# $f(z) = \sum_{n=0}^\infty a_nz^n$ converges in the unit disk and $|f(z)| < 1$. Show that $|a_0|^2 + |a_1| \leq 1$.

The series $\sum_{n=0}^\infty a_nz^n$ converges in the unit disk $|z| < 1$ and defines a function mapping the unit disk into itself. Show that $|a_0|^2 + |a_1| \leq 1$.

Only thing I've thought to try so far has been to choose $S$ an automorphism of the unit disk sending $f(0) = a_0$ to $0$, applying Schwarz's lemma to get $|Sf(z)| \leq |z|$ on the unit disk, and doing some algebra, but it isn't quite coming out.

Is there a better approach?

• Do you know the Schwarz-Pick lemma? – Daniel Fischer Aug 19 '14 at 19:04
• Ah, proving the Schwarz-Pick lemma is an exercise in Ahlfors, though not by name. I had forgotten about it, it makes this exercise trivial. I've added it to my toolset. Thank you, @DanielFischer. – user169845 Aug 19 '14 at 19:10

The Schwarz-Pick lemma says that for a holomorphic $f\colon \mathbb{D}\to \mathbb{D}$ we have
$$\left\lvert \frac{f(z)-f(w)}{1-\overline{f(w)}f(z)}\right\rvert \leqslant \left\lvert \frac{z-w}{1-\overline{w}z}\right\rvert$$
for all $z,w \in \mathbb{D}$. Dividing by $\lvert z-w\rvert$ and letting $w\to z$, the "infinitesimal version"
$$\frac{\lvert f'(z)\rvert}{1-\lvert f(z)\rvert^2} \leqslant \frac{1}{1-\lvert z\rvert^2}$$
for all $z\in\mathbb{D}$ follows. Insert $z = 0$.