Showing that $|f^{(n)}| \le n!n^n$ and then making this result sharper

Question

Ahlfors: Show that the successive derivatives of an analytic function at a point can never satisfy $|f^{(n)}(z)| > n!n^n$. Formulate a sharper theorem of the same kind.

Attempt for Part One:

1. Let $\Delta$ be a neighborhood around $z$ of radius $r$ small enough so that $f$ is analytic on $\Delta$. Let $C$ be a circle around $z$ of radius $r$.

2. Cauchy's integral formula then yields that

   $$ f^{(n)}(z) = {n! \over 2 \pi i}\int_{C} {f(\zeta) \over (\zeta - z)^{n+1}}\ d\zeta $$

3. Since $\mathbb{R}$ is complete, we have that $M = \max\{|f(\zeta)| : |\zeta - z| \le r \} \in \mathbb{R}$ exists. Furthermore, we have that $|\zeta - z| \le r$ for all $\zeta$ within the perimeter of $C$. Hence we have

   $$ |f^{(n)}(z)| \le \left|{n! \over 2 \pi i}\right|\int_{C} {|M| \over |r^{n+1}|}\ |d\zeta| \le {n! \over 2 \pi} {M 2 \pi r \over r^{n+1}} = {n!M \over r^{n}} $$

4. Hence we have Cauchy's estimate:

   $$ |f^{(n)}(z)| \le {n! M \over r^n} $$

5. We may further assume that $M > 1$ above (it doesn't affect any of the the inequality reasoning).

6. Consider that there is a point where $n$ is large enough s.t. $n^n\ge {M \over r^n}$ (indeed, it is when $n \ge {M \over r}$). At such a point, we have that:

   $$ n^n \ge {M \over r^n} \implies n^n r^n \ge M \implies r^n \ge {M \over n^n} \implies {1 \over r^n} \le {n^n \over M} $$

7. Then assuming $n^n \ge {M \over r^n}$, we have that

   $$ \underbrace{|f^{(n)}(z)| \le {n!M \over r^n}}_{\text{Cauchy's estimate}} = {n!M} \cdot \left({1 \over r^n}\right) \le \underbrace{{n!M}\cdot \left({n^n \over M}\right)}_{\text{since }{1 \over r^n} \le {n^n \over M}} = n! n^n $$

   as desired.

8. We have thus far shown that if $n \ge {M \over r}$, then

   $$ |f^{(n)}(z)| \le n!n^n $$

Question: In what way could we use this result to make a sharper theorem of the same kind?

Answer 1

The inequality you derive in the 4th point $$ \lvert f^{(n)}(z) \rvert \le \frac{n! M}{r^n} $$ is enough to substantially improve the result:

if $g(n)$ is any increasing function that increases to $\infty$, no matter how slowly, then chose $n_0$ such that $f(z)$ is analytic in the circle centered in $z$ of radius $$ \frac{2}{g(n_0)} $$ and such that $2^{n_0} > M$ then for any $n>n_0$ using your inequality in the circle of radius $$ r=\frac{2}{g(n)} < \frac{2}{g(n_0)}$$ gives $$ \lvert f^{(n)}(z) \rvert \le \frac{n!M}{r^n} =n!M2^{-n}g(n)^n \leq n! 2^{n_0} 2^{-n}g(n)^n \leq n! 2^{n} 2^{-n} g(n)^n =g(n)^nn!. $$

This result is essentially best possible, to see it, suppose that the inequality $$ \lvert f^{(n)}(z) \rvert \le H(n)^n n! $$ is true for certain bounded increasing function $H(n)$ and for all the analytic functions $f(z)$, then if $C > H(n)$ for all $n$ then consider the funcion $$ h(z) = \sum_{n=0}^\infty (2C)^n z^n $$ this function is analytic inside the circle of radius $1/2C$, but we have $$ \lvert h^{(n)}(0) \rvert = (2C)^n n! > H(n)^n n! $$ a contradiction.