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This article is cited in 1 scientific paper (total in 1 paper) New estimates for short Kloosterman sums with weights N. K. Semenova Lomonosov Moscow State University, Faculty of Mechanics and Mathematics
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    § 1. IntroductionIn 1926, Kloosterman [1] considered trigonometric sums of the form
$$
\begin{equation*}
S(m;a, b)=\sum_{\substack{\nu=1\\ (\nu, m)=1}}^{m}{e_m(a\overline{\nu}+b\nu)}, \qquad e_m(\nu)=e^{2\pi i \nu/m},
\end{equation*}
\notag
$$
where $m$, $a$, and $b$ are integers and $(a, m)\,{=}\,(\nu, m)\,{=}\,1$. We denote by $\overline{\nu}$ the residue inverse to $\nu$ modulo $m$, $\nu \overline{\nu} \equiv 1 \pmod{m}$. These sums were later called Kloosterman sums. By an incomplete Kloosterman sum we mean a sum of the form
$$
\begin{equation}
\sum_{\substack{\nu \leqslant x\\ (\nu, m)=1}}{e_m(a\overline{\nu}+b\nu)}, \qquad 1<x<m.
\end{equation}
\tag{1.1}
$$
When $x \geqslant m^{1/2+\varepsilon}$, a non-trivial estimate for sums of this kind follows from classical results of Weil (see [2] and [3]). For estimates of “short” sums satisfying $x \leqslant \sqrt{m}$, Karatsuba proposed a fundamentally new method in the early 1990s (see [4]–[6]), which was further developed in papers of Bourgain and Garaev [7] and Korolev (see [8]–[11]). For a sufficiently detailed survey of studies on this topic, see [12]. Along with the sums (1.1), so-called incomplete Kloosterman sums with weights are considered, that is, sums of the form
$$
\begin{equation}
S(x)=S(m,x;f,a,b)=\sum_{\substack{\nu \leqslant x\\ (\nu, m)=1}}{f(\nu)e_m(a\overline{\nu}+ b\nu)},
\end{equation}
\tag{1.2}
$$
where $f(\nu)$ is an arithmetic function. As in the case of the sums (1.1), a bound for the sums (1.2) when $x \geqslant m^{1/2+\varepsilon}$ can be derived from Weil’s bound (see [2]). The case of a short sum ($x \leqslant \sqrt{m}$) was first considered in [9]. In the present paper, using the ideas and techniques in [5]–[9] and [13], we refine bounds obtained in [9] for the sums (1.2). § 2. Statement of main resultsLet us introduce some notation needed below.
The main result for “homogeneous” sums (that is, those that satisfy the condition $b\equiv0 \pmod{m}$) is the following theorem. Theorem 1. Let $m\,{\geqslant}\, m_0$ be a sufficiently large prime, $(a,m)\,{=}\,1$, and let The following theorem holds in the case of inhomogeneous short sums with weights ($b\not\equiv0 \pmod{m}$). Theorem 2. Let $m\,{\geqslant}\, m_1$ be a sufficiently large prime, $(a,m)\,{=}\,1$, and let Remark 1. The bound obtained in [9] for the sum (1.2) has a decreasing factor of the form Remark 2. In the special case $x=m^\varepsilon$, $f(\nu)=\tau_r(\nu)$, where $0<\varepsilon \leqslant 0.5$, $r \geqslant 2$, are fixed numbers, the bounds in Theorems 1 and 2 take the form Using Theorems 1 and 2, we can show that the following assertions hold. Theorem 3. Let $m\,{\geqslant}\, m_0$ be a sufficiently large prime, $(a,m)\,{=}\,1$, and let Theorem 4. Let $m\,{\geqslant}\, m_0$ be a sufficiently large prime, $(a,m)\,{=}\,1$, and let Remark 3. The following asymptotic formulae hold for the sums in the right-hand sides of the equations of Theorems 3 and 4: § 3. Homogeneous short Kloosterman sums with weights3.1. Auxiliary assertionsWe need several auxiliary assertions. Lemma 1. Let $a_n, b_n \geqslant 0$, $n=1,\dots, N$, and let $k$ be an integer, $k \geqslant 2$. Then the following inequalities hold: Lemma 2. Let $N<m$ and let $J_{2k}(N)$ be the number of solutions of the congruence in primes $p_1, \dots, p_{2k}$ such that $p_{1}, \dots , p_{2k} \leqslant N $. Then This is Theorem 6 in [7]. Corollary 1. Let $m$ be a prime, $k$ an integer, $2 \leqslant k<X<X_1 \leqslant 2X$, and $I_k(X)$ the number of solutions of the congruence in primes $p_1, \dots, p_{2k}$ such that $X<p_1, \dots, p_{2k} \leqslant X_{1}$. Then Lemma 3. Let $m$ be a prime, and $k, s \geqslant 2$ integers with $ k<P<P_1 \leqslant 2P$ and $s<Q<Q_1 \leqslant 2Q$. Further, let $\xi(p)$ and $\eta(q)$ be arbitrary arithmetic functions defined on the primes in the intervals $P<p \leqslant P_1$ and $Q<q \leqslant Q_1$, respectively, where Proof. We follow the reasoning in [13]. First of all, we have
$$
\begin{equation*}
|W_1|=\biggl| \sum_q \eta(q) \sum_{p} \xi(p) e_{m}(a\overline{p}\, \overline{q})\biggr| \leqslant \sum_q |\eta(q)| \biggl| \sum_p \xi(p) e_{m}(a\overline{p}\, \overline{q}) \biggr|,
\end{equation*}
\notag
$$
where $\sum_p$ and $\sum_q$ stand for sums over the primes in the intervals $P<p \leqslant P_1$ and $Q<q \leqslant Q_1$, respectively. Taking the modulus of $W_1$ to a power $k$ and applying the second inequality in Lemma 1, we obtain
$$
\begin{equation*}
\begin{aligned} \, |W_1|^k &\leqslant \biggl(\sum_q|\eta(q)| \biggr)^{k-1}\sum_q |\eta(q)| \biggl| \sum_p \xi(p) e_{m}(a\overline{p}\, \overline{q}) \biggr|^k \\ &\leqslant \eta_{0}^{k-1}\eta \sum_q \biggl| \sum_p \xi(p) e_{m}(a\overline{p}\, \overline{q}) \biggr|^k \\ &= \eta_{0}^{k-1}\eta \sum_q \biggl| \sum_{p_1, \dots, p_k} \xi(p_1)\cdots \xi(p_k) e_{m}(a \overline{q}(\overline{p}_1+\dots+ \overline{p}_k)) \biggr| \\ &= \eta_{0}^{k-1}\eta \sum_q \Biggl| \sum_{\lambda=0}^{m-1} e_{m}(a \overline{q} \lambda) \sum_{\substack{p_1,\dots, p_k\\\overline{p}_1+\dots+ \overline{p}_k \equiv \lambda \ (\operatorname{mod} m)}} \xi(p_1)\cdots\xi(p_k) \Biggr|. \end{aligned}
\end{equation*}
\notag
$$
Writing
$$
\begin{equation*}
A_k(\lambda)=\sum_{\substack{p_1,\dots, p_k\\\overline{p}_1+\dots+ \overline{p}_k \equiv \lambda \ (\operatorname{mod} m)}} \xi(p_1)\cdots\xi(p_k),
\end{equation*}
\notag
$$
we have
$$
\begin{equation*}
|W_1|^k \leqslant \eta_{0}^{k-1}\eta \sum_q \biggl| \sum_{\lambda=0}^{m-1} e_{m}(a \overline{q} \lambda) A_k(\lambda) \biggr|.
\end{equation*}
\notag
$$
Let $\theta(q)$ be the argument of the inner sum. Then
$$
\begin{equation*}
|W_1|^k \leqslant \eta_{0}^{k-1}\eta \sum_q e^{-i \theta(q)}\!\sum_{\lambda=0}^{m-1} A_k(\lambda) e_{m}(a \overline{q} \lambda) =\eta_{0}^{k-1}\eta \!\sum_{\lambda= 0}^{m-1}A_k(\lambda) \sum_q e^{-i \theta(q)} e_m(a\overline{q}\lambda),
\end{equation*}
\notag
$$
and hence
$$
\begin{equation*}
|W_1|^k \leqslant \eta_{0}^{k-1}\eta \sum_{\lambda=0}^{m-1} |A_k(\lambda)| \biggl| \sum_q e^{-i \theta(q)} e_m(a\overline{q}\lambda) \biggr|.
\end{equation*}
\notag
$$
Raising this inequality to the power $s$ and applying Lemma 1 again, we obtain
$$
\begin{equation*}
|W_1|^{ks} \leqslant \eta_{0}^{s(k-1)}\eta^{s} \biggl( \sum_{\lambda=0}^{m-1} |A_k(\lambda)| \biggr)^{s-1} \sum_{\lambda=0}^{m-1} |A_k(\lambda)| \biggl| \sum_q e^{-i \theta(q)} e_m(a\overline{q}\lambda) \biggr|^s.
\end{equation*}
\notag
$$
We estimate the first sum:
$$
\begin{equation*}
\begin{aligned} \, \sum_{\lambda=0}^{m-1} |A_k(\lambda)| &\leqslant \sum_{\lambda=0}^{m-1} \sum_{\substack{p_1, \dots, p_k\\\overline{p}_1+\dots+ \overline{p}_k \equiv \lambda \ (\operatorname{mod} m) }} |\xi(p_1)| \cdots |\xi(p_k)| \\ &= \sum_{p_1, \dots, p_k} |\xi(p_1)| \cdots |\xi(p_k)|=\biggl( \sum_p |\xi(p)| \biggr)^k= \xi_0^k. \end{aligned}
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
|W_1|^{ks} \leqslant \eta_{0}^{s(k-1)}\eta^{s} \xi_0^{k(s-1)} \sum_{\lambda=0}^{m-1} |A_k(\lambda)| \biggl| \sum_q e^{-i \theta(q)} e_m(a\overline{q}\lambda) \biggr|^s.
\end{equation*}
\notag
$$
Squaring this inequality and using Lemma 1, we see that
$$
\begin{equation*}
|W_1|^{2ks} \leqslant \eta_{0}^{2s(k-1)} \eta^{2s} \xi_0^{2k(s-1)} BC,
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
B=\sum_{\lambda=0}^{m-1} |A_k(\lambda)|^2, \qquad C=\sum_{\lambda=0}^{m-1} \biggl| \sum_q e^{-i \theta(q)} e_m(a\overline{q}\lambda) \biggr|^{2s}.
\end{equation*}
\notag
$$
Estimating the sum $B$, we have
$$
\begin{equation*}
\begin{aligned} \, B &=\sum_{\lambda=0}^{m-1} |A_k(\lambda)|^2=\sum_{\lambda=0}^{m-1} \Biggl| \sum_{\substack{p_1, \dots, p_k\\\overline{p}_1+ \dots+\overline{p}_k \equiv \lambda \ (\operatorname{mod} m)}} \xi(p_1) \cdots \xi(p_k) \Biggr|^2 \\ &\leqslant \sum_{\lambda=0}^{m-1} \sum_{\substack{p_1,\dots, p_{2k}\\ \overline{p}_1+\dots+ \overline{p}_k \equiv \lambda \equiv \overline{p}_{k+1}+\dots+ \overline{p}_{2k} \ (\operatorname{mod}m)}} |\xi(p_1)| \cdots |\xi(p_{2k})| \\ &\leqslant \xi^{2k} \sum_{\lambda=0}^{m-1} \sum_{\substack{p_1, \dots, p_{2k}\\ \overline{p}_1+\dots+ \overline{p}_k \equiv \lambda \equiv \overline{p}_{k+1}+\dots+ \overline{p}_{2k} \ (\operatorname{mod}m)}} 1 \\ &=\xi^{2k} \sum_{\substack{p_1, \dots, p_{2k}\\ \overline{p}_1+\dots+ \overline{p}_k \equiv\overline{p}_{k+1}+\dots+ \overline{p}_{2k} \ (\operatorname{mod}m)}} 1=\xi^{2k} I_k(P). \end{aligned}
\end{equation*}
\notag
$$
Estimating the sum $C$, we have
$$
\begin{equation*}
\begin{aligned} \, C&=\sum_{\lambda=0}^{m-1} \sum_{Q<q_1, \dots, q_{2s} \leqslant Q_1} e^{-i (\theta(q_1)+ \dots -\theta(q_{2s}))} e_m(a\lambda (\overline{q}_1 +\dots -\overline{q}_{2s})) \\ &=\sum_{Q<q_1, \dots, q_{2s} \leqslant Q_1} e^{-i (\theta(q_1)+\dots -\theta(q_{2s}))} \sum_{\lambda=0}^{m-1}e_m(a\lambda (\overline{q}_1 +\dots -\overline{q}_{2s})), \\ |C| &\leqslant \sum_{\substack{Q<q_1, \dots, q_{2s} \leqslant Q_1 \\ \overline{q}_1+\dots+ \overline{q}_s \equiv \overline{q}_{s+1}+\dots+ \overline{q}_{2s} \ (\operatorname{mod}m)}} m=m I_s(Q). \end{aligned}
\end{equation*}
\notag
$$
Thus, the following inequality holds:
$$
\begin{equation*}
|W_1|^{2ks} \leqslant \eta_0^{2s(k-1)}\eta^{2s}\xi_0^{2k(s-1)} \xi^{2k} m I_k(P) I_s(Q).
\end{equation*}
\notag
$$
Using Lemma 2, we obtain
$$
\begin{equation*}
\begin{aligned} \, &|W_1|^{2ks} {\leqslant}\, (\eta_0 \xi_0)^{2ks}(\eta_0^{-1} \eta)^{2s} (\xi_0^{-1}\xi)^{2k} m (16k)^k (16s)^s P^k Q^s \biggl( 1{\kern1pt}{+}{\kern1pt}\frac{P^{2k-1}}{m} \biggr) \biggl( 1{\kern1pt}{+}{\kern1pt}\frac{Q^{2s-1}}{m} \biggr) \\ &\quad = (\eta_0 \xi_0)^{2ks}(4 \eta_0^{-1} \eta \sqrt{s})^{2s} (4\xi_0^{-1}\xi \sqrt{k})^{2k} P^{2k} Q^{2s} \biggl( \frac{\sqrt{m}}{P^k}+\frac{P^{k-1}}{\sqrt{m}} \biggr) \biggl( \frac{\sqrt{m}}{Q^s}+\frac{Q^{s-1}}{\sqrt{m}} \biggr). \end{aligned}
\end{equation*}
\notag
$$
Hence where
$$
\begin{equation*}
\Delta\,{=}\,(4\xi \sqrt{k})^{1/s} (4\eta \sqrt{s})^{1/k} \biggl( \frac{P}{\xi_0}\biggr)^{1/s} \!\biggl( \frac{Q}{\eta_0}\biggr)^{1/k} \! \biggl( \biggl( \frac{\sqrt{m}}{P^k}+\frac{P^{k-1}}{\sqrt{m}} \biggr) \biggl( \frac{\sqrt{m}}{Q^s}+\frac{Q^{s-1}}{\sqrt{m}} \biggr) \biggr)^{1/(2ks)}\! .
\end{equation*}
\notag
$$
This completes the proof of the lemma. Corollary 2. In the case when $\xi(\nu)=\eta(\nu)=f(\nu)$, the sum $W_1$ satisfies the bound $|W_1| \ll PQ \Delta_1$, where Proof. We write the bound in Lemma 3 in the form
$$
\begin{equation*}
\begin{aligned} \, |W_1| &\leqslant \xi_0^{1-1/s} (\xi P)^{1/s} \eta_0^{1-1/k} (\eta Q)^{1/k} (4\sqrt{s})^{1/k} (4\sqrt{k})^{1/s} \\ &\qquad\times \biggl( \biggl( \frac{\sqrt{m}}{P^k}+\frac{P^{k-1}}{\sqrt{m}} \biggr) \biggl( \frac{\sqrt{m}}{Q^s}+\frac{Q^{s-1}}{\sqrt{m}} \biggr) \biggr)^{1/(2ks)}. \end{aligned}
\end{equation*}
\notag
$$
Consider the case $\xi(\nu)=\eta(\nu)=\tau_r(\nu)$. Let a constant $x_0$ be chosen in such a way that the inequality holds for all $x \geqslant x_0$. Then, noting that $\xi=\eta=r$ and
$$
\begin{equation*}
\xi_0=r (\pi (P_1)-\pi(P)) \leqslant r \pi_1(P), \qquad \eta_0 \leqslant r \pi_1(Q),
\end{equation*}
\notag
$$
we have, provided that $P, Q \geqslant \max(x_0, e^{2r^2})$,
$$
\begin{equation*}
\begin{aligned} \, \xi_0 (\xi_0^{-1} \xi )^{1/s} &=\xi_0^{1-1/s} \xi^{1/s} \leqslant \bigl(r\pi_1(P)\bigr)^{1-1/s} r^{1/s}= r \bigl( \pi_1(P)\bigr)^{1-1/s} \\ &\leqslant r \biggl( \frac{2P}{\ln{P}} \biggr)^{1-1/s}=r \biggl( \frac{2}{\ln{P}} \biggr)^{1-1/s} P^{1-1/s} \leqslant r \sqrt{\frac{2}{\ln P}}\, P^{1-1/s} \leqslant P^{1-1/s}. \end{aligned}
\end{equation*}
\notag
$$
Similarly, $\eta_0 (\eta_0^{-1} \eta )^{{1}/{k}} \leqslant Q^{1-{1}/{k}}$. Therefore, the final bound for $W_1$ is as follows:
$$
\begin{equation*}
\begin{aligned} \, |W_1| &\leqslant PQ \cdot 4^{1/{s}+{1}/{k}} k^{{1}/({2s})} s^{1/({2k})} \biggl( \biggl( \frac{\sqrt{m}}{P^k}+\frac{P^{k-1}}{\sqrt{m}} \biggr) \biggl( \frac{\sqrt{m}}{Q^s}+\frac {Q^{s-1}}{\sqrt{m}} \biggr) \biggr)^{{1}/({2ks})} \\ &\leqslant PQ \cdot 4 k^{1/({2s})} s^{1/({2k})} \biggl( \biggl( \frac{\sqrt{m}}{P^k}+\frac {P^{k-1}}{\sqrt{m}} \biggr) \biggl( \frac{\sqrt{m}}{Q^s}+\frac{Q^{s-1}}{\sqrt{m}} \biggr) \biggr)^{1/({2ks})}. \end{aligned}
\end{equation*}
\notag
$$
In the remaining cases, the validity of the desired equation is established similarly. This completes the proof of the corollary. We also need the following lemmas. This is Lemma 1 in [9]. Lemma 5. Let $x>x_0$ and put Then the following bound holds for $N(x)$: This is Lemma 2 in [9]. Lemma 6. Let $y_0<y<x$ and let $G(x, y)$ be the sum of the values of $f(\nu)$ over the square-free numbers $\nu\leqslant x$ all of whose prime divisors exceed $y$, that is, Then the following inequality holds: Proof. When $f(n)=\mu^2(n)$ and $f(\nu)=\mu(n)$, the desired inequality follows from the classical bound for the number of integers $n \leqslant x$ satisfying the condition $P^{-}(n)> y$ (see, for example, Ch. I, § 4.2, Theorem 2 of [14]). When $f(n)=\tau_r(n)$, the assertion follows from Lemma 6.1 in [15]. Finally, when $f(n)=b(n)$, the desired inequality follows from Theorem 2 in [16]. This completes the proof of the lemma. 3.2. Proof of Theorem 1The sum $S(x)$ is estimated as follows. The terms corresponding to numbers $\nu$, $1 \leqslant \nu \leqslant x$, having no prime divisors belonging to special intervals are estimated trivially. The remaining terms can be grouped into sums to which the bound in the corollary to Lemma 3 can be applied. Step 1. We write $H=\exp (2\sqrt{\ln x}) $. Let $S_1$ be the part of $S(x)$ over all $\nu$ of the form $\nu=c^2 d$, where $c>H$. By Lemma 4, $S_1 \ll F(x)e^{-\sqrt{\ln x}}$. We note that all the remaining $\nu$ have the form $\nu=c^{2} d$, where $c\leqslant H$, $d$ is square-free and $\mu(d)\neq 0$. Before passing to Step 2, we introduce the following parameters. Let $ R=\exp (\ln x/\ln \ln x)$ and define an integer $l$ by the condition $m^{1/(2l)} \leqslant R<m^{1/(2l-2)}$. Let $n$ satisfy the condition $3l \leqslant n \leqslant (1/4)\sqrt{\ln m}$ (the exact value of $n$ will be chosen below), and let $\delta=1/(4n)$. For an integer $k$, $l \leqslant k \leqslant n$, we write
$$
\begin{equation*}
X_{k}=m^{(1-\delta)/(2k) }, \qquad Y_{k}=m^{(1+\delta)/(2k) }.
\end{equation*}
\notag
$$
We note that $X_{k-1}>2Y_k$ for all $k$ under consideration. Indeed,
$$
\begin{equation*}
\begin{aligned} \, \frac{X_{k-1}}{Y_k} &=m^{(1-\delta)/(2k-2)-(1+\delta)/(2k)}= m^{(1-(2k-1)\delta)/(2k(k-1))} \\ &> m^{(1-2k\delta)/(2k^2)}\geqslant m^{(1-2n\delta)/(2n^2)}=m^{1/(4n^2)}, \end{aligned}
\end{equation*}
\notag
$$
and it suffices to prove that $2^{4n^2}<m$. However, $2^{4n^2} \leqslant 2^{\ln m/4}<m^{7/40}<m$ since $n \leqslant (1/4) \sqrt{\ln m}$. Step 2. We denote by $S_2$ the sum of the remaining terms $S(x)$ that correspond to numbers $\nu$ all of whose prime divisors do not exceed $m^{1/(2l)}$. By the choice of $l$ and according to Lemma 5, we have the bound $S_2 \ll F(x) \exp (-(1/5)(\ln \ln x ) \ln \ln \ln x)$. We note that all the remaining $\nu$ have at least one prime divisor $q$ such that $m^{1/(2s)}<q \leqslant m^{1/(2s-2)}$, where $s \leqslant l$, and denote by $J$ the set of all primes in the union of the intervals $m^{1/(2s)}<q \leqslant m^{1/(2s-2)}$, where $s \leqslant l$. Step 3. We denote by $I$ the set of primes in the union of the intervals $(Y_k, X_{k-1}]$, $l<k\leqslant n$, and denote by $S_3$ the sum of the remaining terms of $S(x)$ that correspond to numbers $\nu$ having no prime divisors in $I$. Let $M=m^{1/(2l)}$, $Y=Y_n$ and $X=X_l$, and denote by $K$ the set of primes in the union of the intervals $(1,Y]$, $(X_k, Y_k]$, where $ l<k< n$, $(X, M]$. Noting that $Y>H$, we can conclude that $\nu$ has the form $\nu=c^2 u v$, where $1\leqslant c \leqslant H$, $\mu(uv)\neq 0$, and all the prime divisors of $v$ belong to $J$ and either $u= 1$ or all the prime divisors of $u$ belong to $K$. We estimate $S_3$ for each of the functions $f(\nu)$ separately. 1) Case $f(\nu)=\mu(n)$ or $f(\nu)=\mu^2(\nu)$. Since $f(\nu) \neq 0 $ when only $c=1$, we have
$$
\begin{equation*}
\begin{aligned} \, |S_3| &\leqslant \sum_{c^2uv\leqslant x} \mu^2(c^2uv) \leqslant \sum_{u\leqslant x/M} \mu^2(u) \sum_{v \leqslant x/u} \mu^2(v) \\ &=\sum_{u\leqslant x / M}\sum_{M \leqslant v \leqslant x/u} 1= \sum_{u\leqslant x/M} G\biggl(\frac{x}{u}; M\biggr), \end{aligned}
\end{equation*}
\notag
$$
where $u$ and $v$ range over the sets indicated above. Thus, using Lemma 6, we obtain
$$
\begin{equation*}
|S_3| \ll \sum_{u \leqslant x/M} \frac{x}{u} \frac{1}{\ln M} \ll \frac{x}{\ln M} \sum_{u \leqslant x} \frac{1}{u} \ll \frac{x}{\ln M} \prod_{p \in K} \biggl(1+\frac{1}{p}\biggr) \ll \frac{x e^{\sigma_1}}{\ln M},
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\sigma_1=\sum_{p \in K} \frac{1}{p}=\sum_{p \leqslant Y} \frac{1}{p}+\sum_{l<k<n}\, \sum_{p \in (X_k, Y_k]} \frac{1}{p} +\sum_{X<p\leqslant M}\frac{1}{p}.
\end{equation*}
\notag
$$
Since
$$
\begin{equation*}
\begin{aligned} \, \sum_{X_k<p \leqslant Y_k} \frac{1}{p} &=\ln\ln Y_k-\ln\ln X_k +O\biggl(\frac{1}{\ln X_k}\biggr) \\ &=\ln \ln m^{(1+\delta)/(2k)}-\ln \ln m^{(1-\delta)/(2k)}+O\biggl( \frac{k}{\ln m}\biggr) \\ &=\ln (1+\delta)-\ln (1-\delta)+O \biggl( \frac{k}{\ln m}\biggr)=2 \delta+O (\delta^3)+O \biggl( \frac{k}{\ln m}\biggr) \\ &=\frac{1}{2n}+O\biggl( \frac{1}{n^3}\biggr)+O \biggl( \frac {k}{\ln m}\biggr), \end{aligned}
\end{equation*}
\notag
$$
it follows that
$$
\begin{equation*}
\sum_{k=l+1}^{n-1}\biggl( \sum_{X_k<p \leqslant Y_k} \frac{1}{p}\biggr) =\frac{n-l-1}{2n}+ O\biggl(\frac{1}{l^2}\biggr)+O \biggl(\frac{n^2}{\ln m}\Big )=O(1).
\end{equation*}
\notag
$$
Hence We finally obtain
$$
\begin{equation*}
|S_3| \ll \frac{x}{\ln M} \frac{\ln Y}{\ln X} \ln M \ll x \frac{\ln Y}{\ln X} \ll x \frac{(\ln m)/n}{(\ln m)/l} \ll x \frac{l}{n} \ll F(x) \biggl(\frac{l}{n}\biggr)^\alpha.
\end{equation*}
\notag
$$
2) Case $f(\nu)=\tau_r(\nu)$. Since $ \tau_r(gh) \leqslant \tau_r(g) \tau_r(h)$ for every $g$ and $h$, we have
$$
\begin{equation*}
\begin{aligned} \, |S_3| &\leqslant \sum_{c^2uv\leqslant x} \tau_r(c^2) \tau_r(u) \tau_r(v) \leqslant \sum_{c \leqslant H} \tau_r(c^2) \sum_{u \leqslant x(c^2M)^{-1}} \tau_r(u) \sum_{M \leqslant v \leqslant x(c^2u)^{-1}} \tau_r(v) \\ &=\sum_{c \leqslant H} \tau_r(c^2) \sum_{u \leqslant x(c^2M)^{-1}} \tau_r(u) G\biggl( \frac{x}{c^2 u}; M\biggr). \end{aligned}
\end{equation*}
\notag
$$
Using Lemma 6, we obtain
$$
\begin{equation*}
\begin{aligned} \, |S_3| &\ll \sum_{c \leqslant H} \tau_r(c^2) \sum_{u \leqslant x(c^2M)^{-1}} \tau_r(u) \frac{x}{c^2u}\frac{1}{(\ln M)^r}\biggl(\ln \frac{x}{c^2 u }\biggr)^{r-1} \\ &\ll \frac{x (\ln x)^{r-1}}{(\ln M)^r} \sum_{c \leqslant H} \frac{\tau_r(c^2)}{c^2} \sum_{u \leqslant x(c^2M)^{-1}} \frac{\tau_r(u)}{u} \ll \frac{x (\ln x)^{r-1}}{(\ln M)^r} \sum_{c =1}^{+\infty} \frac{\tau_r(c^2)}{c^2} \sum_{u \leqslant x} \frac{\tau_r(u)}{u} \\ &\ll \frac{x (\ln x)^{r-1}}{(\ln M)^r} \prod_{p \in K} \biggl( 1+\frac{r}{p}\biggr) \ll \frac{x (\ln x)^{r-1}}{(\ln M)^r} e^{\sigma_2}, \end{aligned}
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\sigma_2=r \sum_{p\in K} \frac{1}{p}=r(\ln\ln Y+\ln\ln M-\ln\ln X)+O(1).
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
|S_3| \ll \frac{x (\ln x)^{r-1}}{(\ln M)^r} \biggl( \frac{\ln Y}{\ln X}\biggr)^r (\ln M )^r \ll x (\ln x)^{r-1} \biggl(\frac{l}{n} \biggr)^r \ll F(x) \biggl(\frac{l}{n} \biggr)^\alpha.
\end{equation*}
\notag
$$
3) Case $f(\nu)=b(\nu)$. Since all the prime divisors of $v$ lie in $J$, it follows that $(v, c^2u)=1$ and $b(c^2uv)=b(c^2u)b(v)$. We claim that $b(c^2 u)=b(c^2)b(u)=b(u)$. If $(c, u)=1$, then this is obvious. Now let $(c,u)>1$. Since $u$ is square-free, it follows that $(c, u)=q_1\cdots q_k$, where the $q_i$ are distinct primes. Then $c=q_1^{\alpha_1}\cdots q_k^{\alpha_k}c_1$ and $u=q_1\cdots q_k u_1$, where $(c_1, u_1)=1$. We have
$$
\begin{equation*}
b(c^2 u)=b(q_1^{2\alpha_1+1}\cdots q_k^{2\alpha_k+1})b(c_1^2)b(u_1) =b(q_1\cdots q_k)b(u_1)=b(u).
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
|S_3| \leqslant \sum_{c \leqslant H }\sum_{u \leqslant x (c^2 M)^{-1}} b(u) \sum_ {M \leqslant v \leqslant(c^2 u)^{-1}} b(v) \leqslant \sum_{c \leqslant H } \sum_{u \leqslant x (c^2 M)^{-1}} b(u) G\biggl( \frac{x}{c^2 u}; M\biggr).
\end{equation*}
\notag
$$
Using Lemma 6, we obtain
$$
\begin{equation*}
\begin{aligned} \, |S_3| &\ll \sum_{c \leqslant H } \sum_{u \leqslant x (c^2 M)^{-1}} b(u) \frac{x}{c^2 u } \frac{1}{\sqrt{\ln(x/(c^2u))}} \frac{1}{\sqrt{\ln M}} \\ &\ll \frac{x}{\ln M} \sum_{c \leqslant H }\frac{1}{c^2} \sum_{u \leqslant x (c^2 M)^{-1}} \frac{b(u)}{u} \ll \frac{x}{\ln M} \sum_{u \leqslant x} \frac{b(u)}{u}. \end{aligned}
\end{equation*}
\notag
$$
Since $u$ is square-free, it follows that $b(u)=1$ if and only if all the prime divisors of $u$ are congruent to 1 modulo 4. Therefore,
$$
\begin{equation*}
\sum_{u \leqslant x} \frac{b(u)}{u} \leqslant \prod_{\substack{p \in K \\ p \equiv 1 \ (\operatorname{mod}4)}} \biggl( 1+\frac{1}{p}\biggr) \leqslant e^{\sigma_3},
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\begin{aligned} \, \sigma_3 &=\sum_{\substack{p \in K\\ p \equiv 1 \ (\operatorname{mod}4)}} \frac{1}{p} =\sum_{\substack{p \leqslant Y\\ p \equiv 1 \ (\operatorname{mod}4)}} \frac{1}{p}+\sum_{l<k<n} \sum_{\substack{p \in (X_k, Y_k]\\ p \equiv 1 \ (\operatorname{mod}4)}} \frac{1}{p} +\sum_{\substack{X<p\leqslant M \\ p \equiv 1 \ (\operatorname{mod}4)}}\frac {1}{p} \\ &=\frac{1}{2} (\ln\ln Y+\ln\ln M -\ln\ln X)+O(1). \end{aligned}
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
\begin{aligned} \, |S_3| &\ll \frac{x}{\ln M} \sqrt{\frac{\ln Y}{\ln X}} \sqrt{\ln M}\ll \frac{x}{\sqrt{\ln M}} \sqrt{\frac{\ln Y}{\ln X}} \ll \frac{x}{\sqrt{\ln x}} \sqrt{\ln \ln x} \sqrt {\frac{l}{n}} \\ &\ll F(x) \biggl(\frac{l}{n} \biggr)^ \alpha (\ln\ln m)^\beta. \end{aligned}
\end{equation*}
\notag
$$
Hence the following bound holds for all the functions $f(\nu)$ under consideration:
$$
\begin{equation*}
|S_3| \ll F(x) \biggl( \frac{l}{n}\biggr)^{\alpha} (\ln \ln m)^\beta.
\end{equation*}
\notag
$$
The sum of the remaining terms is denoted by $S_4$. All the $\nu$ corresponding to these terms have at least one prime divisor $p$ in $I$ and at least one prime divisor $q$ in $J$. Thus, all such $\nu$ have the form $\nu=c^2uvw$, where $1\leqslant c \leqslant H$, $\mu(uvw)\,{\neq}\,0$, all the prime divisors of $u$ belong to $I$, all the prime divisors of $v$ belong to $J$, and $w$, in contrast, has no prime divisors in $I$ and $J$. Let us group together the terms of $S_4$ for which $u$ and $v$ consist of $\mu$ and $\lambda$ factors, respectively:
$$
\begin{equation*}
S_4=\sum_{\mu\geqslant 1} \sum_{\lambda \geqslant 1} S_4(\mu, \lambda).
\end{equation*}
\notag
$$
We write $\Phi(n)=e_m(a \overline{n})$. Then, since $f(c^2 u v w)=f(c^2w)f(u)f(v)$, we obtain
$$
\begin{equation*}
\begin{aligned} \, |S_4(\mu, \lambda)| &=\biggl| \sum_{c^2uvw \leqslant x}f(c^2uvw) \Phi (c^2uvw) \biggr| \\ &\leqslant \sum_{c \leqslant H} f(c^2) \sum_{w}f(w) \biggl| \sum_{uv \leqslant x(c^2w)^{-1}}f(u)f(v) \Phi(c^2uvw)\biggr|, \end{aligned}
\end{equation*}
\notag
$$
where $w$ ranges over an increasing sequence of numbers that have no prime divisors in $I$ or $J$ and $w \leqslant x(c^2 Y^\mu M^\lambda )^{-1}$. Let $c$ and $w$ be fixed. We compare the inner sum $S_4(\mu, \lambda; w )$ over $u$ and $v$ with the sum
$$
\begin{equation*}
S' (\mu, \lambda;w)=\frac{1}{\mu\lambda} \sum_{u_1,v_1}\sum_{p,q}{f(u_1)f(p)f(v_1)f(q) \Phi(c^2u_{1}pv_{1}qw)},
\end{equation*}
\notag
$$
where $u_1$ and $v_1$ range independently over increasing sequences of numbers that are products of $\mu-1$ and $\lambda-1$ different prime factors in $I$ and $J$, respectively, while $p$ and $q$ take values in the primes in $I$ and $J$, and the condition $u_{1}v_{1}pq \leqslant x(c^2w)^{-1}$ is satisfied. The numbers $u$ and $v$ corresponding to the terms in $S_4(\mu, \lambda; w)$ are represented in the form $u= u_{1}p$ and $v=v_{1}q$, where $(u_{1},p)=1$ and $(v_{1},q)=1$, in $\mu$ and $\lambda$ ways, respectively. Therefore, any term in this sum occurs in $S'(\mu, \lambda; w)$ with coefficient 1. Moreover, $S'(\mu, \lambda; w)$ contains summands for which at least one of the conditions $(u_{1},p)=1$ and $(v_{1},q)=1$ is violated. Denoting their contribution by $S''(\mu, \lambda; w)$ and writing $u_{1}=u_{2}p$ and $v_1=v_{2}q$, we obtain
$$
\begin{equation*}
\begin{aligned} \, &| S'' (\mu, \lambda; w)| \leqslant \frac{1}{\mu\lambda} \biggl(\sum_{p^{2}qu_{2}v_{1} \leqslant x(c^2w)^{-1}}f(pu_{2})f(p)f(v_1)f(q) \\ &\quad +\sum_{pq^{2}u_{1}v_{2} \leqslant x(c^2w)^{-1}}\!\! f(u_1)f(p)f(v_{2}q)f(q) \,{+}\!\!\sum_{p^{2}q^{2}u_{2}v_{2} \leqslant x(c^2w)^{-1}} \!\! f(u_2p)f(p)f(v_{2}q)f(q) \biggr) \\ &\enspace = \frac{1}{\mu\lambda} (S_1'' +S_2''+S_3''). \end{aligned}
\end{equation*}
\notag
$$
Let us estimate the sum $S_1''$. Since $pu_{2}$ is square-free, it follows that
$$
\begin{equation*}
S_1''=\sum_{p^2qu_{2}v_{1} \leqslant x(c^2w)^{-1}} f(pu_2) f(p)f(v_1)f(q) =\sum_{p^2qu_{2}v_{1} \leqslant x(c^2w)^{-1}}f^2(p)f(v_1 u_2 q).
\end{equation*}
\notag
$$
We write $d=v_1 u_2 q$. The number of representations of $d$ in this form does not exceed $\lambda$. The following inequality holds:
$$
\begin{equation*}
S_1'' \leqslant \lambda \sum_{p \in I} f^2(p) \sum_{p^2d \leqslant x(c^2w)^{-1}} f(d) \ll \ln x \sum_{p \in I} f^2(p) F\biggl( \frac{x}{p^2c^2w} \biggr).
\end{equation*}
\notag
$$
Since $F(x/t) \ll (F(x)/t)\ln x$ when $1\leqslant t \leqslant x/2$, we obtain the bound
$$
\begin{equation*}
S_1'' \ll \frac{F(x)}{c^2w} ( \ln x)^2 \sum_{p \in I} {\frac{f^2(p)}{p^2}} \ll \frac{F(x)}{c^2w} ( \ln x)^2 \sum_{p \geqslant Y} {\frac{1}{p^2}} \ll \frac{F(x)}{c^2w} \frac{(\ln x)^2}{Y}.
\end{equation*}
\notag
$$
Similar bounds hold for $S_2''$ and $S_3''$. Thus,
$$
\begin{equation*}
\begin{gathered} \, |S''(\mu, \lambda; w)| \ll \frac{1}{\mu\lambda} \frac{F(x)}{c^2w} \frac{(\ln x)^2}{Y}, \\ | S_4(\mu, \lambda;w)| \ll |S'(\mu, \lambda; w)|+|S''(\mu, \lambda; w)| \ll |S'(\mu, \lambda; w)|+\frac{1}{\mu\lambda} \frac{F(x)}{c^2w} \frac{(\ln x)^2}{Y}, \end{gathered}
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
S'(\mu, \lambda; w)=\frac{1}{\mu\lambda}\sum_{u_1}\sum_{v_1} f(u_1)f(v_1) \widetilde{S},
\end{equation*}
\notag
$$
and for fixed $\mu, \lambda, w, u_1, v_1$ we have
$$
\begin{equation*}
\widetilde{S}=\sum_{pq \leqslant Z} {f(p)f(q)\Phi(c^2 u_1 v_1 w pq)}, \qquad Z=\biggl[ \frac{x}{c^2 u_1 v_1 w} \biggr].
\end{equation*}
\notag
$$
Let us represent $\widetilde{S}$ in the form
$$
\begin{equation*}
\widetilde{S}=\sum_{pq\leqslant Z} f(p)f(q)e_m(a_1\overline{p}\, \overline{q}),
\end{equation*}
\notag
$$
where $a_1 \equiv a(\overline{c^2 u_1 v_1}) \pmod m$. Further, we split the domains over which $p$ and $q$ vary into intervals of the form $P<p \leqslant P_1 \leqslant 2P$ and $Q<q \leqslant Q_1 \leqslant 2Q$ and choose the quantities $P$, $P_1$, $Q$, $Q_1$ in such a way that every interval $(P, P_1]$ is entirely contained in some interval $(Y_k, X_{k-1}]$, $l<k\leqslant n$, and every interval $(Q, Q_1]$ is entirely contained in some interval $(m^{1/(2s)}, m^{1/(2s-2)}]$, $s \leqslant l$. Thus, $\widetilde{S}$ splits into at most $(\ln m)^2$ sums of the form
$$
\begin{equation*}
S(P, Q)=\sum_{P <p\leqslant P_1} \sum_{\substack{Q< q\leqslant Q_1\\ pq \leqslant Z}} f(p)f(q)e_m(a_1 \overline{p}\, \overline{q}).
\end{equation*}
\notag
$$
We consider one of these sums. We will get rid of the condition $pq \leqslant Z$. Since $p \leqslant Z/Q$, it follows that $ P<p \leqslant P_2$, where $P_2=\min (P_1, Z/Q)$. Hence $Q< q\leqslant Q_2$, where $Q_2=\min (Q_1, Z/p)$. We transform the sum to make the upper limit of $q$ independent of $p$:
$$
\begin{equation*}
\begin{aligned} \, S(P, Q) &=\sum_{P <p\leqslant P_2} \sum_{Q< q\leqslant Q_{1}} \biggl( \frac{1}{m} \sum_{|d|<m/2}\, \sum_{Q< \xi \leqslant Q_2} e_m\bigl( d(q-\xi)\bigr) \biggr) f(p)f(q)e_m(a_1\overline{p}\, \overline{q}) \\ &=\sum_{|d|<m/2} {\frac{1}{|d|+1}} \sum_{P<p\leqslant P_2} \frac{|d|+1}{m} \biggl(\sum_{Q<\xi \leqslant Q_2}{e_m(-d\xi)}\biggr) \\ &\qquad\times\sum_{Q<q\leqslant Q_1} e_m(dq)f(p)f(q)e_m(a_1 \overline{p}\, \overline{q}) =\sum_{|d|<m/2}{\frac{T(d)}{|d|+1}}, \end{aligned}
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\begin{gathered} \, T(d)=\sum_{P<p\leqslant P_2} \sum_{Q<q\leqslant Q_1} {\alpha(p) \beta(q) e_m(a_1 \overline{p}\, \overline{q})}, \\ \alpha(p)=\frac{|d|+1}{m} \biggl(\sum_{Q<\xi \leqslant Q_2}{e_m(-d\xi)} \biggr) f(p), \qquad \beta(q)=e_m(dq) f(q). \end{gathered}
\end{equation*}
\notag
$$
Since $|\alpha(p)| \leqslant |f(p)|$, $|\beta(q)| \leqslant |f(q)|$, it follows that $|T(d)| \ll PQ\Delta_1$ by the corollary to Lemma 3, where
$$
\begin{equation*}
\Delta_1=k^{1/(2s)} s^{1/(2k)} \biggl( \frac{\sqrt{m}}{P^k}+\frac{P^{k-1}}{\sqrt{m}} \biggr)^{1/(2ks)} \biggl( \frac{\sqrt{m}}{Q^s}+\frac{Q^{s-1}}{\sqrt{m}}\biggr)^{1/(2ks)}.
\end{equation*}
\notag
$$
It follows from the conditions on $P$ and $Q$ that
$$
\begin{equation*}
\frac{\sqrt{m}}{P^k} \leqslant \frac{\sqrt{m}}{m^{(1+\delta)/2}}= m^{-\delta/2} , \qquad \frac{P^{k-1}}{\sqrt{m}} \leqslant \frac{m^{(1-\delta)/2}}{\sqrt{m}} \leqslant m^{-\delta/2} , \qquad \frac{\sqrt{m}}{Q^s}+\frac{Q^{s-1}}{\sqrt{m}} \leqslant 2.
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
\begin{aligned} \, \Delta_1 &\ll k^{1/(2s)}s^{1/(2k)} m^{-{\delta}/{(4 ks)}} \ll l^{1/(2k)} n^{1/(2s)}m^{-1/(16ln^2)} \ll l^{1/(2l)} n^{1/ 2} m^{-1/(16ln^2)} \\ &\ll \sqrt{n}\, m^{-1/(16ln^2)}\ll {(\ln m)}^{1/4}m^{-1/(16ln^2)}=\Delta_2 {(\ln m)}^{1/4}. \end{aligned}
\end{equation*}
\notag
$$
Therefore, $ T(d) \ll PQ\Delta_2 (\ln m)^{1/4}$, and hence
$$
\begin{equation*}
S (P, Q) \ll PQ\Delta_2 (\ln m)^{5/4} \ll Z \Delta_2 (\ln m)^{5/4}, \qquad \widetilde{S}\ll Z \Delta_2 (\ln m)^{13/4}.
\end{equation*}
\notag
$$
We proceed to estimate $S_4$. We obtain
$$
\begin{equation*}
\begin{aligned} \, S_4(\mu, \lambda; w) &\ll \frac{1}{\mu \lambda} \sum_{u_1} \sum_{v_1} f(u_1)f(v_1) \frac{x\Delta_2 (\ln m)^{13/4}}{c^2 u_{1} v_{1}w}+\frac{1}{\mu\lambda} \frac{F(x)}{c^2w} \frac{(\ln x)^2}{Y} \\ &\ll \frac{x\Delta_2}{\mu \lambda} \frac{(\ln m)^{13/4}}{c^2w} \sum_{u_1} \frac{f(u_1)}{u_1} \sum_{v_1} \frac{f(v_1)}{v_1}+ \frac{1}{\mu\lambda} \frac{F(x)}{c^2w} \frac{(\ln x)^2}{Y}. \end{aligned}
\end{equation*}
\notag
$$
Further,
$$
\begin{equation*}
\begin{aligned} \, S_4(\mu, \lambda) &\ll \frac{x\Delta_2 (\ln m)^{13/4}}{\mu \lambda} \sum_{c\leqslant H} \frac{f(c^2)}{c^2} \sum_{w} \frac{f(w)}{w} \sum_{u_1}\frac{f(u_1)}{u_1} \sum_{v_1}\frac{f(v_1)}{v_1} \\ &\qquad+ \frac{1}{\mu\lambda} \frac{F(x)(\ln x)^2}{Y}\sum_{c\leqslant H} \frac{1}{c^2} \sum_{w}\frac{1}{w}. \end{aligned}
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
\begin{aligned} \, S_4 &\ll \sum_{\mu, \lambda \geqslant 1} {|S(\mu, \lambda)|} \\ &\ll x \Delta_2 (\ln m)^{13/4} \sum_{\mu \geqslant 1} \sum_{\lambda \geqslant 1} \frac{1}{\mu \lambda} \sum_{c\leqslant H} \frac{f(c^2)}{c^2} \sum_{w\leqslant x}\frac{f(w)}{w} \sum_{u_1 \leqslant x}\frac{f(u_1)}{u_1} \sum_{v_1 \leqslant x}\frac{f(v_1)}{v_1} \\ &\qquad+ \frac{F(x) (\ln x)^2}{Y} \sum_{\mu \geqslant 1} \sum_{\lambda \geqslant 1} \frac{1}{\mu \lambda}\sum_{c\leqslant H} \frac{1}{c^2} \sum_{w\leqslant x} \frac{1}{w} \ll x \Delta_2 (\ln m)^{13/4} D+\frac{F(x)}{H} (\ln x)^5, \end{aligned}
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
D =\sum_{\mu \geqslant 1} \sum_{\lambda \geqslant 1} \frac{1}{\mu \lambda} \sum_{c\leqslant H} \frac{f(c^2)}{c^2} \sum_{w\leqslant x} \frac{f(w)}{w} \sum_{u_1 \leqslant x} \frac{f(u_1)}{u_1} \sum_{v_1 \leqslant x} \frac{f(v_1)}{v_1}.
\end{equation*}
\notag
$$
Estimating the sum $D$, we obtain
$$
\begin{equation*}
\begin{aligned} \, D &\ll \biggl( \sum_{c=1}^{+\infty} \frac{f(c^2)}{c^2}\biggr) \biggl( \sum_{w \leqslant x} \frac{f(w)}{w}\biggr) \biggl( \sum_{\mu \geqslant 1} \sum_{u_1} \frac{f(u_1)}{u_1}\biggr) \biggl( \sum_{\lambda \geqslant 1} \sum_{v_1} \frac{f(v_1)}{v_1}\biggr) \\ &\ll \biggl( \sum_{w \leqslant x} \frac{f(w)}{w}\biggr) \sum_{u} \frac{f(u)}{u} \sum_{v} \frac{f(v)}{v}, \end{aligned}
\end{equation*}
\notag
$$
where $u$ and $v$ range over increasing sequences of square-free numbers all of whose prime divisors belong to the sets $I$ and $J$, respectively, and $w$ ranges over square-free numbers having no prime divisors in $I$ and $J$. We also see that
$$
\begin{equation*}
D \ll \!\! \prod_{\substack{p \leqslant x\\ p \notin I,\, p \notin J }}\biggl( 1+ \frac{f(p)}{p}\biggr) \prod_{p \in I}\biggl( 1+ \frac{f(p)}{p}\biggr) \prod_{p \in J}\biggl( 1+ \frac{f(p)}{p}\biggr) \ll \prod_{p \leqslant x}\biggl( 1+ \frac{f(p)}{p}\biggr) \ll (\ln x)^\alpha.
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
S_4 \ll x \Delta_2 (\ln m)^{13/4} (\ln x)^\alpha+\frac{F(x)}{H} (\ln x)^5 \ll F(x) \Delta_2 (\ln m)^{17/4}+\frac{F(x)}{\sqrt{H}},
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
\begin{aligned} \, S &\ll F(x) \exp \biggl( -\frac{1}{5} (\ln\ln x) \ln\ln\ln x \biggr) \\ &\qquad+ F(x)\biggl(\frac{l}{n}\biggr)^\alpha (\ln\ln x)^\beta+ \frac{F(x)}{\sqrt{H}}+F(x)\Delta_2 (\ln m)^{17/4} \\ &\ll F(x)\biggl( \biggl(\frac{l}{n}\biggr)^\alpha (\ln\ln x)^\beta+(\ln m)^{17/4} m^{-1/(16ln^2)}\biggr). \end{aligned}
\end{equation*}
\notag
$$
Let us now choose $n$ in such a way that where $C \geqslant \alpha+5$ is a constant. Noting that
$$
\begin{equation*}
\frac{1}{2}\, \frac{\ln m}{\ln x} \ln\ln x \leqslant l<\frac{1}{2}\, \frac{\ln m}{\ln x} \ln\ln x+ 1<\frac{3}{4}\, \frac{\ln m}{\ln x} \ln\ln m,
\end{equation*}
\notag
$$
we obtain
$$
\begin{equation*}
\begin{gathered} \, \frac{1}{16ln^2}>\frac{1}{16n^2} \, \frac{4 \ln x}{3 \ln m}\, \frac{1}{\ln\ln m} \geqslant \frac{1}{12n^2}\, \frac{\ln x}{(\ln m) (\ln\ln m)}, \\ m^{1/(16ln^2)}> \exp \biggl( \frac{1}{12n^2}\, \frac{\ln x}{\ln\ln m }\biggr). \end{gathered}
\end{equation*}
\notag
$$
Hence, the condition (3.1) is satisfied if $n$ is chosen in such a way that
$$
\begin{equation*}
\frac{1}{12n^2}\, \frac{\ln x}{\ln\ln m } \geqslant C \ln\ln m, \quad \text{that is,} \quad n^2 \leqslant \frac{1}{12C}\, \frac{\ln x}{(\ln\ln m)^2}.
\end{equation*}
\notag
$$
We write
$$
\begin{equation*}
n=\biggl[ \frac{1}{\sqrt{12C}} \, \frac{\sqrt{\ln x}}{\ln\ln m}\biggr].
\end{equation*}
\notag
$$
Since the inequality $ n \leqslant (1/4)\sqrt{\ln m}$ is obvious, we only have to prove that $3l \leqslant n$, to this end it suffices to establish that
$$
\begin{equation*}
\frac{2 \ln m}{\ln x} \ln\ln m<\frac{1}{4\sqrt{C}}\, \frac{\sqrt{\ln x}}{\ln \ln m}
\end{equation*}
\notag
$$
or, equivalently, $(\ln x )^{3/2}>8\sqrt{C} (\ln m) (\ln\ln m)^2$. However, the last inequality follows from the hypotheses of the theorem. Thus,
$$
\begin{equation*}
S(x) \ll F(x) \Delta, \qquad \Delta=\biggl(\frac{\ln m}{(\ln x)^{3/2}} (\ln\ln m)^2 \biggr)^\alpha (\ln\ln m)^\beta.
\end{equation*}
\notag
$$
This completes the proof of Theorem 1. Remark 4. The following bound holds in the case of a fixed $\varepsilon$, $0<\varepsilon \leqslant 1/2$, and $x=m^\varepsilon$: § 4. Inhomogeneous short Kloosterman sums with weights4.1. Auxiliary assertionsTo prove the main theorem in the case of inhomogeneous sums, we need a bound similar to that in Lemma 3. Lemma 7. Let the conditions of Lemma 3 hold. Then the sum Proof. Following [13] and using the notation in Lemma 3 together with the inequalities in Lemma 1, we see that
$$
\begin{equation*}
\begin{aligned} \, |W_2|^s &\leqslant \biggl( \sum_{p} |\xi(p)| \biggl|\sum_{q}{\eta(q) e_{m}(a\overline{p}\, \overline{q}+bpq)} \biggr|\biggr)^s \\ &\leqslant \biggl( \sum_{p} |\xi(p)|\biggr)^{s-1} \sum_{p}|\xi(p)| \biggl| \sum_{q} \eta(q) e_{m}(a\overline{p}\, \overline{q}+bpq)\biggr|^s \\ &= \xi_{0}^{s-1} \sum_{p} |\xi(p)| \biggl| \sum_{q} {\eta(q) e_{m}(a\overline{p}\, \overline{q}+bpq)}\biggr|^s \\ &\leqslant \xi_{0}^{s-1} \xi \sum_{p}\biggl| \sum_{q_1,\dots, q_s} \eta(q_1)\cdots \eta(q_s) e_{m}\bigl(a\overline{p}(\overline{q}_1+\dots+\overline{q}_s)+bp({q_1} +\dots+{q_s})\bigr) \biggr| \\ &= \xi_{0}^{s-1} \xi \sum_{p} \biggl| \sum_{\lambda=1}^{m} \sum_{sQ< \mu \leqslant sQ_1} e_m(a\overline{p}\lambda+bp\mu) A_s(\lambda, \mu) \biggr|, \end{aligned}
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
A_s(\lambda, \mu)=\sum_{\substack{q_1, \dots, q_s\\\overline{q}_1+\dots+ \overline{q}_s \equiv \lambda \ (\operatorname{mod}m) \\ q_1+\dots+q_s \equiv \mu \ (\operatorname{mod}m)}}\eta(q_1)\cdots \eta(q_s).
\end{equation*}
\notag
$$
Let $\theta(p)$ be the argument in the inner sum. Then
$$
\begin{equation*}
\begin{aligned} \, |W_2|^s &\leqslant \xi_{0}^{s-1} \xi \sum_{p} e^{-i\theta(p)} \sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1 }} {A_s(\lambda, \mu) e_m(a\overline{p}\lambda+bp\mu)} \\ &= \xi_{0}^{s-1} \xi \sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} A_s(\lambda, \mu) \sum_{p}{e^{-i\theta(p)} e_m(a\overline{p}\lambda+bp\mu)}, \end{aligned}
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
|W_2|^s \leqslant \xi_{0}^{s-1} \xi \sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} |A_s(\lambda, \mu)| \biggl| \sum_{p} e^{-i\theta(p)} e_m(a\overline{p}\lambda+bp\mu)\biggr|.
\end{equation*}
\notag
$$
Taking this inequality to the power $k$ and applying Lemma 1,
$$
\begin{equation*}
\begin{aligned} \, |W_2|^{ks} &\leqslant \xi_{0}^{k(s-1)} \xi^{k} \Biggl( \sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ< \mu \leqslant sQ_1}} |A_s(\lambda, \mu)|\Biggr)^{k-1} \\ &\qquad\times \sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} |A_s(\lambda, \mu)|\biggl| \sum_{p} {e^{-i\theta(p)} e_m(a\overline{p}\lambda+bp\mu)}\biggr|^k \\ &\leqslant \xi_{0}^{k(s-1)} \xi^{k} \eta_0^{s(k-1)} \sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} |A_s(\lambda, \mu)| \biggl| \sum_{p} {e^{-i\theta(p)} e_m(a\overline{p}\lambda+ bp\mu)}\biggr|^k. \end{aligned}
\end{equation*}
\notag
$$
Squaring and using Lemma 1, we obtain
$$
\begin{equation*}
|W_2|^{2ks} \leqslant \xi_0^{2k(s-1)} \xi^{2k} \eta_0^{2s(k-1)} BC,
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
B=\sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}}|A_s(\lambda, \mu)|^2, \qquad C=\sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} \biggl| \sum_{p} {e^{-i\theta(p)} e_m(a\overline{p}\lambda+bp\mu)}\biggr|^{2k}.
\end{equation*}
\notag
$$
Estimating the sum $B$, we see that
$$
\begin{equation*}
\begin{aligned} \, &\sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}}|A_s(\lambda, \mu)|^2 = \sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} \Biggl|\sum_{\substack{q_1, \dots, q_s\\ \overline{q}_1+\dots+ \overline{q}_s \equiv \lambda \ (\operatorname{mod}m) \\ q_1+\dots+q_s \equiv \mu \ (\operatorname{mod}m)}}{\eta(q_1)\cdots \eta(q_s)}\Biggr|^2 \\ &\ \leqslant \sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} \sum_{\substack{q_1, \dots, q_{2s}\\ \overline{q}_1+\dots+ \overline{q}_s \equiv \lambda \equiv \overline{q}_{s+1}+\dots+ \overline{q}_{2s} \ (\operatorname{mod}m) \\ q_1+\dots+q_s \equiv \mu \equiv q_{s+1}+\dots+q_{2s} \ (\operatorname{mod}m)}}|\eta(q_1)|\cdots |\eta(q_{2s})| \\ &\ \leqslant \eta^{2s} \sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} \sum_{\substack{q_1,\dots, q_{2s}\\ \overline{q}_1+\dots+ \overline{q}_s \equiv \lambda \equiv \overline{q}_{s+1}+\dots+ \overline{q}_{2s} \ (\operatorname{mod}m) \\ q_1+\dots+q_s \equiv \mu \equiv q_{s+1}+\dots+q_{2s} \ (\operatorname{mod}m)}} 1 \\ &\ = \eta^{2s} \sum_{\substack{q_1, \dots, q_{2s}\\ \overline{q}_1+\dots+ \overline{q}_s \equiv \overline{q}_{s+1}+\dots+ \overline{q}_{2s} \ (\operatorname{mod}m) \\ q_1+\dots+q_s \equiv q_{s+1}+\dots+ q_{2s} \ (\operatorname{mod}m)}}\!\!\!\!\!1\leqslant \eta^{2s} \sum_{\substack{q_1, \dots, q_{2s}\\ \overline{q}_1+\dots+ \overline{q}_s \equiv \overline{q}_{s+1}+\dots+ \overline{q}_{2s} \ (\operatorname{mod}m)}}\!\!\!\!\!1 \\ &\ =\eta^{2s} I_s(Q). \end{aligned}
\end{equation*}
\notag
$$
Estimating the sum $C$, we obtain
$$
\begin{equation*}
\begin{aligned} \, C &= \!\!\sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} \!\!\sum_{p_1,\dots, p_{2k}} \!e^{-i(\theta(p_1)+\dots-\theta(p_{2k}))} e_m\bigl(a\lambda(\overline{p}_1+\dots- \overline{p}_{2k})\,{+}\,b\mu (p_1+\dots -p_{2k})\bigr) \\ &= \!\!\sum_{p_1,\dots, p_{2k}} e^{-i(\theta(p_1)+\dots-\theta(p_{2k}))} \!\!\!\sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} \!e_m\bigl(a\lambda(\overline{p}_1+\dots- \overline{p}_{2k})\,{+}\,b\mu (p_1+\dots -p_{2k})\bigr) \\ &\leqslant \sum_{p_1,\dots, p_{2k}} \Biggl| \sum_{\substack{1\leqslant \lambda \leqslant m \\ sQ<\mu \leqslant sQ_1}} e_m\bigl(a\lambda(\overline{p}_1+\dots-\overline{p}_{2k})+b\mu (p_1+\dots -p_{2k})\bigr)\Biggr| \\ &=\sum_{\sigma=1}^{m} \sum_{|\tau| \leqslant kP} j_k(\sigma, \tau) \biggl| \sum_{sQ< \mu \leqslant sQ_1} \sum_{\lambda=1}^{m} e_m(a\lambda\sigma+b\mu \tau)\biggr|, \end{aligned}
\end{equation*}
\notag
$$
where $j_k(\sigma, \tau)$ is the number of solutions of the system under the given conditions on the $p_j$, $j= 1,\dots, 2k$,
$$
\begin{equation*}
\begin{cases} \overline{p}_1+\dots+\overline{p}_k \equiv \overline{p}_{k+1}+\dots+\overline{p}_{2k}+ \sigma \pmod{m}, \\ p_1+\dots +p_k \equiv p_{k+1}+\dots +p_{2k}+\tau \pmod{m}. \\ \end{cases}
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
\begin{aligned} \, C &\leqslant \sum_{\sigma=1}^{m} \sum_{|\tau| \leqslant kP} j_k(\sigma, \tau) \sum_{sQ< \mu \leqslant sQ_1} \biggl| \sum_{\lambda=1}^{m} e_m(a\lambda\sigma)\biggr| \\ &\leqslant sQ \sum_{\sigma=1}^{m} \biggl(\sum_{|\tau| \leqslant kP} j_k(\sigma, \tau)\biggr) \biggl| \sum_{\lambda=1}^{m} e_m(a\lambda\sigma) \biggr| =sQm \sum_{|\tau| \leqslant kP} j_k(0, \tau) \leqslant smQ I_k(P). \end{aligned}
\end{equation*}
\notag
$$
We thus obtain
$$
\begin{equation*}
|W_2|^{2ks} \leqslant \xi_0^{2k(s-1)} \xi^{2k} \eta_0^{2s(k-1)} \eta^{2s} smQ I_s(Q)I_k(P).
\end{equation*}
\notag
$$
Applying Lemma 2 to $I_k(P)$ and $I_s(Q)$, we arrive at the desired assertion. This completes the proof Lemma 7. Corollary 3. When $\xi(\nu)=\eta(\nu)=f(\nu)$, the sum $W_2$ satisfies the bound $|W_2| \ll PQ \Delta_1$, where 4.2. Proof of Theorem 2We follow the same scheme of reasoning as in the proof of Theorem 1. Step 1. We write $H=\exp(2\sqrt{\ln x})$. Let $S_1$ be the part of the sum $S(x)$ over those $\nu$ of the form $\nu=c^2d$, $c>H$. By Lemma 4, $S_1 \ll F(x)e^{-\sqrt{\ln x}}$. The other $\nu$ have the form $\nu=c^2d$, where $c \leqslant H$, $d$ is square-free and $\mu(d) \neq 0$. Step 2. We write $R=\exp(\ln x/\ln\ln x)$ and define the integer $l_1$ by the condition $m^{1/(2l_1)} \leqslant R<m^{1/(2l_1-2)}$. We denote by $S_2$ the sum of the remaining terms corresponding to numbers $\nu$ all of whose prime divisors do not exceed $M= m^{1/(2l_1)}$. Then by Lemma 5,
$$
\begin{equation*}
S_2 \ll F(x)\exp\biggl(-\frac{1}{5} (\ln\ln x) \ln{\ln\ln x} \biggr).
\end{equation*}
\notag
$$
All the remaining $\nu$ have at least one prime divisor exceeding $M$. Before passing to Step 3, we introduce the following parameters. Let $n_1$ satisfy the condition $3l_1 \leqslant n_1 \leqslant (1/60)\sqrt{\ln m}$ (exact value will be chosen below) and put $\delta=1/(4n_1)$. For an integer $k$, $l_1 \leqslant k \leqslant n_1$, we write
$$
\begin{equation*}
X_k=m^{(1-\delta)/({2k})}, \qquad Y_k=m^{(1+\delta)/({2k})}.
\end{equation*}
\notag
$$
Just as was done above, we can readily see that $X_{k-1}>2Y_k$ for all $k$ under consideration. Let $l_2$ and $n_2$ satisfy the inequalities $l_2 \geqslant 5n_1$, $n_2 \geqslant 3 l_2$ and $n_2 \leqslant (1/4) \sqrt {\ln m}$ (we will specify the exact values of $l_2$ and $n_2$ below). We write We denote by $J$ the set of primes in $(U, V]$, and by $I$ the set of primes in the union of the $(Y_k, X_{k-1}]$, $l_1<k\leqslant n_1$. Step 3. We let $S_3$ denote the sum of the remaining terms of $S(x)$ which correspond to numbers $\nu$ that have no prime divisors in $I$. Repeating the arguments in Step 3 of Theorem 1 almost verbatim, we arrive at the bound
$$
\begin{equation*}
|S_3| \ll F(x) \biggl( \frac{l_1}{n_1} \biggr)^\alpha (\ln\ln x)^\beta.
\end{equation*}
\notag
$$
Step 4. We let $S_4$ denote the sum of the terms that remain in $S(x)$ and correspond to numbers $\nu$ having no prime divisors in $J$. As in Step 3, we obtain the bound
$$
\begin{equation*}
|S_4| \ll F(x) \biggl( \frac{l_2}{n_2} \biggr)^\alpha (\ln\ln x)^\beta.
\end{equation*}
\notag
$$
The sum of the terms not entering any of the above sums we denote by $S_5$. All the numbers $\nu$ corresponding to them have at least one prime divisor in $I$ and at least one in $J$. Thus, all these $\nu$ have the form $\nu=c^2 uvw$, where $1\leqslant c\leqslant H$, $\mu(uvw) \neq 0$, all the prime divisors of $u$ are in $I$, all prime divisors of $v$ are in $J$, while $w$, in contrast, has no prime divisors in the union of $I$ and $J$. We take integers $\lambda$, $\mu \geqslant 1$ and denote by $S_5(\mu, \lambda)$ the sum of the terms in $S_5$ that correspond to the numbers $\nu=c^2 u v w$ in which $u$ and $v$ consist of $\mu$ and $\lambda$ prime factors, respectively. In this case,
$$
\begin{equation*}
S_5=\sum_{\mu \geqslant 1} \sum_{\lambda \geqslant 1} S_5(\mu, \lambda).
\end{equation*}
\notag
$$
We write $\Phi(n)=e_m(a\overline{n}+bn)$. Then
$$
\begin{equation*}
\begin{aligned} \, |S_5(\mu, \lambda)| &=\biggl| \sum_{c^2uvw \leqslant x}{f(c^2uvw)\Phi(c^2uvw)}\biggr| \\ &\leqslant \sum_{c\leqslant H}{f(c^2)} \sum_{w}{f(w)} \biggl| \sum_{uv \leqslant x(c^2w)^{-1}}{f(u)f(v)\Phi(c^2uvw)}\biggr|, \end{aligned}
\end{equation*}
\notag
$$
where $w$ ranges over an increasing sequence of numbers that have no prime divisors in the union of the intervals $I$ and $J$, and $w \leqslant x (c^2 X^{\mu} U^{\lambda})^{-1}$. Let $c$ and $w$ be fixed. We compare the inner sum in $S_5(\mu, \lambda; w)$ over $u$, $v$ with the sum
$$
\begin{equation*}
S'(\mu, \lambda; w)=\frac{1}{\mu\lambda} \sum_{u_1, v_1} \sum_{p, q} {f(u_1)f(p)f(v_1)f(q) \Phi(c^2u_1pv_1qw)},
\end{equation*}
\notag
$$
where $u_1$, $v_1$ range independently over increasing sequences of numbers that are products of $\mu\,{-}\,1$ and $\lambda\,{-}\,1$ different prime factors in $I$ and $J$, respectively, while $p$ and $q$ take the values of the primes in $I$ and $J$, and the condition $u_1 v_1p q \leqslant x(c^2w)^{-1}$ holds. The numbers $u$ and $v$ corresponding to terms in $S_5(\mu, \lambda; w)$ can be represented in the form $u= u_{1}p$ and $v=v_{1}q$, where $(u_{1},p)=1$ and $(v_{1},q)=1$, in exactly $\mu$ and $\lambda$ ways. Therefore, every term in $S_5(\mu, \lambda; w)$ enters $S'(\mu, \lambda; w)$ with the coefficient 1. Further, there are terms in $S'(\mu, \lambda; w)$ for which at least one of the conditions $(u_{1},p)=1$ and $(v_{1},q)=1$ is violated. We denote by $S''(\mu, \lambda; w)$ the sum over these terms. Setting $u_{1}=u_{2}p$ and $v_1=v_{2}q$, we see that
$$
\begin{equation*}
\begin{aligned} \, &|S'' (\mu, \lambda; w)| \leqslant \frac{1}{\mu\lambda} \biggl(\sum_{p^{2}qu_{2}v_{1} \leqslant x(c^2w)^{-1}} f(pu_{2})f(p)f(v_1)f(q) \\ &+\!\sum_{pq^{2}u_{1}v_{2} \leqslant x(c^2w)^{-1}} f(u_1)f(p)f(v_{2}q)f(q)+ \! \!\sum_{p^{2}q^{2}u_{2}v_{2} \leqslant x(c^2w)^{-1}} f(u_2p)f(p)f(v_{2}q)f(q) \biggr). \end{aligned}
\end{equation*}
\notag
$$
Estimating $S''(\mu, \lambda; w)$ as was done in Theorem 1, we obtain the bound
$$
\begin{equation*}
|S''(\mu, \lambda; w)| \ll \frac{1}{\mu\lambda} \frac{F(x)}{c^2w} \frac{(\ln x)^2}{H}.
\end{equation*}
\notag
$$
Thus, $|S_5(\mu, \lambda; w)|\leqslant |S'|+|S''|$, where
$$
\begin{equation*}
S'(\mu, \lambda; w)=\frac{1}{\mu\lambda} \sum_{u_1} \sum_{v_1}f(u_1)f(v_1) \widetilde{S},
\end{equation*}
\notag
$$
and for fixed $\mu, \lambda, w, u_1, v_1$ we have
$$
\begin{equation*}
\widetilde{S}=\sum_{pq \leqslant Z} f(p)f(q)\Phi(c^2 u_1 v_1 w pq), \qquad Z=\biggl[ \frac{x}{c^2 u_1 v_1 w} \biggr].
\end{equation*}
\notag
$$
We represent $\widetilde{S}$ in the form
$$
\begin{equation*}
\widetilde{S}=\sum_{pq\leqslant Z} f(p)f(q)e_m(a_1\overline{p}\, \overline{q}+b_1pq),
\end{equation*}
\notag
$$
where $a_1 \equiv a(\overline{c^2 u_1 v_1}) \pmod{m}$, $b_1 \equiv bc^2 u_1 v_1 w \pmod{m}$. Further, we divide the ranges of $p$ and $q$ into intervals of the form $P\,{<}\,p \,{\leqslant}\, P_1 \,{\leqslant}\, 2P$ and $Q\,{<}\, q \,{\leqslant}\, Q_1\,{\leqslant}\,2Q$ and choose $P$, $P_1$, $Q$, $Q_1$ in such a way that every interval $(P, P_1]$ is contained entirely in some interval $(Y_k, X_{k-1}]$, $l_1<k\leqslant n_1$, and every interval $(Q, Q_1]$ is contained entirely in some interval $(m^{1/(2s)}, m^{1/(2s-2)}]$, $ l_2<s \leqslant n_2$. Thus, $\widetilde{S}$ splits into at most $(\ln m)^2$ sums of the form
$$
\begin{equation*}
S(P, Q)=\sum_{P <p\leqslant P_1} \sum_{\substack{Q< q\leqslant Q_1\\ pq \leqslant Z}} f(p)f(q)e_m(a_1 \overline{p}\, \overline{q}+b_1pq).
\end{equation*}
\notag
$$
We consider one of these sums. Getting rid of the condition $pq \leqslant Z$, as was done in Theorem 1, we see that
$$
\begin{equation*}
S(P, Q)=\sum_{|d|<m/2}\frac{T(d)}{|d|+1}, \qquad T(d)=\sum_{P<p\leqslant P_2}\sum_{Q<q \leqslant Q_1}\alpha(p) \beta(q) e_m(a_1\overline{p}\, \overline{q}+b_1pq),
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\alpha(p)=\frac{|d|+1}{m} \biggl(\sum_{Q<\xi \leqslant Q_2} e_m(d\xi) \biggr)f(p), \qquad \beta(q)= e_m(dq)f(q),
\end{equation*}
\notag
$$
and $|\alpha(p)| \leqslant |f(p)|$ and $|\beta(q)| \leqslant |f(q)|$. By the corollary to Lemma 7, $T(d)$ satisfies the bound $|T(d)| \ll PQ \Delta_1$, where
$$
\begin{equation*}
\Delta_1=k^{1/(2s)} s^{1/(2k)}Q^{1/(2ks)} \biggl(\frac{P^{k-1}}{\sqrt{m}} +\frac{\sqrt{m}}{P^{k}}\biggr)^{1/(2ks)} \biggl(\frac{Q^{s-1}}{\sqrt{m}} +\frac{\sqrt{m}}{Q^s}\biggr)^{1/(2ks)}.
\end{equation*}
\notag
$$
It follows from the conditions on $P$ and $Q$ that
$$
\begin{equation*}
\frac{\sqrt{m}}{P^k} \leqslant \frac{\sqrt{m}}{m^{(1+\delta)/2}}=m^{-\delta/2}, \qquad \frac{P^{k-1}}{\sqrt{m}}\leqslant \frac{m^{(1-\delta)/2}}{\sqrt{m}} \leqslant m^{-\delta/2}, \qquad \frac{\sqrt{m}}{Q^s}+\frac{Q^{s-1}}{\sqrt{m}} \leqslant 2.
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\begin{aligned} \, \Delta_1 &\ll k^{1/(2s)} s^{1/(2k)} (m^{1/(2l_2)} m^{-\delta/2})^{1/(2ks)} \\ &\ll n_2^{1/(2l_1)} n_1^{1/(2l_2)} (m^{1/(10n_1)-1/(8n_1)})^{1/(2ks)} \\ &\ll n_2^{1/(2l_1)} \biggl(\frac{1}{5} l_2\biggr)^{1/(2l_2)} m^{-1/({80n_1ks})} \ll (\ln m)^{1/4} m^{-1/(80n_1^2 n_2)}= \Delta_2 (\ln m)^{1/4}. \end{aligned}
\end{equation*}
\notag
$$
Thus, $T(d) \ll PQ\Delta_2 (\ln m)^{1/4}$, and hence
$$
\begin{equation*}
S(P,Q) \ll PQ \Delta_2(\ln m)^{5/4} \ll Z\Delta_2 \ln m, \qquad \widetilde{S} \ll Z\Delta_2 (\ln m)^{13/4}.
\end{equation*}
\notag
$$
We pass to a bound for $S_5$. We obtain
$$
\begin{equation*}
\begin{aligned} \, S_5(\mu, \lambda;w) &\ll \frac{1}{\mu\lambda} \sum_{u_1}\sum_{v_1} f(u_1)f(v_1) \frac{x \Delta_2 (\ln m)^3}{c^2u_1v_1w}+\frac{1}{\mu\lambda}\, \frac{F(x)}{c^2w}\,\frac{(\ln x)^2}{H} \\ &=\frac{x\Delta_2}{\mu\lambda}\, \frac{(\ln m)^3}{c^2 w}\sum_{u_1} \frac{f(u_1)}{u_1} \sum_{v_1} \frac{f(v_1)}{v_1}+\frac{1}{\mu\lambda}\, \frac{F(x)}{c^2w}\,\frac{(\ln x)^2}{H}. \end{aligned}
\end{equation*}
\notag
$$
Further,
$$
\begin{equation*}
S_5(\mu, \lambda) \ll \frac{x \Delta_2 (\ln m)^3}{\mu \lambda} \sum_{c\leqslant H}{\frac{f(c^2)}{c^2}} \sum_{w}{\frac{f(w)}{w}} \sum_{u_1}{\frac{f(u_1)}{u_1}} \sum_{v_1}{\frac{f(v_1)}{v_1}}+\frac{1}{\mu\lambda}\, \frac{F(x)(\ln x)^3}{H}.
\end{equation*}
\notag
$$
Using arguments similar to those in Theorem 1, we see that
$$
\begin{equation*}
S_5 \ll x \Delta_2(\ln m)^{13/4} (\ln x)^\alpha+\frac{F(x)}{\sqrt{H}}.
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
\begin{aligned} \, S(x) &\ll F(x) \biggl( \frac{l_1}{n_1}\biggr)^\alpha (\ln\ln x)^\beta+F(x) \biggl( \frac{l_2}{n_2}\biggr)^\alpha(\ln\ln x)^\beta \\ &\qquad+\frac{F(x)}{\sqrt{H}}+F(x) (\ln m)^{17/4} m^{-{1}/{(80n_1^2 n_2)}} \\ &\ll F(x) \biggl( \biggl( \frac{l_1}{n_1}\biggr)^\alpha (\ln\ln x)^\beta+\biggl( \frac{l_2}{n_2}\biggr)^\alpha(\ln\ln x)^\beta+(\ln m)^{17/4} m^{-{1}/{(80n_1^2 n_2)}}\biggr). \end{aligned}
\end{equation*}
\notag
$$
We now choose $n_1$ and $n_2$ in such a way that where $C\geqslant \alpha+ 5$ is a constant. This condition is satisfied if $n_2$ satisfies
$$
\begin{equation*}
\frac{1}{ 80 n_1^2 n_2} \ln m \geqslant C \ln\ln m, \quad \text{that is,} \quad n_2 \leqslant \frac{1}{80 C} \frac{\ln m}{ n_1^2 \ln\ln m}.
\end{equation*}
\notag
$$
We write
$$
\begin{equation*}
n_2=\biggl[ \frac{1}{80C}\, \frac{\ln m}{n_1^2 \ln\ln m}\biggr].
\end{equation*}
\notag
$$
Noting that
$$
\begin{equation*}
n_2 \geqslant \frac{\ln m}{80C n_1^2 \ln\ln m}-1 \geqslant \frac{\ln m}{102C n_1^2 \ln\ln m},\quad \text{that is,}\quad \frac{1}{n_2} \leqslant \frac{102C n_1^2 \ln\ln m}{\ln m}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\frac{1}{2}\,\frac{\ln m}{\ln x} \ln\ln x \leqslant l_1 \leqslant \frac{1}{2}\, \frac{\ln m}{\ln x} \ln\ln x+1<\frac{3}{4}\, \frac{\ln m}{\ln x} \ln\ln m,
\end{equation*}
\notag
$$
we see that
$$
\begin{equation*}
\biggl( \frac{l_1}{n_1}\biggr)^\alpha+\biggl( \frac{l_2}{n_2}\biggr)^ \alpha \ll \biggl( \frac{3}{4}\, \frac{\ln m}{n_1 \ln x} \ln\ln m \biggr)^\alpha+\biggl( 102C \frac{l_2 n_1^2 }{\ln m} \ln\ln m \biggr)^\alpha.
\end{equation*}
\notag
$$
We define $n_2$ and $l_2$ as follows:
$$
\begin{equation*}
n_1=\biggl[ \frac{1}{\sqrt[4]{680C}}\, \frac{\sqrt{\ln m}}{(\ln x)^{1/4}}\biggr], \qquad l_2=5n_1.
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
\biggl( \frac{l_1}{n_1}\biggr)^\alpha+\biggl( \frac{l_2}{n_2}\biggr)^\alpha \ll \biggl( \frac{\sqrt{\ln m}}{(\ln x)^{3/4}}\ln\ln m \biggr)^\alpha.
\end{equation*}
\notag
$$
Since the inequalities $l_2 \geqslant 5n_1$, $n_2 \leqslant (1/4)\sqrt{\ln m}$ and $n_2\leqslant (1/60) \sqrt{\ln m}$ are obvious, it follows that we only need to show that $3l_1\leqslant n_1$ and $3l_2 \leqslant n_2$. In turn, to satisfy the first inequality, it suffices to prove that
$$
\begin{equation*}
2\frac{\ln m}{\ln x} \ln \ln m<\frac{1}{6\sqrt[4]{C}}\, \frac{\sqrt{\ln m}}{(\ln x)^{1/4}}
\end{equation*}
\notag
$$
or, equivalently,
$$
\begin{equation*}
(\ln x)^{3/4}>12 \sqrt[4]{C}\, (\ln m)^{1/2} \ln\ln m.
\end{equation*}
\notag
$$
However, the last inequality follows from the hypothesis of the theorem. The validity of the other inequality follows from the bound
$$
\begin{equation*}
15 n_1 \leqslant \frac{\ln m}{102C n_1^2 \ln\ln m}, \quad \text{that is,} \quad \frac{5}{9\sqrt[4]{C^3}}\, \frac{(\ln m)^{3/2}}{(\ln x)^{3/4}} \leqslant \frac{1}{102 C}\, \frac{\ln m}{\ln\ln m},
\end{equation*}
\notag
$$
or, equivalently,
$$
\begin{equation*}
(\ln x)^{3/4} \geqslant \frac{170}{3} \sqrt[4]{C}\, (\ln m)^{1/2} \ln\ln m.
\end{equation*}
\notag
$$
However, the last relation follows from the hypothesis of the theorem. Thus,
$$
\begin{equation*}
S(x) \ll F(x) \Delta, \qquad \Delta=\biggl( \frac{\sqrt{\ln m}}{(\ln x)^{3/4}} \ln\ln m\biggr)^\alpha (\ln\ln m)^\beta.
\end{equation*}
\notag
$$
This completes the proof of Theorem 2. Remark 5. The following bound holds in the case of a fixed $\varepsilon$, $0<\varepsilon \leqslant 1/2$, and $x=m^\varepsilon$: The author is deeply grateful to M. A. Korolev for suggesting the problem and for his continuing interest in the work. 
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