TITLE: Locally Lipschitz implies Lipschitz under equivalent metrics?
QUESTION [7 upvotes]: Bonjour.
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 $(U_x, d_1) \to (X_2, d_2)$.

REPLY [5 votes]: 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.