Guanghua JI Haiwei SUN

1 Introduction

An important problem in number theory is to study the moments of central values ofL-functions.Many authors considered this problem for several families ofL-functions(see[1,3,6,8,10,16–17]etc).Among them,the family of twistingL-functions has received much attention in recent years.The aim of this paper is to consider the moments ofL-functions attached to the twist of the modular form by Dirichlet characters.

Letf(z)be a holomorphic cusp form of weightκwith respect to the full modular groupSL2(Z).Moreover,we assume thatf(z)is a normalized eigenfunction for all Hecke operators.In this casef(z)has the following Fourier series expansion:

withλf(1)=1.

Such anfis called a holomorphic Hecke eigenform.Associated with each Hecke eigenformf,there exists anL-functionL(s,f),which is defined as

for Res>1.

Letqbe a positive integer andχbe a Dirichlet character moduloq.For Res>1,the automorphicL-functionsL(s,f?χ)are defined by

For any positive real numberk,we define

In this paper,we will estimate the upper and lower bounds ofMk(q,f).

Recently,Heath-Brown[6]proved that

fork=withn∈N,and under the GRH the estimate also holds for all positive real numbers 0

Using the method of[6],we hope to get the same upper bounds forMk(q,f)withunconditionally and 0

where(see Section 2 for the details)

andσ≤1.In[13],Pi considered the last average for any real numberk(see Lemma 2.1).Using this lemma and the upper bounds of corresponding integrals ofMk(q,f),we will prove the following theorem.

Theorem 1.1Forwith n∈N,we have

and the estimate also holds for any real number0

To consider the lower bounds we will use the method of[15].In[15],Rudnick and Soundararajan considered the lower bounds ofMk(q)and proved that unconditionally

for any rational numberk≥1,at least whenqis prime.

In Section 3,we will establish different constitutions of the two sums and use the average estimates of Lemma 2.1,to prove the lower bound ofMk(q,f).But here we can not get unconditionally the lower bounds,since unconditionally the analytic continuation ofL(s,f?χ)kand the estimates of error terms are not good enough(see Section 3).So we assume that the GRH is ture forL(s,f?χ),and under this condition we can get the lower bounds for a more generalk.

Theorem 1.2Let k be a fixed positive real number and q be any large prime,and under GRH for L(s,f?χ),we have

2 Proof of Theorem 1.1

2.1 Introduction

It is known that for a primitiveχ,L(s,f?χ)admits analytic continuation to C as an entire function and satisfies the functional equation(see Proposition 14.20 of[7])

where

is the completeL-function,τ(χ)is the Gauss sum,andηfis the eigenvalue offfor the operatorwith|ηf|=1.In addition,L(s,f?χ)has an Euler product of degree 2,which is

for Res>1,whereαf(p),βf(p)∈C.For Res>1,we have

where

and

For the definition of functiondk(·)we can refer to[5].If we assume the generalized Riemann hypothesis hold,then there exist no zeros forσ>,so that one can define a holomorphic extension of

in the half-planeσ>.

For the proof of Theorem 1.1,we will use the following integral:

where the weight functionW(s)is defined by

withδ>0 to be specified later.In addition toJ(σ,f?χ)we will also consider its average over non-principal characters

In this section,we are mainly to estimate the upper bound ofJ(σ),i.e.,Lemma 2.6.Using this lemma we can prove Theorem 1.1.

2.2 Necessary lemmas

In[13],the author gave the estimate of the average(1.2),which we state as the following lemma whose proof is included for completeness.

Lemma 2.1Let<σ≤1.For any real number k>0we have

and

ProofFirstly,we use the result of[18]to prove the following asymptotic formula:

From[18]we know that if a multiplicative and nonnegative functionλ(n)satisfies the following three conditions:

(i)For some constantτ>0,

(ii)For some constantG>0 and any primep,

(iii)

then

Hence we just need to show thatAf(n)satisfies the last three conditions.

To check the condition(i),we apply(2.3)–(2.4)to see that

where we use Deligne’s estimate(see[2]):

From the following estimate(see[11]or[14])

and the partial summation formulae we have

from which the condition(ii)follows.From(2.4)and Deligne’s estimate we have

and now we can easily check that the condition(iii)is also true:

Then from(2.6)–(2.8)we get the asymptotic formula(2.5).So now we just need to estimate the asymptotic formula.From the above estimates we can see that

Now from(2.5)we get that

with some constants 0

The following lemma is a convexity estimate(see[4]or[6]),which plays an important role in the proof.

Lemma 2.2Let f and g be complex-valued functions which are regular in the strip{s∈C:α<σ<β},and continuous in the closed strip{s=σ+it:α≤σ≤β}.Let b and c be positive real numbers.Suppose that|f(s)|b|g(s)|cand|g(s)|tend to zero as t→∞,uniformly in{s=σ+it:α≤σ≤β}.Set

Then,for α≤γ≤β,

The following lemma gives the upper bounds ofJ(σ).In the proof,we use some properties and a mean value estimate ofL(s,f?χ),which are similar to the case ofL(s,χ).We just give the outline of the proof.For the details one can refer to Lemma 4 in[6].

Lemma 2.3Let≤σ≤1and1?σ≤γ≤σ.We have

ProofBy Lemma 2.2,we get that if≤σ≤and 1?σ≤γ≤σ,

By Hlder’s inequality,we get

Firstly,we estimateJ(1?σ).Following the argument of Lemma 4 in[6],we can get the similar result

with

Note that the exponent ofqis 2k(2σ?1)but notk(2σ?1),becauseL(s,f?χ)is of degree 2 and the Stirling formula gives a doubled exponent.

Then it is sufficient to prove the following inequality:

From Lemma 4 of[6],we know that a mean value estimate is required.The following estimate(see[9])is the result corresponding to[12]for the fourth power moment ofL(s,χ),which is

forindicates that only primitive characters are to be considered.From this mean value estimate we can prove(2.10)following the argument in[6].Therefore we prove that the result is true.

In the following lemma,we will consider two other integrals.For the proof of the unconditional result of Theorem 1.1 we use

forσ>Under the GRH we will employ

forσ>Their averages over non-principal characters are

Note that forH(σ),here we get an upper bound which is different from the result of Lemma 5 in[6].This improvement is the main point to get the unconditional result in our theorem.ForG(σ),the proof is exactly the same as Lemma 5 in[6].So we only give the proof forH(σ)here.

Lemma 2.4LetUnder the GRH we have

and unconditionally

ProofWe only considerH(σ,f?χ).By Lemma 2.2 we have

Therefore

Forwe follow the argument in[6]and get that

Now we estimate By Hlder’s inequality we have

The first integral on the right is triviallyO(q6δ).Moreover,

with certain coefficientsλf(n)n?.The argument then proceeds as[6],noting that

It follows that

Then from(2.11)we deduce that

Now we consider the last integral.Let

where

Let

be the average over non-principal characters.From Lemma 2.1 we can estimate the upper bound ofK(σ).

Lemma 2.5LetWe have

and

ProofBy the definition we get

where

and

Evaluating the sumS(m,n)we find that

By Lemma 2.1 and observing that

we can easily complete the lemma.

From the above results,we can prove our main Lemma.

Lemma 2.6Letwith some constant c>0.For1?σ0≤γ≤σ0we have

ProofBy the definitions ofG(σ,f?χ)andH(σ,f?χ),we have

under the GRH,and

unconditionally.In view of Lemma 2.4 we have

and

respectively.However we also have

and

and therefore

and

in the two cases,respectively.

Using Lemma 2.5,we find that

under the GRH,since

for 0

unconditionally.

Finally we take then apply Lemma 2.3 withγ=and use(2.12)again,to deduce that under GRH

and unconditionally

We are now ready to choose the value ofδ.We writeck,1andck,2for the implied constants in the last two estimates respectively,and note that they depend only onk.

Under the GRH,we take

which ensure that

and hence imply that

Unconditionally,we take

which ensure that

and hence also imply that

At last using Lemma 2.3,we can easily prove our result.

2.3 Proof of the theorem

Now following the argument in[6],we can extract the sumMk(q,f)from the integralJ(γ)and prove Theorem 1.1.Since|L(s,f?χ)|2kis subharmonic we have

We now multiply byrand integrate for 0≤r≤Rto show that

whereD=D(0,R)is the disc of radius ofRabout the origin,and dAis the measure of area.

We take

so that ifz∈DthenandIt follows that

whence

Since Meas(D)(logq)?2we now deduce from Lemma 2.6 that

as required.

3 Proof of Theorem 1.2

In this section,we will give the lower bound ofMk(q,f).Letxbe a small power ofq,and set

We will evaluate

and show thatThen Theorem 1.2 follows from Hlder’s inequality:

We start withS2.By Lemma 2.1 we getand thus

Sincex

Then using Lemma 2.1 we find thatS2φ(q)(logq)k2.

We now turn toS1.If Re(s)>1,integration by parts gives

whereX≥x.To deal with the sum in the second integration we use Perron’s formula and get

forThen we shift the path of integration to Re(s)=σ1>0.SinceL(s,f?χ)khas no pole,we have

where

Note that under GRH we have

where?is a positive number.Hence we have

and

If we takethen we conclude that under GRH

Ifthen(3.1)furnishes an analytic continuation ofThus we have

Therefore

Obviously the main term is

Then using the orthogonality relation for characters we conclude that

By Hlder’s inequality we have

and

So we get

Now we first estimate the contribution of the off-diagonal terms. Here we may writen=m+ql,whereThe contribution of these of f-diagonal terms is

Therefore

From Lemma 2.5,we know that

At last takingX=qandx=we deduce that

This proves the theorem.

AcknowledgementThe authors would like to express their thanks to the referees for many useful suggestions and comments on the manuscript.

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