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Let $S^1$ the unit circle in $\mathbb{R}^2$ and

$$d: S^1\times S^1\to\mathbb{R}$$

$$d(\theta_1,\theta_2) = \left\{ \begin{array}{ll} |\theta_1-\theta_2| & \mbox{if } |\theta_1-\theta_2|\le \pi \\ 2\pi-|\theta_1-\theta_2| & \mbox{else} \end{array} \right.$$

I'm trying to prove that this function satisfies the triangle inequality

$$d(\theta_1,\theta_3)\le d(\theta_1,\theta_2)+d(\theta_2,\theta_3)$$

There are three possible cases:

  1. $|\theta_1-\theta_2|, |\theta_2-\theta_3|\le\pi$
  2. $|\theta_1-\theta_2|\le\pi, |\theta_2-\theta_3|>\pi$ or $|\theta_2-\theta_3|\le\pi, |\theta_1-\theta_2|>\pi$
  3. $|\theta_1-\theta_2|, |\theta_2-\theta_3|>\pi$

I proved the first two ones using the triangle inequality for the absolute value, but I'm stuck for the third.

It is equivalent to prove that

$$|\theta_1-\theta_3|+|\theta_1-\theta_2|+|\theta_2-\theta_3|\le 4\pi\text{ if }|\theta_1-\theta_3|\le\pi$$ and $$|\theta_1-\theta_2|+|\theta_2-\theta_3|-|\theta_1-\theta_3|\le 2\pi\text{ if }|\theta_1-\theta_3|>\pi$$

Could you give me a hint ? Or did I misdefine $d$ or any other error ?

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4 Answers 4

up vote 1 down vote accepted

I would say that w.l.o.g. $\theta_1 \leq \theta_3$ and break it up into the cases of $\theta_1 \leq \theta_2 \leq \theta_3$, $\theta_2 \leq \theta_1 \leq \theta_3$, and $\theta_1 \leq \theta_3 \leq \theta_2$. That allows you to use more than just the triangle inequality on the real line.

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Thank you for the answer :) It is then easy to prove the first equality and show that the second case never happens. I don't know if it is possible to proceed without distinguishing those cases (more elegantly ;)) –  Klaus Nov 20 '10 at 20:12

Apologies: this should be a comment, but I am not "reputable" enough to leave comments.

If you proved 1, then you should have no problems with 3. The key to the exercise is to understand this metric geometrically. Note that if you draw two radii on the unit circle, then there are two obvious angles between them, that you could measure. The two-case-rule just says "always measure the smaller one of the two angles". In other words, the distance between two angles is always at most $\pi$, which should allow you to easily complete the proof.

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Thank you for the answer. I understand the metric geometrically well, that's how I defined it at first. Geometrically, the triangle inequality is by the way obvious, but I cannot obtain the wanted inequalities above to get the last case :/ –  Klaus Nov 20 '10 at 13:32
    
Ok, here is another hint: the second of the two cases that you list after "it is equivalent to..." doesn't occur. Why? Your geometric intuition should help you here, as well. –  Alex B. Nov 20 '10 at 13:55

This is not a formal proof, but gives, I think, the geometric intuition:

alt text

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Thank you for the drawing, that's very nice of you :) However, as I said to Alex, I understood how it worked geometrically, but I'm unable to conclude the formal proof above. Maybe should I have considered the relative positions of the points (order relations between angles) rather than the values of $|\theta_i-\theta_j|$ ? –  Klaus Nov 20 '10 at 13:38
    
Your idea seems good to me, having in mind that as stated by Alex "the distance between two angles is always at most $\pi$". The angles are to be measured as in trigonometry. In this respect angle $2\pi-\theta_3$ on the right should be exchanged with $\theta_3$. On the left it is correct. –  Américo Tavares Nov 20 '10 at 13:50

If the three points $z_1$, $z_2$, $z_3\in S^1$ can be covered by a semicircle $H$ then the restriction $d\restriction H$ acts on the $z_i$ like the usual metric on $\mathbb R$, if not, then the sum of any two distances between the $z_i$ is $>\pi$ and the third distance is $<\pi$.

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