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The third-power moment of the Riesz mean error term of symmetric square L-function

  • Let f(z) be a holomorphic Hecke eigenform of weight k with respect to the full modular group SL(2,Z), and let Δρ(x;sym2f) be the error term of the Riesz mean of the symmetric square L-function L(s,sym2f). In this paper, using a Voronoï type formula for Δρ(x;sym2f), we consider the third-power moment of Δρ(x;sym2f) and derive the asymptotic formula for

    T1Δ3ρ(x;sym2f)dx.

    Citation: Rui Zhang, Xiaofei Yan. The third-power moment of the Riesz mean error term of symmetric square L-function[J]. AIMS Mathematics, 2021, 6(9): 9436-9445. doi: 10.3934/math.2021548

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  • Let f(z) be a holomorphic Hecke eigenform of weight k with respect to the full modular group SL(2,Z), and let Δρ(x;sym2f) be the error term of the Riesz mean of the symmetric square L-function L(s,sym2f). In this paper, using a Voronoï type formula for Δρ(x;sym2f), we consider the third-power moment of Δρ(x;sym2f) and derive the asymptotic formula for

    T1Δ3ρ(x;sym2f)dx.



    Let H be the upper half plane, i.e., H={z=x+iy| x,yR,and y>0}. Let Sk(Γ) be the space of holomorphic cusp forms of even weight k2 for the full modular group Γ=SL(2,Z). Then for fSk(Γ) and zH, we have the Fourier expansion

    f(z)=n=1af(n)e(nz),

    where af(n) denotes the n-th Fourier coefficient, and e(x)=e2πix. Set λf(n)=af(n)nk12, then λf(n) is multiplicative. The analytic properties of λf(n) were studied by many authors (see [5,7,8,9,10,12,13,14,16,20,22,23,24,25]). The symmetric square L-function attached to f can be written as

    L(s,sym2f)=ζ(2s)n=1λf(n2)ns=n=1cnns

    for s>1. And it can be continued to an entire function on the whole complex plane. Then it is known that for any ϵ>0,

    |cn|d3(n)nϵ,

    where d3(n) is the number of ways to write n as a product of three factors.

    Hafner [3] considered the Riesz mean of the form

    Dρ(x;sym2f)=1Γ(ρ+1)nx(xn)ρcn,

    where ρ0 is a fixed number and indicates that cn is replaced by cn/2 if ρ=0. Also, the Riesz mean can be represented as a sum of "residue function" and "error term":

    Dρ(x;sym2f)=L(0,sym2f)Γ(ρ+1)xρ+Δρ(x;sym2f).

    Then let Δρ(x;sym2f) be the error term of the Riesz mean of the symmetric square L-function. Hafner [3] gave that for ρ=0, one has

    nxcn=L(0,sym2f)+Δ0(x;sym2f).

    The symmetric square L-function and the Riesz mean error term have been studied by many authors, for example, see [1,4,6,11,15,18,21,25]. Fomenko [2] considered Δρ(x;symmf) when m=2 and obtained a truncated Voronoï type formula, and we give it in Lemma 2.1. Let N=x12 in Lemma 2.1, one obtain

    Δρ(x;sym2f)x1+ρ2.

    Using the Voronöi formula, Fomenko got

    X1Δ2ρ(x;sym2f)dx=CX4ρ+53+O(Xρ+53+ϵ),

    where

    C=22ρ1π2ρ2(4ρ+5)1n=1c2nn2ρ+43.

    Wang [19] generalized the truncated Voronöi type formula to the case m3 under the hypothesis Nice(m,f):

    Δρ(x;symmf)=ϵ(symmf)2ρπρ1(m+1)12xmm+1ρ+m2(m+1)×nNcnn1m+1ρm+22(m+1)cos(2π(m+1)(nx)1m+1(A(m,k)+ρ2)π)+O(xmm+1ρ+m12(m+1)Nρm+1ρ+m12(m+1)+ϵ)+O(xmm+1(ρ+1)+ϵNρ+1m+1),

    where

    A(m,k)={m(m+2)(k1)8+12,   m0(mod4);m(m+2)(k1)8+1,   m2(mod4);(m+1)2(k1)8+34,   m is odd,
    ϵ(symmf)={+1, if  m  even ;ϵ(k,m), otherwise ,

    with

    ϵ(k,m):=i(m+12)2(k1)+m+12={ik, if  m 1 (mod 8),1, if  m 3 (mod 8),ik, if  m 5 (mod 8),+1, if  m 7 (mod 8).

    Liu and Wang [15] studied the higher-power moments of Δρ(x;sym2f). If exists a real number A0=A0(ρ)>3 such that

    T1ΔA0ρ(x;sym2f)dxT1+2ρ+13A0+ϵ,

    then for any integer 3h<A0, they had the following asymptotic formula

    T1Δhρ(x;sym2f)dx=6Bρ(h,c)(3+(2ρ+1)h)(2π)(ρ+1)h3h2T1+2ρ+13h+O(T1+2ρ+13h+ϵ(Tδρ(h,A0)+Tρ3)),

    where

    Bρ(h;f):=h1l=1(h1l)sρ(l,h;f)cos(πρ2(h2l))

    with

    sρ(l,h;f):=3n1++3nl=3nl+1++3nhf(n1)f(nh)(n1nh)(ρ+2)/3(1l<h),

    and

    δρ(h,A0):=σρ(h,A0)min(2ρ2ρ+1,13bρ(H0))

    with

    σρ(h,A0):=(2ρ+1)(A0h)3(A02),3h<A0,
    bρ(h):=3h213+(1ρ)h3

    and H0 is the least even integer such that nA0. They also proved the result for ρ=12, h=3,4,5,

    T1Δhρ(x;sym2f)dx=6B12(h,c)(3+2h)(2π)32h3h2T1+23h+O(T1+23hλ1/2(h,6)+ϵ),

    where λ1/2(3,6)=122, λ1/2(4,6)=187, λ1/2(5,6)=1498. Zhang, Han and Zhang [26] studied the power moment of the Riesz mean error term in short intervals. For k3, T23+2ϵdρ,k(2ρ+1)δHT and ϵ>0, they got

    T+HTHΔkρ(x;sym2f)dx=BkT+HTHx2ρ+13kdx+O(HT2ρ+13k+ϵ(HT23)(2ρ+1)δdρ,k),

    where

    Bk=2kρk+1πkρk3k2ck, dρ,k=(3k14+kρk)(k+δ2)+(2ρ+1)δ.

    Their results improved the results in [15] when ρ=12, k=3,4,5 and δ>5031.

    In this paper, we will use the Voronöi type formula for Δρ(x;sym2f) to study the third-power moment estimates of Δρ(x;sym2f) and obtain the following theorem.

    Theorem 1. For any fSk(Γ)+, x>1 and 612181<ρ<23, we have

    T1Δ3ρ(x;sym2f)dx=(2ρπρ+13)338ρ+8cos(πρ2)AT2ρ+2+O(T6ρ2+21ρ+168+3ρ+ϵ),

    where

    A=α, β=1(αβ(α+β))ρ2h=1h is cubefreehρ2cα3hcβ3hc(α+β)3h.

    Particularly, for ρ=12, we can get the following result.

    Corollary 2. Let ρ=12. For any ϵ>0, we have

    T1Δ3ρ(x;sym2f)dx=3π92144BT3+O(T5619+ϵ),

    where

    B=α, β=1(αβ(α+β))32h=1h is cubefreeh32cα3hcβ3hc(α+β)3h.

    Remark. Note that our results improve the results in [15].

    To prove our theorem, we need the following lemmas.

    Lemma 2.1. Let x>1, N1. Then for any fixed ρ, 0ρ1, we have

    Δρ(x;sym2f)=2ρπρ1312Σ(x)+O(x4ρ+16N2ρ+16+ϵ)+O(x2ρ+23+ϵNρ+13)+O(xϵ), (2.1)

    where

    Σ(x)=nNcnnρ+23x2ρ+13cos(6π3nxπρ2).

    Proof. See Theorem 1.1 in [2].

    Lemma 2.2. Let h3, (i1,,ih1){0,1}h1 such that

    3n1+(1)i13n2+(1)i23n3++(1)ih13nh0.

    Then

    |3n1+(1)i13n2++(1)ih13nh|max(n1,,nh)(3h231).

    Proof. See for example Lemma 2.3 in [21].

    Lemma 2.3. If g(x) and h(x) are continuous real-valued functions of x and g(x) is monotonic, then

    bag(x)h(x)dx(maxaxb|g(x)|)(maxau<vb|vuh(x)dx|).

    Proof. See the Lemma 1 in [17].

    Lemma 2.4. Suppose k3, (i1,,ik1){0,1}k1, (i1,,ik1)(0,,0) and

    N1,,Nk>1,  0<ΔE13,  E=max(N1,,Nk).

    Let

    A=A(N1,,Nk;i1,,ik1;Δ)

    denote the number of solutions of the inequality

    |3n1+(1)i13n2++(1)ik13nk|<Δ (2.2)

    with Nj<nj2Nj, 1jk. Then

    AΔE13N1Nk+E1N1Nk.

    Proof. It can be proved similarly as Lemma 2.4 of [21]. Without loss of generality, suppose E=Nk. If n1,,nk satisfy (2.2), then

    |3n1+(1)i13n2++(1)ik23nk1|=(1)ik1+13nk+θΔ

    for some |θ|<1. We can get

    (3n1+(1)i13n2++(1)ik23nk1)3=(1)ik1+1nk+O(ΔN23k).

    Hence for fixed n1,,nk1, the number of nk is 1+ΔN23k and thus

    AΔN23kN1Nk1+N1Nk1.

    Throughout this paper, n, m and k denote natural numbers. The constants implied in the symbols and O may depend on ϵ.

    By (2.1), we know that to establish the asymptotic formula in our theorem, we shall prove, for H1, that

    2HHΣ(x)2dx=12n=1c2nn2ρ+432HHx4ρ+23dx+O(H4ρ+53+ϵN2ρ+13), (3.1)

    and take N=H6ρ+8(8+3ρ)(ρ+1),

    2HHΣ(x)3dx=34cos(πρ2)A2HHx2ρ+1dx+O(H6ρ2+21ρ+168+3ρ+ϵ). (3.2)

    Then, since (a+b)3=a3+O(|b|a2+|b|3), it follows from (2.1) that

    2HHΔ3ρ(x;sym2f)dx=(2ρπρ+13)32HHΣ(x)3dx+O(H2ρ+23+ϵNρ+132HHΣ(x)2dx+H2ρ+3+ϵNρ1)=(2ρπρ+13)33A4cos(πρ2)2HHx2ρ+1dx+O(H6ρ2+21ρ+168+3ρ+ϵ) (3.3)

    by (3.1) and (3.2). Adding this for H=T/2,T/22,, we see that the asymptotic formula in our theorem follows.

    We shall now give the details of the proof of (3.2). The proof of (3.1), which we shall omit, employs similar arguments and is simpler.

    For simplicity, put

    r=r(n,m,k):=(nmk)ρ+23cncmck,     n,m,kN, (3.4)

    and r=0 otherwise. Taking the third power of both sides of Σ(x) and for each element i=(i1,i2)I2,I={0,1}, using the elementary formula

    cosa1cosa2cosa3=14iI2cos(a1+(1)i1a2+(1)i2a3),

    we can write

    Σ(x)3=34rx2ρ+1cos(6π(3n+3m3k)3xπρ2)+14rx2ρ+1cos(6π(3n+3m+3k)3x3πρ2)=S0(x)+S1(x)+S2(x), (3.5)

    where

    S0(x):=34cos(πρ2)3n+3m=3krx2ρ+1, (3.6)
    S1(x):=343n+3m3krx2ρ+1cos(6π(3n+3m3k)3xπρ2), (3.7)
    S2(x):=14rx2ρ+1cos(6π(3n+3m+3k)3x3πρ2). (3.8)

    The main term in the right-hand side of (3.2) comes from S0(x). Indeed, integrating both sides of (3.6) with respect to x over the interval (H,2H), we have

    2HHS0(x)dx=34cos(πρ2)3n+3m=3kr2HHx2ρ+1dx. (3.9)

    For natural numbers n, m and k, the condition 3n+3m=3k holds if and only if n, m and k all have the same cube-free part h, such that n=α3h, m=β3h, k=γ3h and α+β=γ. Hence

    3n+3m=3kr=hNh is cubefreehρ2α+β3Nh(αβ(α+β))ρ2cα3hcβ3hc(α+β)3h.

    Since cuvcucv and cwwϵ, we have

    α+β>3Nh(αβ(α+β))ρ2cα3hcβ3hc(α+β)3hc3hα>123Nhβα(αβ(α+β))ρ2cα3cβ3c(α+β)3c3hα>123Nhα2ρ4+ϵβαβρ2c3βc3h(Nh)2ρ+33+ϵ.

    Thus

    3n+3m=3kr=hNh is cubefreehρ2α,β=1(αβ(α+β))ρ2cα3hcβ3hc(α+β)3h+O(hNhρ2c3h(Nh)2ρ+33+ϵ)=α,β=1(αβ(α+β))ρ2h=1h is cubefreehρ2cα3hcβ3hc(α+β)3h+O(α,β=1(αβ(α+β))ρ2cα3cβ3c(α+β)3h>Nhρ2c3h)+O(N2ρ+33+ϵhNhρ+33)=A+O(Nρ1+ϵ).

    Substituting this into (3.9), we have

    2HHS0(x)dx=34cos(πρ2)A2HHx2ρ+1dx+O(H2ρ+2Nρ1+ϵ). (3.10)

    This yields the main term in (3.2) and hence that in the formula in our theorem.

    We now proceed to show that the contributions from S1(x) and S2(x) are bounded by H6ρ2+21ρ+168+3ρ+ϵ. Applying the Lemma 2.3 we find that for any real numbers p(0) and q,

    2HHx2ρ+1cos(p3x+q)dx=2HH3p1x2ρ+53(p3x23cos(p3x+q))dxH2ρ+53|p|1|vu(p3x23cos(p3x+q))dx|H2ρ+53|p|1. (3.11)

    Let us examine S2(x) first. Integrating both sides of (3.8) and then using (3.11) and (3.4), we have

    2HHS2(x)dxrH2ρ+53(3n+3m+3k)1H2ρ+53nmkNrk13H2ρ+53+ϵnmkNnρ+23mρ+23kρ+33H2ρ+53+ϵN43ρ. (3.12)

    Next we consider S1(x). For convenience, let

    Δ=3n+3m3k.

    Using the trivial bound

    2HHx2ρ+1cos(6πΔ3xπρ2)dxH2ρ+2,

    where |Δ|H824+9ρ and applying (3.11) when |Δ|>H824+9ρ, we deduce from (3.7) that

    2HHS1(x)dxH2ρ+20<|Δ|H824+9ρnmr+H2ρ+53|Δ|>H824+9ρnmr|Δ|1:=H2ρ+2W1+H2ρ+53W2. (3.13)

    Now we consider first the sum W1. For given nmN there is most one natural number k for which |Δ|H824+9ρ, since Δ=3n+3m3k and |Δ3N2|H824+9ρ3N2=o(1). Such k, if it exists, must be greater than m. By Lemma 2.2, 0<|Δ|H824+9ρ implies m83H824+9ρ, so mH18+3ρ. Hence, by (3.4),

    W1HϵnmNmH18+3ρ(nm)ρ+23mρ+23HϵmH18+3ρmρ1Hρ8+3ρ+ϵ.

    Similarly, in the sum W2, we have k=(3n+3m)3+O(|Δ|3m2)>m and there are 1+|Δ|3m2 such k. Hence

    W2HϵnmN|Δ|>H824+9ρnρ+23mρ+23|Δ|1(1+|Δ|3m2)mρ+23HϵnmNnρ+23m2ρ+43H824+9ρ+HϵnmNnρ+23m2ρ+23H824+9ρ+ϵ+HϵN23ρH824+9ρ+ϵ.

    Collecting these estimates we find that

    2HHS1(x)dxH6ρ2+21ρ+168+3ρ+ϵ.

    Combining this with (3.10) and (3.12), we deduce from (3.2) the formula (3.5). Then according to (3.3), our theorem is proved.

    Zhang is supported by the National Natural Science Foundation of China [Grant No. 11771256]. Yan is supported by the National Natural Science Foundation of China [Grant No. 11801327] and Natural Science Foundation of Shandong Province [Grant No. ZR201709280100].

    The authors declare no conflict of interest.



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