I am trying to solve the following problem:
Let $f$ be analytic on $\mathbb{D} = \{z \in \mathbb{C}: |z|<1\}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$ for $|z|<1$. Prove $|f'(0)| \leq 4$.
I have an idea of how to do it, but I am stuck at one of the steps.
I'd like to use Schwarz' lemma. So I wish to find a function $h = g \circ f, $ such that $h(0) = 0$, and where $g: f(\mathbb{D}) \rightarrow \mathbb{D}$. Then $h$ would be a function from $\mathbb{D}$ to $\mathbb{D}$, and Schwarz' lemma would apply, so that $|h'(0)| = |g'(f(0))f'(0)| \leq 1.$ If I have then managed to find a $g$ such that $g'(-1) = 1/4$, I believe I would be done with the proof.
Right now, however, I am stuck on what $f(\mathbb{D})$ is. I suppose I should use the assumption that $|1+f(z)| < 1 + |f(z)|$. Simply by drawing a picture I've been able to conclude that if $Im(z) = 0$, then $Re(z)$ cannot be in $\left[0, \infty\right].$ But that's all I have at this point...
$|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| \le 1+|f(z)|\, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words: $$ f(z) \in \Bbb C \setminus [0, \infty) \, . $$ It is a bit more convenient to consider $$ -f(z) \in \Bbb C \setminus (-\infty, 0] = U \, . $$ The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = \frac{w-1}{w+1}$.
Then $$ h(z) = \frac{\sqrt{-f(z)}-1}{\sqrt{-f(z)}+1} $$ maps $\Bbb D$ into $\Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| \le 1$.
Reversing the compositions we get $$ f(z) = - \left( \frac{1+h(z)}{1-h(z)} \right)^2 \\ f'(z) = - 4 \frac{1+h(z)}{(1-h(z))^3} h'(z) \\ f'(0) = -4 h'(0) $$ and the desired conclusion $| f'(0) | \le 4$ follows. Equality holds if and only if $$ f(z) = - \left( \frac{1+\lambda z}{1-\lambda z} \right)^2 \\ $$ for some $\lambda \in \Bbb C$ with $|\lambda| = 1$.