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Hi!

I just came across the Ostroski-Hadamard gap theorem, and while I can understand the proofs as well as the principle that the series $\sum_{n=0}^\infty z^{2^n}$ ought to have a singularity at every $2^n$-th root of unity for every $n$, I feel I'm missing some intuition into what exactly is going on.

Specifically, there is certainly the intuition that the faster a power series' coefficients decrease, the larger the radius of convergence will be - say, comparing the geometric series with the exponential power series. When contrasted with lacunary series, this seems to fail: the coefficients seem to be increasingly "smaller", at least in an average sense, but the function becomes terribly ill-behaved. (One could try and argue that in the Cesàro sense the coefficients do tend to zero: if $\sum_{n=0}^\infty z^{2^n}=\sum_{k=0}^\infty a_k z^k$, then $\frac{1}{n}\sum_{k=0}^n a_k\approx\frac{\lfloor\log_2(n)\rfloor}{n}\rightarrow0$ as $n\rightarrow\infty$. On the other hand, the power series $\sum_{k=0}^\infty \frac{z^k}{k}$, while having the same radius of convergence, can easily, if non-uniquely, be analytically extended to the whole complex plane; I'd expect the same of any series of the form $\sum_{k=0}^\infty \frac{\log(k)}{k}z^k$.)

Can anyone share some insight?

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    $\begingroup$ I blame the parents. $\endgroup$
    – Will Jagy
    Mar 5, 2012 at 1:26
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    $\begingroup$ For power series with integer coefficients, there is Carlson's result that if such a power series has radius of convergence one, then either it is a rational function or the unit circle is a natural boundary. See Enumerative Combinatorics, vol..1, second ed., Exercise 4.46(c). $\endgroup$ Dec 2, 2019 at 0:26

5 Answers 5

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You can read more about this in the excellent survey

J.-P. Kahane: A century of interplay between Taylor series, Fourier series and Brownian motion, Bull. London Math. Soc. 29(1997), 257-279

In particular you can learn from this survey that the phenomenon you mentioned is rather typical. It's definitely worth having a look at it.

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Maybe your question is backwards. Natural boundary at the radius of convergence is the usual thing, and analytic continuation outside the circle of convergence is the fluke. Only VERY SPECIAL series have continuations.

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    $\begingroup$ Can this veryspecialness be more precise? $\endgroup$ Mar 5, 2012 at 23:49
  • $\begingroup$ @Mariano: Yes. A nice place to see how (historically) this came to be is the paper by Kahane mentioned in Liviu Nicolaescu's answer. Around the time of his dissertation, Borel came to the realization that Taylor series are, in general, not continuable. But he didn't have a way of making his intuition precise as randomness and probability where not yet formally developed. This was addressed in 1929 by Steinhaus. $\endgroup$ Mar 7, 2012 at 4:14
  • $\begingroup$ See also math.stackexchange.com/questions/318009/… $\endgroup$ Mar 2, 2013 at 2:40
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    $\begingroup$ @‍LiviuNicolaescu's answer mentioned by @AndrésE.Caicedo. $\endgroup$
    – LSpice
    Dec 2, 2019 at 15:34
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The mentioned gap theorem was generalized by Fabry (Acta Math. 1899, pp. 65-87): if the power series $f(z)=\sum_n a_n z^{\lambda_n}$ has radius of convergence $1$, and the exponents $\lambda_n\in\mathbb{N}$ satisfy $\lambda_n/n\to\infty$, then the unit circle is a natural boundary for $f(z)$.

Turán (Acta Math. Hung. 1947, pp. 21-29) gave a simple proof which might provide some insight into the phenomenon. His main inequality, from which he deduces the result, reads as follows:

$$ \max_{0\leq x\leq 2\pi}\ \left| \sum_{n=1}^N a_n e^{i\lambda_n x} \right| \leq \left(\frac{48\pi}{\delta}\right)^N \max_{a\leq x\leq a+\delta}\ \left| \sum_{n=1}^N a_n e^{i\lambda_n x} \right| $$

In other words, the key feature seems to be that on every arc of the unit circle, the partial sums are considerably bounded away from zero. For more details I would recommend to study Turán's paper.

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    $\begingroup$ I love the phrasing «considerably bounded away from zero.» :) $\endgroup$ Mar 4, 2012 at 21:03
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    $\begingroup$ @Mariano: Well, what can we do... I should add that this approach to Fabry's gap theorem made quite a revolution in complex analysis and analytic number theory. Two closely related papers of Turán are Acta Sci. Szeged 1952, pp. 209-218; Rev. Math. Pures Appl. 1956, pp. 27-32 $\endgroup$
    – GH from MO
    Mar 4, 2012 at 21:37
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    $\begingroup$ @GH, it was a compliment, really! Math writing, its wordings and style are, very often, like an uneventful tetris game, in which pieces more or less fall in their natural places—yet sometimes, just as in tetris, one finds little gems of phrasing like yours. $\endgroup$ Mar 6, 2012 at 22:13
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    $\begingroup$ @Mariano: Thank you! I should have put a smiley after "Well, what can we do..." :-) $\endgroup$
    – GH from MO
    Mar 6, 2012 at 23:35
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"Objection, the question assumes facts not in evidence!"

Talking about the general question as in the title, I wonder in what measure can we say that lacunary series are particularly badly behaved. Maybe the point is just that a lacunary form makes it easier to construct badly behaved series, which is slightly different. An example: we know that a real entire function $f$, say with real coefficients, may grow as fast as any given increasing function on $g:\mathbb{R}\to\mathbb{R}$, and building an example is easy by means of lacunary series. But $f(z+1)$ grows even faster, although the translation destroys the lacunary form.

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  1. The intuition is simple. Consider the example $f(z)=\sum_{n=0}^\infty z^{2^n}.$ This function satisfies the functional equation $f(z^2)=z^2+f(z)$. On the positive ray, we evidently have $f(r)\to\infty$ as $r\to 1$, and the functional equation shows that the same must happen on all rays $\{ re^{i\theta}:0<r<1\}$ for $\theta=k/2^n$. Since these rays are dense, all boundary points of the unit disk must be singular. Hadamard noticed that the special arithmetic nature of the sequence $2^n$ is irrelevant here it is enough to assume that $ \liminf m_{n+1}/m_n>1$. This is an implementation of the general principle: a lacunary series behaves in the same way in all directions. So if the radius of convergence is $R<\infty$ then all points $Re^{i\theta}$ must be singular.

  2. The final form of this theorem is due to E. Fabry, and it is called "Fabry's gap theorem", which implies, for example that $\sum_{n=0}^\infty z^{n^2}$ is singular at every boundary point of the circle of convergence. This gap theorem is in turn is a very special case of "Fabry's General Theorem". The best source for all of this is the German book of L. Bieberbach, Analytische Fortsetzung, Springer 1955, except that it is somewhat out of date. He describes the story of Fabry's General Theorem and related results in great detail. For a modern exposition of Fabry's theorems in English, I recommend my papers

MR2595767 Eremenko, Alexandre, Densities in Fabry's theorem. Illinois J. Math. 52 (2008), no. 4, 1277–1290, and

MR2431054 Eremenko, Alexandre, A version of Fabry's theorem for power series with regularly varying coefficients. Proc. Amer. Math. Soc. 136 (2008), no. 12, 4389–4394.

Remark. Alexandr Ostrowski was 5 years old when Fabry published his general theorem. So it is unclear why the Wikipedia author calls it "Hadamard-Ostrowski".

Remark 2. A complex analyst will not describe such behavior as "badly behaved". Anyway, this behavior is typical for analytic functions, both in the sense of Baire category and in the sense of measure.

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  • $\begingroup$ Thanks. I corrected. $\endgroup$ Nov 28, 2019 at 12:57
  • $\begingroup$ Best answer here. $\endgroup$
    – GH from MO
    Jun 1, 2021 at 23:17

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