A Diophantine approximation problem with two primes and one k-th power of a prime
Alessandro Gambini, Alessandro Languasco, Alessandro Zaccagnini
aa r X i v : . [ m a t h . N T ] F e b A Diophantine approximation problem with two primes and one k -th power of a prime Alessandro Gambini a, ∗ , Alessandro Languasco b , Alessandro Zaccagnini a a Università di Parma, Dipartimento di Scienze Matematiche, Fisiche e Informatiche, Parco Area delle Scienze 53/a,43124 Parma, Italy. b Università di Padova, Dipartimento di Matematica “Tullio Levi-Civita”, Via Trieste 63, 35121 Padova, Italy.
Abstract
We refine a result of the last two Authors of [8] on a Diophantine approximation problem with twoprimes and a k -th power of a prime which was only proved to hold for 1 < k < /
3. We improve the k -range to 1 < k ≤ L -norm of the relevant exponential sum over primes. Keywords:
Diophantine inequalities, Goldbach-type problems, Hardy-Littlewood method primary secondary 11J25, 11P32, 11P55
1. Introduction
This paper deals with an improvement of the result contained in [8], which is due to the last twoAuthors: we refer to its introduction for a more thorough description of the general Diophantineproblem with prime variables. Here we just recall that the goal is to prove that the inequality | λ p k + · · · + λ r p k r r − ω | ≤ η, where k , . . . , k r are fixed positive numbers, λ , . . . , λ r are fixed non-zero real numbers and η > is arbitrary, has infinitely many solutions in prime variables p , . . . , p r for any given real number ω , under as mild Diophantine assumptions on λ , . . . , λ r as possible. In some cases, it is evenpossible to prove that the above inequality holds when η is a small negative power of the largestprime occurring in it, usually when / k + · · · + / k r is large enough.The problem tackled in [8] had r = , k = k = , k = k ∈ ( , / ) . Assuming that λ / λ is irrational and that the coefficients λ j are not all of the same sign, the last two Authors proved ∗ Corresponding author
Email addresses: [email protected] (Alessandro Gambini), [email protected] (Alessandro Languasco), [email protected] (Alessandro Zaccagnini)
Preprint submitted to Elsevier September 19, 2018
OUTLINE OF THE PROOF η = (cid:0) max { p , p , p k } (cid:1) − φ ( k ) + ε for any fixed ε > , where φ ( k ) = ( − k )/( k ) .Our purpose in this paper is to improve on this result both in the admissible range for k and inthe exponent, replacing φ ( k ) by a larger value in the common range. More precisely, we prove thefollowing Theorem. Theorem 1.
Assume that < k ≤ , λ , λ and λ are non-zero real numbers, not all of the samesign, that λ / λ is irrational and let ω be a real number. The inequality | λ p + λ p + λ p k − ω | ≤ (cid:0) max { p , p , p k } (cid:1) − ψ ( k ) + ε (1) has infinitely many solutions in prime variables p , p , p for any ε > , where ψ ( k ) = ( − k )/( k ) if < k ≤ , / if < k ≤ , ( − k )/( k ) if < k < , / if k = . (2)We point out that in the common range < k < / we have ψ ( k ) > φ ( k ) . We also remark thatthe strong bounds for the exponential sum S k , defined in (3) below, that recently became availablefor 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 the1940’s by Davenport & Heilbronn [4], where the integration on a circle, or equivalently on theinterval [ , ] , is replaced by integration on the whole real line. Our improvement is due to the useof the Harman technique on the minor arc and to the fourth-power average for the exponential sum S k for k ≥ .We thank the anonymous referee for an extremely careful reading of a previous version of thispaper.
2. Outline of the proof
Throughout this paper p i denotes a prime number, k ≥ is a real number, ε is an arbitrarily smallpositive number whose value may vary depending on the occurrences and ω is a fixed real number.In order to prove that (1) has infinitely many solutions, it is sufficient to construct an increasingsequence X n that tends to infinity such that (1) has at least one solution with max { p , p , p k } ∈[ δ X n , X n ] , with a fixed δ > which depends only on the choice of λ , λ and λ . Let q be adenominator of a convergent to λ / λ and let X n = X (dropping the suffix n ) run through thesequence X = q . The main quantities we will use are: S k ( α ) = Õ δ X ≤ p k ≤ X log p e ( p k α ) , U k ( α ) = Õ δ X ≤ n k ≤ X e ( n k α ) and T k ( α ) = ∫ X / k ( δ X ) / k e ( t k α ) d t , (3) OUTLINE OF THE PROOF e ( α ) = e π i α . We will approximate S k with T k and U k . By the Prime Number Theorem andfirst derivative estimates for trigonometric integrals we have S k ( α ) ≪ k ,δ X / k , T k ( α ) ≪ k ,δ X / k − min { X , | α | − } . (4)Moreover the Euler summation formula implies that T k ( α ) − U k ( α ) ≪ k ,δ + | α | X . (5)We also need a continuous function we will use to detect the solutions of (1), so we introduce b K η ( α ) : = max { , η − | α |} , where η > , which is the Fourier transform of the function K η defined by K η ( α ) = (cid:16) sin ( παη ) πα (cid:17) for α , and, by continuity, K η ( ) = η . A well-known estimate is K η ( α ) ≪ min { η , | α | − } . (6)Let now P ( X ) = {( p , p , p ) : δ X < p , p ≤ X , δ X < p k ≤ X } and I ( η, ω, X ) = ∫ X S ( λ α ) S ( λ α ) S k ( λ α ) K η ( α ) e (− ωα ) d α, where X is a measurable subset of R . From (3) and using the Fourier transform of K η ( α ) , we get I ( η, ω, R ) = Õ ( p , p , p )∈ P ( X ) log p log p log p max (cid:8) , η − | λ p + λ p + λ p k − ω | (cid:9) ≤ η ( log X ) N ( X ) , where N ( X ) actually denotes the number of solutions of the inequality (1) with ( p , p , p ) ∈ P ( X ) .In other words I ( η, ω, R ) provides a lower bound for the quantity we are interested in; therefore itis sufficient to prove that I ( η, ω, R ) > .We now decompose R into subsets such that R = M ∪ M ∗ ∪ m ∪ t where M is the major arc, M ∗ is the intermediate arc (which is non-empty only for some values of k , see section 6), m is theminor arc and t is the trivial arc. The decomposition is the following: if < k < / we consider M = [− P / X , P / X ] , M ∗ = Ø , m = [ P / X , R ] ∪ [− R , − P / X ] , t = R \ ( M ∪ M ∗ ∪ m ) , (7) LEMMAS / ≤ k ≤ , we set M = [− P / X , P / X ] , M ∗ = [ P / X , X − / ] ∪ [− X − / , − P / X ] , m = [ X − / , R ] ∪ [− R , − X − / ] , t = R \ ( M ∪ M ∗ ∪ m ) , (8)where the parameters P = P ( X ) > and R = R ( X ) > / η are chosen later (see (15) and (16))as well as η = η ( X ) , that, as we explained before, we would like to be a small negative power of max { p , p , p k } (and so of X ). We have to distinguish two cases in the previous decomposition ofthe real line in order to avoid a gap between the end of the major arc and the beginning of the minorarc, where we can prove Lemma 12 in the form that we need: see the comments at the beginningof section 6 and just before the statement of Lemma 12. As we will see later in section 6, we needto introduce intermediate arc only for k ≥ / .The constraints on η are in (18), (20) and (21), according to the value of k . In any case, we have I ( η, ω, R ) = I ( η, ω, M ) + I ( η, ω, M ∗ ) + I ( η, ω, m ) + I ( η, ω, t ) . We expect that M provides themain term with the right order of magnitude without any special hypothesis on the coefficients λ j .It is necessary to prove that I ( η, ω, M ∗ ) , I ( η, ω, m ) and I ( η, ω, t ) are o (cid:0) I ( η, ω, M ) (cid:1) as X → + ∞ on the particular sequence chosen: we show that the contribution from trivial arc is “tiny” withrespect to the main term. The main difficulty is to estimate the minor arc contribution; in this casewe will need the full force of the hypothesis on the coefficients λ j and the theory of continuedfractions. Remark : from now on, anytime we use the symbol ≪ or ≫ we drop the dependence of theapproximation from the constants λ j , δ and k . We use the notation f = ∞( g ) for g = o ( f ) .
3. Lemmas
In their original paper [4] Davenport and Heilbronn approximate directly the difference | S k ( α ) − T k ( α )| estimating it with O ( ) . The L -norm estimation approach (see Brüdern, Cook & Perelli[3] and [8]) improves on this taking the L -norm of | S k ( α ) − T k ( α )| : this leads to the possibility ofhaving a wider major arc compared to the original approach. We introduce the generalized versionof the Selberg integral J k ( X , h ) = ∫ XX (cid:0) θ (( x + h ) / k ) − θ ( x / k ) − (( x + h ) / k − x / k ) (cid:1) d x , where θ ( x ) = Í p ≤ x log p is the usual Chebyshev function. We have the following lemmas. Lemma 1 ([7], Theorem 3.1).
Let k ≥ be a real number. For < Y ≤ / we have ∫ Y − Y | S k ( α ) − U k ( α )| d α ≪ X / k − log XY + Y X + Y J k (cid:16) X , Y (cid:17) . LEMMAS Lemma 2 ([7], Theorem 3.2).
Let k ≥ be a real number and ε be an arbitrarily small positiveconstant. There exists a positive constant c ( ε ) , which does not depend on k , such that J k ( X , h ) ≪ h X / k − exp (cid:16) − c (cid:16) log X log log X (cid:17) / (cid:17) uniformly for X − /( k ) + ε ≤ h ≤ X . In order to prove our crucial Lemma 4 on the L -norm of S k ( α ) , we need the following technicalresult. Lemma 3.
Let ε > , k > and γ > . Let further B ( X / k ; k ; γ ) denote the number of solutionsof the inequalities | n k + n k − n k − n k | < γ, X / k < n , n , n , n ≤ X / k . Then B ( X / k ; k ; γ ) ≪ (cid:0) X / k + γ X / k − (cid:1) X ε . Proof.
This is an immediate consequence of Theorem 2 of Robert & Sargos [9]; we just need tochoose M = X / k , α = k and γ = δ M k there. (cid:3) Lemma 4.
Let ε > , δ > , k > , n ∈ N and τ > . Then we have ∫ τ − τ | S k ( α )| d α ≪ (cid:0) τ X / k + X / k − (cid:1) X ε and ∫ n + n | S k ( α )| d α ≪ (cid:0) X / k + X / k − (cid:1) X ε . Proof.
A direct computation gives ∫ τ − τ | S k ( α )| d α = Õ δ X < p k , p k , p k , p k ≤ X ( log p ) · · · ( log p ) ∫ τ − τ e (( p k + p k − p k − p k ) α ) d α ≪ ( log X ) Õ δ X < p k , p k , p k , p k ≤ X min ( τ, | p k + p k − p k − p k | ) ≪ ( log X ) Õ δ X < n k , n k , n k , n k ≤ X min ( τ, | n k + n k − n k − n k | ) ≪ U τ ( log X ) + V ( log X ) , (9)where U : = Õ δ X < n k , n k , n k , n k ≤ X | n k + n k − n k − n k |≤ / τ , and V : = Õ δ X < n k , n k , n k , n k ≤ X | n k + n k − n k − n k | > / τ | n k + n k − n k − n k | , LEMMAS U we get U ≪ B ( X / k ; k ; 1 / τ ) ≪ (cid:16) X / k + τ X / k − (cid:17) X ε . (10)Concerning V , by a dyadic argument we get V ≪ log X (cid:16) max / τ< W ≪ X Õ δ X < n k , n k , n k , n k ≤ XW < | n k + n k − n k − n k |≤ W | n k + n k − n k − n k | (cid:17) ≪ log X max / τ< W ≪ X (cid:16) W B ( X / k ; k ; 2 W ) (cid:17) ≪ max / τ< W ≪ X (cid:16) X / k − + X / k W (cid:17) X ε ≪ ( τ X / k + X / k − ) X ε . (11)Combining (9)-(11), the first part of the lemma follows. The second part can be proved in a similarway. (cid:3) We need the following result in the proof of Lemma 9 and also when dealing with M ∗ ; seesection 6. Lemma 5.
Let δ > , k > , n ∈ N and τ > . Then ∫ τ − τ | S k ( α )| d α ≪ (cid:0) τ X / k + X / k − (cid:1) ( log X ) and ∫ n + n | S k ( α )| d α ≪ X / k ( log X ) . Proof.
It follows directly from the proof of Lemma 7 of Tolev [10] by letting c = k and using X / k instead of X there. We explicitly remark that the condition c ∈ ( , / ) in Tolev’s originalversion of this lemma depends on other parts of his paper; in fact the proof of Lemma 7 of [10]holds for every c > . (cid:3) 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 α be a real number and a , q be positive integerssatisfying ( a , q ) = and | α − a / q | < / q . Then S ( α ) ≪ (cid:16) X √ q + p X q + X / (cid:17) ( log X ) . Lemma 7.
Let X − ≪ | α | ≪ X − / . Then S ( α ) ≪ X / | α | − / ( log X ) . Proof.
It follows immediately from Lemma 6 by choosing q = ⌊ / α ⌋ and a = . (cid:3) LEMMAS Lemma 8.
Let λ ∈ R \ { } , X ≥ Z ≥ X / ( log X ) and | S ( λα )| > Z . Then there are coprimeintegers ( a , q ) = satisfying ≤ q ≪ (cid:16) X ( log X ) Z (cid:17) , | q λα − a | ≪ X ( log X ) Z . Proof.
Let Q be a parameter that we will choose later. By Dirichlet’s theorem there exist coprimeintegers ( a , q ) = such that ≤ q ≤ Q and | q λα − a | ≪ Q − ≤ q − . The choice Q = Z X ( log X ) allows us to prove the second part of the statement and to neglect some terms in the estimations of | S ( λα )| . Using Lemma 6, knowing that Z ≥ X / ( log X ) and | S ( λα )| > Z , we can rewrite thebound for | S ( λα )| neglecting the term X / : Z < | S ( λα )| ≪ ( X q − / + X / q / )( log X ) . The condition q ≤ Q allows us to neglect the term X / q / and deal with small values of q ; in fact,if q > X / then we would have a contradiction Z < | S ( λα )| ≪ X / q / ( log X ) ≤ X / ZX / ( log X ) ( log X ) = o ( Z ) . Then q ≤ min { X / , Q } = X / , since Z = X / ( log X ) > X / ( log X ) . Moreover, we can rewritethe inequality on | S ( λα )| as Z < | S ( λα )| ≪ X q − / ( log X ) and finally we get q / Z ≪ X ( log X ) , which completes the proof. (cid:3) The optimizations in section 7 depend either on L or on L averages of S k , according to thevalue of k ; these are provided by the following Lemmas. For brevity, we skip the proof of the firstone, remarking that it requires Lemma 5. Lemma 9 (Lemma 5 of [8]).
Let λ ∈ R \ { } , k > , < η < , R > / η and < P < X . Wehave ∫ RP / X | S ( λα )| K η ( α ) d α ≪ η X log X and ∫ RP / X | S k ( λα )| K η ( α ) d α ≪ η X / k ( log X ) . Lemma 10.
Let λ ∈ R \ { } , ε > , k > , < η < , R > / η and < P < X . Then ∫ RP / X | S k ( λα )| K η ( α ) d α ≪ η max { X / k , X / k − } X ε . THE MAJOR ARC Proof.
Using (6) we obtain ∫ RP / X | S k ( λα )| K η ( α ) d α ≪ η ∫ / η P / X | S k ( λα )| d α + ∫ R / η | S k ( λα )| d αα = A + B , (12)say. By Lemma 4, we immediately get A ≪ η ∫ | λ |/ η −| λ |/ η | S k ( α )| d α ≪ η max { X / k , η X / k − } X ε . (13)Moreover, again by Lemma 4, we have that B ≪ ∫ + ∞| λ |/ η | S k ( α )| d αα ≪ Õ n ≥| λ |/ η ( n − ) ∫ nn − | S k ( α )| d α ≪ η max { X / k , X / k − } X ε . (14)Combining (12)-(14) and using < η < , the lemma follows. (cid:3) 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 thatenables us to give a non-trivial result also when k = . Lemma 11.
Let λ ∈ R \ { } , ε > , < η < , R > / η and < P < X . Then ∫ RP / X | S ( λα )| K η ( α ) d α ≪ η X / + ε . Proof.
Inserting Hua’s estimate in [6], i.e. ∫ | S ( α )| d α ≪ X / + ε , in the body of the proof ofLemma 10 and exploiting the periodicity of S ( α ) , the result follows immediately. (cid:3) 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]: I ( η, ω, M ) = ∫ M S ( λ α ) S ( λ α ) S k ( λ α ) K η ( α ) e (− ωα ) d α = ∫ M T ( λ α ) T ( λ α ) T k ( λ α ) K η ( α ) e (− ωα ) d α + ∫ M ( S ( λ α ) − T ( λ α )) T ( λ α ) T k ( λ α ) K η ( α ) e (− ωα ) d α THE MAJOR ARC + ∫ M S ( λ α )( S ( λ α ) − T ( λ α )) T k ( λ α ) K η ( α ) e (− ωα ) d α + ∫ M S ( λ α ) S ( λ α )( S k ( λ α ) − T k ( λ α )) K η ( α ) e (− ωα ) d α = J + J + J + J , say. J As the reader might expect the main term is given by the summand J .Let H ( α ) = T ( λ α ) T ( λ α ) T k ( λ α ) K η ( α ) e (− ωα ) so that J = ∫ R H ( α ) d α + O (cid:16)∫ + ∞ P / X | H ( α )| d α (cid:17) . Using (6) and (4), we get ∫ + ∞ P / X | H ( α )| d α ≪ η X / k − ∫ + ∞ P / X d αα ≪ η X + / k P = o (cid:0) η X + / k (cid:1) , provided that P → + ∞ . Let now D = [ δ X , X ] × [( δ X ) / k , X / k ] . We obtain ∫ R H ( α ) d α = ∭ D ∫ R e (( λ t + λ t + λ t k − ω ) α ) K η ( α ) d α d t d t d t = ∭ D max { , η − | λ t + λ t + λ t k − ω )|} d t d t d t . Apart from trivial changes of sign, there are essentially two cases:1. λ > , λ > , λ < λ > , λ < , λ < .We deal with the first one. We warn the reader that here it may be necessary to adjust the valueof δ in order to guarantee the necessary set inclusions. After a suitable change of variables, letting D ′ = [ δ X , ( − δ ) X ] , we find that ∫ R H ( α ) d α ≫ ∭ D ′ max { , η − | λ u + λ u + λ u )|} u / k − d u d u d u ≫ X / k − ∭ D ′ max { , η − | λ u + λ u + λ u )|} d u d u d u . THE MAJOR ARC λ j : weassume, as we may, that | λ | ≥ max { λ , λ } , the other cases being similar. Now, for j = , let a j = δ | λ || λ j | , b j = a j and D j = [ a j X , b j X ] ; if u j ∈ D j for j = , then λ u + λ u ∈ [ | λ | δ X , | λ | δ X ] so that, for every choice of ( u , u ) the interval [ a , b ] with endpoints ± η /| λ | + ( λ u + λ u )/| λ | iscontained in [ δ X , ( − δ ) X ] . In other words, for u ∈ [ a , b ] the values of λ u + λ u + λ u coverthe whole interval [− η, η ] . Hence for any ( u , u ) ∈ D × D we have ∫ ( − δ ) X δ X max { , η − | λ u + λ u + λ u |} d u = | λ | − ∫ η − η max { , η − | u |} d u ≫ η . Summing up, we get J ≫ η X / k − ∬ D × D d u d u ≫ η X / k − X = η X + / k , which is the expected lower bound. J , J and J The computations for J and J are similar to and simpler than the corresponding one for J ;moreover the most restrictive condition on P arises from J ; hence we will skip the computationfor both J and J . Using the triangle inequality and (6), J ≪ η ∫ M | S ( λ α )|| S ( λ α )|| S k ( λ α ) − T k ( λ α )| d α ≤ η ∫ M | S ( λ α )|| S ( λ α )|| S k ( λ α ) − U k ( λ α )| d α + η ∫ M | S ( λ α )|| S ( λ α )|| U k ( λ α ) − T k ( λ α )| d α = η ( A + B ) , say, where U k ( λ α ) and T k ( λ α ) are defined in (3). Using the Cauchy-Schwarz inequality, Lemmas1-2 and trivial bounds yields, for any fixed A > , A ≪ X (cid:16)∫ M | S ( λ α )| d α (cid:17) / (cid:16)∫ M | S k ( λ α ) − U k ( λ α )| d α (cid:17) / ≪ X + / k ( log X ) ( − A )/ = o (cid:0) X + / k (cid:1) THE TRIVIAL ARC A > , provided that P ≤ X /( k )− ε . Using again the Cauchy-Schwarz inequality, (5) andtrivial bounds, we see that B ≪ ∫ / X | S ( λ α )|| S ( λ α )| d α + X ∫ P / X / X α | S ( λ α )|| S ( λ α )| d α ≪ X + P (cid:16)∫ P / X / X | S ( λ α )| d α ∫ P / X / X | S ( λ α )| d α (cid:17) / ≪ PX log X . Taking P = o (cid:0) X / k ( log X ) − (cid:1) we get η B = o (cid:0) η X + / k (cid:1) . We may therefore choose P = X /( k )− ε . (15)
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 | I ( η, ω, t )| ≪ ∫ + ∞ R | S ( λ α ) S ( λ α ) S k ( λ α )| K η ( α ) d α ≪ X / k (cid:16)∫ + ∞ R | S ( λ α )| K η ( α ) d α (cid:17) / (cid:16)∫ + ∞ R | S ( λ α )| K η ( α ) d α (cid:17) / ≪ X / k C / C / , say. Using the PNT and the periodicity of S ( α ) , for every j = , we have that C j = ∫ + ∞ R | S ( λ j α )| d αα ≪ ∫ + ∞| λ j | R | S ( α )| d αα ≪ Õ n ≥| λ j | R ( n − ) ∫ nn − | S ( α )| d α ≪ X log X | λ j | R . Hence, recalling that | I ( η, ω, t )| has to be o ( η X + / k ) , the choice R = η − ( log X ) / (16)is admissible.
6. The intermediate arc: / ≤ k ≤ 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 and a appearing in (22) below do not vanish; in fact, if a = , say, then α is very small ( α ≪ X − / ) and,according to our definitions above, it belongs to M ∪ M ∗ . THE MINOR ARC k we do not need an intermediate arc, because the major arc is wide enough to ruleout the possibility that a a = for α ∈ m . For larger values of k , the constraint (15) implies thatthere 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 M ∗ , we are able to cover more than needed.Let / ≤ k ≤ : we now show that the contribution of M ∗ is negligible. Using (6), Lemma 7,the Cauchy-Schwarz inequality and (15) we get I ( η, ω, M ∗ ) ≪ η ∫ X − / P / X | S ( λ α )|| S ( λ α )|| S k ( λ α )| d α ≪ η X ( log X ) ∫ X − / P / X | S k ( λ α )| d αα ≪ η X ( log X ) (cid:16)∫ X − / − X − / | S k ( λ α )| d α (cid:17) / (cid:16)∫ X − / P / X d αα (cid:17) / ≪ η X ( X / k − / ) / ( X − /( k ) ) / X ε ≪ η X / + /( k ) + ε , where we also used Lemma 5 with τ = X − / and the fact that k ≥ / . The last estimate is o (cid:0) η X + / k (cid:1) for every / ≤ k < / .
7. The minor arc
Here we use Harman’s technique as described in [5]. The minor arc m is defined in (7) and (8),according to the value of k . In view of using Lemma 8, we now split m into subsets m , m and m ∗ = m \ ( m ∪ m ) , where m i = { α ∈ m : | S ( λ i α )| ≤ X / ( log X ) } for i = , . In order to obtain the optimization, we chose to split the range for k into two intervals in which totake advantage of the L -norm of S k ( α ) in one case (Lemma 9) and the L -norm of S k ( α ) in theother one (Lemma 10). The same choice will be made later in the discussion of the arc m ∗ . Wewill see that it is not possible to split the minor arc in another way in order to get a better result, inthe present state of knowledge on exponential sums. m ∪ m Using Hölder’s inequality and Lemma 9, for < k ≤ / we obtain | I ( η, ω, m i )| ≪ ∫ m i | S ( λ α )|| S ( λ α )|| S k ( λ α )| K η ( α ) d α ≪ (cid:16) max α ∈ m i | S ( λ α )| (cid:17) (cid:16)∫ m i | S ( λ α )| K η ( α ) d α ) (cid:17) / THE MINOR ARC × (cid:16)∫ m i | S k ( λ α )| K η ( α ) d α ) (cid:17) / ≪ X / ( log X ) ( η X log X ) / ( η X / k ( log X ) ) / ≪ η X / + /( k ) + ε . (17)The estimate in (17) should be o ( η X + / k ) ; hence this leads to the constraint η = ∞( X / − /( k ) + ε ) , (18)where f = ∞( g ) means g = o ( f ) .Using Hölder’s inequality and Lemmas 9 and 10, for / < k < we obtain | I ( η, ω, m i )| ≪ ∫ m i | S ( λ α )|| S ( λ α )|| S k ( λ α )| K η ( α ) d α ≪ (cid:16) max α ∈ m i | S ( λ α )| / (cid:17) (cid:16)∫ m i | S ( λ α )| K η ( α ) d α ) (cid:17) / × (cid:16)∫ m i | S k ( λ α )| K η ( α ) d α ) (cid:17) / (cid:16)∫ m i | S ( λ α )| K η ( α ) d α ) (cid:17) / ≪ X / ( log X ) / ( η X log X ) / ( η max { X / k , X / k − }) / ( η X log X ) / ≪ η max (cid:8) X / + /( k ) , X / + / k (cid:9) X ε . (19)The estimate in (19) should be o ( η X + / k ) ; hence this leads to η = ∞ (cid:0) max { X / − /( k ) + ε , X − / + ε } (cid:1) . (20)If k = we use Lemmas 9 and 11 thus getting | I ( η, ω, m i )| ≪ ∫ m i | S ( λ α )|| S ( λ α )|| S ( λ α )| K η ( α ) d α ≪ (cid:16) max α ∈ m i | S ( λ α )| / (cid:17) (cid:16)∫ m i | S ( λ α )| K η ( α ) d α ) (cid:17) / × (cid:16)∫ m i | S ( λ α )| K η ( α ) d α ) (cid:17) / (cid:16)∫ m i | S ( λ α )| K η ( α ) d α ) (cid:17) / ≪ η X / + ε . This bound leads to the constraint η = ∞ (cid:0) X − / + ε (cid:1) , (21)which justifies the last line of (2). THE MINOR ARC m ∗ We recall our definitions in (7) and (8). It remains to discuss the set m ∗ where the followingbounds hold simultaneously | S ( λ α )| > X / ( log X ) , | S ( λ α )| > X / ( log X ) , T ≤ | α | ≤ η − ( log X ) / = R , where T = P / X = X /( k )− − ε by our choice in (15) if k < / , and T = X − / otherwise. Using adyadic dissection, we split m ∗ into disjoint sets E ( Z , Z , y ) in which, for α ∈ E ( Z , Z , y ) , we have Z i < | S ( λ i α )| ≤ Z i , y < | α | ≤ y , where Z i = k i X / ( log X ) and y = k X /( k )− − ε for some non-negative integers k , k , k .It follows that the number of disjoint sets is, at most, ≪ ( log X ) . Let us write A as a shorthandfor the set E ( Z , Z , y ) . We need an upper bound for the Lebesgue measure of A . In the followingLemma, it is crucial that both the integers a and a appearing in (22) below do not vanish; in fact,if a = , say, then q = and α is so small that it can not belong to m . If k is large, we treatthe range [ P / X , X − / ] and its symmetrical by means of the argument in section 6: this is neededbecause, in this case, the inequalities (22) below do not rule out the possibility that a a = , unless | α | is large enough. Lemma 12.
Let ε > . We have that µ ( A ) ≪ y X / + ε Z − Z − , where µ (·) denotes the Lebesguemeasure. Proof. If α ∈ A , by Lemma 8 there are coprime integers ( a , q ) and ( a , q ) such that ≤ q i ≪ (cid:16) X ( log X ) Z i (cid:17) , | q i λ i α − a i | ≪ X ( log X ) Z i . (22)We remark that a a , otherwise we would have α ∈ M ∪ M ∗ . In fact, if a a = , recallingthe definitions of Z i and (22), α ≪ q − i X ( log X ) Z − i ≪ X − / .Now, we can further split m ∗ into sets I = I ( Z , Z , y , Q , Q ) where, on each set, Q j ≤ q j ≤ Q j .Note that a i and q i are uniquely determined by α ; in the opposite direction, for a given quadruple a , q , a , q , the inequalities (22) define an interval of α of length ≪ min ( X ( log X ) Q Z , X ( log X ) Q Z ) ≪ X ( log X ) Q / Q / Z Z , by taking the geometric mean.Now we need a lower bound for Q Q : by (22) we obtain (cid:12)(cid:12)(cid:12) a q λ λ − a q (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) a λ α ( q λ α − a ) − a λ α ( q λ α − a ) (cid:12)(cid:12)(cid:12) THE MINOR ARC ≪ q | q λ α − a | + q | q λ α − a |≪ Q X ( log X ) Z + Q X ( log X ) Z . Recalling that Q i ≪ ( X ( log X ) / Z i ) and that Z i ≫ X / ( log X ) , we have (cid:12)(cid:12)(cid:12) a q λ λ − a q (cid:12)(cid:12)(cid:12) ≪ (cid:16) X ( log X ) X / ( log X ) (cid:17) (cid:16) X / ( log X ) X / ( log X ) (cid:17) ≪ X − / ( log X ) − < q . (23)We recall that q = X / is a denominator of a convergent of λ / λ . Hence by (23), Legendre’slaw of best approximation for continued fractions implies that | a q | ≥ q and by the same token,for any pair α , α ′ having distinct associated products a q , | a ( α ) q ( α ) − a ( α ′ ) q ( α ′ )| ≥ q ; thus, by the pigeon-hole principle, there is at most one value of a q in the interval [ r q , ( r + ) q ) forany positive integer r . Furthermore a q determines a and q to within X ε / possibilities (fromthe bound for the divisor function) and consequently also a q determines a and q to within X ε / possibilities from (23).Hence we got a lower bound for q q , since, using Q j ≤ q j ≤ Q j , we get q q = a q q a ≫ r q | α | ≫ r q y − . for the quadruple under consideration.As a consequence we obtain that the total length of the part of I ( Z , Z , y , Q , Q ) with a q ∈[ r q , ( r + ) q ) is ≪ X + ε / ( log X ) Z − Z − r − / q − / y / . Now we need a bound for r : since a q ∈ [ r q , ( r + ) , q ) , we have r q ≤ | a q | ≪ q q | α | ≪ y (cid:16) X ( log X ) Z (cid:17) (cid:16) X ( log X ) Z (cid:17) ≪ y X ( log X ) Z Z and hence we get r ≪ q − y X ( log X ) Z − Z − . Next, we sum on every interval to get an upper bound for the measure of A : we get µ ( A ) ≪ X + ε / y / ( log X ) Z Z q / Õ ≤ r ≪ q − y X ( log X ) Z − Z − r − / . Standard estimates imply that the sum on the right is ≪ ( q − y X ( log X ) Z − Z − ) / , and recallingthat q = X / we can finally write µ ( A ) ≪ y X + ε / ( log X ) Z − Z − q − ≪ y X / + ε Z − Z − . This proves the lemma. (cid:3)
CONCLUSION
8. Conclusion
Here we finally justify the choice of the function ψ in the statement of the main Theorem. UsingLemmas 9-10-12 we are now able to estimate I ( η, ω, A ) for < k ≤ . For k ≥ , we also needthe result in section 6.If < k ≤ / we proceed as follows: | I ( η, ω, A )| ≪ ∫ A | S ( λ α )|| S ( λ α )|| S k ( λ α )| K η ( α ) d α ≪ (cid:16)∫ A | S ( λ α ) S ( λ α )| K η ( α ) d α (cid:17) / (cid:16)∫ A | S k ( λ α )| K η ( α ) d α (cid:17) / ≪ (cid:0) min (cid:8) η , y − (cid:9)(cid:1) / (cid:0) ( Z Z ) µ ( A ) (cid:1) / (cid:0) η X / k + ε (cid:1) / ≪ (cid:0) min (cid:8) η , y − (cid:9)(cid:1) / Z Z (cid:0) y X / + ε Z − Z − (cid:1) / η / X /( k ) + ε / ≪ η X / + /( k ) + ε . Hence we need η = ∞ (cid:0) X / − /( k ) + ε (cid:1) , which is the same condition we got in (18).If / < k < , | I ( η, ω, A )| ≪ ∫ A | S ( λ α )|| S ( λ α )|| S k ( λ α )| K η ( α ) d α ≪ (cid:16)∫ A | S ( λ α ) S ( λ α )| / K η ( α ) d α (cid:17) / (cid:16)∫ A | S k ( λ α )| K η ( α ) d α (cid:17) / ≪ (cid:0) min (cid:8) η , y − (cid:9)(cid:1) / (cid:0) ( Z Z ) / µ ( A ) (cid:1) / (cid:0) η max { X / k , X / k − } X ε (cid:1) / ≪ (cid:0) min (cid:8) η , y − (cid:9)(cid:1) / Z Z (cid:0) y X / + ε Z − Z − (cid:1) / η / max { X /( k ) , X / k − / } X ε / ≪ η Z − / Z − / X + ε max { X /( k ) , X / k − / }≪ η max { X / + /( k ) , X / + / k } X ε . Hence we need η = ∞ (cid:0) max { X / − /( k ) + ε , X − / + ε } (cid:1) , which is the same condition we got in (20).If k = , using Lemmas 11 and 12 we obtain | I ( η, ω, A )| ≪ ∫ A | S ( λ α )|| S ( λ α )|| S ( λ α )| K η ( α ) d α ≪ (cid:16)∫ A | S ( λ α ) S ( λ α )| / K η ( α ) d α (cid:17) / (cid:16)∫ A | S ( λ α )| K η ( α ) d α (cid:17) / ≪ η Z − / Z − / X / + / + ε ≪ η X / + ε . This leads to the same constraint for η that we had in (21). EFERENCES ReferencesReferences [1] J. Bourgain,
On the Vinogradov mean value , Tr. Mat. Inst. Steklova (2017), Analitich-eskaya i Kombinatornaya Teoriya Chisel, 36–46.[2] J. Bourgain, C. Demeter, and L. Guth,
Proof of the main conjecture in Vinogradov’s meanvalue theorem for degrees higher than three , Ann. Math. , (2016), 633–682.[3] J. Brüdern, R. J. Cook, and A. Perelli,
The values of binary linear forms at prime arguments ,Proc. of Sieve Methods, Exponential sums and their Application in Number Theory (G. R.H. Greaves et al , ed.), Cambridge University Press, 1997, pp. 87–100.[4] H. Davenport and H. Heilbronn,
On indefinite quadratic forms in five variables , J. LondonMath. Soc. (1946), 185–193.[5] G. Harman, Diophantine approximation by prime numbers , J. London Math. Soc. (1991),218–226.[6] L. K. Hua, Some results in the additive prime number theory , Quart. J. Math. Oxford (1938),68–80.[7] A. Languasco and A. Zaccagnini, On a ternary Diophantine problem with mixed powers ofprimes , Acta Arith. (2013), 345–362.[8] A. Languasco and A. Zaccagnini,
A Diophantine problem with prime variables , in V. KumarMurty, D. S. Ramana, and R. Thangadurai, editors,
Highly Composite: Papers in NumberTheory, Proceedings of the International Meeting on Number Theory, celebrating the 60thBirthday of Professor R. Balasubramanian (Allahabad, 2011) , volume 23, pages 157–168.RMS-Lecture Notes Series, 2016.[9] O. Robert and P. Sargos,
Three-dimensional exponential sums with monomials , J. reine angew.Math. (2006), 1–20.[10] D. I. Tolev,