Equivalence of Definitions of Lipschitz Equivalent Metrics

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

\(\ds \exists h \in \R_{>0}: \forall x, y \in A: \, \) \(\ds h \map {d_2} {x, y}\) \(\le\) \(\ds \map {d_1} {x, y}\) Definition 1 of Lipschitz Equivalent Metrics
\(\ds \leadsto \ \ \) \(\ds \map {d_2} {x, y}\) \(\le\) \(\ds \dfrac 1 h \map {d_1} {x, y}\)
\(\ds \) \(=\) \(\ds K_1 \map {d_1} {x, y}\)

and:

\(\ds \exists k \in \R_{>0}: \forall x, y \in A: \, \) \(\ds \map {d_1} {x, y}\) \(\le\) \(\ds k \map {d_2} {x, y}\) Definition 1 of Lipschitz Equivalent Metrics
\(\ds \) \(=\) \(\ds K_2 \map {d_2} {x, y}\)

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

\(\ds \exists K_1 \in \R_{>0}: \forall x, y \in A: \, \) \(\ds \map {d_2} {x, y}\) \(\le\) \(\ds K_1 \map {d_1} {x, y}\) Definition 2 of Lipschitz Equivalent Metrics
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 {K_1} \map {d_2} {x, y}\) \(\le\) \(\ds \map {d_1} {x, y}\)
\(\ds \) \(=\) \(\ds h \map {d_1} {x, y}\)


and:

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

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

$\blacksquare$

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