# $f$ is analytic and maps the unit disk to itself. Prove that $|f'(0)|\leq1- |f(0)|^2$

I am having difficulties with the following problem:

$\bf Given$: $f$ is analytic and maps from unit disk to itself. $\bf Prove:$ $|f'(0)|\leq1- |f(0)|^2$.

For some reason (unclear to me) it is hinted to consider: $h(z) = \frac{f(z)-\alpha}{\alpha^* f(z)-1}$ and to let $\alpha =f(0)$.

(note: $\alpha^*$=complex conjugate of $\alpha$.)

• Presumably, $f$ is assumed to be analytic? If so, you should edit that into the question. Jan 19, 2014 at 19:49
• If you look closely at the function $h$, and remember the Schwarz lemma you asked about previously, do you get an idea how to deduce it? Jan 19, 2014 at 19:56
• @DanielFischer Not really. I guess it has something to do with taking a derivative again... But I don't really see a similarity Jan 19, 2014 at 19:59
• How much of a connection do you see? Jan 19, 2014 at 20:04
• @DanielFischer I've been looking at it for a while now. But I don't see a relevant connection. Can you please help me? Jan 19, 2014 at 21:05

We know that for any $w$ with $\lvert w\rvert < 1$, the map

$$T_w \colon z \mapsto \frac{z-w}{1-\overline{w}\cdot z}$$

is an automorphism of the unit disk $\mathbb{D}$.

So if $f \colon \mathbb{D} \to \mathbb{D}$ is a holomorphic mapping, then $h = T_{f(0)} \circ f$ also is a holomorphic map from the unit disk to itself. And we have $h(0) = 0$ by construction. So $h$ is a map satisfying the preises of Schwarz' lemma, whence we have

$$\lvert h(z)\rvert \leqslant \lvert z\rvert\tag{1}$$

for all $z\in\mathbb{D}$. Now we can look at the form of $h$, as far as we know it. Expanding $T_{f(0)}$, the inequality $(1)$ becomes

$$\left\lvert \frac{f(z)-f(0)}{1 - \overline{f(0)}\cdot f(z)}\right\rvert \leqslant \lvert z\rvert \tag{2}$$

for all $z\in\mathbb{D}$. Rearranging $(2)$ yields

$$\left\lvert \frac{f(z)-f(0)}{z}\right\rvert \leqslant \lvert 1 - \overline{f(0)}\cdot f(z)\rvert.$$

The last step should not be difficult to find.