Suppose that $(X,d_X)$ and $(Y,d_Y)$ are complete doubling metric spaces and let $f:X\rightarrow Y$ be a non-constant Lipschitz map. Then can does there exist a lsc function $\rho:(0,\infty)\rightarrow (0,\infty)$ with the properties that

  • $ \lim\limits_{t \uparrow \infty}\rho(t) = \infty $
  • $ \lim\limits_{t \downarrow 0}\rho(t) = 0, $

Such that for every $x,y \in X$, $$ \rho(d_X(x,y)) \leq d_Y(f(x),f(y)) \leq Lip(f) d_X(x,y) , $$ where $Lip(f)$ is the best Lipschitz constant of $f$.


1 Answer 1


Of course, there might be such a function $\rho$ for a specific $f$, but there need not be one in general.

Let $X$ be $\mathbb{R}^2$, $Y$ be $\mathbb{R}$ and let $f(x,y)$ be $x$ if $x<0$ and $0$ if $x\geq 0$. Then, if I understood correctly, such a $\rho$ cannot exist. After all, $f$ identifies pairs of points with arbitrary distances.

  • $\begingroup$ Yes this makes perfect sense. Are there "reasonable" sufficient conditions for its existence. Suppose that I also require that $f$ is injective so this pathology doesn't occur. $\endgroup$
    – AIM
    Jun 25, 2019 at 13:52
  • $\begingroup$ I am sure that injectivity will not save you, since one could could easily modify my example to make an injective function such that, for arbitrarily large values of $t$, one still has $\inf\{|f(x)-f(y)|:d(x,y)=t\}=0$. I am personally not sure what kind of condition one could hope for that is not a trivial rewriting of the one you wrote down. $\endgroup$
    – user130862
    Jun 25, 2019 at 14:00
  • $\begingroup$ Ah ok, so this version of the problem is non-trivial (or totally trivial). Thanks, this helped a lot $\endgroup$
    – AIM
    Jun 25, 2019 at 14:02

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