One example is that if $d$ is a metric, then so is $\frac{d}{1+d}.$ I'm wondering if there are broader generalities than this, that if we can broaden and classify the set of all functions $f$ for which $f(d)$ is also a metric.
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3Clearly $$f(x)=0\iff x=0$$ is necessary, and beyond that only the triangle inequality is worth attention. To preserve the triangle inequality, it's sufficient that we have $$\alpha+\beta\ge\delta\implies f(\alpha)+f(\beta)\ge f(\delta)$$ (and I suspect this will be necessary as well). – Noah Schweber Oct 01 '22 at 20:26
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This has been investigated in Dovgoshey, Oleksiy & Martio, Olli. (2009). Functions transferring metrics to metrics. Beiträge zur Algebra und Geometrie / Contributions to Algebra and Geometry. 54. 1-25. 10.1007/s13366-011-0061-7.
The authors define $\cal F$ as the class of all functions $f:\Bbb R_{\ge 0} \to \Bbb R_{\ge 0}$ with the property that for every metric space $(X, d)$, $f \circ d$ is also a metric on $X$.
They prove
Theorem 1.1. A function $f:\Bbb R_{\ge 0} \to \Bbb R_{\ge 0}$ is in $\cal F$ if and only if the following conditions hold:
- $f(0) = 0$ and $f(t) > 0$ for $t > 0$,
- $2 \max(f(a), f(b), f(c)) \le f(a) + f(b) + f(c)$ for all $a, b, c \ge 0$ with $2 \max(a, b, c) \le a + b + c$.
They show that a function $f \in \cal F$ is necessarily subadditive, and that subadditivity is also sufficient if $f$ is assumed to be increasing:
Theorem 4.1. An increasing function $f:\Bbb R_{\ge 0} \to \Bbb R_{\ge 0}$ is in $\cal F$ if and only if the following conditions hold:
- $f(0) = 0$ and $f(t) > 0$ for $t > 0$, and
- $f(x+y) \le f(x) + f(y)$ for all $x, y \ge 0$, i.e. $f$ is subadditive.
This is for example satisfied if $f$ is strictly increasing and concave with $f(0) = 0$.
Martin R
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