Divisor Problems Related to Hecke Eigenvalues in Three Dimensions

Received 26 March 2021; Revised 4 April 2021; Accepted 7 April 2021; Published 20 April 2021 Academic Editor: Jie Wu Copyright © 2021 Jing Huang and Huafeng Liu. +is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in anymedium, provided the original work is properly cited. In this paper, we consider divisor problems related to Hecke eigenvalues in three dimensions. We establish upper bounds and asymptotic formulas for these problems on average.


Introduction
Let Γ � SL 2 (Z) be the full modular group. Let H * k denote the set of primitive holomorphic forms with even integral weight k ≥ 2 for Γ.
en H * k consists of common eigenfunctions f of all Hecke operators T n . e Hecke eigenfunction f has the following Fourier series expansion: where λ f (n) denotes the n-th normalized eigenvalue. It is known that λ f (n), as a function of n, is real-valued and multiplicative. Moreover, for all integers n ≥ 1, Deligne [1] showed that where d(n) is the divisor function. Also, for every prime p, Now we introduce some specific automorphic L-functions. Define the Hecke L-function attached to f as for Rs > 1. Moreover, the Rankin-Selberg L-function attached to f can be defined as en, L(s, f × f) can be rewritten as λ f×f (n) n s . (6) As usual, ζ(s) denotes the Riemann zeta-function. e symmetric square L-function attached to f can be defined as, for Rs > 1, which can also be expressed in the Dirichlet series e symmetric square L-function L(s, sym 2 f) has a functional equation and could be analytic continued to an entire function over the whole complex plane. We refer to works of Hecke in [2], Gelbert and Jacquet [3], Kim [4], and Kim and Shahidi [5,6] for these properties. erefore, the symmetric square L-function L(s, sym 2 f) could be seen as a general L-function in the sense of Perelli [7].
In number theory, considering the properties and average behaviors of the Fourier coefficients is meaningful and interesting. Some classical problems concern mean values of these Fourier coefficients and related problems with the corresponding automorphic L-functions (see [1,, etc.). Here, we just give a brief history for general divisor problems related to these Fourier coefficients.
In particular, we have λ 1,f (n) � λ f (n) and λ 1,f×f (n) � λ f×f (n). In 1927, Hecke [30] showed that Subsequently, this upper bound was improved by many scholars (for example, see [12,21,24]). In this direction, the best result so far was obtained by Wu [24] who showed that Rankin [20] and Selberg [22] established where C f is a positive constant depending on f. Later, Kanemitsu, Sankaranarayanan, and Tanigawa [31] studied general divisor problems and proved where ω is an integer and ω ≥ 2; M ω (x) comes from a residue and takes the form xP ω− 1 (log x); P ω− 1 (t) denotes a polynomial of t with degree ω − 1. Fomenko [32] also proved the same result for the sum n≤x λ ω,f (n). After that, Kanemitsu, Sankaranarayanan, and Tanigawa's result was improved by Lü [33], and some general results were obtained (see [34][35][36][37], etc.). In this paper, we consider divisor problems related to Hecke eigenvalues in three dimensions motivated by the above results and Ivić's work on three-dimensional divisor problems (see, e.g., [38]). We first introduce some notation. For any fixed integer 1 < a < b < c, we define We are interested in studying the average behaviors of sums which can be seen as divisor problems related to Hecke eigenvalues in three dimensions. We establish the following results.
Theorem 1. Let a, b, and c be fixed integers satisfying 1 < a < b < c. en, for any ϵ > 0, one has Theorem 2. Let a, b, and c be fixed integers satisfying 1 < a < b < c. en, for any ϵ > 0, one has where

Some Lemmas
To prove our theorems, we need to quote some lemmas, which include the individual and averaged subconvexity bounds for the Riemann zeta-function and the symmetric square L-function. And from the following Lemma 1, we know that the Rankin-Selberg L-function L(s, f × f) could be decomposed into the product of the Riemann zetafunction and the corresponding symmetric square L-function.
Proof. By comparing the Euler products of two sides of (20) and applying (3), we can get this lemma. is lemma can also be found in [33].
□ Lemma 2. For any ϵ > 0, one has uniformly for T ≥ 1, and the subconvexity bound in the critical strip where |t| ≥ 1.
Proof. ese results are proved by Good [11].
□ Lemma 3. For any ε > 0, one has uniformly for T ≥ 1, and the subconvexity bound in the critical strip where |t| ≥ 1.
uniformly for T ≥ 1, and the subconvexity bound in the critical strip where |t| ≥ 1.

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Proof.
e first result follows from Perelli's mean value theorem with L(s, sym 2 f) (see [7]), and the second one is due to Nunes [19].

Proof of Theorem 1
In this section, we shall complete the proof of eorem 1. Note that en, by (27) and Perron's formula (see Proposition 5.54 in [41]), we can obtain where s � σ + it, η � (1/a) + ε and 1 ≤ T ≤ x is a parameter to be chosen later.
We shift the line of the integral of (28) to the line Rs � (1/2a). en, Cauchy's residue theorem shows that e following work is to estimate G 1 , G 2 , and G 3 . e estimates for the integrals over the horizontal segments are similar, so we handle G 2 and G 3 first. To get this goal, we consider three cases We first consider the case c ≤ 2a. To estimate G 2 and G 3 , we divide the integral interval into the following eight intervals I 1 , . . . , I 4 , some of which may be empty, and use Lemma 2.
Interval I 1 : In this interval, we have Interval I 2 : In this interval, we have Interval I 3 : is interval is an empty set noting that (1/b) > (1/c). Interval I 4 : In this interval, we have 4 Journal of Mathematics us, by (31)-(36), we have Next, we handle G 1 . We have en, by Lemma 2 and applying Cauchy's inequality, we can obtain According to (29), (37), and (39), we have By taking T � x (3/2(7a− b− c)) in (40), we can obtain which proves the first result of eorem 1. For the case b < 2a < c, to estimate G 2 and G 3 , we still divide the integral interval into four corresponding short intervals I 1 ″ , . . . , I 4 ″ , which are different from ones for the case c ≤ 2a. In fact, the corresponding short intervals I 1 ″ , I 3 ″ become empty sets in the current situation. However, we still can estimate G 2 + G 3 similar to the corresponding parts of the case c ≤ 2a and get e estimate of G 1 becomes the following at the current case by noting (c/2a) > 1.
For the case 2a ≤ b, to estimate G 2 and G 3 , we also divide the integral interval into four corresponding short intervals I 1 ′ , . . . , I 4 ′ , which are different from ones for the case c ≤ 2a. In fact, the corresponding short intervals I 1 ′ , I 2 ′ , I 3 ′ become empty sets in the current situation. However, we still can estimate G 2 + G 3 similarly to the corresponding parts of the case c ≤ 2a and get e estimate of G 1 becomes the following by noting (c/2a) > (b/2a) > 1.
Recalling (29), we have By taking T � x 1/2a in (48), we can get which proves the third result of eorem 1.

Proof of Theorem 2
In this section, we shall prove eorem 2, the process of which is more complicated than eorem 1. Note that en, by (50) and Perron's formula (see Proposition 5.54 in [41]), we have Journal of Mathematics where s � σ + it, η � (1/a) + ε and 1 ≤ T ≤ x is a parameter which will be decided later.
In view of (20), the points s � (1/a), s � (1/b) and s � (1/c) are the only three possible simple poles of the integrand in the rectangle I T � s � σ + it: e corresponding possible residues at s � (1/a), s � (1/b), and s � (1/c) are equal to respectively. We move the integral line of the integral in (28) to the parallel segment with Rs � (1/2a). We first consider the case c ≤ 2a. In this situation, the points s � (1/a), s � (1/b), and s � (1/c) are all poles of the integrand in the rectangle I T . erefore, by Cauchy's residue theorem, we can obtain Now, the remaining work is to handle these three terms J 1 , J 2 , and J 3 . In addition, the estimates of these integrals on the horizontal parts are analogous, and so we deal with J 2 and J 3 firstly. To estimate J 2 and J 3 , similarly to the method estimating G 2 and G 3 , we still divide the integral interval into the following four short intervals I * 1 , . . . , I * 4 and apply Lemmas 3 and 4.

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In this interval, we have (57) Interval I * 3 : is interval is an empty set noting that (1/b) > (1/c).
Interval I * 4 : In this interval, we have From (55)-(60), we can obtain Now, we turn to estimate J 1 , and we have 8 Journal of Mathematics en, using Lemmas 3 and 4 and Hölder's inequality, we obtain By putting (53), (61), and (63) together, we have To estimate J 2 ″ + J 3 ″ , we still divide the integral interval into four short intervals I * * 1 , . . . , I * * 4 , which are different from ones for the case c ≤ 2a. In fact, the corresponding short intervals I * * 1 , 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 c ≤ 2a and get us, recalling (67), we have By taking T � x (84/486a− 131b) in (69), we have To estimate J 2 ′ + J 3 ′ , we still divide the integral interval into four short intervals I * * * (72) e estimate of J 1 ′ becomes the following by noting (c/2a) > (b/2a) > 1.
which proves the third result of eorem 2.
Data Availability e data supporting the findings of this study are included within the article.

Conflicts of Interest
e authors declare that they have no conflicts of interest.