# Holomorphic function's derivative's upper bound is 1

Let $f: \mathbb{D} \to \mathbb{D}$ be a holomorphic function. Then $|f'(0)| \le 1$. How do I prove this?

My thought so far were to write $f(z) = \sum_{n=0}^\infty a_n z^n = a_0 + g(z)$ with $g(0) = 0$. I rewrote this to $g(z) = f(z) - a_0$ and tried using the Schwarz Lemma. The problem I have is that $g$ is not a function that maps $\mathbb{D} \to \mathbb{D}$.

Edit: Let $g: \mathbb{D} \to \mathbb{D}$ be a holomorphic function so that $g^{-1}: \mathbb{D} \to \mathbb{D}$ is holomorphic too. I want $g$ to map $f(0)$ to $0$, that means I have $g(f(0)) = 0$ and $g^{-1}(0) = f(0)$. From there on, I don't really know how to go on. I can't see how this is going to help me, yet. Wouldn't I need $g(0) = 0$ too so that I can apply Schwarz Lemma?

Edit#2: A function $g(z) = \frac{az+b}{cz+d}$ is a linear fractional transformation if $ad-bc \neq 0$. Let's say $a = 1/k$, $b=-1$, $c=0$, $d=1$ and $f(0)=k$. Since $f : \mathbb{D} \to \mathbb{D}$, $|f| < 1$. So our function $g(z) = \frac{1}{k}z - 1$ is a linear fraction transformation.

I get $g(f(0)) = 0$. Let's look at the derivative: $g(f(z))' = \frac{1}{k} \cdot f'(z)$ for $z=0$, I get $1 \ge |g(f(0))'| = |g'(f(0)) \cdot f'(0)| = |\frac{1}{k}| |f'(0)| \Leftrightarrow |f'(0)| \le |k| \le 1$

• Hint: Use composition of functions, instead, composing with a biholomorphic map $\Bbb D\to\Bbb D$ that moves $f(0)$ to $0$. – Ted Shifrin Jul 27 '17 at 17:37
• Thanks for your hint, but I sadly still can't seem to solve the problem. – Limechime Jul 27 '17 at 18:09
• Why not? What happens when you do this? Perhaps you could edit your question to include the details when you do try this? – Ted Shifrin Jul 27 '17 at 18:12
• You need a specific formula for $g(z)$. Call $f(0)=c$. These are famous linear fractional transformations that map $\Bbb D$ to $\Bbb D$. – Ted Shifrin Jul 27 '17 at 18:24
• Does my proof look alright or did I miss something? – Limechime Jul 27 '17 at 18:57

Is $\mathbb{D}$ the unit disk with center at the origin? If it is then we can use Cauchy 's Integral formula on path $\alpha:[0,1]\to \mathbb{D}, \ \alpha(t)=re^{it}$ where $0<r<1$ to get that $|2\pi if'(0)|=\int_\alpha\frac{f(z)}{z^2}dz\le \max |\frac{f(z)}{z^2}|2\pi r\le \frac{2\pi}{r}$ because $|f(z)|\le 1$ for all $z\in \mathbb{D}$. This implies that $|f'(0)|\le\frac{1}{r}$ for all $0<r<1$ which means $|f'(0)|\le 1$
• After rethinking the solution, I didn't really understand the last implication. If $|f'(0)| \le 1/r$ for all $0<r<1$, how can you say that $|f'(0)| \le 1$? Doesn't it have to be $|f'(0)| \le \infty$? – Limechime Jul 28 '17 at 9:35
• Notice that $\underset{r\to 1^-}{\lim}\frac{1}{r}=1$ – user341124 Jul 28 '17 at 13:16