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Let $X$ and $Y$ be metric spaces and let $f:X\rightarrow Y$ be such that: for every compact subset $K$ of $X$ the restricted map $f|_K:K\rightarrow Y$ defined by $f|_K(x)=f(x)$ is bi-Lipschitz (with constants depending on $K$).

What are such functions/what is this property called?

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  • $\begingroup$ "If $X$ is locally compact, then this property is equivalent to $f$ being locally Lipschitz." -- This is not true If by "this property" you mean being bi-Lipschitz on compact subsets. For instance, constant functions on $\mathbb R$ are locally Lipschitz but not bi-Lipschitz on $[0,1]$. Your quoted statement is not true even for bijective functions. E.g., if $f(x)=x^3$ for real $x$, then $f$ is locally Lipschitz but not bi-Lipschitz on $[0,1]$. $\endgroup$
    – Iosif Pinelis
    Commented Jun 15, 2022 at 14:41
  • $\begingroup$ @IosifPinelis You're right. Very nice counter example. Still, does this relaxation of the bi-Lipschitz property have a name? $\endgroup$
    – ABIM
    Commented Jun 15, 2022 at 14:46
  • $\begingroup$ Also, I think you are using the term "etymology" incorrectly. In this context, it would mean "the origin and historical development of a certain mathematical term". But you do not have any certain term; rather, you seem to be looking for one. $\endgroup$
    – Iosif Pinelis
    Commented Jun 15, 2022 at 14:46
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    $\begingroup$ I think "locally bi-Lipschitz" would do, even though I have not encountered this term. $\endgroup$
    – Iosif Pinelis
    Commented Jun 15, 2022 at 14:49
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    $\begingroup$ I see. Maybe then "compactly bi-Lipschitz"; cf. "converge compactly" at en.wikipedia.org/wiki/…. $\endgroup$
    – Iosif Pinelis
    Commented Jun 15, 2022 at 15:07

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