Real zeros of random Dirichlet series

TL;DR

This paper investigates the distribution of real zeros of a random Dirichlet series with random signs, establishing conditions under which the series almost surely has no or infinitely many real zeros.

Contribution

It provides a rigorous analysis of the zero distribution of random Dirichlet series based on the convergence of the sum of reciprocals of the sequence.

Findings

If the sum of reciprocals converges, the series has no real zeros with positive probability.

If the sum diverges, the series almost surely has infinitely many real zeros.

Abstract

Let $F(\sigma)$ be the random Dirichlet series $F(\sigma)=\sum_{p\in\mathcal{P}} \frac{X_p}{p^\sigma}$, where $\mathcal{P}$ is an increasing sequence of positive real numbers and $(X_p)_{p\in\mathcal{P}}$ is a sequence of i.i.d. random variables with $\mathbb{P}(X_1=1)=\mathbb{P}(X_1=-1)=1/2$. We prove that, for certain conditions on $\mathcal{P}$, if $\sum_{p\in\mathcal{P}}\frac{1}{p}<\infty$ then with positive probability $F(\sigma)$ has no real zeros while if $\sum_{p\in\mathcal{P}}\frac{1}{p}=\infty$, almost surely $F(\sigma)$ has an infinite number of real zeros.

Equations

$$(P1)\mathcal{P}\cap [0,1)=\varnothing ,$$

$$(P2)\sum _{p\in \mathcal{P}}\frac{1}{p^{\sigma }}\text{mbox}hasabcissaofconvergence{\sigma }_c=1.$$

$$\pi (x):=|\{p\le x:p\in \mathcal{P}\}|.$$

$$F(s)=s{\int }_1^{\infty }A(x)\frac{dx}{x^{1+s}},\text{mbox}foralls\in \text{mathds}{H}_{{\sigma }_c}.$$

$$F(\sigma+\epsilon)=\sum_{p\leq U}\frac{X_{p}}{p^{\sigma+\epsilon}}+O\bigg{(}U^{-\epsilon}\sup_{x>U}\bigg{|}\sum_{U<p\leq x}\frac{X_{p}}{p^{\sigma}}\bigg{|}\bigg{)},$$

$$\sum _{p>U}\frac{X_p{p}^{-\sigma }}{p^{ϵ}}$$

$$\displaystyle\ll\sup_{x>U}\bigg{|}\sum_{U<p\leq x}\frac{X_{p}}{p^{\sigma}}\bigg{|}\int_{U}^{\infty}\frac{\epsilon}{x^{1+\epsilon}}dx=U^{-\epsilon}\sup_{x>U}\bigg{|}\sum_{U<p\leq x}\frac{X_{p}}{p^{\sigma}}\bigg{|}.$$

$$\mathbb{P}\bigg{(}\max_{1\leq m\leq n}\bigg{|}\sum_{k=1}^{m}X_{k}\bigg{|}\geq t\bigg{)}\leq 3\max_{1\leq m\leq n}\mathbb{P}\bigg{(}\bigg{|}\sum_{k=1}^{m}X_{k}\bigg{|}\geq\frac{t}{3}\bigg{)}.$$

$$\mathbb{P}\bigg{(}\sum_{k=1}^{n}a_{k}X_{k}\geq\lambda\bigg{)}\leq\exp\bigg{(}-\frac{\lambda^{2}}{2\sum_{k=1}^{n}a_{k}^{2}}\bigg{)}.$$

$$\sup_{x>U}\bigg{|}\sum_{U<p\leq x}\frac{X_{p}}{\sqrt{p}}\bigg{|}<\frac{1}{10}.$$

$$F(1/2+ϵ)\ge \sum _{p\le U}\frac{1}{p^{1/2+ϵ}}-\frac{1}{10U^{ϵ}}\ge \frac{\pi (U)}{U^{1/2+ϵ}}-\frac{1}{10U^{ϵ}},$$

$$\underset{U\to \infty }{\mathrm{limsup}}\frac{\pi (U)}{U^{1-\delta }}=\infty .$$

$$\sum _{p\le U}\frac{1}{p^{1-ϵ}}$$

$$\le \frac{c{U}^{1-\delta }}{U^{1-ϵ}}+1+(1-ϵ){\int }_1^U\frac{c{x}^{1-\delta }}{x^{2-ϵ}}dx\ll 1+{\int }_1^U\frac{1}{x^{1+(\delta -ϵ)}}dx\ll 1,$$

$$F(1/2+\epsilon)\geq\frac{U^{1-1/4}}{U^{1/2+\epsilon}}-\frac{1}{10U^{\epsilon}}=\frac{1}{U^{\epsilon}}\bigg{(}U^{1/4}-\frac{1}{10}\bigg{)}>0.$$

$$\mathbb{P}(B_U^c)$$

$$\displaystyle\leq 6\lim_{n\to\infty}\max_{U<x\leq n}\mathbb{P}\bigg{(}\sum_{U<p\leq x}\frac{X_{p}}{\sqrt{p}}\geq\frac{1}{30}\bigg{)}\leq 6\lim_{n\to\infty}\exp\bigg{(}\frac{-1/30^{2}}{2\sum_{U<p\leq n}\frac{1}{p}}\bigg{)}$$

$$\displaystyle\leq 6\exp\bigg{(}-\frac{1}{2\cdot 30^{2}\sum_{p>U}\frac{1}{p}}\bigg{)}.$$

$$\pi (x)\ll \frac{x}{\log x}.$$

$$\frac{{\sum }_{p\in \mathcal{P}}\frac{X_p}{p^{\sigma }}}{\sqrt{{\sum }_{p\in \mathcal{P}}\frac{1}{p^{2\sigma }}}}{\to }_d\mathcal{N}(0,1),\text{mbox}as\sigma \to {\frac{1}{2}}^{+},$$

$$\underset{\sigma \to {\frac{1}{2}}^{+}}{\mathrm{limsup}}\frac{{\sum }_{p\in \mathcal{P}}\frac{X_p}{p^{\sigma }}}{\sqrt{{\sum }_{p\in \mathcal{P}}\frac{1}{p^{2\sigma }}}}\ge L$$

$$\underset{\sigma \to {\frac{1}{2}}^{+}}{\mathrm{liminf}}\frac{{\sum }_{p\in \mathcal{P}}\frac{X_p}{p^{\sigma }}}{\sqrt{{\sum }_{p\in \mathcal{P}}\frac{1}{p^{2\sigma }}}}\le -L$$

$$\mathbb{P}\bigg{(}\bigg{|}\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}-\sum_{p\leq y}\frac{X_{p}}{\sqrt{p}}\bigg{|}\geq 2\lambda\bigg{)}\leq 4\exp(-d\lambda^{2}).$$

$$\displaystyle\mathbb{P}\bigg{(}\bigg{|}\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}-\sum_{p\leq y}\frac{X_{p}}{\sqrt{p}}\bigg{|}\geq 2\lambda\bigg{)}\leq$$

$$+$$

$$\displaystyle\leq\exp\bigg{(}-\frac{\lambda^{2}}{2V_{y}}\bigg{)}+\exp\bigg{(}-\frac{\lambda^{2}}{2W_{y}}\bigg{)},$$

$$\frac{1}{p^{\sigma }}-\frac{1}{\sqrt{p}}=(\sigma -1/2)\frac{\log p}{p^{\theta }},\text{mbox}forsome\theta =\theta (p,\sigma )\in [1/2,\sigma ].$$

$$V_y$$

$$\displaystyle=(\sigma-1/2)^{2}\bigg{(}\frac{\pi(y)\log^{2}y}{y}-\int_{1^{-}}^{y}\pi(t)\frac{2\log t-\log^{2}t}{t^{2}}dt\bigg{)}$$

$$\displaystyle\ll(\sigma-1/2)^{2}\bigg{(}\log y+\int_{1^{-}}^{y}\frac{\log t}{t}dt\bigg{)}\ll(\sigma-1/2)^{2}\log^{2}y.$$

$$U_y$$

$$\ll \frac{1}{y^{2\sigma -1}\log y}+2\sigma {\int }_y^{\infty }\frac{1}{t^{2\sigma }\log t}dt\ll \frac{1}{y^{2\sigma -1}\log y}+\frac{2\sigma }{(2\sigma -1)y^{2\sigma -1}\log y}$$

$$\ll \frac{1}{(2\sigma -1)y^{2\sigma -1}\log y}.$$

$$\underset{k\to \infty }{\mathrm{limsup}}\frac{{\sum }_{p\le y_k}\frac{X_p}{\sqrt{p}}}{\sqrt{{\sum }_{p\le y_k}\frac{1}{p}}}=\infty ,$$

$$\underset{k\to \infty }{\mathrm{liminf}}\frac{{\sum }_{p\le y_k}\frac{X_p}{\sqrt{p}}}{\sqrt{{\sum }_{p\le y_k}\frac{1}{p}}}=-\infty .$$

$$\underset{y\to \infty }{\lim }\mathrm{Var}\sum _{p\le y}\frac{X_p}{\sqrt{p}}=\underset{y\to \infty }{\lim }\sum _{p\le y}\frac{1}{p}=\infty ,$$

$$\mathbb{P}\bigg{(}\sum_{p\leq y}\frac{X_{p}}{\sqrt{p}}\geq L\sqrt{\sum_{p\leq y}\frac{1}{p}}\bigg{)}=\mathbb{P}\bigg{(}\sum_{p\leq y}\frac{X_{p}}{\sqrt{p}}\leq-L\sqrt{\sum_{p\leq y}\frac{1}{p}}\bigg{)}\geq\delta.$$

Full text

1. Introduction.

A Dirichlet series is an infinite sum of the form $F(\sigma):=\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}$, where $\mathcal{P}$ is an increasing sequence of positive real numbers and $(X_{p})_{p\in\mathcal{P}}$ is any sequence of complex numbers. If $F(\sigma)$ converges then $F(s)$ converges for all $s\in\mathds{C}$ with real part greater than $\sigma$ (see [4] Theorem 1.1). The abscissa of convergence of a Dirichlet series is the smallest number $\sigma_{c}$ for which $F(\sigma)$ converges for all $\sigma>\sigma_{c}$.

The problem of finding the zeros of a Dirichlet series is classical in Analytic Number Theory. For instance, the Riemann hypothesis states that the zeros of the analytic continuation of the Riemann zeta function $\zeta(\sigma):=\sum_{k=1}^{\infty}\frac{1}{k^{\sigma}}$ in the half plane $\{\sigma+it\in\mathds{C}:\sigma>0\}$ all have real part equal to $1/2$. This analytic continuation can be described in terms of a convergent Dirichlet series – The Dirichlet $\eta$-function $\eta(s)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{s}}$ satisfies $\eta(s)=(1-2^{1-s})\zeta(s)$, for all complex $s$ with positive real part. Thus, to find zeros of $\eta(s)$ for $0<Re(s)<1$ is the same as finding non-trivial zeros of $\zeta$.

In this paper we are interested in the real zeros of the random Dirichlet series $F(\sigma):=\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}$, where the coefficients $(X_{p})_{p\in\mathcal{P}}$ are random and $\mathcal{P}$ satisfies:

[TABLE]

For instance, $\mathcal{P}$ can be the set of the natural numbers. The conditions $(P1-P2)$ imply, in particular, that the series $\sum_{p\in\mathcal{P}}\frac{1}{p^{2\sigma}}$ converges for each $\sigma>1/2$. Therefore, if $(X_{p})_{p\in\mathcal{P}}$ is a sequence of i.i.d. random variables with $\mathbb{E}X_{p}=0$ and $\mathbb{E}X_{p}^{2}=1$, then, by the Kolmogorov one-series Theorem, the series $F(\sigma)=\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}$ has a.s. abscissa of convergence $\sigma_{c}=1/2$. Moreover, the function of one complex variable $\sigma+it\mapsto F(\sigma+it)$ is a.s. an analytic function in the half plane $\{\sigma+it\in\mathds{C}:\sigma>1/2\}$. In the case $X_{p}=\pm 1$ with equal probability, the line $\sigma=\sigma_{c}$ is a natural boundary for $F(\sigma+it)$, see [2] (pg. 44 Theorem 4).

Our main result states:

Theorem 1.1**.**

*Assume that $\mathcal{P}$ satisfies $P1$-$P2$ and let $(X_{p})_{p\in\mathcal{P}}$ be i.i.d and such that $\mathbb{P}(X_{p}=1)=\mathbb{P}(X_{p}=-1)=1/2$. Let $F(\sigma)=\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}$.

i. If $\sum_{p\in\mathcal{P}}\frac{1}{p}<\infty$, then with positive probability $F$ has no real zeros;

ii. If $\sum_{p\in\mathcal{P}}\frac{1}{p}=\infty$, then a.s. $F$ has an infinite number of real zeros.*

It follows as corollary to the proof of item i. that in the case $\sum_{p\in\mathcal{P}}\frac{1}{p}=\infty$, with positive probability $F(\sigma)$ has no zeros in the interval $[1/2+\delta,\infty)$, for fixed $\delta>0$.

Since a Dirichlet series $F(s)=\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{s}}$ is a random analytic function, it can be viewed as a random Taylor series $\sum_{k=0}^{\infty}Y_{k}(s-a)^{k}$, where $a>\sigma_{c}$ and $(Y_{k})_{k\in\mathds{N}}$ are random and dependent random variables. The case of random Taylor series and random polynomials where $(Y_{k})_{k\in\mathds{N}}$ are i.i.d. has been widely studied in the literature, for an historical background we refer to [3] and [5] and the references therein.

2. Preliminaries

2.1. Notation.

We employ both $f(x)=O(g(x))$ and Vinogradov’s $f(x)\ll g(x)$ to mean that there exists a constant $c>0$ such that $|f(x)|\leq c|g(x)|$ for all sufficiently large $x$, or when $x$ is sufficiently close to a certain real number $y$. For $\sigma\in\mathds{R}$, $\mathds{H}_{\sigma}$ denotes the half plane $\{z\in\mathds{C}:Re(z)>\sigma\}$. The indicator function of a set $S$ is denoted by $\mathds{1}_{S}(s)$ and it is equal to $1$ if $s\in S$, or equal to [math] otherwise. We let $\pi(x)$ to denote the counting function of $\mathcal{P}$:

[TABLE]

2.2. The Mellin transform for Dirichlet series

In what follows $\mathcal{P}=\{p_{1}<p_{2}<...\}$ is a set of non-negative real numbers satisfying $P1$-$P2$ above. A generic element of $\mathcal{P}$ is de noted by $p$, and we employ $\sum_{p\leq x}$ to denote $\sum_{p\in\mathcal{P};p\leq x}$. Let $A(x)=\sum_{p\leq x}X_{p}$ and $F(s)=\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{s}}$. Let $\sigma_{c}>0$ be the abscissa of convergence of $F(\sigma)$. Then $F$ can be represented as the Mellin transform of the function $A(x)$ (see, for instance, Theorem 1.3 of [4]):

[TABLE]

In particular, we can state:

Lemma 2.1**.**

Let $F(s)=\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{s}}$ be such that $F(1/2)$ is convergent. Then for each $\sigma\geq 1/2$ and all $\epsilon>0$, for all $U>1$:

[TABLE]

where the implied constant in the $O(\cdot)$ term above can be taken to be $1$.

Proof.

Put $\displaystyle A(x)=\sum_{p\leq x}\mathds{1}_{(U,\infty)}(p)\frac{X_{p}}{p^{\sigma}}$. By (1) it follows that

[TABLE]

∎

2.3. A few facts about sums of independent random variables

In what follows we use

Levy’s maximal inequality: Let $X_{1},...,X_{n}$ be independent random variables. Then

[TABLE]

Hoeffding’s inequality: Let $X_{1},...,X_{n}$ be i.i.d. with $\mathbb{P}(X_{1}=1)=\mathbb{P}(X_{1}=-1)=1/2$. Let $a_{1},...,a_{n}$ be real numbers. Then for any $\lambda>0$

[TABLE]

3. Proof of the main result

Proof of item i.

Since $\sum_{p\in\mathcal{P}}\frac{1}{p}<\infty$ we have by the Kolmogorov one series theorem that the series $\sum_{p\in\mathcal{P}}\frac{X_{p}}{\sqrt{p}}$ converges almost surely. In what follows $U>0$ is a large fixed number to be chosen later, $A_{U}$ is the event in which $X_{p}=1$ for all $p\leq U$ and $B_{U}$ is the event in which

[TABLE]

We claim that for sufficiently large $U$ on the event $A_{U}\cap B_{U}$ the function $F(s)=\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{s}}$ does not vanish for all $s\geq\frac{1}{2}$. Further for sufficiently large $U$ we will show that $\mathbb{P}(A_{U}\cap B_{U})>0$.

On the event $A_{U}\cap B_{U}$ we have by lemma 2.1 that

[TABLE]

where $\pi(U)=\#\{p\leq U:p\in\mathcal{P}\}$. We claim that for each $\delta>0$ we have that

[TABLE]

In fact, this is a consequence from P2: For any $\delta>0$ the series diverges $\sum_{p\in\mathcal{P}}\frac{1}{p^{1-\delta}}=\infty$. To show that this is true we argue by contraposition: Assume that for some fixed $\delta>0$ $\limsup_{U\to\infty}\frac{\pi(U)}{U^{1-\delta}}<\infty$ and hence that there exists a constant $c>0$ such that for all $U>0$, $\pi(U)\leq cU^{1-\delta}$. In that case we have for $0<\epsilon<\delta$

[TABLE]

and hence that the series $\sum_{p\in\mathcal{P}}\frac{1}{p^{1-\epsilon}}$ converges. Therefore, we showed that $\limsup_{U\to\infty}\frac{\pi(U)}{U^{1-\delta}}<\infty$ implies that $\sum_{p\in\mathcal{P}}\frac{1}{p^{\sigma}}$ has abscissa of convergence $\sigma_{c}\leq 1-\delta$.

Now we may select arbitrarily large values of $U>1$ for which $\pi(U)\geq U^{1-1/4}$ and $\sum_{p\leq U}\frac{1}{\sqrt{p}}>\frac{1}{10}$, and hence, by (4), for all $\epsilon>0$ we obtain that

[TABLE]

This proves that on the event $A_{U}\cap B_{U}$ we have that $F(s)\neq 0$ for all $s\in[1/2,\infty)$.

Observe that $A_{U}$ and $B_{U}$ are independent and $A_{U}$ has probability $\frac{1}{2^{\pi(U)}}>0$. Now we will show that the complementary event $B_{U}^{c}$ has small probability. Indeed, by applying the Levy’s maximal inequality and the Hoeffding’s inequality, we obtain:

[TABLE]

Since $\sum_{p\in\mathcal{P}}\frac{1}{p}$ is convergent, the tail $\sum_{p>U}\frac{1}{p}$ converges to [math] as $U\to\infty$. Therefore, for sufficiently large $U$ we can make $\mathbb{P}(B_{U}^{c})<1/2$. ∎

Now we are going to prove Theorem 1.1 part $ii$. We present two different proofs. In the first proof we assume that the counting function of $\mathcal{P}$

[TABLE]

In this case, for instance, $\mathcal{P}$ can be the set of prime numbers. In this proof we show that, for $\sigma$ close to $1/2$, the infinite sum $\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}$ can be approximated by the partial sum $\sum_{p\leq y}\frac{X_{p}}{\sqrt{p}}$ for a suitable choice of $y$ (Lemma 3.1). Then we show that these partial sums change sign for an infinite number of $y$, and hence, $F(\sigma)=\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}$ changes sign for an infinite number of $\sigma\to 1/2^{+}$.

The case in which $\mathcal{P}$ is the set of natural numbers, the infinite sum $\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}$ can not be approximated by the finite sum $\sum_{p\leq y}\frac{X_{p}}{\sqrt{p}}$, i.e, Lemma 3.1 fails in this case. Thus, our approach is different in the general case. First we show (Lemma 3.3) that $\sum_{p\in\mathcal{P}}\frac{1}{p}=\infty$ implies that

[TABLE]

and second, for each $L>0$, the event

[TABLE]

is a tail event, and by (6), it has positive probability. Similarly,

[TABLE]

also is a tail event and has positive probability. Thus, by the Kolmogorov $0-1$ Law, with probability $1$, $\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}$ changes sign for an infinite number of $\sigma\to 1/2^{+}$.

3.1. Proof of Theorem 1.1 (ii) in the case $\pi(x)\ll\frac{x}{\log x}$

Lemma 3.1**.**

Assume that $\mathcal{P}$ satisfies $P1$-$P2$ and that $\sum_{p\in\mathcal{P}}\frac{1}{p}=\infty$. Further, assume that $\pi(x)\ll\frac{x}{\log x}$. Let $\sigma>1/2$ and $y=\exp((2\sigma-1)^{-1})\geq 10$. Then there is a constant $d>0$ such that for all $\lambda>0$

[TABLE]

Proof.

If $|a+b|\geq 2\lambda$ then either $|a|\geq\lambda$ or $|b|\geq\lambda$. This fact combined with the Hoeffding’s inequality allows us to bound:

[TABLE]

where V_{y}=\sum_{p\leq y}\bigg{(}\frac{1}{p^{\sigma}}-\frac{1}{\sqrt{p}}\bigg{)}^{2} and $U_{y}=\sum_{p>y}\frac{1}{p^{2\sigma}}$. To complete the proof we only need to estimate these quantities. By the mean value theorem

[TABLE]

Therefore

[TABLE]

In particular, the choice $y=\exp((2\sigma-1)^{-1})$ implies that both variances $V_{y}$ and $U_{y}$ are $O(1)$. ∎

The simple random walk $S_{n}=\sum_{k=1}^{n}X_{n}$ where $(X_{n})_{n\in\mathds{N}}$ is i.i.d with $X_{1}=\pm 1$ with probability $1/2$ each, satisfies a.s. $\limsup_{n\to\infty}S_{n}=\infty$ and $\liminf_{n\to\infty}S_{n}=-\infty$. We follow the same line of reasoning as in the proof of this result ([6] pg. 381, Theorem 2) to prove:

Lemma 3.2**.**

Assume that $\sum_{p\in\mathcal{P}}\frac{1}{p}=\infty$. Let $y_{k}$ be a increasing sequence of positive real numbers such that $\lim y_{k}=\infty$. Then it a.s. holds that:

[TABLE]

Proof.

We begin by observing that $(X_{p}/\sqrt{p})_{p\in\mathcal{P}}$ is a sequence of independent and symmetric random variables that are uniformly bounded by $1$. It follows that

[TABLE]

and hence this sequence satisfies the Lindenberg condition. By the Central Limit Theorem it follows that for each fixed $L>0$ there exists a $\delta>0$ such that for sufficiently large $y>0$

[TABLE]

Next observe that the event in which $\limsup_{k\to\infty}\frac{\sum_{p\leq y_{k}}\frac{X_{p}}{\sqrt{p}}}{\sqrt{\sum_{p\leq y_{k}}\frac{1}{p}}}\geq L$ is a tail event, and hence by the Kolmogorov zero or one law it has either probability zero or one. Since

[TABLE]

it follows that for each fixed $L>0$ $\limsup_{k\to\infty}\frac{\sum_{p\leq y_{k}}\frac{X_{p}}{\sqrt{p}}}{\sqrt{\sum_{p\leq y_{k}}\frac{1}{p}}}\geq L$, a.s. Similarly, we can conclude that for each fixed $L>0$ $\liminf_{k\to\infty}\frac{\sum_{p\leq y_{k}}\frac{X_{p}}{\sqrt{p}}}{\sqrt{\sum_{p\leq y_{k}}\frac{1}{p}}}\leq-L$, a.s. ∎

Proof of item ii.

Take $\lambda=\lambda(y)=\sqrt{\sum_{p\leq y}\frac{1}{p}}$ in Lemma 3.1 and let $y=\exp((2\sigma-1)^{-1})$. Since $\lim_{y\to\infty}\lambda(y)=\infty$, it follows that there is a subsequence $y_{k}\to\infty$ for which $\sum_{k=1}^{\infty}\exp(-d\lambda^{2}(y_{k}))<\infty$ and hence, by the Borel-Cantelli Lemma, it a.s. holds that

[TABLE]

where $y_{k}=\exp((2\sigma_{k}-1)^{-1})$. This combined with Lemma 3.2 gives a.s.

[TABLE]

Similarly, we conclude that $\liminf_{\sigma\to 1/2^{+}}\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}=-\infty$, a.s. Since $F(\sigma)$ is a.s. analytic, it follows that there is an infinite number of $\sigma>1/2$ for which $F(\sigma)=0$. ∎

3.2. Proof of Theorem 1.1 (ii), the general case

The following Lemma is an adaptation of [1], Theorem 1.2:

Lemma 3.3**.**

Assume that $\mathcal{P}$ satisfies P1-P2 and that $\sum_{p\in\mathcal{P}}\frac{1}{p}=\infty$. Then

[TABLE]

Proof.

Let $V(\sigma)=\sqrt{\sum_{p\in\mathcal{P}}\frac{1}{p^{2\sigma}}}$. Observe that $V(\sigma)\to\infty$ as $\sigma\to 1/2^{+}$: For each fixed $y>0$

[TABLE]

Thus, by making $y\to\infty$ in the equation above, we obtain the desired claim.

For each fixed $\sigma>1/2$, by the Kolmogorov one series Theorem, we have that $\sum_{p\leq y}\frac{X_{p}}{p^{\sigma}}$ converges almost surely as $y\to\infty$. Since $(X_{p})_{p\in\mathcal{P}}$ are independent, by the dominated convergence theorem:

[TABLE]

We will show that for each fixed $t\in\mathds{R}$, $\varphi_{\sigma}(t)\to\exp(-t^{2}/2)$ as $\sigma\to 1/2^{+}$. Observe that $\varphi_{\sigma}(t)=\varphi_{\sigma}(-t)$, so we may assume $t\geq 0$. Thus, for each fixed $t\geq 0$ we may choose $\sigma>1/2$ such that $0\leq\frac{t}{V(\sigma)p^{\sigma}}\leq\frac{1}{100}$ and 0\leq 1-\cos\big{(}\frac{t}{V(\sigma)p^{\sigma}}\big{)}\leq\frac{1}{100}, for all $p\in\mathcal{P}$.

For $|x|\leq 1/100$, we have that $\log(1-x)=-x+O(x^{2})$ and $\cos(x)=1-\frac{x^{2}}{2}+O(x^{4})$. Further, $1-\cos(x)=2\sin^{2}(x/2)\leq\frac{x^{2}}{2}$. Thus, we have:

[TABLE]

We conclude that $\varphi_{\sigma}(t)\to\exp(-t^{2}/2)$ as $\sigma\to 1/2^{+}$. ∎

Proof of item ii.

Let $V(\sigma)$ be as in the proof of Lemma 3.3. Since $V(\sigma)\to\infty$ as $\sigma\to 1/2^{+}$, we have, for each fixed $y>0$

[TABLE]

Thus, for each fixed $L>0$,

[TABLE]

is a tail event. By Lemma 3.3, $\frac{1}{V(\sigma)}\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}\to_{d}\mathcal{N}(0,1)$, as $\sigma\to 1/2^{+}$. Thus, this tail event has positive probability (see the proof of Lemma 3.2). By the Kolmogorov zero or one Law, a.s.:

[TABLE]

Similarly, a.s.:

[TABLE]

Since $F(\sigma)=\sum_{p\in\mathcal{P}}\frac{X_{p}}{p^{\sigma}}$ is a.s. an analytic function, with probability $1$ we have that $F(\sigma)=0$ for an infinite number of $\sigma\to 1/2^{+}$. ∎

Acknowledgments. The proof of Theorem 1.1 item ii was initially presented only in the case $\pi(x)\ll\frac{x}{\log x}$. I would like to thank the anonymous referee for pointing that this should not be a necessary condition for the existence of an infinite number of real zeros, and for pointing the reference [1], from which I could adapt their Theorem 1.2 for random Dirichlet series (Lemma 3.3).