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Take some power series $f(z) = \sum a_n z^n$ with a finite non-zero radius of convergence. I can rearrange the terms of the series, say, to get a different infinite series $$f_{\sigma}(z) = \sum_{n=0}^{\infty} a_{\sigma(n)} z^{\sigma(n)}$$ where $\sigma$ is a bijection from the nonnegative integers $\mathbb{N}$ to $\mathbb{N}$. Doing this rearrangement doesn't change anything in the interior of the disk of convergence, since the series converges absolutely inside the disk.

But suppose the original power series had conditional convergence at some points on the boundary of the disk. (The series $$f(z) = \sum_{n=1}^{\infty} \frac{z^n}{n}$$ is one example.) What is the space of possible functions $f_{\sigma}$ that could result from rearrangements? I'm thinking that we would have something at least vaguely akin to Riemann's rearrangement theorem, but that theorem, as far as I know it, can only deal with series of numbers, not series of functions.

echinodermata
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    Great question. – Andrés E. Caicedo Jan 19 '14 at 06:12
  • Hi. It has been suggested that you repost your question on MathOverflow, including a link to the version here indicating that no answers have yet been received. It would increase the exposure of the question with the experts that may have something to say about it. As I said, I doubt a full answer is known, but even partial results may be interesting and shed some light on the key difficulties. – Andrés E. Caicedo Feb 13 '14 at 14:45
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    By OP request + in regards to @AndresCaicedo's comment above, migrating to MO. – Willie Wong Feb 14 '14 at 09:54
  • Some rearangements will diverge at some points. When you say "a space of functions" do you consider only those rearrangements that converge everywhere? Or almost everywhere? – Alexandre Eremenko Dec 07 '14 at 16:38
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    @AlexandreEremenko some rearrangements could converge on different sets than others, and some could have a very very nasty set of convergence; in its full generality, the problem asks you to consider them all, with no restrictions whatsoever on converging everywhere, almost everywhere, etc. But since that seems difficult, feel free to restrict the scope of the problem to try to make progress. – echinodermata Dec 08 '14 at 10:24
  • Before we decide whether it is difficult or not, state your question precisely. I do not understand what is a "space of functions" in which each function is defined on its own set. – Alexandre Eremenko Dec 08 '14 at 15:32
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    @AlexandreEremenko It is difficult. The question is stated precisely. If the word "space" is confusing you, just read "collection" instead. – Andrés E. Caicedo Jan 03 '15 at 08:10
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    I'm a bit confused. For all permutations $\sigma$, $f_{\sigma}$ gives the same function inside the disc of convergence. So, by identity theorem, there is only one analytic function with that power series. Therefore, the set of possible functions contains only one analytic function. Is this correct ? – Srinivas K Mar 26 '15 at 05:03
  • How does this address the original question? – Yemon Choi Mar 26 '15 at 02:57
  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. – Ryan Budney Mar 26 '15 at 05:07

1 Answers1

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This question is not very well defined, so

there is a general science on rearragment of series in Banach spaces (see e.g. here). Now there are many ways to understand your question. For example, you can think about the function on the boundary as an element of $L_1(\mathbb{T})$ . Then because the functionals of scalar products with $z^n$ form a total system, the set of possible resulting functions is single point. (Note, that single point in $L_1$ is not a single function in ``pointwise'' sense)

However I don't know how to approach the problem when there is no natural Banach space of functions in which you want your resulting sum to lie in.

user68061
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  • The question is not very well defined, because I don't quite understand what is understood by the resulting ``function'' in the case when, let's say, series is nowhere convergent? Your link is very interesting, maybe you should add it at least as a comment to the question? The problem for all "partially defined" functions seem to monstrous)) – user68061 Jan 19 '14 at 06:36
  • Let $E_{\sigma}$ be the set of points $z$ where the series $f_{\sigma}(z)$ converges; each $f_{\sigma}$ is a function with domain $E_{\sigma}$. If it's too painful to consider a set of functions that don't even have the same domain, you could give up some generality and use some sort of "aggregate": Say, a function $F:\mathbb{C} \rightarrow \mathcal{P}(\mathbb{C})$ which takes each value of $z$ and returns the (possibly empty) set of convergent values of the series at $z$ over all rearrangements. In any case, you don't have to take "space" too seriously in the vein of a typical function space. –  Jan 20 '14 at 12:42
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    (Reposting comment lost on migration.) The question seems well defined to me; but it is a pointwise problem related to a known open problem, so sadly I doubt there is much current literature on it. Rearranging the series for f gives us a new series that converges on some points of the boundary. The domain of this rearrangement (on the boundary) may be different from the domain of the original series. Currently, I do not see clearly how to even describe the collection of sets obtainable as domains of a given f. The problem being asked is harder. – Andrés E. Caicedo Feb 14 '14 at 19:44