Inequality using coefficients of Taylor development of bounded holomorphic funciton

Question

Let $f: D(0,1) \rightarrow \mathbb{C}$ be a holomorphic function on $D(0,1)$ and let $\sum_{n=0}^{+\infty} a_n z^n$ be its Taylor series at $0$.

We suppose that the Taylor development converges in $D(0,1)$, that $f(0) \geq 0$ and that $\forall z\in D: |f(z)| \leq 1$.

I try to prove that: $$\sum_{n=0}^{+\infty} \frac{|a_n|}{ 3^n} \leq 1 \, .$$

What I did:

Since $\frac{1}{3} \in D(0,1)$, $|f(\frac{1}{3}) | \leq 1$ which means $|\sum_{n=0}^{+\infty} \frac{a_n}{ 3^n} |\leq 1 $. However what I need to prove is stronger since $|\sum_{n=0}^{+\infty} \frac{a_n}{ 3^n}| \leq \sum_{n=0}^{+\infty} \frac{|a_n|}{ 3^n} $.

Could you please help with any technique or theorems I could use?

Answer 1

This is “Bohr's theorem on power series” and the number $\frac 13$ is the so-called “Bohr radius,” after the Danish mathematician (and soccer player) Harald Bohr:

Let $f$ be holomorphic in the unit disk $\Bbb D$ with power series $f(z) = \sum_{n=0}^\infty a_n z^n $, and $|f(z)| < 1$ for all $|z| \in \Bbb D$. Then $$ \sum_{n=0}^\infty |a_n z^n| \le 1 \quad \text{for } |z| \le \frac 13 \, . $$ (The number $\frac 13$ cannot be replaced by any larger number.)

The following proof is taken from Why Bohr got interested in his radius and what it has led to, which also contains more information about the history of this theorem and further developments.

The Schwarz-Pick theorem states that $$ \frac{|f'(z)|}{1-|f (z)|^2} \le \frac{1}{1-|z|^2} \, . $$ In particular for $z=0$ it follows that $$ |a_1| \le 1 - |a_0|^2 \, . $$

The crucial idea now is that the same relationship holds for higher derivatives as well: $$ \tag{*} |a_n| \le 1 - |a_0|^2 \quad \text{for } n = 1, 2, 3, \ldots $$ Then $$ \sum_{n=0}^\infty \frac{|a_n|}{3^n} \le |a_0| + (1 - |a_0|^2) \sum_{n=1}^\infty \frac{1}{3^n} = |a_0| + \frac{1 - |a_0|^2}{2} \\ = 1 -\frac{(1- |a_0|)^2}{2} \le 1 \, . $$

So it remains to prove $(*)$: For fixed $n\ge 1$, let $\omega$ be a $n^{\text{th}}$ root of unity and set $$ g(z) = \frac{1}{n} \sum_{j=0}^{n-1} f(\omega^j z) = a_0 + a_n z^n + a_{2n} z^{2n} + \ldots = h(z^n) $$ where $$ h(z) = \sum_{k=0}^{\infty} a_{kn} z^k $$ is holomorphic in the unit disk, and also $|h(z)| \le 1$. Therefore we can apply the Schwarz-Pick theorem to $h$ and conclude that $$ |a_n| = |h'(0)| \le 1 - |h(0)|^2 = 1 - |a_0|^2 \, . $$ This finishes the proof.

Remark: The condition $f(0) \ge 0$ is not needed.