Moments and Oscillations of Exponential Sums Related to Cusp Forms

Transcription

1 oments and Oscillations of Exponential Sums Related to Cusp Forms Esa V. Vesalainen arxiv: v1 [math.nt] 1 Feb 014 Abstract We consider large values of long linear exponential sums involving Fourier coefficients of holomorphic cusp forms. The sums we consider involve rational linear twists enh/) with sufficiently small denominators. We prove both pointwise upper bounds and bounds for the frequency of large values. In particular, the -aspect is treated. As an application we obtain upper bounds for all the moments of the sums in question. We also give the asymptotics with the right main term for fourth moments. We also consider the mean square of very short sums, proving that on average short linear sums with rational additive twists exhibit square root cancellation. This result is also proved in a slightly sharper form. Finally, the consideration of moment estimates for both long and short exponential sums culminates in a result concerning the oscillation of the long linear sums. Essentially, this result says that for positive proportion of time, such a sum stays in fairly long intervals, where its order of magnitude does not drop below the average order of magnitude and where its argument is in a given interval of length 3π/ and so can not wind around the origin. 1 Introduction 1.1 Linear exponential sums related to cusp forms Let F be a fixed holomorphic cusp form of even weight κ Z + for the full modular group SL, Z). Then f has the usual normalized Fourier expansion Fz) = an)n κ 1)/ enz), n=1 where z C with Iz > 0. It is interesting to study the linear exponential sums A,α) = an)enα), n and more generally A, ;α) = an)enα), n + where R + is large, [1,], and α R. These sums provide an interesting window into the life of the Fourier coefficients. They are also connected 1

2 to various other problems; as an example we mention second moments of the corresponding L-function see e.g. the introduction to [17]). When is small compared to, the resulting short sums provide a natural analogue of the classical problems of analytic number theory studying various error terms in short intervals, but good estimates for short sums also provide a practical tool for reducing smoothing error thereby potentially leading to better estimates as examples we mention [15, ] and the proof of Theorem 1 below). Pointwise bounds for short sums have been obtained by Jutila [15], and Ernvall- Hytönen and Karppinen [5, ]. The first estimate for long linear exponential sums involving holomorphic cusp form coefficients was proved by Wilton [8] in the course of proving an analogue of Voronoi s summation formula for cusp form coefficients see Theorem 1 below) and that the L-function connected to the Ramanujan τ-function has infinitely many zeros on the critical line. Wilton s estimate is that the long linear sum is 1/ log, uniformly for α R. Ranin [1] and Selberg [3] famously proved that 1 an) enα) dα = an) = A +O 3/5 ), 0 n n where A is a certain positive real constant depending on F. This implies that only the logarithm might possibly be removed from Wilton s estimate, and this was done by Jutila [15]. Indeed, we have an)enα) 1/, uniformly for α R. n 1. Linear exponential sums with rational twists In the study of linear exponential sums, the case in which α is near a fraction with a small denominator is often different in character from the one in which α is not close to a such a fraction. Furthermore, the behavior near such rational values is often strongly lined to the behavior at such rational points, and it is this latter behavior that we are concerned with in this paper. For rational values of α the study of the linear exponential sums is more ain to the classical problems of understanding the error terms in the Dirichlet divisor problem and the circle problem. Indeed, the analogies go rather deep see e.g. [14]), largely due to the fact that the divisor and circle problems, too, have modular origins. However, the cusp form problems are in some respects more challenging. For example, the sums related to cusp forms have not been connected to such theories as the machinery of exponential pairs. any of the best estimates are weaer than the best results on the analogous classical problems. An example of this is provided by the case α = 0: for the sums of Fourier coefficients, the best estimate to date, due to Ranin [], is A,0) 1/3 log δ, for a certain small positive δ R +. For, say, the error term in the Dirichlet divisor problem instead, the 0th century saw a long string of upper bounds improving the cubic root cancellation first obtained by Voronoi [7]. For more on the

3 rather extensive literature Dirichlet divisor problem, we recommend Tsang s survey [6] and references therein, and the Chapter 13 in the boo [8] of Ivić. 1.3 What we do in this paper On average the long linear sum with rational twist α = h/ has the size 1/ 1/4, which is Theorem 1. in [14], and recalled in Theorem 11 below. However, the pointwise upper bounds are far from this. The upper bound /3 1/3+ε is Corollary on p. 30 in [14], and it is difficult to improve upon this, but we give a slight improvement for 1/10 3/8 by arguing from the Voronoi-type summation formula along the lines of Ivić s paper [9] and reducing the smoothing error using the estimates for short exponential sums by Ernvall-Hytönen and Karppinen [5]. Using arguments of Ivić s paper [7], we prove in Theorem that, in a certain sense, a long linear exponential sum can not exceed its average value 1/ 1/4 too often. These estimates will lead, as in [7], to rather general moment estimates for such sums. In particular, the sums of coefficients will exhibit in every moment cancellation beyond the cubic root cancellation. We also tae a closer loo at the fourth moment of those sums, obtaining the main term of the asymptotics, following Ivić and Sargos [11]. The main term for the fourth moment for the divisor problem was first obtained by Tsang [5]. In our result, the main term will be larger than the error terms when 1/6 ε. We shall also consider the mean square of short sums with rational additive twists, proving essentially the expected square root cancellation on average: for 1/ ε, and max 0 U x n x+ x n x+u ) nh an)e dx ) nh an)e dx log for 1/4 ε. For sums of coefficients, the first of these essentially follows from the arguments of Jutila [13], whose arguments we will follow. The mean square of square root length sums has been considered by Ernvall-Hytönen [3] who obtained square root cancellation with rational additive twists. The second estimate is proved following Heath-Brown and Tsang as in the proof of Lemma of [6], and following the proof of Theorem in [1]. We would also lie to mention that recently Ernvall-Hytönen [4] has considered the mean square of short exponential sums for which is larger than 1/. Our final Theorem 8 will be an analogue of the main theorem of Heath- Brownand Tsangin [6], and it is the originalmotivation for the variousmoment estimates of this paper. Combining the various moment estimates for both long and short sums, we will see that, if 1/8 ε, as the increases, the long linearsum with arationaltwist spends apositiveproportionofits time in nearly square root length intervals in which its absolute value is at least of the average size 1/ 1/4 and where its argument is in a given fixed interval of length 3

4 3π/. Thus, essentially the result says that for a positive proportion of time, such sums do not exhibit much oscillation: the absolute value stays at least at the average level and the value does not wind around the origin. 1.4 Notation The cuspformfz) willbe a fixedholomorphiccusp formofevenweightκ Z + for the full modular group SL, Z). Its Fourier expansion will be normalized as follows: Fz) = an)n κ 1)/ enz). n=1 The symbols,, and O are used for the usual asymptotic notation: for complex valued functions f and g in some set Ω, the notation f g means that fx) C gx) for all x Ω for some implied constant C R +. When the implied constant depends on some parameters α,β,..., we use α,β,... instead of mere. The notation g f means f g, and f g means f g f. The implied constants are allowed to depend on the cusp form F under study and ε. The symbol ε denotes an arbitrarily small positive real constant, and its value can and will change from one instance to the next. We emphasize that in the exponents of assumptions and conclusions it may be chosen arbitrarily small. The only special notation used below will be the following: If f is a realvalued function, then f + = max{f,0}. If f is complex-valued, then we will write The results f ++ = Rf) + +iif) +. The classical upper bound [14, Cor., p. 30] which follows from the truncated Voronoi identity by taing absolute values is ) nh an)e /3 1/3+ε, n where h and are coprime integers with 1. On the other hand, the estimates from [15] give the upper bound 1/, for arbitrary. It turns out that in the range 1/10 3/8, these two estimates can be improved upon by combining the argument of [9] with the estimates for short sums from [5]. Theorem 1. Let [1, [, and let h and be coprime integers with 1 1/. We have /3 1/3 log 1/3 when 1/, ) nh 1/4 3/8+ε when 1/10 1/4, an)e /3 13/48+ε when 1/4 1/, 15/3+ε when 19/64 1/64, n /3 1/4+ε when 1/64 1/, and 1/ when 1/. 4

5 The following theorem estimates the rarity of large values a long linear sum can tae. Theorem. Let [1, [, and let h and be coprime integers with 1. Let 1 t 1 < t <... < t R and t r t s V for r s R. Fix an exponent pair p,q. If A t r, h ) V for r {1,,...,R}, where then V 1+q p) q p) q p+ε and q p, R 1+ε V 3 + q/p 1+q/p+ε V 1+q)/p. The large value result can be turned into a general moment estimate: Theorem 3. Let [1, [, fix an exponent pair p,q, and let h and be coprime integers with 1 q p. Furthermore, let α [0, [ and β [0, [ be exponents so that, restricting to some suitable interval, ) nh an)e α x β+ε, n x for sufficiently large x. Then, for the same range of, 1 x, A h ) A dx A/ A/4+1+ε +Φ+Ψ, where and Φ = { αa+1 α) βa+1 β)+ε if A, A/+1 A/4+1/+ε if A, { αa α α/p+1 α)q/p βa+1 β β/p+1 β)q/p+ε if A 1+ 1+q p Ψ =, A/ 1/ 1/p)+q/p A/4+3/4 1/4p)+q/p)+ε if A 1+ 1+q p. This theorem of course admits many special cases. We will mention only a few. Using the classical upper bound n x an)e ) nh /3 x 1/3+ε, the bounds for Φ and Ψ simplify to { Φ = A/3 A/3+1/3+ε if A, A/+1 A/4+1/+ε if A, and { A/3 /3 /3p) q/p A/3+/3+q/3p) 1/3p)+ε if A 1+ 1+q p Ψ =, A/ 1/ 1/p)+q/p A/4+3/4 1/4p)+q/p)+ε if A 1+ 1+q p. When 1/10 1/4, we can use the better exponents α = 1/4 and β = 3/8 coming from Theorem 1, and Φ and Ψ can be estimated as { Φ = A/4+3/ 3A/8+1/4+ε if A, A/+1 A/4+1/+ε if A, 5

6 and { A/4 1/4 1/4p+3q/p) 3A/8+5/8 5/8p+q/4p)+ε if A 1+ 1+q p Ψ =, A/ 1/ 1/p)+q/p A/4+3/4 1/4p)+q/p)+ε if A 1+ 1+q p. The third moment case A = 3 is of particular interest to us since we shall use it in the proof of Theorem 5. We choose the exponent pair p,q = 75/1,57/106, which can be found from Table 1 on page 58 in [19]. For this pair q p is slightly larger than 1/6 and 1+ 1+q p is nearly 7. In the range 1/10 we then obtain and Φ = 4/3+ε = 3/ 1/ 4/3+ε 3/ 7/4+ε, Ψ = 83/75 33/150+ε 3/ 7/4+ε. Similarly, in the range 1/10 1/6 we obtain and Φ = 9/4 11/8+ε = 3/ 1/4 11/8+ε 3/ 7/4+ε, Thus, combining these we obtain Corollary 4. For 1/6, we have Ψ = 83/75 33/150+ε 3/ 7/4+ε. 1 x, A h ) 3 dx 3/ 7/4+ε. Apart from the ε and the range of, this is essentiallly best possible. Because of the ε in the exponents, these moment estimates are not quite strong enough that we could see that the oscillation result Theorem 8 holds for a positive proportion of time. Thus, we need a higher moment estimate where the ε has been shaved away. For this purpose, as well as for its intrinsic interest, we derive a reasonably sharp fourth moment estimate. Theorem 5. For [1, [ and for coprime integers h and with 1 1/6 ε, we have 1 x, A h ) 4 dx = C F +O 11/4 15/8+ε )+O 13/6 3/1+ε ), where the constant coefficient C F is 3 3/4 a+b c d)h aa)ab)ac)ad)e 64π 4 abcd) a,b,c,d=1 a+ b= c+ d The constant C F is a finite positive real number by Deligne s estimate an) dn) and the estimates 3.6) in [5]. We remar that Theorem 5 immediately implies that Corollary 4 holds without the ε in the upper bound when 1/6 ε. Next, we turn to the mean square estimates for short linear sums. ). 6

7 Theorem 6. Let 1, let 1, let h and be coprime integers with 1 1/ ε, and, finally, let 1+ε 1 Ξ. Then +Ξ ) nh an)e dx Ξ. x n x+ Furthermore, if 1/ ε, and if the underlying cusp form does not vanish identically, then the second moment is Ξ. Next, a slightly different upper bound which treats sums of many different lengths at once. Theorem 7. Let 1, let ε, let h and be coprime integers with 1 1/4 ε. Then ) max nh 0 U an)e dx log. x n x+u Finally, we arrive at our final theorem: Theorem 8. Let R + be sufficiently large, and let h and be coprime integers with 1 1/8 ε. Then there are 1/ log subintervals I of [,] of length 1/ log such that, for x I, ) nh an)e 1/ 1/4, n x andwemayevendemand thatthe argumentbelongstoagivenintervaloflength 3π/ for x I. Here, of course, the intervals will depend on, h and, but the implicit constants will not. 3 Various lemmas and well-nown theorems We collect in this section the pre-existing results and some of their consequences required in the proofs. 3.1 Results on holomorphic cusp forms Deligne s famous wor on Weil s conjectures gives as an application [1] an estimate for individual Fourier coefficients of a holomorphic cusp form: an) dn), and in fact, an) dn), if the cusp form in question is a Hece eigenform. This is often combined with Shiu s estimate [4] for the divisor function: dn) y logx, x n x+y for x ε y x. Ranin [1] and Selberg [3] obtained the main term for the mean square of the Fourier coefficients: 7

8 Theorem 9. We have an) = Cx+O x 3/5+ε), n x where C R + only depends on the cusp form in question. Through integration by parts, this can be used in many forms. For example, for x [1, [, α ]0,1[ and β ]1, [, we have an) α x 1 α and n x n α an) n β β x 1 β. ost of our proofs depend on the important truncated Voronoi identity, a good presentation of which is given in Chapter 1 of Jutila s monograph [14], where this identity appears in Theorem 1.1. Theorem 10. Let x [1, [, let N [1, [, assume that N x, let h be an integer, let be a positive integer, and assume that h,) = 1. Then ) nh an)e = 1/ x 1/4 π ) nh an)n 3/4 e n x n N 4π nx cos π ) +O x 1/+ε N 1/). 4 In Theorem 1. of the same chapter, Jutila also givesas an application the mean square estimate for long linear sums. Theorem 11. Let 1,andlethand becoprimeintegerswith1. Then where 1 x>n x, A h ) dx = C 3/ +O 1+ε) +O 3/ 5/4+ε), C = 1 6π n=1 an) n 3/. The proof of Theorem 1 requires the full Voronoi type summation formula for holomorphic cusp forms. Theorem 1. For positive real numbers a and b with a < b, and for coprime integers h and with 1, and a continuously differentiable function f: R + R +, we have a n b ) nh an)e fn) = 1) κ/ π ) b nh an)e n=1 a ) 4π nx fx)j κ 1 dx. Here means that if a or b) is an integer, then the term fa) or fb)) should be halved. 8

9 Again, the presentation of Chapter 1 of [14] is recommended; Theorem 1.7 there gives the above summation formula. 3. Tools for exponential sums without Fourier coefficients The following lemma due to Bombieri, which is Lemma 5.1 in [18], is not strictly speainganexponentialsum results, but we willuse it in the proofoftheorem to separate the Fourier coefficients from the exponential sums, leaving only exponential sums which then can be estimated using the theory of exponent pairs. Theorem 13. Let H be a Hilbert space. Denote its inner product by and its norm by. Also, let R Z +, and let ξ, ϕ 1, ϕ,..., ϕ R be vectors of H. Then R R ξ ϕ r ξ max ϕ r ϕ s. r=1 1 r R s=1 We will actually only apply this for H = C N for some N Z +, and the inner product will be the usual one so that for z = z 1,...,z N C N and w = w 1,...,w N C N we have z w = z 1 w z N w N, and z = z z N. Discussionsofthebasictheoryofexponentpairscanbefoundine.g. Chapter of [8], or in Chapter 3 of [19]. For the purposes of the proof of Theorem, it is enough to state here that if the pair p,q [ [ 0, ] 1 1,1] happens to be an exponent pair, then we may estimate ea n) A p q p/ +A 1 1/, n + for nonzero real constants A. The square root will appear from the right-hand side of the truncated Voronoi identity. 3.3 Lemmas on exponential integrals Lemma 14. Let 1, let 0 Ξ, and let and n be positive integers. Then +Ξ 4π nx x 1/4 cos π ) dx T 3/4. 4 n This is a typical consequence of the first derivative test, which is discussed e.g. in Section.1 of [8]. ore precisely, the claim follows from Lemma.1 of [8] and integration by parts. Lemma 15. Let 1, 0 U, 0 T, 0 Ξ, and let m, n and be positive integers. Then +Ξ x 1/ e ± mx+t) This follows immediately from the first derivative test. )) + nx+u) dx. m+ n 9

10 Lemma 16. Let 1, 0 U T, 0 Ξ, and let m, n and be positive integers with m < n. Then +Ξ x 1/ )) mx+t) e ± nx+u) dx n n m. This also follows immediately from the first derivative test given the simple observation that n n m n m. Lemma 17. Let 1, 0 < T, 0 Ξ, and let m, n and be positive integers with m < n and m T. Then +Ξ x 1/ e ± mx )) nx+t) dx n n m. This is once more a consequence of the first derivative test. The relevant estimate is d ) nx+t) mx n m = dx x+t x nx mx mt = n m) nx+ ) xx+t) mx+t) n+ m) n m n. Finally, one more corollary of the first derivative test: Lemma 18. Let [1, [, let Z +, and let R +. Then xe ± x )dx 3/. 3.4 Lemmas on the spacing of square roots The following three lemmas are Lemmas, 5 and 6 in [11], and they will be used in the consideration of the fourth moment estimate in Section 7. Lemma 19. If a, b, c and d are positive integers with a c, b c and a+ b± c d, then a+ b± c d c abc) 1/. Lemma 0. Let A, B, C, D and δ be positive real numbers with 1 A C, 1 B C and 1 D C. Then the number of quadruples of positive integers a, b, c and d with A < a A, B < b B, C < c C, D < d D and a+ b c d δ C is both and ABCD δ +C 3/+ε) +Cmin{A,B,D} ABCD δc +ABCD) 1/) C ε. 10

11 Lemma 1. Let A, B, C, D and δ be positive real numbers with 1 A C and 1 B C. Then the number of quadruples of positive integers a, b, c and d with A < a A, B < b B, C < c C, D < d D and 0 < a+ b+ c d δ C is both and ABCD δ +C 3/+ε) ABCD δc +ABCD) 1/) C ε. 4 Proof of Theorem 1 Theorem 1 will follow by introducing a smooth weight function and using the following weighted result: Lemma. Let [1, [, let [1,], let h and be coprime integers with 1 1/, and let X [1, [. Also, let U [1, /3] and let w C R) be supported in [, + ] and satisfy w 1, w U 1 and w U, as well as wx) = 1 for x [ +U, + U]. Then ) nh an)e wn) 1/ 1/4 X 1/4 + 3/ X 1/4 U 1 3/4. n + Remar: By examining the proof of the lemma, one sees that the first term can be omitted with the upper bound then being 3/ U 1 3/4. For example, if U 3/ 3/4+ε, then the weighted sum in question is o1) as. Proof of Theorem 1. ThelastestimateofTheorem1followsfromthe results in [15]. The other estimates follow by first smoothing the sum with a weight function w satisfying the assumptions of Lemma, and then applying Lemma tothe smoothsum. However,the different rangesof will behandled slightly differently. First, if 1/10 1/4, then we shall choose U = 3/ 1/4. It is easy to chec that 1/10 1/4 implies that /5 U 5/8, so that Theorem 5.5 from [5] applies and the smoothing error will be U 1/6 1/3+ε by a simple integration by parts argument. Combining this with Lemma gives ) nh an)e U 1/6 1/3+ε + 1/ 1/4 X 1/4 + 3/ X 1/4 U 1 3/4. n Choosing now X = U this simplifies to 1/4 3/8+ε, as required. Similarly, if 1/4, we choose U = /3 11/4 5/8, so that Theorem 5.7 from [5] gives the smoothing error U 3/16+ε. Choosing then X = /3 1/1 in Lemma gives the upper bound /3 13/48+ε. In the same vein, if 19/64 1/64, then we may choose U = 3/3 and use Theorem 5.16 from [5] to get the smoothing error 15/3+ε. Then choosing X = 7/8 in Lemma gives only terms which are 15/3+ε. 11

12 In the range 1/64 3/4 we choose U = /3 1/3 and apply Theorem5.16from[5]togetthesmoothingerror U 1/4+ε. Then, choosing X = /3 in Lemma gives the upper bound /3 1/4+ε. Finally, for, we may choose U = /3 1/3 log /3 and X = /3 1/3 log 4/3, and the smoothing error can be estimated by Deligne s estimate and Shiu s theorem by /3 1/3 log 1/3, which is also the upper bound that Lemma gives for the smooth sum. Proof of Lemma. An application of the additively twisted Voronoi-type summation formula for holomorphic cusp forms gives n + ) nh an)e wn) = 1) κ/π nh an)e n=1 ) + J κ 1 4π nx ) wx) dx. We shall consider the series in two parts n X + n>x. In the terms with n X we first apply the asymptotics of the Bessel J- function J ν x) = πx cos x ν 4 1 )+O ν x 3/), 8 and then apply the first derivative test on the integrals involving the main terms of the J-asymptotics and estimates by absolute values on the integrals involving the error term of the J-asymptotics: + J κ 1 4π nx ) wx) dx + = 1/ 4π nx π n 1/4 x 1/4 cos +O 3/ n 3/4 3/4). κ 1 8 ) dx 1/ n 1/4 1/4 1/ n 1/ 1/ + 3/ n 3/4 3/4 3/ n 3/4 1/4. Now the sum over n X, using the Ranin Selberg estimate with the Cauchy Schwarz inequality, gives the first term 1/ X 1/4 1/4. Next, we treat the terms with n > X by integrating by parts twice using the fact that d dx xν J ν x)) = x ν J ν 1 x), 1

13 and estimating by absolute values, and get + J κ 1 4π nx = 4π n + ) wx) dx xj κ+1 4π nx 5/ n 5/4 3/4 U 1. ) w x)dx It is important to observe here that we really get just U 1 instead of U the reason being that after integrations by parts the resulting integrands vanish outside the support of w x) which is of length U. Thus the series over n > X gives the second term 3/ X 1/4 3/4 U 1. 5 Proof of Theorem From the truncated Voronoi identity, we see that for x and N x, ) nh an)e n x = 1/ x 1/4 π ) nh 4π nx an)n 3/4 e cos π ) 4 n N +O x 1/+ε N 1/). ) ) 1/ x 1/4 nh nx an)n 3/4 e e n N ) + 1/ x 1/4 nh an)n 3/4 e e ) nx n N +x 1/+ε N 1/. Thesecondtermwillhavethesameupperboundsasthefirstones; notforgetting this, we may simplify the expressions that appear below. We shall consider the interval [,] and divide it into subintervals of length not exceeding 0 V. We shall consider the number R 0 of points in a single subinterval, and we shall call these points t 1, t,..., t R0. The estimate for R will be obtained from that of R 0 via R R ). Now, for N we have R 0 V 1/ ) ) nh 1/4 an)n 3/4 ntr e e n N r R 0 + R 0 1+ε N 1. 13

14 The error term coming from the truncated Voronoi identity may be absorbed to the left-hand side provided that N V 1+ε. This requirement implies that ε V, which is far weaer than the lower bound for V will be. R 0 V r R 0 1/ ) ) nh 1/4 an)n 3/4 ntr e e n N ) ) 1/ nh an)n 3/4 ntr e e r R 0 U N/U<n U dyadic 1/ logn ) ) nh an)n 3/4 ntr e e r R 0 1/ log N max 1/+ε max max r R 0 s R 0 1/+ε N 1/ U N/ dyadic U<n U U N/ r R 0 U<n U U N/ 1/ U<n U + 1/+ε max max r R 0 e U N/ U 1/ s R 0 s r U<n U n tr t s ) e ) ) nh an)n 3/4 ntr e e ) n tr t s ) We pic an exponent pair p,q. We have t r t s t s dt t 1/ t r t s. 1/ t r This is both 0 1/ and V 1/. Therefore we have the estimate n tr )) t s e p p 0 p/ U q p/ +U 1/ 1/ V 1. U<n U Substituting this bac to the previous estimates gives R 0 V 1/+ε N 1/ + 1+ε V 1 ) + 1 p R 0 p 0 1/ p/ N q p/ 1/. 14

15 Choosing N = 1+ε V allows us to merge the first two terms on the right-hand side giving R 0 1+ε V 3 + q p) R 0 p 0 q p+ε V p q 1. In order to absorb the last term to the left-hand side, the choice of 0 should be such that q p) p 0 q p+ε V p q 1 1, with a sufficiently small implicit constant, of course. We choose p 0 to be as large as possible, so that Finally, we obtain p 0 p q) p q ε V 1+q p. R 0 1+ε V 3. This, of course, is conditional to the requirement 0 V, meaning that which simplifies to V p p q) p q ε V 1+q p V 1+q p) q p) q p+ε. Combining the above considerations leads to the following estimate: R 1+ε V ε V 3 1+ε V 3 + q/p 1+q/p+ε V 1+q)/p. We still have to see what the allowed range for is. The only requirement is that ε V. This will hold, if 1+q p) q p) q p, or more simply, if q p. 6 Proof of Theorem 3 The idea is following: We shall consider the integral / x, A h ) A dx. Thepartswheretheintegrandis A/ A/4 areestimatedby A/ A/4+1. The remaining value range [ A/ 1/4, α β+ε] is divided dyadically into intervals of the shape [V, V]. This might require extending two of the subintervals, but this doesn t matter. The number of such subintervals is log. For each interval, we choose RV) points in [,], at which the size of A x, ) h is in [V,V]. We choose the points so that they are spaced with distance at least V between each pair of them. 15

16 The integral in the regions, where the integrand is large, will be V V V RV)V A 1+ε V 3 + q/p 1+q/p+ε V 1+q)/p) V A+1. The result now follows easily: the first term in the parantheses gives rise to Φ, the second to Ψ. The upper bounds of Φ and Ψ are then obtained just by estimatingv appropriatelyfrombelowby 1/ 1/4 orfromaboveby α β+ε. 7 Proof of Theorem 5 Let 1 N be arbitrary for the present time. The truncated Voronoi identity gives x, A h ) 4 dx = 4π 4 n N ) 4) 4 x nh 4π nx an)e cos π dx+error, n N where the error is ) 4) nh 4π nx an)e cos π 3 x 1/+ε N 1/ dx 1/+ε N 1/ 5/ 9/4+ε N 1/ + 3 N. n, A h ) 3 dx+ 1/+ε N 1/) 4 To treat the main term, we will expand the fourth power as Σ 4 = Σ Σ, exchange the order of integration and summation, and write the cosines in terms of exponential functions. This, of couse, leads to a large number of terms of the form ) a+b c d)h 64π 4 aa)ab)ac)ad)abcd) 3/4 e e ) ) xe ± ax ± bx ± cx ± dx dx, 8 where a, b, c and d are positive integers from the interval [1,N], where the signs correspond to the ±, and where otherwise all possible choices of the ±-signs appear. The main term of the fourth moment comes from the terms with two plus signs and two minus signs and in which a+ b = c+ d. In this case the 16

17 integral reduces to 3 /. The sum can be extended from a,b,c,d N to with an error a,b,c,d=1 +ε N 1/4, as in 3.6) in [5]. The terms with four plus signs or four minus signs are estimated by absolute values by Lemma 18 giving a N aa)ab)ac)ad) abcd) 3/4 3/ a+ b+ c+ d b N c N d N 4 3 3/ an) n 7/8 3 3/ N 1/8 ) 4 = 3 3/ N 1/. n N Let us next consider the terms involving three signs of one ind and one of the other ind. Without loss of generality we may consider the signs +, +, + and. We first observe that those terms in which = a + b + c d vanishes cancel simply because for them e ) ) x e +e ) ) x e = 0. 8 Next we split the sums a b c d dyadically in each variable into log 4 N ε subsums of the form, A<a AB<b B C<c C D<d D where naturally A, B, C and D are all N. It turns out, that by symmetry, it is enough to consider the terms in which A C and B C. In particular, we do these simplifications in order to split the subsums further in terms of a dyadic decomposition of the range of and apply Lemma 1 to count the number of terms in each subsubsum. We define δ to denote C 1/. The rest of estimating the +++ )-terms is divided into three cases according to whether δ 1, 1/C δ 1 or δ 1/C. The terms with δ 1 are the easiest to dispose of. Namely, there are trivially at most ABCD such terms, and so by Lemma 18, they contribute ABCDABCD) 3/4 3/+ε δ C 3 3/+ε ABCD) 1/4 C 1/ 3 3/+ε CD) 1/4 3 3/+ε N 1/. Next, let us consider the terms with 1/C δ 1. Since δ C = a+ b+ c d ) = a+ b d ) c+ a+ b+ c d ) c, we have c = a+ b d ) +OδC), 17

18 andsothereis δc possiblevaluesofcforanygiventriple a,b,d. Thus, there are ABδCD terms with 1/C δ 1, and by Lemma 18, they contribute ABδCDABCD) 3/4 3/+ε δ C 3 3/+ε N 1/. We next consider the third case in which δ 1/C. This case will be split into two subcases depending on whether C Γ or C Γ, where Γ is some positive real number whose value will be set later when its impact on the final error terms is easier to see. Let first C Γ. We split our terms into further subsums by performing a dyadic division of the value range of δ. Since by Lemma 19 we have δ = C 1/ C 5/ ABC) 1/ N 4, there will be logn ε such subranges of δ to consider. Let us consider the terms where δ lies in one of these. Let us observe that D C. By the first upper bound given by Lemma 1 there are ABCDδ + C 3/+ε ) corresponding terms, which then contribute, estimating the integral either by absolute values or by Lemma 18, ε ABCD δ +C 3/+ε) } ABCD) 3/4 min {, 3/ δ C 3 3/+ε ABCD) 1/4 C 1/ + +ε ABCD) 1/4 C 3/+ε 3 3/+ε N 1/ + +ε Γ 1/. Next, let C Γ. We again perform the same dyadic division of the value range of δ. In one such subrange, the second upper bound of Lemma 1 says that there ABCD δc +ABCD) 1/) C ε corresponding terms which then contribute, estimating the integrals by Lemma 18 and 1/δ by Lemma 19, ε ABCD δc +ABCD) 1/) C ε 3/4 3/+ε ABCD) δ C 3/+ε C 5/ +C 3/ ABCD) 1/4) 3/+ε Γ 5/. This concludes our treatment of the +++ )-terms. The last group of signs to consider are those with two plus signs and two minussigns. Thosetermsin which = a+ b c dvanisheswerealready consideredinthe derivationofthemainterm, andsowemayassumethroughout that 0. We perform again the dyadic division of the summations in a, b, c and d, and by symmetry, we may focus on the terms with A C, B C and D C. Wemayagaindefine δ = C 1/, andthe regionsδ 1and1/C δ 1 are handled by the same estimates as in the +++ )-terms. The case δ 1/C is also similar, and split into subcases depending on whether C Γ or C Γ. The latter wors out in exactly the same as for the +++ )-terms, except that Lemma 0 is to be used instead of Lemma 1. The only new complication arising from the case C Γ is that the first bound of Lemma 0 has an extra term which does not appear in Lemma 1. Since C 1/, at least one of A and B must be C. Let us suppose that 18

19 B C. The contribution from the extra term is, estimating the integrals by absolute values and AD 1, ε Cmin{A,B,D}ABCD) 3/4 +ε C 1/4 AD) 1/ ABD) 3/4 +ε C 1/4 B 3/4 AD) 1/4 +ε C 1/ +ε Γ 1/. Finally, we only need to collect all the error terms and choose suitable values for N and Γ. We have established that x, A h ) 4 dx C F 5/ 9/4+ε N 1/ + 3 N + +ε N 1/ /+ε N 1/ + 3 3/+ε Γ 5/ + +ε Γ 1/. Choosing Γ = 1/3 1/6 optimizes the last two terms to 13/6 3/1+ε. ThevalueofN canbechosenbyoptimizingthefirstandfourthterms; wechoose N = 1/ 3/4 and the first and fourth terms simplify to 11/4 15/8+ε. The second and third terms will become 1/+ε + 7/8 9/16+ε. This is easily seen to be 13/6 3/1+ε + 11/4 15/8+ε, and we are done. 8 ore moment estimates In the proof of Theorem 8, we will need a lower bound for the mean square of A ++. We will prove this following Ivić and Zhai [1]. Proposition 3. Let 1, let 0 Ξ, and let h and be coprime integers with 1. Then +Ξ +Ξ A x, h ) dx 3/ 3/4 +Ξ ε. Using the truncated Voronoi identity and Lemma 14 we obtain A n, h ) dx = 1/ π + = 1/ π +Ξ +Ξ x ) nh 4π nx 1/4 an)n 3/4 e cos π ) dx 4 n O ε )dx nh an)n 3/4 e n n ) +Ξ 4π nx x 1/4 cos π ) dx+oξ ε ) 4 1/ an) n 3/4 3/4 +Ξ ε 3/ 3/4 +Ξ ε. n 19

20 Proposition 4. Let 1, and let h and be coprime integers with 1 1/6 ε. Then n, A h dx ) 1/ 5/4. Using Theorem 11, the Cauchy Schwarz inequality, and Corollary 4 remembering that Theorem 5 removes the ε) we get 3/ and the claim follows easily. n, A h ) dx n, A h ) 3 dx 3/ 7/4 n, A h dx, ) n, A h dx ) Lemma 5. Let 1, and let h and be coprime integers with 1 1/6 ε. Then A ++ x, h ) dx 3/. By Proposition 4 and Proposition 3, we have 1/ 5/4 = x, A h dx ) x, RA h dx+ ) RA x, h )) RA x, h + + x, IA h dx ) IA x, h )) ) dx + ) +IA x, h )) dx For sufficiently large, the last integral, which is 3/ 3/4 + 1+ε, may be absorbed to the left-hand side, and we may continue the argument with the 0

21 Cauchy Schwarz inequality: 1/ 5/4 and the result follows easily. RA x, h )) + A ++ x, h dx ) + IA x, h )) ) dx + A ++ x, h ) dx, 9 Proof of Theorem 6 Choosing N x in the truncated Voronoi identity gives ) nh an)e x n x+ = 1/ π ) nh an)n 3/4 e n 4π ) nx+ ) cos π x+ ) 1/4 4 ) 4π nx cos π )x 1/4 +Ox ε ) 4 = 1/ x 1/4 π ) nh an)n 3/4 e n 4π ) nx+ ) cos π 4π nx cos π 4 4) ) +O ε ). Here the last estimate follows from estimates by absolute values and the simple observation that x+ ) 1/4 x 1/4 = 1 4 x+ x t 3/4 dt 3/4. Plugging this output of the truncated Voronoi formula into the mean square expression gives +Ξ = π x n x+ +Ξ x 1/ n ) nh an)e dx ) nh an)n 3/4 e cos...) cos )) dx 1

22 + 1/ R +Ξ x ) nh 1/4 an)n 3/4 e cos...) cos...))o ε )dx n +O Ξ ε). The last term is Ξ ε, and once we have proved that the first term is Ξ, an application of the Cauchy Schwarz inequality gives the conclusion that the second term is Ξ Ξ ε = Ξ 1/ ε Ξ ε. The rest of the proof consists basically of expanding the square, applying the first derivative test and collecting terms. The main contribution comes from the low-frequency diagonal terms, which are easy enough to handle. However, one must be careful in order to avoid resonances in the off-diagonal terms. First, let us introduce some notation: for λ R, we shall write s λ = n an)n 3/4 cos πnh e nx+λ) where λ will ultimately be either 0 or. Similarly, we shall write s λ for the same sum with cos πnh/ ) replaced by sin πnh/ ). Thus we shall have +Ξ x n x+ = π ) nh an)e dx +Ξ + π x 1/ Rs Rs 0 ) dx +Ξ 1 8 x 1/ R s R s 0 ) dx+o Ξ ε). The second integral on the right-hand side can be treated in exactly the same way as the first one. In the end the factors cos πnh/ ) and sin πnh/ ) will go away due to the fact that cos α+sin α = 1. In the following we shall, for the sae of simpler notation, to assume that the coefficients an) are real. If this is not the case, then we can just replace an)cosπnh/ and an)sinπnh/ by the real and imaginary parts of an)enh/). The integrand Rs Rs 0 ) will be exchanged for a handful of other terms which lead to exponential integrals which can be fed to the first derivative test. The first step will be splitting the sums into low-frequency and high-frequency terms: s λ =..., s > λ =..., n N 0 N 0<n where N 0 = / ). Furthermore, we shall denote the corresponding real parts by σ λ and σ> λ, and the imaginary parts by t λ and t> λ. ),

23 The low- versus high-frequency split: ) Rs Rs 0 ) = σ σ 0 )+σ> σ> 0 ) = σ σ 0 ) +σ > σ> 0 ) +σ σ 0 )σ> σ> 0 ). These three terms and their integrals will be handled separately. We start with the last one. The last term can be written in terms of the exponential function: σ σ 0 )σ> σ> 0 ) s s 0 )σ> σ> 0 ) = s s 0 )s> s> 0 )+s s 0 )s> s> 0 ). By Lemmas 16 and 17, the contribution from the integral of the first term will be +Ξ x 1/ s s 0 )s> s> 0 )dx m N 0 N 0<n 1/ ε Ξ ε. am)an) mn) 3/4 n n m The contribution from the other term, the one without the complex conjugation, can be estimated in the same way using Lemma 15. The middle term can be estimated by two mean squares: σ > σ> 0 ) = σ > ) +σ > 0 ) σ > σ> 0 σ> ) +σ > 0 ) s > + s > 0. The contribution from the integral of either of these is obtained by expanding the square, considering the diagonal and the non-diagonal terms separately, estimating the diagonal terms by absolute values, and estimating the non-diagonal terms using Lemma 16, giving π +Ξ x 1/ s > λ dx N 0<n + an) n 3/ Ξ 1/ N 0<m m<n Ξ 1/ N 1/ ε Ξ 1/ +Ξ ε Ξ ε. am)an) mn) 3/4 n n m The first term will give the main contribution. First, we split it into two parts, the second of which will give rise to only oscillating integrals: σ σ 0 ) = 1 σ σ 0 ) +t t 0 )) + 1 σ σ 0 ) t t 0 )) = 1 s s Rs s 0 ) = 1 s s 0 +O s s 0 )). 3

24 The integral of the O-term can be estimated by expanding the square and estimating the resulting exponential integrals by Lemma 15: π +Ξ x 1/ s s 0 ) dx m N 0 n N 0 am)an) mn) 3/4 1+ε Ξ ε. The last integral we have to consider is 4π +Ξ x 1/ s s 0 dx. n+ m Here we again expand the square and consider the diagonal and the non-diagonal terms separately. The non-diagonal terms are estimated as before using Lemma 17 giving 3/4 am)an) mn) n m 1+ε Ξ ε. m<n N 0 m N 0 The diagonal terms give the main contribution; they are an) 4π cos πnh +Ξ )) n x+ x x 1/ e 1 n 3/ dx. n N 0 The integrals involving s λ give the same main terms with cos replaced by sin, and so these trigonometric factors cancel away leaving only +Ξ an) )) n x+ x 4π x 1/ e 1 n 3/ dx. n N 0 This sum is split into low-frequence terms and high-frequency terms according to whether n N 1 or N 1 < n N 0, where N 1 = 1 4. The high-frequency terms give a nonnegative contribution which is at most an) Ξ 1/ N 1/ n 3/ 1 Ξ 1/ Ξ. N 1<n N 0 In the low-frequency terms with n N 1, we have 0 < n x+ x ) n = x+ x dt t where x [, +Ξ]. Thus, the low-frequency terms are n 1, an) 1/ n Ξ n 3/ 1 Ξ 1/ N 1/ 1 Ξ, n N 1 and we are done. By inspecting the last line, we also observe that, if ε 1/ ε, then N 1 ε and the low-frequency terms actually are Ξ, so that we get the second conclusion of the theorem. 4

25 10 Proof of Theorem 7 Let λ Z + be such that λ 1/ < λ+1, and write b = λ. Then b λ 1/. The idea of the proof is to consider subsums of length b instead of individual terms. The relevant observation here is that max 1 U x n x+u ) nh an)e max 1 j< λ x<n x+jb ) nh an)e +O b log ), where j taes only integral values and we have used Deligne s and Shiu s estimates. The maximum is certainly attained for some value particular value of j which we shall call j 0. The remainder of the argument is really just a matter of estimating the sum of length j 0 b so that the dependence on j 0 goes away, and then finishing off with Theorem 6. We use the binary representation of j 0 to dyadically dissect [x,x+j 0 b]. Write j 0 = λ1 + λ λn, where N Z + and Writing also 0 λ N <... < λ < λ 1 < λ. Λ 0 = 0, Λ 1 = λ1, Λ = λ1 + λ,..., Λ N = λ1 + λ λn, we estimate x<n j 0b ) nh an)e N N =1 x+λ 1 b<n x+λ b ) nh an)e. Writing next Λ = ν λ for {0,1,...,N}, where we set λ 0 = 0, we have 0 ν < λ λ, and we may estimate x+λ 1 b<n x+λ b = ) nh an)e x+ν 1 λ 1b<n x+ν 1 λ 1 1 b+ λ b λ λ 1 1 ν=0 x+ν λ 1b<n x+ν λ 1b+ λ b ) nh an)e ) nh an)e. 5

26 Combining this with max 0 U N λ λ x n x+u N =1 λ l=1 λ l=1 λ λ 1 1 ν=0 λ l+1 1 ν=0 λ l+1 1 ν=0 and applying Theorem 6 gives ) nh an)e dx x+ν λ 1b<n x+ν λ 1b+ λ b x+ν l+1 b<n x+ν l+1 b+ l b l b λ 11 Proof of Theorem 8 ) nh an)e dx ) nh an)e dx λ λ l l b λ λ b log. l=1 We will only prove the case in which at least one of the real and imaginary parts of A ) x, nh is required to be positive. The more general case follows from multiplying the underlying cusp form by a suitable unimodular constant. We write = c 1/ log, and define a function ωx) = A ++ x, h ) 4 max 0 U A x+u, nh ) A x, nh ) cx 1/ for x [,]. Here c is a small positive real constant. The point of this definition is that when ωx) > 0 then x must lie in a subinterval of length in which A ) x, nh 1/ 1/4 and A ) x, nh has a positive real or positive imaginary part. If c is small enough, then we may estimate ωx)dx 3/ O c log ) O c 3/) 3/, and so ωx)dx will be positive and 3/ for sufficiently large. Finally, let S be the set of x [,] for which ωx) > 0, and write l for the measure of S. Then 3/ ωx)dx ωx)dx A ++ 1 dx so that l, and we are done. S S S x, h ) dx x, A h ) 4 dx l 1/, 6

27 Acnowledgements The author would lie to express his gratitude for the valuable advice of A.-. Ernvall-Hytönen as well as for an insightful conversation with K.-. Tsang on matters related to Voronoi-type formulae. The author first learned about the sign change results for the divisor problem and related problems at the excellent conference Elementare und analytische Zahlentheorie, held in Schloss Schney, August 13 18, 01. The author is grateful for the generous support of the organizers. This research was funded by Finland s inistry of Education through the Graduate School of Inverse Problems, the Academy of Finland through the Finnish Centre of Excellence of Inverse Problems Research, and the Foundation of Vilho, Yrjö and Kalle Väisälä. References [1] P. Deligne: La conjecture de Weil: I, Publ. ath. Inst. Hautes Études Sci., ), [] A.-. Ernvall-Hytönen: On the error term in the approximate functional equation for exponential sums related to cusp forms, Int. J. Number Theory, 4 008), [3] A.-. Ernvall-Hytönen: On the mean square of short exponential sums related to cusp forms, Funct. Approx. Comment. ath., ), [4] A.-. Ernvall-Hytönen: ean square estimate for relatively short exponential sums involving Fourier coefficients of cusp forms, preprint, arxiv: [math.nt]. [5] A.-. Ernvall-Hytönen and K. Karppinen: On short exponential sums involving Fourier coefficients of holomorphic cusp forms, Int. ath. Res. Not., article ID rnn0 008), [6] D. R. Heath-Brown and K.-. Tsang: Sign changes of ET), x), and Px), J. Number Theory, ), [7] A. Ivić: Large values of the error term in the divisor problem, Invent. ath., ), [8] A. Ivić: The Riemann Zeta-Function: Theory and Applications, Dover Publications, 003. [9] A. Ivić: The circle and divisor problem, Bull. Cl. Sci. ath. Nat. Sci. ath., 9 004), [10] A. Ivić: On the divisor function and the Riemann zeta-function in short intervals, Ramanujan J., ), [11] A. Ivić and P. Sargos: On the higher moments of the error term in the divisor problem, Illinois J. ath., ),

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