Power series without analytic continuation

Question

Given a formal power series $\sum a_n z^n$ and a radius of convergence $R>0$, there are various ways to extend the function to the boundary such as

Abel's theorem
Fatou's lemma
$H^\infty$ theorem.

What is an example of a function that does (almost) nowhere to the boundary? Which power series are proven to not possess an analytic continuation beyond the radius of convergence.

The craziest things what I can construct is a finite number of essential singularities using an entire function $f$, which is not a polynomial, and looking at something like $z \mapsto f(1/z)$.

Well, this is an extremely well-studied problem and it is rather subtle. As Andrés Caicedo's fantastic answer in this MO-thread shows, quite a bit is known (and quite a bit unknown) about the subsets of the circle of convergence that may arise as sets of divergence of an analytic function. — t.b., Jun 22 '11 at 13:28
The examples given by AmbroseH are the canonical ones, besides the theta function $\vartheta(1;q) = 1 + 2\sum_{n \in \mathbb{Z}} q^{n^2}$ (which is a bit more subtle). — t.b., Jun 22 '11 at 13:35
By the way: Remmert's books Theory of complex functions and Classical topics in complex function theory contain a very detailed and lucid discussion of these topics, with lots of historical references. — t.b., Jun 22 '11 at 13:38

score 10 · Accepted Answer · answered Jun 22 '11 at 13:21

10

This is problem 2 in Chapter 2 of Stein & Shakarchi's Complex Analysis. They give two examples:

$f(z) = \sum_{n=0}^\infty z^{2^n},$

and, for $0 < \alpha < \infty,$

$f(z) = \sum_{n = 0}^\infty 2^{-n \alpha} z^{2^n}.$

The latter can in fact be extended continuously but not analytically to the boundary circle.

answered Jun 22 '11 at 13:21

Robert Haraway

1,475

This answers the 2nd question: Which powerseries are proven to not posess an analytic continuation beyond the radius of convergence? Thanks! – Marc Palm Jun 22 '11 at 13:28
Actually, the former series (I think) answers your first question. Stein and Shakarchi's hint for that problem is to consider $\theta = \frac{2\pi p}{2^k}$ for positive integers $p,k,$ and consider $z = r e^{i \theta},$ and – Robert Haraway Jun 22 '11 at 13:41
2

show that $|f(r e^{i\theta})| \to \infty$ as $r \to 1.$ – Robert Haraway Jun 22 '11 at 13:41
I apologize for not specifying that in my answer. – Robert Haraway Jun 22 '11 at 13:43
1

It's useful to notice that $f(z^{2^{n}}) = f(z) - (z + z^2 + \cdots + z^{2^{n-1}})$. – t.b. Jun 22 '11 at 13:49

score 9 · Answer 2 · edited Jun 22 '11 at 15:00

9

The sum $f(z) = \sum_{n = 0}^\infty z^{n!}.$ extends nowhere outside of $|z|\lt1$; there are infinitely many points in the set of n-th roots of unity in the boundary each of which blow up to infinity; the set of n-th roots is dense in the boundary $|z|=1$, so the series cannot squeeze out anywhere.

edited Jun 22 '11 at 15:00

t.b.

81,476

answered Jun 22 '11 at 14:57

gary

4,187

@Theo: Thanks again; hope editing this one was not as much work as the last one, which netted me some 60 points--I should give you a few of mine. – gary Jun 22 '11 at 15:15
2

you're welcome. Don't worry about the points, you earned them :) The only thing I did here was to change z^n! into z^{n!} to change $z^n!$ into $z^{n!}$ and some minor details. – t.b. Jun 22 '11 at 15:21
Hey. How do you prove this? – Gonzalo A. Benavides Jul 08 '22 at 00:22
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@GonzaloBenavides : Basically, you parametrize points i unit circle using Polars, i.e., $ z=e^{i \theta}; 0<\theta < 2\pi$. Then you use that Rationals are dense on the boundary, so that $z^{n!} $ equals $1 $ infinitely often, so that $\Sigma_{n=1}^{\infty} z^{n!}$ diverges, since n! cancels out the denominator. Because the set of Rational points $p/q$ are dense in $[0, 2\pi]$. – MSIS Jul 28 '22 at 00:03

Power series without analytic continuation

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