How could I go about proving that a bi-Lipschitz function has an inverse that is a Lipschitz function.
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 $K>0$ such that for all $x_1,x_2 \in X$, we have that
$$\frac{1}{K} d_X(x_1,x_2)\leq d_Y(f(x_1),f(x_2))\leq K d_X(x_1,x_2)$$
Definition 2: Lipschitz function. Given metric spaces $(X,d_X)$, $(Y,d_Y)$, a function $f:X \to Y$ is called Lipschitz if there exists a constant $K>0$ such that for all $x_1,x_2 \in X$, we have that
$$d_Y(f(x_1),f(x_2))\leq K d_X(x_1,x_2)$$
$\endgroup$ 21 Answer
$\begingroup$First, note that since $f$ is Lipschitz, then $Kd_x(x_1,x_2) \geq d_y(f(x_1),f(x_2))$, therefore, $f(x_1) = f(x_2) \iff x_1 = x_2$, so $f$ is one-to-one. Now, let $f^{-1}$ be the inverse map of $f$ mapping $f(X)$ one-to-one onto $X$.
Hence, for all $y_1,y_2 \in f(X)$, there is a unique $x_1,x_2 \in X$ such that $f^{-1}(y_1) = x_1$ and $f^{-1}(y_2) = x_2$. Now we can just substitute in the original equation:$$\frac{1}{k}d_x(x_1,x_2) \leq d_y(f(x_1),f(x_2)) \leq K d_x(x_1,x_2)$$We get:$$\frac{1}{k}d_x(f^{-1}(y_1),f^{-1}(y_2)) \leq d_y(y_1,y_2)$$And,$$d_y(y_1,y_2) \leq K d_x(f^{-1}(y_1),f^{-1}(y_2))$$
So the inverse is also bi-Lipschitz.
$\endgroup$ 5
