
$$\frac{1}{K_{1}} d_X(x_1,x_2)\leq d_Y(f(x_1),f(x_2))\leq K_{1} d_X(x_1,x_2)$$
Definition 2: bi-Lipschitz function. Given metric spaces $(X,d_X)$, $(Y,d_Y)$, a function $f:X \to Y$ is called bi-Lipschitz if $f$ is Lipschitz and there exists a constant $K_{2}>0$ such that for all $x_1,x_2 \in X$, we have that
$$ K_{2} \; d_X(x_1,x_2) \leq d_Y(f(x_1),f(x_2))$$
Definition 3: bi-Lipschitz function. Given metric spaces $(X,d_X)$, $(Y,d_Y)$, a function $f:X \to Y$ is called bi-Lipschitz if $f$ is Lipschitz and has an inverse mapping $f^{-1}: f(\mathcal{X}) \rightarrow \mathcal{X}$ which is also Lipschitz.
Based on definition 1, can we say that the Lipschitz constant of $f^{-1}$ is the inverse of the Lipschitz constant of $f$?
Best Answer
No.
Here is an example to illustrate. Let $\epsilon > 0$, $X = [\epsilon,1]$, $y = [\epsilon^2,1]$, and $f(x) = x^2$.
You can easily verify that $f : X \to Y$ is bi-Lipschitz with Lipschitz constant $2$. On the other hand, the Lipschitz constant of $f^{-1}$ is $\dfrac{1}{2\sqrt{\epsilon}}$ which can be made arbitrarily large yet finite by adjusting the domain $X$.