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Research article

The short interval results for power moments of the Riesz mean error term

  • Received: 01 March 2023 Revised: 26 June 2023 Accepted: 09 August 2023 Published: 31 August 2023
  • 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 (3k5) 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 (3k5) 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 nN, where T(n) is the Hecke operator of order n. Let cn be the convolution function defined by

    cn=n1κm2|nm2(κ1)|a(nm2)|2.

    In 1974, Deligne [1] proved the estimate |a(n)|nκ12d(n), where d(n) is the Dirichlet divisor function, which implies cnεnε. 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):=nxcnCx, (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)nx(xn)ρcn

    for any fixed ρ0 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)κ1Γ(κ+1)Fyκ2|φ(z)|2dxdy,Z(s)=n=1cnns, s>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 α0. They also proved that

    Δ1(x;φ)=O(x65)

    and

    T1Δ21(x;φ)dx=213(2π)4(n=1c2nn7/4)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(T358136+ε),T1Δ41(x;φ)dx=B4(c)11264π8T112+O(T1121221+ε),T1Δ51(x;φ)dx=B5(c)108544π10T538+O(T53811731+ε), (1.4)

    where

    Bk(f):=k1l=1(k1l)sk;l(f)cosπ(k2l)4,sk;l:=4n1++4nl=4nl+1++4nkf(n1)f(nk)(n1nk)7/8, 1lk.

    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 k3 be a fixed integer. For any sufficiently small ε>0, let δk:=(k1)(4k1+k102)+3, 0<δ<1 be a fixed constant, which satisfies 83δkε<δ, T and H are two large positive real number, which satisfies

    T+HTH|Δ1(x;φ)|k+δdxHT9(k+δ)/8+ε (1.5)

    and T3/4+2δkε/(3δ)HT. Then, we have

    T+HTHΔk1(x;φ)dx=Bk(c)2k1(2π)2kT+HTHx9k/8dx+O(HT9k8+ε(HT3/4)3δδk), (1.6)

    where

    Bk(c):=k1l=1(k1l)sk;l(c)cosπ(k2l)4sk;l(c):=4n1++4nl=4nl+1++4nkcn1cnk(n1nk)7/8, 1l<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 3k5, 1/8<θ<1/5 is a real number. Let Δ1(x;φ)xθ+1. Then, we have asymptotic formula

    T+HTHΔk1(x;φ)dx=Bk(c)2k1(2π)2kT+HTHx9k/8dx+O(HT9k8ε), (1.7)

    when T1+(k2)θk/8+εHT.

    Corollary 1. For 3k5, if T(k2)/5+1k/8+εHT, asymptotic formula (1.7) is true.

    Theorem 3. Suppose k3 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/2HT.

    Remark 2. By variable substitution, it is easy to see that

    T+HTHΔk1(x;φ)dx=4T+HTHΔk1(x4;φ)x3dx,

    here

    T:=(T+H)1/4+(TH)1/44T1/4, H:=(T+H)1/4(TH)1/44H/T3/4.

    If the conjecture Δ1(x;φ)=O(X9/8+ε) is true and H=T3/4+ε/2, then we have HTε/2. Thus, Theorem 3 contains the integral T+GTGΔk1(x4;φ)dx has asymptotic formula for G=Tε/2. Thus, the constant 3/4 in Theorem 3 is probably the best.

    Lemma 1. Suppose x>1 is a real number. For 1Nx2 a parameter we have

    Δ1(x;φ)=1(2π)2R(x;N)+O(x1+ε+x3/2+εN1/2), (2.1)

    where

    R:=R(x,N)=x9/8nNcnn7/8cos(8π4nxπ4).

    Proof. This is [9, Lemma 2.1].

    Lemma 2. Suppose k3, (i1,,ik1){0,1}k1 such that

    4n1+(1)i14n2+(1)i24n3++(1)ik14nk0.

    Then, we have

    |4n1+(1)i14n2+(1)i24n3++(1)ik14nk|max(n1,,nk)(4k241).

    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(maxaxb|f(x)|)(maxau<vb|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(p4x+q)dx=2TT4p1x33/8(p4x3/4cos(p4x+q))dxT33/8|p|1|vup4x3/4cos(p4x+q)dx|T33/8|p|1.

    Lemma 4. Suppose k3, (i1,,ik1){0,1}k1, (i1,,ik1)(0,,0), N1,,Nk>1, 0<ΔE1/4, E=max(N1,,Nk). Let A denote the number of solutions of the inequality

    |4n1+(1)i14n2+(1)i24n3++(1)ik14nk|<Δ (2.2)

    with Nj<nj2Nj, 1jk, where

    A=A(N1,,Nk;i1,,ik1;Δ).

    Then, we have

    AΔE1/4N1Nk+E1N1Nk.

    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

    4n1+(1)i14n2+(1)i24n3++(1)ik24nk1=(1)ik14nk+θΔ.

    Thus, we have

    4n1+(1)i14n2+(1)i24n3++(1)ik24nk1=(1)ik14nk+θΔ,
    (4n1+(1)i14n2+(1)i24n3++(1)ik24nk1)4=nk+O(ΔN3/4k).

    Therefore, for fixed (n1,,nk1), the number of nk is 1+ΔN3/4k and so

    AΔN3/4kN1Nk1+N1Nk1.

    If T/2HT, it is easily to get Theorem 1. Suppose HT2 and y is a parameter such that Tε<yT1/3. For any Tx2T, we define

    R=x9/8(2π)2nycnn7/8cos(8π4nxπ4),R1=R1(x,y):=Δ1(x;φ)R.

    We will prove that the higher-power moment of R1 is small, so the integral T+HTHΔk1(x;φ)dx can be well approximated by T+HTHRkdx, which is easy to evaluate.

    First, by the elementary formula

    cosb1cosbk=12k1(i1,,ik1){0,1}k1cos(b1+(1)i1b2+(1)i2b3++(1)ik1bk),

    we can write

    Rk=(2π)2kx9k/8n1ynkycn1cnk(n1nk)7/8kj=1cos(8π4njxπ/4)=(2π)2kx9k/8(i1,,ik1){0,1}k1n1ynkycn1cnk(n1nk)7/8×cos(8π4xα(n1,,nk;i1,,ik1)π/4β(i1,,ik1)),

    where

    α(n1,,nk;i1,,ik1):=4n1+(1)i14n2+(1)i24n3++(1)ik14nk,β(i1,,ik1):=1+(1)i1+(1)i2++(1)ik1.

    Therefore, we have

    Rk=12k1(2π)2kx9k/8(i1,,ik1){0,1}k1cos(πβ4)njy,1jkα=0cn1cnk(n1nk)7/8+12k1(2π)2kx9k/8(i1,,ik1){0,1}k1njy,1jkα0cn1cnk(n1nk)7/8cos(8πα4xπβ4)=12k1(2π)2k(S3(x,k)+S4(x,k)), (3.1)

    where

    α:=α(n1,,nk;i1,,ik1), β:=β(i1,,ik1).

    Consider S3(x,k). We have

    T+HTHS3(x,k)dx=(i1,,ik1){0,1}k1cos(πβ4)njy,1jkα=0cn1cnk(n1nk)7/8T+HTHx9k/8dx. (3.2)

    By (4.3) in [9], we know that

    12k1(2π)2kT+HTHS3(x,k)dx=Bk(c)2k1(2π)2kT+HTHx9k/8dx+O(HT9k/8+εy3/4). (3.3)

    Now we consider S4(x,k). By the first derivative of van der Corput method, one has

    T+HTHS4(x,k)dxT3/4+9k/8(i1,,ik1){0,1}k1njy,1jkα0cn1cnk(n1nk)7/8|α|. (3.4)

    For fixed (i1,,ik1){0,1}k1, we write

    (y;i1,,ik1)=njy,1jkα0cn1cnk(n1nk)7/8|α|.

    If (i1,,ik1)=(0,,0), then we have

    (y;0,,0)njy,1jkcn1cnk(n1nk)7/8(4n1++4nk)njy,1jkcn1cnk(n1nk)7/8+1/(4k)y(k2)/8logky.

    For (i1,,ik1)(0,,0), by a splitting argument we deduce that there exist a collection of numbers 1<N1,,Nk<y such that

    (y;i1,,ik1)Nj<nj2Nj,1jkα0cn1cnk(n1nk)7/8|α|logky.

    Suppose N1Nky. According to Lemma 2, we know that |α|N(4k241)k. For some N(4k241)kΔ<y1/4, by using Lemma 4 we can get

    Nj<nj2Nj,1jkα0cn1cnk(n1nk)7/8|α|yε(N1Nk)7/8ΔA(N1,,Nk;i1,,ih1;Δ)yε(N1Nk)7/8Δ(ΔN3/4kN1Nk1+N1Nk1)yε(Nk28k+N4k2+k108k)yb(k)+ε, (3.5)

    where b(k)=4k2+k108. Therefore we have

    T+HTHS4(x,k)dxH3/4T9k/8+εyb(k). (3.6)

    According to (3.1)-(3.6), we have

    T+HTH|R|kdx=Bk(c)2k1(2π)2kT+HTHx9k/8dx+O(HT9k/8+εy3/4+T3/4+9k/8+εyb(k)) (3.7)

    From the definition of K0 (K0:=min{nN:nA0,2|n} in [9] we write

    K0={k+1,k is odd,k+2,otherwise.

    Suppose that y(HT3/4)1/b(K0). Let N=T in the formula (2.1) of Lemma 1. Then, we can obtain

    R1=(2π)2x9/8y<nTcnn7/8cos(8π4nxπ/4)+O(T1+ε)|x9/8y<nTcnn7/8e(44nx)|+T1+ε.

    Thus, we have

    T+HTHR21dxHT2+ε+T+HTH(x9/8y<nTcnn7/8e(44nx))2dxHT2+ε+HT9/4y<nTc2nn7/4+T3y<m<nTcncm(mn)7/8(4n4m)HT2+ε+T3+ε+HT9/4log3Ty3/4HT9/4log3Ty3/4. (3.8)

    From (3.7) we can obtain that

    T+HTH|R|K0dxHT9K0/8+ε.

    By Hölder's inequality and the above formula we have

    T+HTH|R|A0dxHT9A0/8+ε (3.9)

    for A0=k+δ, here 0<δ<1 is a fixed constant. From (1.5) and (3.9) we get

    T+HTH|R1|A0dxT+HTH(|Δ(x;φ)|A0+|R|A0)dxHT9A0/8+ε. (3.10)

    For any 2<A<A0, by (3.8), (3.10) and Hölder inequality we have

    T+HTH|R1|Adx=T+HTH|R1|2(A0A)A02+A0(A2)A02dx(T+HTHR21dx)(A0A)A02(T+HTH|R1|A0dx)(A2)A02HT9A8+εy3(A0A)4(A02). (3.11)

    Thus we have

    Lemma 5. Suppose Tεy(HT3/4)1/b(K0), 2<A<A0. Then, we have

    T+HTH|R1|AdxHT9A8+εy3(A0A)4(A02). (3.12)

    Suppose 3k<A0. By the elementary formula (a+b)kak|b|k+|ak1b| we have

    T+HTHΔk1(x;φ)dx=T+HTHRkdx+O(T+HTH|Rk1R1|dx)+O(T+HTH|R1|kdx). (3.13)

    By (3.9), Lemma 5 and Hölder inequality we have

    T+HTH|Rk1R1|dx(T+HTH|RA0|dx)k1A0(T+HTH|RA0A0k+11|dx)A0k+1A0HT9k8+εy3(A0k)4(A02). (3.14)

    Taking y=(HT3/4)1b(K0)+3δ4(k+δ2). From (3.13), (3.14), (3.7) and Lemma 5, we can obtain

    T+HTHΔk1(x;φ)dx=Bk(c)2k1(2π)2kT+HTHx9k/8dx+O(T3/4+9k/8+εyb(K0)+HT9k8+εy3(A0k)4(A02))=Bk(c)2k1(2π)2kT+HTHx9k/8dx+O(HT9k8+ε(HT3/4)3δ4b(K0)(k+δ2)+3δ)=Bk(c)2k1(2π)2kT+HTHx9k/8dx+O(HT9k8+ε(HT3/4)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 THx1<<xRT+H satisfies |xrxs|V (rsR), T1/8VTθ and |Δ1(xr;φ)|VT (r=1,,R). Dividing interval [TH,T+H] into the subinterval of length not exceeding T0 (T0V). Let R0 denotes the number of xr which exists in the subinterval of length not exceeding T0. Then, we have

    RR0(1+H/T0).

    Tanigawa, Zhai and Zhang [9] proved that

    RTV3L5+HT3V25L39, (4.1)

    if T0=V22T2L34, R0T1+4εV2.

    Now we consider T+HTH|Δ1(x;φ)|Adx, where 2<A<24 be a fixed real number. According to [9], we have

    T+HTH|Δ1(x;φ)|AdxHT9A/8+ε+VVrRV|Δ1(xr;φ)|A, (4.2)

    where T1/8V=2mTθ, VT<|Δ1(xr;φ)|2VT (r=1,,RV) and |xrxs|V for rsRRV. By (4.1) we have

    VrRV|Δ1(xr;φ)|ARVTAVA+1L5T1+AVA2+L39HT3+AVA24T1+AVA2L5+HT9A/8L39. (4.3)

    Combining (4.2) and (4.3), we get

    T+HTH|Δ1(x;φ)|AdxHT9A/8+ε+T1+A+θ(A2)+ε.

    Now suppose 2<A<2θ/(θ1/8). If HT1+θ(A2)A/8, then we have

    T+HTH|Δ1(x;φ)|AdxHT9A/8+ε.

    We notice that 2θ/(θ1/8)=16/3=5.33, if θ=1/5. In Theorem 1, taking δ=4/3δkε for 3k5, we can get Theorem 2.

    Suppose Δ1(x;φ)x9/8+ε. Then,

    T+HTH|Δ1(x;φ)|k+δdxHT9(k+δ)/8+ε,

    where ε=(k+1)ε. From Theorem 1, we can get asymptotic formula

    T+HTHΔk1(x;φ)dx=Bk(c)2k1(2π)2kT+HTHx9k/8dx+O(HT9k8+ε(HT3/4)3δ/δk)

    for T2δkε3δ+34HT. 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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