A Diophantine approximation problem with two primes and one $k$-th power of a prime

TL;DR

This paper improves bounds on a Diophantine approximation problem involving two primes and a prime's k-th power, extending the range of k and providing stronger approximation bounds.

Contribution

The authors extend the range of k from 1<k<4/3 to 1<k≤3 and enhance approximation bounds using harmonic analysis techniques.

Findings

Extended the k-range to 1<k≤3

Combined Harman's technique with new estimates for exponential sums

Provided stronger bounds for Diophantine approximation

Abstract

We refine a result of the last two Authors of [8] on a Diophantine approximation problem with two primes and a $k$-th power of a prime which was only proved to hold for $1<k<4/3$. We improve the $k$-range to $1<k\le 3$ by combining Harman's technique on the minor arc with a suitable estimate for the $L^4$-norm of the relevant exponential sum over primes $S_k$. In the common range we also give a stronger bound for the approximation.

Equations

$$|{\lambda }_1{p}_1^{k_1}+\cdots +{\lambda }_r{p}_r^{k_r}-\omega |\le \eta ,$$

$$|\lambda_{1}p_{1}+\lambda_{2}p_{2}+\lambda_{3}p_{3}^{k}-\omega|\leq\bigl{(}\max\{p_{1},p_{2},p_{3}^{k}\}\bigr{)}^{-\psi(k)+\varepsilon}$$

$$\psi (k)=\{\begin{array}{ll}(3-2k)/(6k)& \text{if 1<k≤56,}\\ 1/12& \text{if 56<k≤2,}\\ (3-k)/(6k)& \text{if 2<k<3,}\\ 1/24& \text{if k=3.}\end{array}$$

$$S_k(\alpha )=\sum _{\delta X\le p^k\le X}\log pe(p^k\alpha ),U_k(\alpha )=\sum _{\delta X\le n^k\le X}e(n^k\alpha )\text{and}{T}_k(\alpha )={\int }_{(\delta X{)}^{1/k}}^{X^{1/k}}e(t^k\alpha )\mathrm{d}t,$$

$$S_k(\alpha ){\ll }_{k,\delta }{X}^{1/k},T_k(\alpha ){\ll }_{k,\delta }{X}^{1/k-1}\min \{X,|\alpha {|}^{-1}\}.$$

$$T_k(\alpha )-U_k(\alpha ){\ll }_{k,\delta }1+|\alpha |X.$$

$${\hat{K}}_{\eta }(\alpha ):=\max \{0,\eta -|\alpha |\},\text{where}\eta >0,$$

$$K_{\eta}(\alpha)=\Bigl{(}\frac{\sin(\pi\alpha\eta)}{\pi\alpha}\Bigr{)}^{2}$$

$$K_{\eta }(\alpha )\ll \min \{{\eta }^2,|\alpha {|}^{-2}\}.$$

$$\mathcal{P}(X)=\{(p_1,p_2,p_3):\delta X<p_1,p_2\le X,\delta X<p_3^k\le X\}$$

$$\mathcal{I}(\eta ,\omega ,\mathfrak{X})={\int }_{\mathfrak{X}}{S}_1({\lambda }_1\alpha )S_1({\lambda }_2\alpha )S_k({\lambda }_3\alpha )K_{\eta }(\alpha )e(-\omega \alpha )\mathrm{d}\alpha ,$$

$$\mathcal{I}(\eta ,\omega ,\mathbb{R})$$

$$\le \eta (\log X{)}^3\mathcal{N}(X),$$

$$\mathfrak{M}$$

$$\mathfrak{m}$$

$$\mathfrak{M}$$

$$\mathfrak{m}$$

$$\mathcal{J}_{k}(X,h)=\int_{X}^{2X}\bigl{(}\theta((x+h)^{1/k})-\theta(x^{1/k})-((x+h)^{1/k}-x^{1/k})\bigr{)}^{2}\,\mathrm{d}x,$$

$$\int_{-Y}^{Y}|S_{k}(\alpha)-U_{k}(\alpha)|^{2}\,\mathrm{d}\alpha\ll\frac{{X^{2/k-2}}\log^{2}X}{Y}+Y^{2}X+Y^{2}\mathcal{J}_{k}\Bigl{(}X,\frac{1}{2Y}\Bigr{)}.$$

$$\mathcal{J}_{k}(X,h)\ll h^{2}X^{2/k-1}\exp\Bigl{(}-c_{1}\Bigl{(}\frac{\log X}{\log\log X}\Bigr{)}^{1/3}\Bigr{)}$$

$$|n_1^k+n_2^k-n_3^k-n_4^k|<\gamma ,X^{1/k}<n_1,n_2,n_3,n_4\le 2X^{1/k}.$$

$$\mathcal{B}(X^{1/k};k;\gamma)\ll\bigl{(}X^{2/k}+\gamma X^{4/k-1}\bigr{)}X^{\varepsilon}.$$

$$\int_{-\tau}^{\tau}|S_{k}(\alpha)|^{4}\,\mathrm{d}\alpha\ll\bigl{(}\tau X^{2/k}+X^{4/k-1}\bigr{)}X^{\varepsilon}\quad\textrm{and}\quad\int_{n}^{n+1}|S_{k}(\alpha)|^{4}\,\mathrm{d}\alpha\ll\bigl{(}X^{2/k}+X^{4/k-1}\bigr{)}X^{\varepsilon}.$$

$${\int }_{-\tau }^{\tau }|S_k(\alpha ){|}^4\mathrm{d}\alpha$$

$$\displaystyle\ll(\log X)^{4}\sum_{\delta X<p_{1}^{k},p_{2}^{k},p_{3}^{k},p_{4}^{k}\leq X}\min\Biggl{\{}\tau,\frac{1}{|p_{1}^{k}+p_{2}^{k}-p_{3}^{k}-p_{4}^{k}|}\Biggr{\}}$$

$$\displaystyle\ll(\log X)^{4}\sum_{\delta X<n_{1}^{k},n_{2}^{k},n_{3}^{k},n_{4}^{k}\leq X}\min\Biggl{\{}\tau,\frac{1}{|n_{1}^{k}+n_{2}^{k}-n_{3}^{k}-n_{4}^{k}|}\Biggr{\}}$$

$$\ll U\tau (\log X{)}^4+V(\log X{)}^4,$$

$$U:=\sum _{\begin{array}{c}\delta X<n_1^k,n_2^k,n_3^k,n_4^k\le X\\ |n_1^k+n_2^k-n_3^k-n_4^k|\le 1/\tau \end{array}}1,\text{and}V:=\sum _{\begin{array}{c}\delta X<n_1^k,n_2^k,n_3^k,n_4^k\le X\\ |n_1^k+n_2^k-n_3^k-n_4^k|>1/\tau \end{array}}\frac{1}{|n_1^k+n_2^k-n_3^k-n_4^k|},$$

$$U\ll\mathcal{B}(X^{1/k};k;1/\tau)\ll\Bigl{(}X^{2/k}+\frac{1}{\tau}X^{4/k-1}\Bigr{)}X^{\varepsilon}.$$

$$V$$

$$\displaystyle\ll\log X\max_{1/\tau<W\ll X}\Bigl{(}\frac{1}{W}\mathcal{B}(X^{1/k};k;2W)\Bigr{)}\ll\max_{1/\tau<W\ll X}\Bigl{(}X^{4/k-1}+\frac{X^{2/k}}{W}\Bigr{)}X^{\varepsilon}$$

$$\ll (\tau X^{2/k}+X^{4/k-1})X^{\epsilon }.$$

$$\int_{-\tau}^{\tau}|S_{k}(\alpha)|^{2}\,\mathrm{d}\alpha\ll\bigl{(}\tau X^{1/k}+X^{2/k-1}\bigr{)}(\log X)^{3}\quad\textrm{and}\quad\int_{n}^{n+1}\!\!\!|S_{k}(\alpha)|^{2}\,\mathrm{d}\alpha\ll X^{1/k}(\log X)^{3}.$$

$$S_{1}(\alpha)\ll\Bigl{(}\frac{X}{\sqrt{q}}+\sqrt{Xq}+X^{4/5}\Bigr{)}(\log X)^{4}.$$

Full text

1 Introduction

This paper deals with an improvement of the result contained in [8], which is due to the last two Authors: we refer to its introduction for a more thorough description of the general Diophantine problem with prime variables. Here we just recall that the goal is to prove that the inequality

[TABLE]

where $k_{1}$, …, $k_{r}$ are fixed positive numbers, $\lambda_{1}$, …, $\lambda_{r}$ are fixed non-zero real numbers and $\eta>0$ is arbitrary, has infinitely many solutions in prime variables $p_{1}$, …, $p_{r}$ for any given real number $\omega$, under as mild Diophantine assumptions on $\lambda_{1}$, …, $\lambda_{r}$ as possible. In some cases, it is even possible to prove that the above inequality holds when $\eta$ is a small negative power of the largest prime occurring in it, usually when $1/k_{1}+\cdots+1/k_{r}$ is large enough.

The problem tackled in [8] had $r=3$, $k_{1}=k_{2}=1$, $k_{3}=k\in(1,4/3)$. Assuming that $\lambda_{1}/\lambda_{2}$ is irrational and that the coefficients $\lambda_{j}$ are not all of the same sign, the last two Authors proved that one can take \eta=\bigl{(}\max\{p_{1},p_{2},p_{3}^{k}\}\bigr{)}^{-\phi(k)+\varepsilon} for any fixed $\varepsilon>0$, where $\phi(k)=(4-3k)/(10k)$. Our purpose in this paper is to improve on this result both in the admissible range for $k$ and in the exponent, replacing $\phi(k)$ by a larger value in the common range. More precisely, we prove the following Theorem.

Theorem 1

Assume that $1<k\leq 3$, $\lambda_{1}$, $\lambda_{2}$ and $\lambda_{3}$ are non-zero real numbers, not all of the same sign, that $\lambda_{1}/\lambda_{2}$ is irrational and let $\omega$ be a real number. The inequality

[TABLE]

has infinitely many solutions in prime variables $p_{1}$, $p_{2}$, $p_{3}$ for any $\varepsilon>0$, where

[TABLE]

We point out that in the common range $1<k<4/3$ we have $\psi(k)>\phi(k)$. We also remark that the strong bounds for the exponential sum $S_{k}$, defined in (3) below, that recently became available for integral $k$ (see Bourgain [1] and Bourgain, Demeter & Guth [2]) are not useful in our problem.

The technique used to tackle this problem is the variant of the circle method introduced in the 1940’s by Davenport & Heilbronn [4], where the integration on a circle, or equivalently on the interval $[0,1]$, is replaced by integration on the whole real line. Our improvement is due to the use of the Harman technique on the minor arc and to the fourth-power average for the exponential sum $S_{k}$ for $k\geq 1$.

We thank the anonymous referee for an extremely careful reading of a previous version of this paper.

2 Outline of the proof

Throughout this paper $p_{i}$ denotes a prime number, $k\geq 1$ is a real number, $\varepsilon$ is an arbitrarily small positive number whose value may vary depending on the occurrences and $\omega$ is a fixed real number. In order to prove that (1) has infinitely many solutions, it is sufficient to construct an increasing sequence $X_{n}$ that tends to infinity such that (1) has at least one solution with $\max\{p_{1},p_{2},p_{3}^{k}\}\in[\delta X_{n},X_{n}]$, with a fixed $\delta>0$ which depends only on the choice of $\lambda_{1}$, $\lambda_{2}$ and $\lambda_{3}$. Let $q$ be a denominator of a convergent to $\lambda_{1}/\lambda_{2}$ and let $X_{n}=X$ (dropping the suffix $n$) run through the sequence $X=q^{3}$. The main quantities we will use are:

[TABLE]

where $e(\alpha)=e^{2\pi i\alpha}$. We will approximate $S_{k}$ with $T_{k}$ and $U_{k}$. By the Prime Number Theorem and first derivative estimates for trigonometric integrals we have

[TABLE]

Moreover the Euler summation formula implies that

[TABLE]

We also need a continuous function we will use to detect the solutions of (1), so we introduce

[TABLE]

which is the Fourier transform of the function $K_{\eta}$ defined by

[TABLE]

for $\alpha\neq 0$ and, by continuity, $K_{\eta}(0)=\eta^{2}$. A well-known estimate is

[TABLE]

Let now

[TABLE]

and

[TABLE]

where $\mathfrak{X}$ is a measurable subset of $\mathbb{R}$. From (3) and using the Fourier transform of $K_{\eta}(\alpha)$, we get

[TABLE]

where $\mathcal{N}(X)$ actually denotes the number of solutions of the inequality (1) with $(p_{1},p_{2},p_{3})\in\mathcal{P}(X)$. In other words $\mathcal{I}(\eta,\omega,\mathbb{R})$ provides a lower bound for the quantity we are interested in; therefore it is sufficient to prove that $\mathcal{I}(\eta,\omega,\mathbb{R})>0$.

We now decompose $\mathbb{R}$ into subsets such that $\mathbb{R}=\mathfrak{M}\,\cup\ \mathfrak{M}^{*}\,\cup\ \mathfrak{m}\,\cup\,\mathfrak{t}$ where $\mathfrak{M}$ is the major arc, $\mathfrak{M}^{*}$ is the intermediate arc (which is non-empty only for some values of $k$, see section 6), $\mathfrak{m}$ is the minor arc and $\mathfrak{t}$ is the trivial arc. The decomposition is the following: if $1<k<5/2$ we consider

[TABLE]

while, for $5/2\leq k\leq 3$, we set

[TABLE]

where the parameters $P=P(X)>1$ and $R=R(X)>1/\eta$ are chosen later (see (15) and (16)) as well as $\eta=\eta(X)$, that, as we explained before, we would like to be a small negative power of $\max\{p_{1},p_{2},p_{3}^{k}\}$ (and so of $X$). We have to distinguish two cases in the previous decomposition of the real line in order to avoid a gap between the end of the major arc and the beginning of the minor arc, where we can prove Lemma 12 in the form that we need: see the comments at the beginning of section 6 and just before the statement of Lemma 12. As we will see later in section 6, we need to introduce intermediate arc only for $k\geq 5/2$.

The constraints on $\eta$ are in (18), (20) and (21), according to the value of $k$. In any case, we have $\mathcal{I}(\eta,\omega,\mathbb{R})=\mathcal{I}(\eta,\omega,\mathfrak{M})+\mathcal{I}(\eta,\omega,\mathfrak{M}^{*})+\mathcal{I}(\eta,\omega,\mathfrak{m})+\mathcal{I}(\eta,\omega,\mathfrak{t})$. We expect that $\mathfrak{M}$ provides the main term with the right order of magnitude without any special hypothesis on the coefficients $\lambda_{j}$. It is necessary to prove that $\mathcal{I}(\eta,\omega,\mathfrak{M}^{*})$, $\mathcal{I}(\eta,\omega,\mathfrak{m})$ and $\mathcal{I}(\eta,\omega,t)$ are o\bigl{(}\mathcal{I}(\eta,\omega,\mathfrak{M})\bigr{)} as $X\to+\infty$ on the particular sequence chosen: we show that the contribution from trivial arc is “tiny” with respect to the main term. The main difficulty is to estimate the minor arc contribution; in this case we will need the full force of the hypothesis on the coefficients $\lambda_{j}$ and the theory of continued fractions.

Remark: from now on, anytime we use the symbol $\ll$ or $\gg$ we drop the dependence of the approximation from the constants $\lambda_{j},\delta$ and $k$. We use the notation $f=\infty(g)$ for $g={o}\left(f\right)$.

3 Lemmas

In their original paper [4] Davenport and Heilbronn approximate directly the difference $|S_{k}(\alpha)-T_{k}(\alpha)|$ estimating it with $\mathcal{O}\left(1\right)$. The $L^{2}$-norm estimation approach (see Brüdern, Cook & Perelli [3] and [8]) improves on this taking the $L^{2}$-norm of $|S_{k}(\alpha)-T_{k}(\alpha)|$: this leads to the possibility of having a wider major arc compared to the original approach. We introduce the generalized version of the Selberg integral

[TABLE]

where $\theta(x)=\sum_{p\leq x}\log p$ is the usual Chebyshev function. We have the following lemmas.

Lemma 1** ([7], Theorem 3.1)**

Let $k\geq 1$ be a real number. For $0<Y\leq 1/2$ we have

[TABLE]

Lemma 2** ([7], Theorem 3.2)**

Let $k\geq 1$ be a real number and $\varepsilon$ be an arbitrarily small positive constant. There exists a positive constant $c_{1}(\varepsilon)$, which does not depend on $k$, such that

[TABLE]

uniformly for $X^{1-5/(6k)+\varepsilon}\leq h\leq X$.

In order to prove our crucial Lemma 4 on the $L^{4}$-norm of $S_{k}(\alpha)$, we need the following technical result.

Lemma 3

Let $\varepsilon>0$, $k>1$ and $\gamma>0$. Let further $\mathcal{B}(X^{1/k};k;\gamma)$ denote the number of solutions of the inequalities

[TABLE]

Then

[TABLE]

Proof 1

This is an immediate consequence of Theorem 2 of Robert & Sargos [9]; we just need to choose $M=X^{1/k}$, $\alpha=k$ and $\gamma=\delta M^{k}$ there. \qed

Lemma 4

Let $\varepsilon>0$, $\delta>0$, $k>1$, $n\in\mathbb{N}$ and $\tau>0$. Then we have

[TABLE]

Proof 2

A direct computation gives

[TABLE]

where

[TABLE]

say. Using Lemma 3 on $U$ we get

[TABLE]

Concerning $V$, by a dyadic argument we get

[TABLE]

Combining (9)-(11), the first part of the lemma follows. The second part can be proved in a similar way. \qed

We need the following result in the proof of Lemma 9 and also when dealing with $\mathfrak{M}^{*}$; see section 6.

Lemma 5

Let $\delta>0$, $k>1$, $n\in\mathbb{N}$ and $\tau>0$. Then

[TABLE]

Proof 3

It follows directly from the proof of Lemma 7 of Tolev [10] by letting $c=k$ and using $X^{1/k}$ instead of $X$ there. We explicitly remark that the condition $c\in(1,15/14)$ in Tolev’s original version of this lemma depends on other parts of his paper; in fact the proof of Lemma 7 of [10] holds for every $c>1$. \qed

We now state some other lemmas which will be mainly useful on the minor and trivial arcs.

Lemma 6** (Vaughan [11], Theorem 3.1)**

Let $\alpha$ be a real number and $a,q$ be positive integers satisfying $(a,q)=1$ and $|\alpha-a/q|<1/q^{2}$. Then

[TABLE]

Lemma 7

Let $X^{-1}\ll|\alpha|\ll X^{-3/5}$. Then $S_{1}(\alpha)\ll X^{1/2}|\alpha|^{-1/2}(\log X)^{4}$.

Proof 4

It follows immediately from Lemma 6 by choosing $q=\lfloor 1/\alpha\rfloor$ and $a=1$. \qed

Lemma 8

Let $\lambda\in\mathbb{R}\setminus\{0\}$, $X\geq Z\geq X^{4/5}(\log X)^{5}$ and $|S_{1}(\lambda\alpha)|>Z$. Then there are coprime integers $(a,q)=1$ satisfying

[TABLE]

Proof 5

Let $Q$ be a parameter that we will choose later. By Dirichlet’s theorem there exist coprime integers $(a,q)=1$ such that $1\leq q\leq Q$ and $|q\lambda\alpha-a|\ll Q^{-1}\leq q^{-1}$. The choice

[TABLE]

allows us to prove the second part of the statement and to neglect some terms in the estimations of $|S_{1}(\lambda\alpha)|$. Using Lemma 6, knowing that $Z\geq X^{4/5}(\log X)^{5}$ and $|S_{1}(\lambda\alpha)|>Z$, we can rewrite the bound for $|S_{1}(\lambda\alpha)|$ neglecting the term $X^{4/5}$:

[TABLE]

The condition $q\leq Q$ allows us to neglect the term $X^{1/2}q^{1/2}$ and deal with small values of $q$; in fact, if $q>X^{1/2}$ then we would have a contradiction

[TABLE]

Then $q\leq\min\{X^{1/2},Q\}=X^{1/2}$, since $Z=X^{4/5}(\log X)^{5}>X^{3/4}(\log X)^{5}$. Moreover, we can rewrite the inequality on $|S_{1}(\lambda\alpha)|$ as

[TABLE]

and finally we get $q^{1/2}Z\ll X(\log X)^{4}$, which completes the proof. \qed

The optimizations in section 7 depend either on $L^{2}$ or on $L^{4}$ averages of $S_{k}$, according to the value of $k$; these are provided by the following Lemmas. For brevity, we skip the proof of the first one, remarking that it requires Lemma 5.

Lemma 9** (Lemma 5 of [8])**

Let $\lambda\in\mathbb{R}\setminus\{0\}$, $k>1$, $0<\eta<1$, $R>1/\eta$ and $1<P<X$. We have

[TABLE]

Lemma 10

Let $\lambda\in\mathbb{R}\setminus\{0\}$, $\varepsilon>0$, $k>1$, $0<\eta<1$ , $R>1/\eta$ and $1<P<X$. Then

[TABLE]

Proof 6

Using (6) we obtain

[TABLE]

say. By Lemma 4, we immediately get

[TABLE]

Moreover, again by Lemma 4, we have that

[TABLE]

Combining (12)-(14) and using $0<\eta<1$, the lemma follows. \qed

As we remarked in the introduction, stronger bounds are now available for larger integral $k$, but they are not useful for our purpose. The next Lemma provides the additional information that enables us to give a non-trivial result also when $k=3$.

Lemma 11

Let $\lambda\in\mathbb{R}\setminus\{0\}$, $\varepsilon>0$, $0<\eta<1$, $R>1/\eta$ and $1<P<X$. Then

[TABLE]

Proof 7

Inserting Hua’s estimate in [6], i.e. $\int_{0}^{1}|S_{3}(\alpha)|^{8}\ \mathrm{d}\alpha\ll X^{5/3+\varepsilon},$ in the body of the proof of Lemma 10 and exploiting the periodicity of $S_{3}(\alpha)$, the result follows immediately. \qed

Another lemma on the minor arc is inserted in the body of section 7.

4 The major arc

We recall the definitions in (7) and (8). The major arc computation is the same as in [8]:

[TABLE]

say.

4.1 Main term: lower bound for $J_{1}$

As the reader might expect the main term is given by the summand $J_{1}$.

Let $H(\alpha)=T_{1}(\lambda_{1}\alpha)T_{1}(\lambda_{2}\alpha)T_{k}(\lambda_{3}\alpha)K_{\eta}(\alpha)e(-\omega\alpha)$ so that

[TABLE]

Using (6) and (4), we get

[TABLE]

provided that $P\rightarrow+\infty$. Let now $D=[\delta X,X]^{2}\times[(\delta X)^{1/k},X^{1/k}]$. We obtain

[TABLE]

Apart from trivial changes of sign, there are essentially two cases:

$\lambda_{1}>0$, $\lambda_{2}>0$, $\lambda_{3}<0$ 2. 2.

$\lambda_{1}>0$, $\lambda_{2}<0$, $\lambda_{3}<0$.

We deal with the first one. We warn the reader that here it may be necessary to adjust the value of $\delta$ in order to guarantee the necessary set inclusions. After a suitable change of variables, letting $D^{\prime}=[\delta X,(1-\delta)X]^{3}$, we find that

[TABLE]

Apart from sign, the computation is essentially symmetrical with respect to the coefficients $\lambda_{j}$: we assume, as we may, that $|\lambda_{3}|\geq\max\{\lambda_{1},\lambda_{2}\}$, the other cases being similar. Now, for $j=1,2$ let $a_{j}=\dfrac{2\delta|\lambda_{3}|}{|\lambda_{j}|}$, $b_{j}=\dfrac{3}{2}a_{j}$ and $\mathcal{D}_{j}=[a_{j}X,b_{j}X]$; if $u_{j}\in\mathcal{D}_{j}$ for $j=1,2$ then

[TABLE]

so that, for every choice of $(u_{1},u_{2})$ the interval $[a,b]$ with endpoints $\pm\eta/|\lambda_{3}|+(\lambda_{1}u_{1}+\lambda_{2}u_{2})/|\lambda_{3}|$ is contained in $[\delta X,(1-\delta)X]$. In other words, for $u_{3}\in[a,b]$ the values of $\lambda_{1}u_{1}+\lambda_{2}u_{2}+\lambda_{3}u_{3}$ cover the whole interval $[-\eta,\eta]$. Hence for any $(u_{1},u_{2})\in\mathcal{D}_{1}\times\mathcal{D}_{2}$ we have

[TABLE]

Summing up, we get

[TABLE]

which is the expected lower bound.

4.2 Bound for $J_{2}$, $J_{3}$ and $J_{4}$

The computations for $J_{2}$ and $J_{3}$ are similar to and simpler than the corresponding one for $J_{4}$; moreover the most restrictive condition on $P$ arises from $J_{4}$; hence we will skip the computation for both $J_{2}$ and $J_{3}$. Using the triangle inequality and (6),

[TABLE]

say, where $U_{k}(\lambda_{3}\alpha)$ and $T_{k}(\lambda_{3}\alpha)$ are defined in (3). Using the Cauchy-Schwarz inequality, Lemmas 1-2 and trivial bounds yields, for any fixed $A>0$,

[TABLE]

as long as $A>1$, provided that $P\leq X^{5/(6k)-\varepsilon}$. Using again the Cauchy-Schwarz inequality, (5) and trivial bounds, we see that

[TABLE]

Taking P=o\bigl{(}X^{1/k}(\log X)^{-1}\bigr{)} we get \eta^{2}B_{4}=o\bigl{(}\eta^{2}X^{1+1/k}\bigr{)}. We may therefore choose

[TABLE]

5 The trivial arc

We recall that the trivial arc is defined in (7) and (8). Using the Cauchy-Schwarz inequality and (4), we see that

[TABLE]

say. Using the PNT and the periodicity of $S_{1}(\alpha)$, for every $j=1,2$ we have that

[TABLE]

Hence, recalling that $|\mathcal{I}(\eta,\omega,\mathfrak{t})|$ has to be $o(\eta^{2}X^{1+1/k})$, the choice

[TABLE]

is admissible.

6 The intermediate arc: $5/2\leq k\leq 3$

In section 7 we apply Harman’s technique to the minor arc, using Lemma 8 as the starting point. We remark that in the course of the proof of Lemma 12 it is crucial that both the integers $a_{1}$ and $a_{2}$ appearing in (22) below do not vanish; in fact, if $a_{1}=0$, say, then $\alpha$ is very small ($\alpha\ll X^{-2/3}$) and, according to our definitions above, it belongs to $\mathfrak{M}\cup\mathfrak{M}^{*}$.

For small $k$ we do not need an intermediate arc, because the major arc is wide enough to rule out the possibility that $a_{1}a_{2}=0$ for $\alpha\in\mathfrak{m}$. For larger values of $k$, the constraint (15) implies that there is a gap between the major arc and the minor arc which we need to fill: see the definition in (8). Using the intermediate arc $\mathfrak{M}^{*}$, we are able to cover more than needed.

Let $5/2\leq k\leq 3$: we now show that the contribution of $\mathfrak{M}^{*}$ is negligible. Using (6), Lemma 7, the Cauchy-Schwarz inequality and (15) we get

[TABLE]

where we also used Lemma 5 with $\tau=X^{-3/5}$ and the fact that $k\geq 5/2$. The last estimate is o\bigl{(}\eta^{2}X^{1+1/k}\bigr{)} for every $5/2\leq k<55/12$.

7 The minor arc

Here we use Harman’s technique as described in [5]. The minor arc $\mathfrak{m}$ is defined in (7) and (8), according to the value of $k$. In view of using Lemma 8, we now split $\mathfrak{m}$ into subsets $\mathfrak{m}_{1}$, $\mathfrak{m}_{2}$ and $\mathfrak{m}^{*}=\mathfrak{m}\setminus(\mathfrak{m}_{1}\cup\mathfrak{m}_{2})$, where

[TABLE]

In order to obtain the optimization, we chose to split the range for $k$ into two intervals in which to take advantage of the $L^{2}$-norm of $S_{k}(\alpha)$ in one case (Lemma 9) and the $L^{4}$-norm of $S_{k}(\alpha)$ in the other one (Lemma 10). The same choice will be made later in the discussion of the arc $\mathfrak{m}^{*}$. We will see that it is not possible to split the minor arc in another way in order to get a better result, in the present state of knowledge on exponential sums.

7.1 Bounds on $\mathfrak{m}_{1}\cup\mathfrak{m}_{2}$

Using Hölder’s inequality and Lemma 9, for $1<k\leq 6/5$ we obtain

[TABLE]

The estimate in (17) should be $o(\eta^{2}X^{1+1/k})$; hence this leads to the constraint

[TABLE]

where $f=\infty(g)$ means $g={o}\left(f\right)$.

Using Hölder’s inequality and Lemmas 9 and 10, for $6/5<k<3$ we obtain

[TABLE]

The estimate in (19) should be $o(\eta^{2}X^{1+1/k})$; hence this leads to

[TABLE]

If $k=3$ we use Lemmas 9 and 11 thus getting

[TABLE]

This bound leads to the constraint

[TABLE]

which justifies the last line of (2).

7.2 Bound on $\mathfrak{m}^{*}$

We recall our definitions in (7) and (8). It remains to discuss the set $\mathfrak{m}^{*}$ where the following bounds hold simultaneously

[TABLE]

where $T=P/X=X^{5/(6k)-1-\varepsilon}$ by our choice in (15) if $k<5/2$, and $T=X^{-3/5}$ otherwise. Using a dyadic dissection, we split $\mathfrak{m}^{*}$ into disjoint sets $E(Z_{1},Z_{2},y)$ in which, for $\alpha\in E(Z_{1},Z_{2},y)$, we have

[TABLE]

where $Z_{i}=2^{k_{i}}X^{5/6}(\log X)^{5}$ and $y=2^{k_{3}}X^{5/(6k)-1-\varepsilon}$ for some non-negative integers $k_{1},k_{2},k_{3}$.

It follows that the number of disjoint sets is, at most, $\ll(\log X)^{3}$. Let us write $\mathcal{A}$ as a shorthand for the set $E(Z_{1},Z_{2},y)$. We need an upper bound for the Lebesgue measure of $\mathcal{A}$. In the following Lemma, it is crucial that both the integers $a_{1}$ and $a_{2}$ appearing in (22) below do not vanish; in fact, if $a_{1}=0$, say, then $q_{1}=1$ and $\alpha$ is so small that it can not belong to $\mathfrak{m}$. If $k$ is large, we treat the range $[P/X,X^{-3/5}]$ and its symmetrical by means of the argument in section 6: this is needed because, in this case, the inequalities (22) below do not rule out the possibility that $a_{1}a_{2}=0$, unless $|\alpha|$ is large enough.

Lemma 12

Let $\varepsilon>0$. We have that $\mu(\mathcal{A})\ll yX^{8/3+\varepsilon}Z_{1}^{-2}Z_{2}^{-2}$, where $\mu(\cdot)$ denotes the Lebesgue measure.

Proof 8

If $\alpha\in\mathcal{A}$, by Lemma 8 there are coprime integers $(a_{1},q_{1})$ and $(a_{2},q_{2})$ such that

[TABLE]

We remark that $a_{1}a_{2}\neq 0$ otherwise we would have $\alpha\in\mathfrak{M}\cup\mathfrak{M}^{*}$. In fact, if $a_{1}a_{2}=0$, recalling the definitions of $Z_{i}$ and (22), $\alpha\ll q_{i}^{-1}X(\log X)^{10}Z_{i}^{-2}\ll X^{-2/3}$.

Now, we can further split $\mathfrak{m}^{*}$ into sets $I=I(Z_{1},Z_{2},y,Q_{1},Q_{2})$ where, on each set, $Q_{j}\leq q_{j}\leq 2Q_{j}$. Note that $a_{i}$ and $q_{i}$ are uniquely determined by $\alpha$; in the opposite direction, for a given quadruple $a_{1}$, $q_{1}$, $a_{2}$, $q_{2}$, the inequalities (22) define an interval of $\alpha$ of length

[TABLE]

by taking the geometric mean.

Now we need a lower bound for $Q_{1}Q_{2}$: by (22) we obtain

[TABLE]

Recalling that $Q_{i}\ll(X(\log X)^{4}/Z_{i})^{2}$ and that $Z_{i}\gg X^{5/6}(\log X)^{5}$, we have

[TABLE]

We recall that $q=X^{1/3}$ is a denominator of a convergent of $\lambda_{1}/\lambda_{2}$. Hence by (23), Legendre’s law of best approximation for continued fractions implies that $|a_{2}q_{1}|\geq q$ and by the same token, for any pair $\alpha$, $\alpha^{\prime}$ having distinct associated products $a_{2}q_{1}$,

[TABLE]

thus, by the pigeon-hole principle, there is at most one value of $a_{2}q_{1}$ in the interval $[rq,(r+1)q)$ for any positive integer $r$. Furthermore $a_{2}q_{1}$ determines $a_{2}$ and $q_{1}$ to within $X^{\varepsilon/2}$ possibilities (from the bound for the divisor function) and consequently also $a_{2}q_{1}$ determines $a_{1}$ and $q_{2}$ to within $X^{\varepsilon/2}$ possibilities from (23).

Hence we got a lower bound for $q_{1}q_{2}$, since, using $Q_{j}\leq q_{j}\leq 2Q_{j}$, we get

[TABLE]

for the quadruple under consideration.

As a consequence we obtain that the total length of the part of $I(Z_{1},Z_{2},y,Q_{1},Q_{2})$ with $a_{2}q_{1}\in[rq,(r+1)q)$ is

[TABLE]

Now we need a bound for $r$: since $a_{2}q_{1}\in[rq,(r+1),q)$, we have

[TABLE]

and hence we get

[TABLE]

Next, we sum on every interval to get an upper bound for the measure of $\mathcal{A}$: we get

[TABLE]

Standard estimates imply that the sum on the right is $\ll(q^{-1}yX^{4}(\log X)^{16}Z_{1}^{-2}Z_{2}^{-2})^{1/2}$, and recalling that $q=X^{1/3}$ we can finally write

[TABLE]

This proves the lemma. \qed

8 Conclusion

Here we finally justify the choice of the function $\psi$ in the statement of the main Theorem. Using Lemmas 9-10-12 we are now able to estimate $\mathcal{I}(\eta,\omega,\mathcal{A})$ for $1<k\leq 3$. For $k\geq\frac{5}{2}$, we also need the result in section 6.

If $1<k\leq 6/5$ we proceed as follows:

[TABLE]

Hence we need \eta=\infty\bigl{(}X^{1/3-1/(2k)+\varepsilon}\bigr{)}, which is the same condition we got in (18).

If $6/5<k<3$,

[TABLE]

Hence we need \eta=\infty\bigl{(}\max\{X^{1/6-1/(2k)+\varepsilon},X^{-1/12+\varepsilon}\}\bigr{)}, which is the same condition we got in (20). If $k=3$, using Lemmas 11 and 12 we obtain

[TABLE]

This leads to the same constraint for $\eta$ that we had in (21).