5
$\begingroup$

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 $M$?

  • 0
    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

1 Answers 1

1

You need to explicitly exclude \psi'(0)=0 (and thus due to the range of $\psi$ it follows $\psi'(0)\gt 0$) or alternatively require \psi' strictly decreasing in any interval $(0,\epsilon)$. Otherwise, $\psi$ might satisfy the starting conditions of your second statement, but can violate $\psi(x)=0\iff x=0$, because if \psi' can be zero on a positive interval starting at $0$, $\psi$ would be constant on that interval.

Then, from the starting conditions, \psi' > 0 on an interval near $0$ and thus $\psi$ strictly increasing, so $\neq 0$ on that interval. Now, from the given range of $\psi$ and \psi' non-increasing it follows that \psi' \ge 0 everywhere - if it was not, then you can derive that $\psi$ must eventually become negative, which is a contradiction.

That in turn implies $\psi$ is strictly increasing near $0$ and (not necessarily strictly) increasing everywhere, and thus the remaining needed precondition of the first statement $\psi(x)=0\iff x=0$ follows.

  • 0
    Fair enough, edited that in.2012-03-23