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Let $f\left(z\right)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be a power series with $0$s and $1$s as its coefficients ($a_{n}\in\left\{0,1\right\}$ for all $n$) with a radius of convergence of $1$. I call such holomorphic functions “digital" functions. I've been investigating the boundary values and limiting behaviors of digital functions, among other things, and have found myself confronted with the following question.

Suppose the unit circle is a natural boundary of $f\left(z\right)$ (meaning that there exists no analytic continuation of $f$ to a domain larger than $\mathbb{D}$). Letting $\xi$ denote any root of unity ($\xi=e^{2\pi i\frac{a}{b}}$ for some co-prime integers $a,b$), does it then follow that the sequence of partial sums: $$S_{N}\left(\xi\right)=\sum_{n=0}^{N}a_{n}\xi^{n}$$ is necessarily unbounded in magnitude (i.e., $\lim_{N\rightarrow\infty}\left|S_{N}\left(\xi\right)\right|=\infty$)? More generally, what, if anything, can be said about the behavior of a digital function $f\left(z\right)$ at a root of unity $\xi$ when it is known that the partial sums of $f$'s power series at $\xi$ are bounded in magnitude (i.e., $\limsup_{N\rightarrow\infty}\left|S_{N}\left(\xi\right)\right|<\infty$). Ex: is this a sufficient condition to imply the existence of an analytic continuation of $f$ to some open neighborhood of $\xi$?

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I think this question is odd because boundedness of $S_N(\xi)$ is a very weak condition whereas existence of an analytic continuation to some open neighborhood of $\xi$ (and even much less nice behavior) is a very strong condition.

A "random" Taylor series has no analytic continuation beyond its circle of convergence. (See Why are lacunary series so badly behaved?, and especially the survey of J.-P. Kahane recommended in the accepted answer, Sec. 7.) Imposing some additional conditions, such that $a_n\in\{0,1\}$ and $|S_N(\xi)|<C$ would not change very much.

Finding explicit (not random) examples might take some ingenuity. Robert Israel's series is not bad. A still more simple one is the theta series $\sum z^{n^2}$ (which is known to be a lacunary series), again at $z=-1$.

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  • $\begingroup$ Okay. Suppose the limit as we approach $\xi$ radially exists (such as in Robert Israel's example, where $f$ "vanishes" as you approach $-1$ from above on the negative real axis) and is finite; does that imply that the magnitudes of the $N$th partial sums of $f$ at $\xi$ are uniformly bounded with respect to $N$? Likewise, would the radial limit having infinite magnitude force the partial sums to be unbounded with respect to $N$? $\endgroup$
    – MCS
    Commented Feb 4, 2018 at 19:47
  • $\begingroup$ If $S_N$ are bounded, then one can prove (using summation by parts) that the function $f$ is bounded on the line segment $[0,\xi)$ but pretty much nothing beyond that. (I think I can conjure examples where it does not have a limit.) So yes, infinite limit does imply that sums are not bounded. If a finite limit exists, it simply means that the series admits Abel summation. In general, it does not imply at all that partial sums are bounded, and probably it is still true in this special case. $\endgroup$
    – Alex Gavrilov
    Commented Feb 5, 2018 at 8:51
  • $\begingroup$ Actually, I figured that out all on my own not long after Robert Israel answered my question. :) Specifically: the limsup of $\left|f\left(\xi x\right)\right|$ as $x$ increases to $0$ is bounded whenever the limsup of the Cesàro means of $f\left(\xi\right)=\sum_{n=0}^{\infty}c_{n}\xi^{n}$ is finite. $\endgroup$
    – MCS
    Commented Feb 10, 2018 at 20:38
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Consider the case $a_n = 1$ if $n$ or $n-1$ is a power of $3$, $0$ otherwise, and $\xi = -1$. Since $f(z) = (1+z) \sum_{k=0}^\infty z^{3^k}$ and $\sum_{k=0}^\infty z^{3^k}$ is a lacunary series, the unit circle is a natural boundary. But for $\xi = -1$ all partial sums are $-1$ or $0$.

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  • $\begingroup$ Huh. What if we require $f$ to not have any non-constant polynomial divisors? I suppose I should explain myself: my intuition is that for such functions, there should be at least one integer $a$ so that the exponents of $z$ are unequally distributed modulo $a$, so as to guarantee the divergence (or, at least, the boundedness away from zero) of the partial sums at $z=e^{2\pi i/a}$). Alternatively: (assuming the non-polynomial divisor condition), might it be possible to guarantee this behavior for at least some root of unity other than $z=1$? $\endgroup$
    – MCS
    Commented Feb 4, 2018 at 17:15
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    $\begingroup$ To avoid the polynomial factor, modify the power series as follows. Let $A$ be an infinite subset of the positive integers containing all powers of $3$ and such that for every interval $(3^k, 3^{k+1}) \cap A$ consists of exactly one number and it is a power of $2$. [It is easy to see that such an $A$ exists.] Now set $a_n = \chi_A (n)$, the latter being the indicator function of $A$. This is lacunary and has the unit circle as natural boundary. Then the sequence of values at $-1$ is alternating, so the partial sums take on only two values. $\endgroup$
    – David Handelman
    Commented Feb 5, 2018 at 0:40

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