Goldfeld and Huang Res Math Sci 61:5)802( https://doi.org/10.1007/s40687-018-0134-4
RESEARCH
Super-positivity of a family of L-functions in the level aspect Dorian Goldfeld1* and Bingrong Huang2,3 * Correspondence:
[email protected] 1 Department of Mathematics, Columbia University, New York, NY 10027, USA Full list of author information is available at the end of the article
Abstract An automorphic self dual L-function has the super-positivity property if all derivatives of the completed L-function at the central point s = 1/2 are nonnegative and all derivatives at a real point s > 1/2 are positive. In this paper, we prove that at least 12% of L-functions associated to Hecke basis cusp forms of weight 2 and large prime level q have the super-positivity property. It is also shown that at least 49% of such L-functions have no real zeros on Re(s) > 0 except possibly at s = 1/2. Keywords: Level, L-function, Mollification, Real zeros, Super-positivity, Zero-density
1 Introduction The notion of super-positivity of self dual L-functions was introduced in [14] with breakthrough applications to the function field analog of the Gross–Zagier formula for higher derivatives of L-functions. We say a self dual L-function has the super-positivity property if all derivatives of its completed L-function (including Gamma factors and power of the conductor) at the central value s = 1/2 are nonnegative, all derivatives at a real point s > 1/2 are positive, and if the k0 th derivative is positive then all (k0 + 2j)th derivatives are positive, for all j ∈ N. See [4] for more details. In this paper, we continue our investigation of super-positivity of L-functions associated to classical modular forms as begun in [4]. Let q be a fixed large prime. Let S2 (q) be the space of holomorphic weight 2 cusp forms of level q, and H2 (q) a basis of S2 (q) that are eigenfunctions of all the Hecke operators and have the first Fourier coefficient af (1) = 1. Our main theorem in this paper is as follows. Theorem 1.1 There are infinitely many modular forms f of weight 2 and prime level such that L(s, f ) has the super-positivity property. In fact, the proportion of f ∈ H2 (q) which have the super-positivity property is ≥ 12%, when q is a sufficiently large prime. In the course of proving Theorem 1.1, we also obtained the following result about real zeros of L-functions as a by-product. Theorem 1.2 There are infinitely many modular forms f of weight 2 and prime level such that L(s, f ) has no nontrivial real zeros except possibly at s = 1/2. In fact, the proportion of f ∈ H2 (q) with this property is ≥ 49%, when q is a sufficiently large prime.
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A key ingredient of the proofs of the above theorems is given in a paper of Stark and Zagier [13] which essentially establishes that super-positivity follows if L(s, f ) has no zeros in the triangle s = σ + it 1/2 < σ < 1, t ∈ R, |t| ≤ σ − 1/2 . In this regard, see [4]. To prove such a zero-free region for a large number of those L-functions, we first use Selberg’s lemma [12, Lemma 14] to convert the count of the number of zeros in this triangle to estimates of the mollified second moments. We will need an asymptotic formula for the mollified second moment of L-functions near the central point 1/2. In Sect. 4, we prove this (with the harmonic weights) by an asymptotic formula for the twisted second moment (see Theorem 3.2) and Perron’s formula. Then we deal with L-functions away from the critical line with the harmonic weights by the convexity principle (see Sect. 5). In order to get a better bound, we also need to improve an estimate of the first moment, see Lemmas 5.3 and 6.7. The harmonic weights appear naturally, since we will use the Petersson trace formula. We will remove those harmonic weights in Sect. 6. The proofs of both Theorems 1.1 and 1.2 depend on numerics, which we didn’t try to optimize. Note that some of the techniques in this paper are motivated by Kowalski–Michel [9,10] and Conrey–Soundararajan [1,2]. Remark 1.3 It is possible to consider even and odd forms separately, and then get a more precise result even for fixed even weight and square-free level. One can also prove a positive proportion result in the weight or spectral aspect by combining the ideas in this paper with [4].
2 Preliminaries 2.1 The approximate functional equation
For f ∈ H2 (q), let f (z) =
∞
n1/2 λf (n)e(nz),
(for z ∈ H)
n=1
be its Fourier expansion. We can associate it with an L-function L(s, f ) :=
∞ λf (n) n=1
ns
,
(for Re(s) > 1).
The functional equation is written in terms of the completed L-function √ s q 1 L(s, f ), s+ (s, f ) = 2π 2 namely (s, f ) = εf (1 − s, f ),
(2.1)
and the sign εf of the functional equation is εf = q 1/2 λf (q) ∈ {±1}.
(2.2)
Let G be a polynomial of degree 2N (large enough) satisfying G(−s) = G(s),
G(0) = 1,
and G(−N ) = · · · = G(−1) = 0.
(2.3)
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Notice that we have G (0) = G (0) = 0. That is G(s) = a0
N
(s2 − k 2 ),
a0 = (−1)N (N !)−2 .
k=1
Define Ht (s) := G(s + it)G(s − it)(s + 1 + it)(s + 1 − it).
(2.4)
By Stirling’s formula, (in a vertical strip) we have
Ht (s) (1 + |t| + |Im(s)|)B · min 1, e−π (|Im(s)|−|t|) , Ht (δ)
(2.5)
for some constant B > 0 depending on N and Re(s). Note that for |δ|
log log q log q ,
we have
Ht (−δ) (−δ + 1 + it)(−δ + 1 − it) = = 1 + O |δ| . Ht (δ) (δ + 1 + it)(δ + 1 − it) By considering the following integral 1 Iδ = 2πi
3+i∞
(1/2 + δ + it + s, f )(1/2 + δ − it + s, f )G(s + it)G(s − it) 3−i∞
ds , s−δ
we can obtain the approximate functional equation of |L(1/2 + δ + it, f )|2 . Lemma 2.1 Let f ∈ H2 (q). Let − logB q ≤ δ ≤ ϑ, δ = 0, and t ∈ R. We have ∞ q −δ λf (n)ηit (n) L(1/2 + δ + it, f )2 = Vδ,t 2 4π n1/2
Here ην (n) :=
a ν
ad=n d
1 Vδ,t (y) := 2πi
3+i∞
3−i∞
n=1
4π 2 n . q
is the generalized divisor function, and for any y > 0,
Ht (s) 2s ζq (1 + 2s)y−s ds Ht (δ) (s − δ)(s + δ)
is real valued, and satisfies the following: Vδ,t (y) = ζq (1 + 2δ)y−δ +
Ht (−δ) ζq (1 − 2δ)yδ + ON yN (1 + |t|)B , Ht (δ)
for 0 < y ≤ 1; and Vδ,t (y) A,N y−A (1 + |t|)B for all A ≥ 1, for y ≥ 1, and B depending on A and N . In particular, for all y ≥ 1/q we have Vδ,t (y) q δ (log q)(1 + |t|)B . Here ζq is the Riemann zeta function with the Euler factor at q removed. Proof See Kowalski–Michel [9, pp. 526–527].
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2.2 The Petersson trace formula
Let H denote the upper half plane. The Fourier coefficients of f ∈ H2 (q) satisfy the relation λf (m)λf (n) =
εq (d)λf
mn
d|(m,n)
d2
(2.6)
,
where εq (·) is the principle Dirichlet character modulo q. In particular, we have λf (q)λf ( ) = λf (q ) for any ∈ Z>0 . The Petersson trace formula is given by the following basic orthogonality relation on H2 (q). Lemma 2.2 Let m, n ≥ 1. Then
ωf · λf (m)λf (n) = δm,n − 2π
f ∈H2 (q)
c≡0(modq)
S(m, n; c) J1 c
√ 4π mn . c
where
(log q)4 ζ (2) 1 1 ωf := = · · 1+O 4π f, f |H2 (q)| L(1, sym2 f ) q
is termed the harmonic weight of f , where ·, · denotes the Petersson inner product on S2 (q),
dx dy f (z) g(z) .
f, g = y2 0 (q)\H
Proof See e.g., Kowalski–Michel [9, pp. 507–509].
By Weil’s bound for Kloosterman sums and the bound J1 (x) x for J -Bessel function, we have (see e.g., [9, eq. (12)])
ωf = 1 + O q −3/2 .
(2.7)
f ∈H2 (q)
We can also consider the number of odd cusp forms. Note that
αf =
f ∈H2 (q) f is odd
1 − εf · αf , 2
(2.8)
f ∈H2 (q)
for any finite set of complex numbers (αf ). By Kowalski–Michel [11, Lemma 1], we have
ωf =
f ∈H2 (q) f is odd
1 1 +O . 2 q
(2.9)
Also, it’s well known that f ∈H2 (q) f is odd
1∼
1 1. 2 f ∈H2 (q)
(2.10)
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3 The twisted second moment near the critical point Recall that H2 (q) denotes a basis for the space of holomorphic Hecke cusp forms of weight k = 2 for 0 (q) ⊂ SL(2, Z). Assume that for all f ∈ H2 (q) there is some uniquely defined αf ∈ C. Consider the set {αf } := {αf }f ∈H2 (q) . The basic objects of study for the rest of this paper are given in the following definition. Definition 3.1 Let R1/2 := β + iγ β ∈ (1/2, 1) and |γ | ≤ β − 1/2 . For q > 1, a large prime number, we define the following sums A {αf }; q := αf ,
A(q) :=
f ∈H2 (q)
M(q) :=
f ∈H2 (q) L(s,f ) = 0 for s ∈ R1/2
1,
f ∈H2 (q)
1,
N (q) :=
1.
f ∈H2 (q) L(s,f ) has at least one zero in R1/2
It follows from Stark–Zagier [13] that it is enough to prove that at least 12% of Lfunctions associated to f ∈ H2 (q) have no zeros in the triangle R1/2 . Furthermore, it is clear that
M(q) + N (q) = A(q). The key strategy for proving Theorem 1.1 is to try to show that M(q) is large compared to A(q). To achieve this goal, we will use the mollification method which leads us to first consider the following twisted second moment of L(s, f ) at the special value s = 1/2+δ+it. 1 both fixed, and t ∈ R. Let Theorem 3.2 Let − logc q ≤ δ ≤ ϑ with c > 0, 0 < ϑ < 100 1−4ϑ
≤q be a positive integer. We have the following asymptotic formula A ωf λf ( )|L(1/2 + δ + it, f )|2 ; q ηit ( ) ηit ( ) Ht (−δ) q −2δ = ζq (1 + 2δ) 1/2+δ + ζq (1 − 2δ) 1/2−δ
Ht (δ) 4π 2 B −1/2+ε + Oε 1 + |t| q .
Remark 3.3 One may use a similar method to obtain the asymptotic formula for the average over even or odd forms, respectively. See Kowalski–Michel [11] for example. Proof We now begin the proof of Theorem 3.2. From the approximate functional equation (Lemma 2.1), we have
A
ωf · λf ( )|L(1/2 + δ + it, f )|2 ωf λf ( )|L(1/2 + δ + it, f )|2 ; q = f ∈H2 (q)
2 ∞ q −δ ηit (n) 4π n = V ωf · λf ( )λf (n). δ,t 2 1/2 4π n q n=1
f ∈H2 (q)
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Applying the Petersson trace formula (Lemma 2.2), we obtain that
A
ωf λf ( )|L(1/2 + δ + it, f )|2 ; q = D + F ,
(3.1)
where we have the diagonal term (noting that < q)
D :=
2 q −δ η ( ) 4π it V , δ,t 2 1/2 4π
q
and the off-diagonal term 2 ∞ q −δ 4π n ηit (n) S( , n; c) F := −2π Vδ,t J1 2 1/2 4π n q c n=1
q|c
√ 4π n . c
3.1 The diagonal term
Recall that ϑ is a fixed positive real number less than 1/100. We first handle the case − logc q ≤ δ ≤ ϑ and δ = 0. Note that for the remaining case δ = 0, we can just view it as the limitation of δ → 0. Introducing the integral defining Vδ,t (y) in Lemma 2.1, we have
D=
2 −s
q −δ η ( ) 1 3+i∞ Ht (s) 2s 4π it ζ ds. (1 + 2s) q 4π 2
1/2 2πi Ht (δ) q (s − δ)(s + δ) 3−i∞
For −1/2 ≤ Re(s) ≤ 3, we have
Re(s) + 1 2 |(s + 1 + it)(s + 1 − it)| 1. ≤ (δ + 1 + it)(δ + 1 − it) (δ + 1)2
Hence by shifting the contour to the line Re(s) = −1/2 + ε, we obtain ηit ( ) ηit ( ) Ht (−δ) q −2δ + ζ (1 − 2δ) q
1/2+δ
1/2−δ Ht (δ) 4π 2 −1/2+ε+i∞ 2 −s
Ht (s) 2s 4π τ ( )q −δ ζ ds +O (1 + 2s) q 1/2
Ht (δ) q (s − δ)(s + δ)
D = ζq (1 + 2δ)
−1/2+ε−i∞
ηit ( ) ηit ( ) Ht (−δ) q −2δ = ζq (1 + 2δ) 1/2+δ + ζq (1 − 2δ) 1/2−δ
Ht (δ) 4π 2 B −1/2+ε + Oε 1 + |t| q .
(3.2)
Here we use the bound (2.5) for Ht (s)/Ht (δ). 3.2 The off-diagonal term
One can show that
F (1 + |t|)B q −1/2+ε .
(3.3)
We won’t give the details of the proof of the above estimate, since a very similar result already appeared in Kowalski–Michel [10, Appendix A], and the techniques needed for the proof can be found in Kowalski–Michel [11, Sect. 2.4]. Combining (3.1)–(3.3), completes the proof of Theorem 3.2.
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4 Mollification near the critical point 4.1 Choosing the mollifier
We will take the same mollifier as in Kowalski–Michel [10, Appendix A]. Let ϑ be a fixed small positive constant, and M ≤ q 1/2−2ϑ . Fix 0 < ϒ < 1. Define ⎧ 1−ϒ , ⎪ ⎪ ⎨ 1, if 0 ≤ x ≤ M log(M/x) Fϒ,M (x) := P log M , if M 1−ϒ ≤ x ≤ M, ⎪ ⎪ ⎩ 0, if x ≥ M, with P(y) = y/ϒ. We define the mollifier for L(s, f ) by ∞ x (s)
λf ( ),
(4.1)
∞ μ( ) μ2 ( n)Fϒ,M ( n) .
s−1/2 n2s
(4.2)
M(s, f ) :=
=1
1/2
where
x (s) :=
n=1
We always write L(s, f )M(s, f ) as LM(s, f ). In this section, we will prove the following theorem. Theorem 4.1 With the notations as above and 0 < ϒ < 1, M ≤ q 1/2−2ϑ , assume that 100 log log q 100 log log q 100 |t| ≤ and that − log . Then we have log q q ≤δ ≤ log q
2
ωf |LM(1/2 + δ + it, f )| = 1 + q
f ∈H2 (q)
−2δ M
− M (1−ϒ)(−2δ+2it) 2 ϒ(−2δ + 2it) log M
−2δ+2it
M −2δ(1−ϒ) − M −2δ −2δ 1 − q (2δϒ log M)2 log log q −2(1−ϒ)δ M . +O log q
+
Proof By (2.6), we have |M(1/2 + δ + it, f )|2 = = =
∞ ∞ x 1 (1/2 + δ + it)x 2 (1/2 + δ − it)
1 =1 2 =1 ∞ ∞ ∞
1 =1 2 =1 d=1 ∞ ∞ ∞
1 =1 2 =1 d=1
1/2 1/2
1 2
λf ( 1 )λf ( 2 )
εq (d) xd 1 (1/2 + δ + it)xd 2 (1/2 + δ − it) λf ( 1 2 ) 1/2 1/2 d
1
2
1 xd 1 (1/2 + δ + it)xd 2 (1/2 + δ − it) λf ( 1 2 ). 1/2 1/2 d
1
(4.3)
2
Here for the last step, we use the fact that the sum over d is actually supported on [1, M] and that M < q and q being prime. Hence we get
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A
ωf |LM(1/2 + δ + it, f )|2 ; q =
∞ ∞ ∞ 1 xd 1 (1/2 + δ + it)xd 2 (1/2 + δ − it) 1/2 1/2 d
1 2
1 =1 2 =1 d=1 × A ωf λf ( 1 2 )|L(1/2 + δ + it, f )|2 ; q .
There are two cases we need to consider: ⎧ 100 log log q 100 −100 ⎪ ⎪ and |δ| ≥ , ⎨ log q ≤ δ ≤ log q log q log log q (I) ⎪ 100 log log q ⎪ ⎩ |t| ≤ ; log q and ⎧ 100 ⎪ ⎪ ⎨ |δ| ≤ log q log log q , (II) 100 1 ⎪ ⎪ ⎩ ≤ |t| ≤ . 100 log q log q We will focus on the first case (I), which we will assume in this section from now on. Note that the other case can be handled by combining the ideas of Conrey and Soundararajan [1, §6]. Define νδ (r) :=
1 μ(d) . r d 1+2δ
(4.4)
d|r
By Theorem 3.2, we obtain
A
ωf |LM(1/2 + δ + it, f )|2 ; q
∞ ∞ ∞ 1 xd 1 (1/2 + δ + it)xd 2 (1/2 + δ − it) ηit ( 1 2 ) · ζq (1 + 2δ) = 1/2 1/2 1/2+δ d ( 1 2 )
1 2
1 =1 2 =1 d=1 ηit ( 1 2 ) Ht (−δ) q −2δ + ζq (1 − 2δ) + Oε (1 + |t|)B Mq −1/2+ε 1/2−δ 2 ( 1 2 ) Ht (δ) 4π = S1 + S2 + Oε (1 + |t|)B q −2ϑ+ε , (4.5)
where 2 ηit ( ) , νδ (r) x (1/2 + δ + it) r 1+δ
r=1
2 ∞ ηit ( ) Ht (−δ) q −2δ S2 := ζq (1 − 2δ) ν−δ (r) xr (1/2 + δ + it) . 2 1−δ Ht (δ) 4π
S1 := ζq (1 + 2δ)
∞
r=1
Here we use the following Möbius inversion formula to separate the variables 1 and 2 , m n ηit (mn) = ηit . μ(d)ηit d d d|(m,n)
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4.2 The treatment of S1
For S1 , we will follow Conrey and Soundararajan’s method in [2] to show M −2(1−ϒ)δ − M −2δ log log q −2(1−ϒ)δ M S1 = 1 + +O . (2δϒ log M)2 log q
(4.6)
One may also use the calculation in Kowalski–Michel [10, Apendix D] to prove it. Let ϒ,M (w) be the Mellin transform of Fϒ,M (x). We have ϒ,M (w) =
M w − M (1−ϒ)w . ϒw2 log M
(4.7)
It has a simple pole at w = 0 with residue 1. By Mellin inversion formula we have 1 Fϒ,M (x) = 2πi
α+i∞
ϒ,M (w)x−w dw,
α−i∞
for any constant α > 0. We have
S1 = ζq (1 + 2δ)
∞ r=1
∞ ηit ( ) μ(r ) μ2 (r n)Fϒ,M (r n) 2 νδ (r)
1+δ (r )δ+it n1+2δ+2it
n=1
∞ ∞ νδ (r) μ(r 1 )ηit ( 1 ) μ2 (r 1 n1 )Fϒ,M (r 1 n1 ) = ζq (1 + 2δ) r 2δ
1+2δ+it n1+2δ+2it 1 1
×
n1 =1
1
r=1
∞ μ(r 2 )ηit ( 2 ) μ2 (r 2 n2 )Fϒ,M (r 2 n2 )
2
1+2δ−it 2
= ζq (1 + 2δ)
1 (2πi)2
n1+2δ−2it 2
n2 =1 α+i∞
α+i∞
G1 (w1 , w2 , δ, t)ϒ,M (w1 )ϒ,M (w2 )dw1 dw2 , α−i∞ α−i∞
where G1 (w1 , w2 , δ, t) :=
∞ r=1
×
νδ (r) 2δ+w 1 +w2 r
μ2 (r 1 n1 )μ(r 1 )ηit ( 1 )μ2 (r 2 n2 )μ(r 2 )ηit ( 2 )
1
n1
2
n2
1 +2δ+it 1+w1 +2δ+2it 1+w2 +2δ−it 1+w2 +2δ−2it
1+w n1
2 n2 1
To deal with G1 (w1 , w2 , δ, t), we will use the following easy lemma. Lemma 4.2 Let f, g, h be three multiplicative functions. Denote F (k) = μ2 (rm)μ2 (rn)h(r)f (m)g(n). k=rmn
Then F is a multiplicative function. Proof We first note that F is supported on cubic-free numbers. Let k = ab2 , where a, b are square-free numbers. We only need to consider the decompostion of k = rmn with b|m and b|n; otherwise the contribution will be zero because of μ2 . Hence h(r)f (m)g(n). F (k) = F (ab2 ) = f (b)g(b)H(a), where H(a) = a=rmn
Now it’s not hard to check that F is multiplicative.
.
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We can rearrange the terms in G1 (w1 , w2 , δ, t) and view it as a summation over k = rmn where m = 1 n1 and n = 2 n2 . Hence by Lemma 4.2, we have ηit (p) ηit (p) 1 − p−1−2δ 1 + 1+2δ+w +w − 1+w +2δ+it − 1+2δ+w −it G1 (w1 , w2 , δ, t) = 1 2 1 2 p p p p + − =:
1 p1+w1 +2δ+2it
+
ηit (p) p2+w1 +w2 +4δ−it
1 p1+w2 +2δ−2it −
+
ηit (p)2 2+w p 1 +w2 +4δ
ηit (p)
+
1 p2+w1 +w2 +4δ
p2+w1 +w2 +4δ+it
ζ (1 + 2δ + w1 + w2 ) H1 (w1 , w2 , δ, t). ζ (1 + w1 + 2δ)ζ (1 + w2 + 2δ)
Here the Euler product defining H1 (w1 , w2 , δ, t) converges absolutely for min Re(w1 ), Re(w2 ), Re(w1 + w2 ) > −1/2. To evaluate the integral in S1 , we shift both contours to the line Re(w1 ) = Re(w2 ) = (2A + 1)/ log q, and truncate the w2 integral at |Im(w2 )| ≤ q with error O(q −1+ε ). Then we shift the w1 integral to the contour C1 given by c C1 := w1 Re(w1 ) = −2δ − , log3/4 2 + |Im(w1 )| where c > 0 is a suitable constant such that ζ (s) has no zero in the region Re(s) ≥ 1 − c log−3/4 2 + |Im(s)| and c log−3/4 (2) < 1/3. If q is sufficiently large, we encounter poles at w1 = 0 and w1 = −2δ − w2 . The first pole at w1 = 0 yields a residue 1 1 ζ (1 + 2δ) 2πi
(2A+1)/
log q+iq
H1 (0, w2 , δ, t)ϒ,M (w2 )dw2 (2A+1)/ log q−iq
=
1 1 + Oε ζ (1 + 2δ)
−(1−ϒ)/2+ε M ,
(4.8)
by moving the w2 -contour to C21 where 2A + 1 1 1 1 1 2A + 1 C21 := − iq, − − iq ∪ − − iq, − + iq ∪ − + iq, + iq , log q 2 2 2 2 log q and estimating the integration trivially. The pole at w1 = −2δ − w2 leaves the residue 1 2πi
(2A+1)/
log q+iq
(2A+1)/ log q−iq
H1 (−2δ − w2 , w2 , δ, t) ϒ,M (−2δ − w2 )ϒ,M (w2 )dw2 . ζ (1 − w2 )ζ (1 + 2δ + w2 )
(4.9)
Note that by (4.7) we have ϒ,M (−2δ − w2 )ϒ,M (w2 ) =
M −2δ + M −2δ(1−ϒ) − M −ϒw2 −2δ − M ϒw2 −2δ(1−ϒ) . w22 (2δ + w2 )2 (ϒ log M)2 (4.10)
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We can see that the contributions from the first three terms on the right-hand side of −2δ −2δ(1−ϒ) −M −ϒw2 −2δ the above identity, that is, M w+M , are small. Indeed, by shifting the 2 (2δ+w )2 (ϒ log M)2 2
2
integration to the contour C22 where 2A + 1 2A + 1 1 1 1 2A + 1 − iq, −i ∪ − i, − i ∪ − i, + i C22 := log q log q log q 4 4 4 1 2A + 1 2A + 1 2A + 1 ∪ + i, +i ∪ + i, + iq , 4 log q log q log q
and using the standard bounds for the Riemann zeta function, one can show that the contributions from the first three terms are bounded by
O
M −2δ(1−ϒ) (log M)2
.
Now we consider the contribution from
−M ϒw2 −2δ(1−ϒ) w22 (2δ+w2 )2 (ϒ log M)2
to (4.9). We move the contour
of integration to the left C23 where 2A + 1 c − iq, −2δ − C23 := − iq log q log3/4 q c c ∪ −2δ − − iq, −2δ − + iq log3/4 q log3/4 q c 2A + 1 ∪ −2δ − + iq , + iq, log q log3/4 q
for the same constant c in the definition of C1 , and pick up contributions from the poles at w2 = 0 and w2 = −2δ. Similarly, we find the integration on C23 can be bounded by 1/4 O M −2δ(1−ϒ) e−c log q , for some c > 0. The pole at w2 = 0 contributes
M −2δ(1−ϒ) 1 H1 (−2δ, 0, δ, t). ζ (1 + 2δ) (2δϒ log M)2
(4.11)
Lastly the pole at w2 = −2δ leaves the residue
−
1 M −2δ H1 (0, −2δ, δ, t). ζ (1 + 2δ) (2δϒ log M)2
(4.12)
The remaining integral, for w1 on C1 , is bounded by 1/4 O M −2δ(1−ϒ) e−c log q , for some c > 0. The proof that M ≤ q 1/2−2ϑ now follows upon combining (4.8), (4.11), and (4.12), and noting that H1 (0, 0, δ, t) = 1.
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4.3 The treatment of S2
We now consider S2 . Following a similar argument as for S1 , we have ∞ ∞ ∞ Ht (−δ) q −2δ μ2 (r) ηit ( 1 )μ( 1 )μ2 (r 1 n1 ) S2 = ζ (1 − 2δ) ν (r) q −δ Ht (δ) 4π 2 r 2δ
1+it n1+2δ+2it 1 1 r=1
1 =1 n1 =1 ×
∞ ∞ ηit ( 2 )μ( 2 )μ2 (r 2 n2 )
Fϒ,M (r 1 n1 )Fϒ,M (r 2 n2 )
1−it n1+2δ−2it 2 2 Ht (−δ) q −2δ = ζq (1 − 2δ) Ht (δ) 4π 2 α+i∞
α+i∞
1 × G2 (w1 , w2 , δ, t)ϒ,M (w1 )ϒ,M (w2 )dw1 dw2 , (2πi)2
2 =1 n2 =1
α−i∞ α−i∞
where G2 (w1 , w2 , δ, t) μ2 (r)ν−δ (r) μ2 (r 1 n1 )μ( 1 )ηit ( 1 )μ2 (r 2 n2 )μ( 2 )ηit ( 2 ) := 1+w1 +it 1+w1 +2δ+2it 1+w2 −it 1+w2 +2δ−2it r 1+2δ+w1 +w2 n1
2 n1 r
1 n1 2 n2 1 ηit (p) ηit (p) 1 1 1 − p−1+2δ = 1 + 1+2δ+w +w − 1+w +it − 1+w −it + 1+w +2δ+2it + 1+w +2δ−2it 1 2 1 2 1 2 p p p p p p ηit (p)2 1 ηit (p) ηit (p) + 2+w +w + 2+w +w +4δ − 2+w +w +2δ−it − 2+w +w +2δ+it p 1 2 p 1 2 p 1 2 p 1 2 =:
ζ (1 + 2δ + w1 + w2 )ζ (1 + w1 + 2δ + 2it)ζ (1 + w2 + 2δ − 2it) H2 (w1 , w2 , δ, t). ζ (1 + w1 )ζ (1 + w1 + 2it)ζ (1 + w2 )ζ (1 + w2 − 2it)
Here the Euler product defining H2 (w1 , w2 , δ, t) converges absolutely for min Re(w1 ), Re(w2 ), Re(w1 + w2 ) > −1/2. To estimate the integral in S2 , we again shift the w1 and w2 contours to the lines Re(w1 ) = Re(w2 ) = 2A+1 log q . We now truncate the w2 contour at Im(w2 ) ≤ q, and shift the w1 contour to C1 given by c C1 := w1 Re(w1 ) = − 3/4 , log 2 + |Im(w1 + 2it)| for the same constant c in the definition of C1 . If q is sufficiently large, we encounter poles at w1 = −2δ − 2it and w1 = −2δ − w2 . The first pole at w1 = −2δ − 2it yields a residue 1 1 · ζ (1 − 2δ)ζ (1 − 2δ − 2it) 2πi
(2A+1)/
log q+iq
(2A+1)/ log q−iq
ζ (1 + w2 + 2δ − 2it) ζ (1 + w2 )
(4.13)
× H2 (0, w2 , δ, t)ϒ,M (−2δ − 2it)ϒ,M (w2 )dw2 . where We now move the w2 -contour to C21 2A + 1 2A + 1 C21 := − iq, α0 − iq ∪ [α0 − iq, α0 + iq] ∪ α0 + iq, + iq log q log q
c c with α0 = max − 3/4 , −2δ − 3/4 , pick up the pole at w2 = −2δ + 2it when q is log q log q large enough, and estimate the remaining integration trivially. By doing so, we find that (4.13) is equal to
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H2 (0, −2δ + 2it, δ, t) ϒ,M (−2δ − 2it)ϒ,M (−2δ + 2it) ζ (1 − 2δ)ζ (1 − 2δ − 2it)ζ (1 − 2δ + 2it) (4.14) −2δ(1−ϒ) −c log1/4 q +O M e , for some constant c > 0 depending on c and ϒ. The pole at w1 = −2δ − w2 leaves the residue 1 2πi
(2A+1)/
log q+iq
(2A+1)/ log q−iq
ζ (1 − w2 + 2it)ζ (1 + w2 + 2δ − 2it)H2 (−2δ − w2 , w2 , δ, t) ζ (1 − 2δ − w2 )ζ (1 − w2 − 2δ + 2it)ζ (1 + w2 )ζ (1 + w2 − 2it)
× ϒ,M (−2δ − w2 )ϒ,M (w2 )dw2 . (4.15) Note that we have (4.10). We can see that the contributions from the first three terms there are small again. Indeed, by the same method as in the treatment of S1 we see that the contributions from the first three terms are M −2δ(1−ϒ) O . (log M)2 ϒw2 −2δ(1−ϒ)
−M Now we consider the contribution from w2 (2δ+w C23
2
2)
2 (ϒ
log M)2
to (4.15). We move the contour
of integration to the left where 2A + 1 c c c C23 := − iq, − 3/4 − iq ∪ − 3/4 − iq, − 3/4 + iq log q log q log q log q c 2A + 1 + iq , ∪ − 3/4 + iq, log q log q for the same constant c in the definition of C1 , and pick up contributions from the poles can be bounded by at w2 = 0 and w2 = −2δ. Similarly, we find the integration on C23 1/4 O M −2δ(1−ϒ) e−c log q , for some c > 0. The pole at w2 = 0 contributes −
M −2δ(1−ϒ) 1 H2 (−2δ, 0, δ, t). ζ (1 − 2δ) (2δϒ log M)2
(4.16)
Lastly the pole at w2 = −2δ leaves the residue 1 M −2δ H2 (0, −2δ, δ, t). ζ (1 + 2δ) (2δϒ log M)2
(4.17)
The remaining integral, for w1 on C1 , is bounded by 1/4 O M −2δ(1−ϒ) e−c log q , for some c > 0. Combining (4.14), (4.16), and (4.17), and noting that H2 (0, 0, 0, 0) = 1, we obtain − M (1−ϒ)(−2δ+2it) 2 S2 = q ϒ(−2δ + 2it) log M −2δ(1−ϒ) −2δ log log q −2(1−ϒ)δ M −M − q −2δ + O M . (2δϒ log M)2 log q
−2δ M
−2δ+2it
Finally, by (4.5), (4.6), and (4.18), this completes the proof of Theorem 4.1.
(4.18)
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5 Mollification away from the critical line In this section, we will prove Theorem 5.1 which is not needed for the proof of our main theorems. However, the results of this section will serve as a prototype for a similar result (without the harmonic weights) which is actually needed in Sect. 6.2. Theorem 5.1 With notations as in Sect. 4, assume s =
1 2
+ δ + it where
1 1 10 log log q ≤ δ ≤ + . log q 2 (1 − ϒ) log q Then there exists an absolute constant B > 0 such that for any 0 < a < 2(1 − ϒ) we have ωf |LM(s, f )|2 − 1 ϒ,ϑ,a,B (1 + |t|)B M −aδ . f ∈H2 (q)
Proof of Theorem 5.1 The proof follows easily from Lemmas 5.2 and 5.3 which are stated and proved below.
Lemma 5.2 With notations as above, assume s = 1/2 + δ + it with δ ≥ 1/(2 log q). Then for any 0 < a < 2(1 − ϒ), we have ωf |LM(s, f ) − 1|2 ϒ,ϑ,a,B (1 + |t|)B M −aδ . f ∈H2 (q)
Proof See Kowalski–Michel [10, Corollary 1]. Lemma 5.3 With notations as above, assume that log1 q ≤ δ ≤ 1/2 + any 0 < a ≤ 2(1 − ϒ), we have ωf LM(s, f ) − 1 ϒ,ϑ,B (1 + |t|)B M −aδ .
10 log log q (1−ϒ) log q .
Then for
f ∈H2 (q)
Remark 5.4 This is an improvement in lemma 7 in Kowalski–Michel [10], where they proved a similar bound for any 0 < a < 1 − ϒ. We will prove the lemma by using the method of Conrey and Soundararajan [2] (see also [4, Lemma 6.6]). The main idea here is to use the approximate functional equation and the Petersson trace formula instead of obtaining the bound by the second moment and the Cauchy–Schwarz inequality. Proof of Lemma 5.3 In the region Re(s) > 1, by (4.1) we may write ⎞ ⎛ ∞ 1 ⎝ λf (a)λf (b)μ(b)μ(bc)2 Fϒ,M (bc)⎠ . LM(s, f ) = ns 2 abc =n
n=1
Using the Hecke relations and the fact that we may assume b < q, we see that ab λf (a)λf (b)μ(b)μ(bc)2 Fϒ,M (bc) = λf μ(b)μ(bc)2 Fϒ,M (bc), 2 d 2 2 abc =n
abc =n d|(a,b)
and setting a = αd, b = βd, and g = cd, this becomes n 2 λf μ(βg) F (βg) μ(βd) = λ (n) μ(β)Fϒ,M (β), ϒ,M f g2 2 2 g |n
αβ=n/g
αβ=n
cd=g
since the terms with g > 1 are easily seen to disappear. Thus LM(s, f ) =
∞ λf (n) n=1
ns
c(n),
where c(n) =
d|n
μ(d)Fϒ,M (d).
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We have c(1) = 1; for 1 < n ≤ M 1−ϒ , we have c(n) = d|n μ(d) = 0; and for n > M 1−ϒ , we have |c(n)| ≤ τ (n). 10 log log q We will first handle the case Re(s) = 1/2 + δ0 where δ0 = 1/2 + (1−ϒ) log q . Put B(s, f ) := LM(s, f ) − 1. We consider 1 2πi
3+i∞
(w)B(w + s, f )X w dw, 3−i∞
where X = q 1−ϑ . We shift the line of integration to Re(w) = −δ0 + δ1 , where δ1 = The pole at w = 0 gives B(s, f ), and so we conclude that −δ0 +δ
1 +i∞
3+i∞
1 (w)B(w + s, f )X dw − 2πi
1 B(s, f ) = 2πi
w
3−i∞
1 log q .
(w)B(w + s, f )X w dw −δ0 +δ1 −i∞
=: T1 (s, f ) − T2 (s, f ). We first estimate the contribution of the T2 (s, f ) terms. By Cauchy’s inequality and Lemma 6.6, for some constant B > 0 we have ! " −δ0 +δ1 A ωf |T2 (s, f )| ; q X A ωf |(w)| · |B(w + s, f )| |dw| ; q (−δ0 +δ1 )
B
(1 + |t|) X
−δ0 +δ1
(1 + |t|)B q −(1−2ϑ)δ0 (1 + |t|)B M −2(1−ϒ)δ0 . # It remains now to estimate the T1 contribution. Since 2π1 i (α) (w)(X/n)w dw = e−n/X , we see that ∞ λf (n)c(n) −n/X λf (n)c(n) −n/X T1 (s, f ) = e = e + O q −B . s s n n 1−ϒ 2 n=2
M
In order to bound A ωf T1 (s, f ) ; q , for M 1−ϒ < n ≤ X(log q)2 , we consider A ωf λf (n) ; q = ωf · λf (n). f ∈H2 (q)
By Lemma 2.2, we arrive at
− 2π
c ≡ 0 (mod q)
S(1, n; c) J1 c
√ √ √ 4π n n n 3/2−ε r −3/2 3/2−ε . c q q r≥1
Thus we see that if n ≤ X(log q)2 then √ n A ωf λf (n) ; q 3/2−ε q −1−ε . q Hence A ωf T1 (s, f ) ; q q −1 , and A ωf (LM(1/2 + δ0 + it, f ) − 1) ; q B,a (1 + |t|)B M −2(1−ϒ)δ0 . This completes the proof for Re(s) = 1 log q
≤δ≤
1 2
+
10 log log q (1−ϒ) log q
1 2
+ δ0 = 1 +
10 log log q (1−ϒ) log q . Furthermore, the result for
now follows from Lemma 5.2 and the convexity argument.
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6 Removing the harmonic weights 6.1 Near the critical point
To get the natural average from the harmonic average, we will use the method of [9]. For x ≥ 1, let ωf (x) :=
λf (d 2 ) . dl 2 2
(6.1)
dl ≤x
We will use the following lemma to remove the harmonic weights. Lemma 6.1 Let {αf } be complex numbers satisfying the conditions A {ωf αf }; q (log q)C ,
(for some absolute constant C > 0),
(6.2)
and max |ωf αf | q − ,
f ∈H2 (q)
(for some > 0),
(6.3)
as the level q (prime) tends to infinity. Let x = q κ for some 0 < κ < 1. There exists an absolute constant ς = ς (κ, ) > 0 such that A {αf }; q 1 = A {ωf · ωf (x)αf }; q + O(q −ς ). A(q) ζ (2)
Proof See [9, Proposition 2].
Using the above lemma and similar calculations as in Sect. 4, we can obtain the following theorem. Theorem 6.2 Let q, M, ϒ be the same as in Theorem 4.1. Let x = q κ where 0 < κ < 1 is a 100 log log q 100 log log q 100 small constant. Assume that |t| ≤ and − log . Then we have log q q ≤δ ≤ log q 1 ωf · ωf (x) · |LM(1/2 + δ + it, f )|2 ζ (2) f ∈H2 (q)
−2δ+2it M − M (1−ϒ)(−2δ+2it) 2 = 1 + q −2δ ϒ(−2δ + 2it) log M M −2δ(1−ϒ) − M −2δ log log q −2(1−ϒ)δ −2δ M + O 1 − q + . (2δϒ log M)2 log q Furthermore, we have −2δ+2it M A {|LM(1/2 + δ + it, f )|2 }; q − M (1−ϒ)(−2δ+2it) 2 = 1 + q −2δ A(q) ϒ(−2δ + 2it) log M log log q −2(1−ϒ)δ M −2δ(1−ϒ) − M −2δ −2δ M + 1 − q +O . (2δϒ log M)2 log q
(6.4)
(6.5)
Proof We will follow [5, §6] closely. By (4.3), (6.1), and the Hecke relations, we have ωf (x)|M(1/2 + δ + it, f )|2 =
∞ ∞ ∞ 1 1 2 dl g 2
dl ≤x
×
1 =1 2 =1 g=1
xg 1 (1/2 + δ + it)xg 2 (1/2 + δ − it)
1/2 1/2
1 2
h|(d 2 , 1 2 )
λf
d 2 1 2 h2
.
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Write h = h1 h22 where h1 and h2 are square-free. Clearly h2 |( 1 , 2 ) and h2 |d. Also h1 |(d, 1 2 ). Shifting the orders of summations, and then making changes of variables (d → dh1 h2 , 1 → 1 h2 , and 2 → 2 h2 ), we obtain ∞ ∞ ∞ 1 1 1 1 ωf · ωf (x)|LM(1/2 + δ + it, f )|2 = ζ (2) ζ (2) 2 dl 2 g
1 =1 2 =1 g=1 h|(d 2 , 1 2 ) f ∈H2 (q) dl ≤x 2 xg 1 (1/2 + δ + it)xg 2 (1/2 + δ − it) d 2 1 2 × ωf · λf L(1/2 + δ + it) 1/2 1/2 2 h
1 2 f ∈H2 (q) ∞ ∞ ∞ 1 1 1 1 xg 1 h2 (1/2 + δ + it)xg 2 h2 (1/2 + δ − it) 1/2 1/2 ζ (2) l2 g h22
=
l
1 × h1 h1 | 1 2
1 =1 2 =1 g=1
d
1
h2
2
2 1 ωf · λf d 2 1 2 L(1/2 + δ + it) . d f ∈H2 (q)
Now introducing our expression for the twisted second moment, i.e., Theorem 3.2, we obtain 1 ωf · ωf (x)|LM(1/2 + δ + it, f )|2 ζ (2) f ∈H2 (q)
=
∞ ∞ ∞ 1 1 1 1 xg 1 h2 (1/2 + δ + it)xg 2 h2 (1/2 + δ − it) 1/2 1/2 ζ (2) l2 g h22
1 2
1 =1 2 =1 g=1 h2 l 1 1 ηit (d 2 1 2 ) ζq (1 + 2δ) 2 × h1 d (d 1 2 )1/2+δ h1 | 1 2 d
The error term above contributes Oε Mq −1/2+ε = Oε q −2ϑ+ε . Note that we may remove the restriction on the sum over d with error x−1/2+ε . Thus 2 1 ωf · ωf (x)LM(1/2 + δ + it, f ) ζ (2) f ∈H2 (q) = T1 + T2 + Oε q −κ/2+ε + q −2ϑ+ε ,
(6.6)
where ∞ ∞ ∞ 1 1 T1 := ζq (1 + 2δ) g h22
=1 =1 g=1 1
2
h2
xg 1 h2 (1/2 + δ + it)xg 2 h2 (1/2 + δ − it) 1 ηit (d 2 1 2 ) × , ( 1 2 )1+δ h1 d 2+2δ h1 | 1 2
d
∞ ∞ ∞ Ht (−δ) q −2δ 1 1 T2 := ζq (1 − 2δ) Ht (δ) 4π 2 g h22
1 =1 2 =1 g=1
h2
xg 1 h2 (1/2 + δ + it)xg 2 h2 (1/2 + δ − it) 1 ηit (d 2 1 2 ) × . ( 1 2 )1−δ h1 d 2−2δ h1 | 1 2
d
To estimate the main term contributions, we will need the following lemma.
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Lemma 6.3 Let 1 and 2 be square-free. For Re(s ± 2ν) > 1 we have ην (d 2 1 2 )
=
ds
d
ην (p) ζ (s)ζ (s + 2ν)ζ (s − 2ν) ζ (2s) 1 + p−s p 1 2 ( 1 , 2 )2
p|( 1 , 2 )
ην (p2 ) − p−s . 1 + p−s
Proof of Lemma 6.3 See Hough [5, lemma 6.3]. By (4.2), we have T1 = ζq (1 + 2δ)
∞ ∞ ∞ 1 1 μ(g 1 h2 )μ(g 2 h2 ) 1 2 g h1 h2 (gh2 )2δ 11+2δ+it 21+2δ−it
1 =1 2 =1 g=1 h h | 2
1
1 2
μ2 (g 1 h2 n1 )Fϒ,M (g 1 h2 n1 ) μ2 (g 2 h2 n2 )Fϒ,M (g 2 h2 n2 ) ηit (d 2 1 2 )
×
n11+2δ+2it
n1
T2 = ζq (1 − 2δ)
n21+2δ−2it
n2
d 2+2δ
d
∞ ∞ ∞ Ht (−δ) q −2δ 1 1 μ(g 1 h2 )μ(g 2 h2 ) 1 Ht (δ) 4π 2 g h1 h22 (gh2 )2δ 1+it
1−it 1 2
1 =1 2 =1 g=1 h h | 2
1
1 2
μ2 (g 1 h2 n1 )Fϒ,M (g 1 h2 n1 ) μ2 (g 2 h2 n2 )Fϒ,M (g 2 h2 n2 ) ηit (d 2 1 2 )
×
,
n11+2δ+2it
n1
n21+2δ−2it
n2
d
d 2−2δ
.
We first consider T1 . By Lemma 6.3, we have
T1 = ζq (1 + 2δ)
1 ζ (2 + 2δ) |ζ (2 + 2δ + 2it)|2 1+2δ ζ (4 + 4δ) g h2+2δ r=gh
×
μ( 1 )μ( 2 )
1 , 2 ,n1 ,n2 ( 1 ,n1 )=1,( 2 ,n2 )=1 ( 1 2 n1 n2 ,r)=1
×
p
1 2 ( 1 , 2 )2
1 1+ p
1+2δ+it
1+2δ−it n1+2δ+2it n1+2δ−2it 1 2 1 2
ην (p) 1 + p−2−2δ
p|( 1 , 2 )
1 1+ p
Fϒ,M ( 1 n1 r)Fϒ,M ( 2 n2 r)
ην (p2 ) − p−2−2δ 1 + p−2−2δ
.
Here the factor (1 + 1/p) in the products is coming from the h1 -sum. It follows from Mellin inversion that ζ (2 + 2δ) |ζ (2 + 2δ + 2it)|2 T1 = ζq (1 + 2δ) ζ (4 + 4δ) α+i∞
α+i∞
1 × G1 (w1 , w2 , δ, t)ϒ,M (w1 )ϒ,M (w2 )dw1 dw2 , (2πi)2 α−i∞ α−i∞
where G1 (w1 , w2 , δ, t) :=
r=gh
1 r 1+2δ+w1 +w2 h ⎛
1 , 2 ,n1 ,n2 ( 1 ,n1 )=1,( 2 ,n2 )=1 ( 1 2 n1 n2 ,r)=1
μ( 1 )μ( 2 ) 1+w1 +2δ+it 1+w2 +2δ−it
1
2
⎞ −1 2 ) − p−2−2δ )(1 + p−2−2δ ) (p 1 (η ν ⎠ × 1+w +2δ+2it 1+w +2δ−2it ⎝ 1+ p ην (p)2 n1 1 n2 2 p|( 1 , 2 ) ⎞⎛ ⎞ ⎛ (p) (p) 1 1 η η ν ν ⎠⎝ ⎠. 1+ 1+ ×⎝ p 1 + p−2−2δ p 1 + p−2−2δ 1
p| 1
p| 2
To write G1 (w1 , w2 , δ, t) as an Euler product, we can use the following lemma.
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Lemma 6.4 Let f1 , f2 , g1 , g2 , h, t be some multiplicative functions. Denote F (k) = μ2 (r 1 n1 )μ2 (r 2 n2 )h(r)f1 ( 1 )f2 ( 2 )g1 (n1 )g2 (n2 ) t(p). k=r 1 n1 2 n2
p|( 1 , 2 )
Then F is a multiplicative function. Proof of Lemma 6.4 Note that F is supported on cubic-free numbers. Let k = ab2 , where a, b are square-free numbers. We only need to consider the decompostion of k = r 1 n1 2 n2 with b| 1 n1 and b| 2 n2 ; otherwise the contribution will be zero because of μ2 . Write a = r 1 n1 2 n2 and b = b1 b1 = b2 b2 , where 1 = b1 1 , 2 = b2 2 , n1 = b1 n1 , and n2 = b2 n2 . Hence we have F (k) = F (ab2 ) = h(r)f1 ( 1 )f2 ( 2 )g1 (n1 )g2 (n2 ) a=r 1 n1 2 n2
×
f1 (b1 )g1 (b1 )f2 (b2 )g2 (b2 )
b=b1 b1 =b2 b2
t(p).
p|(b1 ,b2 )
Since b is square-free, it’s not hard to check that F is multiplicative. Now by Lemma 6.4, we have
G1 (w1 , w2 , δ, t) 1 + 1/p 1 + 1/p ηit (p) ηit (p) 1 + 1/p = − 1+2δ−it+w 1 + 1+2δ+w +w − 1+2δ+it+w −2−2δ 1 2 1 2 p 1+p 1 + p−2−2δ p p p + −
1 p1+2δ+2it+w1
+
ηit (p) 2+4δ−it+w 1 +w2 p
1 p1+2δ−2it+w2
+
1 p2+4δ+w1 +w2
ηit (p2 ) − p−2−2δ 1 + 1/p p2+4δ+w1 +w2 1 + p−2−2δ 1 + 1/p . 1 + p−2−2δ
+
1 + 1/p ηit (p) − 2+4δ+it+w +w −2−2δ 1 2 1+p p
Hence we get
G1 (w1 , w2 , δ, t)
ζ (4 + 4δ) 1 1 1 1 = 1 + 2+2δ + 1+2δ+w +w + 3+4δ+w +w + 2+2δ+w +w 1 2 1 2 1 2 ζ (2 + 2δ) p p p p p +
1 p4+4δ+w1 +w2
−
ηit (p) 1+2δ+it+w 1 p
−
ηit (p) 2+2δ+it+w 1 p
−
ηit (p) 1+2δ−it+w 2 p
−
ηit (p) 2+2δ−it+w 2 p
1 1 1 1 1 + 3+4δ+2it+w + 1+2δ−2it+w + 3+4δ−2it+w + 2+4δ+w +w 1 2 1 2 2 p p1+2δ+2it+w1 p p p 2 −2−2δ 2 −2−2δ 1 ηit (p ) − p ηit (p ) − p ηit (p) + 4+6δ+w +w + + − 2+4δ−it+w +w 1 2 1 2 p2+4δ+w1 +w2 p3+4δ+w1 +w2 p p ηit (p) ηit (p) ηit (p) − 3+4δ−it+w +w − 2+4δ+it+w +w − 3+4δ+it+w +w 1 2 1 2 1 2 p p p +
ζ (1 + 2δ + w1 + w2 ) H1 (w1 , w2 , δ, t). ζ (1 + 2δ + w1 )ζ (1 + 2δ + w2 ) Here the Euler product defining H1 (w1 , w2 , δ, t) converges absolutely for min Re(w1 ), Re(w2 ), Re(w1 + w2 ) > −1/2. =:
We find that
H1 (0, 0, δ, t) =
ζ (4 + 4δ) . ζ (2 + 2δ)|ζ (2 + 2δ + 2it)|2
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Hence, by exactly the same argument as in Sect. 4.2, we can prove that
T1 = 1 +
M −2(1−ϒ)δ − M −2δ log log q −2(1−ϒ)δ M + O . (2δϒ log M)2 log q
(6.7)
And similarly, we can handle T2 , getting −2δ+2it M − M (1−ϒ)(−2δ+2it) 2 T2 = q −2δ ϒ(−2δ + 2it) log M M −2δ(1−ϒ) − M −2δ log log q −2(1−ϒ)δ M − q −2δ + O . (2δϒ log M)2 log q
(6.8)
By (6.6)–(6.8), we prove (6.4) in the theorem. To prove (6.5), we will apply Lemma 6.1. Note that condition (6.2) is a consequence of the trivial bound |M(1/2 + δ + it, f )| < M 1/2+ε , the convexity bound L(1/2 + δ + it, f ) q 1/4+ε (1 + |t|)1/2+ε , and the fact ωf (log q)q −1 . The result in Theorem 4.1 guarantees condition (6.3) under our choice of the parameters. The proof is completed upon applying Lemma 6.1.
6.2 Away from the critical line
Now, we want to remove the harmonic weights that appear in Sect. 5. Our main result in this subsection is the following theorem. Theorem 6.5 With notations as in Sect. 4, assume s =
1 2
+ δ + it where
1 1 10 log log q ≤ δ ≤ + . log q 2 (1 − ϒ) log q Then there exists an absolute constant B > 0, such that for any 0 < a < 2(1 − ϒ), we have 1 |LM(s, f )|2 − 1 ϒ,ϑ,a,B (1 + |t|)B M −aδ . A(q) f ∈H2 (q)
Proof of Theorem 6.5 As in Sect. 5, the proof follows easily from Lemmas 6.6, 6.7, stated and proved below.
Lemma 6.6 With notations as above, assume s = 12 + δ + it where δ ≥ any 0 < a < 2(1 − ϒ) we have 1 |LM(s, f ) − 1|2 ϒ,ϑ,a,B (1 + |t|)B M −aδ . A(q)
1 (2 log q) .
Then for
f ∈H2 (q)
Proof Together with Theorem 6.2 and the fact that ωf log q, the proof of the above lemma is similar to Kowalski–Michel [10, Corollary 1].
Lemma 6.7 With notations as above, assume that log1 q ≤ δ ≤ any 0 < a < 2(1 − ϒ) we have 1 LM(s, f ) − 1 ϒ,ϑ,B (1 + |t|)B M −aδ . A(q) f ∈H2 (q)
1 2
+
10 log log q (1−ϒ) log q .
Then for
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Proof of Lemma 6.7 We will first handle the case Re(s) = 12 + δ0 where we set 10 log log q δ0 = 12 + (1−ϒ) log q . Put B(s, f ) = LM(s, f )−1. As in the proof of Lemma 5.3, for Re(s) > 1, we have ∞ λf (n) c(n), where c(n) = μ(d)Fϒ,M (d), LM(s, f ) = ns n=1
d|n
and B(s, f ) = T1 (s, f ) − T2 (s, f ). We first estimate the contribution of the T2 (s, f ) terms. By Cauchy’s inequality and Lemma 6.6, for some constant B > 0 we have
# A |(w)||B(w + s, f )||dw| ; q A |T2 (s, f )| ; q (−δ0 +δ1 ) X −δ0 +δ1 A(q) A(q) (1 + |t|)B X −δ0 +δ1 (1 + |t|)B q −(1−2ϑ)δ0 (1 + |t|)B M −2(1−ϒ)δ0 . It remains now to estimate the T1 contribution. As before, we have λf (n)c(n) −n/X −B e + O q . T1 (s, f ) = ns 1−ϒ 2 M
In order to bound A1(q) A T1 (s, f ) ; q , for M 1−ϒ < n ≤ X(log q)2 , we consider A λf (n) ; q 1 (log q)5 2 ωf · L(1, sym f )λf (n) + O = . A(q) ζ (2) q f ∈H2 (q)
We can replace L(1, sym2 f ) by the following Dirichlet series (see e.g., Iwaniec–Michel [8, Lemma 3.1]) λf (l 2 ) l V1 L(1, sym2 f ) = l q 1+ε l
w+k−1 )( w+k where L∞ (w, sym2 f ) := π −3w/2 ( w+1 2 )( 2 2 ) and
L∞ (w + u, sym2 f ) (q) 1 du Vw (y) := ζ (2w + 2u) . y−u 2πi L∞ (w, sym2 f ) u (2)
Here ζ (q) (w) stands for the partial zeta function with local factors at primes of q removed. Thus A λf (n) ; q 1 V1 (l/q) ωf · λf (l 2 )λf (n) = A(q) ζ (2) l 1+ε f ∈H2 (q)
l
L∞ V0 (l/q) ωf · λf (l 2 )λf (n) 2 qζ (2)L∞ (1, sym f ) f ∈H2 (q) l
(0, sym2 f )
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We shall only deal with the first term above, since the second one can be handled similarly. For l < M 1−ϒ , by applying Lemma 2.2, we arrive at √ V1 (l/q) 4πl n 1 S(l 2 , n; c) δn,l 2 − 2π J1 . ζ (2) l c c 1−ϒ c ≡ 0 (mod q)
l
The contribution from the first term above to
qε
M (1−ϒ)/2
1
A(q)
A T1 (s, f ) ; q is
1 M −(1−ϒ)+ε . l3
By the Weil bound for the Kloosterman sums, we know that the contribution from the second term above to A1(q) A T1 (s, f ) ; q is √ 1 1 (c, n, l 2 )1/2 τ (c) l n ε q · l 1−ϒ n c1/2 c 1−ϒ 2 l
M
M
1−ϒ −1+ε
q
M
−(1−ϒ)
c≡0(q)
.
Hence we get that the contribution from l < M 1−ϒ to
1
A(q)
A T1 (s, f ) ; q is
M −(1−ϒ)+ε . Now we are going to treat the case M 1−ϒ ≤ l < q 1+ε . We will use the following two large sieve inequalities (see [3, Theorem 1]) 2 ωf λf (n)an q ε (N /q + 1) |an |2 , f ∈H2 (q)
n≤N
n≤N
and (see [8, Theorem 5.1]) 2 l ωf λf (l 2 )g q ε (X 2 /q + X). L f ∈H2 (q)
l
Here g is a smooth function supported in [1, 2] which satisfies g (j) (x) 1. By applying a smooth partition of unity to the l-sum, we know that the contribution of terms with M 1−ϒ ≤ l < q 1+ε is λf (n)an 1 l ωf λf (l 2 )g sup L ns M 1−ϒ Lq 1+ε L 1−ϒ 2 f ∈H2 (q)
1
sup M 1−ϒ Lq 1+ε
×
f ∈H2 (q)
ωf
L
f ∈H2 (q)
l
M
2 1/2 l ωf λf (l 2 )g L l
M 1−ϒ
λf (n)an 2 1/2 ns
M −(1−ϒ)+ε . Hence
A T1 (s, f ) ; q M −(1−ϒ)+ε , A(q)
and
A (LM(1/2 + δ0 + it, f ) − 1) ; q B,a (1 + |t|)B M −(2(1−ϒ)−ε)δ0 . A(q)
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This completes the proof for Re(s) = 1/2 + δ0 = 1 +
10 log log q (1−ϒ) log q .
The result for
1 10 log log q ≤ δ ≤ 1/2 + log q (1 − ϒ) log q
follows from Lemma 6.6 and the convexity argument.
7 Proof of Theorem 1.2 The proofs depend on Selberg’s lemma in [12, Lemma 14] (see also [1, Lemma 2.1]) and are similar to those of [4, theorems 1.8 and 1.5]. One can see the similarity by replacing K in [4] with q 1/2 , so we don’t give the details. In this section, we will prove that for at least 49% of the modular L-functions of weight 2 and level q we have L(σ , f ) > 0 for σ ∈ (1/2, 1]. Let M = q (1−5ϑ)/2 . We apply Selberg’s lemma with the choices H=
S , log q
R 1 − , 2 log q
W0 =
W1 = 1 +
10 log log q , (1 − ϒ) log q
φ(s) = LM(s, f ),
where R and S are fixed positive parameters which will be chosen later. It follows that 4S
π R + (β − 1/2) log q π γ log q sinh cos 2S 2S
β≥ 12 − logR q 2S 0≤γ ≤ 3 log q L(β+iγ ,f )=0
≤ I1 (f ) + I2 (f ) + I3 (f ),
(7.1)
where
S I1 (f ) := −S
πt cos 2S
2 u S 1 π(u + R) log LM + +i , f du, sinh 2S 2 log q log q
(W1 −1/2)
log q
I2 (f ) := −R
S I3 (f ) := −Re −S
R t 1 log LM − +i , f dt, 2 log q log q
(W1 − 1/2) log q − R + it t cos π log LM W1 + i , f dt. 2iS log q
Now, in the sum over zeros on the left-hand side of (7.1), the weight cos replaced by 1 (see [4, eq. (7.3)]). It follows that
4S
β≥ 12 − logR q 2S 0≤γ ≤ 3 log q L(β+iγ ,f )=0
π R + (β − 1/2) log q ≤ I1 (f ) + I2 (f ) + I3 (f ). sinh 2S
(7.2) πγ 2H
can be
(7.3)
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Claim 7.1 We have
⎧ ⎪ 4S sinh π2SR , for all odd forms f ∈ H2 (q), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ for all odd forms f ∈ H2 (q) and ⎪ ⎪ ⎪ ⎪ π R ⎪ ⎪ 12S sinh 2S , if L(s, f ) has a zero ρ = β + iγ ⎪ ⎨ 2S I1 (f ) + I2 (f ) + I3 (f ) ≥ with β ∈ (1/2, 1] and |γ | ≤ 3 log q, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ for all even forms f ∈ H2 (q) and ⎪ ⎪ ⎪ ⎪ π R ⎪ ⎪ if L(s, f ) has a zero ρ = β + iγ ⎪8S sinh 2S , ⎪ ⎪ ⎪ ⎩ with β ∈ (1/2, 1] and |γ | ≤ 2S . 3 log q
Proof of Claim This is analogous to [4, Claim 7.4]. Note that for odd forms f ∈ H2 (q), we have L(1/2, f ) = 0 for all these f.
Let us now define
Nr (q) :=
1.
f ∈H2 (q) L(β+iγ ,f ) = 0 2S for some β∈(1/2,1], |γ |≤ 3 log q
By (2.10) and claim 7.1, we have A I1 (f ) + I2 (f ) + I3 (f ) ; q 1 A(q) + Nr (q) ≤ . 4 8S sinh π R 2S
That is A I1 (f ) + I2 (f ) + I3 (f ) ; q Nr (q) 1 ≤ − . A(q) 4 8S sinh π R A(q) 2S
It follows from Theorems 6.2 and 6.5, with the same argument as in [4, §7], that
Nr (q) 1 ≤ A(q) 8S sinh π2SR
∞ +
S
πu
πt cos 2S
log V (−R, t) dt
0
1 log V (u − R, S) du − + Oc (log q)−c , sinh 2S 4
(7.4)
0
for some constant c > 0, where (−u+iv)(1−5ϑ) − e(1−ϒ)(−u+iv)(1−5ϑ) 2 −2u e V (u, v) := 1 + e ϒ(−u + iv)(1 − 5ϑ) −2u
+ (1 − e
e−u(1−ϒ)(1−5ϑ) − e−u(1−5ϑ) ) . (uϒ(1 − 5ϑ))2
(7.5)
π . Then, by taking ϒ = 0.48 and R = 7, and Here we set ϑ = 10−10 and S = 2(1−ϒ)(1−20ϑ) a computer calculation of the integrals on the left-hand side of (7.4), we get
Nr (q) ≤ 0.5041, A(q)
(7.6)
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when q is sufficiently large. Hence the theorem follows easily.
8 Proof of Theorem 1.1 π In this section, we will take ϑ = 10−10 , ϒ = 0.64, R = 4.6 and S = 2(1−ϒ)(1−20ϑ) . Let J := [C log log q], where C is a large constant and [x] means the largest integer less than x. Set d := 2S/3 and define the regions: ⎧
⎪ β + iγ β > 12 , |γ | ≤ logd q , if j = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
jd (j+1)d Rj := if 1 ≤ j ≤ J − 1, β + iγ β ≥ 12 + log q , |γ | ≤ log q , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎩ β + iγ β ≥ 1 + Jd , |γ | ≤ 1 , if j = J, 2 log q and the zero counting sum Nj (q) :=
(where 0 ≤ j ≤ J ).
1,
f ∈H2 (q) L(s,f ) has at least one zero in Rj
For j = 0, with our new choice of R and ϒ, by the method in Sect. 7, we have
N0 (q) ≤ 0.60934. A(q)
(8.1)
For 1 ≤ j ≤ J − 1, let Bj be the rectangular box with vertices W0,j ± Hj and W1 ± Hj , where W0,j :=
1 jd/2 + , 2 log q
Hj :=
3(j + 1)d/2 . log q
(8.2)
By Selberg’s lemma, and the argument in Sect. 7, we have ⎡
Nj (q) 1 ⎢ ⎣ ≤ πj A(q) 6(j + 1)d sinh 6(j+1)
∞
πu + sinh 3(j + 1)d 0 + Oc (log q)−c ,
3 2 (j+1)d
0
πt cos 3(j + 1)d
log V (jd/2, t) dt
⎤ log V (u + jd/2, 3(j + 1)d/2) du⎦ (8.3)
for some constant c > 0. By a computer calculation of the integrals on the right-hand side of (8.3), we obtain the following bounds:
N1 (q) ≤ 0.21032, A(q)
N2 (q) ≤ 0.03758, A(q)
20 Nj (q) ≤ 0.00528, A(q) j=4
provided q is sufficiently large.
N3 (q) ≤ 0.00995, A(q) (8.4)
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To obtain similar bounds for 21 ≤ j ≤ J − 1, we will, instead, use the following trivial bound for u ≥ 20 1 ≤ V (u, v) ≤ 1 + e−0.35u . In this case 21≤j≤J −1
Nj (q) ≤ 0.001 A(q)
(8.5)
for suitable choice of C. Note that by [9, Theorem 4] we also have
NJ (q) (log q)−c , A(q)
(8.6)
when C is large enough. Thus by (8.1), (8.4), (8.5), and (8.6), we get J Nj (q) N (q) ≤ ≤ 0.88, A(q) A(q) j=0
for sufficiently large q. It immediately follows from the above that
M(q) = A(q) − N (q) ≥ 0.12 · A(q).
(8.7)
This completes the proof of our theorem. Author details 1 Department of Mathematics, Columbia University, New York, NY 10027, USA, 2 School of Mathematics, Shandong University, Jinan 250100, Shandong, China, 3 Present address: School of Mathematical Sciences, Tel Aviv University, 69978 Tel Aviv, Israel. Acknowledgements The second author would like to thank Professors Jianya Liu and Zeév Rudnick for their valuable advice and constant encouragement. Dorian Goldfeld is partially supported by NSA Grant H98230-16-1-0009, Bingrong Huang is supported by the European Research Council, under the European Union’s Seventh Framework Programme (FP7/2007-2013)/ERC Grant Agreement No. 320755. For Don Zagier on his 65th birthday.
Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 8 October 2017 Accepted: 27 February 2018
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