Question

In a comment on question 110345 I made a claim that might be incorrect. I claimed that if f(z) is a non-constant analytic function defined by a power series whose circle of convergence C has a positive radius, then f(z) cannot be bounded at all points in the interior of C. But is this really true? Or am I just imagining that I learned it somewhere. I could not come up with any simple counter-examples. It sounds like some weird generalization of Liouville's theorem.

Answer 1

This is not true. For a counterexample take

$$\sum_{n=1}^{\infty} \frac{x^n}{n^2} $$

The radius of convergence is one and this is bounded by

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6 $$

You are likely confusing this with the maximum modulus principle saying that there must be no local maximu (in absolute value) in the interior if the function is nonconstant.

Answer 2

You are probably misunderstanding the following folk theorem: If $D $ is the convergence disc of a power series converging to $f$, then there must be some singularity of $f$ on $\partial D$. In other words, you cannot continue $f$ analytically onto a larger disc. A counterexample that is more explicit than quids example is the power series expansion of $\sqrt{1+z}$ around $z=0$, which has convergence radius $1$. The singularity at $z=-1$ is not a pole.

A hint to the proof: if $D \subset U$ is a disc in the domain of definition of a holomorphic function $f$, then the Taylor expansion around the midpoint of $D$ converges in $D$.