let $f:\mathbb {H}\to\mathbb {H}$ holomorphic, show $\Bigg|\frac{f(w)-f(z)}{f(w)-\overline{f(z)}}\Bigg|\leq\bigg|\frac{w-z}{w-\overline{z}}\bigg|$?

Question

let $f:\mathbb {H}\to\mathbb {H}$ be holomorphic. Show that

$\Bigg|\frac{f(w)-f(z)}{f(w)-\overline{f(z)}}\Bigg|\leq\bigg|\frac{w-z}{w-\overline{z}}\bigg|$ for all $w,z$ in $\mathbb{H}=\big \{z\in\mathbb{C}:Im z>0\big \}$

when does equality hold??

i am thinking of composing $f$ with some automorphism of unit disk .

Answer 1

This wikipedia page explains what you are looking for with details. Essentialy you need to use the Caley transform to reduce the problem to the unit disk and there you use Schwarz's lemma to prove it.

For the equality case you just need to keep track of the compositions of the maps of the previous part and exploit the fact that you know when equality holds for Schwarz's lemma.

Answer 2

Hint. Consider the Cayley transformation $w=\dfrac{z-i}{z+i}$ which maps the upper-half plane $\mathbb{H}$ onto the unit disk $\mathbb{D}=\left\{|w|<1\right\}$ and use the Schwarz–Pick theorem.

Answer 3

For every $a \in \mathbb{H}$ the map $$\phi_{a} \colon z\mapsto \frac{z-a}{z- \bar{a}}$$ is an diffeo from $\mathbb{H}$ to $D$. Therefore, for any $f\colon \mathbb{H} \to \mathbb{H}$ and any $z \in \mathbb{H}$ the map $$\phi_{ f(w)} \circ f \circ \phi_{w}^{-1}\colon D \to D$$ takes any $\phi_w(z)$ to $\phi_{f(w)}(f(z))$. Now, for $z = w$ this means $f$ takes $0 = \phi_{w}(w)$ to $\phi_{f(w)}( f(w)) = 0$. We can apply now the Schwarz lemma and conclude that $$|\phi_{f(w)}(f(z)) | \le |\phi_w(z)|$$ for all $z\in \mathbb{H}$.

If we have equality for some distinct $z$, $w$ then we have equality for all $w$, $z$ in $\mathbb{H}$, and, moreover, $f$ is a diffeo of $\mathbb{H}$.