Let Δ1(x;φ) denote the error term in the classical Rankin-Selberg problem. In this paper, our main results are getting the k-th (3≤k≤5) power moments of Δ1(x;φ) in short intervals and its asymptotic formula by using large value arguments.
Citation: Jing Huang, Qian Wang, Rui Zhang. The short interval results for power moments of the Riesz mean error term[J]. Electronic Research Archive, 2023, 31(9): 5917-5927. doi: 10.3934/era.2023300
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Let Δ1(x;φ) denote the error term in the classical Rankin-Selberg problem. In this paper, our main results are getting the k-th (3≤k≤5) power moments of Δ1(x;φ) in short intervals and its asymptotic formula by using large value arguments.
Let φ(z) be a holomorphic form of weight κ with respect to the full modular group SL2(Z) and denote by a(n) the n-th Fourier coefficient of φ(z). We assume that φ(z) is normalized such that a(1)=1 and T(n)φ=a(n)φ for every n∈N, where T(n) is the Hecke operator of order n. Let cn be the convolution function defined by
cn=n1−κ∑m2|nm2(κ−1)|a(nm2)|2. |
In 1974, Deligne [1] proved the estimate |a(n)|≤nκ−12d(n), where d(n) is the Dirichlet divisor function, which implies cn≪εnε. Here and in what follows, ε denotes an arbitrarily small positive number which is not necessarily the same at each occurrence. The classical Rankin-Selberg problem is to estimate the upper bound of the error term
Δ(x;φ):=∑n≤xcn−Cx, | (1.1) |
where C is an explicit constant. In 1939, Rankin [2] proved that
Δ(x;φ)=O(x35), | (1.2) |
which was stated by Selberg [3] again without proof. However, no improvement of (1.2) has been obtained after Rankin and Selberg. In [4], Ivić obtained that Δ(x;φ)=Ω±(x3/8) and conjectured that Δ(x;φ)=O(x3/8+ε).
Ivić, Matsumoto and Tanigawa [5] considered the Riesz mean of the type
Dρ(x;φ):=1Γ(ρ+1)∑n≤x(x−n)ρcn |
for any fixed ρ≥0 and define the error term Δρ(c;φ) by
Dρ(x;φ)=π2κR06Γ(ρ+2)xρ+1+Z(0)Γ(ρ+1)xρ+Δρ(x;φ), | (1.3) |
where
R0=12(4π)κ−1Γ(κ+1)∬Fyκ−2|φ(z)|2dxdy,Z(s)=∞∑n=1cnn−s, ℜs>1, |
where Z(s) can be continued to the whole plane and the integral being taken over a fundamental domain F of SL2(Z). They considered the relation between Δ(x;φ) and Δ1(x;φ) and proved that Δ(x;φ)=O(xα/2) if Δ1(x;φ)=O(xα) holds for some α≥0. They also proved that
Δ1(x;φ)=O(x65) |
and
∫T1Δ21(x;φ)dx=213(2π)−4(∞∑n=1c2nn−7/4)T13/4+O(T3+ε). |
Since this kind sums play a very important role in the study of analytic number theory, many number theorists and scholars have obtained a series of meaningful research results (for example see [6,7,8,9,11,12,13], etc.). In particular, in [9], Tanigawa, Zhai and Zhang studied the third, fourth and fifth power moments of Δ1(x;φ) and proved that
∫T1Δ31(x;φ)dx=B3(c)1120π6T358+O(T358−136+ε),∫T1Δ41(x;φ)dx=B4(c)11264π8T112+O(T112−1221+ε),∫T1Δ51(x;φ)dx=B5(c)108544π10T538+O(T538−11731+ε), | (1.4) |
where
Bk(f):=k−1∑l=1(k−1l)sk;l(f)cosπ(k−2l)4,sk;l:=∑4√n1+⋯+4√nl=4√nl+1+⋯+4√nkf(n1)⋯f(nk)(n1⋯nk)7/8, 1≤l≤k. |
In this paper we shall prove that the k-th power moment of Δ1(x;φ) in short intervals for k=3,4,5, the theorem is as follows.
Theorem 1. Let k≥3 be a fixed integer. For any sufficiently small ε>0, let δk:=(k−1)(4k−1+k−102)+3, 0<δ<1 be a fixed constant, which satisfies 83δkε<δ, T and H are two large positive real number, which satisfies
∫T+HT−H|Δ1(x;φ)|k+δdx≪HT9(k+δ)/8+ε | (1.5) |
and T3/4+2δkε/(3δ)≤H≤T. Then, we have
∫T+HT−HΔk1(x;φ)dx=Bk(c)2k−1⋅(2π)2k∫T+HT−Hx9k/8dx+O(HT9k8+ε(HT−3/4)−3δδk), | (1.6) |
where
Bk(c):=k−1∑l=1(k−1l)sk;l(c)cosπ(k−2l)4sk;l(c):=∑4√n1+⋯+4√nl=4√nl+1+⋯+4√nkcn1⋯cnk(n1⋯nk)7/8, 1≤l<k. |
Remark 1. If we take H=T, δ is larger (for example δ=12), then Theorem 1 implies asymptotic formula (1.4).
As corollaries, we have the following Theorems 2 and 3. Theorem 3 implies the best possible result.
Theorem 2. Suppose 3≤k≤5, 1/8<θ<1/5 is a real number. Let Δ1(x;φ)≪xθ+1. Then, we have asymptotic formula
∫T+HT−HΔk1(x;φ)dx=Bk(c)2k−1⋅(2π)2k∫T+HT−Hx9k/8dx+O(HT9k8−ε), | (1.7) |
when T1+(k−2)θ−k/8+√ε≤H≤T.
Corollary 1. For 3≤k≤5, if T(k−2)/5+1−k/8+√ε≤H≤T, asymptotic formula (1.7) is true.
Theorem 3. Suppose k≥3 be a any fixed integer and conjecture Δ1(x;φ)=O(x9/8+ε) is true. Then, asymptotic formula (1.7) is true if T3/4+√ε/2≤H≤T.
Remark 2. By variable substitution, it is easy to see that
∫T+HT−HΔk1(x;φ)dx=4∫T′+H′T′−H′Δk1(x4;φ)x3dx, |
here
T′:=(T+H)1/4+(T−H)1/44≍T1/4, H′:=(T+H)1/4−(T−H)1/44≍H/T3/4. |
If the conjecture Δ1(x;φ)=O(X9/8+ε) is true and H=T3/4+√ε/2, then we have H′≍T√ε/2. Thus, Theorem 3 contains the integral ∫T+GT−GΔk1(x4;φ)dx has asymptotic formula for G=T√ε/2. Thus, the constant 3/4 in Theorem 3 is probably the best.
Lemma 1. Suppose x>1 is a real number. For 1≪N≪x2 a parameter we have
Δ1(x;φ)=1(2π)2R(x;N)+O(x1+ε+x3/2+εN−1/2), | (2.1) |
where
R:=R(x,N)=x9/8∑n≤Ncnn7/8cos(8π4√nx−π4). |
Proof. This is [9, Lemma 2.1].
Lemma 2. Suppose k≥3, (i1,⋯,ik−1)∈{0,1}k−1 such that
4√n1+(−1)i14√n2+(−1)i24√n3+⋯+(−1)ik−14√nk≠0. |
Then, we have
|4√n1+(−1)i14√n2+(−1)i24√n3+⋯+(−1)ik−14√nk|≫max(n1,⋯,nk)−(4k−2−4−1). |
Proof. This is [9, Lemma 2.3].
Lemma 3. If f(x) and g(x) are continuous real-valued functions of x and f(x) is monotonic, then we have
∫baf(x)g(x)dx≪(maxa≤x≤b|f(x)|)(maxa≤u<v≤b|∫vug(x)dx|). |
Proof. This follows from the second mean value theorem.
For any real numbers p(≠0) and q, by using this lemma we can obtain
∫2TTx27/8cos(p4√x+q)dx=∫2TT4p−1x33/8(p4x3/4cos(p4√x+q))dx≪T33/8|p|−1|∫vup4x3/4cos(p4√x+q)dx|≪T33/8|p|−1. |
Lemma 4. Suppose k≥3, (i1,⋯,ik−1)∈{0,1}k−1, (i1,⋯,ik−1)≠(0,⋯,0), N1,⋯,Nk>1, 0<Δ≪E1/4, E=max(N1,⋯,Nk). Let A denote the number of solutions of the inequality
|4√n1+(−1)i14√n2+(−1)i24√n3+⋯+(−1)ik−14√nk|<Δ | (2.2) |
with Nj<nj≤2Nj, 1≤j≤k, where
A=A(N1,⋯,Nk;i1,⋯,ik−1;Δ). |
Then, we have
A≪ΔE−1/4N1⋯Nk+E−1N1⋯Nk. |
Proof. The proof of this lemma is similar to the proof of [10, Lemma 2.4]. Suppose E=Nk. If (n1,⋯,nk) satisfies (2.2), then for some |θ|<1, we can obtain
4√n1+(−1)i14√n2+(−1)i24√n3+⋯+(−1)ik−24√nk−1=(−1)ik−14√nk+θΔ. |
Thus, we have
4√n1+(−1)i14√n2+(−1)i24√n3+⋯+(−1)ik−24√nk−1=(−1)ik−14√nk+θΔ, |
(4√n1+(−1)i14√n2+(−1)i24√n3+⋯+(−1)ik−24√nk−1)4=nk+O(ΔN3/4k). |
Therefore, for fixed (n1,⋯,nk−1), the number of nk is ≪1+ΔN3/4k and so
A≪ΔN3/4kN1⋯Nk−1+N1⋯Nk−1. |
If T/2≤H≤T, it is easily to get Theorem 1. Suppose H≤T2 and y is a parameter such that Tε<y≤T1/3. For any T≤x≤2T, we define
R=x9/8(2π)2∑n≤ycnn7/8cos(8π4√nx−π4),R1=R1(x,y):=Δ1(x;φ)−R. |
We will prove that the higher-power moment of R1 is small, so the integral ∫T+HT−HΔk1(x;φ)dx can be well approximated by ∫T+HT−HRkdx, which is easy to evaluate.
First, by the elementary formula
cosb1⋯cosbk=12k−1∑(i1,⋯,ik−1)∈{0,1}k−1cos(b1+(−1)i1b2+(−1)i2b3+⋯+(−1)ik−1bk), |
we can write
Rk=(2π)−2kx9k/8∑n1≤y⋯∑nk≤ycn1⋯cnk(n1⋯nk)7/8k∏j=1cos(8π4√njx−π/4)=(2π)−2kx9k/8∑(i1,⋯,ik−1)∈{0,1}k−1∑n1≤y⋯∑nk≤ycn1⋯cnk(n1⋯nk)7/8×cos(8π4√xα(n1,⋯,nk;i1,⋯,ik−1)−π/4β(i1,⋯,ik−1)), |
where
α(n1,⋯,nk;i1,⋯,ik−1):=4√n1+(−1)i14√n2+(−1)i24√n3+⋯+(−1)ik−14√nk,β(i1,⋯,ik−1):=1+(−1)i1+(−1)i2+⋯+(−1)ik−1. |
Therefore, we have
Rk=12k−1(2π)2kx9k/8∑(i1,⋯,ik−1)∈{0,1}k−1cos(−πβ4)∑nj≤y,1≤j≤kα=0cn1⋯cnk(n1⋯nk)7/8+12k−1(2π)2kx9k/8∑(i1,⋯,ik−1)∈{0,1}k−1∑nj≤y,1≤j≤kα≠0cn1⋯cnk(n1⋯nk)7/8cos(8πα4√x−πβ4)=12k−1(2π)2k(S3(x,k)+S4(x,k)), | (3.1) |
where
α:=α(n1,⋯,nk;i1,⋯,ik−1), β:=β(i1,⋯,ik−1). |
Consider S3(x,k). We have
∫T+HT−HS3(x,k)dx=∑(i1,⋯,ik−1)∈{0,1}k−1cos(−πβ4)∑nj≤y,1≤j≤kα=0cn1⋯cnk(n1⋯nk)7/8∫T+HT−Hx9k/8dx. | (3.2) |
By (4.3) in [9], we know that
12k−1⋅(2π)2k∫T+HT−HS3(x,k)dx=Bk(c)2k−1⋅(2π)2k∫T+HT−Hx9k/8dx+O(HT9k/8+εy−3/4). | (3.3) |
Now we consider S4(x,k). By the first derivative of van der Corput method, one has
∫T+HT−HS4(x,k)dx≪T3/4+9k/8∑(i1,⋯,ik−1)∈{0,1}k−1∑nj≤y,1≤j≤kα≠0cn1⋯cnk(n1⋯nk)7/8|α|. | (3.4) |
For fixed (i1,⋯,ik−1)∈{0,1}k−1, we write
∑(y;i1,⋯,ik−1)=∑nj≤y,1≤j≤kα≠0cn1⋯cnk(n1⋯nk)7/8|α|. |
If (i1,⋯,ik−1)=(0,⋯,0), then we have
∑(y;0,⋯,0)≪∑nj≤y,1≤j≤kcn1⋯cnk(n1⋯nk)7/8(4√n1+⋯+4√nk)≪∑nj≤y,1≤j≤kcn1⋯cnk(n1⋯nk)7/8+1/(4k)≪y(k−2)/8logky. |
For (i1,⋯,ik−1)≠(0,⋯,0), by a splitting argument we deduce that there exist a collection of numbers 1<N1,⋯,Nk<y such that
∑(y;i1,⋯,ik−1)≪∑Nj<nj≤2Nj,1≤j≤kα≠0cn1⋯cnk(n1⋯nk)7/8|α|logky. |
Suppose N1≤⋯≤Nk≤y. According to Lemma 2, we know that |α|≫N−(4k−2−4−1)k. For some N−(4k−2−4−1)k≪Δ<y1/4, by using Lemma 4 we can get
∑Nj<nj≤2Nj,1≤j≤kα≠0cn1⋯cnk(n1⋯nk)7/8|α|≪yε(N1⋯Nk)7/8ΔA(N1,⋯,Nk;i1,⋯,ih−1;Δ)≪yε(N1⋯Nk)7/8Δ(ΔN3/4kN1⋯Nk−1+N1⋯Nk−1)≪yε(Nk−28k+N4k−2+k−108k)≪yb(k)+ε, | (3.5) |
where b(k)=4k−2+k−108. Therefore we have
∫T+HT−HS4(x,k)dx≪H3/4T9k/8+εyb(k). | (3.6) |
According to (3.1)-(3.6), we have
∫T+HT−H|R|kdx=Bk(c)2k−1⋅(2π)2k∫T+HT−Hx9k/8dx+O(HT9k/8+εy−3/4+T3/4+9k/8+εyb(k)) | (3.7) |
From the definition of K0 (K0:=min{n∈N:n≥A0,2|n} in [9] we write
K0={k+1,k is odd,k+2,otherwise. |
Suppose that y≤(HT−3/4)1/b(K0). Let N=T in the formula (2.1) of Lemma 1. Then, we can obtain
R1=(2π)−2x9/8∑y<n≤Tcnn7/8cos(8π4√nx−π/4)+O(T1+ε)≪|x9/8∑y<n≤Tcnn7/8e(44√nx)|+T1+ε. |
Thus, we have
∫T+HT−HR21dx≪HT2+ε+∫T+HT−H(x9/8∑y<n≤Tcnn7/8e(44√nx))2dx≪HT2+ε+HT9/4∑y<n≤Tc2nn7/4+T3∑y<m<n≤Tcncm(mn)7/8(4√n−4√m)≪HT2+ε+T3+ε+HT9/4log3Ty3/4≪HT9/4log3Ty3/4. | (3.8) |
From (3.7) we can obtain that
∫T+HT−H|R|K0dx≪HT9K0/8+ε. |
By Hölder's inequality and the above formula we have
∫T+HT−H|R|A0dx≪HT9A0/8+ε | (3.9) |
for A0=k+δ, here 0<δ<1 is a fixed constant. From (1.5) and (3.9) we get
∫T+HT−H|R1|A0dx≪∫T+HT−H(|Δ(x;φ)|A0+|R|A0)dx≪HT9A0/8+ε. | (3.10) |
For any 2<A<A0, by (3.8), (3.10) and Hölder inequality we have
∫T+HT−H|R1|Adx=∫T+HT−H|R1|2(A0−A)A0−2+A0(A−2)A0−2dx≪(∫T+HT−HR21dx)(A0−A)A0−2(∫T+HT−H|R1|A0dx)(A−2)A0−2≪HT9A8+εy−3(A0−A)4(A0−2). | (3.11) |
Thus we have
Lemma 5. Suppose Tε≤y≤(HT−3/4)1/b(K0), 2<A<A0. Then, we have
∫T+HT−H|R1|Adx≪HT9A8+εy−3(A0−A)4(A0−2). | (3.12) |
Suppose 3≤k<A0. By the elementary formula (a+b)k−ak≪|b|k+|ak−1b| we have
∫T+HT−HΔk1(x;φ)dx=∫T+HT−HRkdx+O(∫T+HT−H|Rk−1R1|dx)+O(∫T+HT−H|R1|kdx). | (3.13) |
By (3.9), Lemma 5 and Hölder inequality we have
∫T+HT−H|Rk−1R1|dx≪(∫T+HT−H|RA0|dx)k−1A0(∫T+HT−H|RA0A0−k+11|dx)A0−k+1A0≪HT9k8+εy−3(A0−k)4(A0−2). | (3.14) |
Taking y=(HT−3/4)1b(K0)+3δ4(k+δ−2). From (3.13), (3.14), (3.7) and Lemma 5, we can obtain
∫T+HT−HΔk1(x;φ)dx=Bk(c)2k−1⋅(2π)2k∫T+HT−Hx9k/8dx+O(T3/4+9k/8+εyb(K0)+HT9k8+εy−3(A0−k)4(A0−2))=Bk(c)2k−1⋅(2π)2k∫T+HT−Hx9k/8dx+O(HT9k8+ε(HT−3/4)−3δ4b(K0)(k+δ−2)+3δ)=Bk(c)2k−1⋅(2π)2k∫T+HT−Hx9k/8dx+O(HT9k8+ε(HT−3/4)−3δδk). | (3.15) |
From this we can easily get Theorem 1.
In this section, we will give the proof of Theorem 2. Let N=T3/5. By (2.1) we can obtain
Δ1(x;φ)≪x6/5. |
Suppose 18<θ<15 be a fixed real number. Let Δ1(x;φ)≪xθ+1.
Suppose T−H≤x1<⋯<xR≤T+H satisfies |xr−xs|≥V (r≠s≤R), T1/8≪V≪Tθ and |Δ1(xr;φ)|≫VT (r=1,⋯,R). Dividing interval [T−H,T+H] into the subinterval of length not exceeding T0 (T0≥V). Let R0 denotes the number of xr which exists in the subinterval of length not exceeding T0. Then, we have
R≪R0(1+H/T0). |
Tanigawa, Zhai and Zhang [9] proved that
R≪TV−3L5+HT3V−25L39, | (4.1) |
if T0=V22T−2L−34, R0≪T1+4εV−2.
Now we consider ∫T+HT−H|Δ1(x;φ)|Adx, where 2<A<24 be a fixed real number. According to [9], we have
∫T+HT−H|Δ1(x;φ)|Adx≪HT9A/8+ε+∑VV∑r≤RV|Δ1(xr;φ)|A, | (4.2) |
where T1/8≤V=2m≤Tθ, VT<|Δ1(xr;φ)|≤2VT (r=1,⋯,RV) and |xr−xs|≥V for r≠s≤R≤RV. By (4.1) we have
V∑r≤RV|Δ1(xr;φ)|A≪RVTAVA+1≪L5T1+AVA−2+L39HT3+AVA−24≪T1+AVA−2L5+HT9A/8L39. | (4.3) |
Combining (4.2) and (4.3), we get
∫T+HT−H|Δ1(x;φ)|Adx≪HT9A/8+ε+T1+A+θ(A−2)+ε. |
Now suppose 2<A<2θ/(θ−1/8). If H≥T1+θ(A−2)−A/8, then we have
∫T+HT−H|Δ1(x;φ)|Adx≪HT9A/8+ε. |
We notice that 2θ/(θ−1/8)=16/3=5.33⋯, if θ=1/5. In Theorem 1, taking δ=4/3δk√ε for 3≤k≤5, we can get Theorem 2.
Suppose Δ1(x;φ)≪x9/8+ε. Then,
∫T+HT−H|Δ1(x;φ)|k+δdx≪HT9(k+δ)/8+ε′, |
where ε′=(k+1)ε. From Theorem 1, we can get asymptotic formula
∫T+HT−HΔk1(x;φ)dx=Bk(c)2k−1⋅(2π)2k∫T+HT−Hx9k/8dx+O(HT9k8+ε′(HT−3/4)−3δ/δk) |
for T2δkε′3δ+34≤H≤T. Taking δ=43δk(k+1)√ε in Theorem 1, we can get Theorem 3.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This work was supported by National Natural Science Foundation of China (grant nos. 12171286).
The authors declare that there are no conflicts of interest regarding the publication of this paper.
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