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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

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

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

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  • $\begingroup$ for the conclusion do you mean to say it is also lipschitz? $\endgroup$ – mq1998 Apr 8 at 21:49
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    $\begingroup$ Well, it is Lipschitz, but we can actually prove a stronger claim that it is bi-Lipschitz. $\endgroup$ – Davi Barreira Apr 9 at 0:12
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    $\begingroup$ This proof is based on the proof in the book Lectures on Real Analysis by Yeh. He proves the inverse is actually bi-Lipschitz. $\endgroup$ – Davi Barreira Apr 9 at 0:14
  • $\begingroup$ consequently I understand the proof. it proves the inverse of a bi-lipschitz function is lipschitz and also bi-lipschitz? $\endgroup$ – mq1998 Apr 9 at 0:23
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    $\begingroup$ Yes, it proves it is bi-Lipschitz, which is Lipschitz. $\endgroup$ – Davi Barreira Apr 9 at 0:25

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