ACTA ARITHMETICA
111.4 (2004)
Lower bound of real primitive L-function at s = 1
by
Chun-Gang Ji (Nanjing) and Hong-Wen Lu (Shanghai)
1. Introduction and main results. Let χ be a real primitive Dirichlet
character modulo k (> 1). It is well known that if L(s, χ) has no zero in
the interval (1 − c1 /log k, 1), then L(1, χ) > c2 /log k, where c1 and c2 are
positive constants and c2 depends upon c1 . If, however, L(s, χ) has a real
zero close to 1, the only non-trivial lower bounds that are known for L(1, χ)
are ineffective. Siegel [9] proved that for any ε > 0,
c(ε)
,
kε
where c(ε) is an ineffective positive constant depending upon ε (see also
Chowla [1], Estermann [2], Goldfeld [3] and Goldfeld and Schinzel [4]).
Tatuzawa [10] proved that if 0 < ε < 1/11.2 and k > e1/ε , then with at
most one exception
0.655ε
L(1, χ) >
.
kε
L(1, χ) >
Hoffstein [6] proved that if 0 < ε < 1/(6 log 10) and k > 106 , then with at
most one exception
1
ε
L(1, χ) > min
,
.
7.735 log k 0.349k ε
In this paper we improve upon the result of Hoffstein. Using Lemma 4
of Hoffstein [5], the upper bound estimate of L(1, χ) of Louboutin [7], [8]
and some arithmetic theory of biquadratic bicyclic number fields, we prove
the following:
Theorem. Let 0 < ε < 1/(6 log 10), and χ be a real primitive Dirichlet character modulo k which is greater than 106 . Then with at most one
2000 Mathematics Subject Classification: Primary 11M20; Secondary 11R11, 11R42.
Key words and phrases: L-function, zeroes, quadratic number fields.
This work was supported by the National Natural Science Foundation of China, grant
number: 10201013 and 19531020.
[405]
406
C. G. Ji and H. W. Lu
exception,
L(1, χ) > min
1
32.260ε
.
,
7.7388 log k
kε
2. Several lemmas
√
Lemma 1 ([6, Lemma 2]). Set c = (3 + 2 2)/2 and let dF denote the
absolute value of the discriminant of a number field F 6= Q. Then ζF (s) has
at most one real zero β with β > 1 − 1/(c log dF ), and if it exists it is a
simple zero.
Lemma 2. Let K be an algebraic number field of degree n > 1 and
assume that for each m ≥ 1 there exists at least one integral ideal of K
of norm m2 (e.g. K is a quadratic or a biquadratic bicyclic number field ).
Assume also that 1/2 < β < 1 and ζK (β) ≤ 0. Then the residue at s = 1 of
the Dedekind zeta function ζK (s) of K satisfies
2
dK ζ n (3/2)(n + 1)!
n+2
β−1 π
κK ≥ (1 − β) x
− 2 3/2
− √
(x ≥ 1).
6
(4n − 3)π n
[ x]
x
Proof. According to the proof of Lemma 4 of Hoffstein [5],
2
π
1
n+2
− √
2n+1 (n + 1)! 6
[ x]
X 1
1
m2
≤ n+1
− (n + 1) 2
2
(n + 1)!
m2
x
√
m≤ x
X 1 1
m4 n+1
≤ n+1
1− 2
2
2
(n + 1)!
x
√ m
m≤ x
≤
X 1 1
m4 n+1
1
−
2β
2n+1 (n + 1)!
x2
√ m
m≤ x
X
1
1
(N (a))2 n+1
≤ n+1
1−
2
(n + 1)!
(N (a))β
x2
N (a)≤x
=
1
2πi
=J+
≤J+
2+i∞
2−i∞
ζK (s + β)xs
ds
s(s + 2) . . . (s + 2n + 2)
κK x1−β
ζK (β)
+
2n+1 (n + 1)! (1 − β)(3 − β) . . . (2n + 3 − β)
κK x1−β
(1 − β)2n+1 (n + 1)!
Lower bound of real primitive L-function at s = 1
407
where
1
J :=
2πi
satisfies
−1/2−β+i∞
−1/2−β−i∞
|J| ≤
ζK (s + β)xs
ds
s(s + 2) . . . (s + 2n + 2)
ζ n (3/2)
dK
x1−β ;
x3/2 (2π)n (4n − 3)
here we use the functional equation satisfied by ζK (s) and the bound |ζK (3/2
+ it)| ≤ ζ n (3/2). This completes the proof of Lemma 2.
√
Corollary 3. Let c = (3 + 2 2)/2 and K be a quadratic number field
with the absolute value of the discriminant dK > 106 . If ζK (β) ≤ 0 for some
β satisfying 1 − 1/(4c log dK ) ≤ β < 1, then κK ≥ 1.5063(1 − β).
Proof. Apply the previous
result with n = 2 and x = d0.95
for which
K
√
≥ exp(−0.95/(6 + 4 2)).
xβ−1
Lemma 4 (see [7] and [8]). (1) If χ is a primitive even Dirichlet character modulo k > 1, then
1
2 + γ − log(4π)
1
log k +
≤ (log k + 0.05).
2
2
2
(2) If χ is a primitive odd Dirichlet character modulo k, then
2 + γ − log π
1
1
≤ (log k + 1.44).
|L(1, χ)| ≤ log k +
2
2
2
Here γ = 0.577215 . . . denotes Euler’s constant.
|L(1, χ)| ≤
3. Proof of Theorem. Let 0 < ε < 1/(6 log 10) and let χ1 be a real
primitive Dirichlet character of least conductor k1 > 106 such that
1
(1)
L(1, χ1 ) ≤
7.7388 log k1
if it exists. By Corollary 3, L(1, χ1 ) has a real zero β1 such that
1
.
(2)
1 − β1 <
4c log k1
Let χ be another real primitive Dirichlet character modulo k > 106 different
from χ1 . By our choice of k1 we can assume
k√ ≥ k1 . Let the Kronecker
√
symbols of quadratic number fields Q( d), Q( d1 ) be χ, χ1 respectively,
where d√and√d1 are fundamental discriminants. Then |d| = k, |d1 |√= k1 . Let
F = Q( d, d1 ). Then it has another quadratic number field Q( d2 ) with
the√fundamental discriminant d2 and d2 | dd1 . Let the Kronecker symbol of
Q( d2 ) be χ2 with conductor k2 = |d2 |. Let
(3)
α = −3/2 − β1 ,
x = dA
F,
A > 0.8,
408
C. G. Ji and H. W. Lu
and notice that dF ≤ (kk1 )2 ≤ k 4 . Applying Lemma
√ 2 with n = 4 and
9.6 and 1−β < 1/(4c log k ) ≤ 1/((6+4 2) log 106 ) < 0.00621,
x = d0.8
>
10
1
1
F
we obtain
(4)
1.60452(1 − β1 ) < κF x1−β1 = L(1, χ)L(1, χ1 )L(1, χ2 )x1−β1 .
If L(s, χ) 6= 0 in the range 1 − 1/(4c log k) < s < 1, then by Corollary 3 we
have
1
1.5063
>
L(1, χ) >
4c log k
7.7388 log k
and the result follows. If L(β, χ) = 0 for some β such that 1 − 1/(4c log k) <
β < 1 then both β and β1 are zeros of ζF (s). Since β > 1 − 1/(4c log k) ≥
1 − 1/(c log dF ), we must have
1
1
(5)
1 − β1 >
≥
c log dF
2c log(kk1 )
by Lemma 1. Let A = 2/(1.5 + β1 ). Then
2A(1 − β1 ) =
4(1 − β1 )
4
1
0.1376
<
·
<
,
1
2.5 − (1 − β1 )
log k1
10c − log k1 log k1
by (3) and k1 ≥ 106 , and we have
(6)
A(1−β1 )
x1−β1 = dF
≤ (kk1 )2A(1−β1 ) < (kk1 )0.1376/log k1 .
From Lemma 4 we have
1
1
(7) L(1, χ2 ) < (log k2 + 1.44) ≤ (log(kk1 ) + 1.44) < 0.5261 log(kk1 )
2
2
for kk1 ≥ 1012 . Combining (1) and (4)–(7) we get
4.04947
(8)
L(1, χ) >
0.1376 .
log k 2
log k1
(kk
)
log k1 1 + log
1
k1
Let
η=
log k
,
log k1
η ≥ 1.
If η ≤ 7.54, then L(1, χ) > 1/(7.7388 log k) by
4.04947η
· 7.7388 > 1,
(η + 1)2 e0.1376(1+η)
1 ≤ η ≤ 7.54.
Hence we may assume η > 7.54; in this case 2 log(η + 1) < 0.5023(η + 1) and
2.13546
L(1, χ) >
.
(log k1 )k 0.6399/log k1
For fixed δ > 0, xeδ/x decreases until it reaches a minimum at x = δ. Let
x = log k1 , δ = 0.6399 log k. Then log k > 7.54 log k1 > (0.6425)−1 log k1 ,
Lower bound of real primitive L-function at s = 1
409
log k1 > 1/ε. By (8), we have
2.13546ε
ε
L(1, χ) > 0.6399ε = ε · 2.13546k 0.3601ε.
k
k
Since
2.13546k 0.3601ε > 2.13546e0.3601ε(7.54 log k1 ) > 2.13546e0.3601·7.54 > 32.260,
we have
L(1, χ) >
This completes the proof of Theorem.
32.260ε
.
kε
Acknowledgments. The authors thank the referee for his/her valuable suggestions and comments. The first author would like to thank for
support from the Morningside Center of Mathematics and the Institute of
Mathematics, Chinese Academy of Sciences.
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
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Department of Mathematics
Nanjing Normal University
Nanjing 210097
People’s Republic of China
E-mail: [email protected]
Department of Applied Mathematics
Tongji University
Shanghai 200092
People’s Republic of China
E-mail: [email protected]
Received on 15.4.2002
and in revised form on 16.4.2003
(4263)