# Equivalence of Definitions of Lipschitz Equivalent Metrics

## Contents

## Theorem

The following definitions of the concept of **Lipschitz Equivalent Metrics** are equivalent:

### Definition 1

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.

Let $\exists h, k \in \R_{>0}$ such that:

- $\forall x, y \in A: h \map {d_2} {x, y} \le \map {d_1} {x, y} \le k \map {d_2} {x, y}$

Then $d_1$ and $d_2$ are described as **Lipschitz equivalent**.

### Definition 2

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.

Let $\exists K_1, K_2 \in \R_{>0}$ such that:

- $(1): \quad \forall x, y \in A: \map {d_2} {x, y} \le K_1 \map {d_1} {x, y}$
- $(2): \quad \forall x, y \in A: \map {d_1} {x, y} \le K_2 \map {d_2} {x, y}$

Then $d_1$ and $d_2$ are described as **Lipschitz equivalent**.

## Proof

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ 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:

- $\exists h, k \in \R_{>0} \forall x, y \in A: h \map {d_2} {x, y} \le \map {d_1} {x, y} \le k \map {d_2} {x, y}$

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

\(\, \displaystyle \exists h \in \R_{>0}: \forall x, y \in A: \, \) | \(\displaystyle h \map {d_2} {x, y}\) | \(\le\) | \(\displaystyle \map {d_1} {x, y}\) | Definition 1 of Lipschitz Equivalent Metrics | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {d_2} {x, y}\) | \(\le\) | \(\displaystyle \dfrac 1 h \map {d_1} {x, y}\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle K_1 \map {d_1} {x, y}\) |

and:

\(\, \displaystyle \exists k \in \R_{>0}: \forall x, y \in A: \, \) | \(\displaystyle \map {d_1} {x, y}\) | \(\le\) | \(\displaystyle k \map {d_2} {x, y}\) | Definition 1 of Lipschitz Equivalent Metrics | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle K_2 \map {d_2} {x, y}\) |

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

$\Box$

### Definition 2 implies Definition 1

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

Then by definition:

- $(1): \quad \exists K_1 \in \R_{>0}: \forall x, y \in A: \map {d_2} {x, y} \le K_1 \map {d_1} {x, y}$
- $(2): \quad \exists K_2 \in \R_{>0}: \forall x, y \in A: \map {d_1} {x, y} \le K_2 \map {d_2} {x, y}$

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

\(\, \displaystyle \exists K_1 \in \R_{>0}: \forall x, y \in A: \, \) | \(\displaystyle \map {d_2} {x, y}\) | \(\le\) | \(\displaystyle K_1 \map {d_1} {x, y}\) | Definition 2 of Lipschitz Equivalent Metrics | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \dfrac 1 {K_1} \map {d_2} {x, y}\) | \(\le\) | \(\displaystyle \map {d_1} {x, y}\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle h \map {d_1} {x, y}\) |

and:

\(\, \displaystyle \exists K_2 \in \R_{>0}: \forall x, y \in A: \, \) | \(\displaystyle \map {d_1} {x, y}\) | \(\le\) | \(\displaystyle K_2 \map {d_2} {x, y}\) | Definition 2 of Lipschitz Equivalent Metrics | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle k \map {d_2} {x, y}\) |

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

$\blacksquare$