Proof Our tool here is the Siegel–Walfisz theorem [Reference Iwaniec and Kowalski26, Equation (5.77)]. Let $ k_0 $ be the square-free part of $ k $ . Then $ k_0 \geq 2 $ . By the multiplicativity of Legendre symbols, we write

$$ \begin{align*} \sum_{\substack{p \in (X, 2X] }} \bigg ( \frac{k}{p}\bigg) = \sum_{\substack{p \in (X, 2X] }} \bigg ( \frac{k_0}{p}\bigg) = V_{1}+V_{2}, \end{align*} $$

where

$$ \begin{align*} V_{1} : &= \sum_{\substack{p \in (X, 2X] \\ p \equiv 1 \bmod 4}} \bigg ( \frac{k_0}{p}\bigg ), \quad V_{2} : = \sum_{\substack{p \in (X, 2X] \\ p \equiv 3 \bmod 4}} \bigg ( \frac{k_0}{p}\bigg). \end{align*} $$

We consider the following cases.

Case 1: $k_0$ odd. Using quadratic reciprocity, we can express $ V_1 $ as

$$ \begin{align*} V_1 &= \sum_{\substack{p \in (X, 2X] \\ p \equiv 1 \bmod 4}} \bigg( \frac{p}{k_0} \bigg) = \sum_{\substack{1 \leq a \leq k_0 \\ (a, k_0) = 1}} \bigg( \frac{a}{k_0} \bigg) \sum_{\substack{p\in (X, 2X] \\ p \equiv 1 \bmod{4} \\ p \equiv a \bmod{k_0}}} 1 = \sum_{\substack{1 \leq a \leq k_0 \\ (a, k_0) = 1}} \bigg( \frac{a}{k_0} \bigg)\sum_{\substack{p \in (X, 2X] \\ p \equiv a' \bmod{4k_0}}} 1. \end{align*} $$

Here, we have noted that $ (p, k_0) = 1 $ and used the Chinese remainder theorem to consolidate the congruence conditions. By the Siegel–Walfisz theorem, for any $B>0$ , we find that

$$ \begin{align*} V_1 &= \frac{\text{Li}(X)}{\varphi(4k_0)}\sum_{\substack{1 \leq a \leq k_0 \\ (a, k_0) = 1}}\bigg( \frac{a}{k_0} \bigg) +O_{A,B}\left( \varphi(k_0)X (\log X)^{-B} \right ). \end{align*} $$

The sum over $ a $ is a complete character sum, which is equal to zero. Hence, by suitably adjusting the parameter B to absorb the factor $\varphi (k_0)$ , we have $V_1 \ll _{A,B} X(\log X)^{-B}$ for any $B>0$ .

Case 2: $k_0$ even. Since $k_0$ is square-free, we must have $k_0=2k_0'$ with $k_0'$ odd. We again apply quadratic reciprocity to write $ V_1 $ as

$$ \begin{align*} V_1 &= \sum_{\substack{p \in (X, 2X] \\ p \equiv 1 \bmod{4}}} \bigg( \frac{2}{p} \bigg) \bigg( \frac{k_0'}{p} \bigg) =\sum_{\substack{p \in (X, 2X] \\ p \equiv 1 \bmod{8}}} \bigg( \frac{k_0'}{p} \bigg)-\sum_{\substack{p \in (X, 2X] \\ p \equiv 5 \bmod{8}}} \bigg( \frac{k_0'}{p} \bigg) \notag \\ &= \sum_{\substack{1 \leq a \leq k_0' \\ (a, k_0') = 1}} \bigg( \frac{a}{k_0'} \bigg) \sum_{\substack{p \in (X, 2X] \\ p \equiv 1 \bmod{8} \\ p \equiv a \bmod{k_0'}}} 1-\sum_{\substack{1 \leq b \leq k_0' \\ (b, k_0') = 1}} \bigg( \frac{b}{k_0'} \bigg) \sum_{\substack{p \in (X, 2X]\\ p \equiv 5 \bmod{8} \\ p \equiv b \bmod{k_0'}}} 1 \notag \\ &= \sum_{\substack{1 \leq a \leq k_0' \\ (a, k_0') = 1}} \bigg( \frac{a}{k_0'} \bigg) \sum_{\substack{p \in (X, 2X] \\ p \equiv a' \bmod{8k_0'}}} 1-\sum_{\substack{1 \leq b \leq k_0' \\ (b, k_0') = 1}} \bigg( \frac{b}{k_0'} \bigg)\sum_{\substack{p \in (X, 2X] \\ p \equiv b' \bmod{8 k_0'}}} 1. \end{align*} $$

Similar to the previous case, using the Siegel–Walfisz theorem, for any $B>0$ , $V_1 \ll _{A,B} X(\log X)^{-B}$ .

The same case-by-case treatment applies to $V_2$ as well, and the desired conclusion follows.

Proof Consider first the case when the exponent is even. Expanding the power, we write

(4.2) $$ \begin{align} \sum_{p \in (X, 2X]}S_{p}(K)^{2r} &= N_{r,K} \left( \pi(2X)-\pi(X) \right) +\underset{ \substack{k_i\ \mathrm{ prime}, \, k_1 \ldots k_{2r} \neq \square}}{\sum_{k_1 \leq K} \sum_{k_2 \leq K } \dots \sum_{k_{2r} \leq K}} \sum_{p\in(X, 2X]} \left( \frac{k_1\ldots k_{2r}}{p} \right), \end{align} $$

where $ N_{r,K} $ denotes the number of tuples $ (k_1, \dots ,k_{2r}) $ such that $ 2 \leq k_i \leq K $ , $ k_i $ is prime, and the product $ k_1 \ldots k_{2r} $ is a perfect square. For such tuples, there are at most r distinct entries, and each must occur an even number of times. The number of ways to choose exactly r distinct primes from $\{2, \ldots , K\}$ is $\left (\frac {1}{r!}\right ) \pi (K)(\pi (K)-1) \dots (\pi (K)-r+1)$ , and the number of ways to arrange these as r pairs is $\frac {(2 r)!}{2^r}=r!(2 r-1) \cdot (2 r-3) \dots 3 \cdot 1$ . Hence, we obtain the inequality

(4.3) $$ \begin{align} N_{r, K} \geq \mu_{2r} \cdot \pi(K)(\pi(K)-1) \dots(\pi(K)-r+1). \end{align} $$

Also, the number of ways of choosing at most r distinct primes from $2, \ldots , K$ is less than or equal to $\pi (K)^{r}$ . When these have been chosen, the number of different ways of arranging them in $2 r$ places (each occurring an even number of times) is at most $(2 r-1)\cdot (2 r-3) \dots 5 \cdot 3 \cdot 1$ . Therefore, we get

(4.4) $$ \begin{align} N_{r, K} \leq \mu_{2r} \cdot \pi(K)^{r}. \end{align} $$

Using the crude estimate,

$$ \begin{align*}& \pi(K)(\pi(K)-1) \dots(\pi(K)-r+1) \\&\quad= \pi(K)^r \exp \left ( \log \left (1-\frac{1}{\pi(K)}\right )+\cdots+\log \left (1-\frac{r-1}{\pi(K)} \right ) \right ) \\ &\quad = \pi(K)^r \left (1+O\left (\frac{r^2}{\pi(K)}\right )\right ), \end{align*} $$

and combining (4.3) and (4.4), we derive

(4.5) $$ \begin{align} N_{r, K}=\mu_{2r} \cdot \pi(K)^r \left (1+O\left (\frac{r^2}{\pi(K)}\right )\right ). \end{align} $$

Now, we bound the contribution from the cases where the product $ k_1 \ldots k_{2r} $ is not a perfect square. By Lemma 4.1, for any $B>0$ ,

(4.6) $$ \begin{align} \sum_{\substack{p \in (X, 2X] }} \bigg ( \frac{k_1\ldots k_{2r}}{p}\bigg) \ll_{r,A,B} X(\log X)^{-B}. \end{align} $$

Substituting (4.5) and (4.6) in (4.2), we obtain

$$\begin{align*}\sum_{p \in (X, 2X]}S_{p}(K)^{2r}&= \mu_{2r} \cdot \pi(K)^r \left( \pi(2X)-\pi(X) \right) \left(1+O\left (\frac{r^2}{\pi(K)}\right )\right )\\ &\quad +O_{r,A,B} \left ( X(\log X)^{-B} \right). \end{align*}$$

This completes the proof for the even exponent case. When the exponent is odd, the proof is similar except that the product $k_1\ldots k_{2r-1}$ is never a perfect square.

Proof By our definition of $S_p(K)$ , the condition $H_p \leq \tfrac {1}{2}( \pi (K)+\lambda \cdot \pi (K)^{\frac {1}{2}} )$ is equivalent to $S_{p}(K) \geq - \lambda \cdot \pi (K)^{\frac {1}{2}}$ . By Lemma 4.2, for every fixed positive integer j, we have

$$ \begin{align*} \frac{1}{\# \{ p \in (X,2X]\ \mathrm{ prime}\}}\sum_{p \in (X, 2X]}\left((\pi(K))^{-\frac{1}{2}} S_{p}(K)\right)^{j} \rightarrow \mu_{j} \quad \text { as } \quad X \rightarrow \infty, \end{align*} $$

where

$$\begin{align*}\mu_{j}= \begin{cases}0, & \text { if } j=2r-1, \\ (2 r-1)\cdot (2 r-3) \dots 5 \cdot 3 \cdot 1, & \text { if } j=2r. \end{cases} \end{align*}$$

The proof now follows using the moment method and an application of the Berry–Esseen theorem. See [Reference Basak, Nath and Zaharescu5, Reference Davenport and Erdös13] for such applications.

Proof We begin by expressing $\log f_p(\tau )$ as

(5.4) $$ \begin{align} \log f_p(\tau) = -\sum_{k \leq K} a_{k,p} \log (1-e^{-k\tau}) = \sum_{\ell=1}^{\infty} \frac{1}{\ell} \sum_{k \leq K} a_{k,p} e^{-k\ell \tau}. \end{align} $$

For $\operatorname {Re}(\tau ) = \sigma>0$ , using the inverse Mellin transform, we find that

$$ \begin{align*} e^{-\tau} = \frac{1}{2 \pi i} \int_{2 - i\infty}^{2+i \infty}\tau^{-s} \Gamma(s) \, d s. \end{align*} $$

Substituting this into (5.4), we obtain

(5.5) $$ \begin{align} \log f_p(\tau)&= \frac{1}{2 \pi i} \int_{2 - i\infty}^{2+i \infty} \tau^{-s} \Gamma(s) \zeta(s+1) \mathcal{D}_p(s) \, d s, \end{align} $$

where

$$ \begin{align*} \mathcal{D}_{p}(s) = \sum_{k \leq K} \frac{a_{k,p}}{k^s}. \end{align*} $$

Our goal now is to estimate the integral on the right-hand side of (5.5). The path of integration in (5.5) is taken along the vertical line $\operatorname {Re}(s) = 2$ . We modify the path for $|t| \leq T$ by moving it leftward and replacing it with a rectangular contour. This new contour connects the points $2 \pm i T$ and $-c \pm i T$ , where $c\in (0,1)$ . The resulting path of integration L consists of the segments $L = \bigcup _{i=1}^5 L_i$ , defined as follows: $L_1 : (2 - i\infty , 2 - iT], L_2 : [2 - iT, -c - iT], L_3 : [-c - iT, -c + iT], L_4 : [-c + iT, 2 + iT]$ , and $L_5 : [2 + iT, 2 + i\infty )$ . Using this modified path, the Cauchy residue theorem gives

(5.6) $$ \begin{align} \log f_p(\tau) = \text{Res} + \frac{1}{2\pi i} \sum_{j=1}^5 I_{j}, \end{align} $$

where $\text {Res}$ accounts for the residues of the integrand’s singularities enclosed by the original and modified paths, and $I_{j}$ represents the integrals along the respective segments $L_j$ . The integrand in (5.5) has a double pole at $s=0$ . Computing the residue, we derive

(5.7) $$ \begin{align} \text{Res} = \mathcal{D}_p'(0)-\mathcal{D}_p(0)\log \tau. \end{align} $$

To estimate the integrals $I_{j}$ , we rely on the following bounds:

(5.8) $$ \begin{align} \sigma &\leq |\tau| \leq \sqrt{2} \sigma, \, \text{for } |\operatorname{Arg}(\tau)| \leq \tfrac{\pi}{4}, \end{align} $$

(5.9) $$ \begin{align} \lvert \Gamma(s) \rvert &\ll_{c} e^{-\frac{\pi}{2} \lvert \operatorname{Im}(s) \rvert}, \, \text{uniformly for } -c \leq \operatorname{Re}(s) \leq 2, \end{align} $$

(5.10) $$ \begin{align} \lvert \zeta(s) \rvert & \ll_{c} (\lvert t\rvert+2)^C, \, \text{uniformly for} -c \leq \operatorname{Re}(s)<1, \end{align} $$

for some constant $C>0$ depending only on c. Since $\lvert \operatorname {Arg}(\tau )\rvert \leq \frac {\pi }{4}$ , the absolute value of $\tau ^{-s}$ can be bounded as follows:

(5.11) $$ \begin{align} \lvert \tau^{-s} \rvert &=\lvert \exp(-\operatorname{Re}(s) \log \lvert\tau \rvert +\operatorname{Im}(s) \operatorname{Arg}(\tau)) \rvert \leq \lvert \tau \rvert^{-\operatorname{Re}(s)} e^{\frac{\pi}{4} \lvert \operatorname{Im}(s) \rvert}. \end{align} $$

Consider first the integral $I_5$ . In our range of integration, we have $\zeta (s+1), \mathcal {D}_p(s) \ll 1$ . Therefore, using the bounds (5.8), (5.9), and (5.11), we find that

$$ \begin{align*} I_{5} \ll_{c} \lvert \tau \rvert^{-2} \int_{T}^{\infty} e^{-\frac{\pi t}{4}} \, d t \ll_{c} \sigma^{-2} e^{-\frac{\pi T}{4}}. \end{align*} $$

An analogous argument holds for $I_{1}$ . So, we obtain

(5.12) $$ \begin{align} I_{1}+I_{5} \ll_{c} \sigma^{-2} e^{-\frac{\pi T}{4}}. \end{align} $$

Now, we consider the horizontal integrals $I_2$ and $I_4$ . Using (5.8)–(5.11), we have

(5.13) $$ \begin{align} I_2 & \ll \int_{2 - iT}^{-c-i T} \tau^{-s} \Gamma(s) \zeta(s+1) \sum_{k \leq K} k^c \, d s \notag \\ &\ll_{c} e^{-\frac{\pi T}{4}} ( T^C + K^{c+1} ) \int_{-c}^{2} \sigma^{-u} \, du \ll_{c} \sigma^{-2} e^{-\frac{\pi T}{4}}\left( T^C+K^{c+1} \right). \end{align} $$

A similar bound holds for $I_4$ . Therefore, we obtain

(5.14) $$ \begin{align} I_2+I_4 &\ll_{c} \sigma^{-2} e^{-\frac{\pi T}{4}}\left( T^C+K^{c+1} \right). \end{align} $$

Finally, it remains to estimate the vertical integral $I_3$ . To this end, using (5.8)–(5.11), we have

(5.15) $$ \begin{align} I_3 &\ll \int_{-c - iT}^{-c+i T} \tau^{-s} \Gamma(s) \zeta(s+1) \sum_{\substack{ k \leq K}} k^c \, d s \notag \\ &\ll_{c} \lvert \tau \rvert^{c} K^{c+1} \int_{-T}^{T} (\lvert t \rvert+2)^C e^{-\frac{\pi \lvert t \rvert}{4}} \, d t \ll_{c} \sigma^{c} K^{c+1}. \end{align} $$

Putting together (5.6), (5.7), (5.12), (5.14), and (5.15), and letting $T \to \infty $ , we arrive at

$$\begin{align*}\log f_p(\tau) = \mathcal{D}_p'(0)-\mathcal{D}_p(0)\log \tau +O_{c,A}( \sigma^{c}K^{c+1}). \end{align*}$$

Noting that $\mathcal {D}_p'(0)= -\log ( \prod _{k \in \mathbb {H}_p} k )$ and $\mathcal {D}_p(0) = H_p$ , we complete the proof.

Proof Define

(6.2) $$ \begin{align} \mathcal{F} := \left\{ p \in (X,2X]\ \mathrm{ prime}: H_p < \tfrac{\pi(K)}{3} \right\}. \end{align} $$

If $H_p < \tfrac {\pi (K)}{3}$ , then $S_{p}(K)> \tfrac {\pi (K)}{3}$ , where $S_{p}(K)$ is as in (4.1). Then by Lemma 4.2, for any $r \in \mathbb {N}$ ,

$$ \begin{align*} \#\mathcal{F} \cdot \frac{\pi(K)^{2r}}{3^{2r}} &\leq \sum_{\substack{p \in \mathcal{F}}} S_{p}(K)^{2r} \leq \sum_{\substack{p \in (X, 2X] \\ p\ \mathrm{ prime} }} S_{p}(K)^{2r} \ll_{\eta,A,r} X(\log X)^{-1} \cdot \pi(K)^{r}. \end{align*} $$

By choosing $ r $ sufficiently large in terms of $ \eta $ , $ A $ , and $ B $ , it follows that $\# \mathcal {F} \ll _{\eta , A, B} X (\log X)^{-B}$ . Therefore, for all primes $ p \in (X, 2X] \setminus \mathcal {F} $ , we have the uniform bound $H_p \geq \tfrac {\pi (K)}{3}$ which implies that $\sigma _p \geq \tfrac {\pi (K)}{3m}$ . Throughout the rest of the proof, we will restrict ourselves to working with primes from the set $(X, 2X] \setminus \mathcal {F}$ . We consider two cases depending on the range of t.

Case 1: $(\log K)^{-\frac {1}{2}} \leq \lvert 2\pi K t \rvert \leq \sqrt {10}$ . By (5.4) and nonnegativity,

(6.3) $$ \begin{align} \operatorname{Re}(\log f_p(\sigma_p+2\pi i t))-\log f_p(\sigma_p) & =\sum_{\ell=1}^{\infty} \frac{1}{\ell} \sum_{k \in \mathbb{H}_p} e^{-k \ell \sigma_p} \left ( \cos(2\pi k \ell t)-1 \right) \notag \\ & \leq \sum_{k \in \mathbb{H}_p} e^{-k \sigma_p} \left ( \cos(2\pi k t)-1 \right)\notag\\& \leq \frac{1}{2}\sum_{k \in \mathbb{H}_p} \left ( \cos(2\pi k t)-1 \right). \end{align} $$

By the prime number theorem and partial summation, for $p \in (X,2X]\setminus \mathcal {F}$ ,

$$\begin{align*}\sum_{\substack{ k \in \mathbb{H}_p}} k^2 \geq c_1K^3(\log K)^{-1}, \end{align*}$$

for some constant $c_1>0$ . Therefore, by Taylor expansion, the right-hand side of (6.3) is

$$ \begin{align*} &\leq \frac{1}{2} \sum_{\substack{k \in \mathbb{H}_p}} \sum_{j=1}^{\infty} \frac{(-1)^j(2\pi kt)^{2j}}{(2j)!} \leq -\frac{1}{200} \sum_{\substack{k \in \mathbb{H}_p}} (2\pi kt)^{2} \leq -c_2 K(\log K)^{-2}, \end{align*} $$

for some constant $c_2>0$ .

Case 2: $ K \sigma _p \leq \lvert 2\pi K t \rvert < (\log K)^{-\frac {1}{2}}$ . In this range, (6.3) is no longer efficient. Indeed, when t is small, the terms with $\ell \asymp \sigma _p^{-1}$ can significantly contribute to the left-hand side of (6.3). Rather than relying on (6.3), we work directly with (5.4). We write

(6.4) $$ \begin{align} \operatorname{Re}(\log f_p(\sigma_p+2\pi i t))-\log f_p(\sigma_p) = -\sum_{k \in \mathbb{H}_p} \log \frac{\lvert 1-e^{-k\tau} \rvert}{(1-e^{-k\sigma_p})}. \end{align} $$

By Taylor expansion, we have

$$ \begin{align*} 1-e^{-k\tau} = k\tau+O(K^2t^2), \end{align*} $$

which implies that

(6.5) $$ \begin{align} \lvert 1-e^{-k\tau} \rvert = \lvert k\tau \rvert (1+O((\log K)^{-1})). \end{align} $$

Another application of the Taylor expansion shows that

(6.6) $$ \begin{align} 1-e^{-k\sigma_p} = k\sigma_p(1+O(K\sigma_p)). \end{align} $$

Substituting (6.5) and (6.6) in (6.4), we see that the right-hand side of (6.4) is

(6.7) $$ \begin{align} -\frac{1}{2}\sum_{k \in \mathbb{H}_p} \log \left (1+\frac{4 \pi^2t^2}{\sigma_p^2} \right)+O(H_p (\log K)^{-1}). \end{align} $$

It follows that in this case,

$$\begin{align*}\operatorname{Re}(\log f_p(\sigma_p+2\pi i t))-\log f_p(\sigma_p) \leq -c_3 K(\log K)^{-1}, \end{align*}$$

for some constant $c_3>0$ .

Combining Cases 1 and 2 and choosing $C_0 = \min \{c_2,c_3 \}$ , we arrive at our desired conclusion.

Proof For $X>0$ sufficiently large, uniformly for all $p \in (X,2X]$ , we have

(6.8) $$ \begin{align} \sum_{\substack{k \leq K \\ k\ \mathrm{ prime}}} e^{-k\sigma_p} (\cos(2 \pi k t)-1) \leq \frac{1}{2}\sum_{\substack{k \leq K \\ k\ \mathrm{ prime}}} (\cos (2\pi k t)-1). \end{align} $$

To handle the right-hand side of (6.8), we will use Vinogradov’s bound on exponential sums over primes [Reference Davenport12, Chapter 25]. If $\lvert t-u/v\rvert \leq 1/v^2$ with $(u,v)=1$ , then

(6.9) $$ \begin{align} \sum_{k \leq K} \Lambda(k)e(kt) \ll (K v^{-\frac{1}{2}}+K^{\frac{4}{5}}+K^{\frac{1}{2}}v^{\frac{1}{2}})(\log K)^4. \end{align} $$

Let $K_1 = K(\log K)^{-10}.$ By Dirichlet’s theorem on Diophantine approximation, for $t \in \mathbb {R}$ , there exists a reduced fraction $u/v$ with $1 \leq v \leq K_1$ such that $\lvert t-u/v \rvert \leq 1/(vK_1) \leq 1/v^2$ . If $v \geq (\log K)^{10}$ , then by (6.9) and partial summation, we have

$$\begin{align*}\sum_{\substack{k \leq K \\ k\ \mathrm{ prime}}}\cos(2\pi k t) \ll K (\log K)^{-2}. \end{align*}$$

When $3 \leq v \leq (\log K)^{10}$ , then writing $t=u/v+\beta $ , we have, for any $B>0$ (see [Reference Davenport12, Chapter 26]),

$$\begin{align*}\sum_{k \leq K} \Lambda(k)e(kt)= \frac{\mu(v)}{\varphi(v)}T(\beta)+O_{B}(K(\log K)^{-B}), \end{align*}$$

where $T(\beta ) = \sum _{k \leq K} e(k \beta )$ . We trivially bound $T(\beta )$ and use partial summation to obtain

$$\begin{align*}\bigg \lvert \sum_{\substack{k \leq K \\ k\ \mathrm{ prime}}} \cos(2\pi k t) \bigg \rvert \leq \frac{2}{3}\pi(K). \end{align*}$$

Finally, we are left with $v=1,2$ . For $v=1$ , since $\sqrt {10}/(2\pi K) < \lvert \beta \rvert \leq K_1^{-1}$ , instead of trivially bounding $T(\beta )$ , we use the sharper bound $\lvert T(\beta ) \rvert \leq \lvert \sin (\pi \beta ) \rvert ^{-1} \leq \pi K/\sqrt {10}.$ When $v=2$ , we have $\lvert \beta \rvert \leq 1/(2K_1)$ , and therefore, the same approach as in the case $v=1$ works if $\sqrt {10}/(2\pi K) < \lvert \beta \rvert \leq 1/(2K_1)$ . The remaining range is $\lvert \beta \rvert \leq \sqrt {10}/(2\pi K)$ , which corresponds to t lying in the interval

$$\begin{align*}\left [-\frac{1}{2}, -\frac{1}{2}+ \frac{\sqrt{10}}{2 \pi K}\right ] \bigcup \left [\frac{1}{2}- \frac{\sqrt{10}}{2 \pi K}, \frac{1}{2}\right ].\end{align*}$$

It’s easy to verify that for such t, $\cos (2\pi k t) \leq 1/\sqrt {2}$ for all primes $3 \leq k \leq \pi K /(8 \sqrt {10})$ . Combining all the ranges, we conclude that there exists a constant $C_1>0$ such that for $\lvert 2 \pi K t \rvert> \sqrt {10}$ ,

$$\begin{align*}\sum_{\substack{k \leq K \\ k\ \mathrm{ prime}}} e^{-k\sigma_p} (\cos(2 \pi k t)-1) \leq \frac{1}{2}\sum_{\substack{k \leq K \\ k\ \mathrm{ prime}}} (\cos (2\pi k t)-1) \leq -C_1 \pi(K), \end{align*}$$

which completes the proof.

Proof Our approach is to study the high moments of $R_p(t)$ , more precisely, the following sum:

$$ \begin{align*} \sum_{\substack{p \in (X, 2X] \\ p\ \mathrm{ prime}} } R_p^{2r}(t), \end{align*} $$

where $r \in \mathbb {N}$ is fixed. Expanding the $2r$ -th exponent and using Taylor expansion, we have

(6.11) $$ \begin{align} \sum_{\substack{p \in (X, 2X]}} R_p^{2r}(t) &= \underset{ k_i\ \mathrm{ prime}}{\sum_{k_1 \leq K} \sum_{k_2 \leq K} \dots \sum_{k_{2r} \leq K}} \prod_{i=1}^{2r} (1-\cos(2\pi k_it)) \notag \\ &\quad \times \sum_{\substack{p \in (X, 2X]}} \left (\frac{k_1 \ldots k_{2r}}{p} \right ) \prod_{i=1}^{2r} \left ( \sum_{j_i=0}^{\infty} \frac{(-1)^{j_i}(k_i \sigma_p)^{j_i}}{j_i!}\right) \notag \\ &=\sum_{j_1,\dots,j_{2r}} \left (-\frac{1}{m} \right)^{\sum_{i}j_i} \frac{1}{j_1! j_2! \ldots j_{2r}!} \notag \\ &\quad \times \sum_{\substack{k_1,\dots, k_{2r} \leq K \\ k_i\ \mathrm{ prime} }} \prod_{i=1}^{2r} k_i^{j_i}(1-\cos(2\pi k_it)) \sum_{\substack{p \in (X, 2X]}} H_p^{\sum_{i}j_i}\left (\frac{k_1 \ldots k_{2r}}{p} \right ). \end{align} $$

Let $J_0>0$ be a large parameter depending only on $\eta $ and r. We consider the following cases.

Case 1: $0 \leq \sum _{i}j_i = J \leq J_0$ . Note that

$$\begin{align*}H_p = \frac{1}{2}\bigg ( \sum_{\substack{\ell \leq K \\ \ell\ \mathrm{ prime}}} 1-\left (\frac{\ell}{p} \right)\bigg). \end{align*}$$

Therefore, the contribution from this case to (6.11) equals

(6.12) $$ \begin{align} \sum_{\substack{j_1, \dots, j_{2r} \\ J \leq J_0}} &\left (-\frac{1}{2m} \right)^{J} \frac{1}{j_1! j_2! \ldots j_{2r}!} \sum_{\substack{k_1, \dots, k_{2r} \\ k_i \leq K\ \mathrm{ prime}}} \prod_{i=1}^{2r} k_i^{j_i}(1-\cos(2\pi k_it)) \notag \\ \quad &\times \sum_{\substack{\ell_1, \dots, \ell_J \\ \ell_i \leq K \ \mathrm{ prime}}} \sum_{\mathcal{U} \subseteq [J]} (-1)^{\# \mathcal{U}}\sum_{\substack{p \in (X, 2X]}} \left ( \frac{\prod_{i=1}^{2r} k_i \cdot \prod_{u\in \mathcal{U}} \ell_u}{p} \right ). \end{align} $$

When the product $\prod _{i=1}^{2r} k_i \cdot \prod _{u\in \mathcal {U}} \ell _u$ is not a perfect square, by Lemma 4.1, for any $B_0>0$ ,

(6.13) $$ \begin{align} \sum_{\substack{p \in (X, 2X]}} \left ( \frac{\prod_{i=1}^{2r} k_i \cdot \prod_{u\in \mathcal{U}} \ell_u}{p} \right ) \ll_{\eta,A,r} X (\log X)^{-B_0}. \end{align} $$

Choosing $B_0$ large depending at most on $\eta ,A$ , and r, the contribution here is $\ll _{\eta ,A,r} X (\log X)^{-1}$ .

Now suppose $\prod _{i=1}^{2r} k_i \cdot \prod _{u\in \mathcal {U}} \ell _u$ is a perfect square. This forces $\#\mathcal {U}$ to be even. Then (6.12) equals

(6.14) $$ \begin{align} (\pi(2X)-\pi(X)) \cdot \sum_{\substack{j_1, \dots, j_{2r} \\ J \leq J_0}} &\left (-\frac{1}{2m} \right)^{J} \frac{1}{j_1! j_2! \ldots j_{2r}!} \sum_{\substack{k_1, \dots, k_{2r} \\ k_i \leq K \ \mathrm{ prime}}} \prod_{i=1}^{2r} k_i^{j_i}(1-\cos(2\pi k_it)) \notag \\ \quad &\times \sum_{\substack{\ell_1, \dots, \ell_J \\ \ell_i \leq K \ \mathrm{ prime}}} \sum_{\mathcal{U} \subseteq [J]} \mathbb{1}_{\prod_{i=1}^{2r} k_i \cdot \prod_{u\in \mathcal{U}} \ell_u =\square}. \end{align} $$

The count of cases where the product is a perfect square is $\ll _{\eta ,A,r} \pi (K)^{r+\#\mathcal {U}/2}$ following the proof of Lemma 4.2. When $\lvert 2 \pi Kt \rvert> \sqrt {10}$ , using the trivial bound $1-\cos (2\pi k_i t) \leq 1$ and $\eta>2$ , (6.14) is

(6.15) $$ \begin{align} &\ll_{\eta,A,r} \frac{X}{\log X} \cdot \pi(K)^{r} \sum_{\substack{j_1, \dots, j_{2r} \\ J \leq J_0}} \frac{1}{j_1! j_2! \ldots j_{2r}!} \left (\frac{K \pi(K)}{m} \right)^{J} \ll_{\eta,A,r} \frac{X}{\log X} \cdot \pi(K)^{r}. \end{align} $$

Case 2: $\sum _{i}j_i = J> J_0$ . Since $H_p \leq \pi (K)$ , the contribution from these cases to (6.11) is

(6.16) $$ \begin{align} &\ll_{\eta,A,r} \frac{X}{\log X} \cdot \pi(K)^{2r}\sum_{\substack{j_1, \dots, j_{2r} \\ J> J_0}} \frac{1}{j_1! j_2! \ldots j_{2r}!} \left (\frac{K \pi(K)}{m} \right)^{J} \ll_{\eta,A,r} \frac{X}{\log X} \cdot \pi(K)^{r}, \end{align} $$

by allowing $J_0 \geq r \cdot (\eta -2)^{-1}$ .

We combine both cases and the bounds (6.15)–(6.16). Choosing $J_0=\lceil r \cdot (\eta -2)^{-1} \rceil $ , we obtain

(6.17) $$ \begin{align} \sum_{\substack{p \in (X, 2X]}} R_p^{2r}(t) \ll_{\eta,A,r} X(\log X)^{-1} \cdot \pi(K)^{r}, \quad \lvert 2 \pi K t \rvert> \sqrt{10}. \end{align} $$

For $t \in [-\tfrac {1}{2},\tfrac {1}{2}]$ with $\lvert 2 \pi K t \rvert> \sqrt {10}$ , define

(6.18) $$ \begin{align} \mathcal{E}(t) : = \{ p \in (X,2X] \ \mathrm{ prime}: \lvert R_p(t) \rvert> \pi(K)^{\frac{3}{4}} \}. \end{align} $$

By (6.17), we have

$$ \begin{align*} \#\mathcal{E}(t) \cdot \pi(K)^{\frac{3r}{2}} &\leq \sum_{\substack{p \in \mathcal{E}(t)}} R_p^{2r}(t) \leq \sum_{\substack{p \in (X, 2X] \\ p \ \mathrm{ prime} }} R_p^{2r}(t) \ll_{\eta,A,r} X(\log X)^{-1} \cdot \pi(K)^{r}. \end{align*} $$

Choosing r sufficiently large in terms of A and B, we see that $\#\mathcal {E}(t) \ll _{\eta ,A,B} X(\log X)^{-B}$ .

Proof As in the proof of Lemma 6.1, we restrict ourselves to primes from the set $(X,2X] \setminus \mathcal {F}$ , where $\mathcal {F}$ is as in (6.2). This ensures the uniform bound $H_p \geq \frac {\pi (K)}{3}$ . By the mean value theorem, if $t_1,t_2 \in [-\tfrac {1}{2}, \tfrac {1}{2}]$ with $\lvert t_1 -t_2 \rvert \leq T^{-1}$ for some $T>0$ , then

(6.19) $$ \begin{align} \lvert R_p(t_1) - R_p(t_2) \rvert &\leq \sum_{\substack{ k \leq K \\ k \ \mathrm{ prime}}} e^{-k\sigma_p}\lvert \cos(2 \pi k t_2)-\cos(2\pi k t_1) \rvert \notag \\ &\leq \frac{2\pi}{T} \sum_{\substack{ k \leq K \\ k \ \mathrm{ prime}}} k < \frac{2 \pi K}{T} \cdot \pi(K). \end{align} $$

We choose $T= \lceil K^2\rceil $ and divide the interval $(\frac {\sqrt {10}}{2 \pi K}, \frac {1}{2}]$ into sub-intervals $I_1,I_2, \dots $ of length $T^{-1}$ . Note that there are at most T such sub-intervals. For each sub-interval $I_j$ , define

$$\begin{align*}\begin{aligned} \mathcal{E}_j:=\left \{ p \in (X, 2X]\setminus \mathcal{F} \ \mathrm{ prime} : \text{There exists } t \in I_j \text{ such that } \lvert R_p(t)\rvert> 2 \cdot \pi(K)^{\frac{3}{4}} \right \}. \end{aligned} \end{align*}$$

Consider a prime $p \in \mathcal {E}_j$ . Let $t_1 \in I_j$ such that $\lvert R_p(t_1)\rvert> 2 \cdot \pi (K)^{\frac {3}{4}}$ . Then by (6.19) and the triangle inequality, for any $t_2 \in I_j$ , we have

$$ \begin{align*} \lvert R_p(t_2) \rvert \geq \lvert R_p(t_1) \rvert - \lvert R_p(t_2) - R_p(t_1) \rvert> \pi(K)^{\frac{3}{4}}. \end{align*} $$

It follows that $p \in \mathcal {E}(t_2)$ , where $\mathcal {E}(t_2)$ is given by (6.18). Therefore, we have $\# \mathcal {E}_j \subseteq \mathcal {E}(t_2)$ and by Lemma 6.3, we obtain the bound $\# \mathcal {E}_j \ll _{\eta ,A,B_0} X (\log X)^{-B_0}$ for any $B_0>0$ . Taking the union over all j’s, we deduce that

$$\begin{align*}\# \cup_{j} \mathcal{E}_j \ll_{\eta,A,B_0} T X(\log X)^{-B_0} \ll_{\eta,A,B} X(\log X)^{-B}, \end{align*}$$

by choosing $B_0$ sufficiently large depending on $\eta ,A$ , and B. A similar argument gives rise to analogous exceptional sets for the interval $[-\frac {1}{2},-\frac {\sqrt {10}}{2 \pi K})$ . Letting $\mathcal {E}$ be the union of these exceptional sets $\mathcal {E}_j$ ’s along with $\mathcal {F}$ , we arrive at the desired conclusion.

Proof Let $\mathcal {E}$ be the exceptional set provided by Lemma 6.4. Since $\mathcal {F} \subseteq \mathcal {E}$ where $\mathcal {F}$ is as in (6.2), we obtain the uniform bound $H_p \geq \frac {\pi (K)}{3}$ for all $p \in (X,2X]\setminus \mathcal {E}$ . Furthermore, by Lemma 6.1, for all $p \in (X,2X]\setminus \mathcal {E}$ , we have

$$\begin{align*}\lvert f_p(\sigma_p+2\pi i t)\rvert \leq \lvert f_p(\sigma_p) \rvert \exp\left ( -C_0 K(\log K)^{-2}\right), \quad \frac{\sigma_p}{2\pi}< \lvert t \rvert \leq \frac{\sqrt{10}}{2\pi K}, \end{align*}$$

for some constant $C_0>0$ . To handle the remaining range, note that by (6.3),

(6.20) $$ \begin{align} \operatorname{Re}(\log f_p(\sigma_p+2\pi i t))-\log f_p(\sigma_p) &\leq \frac{1}{2}\sum_{\substack{k \leq K \\ k \ \mathrm{ prime}}} e^{-k\sigma_p} (\cos(2 \pi k t)-1)+\frac{1}{2} R_p(t), \end{align} $$

where $R_p(t)$ is given by (6.10). By Lemmas 6.2 and 6.4, for all $p \in (X,2X]\setminus \mathcal {E}$ , the right-hand side of (6.20) is $\leq -\frac {C_1}{2} \pi (K)$ , where $ C_1> 0 $ is the constant from Lemma 6.2. Hence, we deduce that

$$\begin{align*}\lvert f_p(\sigma_p+2\pi i t)\rvert \leq \lvert f_p(\sigma_p) \rvert \exp\left ( -\tfrac{C_1}{2} \pi(K)\right), \quad \frac{\sqrt{10}}{2\pi K}< \lvert t \rvert \leq \frac{1}{2}. \end{align*}$$

Combining the two ranges and choosing $C_2 = \min \{ C_0,C_1/2 \}$ , we arrive at the desired result.

Proof The proof follows the arguments in Lemmas 4.2 and 4.3.

Proof The proof follows the arguments in Lemma 5.1.

Proof The range $\frac {\sigma _{p,q}}{2\pi }< \lvert t \rvert \leq \frac {\sqrt {10}}{2 \pi K}$ can be handled precisely following the arguments of Lemma 6.1. In particular, there exists a constant $C_0>0$ such that outside an exceptional set of primes $\mathcal {F} \subseteq (X, 2X]$ given by

(8.8) $$ \begin{align} \mathcal{F} := \left\{ p \in (X,2X] \ \mathrm{ prime}: H_{p,q} < \tfrac{\pi(K)}{3} \right\}, \end{align} $$

which depends at most on $ X, \eta ,A$ , and $q $ , and satisfies

$$\begin{align*}\#\mathcal{F} \ll_{\eta,A,B,q} X(\log X)^{-B}, \end{align*}$$

we have the uniform bounds $H_{p,q} \geq \frac {\pi (K)}{3}$ and

$$ \begin{align*} \lvert f_{p,q}(\sigma_{p,q}+2\pi i t)\rvert \leq \lvert f_{p,q}(\sigma_{p,q}) \rvert \exp\left ( -C_0 K(\log K)^{-2}\right), \quad \frac{\sigma_{p,q}}{2\pi}< \lvert t \rvert \leq \frac{\sqrt{10}}{2 \pi K}. \end{align*} $$

We now present an alternative approach toward handling the remaining range $ \frac {\sqrt {10}}{2 \pi K} < \lvert t \rvert \leq \frac {1}{2} $ . Instead of estimating high moments and applying the discretization method as in Section 6, we use the Erdős–Turán inequality [Reference Montgomery33, Chapter 1, Corollary 1.1]. Throughout the remainder of the proof, we restrict ourselves to working with primes from the set $(X,2X]\setminus \mathcal {F}$ . Let $\varepsilon _0>0$ be an arbitrarily small absolute constant. For t satisfying $ \frac {\sqrt {10}}{2 \pi K} < \lvert t \rvert \leq \frac {1}{2} $ , let

$$\begin{align*}\mathbb{H}_{p,q}(t) = \{ k \in \mathbb{H}_{p,q} : \cos(2\pi k t)>1-\varepsilon_0 \} \quad \text{and} \quad H_{p,q}(t) = \# \mathbb{H}_{p,q}(t). \end{align*}$$

If $H_{p,q}(t) < \tfrac {\pi (K)}{4}$ for all t in our range and all $p \in (X,2X]\setminus \mathcal {F}$ , then we have

(8.9) $$ \begin{align} \sum_{\substack{k \in \mathbb{H}_{p,q}}} e^{-k\sigma_{p,q}} (\cos(2 \pi k t)-1) \leq \frac{1}{2}\sum_{\substack{k \in \mathbb{H}_{p,q}(t)^c}} (\cos (2\pi k t)-1) \leq -C_1 \pi(K), \end{align} $$

for some constant $C_1>0$ depending only on $\varepsilon _0$ . This implies (8.7). So we may assume that for some t in our range and for some prime $p \in (X,2X]\setminus \mathcal {F}$ , $H_{p,q}(t) \geq \tfrac {\pi (K)}{4}$ . We now proceed along the following steps to arrive at a contradiction to this assumption. The key argument is that $\pi (K)/4$ many primes cannot have $\cos (2\pi kt)$ so close to $1$ .

Step 1: Consider the sequence $ \{ k t \bmod 1 \}$ , as k varies over the primes up to K. Denote by $\mathscr {D}$ , the discrepancy of this sequence, which is defined by

$$\begin{align*}\mathscr{D} : = \sup_{[\alpha_1, \alpha_2] \subseteq [-\frac{1}{2},\frac{1}{2}]} \lvert \# \{ k \leq K : k \text{ prime and } k t \bmod 1 \in [\alpha_1,\alpha_2 ] \} - \pi(K)(\alpha_2-\alpha_1) \rvert. \end{align*}$$

We choose $\alpha _1=-(2 \pi )^{-1}\cos ^{-1}(1-\varepsilon _0)$ and $\alpha _2 =(2 \pi )^{-1}\cos ^{-1}(1-\varepsilon _0)$ . It follows that

(8.10) $$ \begin{align} \mathscr{D} \geq H_{p,q}(t) - \pi(K)(\alpha_2-\alpha_1) \geq \frac{\pi(K)}{5}, \end{align} $$

by choosing $\varepsilon _0$ small enough. On the other hand, by the Erdős–Turán inequality, for any $H \in \mathbb {N}$ ,

(8.11) $$ \begin{align} \mathscr{D} \leq \frac{\pi(K)}{H+1}+3 \sum_{h=1}^{H} \frac{1}{h} \bigg \lvert \sum_{\substack{k \leq K \\ k \ \mathrm{ prime}}} e(k h t) \bigg \rvert \end{align} $$

for each $1 \leq h \leq H$ , there exists a reduced fraction $u_h/v_h$ with $0 \leq u_h < v_h$ and $1 \leq v_h \leq K_1$ such that $\lvert \{ht\} -u_h/v_h\rvert \leq 1/(v_hK_1) \leq 1/v_h^2$ . If $v_h \geq (\log K)^{10}$ for some h, then by (6.9) and partial summation, we find that

(8.12) $$ \begin{align} \sum_{\substack{k \leq K \\ k \ \mathrm{ prime}}} e( k h t) \ll K (\log K)^{-2}. \end{align} $$

On the other hand, if $300 < v_h \leq (\log K)^{10}$ for some h, then writing $\{ht\}=u_h/v_h+\beta _h$ , we have

$$\begin{align*}\sum_{k \leq K} \Lambda(k)e(kht) = \frac{\mu(v_h)}{\varphi(v_h)}T(\beta_h)+O_{B}(K(\log K)^{-B}), \end{align*}$$

for any $B>0$ , where $T(\beta _h) = \sum _{k \leq K} e(k \beta _h)$ . Trivially bounding $T(\beta _h)$ and using partial summation,

(8.13) $$ \begin{align} \bigg \lvert \sum_{\substack{k \leq K \\ k \ \mathrm{ prime}}} e(k h t) \bigg \rvert \leq \frac{1}{70} \pi(K). \end{align} $$

Finally, if $1 \leq v_h \leq 300$ but $36/K \leq \lvert \beta _h \rvert \leq 1/(v_h K_1) $ , then similar to the proof of Lemma 6.2, instead of trivially bounding $T(\beta _h)$ , we use the sharper bound $\lvert T(\beta _h) \rvert \leq \lvert \sin (\pi \beta _h) \rvert ^{-1}$ to obtain

(8.14) $$ \begin{align} \bigg \lvert \sum_{\substack{k \leq K \\ k \ \mathrm{ prime}}} e(k h t) \bigg \rvert \leq \frac{1}{70} \pi(K). \end{align} $$

In conclusion, if t is such that for all $1 \leq h \leq H$ , $v_h>300$ , or $36/K \leq \lvert \beta _h \rvert \leq 1/(v_h K_1)$ , then substituting the bounds (8.12)–(8.14) in (8.11), and evaluating the harmonic sum, we see that $\mathscr {D} < \frac {\pi (K)}{5}$ , which contradicts (8.10), and thereby, contradicts the assumption that $H_{p,q}(t) \geq \tfrac {\pi (K)}{4}$ .

Step 2: Following Step 1, we may now assume that for some $1 \leq h \leq H$ , we have $1 \leq v_h \leq 300$ and $\lvert \beta _h \rvert \leq 36/K$ . Note that

(8.15) $$ \begin{align} t = \frac{\lfloor ht \rfloor}{h} +\frac{u_h}{hv_h}+\frac{\beta_h}{h} = \frac{u_h^{\prime}}{v_h^{\prime}}+\beta_h^{\prime}, \end{align} $$

where $\beta _h^{\prime } = \beta _h/h$ and $(u_h^{\prime }, v_h^{\prime })=1$ . For $1 \leq b \leq v_h^{\prime }$ with $(b,v_h^{\prime })=1$ , consider the set $\{ k \in \mathbb {H}_{p,q}(t) : k \equiv b \bmod v_h^{\prime }\}.$ Then our assumption $H_{p,q}(t) \geq \tfrac {\pi (K)}{4}$ implies that there exists some b for which

(8.16) $$ \begin{align} \# \{ k \in \mathbb{H}_{p,q}(t) : k \equiv b \bmod v_h^{\prime}\} \geq \frac{1}{5 } \cdot \frac{\pi(K)}{\varphi(v_h^{\prime})}. \end{align} $$

We will contradict (8.16). To see this, let $k \in \mathbb {H}_{p,q}(t)$ with $k \equiv b \bmod v_h^{\prime }$ . Using (8.15), we have

(8.17) $$ \begin{align} \varepsilon_0 &>1-\cos\left (2 \pi \left ( \frac{bu_h^{\prime}}{v_h^{\prime}}+k \beta_h^{\prime}\right) \right) \geq \frac{1}{4} \cdot \bigg \lVert \frac{bu_h^{\prime}}{v_h^{\prime}}+k \beta_h^{\prime} \bigg\rVert^2. \end{align} $$

Writing $n_{k,h}$ to be the closest integer to $bu_h^{\prime }/v_h^{\prime }+k \beta _h^{\prime }$ , it follows that

$$\begin{align*}\lvert n_{k,h} \rvert \leq \lvert u_h^{\prime} \rvert+K \lvert \beta_h^{\prime} \rvert+2 \sqrt{\varepsilon_0} \leq c_0 \end{align*}$$

for some constant $c_0>0$ . Hence, we deduce that

(8.18) $$ \begin{align} k \in \bigcup_{\lvert n_{k,h} \rvert \leq c_0} \bigg [ \frac{1}{\beta_h^{\prime}} \left ( n_{k,h}-\frac{bu_h^{\prime}}{v_h^{\prime}}-2\sqrt{\varepsilon_0} \right), \frac{1}{\beta_h^{\prime}}\left ( n_{k,h}-\frac{bu_h^{\prime}}{v_h^{\prime}}+2\sqrt{\varepsilon_0} \right) \bigg ]. \end{align} $$

To conclude the argument, first consider the range $1/(3Kv_h^{\prime }) \leq \lvert \beta _h^{\prime } \rvert \leq 36/(Kh)$ . By (8.18) and the Siegel–Walfisz theorem, we have

$$\begin{align*}\# \{ k \in \mathbb{H}_{p,q}(t) : k \equiv b \bmod v_h^{\prime}\} \leq 10^{6} \cdot c_0 \sqrt{\varepsilon_0} \cdot \pi(K). \end{align*}$$

By choosing $\varepsilon _0$ sufficiently small, this contradicts (8.16). On the other hand, if $\lvert \beta _h^{\prime } \rvert <1/(3Kv_h^{\prime })$ , then choosing $\varepsilon _0$ small enough such that $\varepsilon _0 < 10 \cdot 5700^{-2} \leq 10 v_h^{\prime -2}$ , we see that no k satisfies (8.17), and we again arrive at a contradiction.

We combine Steps 1 and 2 to conclude that (8.9) holds for all t in our range and all $p \in (X,2X]\setminus \mathcal {F}$ . From here on, we can follow the argument in Corollary 6.5 to arrive at our desired conclusion.

Proof The proof follows the arguments in Lemma 5.1.

Proof The proof precisely follows the arguments in Lemma 6.1. The only key difference is, we use the Siegel–Walfisz theorem instead of the prime number theorem.

Proof By Dirichlet’s theorem, for any $t \in \mathbb {R}$ and for any $V\geq 1$ , there exists a reduced fraction $u_1/v_1$ , where $1 \leq v_1 \leq V$ , and $(u_1,v_1)=1$ , such that $\lvert t-u_1/v_1 \rvert \leq 1/(v_1V)\leq 1/v_1^2$ . Let t be as in the statement of the lemma and $V=2v$ . Then $1\leq v_1\leq 2v$ and $\lvert t-u_1/v_1 \rvert \leq 1/(2vv_1) \leq 1/v_1^2$ . By (6.9), we have

(9.7) $$ \begin{align} \sum_{ k \leq K} \Lambda(k)e(kt) \ll (K v_1^{-\frac{1}{2}} +K^{\frac{4}{5}} +K^{\frac{1}{2}}v_1^{\frac{1}{2}})(\log K)^4. \end{align} $$

In the special case, when $u/v=u_1/v_1$ , we must have $u_1=u$ and $v_1=v$ . Therefore, (9.7) implies that

(9.8) $$ \begin{align} \sum_{ k \leq K} \Lambda(k)e(kt) \ll (K v^{-\frac{1}{2}} +K^{\frac{4}{5}} +K^{\frac{1}{2}}v^{\frac{1}{2}})(\log K)^4. \end{align} $$

Otherwise, assume that $u/v\neq u_1/v_1$ . Then we have

(9.9) $$ \begin{align} \frac{1}{vv_1}&\leq \left\lvert\frac{u_1}{v_1}-\frac{u}{v}\right\rvert\leq \left\lvert\frac{u_1}{v_1}-t\right\rvert+\left\lvert t-\frac{u}{v}\right\rvert\leq \frac{1}{2vv_1}+\frac{\gamma}{v^2}. \end{align} $$

This implies that $1/(2vv_1)\leq \gamma /v^2$ , which is equivalent to $v_1^{-\frac {1}{2}} \leq (2\gamma )^{\frac {1}{2}} v^{-\frac {1}{2}}.$ It follows that

(9.10) $$ \begin{align} \sum_{k \leq K} \Lambda(k)e(kt) \ll ( \gamma^{\frac{1}{2}}K v^{-\frac{1}{2}} +K^{\frac{4}{5}} +K^{\frac{1}{2}}v^{\frac{1}{2}})(\log K)^4. \end{align} $$

Putting together (9.8) and (9.10), we get our desired result.

Proof By partial summation, it suffices to study the sum

$$\begin{align*}\sum_{\substack{k \leq K \\ k \equiv b \bmod D}}\Lambda(k) e(kt). \end{align*}$$

By orthogonality of additive characters, we write

(9.12) $$ \begin{align} \sum_{\substack{k \leq K \\ k \equiv b \bmod D}}\Lambda(k) e(kt) = \frac{1}{D}\sum_{\ell=0}^{D-1} e\bigg (-\frac{b\ell}{D} \bigg) \sum_{k \leq K} \Lambda(k) e(k t_{\ell}), \end{align} $$

where $t_{\ell }= t+\ell /D$ . Let $K_1 = K(\log K)^{-10}.$ By Dirichlet’s theorem on Diophantine approximation, there exists a reduced fraction $u/v$ with $1 \leq v \leq K_1$ and $\lvert t-u/v \rvert \leq 1/(vK_1) \leq 1/v^{2}$ . We write

(9.13) $$ \begin{align} \frac{u}{v}+\frac{\ell}{D} = \frac{uD+\ell v}{vD} = \frac{u_{\ell}}{v_{\ell}}, \end{align} $$

where $(u_{\ell },v_{\ell })=1$ . Then $v_{\ell } \geq v/(v,D)$ . Furthermore, we have $\lvert t_{\ell } -u_{\ell }/v_{\ell } \rvert \leq 1/v^2 \leq D^2/v_{\ell }^2$ . Therefore, if $v \geq (\log K)^{10}$ , we apply Lemma 9.3 with $\gamma = D^2$ to obtain

$$\begin{align*}\sum_{k \leq K} \Lambda(k) e(k t_{\ell})\ll D^{\frac{3}{2}} K (\log K)^{-1}, \end{align*}$$

for all $0 \leq \ell \leq D-1$ . Substituting this in (9.12) and using partial summation, we arrive at (9.11). So in what follows, we assume $1 \leq v < (\log K)^{10}$ .

Let $\beta = t-u/v$ . We write

(9.14) $$ \begin{align} \sum_{k \leq K} \Lambda(k) e(k t_{\ell}) & =\sum_{\substack{ k \leq K \\ (k, v_{\ell})=1}} \Lambda(k) e\left(\frac{k u_{\ell}}{v_{\ell}}\right) e(k \beta)+\sum_{\substack{k \leq K \\ (k, v_{\ell})>1}} \Lambda(k) e\left(\frac{k u_{\ell}}{v_{\ell}}\right) e(k \beta). \end{align} $$

We detect the exponential phase in (9.14) using the following identity:

$$ \begin{align*}\frac{1}{\varphi(v_{\ell})} \sum_{\chi \bmod v_{\ell}} \tau(\overline{\chi}) \chi(h)= \begin{cases}e(h / v_{\ell}), & \text { if }(h, v_{\ell})=1, \\ 0, & \text { if }(h, v_{\ell})>1 ,\end{cases} \end{align*} $$

where $\tau (\chi )$ is the Gauss sum. This yields

$$ \begin{align*} \sum_{k \leq K} \Lambda(k) e(k t_{\ell}) & =\frac{1}{\varphi(v_{\ell})} \sum_{\chi \bmod v_{\ell}} \tau(\overline{\chi}) \chi(u_{\ell}) \sum_{\substack{ k \leq K}} \Lambda(k) \chi(k) e(k \beta)\\&\quad +O\bigg(\sum_{\substack{w \mid v_{\ell} \\ w \ \mathrm{ prime}}} \sum_{r \leqslant \log _w (K)} \Lambda\left(w^r\right)\bigg). \end{align*} $$

Following [Reference Davenport12, Chapter 26], by partial summation and the Siegel–Walfisz theorem, we see that only the principal character contributes to the main term. In particular, we deduce that for any $B>0$ ,

$$\begin{align*}\sum_{k \leq K} \Lambda(k) e(k t_{\ell}) = \frac{\mu(v_{\ell})}{\varphi(v_{\ell})} \cdot T(\beta)+O_B(K (\log K)^{-B}) \end{align*}$$

where $T(\beta ) = \sum _{k \leq K} e(k \beta )$ . Substituting this in (9.12), we arrive at

(9.15) $$ \begin{align} \sum_{\substack{k \leq K \\ k \equiv b \bmod D}}\Lambda(k) e(kt) = \frac{1}{D}\sum_{\ell=0}^{D-1} e\bigg (-\frac{b\ell}{D} \bigg) \frac{\mu(v_{\ell})}{\varphi(v_{\ell})} \cdot T(\beta) + O_B(K (\log K)^{-B}). \end{align} $$

We now consider the following two cases.

Case 1: $ (v,D)=1$ . In this case, $v_{\ell } = v$ if $\ell =0$ and $vD$ otherwise. This implies that

$$\begin{align*}\frac{1}{D}\sum_{\ell=0}^{D-1} e\bigg (-\frac{b\ell}{D} \bigg) \frac{\mu(v_{\ell})}{\varphi(v_{\ell})} = \frac{\mu(v)}{\varphi(v)} \cdot \frac{1}{\varphi(D)}. \end{align*}$$

Case 2: $ (v,D)=D$ . If $D^2$ divides v, then $\mu (v_{\ell })=0$ for all $0 \leq \ell \leq D-1$ , and (9.11) follows trivially. So we may assume that D divides v but $D^2$ doesn’t. In this case, $v_{\ell } = v/D$ if $\ell \equiv -u \bmod D$ and v otherwise. This implies that

$$ \begin{align*} \frac{1}{D}\sum_{\ell=0}^{D-1} e\bigg (-\frac{b\ell}{D} \bigg) \frac{\mu(v_{\ell})}{\varphi(v_{\ell})}= \frac{\mu(v/D)}{\varphi(v/D)} \cdot \frac{1}{\varphi(D)} \cdot e\bigg (\frac{bu}{D} \bigg). \end{align*} $$

We now combine the two cases. If $v \geq 3$ in Case 1 or $v/D \geq 3$ in Case 2, then trivially bounding $T(\beta )$ in (9.15) and using partial summation, we obtain (9.11). Otherwise, if $\lvert \beta \rvert>\sqrt {10}/(2\pi K)$ , we use the more efficient bound $\lvert T(\beta ) \rvert \leq \pi K/\sqrt {10}.$ Finally, we are left with the situation when $v=1,2$ in Case 1 or $v=D,2D$ in Case 2. These can be handled following the arguments in Lemma 6.2.

Proof We first consider the second moment. Expanding the square and using Lemma 4.1, we have

(10.4) $$ \begin{align} \sum_{p \in (X, 2X]}S_{p,\mu}(K)^{2} & = \sum_{k_1 \leq K} \sum_{k_2 \leq K} \mu(k_1)\mu(k_2) \sum_{p \in (X,2X]} \left ( \frac{k_1k_2}{p} \right) \notag \\ &= (\pi(2X)-\pi(X)) \cdot \underset{ \substack{k_1 k_{2}= \square}}{\sum_{k_1 \leq K} \sum_{k_2 \leq K }} \mu(k_1) \mu(k_2) +O_{A,B}\left( X(\log X)^{-B} \right). \end{align} $$

Observe that when $k_1,k_2$ are square-free and their product $k_1k_2$ is a perfect square, we must have $k_1=k_2$ . The desired conclusion follows immediately.

We now focus on the general r-th moment for $r \geq 3$ . Here, we will use a result of Harper–Nikeghbali–Radziwiłł [Reference Harper, Nikeghbali and Radziwiłł23] on the moments of a Rademacher random multiplicative function. Expanding the r-th power and using Lemma 4.1, we arrive at

(10.5) $$ \begin{equation} \begin{aligned}\sum_{p \in (X, 2X]}S_{p,\mu}(K)^{r} & = \sum_{k_1 \leq K} \dots \sum_{k_r \leq K} \prod_{i=1}^{r}\mu(k_i) \sum_{p \in (X,2X]} \left ( \frac{k_1\ldots k_r}{p} \right) \notag \\&= (\pi(2X)-\pi(X)) \cdot \underset{ \substack{k_1 \ldots k_r= \square}}{\sideset{}{^\mathfrak{b}}\sum_{k_1 \leq K} \dots \sideset{}{^\mathfrak{b}} \sum_{k_i \leq K}} 1 +O_{A,B}\left( X(\log X)^{-B} \right).\end{aligned}\end{equation} $$

The asymptotics for the sum over $k_i$ ’s were established in the proof of [Reference Harper, Nikeghbali and Radziwiłł23, Theorem 1.4]. It is shown that for some constant $C_r>0$ depending on r, we have

(10.6) $$ \begin{align} \underset{ \substack{k_1 \ldots k_r= \square}}{\sideset{}{^\mathfrak{b}}\sum_{k_1 \leq K} \dots \sideset{}{^\mathfrak{b}} \sum_{k_i \leq K}} 1 =(1+o(1)) C_{r} K^{\frac{r}{2}} (\log K)^{\binom{r}{2}-r}. \end{align} $$

Substituting (10.6) in (10.5), we get the desired result.

Proof The proof follows the arguments in Lemma 5.1.

Proof See [Reference Hooley24, Theorem 3].

Proof We have

(10.9) $$ \begin{align} \sum_{k \leq K} \mu^2(k) e(kt) = \sum_{d \mid v} \sum_{\substack{ k \leq K \\ (k,v)=d}} \mu^2(k) e\left (\frac{ku}{v} \right) e(k\beta). \end{align} $$

We write $k = \widetilde {k} d$ and $v=\widetilde {v} d$ which implies that $(\widetilde {k}, \widetilde {v})=1$ . Therefore, the right-hand side of (10.9) is

(10.10) $$ \begin{align} &\sum_{d \mid v} \mu^2(d)\sum_{\substack{ \widetilde{k} \leq K/d \\ (\widetilde{k}, \widetilde{v})=1, (\widetilde{k}, d)=1 }} \mu^2(\widetilde{k}) e\bigg (\frac{\widetilde{k}u}{\widetilde{v}} \bigg) e(\widetilde{k}d \beta) \nonumber\\&\quad = \sum_{d \mid v} \mu^2(d) \sum_{(b,[\widetilde{v},d])=1} e\bigg (\frac{bu}{\widetilde{v}} \bigg) \sum_{\substack{ \widetilde{k} \leq K/d \\ \widetilde{k} \equiv b \bmod [\widetilde{v},d] }} \mu^2(\widetilde{k}) e(\widetilde{k}d \beta). \end{align} $$

Set

$$\begin{align*}C_{d} = \frac{6}{\pi^2} \cdot \frac{1}{[\widetilde{v},d]} \prod_{\substack{w \mid [\widetilde{v},d] \\ w \ \mathrm{ prime}}} \left (1 - \frac{1}{w^2} \right)^{-1}. \end{align*}$$

By partial summation and Lemma 10.3, we have

(10.11) $$ \begin{align} \sum_{\substack{ \widetilde{k} \leq K/d \\ \widetilde{k} \equiv b \bmod [\widetilde{v},d] }} \mu^2(\widetilde{k}) e(\widetilde{k}d \beta) &= C_{d} \cdot \bigg \lfloor \frac{K}{d} \bigg\rfloor \cdot e\bigg ( \bigg \lfloor \frac{K}{d} \bigg\rfloor \cdot d\beta \bigg)- 2\pi i d \beta \int_{1}^{ \lfloor K/d \rfloor} C_{d} \lfloor u \rfloor e(u d \beta) \, du \notag \\ &\quad +O_{\varepsilon} \bigg ( (1+K \lvert \beta \rvert) \bigg( \bigg (\frac{K}{d [\widetilde{v},d]} \bigg)^{\frac{1}{2}}+[\widetilde{v},d]^{\frac{1}{2}+\varepsilon} \bigg) \bigg). \end{align} $$

The contribution from the error term in (10.11) to (10.10) is $\ll _{\varepsilon } K^{\frac {1}{2}+\frac {\theta _1}{2}+\theta _2+\varepsilon }+ K^{\frac {3\theta _1}{2}+\theta _2+\varepsilon }.$ On the other hand, by partial summation, we obtain

(10.12) $$ \begin{align} \sum_{\widetilde{k} \leq K/d} e(\widetilde{k} d \beta) = \bigg \lfloor \frac{K}{d} \bigg\rfloor \cdot e\bigg ( \bigg \lfloor \frac{K}{d} \bigg\rfloor \cdot d\beta \bigg)- 2\pi i d \beta \int_{1}^{ \lfloor K/d \rfloor} \lfloor u \rfloor e(u d \beta) \, du. \end{align} $$

Therefore, putting together (10.11) and (10.12), we find that (10.10) is equal to

(10.13) $$ \begin{align} \sum_{d \mid v} \mu^2(d) \cdot C_{d} \cdot \sum_{\widetilde{k} \leq K/d} e(\widetilde{k} d \beta) \sum_{(b,[\widetilde{v},d])=1} e\bigg (\frac{bu}{\widetilde{v}} \bigg)+O_{\varepsilon} (K^{\frac{1}{2}+\frac{\theta_1}{2}+\theta_2+\varepsilon}+ K^{\frac{3\theta_1}{2}+\theta_2+\varepsilon}). \end{align} $$

The sum over b is a Ramanujan sum which is $\leq d/(\widetilde {v},d).$ As in the proof of Lemma 6.2, we have

(10.14) $$ \begin{align} \lvert \sum_{\widetilde{k} \leq K/d} e(\widetilde{k} d \beta) \rvert \leq \min \bigg \{\frac{K}{d}, \frac{1}{2 \lVert d \beta \rVert } \bigg \} \leq \frac{1}{d} \cdot \min \{K, \frac{1}{2} \lvert\beta \rvert^{-1} \}. \end{align} $$

Finally, we note that $C_{d} \leq [\widetilde {v},d]^{-1}$ . Putting these together, we see that

(10.15) $$ \begin{align} \bigg \lvert \sum_{d \mid v} \mu^2(d) \cdot C_{d} \cdot \sum_{\widetilde{k} \leq K/d} e(\widetilde{k} d \beta) \sum_{(b,[\widetilde{v},d])=1} e\bigg (\frac{bu}{\widetilde{v}} \bigg) \bigg \rvert &\leq \min \{K, \frac{1}{2} \lvert\beta \rvert^{-1} \} \cdot \frac{1}{v}\sum_{d \mid v} 1 \notag \\ &= \min \{K, \frac{1}{2} \lvert\beta \rvert^{-1} \} \cdot \frac{\tau(v)}{v}. \end{align} $$

We now combine (10.13) and (10.15) to arrive at our desired conclusion.

Proof Let $\varepsilon>0$ be given. For any $\varepsilon '>0$ , using Lemma 10.4 and the conditions on $\theta _1,\theta _2$ , we have for $K \geq 1$ sufficiently large,

$$\begin{align*}C_{\varepsilon'} ( K^{\frac{1}{2}+\frac{\theta_1}{2}+\theta_2+\varepsilon'}+ K^{\frac{3\theta_1}{2}+\theta_2+\varepsilon'}) \leq \varepsilon K \cdot \frac{\tau(v)}{v}. \end{align*}$$

Choosing $\varepsilon '$ sufficiently small depending on $\theta _1,\theta _2$ , and $\varepsilon $ , the proof is complete.

Proof The proof follows the arguments in Lemma 8.3. The only difference is that we use Lemmas 10.4 and 10.5 to obtain the bounds corresponding to Step 1 in 8.3.