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Suppose $(M,d)$ is metric. I have proven that if $\psi\colon[0,\infty)\to[0,\infty)$ is non-decreasing, subadditive and satisfies $\psi(x)=0\iff x=0$ for $x\ge0$, then $$\rho(x,y)=\psi(d(x,y))$$ is a metric on $M$.

But I want to `tweak' the restriction of this statement, I want to use differentiability of $\psi$. I found the following:

Suppose $(M,d)$ is metric and $\psi\colon[0,\infty)\to[0,\infty)$ is differentiable with continuous non-increasing derivative $\psi'$ and $\psi(0)=0$. Then $\psi$ is non-decreasing and subadditive.

I have also proven this statement, but is $\rho=\psi(d(x,y))$ a metric in this situation? And if not, what is a sufficient condition on the derivative $\psi'$ to turn $\rho$ into a metric on $M$?

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    If $\psi$ is differentiable on $(0, \infty)$, and the derivative is everywhere positive, then $\psi$ is strictly increasing. If in addition $\psi(0) = 0$, then you're in the earlier situation and $\psi\circ d$ is again a metric.2012-03-22
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    @WNY: Something's not quite right. If $\psi(x) = x^2$, for example, the $\psi\circ d$ isn't a metric for the usual $d$ on $\mathbb{R}$. Consider the three points $0$, $1$ and $2$ in $\mathbb{R}$. Then $\psi(d(0,2)) = 4$, while $\psi(d(0,1)) + \psi(d(1,2)) = 1+1 < 4$, so the triangle inequality fails. You really need to know something about $\psi$s derivative, that the derivative is non-increasing.2012-03-22
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    See also: [How do you prove triangle inequality for this metric?](https://math.stackexchange.com/q/1862037) and [Is there a continuous, strictly increasing function $f: [0,\infty)\to [0,\infty)$ with $f(0) = 0$ such that $\tilde d = f\circ d$ is not a metric?](https://math.stackexchange.com/q/715293)2017-07-26

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