TITLE: Analytic function $f$ in $\overline{\mathbb{D}}$ satisfying $\left\lvert\,f'(\tfrac{1}{2})\,\right \rvert\leq 8.$
QUESTION [10 upvotes]: Let $f$ be an analytic function on the closed unit disk $\overline{\mathbb{D}}$.
On its boundary $\partial \mathbb{D}$ it holds that $\lvert\,f(z) -z\rvert < \lvert z\rvert$.
I now have to show that 
$$
\left\lvert\,\, f'\left(\frac{1}{2}\right)\right\rvert \leq 8.
$$
I already figured out, that there cannot be a $\,z_0 \in \partial \mathbb{D}$, such that $\,f(z_0) =0,\,$ since that would mean that $\lvert 0-z_0\rvert < \lvert z_0\rvert\,$ 
which produces a contradiction, since the inequality is strict.
I also know, that $f$ takes its maximum on $\partial \mathbb{D}$ according to the maximum modulus principle.
My assumption is that i should get $\,\lvert\,f'(z)\rvert < \lvert z^{-3}\rvert\,$ by looking at the numbers, which seem a bit random to me.
But now I'm stuck. Any help would be greatly appreciated!

REPLY [7 votes]: Cauchy's Integral Formula provides that
$$
f'\Big(\frac{1}{2}\Big)=\frac{1}{2\pi i}\int_{|z|=1}\!\frac{f(z)\,dz}{\big(z-\frac{1}{2}\!\big)^2},
$$
and hence
\begin{align}
\left\lvert\, f'\Big(\frac{1}{2}\Big)\right|&\le 
\frac{\max_{|z|=1}\lvert\,f(z)|}{(1/2)^2}=4\max_{|z|=1}\lvert \,f(z)|\le 
4\max_{|z|=1}\big(\lvert\, f(z)-z|+|z|\big)
\\&\le 4\max_{|z|=1}\big(|z|+|z|\big)=8.
\end{align}