Is this a "folk theorem" about analytic functions of a complex variable? - MathOverflow

Is this a "folk theorem" about analytic functions of a complex variable?

2012-10-24T15:56:37Z

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 by quid for Is this a "folk theorem" about analytic functions of a complex variable?

2012-10-24T16:09:48Z

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 by Johannes Ebert for Is this a "folk theorem" about analytic functions of a complex variable?

2012-10-24T19:07:29Z

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 $D$.