# Analytic function $f$ in $\overline{\mathbb{D}}$ satisfying $\left\lvert\,f'(\tfrac{1}{2})\,\right \rvert\leq 8.$

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!

• Are you sure about your conditions? We can get a much smaller bound with the given conditions. – Daniel Fischer Feb 1 '14 at 19:56
• Yes, that's how it stands on my sheet. As said, I'm slightly confused by the explicit choice of numbers. – stebu92 Feb 1 '14 at 20:00

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}