Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A\subset\Bbb C$ be a subset of the unit circle. Consider the following condition on $A$.

Cond. There exists a sequence $\{a_i\}_{i=1}^\infty$ of complex numbers such that $$\sum_{n=1}^\infty a_nz^n$$ is a power series with radius of convergence $1,$ and $A$ is exactly the subset of the unit circle in which the series converges.

Are there any interesting conditions on a subset $A$ of the unit circle which imply, are implied by or are equivalent to Cond.? I think all finite subsets of the circle have this property. What about the countable subsets? Does it have anything to do with measurability?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Try this POST or, indeed, the whole thread. A quote:

The convergence set has to be F_sigma_delta, since the (pointwise) convergence set for any sequence of continuous functions is F_sigma_delta.

Herzog and Piranian (together) proved in 1949 that any F_sigma subset of |z| = 1 can be the convergence set of some power series with radius of convergence 1. Lukasenko proved in 1978 that some G_delta subsets of |z| = 1 cannot be the convergence set of any power series with radius of convergence 1. For a fairly elementary survey of the problem of characterizing the convergence set for a power series in C (complex numbers), see Thomas W. Korner, "The behavior of power series on their circle of convergence" [pp. 56-94 in "Banach Spaces, Harmonic Analysis, and Probability Theory", Springer Lecture Notes in Mathematics 995, Springer-Verlag, 1983]. This is a beautifully written paper that contains detailed proofs of virtually everything and is pitched at the level of a beginning graduate student in math.

Dave L. Renfro

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.