Consecutive Primes in Short Intervals

Abstract

We obtain a lower bound for $\#\{x/2<p_n\leq x \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a\pmod{q}$, $p_{n+m} - p_n\leq y\}$, where $p_n$ is the $n$th prime.

1. Introduction

Let $p_n$ denote the $n$th prime. We prove the following result.

Theorem 1.1.

There are positive absolute constants $c$ and $C$ such that the following holds. Let $\varepsilon$ be a real number with $0<\varepsilon<1$. Then there is a number $c_0(\varepsilon)>0,$ depending only on $\varepsilon,$ such that if $x\in{\mathbb R},$ $y\in{\mathbb R},$ $m\in{\mathbb Z},$ $q\in{\mathbb Z},$ and $a\in{\mathbb Z}$ satisfy the conditions

$$c_0(\varepsilon) \leq y\leq \ln x,\qquad 1\leq m \leq c \kern1pt \varepsilon\ln y,\qquad 1\leq q \leq y^{1-\varepsilon},\qquad (a,q)=1,$$

then

$$\#\Bigl\{\frac x2< p_n\leq x \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m}-p_n\leq y\Bigr\} \geq\pi(x)\Bigl( \frac{y}{2q\ln x} \Bigr)^{\exp(Cm)}.$$

Theorem 1.1 extends a result of Maynard [5, Theorem 3.3], who established the same result but with $y=\varepsilon\ln x$.

From Theorem 1.1 we obtain

Corollary 1.1.

There is an absolute constant $C>0$ such that if $m$ is a positive integer and $x$ and $y$ are real numbers satisfying $\exp(Cm)\leq y \leq \ln x,$ then

$$\#\Bigl\{\frac x2< p_n\leq x \colon\, \, p_{n+m}-p_n\leq y\Bigr\} \geq \pi(x)\Bigl( \frac{y}{2\ln x} \Bigr)^{\exp(Cm)}.$$

Let us introduce necessary notation. The expression $b\mid a$ means that $b$ divides $a$. For a fixed $a$ the sum $\sum_{b\mid a}$ and the product $\prod_{b\mid a}$ should be interpreted as being over all positive divisors of $a$.

We will use I. M. Vinogradov’s notation: $A\ll B$ means that $|A|\leq cB$ with a positive absolute constant $c$.

We reserve the letter $p$ for primes. In particular, the sum $\sum_{p\leq K}$ should be interpreted as being over all prime numbers not exceeding $K$.

We will also use the following notation:

$\#A$ is the number of elements of a finite set $A$;

${\mathbb N}$, ${\mathbb Z}$, ${\mathbb R}$, and ${\mathbb C}$ are the sets of all positive integers, integers, real numbers, and complex numbers;

$\mathbb{P}$ is the set of all prime numbers;

$[x]$ is the integer part of a number $x$; i.e., $[x]$ is the largest integer $n$ such that $n\leq x$;

$\{x\}$ is the fractional part of a number $x$; i.e., $\{x\}=x-[x]$;

$\lceil x\rceil$ is the smallest integer $n$ such that $n\geq x$;

$\operatorname{Re} s$ and $\operatorname{Im} s$ are the real and imaginary parts of a complex number $s$;

$(a_1,\dots,a_n)$ is the greatest common divisor of integers $a_1,\dots,a_n$;

$[a_1,\dots,a_n]$ is the least common multiple of integers $a_1,\dots,a_n$;

$\varphi (n)$ is the Euler totient function: $\varphi (n)= \#\{1\leq m \leq n \colon\, (m,n)=1\}$;

$\mu(n)$ is the Möbius function, which is defined as follows:

• (i) $\mu(1)=1$,

• (ii) $\mu(n)=0$ if there is a prime $p$ such that $p^2\mid n$, and

• (iii) $\mu(n)=(-1)^s$ if $n=q_1\dots q_s$, where $q_1<\dots<q_s$ are primes;

$\Lambda (n)$ is the von Mangoldt function:

$$\Lambda (n)= \begin{cases} \displaystyle \ln p & \text{ if }\, \displaystyle n=p^k,\\ \displaystyle 0 & \text{ if }\, \displaystyle n\neq p^k; \end{cases}$$

$P^-(n)$ is the least prime factor of $n>1$ (by convention $P^-(1)=+\infty$);

$\binom{n}{k} = n!/(k!\,(n-k)!)$ is the binomial coefficient.

For real numbers $a$ and $b$ we use $(a,b)$ and $[a,b]$ to denote, respectively, the open and closed intervals with endpoints $a$ and $b$. By $(a_1,\dots,a_n)$ we also denote a vector; the meaning of the notation should be clear from the context.

By definition, we put

$$\sum_{\varnothing} = 0\qquad\text{and}\qquad \prod_{\varnothing}=1.$$

We define

$${\mathcal M} = \{n\in{\mathbb N} \colon\, \, \mu(n)\neq 0\}.$$

We will use the following functions:

$$\begin{gathered} \, \operatorname{li} (x)=\intop_2^x\!\frac{dt}{\ln t},\qquad \Phi(x,z)=\#\bigl\{1\leq n\leq x \colon\, \, P^-(n)>z\bigr\},\\[4pt] \pi(x)=\sum_{p\leq x} 1,\qquad \theta(x)=\sum_{p\leq x} \ln p,\qquad \psi(x)=\sum_{n\leq x} \Lambda (n),\\[4pt] \pi(x;q,a)=\sum_{p\leq x,\; p\equiv a\pmod{q}} 1,\qquad \psi(x;q,a)=\sum_{n\leq x,\; n\equiv a\pmod{q}} \Lambda (n). \end{gathered}$$

Let $m>1$ and $a$ be integers. If $(a,m)=1$, then $a^{ \varphi (m)}\equiv 1\pmod{m}$ (the Fermat–Euler theorem). Let $d$ be the smallest positive value of $\gamma$ for which $a^{\gamma}\equiv 1\pmod{m}$. We call $d$ the order of $a\pmod{m}$ and say that $a$ belongs to $d\pmod{m}$.

Let $q$ be a positive integer. We recall that a Dirichlet character modulo $q$ is a function $\chi \colon\, {\mathbb Z}\to{\mathbb C}$ such that

• (1) $\chi(n+q)=\chi(n)$ for all $n\in{\mathbb Z}$ (i.e., $\chi$ is a periodic function with period $q$);

• (2) $\chi(mn)=\chi(m)\chi(n)$ for all $m, n\in{\mathbb Z}$ (i.e., $\chi$ is a totally multiplicative function);

• (3) $\chi(1) =1$;

• (4) $\chi(n)= 0$ for all $n\in{\mathbb Z}$ such that $(n,q)>1$.

By $X_q$ we denote the set of all Dirichlet characters modulo $q$. We recall that $\# X_q = \varphi (q)$ and that the principal character modulo $q$ is

$$\chi^{}_0(n):= \begin{cases} \displaystyle 1 & \text{ if }\, \displaystyle (n,q)=1,\\ \displaystyle 0 & \text{ if }\, \displaystyle (n,q)>1. \end{cases}$$

Let $\chi\in X_q$. We say that the character $\chi$ restricted by $(n,q)=1$ has period $q_1$ if it has the property that $\chi(m)=\chi(n)$ for all $m,n\in{\mathbb Z}$ such that $(m,q)=1$, $(n,q)=1$ and $m\equiv n\pmod{q_1}$. Let $c(\chi)$ denote the conductor of $\chi$, which is the least positive integer $q_1$ such that $\chi$ restricted by $(n,q)=1$ has period $q_1$. We say that $\chi$ is primitive if $c(\chi)=q$, and imprimitive if $c(\chi)<q$. By $X^*_q$ we denote the set of all primitive characters modulo $q$. We observe that the principal character modulo $1$ is primitive. On the other hand, any principal character modulo $q>1$ is imprimitive, since its conductor is clearly $1$. For $\chi\in X_q$ we put

$$\begin{gathered} \, E_{\chi^{}_0}(\chi):={ \begin{cases} \displaystyle 1 & \text{ }\, \displaystyle \text{if } \chi \text{ is the principal character modulo } q,\\ \displaystyle 0 & \text{ }\, \displaystyle \text{otherwise}, \end{cases}} \\[5pt] \psi(x,\chi)=\sum_{n\leq x} \Lambda (n)\chi(n),\qquad \psi'(x,\chi)=\psi(x,\chi)-E_{\chi^{}_0}(\chi)x. \end{gathered}$$

A character $\chi$ is said to be real if $\chi(n)\in{\mathbb R}$ for all $n\in{\mathbb Z}$. A character $\chi$ is said to be complex if there is an integer $n$ such that $\operatorname{Im}(\chi(n))\neq 0$.

We say that characters $\chi^{}_1$ and $\chi^{}_2$ (modulo $q_1$ and modulo $q_2$, respectively) are equal and write $\chi^{}_1 = \chi^{}_2$ if $\chi^{}_1(n)=\chi^{}_2(n)$ for any integer $n$. Otherwise, we say that characters $\chi^{}_1$ and $\chi^{}_2$ are not equal and write $\chi^{}_1 \neq \chi^{}_2$.

Let $\chi$ be a Dirichlet character modulo $q$. The corresponding $L$-function is defined by the series

$$L(s,\chi)=\sum_{n=1}^{\infty}\frac{\chi(n)}{n^s}$$

for $s\in{\mathbb C}$ with $\operatorname{Re} s>1$. It is well known that if $\chi$ is not the principal character modulo $q$, then $L(s,\chi)$ can be analytically continued to ${\mathbb C}$. If $\chi$ is the principal character modulo $q$, then $L(s,\chi)$ can be analytically continued to ${\mathbb C}\setminus \{1\}$ with a simple pole at $s=1$.

We say that two linear functions $L_1(n)=a_1n+b_1$ and $L_2(n)=a_2n+b_2$ with integer coefficients are equal and write $L_1 = L_2$ if $a_1=a_2$ and $b_1=b_2$. Otherwise, we say that the linear functions $L_1$ and $L_2$ are not equal and write $L_1 \neq L_2$.

Let ${\mathcal L} =\{L_1,\dots,L_k\}$ be a set of $k$ linear functions with integer coefficients:

$$L_i(n)=a_i n+b_i,\qquad i=1,\dots,k.$$

For $L(n)=a n+b,$ $a,b\in{\mathbb Z}$, we define

$$\Delta_L=|a|\prod_{i=1}^k|a b_i - b a_i|.$$

We say that $L(n)=an+b$ belongs to ${\mathcal L}$ ($L\in {\mathcal L}$) if there is an $i$, $1 \leq i \leq k$, such that $L=L_i$. Otherwise, we say that $L(n)=an+b$ does not belong to ${\mathcal L}$ ($L\notin {\mathcal L}$).

This paper is organized as follows. In Sections 2–4 we give necessary lemmas. In Section 5 we prove Theorem 1.1 and Corollary 1.1.

2. Preparatory lemmas

In this section we present some well-known lemmas which will be used in the following sections.

Lemma 2.1 (see, for example, [6, Ch. 1]).

Let $x$ be a real number with $x\geq 2$. Then

$$b_1\ln x \leq \prod_{p\leq x} \biggl( 1-\frac{1}{p} \biggr)^{\!-1}\leq b_2 \ln x \qquad\textit{and}\qquad b_3 \ln x\leq \prod_{p\leq x} \biggl( 1+\frac{1}{p} \biggr)\leq b_4 \ln x,$$

where $b_i,$ $i=1,\dots,4,$ are positive absolute constants.

Lemma 2.2 (see, for example, [4, Chs. 1, 2]).

The limits $\lim_{x\to+\infty}\psi(x)/x,$ $\lim_{x\to+\infty}\theta(x)/x,$ $\lim_{x\to+\infty}\pi(x)/(x/ \ln x),$ and $\lim_{n\to+\infty} p_n/(n\ln n)$ exist and

$$\lim_{x\to +\infty} \frac{\psi(x)}{x}= \lim_{x\to +\infty} \frac{\theta(x)}{x}=\lim_{x\to +\infty}\frac{\pi(x)}{x/ \ln x}=1,\qquad \lim_{n\to +\infty} \frac{p_n}{n\ln n}=1.$$

From Lemma 2.2 we obtain

Lemma 2.3.

It holds that

$$ b_5 x \leq \psi(x) \leq b_6 x,\qquad b_7 x \leq \theta(x) \leq b_8 x\qquad \textit{for}\quad x\geq 2,$$

(2.1)

$$\begin{gathered} \, b_9\frac{x}{\ln x} \leq \pi(x) \leq b_{10}\frac{x}{\ln x}\qquad\textit{for}\quad x\geq 2, \\[4pt] b_{11} n\ln(n+2) \leq p_n \leq b_{12}n\ln(n+2)\qquad\textit{for}\quad n\geq 1,\nonumber \end{gathered}$$

(2.2)

where $b_i,$ $i=5,\dots,12,$ are positive absolute constants.

Lemma 2.4 (see, for example, [7, Ch. 2]).

Let $n$ be an integer with $n>1$. Then

$$\varphi (n)=n\prod_{p\mid n}\biggl( 1-\frac{1}{p} \biggr).$$

From Lemma 2.4 we readily obtain the following two lemmas.

Lemma 2.5.

Let $m$ and $n$ be integers with $m\geq 1$ and $n\geq 1$. Then

$$\varphi (mn) \geq \varphi (m) \varphi (n).$$

Lemma 2.6.

Let $m$ and $n$ be integers with $m\geq 1,$ $n\geq 1,$ and $(m,n)=1$. Then

$$\varphi (mn) = \varphi (m) \varphi (n).$$

Lemma 2.7.

Let $n$ be an integer with $n\geq 1$. Then

$$ \frac{n}{ \varphi (n)} = \sum_{d\mid n} \frac{\mu^2(d)}{ \varphi (d)}.$$

(2.3)

Proof.

For $n=1$, equality (2.3) holds. Let $n>1$. Let us express $n$ in the standard form $n=q_1^{\alpha_1}\dots q_r^{\alpha_r}$, where $q_1<\dots<q_r$ are prime numbers. Applying Lemmas 2.4 and 2.6, we have

$$\begin{aligned} \, \frac{n}{ \varphi (n)}&=\prod_{p\mid n}\biggl( 1-\frac{1}{p} \biggr)^{\!-1} =\prod_{p\mid n}\biggl( 1+\frac{1}{p-1} \biggr) =\biggl( 1+\frac{1}{q_1-1} \biggr)\dots \biggl( 1+\frac{1}{q_r-1} \biggr) \\[2pt] &=\biggl( 1+\frac{1}{ \varphi (q_1)} \biggr)\dots \biggl( 1+\frac{1}{ \varphi (q_r)} \biggr) =1+ \sum_{s=1}^r\,\sum_{1\leq i_1<\dots< i_s\leq r} \frac{1}{ \varphi (q_{i_1})\dots \varphi (q_{i_s})} \\[2pt] &=1+ \sum_{s=1}^r\,\sum_{1\leq i_1<\dots< i_s\leq r} \frac{1}{ \varphi (q_{i_1}\dots q_{i_s})} =\sum_{d\mid n,\; d\in {\mathcal M} }\frac{1}{ \varphi (d)}= \sum_{d\mid n}\frac{\mu^2(d)}{ \varphi (d)}. \ \quad\Box\end{aligned}$$

Lemma 2.8 (see, for example, [6, Ch. 1]).

Let $n$ be an integer with $n\geq 1$. Then

$$\varphi (n)\geq c \kern1pt \frac{n}{\ln\ln(n+2)},$$

where $c>0$ is an absolute constant.

Lemma 2.9 (see, for example, [1, Ch. 28]).

Let $x$ be a real number with $x \geq 2$. Then

$$\sum_{1 \leq n \leq x}\frac{1}{ \varphi (n)} \leq c\ln x,$$

where $c>0$ is an absolute constant.

Lemma 2.10 (see, for example, [2, Ch. 5]).

Let $n$ be an integer with $n\geq 1$. Then

$$\sum_{p\mid n}\frac{\ln p}{p}\leq c \ln\ln(3n),$$

where $c>0$ is an absolute constant.

Lemma 2.11.

Let $a,$ $b,$ and $c$ be integers such that $(a,b)\mid c$. Then the equation

$$ a x+b y = c$$

(2.4)

has a solution in integers.

Proof.

We put $d=(a,b)$. Then $c=dl$ for some $l\in{\mathbb Z}$. It is well known (see, for example, [7, Ch. 1, Exercise 1]) that the equation

$$ ax+by=d$$

(2.5)

has a solution in integers. Let $x_0\in{\mathbb Z}$ and $y_0\in{\mathbb Z}$ be a solution of (2.5). Then the integers $lx_0$ and $ly_0$ satisfy (2.4). $\quad\Box$

Lemma 2.12.

Let $n$ and $k$ be integers such that $1 \leq k \leq n$. Then

$$ \binom{n}{k} \geq k^{-k} (n-k)^k.$$

(2.6)

Proof.

For $k=n$ inequality (2.6) holds. Let $1\leq k < n$. Then

$$\binom{n}{k}=\frac{n!}{k!\,(n-k)!}=\frac{n(n-1)\dots(n-k+1)}{k!}\geq \frac{(n-k)^k}{k!}\geq k^{-k}(n-k)^k.\ \quad\Box$$

Lemma 2.13 (see [3, Ch. 0]).

Let $x$ and $z$ be real numbers such that $2 \leq z \leq x/2$. Then

$$\Phi (x,z) \geq c_0 \frac{x}{\ln z},$$

where $c_0>0$ is an absolute constant.

3. Lemmas on Dirichlet characters

In this section we give some well-known lemmas on Dirichlet characters which will be used in the following sections.

Lemma 3.1.

Let $a,$ $b,$ and $n$ be integers such that $1\leq a<b,$ $a\mid b,$ and $(n,a)=1$. Then there is an integer $t$ such that $(n+ta,b)=1$.

Proof.

If $(n,b)=1$, we take $t=0$. Let $(n,b)>1$. Then the set $\Omega = \{ p\mid b \colon\, p \nmid a\}$ is nonempty. Let $\Omega = \{ q_1,\dots,q_r\}$ with $q_1<\dots<q_r$. Let $1 \leq i \leq r$. Since $(a,q_i)=1$, the congruence

$$n + t a\equiv 1\pmod{q_i}$$

has a solution; i.e., there is an integer $m_i$ such that $n+a m_i \equiv 1 \pmod{q_i}$. Consider the system

$$ \begin{cases} \displaystyle t\equiv m_1\pmod{q_1}, \\ \displaystyle \dots\dots\dots\dots\dots\dots \\ \displaystyle t\equiv m_r\pmod{q_r}. \end{cases}$$

(3.1)

Since the numbers $q_1,\dots,q_r$ are coprime, the system has a solution. Let an integer $t_0$ satisfy system (3.1). We claim that $t_0$ is a desired number, i.e., that $(n+t_0a,b)=1$. Assume the contrary: $(n+t_0a,b)>1$. Then there is a prime $p$ such that $p\mid b$ and $p\mid (n+t_0a)$. If $p\nmid a$, then $p\in\Omega$, i.e., $p=q_i$ for some $1 \leq i \leq r$. However,

$$n+t_0a \equiv 1\pmod{q_i}$$

and hence $p\nmid (n+t_0a)$. We arrive at a contradiction. Thus this case is impossible. Hence, $p\mid a$. Since $p\mid (n+t_0a)$, we see that $p\mid n$. Hence, $(n,a)>1$. This contradicts the hypothesis of the lemma. Therefore, the assumption $(n+t_0a,b)>1$ is false. Hence, $(n+t_0a,b)=1$. $\quad\Box$

Lemma 3.2.

Let $q\ge 2$ be an integer and $\chi\in X_q$. Suppose that $\chi$ restricted by $(n,q)=1$ has period $q_1$. Then $\chi$ restricted by $(n,q)=1$ also has period $(q,q_1)$.

Proof.

We put $\delta = (q,q_1)$. Let $m$ and $n$ be integers such that $(m,q)=1$, $(n,q)=1$, and $m\equiv n\pmod{\delta}$. We need to prove that $\chi(m) = \chi(n)$. By Lemma 2.11, there are integers $k$ and $l$ such that

$$m+ q_1 k= n + q l.$$

We put $A=m+ q_1 k= n+ql$. Since $(n,q)=1$, we have $(n+ql, q)=1$. Hence, $(A,q) =1$. Since $\chi$ has period $q$, it follows that

$$\chi(A)=\chi(n+ql)=\chi(n).$$

Since $(A,q)=1$, $(m,q)=1$, and $A\equiv m\pmod{q_1}$, we have $\chi(A) = \chi(m)$. Hence, $\chi(m)= \chi(n)$. $\quad\Box$

Lemma 3.3.

Let $q\geq 1$ and $\chi\in X_q$. Then $c(\chi)$ divides $q$.

Proof.

If $q=1$, then $c(\chi)=1$ and the statement is obvious. Let $q\geq 2$. By Lemma 3.2, $\chi$ restricted by $(n,q)=1$ has period $\delta = (c(\chi), q)$. If $c(\chi)$ is not a divisor of $q$, then $\delta< c(\chi)$, which contradicts the definition of the conductor. $\quad\Box$

Lemma 3.4.

Let $q\geq 1$ and $\chi\in X_q$. Then there exists a unique Dirichlet character $\chi^{}_1\in X_{c(\chi)}$ such that

$$ \chi(n)= \begin{cases} \displaystyle \chi^{}_1(n) & \textit{ if }\, \displaystyle (n,q)=1,\\ \displaystyle 0 & \textit{ if }\, \displaystyle (n,q)>1. \end{cases}$$

(3.2)

Furthermore, $\chi^{}_1$ is primitive.

We say that $\chi^{}_1$ induces $\chi$.

Proof of Lemma 3.4 .

I. $\,$Let $q=1$. Then $c(\chi)=1$, $\# X_1=1$, and $\chi^{}_1 = \chi$, so the statement is obvious.

II. $\,$Let $q \geq 2$ and $\chi$ be a primitive character modulo $q$. Then $c(\chi)=q$ and we can take $\chi^{}_1=\chi$. Let us prove the uniqueness. Suppose that there are two different characters $\chi^{}_1, \chi^{}_2\in X_q$ satisfying (3.2). Then for any $n$ such that $(n,q)>1$ we have $\chi^{}_1(n)=0=\chi^{}_2(n)$. For any $n$ such that $(n,q)=1$, we have $\chi^{}_1(n)=\chi(n)=\chi^{}_2(n)$. Therefore, $\chi^{}_1(n)=\chi^{}_2(n)$ for any integer $n$; i.e., $\chi^{}_1=\chi^{}_2$, a contradiction.

III. $\,$Let $q\geq 2$ and $\chi$ be an imprimitive character modulo $q$. Then $1\leq c(\chi) < q$ and by Lemma 3.3 we have $c(\chi)\mid q$. We define $\chi^{}_1$. Let $n\in{\mathbb Z}$. Consider several cases.

If $(n,c(\chi))>1$, then we put $\chi^{}_1(n)=0$.

If $(n,c(\chi))=1$, then by Lemma 3.1 there is an integer $t$ such that $(n+tc(\chi),q)=1$. We put

$$\chi^{}_1(n) = \chi(n+tc(\chi)).$$

The choice of $t$ subject to the indicated condition is immaterial, since $\chi$ restricted by $(n,q)=1$ has period $c(\chi)$. Thus, $\chi^{}_1(n)$ is defined for any integer $n$. We claim that $\chi^{}_1$ is a character modulo $c(\chi)$. By construction,

$$\chi^{}_1(n)=0 \qquad\text{for any }\, n\in{\mathbb Z} \,\text{ such that }\, (n,c(\chi))>1.$$

By Lemma 3.1, there is an integer $t$ such that $(1+tc(\chi),q)=1$. Since the choice of such a $t$ is immaterial, we take $t=0$. We have $\chi^{}_1(1)=\chi(1)=1$. Now we prove that

$$ \chi^{}_1(n+c(\chi))= \chi^{}_1(n) \qquad\text{for all} \quad n\in{\mathbb Z}.$$

(3.3)

If $(n,c(\chi))>1$, then we have $(n+c(\chi),c(\chi))>1$. Hence,

$$\chi^{}_1(n+c(\chi))=0=\chi^{}_1(n).$$

Let $(n,c(\chi))=1$. Then we have $(n+c(\chi), c(\chi))=1$. By Lemma 3.1, there are integers $t_1$ and $t_2$ such that $(n+t_1c(\chi),q)=1$ and $(n+c(\chi)+t_2c(\chi),q)=1$. By construction, we have

$$\chi^{}_1(n) = \chi(n+t_1 c(\chi)) \qquad\text{and}\qquad \chi^{}_1(n+c(\chi)) = \chi(n+c(\chi)+t_2 c(\chi)).$$

Since $\chi$ restricted by $(n,q)=1$ has period $c(\chi)$, we have $\chi(n+t_1 c(\chi))= \chi(n+c(\chi)+t_2 c(\chi))$. Hence, $\chi^{}_1(n) = \chi^{}_1(n+c(\chi))$ and (3.3) is proved. Now we prove that

$$ \chi^{}_1(m n)=\chi^{}_1(m) \chi^{}_1(n) \qquad \text{for all}\quad m, n\in{\mathbb Z}.$$

(3.4)

If $(m,c(\chi))>1$, then we have $(mn, c(\chi))>1$. Hence, $\chi^{}_1(mn)=0$ and $\chi^{}_1(m)=0$. Therefore, relation (3.4) holds. Similarly, (3.4) holds if $(n,c(\chi))>1$. Let $(m,c(\chi))=1$ and $(n,c(\chi))=1$. Then $(mn,c(\chi))= 1$. By Lemma 3.1, there are integers $t_1$, $t_2$, and $t_3$ such that $(m+t_1c(\chi),q)=1$, $(n+t_2c(\chi),q)=1$, and $(mn+t_3c(\chi),q)=1$. We put $m_1 = m+t_1c(\chi)$, $n_1 = n+t_2c(\chi)$, and $u = mn+t_3c(\chi)$. By construction,

$$\chi^{}_1(mn)=\chi(u),\qquad \chi^{}_1(m)=\chi(m_1),\qquad\text{and}\qquad \chi^{}_1(n)=\chi(n_1).$$

Since $\chi$ is a totally multiplicative function, it follows that

$$\chi^{}_1(m) \chi^{}_1(n)=\chi(m_1)\chi(n_1)=\chi(m_1 n_1).$$

Since $(m_1,q)=1$ and $(n_1,q)=1$, we have $(m_1 n_1,q)=1$. It is clear that $m_1 n_1 \equiv u\pmod{c(\chi)}$. Since $\chi$ restricted by $(n,q)=1$ has period $c(\chi)$, we find that $\chi(u)=\chi(m_1 n_1)$. Therefore, $\chi^{}_1(mn)= \chi^{}_1(m)\chi^{}_1(n)$ and (3.4) is proved. Thus, we have proved that $\chi^{}_1$ is a character modulo $c(\chi)$, i.e., $\chi^{}_1 \in X_{c(\chi)}$.

Now we prove that $\chi^{}_1$ satisfies (3.2). It suffices to show that

$$ \chi^{}_1(n)=\chi(n) \qquad\text{if}\quad (n,q)=1.$$

(3.5)

Since $(n,q)=1$, we have $(n,c(\chi))=1$ (see Lemma 3.3). By Lemma 3.1, there is an integer $t$ such that $(n+t c(\chi),q)=1$. By construction $\chi^{}_1(n) = \chi(n+t c(\chi))$. Since $(n+t c(\chi),q)=1$, $(n,q)=1$, and $n+tc(\chi) \equiv n\pmod{c(\chi)}$, we have $\chi(n+t c(\chi))=\chi(n)$. Hence, $\chi^{}_1(n)=\chi(n)$ and (3.5) is proved.

Now we prove that $\chi^{}_1$ is a primitive character. Suppose that there is a positive integer $q_2$ such that $\chi^{}_1$ restricted by $(n,c(\chi))=1$ has period $q_2$. Let $m$ and $n$ be integers such that $(m,q)=1$, $(n,q)=1$, and $m\equiv n\pmod{q_2}$. By Lemma 3.3, we have $(m,c(\chi))=1$ and $(n,c(\chi))=1$. Then (see (3.5))

$$\chi(m)=\chi^{}_1(m)=\chi^{}_1(n)=\chi(n).$$

Hence, $\chi$ restricted by $(n,q)=1$ has period $q_2$. From the definition of a conductor it follows that $q_2 \geq c(\chi)$. Hence, $\chi^{}_1$ is a primitive character.

Now we prove the uniqueness. Suppose that there are two different characters $\chi^{}_1, \chi^{}_2 \in X_{c(\chi)}$ satisfying (3.2). If $(n,c(\chi))>1$, then $\chi^{}_1(n)=0=\chi^{}_2(n)$. Let $(n,c(\chi))=1$. By Lemma 3.1, there is an integer $t$ such that $(n+t c(\chi),q)=1$. Since $\chi^{}_1$ and $\chi^{}_2$ are periodic functions with period $c(\chi)$, we have

$$\chi^{}_1(n) = \chi^{}_1(n+tc(\chi))=\chi(n+tc(\chi))=\chi^{}_2(n+tc(\chi))=\chi^{}_2(n).$$

Thus, $\chi^{}_1(n)=\chi^{}_2(n)$ for any $n\in{\mathbb Z}$, and so $\chi^{}_1=\chi^{}_2$. We obtain a contradiction. The uniqueness is proved. $\quad\Box$

Lemma 3.5.

Let $q>1$ be an integer expressed in the standard form as $q=q_1^{\alpha_1}\dots q_r^{\alpha_r},$ where $q_1<\dots<q_r$ are primes and $\alpha_1,\dots,\alpha_r$ are positive integers. Let $\chi$ be a Dirichlet character modulo $q$. Then there exist unique characters $\chi^{}_i$ modulo $q_i^{\alpha_i},$ $i=1,\dots,r,$ such that

$$ \chi(n)=\chi^{}_1(n)\dots \chi^{}_r(n)\qquad \textit{for all}\quad n.$$

(3.6)

Furthermore, if the character $\chi$ is real, then all characters $\chi^{}_i,$ $i=1,\dots,r,$ are real. If the character $\chi$ is primitive, then all characters $\chi^{}_i,$ $i=1,\dots,r,$ are primitive.

Proof.

For any $1\leq i\leq r$ we take $A_i$ such that

$$ A_i\equiv 1\pmod{q_i^{\alpha_i}}\qquad\text{and}\qquad A_i\equiv 0\pmod{q_j^{\alpha_j}}\quad \text{for any }\ j\neq i,\ \ 1\leq j\leq r.$$

(3.7)

Since the moduli of these congruences are coprime, the system has a solution (see, for example, [7, Ch. 4]). Thus, integers $A_1,\dots,A_r$ are defined.

Let $1\leq i \leq r$ and $n\in{\mathbb Z}$. We put

$$ \chi^{}_i(n)=\chi\Biggl(nA_i+\sum_{1\leq j\leq r,\; j\neq i}A_j\Biggr).$$

(3.8)

It is easy to show that $\chi^{}_i$ is a Dirichlet character modulo $q_i^{\alpha_i}$.

Now we prove that (3.6) holds. Let $n\in{\mathbb Z}$. Setting

$$n_i=nA_i+\sum_{1\leq j \leq r,\; j\neq i}A_j,\qquad i=1,\dots,r,$$

we have

$$\chi^{}_1(n)\dots\chi^{}_r(n)=\chi(n_1)\dots\chi(n_r)=\chi(n_1\dots n_r).$$

From (3.7) we obtain

$$n_1\dots n_r \equiv n\pmod{q_s^{\alpha_s}}\qquad \text{for any}\quad 1\leq s \leq r.$$

Hence, $n_1\dots n_r - n$ is divisible by $q$, i.e.,

$$n_1\dots n_r\equiv n\pmod{q}.$$

Hence, $\chi(n_1\dots n_r)=\chi(n)$ and (3.6) is proved.

Now we prove the uniqueness of the representation of $\chi$ in the form (3.6). Suppose that

$$ \chi(n)= \widetilde{\chi}{} _1(n)\dots \widetilde{\chi}{} _r(n),$$

(3.9)

where $\widetilde{\chi}{} _i$ is a Dirichlet character modulo $q_i^{\alpha_i}$, $i= 1,\dots,r$. Let $1 \leq i \leq r$ and $n\in{\mathbb Z}$. We have (see (3.7))

$$nA_i+\sum_{1\leq j\leq r,\; j\neq i}A_j\equiv 1\pmod{q_s^{\alpha_s}}\qquad \text{for any}\quad 1\leq s \leq r,\quad s\neq i,$$

and

$$nA_i+\sum_{1\leq j\leq r,\; j\neq i}A_j\equiv n\pmod{q_i^{\alpha_i}}.$$

Hence,

$$\widetilde{\chi}{} _s\Biggl(nA_i+\sum_{1\leq j\leq r,\; j\neq i}A_j\Biggr)=1\qquad \text{for any }\quad 1 \leq s \leq r,\quad s\neq i,$$

and

$$\widetilde{\chi}{} _i\Biggl(nA_i+\sum_{1\leq j\leq r,\; j\neq i}A_j\Biggr)= \widetilde{\chi}{} _i(n).$$

From (3.9) we obtain

$$\chi\Biggl(nA_i+\sum_{1\leq j\leq r,\; j\neq i}A_j\Biggr)= \widetilde{\chi}{} _i(n).$$

Therefore (see (3.8)), $\widetilde{\chi}{} _i(n)= \chi^{}_i(n)$. Since this equation holds for any $n\in{\mathbb Z}$, we have $\widetilde{\chi}{} _i = \chi^{}_i$, $i=1,\dots,r$. Thus, the uniqueness of the representation of $\chi$ in the form (3.6) is proved.

We see from (3.8) that if the character $\chi$ is real, then all characters $\chi^{}_i$, $i=1,\dots,r$, are real. We claim that if the character $\chi$ is primitive, then all characters $\chi^{}_i$, $i=1,\dots,r$, are primitive. Assume the contrary: there is an $i$, $1\leq i \leq r$, such that the character $\chi^{}_i$ is imprimitive. Then $c(\chi^{}_i)<q_i^{\alpha_i}$. Since $c(\chi^{}_i)\mid q_i^{\alpha_i}$ (see Lemma 3.3), we have

$$c(\chi^{}_i)=q_i^{\beta},\qquad \beta < \alpha_i.$$

We put

$$\widetilde{q}{} =q_i^{\beta}\prod_{1\leq j \leq r,\; j\neq i}q_j^{\alpha_j}.$$

Let us show that the character $\chi$ restricted by $(n,q)=1$ has period $\widetilde{q}{}$. Take integers $m$ and $n$ such that $(m,q)=(n,q)=1$ and $m\equiv n\pmod{ \widetilde{q}{} \,}$. Let $1\leq j \leq r$, $j\neq i$. Since

$$m\equiv n\pmod{q_j^{\alpha_j}},$$

we have $\chi^{}_j(m)=\chi^{}_j(n)$. Since $(m,q_i^{\alpha_i})=(n,q_i^{\alpha_i})=1$,

$$m \equiv n\pmod{q_i^{\beta}},$$

and $\chi^{}_i$ restricted by $(n,q_i^{\alpha_i})=1$ has period $q_i^{\beta}$, we have $\chi^{}_i(m)=\chi^{}_i(n)$. This implies

$$\chi(m)=\chi^{}_i(m)\prod_{1\leq j \leq r,\; j \neq i}\chi^{}_j(m)=\chi^{}_i(n)\prod_{1\leq j \leq r,\; j \neq i}\chi^{}_j(n)=\chi(n).$$

We have proved that $\chi$ restricted by $(n,q)=1$ has period $\widetilde{q}{}$. But then $c(\chi)\leq \widetilde{q}{} <q$. This contradicts the fact that the character $\chi$ is primitive. Hence, all characters $\chi^{}_i$, $i=1,\dots,r$, are primitive. Lemma 3.5 is proved. $\quad\Box$

Lemma 3.6.

Let $q$ be a positive integer such that there exists a real primitive character $\chi$ modulo $q$. Then the number $q$ is of the form $2^\alpha k,$ where $\alpha\in\{0,\dots,3\}$ and $k\geq 1$ is an odd square-free integer.

Proof.

Modulo $q=1$ there exists a real primitive character; namely, $\chi(n)=1$ for all $n\in{\mathbb Z}$. The number $1$ is of the form $2^\alpha k$; namely, $\alpha = 0$ and $k=1$.

Let $q>1$ be an integer such that there exists a real primitive character $\chi$ modulo $q$. Suppose that $q=p^rs$, where $p\geq 3$ is a prime number, $(p,s)=1$, and $r\geq 2$. Let $\widetilde{q}{} =p^{r-1}s$. We claim that the character $\chi$ restricted by $(n,q)=1$ has period $\widetilde{q}{}$. Let $m$ and $n$ be integers such that $(m,q)=(n,q)=1$ and $m\equiv n\pmod{ \widetilde{q}{} \,}$. We have $m=n+ \widetilde{q}{} \kern1pt t$, $t\in{\mathbb Z}$, and

$$ m^{p^{r-1}}=(n+ \widetilde{q}{} \kern1pt t)^{p^{r-1}}=n^{p^{r-1}}+\sum_{i=1}^{p^{r-1}}\binom{p^{r-1}}{i}( \widetilde{q}{} \kern1pt t)^i n^{p^{r-1}-i} =n^{p^{r-1}}+\sum_{i=1}^{p^{r-1}}A_it^i n^{p^{r-1}-i},$$

(3.10)

where

$$A_i = \binom{p^{r-1}}{i}( \widetilde{q}{} \,)^i.$$

Let $2\leq i \leq p^{r-1}$. Then

$$A_i = \binom{p^{r-1}}{i}(p^{r-1}s)^i=p^rs\binom{p^{r-1}}{i}p^{(i-1)r-i}s^{i-1}.$$

It is clear that $i-1\geq 1$. We claim that

$$ (i-1)r-i\geq 0$$

(3.11)

or, which is equivalent, $i(r-1)\geq r$. Indeed, since $i\geq 2$ and $r\geq 2$, we have

$$i(r-1)\geq 2(r-1)\geq r.$$

Hence, $A_i=p^rsN$, where $N\in{\mathbb N}$. Thus, for any $2\leq i \leq p^{r-1}$,

$$A_i\equiv 0\pmod{q}.$$

We have $A_1=p^{r-1}(p^{r-1}s)=p^rs p^{r-2}$. Since $r\geq 2$, we obtain

$$A_1\equiv 0\pmod{q}.$$

Hence (see (3.10)),

$$m^{p^{r-1}}\equiv n^{p^{r-1}}\pmod{q}.$$

Using the properties of a character, we obtain

$$(\chi(m))^{p^{r-1}}=(\chi(n))^{p^{r-1}}.$$

Since $(m,q)= (n,q)=1$ and the character $\chi$ is real, we have $\chi(m),\chi(n)\in\{-1,1\}$. Since $p\geq3$ is a prime number and $r\geq 2$ is an integer, it follows that $p^{r-1}$ is an odd positive integer. Therefore, if $\chi(m)=1$, then $\chi(n)=1$, while if $\chi(m)=-1$, then $\chi(n)=-1$ as well. Thus, $\chi(m)=\chi(n)$. We have proved that the character $\chi$ restricted by $(n,q)=1$ has period $\widetilde{q}{}$. Consequently,

$$c(\chi) \leq \widetilde{q}{} < q.$$

This contradicts the fact that $\chi$ is a primitive character. Hence, the number $q$ is of the form $2^\alpha k$, where $\alpha \geq 0$ is an integer and $k\geq 1$ is an odd square-free integer.

We claim that $\alpha\leq 3$. Assume the contrary: $\alpha \geq 4$. Let $k=q_1\dots q_r$, where $q_1<\dots<q_r$ are odd primes. By Lemma 3.5, we have

$$ \chi(n)=\chi^{}_1(n)\chi^{}_2(n)\dots\chi^{}_{r+1}(n),$$

(3.12)

where $\chi^{}_1$ is a real primitive character modulo $2^\alpha$ and $\chi^{}_i$ is a real primitive character modulo $q_{i-1}$, $i=2,\dots,r+1$ (if $k=1$, then $\chi^{}_2,\dots,\chi^{}_{r+1}$ are omitted in (3.12)). It is well known (see, for example, [7, Ch. 6]) that if numbers $\nu$ and $\gamma$ run independently through the sets $\{0,1\}$ and $\{0,\dots,2^{\alpha-2}-1\}$ respectively, then $(-1)^{\nu}\cdot5^{\gamma}$ runs (without repetitions) through a reduced residue system modulo $2^\alpha$. Hence, for any $n$ with $(n,2) = 1$ there are unique numbers $\nu(n) \in \{0,1\}$ and $\gamma(n) \in \{0,\dots,2^{\alpha-2}- 1\}$ such that

$$ n \equiv (-1)^{\nu(n)}\cdot 5^{\gamma(n)}\pmod{2^\alpha}.$$

(3.13)

Since $(-1)^2=1$, we have $(\chi^{}_1(-1))^2=1$. Thus,

$$\chi^{}_1(-1)= (-1)^a,\qquad a\in\{0,1\}.$$

It is well known (see, for example, [7, Ch. 6]) that the number $5$ belongs to $2^{\alpha-2}\pmod{2^\alpha}$; in particular, $5^{2^{\alpha-2}}\equiv 1\pmod{2^\alpha}$. Hence,

$$(\chi^{}_1(5))^{2^{\alpha-2}}=1.$$

We obtain

$$\chi^{}_1(5) = \exp\biggl(2\pi i\frac{b}{2^{\alpha-2}} \biggr),\qquad b\in\{0,\dots,2^{\alpha-2}-1\}.$$

We see from (3.13) that if $n$ is such that $(n,2)=1$, then

$$ \chi^{}_1(n)=(-1)^{a\nu(n)}\exp\biggl(2\pi i\frac{b\gamma(n)}{2^{\alpha-2}} \biggr).$$

(3.14)

We claim that $(b,2)=1$. Indeed, assume the contrary: $(b,2)>1$. We show that then $\chi^{}_1$ restricted by $(n,2^\alpha)=1$ has period $2^{\alpha-1}$. Let $m$ and $n$ be integers such that $(m,2^\alpha)=(n,2^\alpha)=1$ and $m\equiv n\pmod{2^{\alpha-1}}$. We have

$$m \equiv (-1)^{\nu(m)}\cdot 5^{\gamma(m)}\pmod{2^\alpha} \qquad\text{and}\qquad n \equiv (-1)^{\nu(n)}\cdot 5^{\gamma(n)}\pmod{2^\alpha}.$$

Since these congruences also hold modulo $2^{\alpha-1}$, we have

$$ (-1)^{\nu(m)}\cdot 5^{\gamma(m)}\equiv (-1)^{\nu(n)}\cdot 5^{\gamma(n)}\pmod{2^{\alpha -1}}.$$

(3.15)

Since $\alpha\geq 4$, we obtain

$$(-1)^{\nu(m)}\cdot 5^{\gamma(m)}\equiv (-1)^{\nu(n)}\cdot 5^{\gamma(n)}\pmod{4}.$$

It is clear that

$$(-1)^{\nu(m)}\cdot 5^{\gamma(m)}\equiv (-1)^{\nu(m)}\pmod{4} \qquad\text{and}\qquad (-1)^{\nu(n)}\cdot 5^{\gamma(n)}\equiv (-1)^{\nu(n)}\pmod{4}.$$

Hence,

$$(-1)^{\nu(m)}\equiv (-1)^{\nu(n)}\pmod{4}.$$

If $\nu(m)=0$, then $\nu(n)=0$; if $\nu(m)=1$, then $\nu(n)=1$. Thus,

$$ \nu(m)=\nu(n).$$

(3.16)

Therefore (see (3.15)),

$$5^{\gamma(m)}\equiv 5^{\gamma(n)}\pmod{2^{\alpha -1}}.$$

Suppose, for definiteness, that $\gamma(m) \geq \gamma(n)$. We have

$$5^{\gamma(n)}\bigl(5^{\gamma(m)-\gamma(n)}-1\bigr)\equiv0\pmod{2^{\alpha -1}}.$$

Since $(5^{\gamma(n)}, 2^{\alpha-1})=1$, we obtain

$$5^{\gamma(m)-\gamma(n)}-1\equiv0\pmod{2^{\alpha -1}}.$$

Hence,

$$5^{\gamma(m)-\gamma(n)}\equiv 1\pmod{2^{\alpha -1}}.$$

Since $5$ belongs to $2^{\alpha-3}\pmod{2^{\alpha-1}}$, we have (see [7, Ch. 6])

$$\gamma(m)-\gamma(n)\equiv 0\pmod{2^{\alpha -3}}.$$

Therefore,

$$ \gamma(m)=\gamma(n)+ 2^{\alpha -3}t,$$

(3.17)

where $t\geq 0$ is an integer. Since $(b,2)>1$, we have

$$ b=2 \widetilde{b}{} ,$$

(3.18)

where $\widetilde{b}{} \geq 0$ is an integer. We obtain (see (3.14) and (3.16)–(3.18))

$$\begin{aligned} \, \chi^{}_1(m)&=(-1)^{a\nu(m)}\exp\biggl(2\pi i\frac{ \widetilde{b}{} \gamma(m)}{2^{\alpha-3}} \biggr) =(-1)^{a\nu(n)}\exp\biggl(2\pi i\frac{ \widetilde{b}{} (\gamma(n)+ 2^{\alpha -3}t)}{2^{\alpha-3}} \biggr) \\[3pt] &=(-1)^{a\nu(n)}\exp\biggl(2\pi i\frac{ \widetilde{b}{} \gamma(n)}{2^{\alpha-3}} \biggr)\exp(2\pi i \widetilde{b}{} t) =(-1)^{a\nu(n)}\exp\biggl(2\pi i\frac{ \widetilde{b}{} \gamma(n)}{2^{\alpha-3}} \biggr)= \chi^{}_1(n). \end{aligned}$$

Thus, we have proved that $\chi^{}_1$ restricted by $(n,2^\alpha)=1$ has period $2^{\alpha-1}$. Hence,

$$c(\chi^{}_1)\leq 2^{\alpha-1}<2^\alpha.$$

This contradicts the fact that $\chi^{}_1$ is a primitive character. Hence, $(b,2)=1$.

For $n=5$ we have $\nu(5)=0$ and $\gamma(5)=1$. Therefore (see (3.14)),

$$\chi^{}_1(5)=\exp\biggl(2\pi i\frac{b}{2^{\alpha-2}} \biggr)= \exp\biggl(\pi i\frac{b}{2^{\alpha-3}} \biggr).$$

Since $\alpha\geq 4$ and $(b,2)=1$, we have $\operatorname{Im}(\chi^{}_1(5))\neq 0$. This contradicts the fact that $\chi^{}_1$ is a real character. Hence, $0 \leq \alpha\leq 3$. Lemma 3.6 is proved. $\quad\Box$

Lemma 3.7.

Let $q_1$ and $q_2$ be positive integers with $q_1 \neq q_2,$ $\chi^{}_1$ be a primitive character modulo $q_1,$ and $\chi^{}_2$ be a primitive character modulo $q_2$. Then $\chi^{}_1 \neq \chi^{}_2$.

Proof.

Assume the contrary: $\chi^{}_1=\chi^{}_2$. Let $m$ and $n$ be integers such that $(m,q_1)=(n,q_1)=1$ and $m\equiv n\pmod{q_2}$. Then

$$\chi^{}_1(m)=\chi^{}_2(m)=\chi^{}_2(n)=\chi^{}_1(n).$$

Hence, $\chi^{}_1$ restricted by $(n,q_1)=1$ has period $q_2$. Hence, $c(\chi^{}_1) \leq q_2$. Since $\chi^{}_1$ is a primitive character modulo $q_1$, we have $c(\chi^{}_1)=q_1$. Thus, $q_1\leq q_2$. Similarly, it can be proved that $q_2 \leq q_1$. Hence, $q_1=q_2$. We have arrived at a contradiction, which means that $\chi^{}_1 \neq \chi^{}_2$. $\quad\Box$

4. Lemmas on $\psi(x,\chi)$

In this section we present some lemmas on $\psi(x,\chi)$. Most of these lemmas are well known. The proof of Lemma 4.6 is based on Maynard’s ideas (see the proof of Theorem 3.2 in [5]). The proof of Lemma 4.9 follows a standard proof of the Bombieri–Vinogradov theorem (see, for example, [1, Ch. 28]).

Lemma 4.1.

Let $u\geq 2$ be a real number, and let $Q\geq 2$ and $W$ be integers with $(W,Q)=1$. Then

$$\psi(u;Q,W) - \frac{u}{ \varphi (Q)} = \frac{1}{ \varphi (Q)} \sum_{\chi \in X_Q} \overline{\chi(W)} \psi'(u,\chi)$$

(the overbar denotes complex conjugation).

Proof.

We define

$$I_{Q,W}(n)= \begin{cases} \displaystyle 1 & \text{ }\, \displaystyle \text{if }\, n\equiv W {\textstyle\pmod{Q}} ,\\ \displaystyle 0 & \text{ }\, \displaystyle \text{otherwise}. \end{cases}$$

Since (see, for example, [1, Ch. 4])

$$\frac{1}{ \varphi (Q)}\sum_{\chi \in X_Q}\overline{\chi(W)}\chi(n)= I_{Q,W}(n),$$

we have

$$\begin{aligned} \, \psi(u;Q,W)&= \sum_{\substack{n\leq u\\[1pt] n\equiv W\pmod{Q}}} \Lambda (n) = \sum_{n \leq u} \Lambda (n) I_{Q,W}(n) =\sum_{n \leq u} \Lambda (n) \frac{1}{ \varphi (Q)}\sum_{\chi \in X_Q}\overline{\chi(W)}\chi(n) \\[3pt] &=\frac{1}{ \varphi (Q)}\sum_{\chi \in X_Q}\overline{\chi(W)}\Biggl(\,\sum_{n\leq u} \Lambda (n) \chi(n)\Biggr) = \frac{1}{ \varphi (Q)}\sum_{\chi \in X_Q}\overline{\chi(W)}\psi(u,\chi). \end{aligned}$$

Let $\chi^{}_0$ be the principal character modulo $Q$. Since $(W,Q)=1$, it follows that $\chi^{}_0(W)=1$. We have

$$\sum_{\chi \in X_Q} \overline{\chi(W)} E_{\chi^{}_0}(\chi) u = \overline{\chi^{}_0(W)} u = u.$$

Hence,

$$\psi(u;Q,W) - \frac{u}{ \varphi (Q)} = \frac{1}{ \varphi (Q)}\sum_{\chi \in X_Q}\overline{\chi(W)} \bigl(\psi(u,\chi) - E_{\chi^{}_0}(\chi) u\bigr) =\frac{1}{ \varphi (Q)}\sum_{\chi \in X_Q} \overline{\chi(W)} \psi'(u,\chi).$$

Lemma 4.1 is proved. $\quad\Box$

Lemma 4.2 (see, for example, [1, Ch. 14]).

There is a positive absolute constant $a>0$ such that if $\chi$ is a complex character modulo $q,$ then $L(s,\chi)$ has no zeros in the region

$$\Omega \colon\, \qquad \sigma \geq \begin{cases} \displaystyle 1- \frac{a}{\ln(q|t|)} & \textit{ if }\, \displaystyle |t|\geq 1,\\ \displaystyle 1- \frac{a}{\ln q} & \textit{ if }\, \displaystyle |t|<1 \end{cases}$$

(here $s=\sigma + i t,$ $\sigma = \operatorname{Re} s,$ and $t=\operatorname{Im} s$). If $\chi$ is a real nonprincipal character modulo $q,$ the only possible zero of $L(s,\chi)$ in this region is a single (simple) real zero. Furthermore, $L(s,\chi)$ can have a zero in the region $\Omega$ for at most one of the real nonprincipal characters $\chi\pmod{q}$.

Remark.

It is easy to see that the constant $a$ can be replaced by any constant $a^*$ such that $0<a^*<a$.

Lemma 4.3 (see [1, Ch. 20]).

Let $\chi$ be a nonprincipal character modulo $q$ and $2\leq T\leq u$. Then

$$\psi(u,\chi)=-\frac{u^{\beta_1}}{\beta_1}+ R_4(u,T),$$

where

$$|R_4(u,T)| \leq C\biggl(u \ln^2(qu)\exp\biggl(-\frac{a\ln u}{\ln(qT)} \biggr)+uT^{-1}\ln^2(qu)+u^{1/4}\ln u\biggr).$$

Here $C>0$ is an absolute constant and $a>0$ is the absolute constant in Lemma 4.2. The term $-u^{\beta_1}/\beta_1$ should be omitted unless $\chi$ is a real character for which $L(s,\chi)$ has a zero $\beta_1$ (which is necessarily unique, real, and simple) satisfying

$$\beta_1 > 1- \frac{a}{\ln q}.$$

Lemma 4.4 (Page’s theorem; see, for example, [1, Ch. 14]).

There are absolute constants $a_1>0$ and $a'_1>0$ such that the following holds. Let $z\geq 3$ be a real number. Then there is at most one real primitive character $\chi$ to a modulus $q_0,$ $3\leq q_0 \leq z,$ for which $L(s,\chi)$ has a real zero $\beta$ satisfying

$$\beta> 1 - \frac{a_1}{\ln z}.$$

If such a character $\chi$ exists, then

$$q_0 \geq \frac{a'_1(\ln z)^2}{(\ln\ln z)^4}.$$

Such a modulus $q_0$ is said to be an exceptional modulus in the interval $[3,z]$.

Lemma 4.5.

Let $z\geq 3$ be a real number. If an exceptional modulus $q_0$ in the interval $[3,z]$ exists, then the number $q_0$ is of the form $2^\alpha k,$ where $\alpha\in\{0,\dots,3\}$ and $k\geq 1$ is an odd square-free integer.

Proof.

Suppose an exceptional modulus $q_0$ in the interval $[3,z]$ exists. In particular, this means that there exists a real primitive character $\chi$ modulo $q_0$. By Lemma 3.6, the number $q_0$ is of the form $2^\alpha k$ with $\alpha\in\{0,\dots,3\}$ and an odd square-free integer $k\geq 1$. $\quad\Box$

Lemma 4.6.

There are positive absolute constants $c_0,$ $c_1,$ $\gamma_0,$ and $C$ such that the following holds. Let $x\geq c_0$ be a real number, $q_0$ be an exceptional modulus in the interval $[3,\exp(2c_1\sqrt{\ln x})],$ $Q$ be an integer such that $3\leq Q\leq \exp(2 c_1 \sqrt{\ln x})$ and $Q\neq q_0$ (the last inequality should be interpreted as follows: if $q_0$ exists, then $Q\neq q_0;$ if $q_0$ does not exist, then $Q$ is any integer in the indicated interval), and $\chi$ be a primitive character modulo $Q$. Then

$$\max_{2\leq u\leq x^{1+\gamma_0/\sqrt{\ln x}}}|\psi(u,\chi)|\leq C x\exp(-3c_1\sqrt{\ln x}).$$

Proof.

We will choose $c_1$ and $\gamma_0$ later. The number $c_0$ depends on $c_1$ and $\gamma_0$ and is large enough, and $x\geq c_0(c_1,\gamma_0)$. We put

$$z=\exp(2 c_1 \sqrt{\ln x}).$$

We have $z\geq 3$ if the number $c_0(c_1,\gamma_0)$ is chosen large enough. By Lemma 4.4, there is at most one real primitive $\chi$ to a modulus $q_0$, $3\leq q_0 \leq z$, for which $L(s,\chi)$ has a real zero $\beta$ satisfying

$$ \beta > 1-\frac{a_1}{\ln z}= 1-\frac{a_1}{2c_1 \sqrt{\ln x}}.$$

(4.1)

If such a character $\chi$ exists, then

$$ q_0 \geq \frac{a'_1(\ln z)^2}{(\ln\ln z)^4}= \frac{a'_1(2c_1\sqrt{\ln x})^2}{((1/2)\ln\ln x+\ln(2c_1))^4} \geq \frac{a'_1c_1^2\ln x}{(\ln\ln x)^4}$$

(4.2)

provided that $c_0(c_1,\gamma_0)$ is chosen large enough. Let $Q$ be an integer such that $3 \leq Q\leq \exp(2 c_1 \sqrt{\ln x})$ and $Q\neq q_0$, and let $\chi$ be a primitive character modulo $Q$. Since $Q>1$, we see that $\chi$ is a nonprincipal character. By Lemma 4.3, if $2 \leq T \leq u$, then

$$ \psi(u,\chi)=-\frac{u^{\beta_1}}{\beta_1}+ R_4(u,T),$$

(4.3)

where

$$ \begin{aligned} \, |R_4(u,T)| &\leq C\biggl(u \ln^2(Qu)\exp\biggl(-\frac{a\ln u}{\ln(QT)} \biggr)+uT^{-1}\ln^2(Qu)+u^{1/4}\ln u\biggr) \\[4pt] &= C(\Delta_1 + \Delta_2 +\Delta_3). \end{aligned}$$

(4.4)

The term $-u^{\beta_1}/\beta_1$ is to be omitted unless $\chi$ is a real character modulo $Q$ for which $L(s,\chi)$ has a zero $\beta_1$ (which is necessarily unique, real, and simple) satisfying

$$\beta_1 > 1- \frac{a}{\ln Q}.$$

Let

$$2 \leq u \leq x^{1+ \gamma_0/\sqrt{\ln x}}.$$

Let $u \geq c_2(c_1)$, where $c_2(c_1)>0$ is a number depending only on $c_1$. We choose

$$ T= \exp(4 c_1 \sqrt{\ln u}).$$

(4.5)

Then $2 \leq T \leq u$ if $c_2(c_1)$ is chosen large enough.

I. $\,$Now we estimate the quantity

$$\Delta_1 = u \ln^2(Qu)\exp\biggl(-\frac{a\ln u}{\ln(QT)} \biggr).$$

If $c_0(c_1,\gamma_0)$ is chosen large enough, then

$$ 1+\frac{\gamma_0}{\sqrt{\ln x}}\leq 2.$$

(4.6)

Hence,

$$\begin{gathered} \, \ln u \leq \biggl( 1+\frac{\gamma_0}{\sqrt{\ln x}} \biggr)\ln x\leq 2 \ln x, \\[4pt] QT \leq \exp\bigl(2 c_1 \sqrt{\ln x} +4 c_1 \sqrt{\ln u}\,\bigr)\leq\exp\bigl(10 c_1 \sqrt{\ln x}\,\bigr),\nonumber\\[4pt] \ln(QT) \leq 10 c_1 \sqrt{\ln x},\qquad -\frac{a \ln u}{\ln(QT)}\leq -\frac{a\ln u}{10 c_1 \sqrt{\ln x}}.\nonumber \end{gathered}$$

(4.7)

If $c_0(c_1,\gamma_0)$ is chosen large enough, then

$$\ln Q \leq 2 c_1 \sqrt{\ln x} \leq \ln x.$$

Therefore,

$$ \ln^2(Qu)\leq 2\bigl(\ln^2 Q + \ln^2 u\bigr)\leq 10 \ln^2 x=10\exp(2\ln\ln x).$$

(4.8)

We have

$$\Delta_1 \leq 10 \kern1pt u\exp\biggl(-\frac{a\ln u}{10c_1\sqrt{\ln x}} + 2\ln\ln x \biggr).$$

Consider two cases.

(1) Let $x^{1/4} \leq u \leq x^{1+ \gamma_0/\sqrt{\ln x}}$. Then

$$\frac{\ln x}{4} \leq \ln u \leq \biggl( 1+ \frac{\gamma_0}{\sqrt{\ln x}} \biggr)\ln x\leq 2 \ln x.$$

Let

$$0<c_1 \leq \sqrt{\frac{a}{160}}\qquad\Rightarrow\qquad -\frac{a}{40 c_1}\leq -4 c_1.$$

Hence,

$$-\frac{a \ln u}{10 c_1 \sqrt{\ln x}} \leq -\frac{(a/4) \ln x}{10 c_1 \sqrt{\ln x}}=-\frac{a \sqrt{\ln x}}{40 c_1}\leq - 4c_1 \sqrt{\ln x}$$

and

$$-\frac{a \ln u}{10 c_1 \sqrt{\ln x}} + 2\ln\ln x \leq -4 c_1 \sqrt{\ln x} + 2\ln\ln x\leq -\frac72 c_1 \sqrt{\ln x}$$

provided that $c_0(c_1,\gamma_0)$ is chosen large enough. If $0<\gamma_0 \leq c_1/2$, then

$$\Delta_1 \leq 10 \kern1pt x^{1+\gamma_0/\sqrt{\ln x}}\exp\biggl(-\frac72 c_1 \sqrt{\ln x}\biggr) =10 \kern1pt x\exp\biggl(-\frac72 c_1 \sqrt{\ln x} + \gamma_0 \sqrt{\ln x}\biggr)\leq 10 \kern1pt x\exp(-3 c_1\sqrt{\ln x}).$$

(2) Let $c_2(c_1) \leq u < x^{1/4}$ (we may assume that $c_0(c_1,\gamma_0)>(c_2(c_1))^4$ and $c_2(c_1)\geq 10$). We have

$$\begin{aligned} \, \Delta_1 &\leq 10 \kern1pt u\exp\biggl(-\frac{a\ln u}{10c_1\sqrt{\ln x}} + 2\ln\ln x\biggr) \leq 10 \kern1pt u\exp(2\ln\ln x) \\[4pt] &\leq 10 \kern1pt x^{1/4}\exp(2\ln\ln x) \leq 10 \kern1pt x\exp(-3 c_1\sqrt{\ln x}) \end{aligned}$$

provided that $c_0(c_1,\gamma_0)$ is chosen large enough.

Thus, if $0<c_1<\sqrt{a/160}$, $0<\gamma_0 \leq c_1/2$, $x \geq c_0(c_1,\gamma_0)$, and $c_2(c_1) \leq u \leq x^{1+\gamma_0/\sqrt{\ln x}}$, then

$$\Delta_1 \leq 10 \kern1pt x\exp(-3 c_1\sqrt{\ln x}).$$

II. $\,$Now we estimate the quantity

$$\Delta_2 = u T^{-1} \ln^2(Qu).$$

From (4.5) and (4.8) we obtain

$$\Delta_2 \leq 10 \kern1pt u\exp\bigl(-4 c_1\sqrt{\ln u} + 2\ln\ln x\bigr).$$

Consider two cases.

(1) Let $x^{9/10} \leq u \leq x^{1+ \gamma_0/\sqrt{\ln x}}$. Then

$$\frac{9}{10} \ln x \leq \ln u \leq \biggl( 1+\frac{\gamma_0}{\sqrt{\ln x}} \biggr)\ln x \leq 2 \ln x,\qquad -4c_1 \sqrt{\ln u}\leq -4 c_1 \sqrt{\frac{9}{10} \ln x}< - \frac{15}{4} c_1 \sqrt{\ln x}.$$

Since $0<\gamma_0 \leq c_1/2$, we have

$$\begin{aligned} \, \Delta_2&\leq 10 \kern1pt x^{1+ \gamma_0/\sqrt{\ln x}}\exp\biggl(-\frac{15}{4} c_1 \sqrt{\ln x} + 2\ln\ln x\biggr) = 10 \kern1pt x\exp\biggl(- \frac{15}{4} c_1 \sqrt{\ln x} + 2\ln\ln x+ \gamma_0 \sqrt{\ln x}\biggr) \\[3pt] &\leq 10 \kern1pt x\exp\biggl(- \frac{13}{4} c_1 \sqrt{\ln x} + 2\ln\ln x\biggr) \leq 10 \kern1pt x\exp(- 3 c_1 \sqrt{\ln x}) \end{aligned}$$

provided that $c_0(c_1,\gamma_0)$ is chosen large enough.

(2) Let $c_2(c_1) \leq u < x^{9/10}$. Then

$$\begin{aligned} \, \Delta_2 &\leq 10 \kern1pt u\exp\bigl(-4 c_1\sqrt{\ln u} + 2\ln\ln x\bigr) \leq 10 \kern1pt u\exp(2\ln\ln x) \\[4pt] &\leq 10 \kern1pt x^{9/10}\exp( 2\ln\ln x) \leq 10 \kern1pt x\exp(-3 c_1 \sqrt{\ln x}) \end{aligned}$$

provided that $c_0(c_1,\gamma_0)$ is chosen large enough.

Thus, if $0<c_1<\sqrt{a/160}$, $0<\gamma_0 \leq c_1/2$, $x \geq c_0(c_1,\gamma_0)$, and $c_2(c_1) \leq u \leq x^{1+\gamma_0/\sqrt{\ln x}}$, then

$$\Delta_2 \leq 10 \kern1pt x\exp(-3 c_1\sqrt{\ln x}).$$

III. $\,$Now we estimate the quantity

$$\Delta_3 = u^{1/4}\ln u.$$

Since (see (4.6) and (4.7))

$$\ln u\leq 2 \ln x \qquad\text{and}\qquad u^{1/4}\leq x^{(1+\gamma_0/\sqrt{\ln x})/4}\leq x^{1/2},$$

we have

$$\Delta_3 \leq 2 x^{1/2}\ln x \leq x\exp(-3 c_1 \sqrt{\ln x})$$

provided that $c_0(c_1,\gamma_0)$ is chosen large enough.

Finally, we obtain the following (see (4.4)): if $0<c_1<\sqrt{a/160}$, $0<\gamma_0 \leq c_1/2$, $x \geq c_0(c_1,\gamma_0)$, and $c_2(c_1) \leq u \leq x^{1+\gamma_0/\sqrt{\ln x}}$, then

$$ |R_4(u,T)| \leq 21 C x\exp(-3 c_1\sqrt{\ln x}),$$

(4.9)

where $C>0$ is an absolute constant.

IV. $\,$Now we estimate the quantity (see (4.3))

$$\Delta_4 = \biggl|-\frac{u^{\beta_1}}{\beta_1}\biggr|.$$

If $\chi$ is not a real character modulo $Q$ for which $L(s,\chi)$ has a zero $\beta_1$ (which is necessarily unique, real, and simple) satisfying

$$\beta_1 > 1- \frac{a}{\ln Q},$$

then the term $-u^{\beta_1}/\beta_1$ in (4.3) is to be omitted, and there is nothing to estimate. Let $\chi$ be such a character. Then $\chi$ is a real primitive character modulo $Q$. Since $Q\neq q_0$, we have (see Lemma 3.7 and (4.1))

$$\beta_1 \leq 1-\frac{a_1}{\ln z}= 1-\frac{a_1}{2c_1 \sqrt{\ln x}}.$$

Hence,

$$|u^{\beta_1}|= u^{\beta_1} \leq u^{1-a_1/ (2c_1 \sqrt{\ln x})}=u\exp\biggl(-\frac{a_1\ln u}{2 c_1 \sqrt{\ln x}} \biggr).$$

By the remark made after Lemma 4.2, we may assume that $0<a < 1/2$. Since $Q\geq 3$, we have

$$\beta_1 > 1 - \frac{a}{\ln Q}> 1-\frac{1}{2\ln 3}> \frac{1}{2}.$$

Hence, $0<1/{\beta_1}\leq 2$. Thus,

$$ \Delta_4 \leq 2 u\exp\biggl(-\frac{a_1\ln u}{2c_1\sqrt{\ln x}} \biggr).$$

(4.10)

Consider two cases.

(1) Let $x^{1/2} \leq u \leq x^{1+\gamma_0/\sqrt{\ln x}}$. We have (see (4.6))

$$\frac{\ln x}{2} \leq \ln u \leq \biggl( 1+\frac{\gamma_0}{\sqrt{\ln x}} \biggr)\ln x \leq 2\ln x.$$

We take

$$0<c_1<\sqrt{\frac{\min\{a,a_1\}}{160}}\qquad\Rightarrow\qquad -\frac{a_1}{4c_1}\leq -\frac72 c_1.$$

Then,

$$-\frac{a_1 \ln u}{2 c_1 \sqrt{\ln x}}\leq -\frac{(a_1/2)\ln x}{2 c_1\sqrt{\ln x}} = -\frac{a_1\sqrt{\ln x}}{4 c_1}\leq -\frac72 c_1 \sqrt{\ln x}.$$

Since $0<\gamma_0 \leq c_1/2$, we obtain (see (4.10))

$$\Delta_4\leq 2 x^{1+\gamma_0/\sqrt{\ln x}}\exp\biggl(-\frac72 c_1 \sqrt{\ln x}\biggr) =2x\exp\biggl(-\frac72 c_1 \sqrt{\ln x}+\gamma_0\sqrt{\ln x}\biggr) \leq 2x\exp(-3 c_1 \sqrt{\ln x}).$$

(2) Let $c_2(c_1) \leq u < x^{1/2}$. Then (see (4.10))

$$\Delta_4 \leq 2 u \leq 2 x^{1/2}\leq 2 x\exp(-3 c_1 \sqrt{\ln x})$$

provided that $c_0(c_1,\gamma_0)$ is chosen large enough. Combining the estimates found at steps I–IV together, we obtain the following (see (4.3) and (4.9)): if $0<c_1<\sqrt{\min\{a,a_1\}/160}$, $0<\gamma_0 \leq c_1/2$, $x \geq c_0(c_1,\gamma_0)$, and $c_2(c_1) \leq u \leq x^{1+\gamma_0/\sqrt{\ln x}}$, then

$$|\psi(u,\chi)| \leq (21 C +2) x\exp(-3 c_1\sqrt{\ln x}),$$

where $C>0$ is an absolute constant.

There is a number $d(c_1)>0$, depending only on $c_1$, such that

$$t\exp(-3 c_1 \sqrt{\ln t})\geq 1\qquad \text{if}\quad t\geq d(c_1).$$

We may assume that $c_0(c_1,\gamma_0) > d(c_1)$. Hence, if $2 \leq u < c_2(c_1)$, then (see (2.1))

$$|\psi(u,\chi)|=\Biggl|\sum_{n\leq u} \Lambda (n)\chi(n)\Biggr|\leq \sum_{n\leq u} \Lambda (n)=\psi(u) \leq b_6 u\leq b_6 c_2(c_1) \leq b_6 c_2(c_1) x\exp(-3 c_1 \sqrt{\ln x}).$$

Thus, if $0<c_1<\sqrt{\min\{a,a_1\}/160}$, $0<\gamma_0 \leq c_1/2$, and $x \geq c_0(c_1,\gamma_0)$, then

$$\max_{2\leq u \leq x^{1+\gamma_0/\sqrt{\ln x}}}|\psi(u,\chi)| \leq \bigl(21 C +2+ b_6 c_2(c_1)\bigr) x\exp(-3 c_1\sqrt{\ln x}),$$

where $C>0$ is an absolute constant. We take

$$c_1=\frac{\sqrt{\min\{a,a_1\}}}{16}\qquad\text{and}\qquad \gamma_0 = \frac{c_1}2= \frac{\sqrt{\min\{a,a_1\}}}{32}.$$

Since $a>0$ and $a_1>0$ are absolute constants, we see that $c_1$, $\gamma_0$, $c_0(c_1,\gamma_0)$ and $c_2(c_1)$ are positive absolute constants. Lemma 4.6 is proved. $\quad\Box$

Lemma 4.7 (see [1, Ch. 19]).

Let $u \geq 2$ be a real number, $Q\geq 2$ be an integer, $\chi \in X_Q,$ and $\chi^{}_1$ be a primitive character modulo $q_1$ inducing $\chi$. Then

$$|\psi'(u,\chi) - \psi'(u, \chi^{}_1)|\leq \ln^2(Qu).$$

Lemma 4.8 (see [1, Ch. 28]).

Let $Q_1,$ $Q_2,$ and $t$ be real numbers such that $1\leq Q_1 < Q_2$ and $t\geq 2$. Then

$$\sum_{Q_1 < Q \leq Q_2} \frac{1}{ \varphi (Q)} \sum_{\chi\in X^*_Q} \max_{2\leq u \leq t} |\psi(u,\chi)| \leq C\ln^4(t Q_2)\biggl( \frac{t}{Q_1}+ t^{5/6} \ln Q_2+ t^{1/2}Q_2\biggr),$$

where $C>0$ is an absolute constant.

Lemma 4.9.

Let $\varepsilon$ and $\delta$ be real numbers such that $0<\varepsilon<1$ and $0<\delta<1/2$. Then there exists a number $c(\varepsilon,\delta)>0,$ depending only on $\varepsilon$ and $\delta,$ such that if $x\in{\mathbb R}$ and $q\in{\mathbb Z}$ satisfy the conditions $x\geq c(\varepsilon,\delta)$ and $1 \leq q \leq (\ln x)^{1-\varepsilon},$ then there is a positive integer $B$ for which the following relations hold:

$$1\leq B \leq \exp(c_1 \sqrt{\ln x}),\qquad 1\leq \frac{B}{ \varphi (B)}\leq 2,\qquad (B,q)=1$$

and

$$\sum_{\substack{1\leq Q\leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\, \max_{2 \leq u \leq x^{1+ \gamma/\sqrt{\ln x}}} \;\max_{W\in{\mathbb Z} \colon\, \, (W,Q)=1}\biggl|\psi(u;Q,W)-\frac{u}{ \varphi (Q)}\biggr| \leq c_2 x\exp(-c_3\sqrt{\ln x}).$$

Here $c_1,$ $\gamma,$ $c_2,$ and $c_3$ are positive absolute constants.

Proof.

Let $c_0$, $c_1$, $\gamma_0$ and $C$ be the positive absolute constants in Lemma 4.6. We will choose $\gamma$ and $c(\varepsilon,\delta)=c(\varepsilon,\delta,\gamma)$ later; they are assumed to be small and large enough, respectively; for now, let $0<\gamma \leq \gamma_0$, $c(\varepsilon,\delta,\gamma) \geq c_0$, and $x \geq c(\varepsilon,\delta,\gamma)$. Let $q_0$ be the exceptional modulus in the interval $[3,\exp(2 c_1 \sqrt{\ln x})]$. If $q_0$ does not exist, then we take $B=1$. If $q_0$ exists, then (see (4.2))

$$q_0 \geq \frac{a'_1c_1^2\ln x}{(\ln\ln x)^4}=\frac{c_4\ln x}{(\ln\ln x)^4},$$

where $c_4>0$ is an absolute constant. We have $q_0\geq 24$ if $c(\varepsilon,\delta,\gamma)$ is chosen large enough. By Lemma 4.5, the number $q_0$ is of the form $2^\alpha k$, where $\alpha\in\{0,\dots,3\}$ and $k\geq 3$ is an odd square-free integer. We put

$$M_1=\frac{q_0}{2^\alpha}\geq \frac{q_0}{8}\geq \frac{c_4\ln x}{8(\ln\ln x)^4}.$$

Let $\tau = (M_1,q)$ and $M_2=M_1/\tau$. Then $(M_2, q)=1$. Since $\tau \leq q \leq (\ln x)^{1-\varepsilon}$, we have

$$M_2=\frac{M_1}{\tau}\geq \frac{M_1}{(\ln x)^{1-\varepsilon}}\geq \frac{c_4\ln x}{8(\ln\ln x)^4 (\ln x)^{1-\varepsilon}} =\frac{c_4(\ln x)^\varepsilon}{8(\ln\ln x)^4}.$$

If $c(\varepsilon,\delta,\gamma)$ is chosen large enough, then $M_2 \geq 3$. Hence, $M_2\geq 3$ is an odd square-free integer. Furthermore, we have $(M_2, q)=1$ and $M_2$ divides $q_0$. Let $B$ be the largest prime divisor of $M_2$. Hence, $B\geq 3$ is a prime number and $B$ divides $q_0$. We have (see Lemma 2.4)

$$\frac{B}{ \varphi (B)} = \frac{B}{B(1-1/B)}=\frac{1}{1-1/B}\leq \frac{1}{1-1/3}=\frac{3}{2}.$$

Thus, $1 \leq B \leq \exp(2c_1 \sqrt{\ln x})$ is an integer, $(B,q)=1$, $1\leq{B}/{ \varphi (B)} \leq 2$, and $B\geq 3$ is a prime divisor of $q_0$ if $q_0$ exists.

Let $u$ be a real number such that $2 \leq u \leq x^{1+\gamma/ \sqrt{\ln x}}$, and let $Q$ and $W$ be integers such that $2\leq Q \leq x^{1/2 - \delta}$, $(Q, B)=1$, and $(W,Q)=1$. By Lemma 4.1, we have

$$\psi(u;Q,W) - \frac{u}{ \varphi (Q)} = \frac{1}{ \varphi (Q)} \sum_{\chi \in X_Q} \overline{\chi(W)} \psi'(u,\chi).$$

Therefore,

$$\biggl|\psi(u;Q,W) - \frac{u}{ \varphi (Q)}\biggr| \leq \frac{1}{ \varphi (Q)} \sum_{\chi \in X_Q}\mathopen|\psi'(u,\chi)|.$$

Since the right-hand side of this inequality does not depend on $W$, we have

$$\max_{W\in{\mathbb Z} \colon\, (W,Q)=1} \kern1pt \biggl|\psi(u;Q,W) - \frac{u}{ \varphi (Q)}\biggr| \leq \frac{1}{ \varphi (Q)} \sum_{\chi \in X_Q}\mathopen|\psi'(u,\chi)|.$$

Let $\chi\in X_Q$, and let $\chi^{}_1$ be a primitive character modulo $q_1$ inducing $\chi$. From Lemma 3.4 and the definition of the inducing character (which is given below Lemma 3.4), we have $q_1 = c(\chi)$, and hence $q_1\mid Q$ (see Lemma 3.3). Applying Lemma 4.7, we find

$$|\psi'(u,\chi)| \leq |\psi'(u, \chi^{}_1)| +\ln^2(Qu).$$

Since $\# X_Q = \varphi (Q)$, we obtain

$$\begin{aligned} \, \max_{W\in{\mathbb Z} \colon\, (W,Q)=1} \kern1pt \biggl|\psi(u;Q,W) - \frac{u}{ \varphi (Q)}\biggr| &\leq \frac{1}{ \varphi (Q)} \sum_{\chi \in X_Q} \bigl(|\psi'(u,\chi^{}_1)|+\ln^2(Qu)\bigr) \\[3pt] &= \ln^2(Qu) + \frac{1}{ \varphi (Q)} \sum_{\chi \in X_Q} \mathopen|\psi'(u,\chi^{}_1)|. \end{aligned}$$

We can assume that

$$ 1+ \frac{\gamma}{\sqrt{\ln x}} \leq 2$$

(4.11)

provided that $c(\varepsilon,\delta,\gamma)$ is chosen large enough. Hence,

$$\begin{gathered} \, 0<\ln u \leq \biggl( 1+ \frac{\gamma}{\sqrt{\ln x}} \biggr) \ln x \leq 2\ln x,\qquad \ln^2u\leq 4\ln^2x,\\[3pt] 0<\ln Q \leq \biggl( \frac{1}{2} - \delta\biggr)\ln x \leq \ln x,\qquad \ln^2Q \leq \ln^2x,\\[3pt] \ln^2(Qu)\leq 2\bigl(\ln^2Q+\ln^2u\bigr)\leq 10 \ln^2x. \end{gathered}$$

We obtain

$$\max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}}\;\max_{W\in{\mathbb Z} \colon\, (W,Q)=1} \kern1pt \biggl|\psi(u;Q,W) - \frac{u}{ \varphi (Q)}\biggr| \leq 10\ln^2x + \frac{1}{ \varphi (Q)} \sum_{\chi \in X_Q} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}}|\psi'(u,\chi^{}_1)|.$$

Therefore,

$$\begin{aligned} \, \nonumber S&=\sum_{\substack{1\leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}} A_Q = A_1 + \sum_{\substack{2\leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}} A_Q \\ \nonumber &\leq A_1 + \sum_{\substack{2\leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\Biggl( 10\ln^2x + \frac{1}{ \varphi (Q)} \sum_{\chi \in X_Q} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi'(u,\chi^{}_1)|\Biggr) \\ \nonumber &\leq 10 \kern1pt x^{1/2 - \delta}\ln^2x+ A_1+ \sum_{\substack{2\leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}} \,\sum_{\chi \in X_Q}\frac{1}{ \varphi (Q)} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi'(u,\chi^{}_1)| \\[2pt] &= 10 \kern1pt x^{1/2 - \delta}\ln^2x+ A_1+ S', \end{aligned}$$

(4.12)

where

$$A_Q:= \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}}\;\max_{W\in{\mathbb Z} \colon\, (W,Q)=1} \kern1pt \biggl|\psi(u;Q,W) -\frac{u}{ \varphi (Q)}\biggr|.$$

Let us estimate the sum $S'$. Let $Q$ be an integer with $2 \leq Q \leq x^{1/2 - \delta}$ and $(Q,B)=1$, let $\chi\in X_Q$, and let $\chi^{}_1$ be the primitive character modulo $q_1$ inducing $\chi$. Since $q_1\mid Q$, we have $1\leq q_1 \leq x^{1/2 - \delta}$ and $(q_1,B)=1$. Hence,

$$\begin{aligned} \, S' &= \sum_{\substack{2\leq Q \leq x^{1/2 - \delta}\\[1pt](Q,B)=1}}\; \sum_{\chi \in X_Q}\frac{1}{ \varphi (Q)} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi'(u,\chi^{}_1)| \\ &\leq \sum_{\substack{1\leq q_1 \leq x^{1/2 - \delta}\\[1pt] (q_1,B)=1}}\; \sum_{\chi^{}_1 \in X^*_{q_1}} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}}|\psi'(u,\chi^{}_1)| \sum_{1 \leq m \leq x^{1/2 - \delta}/q_1}\frac{1}{ \varphi (m q_1)}. \end{aligned}$$

Applying Lemmas 2.5 and 2.9, we obtain

$$\sum_{1 \leq m \leq x^{1/2 - \delta}/q_1}\frac{1}{ \varphi (m q_1)}\leq \frac{1}{ \varphi (q_1)} \sum_{1 \leq m\leq x^{1/2 - \delta}/q_1} \frac{1}{ \varphi (m)} \leq \frac{1}{ \varphi (q_1)} \sum_{1 \leq m \leq x^{1/2}}\frac{1}{ \varphi (m)}\leq \frac{1}{ \varphi (q_1)} C\ln x,$$

where $C>0$ is an absolute constant. We have

$$S' \leq C\ln x \sum_{\substack{1\leq q_1 \leq x^{1/2 - \delta}\\[1pt] (q_1,B)=1}}\frac{1}{ \varphi (q_1)} \sum_{\chi^{}_1 \in X^*_{q_1}} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi'(u,\chi^{}_1)|.$$

Redenoting $q_1$ by $Q$ and $\chi^{}_1$ by $\chi$, we find

$$S' \leq C\ln x \sum_{\substack{1\leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\frac{1}{ \varphi (Q)} \sum_{\chi\in X^*_Q} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi'(u,\chi)| =C\ln x\bigl(S'_1 + S'_2 + S'_3 \bigr), $$

(4.13)

where

$$\begin{gathered} \, S'_1=\sum_{\substack{1\leq Q \leq \ln x\\[1pt] (Q,B)=1}} R_Q,\qquad S'_2=\sum_{\substack{\ln x < Q \leq \exp(c_1\sqrt{\ln x})\\[1pt] (Q,B)=1}}R_Q,\qquad S'_3=\sum_{\substack{\exp(c_1\sqrt{\ln x}) < Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}} R_Q, \\[4pt] R_Q:= \frac{1}{ \varphi (Q)} \sum_{\chi\in X^*_Q} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}}|\psi'(u,\chi)|, \end{gathered}$$

and $c_1>0$ is the absolute constant in Lemma 4.6.

I. $\,$Now we estimate $S'_1$. We have

$$ S'_1 = \sum_{\substack{1\leq Q \leq \ln x\\[1pt] (Q,B)=1}} R_Q \leq R_1 + \sum_{2\leq Q \leq \ln x} R_Q = R_1 + S'_4.$$

(4.14)

(1) Let us estimate $R_1$. Since $\# X^*_1=1$, we have

$$R_1 = \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi'(u,\chi)|,$$

where $\chi\in X^*_1$, i.e., $\chi(n)=1$ for any $n\in{\mathbb Z}$. Since $\chi$ is the principal character modulo $1$, it follows that

$$\psi'(u,\chi) = \psi(u,\chi) - u.$$

We have

$$\psi(u,\chi) = \sum_{n\leq u} \Lambda (n)\chi(n)=\sum_{n\leq u} \Lambda (n)=\psi(u),\qquad \psi'(u,\chi) = \psi(u) - u.$$

It is well known (see, for example, [1, Ch. 18]) that

$$ |\psi(u) - u| \leq C u\, \exp(-c\sqrt{\ln u}),\qquad u \geq 2,$$

(4.15)

where $C>0$ and $c>0$ are absolute constants. Consider two cases.

(i) Let $x^{1/4} \leq u \leq x^{1+ \gamma/\sqrt{\ln x}}$ (we may assume that $c(\varepsilon,\delta,\gamma)>16$). Then (see (4.11))

$$\frac{1}{4} \ln x \leq \ln u \leq \biggl( 1+ \frac{\gamma}{\sqrt{\ln x}} \biggr)\ln x \leq 2 \ln x,\qquad -c\sqrt{\ln u} \leq -\frac{c}{2} \sqrt{\ln x}.$$

Hence,

$$\begin{aligned} \, |\psi'(u,\chi)|&\leq C u\, \exp(-c\sqrt{\ln u})\leq C x^{1+\gamma/\sqrt{\ln x}} \exp\Bigl(-\frac c2\sqrt{\ln x} \kern1pt \Bigr) \\[3pt] &= Cx\exp\Bigl(\Bigl(\gamma-\frac c2\Bigr)\sqrt{\ln x} \kern1pt \Bigr) \leq Cx\exp \Bigl(-\frac c4\sqrt{\ln x} \kern1pt \Bigr) \end{aligned}$$

provided that $0<\gamma \leq c/4$.

(ii) Let $2 \leq u < x^{1/4}$. Then

$$|\psi'(u,\chi)|\leq C u\exp(-c\sqrt{\ln u})\leq Cu \leq C x^{1/4}\leq Cx\exp \Bigl(-\frac c4\sqrt{\ln x} \kern1pt \Bigr)$$

provided that $c(\varepsilon,\delta,\gamma)$ is chosen large enough.

We obtain

$$ R_1 = \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi'(u,\chi)|\leq Cx\exp\Bigl(-\frac c4\sqrt{\ln x} \kern1pt \Bigr).$$

(4.16)

(2) Now we estimate

$$ S'_4 = \sum_{2\leq Q \leq \ln x} \frac{1}{ \varphi (Q)} \sum_{\chi\in X^*_Q} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi'(u,\chi)|.$$

(4.17)

Let $Q$ be an integer such that $2 \leq Q \leq \ln x$, and let $\chi \in X^*_Q$. Then $\chi$ is a nonprincipal character modulo $Q$, and hence $\psi'(u,\chi) = \psi(u,\chi)$. Consider two cases.

(i) Let $x^{1/4} \leq u \leq x^{1+ \gamma/\sqrt{\ln x}}$. Then (see (4.11))

$$\frac{1}{4} \ln x \leq \ln u\leq \biggl( 1+\frac{\gamma}{\sqrt{\ln x}} \biggr)\ln x \leq 2 \ln x.$$

We may assume that $c(\varepsilon,\delta,\gamma) \geq e^{16}$. Hence, $\ln u \geq (\ln x)/4\geq 4$. We have

$$2 \leq Q \leq \ln x \leq 4 \ln u \leq \ln^2 u.$$

Therefore (see, for example, [1, Ch. 22]),

$$|\psi(u,\chi)| \leq C u\exp\bigl(-c(2) \sqrt{\ln u}\bigr),$$

where $C>0$ and $c(2)>0$ are absolute constants. We have

$$-c(2)\sqrt{\ln u} \leq -\frac{c(2)}{2}\sqrt{\ln x}$$

and

$$\begin{aligned} \, |\psi(u,\chi)| &\leq C u\exp\biggl(-\frac{c(2)}2 \sqrt{\ln x}\biggr) \leq C x^{1+\gamma/\sqrt{\ln x}}\exp\biggl(-\frac{c(2)}2 \sqrt{\ln x}\biggr) \\[3pt] &= C x\exp\biggl(\!\biggl(\gamma-\frac{c(2)}2 \biggr) \sqrt{\ln x}\biggr) \leq Cx\exp\biggl(-\frac{c(2)}4 \sqrt{\ln x}\biggr) \end{aligned}$$

provided that $0<\gamma \leq c(2)/4$.

(i) Let $2 \leq u < x^{1/4}$. Then (see (2.1))

$$\psi(u,\chi) = \sum_{n \leq u} \Lambda (n) \chi(n)$$

and

$$|\psi(u,\chi)|\leq \sum_{n \leq u} \Lambda (n) = \psi(u)\leq b_6 u \leq b_6 x^{1/4} \leq Cx\exp\biggl(- \frac{c(2)}4\sqrt{\ln x}\biggr)$$

provided that $c(\varepsilon,\delta,\gamma)$ is chosen large enough. Hence,

$$\max_{2 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi(u,\chi)| \leq Cx\exp\biggl(- \frac{c(2)}4\sqrt{\ln x}\biggr).$$

Substituting this estimate into (4.17) and using the fact that $\#X^*_Q\leq \#X_Q= \varphi (Q)$, we obtain

$$\begin{aligned} \, S'_4 &\leq Cx\exp\biggl(- \frac{c(2)}4\sqrt{\ln x}\biggr)\ln x = Cx\exp\biggl(- \frac{c(2)}4\sqrt{\ln x} + \ln\ln x\biggr) \nonumber\\[3pt] &\leq Cx\exp\biggl(- \frac{c(2)}8\sqrt{\ln x}\biggr) \end{aligned}$$

(4.18)

provided that $c(\varepsilon,\delta,\gamma)$ is chosen large enough.

Substituting (4.16) and (4.18) into (4.14), we find

$$ S'_1 \leq Cx\exp(-c\sqrt{\ln x}),$$

(4.19)

where $C>0$ and $c>0$ are absolute constants.

II. $\,$Now we estimate the quantity

$$S'_3 = \sum_{\substack{\exp(c_1\sqrt{\ln x}) < Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}} \frac{1}{ \varphi (Q)} \sum_{\chi\in X^*_Q} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}}|\psi'(u,\chi)|.$$

Let $Q$ be an integer with $\exp(c_1\sqrt{\ln x}) < Q \leq x^{1/2 - \delta}$ and $(Q,B)=1$, and let $\chi\in X^*_Q$. Since $Q>1$, we see that $\chi$ is a nonprincipal character modulo $Q$. Hence,

$$\psi'(u,\chi) = \psi(u,\chi).$$

We have

$$\begin{aligned} \, S'_3 &= \sum_{\substack{\exp(c_1\sqrt{\ln x}) < Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}} \frac{1}{ \varphi (Q)} \sum_{\chi\in X^*_Q} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi(u,\chi)| \\ &\leq \sum_{\exp(c_1\sqrt{\ln x}) < Q \leq x^{1/2 - \delta}} \frac{1}{ \varphi (Q)} \sum_{\chi\in X^*_Q} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi(u,\chi)|. \end{aligned}$$

Applying Lemma 4.8 with $Q_1 = \exp(c_1\sqrt{\ln x})$, $Q_2= x^{1/2 - \delta}$, and $t=x^{1+\gamma/\sqrt{\ln x}}$, we obtain

$$S'_3 \leq C\ln^4\bigl(x^{3/2 - \delta +\gamma/\sqrt{\ln x}}\bigr)\Bigl(x\exp\bigl((\gamma - c_1)\sqrt{\ln x} \kern1pt \bigr) + x^{(5/6)(1+\gamma/\sqrt{\ln x})} \ln(x^{1/2 - \delta})+ x^{1-\delta+ \gamma/(2\sqrt{\ln x})} \Bigr).$$

We can assume that

$$\frac{\gamma}{\sqrt{\ln x}} \leq \delta \qquad\text{and}\qquad \frac{5}{6}\biggl( 1+\frac{\gamma}{\sqrt{\ln x}} \biggr)\leq \frac9{10}$$

if $c(\varepsilon,\delta,\gamma)$ is chosen large enough. Increasing $C$ if necessary, we have

$$S'_3 \leq C\ln^4 x\,\bigl(x\exp\bigl((\gamma - c_1)\sqrt{\ln x} \kern1pt \bigr)+ x^{9/10} \ln x+ x^{1-\delta/2}\bigr).$$

Then

$$(\gamma - c_1)\sqrt{\ln x} \leq -\frac{c_1}{2}\sqrt{\ln x}$$

provided that $0<\gamma\leq c_1/2$. We obtain

$$\begin{gathered} \, x\exp\bigl((\gamma - c_1)\sqrt{\ln x} \kern1pt \bigr)\ln^4x \leq x\exp\Bigl(-\frac{c_1}2 \sqrt{\ln x}+ 4\ln\ln x\Bigr) \leq x\exp\Bigl(-\frac{c_1}4 \sqrt{\ln x} \kern1pt \Bigr),\\[4pt] x^{9/10}\ln^5x\leq x\exp\Bigl(-\frac{c_1}4\sqrt{\ln x} \kern1pt \Bigr),\qquad x^{1-\delta/2}\ln^4 x \leq x\exp\Bigl(-\frac{c_1}4\sqrt{\ln x} \kern1pt \Bigr) \end{gathered}$$

provided that $c(\varepsilon,\delta,\gamma)$ is chosen large enough. Redenoting $3C$ by $C$ and $c_1/4$ by $c$, we arrive at

$$ S'_3 \leq Cx\exp(-c\sqrt{\ln x}),$$

(4.20)

where $C>0$ and $c>0$ are absolute constants.

III. $\,$Now we estimate the quantity

$$S'_2= \sum_{\substack{\ln x < Q \leq \exp(c_1\sqrt{\ln x})\\[1pt] (Q,B)=1}}\frac{1}{ \varphi (Q)} \sum_{\chi\in X^*_Q} \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}}|\psi'(u,\chi)|.$$

Let $Q$ be an integer with $\ln x < Q \leq \exp(c_1\sqrt{\ln x})$ and $(Q,B)=1$, and let $\chi\in X^*_Q$. Since $Q>1$, we see that $\chi$ is a nonprincipal character modulo $Q$, and hence $\psi'(u,\chi)= \psi(u,\chi)$. We recall that if an exceptional modulus $q_0$ in the interval $[3,\exp(2c_1\sqrt{\ln x})]$ does not exist, then $B=1$; if $q_0$ exists, then $B \geq3$ is a prime divisor of $q_0$, and so $Q\neq q_0$. Since $0<\gamma \leq \gamma_0$ and $c(\varepsilon,\delta,\gamma) \geq c_0$, we see from Lemma 4.6 that

$$\max_{2 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}} |\psi(u,\chi)| \leq Cx\exp(-3c_1\sqrt{\ln x}).$$

Since $\#X^*_Q\leq \#X_Q= \varphi (Q)$, we obtain

$$\begin{aligned} \, S'_2 &\leq \sum_{\substack{\ln x < Q \leq \exp(c_1\sqrt{\ln x})\\[1pt] (Q,B)=1}} Cx\exp(-3c_1\sqrt{\ln x}) \leq Cx\exp(-3c_1\sqrt{\ln x}) \exp(c_1\sqrt{\ln x}) \nonumber\\[2pt] &= Cx\exp(-2c_1\sqrt{\ln x}). \end{aligned}$$

(4.21)

From (4.19)–(4.21) we find

$$ S'_1 + S'_2+ S'_3 \leq \widetilde{C}{} x\exp(- \widetilde{c}{} \sqrt{\ln x}),$$

(4.22)

where $\widetilde{C}{} >0$ and $\widetilde{c}{} > 0$ are absolute constants. Substituting (4.22) into (4.13), we obtain

$$S' \leq C'x\exp\bigl(- \widetilde{c}{} \sqrt{\ln x} + \ln\ln x\bigr)\leq C'x\exp\Bigl(-\frac{ \widetilde{c}{} }2\sqrt{\ln x} \kern1pt \Bigr)$$

provided that $c(\varepsilon,\delta,\gamma)$ is chosen large enough. Redenoting $C'$ by $C$ and $\widetilde{c}{} /2$ by $c$, we arrive at

$$ S' \leq Cx\exp(-c\sqrt{\ln x}),$$

(4.23)

where $C>0$ and $c>0$ are absolute constants.

IV. $\,$We have

$$ x^{1/2 - \delta}\ln^2x\leq x^{1/2}\ln^2x \leq x\exp(-c\sqrt{\ln x})$$

(4.24)

provided that $c(\varepsilon,\delta,\gamma)$ is chosen large enough (here $c>0$ is the absolute constant in (4.23)).

V. $\,$Now we estimate the quantity

$$A_1= \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}}\;\max_{W\in{\mathbb Z}}|\psi(u;1,W) - u|.$$

Let $W\in{\mathbb Z}$. We have

$$\psi(u;1,W)=\sum_{n\leq u,\ n\equiv W\pmod{1}} \Lambda (n) = \sum_{n\leq u} \Lambda (n) = \psi(u).$$

Hence,

$$A_1 = \max_{2\leq u \leq x^{1+\gamma/\sqrt{\ln x}}} | \psi(u) - u|.$$

Using (4.15) and arguing as in cases I(1), (i) and I(1), (ii), we obtain

$$ A_1 \leq Cx\exp(-c\sqrt{\ln x}),$$

(4.25)

where $C>0$ and $c>0$ are absolute constants.

Substituting (4.23)–(4.25) into (4.12), we find

$$S\leq Cx\exp(-c\sqrt{\ln x}),$$

where $C>0$ and $c>0$ are absolute constants. Thus, if $\gamma$ is a sufficiently small positive absolute constant, $x \geq c(\varepsilon,\delta,\gamma)$ is a real number, and $q$ is an integer such that $1\leq q \leq (\ln x)^{1-\varepsilon}$, then there is an integer $B$ such that

$$1\leq B \leq \exp(2 c_1\sqrt{\ln x}),\qquad 1\leq \frac{B}{ \varphi (B)}\leq 2,\qquad (B,q)=1$$

and

$$\sum_{\substack{1\leq Q\leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}} \,\max_{2 \leq u \leq x^{1+ \gamma/\sqrt{\ln x}}} \;\max_{W\in{\mathbb Z} \colon\, (W,Q)=1} \kern1pt \biggl|\psi(u;Q,W)-\frac{u}{ \varphi (Q)}\biggr| \leq C x\exp(-c\sqrt{\ln x}),$$

where $c_1$, $C$, and $c$ are positive absolute constants. Let us redenote $2c_1$ by $c_1$, $C$ by $c_2$, and $c$ by $c_3$. Since $\gamma$ is an absolute constant, we see that the positive number $c(\varepsilon,\delta,\gamma)= c(\varepsilon,\delta)$ depends only on $\varepsilon$ and $\delta$. Lemma 4.9 is proved. $\quad\Box$

Lemma 4.10.

Let $\varepsilon$ and $\delta$ be real numbers such that $0<\varepsilon<1$ and $0<\delta<1/2$. Then there is a number $c(\varepsilon,\delta)>0,$ depending only on $\varepsilon$ and $\delta,$ such that if $x\in{\mathbb R}$ and $q\in{\mathbb Z}$ satisfy the conditions $x\geq c(\varepsilon,\delta)$ and $1 \leq q \leq (\ln x)^{1-\varepsilon},$ then there is a positive integer $B$ for which the following relations hold:

$$1\leq B \leq \exp(c_1 \sqrt{\ln x}),\qquad 1\leq \frac{B}{ \varphi (B)}\leq 2,\qquad (B,q)=1$$

and

$$\sum_{\substack{1\leq Q\leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\, \max_{2 \leq u \leq x^{1+ \gamma/\sqrt{\ln x}}} \;\max_{W\in{\mathbb Z} \colon\, (W,Q)=1} \kern1pt \biggl|\pi(u;Q,W)-\frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr| \leq c_2 x\exp(-c_3\sqrt{\ln x}).$$

Here $c_1,$ $\gamma,$ $c_2,$ and $c_3$ are positive absolute constants.

Proof.

We will choose the number $\widetilde{c}{} (\varepsilon,\delta)$ later; it is assumed to be large enough. Let $\widetilde{c}{} (\varepsilon,\delta) \geq c(\varepsilon,\delta)$, where $c(\varepsilon,\delta)$ is the number in Lemma 4.9. Let $x\in{\mathbb R}$ and $q\in{\mathbb Z}$ be such that $x \geq \widetilde{c}{} (\varepsilon,\delta)$ and $1 \leq q \leq (\ln x)^{1-\varepsilon}$. Then, by Lemma 4.9, there is a positive integer $B$ such that

$$ 1\leq B \leq \exp(c_1 \sqrt{\ln x}),\qquad 1\leq \frac{B}{ \varphi (B)}\leq 2,\qquad (B,q)=1$$

(4.26)

and

$$ \sum_{\substack{1\leq Q\leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\, \max_{2 \leq u \leq x^{1+ \gamma/\sqrt{\ln x}}} \;\max_{W\in{\mathbb Z} \colon\, (W,Q)=1} |R(u;Q,W)| \leq c_2 x\exp(-c_3\sqrt{\ln x}),$$

(4.27)

where

$$R(u;Q,W):= \psi(u;Q,W) - \frac{u}{ \varphi (Q)}$$

and $c_1$, $\gamma$, $c_2$, and $c_3$ are positive absolute constants.

We put

$$ R_1(u;Q,W) := \pi(u;Q,W) - \frac{ \operatorname{li} (u)}{ \varphi (Q)}.$$

(4.28)

Let $Q\in{\mathbb Z}$, $W\in{\mathbb Z}$, and $u\in{\mathbb Z}$ be such that $1 \leq Q \leq x^{1/2 - \delta}$, $(Q,B)=1$, $(W,Q)=1$, and $3\leq u \leq x^{1+\gamma/\sqrt{\ln x}}$. We claim that

$$ |R_1(u;Q,W)|\leq C_1 u^{1/2} + |R(u;Q,W)|+ \sum_{2 \leq n \leq u-1} \frac{|R(n;Q,W)|}{n\ln^2n},$$

(4.29)

where $C_1>0$ is an absolute constant. We define

$$\alpha(n) = \begin{cases} \displaystyle 1 & \text{ }\, \displaystyle \text{if }\, n\equiv W {\textstyle\pmod{Q}} ,\\ \displaystyle 0 & \text{ }\, \displaystyle \text{otherwise} \end{cases}\qquad\text{and}\qquad \pi_1(u;Q,W) = \sum_{n\leq u}\frac{ \Lambda (n) \alpha(n)}{\ln n}.$$

Let us show that

$$ \pi(u;Q,W) = \pi_1(u;Q,W) + \widetilde{R}{} (u;Q,W),\qquad | \widetilde{R}{} (u;Q,W)| \leq C u^{1/2},$$

(4.30)

where $C>0$ is an absolute constant. Let $u \geq 8$. Then

$$\begin{aligned} \, \pi_1(u;Q,W) &= \sum_{p^m \leq u} \frac{\alpha(p^m)\ln p}{m\ln p} =\sum_{1 \leq m \leq \ln u/ \ln 2}\; \sum_{p \leq u^{1/m}} \frac{\alpha(p^m)}{m} \\[3pt] &=\sum_{p \leq u} \alpha(p) + \sum_{2 \leq m \leq \ln u/ \ln 2}\frac{1}{m} \sum_{p \leq u^{1/m}}\alpha(p^m)= S_1 + S_2. \end{aligned}$$

We have

$$S_1 = \sum_{p\leq u,\ p\equiv W\pmod{Q}} 1 = \pi(u;Q,W)$$

and

$$\begin{aligned} \, S_2&\leq \sum_{2 \leq m \leq \ln u/ \ln 2} \frac{u^{1/m}}{m} = \frac{1}{2} u^{1/2} + \sum_{3 \leq m \leq \ln u/ \ln 2} \frac{u^{1/m}}{m} \leq \frac{1}{2} u^{1/2} + \frac{1}{3} u^{1/3}\frac{\ln u}{\ln 2} \\[3pt] &\leq u^{1/2} + u^{1/3} \ln u \leq C' u^{1/2}, \end{aligned}$$

where $C'>0$ is an absolute constant. If $3 \leq u <8$, then

$$\Biggl|\sum_{2 \leq m \leq \ln u/ \ln 2}\frac{1}{m} \sum_{p \leq u^{1/m}}\alpha(p^m)\Biggr| \leq \frac{1}{2} \sum_{p \leq 8^{1/2}}1 + \frac{1}{3} \sum_{p \leq 8^{1/3}}1 = C'' \leq C'' u^{1/2}.$$

Thus, (4.30) is proved.

Since

$$\psi(x;Q,W) = \sum_{m\leq x} \Lambda (m)\alpha(m),$$

we have

$$\begin{aligned} \, \pi_1(u;Q,W) &= \sum_{2 \leq n \leq u} \frac{\psi(n;Q,W) - \psi(n-1;Q,W)}{\ln n} \\[2pt] &=\sum_{2 \leq n \leq u-1}\psi(n;Q,W)\biggl( \frac{1}{\ln n} - \frac{1}{\ln(n+1)} \biggr)+\frac{\psi(u;Q,W)}{\ln u} \\[2pt] &= \sum_{2 \leq n \leq u-1}\biggl( \frac{n}{ \varphi (Q)}+ R(n;Q,W)\!\biggr) \biggl( \frac{1}{\ln n}-\frac{1}{\ln(n+1)} \biggr)+\frac{u}{ \varphi (Q)\ln u}+ \frac{R(u;Q,W)}{\ln u}. \end{aligned}$$

Further,

$$\begin{aligned} \, \sum_{2 \leq n \leq u-1}\frac{n}{ \varphi (Q)}\biggl( \frac{1}{\ln n} - \frac{1}{\ln(n+1)} \biggr) &=\sum_{2 \leq n \leq u-1}\frac{n}{ \varphi (Q)}\intop_n^{n+1}\frac{dt}{t\ln^2t} =\frac{1}{ \varphi (Q)}\sum_{2 \leq n \leq u-1}\intop_n^{n+1}\frac{t - \{t\}}{t\ln^2t}\,dt \\[3pt] &=\frac{1}{ \varphi (Q)} \left(\intop_2^u \frac{dt}{\ln^2t} - \intop_2^u \frac{\{t\}\,dt}{t\ln^2t} \right). \end{aligned}$$

Since

$$\intop_2^u \frac{dt}{\ln^2t}=\intop_2^ut\,d\biggl(-\frac{1}{\ln t} \biggr)= -\frac{t}{\ln t}\biggr|_2^u+\intop_2^u \frac{dt}{\ln t} =-\frac{u}{\ln u} + \frac{2}{\ln 2} + \operatorname{li} (u),$$

we obtain

$$\begin{aligned} \, \pi_1(u;Q,W) &= \frac{u}{ \varphi (Q)\ln u}+ \frac{R(u;Q,W)}{\ln u} -\frac{u}{ \varphi (Q)\ln u} + \frac{2}{ \varphi (Q)\ln 2} + \frac{ \operatorname{li} (u)}{ \varphi (Q)} \\[3pt] &\phantom{={}}{}- \frac{1}{ \varphi (Q)}\intop_2^u \frac{\{t\}}{t\ln^2t}\,dt+ \sum_{2 \leq n \leq u-1}R(n;Q,W) \biggl( \frac{1}{\ln n} - \frac{1}{\ln(n+1)} \biggr). \end{aligned}$$

We have (see (4.30))

$$\pi(u;Q,W) = \frac{ \operatorname{li} (u)}{ \varphi (Q)}+ R_1(u;Q,W),$$

where

$$\begin{aligned} \, R_1(u;Q,W) &= \frac{2}{ \varphi (Q)\ln 2} - \frac{1}{ \varphi (Q)}\intop_2^u \frac{\{t\}}{t\ln^2t}\,dt + \widetilde{R}{} (u;Q,W) + \frac{R(u;Q,W)}{\ln u} \\[3pt] &\phantom{={}}{} +\sum_{2 \leq n \leq u-1}R(n;Q,W)\biggl( \frac{1}{\ln n} - \frac{1}{\ln(n+1)} \biggr). \end{aligned}$$

We can estimate this quantity as

$$ \begin{aligned} \, |R_1(u;Q,W)| &\leq \frac{2}{\ln 2}+ \left|\intop_2^u\frac{\{t\}}{t\ln^2t}\,dt\right| + | \widetilde{R}{} (u;Q,W)| + \frac{|R(u;Q,W)|}{\ln u} \\[3pt] &\phantom{={}}{} +\sum_{2 \leq n \leq u-1}|R(n;Q,W)|\biggl( \frac{1}{\ln n} - \frac{1}{\ln(n+1)} \biggr). \end{aligned}$$

(4.31)

Since $u \geq 3$, we have

$$ \frac{|R(u;Q,W)|}{\ln u} \leq |R(u;Q,W)|.$$

(4.32)

Since

$$\left|\intop_2^u \frac{\{t\}}{t\ln^2t}\,dt\right|\leq \intop_2^u \frac{dt}{t\ln^2t} =-\frac{1}{\ln t}\biggr|_2^u=\frac{1}{\ln 2} - \frac{1}{\ln u}\leq \frac{1}{\ln 2},$$

it follows (see (4.30)) that

$$ \frac{2}{\ln 2}+ \left|\intop_2^u \frac{\{t\}}{t\ln^2t}\,dt\right| + | \widetilde{R}{} (u;Q,W)| \leq \frac{3}{\ln 2} + C u^{1/2}\leq \biggl(C+\frac{3}{\ln 2} \biggr)u^{1/2}.$$

(4.33)

Let $f(x)=-\ln^{-1}x$ and $n \geq 2$ be an integer. By the mean value theorem, there is a $\xi\in (n,n+1)$ such that

$$ \frac{1}{\ln n} - \frac{1}{\ln(n+1)} = f(n+1) - f(n) = f'(\xi)=\frac{1}{\xi \ln^2\xi}\leq \frac{1}{n\ln^2n}.$$

(4.34)

Substituting (4.32)–(4.34) into (4.31), we obtain (4.29). Hence,

$$\begin{aligned} \, \nonumber &\sum_{\substack{1 \leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\,\max_{\substack{3 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}\\[1pt] u\in{\mathbb Z}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}}|R_1(u;Q,W)| \\[2pt] \nonumber &\ \leq \sum_{\substack{1 \leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\,\max_{\substack{3 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}\\[1pt] u\in{\mathbb Z}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}} |R(u;Q,W)| +\sum_{\substack{1 \leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\,\max_{\substack{3 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}\\[1pt] u\in{\mathbb Z}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}} \bigl|C_1 u^{1/2}\bigr| \\[2pt] \nonumber &\ \phantom{={}}{} + \sum_{\substack{1 \leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\,\max_{\substack{3 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}\\[1pt] u\in{\mathbb Z}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}}\, \sum_{2 \leq n \leq u-1} \frac{|R(n;Q,W)|}{n\ln^2n} \\[4pt] &\ = S_1 + S_2 + S_3. \end{aligned}$$

(4.35)

I. $\,$Now we estimate $S_1$. We have (see (4.27))

$$ S_1 \leq \sum_{\substack{1\leq Q\leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\, \max_{2 \leq u \leq x^{1+ \gamma/\sqrt{\ln x}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}} |R(u;Q,W)|\leq c_2 x\exp(-c_3\sqrt{\ln x}).$$

(4.36)

II. $\,$Let us estimate $S_2$. We can assume that

$$\frac{\gamma}{\sqrt{\ln x}} \leq \delta$$

provided that $\widetilde{c}{} (\varepsilon,\delta)$ is chosen large enough. We have

$$ S_2 \leq C_1 x^{1- \delta + \gamma/(2\sqrt{\ln x})}\leq C_1 x^{1- \delta/2} \leq x\exp(-c_3\sqrt{\ln x})$$

(4.37)

provided that $\widetilde{c}{} (\varepsilon,\delta)$ is chosen large enough.

III. $\,$Now we estimate $S_3$. Let $Q$, $W$, $u$, and $n$ be integers such that $1 \leq Q \leq x^{1/2 - \delta}$, $(Q,B)=1$, $(W,Q)=1$, $3 \leq u \leq x^{1+ \gamma/\sqrt{\ln x}}$, and $2\leq n \leq u-1$. Then

$$|R(n;Q,W)| \leq \max_{2 \leq m \leq x^{1+\gamma/\sqrt{\ln x}}}\max_{\substack{V\in{\mathbb Z}\\[1pt] (V,Q)=1}}|R(m;Q,V)|.$$

Hence,

$$\begin{aligned} \, \sum_{2 \leq n \leq u-1} \frac{|R(n;Q,W)|}{n\ln^2n} &\leq \max_{2 \leq m \leq x^{1+\gamma/\sqrt{\ln x}}}\max_{\substack{V\in{\mathbb Z}\\[1pt] (V,Q)=1}}\mathopen|R(m;Q,V)| \sum_{2 \leq n \leq u-1} \frac{1}{n\ln^2n} \\[2pt] &\le c_0 \max_{2 \leq m \leq x^{1+\gamma/\sqrt{\ln x}}} \max_{\substack{V\in{\mathbb Z}\\[1pt] (V,Q)=1}}\mathopen|R(m;Q,V)|,\qquad\text{where}\quad c_0:= \sum_{n=2}^{\infty}\frac{1}{n\ln^2n}<+\infty. \end{aligned}$$

We have

$$\max_{\substack{3 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}\\ u\in{\mathbb Z}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}} \,\sum_{2 \leq n \leq u-1} \frac{|R(n;Q,W)|}{n\ln^2n} \leq c_0 \max_{2 \leq m \leq x^{1+\gamma/\sqrt{\ln x}}}\max_{\substack{V\in{\mathbb Z}\\[1pt] (V,Q)=1}}\mathopen|R(m;Q,V)|.$$

Therefore (see (4.27)),

$$ S_3 \leq c_0 \sum_{\substack{1 \leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\,\max_{2 \leq m \leq x^{1+\gamma/\sqrt{\ln x}}} \max_{\substack{V\in{\mathbb Z}\\[1pt] (V,Q)=1}}\mathopen|R(m;Q,V)| \leq c_0c_2x\exp(-c_3 \sqrt{\ln x}).$$

(4.38)

Substituting (4.36)–(4.38) into (4.35), we obtain (see (4.28))

$$ \sum_{\substack{1\leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\,\max_{\substack{3 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}\\ u\in{\mathbb Z}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}}\biggl|\pi(u;Q,W)- \frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr| \leq c_4x\exp(-c_3 \sqrt{\ln x}),$$

(4.39)

where $c_4 = c_2+1+ c_0 c_2> 0$ is an absolute constant.

Let $Q$ and $W$ be integers such that $1 \leq Q \leq x^{1/2 - \delta}$, $(Q,B)=1$, and $(W,Q)=1$, and let $u$ be a real number with $2 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}$. Consider two cases.

(1) Let $2 \leq u \leq 3$. Then

$$|\pi(u;Q,W)| \leq \pi(u) \leq 2,\qquad \biggl|\frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr|\leq \operatorname{li} (u) \leq \operatorname{li} (3),$$

and so

$$\biggl|\pi(u;Q,W) - \frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr| \leq |\pi(u;Q,W)| + \biggl|\frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr| \leq 2+ \operatorname{li} (3). $$

(4.40)

(2) Let $3< u \leq x^{1+\gamma/\sqrt{\ln x}}$. Then

$$\biggl|\frac{ \operatorname{li} (u) - \operatorname{li} ([u])}{ \varphi (Q)}\biggr| \leq \intop_{[u]}^{[u]+1}\frac{dt}{\ln t}\leq \intop_2^3 \frac{dt}{\ln t} = \operatorname{li} (3).$$

Hence,

$$\begin{aligned} \, \nonumber &\biggl| \pi(u;Q,W) - \frac{ \operatorname{li} (u)}{ \varphi (Q)} \biggr| = \biggl| \pi([u];Q,W) - \frac{ \operatorname{li} (u)}{ \varphi (Q)} - \frac{ \operatorname{li} ([u])}{ \varphi (Q)} + \frac{ \operatorname{li} ([u])}{ \varphi (Q)} \biggr| \\[4pt] &\qquad\leq \biggl| \pi([u];Q,W) - \frac{ \operatorname{li} ([u])}{ \varphi (Q)} \biggr| + \biggl|\frac{ \operatorname{li} (u)- \operatorname{li} ([u])}{ \varphi (Q)} \biggr| \leq \operatorname{li} (3)+ \biggl| \pi([u];Q,W) - \frac{ \operatorname{li} ([u])}{ \varphi (Q)} \biggr|. \end{aligned}$$

(4.41)

From (4.40) and (4.41) we obtain

$$\begin{aligned} \, \nonumber \max_{2 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}}\biggl|\pi(u;Q,W) - \frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr| &\leq \max_{\substack{3 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}\\ u\in{\mathbb Z}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}} \biggl|\pi(u;Q,W) - \frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr| \\[4pt] &\phantom{={}}{} +2 \operatorname{li} (3)+ 2. \end{aligned}$$

(4.42)

We can assume that

$$ x^{1/2} \leq x\exp(-c_3\sqrt{\ln x})$$

(4.43)

provided that $\widetilde{c}{} (\varepsilon,\delta)$ is chosen large enough. From (4.39), (4.42), and (4.43) we obtain

$$\begin{aligned} \, \nonumber &\sum_{\substack{1 \leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\,\max_{2 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}}\biggl|\pi(u;Q,W) - \frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr| \\ \nonumber &\qquad\leq \sum_{\substack{1 \leq Q \leq x^{1/2 - \delta}\\[1pt] (Q,B)=1}}\,\max_{\substack{3 \leq u \leq x^{1+\gamma/\sqrt{\ln x}}\\ u\in{\mathbb Z}}} \max_{\substack{W\in{\mathbb Z}\\[1pt] (W,Q)=1}}\biggl|\pi(u;Q,W) - \frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr|+ (2 \operatorname{li} (3)+2) x^{1/2} \\[3pt] &\qquad\leq \bigl(c_4 + 2 \operatorname{li} (3)+2\bigr)x\exp(-c_3\sqrt{\ln x}). \end{aligned}$$

(4.44)

Thus, if $x \geq \widetilde{c}{} (\varepsilon,\delta)$ is a real number and $q$ is an integer such that $1 \leq q \leq (\ln x)^{1-\varepsilon}$, then there is a positive integer $B$ for which (4.26) and (4.44) hold. Let us redenote $\widetilde{c}{} (\varepsilon,\delta)$ by $c(\varepsilon,\delta)$ and $c_4 + 2 \operatorname{li} (3)+2$ by $c_2$. Lemma 4.10 is proved. $\quad\Box$

5. Proof of Theorem 1.1 and Corollary 1.1

Let us introduce some additional notation. Let ${\mathcal A}$ be a set of integers, ${\mathcal P}$ a set of primes, and $L(n)=l_1 n + l_2$ a linear function with integer coefficients. We define

$$\begin{gathered} \, {\mathcal A} (x)=\bigl\{n\in {\mathcal A} \colon\, \, x\leq n< 2 x\bigr\},\qquad {\mathcal A} (x;q,a)=\bigl\{n\in {\mathcal A} (x) \colon\, \, n\equiv a {\textstyle\pmod{q}} \bigr\},\\[4pt] L( {\mathcal A} )=\{L(n) \colon\, \, n\in {\mathcal A} \},\qquad {\mathcal P} _{L, {\mathcal A} }(x)= L( {\mathcal A} (x))\cap {\mathcal P} ,\qquad {\mathcal P} _{L, {\mathcal A} }(x;q,a)= L( {\mathcal A} (x;q,a))\cap {\mathcal P} ,\\[4pt] \varphi _L(q)=\frac{ \varphi (|l_1|q)}{ \varphi (|l_1|)}. \end{gathered}$$

Let ${\mathcal L} =\{L_1,\dots,L_k\}$ be a set of distinct linear functions $L_i (n) = a_i n+b_i$, $i=1,\dots,k$, with positive integer coefficients. We say such a set is admissible if for every prime $p$ there is an integer $n_p$ such that $\bigl(\prod_{i=1}^kL_i(n_p),p\bigr)=1$.

We focus on sets satisfying the following hypothesis, which is given in terms of $( {\mathcal A} , {\mathcal L} , {\mathcal P} ,B,x,\theta)$, where ${\mathcal L}$ is an admissible set of linear functions, $B\in{\mathbb N}$, $x$ is a large real number, and $0<\theta<1$.

Hypothesis 1.

For $( {\mathcal A} , {\mathcal L} , {\mathcal P} ,B,x,\theta)$ and $k=\# {\mathcal L}$, the following holds.

(1) ${\mathcal A}$ is well distributed in arithmetic progressions:

$$\sum_{1\leq q \leq x^{\theta}}\max_{a\in{\mathbb Z}}\biggl|\# {\mathcal A} (x;q,a) - \frac{\# {\mathcal A} (x)}{q}\biggr|\ll \frac{\# {\mathcal A} (x)}{(\ln x)^{100 k^2}}.$$

(2) The primes in $L( {\mathcal A} )\cap {\mathcal P}$ are well distributed in most arithmetic progressions: for any $L\in {\mathcal L}$ we have

$$\sum_{\substack{1\leq q\leq x^{\theta}\\[1pt] (q,B)=1}} \,\max_{\substack{a\in{\mathbb Z}\\[1pt] (L(a),q)=1}} \biggl|\# {\mathcal P} _{L, {\mathcal A} }(x;q,a) - \frac{\# {\mathcal P} _{L, {\mathcal A} }(x)}{ \varphi _L(q)}\biggr|\ll \frac{\# {\mathcal P} _{L, {\mathcal A} }(x)}{(\ln x)^{100 k^2}}.$$

(3) ${\mathcal A}$ is not too concentrated in any arithmetic progression: for any $1\leq q< x^{\theta}$ we have

$$\max_{a\in{\mathbb Z}}\# {\mathcal A} (x;q,a) \ll \frac{\# {\mathcal A} (x)}{q}.$$

Maynard proved the following result (see [5, Proposition 6.1]).

Proposition 5.1.

Let $\alpha$ and $\theta$ be real numbers such that $\alpha>0$ and $0<\theta <1$. Let ${\mathcal A}$ be a set of integers, ${\mathcal P}$ a set of primes, and ${\mathcal L} =\{L_1,\dots,L_k\}$ an admissible set of $k$ linear functions, and let $B$ and $x$ be integers. Let the coefficients of $L_i(n)= a_in+b_i\in {\mathcal L}$ satisfy $1\leq a_i,b_i\leq x^\alpha$ for all $1 \leq i \leq k,$ and let $k \leq (\ln x)^{1/5}$ and $1 \leq B \leq x^\alpha$. Let $x^{\theta/10}\leq R \leq x^{\theta/3}$. Let $\rho$ and $\xi$ satisfy $k(\ln \ln x)^2/ \ln x \leq \rho, \xi \leq \theta/10,$ and define

$${\mathcal S} (\xi;D) = \bigl\{n\in{\mathbb N} \colon\, \, p\mid n\ \Rightarrow\; (p>x^{\xi}\textit{ or }\, p\mid D)\bigr\}.$$

Then there is a number $C>0$ depending only on $\alpha$ and $\theta$ such that the following holds. If $k \geq C$ and $( {\mathcal A} , {\mathcal L} , {\mathcal P} ,B,x,\theta)$ satisfy Hypothesis 1, then there exist nonnegative weights $w_n=w_n( {\mathcal L} )$ satisfying

$$ w_n \ll (\ln R)^{2 k}\prod_{i=1}^k\;\prod_{p\mid L_i(n),\; p \kern1pt \nmid B}4$$

(5.1)

such that the following statements hold.

(1) We have

$$ \sum_{n\in {\mathcal A} (x)}w_n = \biggl( 1+ O\biggl( \frac{1}{(\ln x)^{1/10}} \biggr)\!\biggr) \frac{B^k}{ \varphi (B)^k}\, {\mathfrak S} _B( {\mathcal L} )\,\# {\mathcal A} (x)(\ln R)^kI_k.$$

(5.2)

(2) For $L(n)=a_Ln+b_L\in {\mathcal L}$ we have

$$\begin{aligned} \, \nonumber \sum_{n\in {\mathcal A} (x)} {\mathbf 1} _{ {\mathcal P} }(L(n))w_n &\geq \biggl( 1+ O\biggl( \frac{1}{(\ln x)^{1/10}} \biggr)\!\biggr) \frac{B^{k-1}}{ \varphi (B)^{k-1}}\, {\mathfrak S} _B( {\mathcal L} ) \frac{ \varphi (a_L)}{a_L}\, \# {\mathcal P} _{L, {\mathcal A} }(x)(\ln R)^{k+1}J_k \\ &\phantom{={}}{} +O\biggl( \frac{B^k}{ \varphi (B)^k}\, {\mathfrak S} _B( {\mathcal L} )\, \# {\mathcal A} (x)(\ln R)^{k-1} I_k\biggr). \end{aligned}$$

(5.3)

(3) For $L(n)=a_0n+b_0\notin {\mathcal L}$ and $D \leq x^\alpha,$ if $\Delta_L\neq 0,$ we have

$$ \sum_{n\in {\mathcal A} (x)} {\mathbf 1} _{ {\mathcal S} (\xi;D)}(L(n))w_n \ll \xi^{-1} \frac{\Delta_L}{ \varphi (\Delta_L)}\frac{D}{ \varphi (D)} \frac{B^k}{ \varphi (B)^k}\, {\mathfrak S} _B( {\mathcal L} )\,\# {\mathcal A} (x) (\ln R)^{k-1} I_k,$$

(5.4)

where

$$\Delta_L=|a_0|\prod_{i=1}^k|a_0b_i - b_0a_i|.$$

(4) For $L\in {\mathcal L}$ we have

$$ \sum_{n\in {\mathcal A} (x)} w_n \sum_{p\mid L(n),\; p<x^\rho,\; p \kern1pt \nmid B} 1 \ll \rho^2 k^4(\ln k)^2 \frac{B^k}{ \varphi (B)^k}\, {\mathfrak S} _B( {\mathcal L} )\,\# {\mathcal A} (x) (\ln R)^kI_k.$$

(5.5)

Here $I_k$ and $J_k$ are quantities depending only on $k,$ and ${\mathfrak S} _B( {\mathcal L} )$ is a quantity depending only on ${\mathcal L} ,$ and these satisfy

$${\mathfrak S} _B( {\mathcal L} ) = \prod_{p \kern1pt \nmid B}\biggl( 1 - \frac{\#\{1\leq n \leq p \colon\, \, p\mid \prod_{i=1}^k L_i(n)\}}{p} \biggr) \biggl( 1-\frac{1}{p} \biggr)^{\!-k} \geq \exp(-ck),$$

(5.6)

$$I_k = \intop_0^{\infty}\dots\intop_0^{\infty} F^2(t_1,\dots,t_k)\,dt_1\dots dt_k\gg (2k\ln k)^{-k},$$

(5.7)

$$J_k = \intop_0^{\infty}\dots\intop_0^{\infty} \left(\intop_0^{\infty} F(t_1,\dots,t_k)\, dt_k\right)^{\!2} dt_1\dots dt_{k-1} \gg \frac{\ln k}{k} I_k$$

(5.8)

for a smooth function $F=F_k \colon\, {\mathbb R}^k\to{\mathbb R}$ depending only on $k$. The implied constants here depend only on $\alpha,$ $\theta,$ and the implied constants from Hypothesis 1. The constant $c$ in inequality (5.6) is positive and absolute.

Proof of Theorem 1.1 .

First we prove the following

Lemma 5.1.

Let $k$ be a positive integer. Let $a,$ $q,$ and $b_1,\dots,b_k$ be positive integers such that $b_1<\dots<b_k$ and $(a,q) =1$. Let $L_i(n) = q n+a+ q b_i,$ $i=1,\dots,k$. Then ${\mathcal L} =\{L_1,\dots,L_k\}$ is an admissible set if and only if for any prime $p$ such that $p\nmid q$ there is an integer $m_p$ with $m_p\not\equiv b_i\pmod{p}$ for all $1\leq i\leq k$.

Proof.

(1) Let ${\mathcal L} =\{L_1,\dots,L_k\}$ be an admissible set. Let $p$ be a prime such that $p \nmid q$. Since ${\mathcal L}$ is an admissible set, there is an integer $n_p$ such that $\bigl(\prod_{i=1}^kL_i(n_p),p\bigr)=1$. Since $(q,p)=1$, there is an integer $q'$ such that $q q' \equiv 1\pmod{p}$. We put $m_p = -(n_p+q'a)$. Let $i$ be an integer with $1 \leq i \leq k$. Since $(q',p)=1$ and $(L_i(n_p),p)=1$, it follows that $(q'L_i(n_p), p)=1$. We have

$$q'L_i(n_p) \equiv -m_p + b_i\pmod{p}.$$

Hence, $m_p \not\equiv b_i\pmod{p}$.

(2) Suppose that for any prime $p$ with $p\nmid q$ there is an integer $m_p$ such that $m_p\not\equiv b_i\pmod{p}$ for all $1 \leq i\leq k$. Let us show that then ${\mathcal L}$ is an admissible set. First we observe that ${\mathcal L} =\{L_1,\dots,L_k\}$ is a set of distinct linear functions $L_i (n) = q n+l_i$, $i=1,\dots,k$, with positive integer coefficients. Thus, we need to prove that for any prime $p$ there is an integer $n_p$ such that $\bigl(\prod_{i=1}^k L_i(n_p),p\bigr)=1$. Let $p$ be a prime number. Consider two cases.

(i) Let $p\mid q$. Since $(a,q)=1$, we have $(a,p)=1$. Let $i$ be an integer with $1 \leq i \leq k$. For any integer $n$ we have

$$L_i(n)\equiv a\pmod{p},$$

and so $L_i(n) \not\equiv 0\pmod{p}$. Hence, $\bigl(\prod_{i=1}^k L_i(n), p\bigr)=1$. Therefore, in this case we may take any integer as $n_p$.

(ii) Let $p\nmid q$. Then $(q,p)=1$, and so there is an integer $c$ such that

$$ qc\equiv a\pmod{p}.$$

(5.9)

By assumption, there is an integer $m_p$ such that $m_p\not\equiv b_i\pmod{p}$ for all $1 \leq i \leq k$. We put $n_p=-m_p - c$. Let $i$ be an integer with $1 \leq i \leq k$. We have

$$n_p+c+ b_i\not\equiv 0\pmod{p}.$$

Since $(q,p)=1$, we obtain

$$qn_p+ qc+ qb_i\not\equiv 0\pmod{p}.$$

In view of (5.9) this yields $L_i(n_p)\not\equiv 0\pmod{p}$. Hence, $(L_i(n_p), p) =1$. Since this holds for all $1 \leq i \leq k$, we have $\bigl(\prod_{i=1}^kL_i(n_p),p\bigr)=1$. Lemma 5.1 is proved. $\quad\Box$

The proof of the following lemma is based on Maynard’s ideas used in the proof of Lemma 8.1 in [5] (the notation $L\in {\mathcal L}$ was explained in the Introduction).

Lemma 5.2.

There are positive absolute constants $c$ and $C$ such that the following holds. Let $x$ and $\eta$ be real numbers with $x\geq c$ and $(\ln x)^{-9/10}\leq \eta \leq 1$. Let $k$ and $a$ be positive integers. Let $b_1,\dots,b_k$ be integers with $1\leq b_i \leq \ln x,$ $i=1,\dots,k$. Let ${\mathcal L} =\{L_1,\dots,L_k\}$ be a set of $k$ linear functions $L_i(n)=an+b_i,$ $i=1,\dots,k$. For $L(n)=an+b,$ $b\in{\mathbb Z},$ we define

$$\Delta_L = a^{k+1}\prod_{i=1}^k|b_i-b|.$$

Then

$$\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }} \frac{\Delta_L}{ \varphi (\Delta_L)}\leq C\eta\ln\ln(a+2)\ln(k+1)\ln x.$$

Proof.

Consider two cases.

(1) Let $k> \ln\ln x$. We can assume that $\ln\ln x \geq 100$ provided that $c$ is chosen large enough. Therefore, $k\geq 100$. Let $b$ be an integer such that $1 \leq b \leq \eta\ln x$ and $L=an+b\notin {\mathcal L}$. Then $\Delta_L\in{\mathbb N}$. Applying Lemma 2.8, we see that

$$ \frac{\Delta_L}{ \varphi (\Delta_L)}\leq c_0\ln\ln(\Delta_L+2),$$

(5.10)

where $c_0>0$ is an absolute constant. Further,

$$\ln \Delta_L = (k+1)\ln a + \sum_{i=1}^k\ln\mathopen|b_i - b|.$$

For any $1 \leq i \leq k$ we have $|b_i - b| \leq \ln x$. Hence,

$$\ln \Delta_L \leq (k+1)\ln a + k \ln\ln x \leq 2 k \ln a + k^2.$$

Since

$$2k\ln a\leq k^2\ln(a+2)\qquad\text{and}\qquad k^2\leq k^2\ln(a+2),$$

we have

$$\ln \Delta_L \leq 2 k^2\ln(a+2).$$

We observe that if $u\geq 2$ and $v\geq 2$ are real numbers, then

$$ u+v\leq uv.$$

(5.11)

Applying (5.11), we obtain

$$\ln(\Delta_L+2)\leq \ln(3\Delta_L)=\ln \Delta_L + \ln 3 \leq 2 k^2\ln(a+2) + 3\leq 6k^2\ln(a+2).$$

Applying (5.11) again, we have

$$\begin{aligned} \, \ln\ln(\Delta_L+2)&\leq \ln 6 + 2\ln k+\ln\ln(a+2) \leq 2+2\ln k+25\ln\ln(a+2) \\[4pt] &\leq 4\ln k+ 25\ln\ln(a+2) \leq 100 \ln k \ln\ln(a+2)\leq 100 \ln(k+1)\ln\ln(a+2). \end{aligned}$$

Substituting this estimate into (5.10), we obtain

$$\frac{\Delta_L}{ \varphi (\Delta_L)} \leq 100 c_0\ln\ln(a+2) \ln(k+1)= c_1 \ln\ln(a+2) \ln(k+1),$$

where $c_1=100 c_0>0$ is an absolute constant. Thus,

$$\begin{aligned} \, \nonumber \sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }} \frac{\Delta_L}{ \varphi (\Delta_L)} &\leq c_1\ln\ln(a+2)\ln(k+1)\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }}1 \leq c_1\ln\ln(a+2)\ln(k+1) [\eta \ln x] \\[3pt] &\leq c_1\eta \ln\ln(a+2)\ln(k+1) \ln x. \end{aligned}$$

(5.12)

(2) Let $1 \leq k \leq \ln\ln x$. For an integer $b$ we define

$$\Delta(b):=\prod_{i=1}^k|b-b_i|.$$

Let $b$ be an integer such that $1\leq b \leq \eta\ln x$ and $L=an+b\notin {\mathcal L}$. Applying Lemmas 2.5 and 2.4, we obtain

$$\frac{\Delta_L}{ \varphi (\Delta_L)} = \frac{a^{k+1}\Delta(b)}{ \varphi (a^{k+1}\Delta(b))} \leq \frac{a^{k+1}}{ \varphi (a^{k+1})}\frac{\Delta(b)}{ \varphi (\Delta(b))} =\frac{a}{ \varphi (a)}\frac{\Delta(b)}{ \varphi (\Delta(b))}.$$

Hence,

$$ S=\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }} \frac{\Delta_L}{ \varphi (\Delta_L)} \leq \frac{a}{ \varphi (a)}\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }} \frac{\Delta(b)}{ \varphi (\Delta(b))} = \frac{a}{ \varphi (a)} \widetilde{S}{} .$$

(5.13)

Applying Lemma 2.7, we have

$$\begin{aligned} \, \nonumber \widetilde{S}{} &=\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }} \frac{\Delta(b)}{ \varphi (\Delta(b))} = \sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }}\,\sum_{d\mid\Delta(b)} \frac{\mu^2(d)}{ \varphi (d)} \\[3pt] &=\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }}\, \sum_{\substack{1\leq d\leq \eta\ln x\\ d\mid\Delta(b)}} \frac{\mu^2(d)}{ \varphi (d)} +\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }}\, \sum_{\substack{d> \eta\ln x\\ d\mid\Delta(b)}}\frac{\mu^2(d)}{ \varphi (d)} = S_1+S_2. \end{aligned}$$

(5.14)

First we estimate the sum $S_2$. Let $b$ and $d$ be positive integers such that $1\leq b \leq \eta\ln x$, $L=an+b\notin {\mathcal L}$, $d> \eta \ln x$, and $d\mid\Delta(b)$. We claim that

$$ \frac{\mu^2(d)}{ \varphi (d)}\leq \frac{\mu^2(d)\sum_{p\mid d}\ln p}{ \varphi (d)\ln(\eta\ln x)}.$$

(5.15)

We can assume that

$$d>\eta \ln x\geq (\ln x)^{1/10}\geq 100$$

provided that $c$ is chosen large enough. If $\mu^2(d)=0$, then inequality (5.15) holds. Let $\mu^2(d)\neq 0$. Then $d$ is square-free. Therefore, $\sum_{p\mid d}\ln p = \ln d$. Inequality (5.15) is equivalent to the inequality

$$\ln(\eta\ln x)\leq \sum_{p\mid d}\ln p = \ln d,$$

which obviously holds. Thus, (5.15) is proved. We have

$$\begin{aligned} \, S_2&= \sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }}\, \sum_{\substack{d> \eta\ln x\\ d\mid\Delta(b)}} \frac{\mu^2(d)}{ \varphi (d)} \leq \sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }}\, \sum_{\substack{d> \eta\ln x\\ d\mid\Delta(b)}} \frac{\mu^2(d)\sum_{p\mid d}\ln p}{ \varphi (d)\ln(\eta\ln x)} \\[2pt] &=\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }}\, \sum_{p\mid \Delta(b)}\frac{\ln p}{\ln(\eta\ln x)} \sum_{\substack{d> \eta\ln x\\ p\mid d,\; d\mid\Delta(b)}}\frac{\mu^2(d)}{ \varphi (d)}. \end{aligned}$$

Let $b\in{\mathbb N}$, $d\in{\mathbb N}$, and $p\in\mathbb{P}$ be such that $1 \leq b \leq \eta\ln x$, $L=an+b\notin {\mathcal L}$, $p\mid \Delta(b)$, $d>\eta\ln x$, $d$ is a multiple of $p$, and $d\mid\Delta (b)$. Then $d=pt$, where $t\in{\mathbb N}$, $t>(\eta\ln x)/p$, and $t\mid\Delta(b)$. We have (see Lemmas 2.5 and 2.4)

$$\varphi (d)= \varphi (pt)\geq \varphi (p) \varphi (t)= (p-1) \varphi (t)\geq \frac{p}{2} \varphi (t).$$

Hence,

$$\frac{\mu^2(d)}{ \varphi (d)}= \frac{\mu^2(pt)}{ \varphi (pt)}\leq \frac{2 \mu^2(pt)}{p \varphi (t)}\leq \frac{2 \mu^2(t)}{p \varphi (t)}.$$

We obtain (see Lemma 2.7)

$$\sum_{\substack{d>\eta\ln x\\ p\mid d,\; d\mid\Delta(b)}} \frac{\mu^2(d)}{ \varphi (d)} \leq \frac{2}{p}\sum_{\substack{t> (\eta\ln x)/p\\ t\mid\Delta(b)}}\frac{\mu^2(t)}{ \varphi (t)} \leq \frac{2}{p}\sum_{t\mid\Delta(b)}\frac{\mu^2(t)}{ \varphi (t)}=\frac{2}{p}\frac{\Delta(b)}{ \varphi (\Delta(b))}.$$

Therefore,

$$S_2\leq \sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }}\, \sum_{p\mid \Delta(b)} \frac{\ln p}{\ln(\eta\ln x)} \frac{2}{p}\frac{\Delta(b)}{ \varphi (\Delta(b))} =\frac{2}{\ln(\eta\ln x)}\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }} \frac{\Delta(b)}{ \varphi (\Delta(b))}\sum_{p\mid \Delta(b)}\frac{\ln p}{p}.$$

Since $\eta\geq (\ln x)^{-9/10}$, we have

$$\frac{2}{\ln(\eta\ln x)}\leq \frac{2}{\ln((\ln x)^{1/10})}=\frac{20}{\ln\ln x}.$$

Thus,

$$ S_2\leq \frac{20}{\ln\ln x}\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }} \frac{\Delta(b)}{ \varphi (\Delta(b))}\sum_{p\mid \Delta(b)}\frac{\ln p}{p}.$$

(5.16)

Let $b$ be an integer such that $1 \leq b \leq \eta\ln x$ and $L=an+b\notin {\mathcal L}$. Applying Lemmas 2.8 and 2.10, we obtain

$$ \frac{\Delta(b)}{ \varphi (\Delta(b))}\leq c_2\ln\ln(\Delta(b)+2)\leq c_2\ln\ln(3\Delta(b)),\qquad \sum_{p\mid \Delta(b)}\frac{\ln p}{p}\leq c_3\ln\ln(3\Delta(b)),$$

(5.17)

where $c_2>0$ and $c_3>0$ are absolute constants. We have

$$ \ln \Delta(b)=\sum_{i=1}^k\ln\mathopen|b_i-b|\leq k \ln\ln x\leq (\ln\ln x)^2.$$

(5.18)

Hence,

$$\ln\ln(3\Delta(b))=\ln(\ln 3 + \ln \Delta(b)) \leq \ln(\ln 3 + (\ln\ln x)^2)\leq 3\ln\ln\ln x $$

(5.19)

provided that $c$ is chosen large enough. It follows from (5.17) and (5.19) that

$$\frac{\Delta(b)}{ \varphi (\Delta(b))}\sum_{p\mid \Delta(b)}\frac{\ln p}{p}\leq 9c_2c_3 (\ln\ln\ln x)^2=c_4(\ln\ln\ln x)^2,$$

where $c_4=9c_2c_3>0$ is an absolute constant. Substituting this estimate into (5.16), we obtain

$$S_2 \leq \frac{20 c_4 (\ln\ln\ln x)^2}{\ln\ln x} \eta\ln x.$$

We can assume that

$$\frac{20 c_4 (\ln\ln\ln x)^2}{\ln\ln x} \leq 1$$

provided that $c$ is chosen large enough. Hence,

$$ S_2 \leq \eta\ln x\leq \frac{1}{\ln 2}\ln(k+1) \eta \ln x\leq 2 \ln(k+1) \eta \ln x.$$

(5.20)

Now we estimate $S_1$. We have

$$\begin{aligned} \, \nonumber S_1 &= \sum_{\substack{1\leq b \leq \eta\ln x\\ L=an+b\notin {\mathcal L} }}\, \sum_{\substack{1\leq d\leq\eta\ln x\\ d\mid\Delta(b)}} \frac{\mu^2(d)}{ \varphi (d)} = \sum_{1\leq d\leq\eta\ln x}\, \sum_{\substack{1\leq b \leq \eta\ln x\\ L=an+b\notin {\mathcal L} \\ d\mid\Delta(b)}} \frac{\mu^2(d)}{ \varphi (d)} =\sum_{1\leq d\leq\eta\ln x}\frac{\mu^2(d)}{ \varphi (d)} \sum_{\substack{1\leq b \leq \eta\ln x\\ L=an+b\notin {\mathcal L} \\ d\mid\Delta(b)}}1 \\ &=\sum_{1\leq d\leq\eta\ln x}\frac{\mu^2(d)}{ \varphi (d)}N_0(d) =\sum_{\substack{1\leq d\leq\eta\ln x\\ d\in {\mathcal M} }}\frac{1}{ \varphi (d)}N_0(d). \end{aligned}$$

(5.21)

Let $d$ be an integer such that $1\leq d \leq \eta\ln x$ and $d\in {\mathcal M}$. We claim that

$$ N_0(d)\leq \frac{2\eta\ln x}{d}\prod_{p\mid d}\min\{p,k\}.$$

(5.22)

If $d=1$, then the inequality is obvious. Let $d>1$. We define

$$R(b)=(b-b_1)\dots(b-b_k).$$

Then $\Delta(b)=|R(b)|$. We have

$$N_0(d)=\sum_{\substack{1\leq b \leq \eta\ln x\\ L=an+b\notin {\mathcal L} \\ d\mid\Delta(b)}}1 =\sum_{\substack{1\leq b \leq \eta\ln x\\ L=an+b\notin {\mathcal L} \\ R(b)\equiv 0\pmod{d}}}1.$$

Let $d$ be expressed in the standard form as $d=q_1\dots q_r$, where $q_1<\dots <q_r$ are prime numbers. It is well known (see, for example, [7, Ch. 4]) that the congruence $R(b)\equiv 0\pmod{d}$ is equivalent to the system of congruences

$$ \begin{cases} \displaystyle R(b)\equiv 0\pmod{q_1}, \\ \displaystyle \dots\dots\dots\dots\dots\dots \\ \displaystyle R(b)\equiv 0\pmod{q_r}. \end{cases}$$

(5.23)

Let $1\leq j \leq r$. Let $\Omega_j$ be the set of numbers of a complete system of residues modulo $q_j$ satisfying the congruence $R(b)\equiv 0\pmod{q_j}$. Since $R(b_1)=0$, we see that $\Omega_j\neq\varnothing$. Since the leading coefficient of the polynomial $R(b)$ is $1$ and the degree of the polynomial $R(b)$ is $k$, we have $\#\Omega_j\leq k$ (see, for example, [7, Ch. 4]). It is also clear that $\#\Omega_j \leq q_j$. Thus,

$$\#\Omega_j \leq \min\{q_j,k\}.$$

System (5.23) is equivalent to the union of $T=\#\Omega_1\dots\#\Omega_r$ systems

$$ \begin{cases} \displaystyle b\equiv \tau_1\pmod{q_1}, \\ \displaystyle \dots\dots\dots\dots\dots\dots \\ \displaystyle b\equiv \tau_r\pmod{q_r}, \end{cases}$$

(5.24)

where $\tau_1\in\Omega_1,\dots,\tau_r\in\Omega_r$. It is well known (see, for example, [7, Ch. 4]) that the system of congruences (5.24) is equivalent to the congruence

$$b\equiv x_0\pmod{d},$$

where $x_0=x_0(\tau_1,\dots,\tau_r)$. It is also known that the numbers $x_0(\tau_1,\dots,\tau_r)$, $\tau_1\in\Omega_1,\dots,\tau_r\in \Omega_r$, are incongruent modulo $d$. Thus,

$$\bigl\{b\in{\mathbb Z} \colon\, \, R(b)\equiv 0 {\textstyle\pmod{d}} \bigr\}=\bigsqcup_{\tau_1\in\Omega_1,\dots,\tau_r\in\Omega_r} \bigl\{x_0(\tau_1,\dots,\tau_r)+ dt \colon\, \, t\in{\mathbb Z}\bigr\}.$$

Let $\tau_1\in\Omega_1,\dots,\tau_r\in\Omega_r$ and $x_0=x_0(\tau_1,\dots,\tau_r)$. We have

$$\begin{aligned} \, \#\bigl\{t\in{\mathbb Z} \colon\, \, 1\leq x_0+dt\leq \eta\ln x\bigr\} &=\biggl[\frac{\eta\ln x-x_0}{d}\biggr]-\biggl\lceil\frac{1-x_0}{d}\biggr\rceil +1 \\[4pt] &=\frac{\eta\ln x-x_0}{d} - \theta_1- \biggl( \frac{1-x_0}{d}+\theta_2\biggr)+1 =\frac{\eta\ln x}{d} + 1 -\theta_1-\theta_2 - \frac{1}{d}, \end{aligned}$$

where $\theta_1$ and $\theta_2$ are real numbers with $0 \leq \theta_1<1$ and $0 \leq \theta_2<1$. Since $1 \leq d \leq \eta\ln x$, we obtain

$$\#\bigl\{t\in{\mathbb Z} \colon\, \, 1\leq x_0+dt\leq \eta\ln x\bigr\}\leq \frac{\eta\ln x}{d} + 1\leq 2\frac{\eta\ln x}{d}.$$

Thus,

$$N_0(d)\leq \frac{2\eta\ln x}{d}T\leq \frac{2\eta\ln x}{d}\prod_{p\mid d}\min\{p,k\}.$$

Inequality (5.22) is proved.

Substituting (5.22) into (5.21), we obtain

$$ S_1\leq \sum_{\substack{1\leq d\leq\eta\ln x\\ d\in {\mathcal M} }}\frac{1}{ \varphi (d)}\frac{2\eta\ln x}{d}\prod_{p\mid d}\min\{p,k\} =2\eta\ln x\sum_{\substack{1\leq d\leq\eta\ln x\\ d\in {\mathcal M} }} \frac{\prod_{p\mid d}\min\{p,k\}}{d \varphi (d)}= 2\eta\ln x\,S_3.$$

(5.25)

Let $d$ be an integer such that $1\leq d \leq \eta\ln x$ and $d\in {\mathcal M}$. We have (see Lemmas 2.6 and 2.4)

$$\begin{gathered} \, d=\prod_{p\mid d}p,\qquad \varphi (d)=\prod_{p\mid d} \varphi (p)=\prod_{p\mid d}(p-1),\\[4pt] \frac{\prod_{p\mid d}\min\{p,k\}}{d \varphi (d)}= \frac{\prod_{p\mid d}\min\{p,k\}}{\prod_{p\mid d}p(p-1)} =\prod_{p\mid d,\; p\leq k}\frac{1}{p-1} \prod_{p\mid d,\; p>k}\frac{k}{p(p-1)}. \end{gathered}$$

Hence,

$$S_3=\sum_{\substack{1\leq d\leq\eta\ln x\\ d\in {\mathcal M} }}\, \prod_{p\mid d,\; p\leq k}\frac{1}{p-1} \prod_{p\mid d,\; p>k}\frac{k}{p(p-1)} \leq \prod_{p\leq k} \biggl( 1+\frac{1}{p-1} \biggr)\prod_{p>k}\biggl( 1+\frac{k}{p(p-1)} \biggr)=AB. $$

(5.26)

We have (see Lemma 2.1)

$$A=\prod_{p\leq k}\biggl( 1+\frac{1}{p-1} \biggr) \leq\prod_{p\leq k+1}\biggl( 1+\frac{1}{p-1} \biggr) =\prod_{p\leq k+1}\biggl( 1-\frac{1}{p} \biggr)^{\!-1} \leq c_5\ln(k+1), $$

(5.27)

where $c_5>0$ is an absolute constant.

Now we estimate $B$. Since $\ln(1+u)\leq u$ and $u>0$, we get

$$\ln B = \sum_{p> k} \ln\biggl( 1+\frac{k}{p(p-1)} \biggr) \leq \sum_{p> k}\frac{k}{p(p-1)} =k\sum_{p\geq k+1}\frac{1}{p(p-1)} \leq k\sum_{n\geq k+1}\frac{1}{n(n-1)}.$$

We define

$$s_m=\sum_{n=k+1}^m\frac{1}{n(n-1)}=\sum_{n=k+1}^m\biggl( \frac{1}{n-1}-\frac{1}{n} \biggr) =\frac{1}{k} - \frac{1}{m},\qquad m\geq k+1.$$

Then,

$$\sum_{n\geq k+1}\frac{1}{n(n-1)}=\lim_{m\to +\infty} s_m=\frac{1}{k}.$$

We obtain $\ln B\leq 1$, i.e.,

$$ B\leq e<3.$$

(5.28)

It follows from (5.26)–(5.28) that $S_3 \leq c_6\ln(k+1)$, where $c_6>0$ is an absolute constant. Substituting this estimate into (5.25), we obtain

$$ S_1 \leq c_7\eta\ln(k+1)\ln x,$$

(5.29)

where $c_7>0$ is an absolute constant. Therefore (see (5.14), (5.20), and (5.29)),

$$\widetilde{S}{} \leq (c_7+2)\eta\ln(k+1)\ln x = c_8\eta\ln(k+1)\ln x,$$

where $c_8=c_7+2>0$ is an absolute constant. We obtain (see (5.13) and Lemma 2.8)

$$ S\leq c_8\frac{a}{ \varphi (a)}\eta\ln(k+1)\ln x\leq c_9\eta\ln\ln(a+2)\ln(k+1)\ln x,$$

(5.30)

where $c_9>0$ is an absolute constant. We put $C=c_1+c_9$, where $c_1$ is the constant in (5.12). Then $C>0$ is an absolute constant and in both cases, $1\leq k \leq \ln\ln x$ and $k>\ln\ln x$, we have

$$\sum_{\substack{1\leq b \leq \eta \ln x\\ L=an+b\notin {\mathcal L} }} \frac{\Delta_L}{ \varphi (\Delta_L)} \leq C \eta\ln\ln(a+2)\ln(k+1)\ln x.$$

Lemma 5.2 is proved. $\quad\Box$

Lemma 5.3.

Let ${\mathcal A} ={\mathbb N},$ ${\mathcal P} =\mathbb{P},$ $\alpha=1/5,$ and $\theta=1/3,$ and let $C_0=C(1/5,1/3)>0$ be the absolute constant in Proposition 5.1. Let $\varepsilon$ be a real number with $0<\varepsilon<1$. Then there is a number $c_0(\varepsilon)>0$ such that the following holds. Let $x\in{\mathbb N},$ $y\in{\mathbb R},$ and $q\in{\mathbb N}$ satisfy the conditions $x\geq c_0(\varepsilon),$ $1\leq y \leq \ln x,$ and $1\leq q \leq y^{1-\varepsilon}$. Then there is a positive integer $B$ such that

$$ 1\leq B \leq \exp(\vartheta\sqrt{\ln x}),\qquad 1\leq\frac{B}{ \varphi (B)}\leq 2,\qquad (B,q)=1.$$

(5.31)

Furthermore, let $k\in{\mathbb N},$ $\rho\in{\mathbb R},$ $\xi\in{\mathbb R},$ $R\in{\mathbb R},$ $\eta\in{\mathbb R},$ and $a\in{\mathbb Z},$ be such that

$$C_0\leq k \leq (\ln x)^{1/5},$$

(5.32)

$$\frac{k(\ln\ln x)^2}{\ln x} \leq \rho \leq \frac{1}{30},\qquad \xi = \rho,$$

(5.33)

$$R=x^{1/9},\qquad 0<\eta \leq \frac{1}{2},$$

(5.34)

$$1\leq a \leq q,\qquad (a,q)=1.$$

(5.35)

Let ${\mathcal L} =\{L_1,\dots,L_k\}$ be an admissible set of $k$ linear functions, where $L_i(n) = q n + a+ q b_i,$ $i=1,\dots,k,$ $b_1,\dots,b_k$ are positive integers with $b_1<\dots<b_k,$ and $q b_k \leq \eta y$. Then the hypothesis of Proposition 5.1 holds and there exist nonnegative weights $w_n= w_n( {\mathcal L} )$ with the properties stated in Proposition 5.1; the implied constants in (5.1)–(5.5) are positive and absolute. In (5.31), $\vartheta>0$ is also an absolute constant.

Proof.

We will choose $c_0(\varepsilon)$ later; this number is assumed to be large enough. We take $\delta =1/10$ and let $c_0(\varepsilon)\geq c(\varepsilon,\delta)=c(\varepsilon,1/10)$, where $c(\varepsilon,\delta)$ is the quantity in Lemma 4.10. Let $x\in{\mathbb N}$, $y\in{\mathbb R}$, and $q\in{\mathbb N}$ be such that $x\geq c_0(\varepsilon)$, $1 \leq y \leq \ln x$, and $1 \leq q \leq y^{1-\varepsilon}$. By Lemma 4.10, there is a positive integer $B$ such that

$$1\leq B \leq \exp(c_1\sqrt{\ln x}),\qquad 1\leq\frac{B}{ \varphi (B)}\leq 2,\qquad (B,q)=1$$

and

$$ \sum_{\substack{1\leq Q\leq x^{2/5}\\[1pt] (Q,B)=1}}\, \max_{2 \leq u \leq x^{1+ \gamma/\sqrt{\ln x}}}\; \max_{W\in{\mathbb Z} \colon\, (W,Q)=1} \biggl|\pi(u;Q,W)-\frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr| \leq c_2 x\exp(-c_3\sqrt{\ln x}),$$

(5.36)

where $c_1$, $\gamma$, $c_2$, and $c_3$ are positive absolute constants. Let (5.32)–(5.35) hold. Let ${\mathcal L} =\{L_1,\dots,L_k\}$ be an admissible set of $k$ linear functions $L_i(n) = q n + a+ q b_i$, $i=1,\dots,k$, where $b_1,\dots,b_k$ are positive integers with $b_1<\dots<b_k$ and $q b_k \leq \eta y$. Let us show that the hypothesis of Proposition 5.1 holds. First we show that the set $( {\mathcal A} , {\mathcal L} , {\mathcal P} ,B,x,1/3)$ satisfies Hypothesis 1.

I. $\,$Let us show that condition (2) of Hypothesis 1 holds. Let $L(n)=l_1 n+l_2\in {\mathcal L}$. Clearly, we have

$$ 1\leq l_1 \leq \ln x \qquad\text{and}\qquad 1\leq l_2 \leq \ln x.$$

(5.37)

Let us show that

$$ S:=\sum_{\substack{1 \leq r \leq x^{1/3}\\[1pt] (r,B)=1}}\,\max_{\substack{b\in{\mathbb Z}\\[1pt] (L(b),r)=1}} \biggl|\# {\mathcal P} _{L, {\mathcal A} }(x;r,b) - \frac{\# {\mathcal P} _{L, {\mathcal A} }(x)}{ \varphi _L(r)}\biggr| \leq \frac{\# {\mathcal P} _{L, {\mathcal A} }(x)}{(\ln x)^{100 k^2}}.$$

(5.38)

It is not hard to see that

$$\begin{aligned} \, {\mathcal P} _{L, {\mathcal A} }(x)&=\bigl\{ l_1x+ l_2 \leq p < 2l_1x+ l_2 \colon\, \, p\equiv l_2 {\textstyle\pmod{l_1}} \bigr\},\\[4pt] {\mathcal P} _{L, {\mathcal A} }(x;r,b)&=\bigl\{l_1x + l_2\leq p < 2l_1x+l_2 \colon\, \, p\equiv l_1b+l_2 {\textstyle\pmod{l_1r}} \bigr\} \end{aligned}$$

and hence

$$\begin{aligned} \, \# {\mathcal P} _{L, {\mathcal A} }(x)&= \pi(2l_1x+l_2-1; l_1, l_2) - \pi(l_1x+l_2-1; l_1, l_2), \\[4pt] \# {\mathcal P} _{L, {\mathcal A} }(x;r,b)&=\pi(2l_1x+l_2-1; l_1r, L(b))- \pi(l_1x+l_2-1; l_1r, L(b)).\nonumber \end{aligned}$$

(5.39)

We obtain

$$\begin{aligned} \, S&=\sum_{\substack{1 \leq r \leq x^{1/3}\\[1pt] (r,B)=1}}\,\max_{\substack{b\in{\mathbb Z}\\[1pt] (L(b),r)=1}} \biggl|\pi(2l_1x+l_2-1; l_1r, L(b))-\pi(l_1x+l_2-1; l_1r, L(b)) \nonumber\\[3pt] &\phantom{={}}{} - \frac{\pi(2l_1x+l_2-1; l_1, l_2) - \pi(l_1x+l_2-1; l_1, l_2)}{ \varphi (l_1r)/ \varphi (l_1)}\biggr| \leq S_1+S_2+S_3+S_4, \end{aligned}$$

(5.40)

where

$$\begin{aligned} \, S_1 &= \sum_{\substack{1 \leq r \leq x^{1/3}\\[1pt] (r,B)=1}}\,\max_{\substack{b\in{\mathbb Z}\\[1pt] (L(b),r)=1}} \biggl| \pi(l_1x+l_2-1; l_1r, L(b)) - \frac{ \operatorname{li} (l_1x+l_2-1)}{ \varphi (l_1r)}\biggr|,\\[2pt] S_2 &= \sum_{\substack{1 \leq r \leq x^{1/3}\\[1pt] (r,B)=1}} \biggl|\frac{\pi(l_1x+l_2-1; l_1,l_2)}{ \varphi (l_1r)/ \varphi (l_1)} -\frac{ \operatorname{li} (l_1x+l_2-1)}{ \varphi (l_1r)} \biggr|, \\[2pt] S_3 &= \sum_{\substack{1 \leq r \leq x^{1/3}\\[1pt] (r,B)=1}}\,\max_{\substack{b\in{\mathbb Z}\\[1pt] (L(b),r)=1}} \biggl| \pi(2l_1x+l_2-1; l_1r, L(b)) - \frac{ \operatorname{li} (2l_1x+l_2-1)}{ \varphi (l_1r)}\biggr|,\\[2pt] S_4 &= \sum_{\substack{1 \leq r \leq x^{1/3}\\[1pt] (r,B)=1}} \biggl|\frac{\pi(2l_1x+l_2-1; l_1,l_2)}{ \varphi (l_1r)/ \varphi (l_1)} - \frac{ \operatorname{li} (2l_1x+l_2-1)}{ \varphi (l_1r)} \biggr|. \end{aligned}$$

Let us show that

$$ (L(b),l_1)=1$$

(5.41)

for any $b\in{\mathbb Z}$. Assume the contrary: there is an integer $b$ such that $(L(b),l_1)>1$. Then there is a prime $p$ such that $p\mid l_1$ and $p\mid L(b)$. Hence $p\mid l_2$, and we see that $p\mid L(n)$ for any integer $n$. Since $L\in {\mathcal L}$, we see that $p\mid L_1(n)\dots L_k(n)$ for any integer $n$. But this contradicts the fact that ${\mathcal L} =\{L_1,\dots,L_k\}$ is an admissible set. Thus, (5.41) is proved. We also observe that since $(B,q)=1$ and $l_1=q$, we have

$$ (B,l_1)=1.$$

(5.42)

Let $r$ be an integer with $1\leq r \leq x^{1/3}$ and $(r,B)=1$. Applying (5.37), we have

$$l_1 r\leq x^{1/3} \ln x \leq x^{2/5},\qquad l_1x+l_2-1\geq l_1x\geq x\geq 2,\qquad l_1x+ l_2 -1 \leq 2 x\ln x\leq x^{1+\gamma/\sqrt{\ln x}}$$

provided that $c_0(\varepsilon)$ is chosen large enough. Hence, we obtain (see (5.41), (5.42) and (5.36))

$$\begin{aligned} \, S_1 &= \sum_{\substack{r \colon\, \, l_1 \leq l_1r \leq l_1x^{1/3}\\[1pt] (l_1r, B)=1}}\;\max_{\substack{b\in{\mathbb Z}\\[1pt] (L(b),l_1r)=1}} \biggl| \pi(l_1x+l_2-1; l_1r, L(b)) - \frac{ \operatorname{li} (l_1x+l_2-1)}{ \varphi (l_1r)}\biggr| \nonumber\\[2pt] &\leq \sum_{\substack{1\leq Q\leq x^{2/5}\\[1pt] (Q,B)=1}}\, \max_{2 \leq u \leq x^{1+ \gamma/\sqrt{\ln x}}}\; \max_{W\in{\mathbb Z} \colon\, (W,Q)=1} \biggl|\pi(u;Q,W)-\frac{ \operatorname{li} (u)}{ \varphi (Q)}\biggr| \leq c_2 x\exp(-c_3\sqrt{\ln x}). \end{aligned}$$

(5.43)

Applying Lemmas 2.5 and 2.9, we get

$$\begin{aligned} \, S_2&= \varphi (l_1)\biggl|\pi(l_1x+l_2-1; l_1, l_2)- \frac{ \operatorname{li} (l_1x+l_2-1)}{ \varphi (l_1)}\biggr| \sum_{\substack{1 \leq r \leq x^{1/3}\\[1pt] (r,B)=1}}\frac{1}{ \varphi (l_1r)} \\[2pt] &\leq \biggl|\pi(l_1x+l_2-1; l_1, l_2)- \frac{ \operatorname{li} (l_1x+l_2-1)}{ \varphi (l_1)}\biggr| \sum_{1 \leq r \leq x^{1/3}}\frac{1}{ \varphi (r)} \\[4pt] &\leq \widetilde{c}{} \ln x \biggl|\pi(l_1x+l_2-1; l_1, l_2)- \frac{ \operatorname{li} (l_1x+l_2-1)}{ \varphi (l_1)}\biggr|, \end{aligned}$$

where $\widetilde{c}{} >0$ is an absolute constant. Since $l_1 x+ l_2-1\geq l_1x\geq x$ (see (5.37)), we obtain

$$1\leq l_1 \leq \ln x \leq \ln(l_1x+l_2 -1).$$

Hence (see, for example, [1, Ch. 22]),

$$\biggl|\pi(l_1x+l_2-1; l_1, l_2) - \frac{ \operatorname{li} (l_1x+l_2-1)}{ \varphi (l_1)}\biggr| \leq C(l_1x+l_2-1)\exp\bigl(-c\sqrt{\ln(l_1x+l_2-1)}\,\bigr), $$

(5.44)

where $C$ and $c$ are positive absolute constants. We have

$$\exp\bigl(-c\sqrt{\ln(l_1x+l_2-1)}\,\bigr) \leq \exp(-c\sqrt{\ln x}),$$

(5.45)

$$l_1x+l_2-1\leq x\ln x + \ln x \leq 2 x\ln x.$$

(5.46)

We can assume that

$$ -c\sqrt{\ln x} + 2\ln\ln x \leq -\frac{c}{2}\sqrt{\ln x}$$

(5.47)

if $c_0(\varepsilon)$ is chosen large enough. Hence,

$$S_2 \leq \widetilde{C}{} x\exp\bigl(-c\sqrt{\ln x} + 2\ln\ln x\bigr) \leq \widetilde{C}{} x\exp\Bigl(-\frac c2\sqrt{\ln x} \kern1pt \Bigr), $$

(5.48)

where $\widetilde{C}{} =2 \widetilde{c}{} \kern1pt C$ is a positive absolute constant. Similarly, it can be shown that

$$ S_3 \leq C x\exp(-c\sqrt{\ln x}) \qquad\text{and}\qquad S_4 \leq C x\exp(-c\sqrt{\ln x}),$$

(5.49)

where $C$ and $c$ are positive absolute constants. Substituting (5.43), (5.48), and (5.49) into (5.40), we obtain

$$ \sum_{\substack{1 \leq r \leq x^{1/3}\\[1pt] (r,B)=1}}\,\max_{\substack{b\in{\mathbb Z}\\[1pt] (L(b),r)=1}} \biggl|\# {\mathcal P} _{L, {\mathcal A} }(x;r,b) - \frac{\# {\mathcal P} _{L, {\mathcal A} }(x)}{ \varphi _L(r)}\biggr| \leq c_4 x\exp(-c_5\sqrt{\ln x}),$$

(5.50)

where $c_4$ and $c_5$ are positive absolute constants. Applying (5.44)–(5.47), we have

$$\pi(l_1x+l_2-1; l_1, l_2) = \frac{ \operatorname{li} (l_1x+l_2-1)}{ \varphi (l_1)} + R_1,\qquad |R_1| \leq C x\exp(-c\sqrt{\ln x}),$$

where $C$ and $c$ are positive absolute constants. Similarly, it can be shown that

$$\pi(2l_1x+l_2-1; l_1, l_2) = \frac{ \operatorname{li} (2l_1x+l_2-1)}{ \varphi (l_1)} + R_2,\qquad |R_2| \leq C x\exp(-c\sqrt{\ln x}),$$

where $C$ and $c$ are positive absolute constants. Therefore (see (5.39)),

$$\# {\mathcal P} _{L, {\mathcal A} }(x) = \frac{ \operatorname{li} (2l_1x+l_2-1) - \operatorname{li} (l_1x+l_2-1)}{ \varphi (l_1)}+ R,$$

(5.51)

$$|R|\leq c_6 x\exp(-c_7\sqrt{\ln x}),$$

(5.52)

where $c_6$ and $c_7$ are positive absolute constants. We have

$$\begin{gathered} \, 2l_1x+l_2 - 1\leq 2x\ln x + \ln x \leq 3 x \ln x,\\[4pt] \ln(2l_1 x+ l_2 -1) \leq \ln x + \ln\ln x + \ln 3\leq 2 \ln x \end{gathered}$$

provided that $c_0(\varepsilon)$ is chosen large enough. Hence,

$$\begin{aligned} \, \frac{ \operatorname{li} (2l_1x+l_2-1)- \operatorname{li} (l_1x+l_2-1)}{ \varphi (l_1)} &=\frac{1}{ \varphi (l_1)}\intop_{l_1x+l_2-1}^{2l_1x+l_2-1}\frac{dt}{\ln t} \geq \frac{l_1 x}{ \varphi (l_1)\ln(2l_1x+l_2-1)} \nonumber\\[4pt] &\geq \frac{l_1 x}{2 \varphi (l_1)\ln x}. \end{aligned}$$

(5.53)

Let us show that

$$ |R| \leq \frac{l_1 x}{4 \varphi (l_1)\ln x}.$$

(5.54)

Since $l_1/ \varphi (l_1)\geq 1$, we see from (5.52) that it is sufficient to show that

$$c_6 x\exp(-c_7\sqrt{\ln x}) \leq \frac{x}{4\ln x}.$$

This inequality holds if $c_0(\varepsilon)$ is chosen large enough. Thus, (5.54) is proved. From (5.51), (5.53), and (5.54) we obtain

$$ \# {\mathcal P} _{L, {\mathcal A} }(x) \geq \frac{l_1 x}{4 \varphi (l_1)\ln x}.$$

(5.55)

Now we prove (5.38). Since $l_1/ \varphi (l_1)\geq 1$, we see from (5.50) and (5.55) that it suffices to establish the estimate

$$ c_4 x\exp(-c_5\sqrt{\ln x})\leq \frac{x}{4 (\ln x)^{100 k^2+1}}.$$

(5.56)

Taking logarithms, we obtain

$$\ln c_4 +\ln x - c_5\sqrt{\ln x} \leq \ln x - \ln 4 - 100k^2\ln\ln x - \ln\ln x$$

or, which is equivalent,

$$100 k^2\ln\ln x \leq c_5\sqrt{\ln x} - \ln\ln x - \ln(4 c_4).$$

Since $k \leq (\ln x)^{1/5}$, we have

$$100 k^2\ln\ln x \leq 100 (\ln x)^{2/5}\ln\ln x.$$

The inequality

$$100 (\ln x)^{2/5}\ln\ln x \leq c_5\sqrt{\ln x} - \ln\ln x - \ln(4 c_4)$$

holds if $c_0(\varepsilon)$ is chosen large enough. Inequality (5.56) is proved. Thus, (5.38) is proved.

II. $\,$Let us show that condition (1) of Hypothesis 1 holds. We show that

$$ S:= \sum_{1 \leq r \leq x^{1/3}}\max_{b\in{\mathbb Z}}\biggl|\# {\mathcal A} (x;r,b) - \frac{\# {\mathcal A} (x)}{r}\biggr|\leq \frac{\# {\mathcal A} (x)}{(\ln x)^{100 k^2}}.$$

(5.57)

Let $1 \leq r \leq x^{1/3}$ and $b\in{\mathbb Z}$. We have

$${\mathcal A} (x)=\{x\leq n < 2x\} \qquad\text{and}\qquad {\mathcal A} (x;r,b) = \bigl\{x \leq n < 2x \colon\, \, n \equiv b {\textstyle\pmod{r}} \bigr\}.$$

Hence,

$$ \# {\mathcal A} (x)=x \qquad\text{and}\qquad \# {\mathcal A} (x;r,b)=\frac{x}{r}+\rho,\quad |\rho|\leq 1.$$

(5.58)

We obtain

$$ \biggl|\# {\mathcal A} (x;r,b) - \frac{\# {\mathcal A} (x)}{r}\biggr|= |\rho| \leq 1.$$

(5.59)

Hence, $S \leq x^{1/3}$. Thus, to prove (5.57), it suffices to show that

$$x^{1/3}\leq \frac{x}{(\ln x)^{100 k^2}}$$

or, which is equivalent, $(\ln x)^{100 k^2} \leq x^{2/3}$. Taking logarithms, we obtain

$$100 k^2\ln\ln x \leq \frac{2}{3}\ln x.$$

Since $k \leq (\ln x)^{1/5}$, we have

$$100 k^2\ln\ln x \leq 100 (\ln x)^{2/5}\ln\ln x.$$

The inequality

$$100 (\ln x)^{2/5}\ln\ln x \leq \frac{2}{3}\ln x$$

holds if $c_0(\varepsilon)$ is chosen large enough. Thus, (5.57) is proved.

III. $\,$Let us show that condition (3) of Hypothesis 1 holds. To this end we show that for any integer $r$ with $1\leq r < x^{1/3}$ we have

$$ \max_{b\in{\mathbb Z}}\# {\mathcal A} (x;r,b) \leq 2\frac{\# {\mathcal A} (x)}{r}.$$

(5.60)

Let $1\leq r < x^{1/3}$ and $b\in{\mathbb Z}$. We may assume that $c_0(\varepsilon)\geq 2$. Hence, $r \leq x^{1/3} \leq x$. Applying (5.58), we obtain

$$\# {\mathcal A} (x;r,b)\leq \frac{x}{r}+1\leq 2\frac{x}{r}= 2\frac{\# {\mathcal A} (x)}{r},$$

and (5.60) is proved. Thus, the set $( {\mathcal A} , {\mathcal L} , {\mathcal P} ,B,x,1/3)$ satisfies Hypothesis 1.

We can assume that

$$\exp(c_1\sqrt{\ln x}) \leq x^{1/5} \qquad\text{and}\qquad \ln x \leq x^{1/5}$$

provided that $c_0(\varepsilon)$ is chosen large enough. Since $1\leq B \leq \exp(c_1\sqrt{\ln x})$, we obtain $1 \leq B \leq x^{1/5}$. Let $L=l_1n+l_2\in {\mathcal L}$. Applying (5.37), we have $1\leq l_1 \leq x^{1/5}$ and $1\leq l_2 \leq x^{1/5}$. Thus, the hypothesis of Proposition 5.1 holds and there are nonnegative weights $w_n= w_n( {\mathcal L} )$ with the properties stated in Proposition 5.1. In that proposition, the implied constants in (5.1)–(5.5) depend only on $\alpha$, $\theta$ and on the implied constants from Hypothesis 1, and in our case these constants are absolute ($\alpha=1/5$, $\theta=1/3$, and estimates (5.38), (5.57), and (5.60) hold). Therefore, in our case the implied constants in (5.1)–(5.5) are positive and absolute. Finally, let us denote $c_1$ by $\vartheta$. Lemma 5.3 is proved. $\quad\Box$

Lemma 5.4.

There are positive absolute constants $c$ and $C$ such that the following holds. Let $\varepsilon$ be a real number with $0<\varepsilon<1$. Then there is a number $c_0(\varepsilon)>0,$ depending only on $\varepsilon,$ such that if $x\in{\mathbb N},$ $y\in{\mathbb R},$ $m\in{\mathbb Z},$ $q\in{\mathbb Z},$ and $a\in{\mathbb Z}$ satisfy the conditions $c_0(\varepsilon) \leq y \leq \ln x,$ $1 \leq m \leq c \kern1pt \varepsilon\ln y,$ $1 \leq q \leq y^{1-\varepsilon},$ and $(a,q)=1,$ then

$$\#\bigl\{qx < p_n \leq 2qx - 5q \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\} \geq \pi(2qx) \biggl( \frac{y}{2q\ln x} \biggr)^{\!\exp(Cm)}.$$

Proof.

Let ${\mathcal A} ={\mathbb N}$, ${\mathcal P} =\mathbb{P}$, $\alpha=1/5$, and $\theta=1/3$, and let $C_0=C(1/5, 1/3)>0$ be the absolute constant in Proposition 5.1. Let $c_0(\varepsilon)$ be the quantity in Lemma 5.3. We will choose $c(\varepsilon)$ later; this number is large enough. Let $c(\varepsilon) \geq c_0(\varepsilon)$. Let $x\in{\mathbb N}$, $y\in{\mathbb R}$, and $q\in{\mathbb Z}$ be such that

$$c(\varepsilon) \leq y \leq \ln x,$$

(5.61)

$$1\leq q \leq y^{1-\varepsilon}.$$

(5.62)

By Lemma 5.3, there is a positive integer $B$ such that (5.31) holds. We assume that

$$ \widetilde{C}{} _0 \leq k \leq y^{\varepsilon/14},$$

(5.63)

where $\widetilde{C}{} _0>0$ is an absolute constant. We will choose $\widetilde{C}{} _0$ later. For now, we assume that $\widetilde{C}{} _0$ is large enough; in particular, $\widetilde{C}{} _0 \geq C_0$. It follows from (5.61) and (5.63) that $k\leq (\ln x)^{1/5}$. Thus, (5.32) holds. Let (5.33)–(5.35) hold. Let ${\mathcal L} =\{L_1,\dots,L_k\}$ be an admissible set of $k$ linear functions $L_i(n) = q n + a+ q b_i$, $i=1,\dots,k$, where $b_1,\dots,b_k$ are positive integers such that $b_1< \dots< b_k$ and $q b_k \leq \eta y$. Then (see Lemma 5.3) the hypothesis of Proposition 5.1 holds and there are nonnegative weights $w_n= w_n( {\mathcal L} )$ with the properties stated in Proposition 5.1; the implied constants in (5.1)–(5.5) are positive and absolute. We write ${\mathcal L} = {\mathcal L} (\textbf{b})$ for such a set defined by $b_1,\dots,b_k$. Denote the class of admissible sets by $\mathrm{AS}$.

Let $m$ be a positive integer. We consider

$$\begin{aligned} \, S&=\sum_{\substack{1\leq b_1<\dots< b_k\\ qb_k\leq \eta y\\[1pt] {\mathcal L} = {\mathcal L} (\textbf{b})\in \mathrm{AS} }} \,\sum_{n\in {\mathcal A} (x)} \! \left(\vphantom{\intop_j}\right.\!\sum_{i=1}^k\textbf{1}_{ {\mathcal P} }(L_i(n)) - m -k\sum_{i=1}^k \sum_{\substack{p\mid L_i(n)\\ p<x^\rho,\; p \kern1pt \nmid B}} 1 -k\sum_{\substack{1\leq b \leq 2\eta y\\ L=qt+b\notin {\mathcal L} }} \textbf{1}_{ {\mathcal S} (\rho; B)}(L(n))\!\left.\vphantom{\intop_j}\right) \!w_n( {\mathcal L} ) \nonumber\\[3pt] &=\sum_{\substack{1\leq b_1<\dots< b_k\\ qb_k\leq \eta y\\[1pt] {\mathcal L} = {\mathcal L} (\textbf{b})\in \mathrm{AS} }} \,\sum_{n\in {\mathcal A} (x)}A_n( {\mathcal L} )w_n( {\mathcal L} ). \end{aligned}$$

(5.64)

Let ${\mathcal L} =\{L_1,\dots,L_k\}$ and $n$ be in the range of summation of $S$ and $A_n( {\mathcal L} )>0$. Then the following statements hold:

• (1) The number of primes among $L_1(n),\dots,L_k(n)$ is at least $m+1$.

• (2) For any $1\leq i \leq k$, $L_i(n)$ has no prime factor $p$ such that $p<x^\rho$ and $p\nmid B$.

• (3) For any linear function $L=qt+b\notin {\mathcal L}$, where $b$ is an integer with $1\leq b \leq 2\eta y$, $L(n)$ has a prime factor $p$ such that $p<x^\rho$ and $p\nmid B$ (we choose $\rho$ so that $x^\rho$ is not an integer; therefore, the conditions $p \leq x^\rho$ and $p<x^\rho$ are equivalent). Since $L(n)> n\geq x>x^\rho$, we see that $L(n)$ is not a prime number.

As a consequence we obtain the following statements:

• (i) None of $n\in {\mathcal A} (x)$ can make a positive contribution to $S$ from two different admissible sets (since if $n$ makes a positive contribution for some admissible set ${\mathcal L} =\{L_1,\dots,L_k\}$, then the numbers $L_1(n),\dots,L_k(n)$ are uniquely determined as the integers in $[qn+1, qn+2\eta y]$ with no prime factors $p$ such that $p<x^\rho$ and $p\nmid B$).

• (ii) If ${\mathcal L} =\{L_1,\dots,L_k\}$ and $n$ are in the range of summation of $S$ and $A_n( {\mathcal L} )>0$, then there can be no primes in the interval $[qn+1, qn+2\eta y]$ apart from possibly $L_1(n),\dots,L_k(n)$, and so the primes counted in this way must be consecutive.

Let ${\mathcal L} =\{L_1,\dots,L_k\}$ and $n$ be in the range of summation of $S$ and $A_n( {\mathcal L} )>0$. Let $1 \leq i \leq k$. If $p\mid L_i(n)$ and $p\nmid B$, then $p \geq x^\rho$. Setting

$$\Omega = \bigl\{p \colon\, \, p\mid L_i(n)\,\text{ and }\, p\nmid B\bigr\},$$

we have

$$x^{\rho\,\#\Omega}\leq \prod_{p\in\Omega} p \leq L_i(n).$$

Since

$$q\leq y^{1-\varepsilon} \leq y\leq \ln x \qquad\text{and}\qquad a+qb_i\leq 2 \eta y\leq \ln x,$$

we obtain

$$L_i(n) =qn + a+qb_i\leq n\ln x+ \ln x \leq 2x\ln x+\ln x \leq x^2$$

provided that $c(\varepsilon)$ is chosen large enough. Hence, $\rho\,\#\Omega \leq 2$, i.e., $\#\Omega \leq {2}/{\rho}$. We have

$$\prod_{\substack{p\mid L_i(n)\\ p \kern1pt \nmid B}} 4 = \prod_{p\in\Omega} 4 = 4^{\#\Omega}\leq 4^{2/\rho}=e^{(2/\rho)\ln 4}\leq e^{4/\rho} \qquad\text{and}\qquad \prod_{i=1}^k\, \prod_{\substack{p\mid L_i(n)\\ p \kern1pt \nmid B}} 4 \leq e^{(4k)/\rho}.$$

Thus, if ${\mathcal L} =\{L_1,\dots,L_k\}$ and $n$ are in the range of summation of $S$ and $A_n( {\mathcal L} )>0$, then (see (5.1))

$$ w_n( {\mathcal L} ) \leq C (\ln R)^{2k} e^{(4k)/\rho},$$

(5.65)

where $C>0$ is an absolute constant.

Let ${\mathcal L} =\{L_1,\dots,L_k\}$ be in the range of summation of $S$. We consider

$$\begin{aligned} \, \widetilde{S}{} ( {\mathcal L} )&= \sum_{n\in {\mathcal A} (x)}\! \left(\vphantom{\intop_j}\right.\! \sum_{i=1}^k\textbf{1}_{ {\mathcal P} }(L_i(n)) - m -k\sum_{i=1}^k \sum_{\substack{p\mid L_i(n)\\ p<x^\rho,\; p \kern1pt \nmid B}} 1 -k\sum_{\substack{1\leq b \leq 2\eta y\\ L=qt+b\notin {\mathcal L} }} \textbf{1}_{ {\mathcal S} (\rho;B)}(L(n))\!\left.\vphantom{\intop_j}\right)\! w_n( {\mathcal L} ) \\ &=S_1 - S_2 - S_3 - S_4. \end{aligned}$$

Our aim is to obtain a lower bound for $\widetilde{S}{} ( {\mathcal L} )$. We write $w_n$ instead of $w_n( {\mathcal L} )$ for brevity. Let $1 \leq i \leq k$. Since $\# {\mathcal A} (x)= x$, we have (see (5.3))

$$\begin{aligned} \, \sum_{n\in {\mathcal A} (x)} {\mathbf 1} _{ {\mathcal P} }(L_i(n))w_n &\geq (1+o(1))\frac{B^{k-1}}{ \varphi (B)^{k-1}} {\mathfrak S} _B( {\mathcal L} )\frac{ \varphi (q)}{q} \,\# {\mathcal P} _{L_i, {\mathcal A} }(x)(\ln R)^{k+1}J_k \\ &\phantom{={}}{} +O\biggl( \frac{B^k}{ \varphi (B)^k} {\mathfrak S} _B( {\mathcal L} ) x(\ln R)^{k-1}I_k\biggr). \end{aligned}$$

Hence,

$$\begin{aligned} \, S_1 &= \sum_{n\in {\mathcal A} (x)}\,\sum_{i=1}^k \textbf{1}_{ {\mathcal P} }(L_i(n))w_n =\sum_{i=1}^k\, \sum_{n\in {\mathcal A} (x)} \textbf{1}_{ {\mathcal P} }(L_i(n))w_n \\[2pt] &\geq (1+ o(1))\frac{B^{k-1}}{ \varphi (B)^{k-1}} {\mathfrak S} _B( {\mathcal L} )\frac{ \varphi (q)}{q}(\ln R)^{k+1}J_k \sum_{i=1}^k\# {\mathcal P} _{L_i, {\mathcal A} }(x) +O\biggl(k\frac{B^k}{ \varphi (B)^k} {\mathfrak S} _B( {\mathcal L} ) x(\ln R)^{k-1} I_k\biggr) \\[2pt] &=(1+ o(1))\frac{B^k}{ \varphi (B)^k} {\mathfrak S} _B( {\mathcal L} )(\ln R)^{k+1}J_k\frac{ \varphi (B)}{B}\frac{ \varphi (q)}{q}\sum_{i=1}^k\# {\mathcal P} _{L_i, {\mathcal A} }(x) +o\biggl( \frac{B^k}{ \varphi (B)^k} {\mathfrak S} _B( {\mathcal L} ) x(\ln R)^k I_k\biggr) \\[3pt] &=S_1'+S_1'', \end{aligned}$$

since

$$0<\frac{k}{\ln R}\leq \frac{(\ln x)^{1/5}}{(1/9)\ln x}\to 0\qquad \text{as}\quad x\to +\infty.$$

We have shown (see (5.55)) that if $x \geq c_0$, where $c_0> 0$ is an absolute constant, then for any $L\in {\mathcal L}$

$$\# {\mathcal P} _{L, {\mathcal A} }(x) \geq \frac{q x}{4 \varphi (q)\ln x}.$$

We may assume that $c(\varepsilon) \geq c_0$. Since $\varphi (B)/B \geq 1/2$ (see (5.31)), we obtain

$$\frac{ \varphi (B)}{B}\frac{ \varphi (q)}{q}\sum_{i=1}^k\# {\mathcal P} _{L_i, {\mathcal A} }(x)\geq \frac{kx}{8\ln x} = \frac{kx}{72\ln R}.$$

We have $|o(1)|\leq 1/2$ in $S_1'$ if $x \geq c'$, where $c'>0$ is an absolute constant. We may assume that $c(\varepsilon)\geq c'$. Since (see (5.8))

$$J_k\geq c''\frac{\ln k}{k}I_k,$$

where $c''>0$ is an absolute constant, we get

$$S_1'\geq \frac{c''}{144}\frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k\ln k.$$

We have

$$|S_1''|\leq \frac{c''}{288}\frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} ) x(\ln R)^k I_k \leq \frac{c''}{288}\frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} ) x(\ln R)^k I_k\ln k$$

provided that $c(\varepsilon)$ is chosen large enough. Therefore,

$$S_1 \geq \frac{c''}{288}\frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} ) x(\ln R)^k I_k\ln k = c\frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} ) x(\ln R)^k I_k\ln k, $$

(5.66)

where $c>0$ is an absolute constant.

We have (see (5.2))

$$S_2 = m\sum_{n\in {\mathcal A} (x)} w_n = m (1+ o(1))\frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k \geq \frac{m}{2} \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k $$

(5.67)

provided that $c(\varepsilon)$ is chosen large enough. Applying (5.5), we obtain

$$S_3 = k \sum_{n\in {\mathcal A} (x)} \,\sum_{i=1}^k\sum_{\substack{p\mid L_i(n)\\ p<x^\rho,\; p \kern1pt \nmid B}}\! w_n =k\sum_{i=1}^k \,\sum_{n\in {\mathcal A} (x)} w_n \sum_{\substack{p\mid L_i(n)\\ p<x^\rho,\; p \kern1pt \nmid B}}\! 1 \leq c_2 \rho^2 k^6(\ln k)^2 \frac{B^k}{ \varphi (B)^k} {\mathfrak S} _B( {\mathcal L} )x (\ln R)^kI_k,$$

where $c_2>0$ is an absolute constant. Let $c_3>0$ be an absolute constant such that

$$c_2c_3^2\leq \frac{1}{12}\qquad\text{and}\qquad \frac{c_3}{j^3\ln j}\leq \frac{1}{30} \quad \text{for any }\, j\geq 2.$$

We choose an arbitrary number $\rho$ in the interval

$$ \Bigl[\frac{c_3}{2k^3\ln k}, \frac{c_3}{k^3\ln k}\Bigr]$$

(5.68)

so that $x^\rho$ is not an integer. It is clear that $\rho \leq 1/30$. Let us show that the first inequality in (5.33) holds. It suffices to show that

$$\frac{k(\ln\ln x)^2}{\ln x}\leq \frac{c_3/2}{k^3\ln k}.$$

This inequality is equivalent to

$$k^4\ln k (\ln \ln x)^2 \leq \frac{c_3}{2} \ln x.$$

Since $k \leq (\ln x)^{1/5}$, we have

$$k^4\ln k (\ln \ln x)^2 \leq \frac{1}{5}(\ln x)^{4/5}(\ln \ln x)^3 \leq \frac{c_3}{2} \ln x$$

provided that $c(\varepsilon)$ is chosen large enough. Thus, the inequalities in (5.33) hold. We have (see (5.67))

$$\begin{aligned} \, \nonumber S_3 &\leq c_2 \frac{c_3^2}{k^6(\ln k)^2} k^6(\ln k)^2\frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x (\ln R)^kI_k \leq \frac{1}{12}\frac{B^k}{ \varphi (B)^k} {\mathfrak S} _B( {\mathcal L} )x (\ln R)^kI_k \\[4pt] &\leq\frac{m}{12}\frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x (\ln R)^kI_k \leq \frac{1}{6}S_2. \end{aligned}$$

(5.69)

Now we estimate the quantity

$$S_4 = k\sum_{n\in {\mathcal A} (x)}\sum_{\substack{1\leq b \leq 2\eta y\\ L=qt+b\notin {\mathcal L} }} \textbf{1}_{ {\mathcal S} (\rho; B)}(L(n))w_n = k \sum_{\substack{1\leq b \leq 2\eta y\\ L=qt+b\notin {\mathcal L} }}\sum_{n\in {\mathcal A} (x)}\textbf{1}_{ {\mathcal S} (\rho; B)}(L(n))w_n.$$

Let $b$ be in the range of summation of $S_4$. Then $L=qt+b\notin {\mathcal L}$ and

$$\Delta_L=q^{k+1}\prod_{i=1}^k|(a+qb_i) - b| \neq 0.$$

Since $1\leq B \leq x^{1/5}$, we have (see (5.4))

$$\sum_{n\in {\mathcal A} (x)}\textbf{1}_{ {\mathcal S} (\rho; B)}(L(n))w_n \leq \frac{c_4}{\rho}\frac{\Delta_L}{ \varphi (\Delta_L)}\frac{B}{ \varphi (B)} \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^{k-1}I_k,$$

where $c_4>0$ is an absolute constant. Since $B/ \varphi (B) \leq 2$ and $\rho$ lies in the interval (5.68), we obtain

$$\begin{aligned} \, \sum_{n\in {\mathcal A} (x)}\textbf{1}_{ {\mathcal S} (\rho; B)}(L(n))w_n &\leq \frac{4c_4}{c_3}k^3\ln k\frac{\Delta_L}{ \varphi (\Delta_L)} \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^{k-1}I_k \\[1pt] &= c_5 k^3\ln k\frac{\Delta_L}{ \varphi (\Delta_L)} \frac{B^k}{ \varphi (B)^k} {\mathfrak S} _B( {\mathcal L} )x(\ln R)^{k-1}I_k. \end{aligned}$$

Hence,

$$ S_4 \leq c_5 k^4\ln k \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^{k-1}I_k \sum_{\substack{1\leq b \leq 2\eta y\\ L=qt+b\notin {\mathcal L} }} \frac{\Delta_L}{ \varphi (\Delta_L)}.$$

(5.70)

We put

$$ c_6=36Cc_5,$$

(5.71)

where $C>0$ is the absolute constant in Lemma 5.2, and

$$ \eta = \frac{1}{12 c_6 k^4(\ln k)^2\ln\ln(q+2)}.$$

(5.72)

Let us show that

$$ (\ln x)^{-9/10}\leq 2\eta\leq 1.$$

(5.73)

The second inequality in (5.73) is equivalent to the inequality

$$6c_6 k^4(\ln k)^2\ln\ln(q+2) \geq 1.$$

We may assume that $\widetilde{C}{} _0\geq 3$; therefore, $\ln k \geq 1$. We have

$$6c_6 k^4(\ln k)^2\ln\ln(q+2) \geq 6c_6(\ln\ln 3) k^4 \geq 6c_6(\ln\ln 3) { \widetilde{C}{} _0}^4 \geq 1$$

provided that $\widetilde{C}{} _0$ is chosen large enough. The first inequality in (5.73) is equivalent to the inequality

$$6c_6 k^4(\ln k)^2\ln\ln(q+2) \leq (\ln x)^{9/10}.$$

Since $q \leq \ln x$ and $k\leq (\ln x)^{1/5}$, we have

$$6c_6 k^4(\ln k)^2\ln\ln(q+2) \leq 6c_6 (\ln x)^{4/5}\frac{1}{25}(\ln \ln x)^2\ln\ln(\ln x+2) \leq (\ln x)^{9/10}$$

provided that $c(\varepsilon)$ is chosen large enough. Thus, (5.73) holds. We can assume that $x \geq c$, where $c$ is the absolute constant in Lemma 5.2, provided that $c(\varepsilon)$ is chosen large enough. Applying Lemma 5.2 and taking into account that $\ln(k+1) \leq 2 \ln k$, we have

$$\sum_{\substack{1\leq b \leq 2\eta y\\ L=qt+b\notin {\mathcal L} }} \frac{\Delta_L}{ \varphi (\Delta_L)} \leq \sum_{\substack{1\leq b \leq 2\eta \ln x\\ L=qt+b\notin {\mathcal L} }} \frac{\Delta_L}{ \varphi (\Delta_L)} \leq 4C\ln\ln(q+2)(\ln k) \eta \ln x = 36C \ln\ln(q+2)(\ln k) \eta \ln R.$$

Substituting this estimate into (5.70), we get (see also (5.71), (5.72), and (5.67))

$$\begin{aligned} \, S_4 &\leq 36Cc_5 k^4(\ln k)^2 \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k\eta \ln\ln(q+2) \nonumber\\[4pt] &= c_6 k^4(\ln k)^2 \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k \ln\ln(q+2) \frac{1}{12 c_6 k^4(\ln k)^2\ln\ln(q+2)} \nonumber\\[4pt] &= \frac{1}{12} \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k \leq \frac{m}{12} \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k \leq \frac{1}{6} S_2. \end{aligned}$$

(5.74)

From (5.69) and (5.74) we obtain

$$\widetilde{S}{} ( {\mathcal L} ) = S_1- S_2-S_3-S_4\geq S_1 - \frac{4}{3}S_2.$$

We have (see (5.2))

$$S_2 = m\sum_{n\in {\mathcal A} (x)} w_n = m (1+ o(1))\frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k \leq \frac{3}{2}m \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k$$

provided that $c(\varepsilon)$ is chosen large enough. Applying (5.66) with $c$ replaced by $3c_1$, we obtain

$$\widetilde{S}{} ( {\mathcal L} )\geq \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k(3c_1\ln k - 2m),$$

where $c_1>0$ is an absolute constant. We put

$$\widetilde{c}{} = \widetilde{C}{} _0+ \frac{1}{c_1},$$

(5.75)

$$k=\lceil \exp( \widetilde{c}{} m)\rceil.$$

(5.76)

It is not hard to see that

$$k\geq \widetilde{C}{} _0\qquad\text{and}\qquad 3c_1\ln k - 2m \geq m.$$

Since $m$ is a positive integer, we see that $3c_1\ln k - 2m \geq 1$. Hence,

$$\widetilde{S}{} ( {\mathcal L} )\geq \frac{B^k}{ \varphi (B)^k} \kern1pt {\mathfrak S} _B( {\mathcal L} )x(\ln R)^kI_k.$$

Since $B^k/ \varphi (B)^k\geq 1$, $\ln R = (1/9)\ln x$, ${\mathfrak S} _B( {\mathcal L} )\geq \exp(-c_2k)$, and $I_k\geq c_3(2k\ln k)^{-k}$, where $c_2$ and $c_3$ are positive absolute constants (see (5.6) and (5.7)), it follows that

$$\widetilde{S}{} ( {\mathcal L} )\geq \frac{1}{9^k}c_3(2k\ln k)^{-k}\exp(-c_2k)x(\ln x)^k \geq \exp(-k^2) x(\ln x)^k$$

provided that $\widetilde{C}{} _0$ is chosen large enough. We obtain

$$\begin{aligned} \, S=\sum_{\substack{1\leq b_1<\dots< b_k\\ qb_k\leq \eta y\\[1pt] {\mathcal L} = {\mathcal L} (\textbf{b})\in \mathrm{AS} }} \widetilde{S}{} ( {\mathcal L} ) \geq \exp(-k^2) x(\ln x)^k \sum_{\substack{1\leq b_1<\dots< b_k\\ qb_k\leq \eta y\\[1pt] {\mathcal L} = {\mathcal L} (\textbf{b})\in \mathrm{AS} }} 1 = \exp(-k^2) x(\ln x)^k S'. \end{aligned}$$

(5.77)

Now we derive a lower bound for $S'$. First let us show that

$$ 2 \leq k \leq \frac12\biggl[\frac{\eta y}{q}\biggr].$$

(5.78)

The first inequality obviously holds, since we may assume that $\widetilde{C}{} _0 \geq 2$. To prove the second inequality, it suffices to show that

$$ 2 k \leq \frac{\eta y}{q}.$$

(5.79)

We have (see (5.62) and (5.72))

$$\frac{\eta y}{q} \geq \eta y^\varepsilon= \frac{c_4 y^\varepsilon}{k^4(\ln k)^2\ln\ln(q+2)},$$

where $c_4>0$ is an absolute constant. Thus, to prove (5.79), it suffices to show that

$$2k^5(\ln k)^2\ln\ln(q+2)\leq c_4 y^\varepsilon.$$

In particular, from (5.62) it follows that $q\leq y$. Applying (5.63), we have

$$2k^5(\ln k)^2\ln\ln(q+2) \leq 2 y^{5\varepsilon/14}\frac{\varepsilon^2}{196}(\ln y)^2\ln\ln(y+2) \leq c_4 y^\varepsilon$$

provided that $c(\varepsilon)$ is chosen large enough. Thus, (5.78) is proved.

We put

$$\Omega = \biggl\{ 1 \leq n \leq \biggl[\frac{\eta y}{q}\biggr] \colon\, \, (n,p)=1\ \,\forall p\leq k\biggr\}.$$

Applying Lemma 2.13, we have

$$\#\Omega = \Phi \biggl(\biggl[\frac{\eta y}{q}\biggr], k\biggr)\geq c_0 \frac{[\eta y/q]}{\ln k},$$

where $c_0>0$ is an absolute constant. In particular, from (5.78) it follows that $\eta y/ q \geq 4$, and so

$$\biggl[\frac{\eta y}{q}\biggr]\geq \frac{\eta y}{q} - 1\geq \frac{\eta y}{2q}.$$

We obtain

$$ \#\Omega \geq c_5 \frac{\eta y}{q\ln k},$$

(5.80)

where $c_5>0$ is an absolute constant. Let us show that

$$ c_5 \frac{\eta y}{q\ln k}\geq 2 k.$$

(5.81)

Applying (5.62) and (5.72), we have

$$c_5 \frac{\eta y}{q\ln k}\geq \frac{c_6 y^\varepsilon}{k^4(\ln k)^3\ln\ln(q+2)},$$

where $c_6>0$ is an absolute constant. Therefore, it suffices to show that

$$2k^5(\ln k)^3\ln\ln(q+2) \leq c_6 y^\varepsilon.$$

Applying (5.63) and taking into account that $q\leq y$, we have

$$2k^5(\ln k)^3\ln\ln(q+2)\leq 2 y^{5\varepsilon/14}\Bigl( \frac{\varepsilon}{14} \Bigr)^3(\ln y)^3\ln\ln(y+2) \leq c_6 y^\varepsilon$$

provided that $c(\varepsilon)$ is chosen large enough. Thus, (5.81) is proved.

Let $b_1 <\dots<b_k$ be positive integers from the set $\Omega$. Let us show that for any prime $p$ with $p\nmid q$ there is an integer $m_p$ such that $m_p\not \equiv b_i\pmod{p}$ for all $1\leq i \leq k$. Let $p$ be a prime with $p\nmid q$. If $p > k$, then the statement is obvious. If $p \leq k$, then we may put $m_p=0$; from the definition of the set $\Omega$ it follows that $b_i\not\equiv 0\pmod{p}$ for all $1\leq i \leq k$. Thus, the statement is proved. By Lemma 5.1, ${\mathcal L} (\textbf{b})$ is an admissible set. Hence (see also Lemma 2.12, (5.80), (5.81), and (5.72)),

$$\begin{aligned} \, S' &\geq \binom{\#\Omega}{k}\geq k^{-k}(\#\Omega - k)^k \geq k^{-k}\biggl(c_5 \frac{\eta y}{q\ln k} - k\biggr)^{\!k} \geq k^{-k}\biggl( \frac{c_5}{2} \frac{\eta y}{q\ln k} \biggr)^{\!k} \\[4pt] &=k^{-k}\biggl(c_6 \frac{y}{q\ln\ln(q+2)k^4(\ln k)^3} \biggr)^{\!k} = \biggl( \frac{y}{q\ln\ln(q+2)} \biggr)^{\!k}\biggl( \frac{c_6}{k^5(\ln k)^3} \biggr)^{\!k}, \end{aligned}$$

where $c_6>0$ is an absolute constant. We have

$$\biggl( \frac{c_6}{k^5(\ln k)^3} \biggr)^{\!k} \geq \exp(-k^2)$$

provided that $\widetilde{C}{} _0$ is chosen large enough. Hence,

$$S' \geq \biggl( \frac{y}{q\ln\ln(q+2)} \biggr)^{\!k} \exp(-k^2).$$

Substituting this estimate into (5.77), we obtain

$$S\geq \exp(-2k^2)x (\ln x)^k \biggl( \frac{y}{q\ln\ln(q+2)} \biggr)^{\!k} \geq \exp(-2k^5)x (\ln x)^k \biggl( \frac{y}{q\ln\ln(q+2)} \biggr)^{\!k}. $$

(5.82)

Now we obtain an upper bound for $S$. Applying (5.64) and (5.65), we get

$$S\leq \sum_{\substack{1\leq b_1<\dots< b_k\\ qb_k\leq \eta y\\[1pt] {\mathcal L} = {\mathcal L} (\textbf{b})\in \mathrm{AS} }} \,\sum_{\substack{n\in {\mathcal A} (x) \colon\, A_n( {\mathcal L} )>0}}A_n( {\mathcal L} )w_n( {\mathcal L} ) \leq Ck (\ln R)^{2k} e^{(4k)/\rho}\sum_{\substack{1\leq b_1<\dots< b_k\\ qb_k\leq \eta y\\[1pt] {\mathcal L} = {\mathcal L} (\textbf{b})\in \mathrm{AS} }} \,\sum_{\substack{n\in {\mathcal A} (x) \colon\, A_n( {\mathcal L} )>0}} 1.$$

We have (see assertions (1)–(3), (i), and (ii) at the beginning of the proof)

$$\begin{aligned} \, &\sum_{\substack{1\leq b_1<\dots< b_k\\ qb_k\leq \eta y\\[1pt] {\mathcal L} = {\mathcal L} (\textbf{b})\in \mathrm{AS} }} \,\sum_{\substack{n\in {\mathcal A} (x) \colon\, A_n( {\mathcal L} )>0}} 1 \\[4pt] &\leq\#\bigl\{ x\leq n < 2x \colon\, \,\exists p_j,p_{j+1},\dots,p_{j+m}\in[qn+1, qn + 2\eta y],\ p_j,p_{j+1},\dots,p_{j+m}\equiv a {\textstyle\pmod{q}} \bigr\} \\[4pt] &\leq\#\bigl\{ x\leq n < 2x \colon\, \,\exists p_j,p_{j+1},\dots,p_{j+m}\in[qn+1, qn + y],\ p_j,p_{j+1},\dots,p_{j+m}\equiv a {\textstyle\pmod{q}} \bigr\}:=N_1. \end{aligned}$$

Hence,

$$S \leq Ck (\ln R)^{2k} e^{(4k)/\rho} N_1.$$

Since $\rho$ lies in the interval (5.68), we have

$$\frac{4k}{\rho} \leq \frac{8 k^4\ln k}{c_3}= c_4k^4\ln k,$$

where $c_4>0$ is an absolute constant. Since $\ln R = (1/9)\ln x$, it follows that

$$Ck (\ln R)^{2k} e^{(4k)/\rho}\leq C\frac{k}{9^{2k}}\exp(c_4k^4\ln k) (\ln x)^{2k}\leq \exp(k^5)(\ln x)^{2k}$$

provided that $\widetilde{C}{} _0$ is chosen large enough. Hence,

$$ S \leq \exp(k^5)(\ln x)^{2k} N_1.$$

(5.83)

From (5.82) and (5.83) we obtain

$$ N_1\geq x \Bigl( \frac{y}{\ln x} \Bigr)^{k}\biggl( \frac{1}{q\ln\ln(q+2)} \biggr)^{\!k}\exp(-3k^5).$$

(5.84)

We define

$$\begin{aligned} \, \Omega_1 &= \bigl\{ x\leq n \leq 2x-1 \colon\, \,\exists p_j,p_{j+1},\dots,p_{j+m}\in[qn+1, qn + y],\ p_j,p_{j+1},\dots,p_{j+m}\equiv a {\textstyle\pmod{q}} \bigr\},\\[5pt] \Omega_2 &= \bigl\{qx+1\leq p_n \leq q(2x-1)+y \colon\, \, p_n\equiv \dots \equiv p_{n+m} \equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\} \end{aligned}$$

and put $N_2=\#\Omega_2$. Since $x$ is a positive integer, we have $N_1=\#\Omega_1$. Let us show that

$$ N_1 \leq (\lceil y\rceil +1)N_2.$$

(5.85)

Let $n\in\Omega_1$. Then there are at least $m+1$ consecutive primes all congruent to $a\pmod{q}$ in the interval $[qn+1, qn + y]$. Let $p$ be the first of them. Then $p\in\Omega_2$. We put

$$\Lambda = \bigl\{j\in{\mathbb Z} \colon\, \, qj+ 1 \leq p \leq qj+y\bigr\}$$

and claim that

$$ \# \Lambda \leq \lceil y \rceil +1.$$

(5.86)

Let $I_j= [qj+1, qj+y]$, $j\in{\mathbb Z}$. Since $p\in I_n$, we have $\Lambda \neq \varnothing$. Let $l$ be the minimal element in $\Lambda$. We put $t=\lceil y\rceil + 1$. Then $t>y$ and

$$q(l+t)+1> q(l+t)=ql+qt\geq ql +t> ql+y\geq p.$$

Hence, $p\notin I_j$ for $j \geq l+t$ and $j\leq l-1$. We obtain $\# \Lambda \leq t$. Thus, (5.86) is proved; (5.85) follows from (5.86). We have $\lceil y\rceil + 1\leq y+2 \leq 2y$ provided that $c(\varepsilon)$ is chosen large enough. Since

$$N_2 \leq \#\bigl\{qx+1\leq p_n \leq 2qx+y \colon\, \, p_n\equiv \dots \equiv p_{n+m} \equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\}=:N_3,$$

we obtain (see (5.84))

$$ N_3\geq \frac{1}{2}\frac{x}{y} \Bigl( \frac{y}{\ln x} \Bigr)^{k} \biggl( \frac{1}{q\ln\ln(q+2)} \biggr)^{\!k}\exp(-3k^5).$$

(5.87)

We put

$$\begin{aligned} \, N_4 &= \#\bigl\{qx< p_n \leq 2qx - 5q \colon\, \, p_n\equiv \dots \equiv p_{n+m} \equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\}, \\[5pt] N_5 &= \#\bigl\{2qx -5q<p_n \leq 2qx + y \colon\, \, p_n\equiv \dots \equiv p_{n+m} \equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\}.\nonumber \end{aligned}$$

(5.88)

Then

$$ N_3= N_4+N_5.$$

(5.89)

Since $q \leq y$, we have

$$ N_5 \leq 5q+ [y]\leq 5q+ y\leq 6 y.$$

(5.90)

Let us show that

$$ y \leq \frac{1}{24}\frac{x}{y} \Bigl( \frac{y}{\ln x} \Bigr)^{k}\biggl( \frac{1}{q\ln\ln(q+2)} \biggr)^{\!k}\exp(-3k^5) :=T_1.$$

(5.91)

Since $q\leq y^{1-\varepsilon}\leq y$ and $k \leq y^{\varepsilon/14}$, we have

$$T_1 \geq \frac{1}{24}\frac{x}{y} \biggl( \frac{y^\varepsilon}{\ln x\ln\ln(y+2)} \biggr)^{\!k}\exp\bigl(-3y^{5\varepsilon/14}\bigr).$$

Therefore, to prove (5.91), it suffices to show that

$$y \leq \frac{1}{24}\frac{x}{y} \biggl( \frac{y^\varepsilon}{\ln x\ln\ln(y+2)} \biggr)^{\!k}\exp\bigl(-3y^{5\varepsilon/14}\bigr).$$

Taking logarithms, we obtain

$$\ln y \leq -\ln 24 + \ln x-\ln y + k\bigl(\varepsilon\ln y-\ln\ln x-\ln\ln\ln(y+2)\bigr) - 3y^{5\varepsilon/14}$$

or, which is equivalent,

$$T_2:= 2\ln y + \ln 24 - \varepsilon k \ln y + k\ln\ln x + k\ln\ln\ln(y+2) + 3y^{5\varepsilon/14} \leq \ln x.$$

Since $y \leq \ln x$ and $0<\varepsilon<1$, we have $k \leq (\ln x)^{\varepsilon/14}\leq (\ln x)^{1/14}$. Then

$$T_2 \leq 2 \ln\ln x + \ln 24 + (\ln x)^{1/14}\ln\ln x+ (\ln x)^{1/14} \ln\ln\ln(\ln x+2) + 3 (\ln x)^{5/14} \leq \ln x$$

provided that $c(\varepsilon)$ is chosen large enough. Thus, (5.91) is proved. From (5.90) and (5.91) it follows that

$$ N_5 \leq \frac{1}{4}\frac{x}{y} \Bigl( \frac{y}{\ln x} \Bigr)^{k}\biggl( \frac{1}{q\ln\ln(q+2)} \biggr)^{\!k}\exp(-3k^5).$$

(5.92)

Applying (5.87), (5.89), and (5.92), we obtain

$$ N_4 \geq \frac{1}{4}\frac{x}{y} \Bigl( \frac{y}{\ln x} \Bigr)^{k}\biggl( \frac{1}{q\ln\ln(q+2)} \biggr)^{\!k}\exp(-3k^5)=: T_3.$$

(5.93)

We have (see (2.2))

$$\pi(2qx) \leq c_1 \frac{2qx}{\ln(2qx)}\leq c_1 \frac{2qx}{\ln x}=c_2\frac{qx}{\ln x},$$

where $c_1>0$ and $c_2=2c_1>0$ are absolute constants. Therefore,

$$\begin{aligned} \, T_3 &= \frac{qx}{\ln x}\frac{\ln x}{qx} \frac{1}{4}\frac{x}{y} \Bigl( \frac{y}{\ln x} \Bigr)^{k} \biggl( \frac{1}{q\ln\ln(q+2)} \biggr)^{\!k}\exp(-3k^5) \\[4pt] &\geq \frac{1}{4c_2} \kern1pt \pi(2qx)\Bigl( \frac{y}{\ln x} \Bigr)^{k-1}\frac{1}{q^{k+1}(\ln\ln(q+2))^k}\exp(-3k^5). \end{aligned}$$

Using the inequality $\ln(1+x)\leq x$, $x>0$, we obtain $\ln\ln(q+2)\leq \ln(1+q)\leq q$. Hence,

$$\frac{1}{q^{k+1} (\ln\ln(q+2))^k}\geq \frac{1}{q^{2k+1}} \geq \frac{1}{q^{3k^5}}.$$

We can assume that $4c_2 \leq 2^{3k^5}$ if $\widetilde{C}{} _0$ is chosen large enough. We have

$$T_3 \geq \pi(2qx)\Bigl( \frac{y}{\ln x} \Bigr)^{k-1}\frac{1}{(2eq)^{3k^5}}.$$

We can also assume that $3k^5\leq k^6$ if $\widetilde{C}{} _0$ is chosen large enough. Hence,

$$\frac{1}{(2eq)^{3k^5}}\geq \frac{1}{(2eq)^{k^6}}.$$

We have $(2e)^{k^6} \leq 2^{k^7}$ if $\widetilde{C}{} _0$ is chosen large enough. It is clear that $q^{k^6} \leq q^{k^7}$. Then

$$\frac{1}{(2eq)^{k^6}} \geq \frac{1}{(2q)^{k^7}}.$$

Further (see (5.61)),

$$0<\frac{y}{\ln x} \leq 1\qquad\Rightarrow\qquad \Bigl( \frac{y}{\ln x} \Bigr)^{k-1} \geq \Bigl( \frac{y}{\ln x} \Bigr)^{k^7}.$$

We obtain

$$T_3 \geq \pi(2qx)\biggl( \frac{y}{2q\ln x} \biggr)^{k^7}.$$

From (5.75) and (5.76) we find

$$ k= \lceil \exp( \widetilde{c}{} m)\rceil \leq \exp( \widetilde{c}{} m) +1 \leq \exp(2 \widetilde{c}{} m)$$

(5.94)

provided that $\widetilde{C}{} _0$ is chosen large enough. Therefore, $k^7 \leq \exp(14 \widetilde{c}{} m)$. Since $\widetilde{C}{} _0$ is a positive absolute constant, we see from (5.75) that $\widetilde{c}{}$ is a positive absolute constant. We have

$$ T_3 \geq \pi(2qx)\biggl( \frac{y}{2q\ln x} \biggr)^{\!\exp(14 \widetilde{c}{} m)}=\pi(2qx)\biggl( \frac{y}{2q\ln x} \biggr)^{\!\exp(Cm)},$$

(5.95)

where $C=14 \widetilde{c}{} >0$ is an absolute constant. Combining (5.88), (5.93), and (5.95) we obtain

$$\begin{aligned} \, \#\bigl\{qx< p_n \leq 2qx - 5q \colon\, \, p_n\equiv \dots \equiv p_{n+m} \equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\} \geq \pi(2qx)\biggl( \frac{y}{2q\ln x} \biggr)^{\!\exp(Cm)}. \end{aligned}$$

Applying (5.94), we see that the inequality $k\leq y^{\varepsilon/14}$ holds if $\exp(2 \widetilde{c}{} m) \leq y^{\varepsilon/14}$. This inequality is equivalent to

$$m \leq \frac{\varepsilon}{28 \kern1pt \widetilde{c}{} } \ln y =c \kern1pt \varepsilon\ln y,$$

where $c= 1/(28 \widetilde{c}{} \kern1pt )>0$ is an absolute constant. Let us redenote $c(\varepsilon)$ by $c_0(\varepsilon)$. Lemma 5.4 is proved. $\quad\Box$

Lemma 5.5.

There are positive absolute constants $c$ and $C$ such that the following holds. Let $\varepsilon$ be a real number with $0<\varepsilon<1$. Then there is a number $c_0(\varepsilon)>0,$ depending only on $\varepsilon,$ such that if $x\in{\mathbb R},$ $y\in{\mathbb R},$ $m\in{\mathbb Z},$ $q\in{\mathbb Z},$ and $a\in{\mathbb Z}$ satisfy the conditions $c_0(\varepsilon)\leq y \leq \ln x,$ $1 \leq m \leq c \kern1pt \varepsilon\ln y,$ $1 \leq q \leq y^{1-\varepsilon},$ and $(a,q)=1,$ then

$$\#\bigl\{qx< p_n\leq 2qx \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\} \geq \pi(2qx)\biggl( \frac{y}{2q\ln x} \biggr)^{\!\exp(Cm)}.$$

Proof.

Let $c$, $C$, and $c_0(\varepsilon)$ be the quantities mentioned in Lemma 5.4. We will choose a quantity $\widetilde{c}{} _0(\varepsilon)$ and an absolute constant $\widetilde{C}{}$ later; they will be large enough. In particular, let $\widetilde{c}{} _0(\varepsilon) \geq c_0(\varepsilon)$ and $\widetilde{C}{} \geq C$. Let $x\in{\mathbb R}$, $y\in{\mathbb R}$, $m\in{\mathbb Z}$, $q\in{\mathbb Z}$, and $a\in{\mathbb Z}$ be such that $\widetilde{c}{} _0(\varepsilon)\leq y \leq \ln x$, $1\leq m \leq c \kern1pt \varepsilon\ln y$, $1\leq q \leq y^{1-\varepsilon}$, and $(a,q)=1$. We put $l=\lceil x\rceil$. Then, by Lemma 5.4, we have

$$\begin{aligned} \, \nonumber N_1&=\#\bigl\{ql< p_n\leq 2ql - 5q \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\} \\[4pt] &\geq \pi(2ql)\biggl( \frac{y}{2q\ln l} \biggr)^{\!\exp(Cm)}=:T_1. \end{aligned}$$

(5.96)

Since $x \leq l < x+1$, we have

$$ql\geq qx \qquad\text{and}\qquad 2ql - 5q\leq 2q(x+1)-5q=2qx - 3q< 2qx.$$

Therefore,

$$N_1\leq\#\bigl\{qx< p_n\leq 2qx \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\}=: N_2. $$

(5.97)

We have $x+1 \leq x^2$ provided that $\widetilde{c}{} _0(\varepsilon)$ is chosen large enough. Hence,

$$\ln l \leq \ln(x+1)\leq 2\ln x.$$

Since $\pi(2ql)\geq \pi(2qx)$, we have

$$T_1 \geq \pi(2qx)\biggl( \frac{y}{4q\ln x} \biggr)^{\!\exp(Cm)} =\pi(2qx)\biggl( \frac{y}{q\ln x} \biggr)^{\!\exp(Cm)} \biggl( \frac{1}{4} \biggr)^{\!\exp(Cm)}.$$

Then

$$2\exp( \widetilde{C}{} m)\leq \exp(2 \widetilde{C}{} m)$$

provided that $\widetilde{C}{}$ is chosen large enough. Since $\widetilde{C}{} \geq C$, we have

$$\biggl( \frac{1}{4} \biggr)^{\!\exp(Cm)} \geq \biggl( \frac{1}{4} \biggr)^{\!\exp( \widetilde{C}{} m)} =\biggl( \frac{1}{2} \biggr)^{\!2\exp( \widetilde{C}{} m)}\geq \biggl( \frac{1}{2} \biggr)^{\!\exp(2 \widetilde{C}{} m)}.$$

Further,

$$0<\frac{y}{q\ln x}\leq 1 \qquad\Rightarrow\qquad \biggl( \frac{y}{q\ln x} \biggr)^{\!\exp(Cm)} \geq \biggl( \frac{y}{q\ln x} \biggr)^{\!\exp(2 \widetilde{C}{} m)}.$$

Hence,

$$ T_1 \geq \pi(2qx)\biggl( \frac{y}{2q\ln x} \biggr)^{\!\exp(2 \widetilde{C}{} m)}.$$

(5.98)

From (5.96)–(5.98) we obtain

$$\#\bigl\{qx< p_n\leq 2qx \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\} \geq \pi(2qx)\biggl( \frac{y}{2q\ln x} \biggr)^{\!\exp(2 \widetilde{C}{} m)}.$$

Let us denote $\widetilde{c}{} _0(\varepsilon)$ by $c_0(\varepsilon)$ and $2 \widetilde{C}{}$ by $C$. Lemma 5.5 is proved. $\quad\Box$

Let us complete the proof of Theorem 1.1. Let $c_0(\varepsilon)$, $c$, $C$ be the quantities in Lemma 5.5. We will choose a quantity $\widetilde{c}{} _0(\varepsilon)$ and an absolute constant $\widetilde{C}{}$ later; they will be large enough. Let $\widetilde{c}{} _0(\varepsilon) \geq c_0(\varepsilon)$ and $\widetilde{C}{} \geq C$.

Let us prove the following statement.

Proposition 5.2.

Let $\varepsilon$ be a real number with $0<\varepsilon<1$. Let $t\in{\mathbb R},$ $y\in{\mathbb R},$ $m\in{\mathbb Z},$ $q\in{\mathbb Z},$ and $a\in{\mathbb Z}$ be such that

$$t\geq 100,\qquad \widetilde{c}{} _0(\varepsilon) \leq y \leq \ln \frac{t}{2\ln t},\qquad 1\leq m \leq c \kern1pt \varepsilon\ln y,\qquad 1\leq q \leq y^{1-\varepsilon},\qquad (a,q)=1.$$

Then

$$\#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\biggr\} \geq \pi(t)\biggl( \frac{y}{2q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)}. $$

(5.99)

Proof.

Indeed, since $t\geq 100$, we have $2\ln t \geq 1$. Hence,

$$y\leq \ln \frac{t}{2\ln t} \leq \ln t.$$

We have $q\leq y^{1-\varepsilon} \leq y \leq \ln t$. Therefore,

$$y\leq \ln \frac{t}{2\ln t} \leq \ln \frac{t}{2q}.$$

We put $x={t}/{2q}$. Then $x\in{\mathbb R}$, $y\in{\mathbb R}$, $m\in{\mathbb Z}$, $q\in{\mathbb Z}$, and $a\in{\mathbb Z}$ are such that

$$c_0(\varepsilon)\leq y \leq \ln x,\qquad 1\leq m \leq c \kern1pt \varepsilon\ln y,\qquad 1\leq q \leq y^{1-\varepsilon},\qquad (a,q)=1.$$

By Lemma 5.5, we have

$$\begin{aligned} \, &\#\bigl\{qx< p_n\leq 2qx \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\bigr\} \\[4pt] &\qquad\geq \pi(2qx)\biggl( \frac{y}{2q\ln x} \biggr)^{\!\exp(Cm)} \geq \pi(2qx)\biggl( \frac{y}{2q\ln x} \biggr)^{\!\exp( \widetilde{C}{} m)} \geq \pi(2qx)\biggl( \frac{y}{2q\ln(2qx)} \biggr)^{\!\exp( \widetilde{C}{} m)}. \end{aligned}$$

Returning to the variable $t$, we obtain (5.99). $\quad\Box$

Let us prove the following statement.

Proposition 5.3.

Let $\varepsilon$ be a real number with $0<\varepsilon<1$. Let $t\in{\mathbb R},$ $m\in{\mathbb Z},$ $q\in{\mathbb Z},$ and $a\in{\mathbb Z}$ be such that

$$t\geq 100,\qquad \widetilde{c}{} _0\Bigl( \frac{\varepsilon}{2} \Bigr) \leq \ln \frac{t}{2\ln t},\qquad 1\leq m \leq \frac{c}{4} \kern1pt \varepsilon\ln\ln t,\qquad 1\leq q \leq (\ln t)^{1-\varepsilon},\qquad (a,q)=1.$$

Then

$$\#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq \ln \frac{t}{2\ln t}\biggr\} \geq \pi(t)\biggl( \frac{1}{4q} \biggr)^{\!\exp( \widetilde{C}{} m)}.$$

Proof.

We need the following

Lemma 5.6.

Let $t$ be a real number with $t \geq 100$. Then

$$2\ln t \leq \sqrt{t},\qquad \ln \frac{t}{2\ln t}\geq \frac{1}{2}\ln t,\qquad \ln\ln \frac{t}{2\ln t}\geq \frac{1}{2}\ln\ln t,\qquad 1-\frac{\ln(2\ln t)}{\ln t} \geq \frac{1}{2}.$$

The proof of lemma 5.6 is a simple exercise in calculus, and we omit it.

We put

$$y=\ln \frac{t}{2\ln t}.$$

Since $t\geq 100$, we have (see Lemma 5.6)

$$\ln y = \ln\ln \frac{t}{2\ln t}\geq \frac{1}{2}\ln\ln t.$$

Therefore,

$$1\leq m\leq c \kern1pt \frac{\varepsilon}{2}\ln y.$$

We may assume that $\widetilde{c}{} _0(\varepsilon) \geq 2^{1/\varepsilon}$. Since $t \geq 100$, we have $t/(2\ln t)\leq t$ and

$$\widetilde{c}{} _0\Bigl( \frac{\varepsilon}{2} \Bigr) \leq \ln \frac{t}{2\ln t} \leq \ln t.$$

Hence,

$$t \geq \exp\Bigl( \widetilde{c}{} _0\Bigl( \frac{\varepsilon}{2} \Bigr) \Bigr) \geq \exp\bigl(2^{2/\varepsilon}\bigr).$$

Therefore,

$$ \frac{1}{2} (\ln t)^{1- \varepsilon/2}\geq (\ln t)^{1-\varepsilon}.$$

(5.100)

From (5.100) and the last inequality in Lemma 5.6 we find

$$\begin{aligned} \, y^{1 - \varepsilon/2} &= \biggl( \ln \frac{t}{2\ln t} \biggr)^{\!1- \varepsilon/2} =(\ln t)^{1- \varepsilon/2} \biggl( 1- \frac{\ln(2\ln t)}{\ln t} \biggr)^{\!1-\varepsilon/2} \geq (\ln t)^{1- \varepsilon/2} \biggl( \frac{1}{2} \biggr)^{\!1-\varepsilon/2} \\[4pt] &\geq \frac{1}{2}(\ln t)^{1-\varepsilon/2} \geq (\ln t)^{1-\varepsilon}. \end{aligned}$$

Since $1 \leq q \leq (\ln t)^{1-\varepsilon}$, we have $1 \leq q \leq y^{1 - \varepsilon/2}$. Applying Proposition 5.2 with $\varepsilon/2$ and the second inequality of Lemma 5.6, we have

$$\begin{aligned} \, &\#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq \ln \frac{t}{2\ln t}\biggr\} \\[4pt] &\qquad\geq \pi(t)\biggl( \frac{\ln(t/(2\ln t))}{2q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)} \geq \pi(t)\biggl( \frac{1}{4q} \biggr)^{\!\exp( \widetilde{C}{} m)}. \end{aligned}$$

The statement is proved. $\quad\Box$

Proposition 5.4.

Let $\varepsilon$ be a real number with $0<\varepsilon<1$. Let $t\in{\mathbb R},$ $y\in{\mathbb R},$ $m\in{\mathbb Z},$ $q\in{\mathbb Z},$ and $a\in{\mathbb Z}$ be such that

$$t\geq 100,\qquad \widetilde{c}{} _0\Bigl( \frac{\varepsilon}{2} \Bigr) \leq \ln \frac{t}{2\ln t} \leq y \leq \ln t,\qquad 1\leq m \leq \frac{c}{4} \kern1pt \varepsilon\ln y,\qquad 1\leq q \leq y^{1-\varepsilon},\qquad (a,q)=1.$$

Then

$$\#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\biggr\} \geq \pi(t)\biggl( \frac{y}{4q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)}.$$

Proof.

Since $y \leq \ln t$, we have

$$1\leq m \leq \frac{c}{4} \kern1pt \varepsilon\ln \ln t \qquad\text{and}\qquad 1\leq q \leq (\ln t)^{1-\varepsilon}.$$

Applying Proposition 5.3, we obtain

$$\begin{aligned} \, &\#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\biggr\} \\[2pt] &\qquad\geq \#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m}-p_n\leq \ln \frac{t}{2\ln t}\biggr\} \\[2pt] &\qquad\geq \pi(t)\biggl( \frac{1}{4q} \biggr)^{\!\exp( \widetilde{C}{} m)} \geq \pi(t)\biggl( \frac{y}{4q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)}. \ \quad\Box\end{aligned}$$

For $0<\varepsilon<1$ we define the quantity $t_0(\varepsilon)$ as follows:

$$t_0(\varepsilon) \geq 100,\qquad \ln \frac{t}{2\ln t} \geq \max\Bigl\{ \widetilde{c}{} _0\Bigl( \frac{\varepsilon}{2} \Bigr), \widetilde{c}{} _0(\varepsilon)\Bigr\}\quad \text{for any }\ t \geq t_0(\varepsilon).$$

Let us prove the following statement.

Proposition 5.5.

Let $\varepsilon$ be a real number with $0<\varepsilon<1$. Let $t\in{\mathbb R},$ $y\in{\mathbb R},$ $m\in{\mathbb Z},$ $q\in{\mathbb Z},$ and $a\in{\mathbb Z}$ be such that

$$t\geq t_0(\varepsilon),\ \quad \max\Bigl\{ \widetilde{c}{} _0\Bigl( \frac{\varepsilon}{2} \Bigr), \widetilde{c}{} _0(\varepsilon)\Bigr\} \leq y \leq \ln t,\ \quad 1\leq m \leq \frac{c}{4} \kern1pt \varepsilon\ln y,\ \quad 1\leq q \leq y^{1-\varepsilon},\ \quad (a,q)=1.$$

Then

$$\#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\biggr\} \geq \pi(t)\biggl( \frac{y}{4q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)}.$$

Proof.

Consider two cases. If

$$\ln \frac{t}{2\ln t} < y \leq \ln t,$$

then $t$, $y$, $m$, $q$, and $a$ satisfy the hypothesis of Proposition 5.4, which yields

$$\#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\biggr\} \geq \pi(t)\biggl( \frac{y}{4q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)}.$$

If

$$y \leq \ln \frac{t}{2\ln t},$$

then $t$, $y$, $m$, $q$, and $a$ satisfy the hypothesis of Proposition 5.2, which yields

$$\begin{aligned} \, \#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\biggr\} &\geq \pi(t)\biggl( \frac{y}{2q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)} \\[2pt] &\geq \pi(t)\biggl( \frac{y}{4q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)}. \ \quad\Box\end{aligned}$$

For $0<\varepsilon<1$ we put

$$\rho(\varepsilon) = \max\Bigl\{ \widetilde{c}{} _0\Bigl( \frac{\varepsilon}{2} \Bigr), \widetilde{c}{} _0(\varepsilon)\Bigr\} + t_0(\varepsilon).$$

Let us prove the following statement.

Proposition 5.6.

Let $\varepsilon$ be a real number with $0<\varepsilon<1$. Let $t\in{\mathbb R},$ $y\in{\mathbb R},$ $m\in{\mathbb Z},$ $q\in{\mathbb Z},$ and $a\in{\mathbb Z}$ be such that

$$\rho(\varepsilon) \leq y \leq \ln t,\qquad 1\leq m \leq \frac{c}{4} \kern1pt \varepsilon\ln y,\qquad 1\leq q \leq y^{1-\varepsilon},\qquad (a,q)=1.$$

Then

$$\#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\biggr\} \geq \pi(t)\biggl( \frac{y}{2q\ln t} \biggr)^{\!\exp(2 \widetilde{C}{} m)}.$$

Proof.

We have

$$\max\Bigl\{ \widetilde{c}{} _0\Bigl( \frac{\varepsilon}{2} \Bigr), \widetilde{c}{} _0(\varepsilon)\Bigr\}\leq y \leq \ln t \qquad\text{and}\qquad t\geq \exp(\rho(\varepsilon))\geq \rho(\varepsilon)\geq t_0(\varepsilon).$$

Applying Proposition 5.5, we obtain

$$\#\biggl\{\frac t2< p_n\leq t \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ \ p_{n+m} - p_n \leq y\biggr\} \geq \pi(t)\biggl( \frac{y}{4q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)}. $$

(5.101)

We may assume that $\widetilde{C}{} \geq 2$. Therefore, $\exp( \widetilde{C}{} m)\geq \widetilde{C}{} m\geq \widetilde{C}{} \geq 2$. Hence, $2\exp( \widetilde{C}{} m)\leq \exp(2 \widetilde{C}{} m)$. We have

$$\biggl( \frac{1}{4} \biggr)^{\!\exp( \widetilde{C}{} m)}=\biggl( \frac{1}{2} \biggr)^{\!2\exp( \widetilde{C}{} m)} \geq\biggl( \frac{1}{2} \biggr)^{\!\exp(2 \widetilde{C}{} m)}.$$

Further,

$$0<\frac{y}{q\ln t}\leq 1\qquad\Rightarrow\qquad \biggl( \frac{y}{q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)} \geq \biggl( \frac{y}{q\ln t} \biggr)^{\!\exp(2 \widetilde{C}{} m)}.$$

We obtain

$$ \biggl( \frac{y}{4q\ln t} \biggr)^{\!\exp( \widetilde{C}{} m)} \geq \biggl( \frac{y}{2q\ln t} \biggr)^{\!\exp(2 \widetilde{C}{} m)}.$$

(5.102)

Relations (5.101) and (5.102) imply the required assertion. $\quad\Box$

Let us denote $\rho(\varepsilon)$ by $c_0(\varepsilon)$, $c/4$ by $c$, and $2 \widetilde{C}{}$ by $C$. Theorem 1.1 is proved. $\quad\Box$

Proof of Corollary 1.1 .

Let $c_0(\varepsilon)$, $c$, and $C$ be the quantities in Theorem 1.1. We put

$$C_1= \max\biggl\{ \frac{2}{c}, c_0\biggl( \frac{1}{2} \biggr), C\biggr\}.$$

Let $m$ be a positive integer. Let $x\in{\mathbb R}$ and $y\in{\mathbb R}$ be such that $\exp(C_1m) \leq y \leq \ln x$. Then

$$y\geq \exp(C_1m) \geq C_1m\geq C_1 \geq c_0\biggl( \frac{1}{2} \biggr)\qquad\text{and}\qquad y\geq \exp(C_1m)\geq \exp\biggl( \frac{2}{c}m \biggr).$$

The last inequality implies

$$m\leq c \kern1pt \frac{1}{2}\ln y.$$

Putting $q=1$ and $a=1$, we have

$$c_0\biggl( \frac{1}{2} \biggr)\leq y \leq \ln x,\qquad 1\leq m \leq c \kern1pt \frac{1}{2}\ln y,\qquad 1\leq q\leq y^{1/2},\qquad (a,q)=1.$$

Applying Theorem 1.1 with $\varepsilon=1/2$, we see that

$$\begin{aligned} \, &\#\Bigl\{\frac x2< p_n\leq x \colon\, \, p_{n+m} - p_n \leq y\Bigr\} \\[4pt] &\qquad=\#\Bigl\{\frac x2< p_n\leq x \colon\, \, p_n\equiv\dots\equiv p_{n+m}\equiv a {\textstyle\pmod{q}} ,\ p_{n+m} - p_n \leq y\Bigr\} \\[4pt] &\qquad\geq \pi(x)\biggl( \frac{y}{2q\ln x} \biggr)^{\!\exp(Cm)} =\pi(x)\Bigl( \frac{y}{2\ln x} \Bigr)^{\exp(Cm)} \geq \pi(x)\Bigl( \frac{y}{2\ln x} \Bigr)^{\exp(C_1m)}. \end{aligned}$$

Let us redenote $C_1$ by $C$. Corollary 1.1 is proved. $\quad\Box$