Real Zeros Of Eisenstein Series And Rankin-selberg L-functions

Acta Math. Hungar., 2007

DOI: 10.1007/s10474-007-5102-1

REAL ZEROS OF EISENSTEIN SERIES AND RANKINSELBERG L-FUNCTIONS C. BAUER1 and Y. WANG2 1

2

∗

Dolby Laboratories, San Francisco, CA, USA e-mail: [email protected]

Department of Mathematics, Capital Normal University, Beijing 100037, P. R. China e-mail: [email protected] or [email protected]

(Received April 30, 2005; revised August 1, 2006; accepted November 2, 2006)

Abstract. We prove that the Eisenstein series E(z, s) have no real zeroes for s ∈ (0, 1) when the value of the imaginary part of z is in the range 15 < Im z < 4.94. For very large and very small values of the imaginary part of z , E(z, s) ¢ ¡ have real zeros in 12 , 1 , i.e. GRH does not hold for the Eisenstein series. Using these properties, we prove that the RankinSelberg L-function attached with the Ramanujan τ -function has no real zeros in the critical strip, except at the central point s = 21 .

1. Introduction Let f ∈ Sk (Γ), where Sk (Γ) is the space of cusp forms of weight k and Γ := SL2 (Z). At the cusp ∞ it has the Fourier series expansion (1)

f (z) =

∞ X

an e(nz).

n=1

∗ The second author is supported by China NSF Grant #10G01034, China 973 Project #2007CB8079002007CB807903 and Morning Side center. Key words and phrases: Eisenstein series, RankinSelberg L-function. 2000 Mathematics Subject Classication: primary 11F03, 11M36, 11M20.

c 2007 Akadémiai Kiadó, Budapest 02365294/$ 20.00 °

2

C. BAUER and Y. WANG

We dene the L-function Lf (s) =

∞ X

an n−s ,

Re s >

n=1

k+1 , 2

and the famous RankinSelberg L-function (2)

Lf ×f¯(s) =

∞ X

|an |2 n−s ,

Re s > k.

n=1

Lf ×f¯(s) is related with the theory of Eisenstein series E(z, s) by the following formula (Rankin [9], or see a proof in [2, p. 72]): (4π)−(s+k−1) π −s ζ(2s)Γ(s)Γ(s + k − 1)Lf ×f¯(s + k − 1) Z ¯ ¯ 2 dx dy =2 E(z, s)¯ f (z)¯ y k 2 , y Γ\H

where E(z, s) is the Eisenstein series dened in (5). First, we prove Theorem 1. For 51 < Im z < 4.94, we have (3)

E(z, s) < 0

for s ∈ (0, 1),

but for any Im z > 90, there exists s0 ∈ ( 12 , 1) such that (4)

E(z, s0 ) = 0.

While Theorem 1 states a new result and is of independent interest, we mainly use it as a tool to prove an interesting property of the RankinSelberg L-function. In particular, we state the following conjecture: Conjecture 1. The RankinSelberg L-function has no real zeros for s ∈ (0, 1) except at s = 12 , i.e., Lf ×f¯(s + k − 1) 6= 0

1 for s ∈ (0, 1) except s = , 2

for holomorphic modular forms f . This conjecture states that Lf ×f¯(s), and in consequence also the associated symmetric square L-function, do not vanish on the real line in the critical strip. The conjecture can be justied numerically by methods outlined in [4]. It is also known that the conjecture does not hold for certain Maass-forms. Acta Mathematica Hungarica, 2007

EISENSTEIN SERIES AND RANKINSELBERG L-FUNCTIONS

3

Compared with the well-known result ζ(s) < 0 for s ∈ (0, 1), which can be easily proved by elementary methods or using the Mellin transform of ζ(s), this conjecture is quite dicult to prove. We know from the article of Hostein and Ramakrishnan [6], that Lf ×f¯(s + k − 1) has no Siegel zeros, i.e. it has no real zeros near its pole at s = 1. On the other side, Lapid and Rallis [8, Theorem 1] proved the nonnegativity of L(1/2, π) for a symplectic cuspidal automorphic representation π of GLn (A) where A is the ring of adeles of a number eld F . As a simple consequence, µ ¶ 1 Lf ×f¯ k − 5 0, 2 ¡ ¢ for f ∈ Sk SL2 (Z) . Both of the above methods do not allow to derive results for the behavior of the RankinSelberg L-function in the whole interval s ∈ (0, 1). In this paper, we use analytic methods to prove Conjecture 1 for the case k = 12, and all k 5 12. Theorem 2. If f (z) = ∆(z) is the Jacobi discriminant function, i.e. k = 12, then s = 12 is a simple zero of L∆×∆ ¯ (s + k − 1) and ¶ µ 1 , L∆×∆ ¯ (s + k − 1) > 0 for s ∈ 0, 2 µ ¶ 1 L∆×∆ ,1 . ¯ (s + k − 1) < 0 for s ∈ 2 Even though Theorem 2 only states a result for the case k = 12, the proof will show that it holds more generally for all k 5 16. The upper bound k 5 16 is due to the fact that Lemma 3, which is needed for the proof of Theorem 2, only holds for k 5 16. With regard to the real zeros of Lf (s), we know from Hostein and Ramakrishnan [5] that Lf (s) has no Siegel zeros. W. Kohnen and D. Zagier [7] have shown that at the central point s = k2 , we have µ ¶ k Lf = 0. 2

However, no general result is known for s ∈ (0, 1). Particular functions have been evaluated numerically by Dokchitser [4]. The method proposed in this paper can only be applied to L-functions that have positive coecients as the RankinSelberg L-function, but not to Lf (s) where the coecients are not necessarily positive. Acta Mathematica Hungarica, 2007

4

C. BAUER and Y. WANG

2. Real zeros of Eisenstein series As above, we set Γ := SL2 (Z). Recall the denition of the Eisenstein series for SL2 (Γ) for Re s > 1, y = Im z > 0 as X ys 1 (5) E(z, s) = π −s Γ(s) 2 |mz + n|2s m,n∈Z (m,n)6=(0,0)

X 1 = π −s Γ(s)ζ(2s) 2

m,n∈Z (m,n)=1

X ys −s Im (γz)s , 2s = π Γ(s)ζ(2s) |mz + n| γ∈Γ \Γ

where

∞

½ Γ∞ =

µ ¶¯ ¾ 1 n ¯¯ ± n∈Z . 1 ¯

The Fourier series expansions of E(z, s) is (see a proof in [2, p. 66]) (6)

E(z, s) = π −s Γ(s)ζ(2s)y s + π s−1 Γ(1 − s)ζ(2 − 2s)y 1−s +4

∞ X

1 √ ns− 2 σ1−2s (n) yKs− 1 (2πny) cos (2πnx), 2

n=1

where σs (n) is the divisor sum function X σs (n) = ds , d|n

and Kv (z) is the Bessel function dened for arbitrary v and | arg z| < Z 1 ∞ −z(t+t−1 )/2 ν−1 Kν (z) = e t dt. 2 0

π 2

by

Denote the last sum in (6) by H(s, z) =

∞ X

1 √ ns− 2 σ1−2s (n) yKs− 1 (2πny) cos (2πnx). 2

n=1

From the functional equation E(z, s) = E(z, 1 − s) the relation (17), and the functional equation of ζ(s), we derive the functional equation of H(s, z) as (7)

H(s, z) = H(1 − s, z).

In the following, we will prove that the contribution of H(s, z) to E(z, s) is very small for s ∈ (0, 1). We rst prove the following lemma. Acta Mathematica Hungarica, 2007

EISENSTEIN SERIES AND RANKINSELBERG L-FUNCTIONS

Lemma 1.

5

For s ∈ [0, 1],

−2πy ¯ ¯ ¯ H(s, z)¯ 5 1 e−2πy + 3 e−4πy + 3 e−6πy 2 − e . 2 4 4 (1 − e−2πy )2

Thus, for y >

√

3 2 ,

¯ ¯ ¯ H(s, z)¯ < 0.0022.

Proof. We prove the lemma for the range s ∈ [ 12 , 1]. Then, the lemma

follows for the whole range [0, 1] from the functional equation (7). Bateman and Grosswald [1, Lemma 3] proved that for y > 0, and 12 5 s 5 1, (4ny)1/2 e2πny Ks− 1 (2πny) 5 1.

(8)

2

Using (8) and the following estimates (see [1, p. 371]), ns−1 σ1−2s (n) 5 σ−1 (n), 3 σ−1 (n) 5 (n − 1), 2

for

1 5 s 5 1, 2

for n = 2,

we bound the sum in H(s, z) starting from its third term: ¯ ¯X ¯ ¯ ∞ s− 1 √ ¯ S3 := ¯ n 2 σ1−2s (n) yKs− 1 (2πny) cos (2πnx)¯¯ 2 n=3

5

∞ X

1 √ ns− 2 σ1−2s (n) y(4ny)−1/2 e−2πny

n=3

5

∞ X 3 n=3

3 2 − e−2πy (n − 1)e−2πny 5 e−6πy . 4 4 (1 − e−2πy )2

Using (8) again, we nd for the rst term of H(s, z): ¯√ ¯ 1 ¯ ¯ S1 = ¯ yKs− 1 (2πy) cos (2πx)¯ 5 e−2πy , 2 2 and for the second term: ¯ ¯ 1 √ ¯ ¯ S2 = ¯2s− 2 σ1−2s (2) yKs− 1 (4πy) cos (4πx)¯ 2

Acta Mathematica Hungarica, 2007

6

C. BAUER and Y. WANG

¯ ¯ ¯ ¯ s− 12 1−2s √ = ¯2 (1 + 2 ) yKs− 1 (4πy) cos (4πx)¯ 2

³ ´ 1 1 3 5 2s− 2 + 2 2 −s 8−1/2 e−4πy 5 e−4πy . 4

Combining the estimates S1 , S2 and S3 , the lemma is proved for s ∈ [ 21 , 1]. Using the functional equation (7), the lemma follows for s ∈ [0, 1]. ¤ For the proof of Theorem 1, we now estimate the main term of E(z, s) as given in (6) and dene for xed y : F (s) := π −s Γ(s)ζ(2s)y s + π s−1 Γ(1 − s)ζ(2 − 2s)y 1−s .

(9) Let

ω(u) =

∞ X

e−n

2 πu

.

n=1

Using the Mellin transform, we have (see [3, p. 62]) Z ∞³ ´ 1 1 −s + us−1 + u−s− 2 ω(u) du, π Γ(s)ζ(2s) = 2s(2s − 1) 1 Z ∞³ ´ 3 1 s−1 π Γ(1 − s)ζ(2 − 2s) = + us− 2 + u−s ω(u) du. (2s − 1)(2s − 2) 1 Hence, ys y 1−s + 2s(2s − 1) (2s − 1)(2s − 2) Z ∞³ Z ∞³ ´ ´ 3 s s−1 −s− 12 1−s +y u +u ω(u) du + y us− 2 + u−s ω(u) du. F (s) =

1

1

=: F1 (s) + y s F2 (s) + y 1−s F3 (s).

We transform the rst expression as follows: ¡ ¢ 1−s s 2 y s + y 1−s + y s−−y 1 ys y 1−s 2 F1 (s) = + = . 2s(2s − 1) (2s − 1)(2s − 2) 8s(s − 1)

Hence, µ

(10)

F (s) = y

s

1 + 4s(s − 1)

Acta Mathematica Hungarica, 2007

Z

∞³

s−1

u 1

−s− 12

+u

¶ ´ ω(u) du

EISENSTEIN SERIES AND RANKINSELBERG L-FUNCTIONS

µ + y 1−s +

1 + 4s(s − 1)

Z

∞³

7

¶ ´ 3 us− 2 + u−s ω(u) du

1

1 y 1−s − y s T (s) =: G(s) + . 1 8s(s − 1) s − 2 8s(s − 1)

For s ∈ (0, 1), we estimate the integrals in the rst two terms of the righthand side of (10) as follows: Z ∞³ ´ 1 us−1 + u−s− 2 ω(u) du 1

Z

∞³

5

u

s−1

−s− 21

+u

¶ ∞ ´µ X −πu −mπu e + e du

1

m=4

¶ Z ∞µ e−4πu 2e−4π 2 −π −πu 52 e + e + < 0.0276, du 5 1 − e−π π 4π(1 − e−π ) 1

and similarly Z

∞³

s− 32

u 1

−s

+u

Z ´ ω(u) du 5 2

∞ ∞µX

1

Note that

¶ e

−n2 πu

du 5 0.0276.

n=1

1 5 −1 for s ∈ (0, 1), 4s(s − 1)

and

1

y s + y 1−s = 2y 2

for y > 0,

which implies ¡ ¢ 1 G(s) 5 (−1 + 0.0276) y s + y 1−s 5 −1.9448y 2 .

Hence, ¡ ¢ F (s) 5 −0.9724 y s + y 1−s +

1 y 1−s − y s . 8s(s − 1) s − 21

Notice that 1

(11)

1

−s 1 y2 y 1−s − y s − y s− 2 2 T (s) = = y s − 12 s − 12

Acta Mathematica Hungarica, 2007

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C. BAUER and Y. WANG

à = −2y

(ln y)3 ln y + 3!

1 2

! µ ¶ µ ¶ 1 2 (ln y)2k+1 1 2k s− + ··· + s− + ··· , 2 (2k + 1)! 2

which follows from the Taylor expansion of Ax . Hence if 0 < y 5 1, T (s) is a 1 convex function with value 2(1 − y) at s = 0, 1 and minimum 2y 2 ln y −1 = 0 at s = 12 . Thus, 1

(12)

1

F (s) 5 −1.9448y 2 + 2

1 y 2 ln y −1 5 −(1.9448 + ln y −1 )y 2 . 8s(s − 1)

We see from Lemma 1 and (12), 1

E(z, s) = F (s) + 4H(s, z) 5 −(1.9448 + ln y −1 )y 2 µ ¶ 1 −2πy 3 −4πy 3 −6πy 2 − e−2πy +4 e + e + e . 2 4 4 (1 − e−2πy )2

(13)

Setting δ0 = 15 , we see from (13) that the Eisenstein series E(z, s) has no real zeros for δ0 < y 5 1. i.e. (14)

E(z, s) < 0 for alls ∈ (0, 1) and1/5 < Im z 5 1.

If y > 1, we apply the modularity of E(z, s),  µ ¶ 1    E(z, s) = E − z , s , i.e. (15) µ ¶  x y   +i 2 ,s , E(x + iy, s) = E − 2 x + y2 x + y2 and nd that for all s ∈ (0, 1) 1 5 Im z 5 4.94, 5 √ since in the fundamental region, if 1 < y 5 4.94 5 6 + 2.5, then E(z, s) < 0 for

5

y x2 +y 2

5

1 y

< 1. This proves the inequality (3) of Theorem 1.

Acta Mathematica Hungarica, 2007

1 5

5

y 1 +y 2 4

9

EISENSTEIN SERIES AND RANKINSELBERG L-FUNCTIONS

2.1. Analysis for large y . In the previous section, we have shown that

E(z, s) has no real zero in the critical strip for 51 < y < 4.94. In this section, we rst show that for all values of y that are very large, i.e., larger than a constant c1 , E(z, s) has a real zero in the critical strip. Second, we use these results to show that E(z, s) also has a real zero in the critical strip for a subset of all z that have a very small imaginary part y . First consider large values of y . We see F

µ ¶ ¡ ¢ 1 = π −s Γ(s)ζ(2s)y s + π s−1 Γ(1 − s)ζ(2 − 2s)y 1−s s= 1 , 2 2

which is a removable pole, and by our formulas (10) and (11), this expression is equal to µ ¶ µ ¶µ ¶ 1 1 1 1 F = y 2 −1 + ln y 2 2 2s(1 − s) s= 1 2

+y

1 2

µZ

∞³

s−1

u

−s− 12

+u

s− 32

+u

−s

+u

¶ ´ ω(u) du

,

s= 21

1

which shows that F ( 12 ) > 0 for y > e2 . We obtain by a similar calculation that µ ¶ 1 > 2 if y > 90. F 2 This is not the sharpest bound possible, but it is sucient for our application. It implies ¯ E(z, s)¯ s= 1 = F 2

µ ¶ µ ¶ 1 1 + 4H , z > 0 for y > 90, 2 2

because H(s, z) is very small for large y by Lemma 1. Hence, we can show that E(z, s) has a real zero s0 ∈ ( 21 , 1). Since (16)

lim E(z, s) = lim

s→1−

s→1−

1 = −∞ < 0, s(s − 1)

there must be an s0 ∈ ( 12 , 1) at which E(z, s) vanishes. This shows that the General Riemann Hypothesis does not hold for E(z, s). Indeed, as the E(z, s) do not have Euler Products, this result could have been expected. This proves formula (4) of Theorem 1. Acta Mathematica Hungarica, 2007

10

C. BAUER and Y. WANG

Remark 1. Let X + denote the fundamental domain of SL2 (Z)\H de-

ned by − 21 5 Re z 5 12 , |z| = 1, and let X− denote the image of X + under the mapping − z1 . If z ∈ X− , and y = Im z is very small, then Im −1 z will be very large. Thus, by using the modularity (15) of E(z, s), we see that there exists a real zero s0 ∈ ( 21 , 1) of E(z, s) for such z . For those z , we see from (9) that F (s) has a very small absolute value, hence in view of (16) and (13), H(s, z) must grow to innity as y → 0.

3. Analysis of the RankinSelberg L-function In this section, we prove Theorem 2. We initially argue for any even k = 12 and then specialize on the case k 5 16 in Lemma 3. Set à ! X0 π|mz + n|2 exp − u , K(u) = K(u, z) = y m,n where the prime in the summation means that we sum only over pairs(m, n) such that (m, n) 6= (0, 0). Using Poisson summation and Mellin transform, Rankin proved that [9] Z ¡ ¢ 1 ∞ 1 (17) E(z, s) = K(u) u−s + us−1 du + , 2 1 2s(s − 1) which implies both the analytic continuation and the functional equation of E(z, s). Note that Z ¡ ¢ 1 1 ∞ + (18) E (z, s) := E(z, s) − = K(u) u−s + us−1 du, 2s(s − 1) 2 1 is a positive and convex function in s, if s ∈ R, since K(u) is positive. ¡ ¢ Let f (z) ∈ Sk SL2 (Z) , i.e. µ ¶ a b f (γz) = (cz + d)k f (z) for γ = ∈ SL2 (Z), c d and f has a Fourier series expansion as in (1). We now proceed to prove that the corresponding RankinSelberg L-function Lf ×f¯(s + k − 1) as dened in (2) has no real zeros for s ∈ (0, 1) with the exception of a simple zero at s = 12 . It is well-known [2, p. 72] that (19)

(4π)−(s+k−1) π −s ζ(2s)Γ(s)Γ(s + k − 1)Lf ×f¯(s + k − 1) Z ¯ ¯ 2 dx dy E(z, s)¯ f (z)¯ y k 2 . =2 y Γ\H

Acta Mathematica Hungarica, 2007

11

EISENSTEIN SERIES AND RANKINSELBERG L-FUNCTIONS

It will be sucient to prove that the right-hand side of (19) is negative. We notice that ζ(2s) has a simple pole at s = 12 , which implies that s = 12 must be a simple zero of Lf ×f¯(s + k − 1). We also notice that ζ(s) < 0 for s ∈ (0, 1), and ζ(s) > 0 for s > 1. Thus Theorem 2 follows. We see from the denition of Γ\H and Theorem 1 that it is sucient to prove that 1

Z2 Z∞

(20)

¯ ¯ 2 dx dy E(z, s)¯ f (z)¯ y k 2 < 0 for s ∈ (0, 1). y

t(s) =: − 12 1

Using (18), we rewrite the integral t(s) in (20) as Z 1 Z ∞ ¯ ¯ 2 1 ¯ f (z)¯ 2 y k dx dy t(s) = 2 2s(s − 1) 1 y − 1 2

Z + Z =

1 2

− 12

Z

1 2

− 12

Z

∞

1

¯ ¯ 2 dx dy E + (z, s)¯ f (z)¯ y k 2 y

 ¯ ¯ 2 k dx dy  R 12 R ∞ + ¯ ¯ y 1 1 E (z, s) f (z) ¯ 2 k dx dy y2 1 −2 ¯ f (z)¯ y .  + 1 ¯ ¯ R 2 R∞ y2 2s(s − 1) ¯ f (z)¯ 2 y k dx dy

∞¯

1

− 12

1

y2

Thus, for the proof of (20), it is sucient to show that ¯ ¯ 2 k dx dy R 12 R ∞ + ¯ ¯ y 1 1 E (z, s) f (z) y2 −2 (21) < 2 for s ∈ (0, 1). ¯ 2 dx dy R 12 R ∞ ¯ k ¯ ¯ f (z) y y2 −1 1 2

For this purpose, we show the following lemma: Lemma 2. For s ∈ [0, 1], Re z ∈ [ − 12 , 12 ] and Im z > 1, we have à !2 −1 e−4πy + −πy −1 −πy −1 0 < E (z, s) < 4.19ye + 2y e + . 1 − e−πy−1 Proof. The left inequality is obvious by (18). For the right side, consider

the integral (22)

Ã

! π|mz + n|2 exp − u u−s du y 1 Z ∞ ¡ ¢ −s+1 =y exp − π|mz + n|2 u u−s du, Z

∞

y −1

Acta Mathematica Hungarica, 2007

12

C. BAUER and Y. WANG

and similarly (23) Z ∞ 1

Ã

! Z ∞ ¡ ¢ π|mz + n|2 exp − u us−1 du = y s exp − π|mz + n|2 u us−1 du. y y −1

Now, we give an upper bound for the expression k(u) := k(u, z) :=

X0

¡ ¢ exp − π|mz + n|2 u ,

m,n

for u > y −1 > 0, Im z > 1. By denition, X0 ¡ ¢ (24) k(u) = exp − π|mz + n|2 u m,n ∞ X

=2

∞ X ¡ ¢ exp − π|mz|2 u + 2 exp(−πn2 u)

m=1

+

X X

n=1

¡ ¢ exp − π|mz + n|2 u =: 2I1 (u) + 2I2 (u) + I3 (u).

m6=0 n6=0

Since y = Im z > 1, I1 (u) =

∞ X

∞ X ¢ exp − π|mz| u 5 exp(−πm2 u) = I2 (u).

¡

2

m=1

m=1

For xed m and x, there exists at most two n such that −1 < mx + n < 1. For all other n, either n + mx = n + [mx] = 1 or n + mx 5 n + [mx] + 1 5 −1. Hence, X X ¡ ¢ I3 (u) = exp − π|mz + n|2 u m6=0 n6=0

=

X X

¡ ¢ exp ( − π |mx + n|2 + (my)2 u)

m6=0 n6=0

=

X

¡ ¢X ¡ ¢ exp − π(my)2 u exp − π|mx + n|2 u

m6=0

54

n6=0 ∞ X

µ ¶ ∞ X 2 exp(−πm u) 1 + exp (−πn u)

m=1

Acta Mathematica Hungarica, 2007

2

n=1

EISENSTEIN SERIES AND RANKINSELBERG L-FUNCTIONS

= 4I2 (u) + 4

µX ∞

13

¶2 exp (−πm u) . 2

m=1

Combining the estimates for I1 (u), I2 (u) and I3 (u), we see from (22)(24) for 0 < s < 1: ¶ µ Z ∞ Z 1 −s+1 ∞ −s s s−1 + k(u)u du + y k(u)u du E (z, s) = y 2 y −1 y −1 Z ∞ Z ∞ Z ∞ 5y k(u) du 5 4y I2 (u) du + y I3 (u) du y −1

Z

∞

5 8y

∞ X

y −1

y −1

∞

µX ∞

y −1

m=1

Z 2

exp(−πm u) du + 4y

y −1 m=1

¶2 exp (−πm u) du 2

∞ 8 X 1 −πm2 y−1 5y e π m2 m=1

∞

µX ∞

y −1

m=1

Z + 4y

¶µ X ¶ ∞ 2 2 exp (−πm u) πm exp (−πm u) du 2

m=1

5 ye−πy

−1

∞ 8 X 1 π m2 m=1

¯Z ¶ µX ¶¯¯ ∞ ∞ ¯ ∞µX ¯ ¯ + 4y ¯ exp (−πm2 u) d exp (−πm2 u) ¯ ¯ y−1 ¯ m=1

−πy −1

5 4.19ye

m=1

+ 2y

µX ∞

2 −1

exp (−πm y

¶2 )

m=1

à −πy −1

5 4.19ye

+ 2y e

−1

−πy −1

e−4πy + 1 − e−πy−1

!2

for y > 1 and s ∈ [0, 1]. ¤ In the following, we suppose k 5 16 and obtain: Lemma 3. Suppose f (z) ∈ Sk (Γ) is a normalized cusp form, i.e., the rst Fourier coecient a1 = 1. Then, ¯ ¯ (25) e−2πy (1 − Be−2πy ) 5 ¯ f (z)¯ 5 e−2πy (1 + Be−2πy ), Acta Mathematica Hungarica, 2007

14

C. BAUER and Y. WANG

for y = 1, where (26)

k+1 2

B=2

+

(−1)K e4π K (2π)

dK dt

µ

¶¯ ¯ e−6πt ¯ , ¯ −2πt 1−e t=1

2π for k 5 16. where K = [ k+2 2 ], and B 5 e

Proof. By the Ramanujan conjecture (proved by P. Deligne for holomorphic modular forms), for the Fourier coecients of f (z) we have

|an | 5 σ0 (n)n

k−1 2

.

For the normalized cusp form f (z), we have (27) −2πy

e

−

µX ∞

¶ σ0 (n)n

k−1 2

e

−2πny

∞ X ¯ ¯ k−1 −2πy ¯ ¯ < f (z) < e + σ0 (n)n 2 e−2πny .

n=2

n=2

Using σ0 (n) 5 n, the last summation is ∞ X

σ0 (n)n

k−1 2

e−2πny 5

n=2

5e

−4πy

∞ X

∞ X

n

k+1 2

e−2πny

n=2

n

k+1 2

e

−2π(n−2)y

5e

−4πy

n=2

∞ X

n

k+1 2

e−2π(n−2) ,

n=2

for y = 1. Denote the sum above by B , and set K = [ k+2 2 ]. Then, B :=

∞ X

n

k+1 2

e

−2π(n−2)

=2

k+1 2

+e

4π

n=2

52

=2

∞ X n=3

k+1 2

k+1 2

+

(−1)K e4π K (2π)

+ e4π

dK dt

(−1)K dK (2π)K dt

µX ∞ n=3

µ

e

−2πnt

n

k+1 2

e−2πn

¶¯¯ ¯ ¯ ¯

¶¯ ¯ e−6πt ¯ 1 − e−2πt ¯

t=1

. t=1

A numerical calculation using Maple shows that B 5 0.75e2π for k 5 16. Indeed k = 16 is the best possible value for this lemma, since we must assure that B 5 e2π , for making the left side of (25) to be positive. But B=2

k+1 2

+3

k+1 2

Acta Mathematica Hungarica, 2007

e−2π > e2π

for k = 18.

¤

EISENSTEIN SERIES AND RANKINSELBERG L-FUNCTIONS

15

Proof of Theorem 2. It is sucient to prove (21). By Lemmas 2

and 3,

R

1 2

− 12

(28)

¯ ¯2 E + (z, s)¯ f (z)¯ y k dxy2dy ¯ R 12 R ∞ ¯ ¯ f (z)¯ 2 y k dx 2dy 1 1 y − R∞ 1

2

µ ¶ ³ −1 ´2 R ∞ −4πy k−1 −1 −1 2 e−4πy −πy −πy y 4.19e +2 e + (1 + Be−2πy ) dy 1 e 1−e−πy−1 < . R∞ −4πy y k−2 (1 − Be−2πy )2 dy e 1

Calculating the integral on the right-hand side of (28) for k 5 16, and B in (26), we obtain R

1 2

− 21

¯ ¯2 E + (z, s)¯ f (z)¯ y k dxy2dy < 1.12 < 2. ¯ 2 dx dy R 12 R ∞ ¯ k ¯ ¯ f (z) y y2 −1 1 R∞ 1

2

Now, (21) and thus Theorem 2 is proved.

¤

Acknowledgement. The authors thank Tonghai Yang for very helpful conversations and also thank the referee for careful reading and useful suggestions. References [1] P. T. Bateman and E. Grosswald, On Epstein's zeta function, Acta Arith., 9 (1964), 365373. [2] D. Bump, Automorphic Forms and Representations, Cambridge Studies in Advanced Mathematics:55, Cambridge University Press (1997). [3] H. Davenport, Multiplicative Number Theory, Springer-Verlag (1980). [4] T. Dokchitser, Computing special values of motivic L-functions, Experiment. Math., 13 (2004), 137149. [5] J. Hostein and D. Ramakrishnan, Siegel zeros and cusp forms, IMRN International Math. Research Notices, 6 (1995), 279308. [6] J. Hosein and P. Lockhart, Coecients of Maass forms and the Siegel zero, Ann. Math., 140 (1994), 161117, and D. Goldfeld, J. Hostein, D. Lieman, D. Ramakrishnan, Appendix: An eective zero-free region, Ann. Math., 140 (1994), 177181. [7] W. Kohnen and D. Zagier, Values of L-series of modular form at the center of critical strips, Invent. Math., 64 (1981), 175198. ¡ ¢ [8] E. Lapid and S. Rallis, On the nonnegativity of L 12 , π for SO2n+1 , Ann. Math., 157 (2003), 891917. [9] R. A. Rankin, Contributions to the theory of Ramanujan's function τ (n) and similar arithmetical functions, I, II, Proc. Camb. Phil. Soc., 35 (1939), 351372. Acta Mathematica Hungarica, 2007