7

I have to show that $d$ is a metric on the real numbers, and the first three axioms are straight forward, the triangle inequality poses a problem. I know we need to get $$ \begin{align*} d(x,y) &= \arctan|x-y| \\ &\le \arctan|x-z|+\arctan|z-y| \\ &= d(x,z)+d(z,y), \end{align*} $$ so what I've tried is $$\arctan|x-y| = \arctan|x-z+z-y| \le \arctan(|x-z|+|z-y|),$$ but I'm not even sure if this accomplishes anything because I don't know how to split it up.

alexwlchan
  • 2,060
Jeff
  • 1,199
  • For intuition, $\arctan$ is strictly increasing for the domain of positive reals but its derivative is a strictly decreasing, so if you put a single positive value into $\arctan$, you should get a lower value than if you were to put split that value apart and put the two values separately into $\arctan$. – ejcqw Jan 25 '15 at 19:37

4 Answers4

10

Lemma. When $d$ is a metric on a set $X$ and $f:\>{\mathbb R}_{\geq0}\to{\mathbb R}_{\geq0}$ is a monotonically increasing function with $f(0)=0$ and $f(t)>0$ for $t>0$ which is sublinear, i.e., $$f(u+v)\leq f(u)+f(v)\qquad\forall u, \>v\geq0\ ,$$ then $d':=f\circ d$ is again a metric on $X$.

Proof. Only the triangle inequality needs verification: $$d'(x,z)=f\bigl(d(x,z)\bigr)\leq f\bigl(d(x,y)+d(y,z)\bigr)\leq f\bigl(d(x,y)\bigr)+f\bigl(d(y,z)\bigr)=d'(x,y)+d'(y,z)\ .$$ Therefore it remains to prove that $t\mapsto \arctan t$ is sublinear for $t\geq0$. This can be seen as follows. $$\arctan(u+v)=\int_0^{u+v}{dt\over 1+t^2}=\int_0^u{dt\over 1+t^2}+\int_0^v{dt'\over1+(u+t')^2}\leq\arctan u+\arctan v\ .$$

Christian Blatter
  • 226,825
6

Completely orthogonal idea. It is trivial to show that $d_1(x,y)=|x-y|$ is a metric. Function $f(x)=\arctan(x)$ is strictly increasing, continuous (actually real-analytic) bounded function positive for $x\geq 0$. Try to prove that $d(x,y)=f(d_1(x,y))$ must be a metric for any such function $f$. Edit: Please see the comment bellow. The conditions I listed originally are not sufficient.

Predrag Punosevac
  • 388
  • 1
  • 11
  • 5
    The important property is that $f$ is convex and has $f(0)=0$ and $f(x)>0$ for $x>0$. Your conditions are not sufficient - for instance, set $$f(x)=\begin{cases}x &&\text{if }0\leq x \leq 1 \ 2x-1 &&\text{if }x\leq 1 \leq 2 \ 3&&\text{if }x\geq 2 \end{cases}$$ Then $f(|x-y|)$ is not a metric since $2=d(0,1)+d(1,2)<d(0,2)=3$. – Milo Brandt Jan 25 '15 at 20:36
  • Good catch! I edit my answer and pointed to your comment. I typed my answer while eating the lunch as I remember having the similar issues 25 years ago. – Predrag Punosevac Jan 25 '15 at 21:15
3

Suppose that $\alpha, \beta \ge 0$.

For $\alpha\beta < 1$, we have

$$\arctan\alpha + \arctan\beta = \arctan\left(\tan\left(\arctan\alpha + \arctan\beta\right)\right)$$ $$=\arctan \frac{\alpha+\beta}{1-\alpha\beta} \ge \arctan (\alpha + \beta)$$

by the angle sum formula for tangent. On the other hand, for $\alpha\beta \ge 1$ with (say) $\alpha > 1$, we find that $\beta \ge \frac{1}{\alpha}$, so $$\arctan \alpha + \arctan\beta \ge \arctan \alpha + \arctan \frac{1}{\alpha} \ge \arctan\alpha + \frac{\pi}{2} - \arctan\alpha$$ $$\ge \frac{\pi}{2} > \arctan(\alpha+\beta)$$

since $\arctan x < \frac{\pi}{2}$ for all $x$.

Substituting $\alpha = |x-z|$, $\beta=|z-y|$, we find

$$\arctan|x-z| + \arctan|z-y| \ge \arctan |x-y|$$

  • Why is $\arctan\frac{|x-z|+|z-y|}{1-|x-z||z-y|} \geq \arctan(|x-z|+|z-y|)$? Wouldn't the denominator render the function negative, and thus, always lower? –  Jan 30 '15 at 04:24
  • @wearyrowboat Good point, I was being sloppy and assuming $|x-z||z-y| < 1$. However, it turns out that if $|x-z||z-y| \ge 1$, then the sum of their arctangents must at least $\frac{\pi}{2}$. I've edited my answer accordingly. –  Jan 30 '15 at 05:55
  • Isn't $\arctan \alpha + \arctan \frac{1}{\alpha}$ necessarily equal to $\frac{\pi}{2}$? I believe that there is an identity which proves this. –  Jan 31 '15 at 01:00
  • @wearyrowboat Yes, that's true: I've used the identity $\arctan\frac{1}{x} = \frac{\pi}{2} - \arctan x$ to show that. The inequality is just to keep the chain of inequalities starting with $\arctan\beta \ge \arctan\frac{1}{\alpha}$ going. –  Jan 31 '15 at 03:25
2

first let $= \tan t_1 = m_1 \ge 0, \tan t_2 = m_2 \ge 0, 0 \le t_1, t_2 < \pi/2$

then $$\tan(t_1+t_2) = \dfrac{\tan t_1 + \tan t_2}{1 - \tan t_1 \tan t_2} = \dfrac{m_1 + m_2}{1 - m_1 m_2} \ge m_1 + m_2 $$ therefore $$\tan^{-1}(m_1 + m_2) \le t_1 + t_2 = \tan^{-1}m_1 + \tan^{-1} m_2$$ now substituting $m_1 = |x-y|, m_2 = |y-z|$ and using $\tan^{-1}$ is an increasing function with the triangle inequality $|x-z| \le |x-y| + |y-z|$

$$ \tan^{-1} |x-z| \le \tan^{-1}(|x-y| + |y-z|) \le \tan^{-1}|x-y| + \tan^{-1} |y-z|$$ and that proves the triangle inequality.

abel
  • 29,170