Holomorphic function on the unit disc

Question

Let $D=\{z \in \mathbb{C} \mid |z| < 1 \}$ be the unit disc. Let $g: D \to D$ be a holomorphic function with $g(0)=0$. Let $h = \dfrac 1{1-g}$, and assume that $$ h(z) = \sum_{k=0}^{\infty} a_k z^k ,|z|<1 $$

Show that $|a_n|<1(n = 1,2,3,\cdots)$.

Equivalently, let $f:D \to F$ be a holomorphic function with $f(0)=1$, where $F = \{ z \in \mathbb{C} \mid \mathrm{Re} z > \dfrac12\}$ and $$ f(z) = \sum_{k=0}^{\infty} a_k z^k ,|z|<1 $$

Show that $|a_n|<1$ $(n=1,2,\ldots)$

Answer 1

Here is a possible proof of the second version:

Let $f(z) = \sum_{n=0}^\infty a_n z^n$ be holomorphic in the unit disk $\Bbb D$ with $f(0) = 1$ and $\operatorname{Re}f(z) > \frac 12$ for all $z \in \Bbb D$. Then $|a_n| \le 1$ for $n = 1, 2, 3, \ldots$.

Proof: $g(z) = \frac{f(z)-1}{f(z)}$ maps $\Bbb D$ into itself with $g(0) = 0$, so we can apply the Schwarz lemma and conclude that $|g'(0)| \le 1$. It follows that $$ |a_1| = |f'(0)| = \left| \frac{g'(0)}{(1-g(0))^2}\right| = |{g'(0)}|\le 1 \, , $$ which is the desired estimate for $n=1$.

In order to prove that the same estimate holds for any $n \ge 2$ as well one can use the same “trick” as in Inequality using coefficients of Taylor development of bounded holomorphic funciton: For fixed $n\ge 2$, let $\omega$ be a $n^{\text{th}}$ root of unity and set $$ F(z) = \frac{1}{n} \sum_{j=0}^{n-1} f(\omega^j z) = a_0 + a_n z^n + a_{2n} z^{2n} + \ldots = \tilde f(z^n) $$ where $$ \tilde f(z) = \sum_{k=0}^{\infty} a_{kn} z^k $$ is holomorphic in the unit disk.

The function $\tilde f$ also satisfies $\tilde f(0) = 1$ and $\operatorname{Re}\tilde f(z) > \frac 12$, so we can apply the above argument to $\tilde f$ instead of $f$, and conclude that $|\tilde f'(0)| \le 1$. That finishes the proof since $a_n = \tilde f'(0)$.