2. Preliminary lemmas

We now state several results we will need throughout the argument. We will need the following bound on primes p such that $p+h$ is also prime and the singular series:

Lemma 2·1. Let $h\le x$ be an even non-zero integer and suppose that $y\geq4$ . The number of primes $p\in(x,x+y]$ such that $p+h$ is also prime is

\begin{equation*}\ll\frac{\mathfrak{S}(h)y}{(\log y)^2}.\end{equation*}

Furthermore, we have that

\begin{equation*}\sum_{h\leq x}\mathfrak{S}(h)\ll x.\end{equation*}

Proof. See [ Reference Montgomery and Vaughan26 , corollary 3·14] and the subsequent exercises.

We will also need Gallagher’s Lemma, which will reduce bounding integrals over the major and minor arcs to studying almost primes in short intervals.

Lemma 2·2 (Gallagher’s Lemma). Let $2<y<X/2$ . For arbitrary complex numbers $a_n$ , we have

(2·1) \begin{equation}\int_{|\beta|\leq\frac{1}{2y}}\left|\sum_{X<n\leq 2X}a_n e(\beta n)\right|^2d\beta\ll\frac{1}{y^2}\int_X^{2X}\left|\sum_{x<n\leq x+y}a_n\right|^2dx+y\left(\max_{X<n\leq 2X}|a_n|\right)^2.\end{equation}

Proof. This lemma is a modification of [ Reference Gallagher8 , lemma 1] (see also [ Reference Mikawa24 , lemma 1]).

Once we have applied Gallagher’s Lemma in the treatment of the major arcs, part of the error term is reduced to a Dirichlet character analogue of a problem on primes in almost all short intervals. We will use the following result adapted from the work of Koukoulopoulos [ Reference Koukoulopoulos16 ] to bound the second term arising in (1·7):

Lemma 2·3. Let $A\geq 1$ and $\varepsilon\in(0,{1}/{3}]$ be fixed. Let $X\geq1$ , $1\leq Q\leq{\Delta}/{X^{1/6+\varepsilon}}$ and $\Delta=X^{\theta}$ with ${1}/{6}+2\varepsilon\leq\theta\leq 1$ . Then we have that

\begin{equation*}\sum_{q\leq Q}\sum_{\chi(q)}\int_X^{2X}\left|\sum_{x<n\leq x+q\Delta}\left(\Lambda(n)\chi(n)-\delta_\chi\right)\right|^2 dx\ll\frac{Q^3\Delta^2X}{\log^AX},\end{equation*}

where we define $\delta_\chi=1$ if $\chi=\chi_0$ and $\delta_\chi=0$ otherwise.

Proof. The proof can be adapted from the proof given in [ Reference Koukoulopoulos16 , section 4]. Once the contribution of the imprimitive characters has been bounded in a standard way, the main difference we need to account for when compared with [ Reference Koukoulopoulos16 , theorem 1·2] is the presence of the square. To account for this, we adjust certain parameters (namely, replace D with $\sqrt{D}$ and adjust some of the powers of logarithms as needed) and do not require as many applications of the Cauchy–Schwarz inequality. We also note that in our argument we do not need the full strength of [ Reference Koukoulopoulos16 , theorem 1·2].

We use the following Parseval-type result to reduce the problem of finding almost primes in short intervals (cf. the first term of (1·7)) to finding cancellation in the mean square of the associated Dirichlet polynomial:

Lemma 2·4 (Parseval Bound). Let $a_n$ be arbitrary complex numbers, and let $2\leq h_1\leq h_2\leq{X}/{T_0^3}$ with $T_0\geq1$ . Define $F(s)\;:\!=\;\sum_{X<n\leq2X}{a_n}/{n^s}$ . Then

(2·2) \begin{equation}\begin{aligned}&\frac{1}{X}\int_X^{2X}\left|\frac{1}{h_1}\sum_{x<n\leq x+h_1}a_n-\frac{1}{h_2}\sum_{x<n\leq x+h_2}a_n\right|^2dx\\[5pt] &\ll\frac{1}{T_0^2}\max_{X<n\leq2X}|a_n|^2+\int_{T_0}^{\frac{X}{h_1}}|F(1+it)|^2dt+\max_{T\geq\frac{X}{h_1}}\frac{X}{Th_1}\int_T^{2T}|F(1+it)|^2dt.\end{aligned}\end{equation}

Proof. This is [ Reference Teräväinen28 , lemma 1], which is a variant of [ Reference Matomäki and Radziwiłł19 , lemma 14].

Finally, we record an exponential sum bound and a related bound on the sum of reciprocal of the distance to the nearest integer function which provide the necessary cancellation in the estimation of the minor arcs.

Lemma 2·5. Let $\beta\in\mathbb{R}$ , then

\begin{equation*}\sum_{n\leq x}e(\beta n)\ll\min\left(x,\frac{1}{\|\beta\|}\right).\end{equation*}

Proof. This is a standard result, see for example [ Reference Iwaniec and Kowalski13 , chapter 8, Equation (8·6)].

Lemma 2·6. If $1<X\leq Y$ and $\alpha\in\mathbb{R}$ satisfies $\alpha=a/q+O(q^{-2})$ with $(a,q)=1$ , then we have

\begin{equation*}\sum_{n\leq X}\min\left(\frac{Y}{n},\frac{1}{\|\alpha n\|}\right)\ll\left(\frac{Y}{q}+X+q\right)\log(qX).\end{equation*}

Proof. This is a standard result, see for example [ Reference Iwaniec and Kowalski13 , chapter 13, Page 346].

3. Applying the circle method

To prove Theorem 1·1, we will apply the Hardy–Littlewood circle method. Let $\varepsilon>0$ be small and fixed. We define $P>0$ according to the size of H as follows:

(3·1) \begin{equation}P\;:\!=\;\begin{cases}\log^{17+\varepsilon}X&\text{ if }\log^{19+\varepsilon}X\leq H\leq \exp\big((\log X)^{\varepsilon^3}\big),\\\exp\big((\log\log X)^2\big),&\text{ if }\exp\big((\log X)^{\varepsilon^3}\big)< H\leq X\log^{-A}X.\end{cases}\end{equation}

It will be more convenient throughout the argument to have a $\log$ weight attached to the indicator function of $E'_{\!\!2}$ as follows:

Definition 3·1. We define the arithmetic function $\varpi_2:\mathbb{N}\rightarrow\mathbb{R}$ to be

\begin{equation*}\varpi_2(n)=\begin{cases}\log p_2,&\text{ if }n=p_1p_2\text{ with }P<p_1\leq P^{1+\delta},\\0,&\text{ otherwise}.\end{cases}\end{equation*}

From now on we fix $\delta>0$ sufficiently small. We will prove the following asymptotic formula, from which Theorem 1·1 follows immediately after applying dyadic decomposition:

Theorem 3·2. Let $\varepsilon>0$ , $A>3$ be fixed and let $\log^{19+\varepsilon}X\leq H\leq X\log^{-A}X$ . Then, there exists some $\eta=\eta(\varepsilon)>0$ such that for all but at most $O(H\log^{-\eta} X)$ values of $0<|h|\leq H$ we have that

(3·2) \begin{equation}\sum_{X<n\leq 2X}\varpi_2(n)\varpi_2(n+h)=\mathfrak{S}(h)X\left(\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)^2+O\left(\frac{X}{\log^{\eta} X}\right),\end{equation}

where $\mathfrak{S}(h)$ is the singular series defined in (1·2).

We consider the integral

(3·3) \begin{equation}\int_{0}^{1}|S(\alpha)|^2e(-h\alpha)d\alpha=\sum_{X<m,n\leq2X}\varpi_2(m)\varpi_2(n)\int_0^1e(\alpha(m-n-h))d\alpha,\end{equation}

where for $\alpha\in(0,1)$ we define the exponential sum

\begin{equation*}S(\alpha)\;:\!=\;\sum_{X<n\leq 2X}\varpi_2(n)e(n\alpha).\end{equation*}

Then, by the integral identity

(3·4) \begin{equation}\int_0^1 e(nx)dx=\begin{cases}1,&\text{if }n=0,\\0,&\text{otherwise},\end{cases}\end{equation}

we have that the integral in (3·3) vanishes unless $m=n+h$ . Thus (3·3) becomes

(3·5) \begin{align}\int_{0}^{1}|S(\alpha)|^2e(-h\alpha)d\alpha&=\sum_{X<n\leq2X-h}\varpi_2(n)\varpi_2(n+h)\nonumber\\[5pt] &=\sum_{X<n\leq2X}\varpi_2(n)\varpi_2(n+h)+O(h\log^2 X).\end{align}

This error term will be negligible by our choice of H. Thus, except for an acceptable error, we can represent the correlation by an integral over the unit circle.

We split the domain of integration into the major and minor arcs. We define the major arcs $\mathfrak{M}$ to be the set of real $\alpha\in(0,1)$ such that

(3·6) \begin{equation}\left|\alpha-\frac{a}{q}\right|\leq\frac{1}{qQ}\text{ for some }1\leq q\leq Q_0,a<q,(a,q)=1\end{equation}

with $Q_0\;:\!=\;\log^{A'}X$ and $Q\;:\!=\;P\log X$ . Here we define $A'>0$ according to the size of H as follows

(3·7) \begin{equation}A'\;:\!=\;\begin{cases}1+\varepsilon^2,&\text{ if }\log^{19+\varepsilon}X\leq H\leq\exp\big((\log X)^{\varepsilon^3}\big),\\[5pt] 3+\varepsilon^2,&\text{ if }\exp\big((\log X)^{\varepsilon^3}\big)< H\leq X\log^{-A}X.\end{cases}\end{equation}

We define the minor arcs $\mathfrak{m}$ to be the rest of the circle, that is, the set of real $\alpha\in(0,1)$ such that

(3·8) \begin{equation}\left|\alpha-\frac{a}{q}\right|\leq\frac{1}{qQ}\text{ for some }Q_0< q\leq Q,a<q,(a,q)=1.\end{equation}

In Section 4, we will prove the following estimate for the integral over the minor arcs:

Proposition 3·5 (Minor arc estimate). Let $A>3$ be fixed and let $\varepsilon>0$ be fixed sufficiently small. Let $Q\log^{1+\varepsilon}X\leq H\leq X\log^{-A}X$ . With $\mathfrak{m}$ defined as in (3·8), for $\alpha\in\mathfrak{m}$ there exists some $\eta=\eta(\varepsilon)>0$ such that

(3·9) \begin{equation}\int_{\mathfrak{m}\cap[\alpha-\frac{1}{2H},\alpha+\frac{1}{2H}]}|S(\theta)|^2d\theta\ll\frac{X}{\log^{1+\eta}X}.\end{equation}

Sections 5-8 will be dedicated to proving the following expression for the integral over the major arcs:

Proposition 3·6 (Major arc integral). Let $A>3$ be fixed and let $\varepsilon>0$ be fixed sufficiently small. Let $\log^{19+\varepsilon}X\leq H\leq X\log^{-A}X$ . With $\mathfrak{M}$ defined as in (3·6) and $\delta>0$ sufficiently small, there exists some $\eta=\eta(\varepsilon)>0$ such that for all but at most $O(HQ_0^{-1/3})$ values of $0<|h|\leq H$ we have that

\begin{equation*}\int_\mathfrak{M} |S(\alpha)|^2e(-h\alpha) d\alpha=\mathfrak{S}(h)X\left(\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)^2+O\left(\frac{X}{\log^\eta X}\right),\end{equation*}

where $\mathfrak{S}(h)$ is the singular series given in (1·2).

Assuming Proposition 3·5 and Proposition 3·6, we can now prove Theorem 3·2.

Proof of Theorem 3·2. Following [ Reference Matomäki, Radziwiłł and Tao21 , section 3], by [ Reference Matomäki, Radziwiłł and Tao21 , proposition 3·1] we have that

\begin{equation*}\begin{aligned}&\sum_{0<|h|\leq H}\left|\sum_{X<n\leq2X}\varpi_2(n)\varpi_2(n+h)-\int_{\mathfrak{M}}|S(\alpha)|^2e(-h\alpha)d\alpha\right|^2\\[5pt] &\ll H\int_\mathfrak{m}|S(\alpha)|^2\int_{\mathfrak{m}\cap[\alpha-\frac{1}{2H},\alpha+\frac{1}{2H}]}|S(\beta)|^2d\beta d\alpha.\end{aligned}\end{equation*}

By Proposition 3·5, there exists some $\eta=\eta(\varepsilon)>0$ such that

\begin{equation*}\sup_{\alpha\in\mathfrak{m}}\int_{\mathfrak{m}\cap[\alpha-\frac{1}{2H},\alpha+\frac{1}{2H}]}|S(\beta)|^2d\beta\ll\frac{X}{\log^{1+\eta}X}.\end{equation*}

Noting that by partial summation and Mertens’ theorem we have the bound

\begin{equation*}\begin{aligned}\int_0^1|S(\alpha)|^2d\alpha&=\sum_{X<n\leq2X}\varpi_2^2(n)\ll \log X\sum_{X<n\leq2X}\varpi_2(n)\ll X\log X\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\\&\ll X\log X\end{aligned}\end{equation*}

we have that

\begin{equation*}\sum_{0<|h|\leq H}\left|\sum_{X<n\leq2X}\varpi_2(n)\varpi_2(n+h)-\int_{\mathfrak{M}}|S(\alpha)|^2e(-h\alpha)d\alpha\right|^2\ll\frac{HX^2}{\log^\eta X}.\end{equation*}

Applying Chebyshev’s inequality and Proposition 3·6 then gives the result.

4. The minor arcs

We first treat the integral over the minor arcs, proving Proposition 3·5 by following the proof of [ Reference Mikawa24 , lemma 8].

Proof of Proposition 3·5. Starting with the minor arc integral (3·9), we make the substitution $\theta=\alpha+\beta$ to see that

\begin{equation*}I\;:\!=\;\int_{\mathfrak{m}\cap[\alpha-\frac{1}{2H},\alpha+\frac{1}{2H}]}|S(\theta)|^2d\theta=\int_{\substack{\alpha+\beta\in\mathfrak{m}\\|\beta|\leq\frac{1}{2H}}}|S(\alpha+\beta)|^2d\beta.\end{equation*}

We apply Lemma 2·2 to the integral to get

\begin{equation*}I\ll\frac{1}{H^2}\int_{X}^{2X}\left|\sum_{x<n\leq x+H}\varpi_2(n)e(n\alpha)\right|^2dx+H\log^2 X.\end{equation*}

The second term is acceptable by our choice of H, so it remains to bound the first term. We first consider the case $H\leq \exp((\log X)^{\varepsilon^3})$ . We apply the Cauchy–Schwarz inequality to the integrand to get

(4·1) \begin{align} &\Bigg|\sum_{\substack{x<p_1p_2\leq x+H\\P<p_1\leq P^{1+\delta}}}(\log p_2)e(\alpha p_1p_2)\Bigg|^2 \nonumber\\[5pt]&\leq \Bigg(\sum_{P<m_1\leq P^{1+\delta}}|\mathbb{1}_{\mathbb{P}}(m_1)|^2\Bigg)\Bigg(\sum_{P<m_2\leq P^{1+\delta}}\Bigg|\sum_{x<m_2p\leq x+H}(\log p)e(\alpha m_2 p)\Bigg|^2\Bigg). \end{align}

The first term is $\ll {P^{1+\delta}}/{\log P}$ , while the second term is equal to

\begin{equation*}\sum_{\substack{x<mp_1,mp_2\leq x+H \\ P<m\leq P^{1+\delta}}} (\log p_1)(\log p_2) e(\alpha m(p_1-p_2)).\end{equation*}

Next, we perform the integration on this sum. We may trivially extend the domain of integration to [0,3X] as the integrand is positive. Define the set $\Omega\;:\!=\;\{x:0\leq x\leq 3X, mp_i-H\leq x<mp_i, i=1,2\}$ . Exchanging the order of integration and summation and noting that $X<x<mp_1,mp_2\leq x+H\leq3X$ , we have that

\begin{equation*}I\ll\frac{P^{1+\delta}}{H^2\log P}\sum_{P<m\leq P^{1+\delta}}\left|\sum_{X<mp_1,mp_2\leq3X}(\log p_1)(\log p_2)e(\alpha m(p_1-p_2))\cdot|\Omega|\right|.\end{equation*}

If $m|p_1-p_2|>H$ , then $|\Omega|=0$ . Since we have that $mp_i-H>X-H>0$ and $mp_i\leq3X$ for $i=1,2$ , the condition $0\leq x\leq 3X$ is weaker than the condition $\max(mp_1,mp_2)-H\leq x<\min(mp_1,mp_2)$ . Therefore, if $m|p_1-p_2|\leq H$ we have that $|\Omega|=H-m|p_1-p_2|$ .

We now split the sum into the diagonal terms, $p_1=p_2$ , and the off-diagonal terms, $p_1\neq p_2$ , denoted by $S_1$ and $S_2$ respectively. The diagonal terms contribute

(4·2) \begin{equation}S_1\ll \frac{P^{1+\delta}}{H\log P}\sum_{P<m\leq P^{1+\delta}}\sum_{\frac{X}{m}<p\leq\frac{3X}{m}}\log^2 p\ll \frac{XP^{1+\delta}\log(X/P)}{H}.\end{equation}

Now we bound the off-diagonal terms. Let $r=|p_1-p_2|$ . Noting that $0<mr\leq H$ , we need to bound

\begin{equation*}S_2\ll\frac{P^{1+\delta}}{H^2\log P}\sum_{0<r\leq H}\sum_{\substack{\frac{X}{P^{1+\delta}}<p_1,p_2\leq\frac{3X}{P}\\p_2=p_1+r}}(\log p_1)(\log p_2)\Bigg|\sum_{\substack{P<m\leq P^{1+\delta}\\0<m\leq H/r}}e(\alpha mr)(H-mr)\Bigg|.\end{equation*}

Noting that $0<m\leq H/r$ and $P<m\leq P^{1+\delta}$ , we have that $0<r\leq H/P$ . We apply partial summation and Lemma 2·5 to the sum over m to see that

\begin{equation*}S_2\ll \frac{P^{1+\delta}}{H\log P}\sum_{0<r\leq \frac{H}{P}}\min\left(\frac{H}{r},\frac{1}{\|\alpha r\|}\right)\sum_{\substack{\frac{X}{P^{1+\delta}}<p_1,p_2\leq\frac{3X}{P}\\p_2=p_1+r}}(\log p_1)(\log p_2).\end{equation*}

By partial summation followed by Lemma 2·1, we have that the sum over $p_1,p_2$ is bounded by

\begin{equation*}\sum_{\substack{\frac{X}{P^{1+\delta}}<p_1,p_2\leq\frac{3X}{P}\\p_2=p_1+r}}(\log p_1)(\log p_2)\ll\log^2X\sum_{\substack{\frac{X}{P^{1+\delta}}<p_1,p_2\leq\frac{3X}{P}\\p_2=p_1+r}}1\ll\frac{\mathfrak{S}(r)X}{P}.\end{equation*}

Therefore the contribution of the off-diagonal terms can be bounded by

\begin{equation*}S_2\ll\frac{XP^{1+\delta}}{HP\log P}\sum_{0<r\leq \frac{H}{P}}\min\left(\frac{H}{r},\frac{1}{\|\alpha r\|}\right)\mathfrak{S}(r).\end{equation*}

We have that $\mathfrak{S}(r)\ll\log\log r$ , so applying partial summation we have that

\begin{equation*}S_2\ll\frac{XP^{\delta}}{H\log P}\log\log X\sum_{0<r\leq \frac{H}{P}}\min\left(\frac{H}{r},\frac{1}{\|\alpha r\|}\right).\end{equation*}

Next, we apply Lemma 2·6 to the sum over r to get

(4·3) \begin{equation}S_2\ll\frac{XP^\delta }{H}\left(\frac{H}{Q_0}+\frac{H}{P}+Q\right)\log\frac{QH}{P},\end{equation}

recalling that since $\alpha\in\mathfrak{m}$ we have that $Q_0\leq q\leq Q$ . Since $H\leq\exp((\log X)^{\varepsilon^3})$ , we note that $(\log{QH}/{P})\ll(\log X)^{\varepsilon^3}$ . Therefore, combining the contributions of the diagonal terms (4·2) and the off-diagonal terms (4·3), we find

\begin{equation*}I\ll XP^\delta\left((\log X)^{\varepsilon^3}\left(\frac{1}{Q_0}+\frac{1}{P}+\frac{Q}{H}\right)+\frac{P\log(X/P)}{H}\right).\end{equation*}

By our choices of $Q_0=\log^{1+\varepsilon^2}X$ , $Q\log^{1+\varepsilon} X=P\log^{2+\varepsilon}X\ll H$ , we have that

\begin{equation*}I\ll\frac{X}{\log^{1+\eta}X}\end{equation*}

for some $\eta=\eta(\varepsilon)>0$ .

Otherwise, if $H>\exp((\log X)^{\varepsilon^3})$ , we split the sum over $P\le p_1\leq P^{1+\delta}$ into dyadic intervals before applying Cauchy–Schwarz in (4·1). We have that $\log({QH}/{P})\ll\log X$ and $Q_0=\log^{3+\varepsilon^2}X$ , so that the total contribution is

\begin{equation*}I\ll X\log\log X\left(\log X\left(\frac{1}{Q_0}+\frac{1}{P}+\frac{Q}{H}\right)\right)+\frac{XP\log(X/P)}{H}\ll\frac{X}{\log^{2+\eta}X}\end{equation*}

for some $\eta=\eta(\varepsilon)>0$ , which is acceptable.

5. The major arcs

We now shift our attention to evaluating the contribution of the integral over the major arcs. We will first expand the exponential sum $S(\alpha)$ in terms of Dirichlet characters and suitably approximate the contribution of the principal character, which will provide the main term. We will then evaluate this main term and the sequel will then be dedicated to bounding the error terms that arise from this expansion.

5·1. Expanding the exponential sum

First, we rewrite the integral over the major arcs by expanding the exponential sum $S(\alpha)$ in terms of Dirichlet characters. Recalling that $\alpha=a/q+\beta$ satisfies (3·6), we first define

\begin{align*}a(\alpha)&\;:\!=\;\frac{\mu(q)}{\varphi(q)}\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\sum_{X<n\leq2X}e(\beta n),\\b(\alpha)&\;:\!=\;\frac{1}{\varphi(q)}\sum_{\chi(q)}\tau(\overline{\chi})\chi(a)\sum_{X<n\leq2X}\left(\varpi_2(n)\chi(n)-\delta_\chi\sum_{P_1<p\leq P^{1+\delta}}\frac{1}{p}\right)e(\beta n),\\A^2(X)&\;:\!=\;\int_{\mathfrak{M}}|a(\alpha)|^2d\alpha,\\B^2(X)&\;:\!=\;\int_{\mathfrak{M}}|b(\alpha)|^2d\alpha,\end{align*}

where $\tau(\chi)$ denotes the Gauss sum as defined in (1·8) and $\delta_\chi=1$ when $\chi=\chi_0$ and is zero otherwise. We will now find the following expression for the integral over the major arcs, once we have expanded the exponential sum:

Lemma 5·1. Let $\mathfrak{M}$ be defined as in (3·6). We have that

\begin{equation*}\int_{\mathfrak{M}}|S(\alpha)|^2 e(-h\alpha)d\alpha=\int_{\mathfrak{M}}|a(\alpha)|^2 e(-h\alpha)d\alpha+O\left(A(X)B(X)+B^2(X)\right).\end{equation*}

Proof. Let $\alpha\in\mathfrak{M}$ , so that $\alpha={a}/{q}+\beta$ with $q\leq Q_0$ , $(a,q)=1$ and $|\beta|\leq{1}/{(qQ)}$ . Then

\begin{equation*}S(\alpha)=\sum_{X<n\leq 2X}\varpi_2(n)e\left(\frac{an}{q}\right)e(\beta n).\end{equation*}

By Definition 3·1, we have that $n=p_1p_2$ with $P<p_1\leq P^{1+\delta}$ . As we have $P>Q_0$ , we must have that $(p_1,q)=(p_2,q)=1$ and therefore that $(n,q)=1$ . We can now rewrite our expression for $S(\alpha)$ by applying the identity

(5·1) \begin{equation}e\left(\frac{a}{q}\right)=\frac{1}{\varphi(q)}\sum_{\chi(q)}\chi(a)\tau(\overline{\chi})\end{equation}

which holds for $(a,q)=1$ . This gives

(5·2) \begin{align}S(\alpha)&=\frac{1}{\varphi(q)}\sum_{\chi(q)}\tau(\overline{\chi})\chi(a)\sum_{X<n\leq2X}\varpi_2(n)\chi(n)e(\beta n)\end{align}

\begin{align}&=\frac{1}{\varphi(q)}\sum_{\chi(q)}\tau(\overline{\chi})\chi(a)\sum_{\substack{X<p_1p_2\leq2X\\P<p_1\leq P^{1+\delta}}}\chi(p_1)\chi(p_2)(\log p_2)e(\beta p_1p_2)\nonumber,\end{align}

where we have applied the definition of $\varpi_2$ in the last line. Now we approximate the contribution of the principal character, which will become the main term. First, note that since we have $q\leq Q_0<P<p_1$ we must have that $(p_1p_2,q)=1$ for $X<p_1p_2\leq2X$ , so we must have $(\log p_2)\chi_0(p_1)\chi_0(p_2)=\log p_2$ in these ranges. By the prime number theorem, we have that

\begin{equation*}\sum_{X<n\leq2X}\varpi_2(n)=\sum_{P<p_1\leq P^{1+\delta}}\sum_{\frac{X}{p_1}<p_2\leq\frac{2X}{p_1}}\log p_2\sim X\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}.\end{equation*}

Therefore we choose to approximate $\sum_{X<n\leq2X}\varpi_2(n)$ by

\begin{equation*}\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\sum_{X<n\leq2X}1=X\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}+O(1).\end{equation*}

Using this and the fact that $\tau(\chi_0)=\mu(q)$ , we approximate the contribution of the principal character to the exponential sum $S(\alpha)$ by

\begin{equation*}\frac{\mu(q)}{\varphi(q)}\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\sum_{X<n\leq2X}e(\beta n).\end{equation*}

Adding and subtracting this approximation in our expression (5·2) for $S(\alpha)$ we have that

\begin{align*}S(\alpha)=&\frac{\mu(q)}{\varphi(q)}\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\sum_{X<n\leq2X}e(\beta n)\\&+\frac{1}{\varphi(q)}\sum_{\chi(q)}\tau(\overline{\chi})\chi(a)\sum_{X<n\leq2X}\left(\varpi_2(n)\chi(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)e(\beta n)\\=&a(\alpha)+b(\alpha).\end{align*}

Finally, expanding the square and applying the Cauchy–Schwarz inequality, we have that

\begin{align*}\int_{\mathfrak{M}}|S(\alpha)|^2 e(-h\alpha)d\alpha&=\int_{\mathfrak{M}}|a(\alpha)+b(\alpha)|^2e(-h\alpha)d\alpha\\&=\int_{\mathfrak{M}}|a(\alpha)|^2 e(-h\alpha)d\alpha+O\left(A(X)B(X)+B^2(X)\right),\end{align*}

as required.

Thus, in order to prove Proposition 3·6 we need to evaluate $\int_{\mathfrak{M}}|a(\alpha)|^2 e(-h\alpha)d\alpha$ (which will also provide a bound for $A^2(X)$ ) and suitably bound $B^2(X)$ .

5·2. Evaluating the main term

In this section we evaluate the integral $\int_{\mathfrak{M}}|a(\alpha)|^2 e(-h\alpha)d\alpha$ , giving the main term of the asymptotic (and a bound for $A^2(X)$ ):

Proposition 5·2. Let $\varepsilon>0$ be fixed sufficiently small. Then for all but at most $O(HQ_0^{-1/3})$ values of $0<|h|\leq H$ we have that

\begin{equation*}\int_{\mathfrak{M}}|a(\alpha)|^2 e(-h\alpha)d\alpha=\mathfrak{S}(h)X\left(\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)^2+O\left(\frac{X}{\log^\eta X}\right),\end{equation*}

for some $\eta=\eta(\varepsilon)>0$ , where we define the singular series $\mathfrak{S}(h)$ as in (1·2).

Before we can prove Proposition 5·2, we need an expression involving the singular series $\mathfrak{S}(h)$ .

Lemma 5·3 (The Singular Series). Let h be a non-zero even integer and $Q_0$ be defined as in (3·7). Then, for all but at most $O(HQ_0^{-1/3})$ values of $0<|h|\leq H$ we have that

\begin{equation*}\sum_{q\leq Q_0}\frac{\mu^2(q)c_q(-h)}{\varphi^2(q)}=\mathfrak{S}(h)+O(Q_0^{-1/3}\log H).\end{equation*}

Proof. For similar results, see [ Reference Matomäki, Radziwiłł and Tao21 , page 39] and [ Reference Vaughan29 , page 35]. Rewriting the sum over q, we have that

\begin{equation*}\sum_{q\leq Q_0}\frac{\mu^2(q)c_q(-h)}{\varphi^2(q)}=\left(\sum_{q=1}^\infty-\sum_{q> Q_0}\right)\frac{\mu^2(q)c_q(-h)}{\varphi^2(q)}.\end{equation*}

The first term can be seen to be equal to $\mathfrak{S}(h)$ by calculating the Euler product. It remains to bound the tail of the sum. By [ Reference Vaughan29 , page 35], we have that

\begin{equation*}\sum_{0<h\leq H}\left|\sum_{q>Q_0}\frac{\mu^2(q)c_q(-h)}{\varphi^2(q)}\right|^2\ll\frac{H\log^2 H}{Q_0}.\end{equation*}

By Chebyshev’s inequality, we have for all but at most $O(HQ_0^{-1/3})$ values of h the bound

\begin{equation*}\sum_{q>Q_0}\frac{\mu^2(q)c_q(-h)}{\varphi^2(q)}\ll Q_0^{-1/3}\log H,\end{equation*}

as claimed.

We are now able to complete the proof of Proposition 5·2.

Proof of Proposition 5·2. Applying the definition of the major arcs (3·6) and expanding the square, we have that

(5·3) \begin{align} \int_{\mathfrak{M}}|a(\alpha)|^2 e(-h\alpha)d\alpha \end{align}

(5·4) \begin{align} & =\sum_{q\leq Q_0}\sum_{\substack{1\leq a\leq q\\(a,q)=1}}\int_{|\beta|\leq\frac{1}{qQ}}\Bigg|\frac{\mu(q)}{\varphi(q)}\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\sum_{X<n\leq2X}e(\beta n)\Bigg|^2e\left(\frac{-ha}{q}-h\beta\right)d\beta\nonumber\\[5pt] &=\left(\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)^2\sum_{q\leq Q_0}\frac{\mu^2(q)c_q(-h)}{\varphi^2(q)}\int_{|\beta|\leq\frac{1}{qQ}}\sum_{X<m,n\leq2X}e(\beta(m-n-h))d\beta\nonumber\\[5pt] &=\left(\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)^2\sum_{q\leq Q_0}\frac{\mu^2(q)c_q(-h)}{\varphi^2(q)}I_1, \end{align}

say. We rewrite the integral $I_1$ as

\begin{align*}I_1&=\left\{\int_0^1-\int_{\frac{1}{qQ}}^{1-\frac{1}{qQ}}\right\}\sum_{X<m,n\leq2X}e(\beta(m-n-h))d\beta\\&\;=\!:\;I_2-I_3,\end{align*}

say. To the first term $I_2$ , we apply the identity (3·4) to get

(5·5) \begin{equation}I_2=\sum_{\substack{X<m,n\leq2X\\m=n+h}}1=X+O(H)\end{equation}

and by our choice of H the error term is acceptable. Now we bound the integral $I_3$ . Note that $\beta$ is never an integer in the domain of integration, so applying Lemma 2·5 to the sums over m and n we have that

\begin{equation*}I_3=\int_{\frac{1}{qQ}}^{1-\frac{1}{qQ}}\sum_{X<m,n\leq2X}e(\beta(m-n-h))d\beta\ll\int_{\frac{1}{qQ}}^{1-\frac{1}{qQ}}\frac{1}{\|\beta\|^2}d\beta\ll qQ.\end{equation*}

Therefore, combining this with (5·5), we have that

\begin{equation*}I_1=X+O\left(qQ+H\right).\end{equation*}

We now substitute this expression for $I_1$ into (5·4) to get

\begin{equation*}\int_{\mathfrak{M}}|a(\alpha)|^2 e(-h\alpha)d\alpha=\sum_{q\leq Q_0}\frac{\mu^2(q)c_q(-h)}{\varphi^2(q)}\Bigg(X\Bigg(\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\Bigg)^2+O(qQ+H)\Bigg).\end{equation*}

To complete the proof, it remains to treat the sum over q. By Lemma 5·3 and our definitions of H and $Q_0$ , we find immediately that for all but at most $O(HQ_0^{-1/3})$ values of $0<|h|\leq H$ we have that

\begin{equation*}\int_{\mathfrak{M}}|a(\alpha)|^2 e(-h\alpha)d\alpha=\mathfrak{S}(h)X\left(\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)^2+O\left(\frac{X}{\log^\eta X}\right),\end{equation*}

for some $\eta=\eta(\varepsilon)>0$ , as claimed.

6. The error term of the major arcs

In order to complete the proof of Proposition 3·6, and therefore the proof of Theorem 3·2, we need to find sufficient cancellation in the error term $B^2(X)$ arising on the major arcs. In this section we prove the following bound for $B^2(X)$ , which immediately completes the proof of Proposition 3·6 when combined with Proposition 5·2:

Proposition 6·1. Let $\varepsilon>0$ be fixed sufficiently small, then there exists some $\eta=\eta(\varepsilon)>0$ such that

\begin{equation*}B^2(X)\ll\frac{X}{\log^{\eta}X}.\end{equation*}

6·1. Reduction of the problem

First, using Gallagher’s Lemma (Lemma 2·2), we will reduce the problem of estimating $B^2(X)$ to understanding almost primes in almost all short intervals. We first define the following: let $\Delta\;:\!=\;X/T_0^3$ with $T_0\;:\!=\;X^{1/100}$ and

(6·1) \begin{align} B_1(X)\;:\!=\;\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_X^{2X}\Bigg| & \frac{2}{qQ}\sum_{x<n\leq x+qQ/2}\Bigg(\chi(n)\varpi_2(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\Bigg) \\ & -\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}\Bigg(\chi(n)\varpi_2(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\Bigg)\Bigg|^2dx, \nonumber \end{align}

(6·2) \begin{align} B_2(X)&\;:\!=\;\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_X^{2X}\Bigg|\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}\Bigg(\chi(n)\varpi_2(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\Bigg)\Bigg|^2dx. \end{align}

Now we are able to state a bound for $B^2(X)$ in terms of $B_1(X)$ and $B_2(X)$ :

Proposition 6·2. We have that $B^2(X)\ll B_1(X)+B_2(X)$ .

Then, if we can prove that $B_i(X)\ll X\log^{-\eta}X$ for $i=1,2$ , we will immediately be able to conclude Proposition 6·1.

Proof. By definition, we have that $B^2(X)$ equals

\begin{equation*}\sum_{q\leq Q_0}\sum_{\substack{1\leq a\leq q\\(a,q)=1}}\int_{|\beta|\leq\frac{1}{qQ}}\Bigg|\frac{1}{\varphi(q)}\sum_{\chi(q)}\tau(\overline{\chi})\chi(a){\sum_{X<n\leq2X}}\bigg(\chi(n)\varpi_2(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\bigg)e(\beta n)\Bigg|^2d\beta.\end{equation*}

Expanding the square, we have that $B^2(X)$ equals

\begin{align*}&\sum_{q\leq Q_0}\frac{1}{\varphi^2(q)}\sum_{\chi,\chi'(q)}\tau(\overline{\chi})\tau(\chi')\sum_{\substack{1\leq a\leq q\\(a,q)=1}}\chi(a)\overline{\chi'}(a)\sum_{X<m\leq2X}\left(\chi(m)\varpi_2(m)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)\\&\times\sum_{X<n\leq2X}\left(\overline{\chi'}(n)\varpi_2(n)-\delta_{\chi'}\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)\int_{|\beta|\leq\frac{1}{qQ}}e(\beta (m-n))d\beta.\end{align*}

Now, using the definition of Dirichlet characters to trivially extend the sum over a to all $1\leq a\leq q$ , we may apply the character orthogonality relation

\begin{equation*}\sum_{a=1}^q\chi(a)\overline{\chi'}(a)=\begin{cases}\varphi(q),&\text{ if }\chi=\chi',\\[5pt]0,&\text{ if }\chi\neq\chi',\end{cases}\end{equation*}

to see that $B^2(X)$ is

\begin{align*}&=\sum_{q\leq Q_0}\frac{1}{\varphi(q)}\sum_{\chi(q)}|\tau(\overline{\chi})|^2\int_{|\beta|\leq\frac{1}{qQ}}\left|\sum_{X<n\leq2X}\left(\chi(n)\varpi_2(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)e(\beta n)\right|^2d\beta\\[5pt] &\ll\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{|\beta|\leq\frac{1}{qQ}}\left|\sum_{X<n\leq2X}\left(\chi(n)\varpi_2(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)e(\beta n)\right|^2d\beta,\end{align*}

where we have used that $\tau(\overline{\chi})\ll q^{1/2}$ in the last line. Now we apply Lemma 2·2 to the integral term to get that $B^2(X)$ is bounded by

(6·3) \begin{equation}\begin{aligned}\ll&\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_X^{2X}\Bigg|\frac{2}{qQ}\sum_{x<n\leq x+qQ/2}\Bigg(\chi(n)\varpi_2(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\Bigg)\Bigg|^2dx\\[5pt]&+Q\log^2X\sum_{q\leq Q_0}\sum_{\chi(q)}\frac{q^2}{\varphi(q)}.\end{aligned}\end{equation}

The second term contributes

\begin{equation*}Q\log^2X\sum_{q\leq Q_0}\sum_{\chi(q)}\frac{q^2}{\varphi(q)}\ll QQ_0^3\log^2X\end{equation*}

to $B^2(X)$ , which is negligible. Let $\Delta=X/T_0^3$ with $T_0=X^{1/100}$ , then we have that $B^2(X)$ is bounded by

\begin{align*}\ll&\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_X^{2X}\Bigg|\frac{2}{qQ}\sum_{x<n\leq x+qQ/2}\Bigg(\chi(n)\varpi_2(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\Bigg)\\[5pt] &\qquad\qquad\qquad\qquad\quad -\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}\Bigg(\chi(n)\varpi_2(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\Bigg)\Bigg|^2dx\\[5pt]&+\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_X^{2X}\Bigg|\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}\Bigg(\chi(n)\varpi_2(n)-\delta_\chi\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\Bigg)\Bigg|^2dx\\=&B_1(X)+B_2(X),\end{align*}

as claimed.

6·2. Bounding $B_2(X)$

First, we prove the the following estimate for $B_2(X)$ , which will be reduced to a Dirichlet character analogue of a problem on primes in almost all short intervals.

Proposition 6·3. Let $C>0$ be fixed, then with $B_2(X)$ as defined in (6·2) we have

\begin{equation*}B_2(X)\ll\frac{X}{\log^C X}.\end{equation*}

Proof. We separate the cases $\chi=\chi_0$ and $\chi\neq\chi_0$ . If $\chi=\chi_0$ , we have that

\begin{align*}&\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}\left(\varpi_2(n)-\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)\\[5pt]&=\frac{2}{q\Delta}\sum_{P<p_1\leq P^{1+\delta}}\sum_{\frac{x}{p_1}<p_2\leq\frac{x+q\Delta/2}{p_1}}\log p_2-\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}+O\left(\frac{1}{q\Delta}\right).\end{align*}

We now apply the prime number theorem in short intervals (see, for example, [ Reference Iwaniec and Kowalski13 , chapter 10·5]), finding that

\begin{equation*}\frac{2}{q\Delta}\sum_{P<p_1\leq P^{1+\delta}}\sum_{\frac{x}{p_1}<p_2\leq\frac{x+q\Delta/2}{p_1}}\log p_2=\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}+O\left(\exp(-c(\log x)^{1/3-\varepsilon})\right).\end{equation*}

Substituting this back into the above, we have that

\begin{equation*}\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}\left(\varpi_2(n)-\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)=O\left(\exp(-c(\log x)^{1/3-\varepsilon})\right).\end{equation*}

Returning to the integral and summing over q, we find that the contribution of the principal character to $B_2(X)$ is

\begin{equation*}\ll X Q_0\exp(-c'(\log X)^{1/3-\varepsilon}),\end{equation*}

which is acceptable by the choice of $Q_0$ .

We now consider the case $\chi\neq\chi_0$ . By the definition of $\varpi_2$ and the Cauchy–Schwarz inequality, we have that

(6·4) \begin{align}B_2(X)&=\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\substack{\chi(q)\\ \chi\neq\chi_0}}\int_X^{2X}\left|\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}\chi(n)\varpi_2(n)\right|^2dx \nonumber \\[5pt] &=\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\substack{\chi(q)\\\chi\neq\chi_0}}\int_X^{2X}\Bigg|\frac{2}{q\Delta}\sum_{P<p_1\leq P^{1+\delta}}\chi(p_1)\sum_{\frac{x}{p_1}<p_2\leq\frac{x+q\Delta/2}{p_1}}\chi(p_2)\log p_2\Bigg|^2dx \nonumber \\&\ll\frac{P^{1+\delta}}{\log P}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{P<p_1\leq P^{1+\delta}}\frac{4}{(q\Delta)^2}\sum_{\substack{\chi(q)\\\chi\neq\chi_0}}\int_X^{2X}\Bigg|\sum_{\frac{x}{p_1}<p_2\leq\frac{x+q\Delta/2}{p_1}}\chi(p_2)\log p_2\Bigg|^2dx.\end{align}

We make the change of variables $u=x/p_1$ to the integral, so that

\begin{equation*}\int_X^{2X}\Bigg|\sum_{\frac{x}{p_1}<p_2\leq\frac{x+q\Delta/2}{p_1}}\chi(p_2)\log p_2\Bigg|^2dx=p_1\int_{X/p_1}^{2X/p_1}\Bigg|\sum_{u<p_2\leq u+\frac{q\Delta}{2p_1}}\chi(p_2)\log p_2\Bigg|^2du.\end{equation*}

First, in the case $H\leq\exp((\log X)^{\varepsilon^3})$ , we now apply Lemma 2·3 to get that

\begin{equation*}B_2(X)\ll\frac{P^{1+\delta}}{\log P}\frac{X Q_0\log\log Q_0}{\log^DX}\sum_{P<p\leq P^{1+\delta}}\frac{1}{p^2}\ll\frac{X}{\log^C X}\end{equation*}

for $C>0$ , as required. In the case $H>\exp((\log X)^{\varepsilon^3})$ , we split the sum over $P<p_1\leq P^{1+\delta}$ in (6·4) into dyadic intervals and again apply Lemma 2·3 to obtain the required bound.

6·3. Bounding $B_1(X)$

It now remains to prove the required bound for $B_1(X)$ . This problem can be reduced to finding cancellation in the mean square of a Dirichlet polynomial.

Proposition 6·4. Let $\varepsilon>0$ be fixed sufficiently small. With $B_1(X)$ as defined in (6·1), there exists some $\eta=\eta(\varepsilon)>0$ such that

\begin{equation*}B_1(X)\ll\frac{X}{\log^\eta X}.\end{equation*}

To prove this result, we will need the following variant of a result of Teräväinen [ Reference Teräväinen28 ] on the mean square of the Dirichlet polynomial

(6·5) \begin{equation}F(s,\chi)\;:\!=\;\sum_{\substack{X<p_1p_2\leq2X\\P<p_1\leq P^{1+\delta}}}\frac{\chi(p_1)\chi(p_2)}{(p_1p_2)^s},\end{equation}

to be proved in Section 8:

Proposition 6·5. Let $\varepsilon>0$ be fixed sufficiently small. Define $T_0=X^{1/100}$ and $F(s,\chi)$ to be the Dirichlet polynomial defined in (6·5), with P and $\delta>0$ as in Section 3. Then, for $T\geq T_0$ , there exists some $\eta=\eta(\varepsilon)>0$ such that

(6·6) \begin{equation}\begin{aligned}B_3(X)&\;:\!=\;\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{T_0}^{T}|F(1+it,\chi)|^2dt\\[5pt] &\ll \frac{1}{Q_0\log^{2+\eta}X} \sum_{q\leq Q_0} \left(\frac{qTP\log X}{X}+\frac{q}{\varphi(q)}\right).\end{aligned}\end{equation}

Proof of Proposition 6·4. assuming Proposition 6·5. First we consider when $\chi=\chi_0$ as we have a different summand in this case. We have

(6·7) \begin{equation}\frac{2}{qQ}\sum_{x<n\leq x+qQ/2}\left(\varpi_2(n)-\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right)-\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}\left(\varpi_2(n)-\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\right).\end{equation}

We first consider the contribution of the second and fourth terms, namely

(6·8) \begin{equation}\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\left(\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}1-\frac{2}{qQ}\sum_{x<n\leq x+qQ/2}1\right)\ll\frac{1}{qQ}+\frac{1}{q\Delta}.\end{equation}

Returning to our expression for $B_1(X)$ , by our choice of $Q_0,Q$ and $\Delta$ we have that (6·8) contributes

\begin{equation*}\ll\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\int_X^{2X}\frac{1}{(qQ)^2}dx\ll\frac{X}{Q^2}\sum_{q\leq Q_0}\frac{1}{q\varphi(q)}\ll\frac{X}{Q^2},\end{equation*}

which is acceptable. Therefore, when considering the principal character $\chi_0$ , we need only to bound

\begin{align*}&\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_X^{2X}\left|\frac{2}{qQ}\sum_{x<n\leq x+qQ/2}\varpi_2(n)-\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}\varpi_2(n)\right|^2dx\\[5pt]=&\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_X^{2X}\left|\frac{2}{qQ}\sum_{x<n\leq x+qQ/2}\varpi_2(n)\chi_0(n)-\frac{2}{q\Delta}\sum_{x<n\leq x+q\Delta/2}\varpi_2(n)\chi_0(n)\right|^2dx,\end{align*}

noting that in the range of summation we must have $(n,q)=1$ , i.e. $\varpi_2(n)\chi_0(n)=\varpi_2(n)$ for each $X<n\leq2X$ . Thus, from now on we are able to unify the treatment of the principal character $\chi_0$ with the rest of the characters modulo q at the cost of a negligible error.

We now apply Lemma 2·4 with $h_1=qQ/2$ and $h_2=q\Delta/2$ to the integral with respect to x to get

\begin{equation*}\begin{aligned}B_1(X)\ll X\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\Bigg(\frac{\log^2 X}{T_0^2}&+\int_{T_0}^{\frac{2X}{qQ}}|F_1(1+it,\chi)|^2dt\\&+\max_{T\geq\frac{2X}{qQ}}\frac{X}{TqQ}\int_T^{2T}|F_1(1+it,\chi)|^2dt\Bigg),\end{aligned}\end{equation*}

with $T_0=X^{1/100}$ and

(6·9) \begin{equation}F_1(s,\chi)\;:\!=\;\sum_{X<n\leq2X}\frac{\varpi_2(n)\chi(n)}{n^s}=\sum_{\substack{X<p_1p_2\leq2X\\P<p_1\leq P^{1+\delta}}}\frac{\chi(p_1)\chi(p_2)\log p_2}{(p_1p_2)^s}.\end{equation}

The choice of $T_0$ ensures that the first term is negligible. Applying partial summation, we have that $B_1(X)$ is bounded by

\begin{equation*}\ll X\log^2 X\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\left(\int_{T_0}^{\frac{2X}{qQ}}|F(1+it,\chi)|^2dt+\max_{T\geq\frac{2X}{qQ}}\frac{X}{TqQ}\int_T^{2T}|F(1+it,\chi)|^2dt\right).\end{equation*}

We now apply Proposition 6·5. Note that we have $P\log X = Q$ , so that the first term in our bound for $B_1(X)$ is bounded by

\begin{equation*}\ll \frac{X}{Q_0\log^{\eta}X} \sum_{q\leq Q_0} \left(\frac{P\log X}{Q}+\frac{q}{\varphi(q)}\right)\ll\frac{X}{\log^\eta X},\end{equation*}

as needed. For the second term, we want to bound

\begin{equation*}\frac{X^2\log^2X}{Q}\sum_{q\leq Q_0}\frac{1}{\varphi(q)}\max_{T\geq \frac{2X}{qQ}}\frac{1}{T}\sum_{\chi(q)}\int_T^{2T}|F(1+it,\chi)|^2dt.\end{equation*}

Applying Proposition 6·5, we have the bound

\begin{equation*}\ll \frac{X^2}{Q_0Q\log^\eta X}\sum_{q\leq Q_0} \max_{T\geq\frac{2X}{qQ}}\left(\frac{P\log X}{X}+\frac{1}{T\varphi(q)}\right)\ll\frac{X}{\log^\eta X},\end{equation*}

again using that $P\log X=Q$ . Overall we have that

\begin{equation*}B_1(X)\ll\frac{X}{\log^\eta X},\end{equation*}

for some $\eta=\eta(\varepsilon)>0$ , as required.

7. Preliminaries on Dirichlet polynomials

Before we can prove Proposition 6·5, we first need the following preliminary lemmas on Dirichlet polynomials.

7·1. Pointwise bound

After we factorise our Dirichlet polynomial, there will be instances where the best we can do is use a pointwise bound. Before we state this bound, we need the following definition of a well-spaced set.

Definition 7·1 (Well-spaced set). We say a set $\mathcal{T}$ is well-spaced if for any $t,u\in\mathcal{T}$ with $t\neq u$ we have that $|t-u|\geq1$ .

Lemma 7·2 (Pointwise bound). Let $\mathcal{S}$ be a set of pairs $(t,\chi)$ with $t\in[-T,T]$ and $\chi$ a Dirichlet character mod q which is well-spaced (i.e. if $(t,\chi),(u,\chi)\in\mathcal{S}$ then $|t-u|\geq1$ ). Suppose that $\min\{|t|:(t,\chi)\in\mathcal{S}\}\gg\log^AN$ for all $A>0$ if $\chi=\chi_0$ . Let

\begin{equation*}P(s,\chi)\;:\!=\;\sum_{\substack{N<p_1\cdots p_k\leq2N\\p_1,\ldots,p_k\geq z}}\frac{\chi(p_1)\cdots \chi(p_k)}{(p_1\cdots p_k)^{1+it}},\end{equation*}

where $z\geq\exp(\log^{9/10}N)$ . Then for any $C>0$ we have

\begin{equation*}|P(1+it,\chi)|\ll\frac{1}{\log^CN}.\end{equation*}

Proof. This is [ Reference Harman11 , lemma 10·7].

Definition 7·3 (Prime-factored polynomial, [ Reference Teräväinen28 ]). Let $M\geq1$ and

\begin{equation*}M(s,\chi)=\sum_{M<m\leq2M}\frac{a_m\chi(m)}{m^s}\end{equation*}

be a Dirichlet polynomial with $|a_m|\ll d_r(m)$ for some fixed r. We say that $M(s,\chi)$ is prime-factored if for each $C>0$ we have

\begin{equation*}\sup_{(t,\chi)\in\mathcal{S}}|M(1+it,\chi)|\ll\frac{1}{\log^CM}\end{equation*}

when $\exp((\log M)^{1/3})\leq t\leq M^{C\log\log M}$ , where $\mathcal{S}$ is as defined in the previous lemma.

7·2. Decomposing Dirichlet polynomials

As in the work of Teräväinen [ Reference Teräväinen28 ] and Matomäki, Radziwiłł [ Reference Matomäki and Radziwiłł19 ], we take advantage of the bilinear structure to factorise our Dirichlet polynomial.

Lemma 7·4 (Factorisation of Dirichlet polynomials). Define

\begin{equation*}F(s)\;:\!=\;\sum_{\substack{X<mn\leq2X\\M\leq m\leq M'}}\frac{a_mb_n}{(mn)^s}\end{equation*}

for some $M'>M\geq2$ and arbitrary complex numbers $a_m,b_n$ . Let $U\geq1$ and define

\begin{equation*}A_v(s)\;:\!=\;\sum_{e^{\frac{v}{U}}\leq m<e^{\frac{v+1}{U}}}\frac{a_m}{m^s},\quad B_v(s)\;:\!=\;\sum_{Xe^{-\frac{v}{U}}<n\leq2Xe^{-\frac{v}{U}}}\frac{b_n}{n^s}.\end{equation*}

Then

(7·1) \begin{equation}F(s)=\sum_{v\in I\cap\mathbb{Z}}A_v(s)B_v(s)+\sum_{\substack{k\in[Xe^{-1/U},Xe^{1/U}]\\\text{or }k\in[2X,2Xe^{1/U}]}}\frac{d_k}{k^s}\end{equation}

where $I=[U\log M,U\log M']$ and

(7·2) \begin{equation}|d_k|\leq\sum_{k=mn}|a_mb_n|.\end{equation}

Proof. This is [ Reference Teräväinen28 , lemma 2] (see also [ Reference Matomäki and Radziwiłł19 , lemma 12]).

In some cases we will use the Heath–Brown identity to decompose a long polynomial into products of shorter polynomials.

Lemma 7·5 (Heath–Brown decomposition). Let $k\geq1$ be a fixed integer, $T\geq2$ and fix $\varepsilon>0$ . Define the Dirichlet polynomial $P(s,\chi)\;:\!=\;\sum_{P\leq p<P'}\chi(p)p^{-s}$ with $P\gg T^\varepsilon$ , $P'\in\left[P+{P}/{\log T},2P\right]$ . Then, there exist Dirichlet polynomials $Q_1(s,\chi),\ldots,Q_L(s,\chi)$ and a constant $C>0$ such that $L\leq\log^{C}X$ and

\begin{equation*}|P(1+it,\chi)|\ll(\log^{C}X)(|Q_1(1+it,\chi)|+\cdots+|Q_L(1+it,\chi)|)\end{equation*}

for all $t\in[-T,T]$ . Here, each $Q_j(s,\chi)$ is of the form

\begin{equation*}Q_j(s,\chi)=\prod_{i\leq J_j}M_i(s,\chi),\quad J_j\leq 2k,\end{equation*}

where each $M_i(s,\chi)$ is a prime-factored Dirichlet polynomial (depending on j) of the form

\begin{equation*}\sum_{M_i<n\leq2M_i}\frac{\chi(n)\log n}{n^s},\;\sum_{M_i<n\leq2M_i}\frac{\chi(n)}{n^s},\text{ or }\sum_{M_i<n\leq2M_i}\frac{\mu(n)\chi(n)}{n^s},\end{equation*}

whose lengths satisfy $M_1\cdots M_J=X^{1+o(1)},M_i\gg\exp\left({\log P}/{\log\log P}\right)$ . Furthermore, if in fact $M_i>X^{1/k}$ , then $M_i(s,\chi)$ is of the form

(7·3) \begin{equation}\sum_{M_i<n\leq2M_i}\frac{\chi(n)\log n}{n^s}\text{ or }\sum_{M_i<n\leq2M_i}\frac{\chi(n)}{n^s}.\end{equation}

Proof. This is the Dirichlet character analogue of [ Reference Teräväinen28 , lemma 10], which follows from the same argument.

7·3. Mean value theorems for Dirichlet polynomials

Now we state two mean value theorems, the first being the classical result:

Lemma 7·6 (Mean Value Theorem). Let $q,X\geq 1$ and let $a_n$ be arbitrary complex numbers with $F(s,\chi)\;:\!=\;\sum_{X<n\leq2X}\frac{a_n\chi(n)}{n^s}$ . Then

\begin{equation*}\sum_{\chi(q)}\int_{-T}^{T}|F(it,\chi)|^2dt\ll\left(\varphi(q)T+\frac{\varphi(q)}{q}X\right)\sum_{\substack{X<n\leq2X\\(n,q)=1}}|a_n|^2.\end{equation*}

Proof. See, for example, [ Reference Montgomery25 , chapter 6, equation (6·14)].

Next we state a variant of the mean value theorem which will allow us to save a $\log X$ in certain parts of the proof.

Lemma 7·7. With the same assumptions as Lemma 7·6, we have that

\begin{equation*}\sum_{\chi(q)}\int_{-T}^{T}|F(it,\chi)|^2dt\ll T\varphi(q)\Bigg(\sum_{\substack{X<n\leq2X\\(n,q)=1}} |a_n|^2+\sum_{\substack{1\leq h\leq\frac{X}{T}\\q\mid h}}\sum_{\substack{X<n\leq2X\\(n(n+h),q)=1}}|a_{n+h}||a_n|\Bigg).\end{equation*}

Proof. This is the Dirichlet character analogue of [ Reference Teräväinen28 , lemma 4], which follows from [ Reference Iwaniec and Kowalski13 , lemma 7·1]. The proof is contained in the proof of [ Reference Matomäki, Radziwiłł and Tao22 , lemma 5·2].

After factorising the Dirichlet polynomial F and splitting the domain of integration according to the size of the factors, there will be cases where the mean value is taken over a well-spaced set. In this case, we will apply the Halász–Montgomery inequality:

Lemma 7·8 (Halász–Montgomery Inequality). Let $T\geq1$ , $q\geq2$ . Let $\mathcal{S}$ be a well-spaced set of pairs $(t,\chi)$ with $t\in[-T,T]$ and $\chi$ a Dirichlet character mod q. With the same assumptions as Lemma 7·6, we have that

\begin{equation*}\sum_{(t,\chi)\in\mathcal{S}}|F(it,\chi)|^2\ll\left(\frac{\varphi(q)}{q}X+|\mathcal{S}|(qT)^{1/2}\right)(\log(2qT))\sum_{\substack{X<n\leq2X\\(n,q)=1}} |a_n|^2.\end{equation*}

Proof. This is [ Reference Klurman, Mangerel and Teräväinen15 , lemma 7·4].

7·4. Large value theorems

There will be subsets of the domain of integration where a short Dirichlet polynomial factor is large, in which case we apply the following large value theorem.

Lemma 7·9 (Large value theorem). Let $P\geq1,V>0$ , $|a_p|\leq1$ and $F(s,\chi)=\sum_{P<p\leq2P}{a_p\chi(p)}/{p^s}$ . Let $\mathcal{S}\subset[-T,T]\times\{\chi\mod q\}$ be a well-spaced set such that $|F(1+it,\chi)|\geq V$ for all $(t,\chi)\in\mathcal{S}$ . Then

\begin{equation*}|\mathcal{S}|\ll (qT)^{\frac{2\log(1/V)}{\log P}}V^{-2}\exp\left((1+o(1))\frac{\log(qT)\log\log(qT)}{\log P}\right).\end{equation*}

Proof. This is the Dirichlet character analogue of [ Reference Teräväinen28 , lemma 6] and [ Reference Matomäki and Radziwiłł19 , lemma 8]. Also see [ Reference Klurman, Mangerel and Teräväinen15 , lemma 7·5].

Remark 7·10. As remarked in [ Reference Teräväinen28 , remark 6], this lemma can still be applied to polynomials with coefficients not only supported on the primes as long as we have $P\gg X^{\varepsilon}$ , as will be the case in our application.

Alternatively, in the case that we have a longer Dirichlet polynomial factor which is large, we will apply a result of Jutila on large values.

Lemma 7·11 (Jutila’s large value theorem). Let $\varepsilon>0$ be fixed, $|a_n|\leq d_r(n)$ for some fixed r and $F(s,\chi)=\sum_{X<n\leq2X}{a_n\chi(n)}/{n^s}$ . Let k be a fixed positive integer and $\mathcal{S}\subset[-T,T]\times\{\chi\mod q\}$ be a well-spaced set such that $|F(1+it,\chi)|\geq V$ for all $(t,\chi)\in\mathcal{S}$ . Then,

\begin{equation*}|\mathcal{S}|\ll\left(V^{-2}+\left(\frac{qTV^{-4}}{X^2}\right)^k+\frac{qTV^{-8k}}{X^{2k}}\right)(qTX)^{\varepsilon}.\end{equation*}

Proof. This is the first bound of the main theorem in [ Reference Jutila14 ].

7·5. Moments of Dirichlet polynomials

After decomposing the Dirichlet polynomial using the Heath–Brown decomposition (Lemma 7·5), we can have a long polynomial which is the partial sum of a Dirichlet L-function (or its derivative). In this case, we will apply the Cauchy–Schwarz inequality to enable us to use the following bound on the twisted fourth moment of such sums:

Lemma 7·12 (Twisted fourth moment estimate). Let $Q_0\leq T^\varepsilon$ , $T^\varepsilon\leq T_0\leq T$ , $1\leq M,N\leq T^{1+o(1)}$ and define the Dirichlet polynomials

\begin{align*}N(s,\chi)&=\sum_{N<n\leq2N}\frac{\chi(n)}{n^s}\text{ or }\sum_{N<n\leq2N}\frac{\chi(n)\log n}{n^s},\\M(s,\chi)&=\sum_{M<m\leq2M}\frac{a_m\chi(m)}{m^s},\end{align*}

with $a_m$ any complex numbers. Then we have that

(7·4) \begin{align}\sum_{q\leq Q_0}\frac{1}{\varphi(q)}\sum_{\chi(q)}\int_{T_0}^{T} & |N(1+it,\chi)|^4|M(1+it,\chi)|^2dt\nonumber\\&\ll\left(\frac{Q_0T}{MN^2}(1+M^2(Q_0T)^{-1/2})+\frac{1}{T_0}\right)(Q_0T)^{\varepsilon}\max_{M<m\leq2M}|a_m|^2.\end{align}

Proof. This is the Dirichlet character analogue of [ Reference Teräväinen28 , lemma 9] and we follow the same argument. In the case $1\leq qt\leq N$ , we use partial summation and the hybrid result of Fujii, Gallagher and Montgomery [ Reference Fujii, Gallagher and Montgomery7 ]

\begin{equation*}\sum_{n\leq N}\chi(n)n^{it}=\frac{\delta_\chi \varphi(q)N^{1+it}}{q(1+it)}+O((q\tau)^{1/2}\log(q\tau)),\end{equation*}

with $\tau\;:\!=\;|t|+2$ in place of the zeta sum bound to get that

\begin{align*}\sum_{q\leq Q_0}&\frac{1}{\varphi(q)}\sum_{\chi(q)}\int_{T_0}^{T}|N(1+it,\chi)|^4|M(1+it,\chi)|^2dt\\&\ll T^{\varepsilon}\max_{M<m\leq2M}|a_m|^2\sum_{q\leq Q_0}\frac{1}{\varphi(q)}\sum_{\chi(q)}\int_{T_0}^T\left(\frac{\varphi(q)}{q(1+|t|)}\right)^4+\frac{\log^4(q\tau)}{(q\tau)^2}dt\\&\ll \frac{T^{\varepsilon}}{T_0}\max_{M<m\leq2M}|a_m|^2,\end{align*}

providing the third term of (7·4).

In the case $N\leq qt\leq Q_0T$ , we apply in place of Watt’s twisted moment result its Dirichlet character analogue [ Reference Harman, Watt and Wong12 , theorem 2] to obtain the first two terms of (7·4).

8. Bounding the mean value of a Dirichlet polynomial

We are now able to prove Proposition 6·5, completing the proof of Proposition 6·1 and consequently Theorem 3·2. We will adapt the argument appearing in [ Reference Teräväinen28 , sections 2-4]. We first factorise the Dirichlet polynomial $F(s,\chi)$ before bounding the contribution of the remainder terms, that is, the second term of (7·1).

Lemma 8·1. Let $\varepsilon>0$ be fixed sufficiently small and $T_0=X^{1/100}$ . Denote

\begin{equation*}G_v(s,\chi)\;:\!=\;\sum_{e^{\frac{v}{U}}<p\leq e^{\frac{v+1}{U}}}\frac{\chi(p)}{p^s},\quad H_v(s,\chi)\;:\!=\;\sum_{Xe^{-\frac{v}{U}}<p\leq2Xe^{-\frac{v}{U}}}\frac{\chi(p)}{p^s},\end{equation*}

then we have the bound

\begin{align*}B_3(X)\ll \sum_{q\leq Q_0}\Bigg(\frac{qU^2\log^2P}{\varphi(q)}\sum_{\chi(q)}\int_{T_0}^T & |G_{v_0}(1+it,\chi)|^2|H_{v_0}(1+it,\chi)|^2dt\\&+\frac{1}{Q_0\log^{2+\eta}X}\left(\frac{qT\log X}{X}+\frac{q}{\varphi(q)}\right)\Bigg),\end{align*}

for some $\eta=\eta(\varepsilon)>0$ , where we take $I=[U\log P,(1+\delta)U\log P]$ , $U\;:\!=\;Q_0^{1+\varepsilon^2}$ and $v_0\in I$ a suitable integer.

Proof. We factorise $F(s,\chi)$ using Lemma 7·4 to get that

\begin{equation*}F(s,\chi)=\sum_{v\in I\cap\mathbb{Z}}G_v(s,\chi)H_v(s,\chi) +\sum_{\substack{k\in[Xe^{-1/U},Xe^{1/U}]\\\text{or }k\in[2X,2Xe^{1/U}]}}\frac{d_k\chi(k)}{k^s},\end{equation*}

where $I=[U\log P,(1+\delta)U\log P]$ , $U\;:\!=\;Q_0^{1+\varepsilon^2}$ and

\begin{equation*}|d_k|\leq\sum_{\substack{k=p_1p_2\\P<p_1\leq P^{1+\delta}}}1.\end{equation*}

Therefore, taking the maximum in the sum over I, the mean square of the Dirichlet polynomial is bounded by

\begin{align*}&\int_{T_0}^{T}|F(1+it,\chi)|^2dt\\[5pt]&\ll\int_{T_0}^T\left|\sum_{v\in I\cap\mathbb{Z}}G_v(1+it,\chi)H_v(1+it,\chi)\right|^2dt+\int_{T_0}^{T}\Bigg|\sum_{\substack{k\in[Xe^{-1/U},Xe^{1/U}]\\\text{or }k\in[2X,2Xe^{1/U}]}}\frac{d_k\chi(k)}{k^{1+it}}\Bigg|^2dt\\[5pt] &\ll|I|^2\int_{T_0}^T|G_{v_0}(1+it,\chi)|^2|H_{v_0}(1+it,\chi)|^2dt+\int_{T_0}^{T}\Bigg|\sum_{\substack{k\in[Xe^{-1/U},Xe^{1/U}]\\\text{or }k\in[2X,2Xe^{1/U}]}}\frac{d_k\chi(k)}{k^{1+it}}\Bigg|^2dt,\end{align*}

where $v_0\in I$ is the integer maximising the right-hand side. Applying Lemma 7·7 to the second integral, we have that

(8·1) \begin{align} & \sum_{\chi(q)}\int_{T_0}^{T}|F(1+it,\chi)|^2dt \nonumber \\\ll& U^2\log^2P\sum_{\chi(q)}\int_{T_0}^T|G_{v_0}(1+it,\chi)|^2|H_{v_0}(1+it,\chi)|^2dt \nonumber\\&+T\varphi(q)\sum_{\substack{k\in[Xe^{-1/U},Xe^{1/U}] \\\text{or }k\in[2X,2Xe^{1/U}]\\(k,q)=1}}\frac{|d_k|^2}{k^2}+T\varphi(q)\sum_{\substack{1\leq h\leq\frac{2Xe^{1/U}}{T}\\q\mid h}}\sum_{\substack{m-n=h\\m,n\in[Xe^{-1/U},Xe^{1/U}]\\\text{or }m,n\in[2X,2Xe^{1/U}]\\(mn,q)=1}}\frac{|d_m||d_n|}{mn}.\end{align}

We now bound the last two terms. We consider only the sums where $k\in[Xe^{-1/U},Xe^{1/U}]$ , with the sums over $k\in[2X,2Xe^{1/U}]$ being treated analogously. For the first sum,

(8·2) \begin{equation}\sum_{\substack{k=p_1p_2\\Xe^{-1/U}\leq k\leq Xe^{1/U}\\P<p_1\leq P^{1+\delta}}}\frac{1}{k^2}\ll\frac{e^{2/U}}{X^2}\sum_{P<p_1\leq P^{1+\delta}}\sum_{\frac{Xe^{-1/U}}{p_1}\leq p_2\leq\frac{Xe^{1/U}}{p_1}}1.\end{equation}

By the Brun–Titchmarsh inequality, we have the bound

\begin{equation*}\sum_{\frac{Xe^{-1/U}}{p_1}\leq p_2\leq\frac{Xe^{1/U}}{p_1}}1\ll\frac{X(e^{1/U}-e^{-1/U})}{p_1\log X}.\end{equation*}

Returning to (8·2), by Mertens’ theorem we have that

\begin{equation*}\sum_{\substack{Xe^{-1/U}\leq p_1p_2\leq Xe^{1/U}\\P<p_1\leq P^{1+\delta}}}\frac{1}{(p_1p_2)^2}\ll\frac{e^{3/U}-e^{1/U}}{X\log X}\sum_{P<p\leq P^{1+\delta}}\frac{1}{p}\ll\frac{1}{XU\log X}.\end{equation*}

We will use Brun’s sieve to bound the second of these sums. We may trivially bound

\begin{equation*}\begin{aligned}&|\{n\leq2X:n=p_1p_2, p_1\in(P,P^{1+\delta}]\}|\\&\ll|\{n\leq2X:n=p_1m, p_1\in(P,P^{1+\delta}], (m,P(z))=1\}|,\end{aligned}\end{equation*}

where we define $P(z)=\prod_{p<z}p$ with $z=X^{1/\beta}$ and $\beta>1$ suitably large. Let $\Pi$ be the product of all primes in $\tilde{I}\;:\!=\;(P,P^{1+\delta}]\cap[1,z)$ and $P'(z)=\prod_{p<z, p\nmid h}p$ . Therefore, we have that

\begin{align*}\sum_{\substack{1\leq h\leq\frac{2Xe^{1/U}}{T} \\ q\mid h}}&\sum_{Xe^{-1/U}\leq m\leq Xe^{1/U}}\frac{|d_m||d_{m+h}|}{m(m+h)}\\&\ll\frac{e^{2/U}}{X^2}\sum_{\substack{1\leq h\leq\frac{2Xe^{1/U}}{T} \\ q\mid h}}\left|\left\{m\in[Xe^{-1/U},Xe^{1/U}]:\left(m(m+h),\frac{P'(z)}{\Pi}\right)=1\right\}\right|.\end{align*}

Brun’s sieve then gives the bound

\begin{align*}&\left|\left\{m \in[Xe^{-1/U},Xe^{1/U}]:\left(m(m+h),\frac{P'(z)}{\Pi}\right)=1\right\}\right|\\&\ll \frac{hX(e^{1/U}-e^{-1/U})}{\varphi(h)}\prod_{\substack{p<z\\p\notin\tilde{I}}}\left(1-\frac{2}{p}\right)\\&\ll\frac{X}{U\log^2z}\frac{h}{\varphi(h)}.\end{align*}

Therefore we have

\begin{equation*}\begin{aligned}\sum_{\substack{1\leq h\leq\frac{2Xe^{1/U}}{T} \\ q\mid h}}\sum_{Xe^{-1/U}\leq m\leq Xe^{1/U}}\frac{|d_m||d_{m+h}|}{m(m+h)}&\ll\frac{e^{2/U}}{XU\log^2z}\sum_{\substack{1\leq h\leq\frac{2Xe^{1/U}}{T} \\ q\mid h}}\frac{h}{\varphi(h)}\\&\ll\frac{1}{\varphi(q)TU\log^2X}.\end{aligned}\end{equation*}

Combining these estimates and applying the definition of U, these two sums contribute

\begin{equation*}\ll\left(\frac{\varphi(q)T\log X}{X}+1\right)\frac{1}{Q_0\log^{2+\eta}X}\end{equation*}

to (8·1) for some $\eta=\eta(\varepsilon)>0$ , as needed.

It remains to estimate the integral appearing in Lemma 8·1. We split the domain of integration $[T_0,T]$ according to the size of the polynomial $G_{v_0}$ . We will first bound the contribution of $\mathcal{S}_1\subset[T_0,T]\times\{\chi\mod q\}$ defined by

(8·3) \begin{equation}\mathcal{S}_1\;:\!=\;\{(t,\chi)\in[T_0,T]\times\{\chi\mod q\}:|G_{v_0}(1+it,\chi)|\leq e^{-\frac{\alpha_1v_0}{U}}\},\end{equation}

where $\alpha_1\;:\!=\;{3}/{34}-\varepsilon'$ and $\varepsilon'>0$ is sufficiently small in terms of $\varepsilon>0$ . We may write

(8·4) \begin{equation}\mathcal{S}_1=\bigcup_{\chi\mod q}\{\chi\}\times\mathcal{T}_{1,\chi}\end{equation}

for some $\mathcal{T}_{1,\chi}\subset[T_0,T]$ .

8·1. The contribution of $\mathcal{S}_1$

We first treat the contribution of the integral over $\mathcal{T}_{1,\chi}$ , where the polynomial $G_{v_0}(1+it,\chi)$ is pointwise small.

Lemma 8·2. Let $\varepsilon>0$ be fixed sufficiently small and $\mathcal{T}_{1,\chi}$ be defined as in (8·4). Then, there exists some $\eta=\eta(\varepsilon)>0$ such that

\begin{equation*}\begin{aligned}U^2\log^2P\sum_{q\leq Q_0}\frac{q}{\varphi(q)}&\sum_{\chi(q)}\int_{\mathcal{T}_{1,\chi}}|G_{v_0}(1+it)|^2|H_{v_0}(1+it)|^2dt\\&\ll\frac{1}{Q_0\log^{2+\eta}X}\sum_{q\leq Q_0} \left(\frac{qPT\log X}{X}+1\right).\end{aligned}\end{equation*}

Proof. First we apply the definition of $\mathcal{T}_{1,\chi}$ , bounding pointwise $|G_{v_0}(1+it,\chi)|\leq e^{-\frac{\alpha_1v_0}{U}}\leq P^{-\alpha_1}$ to bound the integral over $\mathcal{T}_{1,\chi}$ by

\begin{equation*}\int_{\mathcal{T}_{1,\chi}}|G_{v_0}(1+it,\chi)|^2|H_{v_0}(1+it,\chi)|^2dt\ll P^{-2\alpha_1}\int_{\mathcal{T}_{1,\chi}}|H_{v_0}(1+it,\chi)|^2dt.\end{equation*}

Applying Lemma 7·7, we have that

\begin{align*}\sum_{\chi(q)}&\int_{\mathcal{T}_{1,\chi}}|G_{v_0}(1+it,\chi)|^2|H_{v_0}(1+it,\chi)|^2dt\\&\ll \frac{P^{-2\alpha_1}T\varphi(q)e^{2v_0/U}}{X^2}\Bigg(\sum_{\frac{X}{e^{v_0/U}}<p\leq\frac{2X}{e^{v_0/U}}}1+\sum_{\substack{1\leq h\leq\frac{X}{Te^{v_0/U}}\\q\mid h}}\sum_{\substack{\frac{X}{e^{v_0/U}}<p_1,p_2\leq\frac{2X}{e^{v_0/U}}\\p_1-p_2=h}}1\Bigg).\end{align*}

By Chebyshev’s inequality, we have that

\begin{equation*}\sum_{\frac{X}{e^{v_0/U}}<p\leq\frac{2X}{e^{v_0/U}}}1\ll\frac{X}{e^{v_0/U}\log X}.\end{equation*}

For the second term, we have by Lemma 2·1 that

\begin{equation*}\sum_{\substack{1\leq h\leq\frac{X}{Te^{v_0/U}}\\q\mid h}}\sum_{\substack{\frac{X}{e^{v_0/U}}<p_1,p_2\leq\frac{2X}{e^{v_0/U}}\\p_1-p_2=h}}1\ll\frac{X}{e^{v_0/U}\log^2X}\sum_{\substack{1\leq h\leq\frac{X}{Te^{v_0/U}} \\ q\mid h}}\mathfrak{S}(h)\ll\frac{X^2}{e^{2v_0/U}qT\log^2X}.\end{equation*}

Combining these estimates, we have that

\begin{equation*}\sum_{\chi(q)}\int_{\mathcal{T}_{1,\chi}}|G_{v_0}(1+it,\chi)|^2|H_{v_0}(1+it,\chi)|^2dt\ll \frac{\varphi(q)P^{\delta-2\alpha_1}}{q\log^2X}\left(\frac{qPT\log X}{X}+1\right).\end{equation*}

Thus the overall contribution to the sum $B_3(X)$ is

\begin{equation*}\frac{U^2P^{\delta-2\alpha_1}\log^2P}{\log^2X}\sum_{q\leq Q_0}\left(\frac{qPT\log X}{X}+1\right).\end{equation*}

Now, by our choices of P, U and the definition of $\alpha_1$ with $\varepsilon'$ sufficiently small in terms of $\varepsilon$ , we have that

\begin{equation*}\frac{U^2P^{\delta-2\alpha_1}\log^2P}{\log^2X}\ll\frac{1}{Q_0\log^{2+\eta}X},\end{equation*}

for some $\eta=\eta(\varepsilon)>0$ as $\delta>0$ is sufficiently small.

8·2. The contribution of the complement of $\mathcal{S}_1$

It remains to consider the contribution of the complement of $\mathcal{S}_1$ . We apply Lemma 7·5 to $H_{v_0}(1+it,\chi)$ with $k=3$ , decomposing this polynomial into

\begin{equation*}|H_{v_0}(1+it,\chi)|\ll(\log^C X)\left(|Q_1(1+it,\chi)|+\cdots+|Q_L(1+it,\chi)|\right),\end{equation*}

where $L\leq \log^CX$ for some $C>0$ . Each $Q_j(s,\chi)$ is of the form $Q_j(s,\chi)=\prod_{i\leq J_j}M_i(s,\chi)$ with $J_j\leq6$ for each $1\leq j\leq L$ , where $M_i(s,\chi)$ are prime-factored Dirichlet polynomials of the form

\begin{equation*}\sum_{M_i<n\leq2M_i}\frac{\chi(n)\log n}{n^s},\;\sum_{M_i<n\leq2M_i}\frac{\chi(n)}{n^s},\text{ or }\sum_{M_i<n\leq2M_i}\frac{\mu(n)\chi(n)}{n^s},\end{equation*}

whose lengths satisfy $M_1\cdots M_{J_j}=X^{1+o(1)}$ , $M_i\gg\exp\left({\log X}/{\log\log X}\right)$ for each i. We will treat the polynomials $Q_j(s,\chi)$ according to the lengths $M_i$ of the factors as follows:

Type II Sums. Suppose we have $Q_j(s,\chi)=\prod_{i\leq J_j}M_i(s,\chi)$ for some $1\leq j\leq L$ with $M_i\leq X^{1/3+\varepsilon'}$ for some $i\leq J_j\leq6$ . Then, we rewrite $Q_j(s,\chi)=M_1(s,\chi)M_2(s,\chi)$ with $\exp\left({\log X}/{\log\log X}\right)\ll M_1\leq X^{1/3+\varepsilon'}$ and $M_2=X^{1+o(1)}/M_1$ . Note we may write $M_1=X^{\nu}$ for some $0<\nu\leq1/3+\varepsilon'$ . Where the coefficient $\log n$ appears, we apply partial summation. The polynomial $M_2(s,\chi)$ is a product of polynomials, and the coefficients are given by convolving coefficients which are one of the sequences $(\mu(n)), (1)$ . Thus the coefficients of the polynomial $M_2(s,\chi)$ are bounded in absolute value by $\ll d_r(n)$ with $r\leq5$ .

Type I Sums. Otherwise, we may write $Q_j(s,\chi)=N_1(s,\chi)N_2(s,\chi)$ , where each $N_i(s,\chi)$ is of the form

\begin{equation*}\sum_{N_i<n\leq2N_i}\frac{\chi(n)\log n}{n^s},\text{ or }\sum_{N_i<n\leq2N_i}\frac{\chi(n)}{n^s},\end{equation*}

with lengths satisfying $N_1N_2=X^{1+o(1)}$ . Note that if in fact only one of the lengths $N_i$ satisfies $N_i>X^{1/3+\varepsilon'}$ , then one of $N_1(s,\chi), N_2(s,\chi)$ can be the constant polynomial $1^{-s}$ . Since we have that $N_1N_2=X^{1+o(1)}$ , without loss of generality we may take that $N_1>X^{1/2-\varepsilon'}$ , so that $X^{1/3+\varepsilon'}< N_2\leq X^{1/2+\varepsilon'}$ .

8·2·1. Type II sums

To treat the contribution of these sums, we split the complement of $\mathcal{S}_1$ according to the size of $M_1(1+it,\chi)$ :

(8·5) \begin{equation}\begin{aligned}\mathcal{S}_2&\;:\!=\;\{(t,\chi)\in[T_0,T]\times\{\chi\mod q\}:|M_1(1+it,\chi)|\leq M_1^{-\alpha_2}\}\setminus \mathcal{S}_1,\\\mathcal{S}&\;:\!=\;([T_0,T]\times\{\chi\mod q\})\setminus (\mathcal{S}_1\cup\mathcal{S}_2),\end{aligned}\end{equation}

with $\alpha_2\;:\!=\;{2}/{17}-\varepsilon'>\alpha_1$ . As before, we may write

(8·6) \begin{equation}\begin{aligned}\mathcal{S}_2&=\bigcup_{\chi\mod q}\{\chi\}\times\mathcal{T}_{2,\chi},\\\mathcal{S}&=\bigcup_{\chi\mod q}\{\chi\}\times\mathcal{T}_\chi,\end{aligned}\end{equation}

for some $\mathcal{T}_{2,\chi},\mathcal{T}_\chi\subset[T_0,T]$ . We first consider the contribution of the integral over $\mathcal{T}_{2,\chi}$ .

Lemma 8·3. Let $\varepsilon>0$ be fixed sufficiently small. Let $\mathcal{T}_{2,\chi}$ be defined as in (8·6) and $M_1(s,\chi),M_2(s,\chi)$ be the prime-factored polynomials defined previously. Then

\begin{equation*}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{\mathcal{T}_{2,\chi}}|G_{v_0}(1+it,\chi)|^2|M_1(1+it,\chi)M_2(1+it,\chi)|^2dt\ll X^{-\varepsilon/1000}.\end{equation*}

Proof. By definition of $\mathcal{T}_{2,\chi}$ , we have that $(|G_{v_0}(1+it,\chi)|P^{\alpha_1})^{2(k-1)}\geq1$ , where $k=\lceil\log M_1/\log P\rceil$ . Therefore, we have

(8·7) \begin{align}\sum_{\chi(q)} & \int_{\mathcal{T}_{2,\chi}}|G_{v_0}(1+it,\chi)|^2|M_1(1+it,\chi)M_2(1+it,\chi)|^2dt\nonumber\\&\ll M_1^{-2\alpha_2}P^{2\alpha_1(k-1)}\sum_{\chi(q)}\int_{\mathcal{T}_{2,\chi}}|G_{v_0}^{k}(1+it,\chi)M_2(1+it,\chi)|^2dt.\end{align}

By the choice of k, we have that

\begin{equation*}P^{2\alpha_1k}\ll \exp\left(\frac{2\alpha_1\log P \log M_1}{\log P}\right)\\= M_1^{2\alpha_1}.\end{equation*}

Therefore (8·7) is bounded by

\begin{equation*}\begin{aligned}&\ll M_1^{2\alpha_1-2\alpha_2}P^{-2\alpha_1}\sum_{\chi(q)}\int_{\mathcal{T}_{2,\chi}}|G_{v_0}^{k}(1+it,\chi)M_2(1+it,\chi)|^2dt\\&\ll M_1^{2\alpha_1-2\alpha_2}P^{-2\alpha_1}\sum_{\chi(q)}\int_{\mathcal{T}_{2,\chi}}|A(1+it,\chi)|^2dt,\end{aligned}\end{equation*}

where we define

\begin{equation*}A(s,\chi)\;:\!=\;\sum_{n\in J}\frac{A_n\chi(n)}{n^s},\end{equation*}

with $J\;:\!=\;(M_2e^{kv_0/U},2M_2e^{k(v_0+1)/U}]$ and the coefficients $A_n$ satisfying

\begin{equation*}|A_n|\leq\sum_{\substack{n=p_1\cdots p_km\\e^{v_0/U}<p_i\leq e^{(v_0+1)/U}\\i=1,\ldots,k\\M_2<m\leq2M_2}}d_r(m),\end{equation*}

where $r\leq5$ , as before. Note that the primes $p_1,\ldots,p_k$ are not necessarily distinct and m may also have prime factors in the range $(e^{v_0/U},e^{(v_0+1)/U}]$ . Applying Lemma 7·6 to the integral, we have that

\begin{equation*}\sum_{\chi(q)}\int_{\mathcal{T}_{2,\chi}}|A(1+it,\chi)|^2dt\ll\left(\varphi(q)T+\frac{\varphi(q)}{q}M_2e^{kv_0/U}(2e^{k/U}-1)\right)\sum_{\substack{n\in J\\(n,q)=1}}\frac{|A_n|^2}{n^2}.\end{equation*}

Following [ Reference Matomäki and Radziwiłł19 , lemma 13], for the coefficients $A_n$ of $A(s,\chi)$ , we have the bound

\begin{equation*}|A_n|\ll M_2^{o(1)}\sum_{\substack{n=p_1\cdots p_km\\e^{v_0/U}<p_i\leq e^{(v_0+1)/U}\\i=1,\ldots,k\\M_2<m\leq2M_2}}1.\end{equation*}

The number of ways we can write $d=p_1\cdots p_k$ with $p_i$ not necessarily distinct is at most $k!$ . Then we have the bound

\begin{equation*}|A_n|\ll M_2^{o(1)}k!\sum_{\substack{n=md\\p\mid d\Rightarrow e^{v_0/U}<p\leq e^{(v_0+1)/U}}}1,\end{equation*}

trivially extending the range of summation for m. We write g(n) for the latter sum, which is multiplicative and satisfies

\begin{equation*}g(p^a)=\begin{cases}a+1,&\text{ if }e^{v_0/U}<p\leq e^{(v_0+1)/U},\\0,&\text{ otherwise.}\end{cases}\end{equation*}

Therefore, we clearly have $g(n)\ll d(n)\ll n^{o(1)}$ and thus

\begin{align*}\sum_{\substack{n\in J\\(n,q)=1}}\frac{|A_n|^2}{n^2}&\ll\frac{X^{o(1)}k!}{e^{kv_0/U}M_2}\sum_{\substack{n\in J\\(n,q)=1}}\frac{|A_n|}{n}\\&\ll\frac{X^{o(1)}k!}{e^{kv_0/U}M_2}\sum_{M_2<m\leq2M_2}\frac{d_r(m)}{m}\left(\sum_{e^{v_0/U}<p_i\leq e^{(v_0+1)/U}}\frac{1}{p}\right)^{k}\\&\ll\frac{X^{o(1)}k!}{e^{kv_0/U}M_2}\frac{k}{v_0},\end{align*}

noting that $(\log M_2)^{r-1}\ll(\log X)^4$ is negligible. By the definition of k and since $v_0\in I$ , we have that

\begin{equation*}\frac{k}{v_0}\ll\frac{\log M_1}{U\log^2P}\ll 1.\end{equation*}

By the definition of k, we also have that

\begin{equation*}k!\ll(\log M_1)^{\frac{\log M_1}{\log P}}\ll\exp\left(\frac{\log\log M_1 \log M_1}{(17+\varepsilon)\log\log X}\right)\\\ll M_1^{\frac{1}{17+\varepsilon}}.\end{equation*}

Therefore, we can bound the integral over $\mathcal{T}_{2,\chi}$ by

\begin{equation*}\sum_{\chi(q)}\int_{\mathcal{T}_{2,\chi}}|A(1+it,\chi)|^2dt\ll\left(\varphi(q)\frac{T}{e^{kv_0/U}M_2}+\frac{\varphi(q)}{q}(2e^{k/U}-1)\right)X^{o(1)}M_1^{\frac{1}{17+\varepsilon}}.\end{equation*}

Since $v_0\in I$ , we have that $e^{kv_0/U}\gg P^{k}\gg M_1$ by the definition of k. We also have that $2e^{k/U}-1\ll1$ and therefore we can bound the above integral by

\begin{equation*}\sum_{\chi(q)}\int_{\mathcal{T}_{2,\chi}}|A(1+it,\chi)|^2dt\ll\left(\varphi(q)\frac{T}{M_1M_2}+\frac{\varphi(q)}{q}\right) X^{o(1)}M_1^{\frac{1}{17+\varepsilon}}\ll \varphi(q)X^{o(1)}M_1^{\frac{1}{17+\varepsilon}},\end{equation*}

as we have $M_1M_2=X^{1+o(1)}$ and $T\leq X^{1+o(1)}$ . Returning to (8·7), we have the bound

\begin{equation*}\begin{aligned}&\sum_{\chi(q)}\int_{\mathcal{T}_{2,\chi}}|G_{v_0}(1+it,\chi)|^2|M_1(1+it,\chi)M_2(1+it,\chi)|^2dt\\&\ll \varphi(q)X^{o(1)}M_1^{\frac{1}{17+\varepsilon}+2(\alpha_1-\alpha_2)}P^{-2\alpha_1}.\end{aligned}\end{equation*}

With our choices of $\alpha_1,\alpha_2$ , we have that $2\alpha_1-2\alpha_2=-1/17$ . Summing over q introduces a factor of $Q_0^2$ . Recalling that we may write $M_1=X^{\nu}$ , for some $0<\nu\leq1/3+\varepsilon'$ , we find that

\begin{equation*}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{\mathcal{T}_{2,\chi}}|G_{v_0}(1+it,\chi)|^2|M_1(1+it,\chi)M_2(1+it,\chi)|^2dt\ll Q_0^2P^{-2\alpha_1}X^{-\frac{\varepsilon\nu}{300}+o(1)},\end{equation*}

and choosing $\varepsilon'>0$ sufficiently small in terms of $\varepsilon>0$ ensures the above is bounded by $X^{-\varepsilon/1000}$ , as needed.

We now treat the contribution of the integral over $\mathcal{T}_\chi$ , applying the Halász–Montgomery inequality and the large value theorems.

Lemma 8·4. Let $\mathcal{T_\chi}$ be defined as in (8·6). Let $E>0$ be fixed sufficiently large. Then, we have that

\begin{equation*}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{\mathcal{T}_\chi}|G_{v_0}(1+it,\chi)M_1(1+it,\chi)M_2(1+it,\chi)|^2dt\ll\frac{1}{\log^EX}.\end{equation*}

Proof. Let $M_1=X^{\nu}$ for some $0<\nu\leq1/3+\varepsilon'$ . We first replace the integral over $\mathcal{T}_\chi$ with a sum over a well-spaced set. For each character $\chi$ mod q, cover $\mathcal{T}_\chi$ with intervals of unit length and from each interval take the point which maximises the integral over that interval. This set is not yet necessarily well-spaced, but we can split it into O(1) well-spaced subsets. Therefore we may write

\begin{align*}\sum_{\chi(q)}&\int_{\mathcal{T}_\chi}|G_{v_0}(1+it,\chi)M_1(1+it,\chi)M_2(1+it,\chi)|^2dt\\[5pt]&\ll\sum_{(t,\chi)\in\mathcal{T}'}|G_{v_0}(1+it,\chi)M_1(1+it,\chi)M_2(1+it,\chi)|^2,\end{align*}

where $\mathcal{T}'$ is the well-spaced subset which maximises the right-hand side. We now apply the prime-factored property ${|M_1(1+it,\chi)|^2\ll\log^{-F'}X}$ with $F'>0$ sufficiently large and then Lemma 7·8 to get that

\begin{align*}&\sum_{(t,\chi)\in\mathcal{T}'}|G_{v_0}(1+it,\chi)M_1(1+it,\chi)M_2(1+it,\chi)|^2\\[5pt]&\ll\log^{-F'}X\sum_{(t,\chi)\in\mathcal{T}'}|G_{v_0}(1+it,\chi)M_2(1+it,\chi)|^2\\[5pt]&\ll\log^{-F'}X\left(\frac{\varphi(q)}{q}X^{1-\nu+\varepsilon'}+|\mathcal{T}'|(qT)^{1/2}\right)\sum_{e^{v_0/U}<p\leq e^{(v_0+1)/U}}\frac{1}{p^2}\sum_{M_2<m\leq2M_2}\frac{d_r^2(m)}{m^2}\\[5pt]&\ll\log^{-F}X\left(\frac{\varphi(q)}{q}+\frac{|\mathcal{T}'|(qT)^{1/2}}{X^{1-\nu+\varepsilon'}}\right),\end{align*}

where $F>0$ is suitably large and $r\leq5$ . If we can show that $|\mathcal{T}'|\ll X^{1/2-\nu-\varepsilon^2}$ , then we will have that

\begin{equation*}\sum_{\chi(q)}\int_{\mathcal{T}_\chi}|G_{v_0}(1+it,\chi)M_1(1+it,\chi)M_2(1+it,\chi)|^2dt\ll\log^{-E'}X\end{equation*}

for some suitable $E'>0$ . Summing over q, we have that

\begin{equation*}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{\mathcal{T}_\chi}|G_{v_0}(1+it,\chi)M_1(1+it,\chi)M_2(1+it,\chi)|^2dt\ll\frac{1}{\log^EX},\end{equation*}

where $E>0$ is sufficiently large.

Thus, it remains to prove that $|\mathcal{T}'|\ll X^{1/2-\nu-\varepsilon^2}$ . Applying Lemma 7·11 with $M_1(1+it,\chi)^l, V=M_1^{-\alpha_2l},k=2,$ and $l\in\{2,3\}$ , we have that

\begin{align*}|\mathcal{T}'|&\ll X^{\varepsilon^3}\left(M_1^{2\alpha_2l}+X^2M_1^{4l(2\alpha_2-1)}+XM_1^{4l(4\alpha_2-1)}\right)\\[5pt]&\ll\begin{cases}X^{\max(\frac{8}{17}\nu,2-\frac{104}{17}\nu,1-\frac{72}{17}\nu)-2\varepsilon^2},&\;(l=2),\\[5pt]X^{\max(\frac{12}{17}\nu,2-\frac{156}{17}\nu,1-\frac{108}{17}\nu)-2\varepsilon^2},&\;(l=3).\end{cases}\end{align*}

We have that $\nu\leq{1}/{3}+\varepsilon'$ . The inequality ${8}\nu/{17}\leq{1}/{2}-\nu$ holds when $\nu\leq{17}/{50}$ and we have that $2-{104}\nu/{17}\geq{8}\nu/{17}$ when $\nu\leq{17}/{56}$ . Note that $2-{104}\nu/{17}\leq{1}/{2}-\nu$ fails if $\nu<{17}/{58}$ , so the inequality with $l=2$ provides the bound $|\mathcal{T}'|\ll X^{1/2-\nu-\varepsilon^2}$ in the range ${17}/{58}\leq\nu\leq{1}/{3}$ .

Similarly, ${12}\nu/{17}\leq{1}/{2}-\nu$ holds for $\nu\leq{17}/{58}$ and $2-{156}\nu/{17}\geq{12}\nu/{17}$ when $\nu\leq{17}/{84}$ . We have that $2-{156}\nu/{17}\leq1/2-\nu$ fails when $\nu<{51}/{278}$ , so the inequality with $l=3$ gives the required bound for $|\mathcal{T}'|$ when ${51}/{278}\leq\nu\leq{17}/{58}$ .

For the remaining range $\nu<{51}/{278}$ , we apply Lemma 7·9 with $V=M_1^{-\alpha_2}$ to get that

\begin{equation*}|\mathcal{T}'|\ll (qT)^{2\alpha_2}X^{2\nu\alpha_2+\varepsilon}\ll X^{\frac{4}{17}(1+\nu)+100\varepsilon}\ll X^{1/2-\nu-\varepsilon^2},\end{equation*}

as required.

8·2·2. Type I sums

It remains to treat the contribution of the sums of form (7·3), applying the Cauchy–Schwarz inequality and a result on the twisted fourth moment of partial sums of Dirichlet L-functions.

Lemma 8·5. Let $\varepsilon>0$ be fixed sufficiently small. With $N_1(s,\chi),N_2(s,\chi)$ as defined above, we have that

\begin{equation*}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{[T_0,T]\setminus\mathcal{T}_{1,\chi}}|G_{v_0}(1+it,\chi)|^2|N_1(1+it,\chi)N_2(1+it,\chi)|^2dt\ll X^{-\varepsilon/2}.\end{equation*}

Proof. We split the domain of integration into dyadic intervals $[T_1,2T_1]$ such that $T_0\leq T_1\leq T$ . As we are in the complement of $\mathcal{S}_1$ , we have that $|G_{v_0}(1+it,\chi)P^{\alpha_1}|^{2(l-1)}\geq 1$ , where we define $l=\lfloor\varepsilon\log X/\log P\rfloor$ . Therefore, we have that

\begin{align*}\sum_{q\leq Q_0}&\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{([T_0,T]\setminus\mathcal{T}_{1,\chi})\cap[T_1,2T_1]}|G_{v_0}(1+it,\chi)|^2|N_1(1+it,\chi)N_2(1+it,\chi)|^2dt\\[5pt]\ll& P^{2\alpha_1(l-1)}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{T_1}^{2T_1}|G_{v_0}(1+it,\chi)|^{2l}|N_1(1+it,\chi)N_2(1+it,\chi)|^2dt.\end{align*}

Applying the Cauchy–Schwarz inequality three times (to the integral and the sums over $\chi$ and q), we have that the sum over q above is bounded by

(8·8) \begin{equation}\begin{aligned}\ll &\left(\sum_{q\leq Q_0}\frac{1}{\varphi(q)}\sum_{\chi(q)}\int_{T_1}^{2T_1}|G_{v_0}(1+it,\chi)|^{4l}|N_1(1+it,\chi)|^4dt\right)^{\frac{1}{2}}\\[5pt]&\times\left(\sum_{q\leq Q_0}\frac{q^2}{\varphi(q)}\sum_{\chi(q)}\int_{T_1}^{2T_1}|N_2(1+it,\chi)|^4dt\right)^{\frac{1}{2}}.\end{aligned}\end{equation}

We apply Lemma 7·6 to the second integral. Noting that $N_2(1+it,\chi)$ either has coefficients 1 or $\log n$ , we find that

\begin{equation*}\begin{aligned}\sum_{q\leq Q_0}\frac{q^2}{\varphi(q)}\sum_{\chi(q)}\int_{T_1}^{2T_1}|N_2(1+it,\chi)|^4dt&\ll X^{\varepsilon/10}\sum_{q\leq Q_0}q^2\left(\frac{T_1+N_2^2/q}{N_2^2}\right)\\[5pt]&\ll X^{\varepsilon/10}\left(\frac{T_1+N_2^2}{N_2^2}\right).\end{aligned}\end{equation*}

We next treat the first term appearing in (8·8). We have that

\begin{equation*}|G_{v_0}(1+it,\chi)|^{4l}=\left|\sum_{e^{v_0/U}<p\leq e^{(v_0+1)/U}}\frac{\chi(p)}{p^{1+it}}\right|^{4l}=\left|\sum_{e^{2lv_0/U}<n\leq e^{2l(v_0+1)/U}}\frac{\chi(n)a(n)}{n^{1+it}}\right|^2,\end{equation*}

where $a(n)=0$ unless n is a product of 2l primes, not necessarily distinct, each lying in the interval $(e^{v_0/U},e^{(v_0+1)/U}]$ . Writing n in terms of its prime factorisation $n=p_1^{a_1}\cdots p_{b}^{a_{b}}$ with $b\leq 2l$ , we have that $a(n)=\binom{2l}{a_1,\ldots,a_{b}}$ when it is non-zero and therefore that $a(n)\ll(2l)!$ . Now we can apply Lemma 7·12 with $M(1+it,\chi)=G_{v_0}^{2l}(1+it,\chi)$ , $M=P^{2l}$ and N corresponding to $N_1$ to get that

\begin{align*}\sum_{q\leq Q_0}\frac{1}{\varphi(q)}\sum_{\chi(q)}&\int_{T_1}^{2T_1}|G_{v_0}(1+it,\chi)|^{4l}|N_1(1+it,\chi)|^4dt\\[5pt]&\ll X^{\varepsilon/10}(2l)!^2\left(\frac{Q_0T_1}{N_1^2P^{2l}}\left(1+P^{4l}(Q_0T_1)^{-1/2}\right)+\frac{1}{T_1}\right)\\[5pt]&\ll X^{\varepsilon/10}(l!)^{4+\varepsilon}\left(\frac{Q_0T_1}{N_1^2P^{2l}}+\frac{1}{T_1}\right),\end{align*}

as the definition of l ensures that $P^{4l}(Q_0T_1)^{-1/2}\ll1$ . Returning to (8·8), we have that

\begin{align*}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}&\sum_{\chi(q)}\int_{(\mathcal{T}\setminus\mathcal{T}_1)\cap[T_1,2T_1]}|G_{v_0}(1+it,\chi)|^2|N_1(1+it,\chi)N_2(1+it,\chi)|^2dt\\[5pt]&\ll P^{2\alpha_1(l-1)}X^{\varepsilon/10}(l!)^{2+\varepsilon}\left(\frac{Q_0T_1}{N_1^2P^{2l}}+\frac{1}{T_1}\right)^{1/2}\left(\frac{T_1+N_2^2}{N_2^2}\right)^{1/2}\\[5pt]&\ll P^{2\alpha_1(l-1)}X^{\varepsilon/10}(l!)^{2+\varepsilon}\left(\frac{Q_0T_1}{N_1^2N_2^2P^{2l}}(T_1+N_2^2)+\frac{1}{N_2^2}+\frac{1}{T_1}\right)^{1/2}.\end{align*}

We have that $N_1N_2=X^{1+o(1)}$ with $N_1\geq X^{1/2-\varepsilon'}$ and $X^{1/3+\varepsilon'}\leq N_2\leq X^{1/2+\varepsilon'}$ . As we also have that $X^{1/100}=T_0\leq T_1\leq T\leq X^{1+o(1)}$ , the above is bounded by

\begin{equation*}\ll P^{2\alpha_1(l-1)}X^{\varepsilon/10}(l!)^{2+\varepsilon}\left(\frac{1}{P^l}+\frac{1}{N_2}+\frac{1}{T_0^{1/2}}\right).\end{equation*}

Summing the contribution of each of the integrals over the dyadic intervals multiplies the above estimate by $\log X$ . By the definition of l, we have that $(l!)^{2+\varepsilon}\ll(\log^2X)^{l(1+\varepsilon)}$ , and so we also have that

\begin{align*} (P^{2\alpha_1-1}\log^2 X)^{l(1+\varepsilon)}&\ll\exp\left((1+\varepsilon)\varepsilon\left(2\alpha_1-1+\frac{2}{17+\varepsilon}\right)\log X \right)\\&\ll X^{-2\varepsilon/3}.\end{align*}

Overall we have the bound

\begin{equation*}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{[T_0,T]\setminus\mathcal{T}_{1,\chi}}|G_{v_0}(1+it,\chi)|^2|N_1(1+it,\chi)N_2(1+it,\chi)|^2dt\ll X^{-\varepsilon/2},\end{equation*}

as needed.

8·3. Completing the proof of Proposition 6·5

We may now combine these estimates to complete the proof of Proposition 6·5.

Proof of Proposition 6·5. By Lemma 8·1, we have that

\begin{align*}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{T_0}^T &|F(1+it,\chi)|^2dt\\\ll \sum_{q\leq Q_0}\Bigg(&\frac{qU^2\log^2P}{\varphi(q)}\sum_{\chi(q)}\int_{T_0}^T|G_{v_0}(1+it,\chi)|^2|H_{v_0}(1+it,\chi)|^2dt\\&+\frac{1}{Q_0\log^{2+\eta}X}\left(\frac{qT\log X}{X}+\frac{q}{\varphi(q)}\right)\Bigg)\\\ll \sum_{q\leq Q_0}\Bigg(&\frac{q U^2\log^2P}{\varphi(q)}\sum_{\chi(q)}\Bigg(\int_{\mathcal{T}_{1,\chi}}|G_{v_0}(1+it,\chi)|^2|H_{v_0}(1+it,\chi)|^2dt\\&+\int_{[T_0,T]\setminus\mathcal{T}_{1,\chi}}|G_{v_0}(1+it,\chi)|^2|H_{v_0}(1+it,\chi)|^2dt\Bigg)\\&+\frac{1}{Q_0\log^{2+\eta}X}\left(\frac{qT\log X}{X}+\frac{q}{\varphi(q)}\right)\Bigg)\end{align*}

for some $\eta=\eta(\varepsilon)>0$ and some suitable integer $v_0\in I$ . We apply Lemma 8·2 to bound the contribution of the integral over $\mathcal{T}_{1,\chi}$ , finding that the above is bounded by

\begin{align*}\ll \sum_{q\leq Q_0} \Bigg(\frac{qU^2\log^2P}{\varphi(q)}\sum_{\chi(q)}\int_{[T_0,T]\setminus\mathcal{T}_{1,\chi}} & |G_{v_0}(1+it,\chi)|^2|H_{v_0}(1+it,\chi)|^2dt\\&+\frac{1}{Q_0\log^{2+\eta}X}\left(\frac{qTP\log X}{X}+\frac{q}{\varphi(q)}\right)\Bigg).\end{align*}

Combining Lemmas 8·3 to 8·5, we bound the contribution of the complement of $\mathcal{S}_1$ by

\begin{equation*}\ll \frac{U^2\log^2P}{\log^EX}\ll\frac{1}{\log^FX}\end{equation*}

for some sufficiently large $F>0$ , which is negligible. Thus, we have that

\begin{equation*}\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\int_{T_0}^T|F(1+it,\chi)|^2dt\ll \frac{1}{Q_0\log^{2+\eta}X} \sum_{q\leq Q_0} \left(\frac{qTP\log X}{X}+\frac{q}{\varphi(q)}\right),\end{equation*}

as required.

9. Proof of theorem 1·3

We now briefly outline how to adjust the argument to prove Theorem 1·2. The problem can be reduced to the set of $E_2$ numbers which factorise in the “typical” way. By Mertens’ theorem, almost all products of exactly two primes $p_1p_2\leq X$ with $p_1\leq p_2$ satisfy

(9·1) \begin{equation}p_1\in\left[\exp\left((\log X)^{\varepsilon(X)}\right),\exp\left((\log X)^{1-\varepsilon(X)}\right)\right]\;=\!:\;[P_1,P_2],\end{equation}

where $\varepsilon(X)=o(1)$ . We define $E''_{\!\!\!2}\;:\!=\;E''_{\!\!\!2}(X)$ to be the set of $E_2$ numbers $n=p_1p_2\in(X,2X]$ which factorise in the typical way. Using a sieve theory argument, we have that

\begin{equation*}\begin{aligned}\frac{1}{X}\sum_{X<n\le2X}\mathbb{1}_{E_2}(n)\mathbb{1}_{E_2}(n+h)-o\left(\frac{\mathfrak{S}(h)(\log\log X)^2}{(\log X)^2}\right)&\le \frac{1}{X}\sum_{X<n\le 2X}\mathbb{1}_{E''_{\!\!\!2}}(n)\mathbb{1}_{E''_{\!\!\!2}}(n+h)\\&\le \frac{1}{X}\sum_{X<n\le2X}\mathbb{1}_{E_2}(n)\mathbb{1}_{E_2}(n+h).\end{aligned}\end{equation*}

Therefore, we can reduce the problem to considering the correlations of $n,n+h\in E''_{\!\!\!2}$ .

We modify every definition featuring $(P,P^{1+\delta}]$ , replacing this interval with $[P_1,P_2]$ . We will once again apply the Hardy–Littlewood circle method and in (3·6) and (3·8) we take

(9·2) \begin{equation}Q_0\;:\!=\;\log^{A'} X, A'>4,\qquad Q\;:\!=\;P_2\log^CX,\qquad H\geq Q\log^DX,\end{equation}

where C is chosen sufficiently large in terms of A ^′ and D is chosen sufficiently large in terms of A ^′ and C. In Lemma 8·1 we instead define $I\;:\!=\;[U\log P_1,U\log P_2]$ where $U\;:\!=\;Q_0^{E}$ , $E>0$ and we define $\alpha_1\;:\!=\;\varepsilon'>0$ sufficiently small in terms of $\varepsilon>0$ .

The applications of the Cauchy–Schwarz inequality to sums over products of exactly two primes in the proofs of Proposition 3·5 and Proposition 6·3 are now too inefficient. To overcome this, we split the sum over $p_1\in[P_1,P_2]$ into dyadic intervals before applying the inequality. We now outline how to modify the proof of Proposition 3·5.

Proposition 9·1. Let $A>3$ , $B>1$ be fixed and $\mathfrak{m}$ be defined as in (3·8) with $Q_0,Q$ as in (9·2). Let $Q\log^DX\leq H\leq X\log^{-A}X$ with $D>0$ sufficiently large. For $\alpha\in\mathfrak{m}$ we have that

\begin{equation*}\int_{\mathfrak{m}\cap[\alpha-\frac{1}{2H},\alpha+\frac{1}{2H}]}|S(\theta)|^2d\theta\ll\frac{X}{\log^BX}.\end{equation*}

Proof. As before, we apply Lemma 2·2 to the minor arc integral so that we need to bound

\begin{equation*}I\ll\frac{1}{H^2}\int_{X}^{2X}\left|\sum_{x<n\leq x+H}\varpi_2(n)e(n\alpha)\right|^2dx+H\log^2 X.\end{equation*}

The second term is acceptable by our choice of H, so it remains to bound the first term. Now before applying Cauchy–Schwarz to the integrand we split the sum over $p_1$ into dyadic intervals [P,2P] with $P_1\leq P\leq P_2$ so that we instead need to integrate

\begin{equation*}\Bigg(\sum_{P<m_1\leq 2P}|\mathbb{1}_{\mathbb{P}}(m_1)|^2\Bigg)\Bigg(\sum_{P<m_2\leq 2P}\Bigg|\sum_{x<m_2p\leq x+H}(\log p)e(\alpha m_2 p)\Bigg|^2\Bigg).\end{equation*}

The first term is $\ll \frac{P}{\log P}$ , while the second term is equal to

\begin{equation*}\sum_{\substack{x<mp_1,mp_2\leq x+H \\ P<m\leq 2P}} (\log p_1)(\log p_2) e(\alpha m(p_1-p_2)).\end{equation*}

Next, we perform the integration on this sum and split into the diagonal ( $p_1=p_2$ ) and off-diagonal terms ( $p_1\neq p_2$ ) as before. The diagonal terms now contribute

(9·3) \begin{equation}S_1\ll \frac{P}{H\log P}\sum_{P<m\leq 2P}\sum_{\frac{X}{m}<p\leq\frac{3X}{m}}\log^2 p\ll \frac{XP\log X}{H\log P}\ll\frac{X}{\log^{C+D-1}X}.\end{equation}

Once again applying Lemma 2·5 followed by Lemma 2·1 and Lemma 2·6, the off-diagonal terms contribute

(9·4) \begin{equation}S_2\ll\frac{X\log\log X \log X}{\log P}\left(\frac{1}{Q_0}+\frac{1}{P}+\frac{Q}{H}\right)\ll\frac{X}{\log^{B'}X}\end{equation}

for $B'>3$ by our choice of $Q_0,P,Q$ and H. Combining the contributions of the dyadic intervals [P,2P] gives that

\begin{equation*}I\ll\frac{X}{\log^BX}\end{equation*}

for $B>1$ , as claimed.

Proposition 9·2. Let $A>3,B>0$ be fixed. Let $\varepsilon>0$ be fixed and $\exp((\log X)^{1-\varepsilon})\leq H\leq X\log^{-A}X$ . Let $\mathfrak{M}$ be defined as in (3·6) with $Q_0,Q$ as in (9·2). Then, for all but at most $O(HQ_0^{-1/3})$ values of $0<|h|\leq H$ we have that

\begin{equation*}\int_\mathfrak{M} |S(\alpha)|^2e(-h\alpha) d\alpha=\mathfrak{S}(h)X\left(\sum_{P_1\leq p\leq P_2}\frac{1}{p}\right)^2+ O\left(\frac{X}{\log^B X}\right).\end{equation*}

Proof. Recalling Lemma 5·1, we have the expansion

(9·5) \begin{equation}\begin{aligned}S(\alpha)=&\frac{\mu(q)}{\varphi(q)}\sum_{P_1<p\leq P_2}\frac{1}{p}\sum_{X<n\leq2X}e(\beta n)\\[5pt]&+\frac{1}{\varphi(q)}\sum_{\chi(q)}\tau(\overline{\chi})\chi(a)\sum_{X<n\leq2X}\left(\varpi_2(n)\chi(n)-\delta_\chi\sum_{P_1\leq p\leq P_2}\frac{1}{p}\right)e(\beta n)\\[5pt]=&a(\alpha)+b(\alpha)\end{aligned}\end{equation}

and following the argument of Section 5 we have that

\begin{equation*}\int_\mathfrak{M} |S(\alpha)|^2e(-h\alpha) d\alpha=\mathfrak{S}(h)X\left(\sum_{P_1\leq p\leq P_2}\frac{1}{p}\right)^2+O\left(\frac{X}{\log^BX}+A(X)B(X)+B^2(X)\right).\end{equation*}

Note that $A^2(X)\ll X(\log\log X)^3$ , so it remains to bound $B^2(X)$ . Following Proposition 6·2, we have that $B^2(X)\ll B_1(X)+B_2(X)$ where $B_1(X)$ is defined in (6·1) and $B_2(X)$ is defined in (6·2) with $(P,P^{1+\delta}]$ replaced with $[P_1,P_2]$ . With our choices of (9·2) and $P_1,P_2$ , following the arguments of Proposition 6·4 and Proposition 6·5 we now have that

\begin{equation*}B_1(X)\ll\frac{X}{\log ^BX}.\end{equation*}

The proof of Proposition 6·3 requires modifying in a similar way to Proposition 3·5. In (6·4) we split the sum over $P_1\leq p_1\leq P_2$ into dyadic intervals $P<p_1\leq 2P$ before applying Cauchy–Schwarz, Lemma 2·3 and then combining the contributions of the dyadic sums.

We are now able to complete the proof of Theorem 1·3.

Proof of Theorem 1·3. By partial summation and Mertens’ theorem we have the bound

\begin{equation*}\begin{aligned}\int_0^1|S(\alpha)|^2d\alpha&=\sum_{X<n\leq2X}\varpi_2^2(n)\ll \log X\sum_{X<n\leq2X}\varpi_2(n)\ll X\log X\sum_{P_1\leq p\leq P_2}\frac{1}{p}\\[5pt]&\ll X\log X \log\log X.\end{aligned}\end{equation*}

Therefore, following the proof of Theorem 1·1, the result may be deduced from combining this bound with Proposition 9·1, an application of Chebyshev’s inequality and Proposition 9·2 followed by an application of partial summation.

10. Proof of Proof of Theorem 1·4

We outline the modifications needed to prove Theorem 1·4. When applying the Hardy–Littlewood circle method, in (3·6) and (3·8) we now choose

(10·1) \begin{equation}Q_0\;:\!=\;\log^{A'}X, A'>6, \qquad Q\;:\!=\;X^{1/6+\varepsilon/2},\qquad H\geq QX^{\varepsilon/2}.\end{equation}

As in Section 9, in Lemma 8·1 we instead define $I\;:\!=\;[U\log P_1,U\log P_2]$ where $U\;:\!=\;Q_0^{E}$ , $E>0$ and we define $\alpha_1\;:\!=\;\varepsilon'>0$ sufficiently small in terms of $\varepsilon>0$ . Analogously to the almost prime case, we may write

\begin{equation*}\sum_{X<n\leq2X}\Lambda(n)\varpi_2(n+h)=\int_0^1 S(\alpha)\overline{S'(\alpha)}e(-h\alpha)d\alpha+O(h\log^2X),\end{equation*}

where for $\alpha\in(0,1)$ we define the exponential sum $S'(\alpha)\;:\!=\;\sum_{X<n\leq 2X}\Lambda(n)e(n\alpha)$ . The error term is acceptable by our choice of H. We have the following result for the major arcs.

Proposition 10·1. Let $A>5,B>0$ be fixed and let $\varepsilon>0$ be fixed sufficiently small. Let $X^{1/6+\varepsilon}\leq H\leq X\log^{-A}X$ . Let $\mathfrak{M}$ be defined as in (3·6) with $Q_0,Q$ as in (10·1). Then, for all but at most $O(HQ_0^{-1/3})$ values of $0<|h|\leq H$ we have that

\begin{equation*}\int_\mathfrak{M} S(\alpha)\overline{S'(\alpha)}e(-h\alpha) d\alpha=\mathfrak{S}(h)X\left(\sum_{P_1\leq p\leq P_2}\frac{1}{p}\right)+ O\left(\frac{X}{\log^B X}\right).\end{equation*}

Proof. We can expand S ^′ in terms of Dirichlet characters (see for example [ Reference Mikawa24 ]):

\begin{align*}S'(\alpha)=&\frac{\mu(q)}{\varphi(q)}\sum_{X<n\leq2X}e(n\beta)+\frac{1}{\varphi(q)}\sum_{\chi(q)}\tau(\bar{\chi})\chi(a)\sum_{X<n\leq2X}(\Lambda(n)\chi(n)-\delta_\chi)e(n\beta)\\&+O(\log^2X)\\=&c(\alpha)+d(\alpha)+O(\log^2X),\end{align*}

say. Therefore, using the expansion (9·5) and Cauchy–Schwarz, we may write the integral over the major arcs as

(10·2) \begin{align}&\int_{\mathfrak{M}}S(\alpha)\overline{S'(\alpha)}e(-h\alpha)d\alpha \nonumber \\=&\int_\mathfrak{M}a(\alpha)\overline{c(\alpha)}e(-h\alpha)d\alpha \nonumber \\&+O\big(A(X)D(X)+B(X)(C(X)+D(X))+(A(X)+B(X))\log^2X\big), \end{align}

where we define $C^2(X)=\int_\mathfrak{M}|c(\alpha)|^2d\alpha$ with D(X) defined analogously. Evaluating $\int_{\mathfrak{M}}a(\alpha)\overline{c(\alpha)}e(-h\alpha)d\alpha$ as in Section 5 gives the required main term and an acceptable error. Mikawa [ Reference Mikawa24 , section 3] proves that $C^2(X)\ll X\log\log X$ and that

\begin{equation*}D^2(X)\ll\sum_{q\leq Q_0}\frac{q}{\varphi(q)}\sum_{\chi(q)}\Bigg(\int_X^{2X}\Bigg|\frac{1}{qQ}\sum_{x<n\leq x+qQ/2}(\Lambda(n)\chi(n)-\delta_\chi)\Bigg|^2dx+qQ\log^2X\Bigg).\end{equation*}

The second term is negligible by the definition of Q. Noting that we have chosen $Q=X^{1/6+\varepsilon/2}$ , we apply Lemma 2·3 to the first term to get

\begin{equation*}D^2(X)\ll\frac{X}{\log^BX}\end{equation*}

for $B>0$ , as required. Combining this with our estimates for A(X),B(X) (from Proposition 9·2) and C(X) we have that the error term in (10·2) is $O(X\log^{-B}X)$ , as required.

Proof of Theorem 1·4. Analogously to the proof of Theorem 3·2, by [ Reference Matomäki, Radziwiłł and Tao21 , proposition 3·1] we have that

\begin{align*}\sum_{0<|h|\leq H}&\left|\sum_{X<n\leq2X}\Lambda(n)\varpi_2(n+h)-\int_{\mathfrak{M}}S(\alpha)\overline{S'(\alpha)}e(-h\alpha)d\alpha\right|^2\\&\ll H\int_\mathfrak{m}|S(\alpha)||S'(\alpha)|\int_{\mathfrak{m}\cap[\alpha-\frac{1}{2H},\alpha+\frac{1}{2H}]}|S(\beta)||S'(\beta)|d\beta d\alpha.\end{align*}

By Cauchy–Schwarz, we have that the above is bounded by

\begin{equation*}\ll H\left(\int_0^1|S'(\alpha)|^2d\alpha\right)\left(\int_0^1|S(\alpha)|^2d\alpha\right)^{1/2}\left(\sup_{\alpha\in\mathfrak{m}}\int_{\mathfrak{m}\cap[\alpha-\frac{1}{2H},\alpha+\frac{1}{2H}]}|S(\beta)|^2d\beta\right)^{1/2}.\end{equation*}

Trivially, we have that

\begin{equation*}\int_0^1|S(\alpha)|^2d\alpha\ll X\log X\log\log X,\qquad\int_0^1|S'(\alpha)|^2d\alpha\ll X\log X,\end{equation*}

so, combining these estimates with Proposition 9·1 (suitably adjusting for the choices of $Q_0,Q,H$ ), we have that

\begin{equation*}\sum_{0<|h|\leq H}\left|\sum_{X<n\leq2X}\Lambda(n)\varpi_2(n+h)-\int_{\mathfrak{M}}S(\alpha)\overline{S'(\alpha)}e(-h\alpha)d\alpha\right|^2\ll\frac{HX^2}{\log^BX}.\end{equation*}

Therefore, applying Chebyshev’s inequality and Proposition 10·1 followed by partial summation gives the result.