# Composition of a function with a metric

Which of the following functions, $f: [0,\infty) \rightarrow [0,\infty)$, can be composed with a metric $d$ to get a new metric $f \circ d$:

a)$\;f(x) = \begin{cases}0 & \text{if$x=0$} \\x+1 & \text{if$x >0$}\end{cases}$

b)$f(x) = x^2, x \ge 0$

c)$f(x) = \arctan(x), x \ge 0$

In order for $f \circ d$ to be a metric we need $f$ to be monotonic and subadditive (and of course evaluate to 0 only when the input is 0) so that we can have the triangle inequality property satisfied. This is what I have so far:

a) $f$ is monotonic (looking at the graph, it is always increasing over $[0,\infty)$)
$\;\;\;f$ is subadditive since $1+(x+y) \le (1+x) + (1+y)$, for all $x,y$

b)$f$ is monotonic(looking at the graph, it is always increasing over $[0,\infty)$)
$\;\;\;f$ is not subadditive since $(x+y)^2 = x^2 + 2xy + y^2 \not\le x^2 + y^2$

c)$f$ is monotonic(looking at the graph, it is always increasing over $[0,\infty)$)
$\;\;\;f$ is or is not subadditive ??

I'm thinking that $\arctan(x)$ is subadditive since I can't think of any case where $\arctan(x+y) \le \arctan(x) + \arctan(y)$ doesn't hold over $[0,\infty)$.

Feedback on what I have so far is appreciated.

• You have a mistake in a), $1 + (x+y) < (1+x) + (1+y) = 2+(x+y)$ for all $x,y$. – Daniel Fischer Oct 23 '14 at 15:20
• For c), it may be a good idea to apply $\tan$ to both sides. – Daniel Fischer Oct 23 '14 at 15:22
• For c), use that $\tan$ is increasing and the double angle formula $\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$ – Christopher Oct 23 '14 at 15:23
• @user92638 I fixed the cases for a), you had $x=0$ and $x>1$ leaving $f$ undefined in $(0,1]$. – AlexR Oct 23 '14 at 15:29
• Of course there are certain metrics $d$ for which b) does give a valid metric. The problem statement might be formulated in a better way to stress that $f\circ d$ should be a metric for all metrics $d$. – Hagen von Eitzen Oct 23 '14 at 15:31

In general, if $$d$$ is a metric, and $$f$$ is strictly increasing, $$f(0)=0$$, and concave on $$[0,\infty)$$, then $$f\circ d$$ is again an metric.
Here is a short proof for triangle inequality. Let $$d_1=d(x,y),d_2=d(y,z),d_3=d(x,z)$$, we have $$d_1\leq d_2+d_3$$. By increasing, we have $$f(d_1)\leq f(d_2+d_3)$$, and we want $$f(d_2+f_3)\leq f(d_2)+f(d_3)$$, which is equivalent to $$\frac{f(d_2+d_3)-f(d_2)}{(d_2+d_3)-d_2}\leq \frac{f(d_3)-f(0)}{d_3-0}$$. But this follows directly from concavity of $$f$$ since $$0\leq d_2\leq d_2+d_3$$.
Note, if $$f$$ is 2nd order differentiable, $$f$$ is concave iff $$f''(x)\leq 0$$ for all $$x\in(0,\infty)$$.
• Actually, it suffices that $f$ is increasing, $f(x) \geq 0 \wedge f(x) = 0 \Leftrightarrow x = 0$ (positive definite) and $f(x+y) \leq f(x) + f(y)$ (subadditive). Then we get the triangle inequality via: $f(d(x, y)) \leq f(d(x, y) + d(y, z)) \leq f(d(x, y)) + f(d(y, z))$\$. – ComFreek Apr 30 '19 at 7:31