The Pseudo-Hyperbolic Metric and Lindelöf's Inequality

Let $\Delta$ denote the unit disc in the complex plane $\mathbb{C}$. For any two points $z,a\in \Delta$, we define the pseudo-hyperbolic metric on $\Delta$ by $$d(z,a)=|\varphi_a(z)|=\bigg|\frac{z-a}{1-\bar a z}\bigg|.$$ Observe that if $f:\Delta\to\Delta$ is holomorphic then by the Schwarz-Pick Theorem $$d(f(z),f(a))=\bigg|\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\bigg|\leq \bigg|\frac{z-a}{1-\bar a z}\bigg|=d(z,a),$$ and if $f\in \text{Aut}(\Delta)$ ($f$ is an automorphism of $\Delta$) then equality holds, i.e. $$d(f(z),f(a))=d(z,a) \qquad \text{for all $z,a\in \Delta$}.$$ In today's post, we prove that $d$ is actually a metric on $\Delta$.  Next time, as an application, we will use a useful inequality concerning $d$ to prove the following theorem:      

Theorem (Lindelöf's Inequality). Every holomorphic function $f:\Delta\to\Delta$ satisfies

$$\frac{|f(0)|-|z|}{1-|f(0)||z|}\leq |f(z)|\leq \frac{|f(0)|+|z|}{1+|f(0)||z|}\quad\text{for all $z\in\Delta$}.$$  

Note: Today's post is purely computational and is simply meant to be a reference (for homework-checking or the like). It's usually the case that calculations like those below are "left as an exercise." For this reason I wanted to include them on the blog just in case a student or two might find them helpful, because, well, occasionally we all be like --> 

Another note: I've tried to make sure there are no typos, but undoubtedly I've missed some. (One can get lost in a forest of inequalities.) Let me know if you spot any, and I'll correct them!

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The Pseudo-Hyperbolic Metric IS a Metric

Proof. Let $z,a\in \Delta$. That $d(z,a)\geq 0$ is clear, as is the fact that $d(z,a)=0$ if and only if $z=a$.  To see that $d(z,a)=d(a,z)$, notice that \begin{align}\label{one} d(z,a)=|\varphi_a(z)|= \bigg|\frac{z-a}{1-\bar a z}\bigg| \qquad \text{and} \qquad d(a,z)=|\varphi_z(a)|=\bigg|\frac{a-z}{1-\bar z a}\bigg|. \end{align} But \begin{align*} |1-\bar a z|^2 &= (1-\bar a z)(1-a\bar z)\\ &= (1-\bar za)(1-z\bar a)\\ &=|1-\bar z a|^2 \end{align*} and so $|1-\bar a z|=|1-\bar z a|$. This implies, referring back to (\ref{one}), that $d(z,a)=d(a,z)$.

It remains to show the triangle inequality, and we claim it suffices to prove \begin{align}\label{suffice} d(t_1,t_2)\leq |t_1|+|t_2|, \quad \text{for all $t_1,t_2\in\Delta$.} \end{align} Indeed for $z,w,a\in \Delta$, the triangle inequality $d(z,w)\leq d(z,a)+d(a,w)$ holds if and only if $d(\varphi_a(z),\varphi_a(w))\leq d(\varphi_a(z),0)+d(0,\varphi_a(w))$ since $\varphi_a\in \text{Aut}(\Delta)$ and $\varphi_a(a)=0$. Letting $t_1=\varphi_a(z)$ and $t_2=\varphi_a(w)$, this becomes $d(t_1,t_2)\leq d(t_1,0)+d(0,t_2)$ which is precisely (\ref{suffice}) since $d(z,0)=|z|$ for any $z\in \Delta$.

Thus our goal is to prove (\ref{suffice}), but in fact we aim to prove a much stronger result, namely:

Proposition. For any $t_1,t_2\in\Delta$, $$d(t_1,t_2)\leq \frac{|t_1|+|t_2|}{1+|t_1||t_2|.}$$

This of course implies (\ref{suffice}) since $1/(1+|t_1||t_2|)\leq 1$. So prove the proposition, let $t_1,t_2\in \Delta$ and observe that \begin{align}\label{here} 1-d(t_1,t_2)^2 \geq \frac{(1-|t_1|^2)(1-|t_2|^2)}{(1+|t_1||t_2|)^2} \end{align} since \begin{align*} 1-d(t_1,t_2)^2 &= 1-\bigg|  \frac{t_1-t_2}{1-t_1\bar t_2} \bigg|^2  \\[5pt] &= \frac{(1-t_1\bar t_2)(1-\bar t_1 t_2)-(t_1-t_2)(\bar t_1-\bar t_2)}{|1-t_1\bar t_2|^2} \\[5pt] &= \frac{(1-|t_1|^2)(1-|t_2|^2)}{|1-t_1\bar t_2|^2} \\[5pt] &\geq \frac{(1-|t_1|^2)(1-|t_2|^2)}{(1+|t_1||t_2|)^2}, \end{align*} where the last line follows from the triangle inequality. We also compute the following: \begin{align*} d(|t_1|,-|t_2|)&=\bigg|  \frac{|t_1|-(-|t_2|)}{1-|t_1|(\overline{-|t_2|})}  \bigg|  \\[5pt] &= \bigg| \frac{|t_1|+|t_2|}{1+|t_1||t_2|}  \bigg| \\[5pt] &=  \frac{|t_1|+|t_2|}{1+|t_1||t_2|} \end{align*} and so \begin{align*} 1-d(|t_1|,-|t_2|)^2 = 1-\left(  \frac{|t_1|+|t_2|}{1+|t_1||t_2|}  \right)^2  = \frac{(1-|t_1|^2)(1-|t_2|^2)}{( 1+|t_1||t_2|)^2}. \end{align*} Comparing the previous line with (\ref{here}) we see that $$ 1- d(t_1,t_2)^2\geq 1 - d(|t_1|,-|t_2|)^2$$ which implies $$d(t_1,t_2) \leq  d(|t_1|,-|t_2|) =   \frac{|t_1|+|t_2|}{1+|t_1||t_2|} .$$ This proves the proposition and hence the claim that $d$ satisfies the triangle inequality. Thus $d$ is indeed a metric.