# Showing $\Bigg\vert \frac{f(z)-f(i)}{f(z)-\overline{f(i)}}\Bigg\vert\le \Bigg\vert \frac{z-i}{z+i}\Bigg\vert$

Let $\mathbb{H}=\lbrace z\in \mathbb{C}: \text{Im}z>0 \rbrace$. If $f:\mathbb{H}\to \mathbb{H}$ is analytic, then $$\Bigg\vert \frac{f(z)-f(i)}{f(z)-\overline{f(i)}}\Bigg\vert\le \Bigg\vert \frac{z-i}{z+i}\Bigg\vert,\forall z\in \mathbb{H} \tag{*}$$

Schwarz's Lemma and Schwarz-Pick Lemma cannot be applied at the moment since $f$ is a map on $\mathbb{H}$, not on the unit disc $B(0,1)$. However, the map $f\circ c^{-1}(z)$ is defined on the unit disc, where $c:\mathbb{H}\to B(0,1)$ is the Cayley transformation. It can be shown that $f\circ c^{-1}(z)=f\bigg(i\frac{1+z}{1-z} \bigg)$. I can obtain $f(i)$ by setting $z=0$, I obtain $f(z)$ by setting $z=\big(-1/2 + i/2\big)(\sqrt 3 - 1)\in \mathbb{H}$.

Since $f\circ c^{-1}(0)=f(i)$ and $f\circ c^{-1}\Big(\big(-1/2 + i/2\big)(\sqrt 3 - 1)\Big)=f(z)$, by Schwarz-Pick we have $$\Bigg \vert \frac{f(z)-f(i)}{1-\overline{f(i)}f(z)}\Bigg \vert\le \Bigg \vert \frac{c^{-1}(0)-c^{-1}\Big(\big(-1/2 + i/2\big)(\sqrt 3 - 1)\Big)}{1-\overline{c^{-1}\Big(\big(-1/2 + i/2\big)(\sqrt 3 - 1)\Big)}c^{-1}(0)}\Bigg\vert$$

but I'm not sure if this is the right way to go since I don't see how to get from the left-hand side of the latest expression to the left-hand side of $(*)$. May I have some insight here?

• $c(z) = \frac {z-i}{z+i}$ maps $\mathbb H$ onto $B(0, 1)$ with $c(i) = 0$, and
• $d(z) = \frac {z-f(i)}{z-\overline{f(i)}}$ maps $\mathbb H$ onto $B(0, 1)$ with $d(f(i)) = 0$.
Then $$g = d \circ f \circ c^{-1} : B(0, 1) \to B(0, 1)$$ satisfies the conditions of Schwarz's Lemma, so that $$|d(f(c^{-1}(w)| \le |w| \text{ for all } w \in B(0, 1) \,$$ and therefore $$|d(f(z))| \le |c(z)| \text{ for all } z \in \mathbb H$$ which is what you wanted to prove.
• @EthanAlvaree: That's almost the same question, only slightly more general with $w$ instead of $i$. And you got already an answer. Commented Apr 22, 2015 at 12:50