How do you prove triangle inequality for this metric?

Question

Let $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ be an increasing concave function such that $f(t) = 0$ if and only if $t = 0$. Let $(X,d)$ be a metric space. Show that $d_f = f \circ d $ defines a metric on $X$.

Attempt: I proved symmetry, and other properties (which are straightforward). But I don't know how to prove the triangle inequality. Let $x,y,z \in X$. We want to prove that $d_f(x,y) \leq d_f(x,z) + d_f(z,y).$ We have $$ d_f(x,y) = f(d(x,y)) \leq f(d(x,z) + d(z,y)). $$ How can I deduce from the fact that $f$ is concave that $$f(d(x,z) + d(z,y)) \leq f(d(x,z)) + f(d(z,y))$$?

If a function is concave and $f(0) ≥ 0$, then $f$ is subadditive. For proof see here. — , Jul 17 '16 at 10:18
See also: Conditions for defining new metrics 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? — Martin Sleziak, Jul 26 '17 at 07:20

score 1 · Answer 1 · answered Jul 17 '16 at 10:17

1

Hint: If $f$ is a concave function with $f(0) \ge 0$ then $f$ is subadditive. ie $$f(a+b) \le f(a) + f(b).$$

Can you prove this? Hope this helps!

answered Jul 17 '16 at 10:17

Zestylemonzi

4,103

How do you prove triangle inequality for this metric?

1 Answers1

Linked