# proving bi-Lipschitz function has an inverse that is lipschitz

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

$$d_Y(f(x_1),f(x_2))\leq K d_X(x_1,x_2)$$

• Start writing the definition of "bi-Lipschitz function". – Giuseppe Negro Apr 5 at 16:42
• Use \to in things like $f:X\to Y$. – Angina Seng Apr 5 at 18:09

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, $$d_y(y_1,y_2) \leq K d_x(f^{-1}(y_1),f^{-1}(y_2))$$