Finding an upperbound for $|f^{(n)}(z)|$ 
Ahlfors: If $f(z)$ is analytic and $|f(z)| \le M$ for $|z| \le R$, find an
  upper bound for $|f^{(n)}(z)|$ in $|z| \le \rho < R$.

Attempt:


*

*Let $z$ satisfy $|z| \le \rho < R$ so that $|f(z)| \le M$ by hypothesis.

*Let $C$ be a circle around the origin of radius $\rho_1$ s.t. $\rho < \rho_1 < R$.

*Cauchy's estimate yields that for $z$ inside this circle, we have that
$$
|f^{(n)}(z)| \le {n!M \over R^n}
$$
so how is this not our upper bound?
 A: *

*Let $z$ satisfy $|z| \le \rho < R$.  Let $C$ be a circle around the origin of radius $R$.  Hence $z$ will be inside the perimeter of this circle.  Since $f$ is analytic on the whole plane, it is in particular analytic inside an open $\Delta$ disc of radius larger than $R$.  Hence Cauchy's Integral Formula yields that
$$
f^{(n)}(z) = {n! \over 2 \pi i} \int_{C} {f(\zeta)d\zeta \over (\zeta - z)^{n+1}}
$$
so that we also have
$$
|f^{(n)}(z)| \le {n! \over 2 \pi } \int_{C} {|f(\zeta)|\ |d\zeta| \over |\zeta - z|^{n+1}}
$$

*Since all points $\zeta$ on $C$ satisfy $|\zeta| \le R$, we have that all such points will satisfy $|f(\zeta)| \le M$ as well.  Furthermore, all of the $\zeta$ in $C$ will satisfy that $|\zeta - z| \ge (R - \rho)$ by construction.  Hence we have 
$$
{n! \over 2 \pi } \int_{C} {|f(\zeta)|\ |d\zeta| \over |\zeta - z|^{n+1}}  \le {n! \over 2 \pi } \int_{C} {M |d\zeta| \over (R - \rho)^{n+1}} = {n! 2\pi M R \over 2 \pi (R - \rho)^{n+1}} = {n!MR \over (R- \rho)^{n+1}}
$$

*Then we have shown that for all $|z| \le \rho < R$, we have that
$$
|f^{(n)}(z)| \le {n! MR \over (R-\rho)^{(n+1)}}
$$
which completes the task of establishing an upper bound for $|f^{(n)}(z)|$ on $|z| \le \rho < R$.
