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?

Solution 1:

(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.