On the zeros of sum from n=1 to 00 of lambda_P(n)/n^s

TL;DR

This paper constructs a Dirichlet series with a unique zero in its half-plane of convergence and explores conditions related to the Generalized Riemann Hypothesis, addressing a question posed by Michel Balazard.

Contribution

It provides a specific Dirichlet series with a single zero and proposes new sufficient conditions for the Generalized Riemann Hypothesis.

Findings

Constructed a Dirichlet series with only one zero in its half-plane

Proposed several sufficient conditions for the Generalized Riemann Hypothesis

Addressed a question of Michel Balazard

Abstract

Our main result is to answer a question of Michel Balazard by giving a Dirichlet series with only one zero in its half-plane of convergence. At the end of the paper we also give several sufficient conditions for the Generalized Riemann Hypothesis and ask if any of them are true.

Equations

$$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\frac{1}{9}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\text{mbox}etc.=0.$$

$$\sum _{n=1}^{\infty }\frac{\lambda (n)}{n}=0.$$

$$\prod _p\frac{1}{1-\frac{1}{p}}=\sum _{n=1}^{\infty }\frac{1}{n}=\infty .$$

$$\sum _{n=1}^{\infty }\frac{\lambda (n)}{n}=\prod _p\frac{1}{1+\frac{1}{p}}=0$$

$$\sum _{n=1}^{\infty }\frac{\mu (n)}{n}=0,$$

$$\sum_{n=1}^{\infty}\frac{\lambda(n)}{n^{s}}\mbox{ vanishes only at $s=1$}.$$

$$F(s)=\sum _{n=1}^{\infty }\frac{f(n)}{n^s},$$

$$\lambda_{\mathcal{P}}(p)=\left\{\begin{array}[]{cl}-1&\mbox{ if $p\in\mathcal{P}$}\\ 0&\mbox{ if $p\not\in\mathcal{P}$}\end{array}\right..$$

$$Z_{\mathcal{P}}=Z\biggl{(}\sum_{n=1}^{\infty}\frac{\lambda_{\mathcal{P}}(n)}{n^{s}}\biggr{)}.$$

$${\sigma }_{\mathcal{P}}\le 1.$$

$${\zeta }_{\mathcal{P}}(s):=\prod _{p\in \mathcal{P}}\frac{1}{1-\frac{1}{p^s}}.$$

$$\zeta_{\mathcal{P}}(s)\neq 0\quad\mbox{ for $\sigma>\sigma_{\mathcal{P}}$}.$$

$$\sigma_{\mathcal{N}}=\left\{\begin{array}[]{cl}-\infty&\mbox{ if $\mathcal{P}=\emptyset$}\\ \max\{\sigma_{\mathcal{P}},0\}&\mbox{ if $\mathcal{P}\neq\emptyset$}\end{array}\right..$$

$$\sum _{n\in \mathcal{N}}\frac{1}{n^s}=\prod _{p\in \mathcal{P}}\frac{1}{1-\frac{1}{p^s}}={\zeta }_{\mathcal{P}}(s),(\sigma >{\sigma }_{\mathcal{N}})$$

$$\sum _{n\in \mathbb{N}}\frac{{\lambda }_{\mathcal{P}}(n)}{n^s}=\sum _{n\in \mathcal{N}}\frac{\lambda (n)}{n^s}=\prod _{p\in \mathcal{P}}\frac{1}{1+\frac{1}{p^s}}=\frac{{\zeta }_{\mathcal{P}}(2s)}{{\zeta }_{\mathcal{P}}(s)},(\sigma >{\sigma }_{\mathcal{N}}).$$

$${\sigma }_c=\frac{1}{2}.$$

$$0<\max\Bigl{\{}\frac{a}{2},a-\frac{19}{40}\Bigr{\}}<b<a<1.$$

$$1-\frac{1}{2^s}.$$

$$\Bigl{(}1-\frac{1}{2^{s}}\Bigr{)}\Bigl{(}1-\frac{1}{2^{s-1}}\Bigr{)}.$$

$$0<\max\Bigl{\{}\frac{a}{2},a-\frac{19}{40}\Bigr{\}}<b<a<1.$$

$$\Re \rho -{\sigma }_c$$

$${\sigma }_a-\Re \rho$$

$$\sum _{n=1}^{\infty }\frac{\chi (n)}{n^s}.$$

$$Z$$

$$\sum _{n=1}^{\infty }\frac{f(n)}{n^s}$$

$$\sum _{n=1}^{\infty }\frac{f(n)}{n^s}=\prod _p\frac{1}{1-\frac{f(p)}{p^s}}\ne 0.$$

$$0\le \Re \rho -{\sigma }_c\le {\sigma }_a-{\sigma }_c\le 1,$$

$$f(p)=-\frac{1}{\log \log p}(p\ge 29)$$

$$\sum _p\frac{f(p)}{p}=-\sum _{p\ge 29}\frac{1}{p\log \log p}=-\infty .$$

$$\sum_{p\geq 29}\frac{1}{p^{\sigma}\log\log p}\asymp\log\log\frac{1}{\sigma-1}\quad\mbox{ $(1<\sigma<1+\frac{1}{100})$}.$$

Full text

2010 AMS Mathematics Subject Classification: 11M41, 11M06, 11M26.

Keywords and phrases: zeros of Dirichlet series, completely multiplicative functions

In memory of Jean-Pierre Kahane

TABLE OF CONTENTS

1.

Introduction

1(a)

From Euler to Landau.

1(b)

Zeros of Dirichlet series.

1(c)

Generalization of $\lambda(n)$ to $\lambda_{\mathcal{P}}(n)$ and $\zeta(s)$ to $\zeta_{\mathcal{P}}(s)$.

1(d)

Easy or known results on the set of zeros $Z_{\mathcal{P}}$.

1(e)

New results on the multiset of zeros of Dirichlet series.

1(f)

Zeros and abscissae of convergence for Dirichlet series with completely multiplicative coefficients.

1(g)

Zeros of Helson’s zeta functions.

2.

Proof of Theorem 3.

3.

Proof of Theorem 0.

4.

Architecture of the proofs of Theorems 1 and 2.

5.

Vocabulary, notations and results for Beurling primes.

6.

Primes in short intervals.

7.

Abscissae of convergence.

8.

Proofs of Theorems 2 and 1.

9.

Open questions related to GRH$\setminus$RH.

**1. Introduction

1(a) From Euler to Landau.

**Euler, in his paper [4] of 1737, writes

[TABLE]

In modern words we define $\lambda(n)$, Liouville’s function, to be the completely multiplicative function which is $-1$ at every prime. What Euler writes is then

[TABLE]

It is in this paper he uses the “Euler” product formula. He applies it first to the completely multiplicative function $\frac{1}{n}$ and obtains

[TABLE]

Then he applies it a second time to the completely multiplicative function $\frac{\lambda(n)}{n}$ and obtains

[TABLE]

due to (1.2), and (1.1) follows.

Let us now read formulas (1.2) and (1.3) with our modern eyes, with our definitions of infinite sums. Since the completely multiplicative function $\frac{1}{n}$ is positive, his proof of (1.2) is valid three centuries later. On the other hand the completely multiplicative function $\frac{\lambda(n)}{n}$ is not positive nor summable. Thus the first equality of (1.3) is not proved.

As a matter of fact, it is Riemann [15] in 1859 for the first part, and de la Vallée-Poussin [17] and Hadamard [5] 37 years later for the second part, by continuing Euler’s $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}}$ as a meromorphic function in $\mathbb{C}$ and proving it does not vanish in the closed half-plane Re $s\geq 1$, who brought the tools to prove (1.1). More precisely, von Mangoldt [13] proved in 1897, just one year after de la Vallée-Poussin and Hadamard, that

[TABLE]

where $\mu(n)$ is the Mobius function. In 1907, Landau [12] deduced (1.1) from (1.4). Thus 180 years separates Euler’s claim and its proof!

**1(b) Zeros of Dirichlet series.

**The series

[TABLE]

(This result is part of Theorem 0 below, and we recall briefly its proof in section 3.) In the situation of any Dirichlet series

[TABLE]

we consider the zeros of $F(s)$ from a naive point of view. We denote by $Z(F)$ the set of complex numbers for which the above series converges and its sum is zero. In particular, we have $Z(F)\subset\{s\in\mathbb{C}:\Re s\geq\sigma_{c}(F)\}$ where $\sigma_{c}(F)$ is the abscissa of convergence of $F$. Notice that the series may converge and have zero sum on the line $\sigma=\sigma_{c}(F)$. In other words, $Z(F)$ may contain points on this line.

**1(c) Generalization of $\lambda(n)$ to $\lambda_{\mathcal{P}}(n)$ and $\zeta(s)$ to $\zeta_{\mathcal{P}}(s)$.

**Let $\mathbb{P}$ denote the set of all primes and let $\mathcal{P}\subset\mathbb{P}$ be a subset. We define the generalized Liouville function associated to $\mathcal{P}$ as the completely multiplicative function defined on primes by

[TABLE]

In this paper we study the set of zeros

[TABLE]

Let $\sigma_{\mathcal{P}}$ denote the abscissa of convergence of the series $\sum_{p\in\mathcal{P}}p^{-s}$. It is easy to see that

[TABLE]

We generalize the usual $\zeta(s)$ by denoting

[TABLE]

Of course $\lambda_{\mathbb{P}}=\lambda$ and $\zeta_{\mathbb{P}}=\zeta$. As in this particular important case $\mathcal{P}=\mathbb{P}$, the general zeta function $\zeta_{\mathcal{P}}(s)$ is a normally convergent Euler product in every fixed closed half-plane $\sigma\geq\max\{\sigma_{\mathcal{P}},0\}+\varepsilon$, for any fixed $\varepsilon>0$. If $\mathcal{P}$ is finite, (1.7) defines a non-vanishing meromorphic function in $\mathbb{C}$ whose multiset of poles is the union of $|\mathcal{P}|$ infinite arithmetic progressions of purely imaginary complex numbers. Except perhaps at $s=0$, all those poles are simple. At $s=0$, the multiplicity is $|\mathcal{P}|$.

Let us now study the case where $\mathcal{P}$ is infinite. Then

[TABLE]

Moreover, as in the basic usual case, if $\zeta_{\mathcal{P}}(s)$ has a meromorphic continuation in some open set across the line $\sigma=\sigma_{\mathcal{P}}$, we continue to denote this continuation by $\zeta_{\mathcal{P}}(s)$.

Let $\mathcal{N}=\{n\in\mathbb{N}:p|n\implies p\in\mathcal{P}\}$ (i.e. all the positive integers formed from the primes in $\mathcal{P}$) and let $\sigma_{\mathcal{N}}$ denote the abscissa of convergence of $\sum_{n\in\mathcal{N}}n^{-s}$. It is easy to prove that

[TABLE]

By introducing this complex parameter $s$ and generalizing to any set $\mathcal{P}$ of primes, we can interpret the two Euler formulas (1.2) and (1.3) by the absolutely convergent Euler products

[TABLE]

and

[TABLE]

**1(d) Easy or known results on the set of zeros $Z_{\mathcal{P}}$.

**

**Theorem 0

**Let $\mathcal{P}$ and $\mathcal{P}^{\prime}$ be sets of primes.

1. (a)

If $\sigma_{c}(\sum_{p\in\mathcal{P}\Delta\mathcal{P}^{\prime}}p^{-s})\leq 0$, then $Z_{\mathcal{P}}=Z_{\mathcal{P}^{\prime}}$; 2. (b)

$\overline{Z_{\mathcal{P}}}=Z_{\mathcal{P}}$; 3. (c)

if $\sigma_{\mathcal{P}}\leq 0$, then $Z_{\mathcal{P}}=\emptyset$; 4. (d)

if $\sigma_{\mathcal{P}}>0$, then $\Re Z_{\mathcal{P}}\subset(0,\sigma_{\mathcal{P}}]\subset(0,1]$; 5. (e)

$1\in Z_{\mathcal{P}}\Leftrightarrow\sum_{p\in\mathcal{P}}\frac{1}{p}=\infty$; 6. (f)

$Z_{\mathbb{P}}=\{1\}$;

Remark 1  Point (a) shows that the function $\mathcal{P}\to Z_{\mathcal{P}}$ is, in a way, locally constant.

**1(e) New results on the multiset of zeros of Dirichlet series.

**We know that the maximal open set where a Dirichlet series $F(s)=\sum_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$ is both convergent and holomorphic is $\mathbb{C}_{\sigma(F)}$ where $\mathbb{C}_{\alpha}=\{s\in\mathbb{C}:$ Re $s>\alpha\}$.

Let $\alpha\in[-\infty,\infty)$. Thanks to Weierstrass ([1], Theorem 3.3.1), we know that a necessary and sufficient condition for a multiset $Z$ of $\mathbb{C}_{\alpha}$ to be the multiset $Z(F)$ of some not identically zero holomorphic function $F$ in $\mathbb{C}_{\alpha}$, is for $Z$ to be locally finite in $\mathbb{C}_{\alpha}$.

Now for the same question where we ask $Z$ to be equal to the set $Z(F)\cap\mathbb{C}_{\alpha}$ for some Dirichlet series $F$ with $\sigma(F)=\alpha$, we are far from knowing the necessary and sufficient condition analogue to the Weierstrass theorem. It is even possible that it is impossible to give such a characterization.

In 2000, Balazard (unsolved problem 24 of [14]) asked the first question for this problem. He asked for an example of a Dirichlet series $F(s)$ for which $Z(F)\cap\mathbb{C}_{\sigma_{c}(F)}$ has only one element. Notice that under the Riemann Hypothesis, we have an example, namely the series in (1.5). For this series we then have $Z=\{1\}$ and

[TABLE]

This last formula (1.11) is classical under RH. Eighteen years later, we are able to provide an unconditional family of examples. More precisely we get the following result.

**Theorem 1

**Let $a$ and $b$ be two real numbers such that

[TABLE]

Then there exists a set $\mathcal{P}_{a,b}\subset\mathbb{P}$ such that $Z_{\mathcal{P}_{a,b}}=\{a\}$ with $a$ being a simple zero and $\sigma_{c}=b,\sigma_{a}=a$, where $\sigma_{c}$ and $\sigma_{a}$ are the abscissa of convergence and absolute convergence of $\sum_{n=1}^{\infty}\lambda_{\mathcal{P}_{a,b}}(n)n^{-s}$ respectively.

*Under the Riemann Hypothesis, we can replace $(1.12)$ by $0<\frac{a}{2}<b<a<1$ or $(a,b)=(1,\frac{1}{2})$.

Let us remark that to provide an example of $\mathcal{P}$ with $|Z_{\mathcal{P}}|=1$, we were obliged to choose the zero real, because of the symmetry of $Z_{\mathcal{P}}$ about the real axis ($\overline{Z_{\mathcal{P}}}=Z_{\mathcal{P}}$).

Before giving the answer to another question, let us begin with some obvious remarks. It is very easy to construct Dirichlet series with at least two zeros. The simple example is

[TABLE]

Now if we want to have at least two zeros with different real part, it is also easy: just choose

[TABLE]

But now we ask the following question:

[TABLE]

As far as we know, (1.13) is an open question. Once again, some Dirichlet series $\frac{\zeta_{\mathcal{P}}(2s)}{\zeta_{\mathcal{P}}(s)}$ with well chosen $\mathcal{P}$ give a family of examples which allow us to answer this question positively.

**Theorem 2

**Let $a$ and $b$ be two real numbers such that

[TABLE]

Then there exists a set $\mathcal{P}_{a,b}\subset\mathbb{P}$ such that $Z_{\mathcal{P}_{a,b}}\supset\{a,b\}$ with $a$ and $b$ being simple zeros and $\sigma_{c}\leq\max\{\frac{3ab}{2(a+b)},a-\frac{19}{40}\},\sigma_{a}=a$, where $\sigma_{c}$ and $\sigma_{a}$ are the abscissae of simple and absolute convergence of $\sum_{n=1}^{\infty}\lambda_{\mathcal{P}_{a,b}}(n)n^{-s}$ respectively.

*Under RH, we can replace $\max\{\frac{a}{2},a-\frac{19}{40}\}$ by $\frac{a}{2}$ in $(1.14)$, and $\max\{\frac{3ab}{2(a+b)},a-\frac{19}{40}\}$ by $\frac{3ab}{2(a+b)}$.

Notice that the conditions in Theorems 1 and 2 are the same. As a matter of fact, the proofs of the two results have similar structure, as will be seen in sections 4 and 8.

**1(f) Zeros and abscissae of convergence for Dirichlet series with completely multiplicative coefficients

**We continue with another question on the set of zeros of a Dirichlet series $F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^{s}}$ with completely multiplicative coefficients $f(n)$. Denote the abscissa of convergence and absolute convergence by $\sigma_{c}$ and $\sigma_{a}$ respectively.

Let $V$ denote the set of values of the difference

[TABLE]

when $f$ varies through the completely multiplicative functions and $\rho\in Z(F)$. What can we say about the set $V$?

By Theorem 1, we know that $V\supset(0,\frac{19}{40})$ which is improved under the Riemann Hypothesis to $V\supset(0,\frac{1}{2})$. But, as a matter of fact, a little more is known unconditionally. Before telling our result, let us consider an analogous question for Dirichlet series with completely multiplicative coefficients.

Let $W$ denote the set of values of

[TABLE]

with $\rho$ as before. Now by Theorem 2, we have $W\supset(0,\frac{19}{40})$ which is improved under the Riemann Hypothesis to $W\supset(0,\frac{1}{2})$. But, in contrast to the case for $V$, we are not able to prove this unconditionally.

**Theorem 3

***(i)  $[0,\frac{1}{2}]\subset V\subset[0,1]$.

(ii)  There exists $v\in[\frac{1}{2},1]$ such that $V=[0,v]$ or $V=[0,v)$.

(iii)  $[0,\frac{19}{40})\cup\{\frac{1}{2}\}\subset W\subset[0,1]$, and under RH, we can replace (iii) by

(iii$)^{\prime}$  $[0,\frac{1}{2}]\subset W\subset[0,1]$.

Notice that these results cannot be extended to the case where $f$ is only multiplicative. To see this, consider $f$ defined by $f(1)=1$, $f(2)=-1$ and zero otherwise. Then $F(s)=1-2^{-s}$ vanishes at 0 and $\sigma_{c}(F)=\sigma_{a}(F)=-\infty$.

**1(g) Zeros of Helson zeta functions

**In [16], Seip studies the multiset of zeros of Helson zeta functions. For each completely multiplicative unimodular function $\chi$, Helson’s zeta function $\zeta_{\chi}(s)$, is the meromorphic continuation (if any) of the Dirichlet series

[TABLE]

We shall call a Helson zeta function admissible if it has a meromorphic continuation to $\mathbb{C}_{\frac{1}{2}}$.

Notice that by the Euler product, $\zeta_{\chi}(s)$ cannot vanish for $\sigma>1$. Moreover ([16], Theorem 2.1) an admissible Helson zeta function has at most one zero on the line $\sigma=1$ and, if it has one, it is simple. The function $\zeta(2s)/\zeta(s)$, which we already discussed, is an example where there is one.

Let us look now in the strip $S:=\{s\in\mathbb{C}:1/2<$ Re$s<1\}$. Let $Z$ be a multiset belonging to $S$. We recall that by Weierstrass’ Theorem ([1], Theorem 3.3.1), a necessary condition for $Z$ to be the multiset of zeros of an admissible zeta function is:

[TABLE]

Seip proves that under the Riemann Hypothsesis this necessary condition is also sufficient: if (1.15) is true, then there exists an admissible Helson zeta function whose multiset of zeros in $S$ is equal to $Z$.

The comparison of this result with ours is impressive. Under RH, Seip has found the necessary and sufficient condition. The main result here is only to make unconditional the (conditional on RH) 100 year-old result that there exists an example with a single zero in the half-plane of convergence!

We do not know if it is possible to use some of Seip’s results and/or tools in [16] to find new examples of sets of zeros of Dirichlet series with completely multiplicative coefficients themselves, and not of their possible meromorphic continuation.

**2. Proof of Theorem 3

Step 1**  We prove that $V\subset[0,1]$. Let $f$ be a completely multiplicative function. Let the Dirichlet series

[TABLE]

have abscissa of convergence and absolute convergence $\sigma_{c}$ and $\sigma_{a}$ respectively, and let $\rho$ be a zero of the series. Then $\sigma_{c}\leq\Re\rho\leq\sigma_{a}$ for the series has to converge at $\rho$ and for $\sigma>\sigma_{a}$,

[TABLE]

Hence

[TABLE]

and $V\subset[0,1]$.

Step 2  We prove that $0\in V$. Define the completely multiplicative function $f$ on primes $p$ by

[TABLE]

and zero otherwise. As $29>e^{e}$, we have $-1\leq f(p)\leq 0$ for all $p$. Moreover, we have

[TABLE]

By applying Theorem 9 of [10], it follows that the Dirichlet series in (2.1) vanishes at $s=1$. To finish the proof, it suffices to verify that the abscissa of the series is 1.

Now

[TABLE]

As

[TABLE]

it follows that $\log F(\sigma)\asymp-\log\log\frac{1}{\sigma-1}$ and

[TABLE]

As $F(1)=0$, $F(s)$ cannot be extended holomorphically to a neighbourhood of 1, and the result follows.

Remark 2  Here we have an example where $\sigma_{c}=\Re\rho=\sigma_{a}$.

Step 3  We prove the following: if $f$ is completely multiplicative and $\beta^{*}+i\gamma$ is a zero of (2.1), then

[TABLE]

To see this, let $f_{1}(n)=f(n)n^{-\sigma_{c}-i\gamma}$. The Dirichlet series for $f_{1}$ now has abscissa of convergence 0 and a real zero at some $\beta\geq 0$. We have to show that

[TABLE]

Note that by (2.2), $\beta\leq 1$. Let $\mathcal{P}=\{p_{1},p_{2},\ldots\}$ where $p_{k}$ is an increasing sequence of prime numbers such that $p_{k}\asymp 2^{k}$ and let $\alpha\in(0,\beta)$. Define a completely multiplicative function $g_{\alpha}$ by

[TABLE]

Let $\sigma>\alpha$. As $f_{1}(n)\ll n^{\alpha}$

[TABLE]

It follows that the product

[TABLE]

is absolutely convergent for $\sigma>\alpha$. Moreover, the Dirichlet series for $f_{1}$ and $g_{\alpha}$ are absolutely convergent for $\sigma>1$ and so, using Euler products, we have for $\sigma>1$

[TABLE]

It follows that the series on the left is actually convergent for $\sigma>\alpha$ and hence the above holds for $\sigma>\alpha$. Thus it is zero at $\beta$ and its abscissa of convergence is at most $\alpha$. However, $g_{\alpha}(p)=p^{\alpha}$ for $p\in\mathcal{P}$, so this abscissa is at least $\alpha$. Hence it equals $\alpha$. It follows that $\beta-\alpha\in V$. But $\alpha$ has been chosen arbitrarily in $(0,\beta)$, so (2.3) follows.

Step 4  $\frac{1}{2}\in V$. This follows immediately from the fact that the $L$-function associated to any non-principal Dirichlet character has zeros on the critical line and abscissa of convergence 0.

These steps prove (i) and (ii) of Theorem 3. Now consider (iii).

By replacing $\Re\rho-\sigma_{c}$ with $\sigma_{a}-\Re\rho$ in (2.2) we obtain $W\subset[0,1]$. Step 2 is also valid in the case of $W$ and shows $0\in W$. Next, the example in Step 4 (with $\sigma_{a}=1$) shows that $\frac{1}{2}\in W$. By Theorem 2, we have $(0,\frac{19}{40})\subset W$, which concludes the proof of part (iii).

a $\Box$

**3. Proof of Theorem 0

**Proof of (a).  Let $\rho=a+ib\in Z_{\mathcal{P}}$. Since for all $s$

[TABLE]

we have $\mathcal{P}\neq\emptyset$. Since $\lim_{n\to\infty}\frac{\lambda_{\mathcal{P}}(n)}{n^{\rho}}=0$, it follows that

[TABLE]

Now we call a completely multiplicative function $h$ CMO if $\sum_{n=1}^{\infty}h(n)=0$. We suppose again that $\rho=a+ib\in Z_{\mathcal{P}}$. On one hand it means that $f(n):=\frac{\lambda_{\mathcal{P}}(n)}{n^{\rho}}$ is CMO. On the other hand, by (3.1), we have $a>0$. Thus for $g(n):=\frac{\lambda_{\mathcal{P}^{\prime}}(n)}{n^{\rho}}$, $g$ is completely multiplicative such that for all primes $p$, $|g(p)|<1$. As $\sigma_{c}(\sum_{p\in\mathcal{P}\Delta\mathcal{P}^{\prime}}p^{-s})\leq 0$, we have also $\sum_{p}|g(p)-f(p)|<\infty$. By using théorème 3 of [10], it follows that $g$ is CMO. In other words, $\rho\in Z_{\mathcal{P}^{\prime}}$, and (a) is proven.

Part (b) comes from the fact that $\lambda_{\mathcal{P}}$ is a real function.

Part (c) follows from the combination of $Z_{\emptyset}=\emptyset$ and part (a).

Part (d) follows from the combination of (3.1), (1.10), (1.9), (1.6) and (1.8).

Part (e) comes from théorème 9 of [10].

Proof of (f).  By (e), we know that $1\in Z_{\mathbb{P}}$.

To show that it contains no other points, let $\rho\in Z_{\mathbb{P}}$. Since the abscissa of convergence of $\sum_{n=1}^{\infty}\frac{\lambda(n)}{n^{s}}$ is at least $\frac{1}{2}$ due to the existence of Riemann zeros, it follows that Re $\rho\geq\frac{1}{2}$. Now, by Abel’s Theorem

[TABLE]

But $\zeta(2s)\neq 0$ for $\sigma\geq\frac{1}{2}$, so $\rho$ must be a pole of $\zeta$; i.e. $\rho=1$.

a $\Box$

**4. Architecture of the proofs of Theorems 1 and 2

**Let $a\in(0,1)$. We begin by recalling the proof in [11] of the existence of a set of primes $\mathcal{P}_{a}$ such that $a$ belongs to $Z_{\mathcal{P}_{a}}$. We know that $1$ is a zero of $F(s):=\frac{\zeta(2s)}{\zeta(s)}$. It follows that $a$ is a zero of $F(s/a)$. But

[TABLE]

where $\zeta_{\mathbb{P}^{1/a}}(s)$ is the zeta function associated to the set of Beurling primes $\mathbb{P}^{1/a}=\{p^{1/a}:p\in\mathbb{P}\}$. This set is sparse. It is the reason why the PNT allows us to approximate $\mathbb{P}^{1/a}$ by a subset $\mathcal{P}_{a}$ of $\mathbb{P}$ such that

[TABLE]

We have

[TABLE]

As for the usual case $a=1$ and $\mathcal{P}_{1}=\mathbb{P}$, we prove that we can deduce the wanted formula

[TABLE]

from the combination of (4.1) and (4.2).

The architecture of the proofs of Theorems 1 and 2 is similar, but we need to introduce two new tools. Under RH, we recall that we know an example which answers Balazard’s question. It is

[TABLE]

for which $s=1$ is the only (simple) zero of $F$ in the open half-plane where $\sigma>\frac{1}{2}$. But we do not know that RH is true. To get an unconditional example, we prove that it is possible to choose a set of $\mathcal{P}$ of primes such that (4.4) is replaced by

[TABLE]

with $\sigma_{c}(F_{\mathcal{P}})<1$, and for which $s=1$ is always a simple zero.

More precisely, instead of working with the usual zeta function, we begin to work with Zhang’s zeta function $\zeta_{\mathcal{R}}(s)$ (see [18] or [3]) associated to an appropriate multiset $\mathcal{R}$ of generalized primes which share some of the properties of $\zeta(s)$ under RH:

• •

$\zeta_{\mathcal{R}}(s)=\prod_{r\in\mathcal{R}}\frac{1}{1-\frac{1}{r^{s}}}$ is normally convergent in every half-plane $\sigma>1+\varepsilon$ with $\varepsilon>0$;

• •

$\zeta_{\mathcal{R}}(s)$ has a non-vanishing meromorphic continuation of finite order to $\sigma>\frac{1}{2}$ with a unique simple pole at 1 with residue 1.

Now, more generally than in [11], we work here with the group of meromorphic functions in some non-empty open vertical half-plane generated by the function $\zeta_{\mathcal{R}}(\lambda s)$ where $\lambda>0$.

To prove Theorem 1 we use the function

[TABLE]

which has a simple zero at $a$ and a simple pole at $b$. To prove Theorem 2, we use the function

[TABLE]

which has two simple zeros at $a$ and $b$. To finish off the proofs, we need to approximate these meromorphic functions by functions of the form

[TABLE]

with $\mathcal{P}_{a,b}\subset\mathbb{P}$ such that this function has the same zeros and poles as described above and such that the formula

[TABLE]

is valid for $\sigma>b$ in the first case and for $s=a,b$ in the second.

To do this approximation, instead of using the PNT as in [11], we use the Baker, Harman and Pintz result on primes in short intervals

[TABLE]

It is this number $\frac{21}{40}$ which explains the number $\frac{19}{40}$ in our results.

**5. Vocabulary, notations and results for Beurling primes

**In 1937, Beurling [2] (see also [3]) had the idea to generalize the usual couple $(\mathbb{P},\mathbb{N})$ formed by the usual sets $\mathbb{P}$ of primes and $\mathbb{N}$ of positive integers in the following way. He considers any multiset $\mathcal{P}$ of $(1,\infty)$ which is locally finite in $[1,\infty)$. The elements of $\mathcal{P}$ are called the generalized primes. He defines $\mathcal{N}$ to be the multiset of $[1,\infty)$ formed by the finite product of elements of $\mathcal{P}$ (the number 1 occurs as the product indexed by the subset $\emptyset$). We will talk of a discrete generalized prime system or just g-prime system for such a couple $(\mathcal{P},\mathcal{N})$.

Let us mention that there are two natural generalizations of g-prime systems which we will not use directly here. We refer the interested reader to [8].

Let $(\mathcal{P},\mathcal{N})$ be a g-prime system. Every element $n$ of the multiset $\mathcal{N}$ has a unique decomposition in generalized primes

[TABLE]

We define the generalized Möbius function $\mu_{\mathcal{P}}$ on the multiset $\mathcal{N}$ by the formula

[TABLE]

When $\mathcal{P}=\mathbb{P}$, $\mu_{\mathcal{P}}$ is the usual Möbius function.

Notice that for any sequence $(a_{n})_{n\in\mathcal{N}}$ of complex numbers defined on $\mathcal{N}$, the function

[TABLE]

is a generalized Dirichlet series. We shall use, without referencing anymore, the definitions and properties of generalized Dirichlet series. (See [6] for the theory of generalized Dirichlet series.)

Notice that in the introduction, we considered the particular example of a g-prime system $(\mathcal{P},\mathcal{N})$ with $\mathcal{P}\subset\mathbb{P}$. The definition of $\zeta_{\mathcal{P}}(s)$ generalizes with no difficulty for any g-prime system. If the series $\sum_{n\in\mathcal{N}}n^{-s}$ has a finite abscissa of convergence $\sigma_{c}$, then

[TABLE]

defines a non-vanishing holomorphic function for $\sigma>\sigma_{c}$. If $\zeta_{\mathcal{P}}(s)$ has a meromorphic continuation to some open set across the line $\sigma=\sigma_{c}$ we continue to write $\zeta_{\mathcal{P}}(s)$ for this continuation.

If $\mathcal{P},\mathcal{P}_{1},\mathcal{P}_{2}$ are multisets of primes and $a>0$, we have

[TABLE]

and

[TABLE]

Let $Q=\{q_{1},q_{2},\ldots\}$ be an infinite multiset of generalized primes, $I:Q\to(1,\infty)$ and $\sigma_{0}\geq 0$. We shall say that

[TABLE]

is a simply absolutely convergent quotient of generalized Euler products in $\sigma>\sigma_{0}$ if for all $s$ in the half-plane where $\sigma>\sigma_{0}$ the infinite product

[TABLE]

is absolutely convergent. Notice that the expression in (5.1) represents a non-vanishing holomorphic function in this half-plane.

We shall denote multisets of generalized integers by calligraphic letters ($\mathcal{A},\mathcal{B},\mathcal{C},\ldots$) with the corresponding counting function by its capital equivalent. eg for such a multiset $\mathcal{A}$, let

[TABLE]

Moreover, if $c>0$, we write

[TABLE]

For $\mathbb{P}$ however, we shall keep the traditional notation $\pi(x)$ and $[x]$ for the counting funtions of the primes and natural numbers.

Definition.  Let $(\mathcal{R},\mathcal{N})$ be a g-prime system. We say it is good if it satisfies the following properties:

[TABLE]

for all $\varepsilon>0$111Here li$(x)$ is the usual logarithmic integral, given by li$(x)=\int_{2}^{x}\frac{1}{\log y}\,dy$.. As such, $\zeta_{\mathcal{R}}(s)$ has a non-vanishing meromorphic continuation to the half-plane $\sigma>\frac{1}{2}$, with exactly one (simple) pole at $1$ with residue 1.

Remark 3  Of course, under the Riemann Hypothesis, the basic g-prime system $(\mathbb{P},\mathbb{N})$ is good.

Remark 4  Comparing to Zhang’s work ([18] or [3]) we changed conditions (5.2) and (5.3) a little.

Now Zhang proved that, unconditionally, good systems exist (see [3], Theorem 17.11 and remark 17.12).

Theorem (Zhang)

*A good g-prime system exists.

**Lemma 5.1

***Let $(\cal{R},\cal{N})$ be a good g-prime system Then for all $\varepsilon>0$, there exists $C>0$ such that *

[TABLE]

This follows from Theorem 2.3 of [9].

**6. Primes in short intervals

**We begin with the usual primes.

Lemma 6.1 (Baker, Harman and Pintz)

We have, for $x^{\frac{21}{40}}\leq y\leq x$ and $x$ large enough

[TABLE]

Proof.  This is a consequence of Theorem 10.8 of [7].

a $\Box$

Lemma 6.2

Let $(\mathcal{R},\mathcal{N})$ be a good g-prime system, $c\in(0,1)$, $h\in(0,1)$ and $\varepsilon>0$. Then

[TABLE]

Proof.  We have

[TABLE]

As $(\mathcal{R},\mathcal{N})$ be a good g-prime system, we have

[TABLE]

a $\Box$

Lemma 6.3

Let $\mathcal{Q}$ and $\mathcal{Q}^{*}$ be two multisets of generalized primes. Let $\delta>0$ and $0<h\leq 1$ be two real numbers such that

[TABLE]

[TABLE]

and for $x$ large enough,

[TABLE]

Then there exists an injection $I:\mathcal{Q}\to\mathcal{Q}^{*}$ such that

[TABLE]

is a simply absolutely convergent quotient of generalized Euler products in $\sigma>\max\{\delta+h-1,0\}$.

Proof.  Let $T_{n}$ be an increasing sequence of positive reals defined by $T_{1}=1$ and

[TABLE]

By (6.2) and (6.3) there exists a positive integer $n_{0}$ such that there is an injection $I_{0}:{\cal{Q}}\cap(1,T_{n_{0}}]\to{\cal{Q}}^{*}\cap(1,T_{n_{0}}]$ and for all $n\geq n_{0}$, there is also an injection

[TABLE]

As the intervals $(T_{n},T_{n+1}]$ are disjoint, we get a global injection $I:\cal{Q}\to\cal{Q}^{*}$ such that $I(q)=q+O(q^{h})$.

For fixed $s$ with $\sigma>0$, we have, uniformly in $q$,

[TABLE]

Thus

[TABLE]

and using (6.1)

[TABLE]

if $\sigma>\delta+h-1$, as required.

a $\Box$

**7. Abscissae of convergence

**Let $(\cal{R},\cal{N})$ be a good g-prime system and $c\in(\frac{1}{2},1)$. Let $\mathcal{Q}_{c}:=\mathcal{R}\cup\mathcal{R}^{1/c}$ and $\mathcal{M}_{c}$ the multiset of generalized integers associated to $\mathcal{Q}_{c}$.

**Lemma 7.1

**Let $(\cal{R},\cal{N})$ be a good g-prime system. Then

[TABLE]

Proof.  These follow from writing

[TABLE]

and using the fact that $N(t)=t+O(t^{\frac{1}{2}+\varepsilon})$. Thus, for $\alpha\neq 1$,

[TABLE]

If $\alpha\leq\frac{1}{2}$, the integral is $O(x^{\frac{1}{2}-\alpha+\varepsilon})$ and the result follows. If $\alpha>\frac{1}{2}$, the integral is

[TABLE]

Now for $\alpha>1$, the integral on the right is just $\zeta_{\mathcal{R}}(\alpha)-\frac{\alpha}{\alpha-1}$. But by analytic continuation, this still holds for $\alpha\in(\frac{1}{2},1)$.

a $\Box$

**Lemma 7.2

**Let $(\cal{R},\cal{N})$ be a good g-prime system and $c\in(\frac{1}{2},1)$. Then for all fixed $\varepsilon>0$, we have

[TABLE]

Proof.  Throughout this proof, $k$ and $\ell$ will implicitly be used to denote generalized integers of $\mathcal{N}$. We use Dirichlet’s hyperbola method. Thus for every positive reals $x$ and $y$, we have

[TABLE]

where

[TABLE]

Now we use the formula $N(t)=t+O(t^{\frac{1}{2}+\varepsilon})$ to estimate these three sums, and at the end we optimize in $y$.

We have

[TABLE]

So, by Lemma 7.1, we obtain

[TABLE]

Next,

[TABLE]

Thus

[TABLE]

Finally,

[TABLE]

Combining these formulas, the different terms in $xy^{c-1}$ disappear and it follows that

[TABLE]

Choosing $y=x^{\frac{1}{1+c}}$ gives the result.

a $\Box$

**Lemma 7.3

**Let $(\cal{R},\cal{N})$ be a good g-prime systems and $c\in(\frac{1}{2},1)$. Then for all fixed $\varepsilon>0$,

[TABLE]

Proof.  By Lemma 7.1, the abscissae of convergence of the series

[TABLE]

are both at most 1. By the first effective Perron formula, it follows that for $n\geq 3$ and $T\geq 1$,

[TABLE]

where

[TABLE]

a $\Box$

**Upper bound for $E$

**We divide $E$ into $E=E_{1}+E_{2}+E_{3}$ where $E_{1},E_{2},E_{3}$ are as below.

[TABLE]

by Lemma 7.2. Next $E_{2}=E_{2,1}+E_{2,2}$ with

[TABLE]

The sum on the right is, using Lemma 7.2,

[TABLE]

Thus

[TABLE]

We prove in the same way that

[TABLE]

Furthermore,

[TABLE]

Putting together these bounds we obtain

[TABLE]

**Upper bound for $I_{1}$

**Using the fact that $(\mathcal{R},\mathcal{N})$ is good, the residue theorem and Lemma 5.1, we get

[TABLE]

**Approximate formula for $I_{-1}$

**This is almost the same calculation but now we pick up a residue at $c$ because of the pole of $\zeta(s/c)$. We have

[TABLE]

Finally, we get

[TABLE]

and

[TABLE]

Choosing $T=x$ concludes the proof.

a $\Box$

**Remark on the use of Perron’s effective formula for Beurling primes

**If we had $M_{c}(x)=[x]$, the counting function of the usual integers, then we would have had

[TABLE]

instead of the upper bound (7.1). But in the case of a general Beurling prime system, we do not always have

[TABLE]

It is the reason why, in order to obtain an estimate for for generalized integers in short intervals, we needed first to compute the asymptotic development of the counting function $M_{c}(x)$ of Lemma 7.2.

**8. Proofs of Theorems 1 and 2

Comments**  The two proofs have similar structure. The theorems will follow easily from the two fundamental formulas (see (8.15) and (8.7)),

[TABLE]

for Theorem 1, and

[TABLE]

for Theorem 2, where in both cases $(\mathcal{R},\mathcal{N})$ is a good prime system and $H(s)$ an absolutely convergent product.

The proof of Theorem 1 is longer, mainly because of the presence of the pole at $s=b$ in (8.1) that deserves a special treatment. It is why we begin with the proof of Theorem 2.

Proof of Theorem 2.  Let $a$ and $b$ be two real numbers satisfying (1.14) and let $(\mathcal{R},\mathcal{N})$ be a good g-prime system. Thanks to Zhang, we know such a system exists. Let $Q:=Q_{a,b}=\mathcal{R}^{1/a}\cup\mathcal{R}^{1/b}$, choose $\varepsilon$ such that $0<\varepsilon<\frac{a}{4}$, and define

[TABLE]

By using Lemma 6.2, (1.14), (8.2) and Lemma 6.1, we have, for $x$ large enough

[TABLE]

Moreover, as $\mathcal{R}$ is good, we also have $Q(x)\ll x^{a}$ and

[TABLE]

Put $\sigma(a)=\max\{\frac{a}{2},a-\frac{19}{40}\}$. By Lemma 6.3, there exists a set $\mathcal{P}:=\mathcal{P}_{a,b}$ of ordinary primes, and a bijection $p:\mathcal{Q}\to\mathcal{P}$ such that

[TABLE]

is an absolutely convergent product for $\sigma>\sigma(a)+2\varepsilon$.

As $\varepsilon>0$ can be chosen as small as we please, it follows that

[TABLE]

where $H(s)$ is again an absolutely convergent product for $\sigma>\sigma(a)$.

We denote by $\mathcal{M}_{a,b}$ the multiset of integers associated to the multiset of primes $\mathcal{Q}=\mathcal{Q}_{a,b}$. With the notation of section 7, we have for any $c\in(\frac{1}{2},1)$,

[TABLE]

Let $c=b/a$. For $\sigma>a$, we have the following formula where the generalized Euler products and the generalized Dirichlet series are normally convergent for $\sigma\geq a+\varepsilon$ for any fixed $\varepsilon>0$.

[TABLE]

Let $A(x)=\sum_{m\leq x}\mu_{\mathcal{Q}_{c}}(m)$ (where $m\in\mathcal{M}_{c}$). For $\sigma>\frac{3c}{2(1+c)}$, we have by Abel summation

[TABLE]

by Lemma 7.3(i). It follows that

[TABLE]

We have, normally for $\sigma\geq a+\varepsilon$ for all fixed $\varepsilon>0$

[TABLE]

Combining (8.6), (8.3), (8.4) and (8.5) gives

[TABLE]

and this proves that the abscissa of absolute convergence of the above series is $a$. By (1.14), we have $\max\{\sigma(a),\frac{3ab}{2(a+b)}\}=\max\{\frac{3ab}{2(a+b)},a-\frac{19}{40}\}<b<a$ and the result follows.

a $\Box$

Proof of Theorem 1.  Let $a$ and $b$ be two real numbers satisfying (1.12) and let $(\mathcal{R},\mathcal{N})$ be a good g-prime system. Thanks to Zhang, we know such a system exists. Choose $\varepsilon$ such that $0<\varepsilon<\frac{a}{4}$, and define

[TABLE]

By using Lemma 6.2, (1.12) and (8.8), we have, for $x$ large enough

[TABLE]

Moreover, as $(\mathcal{R},\mathcal{N})$ is good, we also have $R^{1/b}(x)\ll x^{b}$ and

[TABLE]

By applying Lemma 6.3, there exists an injection $I:\mathcal{R}^{1/b}\to\mathcal{R}^{1/a}$ such that

[TABLE]

is an absolutely convergent product for $\sigma>\max\{b-\frac{a}{2}+2\varepsilon,0\}$.

Let $\mathcal{Q}=\mathcal{R}^{1/a}\setminus{\rm Im}(I)$ and define a new $h$ by

[TABLE]

By using Lemma 6.2, (1.12), (8.10) and finally Lemma 6.1, we have, for $x$ large enough

[TABLE]

Moreover, as $\mathcal{R}$ is good, we also have $Q(x)\ll x^{a}$ and

[TABLE]

Recall the notation $\sigma(a)=\max\{\frac{a}{2},a-\frac{19}{40}\}$. By Lemma 6.3, there exists a set $\mathcal{P}=\mathcal{P}_{a,b}$ of ordinary primes and a bijection

[TABLE]

such that the function

[TABLE]

is an absolutely convergent product for $\sigma>\sigma(a)+2\varepsilon$. We have $\sigma(a)\geq a/2>b-a/2$ by (1.12). By (8.9), it follows that

[TABLE]

where $H^{*}(s)$ is an absolutely convergent product for $\sigma>\sigma(a)+2\varepsilon$. As $\varepsilon>0$ can be chosen as small as we please, it follows that

[TABLE]

where

[TABLE]

is again an absolutely convergent product for $\sigma>\sigma(a)$.

Let $c=b/a$. We have the following formula where both sides converge normally for $\sigma\geq a+\varepsilon$ for any fixed $\varepsilon>0$.

[TABLE]

Let

[TABLE]

By Abel summation, for $\sigma>c$,

[TABLE]

by Lemma 7.3(ii). It follows that the abscissa of convergence of the series in (8.12) is at most $b$. As $b$ is a pole, the abscissa is indeed $b$. We have, normally for $\sigma\geq a+\varepsilon$ for all fixed $\varepsilon>0$

[TABLE]

Combining (8.14), (8.11), (8.12) and (8.13) we get

[TABLE]

and this proves that the abscissa of absolute convergence of the above series is $a$.

But by (1.12), we have $b>\sigma(a)$. Thus (8.15) is actually true for $\sigma>b$. It follows that $s=a$ is the only zero in $\mathbb{C}_{b}$ and this zero is simple. As $b$ is a pole it follows that the abscissa of convergence is $b$.

Moreover, for $t\in\mathbb{R}$, we have

[TABLE]

By Abel’s Theorem, it follows that if the series in (8.15) converges at $b+it$, the sum cannot be 0. Thus we have $Z_{\mathcal{P}}=\{a\}$, which concludes the proof of Theorem 1.

a $\Box$

**9. Open questions related to GRH-RH

**Let us recall that one of the classical statements equivalent to the Generalized Riemann Hypothesis (GRH) is the following: *for every Dirichlet character $\chi$, the meromorphic function $L_{\chi}(s)$ does not vanish in $\mathbb{C}_{\frac{1}{2}}$. *

The Dirichlet series defining $\zeta(s)$ and more generally $L_{\chi}(s)$ with $\chi$ a principal Dirichlet character are not convergent in the critical strip. As only the zeros of Dirichlet series themselves (and not of their meromorphic continuation) are studied here, it leads us to introduce GRH$\setminus$RH: *for every non-principal Dirichlet character $\chi$, the Dirichlet series $L_{\chi}(s)$ does not vanish in $\mathbb{C}_{\frac{1}{2}}$.

Let us recall the sets $V$ and $W$ of section 1(f). Theorem 3 says $V\supset[0,\frac{1}{2}]$ and, under RH, $W\supset[0,\frac{1}{2}]$. We wonder if these inclusions are equalities. Write

[TABLE]

and

[TABLE]

We shall see that either of these implies GRH$\setminus$RH.

Let us also recall the two statements mentioned in [10] which imply GRH$\setminus$RH.

[TABLE]

Now, let $\sigma_{n}(f)$ and $\sigma_{p}(f)$ denote the abscissa of convergence of $\sum_{n=1}^{\infty}\frac{f(n)}{n^{s}}$ and $\sum_{p}\frac{f(p)}{p^{s}}$ respectively. Write

[TABLE]

These five statements are related in the following way:

[TABLE]

Proof.  $A\implies B_{c}$

Let us suppose $B_{c}$ is false. Then there exists a completely multiplicative function $f(n)$ and a zero $\beta+i\gamma$ of $\sum_{n=1}^{\infty}\frac{f(n)}{n^{s}}$ (with abscissa of convergence $\sigma_{c}$) such that $\beta-\sigma_{c}>\frac{1}{2}$. By writing $f_{1}(n)=f(n)n^{-\sigma_{c}-i\gamma}$ we have that

[TABLE]

has abscissa of convergence zero, and the series vanishes at $s=\beta>\frac{1}{2}$. Defining $A(x)=\sum_{n\leq x}f_{1}(n)$, we have $A(x)\ll x^{\varepsilon}$ for all $\varepsilon>0$ and by Abel summation that

[TABLE]

As $\sum_{n=1}^{\infty}\frac{f_{1}(n)}{n^{\beta}}=0$, we also have

[TABLE]

Thus $g(n):=f_{1}(n)n^{-\beta}$ is completely multiplicative but does not satisfy $A$.

$B_{c}\implies$ GRH$\setminus$RH

Suppose GRH$\setminus$RH is false. Then there exists a non-principal character $\chi$ and a zero $\beta+i\gamma$ of $L_{\chi}(s)$ with $\beta>\frac{1}{2}$. As $L_{\chi}(s)$ has $\sigma_{c}=0$, it follows that $\beta\in V$ and $B_{c}$ is false.

$B_{a}\implies$ GRH$\setminus$RH

The proof is similar to the above. Suppose GRH$\setminus$RH is false. Then there exists a non-principal character $\chi$ and a zero $\beta+i\gamma$ of $L_{\chi}(s)$ with $\beta<\frac{1}{2}$. As $L_{\chi}(s)$ has $\sigma_{a}=1$, it follows that $1-\beta\in W$ and, since $1-\beta>\frac{1}{2}$, $B_{a}$ is false.

$C\implies$ GRH$\setminus$RH

Suppose $C$ is true. Let $\chi$ be a non-principal character. Note that $\sigma_{n}(\chi)=0$. By $C$, we have

[TABLE]

But this is an equivalent form of RH for $L_{\chi}(s)$; i.e. GRH$\setminus$RH follows.

a $\Box$

Question  Among the four statements $A,B_{c},B_{a}$ and $C$, which are true and which are false?

**Acknowledgements

**We had stimulating discussions on the mathematics in and around this paper with Kristian Seip. We thank him for them.