Given a function $f: [0, \infty) \rightarrow [0, \infty)$ that's concave with $f(0) = 0$ and $f(x) > 0, \forall x \in [0, \infty)$, so $f(tx + (1-t)y) \geq tf(x) + (1-t)f(y)$, how do I show that if $d(x, y)$ is a metric on the metric space $M$, then $f(d(x, y))$ is also a metric for $M$?

I know that since $d$ is a metric, $d(x, y) \leq d(x, z) + d(z, y)$ by the triangle inequality. Is it enough to use the triangle inequality like this:

\begin{align*} d(x, y) &\leq d(x, z) + d(z, y) \\ \Rightarrow f(d(x, y)) & \leq f(d(x, z) + d(z, y)) \\ &\leq f(d(x, z)) + f(d(z, y)) \end{align*}

because concave functions are subadditive (which I can prove). Is that enough? The implication in the second step assumes that the function is increasing, or at least not decreasing (I'm not sure how to prove that), but is that really all I have to do?

This question discusses how this relates to subadditivity, but also makes the assumption of the function being increasing. From the given information, I don't think I have that assumption.


Monotonicity of $f$ follows from your assumptions.

Assume that $f$ is not monotonically increasing. Then there are $x<y$ such that $f(x)\ge f(y) +\delta$, $\delta>0$.

This implies that $f(z)<0$ for some large $z$: Let $z>y$. Then $$ y = x + \frac{y-x}{z-x}(z-x) = \frac{z-y}{z-x}x + \frac{y-x}{z-x}z, $$ and by concavity $$ f(y) \ge \frac{z-y}{z-x} f(x) + \frac{y-x}{z-x}f(z). $$ Since $f(x)\ge f(y)+\delta$, we get $$ f(x) \ge \frac{z-y}{z-x} f(x) + \frac{y-x}{z-x}f(z) + \delta, $$ which is equivalent to $$ f(x)- \frac{z-x}{y-x}\delta\ge f(z). $$ Now for $z\to\infty$ the left-hand side tends to $-\infty$, thus there is $z$ such that $f(z)<0$. Contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.