In a comment on question 110345 I made a claim that might be incorrect. I claimed that if f(z) is a nonconstant 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 counterexamples. It sounds like some weird generalization of Liouville's theorem.
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.
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 $D$.


$\begingroup$ similarly, any polynomial map of degree ≥ 2 admits bounded local inverses in any disc D whose interior does not contain a branch point. if the boundary circle of D does contain a branch point, there is such an inverse which does not converge on any larger disc. $\endgroup$ – roy smith Oct 24 '12 at 21:47

$\begingroup$ +1 and congratulations on 10k. $\endgroup$ – user9072 Oct 24 '12 at 23:32

$\begingroup$ Thanks for the illuminating counterexamples. I wonder if the following modification of my statement might be true. Let D be a closed disk in the complex plane having a positive radius. Let f(z) and g(z) be single valued functions of a complex variable such that (1) f(z) is analytic at all interior points of D. (2) g(z) agrees with f(z) at all interior points of D and is continuous at all points of D. Then f(z) can be extended to a function that is singlevalued and analytic in a connected open set containing D as a subset. $\endgroup$ – Garabed Gulbenkian Oct 26 '12 at 20:10

$\begingroup$ 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. $\endgroup$ – Johannes Ebert Oct 26 '12 at 20:47