For $i=1,2$ let $X_i$ be a non-empty set and $d_i$ a metric $X_i^2 \to \mathbb{R}$. Suppose $f$ is a locally Lipschitz (*) function $(X_1, d_1) \to (X_2, d_2)$.

Question. Do there exist metrics $\delta_1: X_1^2 \to \mathbb{R}$ and $\delta_2: X_2^2 \to \mathbb{R}$ such that i) $\delta_i$ is (topologically) equivalent to $d_i$ (for $i=1,2$) and ii) $f$ is a Lipschitz function $(X_1, \delta_1) \to (X_2, \delta_2)$?

(*) For each $x \in X_1$, there exists a neighborhood $U_x$ of $x$ in $(X_1, d_1)$ such that the restriction of $f$ to $U_x$ is a Lipschitz function $(U_x, d_1) \to (X_2, d_2)$.

  • $\begingroup$ This question does not make sense to me. You can simply choose $\delta_i = d_i$. Your heading seems to imply that you want to know whether $f$ will remain locally Lipschitz if the $d_i$ are replaced by arbitrary (topologically) equivalent metrics $\delta_i$. While I don't know a counterexample, I doubt this is true, though, cause topological equivalence does not let you control the size of neighbourhoods in the metric sense. $\endgroup$
    – user20266
    Dec 21, 2011 at 11:11
  • $\begingroup$ @Thomas: the OP wants to see if one can find a different metric generating the same topology such that the locally Lipschitz function is now globally Lipschitz. For example, let $f:\mathbb{R} \to \mathbb{R}_{\geq 0}$ be $x\mapsto x^2$. This is locally but not globally Lipschitz. But if you take the diffeomorphism $\mathbb{R}_{\geq 0}\mapsto\mathbb{R}_{\geq 0}$ given by $x\mapsto \sqrt{x}$, you pull back a Riemannian metric (and hence a distance function) w.r.t. which the map $f$ is now Lipschitz globally with Lipschitz constant 1. $\endgroup$ Dec 21, 2011 at 11:32

1 Answer 1


Yes — assuming that “$d_1$ is topologically equivalent to $\delta_1$” means that $d_1$ and $\delta_1$ induce the same topologies.

It is sufficient to require $f$ to be continuous, and only the metric on $X_1$ needs to be modified.

Put $\delta_1(x,y) = d_1(x,y) + d_2(f(x),f(y))$.

Then $\delta_1$ is a metric which is topologically equivalent to $d_1$: Indeed, since $d_1 \leq \delta_1$ we have that $x_n \to x$ with respect to $\delta_1$ implies $x_n \to x$ with respect to $d_1$; conversely, continuity of $f$ ensures that $d_1$-convergence implies $\delta_1$-convergence. Thus, both metrics describe the same closed sets on $X_1$, so they induce the same topologies.

As $d_2(f(x),f(y)) \leq \delta_1(x,y)$ we see that $f:(X_1,\delta_1) \to (X_2,d_2)$ is $1$-Lipschitz.

  • $\begingroup$ Is the codomain $(X,\delta_2)$ or $(X,d_2)$? $\endgroup$ Dec 21, 2011 at 12:04
  • 2
    $\begingroup$ @Jesse: the answer to your question is "yes". (t.b. set $d_2 = \delta_2$.) $\endgroup$ Dec 21, 2011 at 12:07

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