Equivalence of Definitions of Lipschitz Equivalent Metrics

Proof
Let $M_1 = \left({A, d_1}\right)$ and $M_2 = \left({A, d_2}\right)$ be metric spaces on the same underlying set $A$.

Definition 1 implies Definition 2
Let $d_1$ and $d_2$ be Lipschitz equivalent by definition 1:

Then by definition:


 * $\exists h, k \in \R_{>0} \forall x, y \in A: h d_2 \left({x, y}\right) \le d_1 \left({x, y}\right) \le k d_2 \left({x, y}\right)$

Let $K_1 = \dfrac 1 h, K_2 = k$.

and:

Thus $d_1$ and $d_2$ are Lipschitz equivalent by definition 2.

Definition 2 implies Definition 1
Let $d_1$ and $d_2$ be Lipschitz equivalent by definition 2.

Then by definition:
 * $(1): \quad \exists K_1 \in \R_{>0}: \forall x, y \in A: d_2 \left({x, y}\right) \le K_1 d_1 \left({x, y}\right)$
 * $(2): \quad \exists K_2 \in \R_{>0}: \forall x, y \in A: d_1 \left({x, y}\right) \le K_2 d_2 \left({x, y}\right)$

Let $h = \dfrac 1 {K_1}, k = K_2$.

and:

Thus $d_1$ and $d_2$ are Lipschitz equivalent by definition 1.