# Is this a “folk theorem” about analytic functions of a complex variable?

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.

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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.

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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 $D$.
I don't think I understand this last remark; probably you want $g$ to be continuous. But you can extend the square root to a continuous function on the whole plane (say by Tietzes theorem). But as I said, there is no analytic extension. –  Johannes Ebert Oct 26 '12 at 20:47