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.

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    $\begingroup$ You have a mistake in a), $1 + (x+y) < (1+x) + (1+y) = 2+(x+y)$ for all $x,y$. $\endgroup$ – Daniel Fischer Oct 23 '14 at 15:20
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    $\begingroup$ For c), it may be a good idea to apply $\tan$ to both sides. $\endgroup$ – Daniel Fischer Oct 23 '14 at 15:22
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    $\begingroup$ For c), use that $\tan$ is increasing and the double angle formula $\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$ $\endgroup$ – Christopher Oct 23 '14 at 15:23
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    $\begingroup$ @user92638 I fixed the cases for a), you had $x=0$ and $x>1$ leaving $f$ undefined in $(0,1]$. $\endgroup$ – AlexR Oct 23 '14 at 15:29
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    $\begingroup$ 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 $d$. $\endgroup$ – 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 $x\in(0,\infty)$.

So by this criterion, a),c) will give you a metric.

  • $\begingroup$ 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))$$. $\endgroup$ – ComFreek Apr 30 '19 at 7:31

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