Here is an old qual problem I'm working on, I have some idea, but I'm not sure if I'm correct or not. I would be happy if anyone could possibly confirm or correct me:
Let $U=\{z\in \mathbb{C} : \frac{1}{2}<|z|<2\}$. Suppose $f:U\rightarrow \mathbb{C}$ is holomorphic. Suppose that for every $n$, there is a holomorphic $g_n:U\rightarrow \mathbb{C}$ with $f(z)=(\frac{d}{dz})^n g_n(z)$, for all $z\in U$. Show that $f$ extends to a holomorphic function $\{z:|z|<2\}\rightarrow \mathbb{C}$.
Yes, my idea was as follows: We know that we can write the Laurent series. Let's say the Laurent series is $\sum_{-\infty}^{\infty} a_n z^n$. I tried to show that $a_n=0$ for all $n<0$. I considered, for example, $a_{-1}$. Assume it's non-zero. But, we know that $f(z)=g_1'(z)$ on $U$ for some $g_1$, which is analytic on $U$. So, these two functions ($f$ and $g_1'$) will also have same Laurent series, by the uniqueness. But, we cannot get any term like $\frac{1}{z}$ by differentiating the Laurent series of $g_1$, which I thought shows that $a_{-1}=0$. Doing the same thing for higher order derivatives repeatedly, I thought I get the result. Is this correct, or do I have some fatal mistake? My doubt is that I haven't used the number $\frac{1}{2}$ in the problem at all, so it doesn't have anything special maybe?
Thanks a lot in advance!
Your idea is spot-on. Being an $n$-th derivative in an annulus implies that the coefficients $a_k$ with $-1 \geqslant k \geqslant -n$ in the Laurent series vanish, so if a function is an $n$-th derivative for all $n$, its Laurent series is a power series, whence it furnishes the continuation to the disk.
The radii of the circles bounding the annulus are completely irrelevant (as long as the inner radius is smaller than the outer radius), so it's no wonder that you didn't use the specific values.