Next Article in Journal
Spontaneous Symmetry Breaking and Its Pattern of Scales
Previous Article in Journal
Relationship between Inter-Limb Asymmetries and Physical Performance in Rink Hockey Players

Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Article

# Power Moments of the Riesz Mean Error Term of Symmetric Square L-Function in Short Intervals

by
Rui Zhang
,
Xue Han
and
Deyu Zhang
*
School of Mathematics and Statistics, Shandong Normal University, Jinan 250300, China
*
Author to whom correspondence should be addressed.
Symmetry 2020, 12(12), 2036; https://doi.org/10.3390/sym12122036
Submission received: 20 November 2020 / Revised: 5 December 2020 / Accepted: 8 December 2020 / Published: 9 December 2020
(This article belongs to the Section Mathematics)

## Abstract

:
Let $f ( z )$ be a holomorphic Hecke eigenform of weight k with respect to $SL ( 2 , Z )$ and let $L ( s , sym 2 f ) = ∑ n = 1 ∞ c n n − s , ℜ s > 1$ denote the symmetric square L-function of f. In this paper, we consider the Riesz mean of the form $D ρ ( x ; sym 2 f ) = L ( 0 , sym 2 f ) Γ ( ρ + 1 ) x ρ + Δ ρ ( x ; sym 2 f )$ and derive the asymptotic formulas for $∫ T − H T + H Δ ρ k ( x ; sym 2 f ) d x$, when $k ≥ 3$.

## 1. Introduction

Let $S k ( Γ )$ be the space of all holomorphic cusp forms
$f ( z ) = ∑ n = 1 ∞ a f ( n ) e ( n z ) , z ∈ H = { z = x + i y : y > 0 }$
of even weight $k ≥ 12$ for $Γ = SL ( 2 , Z )$; $S k ( Γ ) +$ denote the set of Hecke eigenforms from this space with $a f ( 1 ) = 1$. Set
$λ f ( n ) = a f ( n ) n k − 1 2 .$
For prime p, we have
$λ f ( p ) = α f ( p ) + β f ( p ) , α f ( p ) β f ( p ) = 1 .$
For $f ∈ S k ( Γ ) +$, we consider the symmetric square $L ( s , sym 2 f )$ of the Hecke L-function $L ( s , f )$:
$L ( s , sym 2 f ) : = ∏ p ( 1 − p − s ) − 1 1 − α f 2 ( p ) p − s − 1 1 − β f 2 ( p ) p − s − 1 , ℜ s > 1$
and have
$L ( s , sym 2 f ) = ζ ( 2 s ) ∑ n = 1 ∞ λ f ( n 2 ) n − s = ∑ n = 1 ∞ c n n − s .$
Then it is known that for any $ϵ > 0$,
$| c n | ≤ d 3 ( n ) ≪ n ϵ ,$
where $d 3 ( n )$ is the number of ways to write n as a product of three factors.
It is known that the function $L ( s , sym 2 f )$ is entire and satisfies the following functional equation:
$Λ ( 1 − s , sym 2 f ) = Λ ( s , sym 2 f ) ,$
where
$Λ ( s , sym 2 f ) : = π − 3 s 2 Γ ( s 2 + 1 2 ) Γ ( s 2 + k − 1 2 ) Γ ( s 2 + k 2 ) L ( s , sym 2 f ) .$
By using Landau’s theorem, Deligne’s result and properties of $L ( s , sym 2 f )$, we deduce the bound
$∑ n ≤ x c n ≪ x 1 2 log 2 x .$
Khan [1] studied the twisted first moment of $L ( s , sym 2 f )$ at any point s on the critical line:
$∑ f ∈ H k L ( 1 2 , sym 2 f ) w f − 1 = log k + C + O ( k − 1 / 20 + ϵ ) ,$
$∑ f ∈ H k L ( 1 2 + i t , sym 2 f ) a f ( m 2 ) w f − 1 = ζ ( 1 + i 2 t ) m 1 / 2 + i t + ζ ( 1 − i 2 t ) m 1 / 2 − i t L ∞ ( 1 / 2 − i t ) L ∞ ( 1 / 2 + i t ) + O ( k − 1 / 20 + ϵ ) .$
It followed that for any point on the critical line and large enough even k, there exists an eigenform f of weight k such that $L ( s , sym 2 f )$ is nonvanishing at that point. Hafner [2] considered the Riesz mean of the form
$D ρ ( x ; sym 2 f ) = 1 Γ ( ρ + 1 ) ∑ n ≤ x ( x − n ) ρ c n ,$
where $ρ ≥ 0$ is a fixed number. Also, the Riesz mean can be represented as a sum of “residue function” and “error term”:
$D ρ ( x ; sym 2 f ) = L ( 0 , sym 2 f ) Γ ( ρ + 1 ) x ρ + Δ ρ ( x ; sym 2 f ) .$
For $ρ = 0$, we have
$∑ n ≤ x c n = L ( 0 , sym 2 f ) + Δ 0 ( x ; sym 2 f ) .$
The symmetric square L-function and the Riesz mean have been studied by many authors, for example, see [3,4,5,6,7,8]. Fomenko [9] considered $Δ ρ ( x ; sym m f )$ when $m = 2$ and obtained a truncated Voronoi type formula. Wang [10] generalized Fomenko’s result to $m ≥ 3$ under the hypothesis $Nice ( m , f )$. Voronoi type formulas examined below allow one to consider exponential sums connected with the coefficients of the Dirichlet series of automorphic L-functions. It seems that no nontrivial estimates for these sums can be obtained. It is of interest to apply the heuristic arguments proposed in [11,12,13,14] to these sums.
Ivić and Sargos [15] studied the higher moments of $Δ ( x )$, which denote the error term in the Dirichlet divisor problem. They got the asymptotic formulas for the cube and fourth integrals of $Δ ( x )$. Liu and Wang [16] studied the higher-power moments of $Δ ρ ( x ; sym 2 f )$ and got
$∫ 1 T Δ ρ A 0 ( x ; sym 2 f ) d x ≪ T 1 + 2 ρ + 1 3 A 0 + ϵ$
when $A 0 = A 0 ( ρ ) > 3$. Then for any integer $3 ≤ h < A 0$, they had the following asymptotic formula
$∫ 1 T Δ ρ h ( x ; sym 2 f ) d x = 6 B ρ ( h , c ) ( 3 + ( 2 ρ + 1 ) h ) ( 2 π ) ( ρ + 1 ) h 3 h 2 T 1 + 2 ρ + 1 3 h + O ( T 1 + 2 ρ + 1 3 h + ϵ ( T − δ ρ ( h , A 0 ) + T − ρ 3 ) ) .$
They also proved the results for $ρ = 1 2$, $h = 3 , 4 , 5$,
$∫ 1 T Δ ρ h ( x ; sym 2 f ) d x = 6 B 1 2 ( h , c ) ( 3 + 2 h ) ( 2 π ) 3 2 h 3 h 2 T 1 + 2 3 h + O ( T 1 + 2 3 h − λ 1 / 2 ( h , 6 ) + ϵ ) ,$
where $λ 1 / 2 ( 3 , 6 ) = 1 22$, $λ 1 / 2 ( 4 , 6 ) = 1 87$, $λ 1 / 2 ( 5 , 6 ) = 1 498$.
As the vibration of $Δ ρ ( x ; sym 2 f )$ is complicable, it is natural to consider the mean value of $Δ ρ ( x ; sym 2 f )$ in short intervals. In this paper, we use the method of large value estimation and exponential sums to study the higher-power moments of the Riesz mean error term $Δ ρ ( x ; sym 2 f )$ in short intervals. Our results are as followed.
Theorem 1.
Let $k ≥ 3$ be fixed integer, T and H be two real numbers, $δ > 0$ be fixed constant. Then for $T 2 3 + 2 ϵ d ρ , k ( 2 ρ + 1 ) δ ≤ H ≤ T$, we have
$∫ T − H T + H Δ ρ k ( x ; sym 2 f ) d x = B k ∫ T − H T + H x 2 ρ + 1 3 k d x + O ( H T 2 ρ + 1 3 k + ϵ ( H T − 2 3 ) − ( 2 ρ + 1 ) δ d ρ , k ) ,$
where
$B k = 2 − k ρ − k + 1 π − k ρ − k 3 − k 2 c k , d ρ , k = ( 3 k − 1 − 4 + k − ρ k ) ( k + δ − 2 ) + ( 2 ρ + 1 ) δ$
and $ϵ > 0$.
In particular, for $ρ = 1 2$, we can prove the following results.
Corollary 1.
For any $ϵ > 0$, $k = 3 , 4 , 5$, we have
$∫ T − H T + H Δ 1 2 3 ( x ; sym 2 f ) d x = B 3 H T 2 + O ( H T 2 + ϵ ( H T − 2 3 ) − 4 δ 13 + 17 δ ) , ∫ T − H T + H Δ 1 2 4 ( x ; sym 2 f ) d x = B 4 H T 8 3 + O ( H T 8 3 + ϵ ( H T − 2 3 ) − 2 δ 50 + 27 δ ) , ∫ T − H T + H Δ 1 2 5 ( x ; sym 2 f ) d x = B 5 H T 10 3 + O ( H T 10 3 + ϵ ( H T − 2 3 ) − 4 δ 477 + 163 δ ) .$
Remark 1.
Please note that our results improve the results in [16] when $k = 3 , 4 , 5$ and $δ > 50 31$.
To derive the above formulas, we start with Voronoi type formula of $Δ ρ ( x ; sym 2 f )$ to get the large value of it, where Halasz–Montgomery inequality and the estimates of exponential sums are used. For higher-power moments of $Δ ρ ( x ; sym 2 f )$, we divide the interval $[ H , 2 H ]$ into subintervals by binary method, and pick the maximal $| Δ ρ ( x r ; sym 2 f ) |$ in r intervals of length $V ρ + 1$ to get the results. Moreover, if $Δ ρ ( x ; sym 2 f )$ has an upper bound, then we can get the following theorems.
Theorem 2.
Let $3 ≤ k < A$ be a fixed integer, $2 ρ + 1 3 ( ρ + 1 ) < θ < 1 2$ be a real number and
$Δ ρ ( x ; sym 2 f ) ≪ x θ ( ρ + 1 ) .$
Then for any $T θ [ A ( ρ + 1 ) − 3 ] + 2 − 2 ρ + 1 3 A + ϵ ≤ H ≤ T$, we have
$∫ T − H T + H Δ ρ k ( x ; sym 2 f ) d x = B k ∫ T − H T + H x 2 ρ + 1 3 k d x + O ( H T 2 ρ + 1 3 k − ϵ ) .$
Theorem 3.
Let $k ≥ 3$ be a fixed integer. If $Δ ρ ( x ; sym 2 f ) ≪ x 2 ρ + 1 3$ is true, then (3) is holds for $T 2 3 + ϵ ≤ H ≤ T$.

## 2. Some Preliminary Lemmas

To prove our theorems, we need the following lemmas.
Lemma 1.
Let $x > 1$, $N ≤ 1$. Then for any fixed ρ, $0 ≤ ρ ≤ 1$, we have
$Δ ρ ( x ; sym 2 f ) = 2 − ρ π − ρ − 1 3 − 1 2 Σ ( x ) + O ( x 4 ρ + 1 6 N − 2 ρ + 1 6 + ϵ ) + O ( x 2 ρ + 2 3 + ϵ N − ρ + 1 6 ) + O ( x ϵ ) ,$
where
$Σ ( x ) = ∑ n ≤ N c n n − ρ + 2 3 x 2 ρ + 1 3 cos ( 6 π n x 3 − π ρ 2 ) .$
Proof.
See Theorem $1.1$ of Fomenko [9]. □
Lemma 2.
Let $h ≥ 3$, $( i 1 , ⋯ , i h − 1 ) ∈ { 0 , 1 } h − 1$ be such that
$n 1 3 + ( − 1 ) i 1 n 2 3 + ( − 1 ) i 2 n 3 3 + ⋯ + ( − 1 ) i h − 1 n h 3 ≠ 0 .$
Then
$| n 1 3 + ( − 1 ) i 1 n 2 3 + ⋯ + ( − 1 ) i h − 1 n h 3 | ≫ max ( n 1 , ⋯ , n h ) − ( 3 h − 2 − 3 − 1 ) .$
Proof.
See for example Lemma $2.3$ in [17]. □
Lemma 3.
If $g ( x )$ and $h ( x )$ are continuous real-valued functions of x and $g ( x )$ is monotonic, then
$∫ a b g ( x ) h ( x ) d x ≪ max a ≤ x ≤ b | g ( x ) | max a ≤ u < v ≤ b | ∫ u v h ( x ) d x | .$
Proof.
See Lemma 1 in [18]. □
Lemma 4.
Suppose $k ≥ 3$, $( i 1 , … , i k − 1 ) ∈ { 0 , 1 } k − 1$, $( i 1 , … , i k − 1 ) ≠ ( 0 , … , 0 )$ and
$N 1 , … , N k > 1 , 0 < Δ ≪ E 1 3 , E = max ( N 1 , … , N k ) .$
Let
$A = A ( N 1 , … , N k ; i 1 , … , i k − 1 ; Δ )$
denote the number of solutions of the inequality
$| n 1 3 + ( − 1 ) i 1 n 2 3 + ⋯ + ( − 1 ) i k − 1 n k 3 | < Δ$
with $N j < n j ≤ 2 N j$, $1 ≤ j ≤ k$. Then
$A ≪ Δ E − 1 3 N 1 … N k + E − 1 N 1 … N k .$
Proof.
It can be proved similarly as Lemma $2.4$ of [19], so we omit the details. □
Lemma 5.
Let S be an inner-product vector space over $C$, $( a , b )$ denote the inner product in S and $∥ a ∥ 2 = ( a , a )$. Suppose that ξ, $φ 1 , ⋯ , φ R$ are arbitrary vectors in S. Then
$∑ l ≤ R | ( ξ , φ l ) | 2 ≤ ∥ ξ ∥ 2 max l 1 ≤ R ∑ l 2 ≤ R | ( φ l l , φ l 2 ) | .$
Proof.
This is the well-known Halasz–Montgomery inequality (see $( A . 40 )$ of [20]). □

## 3. The Higher-Power Moments of $Δ ρ ( x , sym 2 f )$ in Short Intervals

The conclusion in Theorem 1 is obvious at $T 2 ≤ H ≤ T$. Let $H ≤ T / 2$ and $T ϵ < y ≤ min ( T 2 ρ 2 ρ + 1 , T 1 2 )$ be a parameter to be determined later. For any $T ≤ x ≤ 2 T$, we define
$R 1 = R 1 ( x , y ) : = C ∑ n ≤ y c n n − ρ + 2 3 x 2 ρ + 1 3 cos ( 6 π n x 3 − π ρ 2 ) ,$
where $C = 2 − ρ π − ρ − 1 3 − 1 2$;
$R 2 = R 2 ( x , y ) : = Δ ρ ( x , sym 2 f ) − R 1 .$
We will show that the higher-power moments of $R 2$ is small, and thus the integral
$∫ T − H T + H Δ ρ k ( x , sym 2 f ) d x$
can be approximated by $∫ T − H T + H R 1 k d x$.

#### 3.1. Evaluation of the Integral $∫ T − H T + H R 1 k d x$

Let $k ≥ 3$ be any fixed integer. For each element $i = ( i 1 , i 2 , … , i k − 1 ) ∈ I k − 1 , I = { 0 , 1 }$, from the elementary formula
$cos a 1 ⋯ cos a k = 1 2 k − 1 ∑ i ∈ I k − 1 cos ( a 1 + ( − 1 ) i 1 a 2 + ( − 1 ) i 2 a 3 + ⋯ + ( − 1 ) i k − 1 a k ) ,$
we have
$R 1 k = C k x 2 ρ + 1 3 k ∑ n 1 ≤ y ⋯ ∑ n k ≤ y c n 1 ⋯ c n k ( n 1 ⋯ n k ) ρ + 2 3 ∏ j = 1 k cos ( 6 π n j x 3 − π ρ 2 ) = C k 2 − k + 1 x 2 ρ + 1 3 k ∑ ( i 1 , … , i k − 1 ) ∈ { 0 , 1 } k − 1 ∑ n 1 ≤ y ⋯ ∑ n k ≤ y c n 1 ⋯ c n k ( n 1 ⋯ n k ) ρ + 2 3 × cos ( 6 π x 3 α ( n 1 , … , n k ; i 1 , … , i k − 1 ) − π ρ 2 β ( i 1 , … , i k − 1 ) ) ,$
where
$α ( n 1 , … , n k ; i 1 , … , i k − 1 ) : = n 1 3 + ( − 1 ) i 1 n 2 3 + ⋯ + ( − 1 ) i k − 1 n k 3 , β ( i 1 , … , i k − 1 ) : = 1 + ( − 1 ) i 1 + ⋯ + ( − 1 ) i k − 1 .$
It follows that
$R 1 k = C k 2 − k + 1 S 1 ( x ) + S 2 ( x ) ,$
where
$S 1 ( x ) : = x 2 ρ + 1 3 k ∑ ( i 1 , … , i k − 1 ) ∈ { 0 , 1 } k − 1 cos ( − π ρ 2 β ) ∑ n j ≤ y , 1 ≤ j ≤ k α = 0 c n 1 ⋯ c n k ( n 1 ⋯ n k ) ρ + 2 3 , S 2 ( x ) : = x 2 ρ + 1 3 k ∑ ( i 1 , … , i k − 1 ) ∈ { 0 , 1 } k − 1 ∑ n j ≤ y , 1 ≤ j ≤ k α ≠ 0 c n 1 ⋯ c n k ( n 1 ⋯ n k ) ρ + 2 3 cos ( 6 π x 3 α − π ρ 2 β ) , α : = α ( n 1 , … , n k ; i 1 , … , i k − 1 ) , β : = β ( i 1 , … , i k − 1 ) .$
First, we consider the contribution of $S 1 ( x )$. We have
$∫ T − H T + H S 1 ( x ) d x = ∑ ( i 1 , … , i k − 1 ) ∈ { 0 , 1 } k − 1 cos ( − π ρ 2 β ) ∑ n j ≤ y , 1 ≤ j ≤ k α = 0 c n 1 ⋯ c n k ( n 1 ⋯ n k ) ρ + 2 3 ∫ T − H T + H x 2 ρ + 1 3 k d x .$
By Lemma $( 2.1 )$ in [16] we have
$∑ ( i 1 , … , i k − 1 ) ∈ { 0 , 1 } k − 1 cos ( − π ρ 2 β ) ∑ n j ≤ y , 1 ≤ j ≤ k α = 0 c n 1 ⋯ c n k ( n 1 ⋯ n k ) ρ + 2 3 = c k + O ( y − 2 ρ + 1 3 + ϵ ) ,$
where $c k$ is a constant that can be expressed explicitly. So, we get
$C k 2 − k + 1 ∫ T − H T + H S 1 ( x ) d x = C k 2 − k + 1 c k ∫ T − H T + H x 2 ρ + 1 3 k d x + O ( H T 2 ρ + 1 3 k + ϵ y − 2 ρ + 1 3 ) = B k ∫ T − H T + H x 2 ρ + 1 3 k d x + O ( H T 2 ρ + 1 3 k + ϵ y − 2 ρ + 1 3 ) .$
Now we consider the contribution of $S 2 ( x )$. By Lemma 3 we get
$∫ T − H T + H S 2 ( x ) d x ≪ T 2 ρ + 1 3 k + 2 3 ∑ ( i 1 , … , i k − 1 ) ∈ { 0 , 1 } k − 1 ∑ n j ≤ y , 1 ≤ j ≤ k α ≠ 0 c n 1 ⋯ c n k ( n 1 ⋯ n k ) ρ + 2 3 | α | .$
Let
$Σ 1 = ∑ n j ≤ y , 1 ≤ j ≤ k α ≠ 0 c n 1 ⋯ c n k ( n 1 ⋯ n k ) ρ + 2 3 | α | ,$
by Lemma 2 we have $| α | ≫ y − ( 3 k − 2 − 3 − 1 )$. Then by the Lemma 4, for some
$y − ( 3 k − 2 − 3 − 1 ) ≪ Δ < y 1 3 ,$
we get
$Σ 1 ≪ y ϵ ( n 1 ⋯ n k ) ρ + 2 3 Δ A ( N 1 , … , N k ; i 1 , … , i k − 1 ; Δ ) ≪ y ϵ ( n 1 ⋯ n k ) ρ + 2 3 Δ ( Δ E − 1 3 N 1 … N k + E − 1 N 1 … N k ) ≪ y b ρ ( k ) + ϵ ,$
where $b ρ ( k ) = 3 k − 2 − 4 3 + ( 1 − ρ ) k 3$. So, we have
$∫ T − H T + H S 2 ( x ) d x ≪ T 2 ρ + 1 3 k + 2 3 + ϵ y b ρ ( k ) .$
Combining (5)–(8) we obtain
Lemma 6.
For any fixed $k ≥ 3$, we have
$∫ T − H T + H R 1 k d x = B k ∫ T − H T + H x 2 ρ + 1 3 k d x + O ( H T 2 ρ + 1 3 k + ϵ y − 2 ρ + 1 3 + T 2 ρ + 1 3 k + 2 3 + ϵ y b ρ ( k ) ) .$

#### 3.2. Higher-Power Moments of $R 2$

Taking $N = T$ in (4), by the definition of $R 2$ we have
$R 2 = R 2 * + O ( T ρ + 1 3 + ϵ ) ,$
where
$R 2 * = C x 2 ρ + 1 3 ∑ y < n ≤ T c n n ρ + 2 3 cos ( 6 π n x 3 − π ρ 2 ) .$
Then we have
$R 2 ≪ x 2 ρ + 1 3 ∑ y < n ≤ T c n n ρ + 2 3 e ( 3 n x 3 ) + T ρ + 1 3 + ϵ ,$
which implies
$∫ T − H T + H R 2 2 d x ≪ H T 2 ρ + 2 3 + ϵ + ∫ T − H T + H x 2 ρ + 1 3 ∑ y < n ≤ T c n n ρ + 2 3 e ( 3 n x 3 ) 2 d x ≪ H T 2 ρ + 2 3 + ϵ + H T 4 ρ + 2 3 ∑ y < n ≤ T c n 2 n 2 ρ + 4 3 + T 4 ρ + 4 3 ∑ y < n < m ≤ T c n c m ( n m ) ρ + 2 3 ( n 3 − m 3 ) ≪ H T 4 ρ + 2 3 y − 1 + 2 ρ 3 .$
Here we have used the following estimate
$∑ n ≤ u c n 2 ≪ u 1 + ϵ , ∑ y < n < m ≤ T c n c m ( n m ) ρ + 2 3 ( n 3 − m 3 ) ≪ T 1 − 2 ρ 3 + ϵ .$
By the Lemma 6 and the Hölder inequality we have
$∫ T − H T + H R 1 A 0 d x ≪ H T 2 ρ + 1 3 A 0 + ϵ ,$
where $A 0 = k + δ$. Using (2) and (12), we have
$∫ T − H T + H R 2 A 0 d x ≪ ∫ T − H T + H | Δ ρ ( x ; sym 2 f ) | A 0 + | R 1 | A 0 d x ≪ H T 2 ρ + 1 3 A 0 + ϵ .$
For any $2 < A < A 0$, by (11), (13) and Hölder inequality we have
$∫ T − H T + H R 2 A d x ≪ ( ∫ T − H T + H R 2 2 d x ) A 0 − A A 0 − 2 ( ∫ T − H T + H R 2 A 0 d x ) A − 2 A 0 − 2 ≪ H T 2 ρ + 1 3 A + ϵ y − ( 2 ρ + 1 ) ( A 0 − A ) 3 ( A 0 − 2 ) .$
Lemma 7.
Suppose $T ϵ ≤ y ≤ T 2 ρ 2 ρ + 1$, $2 < A < A 0$, then
$∫ T − H T + H R 2 A d x ≪ H T 2 ρ + 1 3 A + ϵ y − ( 2 ρ + 1 ) ( A 0 − A ) 3 ( A 0 − 2 ) .$

#### 3.3. Proof of Theorem 1

Proof.
Suppose $3 ≤ k < A 0$ and $T ϵ ≤ y ≤ T 2 ρ 2 ρ + 1$. By the elementary formula
$( a + b ) k = a k + k a k − 1 b + O ( | a k − 2 b 2 | + | b | k ) ,$
we get
$∫ T − H T + H Δ ρ k ( x , sym 2 f ) d x = ∫ T − H T + H R 1 k d x + O ( ∫ T − H T + H | R 1 k − 1 R 2 | d x ) + O ( ∫ T − H T + H R 2 k d x ) .$
By (12), Lemma 7 and Hölder inequality we get
$∫ T − H T + H | R 1 k − 1 R 2 | d x ≪ ∫ T − H T + H | R 1 | A 0 d x k − 1 A 0 ∫ T − H T + H | R 2 | A 0 A 0 − k + 1 d x A 0 − k + 1 A 0 ≪ H T 2 ρ + 1 3 k + ϵ y − ( 2 ρ + 1 ) ( A 0 − k ) 3 ( A 0 − 2 ) .$
Now take $y = ( H T − 2 3 ) 1 b ρ ( k ) + ( 2 ρ + 1 ) ( A 0 − k ) / ( 3 A 0 − 6 )$. By Lemma 6, Lemma 7 and (14)–(15) we finally get
$∫ T − H T + H Δ ρ k ( x , sym 2 f ) d x = B k ∫ T − H T + H x 2 ρ + 1 3 k d x + O ( T 2 ρ + 1 3 k + 2 3 + ϵ y b ρ ( k ) + H T 2 ρ + 1 3 k + ϵ y − ( 2 ρ + 1 ) ( A 0 − k ) 3 ( A 0 − 2 ) ) = B k ∫ T − H T + H x 2 ρ + 1 3 k d x + O ( H T 2 ρ + 1 3 k + ϵ ( H T − 2 3 ) − ( 2 ρ + 1 ) ( A 0 − k ) b ρ ( k ) ( 3 A 0 − 6 ) + ( 2 ρ + 1 ) ( A 0 − k ) ) = B k ∫ T − H T + H x 2 ρ + 1 3 k d x + O ( H T 2 ρ + 1 3 k + ϵ ( H T − 2 3 ) − ( 2 ρ + 1 ) δ d ρ , k ) .$
Theorem 1 follows immediately. □

## 4. Proof of Theorem 2 and Theorem 3

#### 4.1. A Large Value Estimate of $Δ ρ ( x ; sym 2 f )$

Theorem 4.
Suppose $N 1 + ϵ ≤ H ≤ x 1 < x 2 < ⋯ < x k ≤ 2 H$ satisfy $| Δ ρ ( x r ; sym 2 f ) | ≥ V ρ + 1 ( r = 1 , 2 , ⋯ , R )$ and $| x j − x i | ≫ V ≫ H 14 39 ( log H ) 30 13 ( i ≠ j )$. Then we have
$R ≪ H 2 V − 4 log 6 H + H 13 2 V − 59 4 log 29 H .$
Proof.
Suppose that $H 0 > V$ is a parameter to be determined later. Let I be any subinterval of $[ H , 2 H ]$ of length not exceeding $H 0$ and let $G = I ∩ { x 1 , x 2 , ⋯ , x R }$. Without loss of generality, we may assume that $G = { x 1 , x 2 , ⋯ , x R 0 }$. Suppose $J = ( 2 + 2 ϵ ) log H − 3 log V log 2$. For any $x r ∈ { x 1 , x 2 , ⋯ , x R }$, we apply Formula (4) with $x = x r$ and $M = 2 J + 1 ≍ H 2 + 2 ϵ V − 3$ to get
$Δ ρ ( x ; sym 2 f ) ≪ H 2 ρ + 1 3 ∑ n ≤ M c n n ρ + 2 3 e ( ± 3 ( n x r ) 1 3 ) + H 1 − 2 ρ 3 ϵ V ρ + 1 ≪ H 2 ρ + 1 3 ∑ j = 0 J ∑ n ∼ 2 j c n n ρ + 2 3 e ( ± 3 ( n x r ) 1 3 ) ,$
if we notice that $| Δ ρ ( x ; sym 2 f ) | ≫ V ρ + 1$. Squaring, summing over the set G and then using the Cauchy inequality, we get that
$∑ r ≤ R 0 Δ ρ 2 ( x ; sym 2 f ) ≪ H 4 ρ + 2 3 ∑ r ≤ R 0 ∑ j = 0 J ∑ n ∼ 2 j c n n ρ + 2 3 e ( ± 3 ( n x r ) 1 3 ) 2 ≪ H 4 ρ + 2 3 log H ∑ r ≤ R 0 ∑ j = 0 J ∑ n ∼ 2 j c n n ρ + 2 3 e ( ± 3 ( n x r ) 1 3 ) 2 ≪ H 4 ρ + 2 3 log 2 H ∑ r ≤ R 0 ∑ n ∼ 2 j 0 c n n ρ + 2 3 e ( ± 3 ( n x r ) 1 3 ) 2$
for some $0 ≤ j 0 ≤ J$.
Let $M 0 = 2 j 0$. Take $ξ ( d ) = { ξ n ( d ) } n = 1 ∞$ with $ξ n ( d ) = c n n − ρ + 2 3$ for $n ∼ M 0$ and zero otherwise. Take $φ r ± ( d ) = { φ r , n ± ( d ) } n = 1 ∞$ with $φ r , n ± ( d ) = e ( ∓ 3 ( n x r ) 1 3 )$ for $n ∼ M 0$ and zero otherwise. Then
$( ξ ( d ) , φ r ± ( d ) ) = ∑ n ∼ M 0 c n n ρ + 2 3 e ( ± 3 ( n x r ) 1 3 ) , ( φ r 1 ± ( d ) , φ r 2 ± ( d ) ) = ∑ n ∼ M 0 e ( ∓ 3 n 1 3 ( x r 1 1 3 − x r 2 1 3 ) ) , ∥ ξ ( d ) ∥ 2 = ∑ n ∼ M 0 c n 2 n 2 ρ + 4 3 ≪ M 0 − 1 + 2 ρ 3 log 3 M 0 ,$
where we have used the estimate
$∑ n ∼ M 0 c n 2 ≪ M 0 log 3 M 0 .$
By (16) and Lemma 5 we get
$R 0 V 2 ρ + 2 ≪ H 4 ρ + 2 3 log 2 H ∑ r ≤ R 0 ∑ n ∼ M 0 c n n ρ + 2 3 e ( ± 3 ( n x r ) 1 3 ) 2 ≪ H 4 ρ + 2 3 M 0 − 1 + 2 ρ 3 log 5 H max r 1 ≤ R 0 ∑ r 2 ≤ R 0 ∑ n ∼ M 0 e ( ∓ 3 n 1 3 ( x r 1 1 3 − x r 2 1 3 ) ) ≪ H 4 ρ + 2 3 M 0 − 1 + 2 ρ 3 log 5 H M 0 + max r 1 ≤ R 0 ∑ r 2 ≤ R 0 r 1 ≠ r 2 ∑ n ∼ M 0 e ( ∓ 3 n 1 3 ( x r 1 1 3 − x r 2 1 3 ) ) .$
Following the calculation on [21], we get, by the Kuzmin–Landau inequality and the exponent pair $( 4 18 , 11 18 )$,
$∑ n ∼ M 0 e ( 3 n 1 3 ( x r 1 1 3 − x r 2 1 3 ) ) ≪ M 0 2 3 | x r 1 1 3 − x r 2 1 3 | − 1 + M 0 − 2 3 · 4 18 | x r 1 1 3 − x r 2 1 3 | 4 18 M 0 11 18 ≪ ( M 0 H ) 2 3 | x r 1 − x r 2 | + | x r 1 − x r 2 | 4 18 ( M 0 H ) 4 27 M 0 11 18 ≪ ( M 0 H ) 2 3 | x r 1 − x r 2 | + H − 4 27 H 0 2 9 M 0 25 54 ,$
where we have used the mean value theorem and the estimate $| x r 1 − x r 2 | ≤ H 0$. Substituting this estimate into (17), we get
$R 0 V 2 ρ + 2 ≪ H 4 ρ + 2 3 M 0 − 1 + 2 ρ 3 log 5 H M 0 + max r 1 ≤ R 0 ∑ r 2 ≤ R 0 r 1 ≠ r 2 ( M 0 H ) 2 3 | x r 1 − x r 2 | + H − 4 27 H 0 2 9 M 0 25 54 ≪ H 4 ρ + 2 3 M 0 − 1 + 2 ρ 3 log 5 H M 0 + M 0 2 3 H 2 3 V − 1 log H + H − 4 27 H 0 2 9 M 0 25 54 R 0 ≪ H 4 ρ + 2 3 M 0 2 − 2 ρ 3 log 5 H + M 0 1 − 2 ρ 3 H 4 + 4 ρ 3 V − 1 log 6 H + H 14 27 + 4 ρ 3 H 0 2 9 M 0 7 54 − 2 ρ 3 R 0 log 5 H ≪ M 0 1 − 2 ρ 3 H 4 + 4 ρ 3 V − 1 log 6 H + H 14 27 + 4 ρ 3 H 0 2 9 M 0 7 54 − 2 ρ 3 R 0 log 5 H ,$
where we use the facts that ${ x r }$ is V-spaced and $M 0 ≪ H 2 + 2 ϵ V − 3$. Take
$H 0 = c H − 7 2 V 43 4 log − 45 2 H$
for some sufficiently small constant c such that we can get for this $H 0$ that
$R 0 V 2 ρ + 2 ≪ H 2 V − 2 + 2 ρ log 6 H ,$
i.e.,
$R 0 ≪ H 2 V − 4 log 6 H .$
It is easy to check that $H 0 ≫ V$ if $V ≫ H 14 39 ( log H ) 30 13$.
Now we divide the interval $[ H , 2 H ]$ into $O ( 1 + H H 0 )$ subintervals of length not exceeding $H 0$. In each interval of this type, the number of $x r$’s is at most $O ( H 2 V − 4 log 6 H )$. So, we have
$R ≪ R 0 1 + H H 0 ≪ H 2 V − 4 log 6 H + H 3 V − 4 H 0 − 1 log 6 H ≪ H 2 V − 4 log 6 H + H 13 2 V − 59 4 log 29 H .$

#### 4.2. Proof of Theorem 2

By (4) Liu and Wang [16] obtained that
$Δ ρ ( x ; sym 2 f ) ≪ x ρ + 1 2 .$
Let $2 ρ + 1 3 ( ρ + 1 ) < θ < 1 2$ be a real number, such that $Δ ρ ( x ; sym 2 f ) ≪ x θ ( ρ + 1 )$. Suppose $T − H ≤ x 1 < x 2 < ⋯ < x k ≤ T + H$ satisfy $| Δ ρ ( x r ; sym 2 f ) | ≥ V ρ + 1 ( r = 1 , 2 , ⋯ , R V )$ and $| x j − x i | ≫ V , ( i ≠ j )$, $T 2 ρ + 1 3 ( ρ + 1 ) ≪ V ≪ T θ$, by Theorem 4 we have
$R ≪ T 2 V − 4 log 6 T + H T 11 2 V − 59 4 log 29 T .$
An argument similar to $( 13.70 )$ of [20] proved that
$∫ T − H T + H Δ ρ A ( x , sym 2 f ) d x ≪ H T 2 ρ + 1 3 A + ∑ V V ∑ r ≤ R V | Δ ρ k ( x r , sym 2 f ) | A ,$
where
$T 2 ρ + 1 3 ( ρ + 1 ) ≪ V ≪ T θ$
and
$V ρ + 1 ≤ | Δ ρ ( x r ; sym 2 f ) | ≤ 2 V ρ + 1 ( r = 1 , 2 , ⋯ , R V ) .$
Therefore, by (18) we have
$V ∑ r ≤ R V | Δ ρ k ( x r , sym 2 f ) | A ≪ R V V A ( ρ + 1 ) + 1 ≪ T 2 + ϵ V A ( ρ + 1 ) − 3 + H T 11 2 + ϵ V A ( ρ + 1 ) − 55 4 ≪ T θ [ A ( ρ + 1 ) − 3 ] + 2 + ϵ + H T 11 2 + 2 ρ + 1 3 A − 55 ( 2 ρ + 1 ) 12 ( ρ + 1 ) + ϵ .$
By (19) and (20) we can get
$∫ T − H T + H Δ ρ A ( x , sym 2 f ) d x ≪ H T 2 ρ + 1 3 A + ϵ + T θ [ A ( ρ + 1 ) − 3 ] + 2 + ϵ ,$
where $1 4 ≤ ρ ≤ 1$. When $2 < A < 9 θ − 3 3 θ ( ρ + 1 ) − 2 ρ − 1$, if $H ≥ T θ [ A ( ρ + 1 ) − 3 ] + 2 − 2 ρ + 1 3 A$, then we have
$∫ T − H T + H Δ ρ A ( x , sym 2 f ) d x ≪ H T 2 ρ + 1 3 A + ϵ .$
Please note that for every $2 < A < 9 θ − 3 3 θ ( ρ + 1 ) − 2 ρ − 1$, taking $δ = 3 d ρ , k ϵ 2 ( 2 ρ + 1 )$ in Theorem 1, we can get the result of Theorem 2.

#### 4.3. Proof of Theorem 3

Suppose $Δ ρ ( x ; sym 2 f ) ≪ x 2 ρ + 1 3$, we can obtain the following estimate
$∫ T − H T + H Δ ρ k + δ ( x , sym 2 f ) d x ≪ H T 2 ρ + 1 3 ( k + δ ) + ϵ ′ ,$
where $ϵ ′ = ( k + 1 ) ϵ$. By Theorem 1 we can get
$∫ T − H T + H Δ ρ k ( x ; sym 2 f ) d x = B k ∫ T − H T + H x 2 ρ + 1 3 k d x + O ( H T 2 ρ + 1 3 k + ϵ ′ ( H T − 2 3 ) − ( 2 ρ + 1 ) δ d ρ , k )$
when $T 2 3 + 2 ϵ ′ d ρ , k ( 2 ρ + 1 ) δ ≤ H ≤ T$.
Let $δ = 3 d ρ , k ( k + 1 ) ϵ 2 ( 2 ρ + 1 )$ in Theorem 1, then we get Theorem 3.

## Author Contributions

Conceptualization, R.Z. and X.H.; methodology, D.Z.; validation, R.Z., X.H. and D.Z.; formal analysis, R.Z.; investigation, R.Z.; resources, R.Z.; data curation, D.Z.; writing–original draft preparation, R.Z.; writing–review and editing, X.H.; visualization, D.Z.; supervision, D.Z.; project administration, D.Z.; funding acquisition, D.Z. All authors have read and agreed to the published version of the manuscript.

## Funding

This research was funded by National Natural Science Foundation of China (Grant No. 11771256).

## Conflicts of Interest

The authors declare no conflict of interest.

## References

1. Khan, R. The first moment of the symmetric-square L-function. J. Number Theory 2007, 124, 259–266. [Google Scholar] [CrossRef] [Green Version]
2. Hafner, J.L. On the representation of the summatory functions of a class of arithmetical functions. Lect. Notes Math. 1981, 899, 148–165. [Google Scholar]
3. Bump, D.; Ginzburg, D. Symmetric square L-functions on GL(r). Ann. Math. 1992, 136, 37–205. [Google Scholar] [CrossRef]
4. Zhang, D.Y.; Lau, Y.; Wang, Y.N. Remark on the paper "On products of Fourier coefficients of cusp forms". Arch. Math. 2017, 108, 263–269. [Google Scholar] [CrossRef]
5. Zhang, D.Y.; Wang, Y.N. Ternary quadratic form with prime variables attached to Fourier coefficients of primitive holomorphic cusp form. J. Number Theory 2017, 176, 211–225. [Google Scholar] [CrossRef]
6. Ichihara, Y. On Riesz mean for the coefficients of twisted Rankin-Selberg L-functions. J. Math. Soc. Jpn. 2003, 55, 81–100. [Google Scholar] [CrossRef]
7. Iwaniec, H.; Michel, P. The second moment of the symmetric square L-functions. Ann. Acad. Sci. Fenn. Math. 2001, 26, 465–482. [Google Scholar]
8. Blomer, V. On the central value of symmetric square L-functions. Math. Z. 2008, 260, 750–777. [Google Scholar] [CrossRef]
9. Fomenko, O.M. The behavior of Riesz means of the Coefficients of a symmetric square L-function. J. Math. Sci. 2007, 143, 3174–3181. [Google Scholar] [CrossRef]
10. Wang, H. On the Riesz means of coefficients of mth symmetric power L-functions. Lith. Math. J. 2010, 50, 474–488. [Google Scholar] [CrossRef]
11. Iwaniec, H.; Luo, W.; Sarnak, P. Low lying zeros of families of L-functions. Publ. IHES 2000, 91, 55–131. [Google Scholar] [CrossRef] [Green Version]
12. Song, P.; Zhai, W.G.; Zhang, D.Y. Power moments of Hecke eigenvalues for congruence group. J. Number Theory 2019, 198, 139–158. [Google Scholar] [CrossRef]
13. Zhang, D.Y.; Zhai, W.G. On the fifth-power moments of Δ(x). Int. J. Number Theory 2011, 7, 71–86. [Google Scholar] [CrossRef]
14. Zhang, D.Y.; Wang, Y.N. Higher-power moments of Fourier coefficients of holomorphic cusp forms for the congruence subgroup Γ0(N). Ramanujan J. 2018, 47, 685–700. [Google Scholar] [CrossRef]
15. Ivić, A.; Sargos, P. On the higher moments of the error term in the divisor problem. Ill. J. Math. 2007, 51, 353–377. [Google Scholar] [CrossRef]
16. Liu, K.; Wang, H.Y. Higher power moments of the Riesz mean error term of symmetric square L-function. J. Number Theory 2011, 131, 2247–2261. [Google Scholar] [CrossRef] [Green Version]
17. Tanigawa, Y.; Zhang, D.Y.; Zhai, W.G. On the Rankin-Selberg problem: Higher power moments of the Riesz mean error term. Sci. China Ser. A 2008, 51, 148–160. [Google Scholar] [CrossRef] [Green Version]
18. Tsang, K.M. Higher-power moments of Δ(x), E(t) and P(x). Proc. Lond. Math. Soc. 1991, 65, 65–84. [Google Scholar] [CrossRef]
19. Zhai, W.G. On higher-power moments of Δ(x) (II). Acta Arith. 2004, 114, 35–54. [Google Scholar] [CrossRef] [Green Version]
20. Ivić, A. The Riemann Zeta Function: Theory and Appllications; Dover: New York, NY, USA, 2003. [Google Scholar]
21. Zhai, W.G. On the error term in Weyl’s law for Heisenberg manifolds. Acta Arith. 2008, 134, 219–257. [Google Scholar] [CrossRef] [Green Version]
 Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Share and Cite

MDPI and ACS Style

Zhang, R.; Han, X.; Zhang, D. Power Moments of the Riesz Mean Error Term of Symmetric Square L-Function in Short Intervals. Symmetry 2020, 12, 2036. https://doi.org/10.3390/sym12122036

AMA Style

Zhang R, Han X, Zhang D. Power Moments of the Riesz Mean Error Term of Symmetric Square L-Function in Short Intervals. Symmetry. 2020; 12(12):2036. https://doi.org/10.3390/sym12122036

Chicago/Turabian Style

Zhang, Rui, Xue Han, and Deyu Zhang. 2020. "Power Moments of the Riesz Mean Error Term of Symmetric Square L-Function in Short Intervals" Symmetry 12, no. 12: 2036. https://doi.org/10.3390/sym12122036

Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here.

Back to TopTop