[Math] Unit circle metric

circlesgeometrymetric-spaces

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 ?

Best Answer

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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