Suppose $f$ is analytic on $D(P,r)$ and unbounded on that disk in that there is no $M$ satisfying $|f(z)| \leq M $ for all $ z \in D(P,r)$. Prove that the radius of convergence of the power series for $f$ about $P$ is equal to $r$.
This question was from my Complex Analysis professor. I am not sure I am understanding the question correctly. Is it saying that because the function $f$ is unbounded, then there can not be a $supremum = R$ that is the radius of convergence(or more specifically that the radius of convergence is $\infty$).
Or is it saying that the radius of convergence is only on the disk $D(P,r)$ where f is analytic?
Thank you for any help.
The statement is that the power series of $f$ cannot converge in a disk $B(0,R)$ for any $R>r$. The reason is if it does converge on a larger disk then the closed disk with radius $r$ around $0$ would be a compact set and the sum of the series is a continuous function this compact set, hence bounded. This forces $f$ to be bounded.