Function f(1/n)=1/n!

Question

I am trying to work on a problem in complex analysis. Although I know how to solve it, I am only stuck at one point.

The problem asks if there exist a holomorphic function $f$ on the unit disk such that $f(\frac{1}{n})=\frac{1}{n!}$.

Here, the approach will be to consider another function $g$ that coincides with $f$ on a discrete set (mainly $\{\frac{1}{n};n \in \mathbb{N}$}) and use the uniqueness theorem to show that they coincide with each other everywhere on the unit ball.

Now, if our function $g$ is not analytic on the unit ball, we will get what we need.

I cannot find such function $g$. It is easy to do it when $f(n)=\frac{1}{n+1}$ but here the factorial is making it a little bit difficult.

Answer 1

The idea is simply that such a function, to exist, would have to be "too flat". More formally: let $f$ be an hypotetical holomorphic function that satisfies your hypothesis. It would then satisty:

$$f(0)=\lim f(\frac 1n)=\lim \frac 1{n!}=0\\ f'(0)=\lim \frac{f(\frac 1n)}{\frac{1}{n}}=\lim \frac{n}{n!}=0\\ f^{(k)}=k!\lim \frac{f(\frac 1n)}{n^k}=k!\lim \frac{n^k}{n!}=0 $$

Thus $f=\sum \frac{f^{(n)}(0)z^n}{n!}=0$, which contradicts $f(\frac 1n)=\frac 1{n!}$

Answer 2

There is no such function. If there was, its Taylor series centered at $0$ would be of the form$$a_kz^k+a_{k+1}z^{k+1}+\cdots,$$for some $k\in\mathbb N$ and $a_k\neq0$. But then$$\lim_{n\to\infty}\frac{\left\lvert f\left(\frac1n\right)\right\rvert}{\left\lvert\frac{a_k}{n^k}\right\rvert}=1.$$In particular$$\lim_{n\to\infty}\frac{\left\lvert f\left(\frac1n\right)\right\rvert}{\frac1{n^k}}\neq0.$$But$$\lim_{n\to\infty}\frac{\frac1{n!}}{\frac1{n^k}}=0.$$