Two-Dimensional Divisor Problems Related to Symmetric L-Functions

In this paper, we study two-dimensional divisor problems of the Fourier coefficients of some automorphic product L-functions attached to the primitive holomorphic cusp form f (z) with weight k for the full modular group SL2(Z). Additionally, we establish the upper bound and the asymptotic formula for these divisor problems on average, respectively.


Introduction
As usual, let H k denote the set of primitive holomorphic cusp forms with even integral weight k ≥ 2 for the full modular group SL 2 (Z); then H k is made up of the common eigenfunctions of all Hecke operators T n . Then the Fourier series expansion of Hecke eigenfunction f at the cusp ∞ has the following form.
where the coefficient λ f (n) denotes the n-th normalized eigenvalue, which is the coefficient divided by n k−1 2 of the Hecke operator T n . Note that λ f (n) is real valued and also multiplicative. Let n be an integer greater than one, Deligne [1] proved that where τ(n) denotes the number of n's positive divisors. For prime p, we have Studying the properties and average behaviors of various sums concerning λ f (n) and λ f × f (n) is a meaningful and interesting problem. In number theory, classical problems are investigate mean value estimates of these Fourier coefficients and related problems with the corresponding automorphic L-functions (for examples, see , etc.). In particular, we give a brief introduction for the general divisor problem.
when ω = 1, we actually have λ 1, f (n) = λ f (n) and λ 1, f × f (n) = λ f × f (n). Hecke [24] proved λ a,b f (n) (4) and For these sums (4) and (5), we establish the upper bound and the asymptotic formula by considering the sizes of a and b, respectively. We refer to Section 3 for our detailed results.
In the following Section 2, we first introduce some specific automorphic L-functions and quote some lemmas. We state our results in Section 3, and show their proofs in Sections 4 and 5. In Sections 6 and 7, we state an application of our results and give a conclusion, respectively.

Some Lemmas
In this section, to prove Theorems 1 and 2 we first introduce some specific automorphic L-functions, which is important for the proof of our results and also help us understand the Fourier coefficients in another way. For Re s > 1, we define the Hecke L-function L(s, f ) attached to f as Moreover, the Rankin-Selberg L-function attached to f could be defined as Then L(s, f × f ) can be rewritten in the following form: The j-th symmetric power L-function attached to f could be defined as follows.
Additionally, the j-th symmetric power L-function attached to f could be expressed in the following Dirichlet series: The j-th symmetric L-function L(s, sym j f ), j = 1, 2, 3, 4 could be analytic continued to an entire function over the whole complex plane C and has a confirmed functional equation. We refer to papers of Hecke [31], Gelbert and Jacquet [32], Kim [33] and Kim and Shahidi [34,35] for these properties of L(s, sym j f ), j = 1, 2, 3, 4. Therefore, we can note that L(s, sym j f ), j = 1, 2, 3, 4 could be recognized as general L-functions in the sense of Perelli [36].
With the help of these automorphic L-functions, we then quote the following lemmas, which include the individual and averaged subconvexity bounds for Riemann zetafunction ζ(s), symmetric square L-function L(s, sym 2 f ) and corresponding Rankin-Selberg L-function L(s, f × f ). From the following Lemma 1 we know that the Rankin-Selberg L-function L(s, f × f ) could be decomposed into the product of Riemann zeta-function ζ(s) and corresponding symmetric square L-function L(s, sym 2 f ).
Proof. By the comparison of Euler products of two sides of (8) and applying Deligne's result (1), we could easily get this lemma. This lemma can also be found in [27,28].

Lemma 2.
For any > 0, one has the mean value estimate uniformly for T ≥ 1 and the upper bounds where |t| ≥ 1.

Lemma 3.
For any > 0, one has the mean value estimate uniformly for T ≥ 1 and the upper bounds where |t| ≥ 1.
Proof. The mean value result (11) follows from analytic properties of L(s, sym 2 f ) and standard arguments in number theory. The upper bounds (12) are due to Nunes [13].

Lemma 4.
For any > 0, one has the mean value estimate uniformly for T ≥ 1 and the upper bounds where |t| ≥ 1.
Proof. The results in this lemma were established by Good [5].

Main Theorems
In this paper, we consider the two-dimensional divisor problems related to the Fourier coefficients λ f (n), λ f × f (n) and establish the following two theorems. To establish these two theorems, we apply some classical methods and instruments, such as Perron's formula, Cauchy's residue theorem, decomposition of the Rankin-Selberg L-function, upper bounds and mean values of specific functions. Theorem 1. Suppose that a and b are any fixed integers with 1 < a < b. Then for any > 0, one has

Theorem 2.
Suppose that a and b are any fixed integers with 1 < a < b. Then for any > 0, one has

Proof of Theorem 1
In this section, we shall complete the proof of Theorem 1. Let s = σ + it and η = 1 + . We have Then, by applying Perron's formula (see the Proposition 5.54 in [39]), we can obtain where T is a parameter which will be decided later.
We shift the line of the integral of (16) to the line Re s = 1 2a . Then Cauchy's residue theorem shows that where The following work is to estimate these three terms, G 1 , G 2 and G 3 . The estimates of these integrals on the horizontal parts are analogous; thus, we always consider G 2 and G 3 firstly in the following parts. To get this goal, we consider two cases b ≤ 2a and b > 2a.
We first consider the case b ≤ 2a. To estimate G 2 and G 3 , we divide the integral interval into the following four short intervals I 1 , · · · , I 4 and apply Lemma 4.

Proof of Theorem 2
We shall prove Theorem 2, the process of which is more complicated than Theorem 1, in this section. Let also s = σ + it and η = 1 + . Note that Then, by applying Perron's formula ( see the Proposition 5.54 in [11] ), we have where T is a parameter which will be decided later. Then we shift the line of the integral of (26) to the line Re s = 1 2a . In view of (8), we know that at the point s = 1, L(s, sym 2 f ) is holomorphic, which was proved by Gelbart-Jacquet [32]. Thus, the points s = 1 a and s = 1 b are the only two possible simple poles of the integrand of (26) in the range R T := {s = σ + it : 1 2a ≤ σ ≤ 1 + , | t |≤ T} depending on the size difference between b and 2a. Thus, we consider two cases, b ≤ 2a and b > 2a.
We first consider the case b ≤ 2a. In this situation, the points s = 1 a and s = 1 b are all simple poles of the integrand of (26) in the range R T . Then, Cauchy's residue theorem gives where x s s ds, and the main terms L(1, the residues of L(as, f × f )L(bs, f × f ) x s s at the simple poles s = 1 a and s = 1 b , respectively. Now the remaining work is to handle these three terms: J 1 , J 2 and J 3 . Additionally, the estimates of these integrals on the horizontal parts are analogous, and thus we deal with J 2 and J 3 firstly. To estimate J 2 and J 3 , similarly to the method of estimating G 2 and G 3 , we also divide the integral interval into the following four short intervals I 1 , · · · , I 4 and apply Lemmas 2 and 3.
For the case b > 2a, we use a similar argument to the corresponding case of Theorem 1. In this situation, the point s = 1 a is the only simple pole of the integrand of (26) in the range R T by noting 1 b < 1 2a . Then Cauchy's residue theorem shows x s s ds where the main term L(1, a derives from the residue of L(as, f × f )L(bs, f × f ) x s s at the simple pole s = 1 a . To estimate J 2 + J 3 we also divide the integral interval into four short intervals I * 1 , · · · I * 4 , which are different from ones for the case b ≤ 2a. In fact, the corresponding short intervals I * 1 and I * 3 become empty sets in this situation. However, we still can estimate J 2 + J 3 by following a similar argument to the corresponding parts of the case b ≤ 2a and get The estimate of J 1 becomes the following at the current case by noting b 2a > 1. Thus, recalling (36) we have Taking T = x 3(2a−1) 8a in (37), we can obtain Note that when a ≥ 3, we have 1 − 3(2a−1) 8a > 1 a . Therefore, we have, recalling b > 2a,