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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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up vote 15 down vote accepted

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 $D$.

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I prefer this example to quid's. –  Gerald Edgar Oct 24 '12 at 21:20
    
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. –  roy smith Oct 24 '12 at 21:47
    
+1 and congratulations on 10k. –  quid Oct 24 '12 at 23:32
    
Thanks for the illuminating counter-examples. 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 single-valued and analytic in a connected open set containing D as a subset. –  Garabed Gulbenkian Oct 26 '12 at 20:10
    
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
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