An equivalent to the Riemann hypothesis in the Selberg class

Abstract

In 2020 S. M. Gonek, S. W. Graham and Y. Lee formulated the Lindelöf hypothesis for prime numbers and proved that it is equivalent to the Riemann Hypothesis. In this note we show that their result holds in the Selberg class of L-functions.

1 Introduction

As usual we denote $s = \sigma + it$, a non-trivial zero of F(s) by $\rho = \beta + i\gamma$ and by p a prime number.

A function F(s) belongs to the Selberg class S if it satisfies the following properties:

• (1) For $\sigma > 1,$ F(s) is an absolutely convergent Dirichlet series

  $$\begin{aligned} F(s) = \sum _{n=1}^\infty \frac{a_n}{n^s}. \end{aligned}$$

• (2) For some integer $m \ge 0,$ $(s-1)^m F(s)$ is an entire function of finite order.

• (3) F(s) satisfies a functional equation of the form

  $$\begin{aligned} \Phi (s) = \omega \overline{\Phi (1 - {\overline{s}})} \end{aligned}$$

  where

  $$\begin{aligned} \Phi (s) = Q^s \prod _{j=1}^r \Gamma (\lambda _j s + \mu _j) F(s), \end{aligned}$$

  with $Q > 0$, $\lambda _j > 0$, $\Re \mu _j \ge 0$ and $|\omega | = 1.$

• (4) (Ramanujan hypothesis) For every $\varepsilon > 0$, $a(n) \ll n^\varepsilon .$

• (5) (Euler product) For $\sigma$ sufficiently large,

  $$\begin{aligned} \log F(s) = \sum _{n=1}^\infty \frac{b_n}{n^s} \end{aligned}$$

  where $b_n = 0$ unless $n = p^k$ for $k \in \mathbb N$ and $b_n \ll n^\theta$ for some $\theta < 1/2.$

The data $Q, \lambda _j, \mu _j$ and $\omega$ does not determine F(s) uniquely, however $d_F = 2 \sum _{j=1}^r \lambda _j$ is an invariant called the degree of F(s). Let $m_F$ be the order of the pole of F(s) at $s=1$.

The zeros of F(s) that come from the poles of the Gamma function in the functional equation are called trivial. We say that F(s) satisfies the Riemann Hypothesis (RH) if all of its non-trivial zeros $\rho =\beta +i\gamma$ have $\beta = 1/2$. More about the Selberg class see Kaczorowski and Perelli [3].

Gonek et al.  [4] proved that the Riemann Hypothesis is equivalent to the following relation

$$\begin{aligned} \sum _{p \le x} p^{-it} = \int _2^x \frac{u^{-it}}{\log u} du + O(x^{1/2}|t|^\varepsilon ), \end{aligned}$$

for all $\varepsilon , B > 0$ and $2 \le x \le |t|^B.$ For further development see Banks [1]. In [2, Corollary 6] a similar equivalent was considered in the case of the Lindelöf hypothesis for the Lerch zeta-function.

In this short note we show that Gonek’s, Graham’s and Lee’s result holds for all functions from S.

Theorem 1

Let $F(s) \in S$ and $d_F\ge 1$. Then F(s) satisfies RH if and only if

$$\begin{aligned} \sum _{n \le x} b_n n^{-it} = m_F \int _2^x \frac{u^{-it}}{\log u} du + O(x^{1/2}|t|^\varepsilon ), \end{aligned}$$

(1)

for all $\varepsilon , B > 0$ and $2 \le x \le |t|^B.$

2 Lemmas and proof of Theorem 1

Let $F(s) \in S$ and denote $\Lambda _F(n) = b_n \log n,$ then

$$\begin{aligned} -\frac{F'}{F}(s) = \sum _{n=1} \frac{\Lambda _F(n)}{n^s}. \end{aligned}$$

Lemma 2

Let $F(s) \in S,$ $\varepsilon > 0$ and let $\mu _F: \mathbb R\rightarrow \mathbb R$ be such that

$$\begin{aligned} F(\sigma + it) \ll |t|^{\mu _F(\sigma ) + \varepsilon }. \end{aligned}$$

Then

$$\begin{aligned} \mu _F(\sigma ) = {\left\{ \begin{array}{ll} 0, \text { if } \sigma > 1, \\ (1/2)d_F(1-\sigma ), \text { if }\,0 \le \sigma \le 1, \\ (1/2 - \sigma )d_F, \text { if }\, 0 < \sigma . \end{array}\right. } \end{aligned}$$

Proof

See Steuding [6, Theorem 6.8]. $\square$

Lemma 3

Let $T>0$ and suppose that $x>0$ is half an odd integer. Then,

$$\begin{aligned} \sum _{n \le x} \Lambda _F(n) n^{-it} = \frac{1}{2 \pi i} \int _{2-iT}^{2+iT} -\frac{F'}{F}(w+it) \frac{x^w}{w} dw + O \left( \frac{x^{2}}{T} + 1\right) . \end{aligned}$$

Proof

Lemma can be proved by using the argumentation presented in the proof of Lemma 3.12 of Titchmarsh [7] by fixing $c=2$ and noticing that

$$\begin{aligned} \sum _{n=1}^\infty \frac{|\Lambda _F(n)|}{n^2} \le \sum _{n=1}^\infty \frac{\log n}{n^{3/2}} \end{aligned}$$

converges. $\square$

Lemma 4

Let $0 \le \delta < 1/4$ be such that $F(s) \in S$ has no trivial zeros with $\sigma = -1+\delta$. Then

$$\begin{aligned} \int _{-1 + \delta - iT}^{-1 + \delta + iT} -\frac{F'}{F}(w)\frac{x^w}{w}dw = O\left( \frac{\log ^2 T}{x^{1-\delta }} \right) \end{aligned}$$

for any $T>0$

Proof

By Hadamard theory we have (see Smajlović [5, proof of Lemma 5.1])

$$\begin{aligned} \frac{F'}{F}(s) = \sum _{|\gamma - T| < 1} \frac{1}{s - \rho } + O(\log T). \end{aligned}$$

(2)

For non-trivial zeros we have $0\le \beta \le 1$ and there are $O(\log T)$ zeros of F(s) with $|\gamma - T| < 1$ (see [3]). Thus,

$$\begin{aligned} \frac{F'}{F}(s) = O(\log t), \end{aligned}$$

when $\sigma = -1 + \delta$. Then,

$$\begin{aligned} \int _{-1 + \delta -iT}^{-1 + \delta +iT} -\frac{F'}{F}(w) \frac{x^w}{w} dw \ll x^{-1 + \delta } \int _{-1 + \delta -iT}^{-1 + \delta +iT} \frac{\log w}{w} dw \ll \frac{\log ^2 (T)}{x^{1 - \delta } } \end{aligned}$$

$\square$

Lemma 5

Let $0 \le \delta < 1/4$ be such that $F(s) \in S$ has no trivial zeros with $\sigma = -1+\delta$. Then

$$\begin{aligned} \int _{-1 + \delta + iT}^{2 + iT} -\frac{F'}{F}(w)\frac{x^w}{w}dw = O\left( \frac{x^2 \log ^2 T}{T} \right) \end{aligned}$$

for any $T>0$ such that it is not an ordinate of a non-trivial zero of F(s).

Proof

Moving by a finite distance we can pick T such that $|T - \gamma | \gg 1 / \log T.$ Then with such a choice of T using (2) we obtain

$$\begin{aligned} \int _{-1 + \delta + iT}^{2 + iT} -\frac{F'}{F}(w)\frac{x^w}{w}dw = O\left( \frac{x^2 \log ^2 T}{ T} \right) . \end{aligned}$$

Moving the line of integration by a bounded amount we may cross at most $O(\log T)$ zeros F(s) (counting with multiplicities) and they will contribute residues of total size at most $O\left( x^2 \log ^2 T/T \right) .$ Hence, noting Lemma 4, we obtain

$$\begin{aligned} \int _{-1 + \delta + iT}^{2 + iT} -\frac{F'}{F}(w)\frac{x^w}{w}dw = O\left( \frac{x^2 \log ^2 T}{ T} \right) \end{aligned}$$

for any $T>0$ such that it is not an ordinate of a non-trivial zero of F(s). $\square$

Proof of Theorem 1

The proof is pretty much the same as the one in Gonek’s, Graham’s and Lee’s paper with additional consideration given for greater generality.

By Abel’s summation formula we see that (1) is equivalent to

$$\begin{aligned} \sum _{n \le x} \Lambda _F(n) n^{-it} = m_F \frac{x^{1-it}}{1-it} + O(x^{1/2}|t|^\varepsilon ) \end{aligned}$$

(3)

for all $\varepsilon , B > 0$ and $2 \le x \le |t|^B.$

Thus, it is enough to prove that RH for F(s) is equivalent to (3)

Suppose F(s) satisfies RH. Let $x \ge 5/2$ be half an odd integer and $T = |t|^C,$ where $C>1$ will be chosen later. By Lemma 3 we have

$$\begin{aligned} \sum _{n \le x} \Lambda _F(n) n^{-it} = \frac{1}{2 \pi i} \int _{2-iT}^{2+iT} -\frac{F'}{F}(w+it) \frac{x^w}{w} dw + O \left( \frac{x^{2}}{T} + 1\right) . \end{aligned}$$

(4)

Choose $0 \le \delta < 1/4$ such that F(s) would have no trivial zeros with $\sigma = -1 + \delta .$ Replacing the line of integration in (4) by one consisting of the three leftmost sides of the rectangle with vertices $2-iT,$ $-1 + \delta -iT,$ $-1 + \delta +iT$ and $2+iT$ and using Lemmas 4 and 5 we see that

$$\begin{aligned} \begin{aligned} \sum _{n \le x} \Lambda _F(n) n^{-it}&= m_F \frac{x^{1-it}}{1-it} - \sum _{|\gamma - t| < T} \frac{x^{\rho - it}}{\rho - it} \\&\quad + O \left( \frac{x^{2} }{T} + \frac{x^2 \log ^2 (|t| + T)}{T} + \frac{\log ^2(|t| + T)}{x^{1-\delta }} + 1\right) . \end{aligned} \end{aligned}$$

Note that the sum is over the non-trivial zeros of F(s). It might happen that we pass over trivial zeros of F(s),  however there are only finitely many of them in $\sigma > -1 + \delta$ and for each of them we have $\beta \le 0,$ thus they contribute a term of size O(1).

By RH, using Abel’s summation formula, we obtain

$$\begin{aligned} \sum _{|\gamma - t|< T} \frac{x^{\rho - it}}{\rho - it} \ll x^{1/2} \sum _{ \genfrac{}{}{0.0pt}2{|\gamma - t| < T}{\beta = 1/2} } \frac{1}{1 + |t-\gamma |} \ll x^{1/2}\log ^2(|t| + T). \end{aligned}$$

Now, suppose that $5/2 \le x \le |t|^B$ and choose $C > \max (1, 3B/2).$ Then $T = |t|^C \ge \max (|t|, x^{3/2})$ and we obtain

$$\begin{aligned} \sum _{n \le x} \Lambda _F(n) n^{-it} = m_F \frac{x^{1-it}}{1-it} + O(x^{1/2}|t|^\varepsilon ). \end{aligned}$$

(5)

We have assumed until now that $x \ge 5/2$ is half an odd integer. If we relax this condition and just assume that $x \ge 2,$ then such x is always within O(1) of half an odd integer. Changing x by this amount in (5) changes the left-hand side by no more than $O(x^{1/2} \log x)$ and the right-hand side by at most $O(|t|^\varepsilon ).$ Since $x^{1/2} \log x \ll x^{1/2}|t|^\varepsilon ,$ (5) holds for $2 \le x \le |t|^B.$

Next we prove that (3) implies RH for F(s). Write

$$\begin{aligned} \psi (x, t) = \sum _{n \le x} \Lambda _F(n) n^{-it} \end{aligned}$$

and

$$\begin{aligned} R(x, t) = \psi (x, t) - m_F \frac{x^{1-it}}{1-it}. \end{aligned}$$

Then by our assumption

$$\begin{aligned} R(x, t) \ll x^{1/2}|t|^\varepsilon \end{aligned}$$

(6)

for $2 \le x \le |t|^B$, where $\varepsilon > 0$ and B is arbitrarily large but fixed.

First we show that for all $s \ne 1$

$$\begin{aligned} \int _1^\infty \frac{R(x, t)}{x^s} dx = - \left( \frac{1}{s-1} \frac{F'}{F}(s+it-1) + \frac{m_F}{(1-it)(s+it-2)} \right) . \end{aligned}$$

(7)

Suppose that $\sigma > 2.$ Then we see that

$$\begin{aligned} \int _1^\infty \frac{\psi (x, t)}{x^s} dx = \sum _{n=2}^\infty \frac{\Lambda _F(n)}{n^{it}} \int _n^\infty x^{-s} dx = - \frac{1}{s-1} \frac{F'}{F} (s+it-1). \end{aligned}$$

Integrating the other term and combining we get (7) for $\sigma > 2,$ the right hand side of which defines a meromorphic continuation of the left hand side which has a simple pole at $s=1$.

Define

$$\begin{aligned} \begin{aligned}&H(s) = \int _1^\infty \frac{R(x, t)}{x^s} dx = \\&\qquad - \left( \frac{(1-it)(s+it-2)F'(s+it-1)+m_F(s-1)F(s+it-1)}{(s-1)(1-it)(s+it-2)F(s+it-1)} \right) . \end{aligned} \end{aligned}$$

Assume, by way of contradiction, that $\rho _0=\beta _0+i\gamma _0$ is a zero of F(s) with $\beta _0 > 1/2.$ Let m be the multiplicity of $\rho _0$, and define

$$\begin{aligned} h(s) = \frac{(s+it-2)F(s+it-1)}{(s+it-\rho _0-1)^m(s+it+1)^{4d_F}}. \end{aligned}$$

For real u,  define

$$\begin{aligned} w(u) = \frac{1}{2\pi i} \int _{3-i\infty }^{3+i\infty } h(s)e^{us} ds \end{aligned}$$

and consider the integral

$$\begin{aligned} \frac{1}{2\pi i} \int _{3-i\infty }^{3+i\infty } h(s)H(s) e^{s \log x} ds = \int _{1}^{\infty } R(y, t) w(\log x - \log y) dy. \end{aligned}$$

(8)

We move the line of integration in the integral of the left hand side to left to $\sigma = 5/4$ and pass two poles at $s=2-it$ and $s=\rho _0+1-it$. The residue at $s=2-it$ is equal to 0 and the other residue is

$$\begin{aligned} -x^{\rho _0+1-it} \frac{(\rho _0 - 1)F^{(m)}(\rho _0)}{(m-1)!(\rho _0-it)(\rho _0+2)^{4d_F}}. \end{aligned}$$

Using the bounds $F(1/4 + iv) \ll (1 + |v|^{1/2d_F})$ and $F'(1/4 + iv) \ll (1+|v|^{1/2d_F})\log (2+|v|),$ the left hand side is

$$\begin{aligned} \begin{aligned}&=x^{\rho _0+1-it} \frac{(\rho _0 - 1)F^{(m)}(\rho _0)}{(m-1)!(\rho _0-it)(\rho _0+2)^{4d_F}} \\&\quad + O\left( x^{5/4} \int _{-\infty }^{\infty } \frac{(1+|t+v|^{1/2d_F})\log (2+|v|)}{(1+|v+t-\gamma _0|)^m(1+|v+t|)^{4d_F})} dv \right) \\&=x^{\rho _0+1-it} \frac{(\rho _0 - 1)F^{(m)}(\rho _0)}{(m-1)!(\rho _0-it)(\rho _0+2)^{4d_F}} + O\left( x^{5/4} \right) . \end{aligned} \end{aligned}$$

Next we estimate w(u). If $u \le 0$ we pull the contour right to $\infty$. Since

$$\begin{aligned} h(s)e^{us} \ll \frac{e^{u\sigma }}{|s+it-\rho _0|^m|s+it|^{4d_F-1}} \end{aligned}$$

for $\sigma \ge 3,$ we see that $w(u) = 0$. If $u>0$, we pull the contour left to $-5/4$. We pass a pole of h(s) at $s=-1-it$ of order $4d_F$ which contributes a residue of size O(1). The integral on the new line is

$$\begin{aligned} \begin{aligned} \int _{-5/4-i\infty }^{-5/4+i\infty } h(s)e^{us}ds \ll \int ^{\infty }_{-\infty } e^{-5/4u} \frac{(1+|v+t|)||F(-9/4+i(v+t))|}{|1+(v+t-\gamma _0)|^m|1+(v+t)|^{4d_F}} dv \\ \ll \int ^{\infty }_{-\infty } e^{-5/4u} \frac{(1+|v+t|)|(1+|v+t|^{23/8d_F})}{|1+(v+t-\gamma _0)|^m|1+(v+t)|^{4d_F}} dv \ll 1. \end{aligned} \end{aligned}$$

Thus,

$$\begin{aligned} w(u) = {\left\{ \begin{array}{ll} 0 \ \text {if} \ u \le 0,\\ O(1) \ \text {if} \ u > 0. \end{array}\right. } \end{aligned}$$

Collecting the estimates in the previous discussion and applying them to (8) we see that for $\rho _0$ fixed

$$\begin{aligned} x^{\beta _0 + 1} \ll _{\rho _0} \int _1^x |R(y, t)| dy + x^{5/4}. \end{aligned}$$

Then, by assumption, setting $B=2/(\beta _0 - 1/2)$ we get

$$\begin{aligned} x^{\beta _0+1} \ll _{\rho _0} x^{3/2}|t|^\varepsilon , \end{aligned}$$

for $2 \le x \le |t|^{2/(\beta _0-1/2)}$. In other words,

$$\begin{aligned} x \ll _{\rho _0} |t|^{(1+\varepsilon )/(\beta _0 - 1/2)}. \end{aligned}$$

This contradiction implies that $\beta _0 = 1/2.$ This completes the proof of the theorem. $\square$