Power moments of automorphic L-functions related to Maass forms for SL3(ℤ)

Jing Huang; Huafeng Liu; Deyu Zhang

doi:10.1515/math-2021-0076

Article Open Access

Power moments of automorphic L-functions related to Maass forms for SL₃(ℤ)

Jing Huang , Huafeng Liu and Deyu Zhang

Published/Copyright: August 31, 2021

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$Open Mathematics$

From the journal Open Mathematics Volume 19 Issue 1

Abstract

Let f be a self-dual Hecke-Maass eigenform for the group S L 3 ( Z ) . For 1 2 < σ < 1 fixed we define m ( σ ) ( ≥ 2 ) as the supremum of all numbers m such that

∫ 1 T ∣ L ( s , f ) ∣ m d t ≪ f , ε T 1 + ε ,

where L ( s , f ) is the Godement-Jacquet L-function related to f . In this paper, we first show the lower bound of m ( σ ) for 2 3 < σ < 1 . Then we establish asymptotic formulas for the second, fourth and sixth powers of L ( s , f ) as applications.

Keywords: power moments; L-function; automorphic form

MSC 2010: 11F03; 11F66

1 Introduction

Let f be a self-dual Hecke-Maass eigenform for the group S L 3 ( Z ) of type ν = ( α , β ) . Then the Langlands’ parameters for f are

μ f ( 1 ) = α + 2 β − 1 , μ f ( 2 ) = α − β , μ f ( 3 ) = 1 − 2 α − β .

It is known that f has the following Fourier-Whittaker expansion:

f ( z ) = ∑ γ ∈ U 2 ( Z ) ⧹ S L 2 ( Z ) ∑ m ≥ 1 ∑ n ≠ 0 A f ( m , n ) m ∣ n ∣ W J M γ 0 0 1 z , ν , ψ 1 , 1 ,

where U 2 = 1 ∗ 0 1 , W J ( z , ν , ψ 1 , 1 ) is the Jacquet-Whittaker function, ψ 1 , 1 is a character of U 3 ( R ) , M = diag ( m ∣ n ∣ , m , 1 ) and A f ( m , n ) are the Fourier coefficients of f . The function W J ( z , ν , ψ 1 , 1 ) represents an exponential decay in y 1 and y 2 for

z = 1 x 12 x 13 1 x 23 1 y 1 y 2 y 1 1 .

From Kim and Sarnak [1] and Sarnak [2] we know that

A f ( m , n ) ≪ ∣ m n ∣ 5 14 + ε .

From [3], the Rankin-Selberg theory shows that

∑ m n 2 ≤ N ∣ A f ( m , n ) ∣ 2 ≪ f N .

Due to A f ( m , n ) = A f ˜ ( n , m ) , then

(1.1) ∑ m 2 n ≤ N ∣ A f ( m , n ) ∣ 2 ≪ f N

also holds, where f ˜ is the contragredient form of f . According to these estimates, we have

(1.2) ∑ m ≤ N ∣ A f ( m , 1 ) ∣ 2 m ≪ log N , ∑ n ≤ N ∣ A f ( 1 , n ) ∣ 2 n ≪ log N .

As in [4] and [5], the Godement-Jacquet L-function associated with f is defined as

L ( s , f ) = ∑ n = 1 ∞ A f ( 1 , n ) n s , for ℜ s > 1 .

This L-function has a standard functional equation and analytic continuation to an entire function on complex plane C . Due to the fact that f is a Hecke eigenform, the Fourier coefficients are multiplicative and the L-function has an Euler product (see [5, pp. 173–174]), for ℜ s > 1 ,

L ( s , f ) = ∏ p ( 1 − A f ( 1 , p ) p − s + A f ( p , 1 ) p − 2 s − p − 3 s ) − 1 .

Then the L-function associated with the dual Maass form f ˜ takes the form

L ( s , f ˜ ) = ∑ n = 1 ∞ A f ( n , 1 ) n s = ∏ p ( 1 − A f ( p , 1 ) p − s + A f ( 1 , p ) p − 2 s − p − 3 s ) − 1 .

We write s = σ + i t and suppose that 1 2 < σ < 1 is fixed. Let m ( σ ) ( ≥ 2 ) be the supremum of all numbers m ( ≥ 2 ) such that

(1.3) ∫ 1 T ∣ L ( s , f ) ∣ m d t ≪ f , ε T 1 + ε ,

where the ≪ -constant may depend on L ( s , f ) and ε . Naturally, we want to seek lower bounds for m ( σ ) , which occurs frequently in applications. In the cases of full modular group S L 2 ( Z ) and the congruence group, many scholars have obtained lot of results (e.g., see [6,7,8, 9,10,11, 12,13,14, 15,16,17, 18,19,20, 21,22,23, 24,25], etc.).

In this paper, we focus our attention on the Hecke-Maass eigenforms for the group S L 3 ( Z ) . In this situation, for one thing, we do not know whether the Ramanujan conjecture is true; for another, the square and fourth mean value estimates of L ( s , f ) are weaker than ones over S L 2 ( Z ) . Our results are as follows.

Theorem 1

Let m ( σ ) for each 2 3 < σ < 1 be defined by (1.3). Then we have

(1.4) m ( σ ) ≥ 4 ( 3 − 2 σ ) 5 ( 4 − 3 σ ) ( 1 − σ ) .

From Theorem 1 we can get the following corollary immediately.

Corollary

We have

m 2 3 ≥ 2 , m 97 − 769 90 ≥ 3 , … , m 103 − 349 90 ≥ 12 , … .

Remark

Due to the fact that L ( s , f ) is an L-function of degree 3, then Perelli’s mean value theorem [26] shows that, for 1 2 ≤ σ ≤ 1 and T ≥ 1 uniformly,

∫ 1 T ∣ L ( σ + i t , f ) ∣ 2 d t ≪ T max ( 3 ( 1 − σ ) , 1 ) + ε ,

which implies

∫ 1 T ∣ L ( σ + i t , f ) ∣ 2 d t ≪ T 1 + ε 2 3 ≤ σ ≤ 1 .

Thus, we restrict the range of σ in Theorem 1 into 2 3 < σ < 1 .

As applications of Theorem 1, we can establish the asymptotic formulas for the second, fourth and sixth powers of L ( s , f ) .

Theorem 2

For any ε > 0 and σ fixed, we have

(1.5) ∫ 1 T ∣ L ( σ + i t , f ) ∣ 2 d t = T ∑ n = 1 ∞ ∣ A f ( 1 , n ) ∣ 2 n − 2 σ + O T 4 − 3 σ 2 + ε ,

(1.6) ∫ 1 T ∣ L ( σ + i t , f ) ∣ 4 d t = T ∑ n = 1 ∞ ∣ A f ∗ A f ( 1 , n ) ∣ 2 n − 2 σ + O T 27 + 69 − 30 σ 2 69 − 6 + ε ,

(1.7) ∫ 1 T ∣ L ( σ + i t , f ) ∣ 6 d t = T ∑ n = 1 ∞ ∣ A f ∗ A f ∗ A f ( 1 , n ) ∣ 2 n − 2 σ + O T 79 + 481 − 90 σ 2 481 − 22 + ε ,

where A f ∗ A f ( 1 , n ) = ∑ n = m l A f ( 1 , m ) A f ( 1 , l ) is the Dirichlet convolution of A f ( 1 , n ) with itself. The asymptotic formulas (1.5), (1.6) and (1.7) follow for 2 3 < σ < 1 , 33 − 69 30 < σ < 1 and 101 − 481 90 < σ < 1 , respectively.

Notation. Throughout this paper, the letter ε stands for a sufficiently small positive number, and the value of ε may change from statement to statement.

2 Some lemmas

In order to prove Theorems 1 and 2, we first introduce some lemmas.

Lemma 2.1

Let T ≤ t ≤ 2 T and k ≥ 1 be a fixed integer. Then for 1 2 < σ < 1 , we have

∣ L ( σ + i t , f ) ∣ k ≪ 1 + log T ∫ − log 2 T log 2 T L ( σ − 1 log T + i t + i v , f ) k e − ∣ v ∣ d v .

Proof

The proof of this lemma is similar to [27, Lemma 7.1], and we just need to use the following functional equation:

G ν ( s ) L ( s , f ) = G ˜ ν ( 1 − s ) L ( 1 − s , f ˜ ) ,

where

G ν ( s ) = π − 3 s 2 Γ s + 1 − 2 α − β 2 Γ s + α − β 2 Γ s − 1 + α + 2 β 2 , G ˜ ν ( s ) = π − 3 s 2 Γ s + 1 − α − 2 β 2 Γ s − α + β 2 Γ s − 1 + 2 α + β 2 ,

in place of the functional equation of ζ ( s ) .□

Lemma 2.2

For m = m ( σ ) ,

(2.1) ∫ 1 T ∣ L ( σ + i t , f ) ∣ m ( σ ) d t ≪ T 1 + ε

is equivalent to

(2.2) ∑ r ≤ R ∣ L ( σ + i t r , f ) ∣ m ( σ ) ≪ T 1 + ε ,

where

(2.3) t r ∈ [ T , 2 T ] for r = 1 , … , R ; ∣ t r − t s ∣ ≥ log 4 T for 1 ≤ r ≠ s ≤ R .

Proof

Let

L ( σ + i t m , f ) = max t ∈ I m ∣ L ( σ + i t , f ) ∣ , I m = [ 2 T − m log 4 T , 2 T − ( m − 1 ) log 4 T ] ,

where m = 1 , 2 , … , [ T log − 4 T ] . Denote by { t r } either of the sets { t 2 m } or { t 2 m − 1 } . Then the t r ’s satisfy (2.3) and

∫ T 2 T ∣ L ( σ + i t , f ) ∣ m ( σ ) d t ≪ ∑ m = 1 [ t log − 4 T ] ∫ 2 T − m log 4 T 2 T − ( m − 1 ) log 4 T ∣ L ( σ + i t m , f ) ∣ m ( σ ) d t ≪ ∑ m = 1 [ t log − 4 T ] ∣ L ( σ + i t m , f ) ∣ m ( σ ) log 4 T ≪ T 1 + ε .

And then replacing T by T 2 , T 2 2 , … and adding we can get (2.1). On the other hand, by Lemma 2.1, we have

∑ r ≤ R L ( σ + i t r , f ) ∣ m ( σ ) d t ≪ R + log T ∑ r ≤ R ∫ − log 2 T log 2 T ∣ L ( σ − 1 log T + i t r + i v , f ) ∣ m ( σ ) d v ≪ R + log T ∑ r ≤ R ∫ t r − log 2 T t r + log 2 T ∣ L ( σ − 1 log T + i t , f ) ∣ m ( σ ) d t ≪ T log − 4 T + log T ∫ 1 2 T + log 2 T ∣ L ( σ − 1 log T + i t , f ) ∣ m ( σ ) d t ≪ T 1 + ε ,

which implies (2.1).□

Lemma 2.3

We suppose that 1 2 < σ < 1 is fixed and

(2.4) R ≪ T 1 + ε V − m ( σ ) ,

where for t r defined by (2.3) we have

(2.5) ∣ L ( σ + i t r , f ) ∣ ≥ V ≥ T ε ( r = 1 , 2 , … , R ) ,

which is equivalent to

(2.6) ∑ r ≤ R ∣ L ( σ + i t r , f ) ∣ m ( σ ) ≪ T 1 + ε .

Proof

We suppose that (2.6) is true and let { t V , 1 , … , t V , R 1 } be the subset of { t r } such that

∣ L ( σ + i t V , j , f ) ∣ ≥ V ( j = 1 , … , R 1 ) .

Then from (2.6) we have

R 1 V m ( σ ) ≤ ∑ r ≤ R ∣ L ( σ + i t r , f ) ∣ m ( σ ) ≪ T 1 + ε ,

thus for R 1 = R , (2.4) holds.

Inversely, we let (2.4) hold and denote by t V , 1 , … , t V , R ( V ) those of the points t 1 , … , t R for which

V ≤ ∣ L ( σ + i t V , j , f ) ∣ ≤ 2 V ( j = 1 , … , R ( V ) ) .

For each V , we have O ( log T ) choices. And from the following Lemma 2.6, we take V = T 5 ( 1 − σ ) 4 , V = 2 − 1 T 5 ( 1 − σ ) 4 , V = 2 − 2 T 5 ( 1 − σ ) 4 , … . Then we can obtain

∑ r ≤ R L ( σ + i t r , f ) ∣ m ( σ ) d t ≪ R T ε + ∑ V ∑ j ≤ R ( V ) ( 2 V ) m ( σ ) ≪ R T ε + ∑ V T 1 + ε ≪ T 1 + ε .

□

Lemma 2.4

Let t 1 < ⋯ < t R be real numbers such that t r ∈ [ T , 2 T ] for r = 1 , … , R ; ∣ t r − t s ∣ ≥ log 4 T for 1 ≤ r ≠ s ≤ R . If

(2.7) T ε ≪ V ≤ ∑ M < n ≤ 2 M a ( n ) n − σ − i t r ,

where ∑ n ≤ M ∣ a ( n ) ∣ 2 ≪ M 1 + ε for 1 ≪ M ≪ T C ( C > 0 ) , then we have

(2.8) R ≪ T ε ( M 2 − 2 σ V − 2 + T V − f ( σ ) ) ,

where

(2.9) f ( σ ) = 2 3 − 4 σ , if 1 2 < σ ≤ 2 3 , 10 7 − 8 σ , if 2 3 ≤ σ ≤ 11 14 , 34 15 − 16 σ , if 11 14 ≤ σ ≤ 13 15 , 98 31 − 32 σ , if 13 15 ≤ σ ≤ 57 62 , 5 1 − σ , if 57 62 ≤ σ ≤ 1 − ε .

Proof

We can get this lemma by following a similar argument to [6, Lemma 8.2] replacing a ( n ) ≪ M ε by ∑ n ≤ M ∣ a ( n ) ∣ 2 ≪ M 1 + ε .□

Lemma 2.5

[27, Theorem 5.2] Let a 1 , … , a N be arbitrary complex numbers. Then

∫ 0 T ∑ n ≤ N a n n i t 2 d t = T ∑ n ≤ N ∣ a n ∣ 2 + O ∑ n ≤ N n ∣ a n ∣ 2 ,

and the above formula remains also valid if N = ∞ , provided that the series on the right hand side of the aforementioned formula converge.

Lemma 2.6

[28, Corollary 1.2] Let 1 2 ≤ σ ≤ 1 be fixed, we have

∣ L ( σ + i t , f ) ∣ ≪ ∣ t ∣ 5 4 ( 1 − σ ) + ε .

Lemma 2.7

For any ε > 0 , we have

∫ 0 T L 2 3 + i t , f 2 d t ≪ T 1 + ε , ∫ 0 T L 2 3 + i t , f 4 d t ≪ T 17 12 + ε .

Proof

The first result is a general result of Perelli [26], which we can also get from Lemma 2.5 with m = 3 and σ = 2 3 in Liu and Zhang [29]. From Lemma 2.6 and the first result, we can easily get the second result.□

Lemma 2.8

For t r satisfying (2.3), we have

∑ r ≤ R L 2 3 + i t r , f 2 d t ≪ T 1 + ε , ∑ r ≤ R L 2 3 + i t r , f 4 d t ≪ T 17 12 + ε .

Proof

Following a similar argument of Lemma 2.2, with the help of Lemma 2.7 we can obtain this lemma.□

Lemma 2.9

[27, Lemma 8.3] Let F ( s ) be regular in the region D : α ≤ σ ≤ β , t ≥ 1 and let F ( s ) ≪ e C t 2 for s ∈ D . Then for any fixed q > 0 and α < γ < β , we have

∫ 2 T ∣ F ( γ + i t ) ∣ q d t ≪ ∫ 1 2 T ∣ F ( α + i t ) ∣ q d t + 1 β − γ β − α ∫ 1 2 T ∣ F ( β + i t ) ∣ q d t + 1 γ − α β − α .

In the following two lemmas, though the definitions of φ k ( m ) and ψ k ( n ) are different from ones in Lemmas 2.11 and 2.12 of [18], we still can get these two lemmas by following similar arguments, respectively.

Lemma 2.10

Let φ k ( n ) be the arithmetic function generated by L ( s , f ) k , that is

(2.10) φ k ( n ) = A f ∗ ⋯ ∗ A f ( 1 , n ) ︸ k times .

Then we have

∑ n ≤ x φ k ( n ) ≪ x 1 + ε , ∑ n ≤ x φ k 2 ( n ) ≪ x 1 + ε .

Lemma 2.11

Let 0 < δ < 1 2 be a fixed constant and

ψ k ( n ) = φ 2 k ( n ) − ∑ n = m l m ≤ T , l ≤ T φ k ( m ) φ k ( l ) , T < n ≤ T 2 , φ 2 k ( n ) , n > T 2 .

Then we have

∑ n ≥ T ψ k 2 ( n ) n − 2 − 2 δ = O ( 1 ) .

3 Proofs of Theorems 1 and 2

3.1 Proof of Theorem 1

In this section, we restrict the range of σ into 2 3 < σ < 1 and shall give lower bounds for m ( σ ) by establishing formulas of type

R ≪ T 1 + ε V − m ( σ ) .

Recalling Mellin’s formula

(3.1) e − x = ( 2 π i ) − 1 ∫ 2 − i ∞ 2 + i ∞ Γ ( ω ) x − ω d ω ( x > 0 ) .

Taking x = n Y and multiplying (3.1) by A f ( 1 , n 1 ) A f ( 1 , n 2 ) n 1 − s n 2 − s , where n = n 1 n 2 and summing over n , we can obtain

(3.2) ∑ n = 1 ∞ ∑ n = n 1 n 2 A f ( 1 , n 1 ) A f ( 1 , n 2 ) e − n Y n − s = ( 2 π i ) − 1 ∫ 2 − i ∞ 2 + i ∞ Y ω Γ ( ω ) L ( s + ω , f ) 2 d ω .

Shifting the line of integration in (3.2) to ℜ ω = 2 3 − σ , we encounter a simple pole at ω = 0 with residue L ( s , f ) 2 and get, as Y → ∞ ,

(3.3) ∑ n ≤ Y log 2 Y ∑ n = n 1 n 2 A f ( 1 , n 1 ) A f ( 1 , n 2 ) e − n Y n − s + o ( 1 ) = L ( s , f ) 2 + ( 2 π i ) − 1 ∫ ℜ ω = 2 3 − σ Y ω Γ ( ω ) L ( s + ω , f ) 2 d ω .

The integral part of (3.3) for which ℑ ω ≥ log 2 T is o ( 1 ) as T → ∞ by Stirling’s formula. Then let s = σ + i t r and thus for each t r we have

(3.4) L ( σ + i t r , f ) 2 ≪ 1 + ∑ n ≤ Y log 2 Y ∑ n = n 1 n 2 A f ( 1 , n 1 ) A f ( 1 , n 2 ) e − n Y n − σ − i t r + ∫ − log 2 T log 2 T Y 2 3 − σ L 2 3 + i t r + i v , f 2 e − ∣ v ∣ d v .

Combining (2.5) with (3.4), we can obtain

(3.5) V 2 ≪ ∑ n ≤ Y log 2 Y ∑ n = n 1 n 2 A f ( 1 , n 1 ) A f ( 1 , n 2 ) e − n Y n − σ − i t r ≪ log T max M ≤ 1 2 Y log 2 Y ∑ M < n ≤ 2 M ∑ n = n 1 n 2 A f ( 1 , n 1 ) A f ( 1 , n 2 ) e − n Y n − σ − i t r

(3.6) V 2 ≪ Y 2 3 − σ L 2 3 + i t r ′ , f 2 ,

where V ≫ T ε and t r ′ is defined as

L 2 3 + i t r ′ , f = max − log 2 T ≤ v ≤ log 2 T L 2 3 + i t r + i v , f .

For convenience, denote by R 1 ′ and R 2 ′ those points which satisfy (3.5) and (3.6), respectively.

Recalling (1.1), we know that Lemma 2.4 is valid. We first consider R 1 ′ . By Lemma 2.4, we have

(3.7) R 1 ′ ≪ log Y × T ε ( M 2 − 2 σ V − 4 + T V − 2 f ( σ ) ) ≪ T ε ( Y 2 − 2 σ V − 4 + T V − 2 f ( σ ) ) .

While for R 2 ′ , by Lemma 2.8, Hölder’s inequality and (3.6), we can obtain

(3.8) R 2 ′ ≪ Y 2 3 − σ V − 2 ∑ r ≤ R 2 ′ L 2 3 + i t r ′ , f 2 ≪ Y 2 3 − σ V − 2 T 1 + ε

and

(3.9) R 2 ′ ≪ Y 2 3 − σ V − 2 ∑ r ≤ R 2 ′ L 2 3 + i t r ′ , f 2 ≪ Y 2 3 − σ V − 2 R 2 ′ 1 2 T 17 24 + ε .

For (3.8), if we take Y = V 6 4 − 3 σ T 3 4 − 3 σ , then we have

(3.10) R ≪ R 1 ′ + R 2 ′ ≪ T ε Y 2 − 2 σ V − 4 + T V − 2 f ( σ ) + Y 2 3 − σ V − 2 T 1 + ε ≪ T ε V − 4 4 − 3 σ T 6 − 6 σ 4 − 3 σ + T V − 2 f ( σ ) .

For (3.9), if we take Y = T 17 8 , then we have

(3.11) R ≪ R 1 ′ + R 2 ′ ≪ T ε Y 2 − 2 σ V − 4 + T V − 2 f ( σ ) + Y 4 3 − 2 σ V − 4 T 17 12 ≪ T ε V − 4 T 17 4 − 17 4 σ + T V − 2 f ( σ ) .

Therefore, combining (3.10) with (3.11) we have

(3.12) R ≪ T ε T V − 2 f ( σ ) + V − 4 4 − 3 σ T 6 − 6 σ 4 − 3 σ + V − 4 T 17 4 − 17 4 σ .

We assume that the second and the third terms in (3.12) do not exceed T V − x and T V − y , for values x and y which can be determined by Lemma 2.6, then we can obtain

x ≤ 4 ( 3 − 2 σ ) 5 ( 1 − σ ) ( 4 − 3 σ ) , y ≤ 7 − 3 σ 5 ( 1 − σ ) .

Thus, we have

R ≪ T 1 + ε V − z

with

z = min 2 f ( σ ) , 4 ( 3 − 2 σ ) 5 ( 1 − σ ) ( 4 − 3 σ ) , 7 − 3 σ 5 ( 1 − σ ) .

For 2 3 < σ ≤ 1 − ε , we always have

4 ( 3 − 2 σ ) 5 ( 1 − σ ) ( 4 − 3 σ ) < 7 − 3 σ 5 ( 1 − σ ) .

Recalling the value of f ( σ ) in Lemma 2.4, we can take

z = 4 ( 3 − 2 σ ) 5 ( 1 − σ ) ( 4 − 3 σ ) , 2 3 < σ ≤ 1 − ε .

Thus, we complete the proof of Theorem 1.

3.2 Proof of Theorem 2

In this section, we give the proof of Theorem 2 by following a similar argument to [6, Theorem 2]. Let σ k ∗ denote the infimum of all numbers σ for which

∫ 1 T ∣ L ( σ + i t , f ) ∣ 2 k d t ≪ T 1 + ε

holds for any ε > 0 , where k ≥ 1 is a fixed integer, 1 2 ≤ σ k ∗ < 1 .

Writing s = σ + i t , we have

(3.13) ∫ 1 T ∣ L ( σ + i t , f ) ∣ 2 k d t = ∫ 1 T ∑ n ≤ T φ k ( n ) n − σ − i t 2 d t + O ∫ 1 T L ( σ + i t , f ) − ∑ n ≤ T φ k ( n ) n − σ − i t 2 d t ,

where φ k ( n ) is given by Lemma 2.10.

Combining Abel’s summation formula with Lemmas 2.5 and 2.10, we can obtain

(3.14) ∫ 1 T ∑ n ≤ T φ k ( n ) n − σ − i t 2 d t = T ∑ n ≤ T φ k 2 ( n ) n − 2 σ + O ∑ n ≤ T φ k 2 ( n ) n 1 − 2 σ = T ∑ n = 1 ∞ φ k 2 ( n ) n − 2 σ + O ( T 2 − 2 σ + ε ) .

Let

F ( σ + i t , f ) = L 2 k ( σ + i t , f ) − ∑ n ≤ T φ k ( n ) n − σ − i t 2 .

And applying Lemma 2.9 with q = 1 , α = σ k ∗ + δ , β = 1 + δ , γ = σ , where 0 < δ < 1 2 is a fixed constant which may be chosen arbitrarily small, for fixed k we have

β − σ β − α = 1 + δ − σ 1 − σ k ∗ ≤ 1 − σ 1 − σ k ∗ + δ 1 2

and

σ − α β − α = σ − σ k ∗ − δ 1 − σ k ∗ ≤ σ − σ k ∗ 1 − σ k ∗ .

Recalling the definition of σ k ∗ , by Lemma 2.5 we have

∫ 1 2 T ∣ F ( α + i t , f ) ∣ d t ≤ ∫ 1 2 T ∣ L ( σ k ∗ + δ + i t , f ) ∣ 2 k d t + ∫ 1 2 T ∣ ∑ n ≤ T φ k ( n ) n − σ k ∗ − δ − i t ∣ 2 d t ≪ T 1 + δ + T 2 − 2 σ k ∗ + ε ≪ T 1 + δ .

Moreover,

F ( β + i t , f ) = ∑ n = 1 ∞ φ 2 k ( n ) n − 1 − δ − i t − ∑ n ≤ T φ k ( n ) n − 1 − δ − i t 2 = ∑ n > T ψ k ( n ) n − 1 − δ − i t ,

where ψ k ( n ) is given by Lemma 2.11.

By Lemma 2.5, Lemma 2.10 and Hölder’s inequality, we can obtain

∫ 1 2 T ∣ F ( β + i t , f ) ∣ d t ≪ T 1 2 ∫ 1 2 T ∣ ∑ n ≥ T ψ k ( n ) n − 1 − δ − i t ∣ 2 d t 1 2 ≪ T 1 2 .

Thus, Lemma 2.9 shows

∫ 1 2 T ∣ F ( σ + i t , f ) ∣ d t ≪ T ( 1 + δ ) 1 − σ 1 − σ k ∗ + δ 1 2 + σ − σ k ∗ 2 − 2 σ k ∗ .

Note that

( 1 + δ ) 1 − σ 1 − σ k ∗ + δ 1 2 + σ − σ k ∗ 2 − 2 σ k ∗ ≤ 2 − σ − σ k ∗ 2 − 2 σ k ∗ + ε

holds for any ε > 0 if δ = δ ( ε ) is sufficiently small. Noting that for the exponent of the O -term in (3.14), we have

2 − 2 σ < 2 − σ − σ k ∗ 2 − 2 σ k ∗ < 1 .

Thus,

∫ 1 T ∣ L ( σ + i t , f ) ∣ 2 k d t = T ∑ n = 1 ∞ φ k 2 ( n ) n − 2 σ + R ( k , σ ; T ) ,

and for fixed σ satisfying σ k ∗ < σ < 1 , we have

R ( k , σ ; T ) ≪ T 2 − σ − σ k ∗ 2 − 2 σ k ∗ + ε .

From Theorem 1 we have

∫ 1 T L 2 3 + i t , f 2 d t ≪ T 1 + ε , ∫ 1 T L 33 − 69 30 + i t , f 4 d t ≪ T 1 + ε , ∫ 1 T L 101 − 481 90 + i t , f 6 d t ≪ T 1 + ε .

Recalling the definition of σ k ∗ , we can take σ 1 ∗ = 2 3 , σ 2 ∗ = 33 − 69 30 and σ 3 ∗ = 101 − 481 90 , from which we can obtain Theorem 2 immediately.

Acknowledgment

The authors are greatly indebted to the reviewers for very beneficial suggestions and comments which led to essential improvement of the original version of this paper. This work was supported by the National Natural Science Foundation of China (Grant Nos. 11771256 and 11801328).

Conflict of interest: Authors state no conflict of interest.

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Received: 2020-09-30

Revised: 2021-06-16

Accepted: 2021-07-12

Published Online: 2021-08-31

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2021-0076

Keywords for this article

power moments; L-function; automorphic form

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