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Is there a continuous, strictly increasing function $f \colon [0,\infty)→ [0,\infty)$ with $f(0) = 0$ such that $\tilde d = f\circ d$ is not a metric? You may take $(X,d)$ to be $\mathbb R$ with the standard metric for this part.

My thoughts are that there isn't any function $\tilde{d}$ which would not be a metric since the distance function $d: X \times X \to \mathbb{R}$ hold the metric space properties. That is, for all $x,y,z\in X$:
(1) $d(x,y) \ge 0$, $d(x,y) = 0$ if and only if $x = y$
(2) $d(x,y) = d(y,x)$
(3) $d(x,y) \le d(x,z) + d(z,y)$

My first thought was if $f$ was a $\ln$ function, then $\tilde{d}$ would not be a metric space, but the question states that $f(0) = 0$, hence $f$ cannot be chosen to be a $\ln$ function anyway.

Just having trouble proving it one way or the other. Any help appreciated.

egreg
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Sam
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1 Answers1

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(Getting this off the Unanswered list.)

You should have no trouble showing that if $f$ is strictly increasing and $f(0)=0$, then $\overline d=f\circ d$ satisfies $(1)$ and $(2)$ of the definition of a metric. Then it all comes down to checking whether $\overline d$ satisfies the triangle inequality. Suppose that $x,y,z\in X$; we want to show (if possible) that

$$\overline d(x,y)\le\overline d(x,z)+\overline d(z,y)\;.$$

Is it possible that we might have, say, $\overline d(x,z)=\overline d(z,y)=1$, but $\overline d(x,y)>2$? That could happen if, for instance, we had $d(x,z)=d(z,y)=1=f(1)$, $d(x,y)=2$, and $f(2)=4$. The function $f(x)=x^2$ would accomplish that — if we can think of a metric space $\langle X,d\rangle$ that has points $x,y$, and $z$ such that $d(x,z)=d(z,y)=1$ and $d(x,y)=2$. But that’s easy, especially since the problem almost suggests that we should look at $X=\Bbb R$ with the usual metric: take $x=0,z=1$, and $y=2$. We now have a counterexample to the triangle inequality for $X=\Bbb R$, $d(x,y)=|x-y|$, and $f(x)=x^2$, so it’s not always true that $\overline d$ is a metric.

Brian M. Scott
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