Until recently, I thought my math teacher had said in his real analysis class that on the boundary of the disc of convergence, a complex power series has at least one point at which it does not converge. But this is not true, so he must have said "uniformly converge", "absolutely converge" or so instead of "converge".

What is the correct statement about the converge of a complex power series on the boundary on the boundary of the disc of convergence?

  • $\begingroup$ Related, if not duplicate $\endgroup$ – The Chaz 2.0 Mar 14 '12 at 3:19
  • $\begingroup$ @TheChaz The issue discussed here is different. $\endgroup$ – Andrés E. Caicedo Mar 14 '12 at 5:06
  • $\begingroup$ @AndresCaicedo: I realized that later (after actually reading this entire question!). Should it be deleted? $\endgroup$ – The Chaz 2.0 Mar 14 '12 at 5:11
  • $\begingroup$ @TheChaz Your comment, you mean? I don't see why. If nothing else, it is natural to think at first both issues are the same. Better leave the comment, as a warning that they are not. $\endgroup$ – Andrés E. Caicedo Mar 14 '12 at 5:17

A series may converge (absolutely and therefore uniformly) at all points of the boundary of convergence, consider $\displaystyle \sum \frac{z^n}{n^2}$.

However, there is at least one point along which the holomorphic function the series defines cannot be analytically continued. Intuitively, think of $\sum z^n$. This series coincides with $1/(1-z)$, which is holomorphic everywhere (except at $z=1$), even though the series only has radius of convergence 1. What this result says is that the "except at $z=1$" is unavoidable.

Let me be more precise. A point on the boundary of the disk $D$ of convergence of a power series (say, about 0) is called regular iff there is an open neighborhood $U$ of that point and an analytic function that coincides with the power series on the intersection of $D$ and $U$. Otherwise, the point is singular.

The theorem your instructor was referring to is probably the following:

Theorem. If a power series has finite radius of convergence, then the set of singular points is a nonempty closed subset of its boundary of convergence.

The idea of the proof that there is at least one singular point is easy. For details, examples, and a discussion of analytic continuation, see for example Chapter 5 of Berenstein-Gay "Complex Variables. An introduction", or a similar textbook.

The thing is, calling $f$ the function defined by the power series, if there are no singular points, we can find around each point $p$ of the boundary, a little disk $D_p$ and an analytic function $f_p$ that coincides with $f$ on $D_p\cap D$.

But, by connectedness, if $D_p$ and $D_{p'}$ intersect, then on their intersection $f_p$ and $f_{p'}$ coincide. This is because they coincide (with $f$) on $D_p\cap D_{p'}\cap D$, which is nonempty if $D_p\cap D_{p'}\ne\emptyset$ to begin with.

Using compactness, we can then use the disks $D_p$ to see that there is a disk concentric with $D$ but slightly larger where there is an analytic function extending $f$. But then, the disk of convergence of $f$ was not $D$ to begin with.


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