Let $f(z)$ be a holomorphic function defined on the disk $|z|\le 2$. Suppose $|f(z)|<1$ for $|z|\le 2$. It looks like there is a constant $c>0$ such that $|f(z)'|<c$ on the disk $|z|\le 1$ (for example, $c=1$?). I wonder if this is true. If yes, is an optimal such constant can be found? 



Bounding the derivative of a holomorphic function on a disk by its absolute value
Yes, this is true. The simple reason is that bounded functions form a normal family. Therefore their derivatives are uniformly bounded on every compact.


To obtain the estimate $|f'(z)|<1$, apply Cauchy theorem:
$$|f'(z)|=\left|\frac{1}{2\pi}\int\_{|\zeta-z|=1}\frac{f(\zeta) d\zeta}{(\zeta-z)^2}\right|\leq 1.$$
Equality in the right inequality can happen only if $f$ is constant, so we always have a strict inequality.


So this estimate is not exact. 
To obtain the exact one, 
follow @Christian Remling's suggestion and use the Schwarz lemma. For convenience, let $g(z)=f(2z)$, then
$g$ maps the unit disk into itself, so 
$$|g'(z)|\leq \frac{1-|g(z)|^2}{1-|z|^2}\leq\frac{1}{1-|z|^2}\leq \frac{4}{3}.$$
Thus the exact estimate us $|f'(z)|=|g'(z)|/2\leq 2/3.$ This is attained on
a Mobius map.


