1
$\begingroup$

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 $1 \ge |g(f(0))'| = |g'(f(0)) \cdot f'(0)| = |\frac{1}{k}| |f'(0)| \Leftrightarrow |f'(0)| \le |k| \le 1$

$\endgroup$
  • 2
    $\begingroup$ Hint: Use composition of functions, instead, composing with a biholomorphic map $\Bbb D\to\Bbb D$ that moves $f(0)$ to $0$. $\endgroup$ – Ted Shifrin Jul 27 '17 at 17:37
  • $\begingroup$ Thanks for your hint, but I sadly still can't seem to solve the problem. $\endgroup$ – Limechime Jul 27 '17 at 18:09
  • $\begingroup$ Why not? What happens when you do this? Perhaps you could edit your question to include the details when you do try this? $\endgroup$ – Ted Shifrin Jul 27 '17 at 18:12
  • $\begingroup$ You need a specific formula for $g(z)$. Call $f(0)=c$. These are famous linear fractional transformations that map $\Bbb D$ to $\Bbb D$. $\endgroup$ – Ted Shifrin Jul 27 '17 at 18:24
  • $\begingroup$ Does my proof look alright or did I miss something? $\endgroup$ – Limechime Jul 27 '17 at 18:57
0
$\begingroup$

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 $|f'(0)|\le 1$

$\endgroup$
  • $\begingroup$ That was a pretty straightforward solution. Thanks! $\endgroup$ – Limechime Jul 27 '17 at 19:44
  • $\begingroup$ 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 $|f'(0)| \le \infty$? $\endgroup$ – Limechime Jul 28 '17 at 9:35
  • $\begingroup$ Notice that $\underset{r\to 1^-}{\lim}\frac{1}{r}=1$ $\endgroup$ – user341124 Jul 28 '17 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.