# Inducing metric spaces

Let $f\colon \mathbb{R}_{\geq0} \to \mathbb{R}_{\geq0}$ be a function. We say that $f$ has the property of inducing metric spaces, whenever for all metric space $(X,d)$, $(X, f \circ d)$ is also a metric space. In other words, for all distances $d$, $f \circ d$ is also a distance. In this question, I am trying to establish more "explicit" and "intrinsic" conditions on $f$ that are equivalent to the property of inducing metric spaces.

EDIT: I have changed what follows, because it was wrong as written before.

What I could prove so far is the following. A sufficient condition for inducing metric spaces is the following.

a) $f^{-1}(0)=0$,

b) for all $a,b,c$ with $a+b=c$, we have $f(a)+f(b) \geq f(c)$ (sublinearity).

c) $f$ is non-decreasing.

However, the converse is not true. For example, the function

$$f= \begin{cases} 0 & x=0 \\ 2& 0<x<1 \\ 1 & 1 \leq x,\end{cases}$$ induces metric spaces, without being non-decreasing. (In general, if the image of $f$ is finite, a condition equivalent to inducing metric spaces is that its maximum is $\leq$ than twice its minimum other than zero).

My question (a bit open-ended and non-mathematical) is: how can one relax conditions a),b),c), so that they actually become equivalent to the property of inducing metric spaces?

• Having $f$ discontinuous in $0$ of course means that the metric $f \circ d$ induces the discrete topology.
– calc
Sep 13, 2013 at 13:33