## 1 Introduction

For an increasing sequence of positive frequencies $$\lambda =\{\lambda _n\}_{n\ge 1}$$, and a generalized Dirichlet series

\begin{aligned} f(s)=\sum \limits _{n\ge 1}a_n e^{-\lambda _ns}, \end{aligned}

the abscissas $$\sigma _c, \sigma _u,$$ and $$\sigma _a$$ of point-wise, uniform, and absolute convergence are defined as in the classical theory of Dirichlet series . In this article we wish to find sets of frequencies such that the analogue of a theorem of Bohr  holds: if $$\sigma _c(f) < \infty$$ and f has a bounded analytic extension to a half-plane $$\{\Re s> \theta \}$$, then $$\sigma _u(f) \le \theta$$. The problem of finding frequencies for which the abscissas of bounded and uniform convergence always coincide, which originated with Bohr and Landau , has recently been revisited [2, 20] with the context of Hardy spaces of Dirichlet series in mind. Indeed, Bohr’s theorem is essentially a necessity for a satisfactory Hardy space theory, see [19, Ch. 6].

An important class of frequencies were introduced by Beurling . Given an arbitrary increasing sequence $$q=\{q_n\}_{n\ge 1}$$, $$1 < q_n\rightarrow \infty$$, such that $$\{\log q_n\}_{n\ge 1}$$ is linearly independent over $${\mathbb {Q}}$$, we will denote by $${\mathbb {N}}_{q}=\{\nu _n\}_{n\ge 1}$$ the set of numbers that can be written (uniquely) as finite products with factors from q, ordered in an increasing manner. The numbers $$q_n$$ are known as Beurling primes, and the numbers $$\nu _n$$ are Beurling integers. The corresponding generalized Dirichlet series are of the form

\begin{aligned} f(s)=\sum \limits _{n\ge 1} a_n\nu _n^{-s}. \end{aligned}

There are a number of criteria to guarantee the validity of Bohr’s theorem for frequencies $$\{\lambda _n\}_{n\ge 1}$$. Bohr’s original condition asks for the existence of $$c_1, c_2>0$$ such that

\begin{aligned} \lambda _{n+1}-\lambda _n \ge c_1e^{-c_2\lambda _{n+1}},\qquad n\in {\mathbb {N}}. \end{aligned}
(1)

Landau relaxed the condition somewhat: for every $$\delta >0$$ there should be a $$c>0$$ such that

\begin{aligned} \lambda _{n+1}-\lambda _n \ge ce^{-e^{\delta \lambda _{n+1}}},\qquad n\in {\mathbb {N}}. \end{aligned}
(2)

Landau’s condition was recently relaxed further by Bayart : for every $$\delta >0$$ there should be a $$C>0$$ such that for every $$n\ge 1$$ it holds that

\begin{aligned} \inf _{m > n} \left( \log \left( \frac{\lambda _{m}+\lambda _n}{\lambda _{m}-\lambda _n}\right) +(m-n) \right) \le Ce^{\delta \lambda _{n}}. \end{aligned}
(3)

For frequencies of Beurling type, $$\lambda _n = \log \nu _n$$, these conditions have natural reformulations. For example, Bohr’s condition (1) is equivalent to the existence of $$c_1, c_2 > 0$$ such that

\begin{aligned} \nu _{n+1} - \nu _n \ge c_1 \nu _{n+1}^{-c_2}. \end{aligned}
(4)

Conditions (1)–(4) are usually very difficult to check for any given Beurling system, since they involve the distances between the corresponding Beurling integers. Furthermore, they often fail. This is especially true if one wants to retain properties of the ordinary integers, such as the asympotic behaviour of the counting function $$N_q(x) = \sum _{\nu _n \le x} 1$$, see e.g.  for the subtleties that arise already when dealing with a finite sequence $$q = (q_1, \ldots , q_N)$$ of Beurling primes.

One motivation for considering Beurling integers is to investigate the properties of the q-zeta function

\begin{aligned} \zeta _q(s)=\sum \limits _{n\ge 1}\nu _n^{-s}=\prod \limits _{n\ge 1}\frac{1}{1-q_n^{-s}}, \end{aligned}

and their interplay with the counting functions

\begin{aligned} N_q(x)=\sum _{\nu _n\le x}1, \qquad \pi _q(x)=\sum _{q_n\le x}1. \end{aligned}

As an example, Beurling  himself showed that the condition

\begin{aligned} N_q(x)=ax+ O\left( \frac{x}{(\log x)^\gamma }\right) ,\qquad \text { for some }\gamma >\frac{3}{2}, \end{aligned}
(5)

implies the analogue of the prime number theorem,

\begin{aligned} \pi _q(x):=\sum _{q_n\le x}1\sim \frac{x}{\log x}. \end{aligned}
(6)

We refer to  for a comprehensive overview of further developments.

In Sect. 2 we begin with a preparatory result which is interesting in its own right. It states that starting with the classical set of primes numbers we can add almost any finite sequence of Beurling primes while retaining the validity of Bohr’s theorem.

### Theorem 1.1

Let $$\{p_n\}_{n\ge 1}$$ be the sequence of ordinary prime numbers and let $$N \ge 1$$. Then Bohr’s condition (1) holds for the Beurling integers generated by the Beurling primes

\begin{aligned} q = \{p_n\}_{n\ge 1}\bigcup \{q_j\}_{j=1}^N, \end{aligned}

for almost every choice $$(q_1,\dots ,q_N)\in (1,\infty )^N$$.

Sequences of Beurling primes of the type considered in Theorem 1.1 previously appeared in .

Our next result requires more careful analysis.

### Theorem 1.2

Let $$q=\{q_n\}_{n\ge 1}$$ be an increasing sequence of Beurling primes such that $$q_1>1$$ and $$\sigma _c(\zeta _q)<\infty$$. Then, for every $$A > 0$$ there exists a sequence of Beurling primes $${\tilde{q}}=\{{\tilde{q}}_n\}_{n\ge 1}$$ for which Bohr’s condition (1) holds and

\begin{aligned} |q_n-{\tilde{q}}_n|\le q_n^{-A},\qquad n\in {\mathbb {N}}. \end{aligned}

Combining our techniques with a probabilistic method from , which refined previous work of Diamond, Montgomery, Vorhauer  and Zhang , we are able to construct a system of Beurling primes that satisfies Bohr’s theorem as well as the analogue of the Riemann hypothesis. Specifically, we say that a Beurling system satisfies RH if the corresponding Beurling zeta function $$\zeta _q(s)$$ has an analytic non-vanishing extension to $$\Re s>1/2$$, except for a simple pole at $$s=1$$, of zero order in any substrip $$\Re s > \sigma _0$$, $$\sigma _0 > 1/2$$. This latter property means that for every $$\varepsilon > 0$$ and $$\sigma > \sigma _0$$,

\begin{aligned} \zeta _q(\sigma + it) = O_{\varepsilon }(|t|^\varepsilon ), \qquad |t| \rightarrow \infty . \end{aligned}

### Theorem 1.3

There exists a system of Beurling primes $$q=\{q_n\}_{n\ge 1}$$ such that:

1. (i)

The system satisfies RH.

2. (ii)

The prime counting function $$\pi _q(x)$$ satisfies

\begin{aligned} \pi _q(x)={{\,\textrm{li}\,}}(x)+O(1), \end{aligned}

where $${{\,\textrm{li}\,}}(x)=\int \limits _{1}^{x}(1-u^{-1})\left( \log u\right) ^{-1}\,du$$.

3. (iii)

The integer counting function $$N_{q}(x)$$ satisfies

\begin{aligned} N_{q}(x) = ax + O_{\varepsilon }(x^{1/2+\varepsilon }), \quad \text {for all } \varepsilon >0, \end{aligned}

for some $$a>0$$.

4. (iv)

The corresponding Beurling integer system satisfies Bohr’s condition.

Note that there are examples of Beurling systems such that $$\zeta _q$$ has a non-vanishing meromorphic extension to $$\Re s > 1/2$$ of infinite order in any substrip $$\Re s > \sigma _0$$, $$1/2< \sigma _0 < 1$$, see . On the other hand, if $$\zeta _q$$ does not vanish and has finite order in every such substrip, then it is actually of zero order, and thus the corresponding Beurling system satisfies RH. Therefore (ii) and (iii) actually imply (i). The latter two statements can both be deduced from [16, Theorem 2.3].

The proofs of our results investigate how well “irrational numbers” may be approximated by fractions of Beurling integers. We will comment further on this kind of Diophantine approximation problems in Sect. 3. In Sect. 3 we will also return to the original idea behind our work. There has been an interest in studying Hardy spaces of generalized Dirichlet series since the 60 s [8, 13, 14]. However, to our knowledge, except for examples that are very closely related to the ordinary integers, there has not been any discussion of the existence of Beurling primes q satisfying the prime number theorem such that Bohr’s theorem holds true for the corresponding Hardy space $${\mathcal {H}}^\infty _q$$. This is in spite of the fact that Bohr’s theorem is crucial for a meaningful theory of the Hardy spaces $${\mathcal {H}}^p_q$$, $$1 \le p \le \infty$$.

Since other aspects of the function theory of Hardy spaces do not depend on the choice q of Beurling primes, the motivation for Theorem 1.3 was to find a canonical Beurling system $${\mathbb {N}}_q$$ which allows us to assume the Riemann hypothesis in the $${\mathcal {H}}^p_q$$-theory. As a specific function theoretic application, we construct an outer function, or synonymously, a cyclic function, $$f \in {\mathcal {H}}^2_q$$ which has a zero in its half-plane of convergence,

\begin{aligned} f(s) = \frac{1}{\zeta _q(s + 1/2 + \varepsilon )}, \qquad 0< \varepsilon < 1/2. \end{aligned}
(7)

The existence of such an f constitutes a counterexample to a conjecture posed by Helson .

### 1.1 Notation

Throughout the article, we will be using the convention that C denotes a positive constant which may vary from line to line. We will write that $$C = C(\Omega )$$ when the constant depends on the parameter $$\Omega$$.

## 2 Proof of the main results

### Lemma 2.1

Suppose that $$\{q_n\}_{n\ge 1}$$ is a Beurling system such that $$d_n := \nu _{n+1} - \nu _n \gg \nu _{n+1}^{-C}$$. Then, for every $$\varepsilon >0$$ and for almost every $$q' >1$$, the Beurling system $$\{q_n\}_{n\ge 1}\cup \{q'\}$$ has a distance function satisfying

\begin{aligned} d'_n = \nu '_{n+1} - \nu '_n \gg \nu _{n+1}^{-C^{'}},\qquad n\in {\mathbb {N}}, \end{aligned}
(8)

where $$C'(q',\,q)=\max \left( C,\,2\sigma _c(\zeta _q)-1+\varepsilon \right)$$.

### Proof

Let $$x_0 > 1$$. First we will prove that the set $${\mathcal {M}}$$ of all numbers $$q' \ge x_0$$ such that there exist infinitely many triples $$(j,n,m)\in {\mathbb {N}}^3$$ with

\begin{aligned} \left| (q')^j-\frac{\nu _n}{\nu _m}\right| \le \nu _n ^{-C_0}\nu _m ^{-C_0},\qquad C_0=\sigma (\zeta _q)+\varepsilon , \end{aligned}

has measure zero. Since

\begin{aligned} \left| q'-\left( \frac{\nu _n}{\nu _m}\right) ^{\frac{1}{j}}\right| \le x_0^{1-j}\left| (q')^j-\frac{\nu _n}{\nu _m}\right| , \end{aligned}

we have that $${\mathcal {M}}\subset \limsup _{m,n,j} \Omega _{m,n,j}$$, where, for j, n, m $$\ge$$ 1,

\begin{aligned} \Omega _{m,n,j}=\left[ \left( \frac{\nu _n}{\nu _m}\right) ^{\frac{1}{j}}-x_0 ^{1-j}\nu _n^{-C_0}\nu _m^{-C_0},\, \left( \frac{\nu _n}{\nu _m}\right) ^{\frac{1}{j}}+x_0 ^{1-j}\nu _n^{-C_0}\nu _m^{-C_0}\right] . \end{aligned}

The Borel–Cantelli lemma thus shows that $$|{\mathcal {M}}|=0$$, since

\begin{aligned} \sum \limits _{m\ge 1}\sum \limits _{n\ge 1}\sum \limits _{j\ge 1}|\Omega _{m,n,j}|\le \frac{2x_{0}}{x_{0}-1}\zeta _q\left( C_0\right) ^2<\infty . \end{aligned}

Fix a number $$q'\in [x_0,\infty )\setminus {\mathcal {M}}$$ such that $$\log q'$$ is not in the (countable) set $${{\,\textrm{span}\,}}_{{\mathbb {Q}}}\{\log q_n\}$$. Note that the set of such numbers has full measure in $$[x_0,\infty )$$, and that $$x_0 > 1$$ is arbitrary. By construction, there are finitely many triples (jnm) such that

\begin{aligned} \left| (q')^j-\frac{\nu _n}{\nu _m}\right| \le \nu _n ^{-C_0}\nu _m ^{-C_0}. \end{aligned}
(9)

For these exceptional triples, the left-hand side is at least positive, since $$\log q'\notin {{\,\textrm{span}\,}}_{{\mathbb {Q}}}\{\log q_n\}$$. Therefore

\begin{aligned} \left| (q')^j-\frac{\nu _n}{\nu _m}\right| \gg \nu _n ^{-C_0}\nu _m ^{-C_0} \end{aligned}

for all $$(j,n,m) \in {\mathbb {N}}^3$$.

Now we consider two arbitrary consecutive Beurling integers generated by the prime system $$\{q_n\}_{n\ge 1}\cup \{q'\}$$,

\begin{aligned} \nu '_{n+1}=(q')^a\nu _m,\qquad \nu '_{n}=(q')^b\nu _l. \end{aligned}

If $$a = b$$, then $$l = m-1$$ and

\begin{aligned}\nu '_{n+1}-\nu '_n \gg \nu _m^{-C} \ge \left( \nu _{n+1}'\right) ^{-C},\end{aligned}

by the hypothesis on the distances $$d_n$$ for the original Beurling system. Otherwise, if, say, $$b < a$$, then

\begin{aligned} \left| \nu '_{n+1}-\nu '_n\right| = (q')^b\nu _m\left| (q')^{a-b}-\frac{\nu _l}{\nu _m}\right| \gg \nu _l ^{-C_0}\nu _m ^{-C_0+1}(q')^b \gg \left( \nu _{n+1}'\right) ^{-C^{'}}. \end{aligned}

where $$C^{'}=2\sigma _c(\zeta _q)-1+\varepsilon$$. $$\square$$

### Proof of Theorem 1.1

The proof is a direct consequence of Lemma 2.1. $$\square$$

In order to prove Bohr’s theorem for more general Beurling systems, we need to control the constant in the distance estimate (8), which comes from the exceptional triples satisfying (9).

### Proof of Theorem 1.2

Fix a small $$\varepsilon > 0$$ and $$x_0\in (1+\varepsilon /2,1+\varepsilon )$$. Consider first any Beurling system $${\mathbb {N}}_\rho = \{\nu _n\}_{n \ge 1}$$ generated by Beurling primes such that $$\rho _1 > 1 + \varepsilon$$ and $$\sigma _c(\zeta _\rho ) < \infty$$. For a number $$\sigma _\infty > \max (2, A)$$ to be chosen in a moment, let

\begin{aligned} {\mathcal {N}} = \bigcup \limits _{m\ge 2}\bigcup \limits _{n\ge 2}\bigcup \limits _{j\ge 1} \Omega _{m, n, j}, \end{aligned}

where $$\Omega _{m, n, j}$$ is defined as in the proof of Lemma 2.1,

\begin{aligned} \Omega _{m,n, j}=\left[ \left( \frac{\nu _n}{\nu _m}\right) ^{\frac{1}{j}}-x_0 ^{1-j}\nu _n^{-\sigma _\infty }\nu _m^{-\sigma _\infty },\, \left( \frac{\nu _n}{\nu _m}\right) ^{\frac{1}{j}}+x_0 ^{1-j}\nu _n^{-\sigma _\infty }\nu _m^{-\sigma _\infty }\right] . \end{aligned}

Then $$|{\mathcal {N}}| \le C(\varepsilon )\left( \zeta _{\rho }(\sigma _\infty )-1\right) ^2.$$ Furthermore, for $$x > 2$$, let

\begin{aligned} I_x=[x-x^{-\frac{\sigma _\infty }{2}},x+x^{-\frac{\sigma _\infty }{2}}]. \end{aligned}

Note that if $$\sigma _\infty$$ is sufficiently large, $$\sigma _\infty \ge C(\varepsilon )$$, then

\begin{aligned} \left( \frac{\nu _n}{\nu _m}\right) ^{\frac{1}{j}}+x_0 ^{1-j}\nu _n^{-\sigma _\infty }\nu _m^{-\sigma _\infty }\le \frac{4}{3}\nu _n\qquad \text {and}\qquad x-x^{-\frac{\sigma _\infty }{2}}\ge \frac{2}{3}x. \end{aligned}

Thus, $$I_x \cap \Omega _{m,n, j} \ne \emptyset$$ only if $$\nu _n \ge x/2$$. Therefore

\begin{aligned}{} & {} \left| I_x\cap {\mathcal {N}}\right| \le \sum \limits _{\begin{array}{c} m \ge 2 \\ j\ge 1\\ \nu _n\ge \frac{x}{2} \end{array}} |\Omega _{m,n,j}|\le C(\varepsilon )\left( \zeta _{\rho }(\sigma _\infty )-1\right) \sum \limits _{\nu _n\ge \frac{x}{2}}\nu _n^{-\sigma _\infty }\\{} & {} \quad \le C(\varepsilon )\left( \zeta _{\rho }(\sigma _\infty )-1\right) \zeta _{\rho }\left( \frac{\sigma _\infty }{4}\right) x^{-\frac{3\sigma _\infty }{4}}. \end{aligned}

We will construct a sequence of Beurling systems such that

\begin{aligned} \left( \zeta _{\rho }(\sigma _\infty )-1\right) \zeta _{\rho }\left( \frac{\sigma _\infty }{4}\right) \le 1 \end{aligned}
(10)

for the number $$\sigma _\infty > 0$$, still to be chosen later. Therefore

\begin{aligned} \left| I_x\cap {\mathcal {N}}\right| \le C(\varepsilon ) x^{-\frac{\sigma _\infty }{4}} |I_x|, \end{aligned}
(11)

We conclude that whenever x is sufficiently large, $$I_x \not \subset {\mathcal {N}}$$.

To include triples where $$\nu _n$$ or $$\nu _m$$ equals one in our considerations, we increase the power $$\sigma _\infty$$. The inequality

\begin{aligned} \left| x^j-\frac{\nu _n}{\nu _m}\right| \le \nu _n^{-3\sigma _\infty } \nu _m^{-3\sigma _\infty } \end{aligned}
(12)

implies, whenever $$x \ge x_0$$, that

\begin{aligned} \left| x - \left( \frac{\nu _n}{\nu _m}\right) ^{\frac{1}{j}}\right| \le x_0^{1-j}\left| x^j-\frac{\nu _n^2\nu _m }{\nu _m^2\nu _n}\right| \le x_0^{1-j} \left( \nu _m^2\nu _n\right) ^{-\sigma _\infty }\left( \nu _n^2\nu _m\right) ^{-\sigma _\infty }. \end{aligned}

Therefore $${\mathcal {M}} \subset {\mathcal {N}}$$, where $${\mathcal {M}}$$ this time denotes the set of all $$x \ge x_0$$ for which there exists an exceptional triple $$(j,n,m) \in {\mathbb {N}}^3$$ such that (12) holds.

Now let q be a sequence of primes in the statement of Theorem 1.2, assuming that $$\varepsilon < q_1 - 1$$. As described, we will only be able to effectively apply (11) when x is sufficiently large, say, $$x \ge B = B(\varepsilon ) = C(\varepsilon )^4 + 2$$, where $$C(\varepsilon )$$ in this instance refers to the same constant as in (11). Let N be such that $$\{q_1, \ldots , q_N\} = (1, B) \cap q$$. Then, as a corollary of Theorem 1.1, we already know that there exists an increasing finite sequence of primes $$\{{\tilde{q}}_1, \ldots {\tilde{q}}_N\}$$, $${\tilde{q}}_1 > 1$$, such that $$|q_j - {\tilde{q}}_j| \le q_j^{-A}$$, $$j=1, \ldots , N$$, and such that Bohr’s condition holds for $$\{\nu _n^{(N)}\}_{n\ge 1}={\mathbb {N}}_{\{{\tilde{q}}_1, \ldots {\tilde{q}}_N\}}$$. Further, we choose $$\sigma _\infty$$ so large that

\begin{aligned} \left| \nu _{n+1}^{(N)}-\nu _n^{(N)}\right| \ge \left( \nu _{n+1}^{(N)}\right) ^{-6\sigma _\infty },\qquad n\in {\mathbb {N}}, \end{aligned}

and

\begin{aligned} \left( \zeta _{q'}(\sigma _\infty )-1\right) \zeta _{q'}\left( \frac{\sigma _\infty }{4}\right) \le 1, \end{aligned}
(13)
\begin{aligned} \textrm{where}\quad q' =\{{\tilde{q}}_1, \ldots {\tilde{q}}_N, q_{N+1} - 1,q_{N+2} - 1, q_{N+3} - 1, \ldots \}.\end{aligned}

This is made possible by the hypothesis that $$\sigma _c(\zeta _q) < \infty$$, since

\begin{aligned} \zeta _{q'}(\sigma )\le \prod \limits _{j\ge 1}\frac{1}{1-(q_j')^{-\sigma }}\le \zeta _q\left( \frac{\sigma }{C}\right) ,\qquad \sigma >0,\qquad C\ge \sup _{n \ge 1}\frac{\log (q_n)}{\log (q'_n)}. \end{aligned}

From here we proceed by induction. Suppose that $${\tilde{q}}_1, \ldots {\tilde{q}}_k$$ have been chosen, where $$k \ge N$$, with corresponding Beurling integers $$\{\nu _n^{(k)}\}_{n\ge 1}={\mathbb {N}}_{\{{\tilde{q}}_{n}\}_{n=1}^k}$$ satisfying that

\begin{aligned} \left| \nu _{n+1}^{(k)}-\nu _n^{(k)}\right| \ge \left( \nu _{n+1}^{(k)}\right) ^{-6\sigma _\infty }. \end{aligned}

We apply the preceding discussion to the Beurling primes $$\rho = \{{\tilde{q}}_1, \ldots {\tilde{q}}_k\}$$ and $$x = q_{k+1}$$, concluding that there exists a number $${\tilde{q}}_{k+1} \in I_{q_{k+1}}$$ such that

\begin{aligned} \left| {\tilde{q}}_{k+1}^j-\frac{\nu ^{(k)}_n}{\nu ^{(k)}_m}\right| \ge \left( \nu _n^{(k)}\right) ^{-3\sigma _\infty } \left( \nu _m^{(k)}\right) ^{-3\sigma _\infty }, \qquad (j, n, m) \in {\mathbb {N}}^3. \end{aligned}

By the same argument as in the last paragraph of the proof of Theorem 1.1 the Beurling system $$\{\nu _n^{(k+1)}\}_{n\ge 1}={\mathbb {N}}_{\{{\tilde{q}}_n\}_{n=1}^{k+1}}$$, then satisfies that

\begin{aligned} \left| \nu _{n+1}^{(k+1)}-\nu _n^{(k+1)}\right| \ge \left( \nu _{n+1}^{(k+1)}\right) ^{-6\sigma _\infty },\qquad n\in {\mathbb {N}}. \end{aligned}

At each step of the construction, () ensures that (10) holds. We hence obtain a sequence $${\tilde{q}}=\{{\tilde{q}}_n\}_{n\ge 1}$$, satisfying that $$|{\tilde{q}}_n-q_n|\le q_n^{-\frac{\sigma _\infty }{2}}$$ as well as Bohr’s condition (1), specifically,

\begin{aligned} \left| {\tilde{\nu }}_{n+1}-{\tilde{\nu }}_{n}\right| \ge \left( {\tilde{\nu }}_{n+1}\right) ^{-6\sigma _\infty },\qquad n\in {\mathbb {N}}. \end{aligned}

where $$\{{\tilde{\nu }}_n\}_{n\ge 1}={\mathbb {N}}_{{\tilde{q}}}$$. $$\square$$

To prove Theorem 1.3, we shall combine the proof of Theorem 1.2 with the probabilistic construction of [6, Theorem 1.2]. Let

\begin{aligned} F(x) = {{\,\textrm{li}\,}}(x) = \int \limits _{1}^{x}\frac{1-u^{-1}}{\log u}\,du \end{aligned}

and set $$x_{n} = F^{-1}(n)$$. We select the nth Beurling prime $$q_{n}$$ randomly from the interval $$[x_{n},x_{n+1}]$$ according to the probability measure $$d{{\,\textrm{li}\,}}(x)|_{[x_{n},x_{n+1}]}$$. That is, we consider a sequence of independent random variables $$Q_{n}$$, representing the coordinate functions $$(q_{1}, q_{2}, \dotsc ) \mapsto q_{n}$$, with cumulative distribution function $$\int \nolimits _{x_{n}}^{x}d{{\,\textrm{li}\,}}(u) = {{\,\textrm{li}\,}}(x) - n$$, $$x_{n} \le x \le x_{n+1}$$. Formally, the probability space is $$X=\prod _{n=1}^{\infty }[x_{n},x_{n+1}]$$, and by appealing to Kolmogorov’s extension theorem, we can equip X with a probability measure dP such that

\begin{aligned} P\biggl (A \times \prod _{n=k+1}^{\infty }[x_{n},x_{n+1}]\biggr ) = \int _{A}d{{\,\textrm{li}\,}}(u_{1})\cdots d{{\,\textrm{li}\,}}(u_{k}), \end{aligned}

$$\text {for every} A \subseteq [x_{1},x_{2}]\times \cdots \times [x_{k},x_{k+1}].$$

### Proof of Theorem 1.3

Let $$A>1$$. We will show the existence of a sequence of Beurling primes $$q=\{q_{n}\}_{n\ge 1}$$ generating integers $${\mathbb {N}}_{q} = \{\nu _{n}\}_{n\ge 1}$$ with the following properties:

1. (a)

The Beurling zeta function $$\zeta _{q}(s)$$ can be written as

\begin{aligned} \zeta _{q}(s) = \frac{se^{Z(s)}}{s-1}, \end{aligned}

where Z(s) is an analytic function in $$\{\Re s > 1/2\}$$ which in every closed half-plane $$\{\Re s \ge \sigma _0\}$$, $$\sigma _{0}>1/2$$, satisfies

\begin{aligned} |Z(s)| \ll _{\sigma _{0}} \sqrt{\log (|t|+2)}, \qquad s = \sigma + it. \end{aligned}
2. (b)

The Beurling integers satisfy $$|\nu _{n}-\nu _{m}| \ge (\nu _{n}\nu _{m})^{-A}$$ whenever $$n\ne m$$.

Suppose that $$\nu , \mu$$ are relatively prime Beurling integers, $$(\nu , \mu ) = 1$$, satisfying $$|\nu -\mu | < (\nu \mu )^{-A}$$. Let $$q_{k}$$ be the largest prime factor of $$\nu \mu$$. Without loss of generality $$q_{k} \mid \nu$$ and $$q_{k}\not \mid \mu$$, and let j be such that $$\nu = q_{k}^{j}\nu '$$ where $$q_{k} \not \mid \nu '$$. Then

\begin{aligned} \left|q_{k}^{j} - \frac{\mu }{\nu '}\right|< \frac{1}{q_{k}^{Aj}(\nu ')^{1+A}\mu ^{A}}, \end{aligned}

so that

\begin{aligned} \left|q_{k} - \Bigl (\frac{\mu }{\nu '}\Bigr )^{1/j}\right|< \frac{1}{q_{k}^{(A+1)j-1}(\nu ')^{1+A}\mu ^{A}} \le \frac{1}{x_{k}^{(A+1)j-1}(\nu ')^{1+A}\mu ^{A}}. \end{aligned}

This motivates the following definitions. For $$k, j \ge 1$$ and

\begin{aligned} q_{1}\in [x_{1}, x_{2}], q_{2}\in [x_{2},x_{3}], \dotsc , q_{k-1}\in [x_{k-1},x_{k}], \end{aligned}

we set

\begin{aligned}{} & {} {\mathcal {M}}_{k,j}(q_{1}, \dotsc , q_{k-1}) \\{} & {} \quad = \bigcup _{\nu , \mu \in {\mathbb {N}}_{(q_{1},\dotsc , q_{k\!-\!1})}}\left[ \Bigl (\frac{\mu }{\nu }\Bigr )^{1/j}\!-\!\frac{1}{x_{k}^{(A+1)j-1}\nu ^{1+A}\mu ^{A}},\!\Bigl (\frac{\mu }{\nu }\Bigr )^{1/j} \!+\! \frac{1}{x_{k}^{(A+1)j-1}\nu ^{1+A}\mu ^{A}}\right] , \end{aligned}

and we consider the events

\begin{aligned} B_{k,j} = \bigl \{(q_{1},q_{2},\dotsc ): q_{k}\in {\mathcal {M}}_{k,j}(q_{1}, \dotsc , q_{k-1})\bigr \}. \end{aligned}

Denoting $${\mathcal {X}} = (x_{1}, x_{2},\dotsc )$$, the Lebesgue measure of $${\mathcal {M}}_{k,j}(q_{1}, \dotsc , q_{k-1})$$ is bounded by $$2\zeta _{{\mathcal {X}}}(1+A)\zeta _{{\mathcal {X}}}(A)x_{k}^{1-(A+1)j}$$. Note that $$\pi _{{\mathcal {X}}}(x) = {{\,\textrm{li}\,}}(x)+O(1)$$, and so $$\zeta _{{\mathcal {X}}}(s)$$ has abscissa of convergence 1. Hence, for the probability of $$B_{k,j}$$ we have

\begin{aligned} P(B_{k,j})&= \int \limits _{x_{1}}^{x_{2}}d{{\,\textrm{li}\,}}(u_{1})\int \limits _{x_{2}}^{x_{3}}d{{\,\textrm{li}\,}}(u_{2})\cdots \int \limits _{x_{k-1}}^{x_{k}}d{{\,\textrm{li}\,}}(u_{k-1})\int \limits _{[x_{k},x_{k+1}] \cap {\mathcal {M}}_{k,j}(u_{1}, \dotsc , u_{k-1})} d{{\,\textrm{li}\,}}(u_{k}) \\&\le \frac{2\zeta _{{\mathcal {X}}}(1+A)\zeta _{{\mathcal {X}}}(A)}{x_{k}^{(A+1)j-1}}. \end{aligned}

In particular, $$\sum _{k=1}^{\infty }\sum _{j=1}^{\infty }P(B_{k,j}) < \infty$$.

We also consider the events

\begin{aligned} A_{k,m} \!=\! \biggl \{(q_{1}, q_{2},\! \dotsc ): \Bigl |\sum _{n=1}^{k}q_{n}^{-i m}\!-\!\int \limits _{x_{1}}^{x_{k}}u^{-im}d{{\,\textrm{li}\,}}(u)\Bigr |\! \ge \! 8\sqrt{\frac{x_{k}}{\log x_{k}}}\bigl (\sqrt{\log x_{k}}\!+\!\sqrt{\log m}\bigr )\biggr \}. \end{aligned}

In the proof of [6, Theorem 1.2] it was shown that also $$\sum _{k=1}^{\infty }\sum _{m=1}^{\infty }P(A_{k,m}) < \infty$$. Hence, by the Borel–Cantelli lemma we have with probability 1 that the sequence $$(q_{1}, q_{2}, \dotsc )$$ is contained in only finitely many of the sets $$A_{k,m}$$ and $$B_{k,j}$$. Take any such sequence q.

Note that by construction, $$\pi _{q}(x) = {{\,\textrm{li}\,}}(x)+O(1)$$. As q is contained in only finitely many $$A_{k, m}$$, we obtain that

\begin{aligned} \left|\sum _{q_{n}\le x}q_{n}^{-it}-\int \limits _{x_{1}}^{x}u^{-it}\,d{{\,\textrm{li}\,}}(u)\right|\ll & {} \sqrt{\frac{x}{\log (x+1)}}\bigl (\sqrt{\log (x+1)}+\sqrt{\log (|t|+1)}\bigr ), \quad \nonumber \\ {}{} & {} \quad x\ge 1, \quad t\in {\mathbb {R}}. \end{aligned}
(14)

For $$x=x_{k}$$ and $$t=m\in {\mathbb {Z}}$$ this is clear. If $$x\in (x_{k}, x_{k+1})$$, then both terms in the absolute value of (14) change by at most O(1) upon replacing x by $$x_{k}$$. Hence the bound also holds for any $$t = m\in {\mathbb {Z}}$$ and arbitrary $$x\ge 1$$. To obtain the bound for $$t\in (m,m+1)$$, we write

\begin{aligned} \sum _{q_{n}\le x}q_{n}^{-it}= & {} \int \limits _{x_{1}}^{x}u^{-i(t-m)}\,d\biggl (\sum _{q_{n}\le u}q_{n}^{-im}\biggr ), \\ \int \limits _{x_{1}}^{x}u^{-it}\,d{{\,\textrm{li}\,}}(u)= & {} \int \limits _{x_{1}}^{x}u^{-i(t-m)}\,d\biggl (\int \limits _{x_{1}}^{u}v^{-im}\,d{{\,\textrm{li}\,}}(v)\biggr ), \end{aligned}

and integrate by parts.

The bound (14) implies (a). Indeed, setting $$\Pi _{q}(x) = \pi _{q}(x) + \frac{\pi _{q}(x^{1/2})}{2}+\cdots$$, we have

\begin{aligned} \log \zeta _{q}(s)= & {} \int \limits _{1}^{\infty }x^{-s}\,d\Pi _{q}(x) = \log \frac{s}{s-1} + \int \limits _{1}^{\infty } x^{-s}\,d\bigl (\pi _{q}(x)-{{\,\textrm{li}\,}}(x)\bigr ) \\{} & {} + \int \limits _{1}^{\infty }x^{-s}\,d\bigl (\Pi _{q}(x)-\pi _{q}(x)\bigr ). \end{aligned}

Let $$\sigma \ge \sigma _{0}>1/2$$. In the second integral we integrate by parts and use (14) to see that it is $$O_{\sigma _{0}}\bigl (\sqrt{\log (|t|+2)}\bigr )$$. The third integral is $$O_{\sigma _{0}}(1)$$ for $$\sigma \ge \sigma _{0}>1/2$$, since $$\Pi _{q}(x)-\pi _{q}(x)$$ is non-decreasing and $$\ll \sqrt{x}/\log x$$.

Now let $$k_{1}, k_{2}, \dotsc , k_{l}$$ be the exceptional integers of the construction. Then $$q\notin B_{k,j}$$ for all j and $$k\ne k_{1}, \dotsc , k_{l}$$. We simply remove the corresponding Beurling primes from the system: $${\tilde{q}} = q {\setminus }\{q_{k_{1}}, \dotsc , q_{k_{l}}\}$$. As $$\zeta _{{\tilde{q}}}(s) = \zeta (s)(1-q_{k_{1}}^{-s})\cdots (1-q_{k_{l}}^{-s})$$, (a) remains valid for $$\zeta _{{\tilde{q}}}(s)$$. Finally, every two distinct Beurling integers $$\nu _{n}\ne \nu _{m}$$ from $${\mathbb {N}}_{{\tilde{q}}}$$ satisfy $$|\nu _{n}-\nu _{m}| \ge (\nu _{n}\nu _{m})^{-A}$$. For if this were not the case, then, by the argument at the beginning of the proof, the largest prime factor $$q_{k}$$ of $$\nu _{n}\nu _{m}/(\nu _{n}, \nu _{m})^{2}$$ would be contained in some $${\mathcal {M}}_{k,j}(q_{1}, \dotsc , q_{k-1})$$, which is impossible by construction.

Every point in Theorem 1.3 has now been proven, except for (iii). However, since $$\zeta _{\tilde{q}}(s)$$ is of zero order in $$\{\Re s > \sigma _0\}$$ for every $$\sigma _0 > 1/2$$ by (a), this follows from a standard application of Perron inversion, see [16, 23]. $$\square$$

## 3 Further discussion

### 3.1 Diophantine approximation and Beurling integers

Using the Borel–Cantelli theorem to study the irrationality of real numbers is a standard technique of Diophantine approximation. The irrationality measure $$\mu (x)$$ of a real number $$x\in {\mathbb {R}}$$ is defined as the infimum of the set

\begin{aligned} R_x=\left\{ r>0:\left| x-\frac{m}{n}\right| <\frac{1}{n^r} \text { for at most finitely many pairs }(m,n)\in {\mathbb {N}}\times {\mathbb {N}}\right\} . \end{aligned}

For a Beurling system $${\mathbb {N}}_{q}=\{\nu _n\}_{n\ge 1}$$, we may also introduce the irrationality measure $$\mu _q(x)$$ of a real number $$x\in {\mathbb {R}}$$ as the infimum of the set

\begin{aligned} R_x=\left\{ r>0:\left| x-\frac{\nu _m}{\nu _n}\right| <\frac{1}{\nu _n ^r} \text { for at most finitely many pairs }(m,n)\in {\mathbb {N}}\times {\mathbb {N}}\right\} . \end{aligned}

Then, by slightly modifying the proof of Lemma 2.1, we obtain the following proposition.

### Proposition 3.1

Let $$q=\{q_n\}_{n\ge 1}$$ be a sequence of Beurling primes with $$\sigma _c(\zeta _q)< \infty$$. Then, for almost every $$x\in {\mathbb {R}}$$, it holds that

\begin{aligned} \mu _q(x)\le 2\sigma _c(\zeta _q). \end{aligned}

In the classical case, Dirichlet’s approximation theorem therefore implies that $$\mu (x)=2$$ for almost every $$x\in {\mathbb {R}}$$. We also recall Roth’s theorem , which states that $$\mu (x)=2$$ for every algebraic irrational number. It would be very interesting to develop corresponding results in the context of Beurling integers.

### 3.2 Hardy spaces of Dirichlet series and a conjecture of Helson

For a sequence q of Beurling primes, we introduce the Hardy space $${\mathcal {H}}_q^2$$ as

\begin{aligned} {\mathcal {H}}_q^2=\left\{ f(s)=\sum _{n\ge 1} a_n \nu _n^{-s} \,:\, \left\Vert f\right\Vert _{{\mathcal {H}}_q^2}^2=\sum _{n\ge 1}|a_n|^2< \infty \right\} . \end{aligned}

More generally, for $$1 \le p < \infty$$, we define $${\mathcal {H}}_q^p$$ as the completion of polynomials (finite sums $$\sum a_n \nu _n^{-s}$$) under the Besicovitch norm

\begin{aligned} \left\Vert P\right\Vert _{{\mathcal {H}}_q^p}:= \left( \lim \limits _{T\rightarrow \infty } \frac{1}{2T} \int \limits _{-T}^{T} |P(it)|^p\,dt\right) ^{\frac{1}{p}}. \end{aligned}

The function theory of these spaces originated with Helson , and was, in the distuingished case where q is the sequence of ordinary primes, continued in very influential papers of Bayart  and Hedenmalm, Lindqvist, and Seip . More generally, there is a developing theory of Hardy spaces of Dirichlet series $$\sum a_n e^{-\lambda _n s}$$ whose frequencies are related to other groups than $${\mathbb {T}}^\infty$$, but we shall restrict our attention to frequencies given by Beurling primes. A cornerstone of the theory is that there is a natural multiplicative linear isometric isomorphism between $${\mathcal {H}}^p_q$$ and the Hardy space $$H^p({\mathbb {T}}^\infty )$$ of the infinite torus [8, 15]. However, more is needed in order to identify $$H^\infty ({\mathbb {T}}^\infty )$$ with $${\mathcal {H}}^\infty _q$$, the space of Dirichlet series $$\sum a_n \nu _n^{-s}$$ which converge to a bounded function in $${\mathbb {C}}_0 = \{\Re s > 0\}$$. In fact, Bohr’s condition is typically used in order to establish this isomorphism .

In identifying $${\mathcal {H}}^p_q$$ with $$H^p({\mathbb {T}}^\infty )$$ one is naturally led to consider twisted Dirichlet series

\begin{aligned} f_\chi (s) = \sum _{n\ge 1} a_n \chi (\nu _n) \nu _n^{-s}, \end{aligned}

where a point $$\chi \in {\mathbb {T}}^\infty$$ is interpreted as the completely multiplicative character $$\chi :{\mathbb {N}}_q \rightarrow {\mathbb {T}}$$ such that $$\chi (q_n) = \chi _n$$. Helson  proved that if $$f \in {\mathcal {H}}^2_q$$ and the associated frequencies satisfy Bohr’s condition, then $$f_\chi (s)$$ converges in $${\mathbb {C}}_0$$ for almost every $$\chi \in {\mathbb {T}}^\infty$$. Helson went on to make a conjecture, which we state only in the special case that the frequencies correspond to a Beurling system. Recall that $$f \in {\mathcal {H}}^2_q$$ is said to be outer (or cyclic) if $$\left\{ fg:\, g\in {\mathcal {H}}_q^\infty \right\}$$ is dense in $${\mathcal {H}}_q^2$$.

### Conjecture

If $${\mathbb {N}}_q$$ is a Beurling system that satisfies Bohr’s condition and f is outer in $${\mathcal {H}}_q^2$$, then $$f_\chi$$ never has any zeros in its half-plane of convergence.

Suppose now that the Beurling primes q are chosen as in Theorem 1.3, so that we have the Riemann hypothesis at our disposal, and consider the Dirichlet series

\begin{aligned} f(s) = \frac{1}{\zeta _q(s + 1/2 + \varepsilon )}, \end{aligned}

for some $$0< \varepsilon < 1/2$$. Through a routine calculation with coefficients, one checks that $$f, f^2, 1/f, 1/f^2 \in {\mathcal {H}}^2_q$$. Therefore, there are polynomials $$p_n$$ which converge to 1/f in $${\mathcal {H}}^4_q$$, so that

\begin{aligned} \Vert 1-p_n f\Vert _{{\mathcal {H}}^2_q} \le \Vert f\Vert _{{\mathcal {H}}^4_q} \Vert 1/f - p_n\Vert _{{\mathcal {H}}^4_q} \rightarrow 0, \qquad n \rightarrow \infty . \end{aligned}

Thus, f is outer. On the other hand, it has a zero at $$s = 1/2 - \varepsilon$$. To disprove Helson’s conjecture it only remains to prove that f converges in $${\mathbb {C}}_0$$.

### Proposition 3.2

The reciprocal $$1/\zeta _q$$ of the Beurling zeta function converges in $$\{\Re s>1/2\}$$.

### Proof

The reciprocal $$1/\zeta _q$$ is of zero order in $$\{\Re s> \sigma _0\}$$, for every $$\sigma _0 > 1/2$$. This is clear from the proof of Theorem 1.3, but it is also well known that this can be deduced from (ii) and (iii) using the Borel–Carathéodory and the Hadamard three circles theorems,

\begin{aligned} \log \zeta _q(\sigma +it)=O\left( (\log |t|)^\alpha \right) ,\qquad \alpha \in (0,1), \end{aligned}
(15)

uniformly for $$\frac{1}{2}<\sigma _0\le \sigma \le 1$$. See for example [16, Theorem 2.3] or [22, Theorem 14.2]. Since we have Bohr’s condition, the standard argument [21, Section 9.44] with Perron’s formula then shows that $$1/\zeta _q$$ is convergent in the half-plane where it is analytic with zero order. $$\square$$