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?
