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I need to prove the triangle inequality for hyperbolic distances. Could someone give me some pointers? I've tried something, but I'm not sure... Is this valid? Could someone look at $\color{red}{(1)}$ and the relevant question at the end.

Definitions:

$$\ell_h(\Gamma) = \int_\Gamma \frac{|\text{d} z|}{1-|z|^2} \qquad \forall \Gamma \subseteq B(0,1)$$

The hyperbolic distance $d_h(z_1,z_2)$ between two points $z_1,z_2 \in B(0,1)$ is the infimum of the hyperbolic length $\ell_h$ of all contours with starting point $z_1$ end endpoint $z_2$.

Proof

Define $f(z) = \frac{1}{1-z^2}$ is holomorphic on $B(0,1)$.

Becase $f$ is holomorpic in this disk $\int_{\Gamma_1} f(z) \text{d}z$ with $\Gamma_1$ an contour connecting $z_1$ to $z_2$ is independent of the path followed.

Choose a path passing through $z_3$ and name the $\Gamma_2$ the contour connecting $z_1$ and $z_3$, $\Gamma_3$ the contour connecting $z_2$ and $z_3$. Then:

$$\int_{\Gamma_1} f(z) \text{d} z = \int_{\Gamma_2} f(z) \text{d} z +\int_{\Gamma_2} f(z) \text{d} z$$

Using the triangle inequality for contours gives:

$$\begin{align} \int_{\Gamma_1} |f(z)| |\text{d} z| &\leqslant \int_{\Gamma_2} |f(z)| |\text{d} z| +\int_{\Gamma_2} |f(z)| |\text{d} z|\\ \int_{\Gamma_1} \frac{|\text{d} z|}{1-|z^2|} &\leqslant \int_{\Gamma_2} \frac{|\text{d} z|}{1-|z^2|} +\int_{\Gamma_2} \frac{|\text{d} z|}{1-|z^2|} \\ \inf \int_{\Gamma_1} \frac{|\text{d} z|}{1-|z^2|} &\leqslant \inf \int_{\Gamma_2} \frac{|\text{d} z|}{1-|z^2|} +\inf \int_{\Gamma_2} \frac{|\text{d} z|}{1-|z^2|} \color{red}{(1)}\\ d_h(z_1,z_2) &= d_h(z_1,z_3) + d_h(z_2,z_3) \end{align}$$

But!

$\color{red}{(1)}$ isn't valid is it? I thought the infimum rule was $$\inf_A (f+g) \geqslant \inf_A f +\inf_A g$$

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  • $\begingroup$ You're right about that infimum rule. $\endgroup$ – kobe Dec 26 '14 at 23:44
  • $\begingroup$ How about $f\leqslant c+g$ with $c$ fixed, that does imply $\inf f \leqslant c+\inf g$ doesn't it? $\endgroup$ – dietervdf Dec 26 '14 at 23:49
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Let $\Gamma_1$ and $\Gamma_2$ be contours in $B(0,1)$ from $z_1$ to $z_2$ and from $z_3$ to $z_2$, respectively. Let $\varepsilon > 0$. There is a contour $\Gamma_3$ in $B(0,1)$ from $z_1$ to $z_3$ such that $$\ell_h(\Gamma_3) < d_h(z_1,z_3) + \varepsilon.$$ Then $$d_h(z_1,z_2) \le \ell_h(\Gamma_1) \le \ell_h(\Gamma_2) + \ell_h(\Gamma_3) < \ell_h(\Gamma_2) + d_h(z_1,z_3) + \varepsilon.$$ Thus $d_h(z_1,z_2) - d_h(z_1,z_3) - \varepsilon < \ell_h(\Gamma_2)$. Since $\Gamma_2$ was arbitrary, $$d_h(z_1,z_2) - d_h(z_1,z_3) - \varepsilon \le d_h(z_3,z_2).$$ Letting $\varepsilon \to 0$, we obtain $$d_h(z_1,z_2) \le d_h(z_1,z_3) + d_h(z_3,z_2).$$

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  • $\begingroup$ Nice! "Since $\Gamma_2$ was arbitrary, ...". I guess you are using the following rule (with the weak inequality) $$c < f\Rightarrow \inf_A c = c \leqslant \inf_A f$$ I guess I could have accomplished 'about' the same in my proof if I would have fixated $\Gamma_2$ and applied the same rule... (?) $\endgroup$ – dietervdf Dec 27 '14 at 0:06
  • $\begingroup$ @dietervdf I used the fact if a nonempty set $S\subset \Bbb R$ is bounded below by a real number $t$, then $\inf S \ge t$. $\endgroup$ – kobe Dec 27 '14 at 0:14
  • $\begingroup$ Why not use the the same $\epsilon$ argument on $\ell_h(\Gamma_2)$? Then $$d_h(z_1,z_2)\leqslant d_h(z_1,z_3)+d_h(z_2,z_3)+2\epsilon$$ Wouldn't that be even simpler? $\endgroup$ – dietervdf Dec 27 '14 at 0:59
  • $\begingroup$ @dietervdf It doesn't really make a difference to me (in terms of difficulty) but I can make those changes if you like. $\endgroup$ – kobe Dec 27 '14 at 6:06

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