On the $L^1$ norm of an exponential sum involving the divisor function

TL;DR

This paper derives bounds on the $L^1$ norm of exponential sums involving the divisor function, contributing to understanding their size and behavior in analytic number theory.

Contribution

It provides new bounds on the $L^1$ norm of exponential sums with the divisor function, advancing previous results in the field.

Findings

Established explicit bounds on the $L^1$ norm for the sum involving $ au(n)$

Improved understanding of the sum's growth and oscillation

Potential applications to problems in analytic number theory

Abstract

In this paper, we obtain bounds on the $L^1$ norm of the sum $\sum_{n\le x}\tau(n) e(\alpha n)$ where $\tau(n)$ is the divisor function.

Equations

$$S(\alpha )=\sum _{n\le x}\tau (n)e(n\alpha ),e(\alpha )=e^{2\pi i\alpha }.$$

$$\sqrt{x}\ll {\int }_0^1|S(\alpha )|d\alpha \ll \sqrt{x}.$$

$$\sqrt{x}\ll {\int }_0^1|S(\alpha )|d\alpha \ll \sqrt{x}\log x.$$

$$\begin{array}{rl}S(\alpha )& =\sum _{n\le x}\left(\sum _{uv=n}1\right)e(n\alpha )\\ & =\sum _{uv\le x}e(uv\alpha )\\ & =2\sum _{u\le \sqrt{x}}\sum _{u<v\le x/u}e(uv\alpha )+\sum _{u\le \sqrt{x}}e(u^2\alpha )\\ & :=2T(\alpha )+V(\alpha ).\end{array}$$

$${\int }_0^1|V(\alpha )|d\alpha \le {\left({\int }_0^1|V(\alpha ){|}^{2}d\alpha \right)}^{\frac{1}{2}}=\sqrt{\lfloor \sqrt{x}\rfloor }\le x^{\frac{1}{4}},$$

$${\int }_0^1|S(\alpha )|d\alpha =2{\int }_0^1|T(\alpha )|d\alpha +O(x^{\frac{1}{4}}).$$

$${\int }_0^1|T(\alpha )|d\alpha \ll \sqrt{x}\log x.$$

$$2\sqrt{x}\le Q\ll \sqrt{x},$$

$${\int }_0^1|T(\alpha )|d\alpha \le \sum _{q\le Q}\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q{\int }_{\frac{a}{q}-1/(2q\sqrt{x})}^{\frac{a}{q}+1/(2q\sqrt{x})}\left|T(\alpha )\right|d\alpha .$$

$$T(\alpha )=F_q(\alpha )+G_q(\alpha )$$

$$F_q(\alpha )=\sum _{\begin{array}{c}u\le \sqrt{x}\\ q|u\end{array}}\sum _{u<v\le x/u}e(uv\alpha )$$

$$G_q(\alpha )=\sum _{\begin{array}{c}u\le \sqrt{x}\\ q\nmid u\end{array}}\sum _{u<v\le x/u}e(uv\alpha ),$$

$$\begin{array}{rl}{\int }_0^1|T(\alpha )|d\alpha & \le \sum _{q\le Q}\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q{\int }_{\frac{a}{q}-1/(2q\sqrt{x})}^{\frac{a}{q}+1/(2q\sqrt{x})}\left|F_q(\alpha )\right|d\alpha +\sum _{q\le Q}\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q{\int }_{\frac{a}{q}-1/(2q\sqrt{x})}^{\frac{a}{q}+1/(2q\sqrt{x})}\left|G_q(\alpha )\right|d\alpha \\ & :=I_F+I_G.\end{array}$$

$$I_F\ll \sqrt{x}.$$

$$I_G\ll \sqrt{x}\log x.$$

$$\alpha =\frac{a}{q}+\beta .$$

$$F_{q}(\alpha)\ll\left\{\begin{array}[]{ll}{\min\left(x,\frac{1}{|\beta|}\right)\frac{\log\frac{2\sqrt{x}}{q}}{q},}&\mbox{if $q\leq\sqrt{x}$, $|\beta|\leq\frac{1}{2q\sqrt{x}}$;}\\ 0,&\mbox{if $q>\sqrt{x}$;}\\ \end{array}\right.$$

$$\begin{array}{rl}{I}_F& \ll \sum _{q\le \sqrt{x}}q{\int }_0^{1/(2q\sqrt{x})}\min \left(x,\frac{1}{|\beta |}\right)\frac{\log \frac{2\sqrt{x}}{q}}{q}d\beta \\ & \ll \sum _{q\le \sqrt{x}}\log \frac{2\sqrt{x}}{q}\left({\int }_0^{1/(2x)}xd\beta +{\int }_{1/(2x)}^{1/(2q\sqrt{x})}\frac{1}{\beta }d\beta \right)\\ & \ll \sum _{q\le \sqrt{x}}{\left(\log \frac{2\sqrt{x}}{q}\right)}^2\ll \sqrt{x}.\end{array}$$

$$F_q(\alpha )=\sum _{j\le \frac{\sqrt{x}}{q}}\sum _{jq\le v\le \frac{x}{jq}}e(jqv\beta ).$$

$$\sum _{N_1<n\le N_2}e(n\alpha )\ll \min \left(N_2-N_1,\frac{1}{\Vert \alpha \Vert }\right)$$

$$F_q(\alpha )\ll \sum _{j\le \frac{\sqrt{x}}{q}}\min \left(\frac{x}{jq},\frac{1}{\Vert jq\beta \Vert }\right).$$

$$F_q(\alpha )\ll \sum _{j\le \frac{\sqrt{x}}{q}}\frac{1}{jq}\min \left(x,\frac{1}{|\beta |}\right)\ll \min (x,\frac{1}{|\beta |})\frac{\log \frac{2\sqrt{x}}{q}}{q}.$$

$$G_q(\alpha )\ll (\sqrt{x}+q)\log q,\text{for}\alpha =\frac{a}{q}+\beta ,|\beta |\le \frac{1}{2q\sqrt{x}},$$

$$\begin{array}{rl}{I}_G& \ll \sum _{q\le Q}q{\int }_0^{1/(2q\sqrt{x})}(\sqrt{x}+q)\log qd\beta \\ & \ll \frac{1}{\sqrt{x}}(Q(\sqrt{x}+Q)\log Q\ll \sqrt{x}\log x\end{array}$$

$$G_q(\alpha )\ll \sum _{\begin{array}{c}u\le \sqrt{x}\\ q\nmid u\end{array}}\min \left(\frac{x}{u},\frac{1}{\Vert u\alpha \Vert }\right).$$

$$\begin{array}{rl}\Vert u\alpha \Vert & \ge \Vert \frac{au}{q}\Vert -\Vert u\beta \Vert \\ & \ge \Vert \frac{au}{q}\Vert -u|\beta |\\ & \ge \Vert \frac{au}{q}\Vert -\frac{u}{2q\sqrt{x}}\\ & \ge \Vert \frac{au}{q}\Vert -\frac{1}{2q}\\ & \ge \frac{1}{2}\Vert \frac{au}{q}\Vert ,\end{array}$$

$$G_q(\alpha )\ll \sum _{\begin{array}{c}u\le \sqrt{x}\\ q\nmid u\end{array}}\frac{1}{\Vert \frac{au}{q}\Vert }.$$

$$\sum _{1\le u\le \frac{q}{2}}\frac{1}{\Vert \frac{au}{q}\Vert }=\sum _{1\le b\le \frac{q}{2}}\frac{q}{b}\ll q\log q.$$

$$\sum _{\begin{array}{c}u\le \sqrt{x}\\ q\nmid u\end{array}}\frac{1}{\Vert \frac{au}{q}\Vert }\le 2\sum _{1\le u\le \frac{q}{2}}\frac{1}{\Vert \frac{au}{q}\Vert }\ll q\log q,$$

$$G_q(\alpha )\ll \frac{\sqrt{x}}{q}(q\log q)\ll \sqrt{x}\log q.$$

Full text

On the $L^{1}$ norm of an exponential sum involving the divisor function

D. A. Goldston and M. Pandey

∗ The first author was in residence at the Mathematical Sciences Research Institute in Berkeley, California (supported by the National Science Foundation under Grant No. DMS-1440140), during the Spring 2017 semester.

1. Introduction

Let $\tau(n)=\sum_{d|n}1$ be the divisor function, and

[TABLE]

In 2001 Brudern [1] considered the $L^{1}$ norm of $S(\alpha)$ and claimed to prove

[TABLE]

However there is a mistake in the proof given there which depends on a lemma which is false. In this note we prove the following result.

Theorem 1**.**

We have

[TABLE]

The upper bound here is obtained by following Brudern’s proof with corrections. The lower bound is based on the method Vaughan introduced to study the $L^{1}$ norm for exponential sums over primes [3], and also makes use of a more recent result of Pongsriiam and Vaughan [2] on the divisor sum in arithmetic progressions. We do not know whether the upper bound or the lower bound reflects the actual size of the $L^{1}$ norm here.

2. Proof of the upper bound

Let $u$ and $v$ always be positive integers. Following Brüdern, we have

[TABLE]

By Cauchy-Schwarz and Parseval

[TABLE]

and therefore by the triangle inequality

[TABLE]

Thus to prove the upper bound in Theorem 1 we need to establish

[TABLE]

We proceed as in the circle method. Clearly in (3) we can replace the integration range $[0,1]$ by $[1/Q,1+1/Q]$. By Dirichlet’s theorem for any $\alpha\in[1/Q,1+1/Q]$ we can find a fraction $\frac{a}{q}$, $1\leq q\leq Q$, $1\leq a\leq q$, $(a,q)=1$, with $|\alpha-\frac{a}{q}|\leq 1/(qQ)$. Thus the intervals $[\frac{a}{q}-\frac{1}{qQ},\frac{a}{q}+\frac{1}{qQ}]$ cover the interval $[1/Q,1+1/Q]$. Taking

[TABLE]

we obtain

[TABLE]

On each interval $[\frac{a}{q}-\frac{1}{2q\sqrt{x}},\frac{a}{q}+\frac{1}{2q\sqrt{x}}]$ we decompose $T(\alpha)$ into

[TABLE]

where

[TABLE]

and

[TABLE]

and have

[TABLE]

The upper bound in Theorem 1 follows from the following two lemmas.

Lemma 2** (Brüdern).**

We have

[TABLE]

Lemma 3**.**

We have

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In what follows we always assume $(a,q)=1$, and define the new variable $\beta$ by

[TABLE]

Proof of Lemma 2.

The proof follows from the estimate

[TABLE]

since this implies

[TABLE]

To prove (6), we first note that the conditions $q|u$ and $u\leq\sqrt{x}$ force $F_{q}(\alpha)=0$ when $q>\sqrt{x}$. Next, when $q\leq\sqrt{x}$ we write $u=jq$ and have

[TABLE]

Making use of the estimate

[TABLE]

we have

[TABLE]

In this sum $jq\leq\sqrt{x}$ so that $|jq\beta|\leq|\beta|\sqrt{x}$, and hence the condition $|\beta|\leq\frac{1}{2q\sqrt{x}}$ implies $|jq\beta|\leq\frac{1}{2q}\leq\frac{1}{2}$. Hence $\|jq\beta\|=jq|\beta|$, and we have

[TABLE]

∎

Proof of Lemma 3.

The proof follows from the estimate,

[TABLE]

since this implies

[TABLE]

by (4). To prove (8), we apply (7) to the sum over $v$ in $G_{q}(\alpha)$ and obtain

[TABLE]

Recalling $\|x\|=\|-x\|$ and the triangle inequality $\|x+y\|\leq\|x\|+\|y\|$, and using the conditions $1\leq u\leq\sqrt{x}$, $q\nmid u$, $|\beta|\leq\frac{1}{2q\sqrt{x}}$, we have

[TABLE]

and therefore

[TABLE]

Here $\left\|\frac{au}{q}\right\|=\frac{b}{q}$ for some integer $1\leq b\leq\frac{q}{2}$ and since the integers $\{au:1\leq u\leq\frac{q}{2}\}$ are distinct modulo $q$ since $(a,q)=1$, we see

[TABLE]

If $q>\sqrt{x}$ then

[TABLE]

while if $q\leq\sqrt{x}$ then the sum bounding $G_{q}(\alpha)$ can be split into $\ll\frac{\sqrt{x}}{q}$ sums of this type and

[TABLE]

∎

3. Proof of the lower bound

Following Brüdern, consider the intervals $|\alpha-\frac{a}{q}|\leq 1/(4x)$ for $1\leq a\leq q\leq Q$, where we take $\frac{1}{2}\sqrt{x}\leq Q\leq\sqrt{x}$. These intervals are pairwise disjoint because for two distinct fractions $|a/q-a^{\prime}/q^{\prime}|\geq 1/(qq^{\prime})\geq 1/x$. (We will see later why these intervals have been chosen shorter than required to be disjoint.) Hence, using (5)

[TABLE]

Next we follow Vaughan’s method [3] and apply the triangle inequality to obtain the lower bound

[TABLE]

Letting

[TABLE]

where

[TABLE]

is the Ramanujan sum, our lower bound may now be written as

[TABLE]

To complete the proof of the lower bound we need the following lemma, which we prove at the end of this section.

Lemma 4**.**

For $q\geq 1$ we have

[TABLE]

where $\gamma$ is Euler’s constant.

Proof of the lower bound in Theorem 1.

For any exponential sum $T(x;\beta)=\sum_{n\leq x}a_{n}e(n\beta)$ we have by partial summation or direct verification

[TABLE]

Taking $T(x;\beta)=U_{q}(x;\beta)$ we thus obtain from (10) and the triangle inequality

[TABLE]

By Lemma 4, with $q\leq\frac{1}{2}\sqrt{x}$,

[TABLE]

Using $|\beta|\leq 1/(4x)$, we have

[TABLE]

We conclude, returning to (12) and making use of Lemma 4 again,

[TABLE]

It is easy to see that the sum above is $\gg\sqrt{x}$ which suffices to proves the lower bound. More precisely, using

[TABLE]

a simple calculation gives the well-known result

[TABLE]

and then by partial summation we find

[TABLE]

∎

Proof of Lemma 4.

Pongsriiam and Vaughan [2] recently proved the following very useful result on the divisor function in arithmetic progressions. For inteqer $a$ and $d\geq 1$ and real $x\geq 1$ we have

[TABLE]

where $\gamma$ is Euler’s constant and $c_{r}(a)$ is the Ramanujan sum. We need the special case when $a=0$ which along with the situation $(a,d)>1$ is explicitly allowed in this formula. Hence we have

[TABLE]

where

[TABLE]

Making use of

[TABLE]

and (13) we have

[TABLE]

We evaluate the sum above using Dirichlet convolution and the identity $1*\mu=\delta$ where $\delta(n)$ is the identity for Dirichlet convolution defined to be 1 if $n=1$ and zero otherwise. Hence

[TABLE]

and Lemma 4 is proved. ∎