# Bound for $f''$ for $f$ unit disc mapping

I have problems with this question: Let $f\colon D(0 ; 1) \to D(0 ; 1)$ be an analytic function such that $f(0) = f'(0) = 0$. Show that $|f''(0)| \leq 2$.

I tried using Schwarz Lemma, but I have no clue how to bound $f'$ first. I tried using Cauchy's Integral Formula: \begin{align*} |f''(0)| &= \left|\frac{1}{\pi i}\int\limits_{\{z\,:\,|z|=1\}} \frac{f(z)}{z^3}\, dz\right| \\ &\leq \frac{1}{\pi}\int\limits_{\{z\,:\,|z|=1\}} \frac{1}{|z|^3}\, dz\\ &= 2 \end{align*} since |z| = 1 on the unit circle, but I am not sure whether you can apply Cauchy's Integral Formula in such a way.

$$|f''(0)| =` \left|\frac{2}{2\pi i} \int_{|z|=1-\epsilon} \frac{f(z)}{z^3} dz\right|\le \frac{1}{\pi}\times \frac{1}{|1-\epsilon|^3}\times 2\pi (1-\epsilon)=\frac{2}{(1-\epsilon)^2}.$$
Note: the assumption $f(0)=f'(0)=0$ is not necessary for this argument to work, and the argument allows to derive bound on derivatives of all orders.
1. If you want to use Schwarz lemma with given assumptions, then observe that for $z\ne 0$, Schwarz lemma gives $|f(z)/z|\le 1$, with equality for some $z$ if and only if $f(z) = e^{i\theta} z$. But since derivative at $0$ is zero, we have $|f(z)|<|z|$. Note also that $\lim_{z\to 0} \frac{f(z)}{z} = f'(0)=0$, therefore the function $g(z) = f(z)/z$ maps ${\mathbb D}$ into ${\mathbb D}$ (with removable singularity at the origin), and $g(0)=0$. By Schwarz lemma, $|g'(0)|\le 1$ with equality if and only if $g(z)=e^{i\theta}z$. By Taylor expansion of $f$, we have $$g(z) = \sum_{n=2}^\infty \frac{f^{(n)}(0)}{n!} z^{n-1}.$$
Hence $g'(0)=f''(0)/2$. Therefore $|f''(0)|\le 2$ with equality if and only if $f(z) = e^{i\theta} z^2$.