Lipschitz Equivalence yields Lipschitz Equivalent Metric

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 $f: M_1 \to M_2$ be a Lipschitz equivalence.

Let $d_3: A^2 \to \R$ be a mapping defined as:

$\map {d_3} {x, y} := \map {d_2} {\map f x, \map f y}$

Then $d_3$ yields a metric on $M_1$ which is Lipschitz equivalent to $d_1$.

Proof

This theorem requires a proof. In particular: Easy when you understand what you're doing You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

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