2.1 A key lemma

In this section, we shall prove a result which will be an essential ingredient in the proof of Theorem 1.1.

Lemma 2.1. Let $P(z)\in \mathbb {Z}[z]$ be a polynomial of degree $d\geq 1$ and $\mathcal {S}$ an infinite set of positive integers. Then there exists a nonzero polynomial $Q(z)\in \mathbb {Z}[z]$ with degree m, say, such that the product $PQ\in \mathbb {Z}[z]$ is a polynomial of the form

$$ \begin{align*} \sum_{\substack{n\in\mathcal{S}(d+m)}}a_nz^n. \end{align*} $$

Proof. Define

$$ \begin{align*} Q(z):=\sum_{i=0}^Lq_iz^i, \end{align*} $$

where L and the $q_i$ values will be chosen later. The polynomial $PQ$ has degree at most $L+d$ and its coefficients are linear forms in $q_0,\ldots , q_L$ . So, it suffices to prove that it is possible to choose the coefficients of $Q(z)$ to eliminate the terms $z^n$ in $(PQ)(z)$ , for which $n\not \in \mathcal {S}(L+d)$ (observe that there are $L+d+1-\#\mathcal {S}(L+d)$ of these terms). By equating the corresponding coefficients of $PQ$ to zero, we obtain a homogeneous linear system with $L+d+1-\#\mathcal {S}(L+d)$ equations in the $L+1$ variables $q_i$ , $i\in [0, L]$ . This system has a nontrivial integer solution provided that

$$ \begin{align*} L+1>L+d+1-\#\mathcal{S}(L+d), \end{align*} $$

that is, $\#\mathcal {S}(L+d)>d.$ Since $\mathcal {S}$ is an infinite set, this inequality holds for all sufficiently large integers L. If we set $m:=\max \{i\leq L: q_i\neq 0\}$ , then

$$ \begin{align*} Q(z)=\sum_{i=0}^mq_iz^i\in\mathbb{Z}[z] \end{align*} $$

is the desired polynomial. This finishes the proof.

2.2 Proof of Theorem 1.1

Let $\{\alpha _1,\alpha _2,\ldots \}$ be an enumeration of $\overline {\mathbb {Q}}\cap B(0,1)$ and let $P_i(z)$ of degree $d_i$ be the minimal polynomial (over $\mathbb {Z}$ ) of the algebraic number $\alpha _i$ . We apply Lemma 2.1 to the polynomials

$$ \begin{align*} P_1(z),\ P_1(z)P_2(z),\ P_1(z)P_2(z)P_3(z),\ \ldots. \end{align*} $$

Since $\mathcal {B}$ is an infinite set, Lemma 2.1 ensures the existence (for any $k\geq 1$ ) of a polynomial $Q_k(z)\in \mathbb {Z}[z]$ of degree $m_k$ such that

$$ \begin{align*} Q_k(z)P_1(z)\ldots P_k(z)=\sum_{\substack{n\in \mathcal{B}(m_k+D_k)}}a_{k,n}z^n, \end{align*} $$

where $D_k:=\sum _{i=1}^kd_i$ . Now, we define recursively the sequence $(t_k)_{k\geq 1}$ by choosing $t_1=\min \mathcal {A}$ and $t_{k+1}\in \mathcal {A}$ satisfying

$$ \begin{align*} t_{k+1}\geq \max\{k(t_{k}+D_{k}+m_{k})+1, \ell(Q_{k+1}P_1\ldots P_{k+1})+(k+1)\}. \end{align*} $$

This choice is possible, because $\mathcal {A}$ is an infinite set of nonnegative integers. As usual, $\ell (P)$ denotes the length of a polynomial P (that is, the sum of the absolute values of its coefficients).

We claim that the function

$$ \begin{align*} f(z):=\sum_{k\geq 1}z^{t_k}Q_k(z)P_1(z)\ldots P_k(z) \end{align*} $$

satisfies the conditions of the theorem. Indeed, first note that by construction, $f(z)$ can be written as $\sum _{n\in \mathcal {S}}a_nz^n$ (using the fact that $t_{k+1}>t_k+D_k+m_k$ ). We also have $f(\overline {\mathbb {Q}}\cap B(0,1))\subseteq \overline {\mathbb {Q}}$ , since for any $i\geq 1$ ,

$$ \begin{align*} f(\alpha_i)=\sum_{k= 1}^{i-1}\alpha_i^{t_k}Q_k(\alpha_i)P_1(\alpha_i)\ldots P_k(\alpha_i) \end{align*} $$

is a finite sum of algebraic numbers. Moreover, $t_{k+1}/(t_k+D_k+m_k)$ tends to infinity as $k\to \infty $ (indeed, $t_{k+1}/(t_k+D_k+m_k)>k$ ), so $f(z)$ is a strongly lacunary series and hence a transcendental function (by the transcendence criterion from [Reference Mahler2, page 40]).

To finish, it remains to show that f is an analytic function in the unit ball. For that, take $R\in (0,1)$ and $z\in \overline {B}(0,R)$ . Since $|P(z)|\leq \ell (P)$ when $|z|\leq 1$ , one infers that

$$ \begin{align*} |z^{t_k}Q_k(z)P_1(z)\ldots P_k(z)| \leq R^{t_k}\ell(Q_kP_1\ldots P_k)<R^{\ell(Q_kP_1\ldots P_k)+k}\ell(Q_kP_1\ldots P_k), \end{align*} $$

since $t_k>\ell (Q_kP_1\ldots P_k)+k$ and $R<1$ . Note now that the maximum value of the function $x\mapsto xR^x$ , for real positive values of x, is attained at $x=1/ |{\log R}|$ and is equal to $e^{-1}/ |{\log R}|$ . This implies that

$$ \begin{align*} R^{\ell(Q_kP_1\ldots P_k)+k}\ell(Q_kP_1\ldots P_k)<\dfrac{e^{-1}}{|\log R|}R^k. \end{align*} $$

Summarising, we see that

$$ \begin{align*} |{z^{t_k}Q_k(z)P_1(z)\ldots P_k(z)}| \leq \frac{e^{-1}}{|{\log R}|}R^k=:M_k, \end{align*} $$

for all $z\in \overline {B}(0,R)$ . Since $\sum _{k\geq 1}M_k$ converges, then by the Weierstrass M-test, the series $\sum _{k\geq 1}z^{t_k}Q_k(z)P_1(z)\ldots P_k(z)$ converges absolutely and uniformly on $\overline {B}(0,R)$ for any $R\in (0,1)$ . Consequently, this series defines an analytic function (namely, $f(z)$ ) in the unit circle $B(0,1)$ .

The function $f(z)=\sum _{n\in \mathcal {S}}a_nz^n$ is analytic in the unit ball and $f(\alpha )\in \overline {\mathbb {Q}}$ for all $\alpha \in \overline {\mathbb {Q}}\cap B(0,1)$ . There is an $\infty $ -ary tree of different possibilities for f. In fact, for any $k\geq 1$ , $t_k$ can be chosen in infinitely many different ways and each choice gives a different function f. Thus, we have constructed uncountably many of these functions. This completes the proof.