Let $X$ be a metric space with metric $d$, and suppose $d$ is unbounded in the sense that for any $N$ there exist points $x$ and $y$ such that $d(x,y) > N$. To replace $d$ with a bounded metric $d^*$, one might try picking a bounded function $f : [0,\infty) \to [0,\infty)$ and defining $d^*(x,y) = f(d(x,y))$. What conditions upon $f$ are necessary for this approach to work?
First we need $d^*$ to define a proper metric, so of course we need $f(0) = 0$ and $f(x) > 0$ for $x > 0$. Then we only need to assure the triangle inequality, i.e.:
$f(d(x,z)) \leq f(d(x,y)) + f(d(y,z))$.
It is sufficient for $f$ to be monotone and subadditive, but is it also necessary?
A similar question is asked here: $d$ is a metric. Under which conditions $f\circ d$ is also a metric? where the author writes that subadditivity is necessary, in the sense that if $f$ is a function such that $f \circ d$ is a metric for any metric $d$ (which is a way to assume no knowledge of the underlying metric $d$), then $f$ must be subadditive. However subadditivity is not necessary in the sense that there exists a metric $d$ and a non-subadditive $f$ such that $f \circ d$ is also a metric, so there is still not a complete answer. As for the monotonicity, there is still no answer either way.
Now I ask about the topological equivalency: let $B_\varepsilon(x) = \{y \in X : d(x,y) < \varepsilon\}$ and $B^*_\varepsilon(x) = \{y \in X : d^*(x,y) < \varepsilon\}$. The metrics are equivalent if these two sets are open with respect to each metric. In other words, it should be true that for any $\varepsilon$ and $x$ there exists a $\delta$ such that $f(d(x,y)) = d^*(x,y) < \delta \implies d(x,y) < \varepsilon$. Also vice versa, that for any $\varepsilon$ and $x$ there exists a $\delta$ such that $d(x,y) < \delta \implies d^*(x,y) = f(d(x,y)) < \varepsilon$.
It is sufficient for $f$ to be monotone in a neighborhood of $0$ (and bounded away from $0$ outside of this neighborhood) and continuous at $0$, but is it also necessary?
