This is taken from an old complex analysis qualifying exam.
Problem
Let $\Delta$ denote the unit disc $\{z\in\mathbb{C}:|z|<1\}$.
Suppose $f:\Delta\rightarrow\Delta$ is holomorphic. Show that
$$\frac{|f(0)|-|z|}{1-|f(0)||z|}\leq|f(z)|\leq\frac{|f(0)|+|z|}{1+|f(0)||z|}$$
for all $|z|<1$.
Attempt
Since $f(0)\in\Delta$, define $\phi\in\operatorname{Aut}(\Delta)$ as
$$\phi(z)=\frac{f(0)-z}{1-\overline{f(0)}z}.$$
Then $\phi\circ f$ maps the unit disc into the unit disc and fixes zero. Thus, by Schwarz' lemma, we have
$$\left|\frac{f(0)-f(z)}{1-\overline{f(0)}f(z)}\right|\leq|z|.$$
But I don't see how this will lead to either of the desired inequalities. Any help would be greatly appreciated.
Best Answer
Here is a proof that uses only Schwarz' lemma and the triangle inequality.
Set $a=f(0)$ and $$\phi_a(z)=\frac{z-a}{1-\overline{a}z}.$$ Then $\phi_a\circ f$ maps the unit disc to the unit disc and fixes zero. Thus, by Schwarz' lemma, $$|\phi_a(f(z))|=\left|\frac{f(z)-a}{1-\overline{a}f(z)}\right|\leq|z|,$$ and so \begin{equation}|f(z)-a|\leq|z||1-\overline{a}f(z)|\leq|z|+|z||a||f(z)|.\tag{1}\end{equation}
Applying the triangle inequality, we arrive at $$|f(z)|\leq|z|+|z||a||f(z)|+|a|.$$ Then $$|f(z)|-|z||a||f(z)|\leq|z|+|a|$$ $$|f(z)|\leq\frac{|z|+|a|}{1-|a||z|}=\frac{|f(0)|+|z|}{1-|f(a)||z|},$$ Where the final equality uses the fact that $f(0)=a$. This is the second desire inequality.
To obtain the first inequality, we begin with $$|a|=|a-f(z)+f(z)|\leq |a-f(z)|+|f(z)|\leq |z|+|z||a||f(z)|+|f(z)|,$$ where the last inequality follows from (1). Then $$|a|-|z|\leq |z||a||f(z)|+|f(z)|$$ $$\frac{|a|-|z|}{|z||a|+1}=\frac{|f(0)|-|z|}{1+|f(0)||z|}\leq |f(z)|.$$