Metrics on Space are Lipschitz Equivalent iff Identity Mapping is Lipschitz Equivalence

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Theorem

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

Let $I_A$ denote the identity mapping on $A$.

Then:

$d_1$ and $d_2$ are Lipschitz equivalent

if and only if:

$I_A: M_1 \to M_2$ is a Lipschitz equivalence.


Proof

By definition of identity mapping:

$\forall x \in A: \map {I_A} x = x$


Sufficient Condition

Let $d_1$ and $d_2$ be Lipschitz equivalent.

Then:

\(\ds \exists h, k \in \R_{>0}: \forall x, y \in A: \, \) \(\ds h \map {d_2} {x, y}\) \(\le\) \(\ds \map {d_1} {x, y} \le k \map {d_2} {x, y}\) Definition of Lipschitz Equivalent Metrics
\(\ds \leadsto \ \ \) \(\ds \exists h, k \in \R_{>0}: \forall x, y \in A: \, \) \(\ds h \map {d_2} {\map {I_A} x, \map {I_A} y}\) \(\le\) \(\ds \map {d_1} {x, y} \le k \map {d_2} {\map {I_A} x, \map {I_A} y}\) Definition of Identity Mapping: $\forall x \in A: \map {I_A} x = x$

That is, by definition:

$I_A: M_1 \to M_2$ is a Lipschitz equivalence.

$\Box$


Necessary Condition

Let $I_A$ be a Lipschitz equivalence.

Then:

\(\ds \exists h, k \in \R_{>0}: \forall x, y \in A: \, \) \(\ds h \map {d_2} {\map {I_A} x, \map {I_A} y}\) \(\le\) \(\ds \map {d_1} {x, y} \le k \map {d_2} {\map {I_A} x, \map {I_A} y}\) Definition of Lipschitz Equivalence (Mapping)
\(\ds \leadsto \ \ \) \(\ds \exists h, k \in \R_{>0}: \forall x, y \in A: \, \) \(\ds h \map {d_2} {x, y}\) \(\le\) \(\ds \map {d_1} {x, y} \le k \map {d_2} {x, y}\) Definition of Identity Mapping: $\forall x \in A: \map {I_A} x = x$

That is, by definition:

$d_1$ and $d_2$ are Lipschitz equivalent metrics.

$\blacksquare$


Sources