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Article

On Fourier Coefficients of the Symmetric Square L-Function at Piatetski-Shapiro Prime Twins

School of Mathematics and Statistics, Shandong Normal University, Jinan 250358, China
*
Author to whom correspondence should be addressed.
Mathematics 2021, 9(11), 1254; https://doi.org/10.3390/math9111254
Submission received: 28 April 2021 / Revised: 23 May 2021 / Accepted: 26 May 2021 / Published: 30 May 2021
(This article belongs to the Special Issue New Developments in Number Theory)

Abstract

:
Let P c ( x ) = { p x | p , [ p c ] are primes } , c R + N and λ s y m 2 f ( n ) be the n-th Fourier coefficient associated with the symmetric square L-function L ( s , s y m 2 f ) . For any A > 0 , we prove that the mean value of λ s y m 2 f ( n ) over P c ( x ) is x log A 2 x for almost all c ε , ( 5 + 3 ) / 8 ε in the sense of Lebesgue measure. Furthermore, it holds for all c ( 0 , 1 ) under the Riemann Hypothesis. Furthermore, we obtain that asymptotic formula for λ f 2 ( n ) over P c ( x ) is p , q p r i m e p x , q = [ p c ] λ f 2 ( p ) = x c log 2 x ( 1 + o ( 1 ) ) , for almost all c ε , ( 5 + 3 ) / 8 ε , where λ f ( n ) is the normalized n-th Fourier coefficient associated with a holomorphic cusp form f for the full modular group.
MSC:
11A41; 11F11; 11F30

1. Introduction

Let k be an even positive integer, f be a holomorphic cusp form of weight k for the full modular group and λ f ( n ) be the normalized n-th Fourier coefficient of f, i.e.,
f ( z ) = n = 1 λ f ( n ) n k 1 2 e ( n z ) .
If we assume that f is an eigenform of all the Hecke operators, then f can be normalized such that λ f ( 1 ) = 1 and λ f ( n ) is real. We define the Hecke L-function associated to f for s > 1 by
L ( s , f ) = n = 1 λ f ( n ) n s = p ( 1 λ f ( p ) p s + p 2 s ) 1 .
For any prime p and all integers ν 0 , we have
λ f ( p ν ) = α f ( p ) ν + α f ( p ) ν 1 β f ( p ) + + β f ( p ) ν ,
where α f ( p ) , β f ( p ) are the local parameters of L ( s , f ) at prime p, satisfying
α f ( p ) + β f ( p ) = λ f ( p ) , α f ( p ) β f ( p ) = 1 .
Then we have
λ f 2 ( p ) = 1 + λ f ( p 2 ) .
Deligne [1,2] proved Ramanujan–Petersson conjecture, i.e.,
| λ f ( n ) | d ( n ) n ε
for all n 1 , where d ( n ) = d | n 1 .
In order to detect the sign changes of λ f ( n ) , many authors have studied the mean value of λ f ( n ) and obtained some good results. For example, see [1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22]. In addition, the sums of λ f ( n ) over primes have also been studied. It is known that (see for example Section 5.6 of Iwaniec and Kowalski [23]) there exists a constant C > 0 such that
p N λ f ( p ) f N exp ( C log N ) .
The upper bound of (1) may reach N 1 2 + ε , assuming the Riemann Hypothesis. Furthermore, we can establish that
p N λ f 2 ( p ) N log N , as N ,
by using the analytic properties of the Rankin–Selberg L-function L ( s , f f ¯ ) .
Another interesting question considered by many authors is the mean value of λ f ( n ) over certain sets of primes. For example, Baier and Zhao [24] studied the distribution of λ f ( n ) at Piatetski–Shapiro primes by considering estimates of exponential sum involving Hecke eigenvalue. Moreover, they conjectured that, for 1 < c < 1 + θ with some suitable θ > 0 , there exists a constant c f > 0 such that
n N [ n c ] R λ f 2 ( [ n c ] ) c f N c log N , as N .
Furthermore, we can define Piatetski–Shapiro prime twins if p , q are primes and q = [ p c ] , for any fixed c R + N . Balog [25] and Dufner [26] proved that
p , q p r i m e p x , q = [ p c ] 1 = x c log 2 x ( 1 + o ( 1 ) ) , x ,
for almost all c ( 0 , 1 ) in the sense of Lebesgue measure. Furthermore, assuming the Riemann Hypothesis of automorphic L-function L ( s , f ) is true, they found that (2) holds for all c ( 0 , 1 ) . Furthermore, Zhang and Zhai [27] studied the mean value of λ f ( n ) over Piatetski–Shapiro prime twins.
Motivated by the above results, we are interested in the distribution of λ f 2 ( n ) at Piatetski–Shapiro prime twins. For the form f, we know the L ( s , s y m 2 f ) is an L-function for some G L ( 3 , Z ) automorphic representation, which is often called the symmetric-square lift of f. The n-th Fourier coefficient of L ( s , s y m 2 f ) satisfies
λ s y m 2 f ( n ) = m l 2 = n λ f ( m 2 ) .
The symmetric square L-function associated to f is defined by
L ( s , s y m 2 f ) = p 0 j 2 1 α f ( p ) 2 j β f ( p ) j p s 1 = n = 1 λ s y m 2 f ( n ) n s
in the half-plane s > 1 . Then, for all n 1 , λ s y m 2 f ( n ) is also multiplicative, real and
| λ s y m 2 f ( n ) | d 3 ( n ) ,
where d 3 ( n ) = m | n d ( m ) . Furthermore, for all primes p, we have
λ s y m 2 f ( p ) = λ f ( p 2 ) .
Many authors studied the mean value of λ s y m 2 f ( n ) . For example, see References [28,29,30,31,32,33,34,35,36]. In this paper, we consider the mean value of Fourier coefficients of symmetric square L-function over Piatetski–Shapiro prime twins and obtain the following results, which imply a result on the distribution of λ f 2 ( n ) at Piatetski–Shapiro prime twins.
Theorem 1.
For almost all c ε , ( 5 + 3 ) / 8 ε and any A > 0 , we have
p , q p r i m e p x , q = [ p c ] λ s y m 2 f ( p ) x log A + 2 x .
Theorem 2.
Assuming the Riemann Hypothesis of symmetric square L-function is true, (3) holds for all c ( 0 , 1 ) .
Corollary 1.
For almost all c ε , ( 5 + 3 ) / 8 ε , we have
p , q p r i m e p x , q = [ p c ] λ f 2 ( p ) = x c log 2 x ( 1 + o ( 1 ) ) , x .
Proof. 
The result follows easily from Theorem 1 and (2), if we notice that λ f 2 ( p ) = 1 + λ f ( p 2 ) .      ☐
Corollary 2.
Assuming the Riemann Hypothesis of symmetric square L-function is true, (4) holds for all c ( 0 , 1 ) .
Proof. 
The result follows from Theorem 2 and (2).      ☐
Notation. 
Throughout the paper, ε always denotes a sufficiently small positive constant. Let δ 1 ( ε ) be sufficiently small and depend on ε. We write f ( x ) g ( x ) , or f ( x ) = O ( g ( x ) ) , to mean that | f ( x ) | C g ( x ) . Let ρ = σ + i η be the nontrivial zero of the symmetric square L-function L ( s , s y m 2 f ) . As usual, Λ ( n ) is the von Mangoldt function.

2. Auxiliary Lemmas

Lemma 1.
Let η run through a countable set of reals, with c ( η ) arbitrary complex such that S ( t ) = η c ( η ) e ( η t ) is absolutely convergent. Let α R , Θ ( 0 , 1 ) , δ = Θ / α . Then
α α | S ( t ) | 2 d t Θ | δ 1 t η t + δ c ( η ) | 2 d t .
Proof. 
This lemma is Lemma 1 of Gallagher [37].      ☐
Lemma 2.
Let T T 0 , where T 0 is a sufficiently large real number. The
T 2 T | L ( α + i t , s y m 2 f ) | 2 d t ( 10 + k + T ) ( log ( 10 + k + T ) ) 17 , i f 2 / 3 α 1 , ( 10 + k + T ) 3 ( 1 α ) ( log ( 10 + k + T ) ) 17 , i f 1 / 3 α 2 / 3 , ( 10 + k + T ) 3 ( 1 2 α ) + 1 ( log ( 10 + k + T ) ) 17 , i f 0 α 1 / 3 .
Proof. 
This lemma is Lemma 3.1 of Lü [33].      ☐
Lemma 3.
For 1 / 2 σ 1 , T 3 , define
N ( σ , T ) = { ϱ = β + i ω : L ( ϱ , s y m 2 f ) = 0 , σ β 1 , | ω | T } .
Then we have
N ( σ , T ) T 5 ( 1 σ ) 3 2 σ + ε , i f 1 / 2 σ 3 / 4 ,
and
N ( σ , T ) T 3 ( 1 σ ) + ε , i f 3 / 4 < σ 1 .
Proof. 
From Lemma 2, we have
T 2 T | L ( 1 / 2 + i t , s y m 2 f ) | 2 d t T 3 / 2 + ε .
This combined with Theorem 1.1, 1.2 of Ye and Zhang [38] and [39] gives this lemma.      ☐
Lemma 4.
For any u 2 , T 2 , we have
n < u Λ ( n ) λ s y m 2 f ( n ) = ρ = σ + i η | η | T u ρ ρ + O log 3 u + u log 3 ( u T ) T .
Proof. 
See, for example, Iwaniec and Kowalski [23]. For convenience of calculation, we reduce the summation range of n from n u to n < u . The contribution of n = u is O ( log 3 u ) , in view of | λ s y m 2 f ( p h ) | d 3 ( p h ) log 2 u .      ☐
Lemma 5.
Let L ( s , f ) be an L-function of degree k such that Rankin–Selberg convolutions L ( s , f f ) and L ( s , f f ¯ ) exist, and the latter has a simple pole at s = α + i t = 1 while the former is entire if f f ¯ . Suppose that the ramified primes | α f ( p ) | p / 2 . There exists an absolute constant c > 0 such that L ( s , f ) has no zeros in the region
α 1 c k 4 log ( q ( f ) | t | + 3 ) ,
where q ( f , s ) is the analytic conductor and q ( f ) = q ( f , 0 ) .
Proof. 
This lemma is Theorem 5.10 of Iwaniec and Kowalski [23].      ☐
Lemma 6.
Let x, T R + , x and
S : = ρ = σ + i η | η | T x σ ,
we have
S x 1 / 2 T 5 / 4 + ε , i f T x 4 / 5 ; x 3 / 2 T 5 / 2 + ε e 5 log M log T , i f x 9 / 20 T < x 4 / 5 ; x 3 / 4 T 5 / 6 + ε + x 1 β T 3 β + ε , i f T < x 9 / 20 .
Proof. 
Note that
S = 1 / 2 1 β x σ d σ N ( σ , T ) ,
where β = β ( T ) is the width of the zero-free region of the symmetric square L-function in Lemma 5. Using integration by parts and Lemma 3, we obtain
S l x 1 / 2 T + 1 / 2 3 / 4 exp ( f 1 ( σ ) ) d σ + 3 / 4 1 β exp ( f 2 ( σ ) ) d σ : = x 1 / 2 T + S 1 + S 2 ,
where l is fulfilled apart from a fixed number of log-factors,
f 1 ( σ ) = 5 ( 1 σ ) 3 2 σ log T + σ log x + ε log T
and
f 2 ( σ ) = 3 ( 1 σ ) log T + σ log x + ε log T .
We estimate S 1 first. The first and second derivatives of f 1 ( σ ) are
f 1 ( σ ) = 5 log T ( 3 2 σ ) 2 + log x
and
f 1 ( σ ) = 20 log T ( 3 2 σ ) 3 < 0 , if σ < 3 / 2 .
Then f 1 ( σ ) is a concave function as σ < 3 / 2 . Let f 1 ( σ 1 ) = 0 , we have σ 1 = 3 2 1 2 5 log T log x . So we have the following three cases.
Case 1. When T x 4 / 5 . In this case, we have σ 1 1 / 2 and f 1 ( σ ) is monotonically decreasing in [ 1 / 2 , 3 / 4 ] . Hence
max σ [ 1 / 2 , 3 / 4 ] f 1 ( σ ) = f 1 1 / 2 = ( 5 log T ) / 4 + ( log x ) / 2 + ε log T .
Case 2. When x 9 / 20 T < x 4 / 5 . In this case, the function f 1 ( σ ) takes the extreme value at σ 1 , which gives
max σ [ 1 / 2 , 3 / 4 ] f 1 ( σ ) = f 1 ( σ 1 ) = ( 5 log T ) / 2 + ( 3 log x ) / 2 5 log T log x + ε log T .
Case 3. When T < x 9 / 20 . In this case, we have σ 1 > 3 / 4 and f 1 ( σ ) is monotonically increasing in [ 1 / 2 , 3 / 4 ] . Hence,
max σ [ 1 / 2 , 3 / 4 ] f 1 ( σ ) = f 1 3 / 4 = ( 5 log T ) / 6 + ( 3 log x ) / 4 + ε log T .
Combining all the above cases, we have
S 1 = 1 2 3 4 exp ( f 1 ( σ ) ) d σ x 1 / 2 T 5 / 4 + ε , if T x 4 / 5 ; x 3 / 2 T 5 / 2 + ε e 5 log x log T , if x 9 / 20 T < x 4 / 5 ; x 3 / 4 T 5 / 6 + ε , if T < x 9 / 20 .
Next, we need to estimate S 2 . It is easy to see that f 2 ( σ ) is linear function in [ 3 / 4 , 1 β ] , hence
f 2 ( σ ) max f 2 3 / 4 , f 2 ( 1 β ) = max ( 3 log T ) / 4 + ( 3 log x ) / 4 , 3 β log T + ( 1 β ) log x + ε log T .
From (6), (7) and (8), we obtain
S l x 1 / 2 T + x 1 / 2 T 5 / 4 + ε + x 3 / 4 T 3 / 4 + ε , if T x 4 / 5 ; x 1 / 2 T + x 3 / 2 T 5 / 2 + ε e 5 log x log T + x 3 / 4 T 3 / 4 + ε , if x 9 / 20 T < x 4 / 5 ; x 1 / 2 T + x 3 / 4 T 5 / 6 + ε + x 1 β T 3 β + ε , if T < x 9 / 20 ,
which proves this lemma.      ☐
Lemma 7.
Let x 0 : = 2 , x v + 1 : = x v + x v log 2 x v . Let γ 0 be a constant and
D ( γ 0 , x v , T ) = γ 0 γ 0 + log 2 x v x v 1 / γ < m x v + 1 1 / γ Λ ( m ) ρ = σ + i η T / 2 < | η | T ( m + 1 ) γ ρ m γ ρ ρ 2 d γ .
If T > x v 1 / γ 0 δ , we have
D ( γ 0 , x v , T ) x v 2 / γ 0 T 3 ρ = σ + i η T / 2 < | η | T x v σ 2 .
If T x v 1 / γ 0 δ , we have
D ( γ 0 , x v , T ) x v 2 / γ 0 T 3 ρ = σ + i η T / 2 < | η | T x v σ 2 T x v 1 / γ 0 2 .
Proof. 
This lemma follows from (13) and (16) of Dufner [26].      ☐

3. Proof of Theorem 2

In this section, we write γ = 1 / c . Then we have
Π 2 , c ( x ) : = p , q p r i m e p x , q = [ p c ] λ s y m 2 f ( p ) = p , q p r i m e p x , q p c < ( q + 1 ) λ s y m 2 f ( p ) = q x c p x q γ p < ( q + 1 ) γ λ s y m 2 f ( p ) .
We split the summation range of q into two parts: q x c 1 and x c 1 < q x c to get
Π 2 , c ( x ) = q x c 1 q γ p < ( q + 1 ) γ λ s y m 2 f ( p ) + x c 1 < q x c q γ p x λ s y m 2 f ( p ) = q x c q γ p < ( q + 1 ) γ λ s y m 2 f ( p ) x c 1 < q x c x < p < ( q + 1 ) γ λ s y m 2 f ( p ) : = E 1 E 2 .
For E 2 , there is at most one prime q x satisfying x c 1 < q x x c , hence
E 2 q x γ < p ( q x + 1 ) γ | λ s y m 2 f ( p ) | ( q x + 1 ) γ q x γ x 1 c ,
if we notice that | λ s y m 2 f ( p ) | d 3 ( p ) 1 .
For E 1 , by the definition of the von Mangoldt function, we have
E 1 = q x c q γ n < ( q + 1 ) γ Λ ( n ) λ s y m 2 f ( n ) log n + O q x c q γ n = p h < ( q + 1 ) γ h 2 λ s y m 2 f ( p h ) log p log p h .
The error term of the above formula contributes
q x c q γ p h < ( q + 1 ) γ h 2 | λ s y m 2 f ( p h ) | = q x c q γ p 2 < ( q + 1 ) γ | λ s y m 2 f ( p 2 ) | + q x c q γ p h < ( q + 1 ) γ h 3 | λ s y m 2 f ( p h ) | q x c q γ n 2 < ( q + 1 ) γ 1 + q x c q γ n h < ( q + 1 ) γ h 3 log 2 x q x c ( q + 1 ) γ / 2 q γ / 2 + q x c ( q + 1 ) γ / 3 q γ / 3 log 3 x q x c q γ / 2 1 + q x c q γ / 3 1 log 3 x x 1 / 2 .
We use the same method to deal with the main term of (13) and get
E 1 = 2 m x c Λ ( m ) log m m γ n < ( m + 1 ) γ Λ ( n ) λ s y m 2 f ( n ) log n + O x 1 c / 2 + O x 1 / 2 ,
where the first O-term comes from
m = q h x c h 2 Λ ( m ) log m m γ n < ( m + 1 ) γ Λ ( n ) λ s y m 2 f ( n ) log n = m = q h x c h 2 log q log q h m γ p r < ( m + 1 ) γ λ s y m 2 f ( p r ) log p log p r m = q h x c h 2 m γ p r < ( m + 1 ) γ | λ s y m 2 f ( p r ) | x 1 c / 2 .
Let
W ( x ) : = 2 m x c Λ ( m ) log m m γ n < ( m + 1 ) γ Λ ( n ) λ s y m 2 f ( n ) log n .
By the range of n and Taylor’s formula, we have
log n = γ log m + O m 1
and
1 log n = 1 γ log m + O 1 m log 2 m .
Then
W ( x ) = γ 1 2 m x c Λ ( m ) log 2 m m γ n < ( m + 1 ) γ Λ ( n ) λ s y m 2 f ( n ) + O ( x 1 c log 3 x ) ,
where the O-term comes from
2 m x c Λ ( m ) m log 3 m m γ n < ( m + 1 ) γ Λ ( n ) λ s y m 2 f ( n ) log 3 x 2 m x c m γ 2 log 3 m x 1 c log 3 x 2 m x c 1 m log 3 m x 1 c log 3 x .
From (14), (15) and partial summation, we obtain
E 1 log 2 x · max x 0 x | Ψ 2 , c ( x 0 ) | + x 1 c log 3 x + x 1 c / 2 + x 1 / 2 ,
where
Ψ 2 , c ( x ) = m x c Λ ( m ) m γ n < ( m + 1 ) γ Λ ( n ) λ s y m 2 f ( n ) .
To get (3), it suffices to prove that
| Ψ 2 , c ( x ) | x log A x , for any A > 0 .
The inequality (17) will be proved from a variant of (17) for short intervals. Let x 0 : = 2 , x v + 1 : = x v + x v log 2 x v , we can see that the inequality
| Δ Ψ 2 , c ( x v ) | = x v c < m x v + 1 c Λ ( m ) m γ n < ( m + 1 ) γ Λ ( n ) λ s y m 2 f ( n ) x v log A 2 x v
implies (17). We use Lemma 4 and get
m γ n < ( m + 1 ) γ Λ ( n ) λ s y m 2 f ( n ) = ρ = σ + i η | η | T v ( m + 1 ) γ ρ m γ ρ ρ + O log 3 m + ( m + 1 ) γ log 3 ( m T ) T .
Taking T v = x v log 2 A 3 x v , then we obtain
| Δ Ψ 2 , c ( x v ) | | Φ ( γ , x v , T v ) | + x v c log 2 A + 4 x v ,
where
Φ ( γ , x v , T v ) = x v c < m x v + 1 c Λ ( m ) ρ = σ + i η | η | T v ( m + 1 ) γ ρ m γ ρ ρ .
Under the Riemann Hypothesis, we have
Φ ( γ , x v , T v ) = x v c < m x v + 1 c Λ ( m ) m γ ( m + 1 ) γ ρ = σ + i η | η | T v t 1 / 2 + i η d t = x v c < m x v + 1 c Λ ( m ) m γ ( m + 1 ) γ ρ = σ + i η | η | T v e ( 1 / 2 + i η ) log t d t .
Making the change of variables u = log t , we deduce that
Φ ( γ , x v , T v ) = x v c < m x v + 1 c Λ ( m ) γ log m γ log ( m + 1 ) e u / 2 ρ = σ + i η | η | T v e i η u d u .
Using the Cauchy–Schwarz inequality twice, we get
| Φ ( γ , x v , T v ) | 2 x v c < m x v + 1 c Λ 2 ( m ) x v c < m x v + 1 c γ log m γ log ( m + 1 ) e u / 2 ρ = σ + i η | η | T v e i η u d u 2 x v c < m x v + 1 c Λ 2 ( m ) x v c < m x v + 1 c m γ 1 γ log m γ log ( m + 1 ) ρ = σ + i η | η | T v e i η u 2 d u x v log 1 x v · log x v γ + log x v + 1 ρ = σ + i η | η | T v e i η u 2 d u .
Making the change of variables t = u ( 2 π ) 1 ( γ + log x v + 1 + log x v ) ( 4 π ) 1 , we deduce that the last integral in (22) can be written as
2 π α α | η | T v c ( η ) e ( η t ) 2 d t ,
where α = ( γ + log x v + 1 log x v ) ( 4 π ) 1 and
c ( η ) = e i η γ + log x v + 1 + log x v / 2 , if | η | T v ; 0 , otherwise .
Applying Lemma 1 with Θ = 1 / 2 and δ = Θ / α 2 π / γ to estimate the integral in (23), we have
δ 1 t η t + δ | η | T v c ( η ) 2 d t = δ 2 t η 1 , η 2 t + δ | η 1 | , | η 2 | T v e i ( η 1 η 2 ) γ + log x v + 1 + log x v / 2 d t δ 2 · δ 2 · T v x v log 2 A 3 x v .
From (20), (22) and (24), we get
| Δ Ψ 2 , c ( x v ) | x v log A 2 x v + x v c log 2 A + 4 x v .
This combined with (16) and (17) gives
E 1 x log A 2 x .
Theorem 2 follows from (11), (12) and (25).

4. Proof of Theorem 1

To prove Theorem 1, we need Lemma 6 and Lemma 7. Furthermore, we only need to estimate Φ ( γ , x v , T v ) unconditionally. Note that
Φ ( γ , x v , T v ) log x v · max 1 T T v | Φ 1 ( γ , x v , T ) | ,
where
Φ 1 ( γ , x v , T ) = x v c < m x v + 1 c Λ ( m ) ρ = σ + i η T / 2 < | η | T ( m + 1 ) ρ γ m ρ γ ρ .
We consider the following integral mean value of Φ 1 ( γ , x v , T ) ,
D ( γ 0 , x v , T ) = γ 0 γ 0 + log 2 x v | Φ 1 ( γ , x v , T ) | 2 d γ
with γ 0 1 ( 5 + 3 ) / 8 ε , 1 ε .
From Lemma 6, we know that the upper bound of S depends on the range of T, so we have the following three cases.
Case 1. When T x v 4 / 5 .
In this case, we have x v 1 / γ 0 δ 1 ( ε ) < T and use (9) to get
D ( γ 0 , x v , T v ) x v 2 / γ 0 T 3 x v 1 / 2 T 5 / 4 + ε 2 = x v 1 + 2 / γ 0 T 1 / 2 + 2 ε x v 1 + 3 / 2 γ 0 + 2 ε / γ 0 + ( 1 / 2 2 ε ) δ 1 ( ε ) .
Note that if 1 / γ 0 5 + 3 / 8 ε , we have
D ( γ 0 , x v , T ) x v 2 ε .
Case 2. When x v 9 / 20 T < x v 4 / 5 .
In case of x v 1 / γ 0 δ 1 ( ε ) < T , we use (9) and get
D ( γ 0 , x v , T v ) x v 2 / γ 0 T 3 x v 3 / 2 T 5 / 2 + ε e 5 log x v log T 2 = x v 3 + 2 / γ 0 T 2 + 2 ε e 2 5 log x v log T = e ( 2 + 2 ε ) log T log x v 2 2 5 log T log x v + 3 + 2 γ 0 log x v = x v ( 2 + 2 ε ) log T log x v 2 2 5 log T log x v + 3 + 2 / γ 0 ,
For convenience, we take t = log T log x v and consider a quadratic function g ( t ) = ( 2 + 2 ε ) t 2 2 5 t + 3 + 2 / γ 0 .
If x v 1 / γ 0 δ 1 ( ε ) x v 9 / 20 T , we have t 9 / 20 , 4 / 5 and g ( t ) is monotonically decreasing in this interval. Hence,
g ( t ) g 9 / 20 = 9 ( 1 + ε ) / 10 + 2 / γ 0 9 / 5 + 9 / 10 ε + 2 δ 1 ( ε )
with 1 / γ 0 9 / 20 + δ 1 ( ε ) . Therefore,
D ( γ 0 , x v , T ) x v 2 ε .
If x v 9 / 20 < x v 1 / γ 0 δ 1 ( ε ) < T < x v 4 / 5 , we get analogously t 1 / γ 0 δ 1 ( ε ) , 4 / 5 and
g ( t ) < g 1 / γ 0 δ 1 ( ε ) = 3 + 4 / γ 0 2 5 1 / γ 0 δ 1 ( ε ) + 2 ε / γ 0 ( 2 + 2 ε ) δ 1 ( ε ) 3 + 4 / γ 0 2 5 / γ 0 + 2 ε / γ 0 + 10 γ 0 2 2 ε δ 1 ( ε ) 2 ε ,
where we choose δ 1 ( ε ) sufficiently small and use the elementary inequality
1 x 1 x / 2 , for x [ 0 , 1 / 2 ] .
Therefore,
D ( γ 0 , x v , T ) x v 2 ε .
In case of T x v 1 / γ 0 δ 1 ( ε ) , we use (10) and get
D ( γ 0 , x v , T ) x v 3 + 2 / γ 0 T 2 + 2 ε e 2 5 log T log x v T x v 1 / γ 0 2 = x v 3 T 4 + 2 ε e 2 5 log T log x v = x v ( 4 + 2 ε ) log T log x v 2 2 5 log T log x v + 3 ,
so we consider a new quadratic function g 1 ( t ) = ( 4 + 2 ε ) t 2 2 5 t + 3 .
If x v 9 / 20 T x v 1 / γ 0 δ 1 ( ε ) < x v 4 / 5 , we get t 9 / 20 , 1 / γ 0 δ 1 ( ε ) and
g 1 ( t ) g 1 1 / γ 0 δ 1 ( ε ) = 4 1 / γ 0 δ 1 ( ε ) 2 5 · 1 / γ 0 δ 1 ( ε ) + 3 + 2 ε / γ 0 2 ε δ 1 ( ε ) 4 / γ 0 2 5 / γ 0 + 3 + 2 ε / γ 0 + 10 γ 0 4 2 ε δ 1 ( ε ) 2 ε .
Therefore,
D ( γ 0 , x v , T ) x v 2 ε .
Case 3. When T < x v 9 / 20 .
If x v 1 / γ 0 δ 1 ( ε ) < T < x v 9 / 20 , we use (9) to get
x v 2 / γ 0 T 3 x v 3 / 4 T 5 / 6 + ε 2 = x v 3 / 2 + 2 / γ 0 T 4 / 3 + 2 ε x v 2 ε
and
x v 2 / γ 0 T 3 x v 1 β T 3 β + ε 2 = x v 2 2 β + 2 / γ 0 T 3 + 6 β + 2 ε x v 2 ε .
Therefore,
D ( γ 0 , x v , T ) x v 2 / γ 0 T 3 x v 3 / 4 T 5 / 6 + ε + x v 1 β T 3 β + ε 2 x v 2 ε .
If T x v 1 / γ 0 δ 1 ( ε ) < x v 9 / 20 or T < x v 9 / 20 x v 1 / γ 0 δ 1 ( ε ) , we use (10) to get
x v 2 / γ 0 T 3 x v 3 / 4 T 5 / 6 + ε 2 T x v 1 / γ 0 2 x v 3 / 2 T 2 / 3 + 2 ε x v 2 ε
and
x v 2 / γ 0 T 3 x v 1 β T 3 β + ε 2 T x v 1 / γ 0 2 x v 2 2 β T 1 + 6 β + 2 ε x v 2 ε .
Therefore,
D ( γ 0 , x v , T ) x v 2 / γ 0 T 3 x v 3 / 4 T 5 / 6 + ε + x v 1 β T 3 β + ε 2 T x v 1 / γ 0 2 x v 2 ε .
Combining all the above cases, we obtain
D ( γ 0 , x v , T ) = γ 0 γ 0 + log 2 x v | Φ 1 ( γ , x v , T ) | 2 d γ x v 2 ε .
Inequality (20) gives us
γ 0 γ 0 + log 2 x v | Ψ 2 , c ( x ) | 2 d γ log 4 x · γ 0 γ 0 + log 2 x v | Δ Ψ 2 , c ( x v ) | 2 d γ log 4 x · γ 0 γ 0 + log 2 x v | Φ ( γ , x v , T v ) | 2 d γ + x v 2 c log 4 A + 10 x log 6 x · max 1 T T v γ 0 γ 0 + log 2 x v | Φ 1 ( γ , x v , T ) | 2 d γ + x v 2 c log 4 A + 10 x x 2 ε .
Let γ 0 : = 1 ( 5 + 3 ) / 8 ε , γ i + 1 : = γ i + 1 log 2 x v and A i : = { γ [ γ i , γ i + 1 ] / | Ψ 2 , c ( x ) | > x log A x } . Then by Tschebytschev’s inequality, we obtain
μ ( A i ) x 2 log 2 A x · γ i γ i + 1 | Ψ 2 , c ( x ) | 2 d γ x ε log 2 A x x ε .

Author Contributions

Conceptualization, X.H. and X.Y.; Methodology, D.Z.; Software, X.H.; Validation, X.H., X.Y. and D.Z.; Formal Analysis, X.H.; Investigation, X.H.; Resources, X.Y.; Data Curation, D.Z.; Writing—Original Draft Preparation, X.H.; Writing—Review & Editing, X.Y.; Visualization, D.Z.; Supervision, D.Z.; Project Administration, D.Z.; Funding Acquisition, X.Y. and 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 and No. 11801327), and Natural Science Foundation of Shandong Province (Grant No. ZR201709280100).

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

The author would like to thank the referees for their many useful comments.

Conflicts of Interest

The authors declare no conflict of interest.

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Han, X.; Yan, X.; Zhang, D. On Fourier Coefficients of the Symmetric Square L-Function at Piatetski-Shapiro Prime Twins. Mathematics 2021, 9, 1254. https://doi.org/10.3390/math9111254

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Han X, Yan X, Zhang D. On Fourier Coefficients of the Symmetric Square L-Function at Piatetski-Shapiro Prime Twins. Mathematics. 2021; 9(11):1254. https://doi.org/10.3390/math9111254

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Han, Xue, Xiaofei Yan, and Deyu Zhang. 2021. "On Fourier Coefficients of the Symmetric Square L-Function at Piatetski-Shapiro Prime Twins" Mathematics 9, no. 11: 1254. https://doi.org/10.3390/math9111254

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