Extension of a holomorphic function $f$ to the disc $D(0,R)$

Question

Let $0<r<R<\infty$ and $A=\{z:r<|z|<R\}$. Let $f \in H(A)$. Prove that if for every positive integer $n$, there corresponds an $F_n \in H(A)$ such that $f=F_n^{(n)}$, then $f$ can be extended to the whole disc $\{z:|z|<R\}$. I think that if I can show that this function satisfies \begin{equation*} \int_\Gamma z^nf(z) dz=0 \end{equation*} for a closed path $\Gamma$ in $A$ and for every $n \ge 0$, then the negative coefficients of the Laurent series of $f$ become zero so that $f \in H(\{z:|z|<R\})$. However, I couldn't show the integral holds. Any good idea?

Answer 1

You are on the right track. Repeated integration by parts shows that $$ \int_\Gamma z^nf(z) dz = \int_\Gamma z^n F_{n+1}^{(n+1)}(z)\, dz = -n \int_\Gamma z^{n-1} F_{n+1}^{(n)}(z)\, dz\\ = \ldots = (-1)^n n! \int_\Gamma F_{n+1}'(z) \, dz = 0 $$ for all $n \ge 0$.

Answer 2

Let $\displaystyle\sum_{n=-\infty}^\infty a_nz^n$ be the Laurent series of $f$ on $A$. As you noticed, you ought to prove that$$n<0\implies a_n=0.\tag1$$You know that there is an analytic function $F_1\colon A\longrightarrow\Bbb C$ such that $(\forall z\in\Bbb C):f(z)=F_1'(z)$. If $\displaystyle\sum_{n=-\infty}^\infty a_{1,n}z^n$ is the Laurent series of $F_1$ on $A$, then\begin{align}\sum_{n=-\infty}^\infty a_nz^n&=f(z)\\&=F_1'(z)\\&=\sum_{n=-\infty}^\infty na_{1,n}z^{n-1}\\&=\sum_{n=-\infty}^\infty(n+1)a_{1,n+1}z^n\end{align}and therefore $a_{-1}=0$ indeed. The same approach allows you to prove that $(1)$ holds.