Joint Value-Distribution of Shifts of the Riemann Zeta-Function

Abstract

We prove that any complex values $z_1,\ldots ,z_n$ can be approximated by the following integral shifts of the Riemann zeta-function $\zeta (s+id_1\tau ),\ldots ,\zeta (s+id_n\tau )$ for infinitely many $\tau$, provided $d_1,\ldots ,d_n\in {\mathbb {Q}}$ are distinct and positive, and s is a fixed complex number lying in the right open half of the critical strip.

1 Introduction

In the 80’s of the last century Bagchi (see [1, 2]) observed that the classical Riemann Hypothesis is equivalent to the fact that for every $\varepsilon >0$ and every compact set $K\subset D:=\{s\in {\mathbb {C}}:1/2<{\text {Re}}(s)<1\}$ with connected complement we have

$$\begin{aligned} \liminf _{T\rightarrow \infty }\frac{1}{T}{\text {meas}}\left\{ \tau \in [0,T]:\max _{s\in K}|\zeta (s+i\tau ) - \zeta (s)|<\varepsilon \right\} >0, \end{aligned}$$

(1)

where ${\text {meas}}$ denotes the Lebesgue measure on ${\mathbb {R}}$. In other words (1) means that the set of real $\tau$ satisfying $\max _{s\in K}|\zeta (s+i\tau ) - \zeta (s)|<\varepsilon$ has a positive lower density. Moreover, let us note that in the language of topological dynamics Inequality (1) says that the Riemann Hypothesis is equivalent to the strong recurrence of the Riemann zeta-function (see [9]).

One implication in Bagchi’s observation is an immediate consequence of the so-called universality theorem due to Voronin [17], which is a generalization of the work of Bohr and his collaborators (see [3,4,5]) on denseness theorems in ${\mathbb {C}}$ of values of the Riemann zeta-function and states that for any non-vanishing and continuous function f(s) on K, analytic in the interior of K, and every $\varepsilon >0$, we have

$$\begin{aligned} \liminf _{T\rightarrow \infty }\frac{1}{T}{\text {meas}}\left\{ \tau \in [0,T]:\max _{s\in K}|\zeta (s+i\tau ) - f(s)|<\varepsilon \right\} >0. \end{aligned}$$

The converse implication is a combination of Rouché’s theorem and the well-known zero density theorem for the Riemann zeta-function.

An interesting generalization of (1) was suggested by Nakamura in [11], where he introduced the problem of finding all $d_1,d_2,\ldots ,d_n$ such that for every compact set $K\subset D$ with connected complement and every $\varepsilon >0$ we have

$$\begin{aligned} \liminf _{T\rightarrow \infty }\frac{1}{T}{\text {meas}}\left\{ \!\tau \in [0,T]:\max _{1\le k\ne \ell \le n}\max _{s\in K}|\zeta (s+id_k\tau ) - \zeta (s+id_\ell \tau )|<\varepsilon \!\right\} >0.\nonumber \!\!\!\!\\ \end{aligned}$$

(2)

Notice that (2) holds for $n=2$ and $d_1=0$, $d_2\ne 0$ if and only if the Riemann Hypothesis is true. Interestingly it turns out (see [11,12,13,14]) that for $n=2$ the case $d_1=0$, $d_2\ne 0$ is the most difficult one as it was shown that (2) holds for $n=2$ and every non-zero real $d_1,d_2$. A common strategy of showing (2) is to prove a joint version of universality for $\zeta (s+id_1\tau ),\ldots ,\zeta (s+id_n\tau )$, namely to prove that for any non-vanishing and continuous functions $f_1(s),\ldots ,f_n(s)$ on K, analytic in the interior of K, and every $\varepsilon >0$ we have

$$\begin{aligned} \liminf _{T\rightarrow \infty }\frac{1}{T}{\text {meas}}\left\{ \tau \in [0,T]:\max _{1\le k\le n}\max _{s\in K}|\zeta (s+id_k\tau ) - f_k(s)|<\varepsilon \right\} >0. \end{aligned}$$

(3)

Obviously, by taking $f_1=f_2=\cdots =f_n$, one can easily see that (3) implies (2). The problem of finding all parameters $d_1,d_2$ satisfying (3) for $n=2$ was also completely solved in the aforementioned series of papers by Nakamura and the author, where it was proved that (3) holds for $n=2$ and any real $d_1,d_2$ with $0\ne d_1,d_2$ and $d_1\ne \pm d_2$. Note that it is indeed a complete answer for $n=2$ as one cannot expect that (3) holds for $d_1=\pm d_2$.

Unfortunately, the situation for $n\ge 3$ is much more complicated and so far we know only (see [11]) that (3) holds if $d_1=1,d_2,\ldots ,d_n$ are algebraic real numbers linearly independent over ${\mathbb {Q}}$, which is an easier case since for such $d_k$’s, in view of Baker’s theorem in transcendental number theory, the set $I(d_1,\ldots ,d_n):=\{d_k\log p: p\text { is prime, }1\le k\le n\}$ is linearly independent over ${\mathbb {Q}}$ and one can follow straightforwardly the proof of Voronin’s theorem in order to get (3). However, if the set $I(d_1,\ldots ,d_n)$ is not linearly independent over ${\mathbb {Q}}$, we need a new approach or at least a significant modification of known proofs of universality, since linear independence of the set $I(d_1,\ldots ,d_n)$ is crucial in several steps of these proofs.

In the paper we focus on considering the following weaker version of (3) in the flavor of denseness theorems due to Bohr, where K is a singleton and $d_1,\ldots ,d_n$ are positive integers. Note that in this case the set $I(d_1,\ldots ,d_n)$ is far from being linearly independent over the field of rational numbers.

Theorem 1.1

Let $d_1<d_2<\cdots <d_n\in {\mathbb {N}}$ and $s=\sigma +it$ with $\sigma \in (1/2,1]$ and $t>0$. Then, for every $z_1,z_2,\ldots ,z_n\in {\mathbb {C}}$ and every $\varepsilon >0$ the set of all positive real $\tau$ satisfying

$$\begin{aligned} \max _{1\le k\le n}|\log \zeta (s+id_k\tau )-z_k|<\varepsilon \end{aligned}$$

has a positive lower density.

Corollary 1.2

Let $0<d_1<d_2<\cdots <d_n\in {\mathbb {Q}}$ and $s=\sigma +it$ with $\sigma \in (1/2,1]$ and $t>0$. Then, for every $z_1,z_2,\ldots ,z_n\in {\mathbb {C}}$ and every $\varepsilon >0$ the set of all positive real $\tau$ satisfying

$$\begin{aligned} \max _{1\le k\le n}|\log \zeta (s+id_k\tau )-z_k|<\varepsilon \end{aligned}$$

has a positive lower density.

Proof

Assume that $d_j=a_j/b_j$ with $a_j,b_j\in {\mathbb {N}}$. Then it suffices to apply Theorem 1.1 with $La_j/b_j\in {\mathbb {N}}$ instead of $d_j$, where $L={{\,\mathrm{lcm}\,}}(b_1,\ldots ,b_n)$, and next make a substitution $L\tau \mapsto \tau$. $\square$

Obviously, the above corollary easily implies that (2) holds for $K=\{s\}$ and arbitrary distinct positive rational numbers $d_1,\ldots ,d_n$.

Moreover, it is worth mentioning that the proof of Theorem 1.1 presented in the paper might be simplified in several places under the assumption that $d_1,\ldots ,d_n$ are integers. However, we decided to make all steps in the proof as general as possible, since it might be interesting for the reader to see that in fact the proof of the above theorem is valid for arbitrary positive real $d_1,\ldots ,d_n$ except only one step where we use the implication

$$\begin{aligned} -\tau \frac{\log p}{2\pi }\text { is close to }\theta _p \text { mod } 1\Longrightarrow -\tau d_k\frac{\log p}{2\pi }\text { is close to }d_k\theta _p\text { mod } 1, \end{aligned}$$

which is trivially true for $d_k\in {\mathbb {Z}}$, but it does not necessary hold if $d_k\notin {\mathbb {Z}}$.

2 Denseness Lemma

In this section we prove the following denseness lemma, which holds for all positive real $d_1,\ldots ,d_n$.

Lemma 2.1

Let $d_1,d_2,\ldots ,d_n$ be real numbers satisfying $0<d_1<d_2<\cdots <d_n$ and $s=\sigma +it$ with $\sigma \in (1/2,1]$ and $t>0$. Then, for every $z_1,z_2,\ldots ,z_n\in {\mathbb {C}}$ and every $\varepsilon >0$, $y>0$, there exist a finite set of prime numbers $M\supset \{p: p<y\}$ and a sequence ${\overline{\theta }} = (\theta _p)_{p\in M}$ such that

$$\begin{aligned} \max _{1\le k\le n}\left| \sum _{p\in M}\frac{e(d_k\theta _p)}{p^s} - z_k\right| <\varepsilon , \end{aligned}$$

where as usual $e(t)=e^{2\pi i t}$.

The proof of the lemma essentially based on the following Pechersky’s [15] generalization of the Riemann rearrangement lemma to Hilbert spaces.

Lemma 2.2

( [15]) Let H be a real Hilbert space with the inner product $(\cdot |\cdot )$ and a sequence $(u_n)_{n\in {\mathbb {N}}}\subset H$ be such that $\sum _{n=1}^\infty \Vert u_n\Vert _H^2<\infty$. Moreover, suppose that for every $e\in H$ with $\Vert e\Vert =1$ the series $\sum _{n=1}^\infty \langle u_n,e\rangle$ are conditionally convergent after suitable permutation of terms. Then, for every $v\in H$ there exists a permutation $(n_k)$ such that $\sum _{k=1}^\infty u_{n_k} = v$.

Moreover, we use the following effective version of Kronecker approximation theorem due to Chen. Henceforth, $\Vert \cdot \Vert$ will denote the distance to the nearest integer.

Lemma 2.3

(Chen, [6, Theorem 1(ii)]) Let $\lambda _1, \dots , \lambda _n$ and $\alpha _1, \dots , \alpha _n$ be real numbers and assume the following property: for all integers $u_1, \dots , u_n$ with $|u_j|\le M$, the assertion

$$\begin{aligned} u_1\lambda _1+ \cdots + u_n\lambda _n \in {\mathbb {Z}} \end{aligned}$$

implies that $u_1\alpha _1+ \cdots + u_n\alpha _n$ is an integer. Then for all positive real numbers $\delta _1, \dots , \delta _n$ and for any integers $T_1<T_2$, we have

$$\begin{aligned} \inf _{t \in [T_1, T_2]\cap {\mathbb {Z}}} \sum _{j=1}^n \delta _j \Vert \lambda _j t -\alpha _j\Vert ^2 \le \frac{ \Delta }{4} \sin ^2\left( \frac{\pi }{2(M+1)}\right) + \frac{\Delta M^n}{8 (T_2-T_1)\Lambda }, \end{aligned}$$

where

$$\begin{aligned} \Delta = \sum _{j=1}^n \delta _j \end{aligned}$$

and

$$\begin{aligned} \Lambda = \min \left\{ \left| u_1\lambda _1+ \cdots + u_n\lambda _n \right| : u_j \in {\mathbb {Z}}, |u_j| \le M, \sum _{j=1}^n \lambda _ju_j \not \in {\mathbb {Z}} \right\} . \end{aligned}$$

Corollary 2.4

Let $\omega \ge 1$, $n\in {\mathbb {N}}$ and $d_1,d_2,\ldots ,d_n\in {\mathbb {R}}$. Then there is an integer $T:=T(d_1,d_2,\ldots ,d_n,n,\omega )>0$ such that for every $a\in {\mathbb {Z}}$ there exists an integer $h\in [a,a+T]$ satisfying

$$\begin{aligned} \sum _{k=1}^n\Vert hd_k\Vert ^2\le \frac{n}{\omega }. \end{aligned}$$

Proof

Applying Chen’s lemma for $T_1=a$, $T_2=a+T$, $(\lambda _1,\ldots ,\lambda _n)=(d_1,d_2,\ldots ,d_n)$, $\alpha _1=\cdots =\alpha _n=0$, $\delta _1=\cdots =\delta _n=1$ and $M:=\pi \omega /4-1$ gives that there is $h\in {\mathbb {Z}}$, $h\in [a,a+T]$ such that

$$\begin{aligned} \sum _{k=1}^n\Vert hd_k\Vert ^2\le \frac{n}{4}\left( \sin ^2\left( \frac{\pi }{2(M+1)}\right) +\frac{M^n}{2\Lambda T}\right) . \end{aligned}$$

Obviously, we have $\sin ^2\left( \frac{\pi }{2(M+1)}\right) \le \frac{2}{\omega }$. Moreover, since M depends only on $\omega$ and $\Lambda$ depends only on M and $d_1,\ldots ,d_n$, there is a sufficiently large T such that $\frac{M^n}{2\Lambda T}\le \frac{2}{\omega }$, which completes the proof. $\square$

The need for the application of Lemma 2.2 forces us to consider the distribution of zeros of functions being linear combinations of exponential functions, which is well understood by the following result due to Wilder.

Lemma 2.5

(Wilder’s theorem) Let $g(z) = \sum _{k=1}^n A_ke^{\omega _kz}$, where $z=x+iy\in {\mathbb {C}}$, ${\mathbb {C}}\ni A_k\ne 0$, $k=1,2,\ldots ,n$, and $\omega _1<\cdots <\omega _n$. Then there exsits $K>0$ such that

• (i) all zeros of g(z) lie in the strip $|x|\le K$;

• (ii) for each pair $(\alpha ,\beta )\in {\mathbb {R}}^2$ with $\beta >0$, we have

  $$\begin{aligned} \left| N_0(\alpha ,\beta ,K) - \beta \frac{\omega _n-\omega _1}{2\pi }\right| \le n-1, \end{aligned}$$

  where $N_0(\alpha ,\beta ,K)$ counts the number of zeros of g(z) in the rectangle $|x|\le K$, $y\in [\alpha ,\alpha +\beta ]$.

Proof

For the proof we refer to [7, Theorem 1]. $\square$

Proof of Lemma 2.1

Let $\theta _{p_m} = \frac{m}{L}$, where $p_m$ denotes the m-th prime number and L is an arbitrary integer satisfying $L>n-1+d_n-d_1$. We shall use Lemma 2.2 for the sequence

$$\begin{aligned} u_m = \left( \frac{e(d_1\theta _{p_m})}{p_m^s},\ldots ,\frac{e(d_n\theta _{p_m})}{p_m^s}\right) \end{aligned}$$

and the real Hilbert space $H={\mathbb {C}}^n$ equipped with the inner product

$$\begin{aligned} ({\mathbf {x}}|{\mathbf {y}}) = {\text {Re}}\sum _{j=1}^n x_j\overline{y_j}\qquad \text {for }\ {\mathbf {x}}=(x_1,\ldots ,x_n),\ {\mathbf {y}} = (y_1,\ldots ,y_n)\in {\mathbb {C}}^n. \end{aligned}$$

Obviously, $\sum _{m=1}^\infty \Vert u_m\Vert _H^2<\infty$, since $\sigma >1/2$. Thus, in order to apply Lemma 2.2, it suffices to show that for every $e=(a_1,\ldots ,a_n)\in {\mathbb {C}}^n$ with $\Vert e\Vert _H =1$ there are two permutations of the series $\sum _m (u_m|e)$ tending to $+\infty$ and $-\infty$, respectively. In fact, we show only the existence of a permutation of the series, which diverges to $+\infty$, since the case $-\infty$ is similar and can be left to the reader.

Let us note that

$$\begin{aligned} (u_m|e) = {\text {Re}}\sum _{k=1}^n \overline{a_k}\frac{e(d_k\theta _{p_m})}{p_m^s} = \frac{1}{p_m^\sigma }{\text {Re}}\left( e^{-it\log p_m}\left( \sum _{k=1}^n\overline{a_k}e(d_k\theta _{p_m})\right) \right) \end{aligned}$$

and consider the function $f(x) = \sum _{k=1}^n\overline{a_k}e(d_k x)$, $x\in {\mathbb {R}}$. Putting $A=\sum _{k=1}^n |a_k|^2$, one can easily deduce from the Cauchy-Schwarz inequality that for every $x\in {\mathbb {R}}$ we have

$$\begin{aligned} |f(x+\tau )-f(x)|&= \left| \sum _{k=1}^n\overline{a_k}e(d_k x)(e(d_k\tau ) - 1)\right| \le 2\pi \sum _{k=1}^n|a_k|\Vert d_k\tau \Vert \\&\le 2\pi \sqrt{A}\left( \sum _{k=1}^n\Vert d_k\tau \Vert ^2\right) ^{1/2}. \end{aligned}$$

Now, from Corollary 2.4, we see that there is a positive integer N such that every interval of length N contains an integer h satisfying $\sum _{k=1}^n\Vert hd_k\Vert ^2\le \frac{\varepsilon ^2}{(2\pi )^2A}$. Therefore, dividing the interval $(0,\infty )$ into disjoint intervals $I_l$, $l\in {\mathbb {N}}$, of length N, one can find integers $\tau _l\in I_l$ satisfying $\sum _{k=1}^n\Vert \tau _ld_k\Vert ^2\le \frac{\varepsilon ^2}{(2\pi )^2A}$. Hence, for every $x\in {\mathbb {R}}$ and $l\in {\mathbb {N}}$, we have

$$\begin{aligned} |f(x+\tau _l)-f(x)|\le \varepsilon . \end{aligned}$$

Now, let us observe that, by Lemma 2.5, the function f(x) has at most $n-1+d_n-d_1$ zeros in the interval [0, 1]. Hence, by the choice of L, there is $m_0$ such that $0\le m_0\le L-1$ and $f(\frac{m_0}{L})=c_0\ne 0$. Thus for every $\tau _l$ we have

$$\begin{aligned} f\left( \frac{m_0+L\tau _l}{L}\right) = f\left( \frac{m_0}{L}+\tau _l\right) = c_0+\xi _l,\qquad \hbox { for suitable}\ |\xi _l|\le \varepsilon , \end{aligned}$$

so each interval $[m_0+kLN,m_0+(k+1)NL]$, $k=1,2,\ldots$, contains an integer $m_k':=m_0+L\tau _l$ satisfying $f(\frac{m_k'}{L})=c_0+\xi _l$ with $|\xi _l|\le \varepsilon$. Assume that $c_0 = |c_0|e^{i\varphi }$. Then for every $m_k'$ and $m\in {\mathbb {N}}$ satisfying

$$\begin{aligned} \frac{2\pi m}{t}-\frac{\pi }{4t}+\frac{\varphi }{t}< \log p_{m_k'}<\frac{2\pi m}{t}+\frac{\pi }{4t}+\frac{\varphi }{t} \end{aligned}$$

we have $\varphi -t\log p_{m'_k} \in (-2\pi m-\pi /4,-2\pi m+\pi /4)$, so ${\text {Re}}(c_0\exp ({-it\log p_{m_k'}})) \ge \frac{c_0}{\sqrt{2}}$. Hence, for sufficiently small $\varepsilon >0$, we have

$$\begin{aligned} {\text {Re}}\left( e^{-it\log p_{m_k'}}f\left( \frac{m_k'}{L}\right) \right) \ge \frac{|c_0|}{2}>0. \end{aligned}$$

Hence it remains to notice that the fact that $\sigma <1$ and the definition of $m_k'$ implies that (as usual, $\lfloor x\rfloor$ denotes the greatest integer less than or equal to x)

$$\begin{aligned}&\sum _{\begin{array}{c} m_k'\\ \frac{2\pi m}{t}-\frac{\pi }{4t}+\frac{\varphi }{t}<\log p_{m_k'}<\frac{2\pi m}{t}+\frac{\pi }{4t}+\frac{\varphi }{t} \end{array}}\frac{1}{p_{m_k'}^\sigma }\\&\quad \ge \frac{1}{e^{\frac{2\pi m}{t}+\frac{\pi }{4t}+\frac{\varphi }{t}}}\sum _{\begin{array}{c} m_k'\\ e^{\frac{2\pi m}{t}-\frac{\pi }{4t}+\frac{\varphi }{t}}<p_{m_k'}<e^{\frac{2\pi m}{t}+\frac{\pi }{4t}+\frac{\varphi }{t}} \end{array}}1\\&\quad \ge \frac{1}{e^{\frac{2\pi m}{t}+\frac{\pi }{4t}+\frac{\varphi }{t}}}\left\lfloor \frac{1}{LN}\sum _{\begin{array}{c} e^{\frac{2\pi m}{t}-\frac{\pi }{4t}+\frac{\varphi }{t}}<p<e^{\frac{2\pi m}{t}+\frac{\pi }{4t}+\frac{\varphi }{t}} \end{array}}1\right\rfloor \gg \frac{1}{m}, \end{aligned}$$

since, for $x=\frac{2\pi m}{t}+\frac{\varphi }{t}$ and $h=\frac{\pi }{4t}$, the prime number theorem gives

$$\begin{aligned} \sum _{\begin{array}{c} e^{x-h}<p<e^{x+h} \end{array}}1&= \pi \left( e^{x+h}\right) -\pi \left( e^{x-h}\right) +O(1)\\&=\frac{e^{x+h}}{x+h}-\frac{e^{x-h}}{x-h} + O_h\left( \frac{e^x}{x^2}\right) \\&\gg _h \frac{e^x}{x} \gg _t \frac{e^{\frac{2\pi m}{t}+\frac{\pi }{4t}+\frac{\varphi }{t}}}{m}. \end{aligned}$$

Thus using the fact that the harmonic series is divergent completes the proof. $\square$

3 Approximation by a Finite Sum Over Primes

In this section we use the following Tsang’s lemma in order to show that the logarithm of the Riemann zeta-function can be approximated by a truncated Dirichlet series over primes for sufficiently many arguments from any vertical line lying in the right open half of the critical strip.

Lemma 3.1

(Tsang [16, Theorem 5.1 and Remark 1]) Let $\sigma \in (1/2,1]$ and $0<\delta <\frac{1}{64}$ be fixed. Then, for sufficiently large T, we have

$$\begin{aligned} \frac{1}{T}\int _T^{2T}\left| \log \zeta (\sigma +it) - \sum _{p\le T^\delta }\frac{1}{p^{\sigma +it}}\right| ^2dt\ll T^{c(1/2-\sigma )}, \end{aligned}$$

where c is a positive constant.

Lemma 3.2

Let $\varepsilon ,\varepsilon '>0$, $0<d_1<d_2<\cdots <d_n$, $s=\sigma +it$ with $\sigma \in (1/2,1], t>0$, $0<\delta <\frac{1}{65}$ and $C(T,\varepsilon )$ denotes the set of $\tau \in [T,2T]$ satisfying

$$\begin{aligned} \max _{1\le k\le n}\left| \log \zeta (s+id_k\tau ) - \sum _{p\le T^\delta }\frac{1}{p^{s+id_k\tau }}\right| <\varepsilon . \end{aligned}$$

Then, for sufficiently large T, we have

$$\begin{aligned} \frac{1}{T}{\text {meas}}C(T,\varepsilon )>1-\varepsilon '. \end{aligned}$$

Proof

First let us observe that using Tsang’s lemma and a suitable version of Hilbert’s inequality due to Montgomery and Vaughan (see [10]) give

$$\begin{aligned}&\int _{T}^{2T}\left| \log \zeta (s+id_k\tau )-\sum _{p\le T^\delta }\frac{1}{p^{s+id_k\tau }} \right| ^2d\tau \\&\quad \ll \int _{d_kT+t}^{2(d_kT+t)}\left| \log \zeta (\sigma +i\tau ) -\sum _{p\le T^\delta }\frac{1}{p^{\sigma +i\tau }} \right| ^2d\tau \\&\quad \ll \int _{d_kT+t}^{2(d_kT+t)}\left| \log \zeta (\sigma +i\tau )\right. \\&\qquad \left. -\sum _{p\le (d_kT+t)^{1/65}}\frac{1}{p^{\sigma +i\tau }} \right| ^2d\tau \\&\qquad +\int _{d_kT+t}^{2(d_kT+t)}\left| \sum _{T^\delta \le p\le (d_kT+t)^{1/65}}\frac{1}{p^{\sigma +i\tau }} \right| ^2d\tau \\&\quad \ll T^{1+c(1/2-\sigma )} + T\sum _{p\le (d_jT+t)^{1/65}}\frac{1}{p^{2\sigma }}\\&\quad \ll T^{1+c'(1/2-\sigma )} \end{aligned}$$

for some $c'>0$ and $k=1,2,\ldots ,n$. Hence for every $\varepsilon _0>0$ and sufficiently large T we have

$$\begin{aligned} \sum _{k=1}^n\frac{1}{T}\int _{T}^{2T}\left| \log \zeta (s+id_k\tau )-\sum _{p\le T^\delta }\frac{1}{p^{s+id_k\tau }} \right| ^2d\tau <\varepsilon _0. \end{aligned}$$

Now let

$$\begin{aligned} A_T'&= \left\{ \tau \in [T,2T]: \sum _{k=1}^n\left| \log \zeta (s+id_k\tau ) - \sum _{p\le T^\delta }\frac{1}{p^{s+id_k\tau }}\right| >\varepsilon _0^{1/3}\right\} ,\\ A_T''&= \left\{ \tau \in [T,2T]: \zeta (s+id_k\tau )=0\text { for some }1\le k\le n\right\} . \end{aligned}$$

Obviously, by the holomorphicity of the Riemann zeta-function, we have ${\text {meas}}(A_T'')=0$ for every T. Moreover,

$$\begin{aligned} \varepsilon _0^{2/3}\frac{{\text {meas}}(A_T')}{T}<n^2\sum _{k=1}^n\frac{1}{T}\int _{A_T'}\left| \log \zeta (s+id_k\tau )-\sum _{p\le T^\delta }\frac{1}{p^{s+id_k\tau }} \right| ^2d\tau \le n^2\varepsilon _0, \end{aligned}$$

so ${\text {meas}}(A_T')\le n^2\varepsilon _0^{1/3}T$. Thus, putting $A_T=A_T'\cup A_T''$ and choosing $\varepsilon _0$ sufficiently small, we get that ${\text {meas}}(A_T)\le \varepsilon 'T$ and

$$\begin{aligned} \sum _{k=1}^n\left| \log \zeta (s+id_k\tau ) - \sum _{p\le T^\delta }\frac{1}{p^{s+id_k\tau }}\right| <\varepsilon \end{aligned}$$

for $\tau \in [T,2T]\setminus A_T$. $\square$

4 Proof of the Main Theorem

In order to prove our main theorem, we need the following crucial lemma, which allows us to connect Lemmas 2.1 and 3.2. The lemma is essentially Lemma 3.6 from [8] where the quantity $\delta (T)$ defined in Lemma 3.6 in [8] has to be estimated in a more subtle way because of the appearance of an extra parameter $d_k$.

Lemma 4.1

Let $d_1,d_2,\ldots ,d_n$ be real numbers satisfying $0<d_1<d_2<\cdots <d_n$ and $s=\sigma +it$ with $\sigma \in (1/2,1]$ and $t>0$. Moreover, suppose that $y>0$ is sufficiently large, $0<\delta <\frac{1}{65}$, $0<d<1/2$, M is a finite set of prime numbers such that $\{p:p\le y\}\subset M$ and $(\theta _p)_{p\in M}$ is a sequence of real numbers. Then for sufficiently large T we have

$$\begin{aligned} \max _{1\le k\le n}\int _{A_d(T)}\left| \sum _{\begin{array}{c} p\le T^\delta \\ p\not \in M \end{array}}\frac{1}{p^{s+id_k\tau }}\right| ^2d\tau \ll T(2d)^{|M|}y^{1-2\sigma }, \end{aligned}$$

where $A_d(T) = \{\tau \in [T,2T]: \max _{p\in M}\Vert -\tau \frac{\log p}{2\pi }-\theta _p\Vert \le d\}$.

Proof

We will closely follow the proof of Lemma 3.6 from [8].

Let $\delta '>0$ be such that $0<d+\delta '<1/2$ and $\xi :{\mathbb {R}}\rightarrow [0,1]$ be a continuous and periodic function with period 1 defined on $[-1/2,1/2]$ by

$$\begin{aligned} \xi (x) = {\left\{ \begin{array}{ll} 1,&{}|x|\le d,\\ -\frac{|x|}{\delta '}+\frac{d+\delta '}{\delta '},&{}d<|x|\le d+\delta ',\\ 0,&{}d+\delta '<|x|\le \frac{1}{2}. \end{array}\right. } \end{aligned}$$

Moreover, we put

$$\begin{aligned} {\tilde{\xi }}(\tau ) = \prod _{p\in M}\xi \left( -\frac{\tau \log p}{2\pi }- \theta _p\right) \qquad \hbox { for}\ \tau \in {\mathbb {R}}. \end{aligned}$$

Obviously $\xi (x) = 1$ if $\Vert x\Vert \le d$. Hence, for $\tau \in A_d(T)$ we have $1={\tilde{\xi }}(\tau ) = {\tilde{\xi }}^2(\tau )$, whereas for arbitrary $\tau \in {\mathbb {R}}$ we have $0\le {\tilde{\xi }}^2(\tau )\le {\tilde{\xi }}(\tau )\le 1$.

Now let us assume that $\varepsilon _0>0$ is small. Then there is a trigonometric polynomial

$$\begin{aligned} P(x) = \sum _{-L\le \ell \le L}c(\ell )e(\ell x),\qquad c(\ell )\in {\mathbb {C}} \end{aligned}$$

such that $|\xi (x)-P(x)|<\varepsilon _0$ for $x\in {\mathbb {R}}$. So let us define

$$\begin{aligned} {\tilde{P}}(\tau ) = \prod _{p\in M}P\left( -\frac{\tau \log p}{2\pi }- \theta _p\right) . \end{aligned}$$

Using the fact that $\left| \prod _{j=1}^{n}a_j-\prod _{j=1}^{n}b_j\right| \le R^{n-1}\sum _{j=1}^n|a_j-b_j|$ for arbitrary complex numbers $a_1,\ldots ,a_n,b_1,\ldots ,b_n$ with $|a_j|,|b_j|\le R$, $j=1,2,\ldots ,n$, we get

$$\begin{aligned} \left| \prod _{p\in M}\xi (x_p)-\prod _{p\in M}P(x_p)\right| \le 2^{|M|-1}|M|\varepsilon _0 \end{aligned}$$

for all real $x_p$, where |M| denotes the number of elements in the set M. Hence $|{\tilde{\xi }}(\tau )-{\tilde{P}}(\tau )|\le 2^{|M|-1}|M|\varepsilon _0$ for $\tau \in {\mathbb {R}}$. Therefore, by the Cauchy–Schwarz inequality, we have

$$\begin{aligned} |{\tilde{\xi }}(\tau )|^2 \le 2|{\tilde{P}}(\tau )|^2+2^{2|M|-1}|M|^2\varepsilon _0^2,\qquad \tau \in {\mathbb {R}} \end{aligned}$$

and

$$\begin{aligned} |{\tilde{P}}(\tau )|^2 \le 2|{\tilde{\xi }}(\tau )|^2+2^{2|M|-1}|M|^2\varepsilon _0^2,\qquad \tau \in {\mathbb {R}}. \end{aligned}$$

Now, let $N'>y$. Then for $\tau \in {\mathbb {R}}$ we have

$$\begin{aligned} \sum _{\begin{array}{c} p\le T^\delta \\ p\not \in M \end{array}}\frac{1}{p^{s+id_k\tau }}= & {} \sum _{\begin{array}{c} p\le N'\\ p\not \in M \end{array}}\frac{1}{p^{s}}e\left( -\frac{d_k\tau \log p}{2\pi }\right) +\sum _{\begin{array}{c} N'<p\le T^\delta \\ p\not \in M \end{array}}\frac{1}{p^{s}}e\left( -\frac{d_k\tau \log p}{2\pi }\right) \\=: & {} S_1(\tau )+S_2(\tau ), \end{aligned}$$

so using the Cauchy–Schwarz inequality again yields

$$\begin{aligned} \int _{A_d(T)}\left| \sum _{\begin{array}{c} p\le T^\delta \\ p\not \in M \end{array}}\frac{1}{p^{s+id_k\tau }}\right| ^2d\tau \le 2\int _{A_d(T)}|S_1(\tau )|^2d\tau +2\int _{A_d(T)}|S_2(\tau )|^2d\tau =: I_1 + I_2. \end{aligned}$$

First let us estimate $I_1$. We have

$$\begin{aligned} I_1&\le \int _T^{2T}|{\tilde{\xi }}(\tau )|^2|S_1(\tau )|^2d\tau \ll \int _T^{2T}|{\tilde{P}}(\tau )|^2|S_1(\tau )|^2d\tau \\&\quad + 2^{2|M|-1}|M|^2\varepsilon _0^2\int _T^{2T}|S_1(\tau )|^2d\tau \end{aligned}$$

and

$$\begin{aligned} {\tilde{P}}(\tau )&= \prod _{p\in M}\left( \sum _{-L\le \ell _p\le L}c(\ell _p)e\left( \ell _p\left( -\frac{\tau \log p}{2\pi }-\theta _p\right) \right) \right) \nonumber \\&=\sum _{-L\le \ell _{p_1},\ldots ,\ell _{p_{|M|}}\le L}{\tilde{c}}(\ell _{p_1},\ldots ,\ell _{p_{|M|}})e\left( -\frac{\tau }{2\pi }\sum _{j=1}^{|M|}\ell _{p_j}\log p_j\right) \end{aligned}$$

(4)

where $p_1,\ldots ,p_{|M|}$ are all elements of M and

$$\begin{aligned} {\tilde{c}}(\ell _{p_1},\ldots ,\ell _{p_{|M|}}) = c(\ell _{p_1})\cdots c(\ell _{p_{|M|}}) e\left( -\sum _{j=1}^{|M|}\ell _{p_j}\theta _{p_j}\right) . \end{aligned}$$

Then

$$\begin{aligned}&\int _T^{2T}|{\tilde{P}}(\tau )|^2|S_1(\tau )|^2d\tau \\&\quad = \int _T^{2T}\left| \sum _{-L\le \ell _{p_1},\ldots ,\ell _{p_{|M|}}\le L}\sum _{\begin{array}{c} p\le N'\\ p\not \in M \end{array}}{\tilde{c}}(\ell _{p_1},\ldots ,\ell _{p_{|M|}})\frac{1}{p^s}e \right. \\&\quad \times \left. \left( -\frac{\tau }{2\pi }\left( d_k\log p+\sum _{j=1}^{|M|}\ell _{p_j}\log p_j \right) \right) \right| ^2 d\tau \end{aligned}$$

Let

$$\begin{aligned} \Omega _k = \left\{ -\left( d_k\log p+\sum _{j=1}^{|M|}\ell _{p_j}\log p_j \right) :p\le N',\ p\not \in M,\ -L\le \ell _{p_1},\ldots ,\ell _{p_{|M|}}\le L\right\} \end{aligned}$$

and

$$\begin{aligned} x&:=x(p,\ell _{p_1},\ldots ,\ell _{p_{|M|}})=-\left( d_k\log p+\sum _{j=1}^{|M|}\ell _{p_j}\log p_j \right) ,\\ y&:=y(q,\ell '_{p_1},\ldots ,\ell '_{p_{|M|}})=-\left( d_k\log q+\sum _{j=1}^{|M|}\ell '_{p_j}\log p_j \right) . \end{aligned}$$

We need to estimate

$$\begin{aligned} \delta _k = \min _{\begin{array}{c} x,y\in \Omega _k\\ (p,\ell _{p_1},\ldots ,\ell _{p_{|M|}})\ne (q,\ell '_{p_1},\ldots ,\ell '_{p_{|M|})} \end{array}}|x-y|, \end{aligned}$$

so we consider the following three cases.

• Case 1: $p=q$. Then $|x-y|>0$ by the fact that the sequence of logarithms of prime numbers are linearly independent over ${\mathbb {Q}}$.

• Case 2: $p\ne q$, $0\ne d_k\in {\mathbb {Q}}$. Similarly to the Case 1, by the unique factorization of integers, we have $x=y\leftrightarrow d_k=0,\ \ell '_{p_i}=\ell '_{p_i}$, so $|x-y|>0$.

• Case 3: $p\ne q$, $d_k\notin {\mathbb {Q}}$. Then $x=y$ means that for suitable $A,B\in {\mathbb {Q}}$ we have

  $$\begin{aligned} d_k\log B = d_k(\log p-\log q) = \sum _{i=1}^{|M|}(\ell '_{p_i}-\ell _{p_i})\log p_i = \log A. \end{aligned}$$

  (5)

Using the Six Exponentials Theorem one can prove (see [13, Corollary 2.3 and Lemma 2.4]) the following result.

Lemma 4.2

For arbitrary irrational number d there exist at most two primes $p_1,p_2$ such that if

$$\begin{aligned} d = \frac{\log a}{\log b},\qquad \hbox { for some} a,b\in {\mathbb {Q}}, \end{aligned}$$

then at least one of $p_1,p_2$ appears in the prime factorization of b.

Now, let us note that $B=p/q$ for some $p,q\notin M$, so $p,q>y$. Hence, taking sufficiently large y and using the above lemma we obtain that (5) never holds, which means that $|x-y|>0$ in Case 3 as well.

Summing up all three cases we get $\delta _k>0$. Hence using again a suitable version of Hilbert’s inequality due to Montgomery and Vaughan (see [10, Corollary 2, Eq. (1.8)]) we see that for some $\theta _k$ with $|\theta _k|\le 1$ and sufficiently large T we have

$$\begin{aligned}&\int _T^{2T}|{\tilde{P}}(\tau )|^2|S_1(\tau )|^2d\tau \\&\quad = (T+2\pi \delta _k^{-1}\theta _k)\sum _{-L\le \ell _{p_1},\ldots ,\ell _{p_{|M|}}\le L}|{\tilde{c}}(\ell _{p_1},\ldots ,\ell _{p_{|M|}})|^2\sum _{\begin{array}{c} p\le N'\\ p\not \in M \end{array}}p^{-2\sigma }\\&\quad \ll Ty^{1-2\sigma }\sum _{-L\le \ell _{p_1},\ldots ,\ell _{p_{|M|}}\le L}|{\tilde{c}}(\ell _{p_1},\ldots ,\ell _{p_{|M|}})|^2. \end{aligned}$$

Similarly one can obtain

$$\begin{aligned} \int _T^{2T}|{\tilde{P}}(\tau )|^2d\tau = (T+2\pi \delta _0^{-1}\theta _0)\sum _{-L\le \ell _{p_1},\ldots ,\ell _{p_{|M|}}\le L}|{\tilde{c}}(\ell _{p_1},\ldots ,\ell _{p_{|M|}})|^2 \end{aligned}$$

for some positive real $\delta _0$ and $\theta _0$ with $|\theta _0|\le 1$. Thus, we have

$$\begin{aligned}&T\sum _{-L\le \ell _{p_1},\ldots ,\ell _{p_{|M|}}\le L}|{\tilde{c}}(\ell _{p_1},\ldots ,\ell _{p_{|M|}})|^2 \\&\quad \ll \int _T^{2T}|{\tilde{P}}(\tau )|^2d\tau \ll \int _T^{2T}{\tilde{\xi }}(\tau )d\tau +T2^{2|M|-1}|M|^2\varepsilon _0^2\\&\quad \le \left| \int _T^{2T}{\tilde{P}}(\tau )\right| +T(2^{|M|-1}|M|\varepsilon _0+2^{2|M|-1}|M|^2\varepsilon _0^2). \end{aligned}$$

Note that from (4) we get

$$\begin{aligned} \int _T^{2T}{\tilde{P}}(\tau )d\tau = c(0)^{|M|}T+O(1). \end{aligned}$$

Moreover, we have

$$\begin{aligned} |c(0)| = \left| \int _{-1/2}^{1/2}P(x)dx\right| \le \int _{-1/2}^{1/2}(|\xi (x)|+\varepsilon _0)dx = 2d+\delta '+\varepsilon _0. \end{aligned}$$

Therefore,

$$\begin{aligned}&\int _T^{2T}|{\tilde{P}}(\tau )|^2|S_1(\tau )|^2d\tau \\&\quad \ll Ty^{1-2\sigma }\left( (2d+\delta '+\varepsilon _0)^{|M|}+2^{|M|-1}|M|\varepsilon _0+2^{2|M|-1}|M|^2\varepsilon _0^2\right) . \end{aligned}$$

Similarly using Hilbert’s inequality one can easily obtain that

$$\begin{aligned} \int _T^{2T}|S_1(\tau )|^2d\tau \ll Ty^{1-2\sigma }, \end{aligned}$$

so

$$\begin{aligned} I_1\ll Ty^{1-2\sigma }\left( (2d+\delta '+\varepsilon _0)^{|M|}+2^{2|M|+1}|M|^2\varepsilon _0\right) . \end{aligned}$$

Next, in order to apply Hilbert’s inequality to estimate $I_2$, we need to use the following estimate

$$\begin{aligned} \delta _k(T) = d_k\min _{\begin{array}{c} N'<p,q\le T^\delta \\ p\ne q\text { and }p,q\notin M \end{array}}|\log p-\log q|\ge d_k\min _{N'<p\le T^\delta }\log \left( 1+\frac{1}{p-1}\right) \ge \frac{d_k}{T^\delta }. \end{aligned}$$

So

$$\begin{aligned} I_2\le \left( T+\frac{2\pi T^\delta }{d_k}\right) \sum _{\begin{array}{c} N'<p\le T^\delta \\ p\not \in M \end{array}}\frac{1}{p^{2\sigma }}\ll TN'^{1-2\sigma }. \end{aligned}$$

Choosing $\delta '$ and $\varepsilon _0$ such that

$$\begin{aligned} \left( 1+\frac{\delta '+\varepsilon _0}{2d}\right) ^{|M|}<\frac{3}{2}\qquad \text {and}\qquad 2^{2|M|+1}|M|^2\varepsilon _0<\frac{1}{2}(2d)^{|M|} \end{aligned}$$

implies that $I_1\ll Ty^{1-2\sigma }(2d)^{|M|}$.

Moreover, if $N'$ is such large that $N'^{1-2\sigma }\le (2d)^{|M|}y^{1-2\sigma }$, then

$$\begin{aligned} \int _{A_d(T)}\left| \sum _{\begin{array}{c} p\le T^\delta \\ p\not \in M \end{array}}\frac{1}{p^{s+id_k\tau }}\right| ^2d\tau \ll T(2d)^{|M|}y^{1-2\sigma }, \end{aligned}$$

which completes the proof. $\square$

Now, we are in the position to complete the proof of our theorem.

Proof of Theorem 1.1

From Lemma 2.1 we know that there exist a finite set of primes $M\supset \{p:p<y\}$ and a sequence $(\theta _p)_{p\in M}$ of real numbers such that

$$\begin{aligned} \max _{1\le k\le n}\left| \sum _{p\in M}\frac{e(d_k\theta _p)}{p^s}-z_k\right| <\varepsilon . \end{aligned}$$

Let $A_d(T) = \{\tau \in [T,2T]: \max _{p\in M}\Vert -\frac{\tau \log p}{2\pi } - \theta _p\Vert \le d\}$ and d is such small that

$$\begin{aligned} \max _{1\le k\le n}\left| \sum _{p\in M}\frac{e(-d_k\frac{\tau \log p}{2\pi })}{p^s}-\sum _{p\in M}\frac{e(d_k\theta _p)}{p^s}\right| <\varepsilon \qquad \text {for any } \tau \in A_d(T). \end{aligned}$$

Let us emphasize that the choice of such d is possible since $d_1,\ldots ,d_n\in {\mathbb {N}}$, and it is the only step in the proof where we need to assume that $d_1,\ldots ,d_n$ are integers.

Let us notice that for $\tau \in A_d(T)$ we have

$$\begin{aligned} \max _{1\le k\le n}\left| \sum _{p\in M}\frac{1}{p^{s+id_k\tau }}-z_k\right| <2\varepsilon . \end{aligned}$$

Next let

$$\begin{aligned} B(T,\varepsilon ) = \left\{ \tau \in A_d(T): \max _{1\le k\le n}\left| \sum _{\begin{array}{c} p\le T^\delta \\ p\not \in M \end{array}}\frac{1}{p^{s+id_k\tau }}\right| <\varepsilon \right\} , \end{aligned}$$

where $0<\delta <\frac{1}{65}$. Then, by Lemma 4.1 and Kronecker’s approximation theorem, for sufficiently large y we have

$$\begin{aligned} \frac{1}{T}{\text {meas}}(B(T,\varepsilon ))>\frac{1}{2}(2d)^{|M|}. \end{aligned}$$

Thus, using the notation from Lemma 3.2, we get that

$$\begin{aligned} \max _{1\le k\le n}|\log \zeta (s+id_k\tau )-z_k|<4\varepsilon \qquad \text {for } \tau \in B(T,\varepsilon )\cap C(T,\varepsilon ), \end{aligned}$$

and, since ${\text {meas}}C(T,\varepsilon )>T(1-\varepsilon ')$ for every $\varepsilon '>0$, we have

$$\begin{aligned}&\frac{1}{T}{\text {meas}}B(T,\varepsilon )\cap C(T,\varepsilon )\\&\quad>\frac{1}{T}{\text {meas}}B(T,\varepsilon )+\frac{1}{T}{\text {meas}}C(T,\varepsilon ) - 1>\frac{1}{2}(2d)^{|M|}-\varepsilon ' \end{aligned}$$

which completes the proof, since the right hand side is positive for sufficiently small $\varepsilon '$. $\square$