zero density for automorphic l-functions

Journal of Number Theory 133 (2013) 3877–3901

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Journal of Number Theory

www.elsevier.com/locate/jnt

Zero density for automorphic L-functions

Yangbo Ye a, Deyu Zhang b,∗

a Department of Mathematics, The University of Iowa, Iowa City, IA 52242-1419, USAb School of Mathematical Sciences, Shandong Normal University, Jinan, Shandong 250014, China

a r t i c l e i n f o a b s t r a c t

Article history:Received 12 April 2013Revised 8 May 2013Accepted 15 May 2013Available online 20 July 2013Communicated by David Goss

MSC:11F6611M2611M41

Keywords:Cusp formMaass formSL2(Z)SL3(Z)Riemann zeta functionAutomorphic L-functionZero density

In this paper, zero density estimates for automorphic L-func-tions L(s, π) for GLn are deduced from a bound for an integralpower moment of L(s, π) on the critical line Re(s) = 1/2. Inparticular for the Riemann zeta function, classical zero den-sity estimates are extended to short vertical strips. For g beinga holomorphic or Maass eigenform for SL2(Z), bounds forzero density for L(s, g) in short strips are proved, which ex-tend Ivić’s results on long strips. For a self-dual Hecke Maasseigenform f for SL3(Z), estimates of zero density for L(s, f)in short and long strips are also proved. The proofs use a zerodetecting argument, a large sieve inequality, a bound for anintegral power moment of L(1/2 + it, π), the Rankin–Selbergtheory, and the Halász–Montgomery–Jutila method.

© 2013 Elsevier Inc. All rights reserved.

1. Introduction

Let

L(s, π) =∞∑

n=1

Aπ(n)ns

* Corresponding author.E-mail addresses: [email protected] (Y. Ye), [email protected] (D. Zhang).

0022-314X/$ – see front matter © 2013 Elsevier Inc. All rights reserved.http://dx.doi.org/10.1016/j.jnt.2013.05.012

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http://dx.doi.org/10.1016/j.jnt.2013.05.012

3878 Y. Ye, D. Zhang / Journal of Number Theory 133 (2013) 3877–3901

be an automorphic (finite-part) L-function with an Euler product

L(s, π) =∏p<∞

m∏j=1

(1 − απ(p, j)p−s

)−1 (1.1)

for Re(s) > 1. Here π is an automorphic representation of the group GLn over Q ora finite algebraic number field. Then L(s, π) has analytic continuation to the wholecomplex plane and a standard functional equation. The non-trivial zeros of L(s, π) are allcontained in the critical strip 0 < Re(s) < 1, while the Generalized Riemann Hypothesis(GRH) predicts that they are all on the critical line Re(s) = 1/2. In the absence of aproof of GRH, it is natural to consider how many zeros of the L-function can have offthe critical line.

To this end, define

Nπ(σ, T1, T2) = #{ρ = β + iγ

∣∣ L(ρ, π) = 0, σ < β < 1, T1 � γ � T2},

where 0 � σ < 1 and T1 < T2. We are interested in zero density estimates for L(s, π)in a strip Nπ(σ, T, T + Tα), where 1/2 � σ < 1 and 0 < α � 1. In this paper, we shalldeduce bounds for Nπ(σ, T, T +Tα) from a bound for its 2�th power moment of the type

T+Tα∫T

∣∣∣∣L(

12 + it, π

)∣∣∣∣2�dt �ε,π T θ+ε (1.2)

for certain θ � α.

Theorem 1.1. Let L(s, π) be an automorphic L-function satisfying (1.2) for certain 0 <

α � 1 and θ � α. For 1/2 � σ < 1 and T � 3, we have

Nπ

(σ, T, T + Tα

)� T

2(α�+θ)(1−σ)�−2σ+2 +ε. (1.3)

Theorem 1.2. Let 0 < α � 1, θ � α, � � 1, and T � 3. Assume (1.2). For 3/4 < σ < 1,we have

Nπ

(σ, T, T + Tα

)� T

2α(1−σ)σ +ε if σ(2α�− 2α− θ) � α�− 2α; (1.4)

� T2θ(1−σ)

�(2σ−1)+2(1−σ)+ε if σ(2α�− 2α− θ) < α�− 2α. (1.5)

We note that for � = 1,

(1.4) is valid for 3/4 < σ � α/θ if 3θ < 4α;

(1.5) is valid for max(3/4, α/θ) < σ < 1.

Y. Ye, D. Zhang / Journal of Number Theory 133 (2013) 3877–3901 3879

For � = 2, we have

(1.4) is valid for 3/4 < σ < 1 if θ � 2α;

(1.5) is valid for 3/4 < σ < 1 if θ > 2α.

For � � 3, (1.4) is valid for

max(

34 ,

α(�− 2)2α�− 2α− θ

)� σ < 1

if � � 3, � > θ/α, and � > 1 + θ/(2α), while (1.5) is valid otherwise.

Theorem 1.3. Let 0 < α � 1, θ � α, � � 1, and T � 3. Assume (1.2). For 3/4 < σ < 1,we have

Nπ

(σ, T, T + Tα

)� T

3α(1−σ)3σ−1 +ε if 6σ(α�− 2α− 2θ) � 3α�− 12α− 4θ; (1.6)

� T4θ(1−σ)

�(2σ−1)+4(1−σ)+ε if 6σ(α�− 2α− 2θ) < 3α�− 12α− 4θ. (1.7)

We note that for � = 1, (1.7) is valid for 3/4 < σ < 1. For � = 2, we have

(1.6) is valid for 3/4 < σ � 3α + 2θ6θ if 5θ � 6α;

(1.7) is valid for max(

34 ,

3α + 2θ6θ

)< σ < 1.

For � � 3, similar ranges can be obtained for (1.6) and (1.7).The proofs of these theorems can be applied to other automorphic L-functions, as long

as there hold good bounds for sums of Fourier coefficients similar to those in Lemma 4.1and

∑n�N

∣∣Aπ(n)∣∣2 �π,ε N

1+ε,∑n�N

|Aπ(n)|2n

�π,ε Nε (1.8)

by the Rankin–Selberg theory. These include products of automorphic L-functions forvarious GLn, Rankin–Selberg L-functions, and certain triple product L-functions. In thenext section, we will consider zero density estimates for some specific L-functions.

Estimation of zero density of automorphic L-functions has a long history. Ingham [9]proved in 1940 for the Riemann zeta function that

Nζ(σ,−T, T ) � T3(1−σ)2−σ log5 T (1.9)


for 1/2 � σ < 1. Note that (1.9) with the power of log T replaced by T ε is contained inour Theorem 1.1 for � = 2 and α = θ = 1. Then in 1972 Huxley [7] improved (1.9) for3/4 � σ < 1

Nζ(σ,−T, T ) � T3(1−σ)3σ−1 log44 T. (1.10)

The same bound with the power of log T replaced by T ε can be deduced from (1.6) with� = 6, α = 1 and θ = 2. Similar bounds as those in (1.9) and (1.10) also hold for DirichletL-functions.

For the L-function L(s, g) attached to a holomorphic cusp form g for SL2(Z), Ivić hasshown in [11] that

Ng(σ,−T, T ) � T4(1−σ)3−2σ +ε for 1/2 � σ < 1; (1.11)

� T2−2σ

σ +ε for 3/4 � σ < 1. (1.12)

Luo [19] in 1995 proved an analogue of Selberg’s density theorem

Ng(σ, 0, T ) � T 1−(σ−1/2)/72 log T

for a holomorphic Hecke eigenform for SL2(Z). In 2007, (1.11) was proved for Maassforms for SL2(Z) by Sankaranarayanan and Sengupta [23]. Recently (1.12) was provedfor Maass forms for SL2(Z) by Zhao Xu [26] and Hengcai Tang [25]. We remark that(1.11) is our (1.3), while (1.12) is our (1.4), both for � = 1 and α = θ = 1.

Possible applications of our results may include sums of primes in short intervalsweighted by the Fourier coefficients of the Maass form (cf. Motohashi [21] for an analo-gous case for SL2(Z)). This is the subject matter of a subsequent paper of the authors.

2. Zero density for specific L-functions

In this section we will consider cases where (1.2) is known and prove bounds for zerodensity for the corresponding L-functions.

For the Riemann zeta function, a bound for the fourth power moment on short inter-vals was proved for 7/8 � δ < 1 by Heath-Brown [6] and for 2/3 � δ < 1 by Iwaniec[12], for any ε > 0:

T+T δ∫T

∣∣∣∣ζ(

12 + it

)∣∣∣∣4dt � T δ+ε. (2.1)

Applying Theorems 1.1 and 1.3 to (2.1) with � = 2, α = δ and θ = δ, we obtain

Nζ

(σ, T, T + T δ

)� T

3δ(1−σ)2−σ +ε for 1 � σ < 1;
2


� T3δ(1−σ)3σ−1 +ε for 3

4 < σ � 56 ;

� T 2δ(1−σ)+ε for 56 < σ � 1.

Note that Theorem 1.2 leads to a bigger bound in this case. Comparing these bounds,we get the following theorem.

Theorem 2.1. Let ζ(s) be the Riemann zeta function. For T � 3 and 2/3 � δ < 1, wehave

Nζ

(σ, T, T + T δ

)� T

3δ(1−σ)2−σ +ε for 1

2 � σ � 34 ; (2.2)

� T3δ(1−σ)3σ−1 +ε for 3

4 < σ � 56 ; (2.3)

� T 2δ(1−σ)+ε for 56 < σ � 1. (2.4)

If δ = 1, we note that the bound in (2.2) is the result (1.9) of Ingham [9] and thebound in (2.3) recovers the result (1.10) of Huxley [7]. Also the bound in (2.4) tells usthat the zero density conjecture holds for 5/6 < σ � 1.

Let g be a self-dual holomorphic or Maass Hecke eigenform for Γ0(N), and χ a real,primitive character mod Q with N |Q. In Ye [27] and Lau, Liu and Ye [17] the followingbound for the integral 4th power moment of L(s, g ⊗ χ) is proved

T+T δ∫T

∣∣∣∣L(

12 + it, g ⊗ χ

)∣∣∣∣4dt � T 1+δ+ε (2.5)

for 1/3 < δ < 1. Applying Theorems 1.1 and 1.2 to (2.5) with � = 2, α = δ and θ = 1+δ,we have θ > 2α. Consequently

Ng⊗χ

(σ, T, T + T δ

)� T

(1+3δ)(1−σ)2−σ +ε for 1

2 � σ < 1;

� T(1+δ)(1−σ)

σ +ε for 34 < σ < 1.

Comparing these two bounds, we proved

Theorem 2.2. Let g be a self-dual holomorphic or Maass Hecke eigenform for Γ0(N), andχ a real, primitive character mod Q with N |Q. For 1/3 < δ < 1 and T � 3, we have

Ng⊗χ

(σ, T, T + T δ

)� T

(1+3δ)(1−σ)2−σ +ε

for 12 � σ <

1 + δ

1 + 2δ if 1/3 < δ � 1/2

or for 1/2 � σ < 3/4 if 1/2 < δ < 1;

(2.6)


� T(1+δ)(1−σ)

σ +ε

for 1 + δ

1 + 2δ � σ < 1 if 1/3 < δ � 1/2

or for 3/4 � σ < 1 if 1/2 < δ < 1.

(2.7)

We also note that the bound in (2.7) exactly extends Ivić’s result in (1.12) to shortstrips.

Now let f be a self-dual Hecke Maass eigenform for the group SL3(Z) of type ν =(ν1, ν2). Then

μf (1) = ν1 + 2ν2 − 1, μf (2) = ν1 − ν2, μf (3) = 1 − 2ν1 − ν2

are the Langlands’ parameters for f . Recall that f has a Fourier Whittaker expansion(cf. Goldfeld [3])

f(z) =∑

γ∈U2(Z)\SL2(Z)

∑m1�1

∑m2 �=0

Af (m1,m2)m1|m2|

WJ

(M

(γ 00 1

)z, ν, ψ1,1

).

Here U2 = {( 1 ∗

0 1

)}, WJ(z, ν, ψ1,1) is the Jacquet–Whittaker function, ψ1,1 is a character

of U3(R),

M = diag(m1|m2|,m1, 1),

and Af (m1,m2) are Fourier coefficients of f . Note that WJ(z, ν, ψ1,1) represents anexponential decay in y1 and y2 for

z =

⎛⎜⎝

1 x12 x13

1 x23

1

⎞⎟⎠

⎛⎜⎝

y1y2

y1

1

⎞⎟⎠ .

It is known that (cf. Kim and Sarnak [16] and Sarnak [24] (21))

Af (m,n) � |mn|5/14+ε.

The Rankin–Selberg theory (cf. [4]) reveals that∑

mn2�N

∣∣Af (m,n)∣∣2 �f N.

Since Af (m2,m1) = Af̃ (m1,m2), where f̃ is the contragredient form of f , there alsoholds

∑2

∣∣Af (m,n)∣∣2 �f N.

m n�N


These estimates lead to

∑m�N

|Af (m, 1)|2m

� logN,∑n�N

|Af (1, n)|2n

� logN.

Let

L(s, f) =∞∑

n=1

Af (1, n)ns

, for Re(s) > 1,

be the Godement–Jacquet L-function attached to f (Godement and Jacquet [2] andGoldfeld [3]). It has a standard functional equation and analytic continuation to an entirefunction on C. Since f is a Hecke eigenform, the Fourier coefficients are multiplicative,and the L-function has an Euler product ([3], pp. 173–174)

L(s, f) =∏p

(1 −Af (1, p)p−s + Af (p, 1)p−2s − p−3s)−1 (2.8)

for Re(s) > 1.Xiaoqing Li [18] proved (1.2) for L(s, f), where f is a fixed self-dual Hecke Maass

eigenform for SL3(Z)

T+T δ∫T

∣∣∣∣L(

12 + it, f

)∣∣∣∣2dt �ε,f T 1+δ+ε (2.9)

for 3/8 < δ � 1/2. For 1/2 < δ � 1, we use a subdivision to decompose [T, T + T δ] intosegments of the form [K,K +

√K]. Then

T+T δ∫T

∣∣∣∣L(

12 + it, f

)∣∣∣∣2dt � T δ

T 1/2 maxT�K�T+T δ

K+√K∫

K

∣∣∣∣L(

12 + it, f

)∣∣∣∣2dt

� T δ−1/2T 3/2+ε

� T 1+δ+ε.

Thus (2.9) holds for 3/8 < δ � 1. Applying Theorems 1.1 and 1.2 to (2.9) with � = 1,α = δ and θ = 1 + δ, we have the following results on its zero density,

Nf

(σ, T, T + T δ

)� T

(2+4δ)(1−σ)3−2σ +ε for 1/2 � σ < 1;

� T 2(1+δ)(1−σ)+ε for 3/4 < σ < 1.

Comparing these bounds, we get the following theorem.


Theorem 2.3. Let f be a self-dual Hecke Maass eigenform for SL3(Z). For 3/8 < δ � 1and T � 3, we have

Nf

(σ, T, T + T δ

)� T

(2+4δ)(1−σ)3−2σ +ε for 1/2 � σ <

2 + δ

2 + 2δ ; (2.10)

� T 2(1+δ)(1−σ)+ε for 2 + δ

2 + 2δ � σ < 1. (2.11)

3. Preliminary lemmas

In the following sections, we will need the following results.

Lemma 3.1. (See Ivić [10], Theorem 5.3.) Let T � t1, t2, . . . , tR � T +Tα be real numberssuch that |tr − ts| � 1 for r �= s � R and let a1, . . . , aN be arbitrary complex numbers.Then

∑r�R

∣∣∣∣∑n�N

ann−itr

∣∣∣∣2�

(Tα

∑n�N

|an|2 +∑n�N

n|an|2)

logN. (3.1)

Lemma 3.2. Suppose (1.2) holds for some 0 < α � 1 and θ � α. Then we have

T+Tα∫T

∣∣∣∣L′(

12 + it, π

)∣∣∣∣2�dt �ε,π T θ+ε.

Proof. It can be derived from (1.2) by the same argument as in the proof of Corollary 1of Kamiya [14]. �Lemma 3.3. Assume (1.2). Let t1, t2, . . . , tR be a sequence of real numbers satisfyingT � tr � T +Tα for each r. Suppose that the sequence {tr} is log2(T ) well-spaced. Then

R∑r=1

∣∣∣∣L(

12 + itr, π

)∣∣∣∣2�

� T θ+ε.

Proof. Recall a well-known estimate

∣∣f(x)∣∣ � 1

b− a

b∫a

∣∣f(u)∣∣ du +

b∫a

∣∣f ′(u)∣∣ du,

for a � x � b. In the above formula, let us take

f(t) = L

(1 + it, π

)2�, a = tr−1, b = tr.
2


Summing over r, 1 � r � R, we get

R∑r=1

∣∣∣∣L(

12 + itr, π

)∣∣∣∣2�

� (log T )−2T+Tα∫T

∣∣∣∣L(

12 + it, π

)∣∣∣∣2�dt

+T+Tα∫T

∣∣∣∣L(

12 + it, π

)∣∣∣∣2�−1∣∣∣∣L′

(12 + it, π

)∣∣∣∣ dt.The first term on the right side is bounded by T θ+ε by (1.2). By Hölder’s inequality,(1.2) and Lemma 3.2, we get

T+Tα∫T

∣∣∣∣L(

12 + it, π

)∣∣∣∣2�−1∣∣∣∣L′

(12 + it, π

)∣∣∣∣ dt

�( T+Tα∫

T

∣∣∣∣L(

12 + it, π

)∣∣∣∣2�dt

) 2�−12�

( T+Tα∫T

∣∣∣∣L′(

12 + it, π

)∣∣∣∣2�dt

) 12�

� T θ+ε.

Lemma 3.3 thus follows. �Lemma 3.4. For L(s, π) we have

Nπ(0, T, T + 1) = m

2π log T(1 + o(1)

).

Proof. By a standard winding number argument on (1.2) as in Davenport [1] and Rud-nick and Sarnak [22]. �Lemma 3.5. Let s be a complex variable. Define

F (s) =2M∑M+1

a(n)n−s, G =2M∑M+1

∣∣a(n)∣∣2.

Suppose that for s = s1, s2, . . . , sR,

∣∣F (s)∣∣ =

∣∣∣∣2M∑M+1

a(n)n−s

∣∣∣∣ � V,

where sr = σr + itr are complex numbers with

T � tr � T + Tα, and |tr − tr′ | � 1 for r �= r′.


Then

R � GV −2M log2 l + G3V −6MTαl11

provided that

V > c G1/2M1/4l2,

where c is an absolute constant and l = log T .

Proof. This is Theorem 1 of Huxley [8] with Q = 1 using the Halász–Montgomery–Jutilamethod [5,20,13]. �4. Proof of Theorem 1.1

We will estimate the number of zeros ρ = β + iγ of L(s, π) in the rectangle

σ < β < 1, T � γ � T + Tα, (4.1)

where σ � 1/2, T � 3 and 0 < α � 1. We define μπ(n) by

1L(s, π) =

∞∑n=1

μπ(n)ns

for Re(s) > 1. By the Euler product of L(s, π) in (1.1), we have

1L(s, π) =

∏p

m∏j=1

(1 − απ(p, j)p−s

).

Consequently μπ(n) is a multiplicative function. We also know that μπ(1) = 1,

μπ

(pk)

= (−1)k∑

1�j1<···<jk�m

k∏ν=1

απ(p, jν)

for k = 1, . . . ,m, and μπ(pk) = 0 for k > m. Then

∑d|n

μπ(d)Aπ

(n

d

)=

{1, if n = 1;0, otherwise.

Lemma 4.1. With notation as above,

∑n�N

∣∣μπ(n)∣∣2 �π,ε N

1+ε,∑n�N

|μπ(n)|2n

�π,ε Nε.


Proof. By the Rankin–Selberg theory,

L(s, π × π̃) =∏p

m∏j=1

m∏i=1

(1 − απ(p, j)απ(p, i)p−s

)−1

converges absolutely for Re(s) > 1. Therefore

∏p

m∏j=1

m∏i=1

(1 +

∣∣απ(p, j)∣∣∣∣απ(p, i)

∣∣p−σ)

=∏p

(1 +

m2∑k=1

1pkσ

∑ k∏ν=1

∣∣απ(p, jν)∣∣∣∣απ(p, iν)

∣∣) (4.2)

converges for σ > 1, where the inner sum on the right side is taken over choices of k

ordered pairs (jν , iν), ν = 1, . . . , k from the set {(j, i) | 1 � j � m, 1 � i � m}. As apartial sum of (4.2) also converges,

∏p

(1 +

m∑k=1

μπ(pk)pks

)=

∏p

(1 +

m∑k=1

1pks

∑1�j1<···<jk�m1�i1<···<ik�m

k∏ν=1

απ(p, jν)απ(p, iν))

converges absolutely for Re(s) > 1. Then Lemma 4.1 follows from a standard Tauberianargument. �

Define

MX(s, π) =∑n�X

μπ(n)ns

.

Set s = σ + it, T � t � T + Tα, 1 � X � Y � TA for some A > 0, X = X(T ), andY = Y (T ), which are parameters to be chosen later. Then

L(s, π)MX(s, π) =∞∑

n=1

bπ(n)ns

, (4.3)

for Re(s) > 1, where

bπ(n) =

⎧⎪⎪⎨⎪⎪⎩

1, if n = 1;0, if 2 � n � X;∑

d|nd�X

μπ(d)Aπ(nd ), if n > X.

Note that |bπ(n)| � n1/2+ε for n > X by the trivial bound for απ(p, j). By Cauchy’sinequality we have


∑n�N

∣∣bπ(n)∣∣2 �

∑n�N

τ(n)∑

n=md

∣∣μπ(d)Aπ(m)∣∣2

� Nε∑d�N

∣∣μπ(d)∣∣2 ∑

m�Nd

∣∣Aπ(m)∣∣2,

where τ(n) � nε is the divisor function. Using (1.6) and Lemma 4.1 it follows that

∑n�N

∣∣bπ(n)∣∣2 � N1+ε

∑d�N

|μπ(d)|2d

� N1+ε. (4.4)

Back to (4.3), for σ > 1 and Re(s) > 0, we get

12πi

∫(σ)

∞∑n=1

bπ(n)ns+w

Γ (w)Y w dw

= 12πi

∫(σ)

L(s + w, π)MX(s + w, π)Γ (w)Y w dw.

By the inverse Mellin transform of Γ (w), the left hand side of the above formula is

12πi

∫(σ)

Γ (w)Y w dw +∑n>X

bπ(n)ns

12πi

∫(σ)

Γ (w)(Y

n

)w

dw

= e−1/Y +∑n>X

bπ(n)ns

e−n/Y .

Hence we get

e−1/Y +∑n>X

bπ(n)ns

e−n/Y = 12πi

∫(σ)

L(s + w, π)MX(s + w, π)Γ (w)Y w dw (4.5)

for σ > 1 and Re(s) > 0. If we move the integration contour to the line Re(w) =1/2 − β < 0, then the integral on the right hand of (4.5) becomes

L(s, π)MX(s, π) + 12πi

∫(1/2−β)

L(s + w, π)MX(s + w, π)Γ (w)Y w dw, (4.6)

where L(s, π)MX(s, π) is the residue of the integrand at w = 0. Taking s equal to anon-trivial zero ρ = β + iγ and setting w = 1/2 − β + iv in (4.5) and (4.6), one has


e−1/Y = −∑n>X

bπ(n)nρ

e−n/Y

+ 12π

+∞∫−∞

L

(12 + iγ + iv, π

)MX

(12 + iγ + iv, π

)

× Γ

(12 − β + iv

)Y 1/2−β+iv dv. (4.7)

Here the absolute convergence of the series is guaranteed by bπ(n) � n1/2+ε. The ab-solute convergence of the integral is given by the convexity bound for L(s + w, π) onIm(w) = v, a trivial bound for MX(s+w, π), and Γ (w) � e−π|v|/2 by Stirling’s formula.

Since |e−1/Y − 1| � 1/3 in (4.7) for Y � 3, a non-trivial zero ρ = β + iγ counted inNπ(σ, T, T + T θ) satisfies either

1 �∣∣∣∣∑n>X

bπ(n)nρ

e−n/Y

∣∣∣∣, (4.8)

or

1 �∣∣∣∣∣

∞∫−∞

L

(12 + iγ + iv, π

)MX

(12 + iγ + iv, π

)

× Γ

(12 − β + iv

)Y 1/2−β+iv dv

∣∣∣∣∣. (4.9)

By the absolute convergence of the integral in (4.9), we can replace it by the same integralfrom − log2 T to log2 T , because

− log2 T∫−∞

++∞∫

log2 T

= o(1).

By Stirling’s formula again, we may further remove the Γ -function from the integrandto get

∣∣∣∣∣log2 T∫

− log2 T

L

(12 + iγ + iv, π

)MX

(12 + iγ + iv, π

)Γ

(12 − β + iv

)Y 1/2−β+iv dv

∣∣∣∣∣

�log2 T∫

2

∣∣∣∣L(

12 + iγ + iv, π

)MX

(12 + iγ + iv, π

)Y 1/2−β+iv

∣∣∣∣ dv.
− log T


Therefore a zero ρ in (4.7) satisfies either (4.8) or

1 � Y 1/2−β

log2 T∫− log2 T

∣∣∣∣L(

12 + iγ + iv, π

)MX

(12 + iγ + iv, π

)∣∣∣∣ dv. (4.10)

Denote by G1 and G2 the numbers of zeros counted in Nπ(σ, T, T + Tα) which satisfy(4.8) and (4.10), respectively. Then

Nπ

(σ, T, T + Tα

)� G1 + G2. (4.11)

Now we divide the short vertical strip (4.1) into consecutive rectangles of height3 log2 T , starting from

σ < β < 1, T � γ < T + 3 log2 T, (4.12)

with the top one being possibly of a shorter height. Number these rectangles startingfrom the first rectangle (4.12). Denote by Gij the number of zeros in all odd-numberedrectangles if j = 1, or in all even-numbered rectangles if j = 2, which satisfy (4.8) ifi = 1, or (4.10) if i = 2. Then (4.11) becomes

Nπ

(σ, T, T + Tα

)� G11 + G12 + G21 + G22. (4.13)

Let’s consider all the zeros counted in Gij . In each rectangle which contains at least onezero, we choose a zero. Then we get a sequence of zeros ρ(ij)

r = β(ij)r +iγ

(ij)r counted in Gij

for r = 1, . . . , Rij . Here Rij is the number of rectangles which contain at least one zerocounted in Gij . By Lemma 3.4, each rectangle contains at most (3m/(2π)) log3 T (1+o(1))zeros. Consequently

Nπ

(σ, T, T + Tα

)� (R11 + R12 + R21 + R22) log3 T. (4.14)

From the construction, we know that zeros counted in Rij are 3 log2 T well-spaced, i.e.,the imaginary parts of any two zeros are different by at least 3 log2 T .

The terms on the left side of (4.8) with n > Y log2 Y contribute o(1) as Y → ∞. Theneach zero ρ

(1j)r = β

(1j)r + iγ

(1j)r counted in R1j , j = 1, 2, satisfies

1 �∣∣∣∣ ∑X<n�Y log2 Y

bπ(n)nβ

(1j)r +iγ

(1j)r

e−n/Y

∣∣∣∣.

Recall that σ < β(1j)r . By partial summation, we have

1 �∣∣∣∣ ∑

2

bπ(n)nβ

(1j)r +iγ

(1j)r

e−n/Y

∣∣∣∣
X<n�Y log Y


�(Y log2 Y

)σ−β(1j)r

∣∣∣∣ ∑X<n�Y log2 Y

bπ(n)nσ+iγ

(1j)r

e−n/Y

∣∣∣∣

+Y log2 Y∫

X

∣∣∣∣ ∑X<n�u

bπ(n)nσ+iγ

(1j)r

e−n/Y

∣∣∣∣uσ−β(1j)r −1 du. (4.15)

Squaring (4.15) we get

1 �∣∣∣∣ ∑X<n�Y log2 Y

bπ(n)nσ+iγ

(1j)r

e−n/Y

∣∣∣∣2

+( Y log2 Y∫

X

∣∣∣∣ ∑X<n�u

bπ(n)nσ+iγ

(1j)r

e−n/Y

∣∣∣∣duu)2

�∣∣∣∣ ∑X<n�Y log2 Y

bπ(n)nσ+iγ

(1j)r

e−n/Y

∣∣∣∣2

+ log(Y log2 Y

) Y log2 Y∫X

∣∣∣∣ ∑X<n�u

bπ(n)nσ+iγ

(1j)r

e−n/Y

∣∣∣∣2du

u(4.16)

by Cauchy’s inequality.Adding (4.16) for all zeros ρ

(1j)r = β

(1j)r + iγ

(1j)r counted in R1j , we get

R1j �R1j∑r=1

∣∣∣∣ ∑X<n�Y log2 Y

bπ(n)nσ+iγ

(1j)r

e−n/Y

∣∣∣∣2

+ log(Y log2 Y

) Y log2 Y∫X

R1j∑r=1

∣∣∣∣ ∑X<n�u

bπ(n)nσ+iγ

(1j)r

e−n/Y

∣∣∣∣2du

u. (4.17)

Now (4.17) can be reduced to bounding

R1j∑r=1

∣∣∣∣ ∑X<n�M

bπ(n)nσ+iγ

(1j)r

e−n/Y

∣∣∣∣2, (4.18)

for M satisfying X < M � Y log2 Y . It follows from Lemma 3.1 that (4.18) is boundedby

O

((Tα

∑ |bπ(n)|2n2σ +

∑ |bπ(n)|2n2σ−1

)e−X/Y logM

).

X<n�M X<n�M


By (4.4) and summation by parts, we get

R1j∑r=1

∣∣∣∣ ∑X<n�M

bπ(n)nσ+iγ

(1j)r

e−n/Y

∣∣∣∣2�

(TαM1−2σ + M2−2σ) logM

�(TαX1−2σ + Y 2−2σ)T ε. (4.19)

Substituting (4.19) into (4.17), we have

R1j �(TαX1−2σ + Y 2−2σ)T ε +

(TαX1−2σ + Y 2−2σ)T ε

Y log2 Y∫X

du

u

�(TαX1−2σ + Y 2−2σ)T ε. (4.20)

Now let us turn to (4.10) for a zero ρ(2j)r = β

(2j)r + iγ

(2j)r , j = 1, 2:

1 � Y 1/2−β(2j)r

log2 T∫− log2 T

∣∣∣∣L(

12 + iγ(2j)

r + iv, π

)MX

(12 + iγ(2j)

r + iv, π

)∣∣∣∣ dv. (4.21)

By the mean value theorem for integration, for r = 1, . . . , R2j there exists t(2j)r = γ(2j)r +v

for some v with − log2 T < v < log2 T such that the integral in (4.21) is equal to

2(log T )2∣∣∣∣L

(12 + it(2j)r , π

)MX

(12 + it(2j)r , π

)∣∣∣∣.Thus (4.21) becomes

1 � Y 1/2−β(2j)r log2 T

∣∣∣∣L(

12 + it(2j)r , π

)MX

(12 + it(2j)r , π

)∣∣∣∣. (4.22)

Raising (4.22) to the power 2�/(� + 1) and adding over all zeros counted in R2j , we get

R2jY2�(σ−1/2)

�+1 T−ε

�R2j∑r=1

∣∣∣∣L(

12 + it(2j)r , π

)MX

(12 + it(2j)r , π

)∣∣∣∣2�

�+1

�( R2j∑

r=1

∣∣∣∣L(

12 + it(2j)r , π

)∣∣∣∣2�) 1

�+1( R2j∑

r=1

∣∣∣∣MX

(12 + it(2j)r , π

)∣∣∣∣2) �

�+1

, (4.23)

by Hölder’s inequality.


From our construction of the zeros counted in R2j , j = 1, 2, the sequence {t(2j)r } islog2 T well-spaced. By Lemma 3.3, the first sum on the right side of (4.23) is boundedby O(T θ+ε). By Lemmas 3.1 and 4.1, the second sum on the right side of (4.23) is

� TαX∑r=1

|μπ(n)|2n

+X∑r=1

∣∣μπ(n)∣∣2 � Tα+ε + X1+ε.

Back to (4.23), now we have

R2j � Y�(1−2σ)

�+1(T

α�+θ�+1 +ε + T

θ�+1+εX

��+1+ε

). (4.24)

Collecting results from (4.20) and (4.24), we deduce from (4.14) that

Nπ

(σ, T, T + Tα

)�

(TαX1−2σ + Y 2−2σ + Y

�(1−2σ)�+1 T

α�+θ�+1 + Y

�(1−2σ)�+1 T

θ�+1X

��+1

)T ε.

Choosing Y = Tα�+θ

�−2σ+2 , we get

Nπ

(σ, T, T + Tα

)�

(TαX1−2σ + T

2(α�+θ)(1−σ)�−2σ+2 + T

�(α�+θ)(1−2σ)(�+1)(�−2σ+2)+

θ�+1X

��+1

)T ε. (4.25)

Now we take X = Tα in (4.25) to get

Nπ

(σ, T, T + Tα

)�

(T 2α(1−σ) + T

2(α�+θ)(1−σ)�−2σ+2

)T ε. (4.26)

We want to prove that the first term in (4.26) is less than or equal to the second term,or equivalently,

2α(1 − σ)(�− 2σ + 2) � 2(α� + θ)(1 − σ). (4.27)

In fact, after canceling 1 − σ from both sides of (4.27), this inequality follows from thefact that α � θ and 1/2 � σ < 1. The inequality (4.27) also shows that X � Y for ourchoice of X and Y . Therefore, (4.27) implies

Nπ

(σ, T, T + Tα

)� T

2(α�+θ)(1−σ)�−2σ+2 +ε

which proves Theorem 1.1. �


5. Proof of Theorem 1.2

To improve our result for σ close to 1, we use the Halász–Montgomery method (Halász[5] and Montgomery [20]) to bound R1j , j = 1, 2. It turns out that for σ close to 1, thisapproach can lead to a better bound for the number of zeros.

Recall that the right side of (4.8) with n > Y log2 Y contributes o(1) as Y → ∞,because of the exponential decay of e−n/Y . Therefore for any zero ρ

(1j)r = β

(1j)r + iγ

(1j)r

counted in R1j , (4.8) can be written as

1 �∣∣∣∣ ∑X<n�Y log2 Y

bπ(n)nρ

(1j)r

e−n/Y

∣∣∣∣. (5.1)

We will use an argument below to shorten the summation in (5.1) to√Y � n � Y 1+ε

by possibly modifying bπ(n). This way we will be able to take X = T ε which will leadto a saving on R2j .

By a dyadic subdivision on the sum in (5.1) we get

1 � · · · +∣∣∣∣ ∑√Y /2�n<

√Y

bπ(n)nρ

(1j)r

e−n/Y

∣∣∣∣ +∣∣∣∣ ∑√Y�n<2

√Y

bπ(n)nρ

(1j)r

e−n/Y

∣∣∣∣ + · · · , (5.2)

where there are O(log(Y log2 Y )) terms. Then there exists Mr = 2νr√Y for some νr ∈ Z

such that

1 � log(Y log2 Y

)∣∣∣∣ ∑Mr�n<2Mr

bπ(n)nρ

(1j)r

e−n/Y

∣∣∣∣. (5.3)

Note that Mr may depend on the zero ρ(1j)r , because for different ρ

(1j)r , different terms

in (5.2) may dominate.For a given ρ

(1j)r , if Mr �

√Y , we set cπ(n) = bπ(n)e−n/Y . If X < Mr <

√Y , then

we can choose an integer k � 2 such that√Y � Mk

r < Y . We observe that

k <log YlogX . (5.4)

Now we raise (5.3) to the kth power to get

1 � logk(Y log2 Y

)∣∣∣∣ ∑Mk

r �n<(2Mr)k

cπ(n)nρ

(1j)r

∣∣∣∣, (5.5)

where

cπ(n) =∑

bπ(n1) · · · bπ(nk)e−n1/Y · · · e−nk/Y . (5.6)
n=n1···nk


We remark that for n > Y log2 Y , cπ(n) may lose the exponential decay exhibited inbπ(n), because e−n1/Y · · · e−nk/Y may be close to 1. By (5.4), however,

(2Mr)k < 2log Y/ log XY = Y log 2/ log XY.

Since logX = ε log T , we know that (2Mr)k < Y 1+ε.We now split the sum in (5.5) by a dyadic subdivision

1 � logk(Y log2 Y

)×

(· · · +

∣∣∣∣ ∑2i−1

√Y�n<2i

√Y

cπ(n)nρ

(1j)r

∣∣∣∣ +∣∣∣∣ ∑2i

√Y�n<2i+1

√Y

cπ(n)nρ

(1j)r

∣∣∣∣ + · · ·). (5.7)

By the argument above, the sums in (5.7) are within the range from√Y to Y 1+ε. Thus

there are O(log Y ) terms in (5.7). By adding possible non-negative terms to (5.7), we get

1 � logk(Y log2 Y

)(∣∣∣∣ ∑√Y�n<2

√Y

cπ(n)nρ

(1j)r

∣∣∣∣ +∣∣∣∣ ∑2√Y�n<22

√Y

cπ(n)nρ

(1j)r

∣∣∣∣ + · · ·)

(5.8)

with a dyadic subdivision on the right side to cover the whole interval of [√Y , Y 1+ε].

Based on (5.4), denote

C =[

log YlogX

]. (5.9)

We note that since X = T ε and Y � TA, C � A/ε. Then we can combine the cases in(5.3) and (5.8) into one formula

1 � logC(Y log2 Y

)(∣∣∣∣ ∑√Y�n<2

√Y

c(r)π (n)nρ

(1j)r

∣∣∣∣ +∣∣∣∣ ∑2√Y�n<22

√Y

c(r)π (n)nρ

(1j)r

∣∣∣∣ + · · ·)

(5.10)

to cover the whole interval of [√Y , Y 1+ε], where c

(r)π (n) = cπ(n) as in (5.6) may depend

on the zero ρ(1j)r with 1 � k � C.

Choosing the largest term from (5.10), we arrive at

1 � logC+1(Y log2 Y)∣∣∣∣ ∑

Mr�n<2Mr

c(r)π (n)nρ

(1j)r

∣∣∣∣, (5.11)

for some Mr in√Y � Mr � Y 1+ε. Note that this Mr may depend on the zero ρ

(1j)r . To

take out the dependence of c(r)π (n) on the zero ρ(1j)r , we add more terms to (5.11) to get

1 �C∑

logC+1(Y log2 Y)∣∣∣∣ ∑ c

[k]π (n)nρ

(1j)r

∣∣∣∣, (5.12)
k=1 Mr�n<2Mr


for Mr in√Y � Mr � Y 1+ε, where c

[k]π (n) is defined to equal the right side of (5.6) for

any specific k within 1 � k � C. If for all 1 � k � C,

∣∣∣∣ ∑Mr�n<2Mr

c[k]π (n)nρ

(1j)r

∣∣∣∣ � 1C

log−(C+2)(Y log2 Y)

for all 1 � k � C, then the right hand side of (5.12) would be � log−1(Y log2 Y ), andhence (5.12) could not hold. Therefore, there is at least one k such that

1C

log−C−2(Y log2 Y)�

∣∣∣∣ ∑Mr�n<2Mr

c[k]π (n)nρ

(1j)r

∣∣∣∣.For any

√Y � M � Y 1+ε and 1 � k � C, let R∗

1j(M,k) be the number of zeros ρ(1j)

such that

1C

log−C−2(Y log2 Y)�

∣∣∣∣ ∑M�n<2M

c[k]π (n)nρ(1j)

∣∣∣∣. (5.13)

Then we have

R1j �∑M

C∑k=1

R∗1j(M,k)

� C(log Y ) max√Y�M�Y 1+ε

1�k�C

R∗1j(M,k). (5.14)

Now we use Lemma 3.5 to estimate R∗1j(M,k) by setting

F (s) =∑

M�n<2M

c[k]π (n)ns

,

V = 1C

log−C−2(Y log2 Y),

G =∑

M�n<2M

|c[k]π (n)|2

n2β(1j)r

. (5.15)

It follows from partial summation on (5.15) that

G �∑

M�n<2M

|c[k]π (n)|2n2σ M2σ−2β(1j)

r +2M∫M

∑M�n<u

|c[k]π (n)|2n2σ u2σ−2β(1j)

rdu

u

�∑

M�n<2M

|c[k]π (n)|2n2σ +

2M∫ ∑M�n<u

|c[k]π (n)|2n2σ

du

u, (5.16)

M


if we notice that β(1j)r � σ. Thus it suffices to bound

∑M�n<u

|c[k]π (n)|2n2σ ,

for M � u � 2M .By Cauchy–Schwarz’s inequality, we have

∑M�n<u

|c[k]π (n)|2n2σ � M−2σ

∑M�n<u

τk(n)∑

n=n1···nk

∣∣bπ(n1)∣∣2 · · · ∣∣bπ(nk)

∣∣2� M−2σ+ε

∑n1�u

∣∣bπ(n1)∣∣2 · · · ∑

nk�u/(n1···nk−1)

∣∣bπ(nk)∣∣2

� M1−2σ+ε,

where we have used (4.4). Back to (5.16) we get

G � M1−2σ+ε � Y 1−2σ+ε.

If 3/4 < σ < 1, the condition

V > c G1/2M1/4l2

can be satisfied. Then by Lemma 3.5 we have

R∗1j � M2−2σ+ε + TαM4−6σ+ε. (5.17)

Recall that√Y � M � Y 1+ε. Then from (5.9) and (5.14) we have

R1j � C(log Y )R∗1j � Y 2−2σ+ε + TαY 2−3σ+ε, (5.18)

for 3/4 < σ < 1.The above arguments allowed us to choose X freely. To estimate R2j , j = 1, 2, we

take X = T ε, so that trivially

MX

(12 + iγ + iv, π

)� T ε (5.19)

for |v| � log2 T . Following our arguments in Section 3, for r = 1, 2, . . . , R2j we chooset(2j)r as before. From (4.22) we get by (5.19) that

1 � T εY 1/2−σ

∣∣∣∣L(

1 + itr, π

)∣∣∣∣. (5.20)
2


Taking the 2�th power of (5.20) and summing it over all zeros ρ(2j)r , we get

R2j � T εY �(1−2σ)R2j∑r=1

∣∣∣∣L(

12 + itr, π

)∣∣∣∣2�

� T θ+εY �(1−2σ) (5.21)

by Lemma 3.3.Comparing (5.18) and (5.21), we get from (4.14) that for 3/4 � σ < 1

Nf

(σ, T, T + Tα

)� (R11 + R12 + R21 + R22) log3 T

� T ε(Y 2−2σ + TαY 2−3σ + T θY �(1−2σ)). (5.22)

Firstly, we set

Y 2−2σ = TαY 2−3σ,

which is equivalent to

Y = Tα/σ.

Thus (5.22) becomes

Nf

(σ, T, T + Tα

)� T ε

(T

2α(1−σ)σ + T θ+α�(1−2σ)

σ

). (5.23)

We compare the two terms in (5.23) to get

Nf

(σ, T, T + Tα

)� T

2α(1−σ)σ +ε, (5.24)

holds for

σ(2α�− 2α− θ) � α�− 2α.

Secondly, we set

Y 2−2σ = T θY �(1−2σ).

Then we have

Y = Tθ

�(2σ−1)+2(1−σ) .

Note that �(2σ− 1) + 2(1−σ) is a linear function of σ with values equal to 1 at σ = 1/2and equal to � at σ = 1. Now we have

Nf

(σ, T, T + Tα

)� T ε

(T

2θ(1−σ)�(2σ−1)+2(1−σ) + Tα+ θ(2−3σ)

�(2σ−1)+2(1−σ)). (5.25)



Nf

(σ, T, T + Tα

)� T

2θ(1−σ)�(2σ−1)+2(1−σ)+ε, (5.26)

holds for

σ(2α�− 2α− θ) < α�− 2α.

The proof of Theorem 1.2 is thus complete. �6. Proof of Theorem 1.3

The method for proving Theorem 1.3 is similar to the proof of Theorem 1.2. The maindifference is the range of M .

We raise (5.3) to the kth power to get (5.5) and choose k such that

Mkr � Y 2 log4 Y < Mk+1

r .

Then we have

T 4/3 log8/3 Y � Mkr � Y 2 log4 Y. (6.1)

Then we split the sum in (5.5) by a dyadic subdivision as in (5.7). By similar argumentswe can get (5.17) for T 4/3 log8/3 Y � M � Y 2+ε. Then we have

R1j � C(log Y )R∗1j � Y 4−4σ+ε + TαY

16−24σ3 +ε, (6.2)

for 3/4 < σ < 1.Comparing (6.2) and (5.21), we get from (4.14) that for 3/4 � σ < 1

Nf

(σ, T, T + Tα

)� (R11 + R12 + R21 + R22) log3 T

� T ε(Y 4−4σ + TαY

16−24σ3 + T θY �(1−2σ)). (6.3)

Firstly, we set

Y 4−4σ = TαY16−24σ

3 ,


Y = T3α

12σ−4 .

Thus (6.3) becomes

Nf

(σ, T, T + Tα

)� T ε

(T

3α(1−σ)3σ−1 + T θ+ 3α�(1−2σ)

12σ−4). (6.4)



Nf

(σ, T, T + Tα

)� T

3α(1−σ)3σ−1 +ε, (6.5)

holds for

σ(6α�− 12α− 12θ) � 3α�− 12α− 4θ.

Secondly, we set

Y 4−4σ = T θY �(1−2σ),


Y = Tθ

�(2σ−1)+4(1−σ) .

Then

Nf

(σ, T, T + Tα

)� T

4θ(1−σ)�(2σ−1)+4(1−σ)+ε + Tα+ 8θ(2−3σ)

3(�(2σ−1)+4(1−σ))+ε.

Now we have

Nf

(σ, T, T + Tα

)� T

4θ(1−σ)�(2σ−1)+4(1−σ)+ε, (6.6)

if

σ(6α�− 12α− 12θ) < 3α�− 12α− 4θ.

The proof of Theorem 1.3 is thus complete. �Acknowledgments

The second author was supported by the National Natural Science Foundation ofChina (Grant No. 11001154) and Natural Science Foundation of Shandong Province(Grant No. ZR2010AQ009). This work was completed when the second author visitedthe University of Iowa, supported by the International Cooperation Program of ShandongProvince. The second author would like to thank Professors Jianya Liu and WenguangZhai for their encouragements. The first author would like to thank ICERM at BrownUniversity for helpful feedback he received during a workshop.


References

[1] H. Davenport, Multiplicative Number Theory, 2nd ed., Springer, New York, 1980.[2] R. Godement, H. Jacquet, Zeta Functions of Simple Algebras, Lecture Notes in Math., vol. 260,

Springer, Berlin, New York, 1972.[3] D. Goldfeld, Automorphic Forms and L-Functions for the Group GL(n,R). With an Appendix by

Kevin A. Broughan, Cambridge Stud. Adv. Math., vol. 99, Cambridge University Press, Cambridge,2006, xiv+493 pp., ISBN 978-0-521-83771-2; 0-521-83771-5.

[4] D. Goldfeld, Xiaoqing Li, The Voronoi formula for GL(n,R), Int. Math. Res. Not. 2008 (2) (2008),Article ID rnm144, 39 pp.

[5] G. Halász, Über die Mittelwerte multiplikativer zahlentheoretischer Funktionen, Acta Math. Acad.Sci. Hungar. 19 (1968) 365–403.

[6] D.R. Heath-Brown, The fourth power moment of the Riemann zeta-function, Proc. Lond. Math.Soc. 38 (1979) 385–422.

[7] M.N. Huxley, On the difference between consecutive primes, Invent. Math. 15 (1972) 164–170.[8] M.N. Huxley, Large values of Dirichlet polynomials, III, Acta Arith. 27 (1975) 435–444.[9] A.E. Ingham, On the estimation of N(σ, T ), Q. J. Math. Oxford Ser. 11 (1940) 291–292.

[10] A. Ivić, The Riemann Zeta Function: The Theory of the Riemann Zeta-Function with Applications,John Wiley, New York, 1985.

[11] A. Ivić, On zeta-functions associated with Fourier coefficients of cusp forms, in: E. Bombieri, et al.(Eds.), Proceedings of the Amalfi Conference on Analytic Number Theory, Università di, Salerno,1992, pp. 231–246.

[12] H. Iwaniec, Fourier coefficients of cusp forms and the Riemann zeta function, Sémin. J. Theor.Nombres Bordeaux 18 (1979–1980) 1–36.

[13] M. Jutila, On a density theorem of H.L. Montgomery for L-functions, Ann. Acad. Sci. Fenn.Ser. A I 520 (1972), 13 pp.

[14] Y. Kamiya, Zero density estimates of L-functions associated with cusp forms, Acta Arith. 3 (1998)209–227.

[15] H. Kim, Functoriality for the exterior square of GL4 and the symmetric fourth of GL2, J. Amer.Math. Soc. 16 (2003) 139–183.

[16] H. Kim, P. Sarnak, Refined estimates towards the Ramanujan and Selberg conjectures, Appendixto Kim [15].

[17] Yuk-Kam Lau, Jianya Liu, Yangbo Ye, A new bound k2/3+ε for Rankin–Selberg L-functions forHecke congruence subgroups, Int. Math. Res. Pap. 2006 (2006) 1–78, Article ID 35090.

[18] Xiaoqing Li, Bounds for GL(3)×GL(2) L-functions and GL(3) L-functions, Ann. of Math. (2) 173(2011) 301–336.

[19] Wenzhi Luo, Zeros of Hecke L-functions associated with cusp forms, Acta Arith. 71 (1995) 139–158.[20] H.L. Montgomery, Mean and large values of Dirichlet polynomials, Invent. Math. 8 (1969) 334–345.[21] Y. Motohashi, On sums of Hecke–Maass eigenvalues squared over primes in short intervals,

arXiv:1209.4140v3 [math.NT], 28 October 2012.[22] Z. Rudnick, P. Sarnak, Zeros of principle L-functions and random matrix theory, Duke Math. J. 81

(1996) 269–322, A celebration of John F. Nash Jr.[23] A. Sankaranarayanan, J. Sengupta, Zero-density estimate of L-functions attached to Maass forms,

Acta Arith. 127 (2007) 273–284.[24] P. Sarnak, Notes on the generalized Ramanujan conjectures, in: Harmonic Analysis, the Trace

Formula, and Shimura Varieties, in: Clay Math. Proc., vol. 4, Amer. Math. Soc., Providence, RI,2005, pp. 659–685.

[25] Hengcai Tang, Zero density of L-functions related to Maass forms, Front. Math. China (2013),http://dx.doi.org/10.1007/s11464-013-0303-0.

[26] Zhao Xu, A new zero-density result of L-functions attached to Maass forms, Acta Math. Sin. Engl.Ser. 27 (6) (2011) 1149–1162.

[27] Yangbo Ye, The fourth power moment of automorphic L-functions for GL(2) over a short interval,Trans. Amer. Math. Soc. 385 (2006) 2259–2268.

http://refhub.elsevier.com/S0022-314X(13)00164-9/bib447670s1

http://refhub.elsevier.com/S0022-314X(13)00164-9/bib47646D4A6371s1

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