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Possible Duplicate:
What operations is a metric closed under?

Let $f: \mathbb R_{\geq 0} \to \mathbb R_{\geq 0}$ be a function and $d : X \times X \to \mathbb R_{\geq 0}$ a metric.

I've been thinking about what properties $f$ has to have for the following to hold: $f(d(x,y))$ is a metric if and only if $f$..., that is, I am looking for a necessary and sufficient condition on $f$ such that $f \circ d$ is a metric.

Clearly, we have to have $f(0) = 0$ since otherwise $d^\prime (x,y) := f(d(x,y))$ doesn't satisfy $d^\prime (x,y) = 0$ if and only if $x=y$. Also, we want that the only point that is mapped to zero by $f$ to be zero.

Since we always have $d^\prime (x,y) = d^\prime (y,x)$, all that remains to think about is the triangle inequality. At first I thought that $f$ had to be decreasing, i.e. $x implies that $f(x) \geq f(y)$ but is that good enough? I tried to come up with an example that breaks it but the examples I tried all seem to work (e.g. $e^{-x}, -x^2+30$)

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    You mean $f \circ d$, not $d \circ f$.2012-05-13
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    Think of conditions $f(0)=0$, and $f$ is convex.2012-05-13
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    This is more or less a duplicate of http://math.stackexchange.com/q/694242012-05-14
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    @NielsDiepeveen But my question is not answered in the thread you linked. My question remains open. What is a necessary and sufficient condition on $f$ for $f \circ d$ to be a metric?2013-02-11
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    I voted to reopen. Then I can set a bounty. I'm still interested in an answer.2013-02-11
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    It needs not to be decreasing. Take for example $f_a:\mathbb{R} \to \mathbb{R}$ defined by $f_a(x)=\min\{a,x\}$, with $a>0$.2016-03-10

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