# On divisor bounded multiplicative functions in short intervals

Yu-Chen Sun  Department of Mathematics and Statistics, University of Turku,
20014 Turku, Finland<EMAIL_ADDRESS>
###### Abstract.

Let $d_{k}(n)=\sum_{n_{1}\cdots n_{k}=n}1$ be the $k$-fold divisor function.
We call a function $f:\mathbb{N}\to\mathbb{C}$ a $d_{k}$-bounded
multiplicative function, if $f$ is multiplicative and $|f(n)|\leq d_{k}(n)$
for every $n\in\mathbb{N}$. In this paper we improve Mangerel’s results which
extend the Matomäki-Radziwiłł theorem to divisor bounded multiplicative
functions.

In particular, we prove that for sufficiently large $X\geq 2$, any
$\epsilon>0$ and $h\geq(\log X)^{k\log k-k+1+\epsilon}$ , we have

$\frac{1}{h}\sum_{x<n\leq x+h}d_{k}(n)-\frac{1}{x}\sum_{x<n\leq
2x}d_{k}(n)=o(\log^{k-1}x)$

for almost all $x\in[X,2X]$. We also demonstrate that the exponent $k\log
k-k+1$ is optimal.

###### Key words and phrases:

multiplicative functions, divisor bounded functions, Matomäki-Radziwiłł
methods, short intervals

##  1\. Introduction

Let $n$ be a positive integer and $d(n)$ denote the number of factors of $n$.
By the elementary argument(hyperbola method), one can show that

$\sum_{n\leq x}d(n)=x\log x+(2\gamma-1)+O(x^{\theta}),$ (1.1)

for any $\theta>1/2$. The classical Dirichlet divisor problem asks whether we
can reduce $\theta$ to $x^{1/4+\delta}$ for any $\delta>0$. The best record
$\theta>131/416$ was given by Huxley [4]. In fact, (1.1) gives information
concerning the sum of the divisor function over shorter intervals.
Specifically, if (1.1) holds, then

$\frac{1}{x^{\theta}}\sum_{x<n\leq
x+x^{\theta}}d(n)\sim\frac{1}{x}\sum_{x<n\leq 2x}d(n)\sim\log x.$

Let $d_{k}(n)=\sum_{n=m_{1}m_{2},\cdots,m_{k}}1$ which is an extension of
$d(n)$. We know that

$\frac{1}{h_{k}}\sum_{x<n\leq
x+x^{\theta_{k}}}d_{k}(n)\sim\frac{1}{x}\sum_{x<n\leq 2x}d_{k}(n)\sim
P_{k-1}(\log x),$

where $0<\theta_{k}<1$ depends on $k$ (see [11]) and $P_{k}(x)$ is a
polynomial of degree $k$ that can be calculated explicitly (see [13, Formula
(5.36) in Section II.5.4]). These results are called results in all intervals.

Another direction is results in almost all intervals which has developed
rapidly in recent years due to the work of Matomäki and Radziwiłł[8].

Let $f:\mathbb{N}\to[-1,1]$ be a multiplicative function. In $2016$, Matomäki
and Radziwiłł in their breakthrough paper [8] proved that, for almost all
$x\in[X,2X]$,

$\frac{1}{h}\sum_{x<n\leq x+h}f(n)-\frac{1}{X}\sum_{X<n\leq 2X}f(n)=o(1),$
(1.2)

provided $h\to\infty$ with $X\to\infty$.

Motivated by the above, we prove the following result for the sum of
$d_{k}(n)$ in almost all intervals.

######  Theorem 1.1.

Let $\epsilon>0$. If $h\geq(\log X)^{k\log k-k+1+\epsilon}$, then

$\frac{1}{h}\sum_{x<n\leq x+h}d_{k}(n)-\frac{1}{x}\sum_{x<n\leq
2x}d_{k}(n)=o(\log^{k-1}x)$

for all but at most $o(X)$ integers $x\in[X,2X]$.

Theorem 1.1 is a special case of Theorem 1.3 below which deals with the
general case for complex valued divisor bounded multiplicative functions.

######  Remark.

Theorem 1.1 improves upon the special case of [7, Theorem 1.7] under the
“almost all” sense. In fact, Mangerel [7] proved that, if $h\geq h_{0}(\log
X)^{(k-1)^{2}}$ and $h_{0}\to\infty$ as $X\to\infty$, then

$\int_{X}^{2X}\left|\frac{1}{h}\sum_{x<n\leq
x+h}d_{k}(n)-\frac{1}{x}\sum_{x<n\leq 2x}d_{k}(n)\right|^{2}=o(\log^{2k-2}x).$

Mangerel’s result is optimal in the $L_{2}$ sense, and one can apply Henriot’s
bound (Lemma 4.2) to prove the fact that if $h=o((\log X)^{(k-1)^{2}})$, then,

$\frac{1}{X}\int_{X}^{2X}\left|\frac{1}{h}\sum_{x<n\leq
x+h}d_{k}(n)-\frac{1}{x}\sum_{x<n\leq
2x}d_{k}(n)\right|^{2}dx=(1+o(1))\frac{1}{h}(\log X)^{k^{2}-1}.$ (1.3)

We will also show in the following theorem the exponent $k\log k-k+1$ in
Theorem 1.1 is optimal.

######  Theorem 1.2.

Let $X$ be sufficiently large. For $\epsilon>0$, and let $1\leq h\leq(\log
X)^{k\log k-k+1-\epsilon}$, then

$\sum_{x<n\leq x+h}d_{k}(n)=o(h(\log x)^{k-1})$

holds for almost all $x\in[X,2X]$.

######  Remark.

Note that Theorem 1.2 was mentioned in [10] without proof. We give the details
of the proof in the end of Section 2. Combining Theorem 1.2 and (1.3), we
observe that when $h\leq(\log X)^{k\log k-k+1-\epsilon}$, the main
contribution to the integral on the left-hand side of (1.3) comes from $o(X)$
intervals $[x,x+h]$. In fact, those intervals contain an integer that has at
least $(1+o(1))k\log\log X$ prime factors.

Recently, Matomäki and Radziwiłł [9] extended their result to $1$-bounded
complex multiplicative functions that might vanish for many primes. For
introducing this extension and our general results, we need the following two
definitions.

######  Definition 1.1.

Let $f:\mathbb{N}\to\mathbb{C}$ be a multiplicative function. We define
$t_{0}=t_{0}(f,X)\in[-X,X]$ to be (one of) $t\in[-X,X]$ which minimizes the
following distance

$\mathbb{D}(f,n^{it};X)^{2}:=\sum_{p\leq X}\frac{|f(p)|-\Re{f(p)p^{-it}}}{p}.$

We say that a multiplicative function $g$ is almost real-valued, if

$\sum_{\begin{subarray}{c}p\in\mathbb{P}\\\
g(p)\not\in\mathbb{R}\end{subarray}}\frac{|g(p)|}{p}<\infty.$

If $g$ is almost real-valued, we let $t_{0}=0$.

######  Definition 1.2.

We call a function $f$ $d_{k}$-bounded if $|f(n)|\leq d_{k}(n)$ for all
$n\in\mathbb{N}$.

Given $\alpha\in(0,1]$ and $\Delta\geq 1$, a $d_{k}$-bounded multiplicative
function $f:\mathbb{N}\to\mathbb{C}$ is said to be $(\alpha,\Delta)$-non-
vanishing if, for any $2\leq w\leq z\leq\Delta$, we have

$\sum_{w<p\leq z}\frac{|f(p)|}{p}\geq k\alpha\sum_{w<p\leq
z}\frac{1}{p}-O\left(\frac{1}{\log w}\right).$

We also define

$P_{f}(x):=\prod_{p\leq x}\left(1+\frac{|f(p)|-1}{p}\right).$

Let $f:\mathbb{N}\to\mathbb{C}$ be a $1$-bounded, $(\alpha,X^{\theta})$-non-
vanishing multiplicative function for some small $\theta>0$ and let $t_{0}$ be
as in Definition 1.1. In recent paper [9, Theorem 1.9], Matomäki and Radziwiłł
proved that, for almost all $x\in[X,2X]$,

$\left|\frac{1}{h}\sum_{x<n\leq
x+h}f(n)-\frac{1}{h}\int_{x}^{x+h}u^{it_{0}}du\cdot\frac{1}{X}\sum_{X<n\leq
2X}f(n)n^{-it_{0}}\right|=o(P_{f}(x)),$

provided $h=h_{0}\prod_{p\leq x}\left(1+\frac{(|f(p)|-1)^{2}}{p}\right)$,
where $h_{0}\to\infty$ with $X\to\infty$.

It is natural to consider whether the Matomäki-Radziwiłł theorem can be
extended to unbounded multiplicative functions, e.g. $d_{k}$-bounded
multiplicative functions.

In $2021$, Mangerel [7] generalized Matomäki-Radziwiłł theorem to the
$d_{k}$-bounded multiplicative functions. Let $f:\mathbb{N}\to\mathbb{C}$ be
$d_{k}$-bounded, $(\alpha,X^{\theta})$-non-vanishing multiplicative function
for some small $\theta>0$ and let $t_{0}$ be as in Definition 1.1. Mangerel
proved that, for almost all $x\in[X,2X]$,

$\left|\frac{1}{h}\sum_{x<n\leq
x+h}f(n)-\frac{1}{h}\int_{x}^{x+h}u^{it_{0}}du\cdot\frac{1}{X}\sum_{X<n\leq
2X}f(n)n^{-it_{0}}\right|=o(P_{f}(x))$

for $h_{0}\prod_{p\leq x}\left(1+\frac{(|f(p)|-1)^{2}}{p}\right)\leq h\leq x$,
whenever $h_{0}\to\infty$ with $X\to\infty$. Note that in [7], his results did
not include that in the almost real-valued case one may take $t_{0}=0$.
However, there is no doubt that one can give the corresponding result in the
almost real-valued case.

In this paper, we will improve Mangerel’s result. Let

$H(f,X,\epsilon):=\frac{(P_{f}(X)\log X)^{(1+\epsilon)\log
k}}{P_{f}(X)}\asymp\prod_{p\leq
X}\left(1+\frac{|f(p)|}{p}\right)^{(1+\epsilon)\log k-1}\log X,$

which is larger than a small power of $\log X$ when $k\geq 2$. We obtain the
following theorem

######  Theorem 1.3.

Let $k\in\mathbb{N}$ and $\alpha\in(0,1]$. For any $\epsilon,\theta>0$, let
$f:\mathbb{N}\to\mathbb{C}$ be $d_{k}$-bounded $(\alpha,X^{\theta})$ non-
vanishing multiplicative function. If $h\in[H(f,X,\epsilon),X]$, then

$\left|\frac{1}{h}\sum_{x<n\leq
x+h}f(n)-\frac{1}{h}\int_{x}^{x+h}u^{it_{0}}du\cdot\frac{1}{X}\sum_{X<n\leq
2X}f(n)n^{-it_{0}}\right|=o(P_{f}(X))$

for all but at most $o(X)$ integers $x\in[X,2X]$.

Letting $f(n)=d_{k}(n)$, we immediately get Theorem 1.1 by replacing
$\epsilon$ by $\frac{\epsilon}{k\log k}$ in Theorem 1.3.

Our arguments rely on the Matomäki-Radziwiłł method, which was first
introduced in [8]. Let us use (1.2) as an example to introduce their method.
For the “relatively long” short intervals, say $h\geq X/(\log X)^{c}$ for some
$c>0$, one can utilize Lipschitz type results to show that (1.2) holds for all
$x\in[X,2X]$. For the “relatively short” short intervals one uses Perron’s
formula and considers $L_{2}$ norm of the left hand side of (1.2). Then the
main terms (integrals around $t=0$) of the “short average” and “long average”
are cancelled, so one only needs to show that the second moment with $t$ far
away form zero is small. Their two powerful tools for controlling the second
moment are Ramaré’s identity and amplifying techniques.

Our new idea is motivated by Matomäki-Radziwiłł-Tao [10], which allows us to
restrict integers into the subset of integers with relatively small number of
prime factors. Let $\Omega(n)$ denote the number of prime factors of $n$. For
instance, for the divisor function $d(n)$, we could consider the set
$\\{n\in[X,2X]:\Omega(n)\leq 2(1+\epsilon)\log\log X\\}$. The reason we can do
this is that

$\sum_{\begin{subarray}{c}n\sim X\\\ |\Omega(n)-2\log\log
X|\geq\epsilon\log\log X\end{subarray}}d(n)=o(X\log X)$

(see Lemma 2.3 below).

More generally, for $d_{k}$-bounded multiplicative functions we consider the
set $\\{n\in[X,2X]:\Omega(n)\leq(1+\epsilon)\log(P_{f}(X)\log X)\\}$.

This is advantageous when we apply the mean-value theorem (see Lemma 4.3
below). In the mean-value theorem, one will get an upper bound consisting of a
diagonal term and an off-diagonal term. The off-diagonal term will be easily
controlled by Henriot’s bound (Lemma 4.2). However, in case of the diagonal
term, e.g.$\sum_{X<n\leq 2X}d_{k}(n)^{2}$, if we directly apply Shiu’s bound,
as Mangerel did, we will get the upper bound $X(\log X)^{k^{2}-1}$ which is
larger than we can afford to have. Fortunately, when we just consider those
integers which have at most $(1+\epsilon)k\log\log X$ prime factors, we have a
pointwise upper bound $d_{k}(n)\leq(\log X)^{(1+\epsilon)k\log k}$. In this
way, we get the upper bound $X(\log X)^{(1+\epsilon)k\log k+k-1}$ for the
diagonal term.

The rest of the paper is organized as follows. In Section 2, we outline the
proof of Theorem 1.3 by reducing Theorem 1.3 to Propositions 2.1 and 2.2,
dealing with the “relatively short” and “relatively long” short intervals
cases separately, and we will prove Theorem 1.2 in the end of this section. In
Section 3, we will collect some Halász type results which will be used in
proving Propositions 2.1 and 2.2, and in the end of Section 3 the almost real-
valued case of Proposition 2.2. Then in Section 4, we will follow Matomäki-
Radziwiłł arguments to prove Proposition 2.1.

######  Acknowledgments.

The author is grateful to his supervisor Kaisa Matomäki for suggesting this
project, drawing his attention to Matomäki-Radziwiłł-Tao [10], and many useful
discussions. He also thanks Joni Teräväinen, Martin Čech and Mengdi Wang for
their useful comments. During the work the author was supported by UTUGS
funding, working in the Academy of Finland project no. 333707

##  2\. Outline of the proof of Theorem 1.3

In this section, we will firstly state Shiu’s bound which we will use many
times and show that we can reduce to the case where $n$ is supported on
certain subset of $[X,2X]$. Then we state two propositions, one is a Matomäki-
Radziwiłł type bound for almost all intervals, the other is a Halász type
bound for all “relatively long” short intervals. In the end of the section we
will give the proofs of Theorems 1.3 and 1.2.

######  Lemma 2.1 (Shiu’s bound).

Let $f:\mathbb{N}\to\mathbb{C}$ be a multiplicative function with
$|f(n)|\ll_{\epsilon}n^{\epsilon}$ for any $\epsilon>0$. Let $\sqrt{X}<Y\leq
X$, $\delta\in(0,1)$ and $Y^{\delta}\leq y\leq Y$. Then

$\sum_{Y-y<n\leq Y}|f(n)|\ll_{k,\delta}yP_{f}(X).$

###### Proof.

See [12, Theorem 1]. ∎

Let $\epsilon_{0}>0$. We define

$\displaystyle P_{1}:=\exp((\log\log X)^{1/2}),\quad Q_{1}:=(\log
X)^{\epsilon_{0}}$ (2.1) $\displaystyle P_{2}:=\exp((\log\log X)^{2}),\quad
Q_{2}:=\exp((\log\log X)^{100})$ (2.2)

Let $A$ denote the set of all $n\in(X,2X]$ having at least one prime factor in
each interval $[P_{j},Q_{j}]$ for $j\in\\{1,2\\}$.

The following lemma says that one can restrict the sums into $n\in A$. This
idea is inspired by the Matomäki-Radziwiłł decomposition (see [8, Section 2]),
but is simpler then theirs.

######  Lemma 2.2.

Let $X>0$ be sufficiently large and $f:\mathbb{N}\to\mathbb{C}$ be a
$d_{k}$-bounded $(\alpha,X^{\theta})$-non-vanishing multiplicative function
for some $\alpha,\theta\in(0,1]$. Let $A$ as the above. For any $h>0$, we have

$\sum_{\begin{subarray}{c}x<n\leq x+h\\\ n\not\in
A\end{subarray}}|f(n)|=o(hP_{f}(X))$

for all but at most $o(X)$ integers $x\in[X,2X]$.

###### Proof.

We first notice that, for $j=1,2$, by the Taylor expansion

$\prod_{p\in[P_{j},Q_{j}]}\left(1-\frac{|f(p)|}{p}\right)=\exp\left(\sum_{p\in[P_{j},Q_{j}]}\log\left(1-\frac{|f(p)|}{p}\right)\right)\ll\exp\left(-\sum_{p\in[P_{j},Q_{j}]}\frac{|f(p)|}{p}\right).$

Since $f$ is $(\alpha,X^{\theta})$-non-vanishing, by the Mertens’ theorem

$\sum_{p\in[P_{j},Q_{j}]}\frac{|f(p)|}{p}\geq
k\alpha\sum_{p\in[P_{j},Q_{j}]}\frac{1}{p}+O\left(\frac{1}{\log
P_{j}}\right)\geq k\alpha\log\left(\frac{\log Q_{j}}{\log
P_{j}}\right)+O\left(\frac{1}{\log P_{j}}\right).$

Thus we have

$\prod_{p\in[P_{j},Q_{j}]}\left(1-\frac{|f(p)|}{p}\right)=o(1).$ (2.3)

By Shiu’s bound (Lemma 2.1) and (2.3),

$\displaystyle\sum_{x\sim X}\sum_{\begin{subarray}{c}x<n\leq x+h\\\ n\not\in
A\end{subarray}}|f(n)|\ll\sum_{\begin{subarray}{c}X<n\leq 2X+h\\\ n\not\in
A\end{subarray}}|f(n)|\sum_{n-h\leq x<n}1\leq
h\sum_{\begin{subarray}{c}X<n\leq 2X+h\\\ n\not\in A\end{subarray}}|f(n)|$
$\displaystyle\ll$ $\displaystyle hX\prod_{p\leq
2X}\left(1+\frac{|f(p)|-1}{p}\right)\sum_{j=1}^{2}\prod_{p\in[P_{j},Q_{j}]}\left(1-\frac{|f(p)|}{p}\right)=o\left(hX\prod_{p\leq
2X}\left(1+\frac{|f(p)|-1}{p}\right)\right).$

∎

The next lemma tells us that one can ignore integers with large number of
prime factors.

######  Lemma 2.3.

Let $X>0$ be large and $\epsilon>0$ sufficiently small. Let $f$ be a
multiplicative function with $|f(n)|\ll_{\epsilon_{0}}n^{\epsilon_{0}}$ for
any $\epsilon_{0}>0$. Let

$B_{\epsilon}=\\{n\in\mathbb{N}:\Omega(n)\leq(1+\epsilon)\log(P_{f}(x)\log
X)\\}.$

For any $\delta>0$, we have

$\sum_{\begin{subarray}{c}n\not\in B_{\epsilon}\\\ X<n\leq
3X\end{subarray}}|f(n)|\ll XP_{f}(X)(P_{f}(X)\log
X)^{-(1+\epsilon)\log(1+\delta)+\delta}.$ (2.4)

Furthermore, for any $h>0$,

$\sum_{\begin{subarray}{c}n\not\in B_{\epsilon}\\\ x<n\leq
x+h\end{subarray}}|f(n)|=o(hP_{f}(X))$ (2.5)

for all but at most $o(X)$ integers $x\in[X,2X]$.

###### Proof.

Since

$\sum_{x\sim X}\sum_{\begin{subarray}{c}n\not\in B_{\epsilon}\\\ x<n\leq
x+h\end{subarray}}|f(n)|\ll\sum_{\begin{subarray}{c}n\not\in B_{\epsilon}\\\
X<n\leq 3X\end{subarray}}|f(n)|\sum_{n-h\leq x<n}1\ll
h\sum_{\begin{subarray}{c}n\not\in B_{\epsilon}\\\ X<n\leq
3X\end{subarray}}|f(n)|.$

(2.5) follows if we can show that

$\sum_{\begin{subarray}{c}n\not\in B_{\epsilon}\\\ X<n\leq
3X\end{subarray}}|f(n)|=o(XP_{f}(X)).$ (2.6)

This follows from (2.4) with $\delta=\frac{\epsilon}{2(1+\epsilon)}$. Hence it
suffices to establish (2.4).

By Shiu’s bound, we have, for any $\delta>0$,

$\displaystyle\sum_{\begin{subarray}{c}n\not\in B_{\epsilon}\\\ X<n\leq
3X\end{subarray}}|f(n)|$ $\displaystyle=\sum_{X<n\leq 3X}|f(n)|{\bf
1}_{\Omega(n)>(1+\epsilon)\log(P_{f}(x)\log X)}$
$\displaystyle\leq\sum_{X<n\leq
3X}|f(n)|(1+\delta)^{\Omega(n)-(1+\epsilon)\log(P_{f}(x)\log X)}$
$\displaystyle\ll X(1+\delta)^{-(1+\epsilon)\log(P_{f}(x)\log X)}\prod_{p\leq
3X}\left(1+\frac{(1+\delta)|f(p)|-1}{p}\right)$ $\displaystyle\ll
XP_{f}(X)(P_{f}(X)\log X)^{-(1+\epsilon)\log(1+\delta)+\delta}.$

Hence our claim follows. ∎

Now we state two propositions, and then combine these with the above lemmas to
prove Theorem 1.3. Let

$\rho_{k,\alpha}:=\frac{k\alpha}{3}-\frac{2k}{3\pi}\sin\left(\frac{\pi\alpha}{2}\right)>0\quad\text{and}\quad\sigma:=\frac{1}{4}\min\left\\{1,\rho_{k,\alpha}\right\\}.$

Recall $t_{0}$ in Definition 1.1.

######  Proposition 2.1 (Matomäki-Radziwiłł type bound).

Let $X>0$ be a large number. Let $f:\mathbb{N}\to\mathbb{C}$ be a
$d_{k}$-bounded $(\alpha,X^{\theta})$-non-vanishing multiplicative function
for some $\alpha,\theta\in(0,1]$. Let $\epsilon>0$, $A$ and $B_{\epsilon/2}$
as in Lemmas 2.2 and 2.3 and $S=A\cap B_{\epsilon/2}$. Let
$H(f,X,\epsilon)\leq h\leq h_{1}=X/(\log X)^{\sigma}$. Then,

$\frac{1}{X}\int_{X}^{2X}\left|\frac{1}{h}\sum_{\begin{subarray}{c}n\in S\\\
x<n\leq
x+h\end{subarray}}f(n)-\frac{1}{h}\int_{x}^{x+h}u^{it_{0}}du\cdot\frac{1}{h_{1}}\sum_{\begin{subarray}{c}n\in
S\\\ x<n\leq
x+h_{1}\end{subarray}}f(n)n^{-it_{0}}\right|^{2}dx=o(P^{2}_{f}(X)).$

######  Proposition 2.2 (Halász bound).

Let $X>0$ be a large number. Let $f:\mathbb{N}\to\mathbb{C}$ be a
$d_{k}$-bounded $(\alpha,X^{\theta})$-non-vanishing multiplicative function
for some $\alpha,\theta\in(0,1]$. One has, for all $h\in[X/(\log
X)^{\sigma},x]$ and $x\in[X,2X]$

$\left|\frac{1}{h}\sum_{x<n\leq x+h}f(n)n^{-it_{0}}-\frac{1}{X}\sum_{X<n\leq
2X}f(n)n^{-it_{0}}\right|\ll\frac{1}{(\log
X)^{\min\\{1,\rho_{k,\alpha}\\}/100}}P_{f}(X)$

and

$\left|\frac{1}{h}\sum_{x<n\leq
x+h}f(n)-\frac{1}{h}\int_{x}^{x+h}u^{it_{0}}du\frac{1}{X}\sum_{X<n\leq
2X}f(n)n^{-it_{0}}\right|\ll\frac{1}{(\log
X)^{\min\\{1,\rho_{k,\alpha}\\}/100}}P_{f}(X).$

In the end of this section, we will prove Theorem 1.3 assuming Propositions
2.1 and 2.2.

###### Proof of Theorem 1.3.

If $h\geq h_{1}=X/(\log X)^{\sigma}$, then Theorem 1.3 follows from the second
estimate in Proposition 2.2. Now we assume that $H(f,X,\epsilon)\leq h<h_{1}$.
By the first estimate in Proposition 2.2, we have, for every $x\in[X,2X]$.

$\displaystyle\left|\frac{1}{h}\sum_{x<n\leq
x+h}f(n)-\frac{1}{h}\int_{x}^{x+h}u^{it_{0}}du\cdot\frac{1}{X}\sum_{X<n\leq
2X}f(n)n^{-it_{0}}\right|$ $\displaystyle=$
$\displaystyle\left|\frac{1}{h}\sum_{x<n\leq
x+h}f(n)-\frac{1}{h}\int_{x}^{x+h}u^{it_{0}}du\cdot\frac{1}{h_{1}}\sum_{X<n\leq
X+h}f(n)n^{-it_{0}}\right|+o(P_{f}(X)).$

Letting $S$ be as in Proposition 2.1 and applying the triangle inequaltiy,
this is

$\displaystyle\leq$
$\displaystyle\left|\frac{1}{h}\sum_{\begin{subarray}{c}x<n\leq x+h\\\ n\in
S\end{subarray}}f(n)-\frac{1}{h}\int_{x}^{x+h}u^{it_{0}}du\cdot\frac{1}{h_{1}}\sum_{\begin{subarray}{c}x<n\leq
x+h_{1}\\\ n\in S\end{subarray}}f(n)n^{-it_{0}}\right|$ $\displaystyle+$
$\displaystyle\frac{1}{h}\sum_{\begin{subarray}{c}x<n\leq x+h\\\ n\not\in
S\end{subarray}}\left|f(n)\right|+\frac{1}{h_{1}}\sum_{\begin{subarray}{c}x<n\leq
x+h_{1}\\\ n\not\in S\end{subarray}}\left|f(n)\right|+o(P_{f}(X))$
$\displaystyle\leq$
$\displaystyle\left|\frac{1}{h}\sum_{\begin{subarray}{c}x<n\leq x+h\\\ n\in
S\end{subarray}}f(n)-\frac{1}{h}\int_{x}^{x+h}u^{it_{0}}du\cdot\frac{1}{h_{1}}\sum_{\begin{subarray}{c}x<n\leq
x+h_{1}\\\ n\in S\end{subarray}}f(n)n^{-it_{0}}\right|$ $\displaystyle+$
$\displaystyle\frac{1}{h}\sum_{\begin{subarray}{c}x<n\leq x+h\\\ n\not\in
A\end{subarray}}\left|f(n)\right|+\frac{1}{h}\sum_{\begin{subarray}{c}x<n\leq
x+h\\\ n\not\in
B_{\epsilon/2}\end{subarray}}\left|f(n)\right|+\frac{1}{h_{1}}\sum_{\begin{subarray}{c}x<n\leq
x+h_{1}\\\ n\not\in
A\end{subarray}}\left|f(n)\right|+\frac{1}{h_{1}}\sum_{\begin{subarray}{c}x<n\leq
x+h_{1}\\\ n\not\in
B_{\epsilon/2}\end{subarray}}\left|f(n)\right|+o(P_{f}(X)).$

Thus, the claim follows by Proposition 2.1, Lemma 2.2 and Lemma 2.3. ∎

Before proving Theorem 1.2, we introduce some properties about $\omega(n)$
which denotes the number of distinct prime factors of $n$.

######  Lemma 2.4.

Let $k$ be a positive integer and $x>0$ be sufficiently large. We have

$\sum_{n\leq x}k^{\omega(n)}=(c_{k}+o(1))x\log^{k-1}x.$

###### Proof.

See [6, Chapter 13, Exercise 13.3]. Here we give a sketch of the proof by a
standard technique.

By Perron’s formula, for example, see [3, Lemma 1], and
$\omega(n)\ll\frac{\log n}{\log\log n}$, for any $\epsilon,T>0$

$\sum_{n\leq x}k^{\omega(n)}=\int_{1+\epsilon-
iT}^{1+\epsilon+iT}F(s)\frac{x^{s}}{s}ds+O\left(\frac{x^{1+\epsilon}}{T}+x^{\epsilon}\right),$
(2.7)

where

$F(s)=\sum_{n=1}^{\infty}\frac{k^{\omega(n)}}{n^{s}}=\prod_{p}\left(1+\frac{k}{p^{s}}+\frac{k}{p^{2s}}+\cdots\right).$

We rewrite $F(s)=\zeta^{k}(s)H(s)$ and note that $H(s)$ is convergent when
$\Re(s)>\frac{1}{2}$. We shift the integral in (2.7) to $\Re(s)=1-\epsilon$,
choose $T=x^{\frac{1}{100k}}$ and apply the upper bound
$|\zeta(\sigma+it)|\ll(1+|t|)^{\frac{1}{3}(1-\sigma+\frac{1}{1000}\epsilon)}$.
Then the integrals on the other three sides are $o(x)$. Then the main term is
$xP(\log x)$ which comes from the residue of $\frac{F(s)x^{s}}{s}$ at $s=1$,
where $P$ is a polynomial with degree $k-1$. ∎

######  Lemma 2.5.

Let $\epsilon>0$, $k\in\mathbb{N}$ and $k\geq 2$. Then

$\sum_{\begin{subarray}{c}n\leq x\\\ |\omega(n)-k\log\log
x|\geq\epsilon\log\log x\end{subarray}}k^{\Omega(n)}=o(x\log^{k-1}x)$

###### Proof.

For the sum over $n$ with $\omega(n)>(k+\epsilon)\log\log x$, we apply Lemma
2.3 (see (2.6)) with $f(n)=k^{\Omega(n)}$, so we have
$P_{f}(X)\ll\log^{k-1}X$. Then we obtain that

$\sum_{\begin{subarray}{c}n\leq x\\\ \omega(n)\geq(k+\epsilon)\log\log
x\end{subarray}}k^{\Omega(n)}\leq\sum_{\begin{subarray}{c}n\leq x\\\
\Omega(n)\geq(k+\epsilon)\log\log
x\end{subarray}}k^{\Omega(n)}=o(x\log^{k-1}x).$

On the other hand, by Shiu’s bound (Lemma 2.1), for any $\frac{1}{k}<A<1$, we
have

$\displaystyle\sum_{\omega(n)<(k-\epsilon)\log\log
x}k^{\Omega(n)}\leq\sum_{n\leq
x}k^{\Omega(n)}A^{\omega(n)}A^{-(k-\epsilon)\log\log x}\ll x(\log
x)^{kA-1-(k-\epsilon)\log A}.$

We choose $A=\frac{k-\epsilon}{k}$ to make $f(A):=kA-1-(k-\epsilon)\log A$
reach the minimum, so the claim follows from $f(A)<f(1)=k-1$. ∎

Combining Lemma 2.4 with Lemma 2.5, one can quickly obtain the following
lemma.

######  Lemma 2.6.

For any sufficiently large $x>0$ and small $\epsilon^{\prime}>0$, we have

$\frac{x}{(\log x)^{(k+\epsilon^{\prime})\log
k-k+1}}\ll\sum_{\begin{subarray}{c}n\leq x\\\ |\omega(n)-k\log\log
x|\leq\epsilon^{\prime}\log\log x\end{subarray}}1\ll\frac{x}{(\log
x)^{(k-\epsilon^{\prime})\log k-k+1}}.$

Now we can prove Theorem 1.2.

###### Proof of Theorem 1.2.

Applying Lemma 2.6 and letting $\epsilon^{\prime}=\frac{\epsilon}{2\log k}$,
we have

$\sum_{X\leq x\leq 2X}\sum_{\begin{subarray}{c}x<n\leq x+h\\\
|\omega(n)-k\log\log X|\leq\epsilon^{\prime}\log\log
X\end{subarray}}1=O\left(h\frac{X}{(\log X)^{k\log k-k+1-\epsilon^{\prime}\log
k}}\right)=o(X),$

which means that for $(1-o(1))X$ intervals $(x,x+h]$ with $x\in[X,2X]$, the
inner sum vanishes.

On the other hand, by Lemma 2.5,

$\sum_{\begin{subarray}{c}x<n\leq x+h\\\ |\omega(n)-k\log\log
X|\geq\epsilon^{\prime}\log\log X\end{subarray}}d_{k}(n)=o(h(\log X)^{k-1}),$

for all but $o(x)$ values $x\in[X,2X]$. The claim follows by combining these
two facts. ∎

##  3\. Some Halász type results

In this section, we will first prove some Halász type results for
$d_{k}$-bounded multiplicative functions. After that we will prove the almost
real-valued cases of Proposition 2.2 in the end of this section.

######  Lemma 3.1.

Let $f:\mathbb{N}\to\mathbb{C}$ be a $d_{k}$-bounded and
$(\alpha,X^{\theta})$-non-vanishing multiplicative function for some
$\alpha,\theta\in(0,1]$. Let $t_{0}$ be as in Definition 1.1, then for any
$|t|\leq X$, and any $0<\rho<\rho_{k,\alpha}$,

$\sum_{p\leq X}\frac{|f(p)|-\Re{f(p)p^{-it}}}{p}\geq\rho\min\\{\log\log
X,3\log(|t-t_{0}|\log X+1)\\}+O_{\rho,\theta,k}(1)$

###### Proof.

Apply [9, Lemma 5.1] with $f/k$ in place of $f$.(Note that [9, Lemma 5.1] does
not use multiplicativity.) ∎

######  Lemma 3.2.

Let $\alpha,\theta\in(0,1]$ and $f:\mathbb{N}\to\mathbb{C}$ be a
$d_{k}$-bounded $(\alpha,X^{\theta})$-non-vanishing multiplicative function.
Let $2\leq P\leq Q\leq\exp\left(\frac{\log X}{\log\log X}\right)$ and
$t_{0},\rho$ be as in Lemma 3.1. Then for any $1\leq Z\leq\log X$,

$\sup_{Z<|t|\leq X}\left|\sum_{\begin{subarray}{c}X<n\leq 2X\\\ p\mid
n\Rightarrow
p\not\in[P,Q]\end{subarray}}\frac{f(n)}{n^{1+it_{0}+it}}\right|\ll_{\theta,\rho,k}\left(\frac{1}{Z^{1/2}}+\left(\frac{\log
Q}{\log P}\right)^{2k}\frac{\log\log X}{(\log X)^{\rho}}\right)P_{f}(X).$

for almost all $x\in[X,2X]$.

###### Proof.

By partial summation, it suffices to show the same upper bound for

$\sup_{Z<|t|\leq X}\frac{1}{X}\left|\sum_{\begin{subarray}{c}X<n\leq y\\\
p\mid n\Rightarrow p\not\in[P,Q]\end{subarray}}f(n)n^{-(it_{0}+it)}\right|,$

for every $y\in[X,2X]$. Our claim follows by applying [7, Lemma 3.3] and
noting that $\rho$ is corresponding to $\sigma$ in [7, Lemma 3.3]. ∎

######  Lemma 3.3.

Let $2\leq P\leq Q=\exp(\log X/(100k^{2}\log\log X))$. Let
$f:\mathbb{N}\to\mathbb{C}$ be a $d_{k}$-bounded $(\alpha,X^{\theta})$-non-
vanishing multiplicative function for some $\alpha,\theta\in(0,1]$,
$t_{0},\rho$ be as in Lemma 3.1 and

$R(s)=\sum_{X<n\leq 2X}\frac{f(n)}{n^{s}}\frac{1}{\\#\\{p\in[P,Q]:p\mid
n\\}+1}.$

Then for any $0<\sigma_{0}<1$, we have that for any $t\in[(\log
X)^{\sigma_{0}},X]$,

$\displaystyle|R(1+it_{0}+it)|$ $\displaystyle\ll\left(\frac{\log Q}{\log
P}\right)^{k}\left(\frac{1}{(\log X)^{\sigma_{0}/2}}+\left(\frac{\log Q}{\log
P}\right)^{2k}\frac{\log\log X}{(\log X)^{\rho}}\right)P_{f}(X).$

###### Proof.

Setting $n=n_{1}n_{2}$, where $n_{1}$ has all prime factors from $[P,Q]$ and
$n_{2}$ has none, we get

$\displaystyle|R(1+it_{0}+it)|$
$\displaystyle=\left|\sum_{\begin{subarray}{c}n_{1}\leq X^{3/4}\\\ p\mid
n_{1}\Rightarrow
p\in[P,Q]\end{subarray}}\frac{f(n_{1})}{n^{1+it_{0}+it}(\omega(n_{1})+1)}\sum_{\begin{subarray}{c}X/n_{1}<n_{2}\leq
2X/n_{1}\\\ p\mid n_{2}\Rightarrow
p\not\in[P,Q]\end{subarray}}\frac{f(n_{2})}{n_{2}^{1+it_{0}+it}}\right|$
$\displaystyle+O\left(\sum_{\begin{subarray}{c}n_{2}\leq X^{1/2}\\\ p\mid
n_{2}\Rightarrow
p\not\in[P,Q]\end{subarray}}\frac{|f(n_{2})|}{n_{2}}\sum_{\begin{subarray}{c}X/n_{2}<n_{1}\leq
2X/n_{2}\\\ p\mid n_{2}\Rightarrow
p\in[P,Q]\end{subarray}}\frac{|f(n_{1})|}{n_{1}}\right)$ $\displaystyle\ll$
$\displaystyle\sum_{\begin{subarray}{c}n_{1}\leq X^{3/4}\\\ p\mid
n_{1}\Rightarrow
p\in[P,Q]\end{subarray}}\frac{d_{k}(n_{1})}{n_{1}}\left|\sum_{\begin{subarray}{c}X/n_{1}<n_{2}\leq
2X/n_{1}\\\ p\mid n_{2}\Rightarrow
p\not\in[P,Q]\end{subarray}}\frac{f(n_{2})}{n_{2}^{1+it_{0}+it}}\right|$
$\displaystyle+$ $\displaystyle\sum_{n_{2}\leq
X^{1/2}}\frac{d_{k}(n_{2})}{n_{2}}\sum_{\begin{subarray}{c}X/n_{2}<n_{1}\leq
2X/n_{2}\\\ p\mid n_{1}\Rightarrow
p\in[P,Q]\end{subarray}}\frac{d_{k}(n_{1})}{n_{1}}$

By applying the Cauchy-Schwarz inequality and an estimate for the number of
$Q$-smooth numbers, the inner sum of the second term is

$\displaystyle\leq$ $\displaystyle\left(\sum_{X/n_{2}\leq n_{1}\leq
2X/n_{2}}\frac{|d_{k}(n_{1})|^{2}}{n^{2}_{1}}\right)^{1/2}\left(\sum_{\begin{subarray}{c}X/n_{2}\leq
n_{1}\leq 2X/n_{2}\\\ p\mid n_{1}\Rightarrow
p\in[P,Q]\end{subarray}}1\right)^{1/2}$ $\displaystyle\ll$ $\displaystyle(\log
X)^{\frac{k^{2}-1}{2}}\exp\left(-\frac{\log X}{3\log Q}\log\frac{\log X}{\log
Q}\right)$

Therefore, the second term is at most

$(\log^{k}X)(\log X)^{\frac{k^{2}-1}{2}}\exp\left(-\frac{\log X}{3\log
Q}\log\frac{\log X}{\log Q}\right)\ll\frac{1}{\log^{C}X},$

for any $C>0$.

In the first term, we apply Lemma 3.2 to the sum over $n_{2}$ and then bound
the sum over $n_{1}$ by $\prod_{p\in[P,Q]}(1-k/p)^{-1}\ll(\frac{\log Q}{\log
P})^{k}$, so the first term is at most

$\left(\frac{\log Q}{\log P}\right)^{k}\left(\frac{1}{(\log
X)^{\sigma_{0}/2}}+\left(\frac{\log Q}{\log P}\right)^{2k}\frac{\log\log
X}{(\log X)^{\rho}}\right)P_{f}(X).$

Hence,

$\displaystyle|R(1+it_{0}+it)|$ $\displaystyle\ll\left(\frac{\log Q}{\log
P}\right)^{k}\left(\frac{1}{(\log X)^{\sigma_{0}/2}}+\left(\frac{\log Q}{\log
P}\right)^{2k}\frac{\log\log X}{(\log X)^{\rho}}\right)P_{f}(X).$

∎

Before proving Proposition 2.2, we will do some preparation. Let
$s\in\mathbb{C}$ and $f$ be a $d_{k}$-bounded multiplicative function. Let

$F(s):=\sum_{n=1}^{\infty}\frac{f(n)}{n^{s}}$

and

$F(s;X):=\prod_{p\leq
X}\left(1+\frac{f(p)}{p^{s}}+\frac{f(p^{2})}{p^{2s}}+\cdots\right).$

If $\Re(s)>1/2$, then we have

$F(s;X)\ll\exp\left(\Re\sum_{p\leq X}\frac{f(p)}{p^{s}}\right).$ (3.1)

######  Lemma 3.4.

Let $f$ be a $d_{k}$-bounded multiplicative function and $F(s)$ and $F(s;X)$
be as above. Then $|F(1+it;X)|\asymp|F(1+1/\log X+it)|$.

###### Proof.

It suffices to show that $\Re\log F(1+it;X)-\Re\log F(1+1/\log X+it)=O(1)$. By
Taylor expansion, we have

$\displaystyle\Re\log F(1+it;X)-\Re\log F(1+1/\log X+it)$
$\displaystyle=\Re\sum_{p\leq
X}\frac{f(p)}{p^{1+it}}-\Re\sum_{p=1}^{\infty}\frac{f(p)}{p^{1+1/\log
X+it}}+O\left(\sum_{p}\frac{1}{p^{2}}\right)$
$\displaystyle\ll\left|\Re\sum_{p\leq X}\frac{(p^{1/\log
X}-1)f(p)}{p^{1+1/\log X+it}}\right|+\sum_{p>X}\frac{|f(p)|}{p^{1+1/\log
X}}+O(1)$ $\displaystyle\ll\frac{1}{\log X}\sum_{p\leq X}\frac{|f(p)|\log
p}{p}+O(1)=O(1).$

∎

######  Lemma 3.5.

Suppose that $\gamma>\frac{1}{\log X}$. Let $f$ be $d_{k}$-bounded
$(\alpha,X^{\theta})$-non-vanishing multiplicative function for some
$\alpha,\theta\in(0,1]$. Let $F(s)$ be as above and $t\in\mathbb{R}$. Then

$F(1+1/\log X+\gamma+it)\ll P_{f}(X)\log X\frac{1}{(\gamma\log X)^{\alpha}}.$

###### Proof.

By Lemma 3.4, we know that

$F(1+1/\log X+\gamma+it)\ll|F(1+\gamma+it;X)|\ll\exp\left({\sum_{p\leq
X}\frac{|f(p)|}{p^{1+\gamma}}}\right).$

By the prime number theorem the contribution of $p\in[e^{1/\gamma},X]$ to the
sum on the right hand side is $O(1)$. Hence, using that and $f$ being
$(\alpha,X^{\theta})$-non-vanishing

$\displaystyle\sum_{p\leq X}\frac{|f(p)|}{p^{1+\gamma}}=\left(\sum_{p\leq
e^{1/\gamma}}\frac{|f(p)|}{p^{1+\gamma}}-\sum_{p\leq
e^{1/\gamma}}\frac{|f(p)|}{p}\right)-\sum_{e^{1/\gamma}<p\leq
X}\frac{|f(p)|}{p}+\sum_{p\leq X}\frac{|f(p)|}{p}+O(1)$ $\displaystyle\leq$
$\displaystyle{\sum_{p\leq X}\frac{|f(p)|}{p}}-\alpha{\sum_{e^{1/\gamma}<p\leq
X}\frac{1}{p}}+O(1).$

Thus, the result follows by noticing that

$\exp\left({\sum_{p\leq X}\frac{|f(p)|}{p}}\right)\ll P_{f}(X)\log X$

and Mertens’ theorem. ∎

Now we are able to prove Proposition 2.2 and the proof essentially follows the
argument of [1, Section 4: Proof of Theorem 1.5].

###### Proof of Proposition 2.2.

If $f$ is not almost real-valued, Proposition 2.2 follows from [7, Theorem
2.2]. Now we deal with the real-valued case, where $t_{0}=0$.

It suffices to show that for all $1\leq w\leq\exp{(\log X)^{2/3}}$ we have

$\left|\frac{1}{X}\sum_{n\leq X}f(n)-\frac{1}{X/w}\sum_{n\leq
X/w}f(n)\right|\ll P_{f}(X)\left(\frac{\log w}{\log
X}\right)^{\min\\{\frac{1}{3},\frac{1}{3}\rho_{k,\alpha}\\}}\log\log X.$ (3.2)

Let $c_{0}=1+1/\log X$, $T=\log^{k+1}X$, $y=\max\\{ew,T^{2}\\}$ and
$\eta=\frac{1}{\log y}$ in [1, Proposition 2.1]. Then by the argument of [1,
Section 4], up to an error term $O((\log y)^{k}/(\log X))$ which is
acceptable, we are able to bound the left-hand-side of (3.2) by

$\ll\int_{0}^{\eta}\int_{0}^{\eta}X^{-\alpha}\min\left\\{\log
X,\frac{1}{\beta}\right\\}\max_{|t|\leq
T}\frac{|F(c_{0}+\beta+it)(1-w^{\alpha+\beta+1-c_{0}-it})|}{|c_{0}+\beta+it|}d\alpha
d\beta,$

where the inner term of the double integral comes from [1, (4.3)]. Hence, the
double integral is

$\displaystyle\ll$
$\displaystyle\int_{0}^{\eta}\int_{0}^{\eta}X^{-\alpha}\min\left\\{\log
X,\frac{1}{\beta}\right\\}\max_{|t|\leq
T}\frac{|F(c_{0}+\beta+it)|(|1-w^{-\beta-it}|+|w^{-\beta-
it}-w^{\alpha+\beta+1-c_{0}-it}|)}{|c_{0}+\beta+it|}d\alpha d\beta$
$\displaystyle\ll$
$\displaystyle\int_{0}^{\eta}\int_{0}^{\eta}X^{-\alpha}\min\left\\{\log
X,\frac{1}{\beta}\right\\}\max_{|t|\leq
T}\frac{|F(c_{0}+\beta+it)||1-w^{-\beta-it}|}{|c_{0}+\beta+it|}d\alpha d\beta$
$\displaystyle+$ $\displaystyle\log(ew)\int_{0}^{\eta}\min\left\\{\log
X,\frac{1}{\beta}\right\\}\max_{|t|\leq
T}\frac{|F(c_{0}+\beta+it)|}{|c_{0}+\beta+it|}\int_{0}^{\eta}X^{-\alpha}\left(\alpha+\beta+\frac{1}{\log
X}\right)d\alpha d\beta$ $\displaystyle\ll$ $\displaystyle\frac{1}{\log
X}\int_{0}^{\eta}\min\left\\{\log X,\frac{1}{\beta}\right\\}\max_{|t|\leq
T}\frac{|F(c_{0}+\beta+it)||1-w^{-\beta-it}|}{|c_{0}+\beta+it|}d\beta$
$\displaystyle+$ $\displaystyle\frac{\log(ew)}{\log
X}\int_{0}^{\eta}\min\left\\{\log X,\frac{1}{\beta}\right\\}\max_{|t|\leq
T}\frac{|F(c_{0}+\beta+it)|}{|c_{0}+\beta+it|}\left(\beta+\frac{1}{\log
X}\right)d\beta$ $\displaystyle\ll$ $\displaystyle\frac{1}{\log
X}\int_{0}^{\eta}\min\left\\{\log X,\frac{1}{\beta}\right\\}\max_{|t|\leq
T}\frac{|F(c_{0}+\beta+it)||1-w^{-\beta-
it}|}{|c_{0}+\beta+it|}d\beta+\frac{\log(ew)}{\log
X}\int_{0}^{\eta}\max_{|t|\leq
T}\frac{|F(c_{0}+\beta+it)|}{|c_{0}+\beta+it|}d\beta$ $\displaystyle=:$
$\displaystyle T_{1}+T_{2}$

Let us estimate $T_{2}$ first. By Lemma 3.4, (3.1) and Lemma 3.5, we obtain
that

$\displaystyle T_{2}$ $\displaystyle\ll\frac{\log(ew)}{\log
X}\int_{0}^{\eta}\max_{|t|\leq
T}\frac{|F(c_{0}+\beta+it)|}{|c_{0}+\beta+it|}d\beta$
$\displaystyle\ll\frac{\log ew}{\log X}\left(\int_{0}^{\frac{1}{\log
X}}\max_{|t|\leq T}|F(c_{0}+\beta+it)|d\beta+\int_{\frac{1}{\log
X}}^{\eta}\max_{|t|\leq T}|F(c_{0}+\beta+it)|d\beta\right)$ $\displaystyle\ll
P_{f}(X)\left(\frac{\log(ew)}{\log X}+\frac{1}{(\log
X)^{\alpha}}(\eta^{1-\alpha}{\bf 1}_{\alpha<1}+\log\log X{\bf
1}_{\alpha=1}+(\log X)^{\alpha-1}{\bf 1}_{\alpha>1}\right)$ $\displaystyle\ll
P_{f}(X)\left(\frac{\log(ew)}{\log X}\right)^{\min\\{1,\alpha\\}}\log\log X$
(3.3)

which is smaller than the right-hand-side of (3.2).

For estimating $T_{1}$, we need the following variant of the maximum modulus
principle which is given by Granville and Soundarajan [2, Lemma 2.2].

######  Lemma 3.6.

Let $a_{n}$ be a sequence of complex number such that
$A(s)=\sum_{n=1}^{\infty}a_{n}n^{-s}$ is absolutely convergent for $\Re(s)\geq
1$. For all real number $T\geq 1$, all $0\leq\gamma\leq 1$ and any $w_{1}\geq
1$ we have

$\max_{|t_{1}|\leq T}|A(1+\gamma+it_{1})(1-w_{1}^{-\gamma-
it_{1}})|\leq\max_{|t_{1}|\leq
2T}|A(1+it_{1})(1-w_{1}^{-it_{1}})|+O\left(\frac{\gamma}{T}\sum_{n=1}^{\infty}\frac{|a_{n}|}{n}\right).$

Recall that $c_{0}=1+1/\log X$. By Lemma 3.4, (3.1) and Lemma 3.1, we have,
for any $|t|\leq T$ and $0<\rho<\rho_{k,\alpha}$

$|F(c_{0}+it)|\ll\log
XP_{f}(X)\left(\max\left\\{\frac{1}{\log^{\rho}X},\frac{1}{(1+|t|\log
X)^{3\rho}}\right\\}\right).$ (3.4)

Hence from Lemma 3.6, with $F(c_{0}+it)$ in place of $A(1+it)$ and $w$ in
place of $w_{1}$, and (3.4) we get that

$\displaystyle\max_{|t|\leq T}\frac{|F(c_{0}+\beta+it)||1-w^{-\beta-
it}|}{|c_{0}+\beta+it|}\ll\max_{|t|\leq 2T}|F(c_{0}+it)(1-w^{-it})|+\eta$
$\displaystyle\ll$ $\displaystyle P_{f}(X)\log X\max_{|t|\leq
2T}|1-w^{-it}|\left(\max\left\\{\frac{1}{\log^{\rho}X},\frac{1}{(1+|t|\log
X)^{3\rho}}\right\\}\right)+\eta$ $\displaystyle\ll$ $\displaystyle
P_{f}(X)\log X\max_{|t|\leq
2T}\left(\max\left\\{\frac{1}{\log^{\rho}X},\frac{1}{(1+|t|\log
X)^{3\rho}}\right\\}\min\\{1,|t|\log w\\}\right)+\eta$ $\displaystyle\ll$
$\displaystyle P_{f}(X)\log X\left(\frac{\log w}{\log
X}\right)^{\min\\{\frac{1}{3},\rho\\}}+\eta.$

Letting $\rho=\frac{1}{3}\rho_{k,\alpha}$ and combining the above with (3.3),
the result follows. ∎

##  4\. Proof of Proposition 2.1

In this section, we will prove Proposition 2.1, which is similar to the
argument in [8, Section 8] with slight modifications.

Before proving Proposition 2.1, Let us first collect some standard mean and
large value results for Dirichlet polynomials.

######  Lemma 4.1.

Let $T\geq 1$ and $N\geq 2$. Let $\\{a_{n}\\}_{n\leq N}$ be a sequence of
complex numbers. Then the following bound holds

$\displaystyle\int_{-T}^{T}|\sum_{n\leq N}a_{n}n^{it}|^{2}dt$
$\displaystyle\ll T\sum_{n\leq N}|a_{n}|^{2}+T\sum_{n\leq N}\sum_{1\leq|m|\leq
n/T}|a_{n}a_{n+m}|$ $\displaystyle\ll T\sum_{n\leq
N}|a_{n}|^{2}+T\sum_{1\leq|m|\leq N/T}\sum_{|n|\leq N}|a_{n}a_{n+m}|.$

###### Proof.

See [9, Lemma 3.1]. ∎

######  Lemma 4.2.

(Henriot) Let $\theta\in(0,1]$ and Let $f:\mathbb{N}\to\mathbb{C}$ be a
multiplicative function with $|f(n)|\ll_{\epsilon}n^{\epsilon}$ for any
$\epsilon>0$. Let $1\leq r_{1},r_{2}\leq x^{3\theta/7}$ be integers. Assume
that $f(pm)=f(p)f(m)$ whenever $p\mid r_{1}r_{2}$. Then, for all $x\geq y\geq
x^{\theta}$ and $K\in[1,x]$, one has

$\sum_{\begin{subarray}{c}0\neq|k|\leq K\\\ (r_{1},r_{2})\mid
k\end{subarray}}\sum_{\begin{subarray}{c}x<n\leq x+y\\\ r_{1}\mid n,r_{2}\mid
n+k\end{subarray}}|f(n)f(n+k)|\ll_{\theta}K\frac{|f(r_{1})f(r_{2})|}{r_{1}r_{2}}y\prod_{p\leq
x}\left(1+\frac{2|f(p)|-2}{p}\right)\prod_{\begin{subarray}{c}p\mid
r_{1}r_{2}\\\ p>K\end{subarray}}\left(1+\frac{1-|f(p)|}{p}\right).$

###### Proof.

See [9, Lemma 3.3]. ∎

The next lemma is a direct corollary of the above lemmas, and will be used
frequently later.

######  Lemma 4.3.

Let $X\geq Y>0$. Let $f:\mathbb{N}\to\mathbb{C}$ be a $d_{k}$-bounded
$(\alpha,X^{\theta})$-non-vanishing multiplicative function for some
$\alpha,\theta\in(0,1]$. Suppose that $a_{n}$ is a complex number for all
$n\in[X,2X]$ with $|a_{n}|\leq|f(n)|$. Then

$\int_{-T}^{T}\left|\sum_{X\leq n\leq
X+Y}\frac{a_{n}}{n^{1+it}}\right|^{2}\ll\frac{T}{X^{2}}\sum_{X<n\leq
X+Y}|a_{n}|^{2}+\frac{Y}{X}P_{f}(X)^{2}.$ (4.1)

In particular, if $B^{\prime}_{\epsilon}\subset B_{\epsilon}$, then

$\int_{-T}^{T}\left|\sum_{X\leq n\leq X+Y}\frac{f(n){\bf
1}_{B^{\prime}_{\epsilon}}(n)}{n^{1+it}}\right|^{2}\ll\left(\frac{TY}{X^{2}}H(f,X,\epsilon)+\frac{Y}{X}\right)P_{f}(X)^{2}=:\mathfrak{S}(T,X,Y,f,\epsilon).$
(4.2)

###### Proof.

To prove the first statement, we apply Lemma 4.1 and use Lemma 4.2 with
$r_{1}=r_{2}=1$ to bound the off-diagonal term.

For proving (4.2), let $a_{n}=f(n){\bf 1}_{B^{\prime}_{\epsilon}}(n)$ in
(4.1). For all $1\leq n\leq X+Y$,

$|a_{n}|\leq k^{(1+\epsilon)\log(P_{f}(X)\log X)}=(P_{f}(X)\log
X)^{(1+\epsilon)\log k}.$

Hence using Shiu’s bound (Lemma 2.1)

$\sum_{n\leq X+Y}|a_{n}|^{2}\leq\max_{X<n\leq X+Y}|a_{n}|\sum_{X<n\leq
X+Y}|f(n)|\ll Y(P_{f}(X)\log X)^{(1+\epsilon)\log k}P_{f}(X).$

Recall that

$H(f,X,\epsilon):=\frac{(P_{f}(X)\log X)^{(1+\epsilon)\log k}}{P_{f}(X)},$

so the lemma follows. ∎

For convenience, we define

$\mathfrak{S}(T,X,f,\epsilon):=\left(\frac{T}{X}H(f,X,\epsilon)+1\right)P_{f}(X)^{2}.$

For the rest of the paper, we say that $\mathcal{T}\subset\mathbb{R}$ is well
spaced if $|t-r|\geq 1$ for all $t,r\in\mathcal{T}$.

######  Lemma 4.4.

Suppose that $\\{a_{n}\\}_{n\leq N}$ is a sequence of complex numbers and
$\mathcal{T}\subset[-T,T]$ is well spaced. Then

$\sum_{t\in\mathcal{T}}\left|\sum_{X<n\leq
2X}\frac{a_{n}}{n^{1+it}}\right|^{2}\ll\min\left\\{\left(1+\frac{T}{X}\right)\log
X,\left(1+|\mathcal{T}|\frac{T^{1/2}}{X}\right)\log
T\right\\}\frac{1}{X}\sum_{X<n\leq 2X}|a_{n}|^{2}.$

###### Proof.

See [5, Theorems 9.4 and 9.6]. ∎

######  Lemma 4.5.

Let $k,T\geq 1$, $P\geq 10$. For all prime $p\in[P,2P]$, let $a_{p}$ be a
complex number with $|a_{p}|\leq k$. Let

$P(s)=\sum_{P<p\leq 2P}\frac{a_{p}}{p^{s}}$

for $s\in\mathbb{C}$. Then for any well spaced $\mathcal{T}\subset[-T,T]$, one
has

  1. (i)

$\sum_{t\in\mathcal{T}}|P(1+it)|^{2}\ll_{k}\frac{1}{\log
P}\left(1+|\mathcal{T}|\exp\left(-\frac{\log P}{(\log
T)^{2/3+\epsilon}}\right)(\log T)^{2}\right).$

  2. (ii)

Let $V\geq 1$. If $\mathcal{T}$ consists only of $t\in[-T,T]$ with
$|P(1+it)|\geq V^{-1}$, then

$|\mathcal{T}|\ll T^{2\frac{\log V}{\log P}}V^{2}\exp\left(2k\frac{\log
T}{\log P}\log\log T\right).$

###### Proof.

The first statement is [8, Lemma 11]. The second one follows from the proof of
[8, Lemma 8] with an extra coefficient $|a_{p}|^{2\lceil\frac{\log T}{\log
P}\rceil}$ which is bouned by $k^{2\lceil\frac{\log T}{\log
P}\rceil}\ll\exp(2\log k\frac{\log T}{\log P})$ ∎

The following lemma is a variant of [8, Lemma 12].

######  Lemma 4.6.

Let $f:\mathbb{N}\to\mathbb{C}$ be a $d_{k}$-bounded and
$(\alpha,X^{\theta})$-non-vanishing multiplicative function for some
$\alpha,\theta\in(0,1]$. Let $H\geq 1$ and $Q\geq P\geq 1$. Let
$B_{\epsilon^{\prime}}$ be as in Lemma 2.3, $a_{m},b_{m}$ and $c_{p}$ be
complex sequences such that $|a_{m}|,|b_{m}|\leq|f(m){\bf
1}_{B_{\epsilon^{\prime}}}(m)|$ and $|c_{p}|\leq|f(p)|$ with
$a_{mp}=b_{m}c_{p}$ whenever $p\nmid m$, and $P\leq p\leq Q$. For $v\in[H\log
P,H\log Q]$, let

$Q_{v,H}=\sum_{\begin{subarray}{c}P\leq p\leq Q\\\ e^{v/H}<p\leq
e^{(v+1)/H}\end{subarray}}\frac{c_{p}}{p^{s}}$

and

$R_{v,H}(s)=\sum_{Xe^{-v/H}<m\leq
2Xe^{-v/H}}\frac{b_{m}}{m^{s}}\frac{1}{\\#\\{P\leq q\leq Q:q\mid
m,q\in\mathbb{P}\\}+1}$

and let $\mathcal{T}\subset[-T,T]\subset[-X,X]$. Then for any $\epsilon>0$

$\displaystyle\int_{\mathcal{T}}\left|\sum_{n\sim
X}\frac{a_{n}}{n^{1+it}}\right|^{2}dt\ll
H\left(\log\left(\frac{Q}{P}\right)+1\right)\times\sum_{j\in\mathcal{I}}\int_{\mathcal{T}}|Q_{j,H}(1+it)R_{j,H}(1+it)|^{2}dt$
$\displaystyle+\left(\frac{1}{H}+\frac{1}{P}+\left(\frac{\log P}{\log
Q}\right)^{k\alpha}\right)\mathfrak{S}(T,X,f,\epsilon^{\prime})$

where $\mathcal{I}$ is the interval $\lfloor H\log P\rfloor\leq j\leq H\log
Q$.

###### Proof.

Since the upper bound in the above lemma looks slightly different from the
upper bounds in [8, Lemma 12], we will give a brief proof which is essentially
same as the proof of [9, Lemma 6.1].

Let $s=1+it$, $\mathcal{P}=\prod_{P\leq p\leq Q}$ and
$\omega(n;P,Q)=\\#\\{P\leq p\leq Q:p\mid n\\}$. By Ramaré’s identity,

$\displaystyle\sum_{X<n\leq 2X}\frac{a_{n}}{n^{s}}$
$\displaystyle=\sum_{v\in\mathcal{I}}Q_{v,H}(s)R_{v,H}(s)+\sum_{X{e^{-1/H}}<m\leq
Xe^{1/H}}\frac{d_{m}}{m^{s}}+\sum_{2X<m\leq 2Xe^{1/H}}\frac{d_{m}}{m^{s}}$
$\displaystyle+\sum_{\begin{subarray}{c}Xe^{-1/H}<mp\leq 2Xe^{1/H}\\\
p\in[P,Q],(m,p)=p\end{subarray}}\frac{e_{m,p}}{(mp)^{s}}+\sum_{\begin{subarray}{c}X<n\leq
2X\\\ (n,\mathcal{P})=1\end{subarray}}\frac{a_{n}}{n^{s}}$
$\displaystyle=:B_{1}(s)+B_{2}(s)+B_{3}(s)+B_{4}(s)+B_{5}(s)$

where

$|d_{m}|\leq|f(m)|{\bf
1}_{B_{\epsilon^{\prime}}}(m)\quad\text{and}\quad|e_{m,p}|\ll|f(mp)|{\bf
1}_{B_{\epsilon^{\prime}}}(mp)+|f(p)f(m)|{\bf 1}_{B_{\epsilon^{\prime}}}(m).$

First, we notice that by the Cauchy-Schwarz inequality

$\int_{\mathcal{T}}|B_{1}(s)|^{2}dt\ll
H\left(\log\left(\frac{Q}{P}\right)+1\right)\times\sum_{v\in\mathcal{I}}\int_{\mathcal{T}}\left|Q_{v,H}(1+it)R_{v,H}(1+it)\right|^{2}dt.$

By Lemma 4.3, it is not difficult to show that

$\int_{\mathcal{T}}|B_{2}(s)|^{2}ds\ll\frac{1}{H}\mathfrak{S}(T,X,f,\epsilon^{\prime})\quad\text{and}\quad\int_{\mathcal{T}}|B_{2}(s)|^{2}ds\ll\frac{1}{H}\mathfrak{S}(T,X,f,\epsilon^{\prime}).$

Now we focus on the integral of $|B_{4}(s)|^{2}$. By Lemma 4.1,

$\displaystyle\int_{\mathcal{T}}|B_{4}(s)|^{2}dt$ $\displaystyle\ll
T\sum_{X/3<n\leq
6X}\frac{1}{n^{2}}\sum_{\begin{subarray}{c}p_{1},p_{2}\in[P,Q]\\\
m_{1},m_{2}\\\
n=p_{1}^{2}m_{1}=p_{2}^{2}m_{2}\end{subarray}}|e_{m_{1}p_{1},p_{1}}e_{m_{2}p_{2},p_{2}}|+T\sum_{0<|k|\leq
6X/T}\sum_{X/3<n\leq 6X}\sum_{\begin{subarray}{c}p_{1},p_{2}\in[P,Q]\\\
m_{1},m_{2}\\\ p_{1}^{2}m_{1}=n\\\
p_{2}^{2}m_{2}=n+k\end{subarray}}\frac{|e_{m_{1}p_{1},p_{1}}e_{m_{2}p_{2},p_{2}}|}{n(n+k)}$

Defining now, for each pair $(p_{1},p_{2})$ and $k\geq 0$, the multiplicative
function $g_{p_{1},p_{2}}$ such that $g_{p_{1},p_{2}}(p^{k})=|f(p^{k})|$ when
$p\not\in\\{p_{1},p_{2}\\}$ and
$g_{p_{1},p_{2}}(p^{k})=\max\\{1,|f(p^{k+1})|,|f(p^{k+2})|\\}$ when
$p\in\\{p_{1},p_{2}\\}$, we obtain the previous expression is

$\displaystyle\ll_{k}$
$\displaystyle\frac{T}{X^{2}}\sum_{p_{1},p_{2}\in[P,Q]}\sum_{\frac{X}{3[p_{1},p_{2}]^{2}}<n\leq\frac{6X}{[p_{1},p_{2}]^{2}}}g_{p_{1},p_{2}}(n)^{2}{\bf
1}_{B_{\epsilon^{\prime}}(n)}$
$\displaystyle+\frac{T}{X^{2}}\sum_{p_{1},p_{2}\in[P,Q]}\sum_{\begin{subarray}{c}0<|k|\leq
6X/T\\\ (p_{1}^{2},p_{2}^{2})\mid
k\end{subarray}}\sum_{\begin{subarray}{c}X/3<n\leq 6X\\\ p_{1}^{2}\mid
n,p_{2}^{2}\mid n+k\end{subarray}}g_{p_{1},p_{2}}(n)g_{p_{1},p_{2}}(n+k).$

Applying now, for each $p_{1},p_{2}$, Lemmas 2.1 and 4.2, we obtain

$\displaystyle\int_{-T}^{T}|B_{4}(s)|^{2}dt$
$\displaystyle\ll\frac{T}{X}\max_{\begin{subarray}{c}X/3<n\leq 6X\\\
p_{3},p_{4}\in[P,Q]\end{subarray}}|g_{p_{3},p_{4}}(n){\bf
1}_{B_{\epsilon^{\prime}}(n)}|\sum_{p_{1},p_{2}\in[P,Q]}\frac{1}{[p_{1},p_{2}]^{2}}\prod_{p\leq
X}\left(1+\frac{|f(p)|-1}{p}\right)$
$\displaystyle+\sum_{p_{1},p_{2}\in[P,Q]}\frac{1}{p_{1}^{2}p_{2}^{2}}\prod_{p\leq
X}\left(1+\frac{2|f(p)|-2}{p}\right)$
$\displaystyle\ll\frac{1}{P}\mathfrak{S}(T,X,f,\epsilon^{\prime}).$

For the case $B_{5}(s)$, we apply Lemma 4.3 with $f=f{\bf
1}_{(n,\mathcal{P})=1}$, which is also multiplicative and Lemma 2.1 and recall
that $f$ is $(\alpha,X^{\theta})$-non-vanishing,

$\displaystyle\int_{\mathcal{T}}|B_{5}(s)|^{2}dt$
$\displaystyle\ll\frac{T}{X^{2}}\sum_{\begin{subarray}{c}X<n\leq 2X\\\
(n,\mathcal{P})=1\end{subarray}}|f{\bf 1}_{B_{\epsilon^{\prime}}}{\bf
1}_{(n,\mathcal{P})=1}(n)|^{2}+(P_{f{\bf 1}_{(n,\mathcal{P})=1}}(X))^{2}$
$\displaystyle\ll\frac{T}{X}H(f,X,\epsilon^{\prime})P_{f}(X)P_{f{\bf
1}_{(n,\mathcal{P})=1}}(X)+(P_{f{\bf 1}_{(n,\mathcal{P})=1}}(X))^{2}$
$\displaystyle\ll\frac{T}{X}H(f,X,\epsilon^{\prime})P_{f}(X)\prod_{p\leq
X}\left(1+\frac{|f(p){\bf 1}_{p\in\mathcal{P}}|-1}{p}\right)+\prod_{p\leq
X}\left(1+\frac{|f(p){\bf 1}_{p\in\mathcal{P}}|-1}{p}\right)^{2}$
$\displaystyle\ll\frac{T}{X}H(f,X,\epsilon^{\prime})P_{f}(X)^{2}\prod_{p\in[P,Q]}\left(1-\frac{|f(p)|}{p}\right)+P_{f}(X)^{2}\prod_{p\in[P,Q]}\left(1-\frac{|f(p)|}{p}\right)^{2}$
$\displaystyle\ll\frac{T}{X}H(f,X,\epsilon^{\prime})\left(\frac{\log P}{\log
Q}\right)^{k\alpha}P_{f}(X)^{2}+\left(\frac{\log P}{\log
Q}\right)^{2k\alpha}P_{f}(X)^{2}$ $\displaystyle\ll\left(\frac{\log P}{\log
Q}\right)^{k\alpha}\mathfrak{S}(T,X,f,\epsilon^{\prime})$

∎

######  Lemma 4.7.

Let $Y_{1},Y_{2}\geq 1$ and consider

$Q(s)=\sum_{p\sim Y_{1}}\frac{c_{p}}{p^{s}}\quad\text{and}\quad
A(s)=\sum_{m\sim X/Y_{2}}\frac{a_{m}}{m^{s}},$

with coefficients $|a_{m}|\leq|f(m){\bf
1}_{B_{\epsilon^{\prime}}}|,|c_{p}|\leq k$. Let $l=\left\lceil\frac{\log
Y_{2}}{\log Y_{1}}\right\rceil$. Then

$\int_{-T}^{T}|Q(1+it)^{l}\cdot
A(1+it)|^{2}\ll\mathfrak{S}(T,X,f,\epsilon^{\prime})k^{2l}(l+1)!^{2}$

###### Proof.

Replace $c_{p}$ by $c_{p}/k$ and $g^{*}(p^{a})=f(p^{a})$ if $p\not\in[Y,2Y]$
in [9, Lemma 7.1], then applying Lemma 4.1, [9, Lemma 3.3] and the proof of
the second statement of Lemma 4.3, the result follows. ∎

Let $S_{f,h}(x)=\sum_{x<m\leq x+h}f(m)$ and $A_{f}(s)=\sum_{x<m\leq
2x}\frac{f(m)}{m^{s}}$. Using Perron’s formula, if $x,x+h\not\in\mathbb{Z}$,
one can rewrite $S_{f,h}(x)$ by a complex integral,

$S_{f,h}(x)=\frac{1}{2\pi
i}\int_{1-i\infty}^{1+i\infty}A_{f}(s)\frac{(x+h)^{s}-x^{s}}{s}ds.$

Recall that $t_{0}$ denotes a point in $[-X,X]$ such that

$\sum_{p\leq X}\frac{|f(p)|-\Re f(p)p^{-it}}{p}$

attains the minimum.

######  Proposition 4.3.

Let $1/\log X\leq\delta<1$ and $|t_{1}|\leq X$ and put
$I_{t_{1},\delta}=[t_{1}-\delta^{-1},t_{1}+\delta^{-1}]$. Let
$\\{a_{n}\\}_{n\leq X}$ be a sequence of complex numbers,

$S_{y}(x)=\sum_{x<n\leq x+y}a_{n}\quad\text{and}\quad A(s)=\sum_{X\leq n\leq
4X}\frac{a_{n}}{n^{s}}.$

Then

$\displaystyle\frac{1}{X}\int_{X}^{2X}\left|\frac{1}{y}S_{y}(x)-\frac{1}{y}\frac{1}{2\pi
i}\int_{I_{t_{1},\delta}}A(1+it)\frac{(x+h)^{1+it}-x^{1+it}}{1+it}dt\right|^{2}dx$
$\displaystyle\ll$ $\displaystyle\int_{[-X/y,X/y]\setminus
I_{t_{1},\delta}}|A(1+it)|^{2}dt+\max_{T\geq
X/y}\frac{X}{yT}\int_{T}^{2T}|A(1+it)|^{2}dt.$ (4.3)

In particular, let $f:\mathbb{N}\to\mathbb{C}$ be a $d_{k}$-bounded
$(\alpha,X^{\theta})$-non-vanishing multiplicative function for some
$\alpha,\theta\in(0,1]$ and $S$ be as in Proposition 2.1. Let

$F(s)=\sum_{\begin{subarray}{c}X\leq n\leq 4X\\\ n\in
S\end{subarray}}\frac{f(n)}{n^{s}}.$

We have

$\displaystyle\frac{1}{X}\int_{X}^{2X}\left|\frac{1}{h}\sum_{\begin{subarray}{c}x<n\leq
x+h\\\ n\in S\end{subarray}}f(n)-\frac{1}{h}\frac{1}{2\pi
i}\int_{-\delta^{-1}}^{\delta^{-1}}F(1+it_{0}+it)\frac{(x+h)^{1+it_{0}+it}-x^{1+it_{0}+it}}{1+it_{0}+it}dt\right|^{2}dx$
$\displaystyle\ll$ $\displaystyle\int_{[-X/h,X/h]\setminus
I_{t_{0},\delta}}|F(1+it)|^{2}dt+\max_{T\geq
X/h}\frac{X}{hT}\int_{T}^{2T}|F(1+it)|^{2}dt.$ (4.4)

Recall $h_{1}=X/(\log X)^{\sigma}$. We have

$\displaystyle\frac{1}{X}\int_{X}^{2X}\left|\frac{1}{h_{1}}\sum_{\begin{subarray}{c}x<n\leq
x+h_{1}\\\ n\in S\end{subarray}}f(n)n^{-it_{0}}-\frac{1}{h_{1}}\frac{1}{2\pi
i}\int_{-\delta^{-1}}^{\delta^{-1}}F(1+it_{0}+it)\frac{(x+h)^{1+it}-x^{1+it}}{1+it}dt\right|^{2}dx.$
$\displaystyle\ll$
$\displaystyle\int_{|t|\in[\delta^{-1},X/h_{1}]}|F(1+it_{0}+it)|^{2}dt+\max_{T\geq
X/h_{1}}\frac{X}{h_{1}T}\int_{T}^{2T}|F(1+it_{0}+it)|^{2}dt.$ (4.5)

###### Proof.

(4.3) follows from the proof of [8, Lemma 14]. Let $a_{n}=f(n){\bf 1}_{S}$,
$y=h$ and $t_{1}=t_{0}$, then (4.4) follows. (4.5) follows by letting
$a_{n}=f(n)n^{-it_{0}}{\bf 1}_{S}$, $y=h_{1}$ and $t_{1}=0$. ∎

Our main tasks is to show that (4.4) and (4.5) are negligible.

######  Lemma 4.8.

Let $x\asymp X$ be sufficiently large and $0<h<h_{1}$ be as in Proposition
2.1. If $\delta^{-1}=o((X/h_{1})^{1/2})$, then

$\displaystyle\left|\int_{-\delta^{-1}}^{\delta^{-1}}F(1+it_{0}+it)\left(\frac{(x+h)^{1+it_{0}+it}-x^{1+it_{0}+it}}{1+it_{0}+it}-\int_{x}^{x+h}u^{it_{0}}du\frac{(x+h_{1})^{1+it}-x^{1+it}}{h_{1}(1+it)}\right)dt\right|$
$\displaystyle=o(hP_{f}(X)).$

###### Proof.

Firstly, by Shiu’s bound (Lemma 2.1), we have the pointwise upper bound

$\max_{|t|\leq\delta^{-1}}|F(1+it_{0}+it)|\ll\frac{1}{X}\sum_{X\leq n\leq
4X}|f(n)|\ll P_{f}(X).$

Secondly, we see that, for $|t|\leq\delta^{-1}$,

$\displaystyle\frac{(x+h)^{1+it_{0}+it}-x^{1+it_{0}+it}}{1+it_{0}+it}-\int_{x}^{x+h}u^{it_{0}}du\frac{(x+h_{1})^{1+it}-x^{1+it}}{h_{1}(1+it)}$
$\displaystyle=\int_{x}^{x+h}u^{it_{0}}\left(u^{it}-\frac{1}{h_{1}}\int_{x}^{x+h_{1}}v^{it}dv\right)du$
$\displaystyle\leq h\max_{\begin{subarray}{c}u\in[x,x+h]\\\
|t|\leq\delta^{-1}\end{subarray}}\left\\{u^{it}-\frac{1}{h_{1}}\int_{x}^{x+h_{1}}v^{it}dv\right\\}.$

By the Taylor expansion we see that for $X\ll u,v\ll X$ with $|u-v|\leq
h_{1}$,

$|v^{it}-u^{it}|=|e^{it\log
u}(1-e^{it\log(v/u)}|\ll|t|\log\left(\frac{X+h_{1}}{X}\right)=O(\delta^{-1}h_{1}/X).$

This is $o(\delta)$ since $\delta^{-1}=o((X/h_{1})^{1/2})$. The claim follows
by combining everything. ∎

From Proposition 4.3, we know that if (4.4) and (4.5) are negligible, then
Proposition 2.1 follows.

Also, by Lemma 4.3 with $f(n)=f{\bf 1}_{S}(n)$, we know that for $\beta>0$ and
$T\geq\frac{X(\log X)^{\beta}}{h}$,

$\frac{X}{h_{1}T}\int_{T}^{2T}|F(1+it_{0}+it)|^{2}dt=o(P_{f}(X)^{2})$

and

$\frac{X}{hT}\int_{T}^{2T}|F(1+it)|^{2}dt=o(P_{f}(X)^{2}).$

Hence, we just need to show that for some $\beta>0$,

$\int_{[-(X/h)(\log X)^{\beta},(X/h)(\log X)^{\beta}]\setminus
I_{t_{0},\delta}}|F(1+it)|^{2}=o(P_{f}(X)^{2})$ (4.6)

and

$\int_{\delta^{-1}}^{(X/h)(\log
X)^{\beta}}|F(1+it+t_{0})|^{2}=o(P_{f}(X)^{2}).$ (4.7)

We will show that (4.6) is true by the following proposition and one can
follow the same argument to show that (4.7) also holds. The proof of
Proposition 4.4 follows the proof of [8, Proposition 1].

######  Proposition 4.4.

Let $\beta=\frac{\alpha\epsilon}{100}$, $\delta=(\log X)^{-\sigma/3}$,
$f,\epsilon,\alpha$ be as in Proposition 2.1 and $F(1+it)$ be as in
Proposition 4.3. Suppose that $H(f,X,\epsilon)\leq h\leq X$, then we have

$\int_{\delta^{-1}}^{(X/h)(\log
X)^{\beta}}|F(1+it_{0}+it)|^{2}=o((P_{f}(X))^{2}).$

###### Proof.

Let $P_{1},Q_{1}$ be as in (2.1) with
$\epsilon_{0}=\frac{\alpha\epsilon}{3}\log k$ and let $P_{2},Q_{2}$ be as in
(2.2). Notice that

$Q_{1}=(\log X)^{\frac{\alpha\epsilon}{3}\log
k}\leq\frac{H(f,X,\epsilon)}{H(f,X,\epsilon/2)}$

Let $\alpha_{1}=\frac{1}{4}-\frac{1}{50}$,
$\alpha_{2}=\frac{1}{4}-\frac{1}{100}$ and $H=P_{1}^{1/6}$. For any
$s\in\mathbb{C}$ and $v\in\mathbb{N}$, let

$Q_{v,j}(s):=\sum_{\begin{subarray}{c}P_{j}\leq p\leq Q_{j}\\\ e^{v/H}\leq
p\leq e^{(v+1)/H}\end{subarray}}\frac{f(p)}{p^{s}},\quad
R_{v,j}(s):=\sum_{\begin{subarray}{c}Xe^{-v/H}\leq m\leq 2Xe^{-v/H}\\\ m\in
S_{j}\end{subarray}}\frac{f(m)}{m^{s}(1+\omega_{[P_{j},Q_{j}]}(m))},$

where $S_{j}$ is the set of those integers which have at most
$(1+\epsilon/2)\log(P_{f}(x)\log X)-1$ prime factors and have at least one
prime factor in every interval $[P_{i},Q_{i}]$ with
$i\in\\{1,2\\}\setminus\\{j\\}$. Note that $Q_{v,j}$ is non-zero only when

$v\in\mathcal{I}_{j}:=\\{v:\lfloor H\log P_{j}\rfloor\leq v\leq H\log
Q_{j}\\}.$

We split $[\delta^{-1},(X/h)(\log X)^{\beta}]=:[T_{0},T]$ into a disjoint
union

$\mathcal{T}_{1}\cup\mathcal{T}_{2}\cup\mathcal{U},$

where

$\displaystyle\mathcal{T}_{1}:=\\{t\in[T_{0},T]:|Q_{v,1}(1+it_{0}+it)|\leq
e^{-\alpha_{1}v/H}\text{ for all }v\in\mathcal{I}_{1}\\}$
$\displaystyle\mathcal{T}_{2}:=\\{t\in[T_{0},T]:|Q_{v,2}(1+it_{0}+it)|\leq
e^{-\alpha_{2}v/H}\text{ for all
}v\in\mathcal{I}_{2}\\}\setminus\mathcal{T}_{1}$
$\displaystyle\mathcal{U}:=[T_{0},T_{1}]\setminus(\mathcal{T}_{1}\cup\mathcal{T}_{2}).$

As arguments in [8, section 8], using Lemma 4.6 with $H=H$, $P=P_{j}$,
$Q=Q_{j}$ and $a_{m}=f(m){\bf 1}_{S}(m)$, $b_{m}=f(m){\bf 1}_{S_{j}}(m)$,
$c_{p}=f(p)$ we have

$\displaystyle\int_{\mathcal{T}_{j}}|F(1+it_{0}+it)|^{2}$ $\displaystyle\ll
H\log\left(\frac{Q_{j}}{P_{j}}\right)\times\sum_{v\in\mathcal{I}_{j}}\int_{\mathcal{T}_{j}}|Q_{v,j}(1+it_{0}+it)R_{v,j}(1+it_{0}+it)|^{2}dt$
$\displaystyle+\left(\frac{1}{H}+\frac{1}{P_{j}}+\left(\frac{\log P_{j}}{\log
Q_{j}}\right)^{k\alpha}\right)\mathfrak{S}(T,X,f,\epsilon/2).$

By our choice of $H,P_{j},Q_{j}$, we can see that the second term contributes
in total to integrals over all $\mathcal{T}_{j}$,

$\ll\sum_{j=1}^{2}\left(\frac{1}{H}+\frac{1}{P_{j}}+\left(\frac{\log
P_{j}}{\log
Q_{j}}\right)^{k\alpha}\right)\mathfrak{S}(X/h,X,f,\epsilon/2)=o((P_{f}(X))^{2}).$

Thus

$\int_{T_{0}}^{T}|F(1+it_{0}+it)|dt\ll
E_{1}+E_{2}+\int_{\mathcal{U}}|F(1+it_{0}+it)|dt+o(P_{f}(X))^{2}).$

where

$E_{j}:=H\log
Q_{j}\cdot\sum_{v\in\mathcal{I}_{j}}\int_{\mathcal{T}_{j}}|Q_{v,j}(1+it_{0}+it)R_{v,j}(1+it_{0}+it)|^{2}dt.$

(1) Bounding $E_{1}$. By the definition of $\mathcal{T}_{1}$ and the mean-
value lemma (Lemma 4.3), we obtain that

$\displaystyle E_{1}$ $\displaystyle\ll H\log
Q_{1}\sum_{v\in\mathcal{I}_{1}}e^{-2\alpha_{1}v/H}\left(\frac{T}{Xe^{-v/H}}H(f,X,\epsilon/2)+1\right)P_{f}(X)^{2}$
$\displaystyle\ll H\log
Q_{1}\times|\mathcal{I}_{1}|\times\max_{v\in\mathcal{I}_{1}}\left\\{e^{-2\alpha_{1}v/H}\left(\frac{e^{v/H}}{H(f,X,\epsilon)}H(f,X,\epsilon/2)+1\right)\right\\}P_{f}(X)^{2}$
$\displaystyle\ll H^{2}(\log
Q_{1})^{2}P_{1}^{-2\alpha_{1}}\left(\frac{Q_{1}}{H(f,X,\epsilon)}H(f,X,\epsilon/2)+1\right)P_{f}(X)^{2}$
$\displaystyle\ll\frac{1}{P_{1}^{1/100}}P_{f}(X)^{2}=o(P_{f}(X)^{2}).$

(2) Bounding $E_{2}$. In this case, we split further

$\mathcal{T}_{2}=\bigcup_{r\in\mathcal{I}_{1}}\mathcal{T}_{2,r},$

where

$\mathcal{T}_{2,r}=\\{t\in\mathcal{T}_{1}:|Q_{r,1}(1+it_{0}+it)|>e^{-\alpha_{1}r/H}\\}.$

Therefore, for some $v\in\mathcal{I}_{2}$ and $r\in\mathcal{I}_{1}$,

$E_{2}\ll H\log Q_{2}\cdot\\#\mathcal{I}_{2}\cdot\\#\mathcal{I}_{1}\cdot
e^{-2\alpha_{2}v/H}\times\int_{\mathcal{T}_{2,r}}|R_{v,2}(1+it_{0}+it)|^{2}dt.$

On $\mathcal{T}_{2,r}$, we have $|Q_{r,{1}}(1+it_{0}+it)|>e^{-\alpha_{1}r/H}$.
Therefore, for any $l\geq 1$, multiplying the right hand side by the term
$(|Q_{r,1}(1+it_{0}+it)|e^{\alpha_{1}r/H})^{2l}\geq 1$, we obtain

$\displaystyle E_{2}$ $\displaystyle\ll(H\log Q_{2})^{3}\cdot
e^{-2\alpha_{2}v/H}\times\exp(2l\cdot\alpha_{1}r/H)$
$\displaystyle\times\int_{\mathcal{T}_{2,r}}|Q_{r,1}(1+it_{0}+it)^{l}R_{v,2}(1+it_{0}+it)|^{2}dt.$

Choosing

$l=l_{v,r}=\left\lceil\frac{v/H}{r/H}\right\rceil\leq\frac{v}{r}+1,$

we get

$\displaystyle E_{2}$ $\displaystyle\ll(H\log
Q_{2})^{3}\cdot\exp(-v/(50H)+(1/2-1/25)r/H)$
$\displaystyle\times\int_{\mathcal{T}_{2,r}}|Q_{r,1}(1+it_{0}+it)^{l}R_{v,2}(1+it_{0}+it)|^{2}dt.$

Now, we use Lemma 4.7 which gives

$\displaystyle\int_{-T}^{T}|Q_{r,1}(1+it_{0}+it)^{l}R_{v,2}(1+it)|^{2}dt$
$\displaystyle\ll$
$\displaystyle(\frac{T}{X}H(f,X,\epsilon/2)+1)P_{f}(X)^{2}k^{2l}(l+1)!^{2}$
$\displaystyle\ll$
$\displaystyle\left(\frac{H(f,X,\epsilon/2)}{H(f,X,\epsilon)}+1\right)P_{f}(X)^{2}\exp(2l\log(kl)).$

By the definition of $l$,

$\displaystyle l\log(kl)$
$\displaystyle\leq\left(\frac{v}{r}+1\right)\log(k(v/r+1))\leq\frac{2v}{H}\frac{\log\log
Q_{2}}{\log P_{1}}\leq\frac{2v}{H}\frac{100\log\log\log X}{(\log\log
X)^{1/2}}.$

Recalling the definition of $P_{j}$ and $Q_{j}$, we obtain that

$\displaystyle E_{2}$ $\displaystyle\ll
P_{f}(X)^{2}\exp\left(-\frac{v}{51H}+\frac{r}{2H}+3\log(H\log Q_{2})\right)$
$\displaystyle\ll P_{f}(X)^{2}\exp\left(-\frac{1}{51}\log
P_{2}+\frac{1}{2}\log Q_{1}+3\log(P_{1}^{1/6}\log Q_{2})\right)$
$\displaystyle\ll P_{f}(X)^{2}\exp\left(-\frac{1}{51}(\log\log
X)^{2}+\frac{1}{2}\log\log X+3\log\log X\right)$
$\displaystyle=o(P_{f}(X)^{2}).$

(3).Bounding. $\int_{\mathcal{U}}|F(1+it_{0}+it)|^{2}dt$. Let $P=\exp((\log
X)^{1-\epsilon_{1}})$ for some small $\epsilon_{1}$ to be chosen later,
$Q=\exp{(\log X/100k^{2}\log\log X)}$ and $H_{0}:=\log Q/\log P$. By Lemma 4.6
the integral is bounded by

$\displaystyle\int_{\mathcal{U}}|F(1+it_{0}+it)|^{2}dt$ $\displaystyle\ll
H_{0}^{2}(\log
X)^{2}\int_{\mathcal{U}}|Q_{v,H_{0}}(1+it_{0}+it)R_{v,H_{0}}(1+it_{0}+it)|^{2}dt$
$\displaystyle+\left(\frac{1}{H_{0}}+\frac{1}{P}+\left(\frac{\log P}{\log
Q}\right)^{k\alpha}\right)\mathfrak{S}(X/h,X,f,\epsilon/2),$

where

$Q_{v,H_{0}}(s):=\sum_{\begin{subarray}{c}P\leq p\leq Q\\\ e^{v/H_{0}}<p\leq
e^{(v+1)/H_{0}}\end{subarray}}\frac{f(p)}{p^{s}},\quad
R_{v,H_{0}}(s):=\sum_{\begin{subarray}{c}Xe^{-v/H_{0}}<m\leq 2Xe^{-v/H_{0}}\\\
m\in S\end{subarray}}\frac{f(m)}{m^{s}(1+\omega_{[P,Q]}(m))},$

and $v\in[\lfloor H_{0}\log P\rfloor,H_{0}\log Q]$. Thus we only need to show
that

$H_{0}^{2}(\log
X)^{2}\int_{\mathcal{U}}|Q_{v,H_{0}}(1+it_{0}+it)R_{v,H}(1+it_{0}+it)|^{2}dt=o((P_{f}(X))^{2}).$

In order to use Halász type results which requires multiplicativity, we will
split the sum over $S=A\cap B_{\epsilon/2}$ as the difference of a sum over
$A$ and a sum over $A\cap\overline{B_{\epsilon/2}}$. Let

$R^{\prime}_{v,H_{0}}(s)=\sum_{\begin{subarray}{c}Xe^{-v/H}\leq m\leq
2Xe^{-v/H}\\\ m\in A\end{subarray}}\frac{f(m)}{m^{s}(1+\omega_{[P,Q]}(m))}.$

and

$R^{\prime\prime}_{v,H_{0}}(s)=\sum_{\begin{subarray}{c}Xe^{-v/H}\leq m\leq
2Xe^{-v/H}\\\ m\in
A\cap\overline{B_{\epsilon/2}}\end{subarray}}\frac{f(m)}{m^{s}(1+\omega_{[P,Q]}(m))}.$

We will prove that both

$\int_{\mathcal{U}}|Q_{v,H_{0}}(1+it_{0}+it)R^{\prime}_{v,H_{0}}(1+it_{0}+it)|^{2}dt\quad\text{and}\quad\int_{\mathcal{U}}|Q_{v,H_{0}}(1+it_{0}+it)R^{\prime\prime}_{v,H_{0}}(1+it_{0}+it)|^{2}dt$

are $o\left(\frac{(P_{f}(X))^{2}}{H_{0}^{2}(\log X)^{2}}\right)$. For the rest
of paper we let $\tilde{R}(s)$ denote $R^{\prime}(s)$ or
$R^{\prime\prime}(s)$. For convenience, we will use a discrete sum to control
the integral over $\mathcal{U}$, i.e. we find a well-spaced set
$\mathcal{T}\subset\mathcal{U}$ such that

$\int_{\mathcal{U}}|Q_{v,H_{0}}(1+it_{0}+it)\tilde{R}_{v,H_{0}}(1+it_{0}+it)|^{2}dt\leq
2\sum_{t\in\mathcal{T}}|Q_{v,H_{0}}(1+it_{0}+it)\tilde{R}_{v,H_{0}}(1+it_{0}+it)|^{2}.$

We first claim that the size of $\mathcal{T}$ is not too large. By the
definition of $\mathcal{U}$, for each $t\in\mathcal{T}$, there is
$v\in\mathcal{I}_{2}$ such that $|Q_{v,H_{0}}(s)|>e^{-\alpha_{2}v/H_{0}}$.
Recall that $P_{2}=\exp((\log\log X)^{2})$. We can apply Lemma 4.5 $(ii)$ to
get that for some small $\eta>0$,

$|\mathcal{T}|\ll X^{1/2-\eta+o(1)}.$

We split $\mathcal{T}$ in order to use a pointwise bound and a large value
result. Let

$\mathcal{T}_{L}=\\{t\in\mathcal{T}:|Q_{v,H_{0}}(1+it_{0}+it)|\geq(\log
X)^{-\frac{1}{2}k^{2}-10}\\}$

and

$\mathcal{T}_{S}=\\{t\in\mathcal{T}:|Q_{v,H_{0}}(1+it_{0}+it)|<(\log
X)^{-\frac{1}{2}k^{2}-10}\\}.$

By Lemma 4.4 and noticing that $P_{f}(X)\ll(\log X)^{k-1}$, we have

$\displaystyle\sum_{t\in\mathcal{T}_{S}}|Q_{v,H_{0}}(1+it_{0}+it)\tilde{R}_{v,H_{0}}(1+it_{0}+it)|^{2}$
$\displaystyle\ll(\log
X)^{-k^{2}-20}\sum_{t\in\mathcal{T}}|\tilde{R}_{v,H_{0}}(1+it_{0}+it)|^{2}$
$\displaystyle\ll(\log X)^{-k^{2}-19}\prod_{p\leq
X}\left(1+\frac{|f(p)|^{2}-1}{p}\right)$ $\displaystyle\ll(\log
X)^{-k^{2}-19}\prod_{p\leq
X}\left(1+\frac{(|f(p)|-1)^{2}}{p}\right)\left(1+\frac{2|f(p)|-2}{p}\right)$
$\displaystyle\ll(\log X)^{-k^{2}-19}(\log X)^{k^{2}-2k+1}P_{f}(X)^{2}\ll(\log
X)^{-19}P_{f}(X).$

Hence, we can focus on the sum over $\mathcal{T}_{L}$. By using Lemma 4.5
$(ii)$ again, we have

$\displaystyle|\mathcal{T}_{L}|$ $\displaystyle\ll\exp\left(2\frac{\log(\log
X)^{\frac{1}{2}k^{2}+10}}{\log P}\log T+2\log(\log
X)^{\frac{1}{2}k^{2}+10}+2k\frac{\log T}{\log P}\log\log T\right)$
$\displaystyle\ll\exp((\log X)^{2\epsilon_{1}}).$

The only difference between $R^{\prime}(s)$ and $R^{\prime\prime}(s)$ is the
following. For $R^{\prime}(s)$, we split $f(n){\bf 1}_{A}(n)$ as the
difference of two multiplicative functions, namely $f(n){\bf
1}_{A}(n)=f(n)(1-{\bf 1}_{(n,[P_{1},Q_{1}])=1})(1-{\bf
1}_{(n,[P_{2},Q_{2}])=1})$, then we apply Lemma 3.3, getting

$\max_{(\log X)^{\sigma/3}\leq|u|\leq
X}|R^{\prime}_{v,H_{0}}(1+it_{0}+it)|\ll\left(\frac{\log Q}{\log
P}\right)^{k}\left(\frac{1}{(\log X)^{\sigma/6}}+\left(\frac{\log Q}{\log
P}\right)^{2k}\frac{\log\log X}{(\log X)^{\rho/3}}\right)P_{f}(X).$

However, for $R^{\prime\prime}(s)$, we apply (2.4) to get that

$|R^{\prime\prime}_{v,H_{0}}(1+it_{0}+it)|\ll\frac{1}{(\log
X)^{k\alpha(-(1+\epsilon/2)\log(1+\delta)+\delta)}}P_{f}(X),$

for some sufficiently small $\delta>0$. Thus by Lemma 4.5 $(i)$ and the above
bound, for any $\epsilon_{1}>0$, we have

$\displaystyle\sum_{t\in\mathcal{T}_{L}}|Q_{v,H_{0}}(1+it_{0}+it)\tilde{R}_{v,H_{0}}(1+it_{0}+it)|^{2}$
$\displaystyle\ll\left(\frac{\log Q}{\log P}\right)^{2k}\left(\frac{1}{(\log
X)^{\sigma/3}}+\left(\frac{\log Q}{\log P}\right)^{4k}\frac{(\log\log
X)^{2}}{(\log X)^{2\rho/3}}+\frac{1}{(\log
X)^{k\alpha(-(2+\epsilon)\log(1+\delta)+\delta)}}\right)$
$\displaystyle\times\left(1+\frac{\exp((\log X)^{2\epsilon_{1}})}{\exp((\log
X)^{1/3-2\epsilon_{1}})}\right)\frac{P_{f}(X)^{2}}{\log P}.$

Combining the above estimates, and choosing $\epsilon_{1}>0$ sufficiently
small depending on $k,\sigma,\rho,\alpha,\epsilon$, we obtain

$H_{0}^{2}(\log
X)^{2}\int_{\mathcal{U}}|Q_{v,H_{0}}(1+it_{0}+it)R_{v,H_{0}}(1+it_{0}+it)|^{2}dt=o((P_{f}(X))^{2}).$

∎

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