I am asked to prove that if $f: \mathbb{C} \rightarrow \mathbb{C}$ is entire and $\lvert f(z) \rvert \leq 1$ on $\overline{B_1(0)}$, then $\lvert f'(z) \rvert \leq 4$ on $\overline{B_\frac{1}{2}(0)}$.
What I did is to choose any $z \in \overline{B_\frac{1}{2}(0)}$. I then easily observed that $\overline{B_\frac{1}{2}(z)} \subseteq \overline{B_1(0)}$. Then: $$ \lvert f'(z) \rvert = \left \lvert \frac{1}{2\pi i} \oint_{\partial K_\frac{1}{2}(z)} \frac{f(w)}{(w-z)^2}~\mathrm{d}w\right \rvert \leq \frac{1}{2\pi} \oint_{\partial K_\frac{1}{2}(z)} \frac{\overbrace{\lvert f(w) \rvert}^{\leq 1}}{\underbrace{\lvert w-z\rvert^2}_{=\frac{1}{4}}}~\mathrm{d}w \leq 2 $$ So I got an even better estimate... . Do you know if I made a mistake? Thank you.
I am also personally interested in what is the sharpest estimation of this kind. I know that $\lvert f' \rvert = 1$ is possible e.g. for $f(z) = z^2$.
Your bound seems alright, but there is a better bound by using the Schwarz-Pick thereom. Assume that $f(z)$ is non-constant, so $|f(z)|=1$ does not occur in the interior of $|z|\leq 1.$ Using that and applying to holomorphic mapping $f(z)$ from unit disk to unit disk, one gets that $$\frac{|f'(z)|}{1-|f(z)|^2}\leq \frac 1{1-|z|^2},$$ for $|z|<1$, which implies that $|f'(z)|\leq \frac 1{1-|z|^2}.$ With $|z|\leq \frac 1 2,$ this shows that $|f'(z)|\leq \frac 4 3.$ The bound is sharp as can be seen from the example $$f(z)=\frac {2z-1}{z-2}=\frac{z-\frac 1 2}{\frac 1 2z-1},$$ which maps unit disk to itself with $|f'(\frac 1 2)|=\frac 4 3.$