Absolute convergence of logarithm of polynomial with positive coefficient ($\ln G(z) = \sum\limits_{i = 0}^\infty {{q_i}{z^i}}$)

Special problem: Let $G(z)$ be a probability generating function(pgf, the $z$ can be seen as real number or complex number), that is $$G(z) = \sum\limits_{i = 0}^\infty {{p_i}{z^i}} ,(\left| z \right| \leqslant 1;\sum\limits_{i = 0}^\infty {{p_i} = 1;} 0 \leqslant {p_i} \leqslant 1).$$ the logarithm of $G(z)$ is $$\ln G(z) = \sum\limits_{i = 0}^\infty {{q_i}{z^i}} ,(\left| z \right| \leqslant 1).$$ If ${p_0} > {p_1} > {p_2} > \cdots$ , then show that ${{q_i}}$ is absolutely convergent, namely $\sum\limits_{i = 0}^\infty {\left| {{q_i}} \right|} < \infty$, or give a counterexample.

I want to use Faa Di Bruno Formula (http://en.wikipedia.org/wiki/Faà_di_Bruno's_formula) to get the expression of $q_i$. then to estimate $\sum\limits_{i = 0}^\infty {\left| {{q_i}} \right|}$.

Remark: The problem above is a special case of my Problem which is connected to probability generating function：

Let $G(z)$ be a probability generating function, that is $$G(z) = \sum\limits_{i = 0}^\infty {{p_i}{z^i}} ,(\left| z \right| \leqslant 1;\sum\limits_{i = 0}^\infty {{p_i} = 1;} 0 \leqslant {p_i} \leqslant 1).$$

Problem 1. Under which necessary and sufficient condition on $p_i$, the logarithm of $G(z)$ $$\ln G(z) = \sum\limits_{i = 0}^\infty {{q_i}{z^i}} ,(\left| z \right| \leqslant 1).$$ is absolutely convergent, namely $\sum\limits_{i = 0}^\infty {\left| {{q_i}} \right|} < \infty$?

My conjecture is that if $G(z)$ has no zeros then $\sum\limits_{i = 0}^\infty {\left| {{q_i}} \right|} < \infty$. If this conjecture is not ture, can somebody give me a counterexample.

The Problem 1 is similar to an open problem：

Problem 2. Let $f(t) = \sum\limits_{k = 0}^\infty {{a_k}{e^{kit}}}$, we use the norm $\left\| f \right\| = \sum\limits_{k = 0}^\infty {\left| {{a_k}} \right| }$. Under which conditions on $f$ is the sequence $\left\| {{f^n}} \right\|$ bounded?

This problem are firstly proposed by Beurling(1938), quoted by Henry(1953). Hedstrom(1966) call the problem "Norms of powers of absolutely convergent Fourier series". It relates to complex-valued probabilities, see Baishanski(1999).

Problem 3. If some $q_i$ are negative, under which necessary and sufficient condition on $q_i$, $$\exp (\sum\limits_{i = 0}^\infty {{q_i}{z^i}} ) = \exp [\sum\limits_{i = 1}^\infty {{q_i}({z^i}} - 1)]$$ is a pgf?

Lévy(1937) prove that $P(z) ={e^{\sum\limits_{i = 1}^m {{q _i}({z^i}-1)} }}{\rm{, }}(\left| z \right| \le 1)$ is a pgf when a term with a sufficiently small negative coefficient is preceded by one term with positive coefficient and followed by at least two terms with positive coefficients as well(see Johnson(2005), p393-394), namely ${q_1} > 0,{q_{m - 1}} > 0,{q_m} > 0$. For $m=4$, van Harn(1987) give four inequalities to ensure $$P(z) = {e^{a(z-1) - b({z^2}-1) + c({z^3}-1) + d({z^4}-1)}}{\rm{, }}(\left| z \right| \le 1)$$ is a pgf, namely $a,b,c,d > 0$ and $b \le \min \{ \frac{{{a^2}}}{3},\frac{c}{a},\frac{{ad}}{{2c}},\frac{{{c^2}}}{{3d}}\}$.

Problem 4. The $\frac{1}{k}($ for each $k \in \mathbb{N})$ power of $G(z)$ is $$\sqrt[k]{{G(z)}} = \sum\limits_{i = 0}^\infty {q_i^{(k)}{z^i}} ,(\left| z \right| \le 1).$$ If some ${q_i^{(k)}}$ are negative, under which necessary and sufficient condition on $p_i$, $\sqrt[k]{{G(z)}}$ is absolutely convergent, namely $\sum\limits_{i = 0}^\infty {\left| {q_i^{(k)}} \right|} < \infty$?

When ${p_0} \ge {p_1} \ge {p_2} \ge \cdots \ge 0$ in Problem 4, this is Szekely's Discrete Convex Theorem (See Theorem 2.3.1 of Kerns(2004), page29). This problem appear firstly in Székely(2005). Székely(2005) discuss the condition that $\sqrt[k]{{G(z)}}$ is absolute convergence, when $G(z)$ is pgf of Bernoulli distibution. It is related to Negative probability(http://en.wikipedia.org/wiki/Negative_probability). Since the absolute convergence of $\sqrt[k]{{G(z)}}$ and $\ln G(z)$ are not equiavalent, so Problem 1 and Problem 4 are different.

Problem 1 can be seen as inverse problem of Problem 3, Problem 4 can be seen as inverse problem of Problem 2.

Baishanski, B. (1999). Norms of powers and a central limit theorem for complex-valued probabilities. In Analysis of Divergence (pp. 523-543). Birkhäuser Boston.

Beurling, A. (1938, August). Sur les intégrales de Fourier absolument convergentes et leur applicationa une transformation fonctionnelle. In Ninth Scandinavian Mathematical Congress (pp. 345-366).

Hedstrom, G. W. (1967). Norms of powers of absolutely convergent Fourier series in several variables. The Michigan Mathematical Journal, 14(4), 493-495.

Helson, H., & Beurling, A. (1953). Fourier-Stieltjes transforms with bounded powers. Mathematica Scandinavica, 1, 120-126.

Kerns, G. J. (2004). Signed Measures in Exchangeability and Infinite Divisibility (Doctoral dissertation, Bowling Green State University).

Johnson, N. L., Kemp, A. W., Kotz S. (2005). Univariate Discrete Distributions, 3ed. Wiley, New Jersey.

Lévy, P. (1937). Sur les exponentielles de polynômes et sur l’arithmétique des produits de lois de Poisson. Annales scientifiques de l’École Normale Supérieure, 54, 231–292.

Székely, G. J. (2005). Half of a coin: negative probabilities. Wilmott Magazine, 66-68.

van Harn K.(1978). Classifying infinitely divisible distributions by functional equations. Amsterdam: Mathematisch.

• There cannot be zeros in $|z|<1$ or even $|z| \leq 1$ because the hypothesis $p_0 > p_1 > p_2 > \cdots > 0$ implies that in the expansion $$(1-z) \, G(z) = p_0 - \sum_{j=1}^\infty (p_j - p_{j-1}) \, z^j$$ all the coefficients $p_j - p_{j-1}$ are positive and their sum telescopes to $p_0$, so $$\left| \sum_{j=1}^\infty (p_j - p_{j-1}) \, z^j \right| < p_0$$ for all $z \neq 1$ such that $|z| \leq 1$. Mar 18 '14 at 3:06
• 14 edits by author, in under 24 hours. Hard to hit a moving target. Mar 18 '14 at 4:46
• Waouh, 24 edits now! This must be a world record.
– abx
Mar 18 '14 at 13:02
• Noam has essentially answered it. All we need to notice is that the trick he mentioned allows one to show that the real part of $(1-z)G(z)$ is positive in the closed unit disk except for the point $z=1$, so $G(z)$ cannot be $0$ or negative real (Except, maybe, at $1$? Nah, it equals $1$ there!). Hence, $\log$ has an analytic brunch in some neighborhood ($\mathbb C\setminus(-\infty,0]$) of the compact $G( \text{Clos}\mathbb D )$, so Wiener's theorem finishes the story in no time. Mar 30 '14 at 9:44
• As to homework, the answer is "Yes", if the person took a decent course on Banach algebras recently and "No" otherwise. Our (or, at least, my) memory is short and rusty and our education is patchy, so, Noah, I guess I can kill you with an elementary for me question and, most likely, you can return the favor :-). Let us, hence, assume (unless it is obvious otherwise) that whoever asks a question asks it in good faith and for a good reason ;-). Mar 30 '14 at 9:53

Not a complete solution, only a remark: assume $G(z)$ has no zeroes in the closed unit disk. If the series for $G(z)$ has radius of convergence >1 the series for $\log(G(z)$ will converge absolutely in $z=1$. (Because $G^\prime(z)/G(z)$ is then regular in a disk slightly larger than the unit disk.) Hence counterexamples (if any) must have a singularity in $z=1$.