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
- $I_A: M_1 \to M_2$ 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
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics