# Convolution of periodic multiplicative functions and the divisor problem

Marco Aymone  Departamento de Matemática, Universidade Federal de Minas
Gerais, Av. Antônio Carlos, 6627, CEP 31270-901, Belo Horizonte, MG, Brazil.
<EMAIL_ADDRESS>,  Gopal Maiti  Institut de Mathématiques de Marseille,
Aix Marseille Université, U.M.R. 7373, Campus de Luminy, Case 907, 13288
MARSEILLE Cedex 9, France<EMAIL_ADDRESS>,  Olivier Ramaré  CNRS /
Institut de Mathématiques de Marseille, Aix Marseille Université, U.M.R. 7373,
Campus de Luminy, Case 907, 13288 MARSEILLE Cedex 9, France
<EMAIL_ADDRESS>and  Priyamvad Srivastav<EMAIL_ADDRESS>
###### Abstract.

We study a certain class of arithmetic functions that appeared in Klurman’s
classification of $\pm 1$ multiplicative functions with bounded partial sums,
c.f., Comp. Math. 153 (8), 2017, pp. 1622-1657. These functions are periodic
and $1$-pretentious. We prove that if $f_{1}$ and $f_{2}$ belong to this
class, then $\sum_{n\leq x}(f_{1}\ast f_{2})(n)=\Omega(x^{1/4})$. This
confirms a conjecture by the first author. As a byproduct of our proof, we
studied the correlation between $\Delta(x)$ and $\Delta(\theta x)$, where
$\theta$ is a fixed real number. We prove that there is a non-trivial
correlation when $\theta$ is rational, and a decorrelation when $\theta$ is
irrational. Moreover, if $\theta$ has a finite irrationality measure, then we
can make it quantitative this decorrelation in terms of this measure.

##  1\. Introduction

###  1.1. Main result and background

A question posed by Erdős in [6], known as the Erdős discrepancy problem,
states that whether for all arithmetic functions $f:\mathds{N}\to\\{-1,1\\}$
we have that the discrepancy

(1) $\sup_{x,d}\left|\sum_{n\leq x}f(nd)\right|=\infty.$

When in addition $f$ is assumed to be completely multiplicative, then this
reduces to whether $f$ has unbounded partial sums.

In 2015, Tao [16] proved that (1) holds for all $f:\mathds{N}\to\\{-1,1\\}$,
and a key point of its proof is that it is sufficient to establish (1) only in
the class of completely multiplicative functions $f$ taking values in the unit
(complex) circle.

When $f:\mathds{N}\to\\{-1,1\\}$ is assumed to be only multiplicative, then
not necessarily $f$ has unbounded partial sums. For example, $f(n)=(-1)^{n+1}$
is multiplicative and clearly has bounded partial sums. In this case,
$f(2^{k})=-1$ for all positive integers $k$. It was observed by Coons [5]
that, for bounded partial sums, this rigidity on powers of $2$ is actually
necessary under suitable conditions on the values that a multiplicative
function $f$ takes at the remaining primes. Later, in the same paper [16], Tao
gave a partial classification of multiplicative functions taking values $\pm
1$ with bounded partial sums: They must satisfy the previous rigidity
condition on powers of $2$ and they must be $1$-pretentious (for more on
pretentious Number Theory we refer reader to [7] by Granville and
Soundararajan), that is,

$\sum_{p}\frac{1-f(p)}{p}<\infty.$

Later, Klurman [10] proved that the only multiplicative functions $f$ taking
$\pm 1$ values and with bounded partial sums are the periodic multiplicative
functions with sum $0$ inside each period, and thus, closing this problem for
$\pm 1$ multiplicative functions.

Building upon the referred work of Klurman, the first author proved in [1]
that if we allow values outside the unit disk, a $M$-periodic multiplicative
function $f$ with bounded partial sums such that $f(M)\neq 0$ satisfies

  1. i.

For some prime $q|M$, $\sum_{k=0}^{\infty}\frac{f(q^{k})}{q^{k}}=0$.

  2. ii.

For each $p^{a}\|M$, $f(p^{k})=f(p^{a})$ for all $k\geq a$.

  3. iii.

For each $\gcd(p,M)=1$, $f(p^{k})=1$, for all $k\geq 1$.

Conversely, if $f:\mathds{N}\to\mathds{C}$ is multiplicative and the three
conditions above are satisfied, then $f$ has period $M$ and has bounded
partial sums. Therefore, these three conditions above give examples of
multiplicative functions with values outside the unit disk with bounded
partial sums, despite of the fact that $f(M)$ is zero or not.

######  Remark 1.1.

It is interesting to observe that when it is assumed that $|f|\leq 1$, the
only way to achieve condition i. is with $q=2$ and $f(2^{k})=-1$ for all
$k\geq 1$.

######  Remark 1.2.

What makes the difference between a multiplicative function $f$ satisfying
i-ii-iii from a non-principal Dirichlet character $\chi$ is that $\chi$
neither satisfies i. nor iii.

Here we are interested in the convolution $f_{1}\ast
f_{2}(n):=\sum_{d|n}f_{1}(d)f_{2}(n/d)$ for $f_{1}$ and $f_{2}$ satisfying
i-ii-iii above. It was proved in [1] that

$\sum_{n\leq x}(f_{1}\ast f_{2})(n)\ll x^{\alpha+\epsilon},$

where $\alpha$ is the infimum over the exponents $a>0$ such that $\Delta(x)\ll
x^{a}$, where $\Delta(x)$ is the classical error term in the Dirichlet divisor
problem defined by:

$\sum_{n\leq x}\tau(n)=x\log x+(2\gamma-1)x+\Delta(x).$

It is widely believed that $\alpha=1/4$ and many results were proven in this
direction. The best upper bound up to date is due to Huxley [8]: $\alpha\leq
131/416\approx 0.315$. Regarding $\Omega$ bounds, Soundararajan [15] proved
that

$\Delta(x)=\Omega\left((x\log x)^{1/4}\frac{(\log\log
x)^{3/4(2^{4/3}-1)}}{(\log\log\log x)^{5/8}}\right).$

It was conjectured in [1] that the partial sums of $f_{1}\ast f_{2}$ obey a
similar $\Omega$-bound for $\Delta(x)$, that is, $\sum_{n\leq x}(f_{1}\ast
f_{2})(n)=\Omega(x^{1/4})$. Here we establish this conjecture.

######  Theorem 1.1.

Let $f_{1}$ and $f_{2}$ be periodic multiplicative functions satisfying i-ii-
iii above. Then $\sum_{n\leq x}(f_{1}\ast f_{2})(n)=\Omega(x^{1/4})$.

######  Example 1.1.

The results from [1] give that for each prime $q$ there exists a unique
$q$-periodic multiplicative function $f$ with bounded partial sums and such
that $f(q)\neq 0$. In the case $q=2$, the corresponding function is
$f(n)=(-1)^{n+1}$. Therefore, in this particular case we have that
$\sum_{n\leq x}(f\ast f)(n)=\Omega(x^{1/4})$. In particular, this establishes
the conjecture in an uncovered case by Proposition 3.1 of [1].

Our proof relies on two ingredients. The second one is a study of a family of
quadratic forms, and is explained in Section 5. The first ingredient is a
generalization of a result of Tong [17] and prove the next theorem.

######  Theorem 1.2.

When $a$ and $b$ are non negative integers, $\lambda=\gcd(a,b)$, $c=a/\lambda$
and $d=b/\lambda$, we have

$\lim_{X\to\infty}\frac{1}{X^{3/2}}\int_{1}^{X}\Delta(x/a)\Delta(x/b)dx=\frac{\tau(cd)}{6\pi^{2}\sqrt{\lambda}cd}\frac{\zeta(3/2)^{4}}{\zeta(3)}\prod_{p^{k}\|cd}\frac{1-\frac{k-1}{(k+1)p^{3/2}}}{1+1/p^{3/2}}.$

Furthermore, when $\theta>0$ is irrational, we have

$\lim_{X\to\infty}\frac{1}{X^{3/2}}\int_{1}^{X}\Delta(x)\Delta(\theta x)dx=0.$

###  1.2. The proof in the large

To prove Theorem 1.1, our starting point is the following formula from [1]: If
$M_{1}$ and $M_{2}$ are the periods of $f_{1}$ and $f_{2}$ respectively, then

(2) $\sum_{n\leq x}(f_{1}\ast f_{2})(n)=\sum_{n|M_{1}M_{2}}(f_{1}\ast
f_{2}\ast\mu\ast\mu)(n)\Delta(x/n),$

where $\mu$ is the Möbius function. Therefore, the partial sums of $f_{1}\ast
f_{2}$ can be written as a finite linear combination of the quantities
$(\Delta(x/n))_{n}$. Apart from the fact that $\Delta(x)=\Omega(x^{1/4})$, we
cannot, at least by a direct argument, prevent a conspiracy among the large
values of $(\Delta(x/n))_{n}$ that would yield a cancellation among a linear
combination of them.

To circumvent this, our approach is inspired by an elegant result of Tong
[17]:

$\int_{1}^{X}\Delta(x)^{2}dx=\frac{(1+o(1))}{6\pi^{2}}\sum_{n=1}^{\infty}\frac{\tau(n)^{2}}{n^{3/2}}\,X^{3/2}.$

By (2), the limit

$\lim_{X\to\infty}\frac{1}{X^{3/2}}\int_{1}^{X}\left|\sum_{n\leq x}(f_{1}\ast
f_{2})(n)\right|^{2}dx$

can be expressed as a quadratic form with matrix $(c_{a,b})_{a,b|M_{1}M_{2}}$
where $c_{a,b}$ is the correlation

$c_{a,b}:=\lim_{X\to\infty}\frac{1}{X^{3/2}}\int_{1}^{X}\Delta(x/a)\Delta(x/b)dx.$

As it turns out, these correlations does not vanish and computed in Theorem
1.2. With that in hand, the matrix correlation-term $c_{a,b}$ can be expressed
as

(3)
$\frac{C}{\sqrt{\gcd(a,b)}}\varphi\left(\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\right),$

for some constant $C>0$ and multiplicative function $\varphi$.

This matrix entanglement is hard to analyze directly. In Section 5 we explore
sufficient conditions for a matrix of the form (3) to be positive definite.
When this happens, this ensures the referred $\Omega$-bound. Thanks to the
Selberg diagonalization process, we show that when $\varphi$ is completely
multiplicative and satisfies other conditions, then this matrix is positive
definite. The main proof somehow reduces to this case; we indeed find a way to
conjugate our original matrix to reach a matrix related to a completely
multiplicative function. With standard linear algebra of Hermitian matrices we
conclude that our matrix $(c_{a,b})_{a,b|M_{1}M_{2}}$ is positive definite. We
ended up with the following result.

######  Theorem 1.3.

Let $f_{1}$ and $f_{2}$ be two periodic multiplicative functions satisfying
i-ii-iii above with periods $M_{1}$ and $M_{2}$ respectively. Let $g=f_{1}\ast
f_{2}\ast\mu\ast\mu$. Then the following limit

$\lim_{X\to\infty}\frac{1}{X^{3/2}}\int_{1}^{X}\left|\sum_{n\leq x}(f_{1}\ast
f_{2})(n)\right|^{2}dx$

is positive and equals to

$\frac{\zeta(3/2)^{4}}{\zeta(3)}\sum_{n,m|M_{1}M_{2}}g(n)\overline{g(m)}\frac{\gcd(n,m)^{3/2}}{nm}\tau\left(\frac{nm}{\gcd(n,m)^{2}}\right)\prod_{p^{k}\|nm/\gcd(n,m)^{2}}\frac{1-\tfrac{(k-1)}{(k+1)}p^{-3/2}}{1+p^{-3/2}}.$

###  1.3. Byproduct study

Motivated by Nyman’s reformulation of the Riemann hypothesis [14], in recent
papers [2, 3, 4] by Balazard, Duarte and Martin, the correlation

$A(\theta):=\int_{0}^{\infty}\\{x\\}\\{\theta x\\}\frac{dx}{x^{2}}$

has been thoroughly studied. Here $\theta>0$ is any real number and $\\{x\\}$
stands for the fractional part of $x$. Several analytic properties for the
function $A(\theta)$ have been shown.

Motivated by this, we studied the “divisor” analogue

$I(\theta)=\lim_{X\to\infty}\frac{1}{X^{3/2}}\int_{1}^{X}\Delta(x)\Delta(\theta
x)dx.$

As stated in Theorem 1.2, when $\theta=p/q$ is a rational number, the limit
above is described by a positive multiplicative function depending on $p$ and
$q$. However and somewhat surprisingly, when $\theta$ is irrational, this
correlation vanishes. The next proposition establishes that this vanishing is
indeed very strong, except maybe at points $\theta$ that are well approximated
by rationals.

######  Proposition 1.1.

Let $\theta>0$ be an irrational number with irrationality measure $\eta$, that
is, for each $\epsilon>0$ there is a constant $C>0$ such that the inequality

$|n-m\theta|\geq\frac{C}{m^{\eta+\epsilon}}$

is violated only for a finite number of positive integers $n$ and $m$. Then,
for every positive $\epsilon$, we have

$\int_{1}^{X}\Delta(x)\Delta(\theta x)dx=O(X^{3/2-1/(18\eta)+\epsilon}).$

In the other cases of irrationals $\theta$, the integral above is
$o(X^{3/2})$.

This shows that we have decorrelation among the values $\Delta(x)$ and
$\Delta(\theta x)$ when $\theta$ is irrational, and moreover, this gives that
the function $I(\theta)$ is continuous at the irrational numbers and
discontinuous at the rationals.

We mention that a similar decorrelation also has been obtained by Ivić and
Zhai in [9]. In this paper they show decorrelation between $\Delta(x)$ and
$\Delta_{k}(x)$, where $\Delta_{k}(x)$ is the error term related to the
$k$-fold divisor function, and $k=3$ or $4$.

##  2\. Notation

###  2.1. Asymptotic notation

We employ both Vinogradov’s notation $f\ll g$ or $f=O(g)$ whenever there
exists a constant $C>0$ such that $|f(x)|\leq C|g(x)|$, for all $x$ in a set
of parameters. When not specified, this set of parameters is $x\in(a,\infty)$
for sufficiently large $a>0$. We employ $f=o(g)$ when $\lim_{x\to
a}\frac{f(x)}{g(x)}=0$. In this case $a$ can be a complex number or
$\pm\infty$. Finally, $f=\Omega(g)$ when $\limsup_{x\to
a}\frac{|f(x)|}{g(x)}>0$, where $a$ is as in the previous notation.

###  2.2. Number-theoretic notation

Here $p$ stands for a generic prime number. We sometimes denote the least
common multiple between $a,b$ as $\operatorname{lcm}(a,b)$. The greatest
common divisor is denoted by $\gcd(a,b)$. The symbol $\ast$ stands for
Dirichlet convolution between two arithmetic functions: $(f\ast
g)(n)=\sum_{d|n}f(d)g(n/d)$.

##  3\. Multiplicative auxiliaries

Our first task is to evaluate $\sum_{n\geq 1}{\tau(cn)\tau(dn)}/{n^{3/2}}$ for
coprime positive integers $c$ and $d$.

######  Lemma 3.1.

Let $c$ be fixed positive number and $f(n)$ be a multiplicative function with
$f(c)\neq 0$. Then $n\mapsto\frac{f(cn)}{f(c)}$ is multiplicative.

###### Proof.

For positive integers $u,v$, we have

$f(u)f(v)=f(\gcd(u,v))f(\operatorname{lcm}(u,v)).$

Let $u=cn$, $v=cm$ with $\gcd(n,m)=1$. Then $f(cn)f(cm)=f(c)f(cnm)$.
Therefore, we obtain

$\frac{f(cm)}{f(c)}\frac{f(cn)}{f(c)}=\frac{f(cnm)}{f(c)}.$

∎

######  Lemma 3.2.

Let $c,d$ be two fixed positive integers with $\gcd(c,d)=1$. Then

$\displaystyle\sum_{n=1}^{\infty}\frac{\tau(cn)\tau(dn)}{n^{s}}=\tau(cd)\frac{\zeta(s)^{4}}{\zeta(2s)}\prod_{p^{k}\|cd}\left(1+p^{-s}\right)^{-1}\left(1-\tfrac{(k-1)}{(k+1)}p^{-s}\right).$

The quantity we compute appears in several places, for instance in [12] by Lee
and Lee of in [13, Theorem 2.4] by Munsch and Shparlinski.

###### Proof.

Note that $\frac{\tau(cn)}{\tau(c)}$ is a multiplicative function in the
variable $n$, and so is $\frac{\tau(cn)\tau(dn)}{\tau(c)\tau(d)}$. Therefore,
for $\Re(s)>1$ we have the following Euler factorization

$\displaystyle\sum_{n=1}^{\infty}\frac{\tau(cn)\tau(dn)}{\tau(c)\tau(d)n^{s}}=\prod_{p\nmid
cd}\left(1+\sum_{\ell=1}^{\infty}\frac{\tau(p^{\ell})^{2}}{p^{\ell
s}}\right)\prod_{p\mid
cd}\left(1+\sum_{\ell=1}^{\infty}\frac{\tau(cp^{\ell})\tau(dp^{\ell})}{\tau(c)\tau(d)p^{\ell
s}}\right).$

For $|x|<1$, we know that

$\sum_{\ell=0}^{\infty}(\ell+1)x^{\ell}=\frac{1}{(1-x)^{2}},\;\;\quad\quad\sum_{\ell=0}^{\infty}(\ell+1)^{2}x^{\ell}=\frac{(1+x)}{(1-x)^{3}},$

from which we also derive that

$\sum_{\ell=0}^{\infty}\ell(\ell+1)x^{\ell}=\frac{2x}{(1-x)^{3}}.$

Now,

$\displaystyle\prod_{p\nmid
cd}\left(1+\sum_{\ell=1}^{\infty}\frac{\tau(p^{\ell})^{2}}{p^{\ell s}}\right)$
$\displaystyle=\prod_{p}\left(1+\sum_{\ell=1}^{\infty}\frac{(\ell+1)^{2}}{p^{\ell
s}}\right)\prod_{p\mid
cd}\left(1+\sum_{\ell=1}^{\infty}\frac{(\ell+1)^{2}}{p^{\ell s}}\right)^{-1}$
$\displaystyle=\prod_{p}\frac{\left(1+p^{-s}\right)}{\left(1-p^{-s}\right)^{3}}\prod_{p|cd}\frac{\left(1-p^{-s}\right)^{3}}{\left(1+p^{-s}\right)}$
$\displaystyle=\frac{\zeta(s)^{4}}{\zeta(2s)}\prod_{p|cd}\frac{\left(1-p^{-s}\right)^{3}}{\left(1+p^{-s}\right)}.$

If $\gcd(c,d)=1$

$\displaystyle\prod_{p\mid
cd}\left(1+\sum_{\ell=1}^{\infty}\frac{\tau(cp^{\ell})\tau(dp^{\ell})}{\tau(c)\tau(d)p^{\ell
s}}\right)$
$\displaystyle=\prod_{p^{k}\|cd}\left(1+\sum_{\ell=1}^{\infty}\frac{(k+1+\ell)(\ell+1)}{(k+1)p^{\ell
s}}\right)$
$\displaystyle=\prod_{p^{k}\|cd}\left(1+\sum_{\ell=1}^{\infty}\frac{(\ell+1)}{p^{\ell
s}}+\frac{1}{k+1}\sum_{\ell=1}^{\infty}\frac{\ell(\ell+1)}{p^{\ell s}}\right)$
$\displaystyle=\prod_{p^{k}\|cd}\left(1-p^{-s}\right)^{-3}\left(1-\tfrac{(k-1)}{(k+1)}p^{-s}\right).$

∎

##  4\. Proof of Theorem 1.2 and of Proposition 1.1

We continue with the proof with the following Lemma.

######  Lemma 4.1.

Let $a,b$ be positive integers, $\lambda=\gcd(a,b)$, $c=a/\lambda$ and
$d=b/\lambda$. Then

$\lim_{X\to\infty}\frac{1}{X^{3/2}}\int_{1}^{X}\Delta(x/a)\Delta(x/b)dx=\frac{1}{6\pi^{2}\sqrt{\lambda}cd}\sum_{n=1}^{\infty}\frac{\tau(cn)\tau(dn)}{n^{3/2}}.$

###### Proof.

Let $N>0$ and $\epsilon>0$ be a small number that may change from line after
line. We proceed with Voronoï’s formula for $\Delta(x)$ in the following form
(see [11])

$\Delta(x)=\frac{x^{1/4}}{\pi\sqrt{2}}\sum_{n\leq
N}\frac{\tau(n)}{n^{3/4}}\cos(4\pi\sqrt{nx}-\pi/4)+R_{N}(x)$

where, for every positive $\epsilon$, we have

$R_{N}(x)\ll x^{\epsilon}+\frac{x^{1/2+\epsilon}}{N^{1/2}}.$

We select $N$ at the end. With this formula we have that in the range $1\leq
x\leq X$,

$\Delta(x/a)=\frac{(x/a)^{1/4}}{\pi\sqrt{2}}\sum_{n\leq
N}\frac{\tau(n)}{n^{3/4}}\cos(4\pi\sqrt{nx/a}-\pi/4)+R_{N}(x/a)=U_{N}(x/a)+R_{N}(x/a)$

say.

Now,

$\displaystyle\int_{1}^{X}\Delta(x/a)\Delta(x/b)dx$
$\displaystyle=\int_{1}^{X}U_{N}(x/a)U_{N}(x/b)dx+\int_{1}^{X}U_{N}(x/a)R_{N}(x/b)dx$
$\displaystyle+\int_{1}^{X}U_{N}(x/b)R_{N}(x/a)dx+\int_{1}^{X}R_{N}(x/a)R_{N}(x/b)dx$
$\displaystyle=\int_{1}^{X}U_{N}(x/a)U_{N}(x/b)dx+O\left(X^{1+1/4+\epsilon}+\frac{X^{1+3/4+\epsilon}}{\sqrt{N}}\right),$

where we used the Cauchy-Schwarz inequality in the last equality. Let now
$\lambda=\gcd(a,b)$, $c=a/\lambda$ and $d=b/\lambda$. By making the change of
variable $u=x/\lambda$, we reach

$\displaystyle\int_{1}^{X}U_{N}(x/a)U_{N}(x/b)dx=\lambda\int_{1}^{X/\lambda}U_{N}(x/c)U_{N}(x/d)dx$
$\displaystyle=\frac{\lambda}{2\pi^{2}(cd)^{1/4}}\sum_{n,m\leq
N}\frac{\tau(n)\tau(m)}{(nm)^{3/4}}\int_{1}^{X/\lambda}x^{1/2}\cos(4\pi\sqrt{nx/c}-\pi/4)\cos(4\pi\sqrt{mx/d}-\pi/4)$
$\displaystyle=\frac{\lambda}{\pi^{2}(cd)^{1/4}}\sum_{n,m\leq
N}\frac{\tau(n)\tau(m)}{(nm)^{3/4}}\int_{1}^{(X/\lambda)^{1/2}}u^{2}\cos(4\pi
u\sqrt{n/c}-\pi/4)\cos(4\pi u\sqrt{m/d}-\pi/4)du,$

where in the last equality above we made a change of variable $u=\sqrt{x}$. We
claim now that the main contribution comes when $n/c=m/d$. Since $c$ and $d$
are coprime, this implies that $m=dk$ and $n=ck$. Therefore the sum over these
$n$ and $m$ can be written as

(4)
$\frac{\lambda}{\pi^{2}cd}\sum_{k=1}^{\infty}\frac{\tau(ck)\tau(dk)}{k^{3/2}}\int_{1}^{(X/\lambda)^{1/2}}u^{2}\cos^{2}(4\pi
u\sqrt{k}-\pi/4)^{2}du+O\left(\frac{X^{3/2+\epsilon}}{\sqrt{N}}\right).$

We recall now that $\cos^{2}(v)=\frac{1+\cos(2v)}{2}$, and hence the integral
above is

(5)
$\int_{1}^{X^{1/2}/\lambda^{1/2}}x^{2}\cos^{2}(4\pi\sqrt{n}x-\pi/4)dx=\frac{X^{3/2}}{6\lambda^{3/2}}+O(X),$

where the big-oh term is uniform in $n$. Now we will show that the sum over
those $n$ and $m$ such that $n/c\neq m/d$ will be $o(X^{3/2})$. With this the
proof will be complete by combining (4) and (5).

We recall the identity $2\cos(u)\cos(v)=\cos(u-v)+\cos(u+v)$. Thus, for
$\sqrt{n/c}\neq\sqrt{m/d}$, we find that

$\displaystyle\int_{1}^{X^{1/2}/\lambda^{1/2}}x^{2}\cos(4\pi\sqrt{n/c}x-\pi/4)\cos(4\pi\sqrt{m/d}x-\pi/4)dx$
$\displaystyle=\int_{1}^{X^{1/2}/\lambda^{1/2}}x^{2}\cos(4\pi(\sqrt{n/c}-\sqrt{m/d})x)dx+\int_{1}^{X^{1/2}/\lambda^{1/2}}x^{2}\sin(4\pi(\sqrt{n/c}+\sqrt{m/d})x)dx$
$\displaystyle\ll\frac{X}{\sqrt{n/c}-\sqrt{m/d}}\ll\frac{\sqrt{n/c}+\sqrt{m/d}}{nd-
mc}X.$

Let $\mathds{1}_{P}(n)$ be the indicator that $n$ has property $P$. We find
that

$\displaystyle\sum_{\begin{subarray}{c}n,m\leq N\\\ nd-mc\neq
0\end{subarray}}\frac{\tau(n)\tau(m)}{(nm)^{3/4}}\int_{1}^{X/\lambda}x^{1/2}\cos(4\pi\sqrt{nx/c}-\pi/4)\cos(4\pi\sqrt{mx/d}-\pi/4)dx$
$\displaystyle\ll XN^{\epsilon}\sum_{\begin{subarray}{c}n,m\leq N\\\ nd-mc\neq
0\end{subarray}}\frac{\sqrt{n/c}+\sqrt{m/d}}{(nm)^{3/4}|nd-mc|}$
$\displaystyle=XN^{\epsilon}\sum_{\begin{subarray}{c}n,m\leq N\\\ nd-mc\neq
0\end{subarray}}\frac{\sqrt{n/c}+\sqrt{m/d}}{(nm)^{3/4}|nd-
mc|}\sum_{\begin{subarray}{c}k=-N\max(c,d)\\\ k\neq
0\end{subarray}}^{N\max(c,d)}\mathds{1}_{nd-mc=k}.$

On calling this sum $S$, we readily continue with

$\displaystyle S$ $\displaystyle\ll
XN^{\epsilon}\sum_{k=1}^{N\max(c,d)}\frac{1}{k}\sum_{m\leq
N}\frac{\sqrt{m}+\sqrt{k}}{((k+mc)m)^{3/4}}$ $\displaystyle\ll
XN^{\epsilon}\left(O(\log N)^{2}+\sum_{k\leq N}\frac{1}{\sqrt{k}}\sum_{m\leq
N}\frac{1}{(m^{2}+mk)^{3/4}}\right)$ $\displaystyle\ll
XN^{\epsilon}\left(O(\log N)^{2}+\sum_{k\leq
N}\frac{1}{\sqrt{k}}\left(\sum_{k\leq m\leq
N}\frac{1}{m^{3/2}}+\frac{1}{k^{3/4}}\sum_{m\leq
k}\frac{1}{m^{3/4}}\right)\right)$ $\displaystyle\ll XN^{\epsilon}(\log
N)^{2}.$

Finally, by selecting $N=X^{2}$, we arrive at

$\int_{1}^{X}\Delta(x/a)\Delta(x/b)dx=\frac{1}{6\pi^{2}\sqrt{\lambda}cd}\left(\sum_{n=1}^{\infty}\frac{\tau(cn)\tau(dn)}{n^{3/2}}\right)X^{3/2}+O(X^{3/2-1/4+\epsilon}),$

where the main contribution in the $O$-term above comes from the usage of
Cauchy-Schwarz in the beginning of the proof.

The proof is complete. ∎

Now we deviate from the main line and prove Proposition 1.1.

###### Proof of Proposition 1.1.

By the proof of Lemma 4.1 we have that

$\displaystyle I_{\theta}(X):=$
$\displaystyle\int_{1}^{X}\Delta(x)\Delta(\theta x)dx$ $\displaystyle=$
$\displaystyle\frac{1}{\pi^{2}}\sum_{n,m\leq
N}\frac{\tau(n)\tau(m)}{(nm)^{3/4}}\int_{1}^{X^{1/2}}x^{2}\cos(4\pi
x\sqrt{n}-\pi/4)\cos(4\pi x\sqrt{m\theta}-\pi/4)dx$
$\displaystyle+O\left(X^{1+1/4+\epsilon}+\frac{X^{1+3/4+\epsilon}}{\sqrt{N}}\right).$

Now, by appealing to the identity $2\cos(u)\cos(v)=\cos(u-v)+\cos(u+v)$, we
reach at

$\displaystyle I_{\theta}(X)=\frac{1}{2\pi^{2}}\sum_{n,m\leq
N}\frac{\tau(n)\tau(m)}{(nm)^{3/4}}\int_{1}^{X^{1/2}}x^{2}\cos(4\pi
x(\sqrt{n}-\sqrt{m\theta}))dx+O\left(X^{1+1/4}+\frac{X^{1+3/4+\epsilon}}{\sqrt{N}}\right).$

On calling the sum above $S_{\theta}(X)$,
$a_{n,m}:=4\pi(\sqrt{n}-\sqrt{m\theta})$, we obtain that

$S_{\theta}(X)=X^{3/2}\sum_{n,m\leq
N}\frac{\tau(n)\tau(m)}{(nm)^{3/4}}\Lambda(a_{n,m}\sqrt{X}),$

where $\Lambda(0):=1/3$ and for $u\neq 0$

$\Lambda(u):=\frac{\sin(u)}{u}+2\frac{\cos(u)}{u^{2}}-2\frac{\sin(u)}{u^{3}}.$

A careful inspection shows that $\Lambda$ is continuous and for large $|u|$,
$\Lambda(u)\ll|u|^{-1}$.

Now, for a large parameter $T$ to be chosen later, we split

$S_{\theta}(X)=X^{3/2}\sum_{\begin{subarray}{c}n,m\leq N\\\
|a_{n,m}\sqrt{X}|\leq
T\end{subarray}}\frac{\tau(n)\tau(m)}{(nm)^{3/4}}\Lambda(a_{n,m}\sqrt{X})+X^{3/2}\sum_{\begin{subarray}{c}n,m\leq
N\\\
|a_{n,m}\sqrt{X}|>T\end{subarray}}\frac{\tau(n)\tau(m)}{(nm)^{3/4}}\Lambda(a_{n,m}\sqrt{X}).$

We call the first sum in the right hand side above diagonal contribution and
the second sum the non-diagonal contribution. We select $T=X^{1/2-\delta}$ and
$N=X^{1/2+\delta}$, for some small $\delta>0$.

The diagonal contribution. We have that

(6) $\displaystyle D(X)$
$\displaystyle=X^{3/2}\sum_{\begin{subarray}{c}n,m\leq N\\\
|a_{n,m}\sqrt{X}|\leq
T\end{subarray}}\frac{\tau(n)\tau(m)}{(nm)^{3/4}}\Lambda(a_{n,m}\sqrt{X})$ (7)
$\displaystyle\ll X^{3/2}N^{\epsilon}\sum_{m\leq
N}\frac{1}{m^{3/4}}\sum_{\begin{subarray}{c}n;\\\
|n-m\theta|\leq\frac{2\sqrt{m\theta}}{X^{\delta}}+\frac{1}{X^{2\delta}}\end{subarray}}\frac{|\Lambda(a_{n,m}\sqrt{X})|}{n^{3/4}}.$

The inner sum above we split accordingly
$\frac{2\sqrt{m\theta}}{X^{\delta}}+\frac{1}{X^{2\delta}}$ is below and above
$1$. In the case that this quantity is great or equal to $1$, we have that
$m\geq((2\theta)^{-1}+o(1))X^{2\delta}$, and hence

$\displaystyle D(X)\ll$ $\displaystyle
X^{3/2}N^{\epsilon}\sum_{((2\theta)^{-1}+o(1))X^{2\delta}\leq m\leq
N}\frac{1}{m^{3/4}}\sum_{\begin{subarray}{c}n;\\\
|n-m\theta|\leq\frac{2\sqrt{m\theta}}{X^{\delta}}+\frac{1}{X^{2\delta}}\end{subarray}}\frac{|\Lambda(a_{n,m}\sqrt{X})|}{n^{3/4}}$
$\displaystyle\ll
X^{3/2}N^{\epsilon}\sum_{((2\theta)^{-1}+o(1))X^{2\delta}\leq m\leq
N}\frac{1}{m^{3/4}}\cdot\frac{1}{m^{3/4}}\frac{\sqrt{m}}{X^{\delta}}$
$\displaystyle\ll X^{3/2-\delta}N^{\epsilon}.$

In the case that $\frac{2\sqrt{m\theta}}{X^{\delta}}+\frac{1}{X^{2\delta}}\leq
1$, we have that $m\leq((2\theta)^{-1}+o(1))X^{2\delta}$, and now the
Diophantine properties of $\theta$ come in to play. If the irrationality
measure of $\theta$ is $\eta$, we have that for each $\epsilon$ there is a
constant $C>0$ such that the inequality

$|n-m\theta|\geq\frac{C}{m^{\eta+\epsilon}}$

is violated only for a finite number of positive integers $n$ and $m$. In our
case, this allows us to lower bound $|a_{n,m}\sqrt{X}|$ for all but a finite
number of $n$ and $m$ such that $1\leq m\ll X^{2\delta}$ and
$1/2\leq\sqrt{n}/\sqrt{m\theta}\leq 2$:

$\displaystyle|a_{n,m}\sqrt{X}|\cdot\frac{\sqrt{n}+\sqrt{m\theta}}{\sqrt{n}+\sqrt{m\theta}}$
$\displaystyle=\sqrt{X}\frac{|n-m\theta|}{\sqrt{n}+\sqrt{m\theta}}$
$\displaystyle\geq\frac{\sqrt{X}}{m^{\eta+\epsilon}(\sqrt{n}+\sqrt{m\theta})}$
$\displaystyle\gg X^{1/2-(2\eta+1)\delta-\epsilon}.$

Observe that the diagonal contribution from those exceptional $n$ and $m$ will
be at most $O(X)$. With these estimates on hand and recalling that
$\Lambda(u)\ll|u|^{-1}$, we obtain

$\displaystyle
X^{3/2}N^{\epsilon}\sum_{m\leq((2\theta)^{-1}+o(1))X^{2\delta}}\frac{1}{m^{3/4}}\sum_{\begin{subarray}{c}n;\\\
|n-m\theta|\leq\frac{2\sqrt{m\theta}}{X^{\delta}}+\frac{1}{X^{2\delta}}\end{subarray}}\frac{|\Lambda(a_{n,m}\sqrt{X})|}{n^{3/4}}$
$\displaystyle\ll
X^{3/2}N^{\epsilon}\sum_{m\leq((2\theta)^{-1}+o(1))X^{2\delta}}\frac{1}{m^{3/2}}\cdot\frac{1}{X^{1/2-(2\eta+1)\delta-\epsilon}}+O(X)$
$\displaystyle\ll X^{1+(2\eta+1)\delta+\epsilon}.$

Therefore, the diagonal contribution is at most

$D(X)\ll X^{1+(2\eta+1)\delta+\epsilon}+X^{3/2-\delta+\epsilon}.$

The non-diagonal contribution. Now, we reach

$\displaystyle X^{3/2}\sum_{\begin{subarray}{c}n,m\leq N\\\
|a_{n,m}\sqrt{X}|>T\end{subarray}}\frac{\tau(n)\tau(m)}{(nm)^{3/4}}\Lambda(a_{n,m}\sqrt{X})$
$\displaystyle\ll\frac{X^{3/2}N^{1/2+\epsilon}}{T}$
$\displaystyle=X^{3/2+1/4+(\delta+\epsilon)/2+\epsilon\delta-1/2+\delta}$
$\displaystyle=X^{1+1/4+3\delta/2+\epsilon/2+\epsilon\delta}.$

We choose $\delta=\frac{1}{3(2\eta+1)}$ and obtain

$I_{\theta}(X)=X^{3/2-1/(18\eta)}.$

The proof of the first part of Proposition 1.1 is complete.

Now we assume that $\theta$ is a Liouville number, i.e., $\theta$ doesn’t have
finite degree of irrationality. We see that the non-diagonal argument does not
depend on the Diophantine properties of $\theta$. Let $\eta>0$ be a large
fixed number, $t>0$ a small number that will tend to $0$. For $D(X)$ as in
(6), by repeating verbatim the estimates above we have that

$D(X)\ll
X^{3/2}\sum_{m\leq((2\theta)^{-1}+o(1))X^{2\delta}}\frac{\tau(m)}{m^{3/4}}\sum_{\begin{subarray}{c}n;\\\
|n-m\theta|\leq\frac{2\sqrt{m\theta}}{X^{\delta}}+\frac{1}{X^{2\delta}}\end{subarray}}\frac{\tau(n)|\Lambda(a_{n,m}\sqrt{X})|}{n^{3/4}}+O(X^{3/2-\delta}N^{\epsilon}).$

Let $\|x\|$ be the distance from $x$ to the nearest integer. We split the sum
over $m$ above into two sums: One over those $m$ such that
$\|m\theta\|>tm^{-\eta}$ and the other over $m$ such that $\|m\theta\|\leq
tm^{-\eta}$.

Repeating the argument above for non-Liouville numbers, we have that the
contribution over those $m$ such that $\|m\theta\|>tm^{-\eta}$ is
$O(t^{-1}X^{1+\delta(2\eta+1)})$. Therefore

$D(X)\ll X^{3/2}\sum_{\begin{subarray}{c}m=1\\\ {\|m\theta\|\leq
tm^{-\eta}}\end{subarray}}^{\infty}\frac{1}{m^{3/2-\epsilon}}+O(t^{-1}X^{1+\delta(2\eta+1)}+X^{3/2-\delta+\epsilon}).$

Combining all these estimates, we see that

$\limsup_{X\to\infty}\frac{1}{X^{3/2}}\left|\int_{1}^{X}\Delta(x)\Delta(\theta
x)dx\right|\ll\sum_{\begin{subarray}{c}m=1\\\ {\|m\theta\|\leq
tm^{-\eta}}\end{subarray}}^{\infty}\frac{1}{m^{3/2-\epsilon}}.$

Since the upper bound above holds for all $t>0$, we have that as $t\to 0^{+}$,
the sum above converges to $0$ and thus implying that the $\limsup$ is $0$.
The proof is complete. ∎

###### Proof of Theorem 1.2.

On combining Lemma 4.1 together with Lemma 3.2, we get the first part of
Theorem 1.2. The second part is a trivial consequence of Proposition 1.1. ∎

##  5\. Quadratic forms auxiliaries

The main proof will lead to considering the quadratic form attached to a
matrix of the form

(8)
$M_{S,\varphi}=\biggl{(}\frac{1}{\sqrt{\gcd(a,b)}}\varphi\biggl{(}\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\biggr{)}\biggr{)}_{a,b\in
S}$

where $S$ is some finite set of integers while $\varphi$ is a _non-negative
multiplicative function such that $\varphi(p^{k})\leq 1$_. So we stray
somewhat from the main line and investigate this situation. Our initial aim is
to find conditions under which the associated quadratic form is positive
definite, but we shall finally restrict our scope. GCD-matrices have received
quite some attention, but it seems the matrices occuring in (8) have not been
explored. We obtain results in two specific contexts.

#### Completely multiplicative case

Here is our first result.

######  Lemma 5.1.

When $\varphi$ is completely multiplicative and $p^{1/4}\varphi(p)\in(0,1]$,
the matrix $M_{S,\varphi}$ is non-negative. When in addition we assume that
$p^{1/4}\varphi(p)\in(0,1)$ and $S$ is divisor closed, this matrix is positive
definite. The determinant in that case is given by the formula

$\det\biggl{(}\frac{1}{\sqrt{\gcd(a,b)}}\varphi\biggl{(}\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\biggr{)}\biggr{)}_{a,b\in
S}=\prod_{d\in S}\varphi(d)^{2}(\mu\ast\psi)(d),$

where $\psi$ is the completely multiplicative function given by
$\psi(p)=1/(\sqrt{p}\varphi(p)^{2})$.

By _divisor closed_ , we mean that every divisor of an element of $S$ also
belongs to $S$.

###### Proof.

We write

$\frac{1}{\sqrt{\gcd(a,b)}}\varphi\biggl{(}\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\biggr{)}=\frac{1}{\sqrt{\gcd(a,b)}}\varphi\biggl{(}\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\biggr{)}=\varphi(a)\varphi(b)\psi(\gcd(a,b))$

where $\psi(n)=1/(\varphi(n)^{2}\sqrt{n})$ is another non-negative
multiplicative function. We introduce the auxiliary function $h=\mu\ast\psi$.
Notice that this function is multiplicative and non-negative, as $\psi(p)\geq
1$. We use Selberg’s diagonalization process to write

$\displaystyle\sum_{a,b\in
S}\frac{1}{\sqrt{\gcd(a,b)}}\varphi\biggl{(}\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\biggr{)}x_{a}x_{b}$
$\displaystyle=\sum_{a,b\in S}\psi(\gcd(a,b))\varphi(a)x_{a}\varphi(b)x_{b}$
$\displaystyle=\sum_{a,b\in
S}\sum_{d|(a,b)}h(d)\varphi(a)x_{a}\varphi(b)x_{b}$
$\displaystyle=\sum_{d}h(d)\biggl{(}\sum_{\begin{subarray}{c}a\in S\\\
d|a\end{subarray}}\varphi(a)x_{a}\biggr{)}^{2}$

from which the non-negativity follows readily. When $\varphi$ verifies the
more stringent condition that $p^{1/4}\varphi(p)\in(0,1)$, we know that both
$\varphi$ and $h$ are strictly positive. Let us define
$y_{d}=\sum_{\begin{subarray}{c}a\in S\\\ d|a\end{subarray}}\varphi(a)x_{a}$.
The variable $d$ varies in the set $D$ of divisors of $S$. We assume that $S$
is divisor closed, so that $D=S$. We can readily invert the triangular system
giving the $y_{d}$’s as functions of the $x_{a}$’s into

$\varphi(a)x_{a}=\sum_{a|b}\mu(b/a)y_{b}$

Indeed, the fact that the mentioned system is triangular ensures that a
solution $y$ is unique if it exists. We next verify that the proposed
expression is indeed a solution by:

$\sum_{\begin{subarray}{c}a\in S\\\
d|a\end{subarray}}\varphi(a)x_{a}=\sum_{\begin{subarray}{c}a\in S\\\
d|a\end{subarray}}\sum_{a|b}\mu(b/a)y_{b}=\sum_{\begin{subarray}{c}b\in S\\\
d|b\end{subarray}}y_{b}\sum_{d|a|b}\mu(b/a)=y_{d}$

as the last inner sum vanishes when $d\neq b$. We thus have a writing as a
linear combination of squares of independant linear forms. In a more
pedestrian manner, if our quadratic form vanishes, then all $y_{d}$’s do
vanish, hence so do the $x_{a}$’s. ∎

Here is a corollary.

######  Lemma 5.2.

When the set $S$ contains solely squarefree integers, the matrix
$M_{S,\varphi}$ is non-negative.

###### Proof.

Simply apply Lemma 5.1 to the completely multiplicative function
$\varphi^{\prime}$ that has the same values on primes as $\varphi$. ∎

Now we recall the Sylvester’s criterion.

######  Lemma 5.3.

A hermitian complex valued matrix $M=(m_{i,j})_{i,j\leq K}$ defines a positive
definite form if and only if all its principal minors $\det(m_{i,j})_{i,j\leq
k}$ for $k\leq K$ are positive.

#### A tensor product-like situation

Lemma 5.2 is enough to solve our main problem when $M_{1}$ and $M_{2}$ are
coprime squarefree integers. We need to go somewhat further. Let $S$ be a
divisor closed set. We consider the quadratic form

(9) $\sum_{a,b\in
S}\varphi\biggl{(}\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\biggr{)}x_{a}x_{b}$

where the variables $x_{a}$’s are also multiplicatively split, i.e.

(10) $x_{a}=\prod_{p^{k}\|a}x_{p^{k}}.$

Let $S(p)$ the subset of $S$ made only of $1$ and of prime powers. We extend
$S$ so that it contains every products of integers from any collection of
distinct $S(p)$***This is not automatically the case, as the example
$S=\\{1,2,3,5,6,10\\}$ shows, since $30$ does not belong to $S$. We then find
that

(11) $\sum_{a,b\in
S}\frac{1}{\sqrt{\gcd(a,b)}}\varphi\biggl{(}\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\biggr{)}x_{a}x_{b}=\prod_{p\in
S}\biggl{(}\sum_{p^{k},p^{\ell}\in
S(p)}\frac{\varphi\bigl{(}p^{\max(k,\ell)-\min(k,\ell)}\bigr{)}}{p^{\min(k,\ell)/2}}x_{p^{k}}x_{p^{\ell}}\biggr{)}.$

We check this identity simply by opening the right-hand side and seeing that
every summand from the left-hand side appears one and only one time.

##  6\. Proof of the main result

###### Proof.

By [1, Theorem 1.4], we have

$S(x)=\sum_{n\leq x}(f_{1}\ast f_{2})(n)=\sum_{a|M_{1}M_{2}}g(a)\Delta(x/a)$

where $g=f_{1}\ast f_{2}\ast\mu\ast\mu$. We infer from this formula that

$\displaystyle\int_{1}^{X}|S(x)|^{2}dx=\sum_{a,b|M_{1}M_{2}}g(a)g(b)\int_{1}^{X}\Delta(x/a)\Delta(x/b)dx$
$\displaystyle=\frac{(1+o(1))}{6\pi^{2}}X^{3/2}\sum_{a,b|M_{1}M_{2}}g(a)g(b)\frac{\gcd(a,b)^{3/2}}{ab}\sum_{n=1}^{\infty}\frac{\tau(an/\gcd(a,b))\tau(bn/\gcd(a,b))}{n^{3/2}}$

by Lemma 4.1. We next use Lemma 3.2 to infer that

$\lim_{X\to\infty}\frac{1}{X^{3/2}}\int_{1}^{X}|S(x)|^{2}dx=\frac{\zeta(3/2)^{4}}{6\pi^{2}\zeta(3)}\sum_{a,b|M_{1}M_{2}}g(a)g(b)\frac{1}{\sqrt{\gcd(a,b)}}\varphi\biggl{(}\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\biggr{)}$

where $\varphi$ is multiplicative and at prime powers:

(12)
$\begin{split}\varphi(p^{k})&=\frac{(k+1)}{p^{k}}\frac{1}{1+p^{-3/2}}\biggl{(}1-\frac{(k-1)}{(k+1)p^{3/2}}\biggr{)}\\\
&=\frac{1}{p^{k}(1+p^{-3/2})}\biggl{(}(k+1)-(k-1)p^{-3/2}\biggr{)}\\\
&=\frac{1}{p^{k}(1+p^{-3/2})}\biggl{(}k(1-p^{-3/2})+(1+p^{-3/2})\biggr{)}\\\
&=\frac{k\beta(p)+1}{p^{k}},\end{split}$

where

$\beta(p)=\frac{1-p^{-3/2}}{1+p^{-3/2}}.$

Now, we can write

$\frac{1}{\sqrt{\gcd(a,b)}}\varphi\left(\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\right)=\frac{1}{(ab)^{1/4}}\left(\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\right)^{1/4}\varphi\left(\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\right).$

Since the terms $a^{-1/4}$ and $b^{-1/4}$ can be absorbed into the variables
$g(a)$ and $g(b)$ of the quadratic form, it is enough to consider the quantity

$\varphi^{*}\left(\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\right),\quad\text{where}\
\,\varphi^{*}(n)=n^{1/4}\varphi(n).$

We note that, at the prime power $p^{k}$, we have

(13) $\varphi^{*}(p^{k})=p^{k/4}\varphi(p^{k})=\frac{k\beta(p)+1}{p^{3k/4}}.$

Due to (11) and the discussion before it, we now restrict to the prime power
case, that is, we look to matrices of the form

$\mathcal{M}_{K}=\bigl{(}\varphi^{*}(p^{|i-j|})\bigr{)}_{i,j\leq K}.$

As we are dealing with a given prime $p$, we shorten $\beta(p)$ in $\beta$.

Since $\varphi^{*}$ is not completely multiplicative, it is not clear how to
handle the matrix $\mathcal{M}_{K}$ directly. So, our aim will be to transform
this into another matrix which, in some way associates with a completely
multiplicative function. So, let us consider

$\mathcal{A}_{K}=\mathcal{U}_{K}^{\top}\mathcal{M}_{K}\,\mathcal{U}_{K},$

where,

(14)
$\mathcal{U}_{K}(i,j)=\begin{dcases}\frac{\mu(p^{|i-j|})}{p^{3(|i-j|)/4}},&\text{when
}i\geq j\,\text{ or }\,(i,j)=(K-1,K),\\\ 0,&\text{otherwise.}\end{dcases}$

Simply speaking, $\mathcal{U}_{K}$ is $1$ on the diagonal and $-p^{-3/4}$ on
all $(i+1,i)$ as well as $(K-1,K)$. Also

$\det(\mathcal{U}_{K})=1-p^{-3/2}.$

We now calculate the entries of the matrix $\mathcal{A}_{K}$. We have the
following:

######  Proposition 6.1.

The matrix $\mathcal{A}_{K}$ above is given by:

$\mathcal{A}_{K}(i,j)=\beta(1-p^{-3/2})\cdot\begin{dcases}p^{-3|i-j|/4},&\text{when
}\,{1\leq i,j\leq K-1}\text{ or }i=j=K,\\\ 0,&\text{otherwise}.\end{dcases}$

We begin with the following lemma:

######  Lemma 6.1.

We have

$\varphi^{*}(p^{m})-p^{-3/4}\varphi^{*}(p^{|m-1|})=p^{-3m/4}\beta,\quad\text{for
all }\,m\geq 0.$

###### Proof.

First, assume $m\geq 1$. We have

$\varphi^{*}(p^{m})-p^{-3/4}\varphi^{*}(p^{m-1})=\frac{m\beta+1}{p^{3m/4}}-p^{-3/4}\frac{(m-1)\beta+1}{p^{3(m-1)/4}}=p^{-3m/4}\beta.$

When $m=0$, we have

$1-p^{-3/4}\varphi^{*}(p)=1-p^{-3/2}(\beta+1)=1-\frac{2p^{-3/2}}{1+p^{-3/2}}=\beta.$

∎

Now, we shall proceed with the proof of the Proposition 6.1.

###### Proof of Proposition 6.1.

Let us first assume that $1\leq i,j\leq K-1$. We have

(15)
$\begin{split}\mathcal{A}_{K}(i,j)&=\sum\limits_{k_{1},k_{2}}\mathcal{U}_{K}^{\top}(i,k_{1})\mathcal{M}_{K}(k_{1},k_{2})\,\mathcal{U}_{K}(k_{2},j)\\\
&=\sum\limits_{\begin{subarray}{c}k_{1}-i\in\\{0,1\\}\\\
k_{2}-j\in\\{0,1\\}\end{subarray}}\frac{\mu(p^{k_{1}-i})}{p^{3(k_{1}-i)/4}}\frac{\mu(p^{k_{2}-j})}{p^{3(k_{2}-j)/4}}\varphi^{*}(p^{|k_{1}-k_{2}|})\\\
&=\biggl{(}\varphi^{*}(p^{|i-j|})\bigl{(}1+p^{-3/2}\bigr{)}-\frac{\varphi^{*}(p^{|i-j+1|})+\varphi^{*}(p^{|i-j-1|})}{p^{3/4}}\biggr{)}.\end{split}$

Here, we do not have the contribution coming from $\mathcal{U}_{K}(K-1,K)$ or
$\mathcal{U}_{K}^{\top}(K,K-1)$ as we have assumed $i,j\leq K-1$. This
assumption is necessary because we are considering the values $k_{1}=i+1$ and
$k_{2}=j+1$ (both of which should remain $\leq K$).

First, let us consider the case $i\geq j$. Letting $i-j=m\geq 0$, (15) becomes

$\begin{split}\mathcal{A}_{K}(i+m,i)&=\varphi^{*}(p^{m})-p^{-3/4}\varphi^{*}(p^{|m-1|})-p^{-3/4}\bigl{(}\varphi^{*}(p^{m+1})-p^{-3/4}\varphi^{*}(p^{m})\bigr{)}\\\
&=p^{-3m/4}\beta-p^{-3/4}p^{-3(m+1)/4}\beta\\\
&=\beta(1-p^{-3/2})p^{-3m/4}.\end{split}$

Similarly, for $j\geq i$, we will obtain the same expression in terms of
$m=j-i$. This proves Proposition 6.1 for $1\leq i,j\leq K-1$.

Next, we consider the case when one of $i$ or $j$ equals $K$.

Claim: $\mathcal{A}_{K}(i,K)=\mathcal{A}_{K}(K,j)=0$, for all $1\leq i,j\leq
K-1$.

We revert to the first line of the expression (15). Letting $m=K-i\geq 1$, we
obtain

$\begin{split}\mathcal{A}_{K}(i,K)&=\sum\limits_{\begin{subarray}{c}k_{1}\in\\{i,i+1\\}\\\
k_{2}\in\\{K-1,K\\}\end{subarray}}\frac{\mu(p^{k_{1}-i})}{p^{3(k_{1}-i)/4}}\frac{\mu(p^{K-k_{2}})}{p^{3(K-k_{2})/4}}\varphi^{*}(p^{|k_{1}-k_{2}|})\\\
&=-p^{-3/4}\varphi^{*}(p^{m-1})+p^{-3/2}\varphi^{*}(p^{|m-2|})+\varphi^{*}(p^{m})-p^{-3/4}g(p^{m-1})\\\
&=-p^{-3/4}\big{(}\varphi^{*}(p^{m-1})-p^{-3/4}\varphi^{*}(p^{|m-2|})\bigr{)}+\varphi^{*}(p^{m})-p^{-3/4}\varphi^{*}(p^{m-1})\\\
&=-p^{-3/4}p^{-3(m-1)/4}\beta+p^{-3m/4}\beta=0.\end{split}$

It similarly follows that $\mathcal{A}_{K}(K,j)=0$ for $1\leq j\leq K-1$,
proving the claim.

Next, we see that

$\begin{split}\mathcal{A}_{K}(K,K)&=\sum\limits_{k_{1},k_{2}\in\\{K-1,K\\}}\frac{\mu(p^{K-k_{1}})}{p^{3(K-k_{1})/4}}\frac{\mu(p^{K-k_{2}})}{p^{3(K-k_{2})/4}}\,\varphi^{*}(p^{|k_{1}-k_{2}|})\\\
&=1-p^{-3/4}\varphi^{*}(p)-p^{-3/4}\bigl{(}\varphi^{*}(p)-p^{-3/4}\bigr{)}\\\
&=\beta-p^{-3/4}\bigl{(}p^{-3/4}\beta\bigr{)}=\beta(1-p^{-3/2}).\end{split}$

This completes the proof of Proposition 6.1. ∎

Now since $n\mapsto n^{-3/4}$ is completely multiplicative, by the proof of
Lemma 5.1, the matrix

$\mathcal{B}_{K}=\left(\left(\frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}\right)^{-3/4}\right)_{a,b\in\\{1,...,p^{K}\\}}$

is positive definite for all $K$. Since the entries $(i,j)$ with $1\leq
i,j\leq K-1$ of $\mathcal{A}_{K}$ coincide with the ones of the matrix
$c\mathcal{B}_{K}$ for some positive constant $c$, and that the entries
$(i,K)$ and $(K,j)$ of $\mathcal{A}_{K}$ are all zero with a single exception
at the entry $(K,K)$, by Sylvester’s criterion (Lemma 5.3), we conclude that
the matrix $\mathcal{A}_{K}$ is positive definite for all $K$.

Since $\mathcal{A}_{K}=\mathcal{U}_{K}^{\top}\mathcal{M}_{K}\mathcal{U}_{K}$,
we have

$\det(\mathcal{A}_{K})=\det(\mathcal{U}_{K})^{2}\,\det(\mathcal{M}_{K})=(1-p^{-3/2})^{2}\,\det(\mathcal{M}_{K}).$

This proves that $\det(\mathcal{M}_{K})>0$ and by induction over $K$ in Lemma
5.3, $\mathcal{M}_{K}$ is positive definite for all $K$.

The factorization (11) completes the proof of Theorem 1.1. ∎

Acknowledgements. This project started while the first author was a visiting
Professor at Aix-Marseille Université, and he is thankful for their kind
hospitality and to CNPq grant PDE no. 400010/2022-4 (200121/2022-7) for
supporting this visit. The first author also is supported by FAPEMIG, grant
Universal no. APQ-00256-23 and by CNPq grant Universal no. 403037/2021-2. The
second and third author are supported by the joint FWF-ANR project Arithrand:
FWF: I 4945-N and ANR-20-CE91-0006. The main bulk of this paper was build when
the fourth author was a guest of the Aix-Marseille Université I2M laboratory.
We thank this place for its support.

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