A ZERO DENSITY ESTIMATE FOR THE SELBERG
CLASS
ANIRBAN MUKHOPADHYAY AND K SRINIVAS
Abstract. It is well known that bounds on moments of a specific
L−function can lead to zero-density result for that L−function.
In this paper, we generalize this argument to all L−functions in
the Selberg class by assuming a certain second power moment. As
an application, it is shown that in the case of symmetric-sqaure
L−function, this result improves the existing one.
To Professor K. Ramachandra with deepest regards
1. Introduction
We begin by recalling that Selberg class S is the class of complex
functions L(s) satisfying the following five axioms:
(i) L(s) can be written as an absolutely convergent Dirichlet series
for σ > 1;
(ii) (s−1)m L(s) is an entire function of finite order for some integer
m ≥ 0;
(iii) L(s) satisfies a functional equation of the form
Φ(s) = ω Φ̄(1 − s),
where
s
Φ(s) = Q
r
Y
Γ(λj s + µj )L(s)
j=1
say, with Q > 0, λj > 0, <µj ≥ 0 and |ω| = 1. Here f¯(s) = f (s̄);
(iv) The Dirichlet coefficients a(n) of L(s) satisfy the Ramanujan
conjecture a(n) n for every > 0;
(v) log L(s) is a Dirichlet series with coefficients b(n) satisfying
b(n) = 0 unless n = pm , for some prime p and integer m ≥ 1
and b(n) nθ with θ < 1/2.
2000 Mathematics Subject Classification. 11E45 (primary); 11M41(secondary).
Key words and phrases. Selberg class, zero-density estimates, symmetric-square
L-functions .
1
2
ANIRBAN MUKHOPADHYAY AND K SRINIVAS
For a survey of basic properties and results on S, see [6]. Condition
(v) implies that L(s) has no zeros in the half-plane σ > 1. From
−n−µ
the functional equation it follows that L(s) = 0 at s = λj j , j =
1, 2, · · · , r; n = 0, 1, 2, · · · . In analogy with the Riemann Zeta-function,
these zeros are termed as trivial zeros and from the theory of entire
functions of finite order, it can be seen that the other zeros ( also
called as non-trivial zeros) lie in the critical strip 0 ≤ σ ≤ 1.
Further, the number of zeros of L ∈ S in the critical strip satisfies a
formula of Riemann-von Mangoldt type. In fact, if we write
NL (T ) =| {ρ : L(ρ) = 0, ρ = β + iγ, 0 ≤ β ≤ 1, 0 ≤ γ ≤ T } |,
then a standard argument shows that for certain constants cL and dL
depending only on L and for T → ∞
dL
T log T + cL T + O(log T )
2π
The number dL is P
called the degree of L, denoted by deg L and is
defined as deg L = 2 rj=1 λj , the λ0j s are the constants appearing in
the functional equation (iii) defined above.
NL (T ) =
For L ∈ S, σ > 12 , let NL (σ, T ) denote the zero-counting function
NL (σ, T ) = |{ρ ∈ C|L(ρ) = 0, ρ = β + iγ, β ≥ σ, 0 ≤ γ ≤ T }|.
For each L ∈ S, one can reasonably assert the analogue of Riemann
Hypothesis (RH) by saying that L(ρ) = 0 ⇒ <ρ = 1/2 in the critical
strip.
In the absence of the truth or falsity of RH for functions in S, results
of the type
(1)
NL (σ, T ) ≤ T A(σ)(1−σ) logB T,
1/2 ≤ σ ≤ 1,
gain importance.
Our objective in this paper is to study (1) for L in S.
When d = 1, the functions in S have been completely characterized
e.g., they are ζ(s) and L(s + iθ, χ), where χ is a primitive Dirichlet
character mod q, q ≥ 2, and θ ∈ R (see [1]). For these functions the
zero-density results have been studied extensively (see chap. 11 of [5]).
For example, for ζ(s), Ingham [3] proved that
(2)
3
Nζ (σ, T ) ≤ T 2−σ (1−σ) log5 T,
uniformly for 1/2 ≤ σ ≤ 3/4 and Huxley [2] proved that
ZERO DENSITY ESTIMATE
(3)
3
3
Nζ (σ, T ) ≤ T 3σ−1 (1−σ) log44 T,
uniformly for 3/4 ≤ σ ≤ 1
For the corresponding zero-density estimates for L− functions, the
reader may refer to (chap. 12, [10] ).
Estimations of NL (σ, T ) depends on the availability of the power
moments
2k
Z T 1
dt T 1+ .
+
it
(4)
L
2
0
For example (3) is obtained from the well-known 4th power moment of
the Riemann zeta-function :
4
Z T
1
ζ( + it) dt < CT (log T )4
2
0
where C > 0 is a constant.
The corresponding 4th power-moments are not known, as of now, for
functions in S when the deg L > 1. In the case of functions of degree
2 the zero-density estimates are obtained by using the corresponding
2nd and 6th power-moments( see [4], [9] ).
In the general setting of Selberg class, Kaczorowski and Perelli [8]
showed that given L and > 0, there exists a constant C > 0 such that
NL (σ, T ) T C(1−σ)+
as T → ∞, uniformly for
was good enough.
1
2
≤ σ ≤ 1. For their purpose, C = 4dL + 12
In this paper, we study the zero-density estimate for L ∈ S when
σ is close to 1, by assuming a certain mean-square upper bound on
the critical line. We use the well known Montgomery zero-detection
method ( see chapter 12 of [10] ) in combination with a lemma of
Halasz and Montgomery (lemma 3 below) to obtain the main result.
The reader may refer to chapter 11 of [5] (as mentioned before) for a
nice account of this method and for various zero-density estimates for
ζ(s) in the critical strip.
More precisely we prove the following
Theorem 1. Let L ∈ S and α > 0. Assume that L satisfies
2
Z T L 1 + it dt T α+ .
2
0
4
ANIRBAN MUKHOPADHYAY AND K SRINIVAS
for any > 0. Then for 2/3 < σ ≤ 1, the following statements hold:
If σ ≥
1
then NL (σ, T ) T 2α(1−σ)+ .
α
Moreover, if 1 ≤ α < 3/2, then for 2/3 < σ < α1 , we have
2
NL (σ, T ) T σ (1−σ)+ .
Remark 1. In [11], Perelli defined a class of L-functions having meromorphic continuation to the complex plane C with at most a (simple)
pole at s = 1, admits a Dirichlet series expansion in some region, has
reasonable growth properties, and has an Euler product and a functional
equation of certain type. Examples of such general L-functions are the
Dirichlet L-series, the Dedekind zeta function, Hecke L-functions, and
zeta functions associated with cusp forms for SL(2, Z) which are eigenforms for the Hecke operators. For this class of functions he obtained
the corresponding mean-square upper bound with α = d/2, where d ≥ 2
denotes the degree of the function in the Selberg class sense. Applying
theorem 1 to this class of functions, we get
N (σ, T ) T d(1−σ)+
for 2/d ≤ σ < 1 and
2
N (σ, T ) T σ (1−σ)+
for 2/3 < σ < 2/d.
Remark 2. In the case of L-functions associated to cusp forms, which
are of degree 2, it is well known that α = 1 is allowed. In this case,
Theorem 1 gives
2
NL (σ, T ) T σ (1−σ)+
for 2/3 < σ < 1. This matches with the result of Ivic (see [4]) in the
range 3/4 < σ < 53/60.
As yet another application of Theorem 1, we consider the symmetricsquare L-function L(sym2 f, s) which is defined as follows:
Let f be a normalized Hecke eigen cusp form of even integral weight
2k for the full modular group. Then f has a Fourier expansion at the
cusp ∞, given by
f (z) =
X
n≥1
λf (n)n(k−1)/2 e(nz)
ZERO DENSITY ESTIMATE
5
where e(α) = e2πiα and λf (1) = 1. For
P <(s) > 1,−sthe L-function
attached to f is defined by L(f, s) =
n≥1 λf (n)n . This has a
functional equation and Euler product given by
Y
L(f, s) =
(1 − αp p−s )−1 (1 − βp p−s )−1
p
where αp + βp = λf (p) and αp βp = 1. We define the symmetric square
L-function by
Y
−1
−1
−1
L(sym2 f, s) =
1 − αp2 p−s
1 − αp βp p−s
1 − βp2 p−s
p
(5)
=
X
λf (n2 )n−s .
n≥1
It is well known that these functions also belong to the Selberg class
and are of degree 3. After suitable normalisation of Dirichlet coefficients, Sankaranarayanan’s result in [12] provides the following meansquare upper bound for L(sym2 f, s)
Lemma 1. Let > 0 and L(sym2 f, s) be as defined in (5). Then
2
Z T
L(sym2 f, 1 + it) dt T 3/2+ .
2
0
Using this mean-square estimate in combination with Montgomery’s
zero-detection method ( see chap.12, [10] ), he obtained the following
zero-density estimate in [13].
5
N (σ, T ) T 3−2σ (1−σ)+
for 2/3 < σ ≤ 1.
A direct application of Theorem 1 along with Lemma 1 immediately
gives the following improvement which we state as:
Corollary 1. Let L(s) = L(sym2 f, s) be as defined in (5) and > 0,
we have
NL (σ, T ) T 3(1−σ)+
uniformly in 2/3 < σ ≤ 1.
2. Proof of the theorem
P
We write L(s) = n≥1 a(n)n−s for <(s) > 1.
Let a0 (n) be the convolution inverse of the Dirichlet
a(n).
P coefficient
0
−s
These a0 (n)’s are defined recursively as 1/L(s) =
a
(n)n
for
n≥1
<(s) > 1.
6
ANIRBAN MUKHOPADHYAY AND K SRINIVAS
We write
P (z) =
Y
p, z ≥ 2.
p≤z
The following lemma on the size of a0 (n) is due to Kaczorowski and
Perelli (see Lemma 1 of [8]).
Lemma 2. Let L ∈ S. Then for every > 0 there exists z = z() such
that a0 (n) n for (n, P (z)) = 1.
Let > 0 be arbitrarily small, X = T a and Y = T b for some
constants a, b > 0 to be chosen later, z = z() be as in the above
lemma, Lp (s) denotes the p-th Euler factor of L(s) and write
Y
X
a0 (n)
F (s, z) =
Lp (s), MX (s, z) =
.
ns
p>z
n≤X,(n,P (z))=1
Note that the zeros of F (s, z) in σ ≥ 21 coincide with those of L(s) in
the same region, since Lp (s) 6= 0 for σ ≥ 21 .
Then we have
X c(n, z, X)
MX (s, z)F (s, z) = 1 +
ns
n>X
where by Lemma 2, the coefficients c(n) = c(n, z, X) satisfy
c(n) n .
From now onward we write F (s) and MX (s) in place of F (s, z) and
MX (s, z) for notational simplicity. Let ρ = β + iγ be a zero of F (s)
with β ≥ σ > 1/2. Consider the Mellin transform
Z 2+i∞
1
−n/Y
Γ(w)Y w n−w dw.
e
=
2πi 2−i∞
This gives
Z 2+i∞
X
1
−1/Y
−ρ −n/Y
e
+
c(n)n e
=
F (ρ+w)MX (ρ+w)Γ(w)Y w dw.
2πi 2−i∞
n>X
We move the line of integration to <(w) = 12 − β. Γ(w) has a simple
pole at w = 0 which is taken care of by the zero of F (ρ + w) at w = 0.
Other possible pole of the integrand is at w = 1 − ρ.
We now recall the well-known Stirling’s formula which states that in
any fixed strip α ≤ σ ≤ β, as t → ∞, we have
|Γ(σ + it)| = (2π)1/2 |t|σ−1/2 e−π|t|/2 (1 + O(1/|t|)) .
ZERO DENSITY ESTIMATE
7
Note that if |γ| ≥ (log T )2 , then by Stirling’s formula the residue at
w = 1 − ρ is o(1). Now we observe, again by Stirling’s formula, that
Z
= o(1)
<(w)=1/2−β
1
+
2πi
Z
(log T )2
F
−(log T )2
1
1
1
+ iγ + iv MX
+ iγ + iv Γ
− β + iv Y
2
2
2
Also
X
c(n)n−ρ e−n/Y = o(1) as Y → ∞,
n>Y (log Y )2
since
X
X
c(n)n−ρ e−n/Y n>Y (log Y )2
n−1/2+ e−n/Y
n>Y (log Y )2
− 12 (log Y )2
X
e
n−1/2+ e−n/2Y
n>Y (log Y )2
1
2
e− 2 (log Y ) Y
1
+
2
(log Y )1+ .
But, e−1/Y → 1 as Y → ∞. Hence each zero ρ of F (s) satisfies one
of the following conditions
X
(6)
c(n)n−ρ e−n/Y 1,
X<n≤Y (log Y )2
1
(7)
2πi
Z
(log T )2
F
−(log T )2
1
1
1
+ iγ + iv MX
+ iγ + iv Γ
− β + iv
2
2
2
1
Y 2 −β+iv dv 1,
(8) |γ| ≤ (log T )2 .
The number of zeros satisfying the third condition is ≤ NF ((log T )2 ) (log T )2 log log T (log T )3 . Let R1 and R2 be the number of zeros
satisfying the condition (6) and condition (7) respectively and having
a distance of (log T )4 between their ordinates. Then
(9)
NL (σ, T ) (R1 + R2 + 1) log5 T.
Using dyadic division of [X, Y (log Y )2 ], we can say that each ρ
counted in R1 satisfies
X
1
(10)
c(n)n−ρ e−n/Y log Y
N <n≤2N
1
−β+iv
2
dv.
8
ANIRBAN MUKHOPADHYAY AND K SRINIVAS
for some N satisfying X < N ≤ Y (log Y )2 /2. If N ≥ Y 1/2 , then put
M = N and if N < Y 1/2 , then choose a positive integer k ≥ 2 such
that
N k < Y (log Y )2 ≤ N k+1 .
So, N k ≥ Y k/k+1 ≥ Y 1/2 . We observe that N k < Y (log Y )2 implies
that k ≤ 2 log Y / log N < 2 log Y / log X = 2b/a, hence k 1 where
the constant depends on as a = is going to be the final choice for a.
Raising (10) to k-th power we get
X
b(n)n−ρ N k <n≤(2N )k
1
(log Y )k
where some of the b(n)’s are 0 and b(n) n .
We again make a dyadic partition of this sum as
X
X
X
=
+
··· +
N k <n≤(2N )k
N k <n≤2N k
2N k <n≤4N k
X
2k−1 N k <n≤(2N )k
to obtain
X
(11)
b(n)n−ρ M <n≤2M
1
(log Y )k+1
for some M of the form 2j N k with 0 ≤ j ≤ k − 1. Hence for a zero
ρ counted in R1 , we can always get an M , Y 1/2 ≤ M = 2j N k ≤
2k−1 N k N k ≤ Y (log Y )2 satisfying
X
1
b(n)n−ρ .
(log Y )D
M <n≤2M
for some D ≥ 1 and b(n) n . We conclude that
X X
−ρ R1 b(n)n (log T )D
ρ
M <n≤2M
for some D ≥ 1. Using partial summation we get
X
X
σ−β
−σ−iγ R1 M
b(n)n
(log T )D .
ρ
M <n≤2M
where β is the real part of ρ. In the preceeding equation n actually
runs over a subset of (M, 2M ], we take b(n) = 0 outside this subset of
ZERO DENSITY ESTIMATE
9
(M, 2M ]. Let t1 , · · · , tR1 be the imaginary parts of the zeros counted
in R1 . Since β ≥ σ, we get from the last inequality
X X
−σ−itr b(n)n
(12)
R1 (log T )D .
r≤R1 M <n≤2M
Now we state a lemma due to Halasz and Montgomery ( see page 494
of [5]).
Lemma 3. Let ξ, φ1 , · · · , φR are arbitrary vectors in an inner product
space over C. Then
!1/2
X
X
|(ξ, φr )| ≤ ||ξ||
|(φr , φs )|
.
r≤R
r,s≤R
Now we use this lemma with ξ = (ξn )n , ξn = b(n)n−σ for M ≤ n ≤
2M and zero otherwise and φr = (φr,n )n , φr,n = n−itr for M ≤ n ≤ 2M
and zero otherwise. We see that
X X |b(n)|2
b(n)n−σ 2 =
||ξ||2 =
T M 1−2σ .
2σ
n
M ≤n≤2M
M ≤n≤2M
Substituiting the above expression in (12), we obtain
!
X X
R12 T M 1−2σ
nitr −its r,s M ≤n≤2M
!!
X X
1−2σ
itr −its = T M
M R1 +
n
(13)
.
r6=s M ≤n≤2M
The following lemma estimates the inner sum of the last expression.
Lemma 4. With the notations as in (13), we have
X
ni(tr −ts ) M |tr − ts |−1 + T 1/2 .
M ≤n≤2M
Proof. We write
X
M <n≤2M
ni(tr −ts ) =
X
eif (n)
M <n≤2M
where f (x) = (tr −ts ) log x. Let us first suppose that 0 < tr −ts < M/2.
Then observing that f 0 (x) is monotonic with f 0 (x) ≥ (tr − ts )/2M > 0,
and f 0 (x) ≤ (tr − ts )/M ≤ 1/2, we get
Z 2M
X
1
M
if (n)
if (x)
e
=
e
dx + O
.
tr −ts
tr − ts
1− M
M
M <n≤2M
10
ANIRBAN MUKHOPADHYAY AND K SRINIVAS
Next, suppose that tr − ts ≥ M/2. Recall the following approximate
functional equation for the Riemann zeta function (see page 97 of [5])
X 1
X 1
1
ζ(s) =
+
χ(s)
+ O(x−σ ) + O(t 2 −σ y σ−1 )
s
1−s
n
n
n≤x
n≤y
t
for 0 ≤ σ ≤ 1, x, y, t > C > 0 and 2πxy = t. Using y = 2πx
,
X 1
X
1
−σ
− 12 1−σ
ζ(s) =
+
χ(s)
+
O(x
)
+
O(t
x ).
s
1−s
n
n
n≤x
n≤t/2πx
Now putting x = 2X and x = X and then subtracting, we obtain
X 1
X
1
−σ
− 12
0=
+
χ(s)
+
O(X
)
+
O(t
X 1−σ ).
s
1−s
n
n
X<n≤2X
t/4πX<n≤t/2πX
Let s = 0 − i(tr − ts ) and X = M . Then
X
X
1
1
1
0=
+χ(−i(t
−t
))
+O((tr −ts )− 2 M ).
r
s
−i(t
−t
)
1+i(t
−t
)
r
s
r
s
n
n
(tr −ts )
(tr −ts )
M <n≤2M
4πM
<n≤
− 21
− 12
We observe that (tr − ts ) M (tr − ts )
1
χ(−i(tr − ts )) (tr − ts ) 2 . Also
X
1
(tr −ts )
(tr −ts )
<n≤ 2πM
4πM
n1+i(tr −ts )
2πM
1
(tr − ts ) = (tr − ts ) 2 and
1.
Hence, in this case,
X
1
1
ni(tr −ts ) (tr − ts ) 2 T 2 .
M <n≤2M
This completes the proof of the lemma.
Now we return to the proof of the theorem. From (13) using the last
lemma we get
X
R12 T R1 M 2−2σ + T M 2−2σ
|tr − ts |−1 + T R12 T 1/2 M 1−2σ .
r6=s
4
Since |tr − ts | (log T ) we have
X
|tr − ts |−1 R1 (log T )−3 .
r6=s
Thus
R12 T R1 M 2−2σ + T M 2−2σ R1 + T R12 T 1/2 M 1−2σ .
If T 1/2 M 1−2σ M − i.e T M 4σ−2− then R12 T R1 M 2−2σ implying R1 T M 2−2σ .
ZERO DENSITY ESTIMATE
11
Note that the above argument is valid for any interval of length atmost T0 = M 4σ−2− in [0, T ]. Therefore, dividing [0, T ] into subintervals
of length atmost T0 and letting R0 to be the representative zeros of R1
in this interval, we obtain R0 T M 2−2σ .
Thus
T
R1 R0 1 +
T M 2−2σ + T M 4−6σ .
T0
Using the fact that Y 1/2 ≤ M ≤ Y (log Y )2 and σ > 2/3, this gives
(14)
R1 T Y 2−2σ + T Y 2−3σ .
Let us turn to estimate R2 . Let γr ; r = 1, · · · , R2 be the imaginary
parts of the zeros counted in R2 . Let
1
1
0
F
F
+ iγr + iv =
max
+
iγ
r + iv −(log T )2 ≤v≤(log T )2
2
2
and tr = γr + v 0 . Then from the condition defining R2 we deduce that
F 1 + itr MX 1 + itr Y 21 −σ 1.
2
2
Putting X = T this gives
1
F ( + itr ) T Y
2
1
−σ
2
1.
We can now conclude that
R2 T Y
2( 21 −σ)
2
X 1
F ( + itr ) .
2
r≤R2
Now we recall the assumption that
2
Z T
1
L( + it) dt T α+ .
2
0
From this we can easily deduce(see page 200 of [5])
2
X 1
F ( + itr ) T α+ .
2
r≤R
2
Hence, putting together we get
(15)
R2 T α+ Y 1−2σ .
From (9), (14) and (15) we get
NL (σ, T ) T Y 2−2σ + T Y 2−3σ + T α Y 1−2σ .
12
ANIRBAN MUKHOPADHYAY AND K SRINIVAS
For σ ≥
1
α
choosing Y = T α , we get
NL (σ, T ) T 2α(1−σ)+ .
If 1 ≤ α < 3/2 then for 2/3 < σ <
1
α
we choose Y = T 1/σ . This gives
2
NL (σ, T ) T σ (1−σ)+ .
This completes the proof of the theorem.
Acknowledgments: The authors wish to thank Professor R. Balasubramanian for some useful discussions.
References
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Institute of Mathematical Sciences, CIT Campus, Tharamani, Chennai 600 113, India
E-mail address, Anirban Mukhopadhyay: [email protected]
E-mail address, K Srinivas: [email protected]