By a "globally bounded $G$-function," following G. Christol, I will mean a solution to a (minimal) linear differential equation on $\mathbb{P}^1$ (then necessarily of the Fuchsian type and with rational exponents), which is regular at $x = 0$, has a Taylor expansion in $O_{K,S}[[x]]$, and has a positive radius of convergence at each place of $K$, where $O_{K,S}$ is the ring of integers of a number field localized at a finite set $S$ of primes. In other words, the coefficients are required to be $S$-integral, excluding the typical examples of $G$-functions with logarithmic branching: the polylogarithms.

**Question.** *May a globally bounded $G$-function have a logarithmic branch point? (I.e., a logarithmic term in the asymptotic expansion near a singular point of the Fuchsian differential equation.)*

For example, algebraic functions are globally bounded by Eisenstein's theorem, and they only admit algebraic branching (Puiseux expansions at every point).

More generally, the diagonals of a rational function in several variables are globally bounded $G$-functions. (They in fact satisfy a Picard-Fuchs differential equation for periods of the complement of an algebraic hypersurface. See, for instance, http://pierre.lairez.fr/objets.html , "Theorem of Christol-Lipschitz.") Christol has a conjecture that all globally bounded $G$-functions are of this form. (Does the question have a negative answer for diagonals of rational functions?)

**Motivation.** If the question has a negative answer, it would imply the following conjecture of Ruzsa: If a mapping $a : \mathbb{N}_0 \to \mathbb{Z}$ preserves congruences (this is to say, it has the divisibility property $n - m \mid a(n)-a(m)$ characteristic of polynomials), and if there is an $A < e$ such that $|a(n)| < A^n$ for all $n \gg 0$, then $a(n)$ is a polynomial. Indeed:

Perelli and Zannier have shown that such an $f_0(x) := \sum_{n \geq 0} a(n)x^n \in \mathbb{Z}[[x]]$ is $D$-finite; it is therefore a globally bounded $G$-function. Assuming as we may that $a(0) = 0$, the divisibility property is easily seen to imply the global boundedness of each of the series of iterated integrals $$ f_{k+1}(x) := - \frac{f_k'(0)}{1-x} + \frac{1}{x} \int_0^x f_k(t) \frac{dt}{t}. $$ For example, the $n$-th coefficient of $f_1$ is $$ b(n) := \frac{a(n+1)}{n+1}-a(1) = \frac{a(n+1)-a(0)}{n+1} -a(1) \in \mathbb{Z}, $$ and satisfies a mildly weaker form of the divisibility property: $$ n-m \mid (n+1)(m+1)(b(n)-b(m)), $$ sufficient to yield $f_2 \in \frac{1}{2} \mathbb{Z}[[x]]$ upon applying it with $m = 0$.

Therefore each $f_k$ is a globally bounded $G$-function. Finally, it is easy to see that unless $f_0$ is meromorphic on all of $\mathbb{P}^1$, regular outside $x = 1$, and vanishing at infinity (in which case the $f_k$ stabilize to $0$ after $\deg{f_0}$ steps; those are exactly the functions predicted by Ruzsa's conjecture), a logarithmic term will enter the iterated integral at each singularity $a \neq 1$ (or at $a = 1$ if the latter is a branch point) as soon as $k \gg_a 0$.

For example, if $a$ is a Laurent pole of order $-m < 0$ of $f_0$, the expansion of $f_m$ at $x = a$ will involve $\log{(x-a)}$. Similar remarks apply to the branch points.

**Added.**

A negative answer to the question would refine the classical theorem of Polya (resp. its generalization by Andre) which states that a globally bounded power series whose derivative is a rational function (resp. an algebraic function) is itself a rational (resp. algebraic) function.

A stronger (contrapositive) question would be the following. If an irreducible linear differential operator $L$ with polynomial coefficients has one globally bounded solution (e.g., recall Apery's differential equation from the proof of the irrationality of $\zeta(3)$), does it follow that $L$ has finite local monodromies?