A sequence $a_1=f'(0),a_2=f''(0),...$

Question

I am working on this problem from my past Qual

"Give a sequence s.t. there is no analytic function $f:D\to \mathbb{C}$ s.t. $a_1=f'(0),a_2=f''(0),...$" where $D$ is the unit disk."

The only thing I can think of the Cauchy's integral formula$$f^{(n)}(0)=\frac{n!}{2\pi i} \int \frac{f(w)}{w^n}dw$$ But that's it. I don't see a relation between these to construct a counterexample. How do I proceed?

Answer 1

Take $a_n=(n!)^{2}$. If such a function exist then $\sum \frac {f^{(n)} (0)} {n!} z^{n}$ would converge for $|z| <1$. But $\sum \frac {f^{(n)} (0)} {n!} z^{n}=\sum (n!)z^{n}$ and this series converges only for $x=0$. [If you prefer you can take $a_n=n^{n}n!$].

Answer 2

Use Cauchy around $\gamma$, the circle of radius $R$. Then, you have $$f^{(n)}(0)=\frac{n!}{2\pi i}\int_\gamma\frac{f(w)}{w^n}dw$$ A trivial bound gives you $$|f^{(n)}(0)|\leq \frac{n!}{2\pi }\frac{M_R}{R^n}2\pi R.$$ i.e. $$|f^{(n)}(0)|\leq n!\frac{M_R}{R^{n-1}}.$$ Where $M_R$ is the maximum of $|f|$ inside the circle of radius $R$. That means that if $f$ is analytic on the disc, $\frac{|f^{(n)}(0)|}{n!}R^{n-1}$ has to be bounded for any $0<R<1$.

Take then $a_n=(n+1)!2^{n-1}$ note that $\frac{a_n}{n!}R^{n-1}$ is not bounded for $R=1/2$.