Overconcentration of Poles on the Circle of Convergence of a Power Series with Bounded Coefficients

Let $$V$$ be an arbitrary set of infinitely many positive integers, and let: $$\varsigma_{V}\left(z\right)\overset{\textrm{def}}{=}\sum_{v\in V}z^{v}$$ Let $$T_{V}$$ denote the set of all $$t\in\left[0,1\right)$$ for which the limit:

$$c_{V}\left(t\right)\overset{\textrm{def}}{=}\lim_{x\uparrow1}\left(1-x\right)\varsigma_{V}\left(e^{2\pi it}x\right)$$ exists and is non-zero. Is $$T_{V}$$ necessarily finite (that is, would an "overconcentration" of radii on which $$\varsigma_{V}\left(z\right)$$ grows like $$\frac{1}{1-\left|z\right|}$$ as $$z$$ tends radially to the unit circle result in a contradiction against the boundedness of $$\varsigma_{V}\left(z\right)$$'s power series coefficients)? Or do there exist $$V$$ for which $$T_{V}$$ is infinite?

Part of the problem is that there is so much literature about the boundary behavior and singularities of power series that trying to find information about something this specific is like looking for the needle in the proverbial haystack. Any assistance would be much appreciated.

The set can be infinite (but only countable). For an example choose any $$t_j$$ linearly independent over $$\mathbb Q$$ and let $$V$$ be the set of all $$v$$ such that $$vt_j\mod 1 \notin (\frac 12-a_j,\frac 12+a_j)$$ for all $$1\le j\le J(v)$$ where $$a_j$$ decrease so fast that $$\sum_j a_j<+\infty$$ and the function $$J$$ increases to $$+\infty$$ so slowly that by the moment $$v_0$$ it reaches $$j_0$$ you may say that $$e^{2\pi ivt_j}$$ for $$v=1,\dots,v_0$$ model the joint distribution of $$j_0$$ independent variables equidistributed over the unit circle with high precision. That (with some small extra care and effort) will give you the existence of non-zero limits even for Cesaro means and the Poisson summation method is stronger.

Edit (about the sum of squares). Let us assume that the limit $$L_j=c(t_j)$$ exists for some finite collection of $$t_j$$. Consider $$L=\lim_{x\to 1}\sum_j \bar L_j(1-x)\sum_{v>0}\chi_V(v)x^ve^{2\pi i t_j v}\,.$$ On the one hand, $$L=\sum|L_j|^2$$. On the other hand, by Cauchy-Schwarz, $$L^2\le \limsup_{x\to 1}\left[(1-x)\sum_{v>0}\chi_V(v)^2x^v\right] \left[(1-x)\sum_{v>0}\left|\sum_j\bar L_j e^{2\pi i t_j v}\right|^2 x^v\right]$$ The expression in the first brackets is at most $$1$$, while, if you open the absolute value and take into account that $$(1-x)\sum_{v>0}e^{2\pi i (t_j-t_k) v} x^v\to 0$$ for $$j\ne k$$, you'll find that the limit of the sum in the second brackets is $$L$$. That gives $$L^2\le L$$, so $$L\le 1$$.

Apologies to everybody for endless edits. Here comes another one.

Consider a very fast increasing sequence $$n_k$$, $$k\ge 1$$ of positive integers (as usual, that means that we will choose them inductively). Let $$N_k=\prod_{m=1}^k(2^{n_m}-1)$$.

Define $$V_k$$ as the set of $$v\ge 1$$ congruent to $$N_{k-1}2^\alpha$$ modulo $$N_k$$ with some $$\alpha\in\{0,1,\dots,n_k-1\}$$ (we define $$N_0=1$$). It is clear that we just take $$n_k$$ possible residues modulo $$N_k$$, so the density of $$V_k$$ is $$n_k/N_k$$.

Put $$V=\cup_{k\ge 1} V_k$$ and $$t_k=N_k^{-1}$$. That's the whole construction. Now it remains to explain why it works.

First, $$V_k$$ are disjoint. Indeed, if $$m, all elements of $$V_k$$ are divisible by $$N_m$$ but no element of $$V_m$$ is.

Second, $$V_k$$ is $$N_k$$-periodic, so even the Cesaro averages $$L_k(z,M)=\frac 1{M}\sum_{v\in V_k,v\le M}z^{v}$$ have limits for all $$z=e^{2\pi it}\in\mathbb T$$ (we will always assume this relation between $$z$$ and $$t$$).

Third, $$\max_M\frac{|V_k\cap[1,M]|}{M}=N_{k-1}^{-1}$$, so as long as the series $$\sum_{k\ge 1}N_{k-1}^{-1}$$ converges, we can safely claim that the limit $$L(z)=\lim_{M\to\infty}L(z,M)=\lim_{M\to\infty}\frac 1{M}\sum_{v\in V,v\le M}z^{v}$$ exists and equals $$\sum_{k\ge 1} L_k(z)$$ ($$L_k(z)$$ are defined similarly with $$V_k$$ instead of $$V$$).

Fourth, each $$V_k$$ satisfies the property that if $$v\in V_k$$, then $$2v\in V_k$$ and, provided that $$v$$ is even, $$\frac v2\in V_k$$. Thus $$\{2v:v\in V_k\}=\{v\in V_k: v\text{ is even}\}\,.$$

Since for every $$M$$, we have $$L_k(z,M)+L_k(-z,M)=\frac 2M\sum_{v\in V_k, v\text{ even}, v\le M}z^v=\frac 2M\sum_{v\in V_k, v\le M/2}z^{2v}=L_k(z^2,M/2)$$, we conclude passing to the limit as $$M\to\infty$$ that $$L_k(z)+L_k(-z)=L_k(z^2)$$ for all $$k$$ and, thereby $$L(z)+L(-z)=L(z^2)$$.

Fifth, if $$t=N_{k}^{-1}$$, then $$L_m(z)=0$$ for $$m (just because $$V_m$$ is $$N_m$$-periodic and $$N_k>N_m$$, but $$L_k(z)=N_k^{-1}\sum_{\alpha=0}^{n_k-1}e^{2\pi i\frac{2^\alpha}{2^{n_k}-1}}\ne 0$$ (all terms are in one half-plane) and $$L_m(z)$$ with $$m>k$$ cannot change that because they are ridiculously small (at most $$n_m/N_m$$) compared to anything depending on $$n_1,\dots, n_k$$ only (it is this condition that forces the fast growth of $$n_k$$). Thus $$L(z)\ne 0$$.

• How would one go about showing that $T_{V}$ must be countable? Also, what would happen if $T_{V}$ is required to contain only rational numbers? Also, also, any links or references you could give me on the matter would be really helpful. Thanks! – MCS Jun 12 at 18:39
• @MCS The sum of squares of limits is finite. For "only rational numbers" I'll have to think for a (possibly long) while. I doubt there is something exactly dealing with your question, but if you look at the literature devoted to almost periodic functions, you'll find all necessary tricks there. – fedja Jun 12 at 18:43
• @MCS Done. Feel free to ask questions if something is unclear. – fedja Jun 13 at 17:38
• Erm... When you state the property the way you did, do you mean that the existence of $R(t)$ implies the existence of $R(t+\frac 12)$ and $R(2t)$ or something weaker or stronger than that? Also, we still assume that $R(t)\ne 0$ for $t\in T_V$, right? – fedja Jun 17 at 5:32
• @MCS Have you read "Alice in Wonderland"? (or was it "Through the looking glass"?). There is a passage there where she recites "Father William" and I like the ending a lot ;-)) That was a joke, of course, but seriously, step back for a while and spend some time understanding what you really want to get and where it is all going. Once you have a clear picture of what you are after, I'll be happy to think of some concise question you may have (start a new thread then) but now I'd rather fry some other fish. Still, you can always ask for clarifications on what has been posted already :-). – fedja Jun 21 at 1:01