Does there exist $f$ holomorphic function in $B(0,1)$ such that $|f^{(n)}(0)|\geq n! n^n$, for all n?

Question

I have a doubt about this question:

Does there exist $f$ holomorphic function in $B(0,1)$ such that $|f^{(n)}(0)|\geq n! n^n$, for all n?

So far, my attempt to solve the problem is as follows:

Take $r\in(0,1)$ then $\overline{B(0,r)}\subseteq B(0,1)$ and by the Cauchy equations we have that

$ \begin{align*} n!n^n & \leq |f^{(n)}(0)| \\\ & = \left | \frac{n!}{2\pi i} \int_{C_r}\frac{f(w)}{w^{n+1}}\right | \\\ & \leq \frac{n!}{r^n} \sup_{w\in C_r}|f(w)| \end{align*} $

Donde $C_r=\partial (B(0,r))$, then

$(nr)^n\leq \sup_{w\in C_r}|f(w)|$, for each $r\in(0,1)$ and each $n$ natural integer.

From here I have tried several ways, however I have not arrived at anything concrete.

Do you have any idea?

Any help would be highly appreciated!

Thank you in advance!

Answer 1

Suppose such $f$ exists.

Let $f(z) =\sum_{n\ge 0}a_nz^n$ be the power series expansion in a neighborhood of $0$.

Then $a_n=\frac{f^n(0) }{n! }$

By the hypothesis $|a_n|=\frac{f^n(0) }{n! }\ge n^n$

Then radius of convergence $R$ is given by

$$\frac{1}{R}=\lim_{n\to\infty}|a_n|^{\frac{1}{n}}=\infty$$

Implies $R=0$