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GAFA, Geom. funct. anal.Vol. 12 (2002) 1296 – 13231016-443X/02/0601296-28
c© Birkhauser Verlag, Basel 2002
GAFA Geometric And Functional Analysis
SUBCONVEXITY FOR RANKIN-SELBERGL-FUNCTIONS OF MAASS FORMS
Jianya Liu and Yangbo Ye
Abstract
In this paper we prove a subconvexity bound for Rankin–Selberg L-functions L(s, f ⊗ g) associated with a Maass cusp form f and a fixedcusp form g in the aspect of the Laplace eigenvalue 1/4 + k2 of f ,on the critical line Re s = 1/2. Using this subconvexity bound, weprove the equidistribution conjecture of Rudnick and Sarnak [RS] onquantum unique ergodicity for dihedral Maass forms, following thework of Sarnak [S2] and Watson [W]. Also proved here is that thegeneralized Lindelof hypothesis for the central value of our L-functionis true on average.
1 Introduction
Let f be a holomorphic Hecke eigenform for the group Γ = SL2(Z) ofeven integral weight k, with Fourier coefficients af (n):
f(z) =∑n>0
af (n)e(nz) .
We normalize f by setting af (1) = 1 and set λf (n) = af (n)/n(k−1)/2. Letg be either a holomorphic Hecke eigenform for Γ of even integral weight lwith
g(z) =∑n>0
ag(n)e(nz)
and λg(n) = ag(n)/n(l−1)/2, or a Maass Hecke eigenform for Γ with ∆g =(1/4+ l2)g. In the latter case, denote the Fourier coefficients of g by λg(n):
g(z) = y1/2∑n �=0
λg(n)Kil
(2π|n|y)e(nx)
Authors partially supported by National Natural Science Foundation (China) grantand National Science Foundation (USA) grant.
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1297
normalized by setting λg(1) = 1. Here λg(n) is then the nth Hecke eigen-value of g, Kil is the modified Bessel function of the third kind, z = x+ iy,and e(x) = e2πix. Consider the finite part of a Rankin–Selberg L-function
L(s, f ⊗ g) = ζ(2s)∞∑
n=1
λf (n)λg(n)ns
. (1.1)
Note that with λf (1) = λg(1) = 1, the L-function defined in (1.1) has itsleading coefficient equal to 1.
Kowalski, Michel, and Vanderkam [KoMV] proved a subconvexity boundfor a Rankin–Selberg L-function L(1/2 + it, f ⊗ g) as f varies over holo-morphic new forms for Γ0(N ) as the level N tends to ∞, with fixed t, g,and the weight of f . In [S2], Sarnak established a subconvexity estimatefor L(1/2 + it, f ⊗ g) in the aspect of the weight of the holomorphic cuspform f . More precisely, for fixed g and t ∈ R and for any ε > 0, he proved
L(
12 + it, f ⊗ g
)�ε,t,g k576/601+ε
as the weight k of f goes to infinity. The precise exponent makes use of therecent bounds towards the Ramanujan conjecture by Kim and Sarnak [KS].
In this paper, we will in turn prove subconvexity bounds for Rankin–Selberg L-functions associated with Maass cusp forms. From now on wewill denote by f a Maass cusp form for Γ = SL2(Z) with ∆f = (1/4+k2)f .We will further assume that f is a Hecke eigenform with Fourier expansion:
f(z) = y1/2∑n �=0
λf (n)Kik
(2π|n|y)e(nx) . (1.2)
We normalize f by setting λf (1) = 1; hence λf (n) is the nth Hecke eigen-value of f . Let g still be either a holomorphic or Maass cusp form. TheRankin–Selberg L-function L(s, f ⊗ g) is then defined by (1.1) again forRe s > 1. If f and g are orthogonal, L(s, f ⊗ g) is indeed holomorphic afteranalytic continuation. Our goal in this paper is to show that a subconvexitybound still holds for L(1/2 + it, f ⊗ g) as k goes to infinity.
The theorems and their proofs below make use of bounds towards theRamanujan conjecture for Maass forms: If π is an automorphic cuspidalrepresentation of GL2(A�) with unitary central character and local Heckeeigenvalues α
(j)π (p) for p < ∞ and µ
(j)π (∞) for p = ∞, j = 1, 2, then∣∣α(j)
π (p)∣∣ ≤ pθ for p at which π is unramified ,∣∣Re
(µ(j)
π (∞))∣∣ ≤ θ if π is unramified at ∞ .
(1.3)
These bounds were proved for θ = 1/4 by Selberg and Kuznetsov [Ku], forθ = 1/5 by Shahidi [Sh4] and Luo, Rudnick, and Sarnak [LRS], for θ = 1/9
1298 JIANYA LIU AND YANGBO YE GAFA
by Kim and Shahidi [KSh], and most recently for θ = 7/64 by Kim andSarnak [KS].Theorem 1.1. Let g be a fixed holomorphic or Maass cusp form forΓ = SL2(Z). Fix ε > 0 and t ∈ R. Then for a Maass Hecke eigenform fwith Fourier expansion (1.2) and λf (1) = 1, we have∑
K−L≤k≤K+L
∣∣L (12 + it, f ⊗ g
)∣∣2 �ε,t,g (KL)1+ε
for K1/2+θ+ε ≤ L ≤ K1−ε if g is holomorphic, and for K(3+2θ)/(5−2θ)+ε ≤L ≤ K1−ε if g is Maass. Here we can take θ = 7/64.
The proof that we will give below applies equally well with Γ = SL2(Z)being replaced by the Hecke congruence subgroup Γ0(N ) for any fixed N .Theorem 1.2. Let g be a fixed holomorphic or Maass cusp form for someΓ0(N ). Fix ε > 0 and t ∈ R. Let f be a Maass Hecke eigenform for Γ0(N )with its first Fourier coefficient λf (1) equal to 1. Then we have
L(
12 + it, f ⊗ g
)�ε,t,g,N k(3+2θ)/4+ε if g is holomorphic ,
�ε,t,g,N k4/(5−2θ)+ε if g is Maass ,
when k → ∞, where θ = 7/64.
Remarks. (i) The convexity bound for L(1/2 + it, f ⊗ g) is k1+ε, whichis deduced from the Phragmen–Lindelof theorem; see [IS]. With θ = 7/64,the bounds in Theorem 1.2 are � k103/128+ε and k128/153+ε, respectively,which break the convexity bound.
(ii) From Theorem 1.2 we can see that any non-trivial θ < 1/2 in (1.3)toward the Ramanujan conjecture gives us a subconvexity bound for ourL-functions.
(iii) Theorem 1.1 shows that the generalized Lindelof hypothesisL(1/2 + it, f ⊗ g) � kε, ε > 0, is true on average for K − L ≤ k ≤ K + Lwith L in the ranges as in Theorem 1.1.
(iv) In Theorems 1.1 and 1.2, we need not assume that the Maass formf is a Hecke eigenform. All we need is a standard normalization and latera normalization for the Kuznetsov formula (cf. Sarnak [S2]).
Let f be a Maass Hecke eigenform for Γ0(N ) with Laplace eigenvalue λ.Normalize f so that µf = |f(z)|2dx dy/y2 is a probability measure onΓ0(N ) \ H, where H is the upper half plane. The equidistribution con-jecture (Rudnick and Sarnak [RS]) predicts that
µf −→ vol(Γ0(N ) \ H
)−1 dx dy
y2
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1299
as λ tends to infinity. According to Sarnak [S1] and Watson [W], thisconjecture would follow from a subconvexity bound for L(1/2, f ⊗ f ⊗ g),with f being as above and g being a fixed Maass Hecke eigenform. If f isa dihedral form corresponding to a representation of the Weil group W� ,i.e. if L(s, f) = L(s, η) for some grossencharacter η on a quadratic num-ber field, then the triple Rankin–Selberg L-function can be factored asL(s, F⊗g)L(s, g⊗χ)L(s, g) for a fixed quadratic character χ of conductor N .Here F is a Maass cusp form with Laplace eigenvalue 1/4 + (2k)2, ifλ = 1/4 + k2. This way the equidistribution conjecture for dihedral Maassforms is reduced to a subconvexity estimate of L(s, F⊗g). Our Theorem 1.2therefore impliesTheorem 1.3. The equidistribution conjecture is true for dihedral Maassforms.
Numerical verification of this theorem has been done by Hejhal andStrombergsson [HS].
We would like to express our heartfelt thanks to Peter Sarnak forhis preprint [S2], helpful discussions, constant encouragement, and care-ful reading of early drafts of this paper. We are also grateful to Frey-doon Shahidi for helpful information and discussion on Rankin–SelbergL-functions.
2 The Approximate Functional Equation
The L-function L(s, f⊗g) satisfies a functional equation proved by Jacquet[J]. The functional equation for Rankin–Selberg L-functions in the generalcase was proved by Shahidi ([Sh1-3], and [Sh5]). When f and g are bothMaass forms, it is also given in Bump [Bu] and Motohashi [M]:
L(s, f ⊗ g) = γ(s)L(1 − s, f ⊗ g)
where (cf. also [Sh3])
γ(s) = (π√
2)4s−2∏
η1,η2=±1
Γ((1 − s + εf,g + ikη1 + ilη2)/2)Γ((s + εf,g + ikη1 + ilη2)/2)
.
Here the product runs over different signs η1, η2 = ±1, and εf,g = 1−εf εg
2 ,where εf and εg are equal to 1 or −1 according to
f(−z) = εff(z) , g(−z) = εgg(z) .
Similar functional equation holds for f being Maass and g being holomor-phic.
1300 JIANYA LIU AND YANGBO YE GAFA
By Stirling’s formula
Γ(z) = e−ze(z−1/2) log z(2π)1/2(1 +
z−1
12+
z−2
288+ O(z−3)
), (2.1)
we can get an asymptotic formula for γ(s). In fact, the factor e−z in (2.1)contributes to γ(s) the following:
exp(−
∑η1,η2=±1
1−s+εf,g+ikη1+ilη2
2+
∑η1,η2=±1
s+εf,g+ikη1+ilη2
2
)= e4s−2.
The factor e(z−1/2) log z contributes
exp( ∑
η1,η2=±1
−s + εf,g + ikη1 + ilη2
2log
1 − s + εf,g + ikη1 + ilη2
2
−∑
η1,η2=±1
−1 + s + εf,g + ikη1 + ilη2
2log
s + εf,g + ikη1 + ilη2
2
)
= exp( ∑
η1,η2=±1
(−s + εf,g + ikη1 + ilη2
2log
εf,g + ikη1 + ilη2
2
− −1 + s + εf,g + ikη1 + ilη2
2log
εf,g + ikη1 + ilη2
2
+−s + εf,g + ikη1 + ilη2
2log(
1 +1 − s
εf,g + ikη1 + ilη2
)− −1 + s + εf,g + ikη1 + ilη2
2log(
1 +s
εf,g + ikη1 + ilη2
)))= exp
( ∑η1,η2=±1
(12−s
)log
εf,g+ikη1+ilη2
2
)exp(2−4s)
(1 + ηk(s)
)where ηk(s) is an error term which is � (1 + |s|)3k−1. Consequently
γ(s) =(π√
2)4s−2
(16∏
η1,η2=±1(εf,g + ikη1 + ilη2)
)s−1/2(1 + ηk(s)
)
=(
64π4∏η1,η2=±1
(εf,g + ikη1 + ilη2)
)s−1/2(1 + ηk(s)
)=(
8π2
(ε2f,g + (k + l)2)1/2(ε2
f,g + (k − l)2)1/2
)2s−1(1 + ηk(s)
).
Let G(s) be an analytic function in −B ≤ Re s ≤ B for a fixed B > 0satisfying
G(0) = 1 , G(s) = G(−s) , |G(s)| � (1 + |s|)−A
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1301
for a fixed large constant A. Then for X ≥ 1, we set
I =1
2πi
∫Re s=2
XsL
(12
+ s, f ⊗ g
)G(s)
ds
s
= L(
12 , f ⊗ g
)+
12πi
∫Re s=−1
XsL
(12
+ s, f ⊗ g
)G(s)
ds
s.
By the functional equation, this becomes
L
(12, f ⊗ g
)+
12πi
∫Re s=−1
Xsγ
(12
+ s
)L
(12− s, f ⊗ g
)G(s)
ds
s
= L
(12, f ⊗ g
)+
12πi
∫Re s=−1
Xs
(8π2
(ε2f,g+(k+l)2)1/2(ε2
f,g+(k−l)2)1/2
)2s
× (1 + ηk
(12 + s
))L(
12 − s, f ⊗ g
)G(s)ds
s .
Changing variables from s to −s, we get
I = L
(12, f ⊗ g
)− 1
2πi
∫Re s=1
X−s
(8π2
(ε2f,g+(k+l)2)1/2(ε2
f,g+(k−l)2)1/2
)−2s
×(
1 + ηk
(12− s
))L
(12
+ s, f ⊗ g
)G(s)
ds
s,
and the integral path can be further shifted to Re s = 2. Set
X = 18π2
(ε2f,g + (k + l)2
)1/2(ε2f,g + (k − l)2
)1/2.
Then
I = L
(12, f ⊗ g
)− 1
2πi
∫Re s=2
Xs
(1 + ηk
(12− s
))L
(12
+ s, f ⊗ g
)G(s)
ds
s.
Consequently
L
(12, f ⊗ g
)=
1πi
∫Re s=2
XsL
(12
+ s, f ⊗ g
)G(s)
ds
s
+ O
(∣∣∣∣ ∫Re s=2
Xsηk
(12− s
)L
(12
+ s, f ⊗ g
)G(s)
ds
s
∣∣∣∣) . (2.2)
For the big O term, we can shift the integral to Re s = 1/2+ε, for someε > 0. Then Xs will contribute O(k1+2ε), and ηk(1/2 − s) will contributeO((1 + |s|)3/k). Recall the well-known fact that for Re s = 1/2 we haveL(1/2 + s, f ⊗ g) �ε kε, which may be proved by Cauchy inequality and
1302 JIANYA LIU AND YANGBO YE GAFA
the techniques in Iwaniec [I2, p. 131]. Therefore, the big O term in (2.2) isOε(kε).
Now let us turn to the main term in (2.2). Recall that
L(s, f ⊗ g) =∑ν≥1
bf⊗g(ν)νs
with bf⊗g(ν) =∑
n2m=ν λf (m)λg(m). Thus
1πi
∫Re s=2
XsL
(12
+ s, f ⊗ g
)G(s)
ds
s= 2
∑n≥1
bf⊗g(n)√n
V( n
X
)= 2
∑b≥1
1b
∑a≥1
λf (a)λg(a)√a
V
(ab2
X
)where
V (y) =1
2πi
∫Re s=2
G(s)y−s ds
s.
Note that
V (y) = 1 +1
2πi
∫Re s=−1
G(s)y−s ds
s
= 1 +y
2π
∫�
G(−1 + it)e−it log y dt
1 + it,
because G(0) = 1. From the assumption that |G(s)| � (1 + |s|)−A in−B ≤ Re s ≤ B, we conclude that limy→0 V (y) = 1. If we write
V (y) =1
2πi
∫Re s=B
G(s)y−s ds
s
=y−B
2π
∫�
G(B + it)e−it log y dt
B + it,
we get V (y) �B (1 + |y|)−B .Consequently, the central value of the L-function on the left side of (2.2)
can be written as
L
(12, f ⊗ g
)= 2
∑n≥1
bf⊗g(n)√n
V( n
X
)+ Oε(kε) .
Using the bound V (y) �B (1 + |y|)−B , we see that we can actually take afinite partial sum above and get an approximation formula for L(1/2, f⊗g).Indeed, we have
L
(12, f ⊗ g
)= 2
∑1≤b≤X1/2+ε
1b
∑a≥1
λf (a)λg(a)√a
V
(ab2
X
)+ Oε(kε) . (2.3)
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1303
Moreover, the estimation of
2∑
1≤b≤X1/2+ε
1b
∑a≥1
λf (a)λg(a)√a
V
(ab2
X
)(2.4)
and hence that of L(1/2, f ⊗ g) can be reduced to estimation of
SY (f) =∑
n
λf (n)λg(n)H(
nY
)for fixed g, where H is a fixed smooth function of compact support con-tained in (1, 2). In fact, we can set Y = X/b2 and have H(a/Y ) in theplace of V (a/(X/b2))
√(X/b2)/a in (2.4). Since (2.4) is indeed the main
term of L(1/2, f ⊗ g), we need the same bound (KL)1+ε for (2.4) as inTheorem 1.2. Note here that X is of the size of k2. We expect a boundfor SY (f) of size
√Y and will see that we actually need this bound in the
case of K2−δ ≤ Y ≤ K2+ε, K − L ≤ k ≤ K + L, and√
K ≤ L ≤ K/4 asK,L → ∞ with K/L also approaching to infinity as a small power of K.
3 Averaging and the Kuznetsov Formula
Our strategy to bound SY (f) is to take a smooth averaging of |SY (f)|2 anduse the Kuznetsov trace formula. By estimating the geometric side of theKuznetsov formula, we hope to obtain savings large enough to offset thewaste from averaging and get a subconvexity bound.
In order to use the Kuznetsov trace formula, we need a different normal-ization of the Maass forms. Let {fj} be an orthonormal basis, consistingof Hecke eigenforms, of the space of Maass cusp forms. Denote by 1/4+ k2
j
the Laplace eigenvalue for fj. Since each fj is normalized by ‖fj‖ = 1, itsleading Fourier coefficient λj(1) in
fj(z) = (y cosh πkj)1/2∑n �=0
λj(n)Kikj
(2π|n|y)e(nx)
is no longer equal to 1. By Iwaniec [I1] and Hoffstein and Lockhart [HoL],however, the leading coefficient is close to 1:(
(1/4 + k2j )N
)−ε �ε
∣∣λj(1)∣∣�ε
((1/4 + k2
j )N)ε
for any ε > 0, where for Γ = SL2(Z), N = 1, and for Γ = Γ0(N ), N is thelevel. Recall that N is fixed. Consequently, by changing normalization fromf to fj, a bound for SY (fj) will come within that for SY (f) by a factor kε.Similarly, a bound for the average of |SY (fj)|2 over K − L ≤ kj ≤ K + Lwill be within that for the average of |SY (f)|2 over K −L ≤ k ≤ K + L bya factor Kε, for any ε > 0.
1304 JIANYA LIU AND YANGBO YE GAFA
Now let L be a number which satisfies√
K ≤ L ≤ K/4. Let h(t) bean even analytic function in | Im t| ≤ 1/2 satisfying h(n)(t) � (1 + |t|)−N
for any N > 0 in this region. Thus h is a Schwartz function on R. Wealso assume that h(t) ≥ 0 for real t. For example, we may simply takeh(t) = 1/ cosh(t). We want to estimate∑
K,L
=∑fj
(h
(kj − K
L
)+ h
(−kj + K
L
)) ∣∣SY (fj)∣∣2
=∑n,m
λg(n)λg(m)H( n
Y
)H(m
Y
)×∑fj
(h
(kj − K
L
)+ h
(−kj + K
L
))λj(n)λj(m) . (3.1)
Recall that 1/4 + k2j is the Laplace eigenvalue for fj. By Weyl’s law, there
are about LK term on the right side of (3.1). We thus expect a boundLKY 1+ε for (3.1). By Kuznetsov’s formula [Ku] (see also Iwaniec [I2]), wecan rewrite the inner sum on the right side of (3.1) as∑
fj
(h
(kj − K
L
)+ h
(−kj + K
L
))λj(n)λj(m)
=δn,m
π2
∫�
tanh(πr)(
h
(r − K
L
)+ h
(−r + K
L
))rdr (3.2)
− 1π
∫�
dir(n)dir(m)(
h
(r − K
L
)+ h
(−r + K
L
))dr
|ζ(1 + 2ir)|2(3.3)
+2iπ
∑c≥1
S(n,m; c)c
(3.4)
·∫�
J2ir
(4π
√nm
c
)(h
(r − K
L
)+ h
(−r + K
L
))rdr
cosh(πr),
where dν(n) =∑
ab=|n|(a/b)ν .
The term in (3.2) will contribute to∑
K,L the following
O
(∑n
∣∣λg(n)∣∣2 ∣∣∣H ( n
Y
)∣∣∣2 ∫�
∣∣∣∣h(r − K
L
)∣∣∣∣ |r|dr
).
By the Rankin–Selberg method, we have∑n
∣∣λg(n)∣∣2 ∣∣H (
nY
)∣∣2 � Y 1+ε. (3.5)
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1305
On the other hand,∫�
∣∣∣∣h(r − K
L
)∣∣∣∣ |r|dr �∫ K+cL
K−cLrdr � KL .
Therefore, (3.2) contributes O(LKY 1+ε) to∑
K,L.
To consider the term in (3.3), we recall that |ζ(1+2ir)|≥c log−2/3(2+|r|)for some c > 0. Using u = (r − K)/L and u = −(r + K)/L, we changevariables in the integral in (3.3). Its contribution to (3.1) becomes
− 2Lπ
∑n,m
λg(n)λg(m)H( n
Y
)H(m
Y
)·∫�
di(uL+K)(n)di(uL+K)(m)h(u)du
|ζ(1 + 2i(uL + K))|2
= −2Lπ
∫�
∣∣∣∣∑n
λg(n)H( n
Y
)di(uL+K)(n)
∣∣∣∣2 h(u)du
|ζ(1 + 2i(uL + K))|2 . (3.6)
Since h(u) is always positive, we may move this term to the side of∑
K,L,which is also positive. Consequently a non-trivial estimation of the contri-bution of (3.4) will give us a bound for
∑K,L.
4 Terms with Kloosterman Sums
4.1 Integrals of Bessel functions. In order to estimate the contribu-tion of (3.4) to
∑K,L, we want to consider the integral
VK,L(x) =∫�
J2ir(x)(
h
(r − K
L
)+ h
(−r + K
L
))rdr
cosh(πr)
=12
∫�
J2ir(x)−J−2ir(x)sinh(πr)
(h
(r−K
L
)+ h
(−r+K
L
))r tanh(πr)dr
for x = 4π√
mn/c. Note that tanh(πr) = sgn(r) + O(e−π|r|) for large |r|.Since the h functions on the right side above isolate r to ±K, we can removetanh(πr) by getting a negligible O(K−N ) for any N > 0. By the Parsevalidentity, we now have
VK,L(x) =12
∫�
(J2ir(x) − J−2ir(x)
sinh(πr)
)∧(y)
×(
h
(r − K
L
)|r| + h
(−r + K
L
)|r|)∧
(−y)dy + O(K−N ) .
1306 JIANYA LIU AND YANGBO YE GAFA
Now (J2ir(x) − J−2ir(x)
sinh(πr)
)∧(y) = −i cos
(x cosh(πy)
)according to Bateman [B, volume 1, p. 59]. Therefore,
VK,L(x) =12i
∫�
cos(x cosh(πy)
)((h
(r−K
L
)+h
(−r+K
L
))|r|)∧
(y)dy.
Note that(h
(r − K
L
)|r| + h
(−r + K
L
)|r|)∧
(y)
=∫�
h
(r − K
L
)|r|e(ry)dr +
∫�
h
(−r + K
L
)|r|e(ry)dr
= e(yK)L(h(u)|uL + K|)∧(yL) + e(−yK)L
(h(u)|uL + K|)∧(−yL) .
As h(u) isolates u to O(1) and L � K1−δ for some δ > 0, we may removethe absolute signs above and get
VK,L(x) = −i
∫�
cos(
x cosh(
πt
L
))e
(tK
L
)(h(u)(uL + K)
)∧(t)dt
=12i(WK,L(x)eix + WK,L(−x)e−ix
)(4.1)
where t = yL and
WK,L(x) =∫�
exp(
2πitK
L+ ix
(cosh(πt/L) − 1
)) (h(u)(uL + K)
)∧(t)dt
=∫�
e
(tK
L+
x
2π
(π2t2
2!L2+
π4t4
4!L4+ · · ·
))(h(u)(uL + K)
)∧(t)dt
for x = ±4π√
mn/c. We will see in Lemma 4.1 that W (x) has only mildoscillations, as its derivatives are not large. The oscillations of VK,L(x) aremainly contained in the exponential functions e±ix in (4.1).
Now we expend the exponential function on the right side from thefourth power of t up to t2N for any N > 0. Then we get a sum of integralsof the form
xµ
L2ν
∫�
e
(tK
L+
πxt2
4L2
)t2ν(h(u)(uL + K)
)∧(t)dt , 0 ≤ 2µ ≤ ν < N ,
(4.2)multiplied by constant coefficients cµ,ν to make the infinite series of themconvergent and bounded, plus
O
( |x|N/2
L2N
∫�
|t|2N∣∣(h(u)(uL + K))∧(t)
∣∣dt
). (4.3)
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1307
4.2 The remainder term of W���(x). We denote a term as in (4.2)by Wµ,ν(x). The term in (4.3) becomes
O
( |x|N/2
L2N−1
∫�
|t|2N∣∣(h(u)u)∧(t)
∣∣dt +|x|N/2K
L2N
∫�
|t|2N∣∣h(t)
∣∣dt
)= O
( |x|N/2
L2N(L + K)
).
Since |x| = 4π√
mn/c ≤ 8πY , Y ≤ K2+ε, and L ∼ K1−δ for a small δ > 0,(4.3) is negligible.
4.3 Main terms of W���(x). As for
Wµ,ν(x) =xµ
L2ν
∫�
e
(tK
L+
πxt2
4L2
)t2ν(h(u)(uL + K)
)∧(t)dt ,
the phase is
φ(t) =tK
L+
πxt2
4L2
with the derivativeφ′(t) =
K
L+
πxt
2L2.
Setting φ′(t) = 0, we get t = −2LK/(πx). Recall that h(u) and hencet2ν(h(u)(uL+K))∧(t) are rapidly decreasing. Thus the contribution to theintegral is only from |t| ≤ Kε for ε > 0 arbitrarily small (i.e. contributionfrom |t| ≥ Kε is O(K−N ) for any large N). Therefore, if LK/|x| ≥ Kε, i.e.if |x| ≤ LK1−ε, then Wµ,ν(x) is negligible. Assume then |x| ≥ LK1−ε. Notethat we still have x = 4π
√mn/c ≤ 8πY and that
√|x|/L ≥ (K/L)1/2K−ε
→ ∞ as we assumed that K/L tends to infinity as a small power of K.Then
Wµ,ν(x) =xµ
(2πiL)2νe
(−K2
πx
)∫�
e
(πx
4L2
(t +
2LK
πx
)2)×(
L
(d2ν
du2ν(uh(u))
)∧(t) + K(h(2ν))∧(t)
)dt . (4.4)
The Fourier transform of the exponential function in the integrand of (4.4)is ∫
�
e
(πx
4L2
(t +
2LK
πx
)2
+ tu
)dt
= e
(−2uLK
πx
)∫�
e
(πxt2
4L2+ tu
)dt
= e
(−2uLK
πx− u2L2
πx
)∫�
e
(πx
4L2
(t +
2L2u
πx
)2)dt
1308 JIANYA LIU AND YANGBO YE GAFA
=L√π|x|
(1 + i sgn(x)
)e
(−2uLK
πx− u2L2
πx
).
By Parseval again
Wµ,ν(x) =xµ
(2πiL)2ν· L√
π|x|(1 + i sgn(x)
)e
(−K2
πx
)×∫�
e
(2uLK
πx− u2L2
πx
)(L
d2ν
du2ν(uh(u)) + Kh(2ν)(u)
)du .
4.4 Asymptotic expansions of main terms. Expanding e(− u2L2
πx
)into Taylor series, we get
Wµ,ν(x) =xµ
(2πiL)2ν· L√
π|x|(1 + i sgn(x)
)e
(−K2
πx
)∫�
e
(2uLK
πx
)×(
Ld2ν
du2ν(uh(u)) + Kh(2ν)(u)
)( ∑0≤k≤N
1k!
(−2iu2L2
x
)k )du
+ O
( |x|µ−1/2
L2ν−1
∫�
∣∣∣∣L d2ν
du2ν
(uh(u)
)+ Kh(2ν)(u)
∣∣∣∣ (u2L2
|x|)N+1
du
)(4.5)
Estimating the integral in (4.5), we get
Wµ,ν(x) =xµ
(2πiL)2ν· L√
π|x|(1+ i sgn(x)
)e
(−K2
πx
) ∑0≤k≤N
1k!
(−2iL2
x
)k
×∫�
u2k
(L
d2ν
du2ν
(uh(u)
)+ Kh(2ν)(u)
)e
(2uLK
πx
)du
+ O
(L2N+3−2ν
|x|(2N+3)/2−µ(L + K)
).
We point out that the integral on the right side above is indeed a sum ofFourier transforms of rapidly decreasing functions:∫
�
u2k
(L
d2ν
du2ν
(uh(u)
)+ Kh(2ν)(u)
)e
(2uLK
πx
)du
= L
(u2k d2ν
du2ν(uh(u))
)∧(2LK
πx
)+ K
(u2kh(2ν)(u)
)∧(2LK
πx
).
This completes the proof of the following results.
Lemma 4.1. (i) For ε > 0 and |x| ≤ LK1−ε, WK,L(x) � K−N for anyN > 0 and hence is negligible.
(ii) Fix N ≥ 1. For LK1−ε ≤ |x| ≤ 8πY , we have
WK,L(x) =∑
0≤2µ≤ν<N
cµ,νWµ,ν(x) ,
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1309
with
Wµ,ν(x) =xµ
(2πiL)2ν· L√
π|x|(1+ i sgn(x)
)e
(−K2
πx
) ∑0≤k≤N
1k!
(−2iL2
x
)k
×(
L
(u2k d2ν
du2ν(uh(u))
)∧(2LK
πx
)+ K
(u2kh(2ν)(u)
)∧(2LK
πx
))+ O
(L2N+3
|x|(2N+3)/2(L + K)
).
4.5 Application of a Voronoi summation formula. Denote by
TK,L(Y ) =∑n,m
λg(n)λg(m)H( n
Y
)H(m
Y
)∑c≥1
S(m,n; c)c
VK,L
(4π
√mn
c
)the terms in
∑K,L corresponding to the expression in (3.4). From
VK,L = 12i
(WK,L(x)eix + WK,L(−x)e−ix
),
the estimation of TK,L(Y ) is reduced to
T±K,L(Y ) =
∑n,m
λg(n)λg(m)H( n
Y
)H(m
Y
)×∑c≥1
S(m,n; c)c
e
(±2
√mn
c
)WK,L
(±4π
√mn
c
).
Using Lemma 4.1 (i) and a non-trivial bound for the Kloosterman sumS(m,n; c), we see that the inner sum may be taken over c ≤ Y/(LK1−ε).We may further reduce the estimation to that of
T±µ,ν(Y ) =
∑n,m
λg(n)λg(m)H( n
Y
)H(m
Y
)×
∑c≤Y/(LK1−ε)
S(m,n; c)c
e
(±2
√mn
c
)Wµ,ν
(±4π
√mn
c
). (4.6)
Now we open the Kloosterman sum
S(m,n; c) =∑
z mod c,(z, c) = 1
e
(mz + nz
c
)
and apply a Voronoi summation formula to the sum with respect to min (4.6). When g is a holomorphic cusp form, this Voronoi formula is givenin Sarnak [S2]:
1310 JIANYA LIU AND YANGBO YE GAFA
∑m≥1
λg(m)H(m
Y
)e(mz
c
)e
(±2
√mn
c
)Wµ,ν
(±4π
√mn
c
)
=2πil
c
∑r≥1
λg(r)e(− zr
c
)×∫ ∞
0H
(t
Y
)e
(±2
√tn
c
)Wµ,ν
(±4π
√tn
c
)Jl−1
(4π
√tr
c
)dt . (4.7)
When g is a Maass cusp form, a similar formula is proved in Kowalski,Michel, and Vanderkam [KoMV]. Since we are evaluating Jl−1(x) at x =4π
√tr/c ≥ LK1−ε/
√Y , we have
Jl−1(x) =1√2πx
ei(x−(2l−1)π/4)∑
0≤j<2N
ijΓ(l + j − 1/2)j!Γ(l − j − 1/2)
(2x)−j
+1√2πx
e−i(x−(2l−1)π/4)∑
0≤j<2N
ijΓ(l + j − 1/2)j!Γ(l − j − 1/2)
(−2x)−j
+ O(x−2N−1/2) . (4.8)
Apply (4.7) and (4.8) to (4.6). We can take N > 0 so that the remainderterms in (4.8) are negligible. For a typical term on the right side of (4.8)
1xj+1/2
e±i(x−(2l−1)π/4 , 0 ≤ j < 2N ,
multiplied by a constant coefficient, we get for (4.6) that
T(η)µ,ν,j(Y ) =
∑c≤Y/(LK1−ε)
1c2
∑n,r≥1
λg(n)λg(r)H( n
Y
) ∑z mod c,(z,c)=1
e
(z(n − r)
c
)
×∫ ∞
0H
(t
Y
)e
(η1
2√
tn
c
)Wµ,ν
(η3
4π√
tn
c
)× cj+1/2
(tr)j/2+1/4eiη2(4π
√tr/c−(2l−1)π/4)dt ,
where η = (η1, η2, η3) and ηi = ±1.
4.6 Asymptotic expansions of T(�)�����(Y ). Changing variables from
t to tY , we get
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1311
T(η)µ,ν,j(Y ) = Y 3/4−j/2
∑c≤Y/(LK1−ε)
cj−3/2∑
n,r≥1
λg(n)λg(r)rj/2+1/4
H( n
Y
)
×∑
z mod c,(z,c)=1
e
(z(n − r)
c
)∫ ∞
0
H(t)tj/2+1/4
e
(η1
2√
tY n
c
)
× Wµ,ν
(η3
4π√
tY n
c
)eiη2(4π
√tY r/c−(2l−1)π/4)dt . (4.9)
By the asymptotic formula of Wµ,ν in Lemma 4.1, (4.9) becomes
T(η)µ,ν,j(Y ) = Y 3/4−j/2
∑c≤Y/(LK1−ε)
cj−3/2∑
n,r≥1
λg(n)λg(r)rj/2+1/4
H( n
Y
)
×∑
z mod c,(z,c)=1
e
(z(n − r)
c
)∫ ∞
0
H(t)tj/2+1/4
e
(η1
2√
tY n
c
)(4π
√tY n/c)µ−1/2
(2πiL)2ν−1
× 1 + η3i
2πie
(−η3
K2c
4π√
tY n
) ∑0≤k≤N
1k!
(−η3
2iL2c
4π√
tY n
)k
×(L
(u2k d2ν
du2ν(uh(u))
)∧(η3LKc
2π2√
tY n
)+K
(u2kh(2ν)(u)
)∧( η3LKc
2π2√
tY n
))× eiη2(4π
√tY r/c−(2l−1)π/4)dt .
Up to a bounded constant coefficient, this can be written as
T(η)µ,ν,j(Y ) =
Y (µ−j+1)/2
L2ν−1
∑0≤k≤N
1k!
(η3L
2
2πi√
Y
)k ∑c≤Y/(LK1−ε)
cj+k−µ−1
×∑
n,r≥1
λg(n)λg(r)n(k−µ)/2+1/4rj/2+1/4
H( n
Y
) ∑z mod c,(z,c)=1
e
(z(n − r)
c
)B
(µ,ν,j)η,Y,c (n, r) ,
where
B(µ,ν,j,k)η,Y,c (n, r) =
∫ ∞
0e
(2√
tY (η1√
n + η2√
r)c
− η3K2c
4π2√
tY n
)×(
L
(u2k d2ν
du2ν(uh(u))
)∧(η3LKc
2π2√
tY n
)+ K
(u2kh(2ν)(u)
)∧( η3LKc
2π2√
tY n
))H(t)
t(j+k−µ+1)/2dt . (4.10)
1312 JIANYA LIU AND YANGBO YE GAFA
To estimate T(η)µ,ν,j(Y ), it suffices to bound
Y (µ−j+1)/2
L2ν−1
∑0≤k≤N
1k!
(L2
2π√
Y
)k ∑c≤Y/(LK1−ε)
cj+k−µ∑|h|<Y
∣∣P (c, h, Y )∣∣ ,
(4.11)where
P (c, h, Y ) =∑
n>max(0,−h)
λg(n)λg(n+h)n(k−µ)/2+1/4(n+h)j/2+1/4
H( n
Y
)B
(µ,ν,j,k)η,Y,c (n, n + h)
for 0 ≤ 2µ ≤ ν < N and 0 ≤ j < 2N .
4.7 The case of the same sign.
Lemma 4.2. If η1 and η2 have the same sign, then B(µ,ν,j,k)η,Y,c (n, n + h) �
K−N for any N > 0, where the implied constant is independent of Y , c, n,and h.
Proof. Change variables in (4.10) from t to t2:
B(µ,ν,j,k)η,Y,c (n, r) =
∫ ∞
0e
(2t√
Y (η1√
n + η2√
r)c
− η3K2c
4π2t√
Y n
)×(
L
(u2k d2ν
du2ν(uh(u))
)∧(η3LKc
2π2t√
Y n
)+ K
(u2kh(2ν)(u)
)∧( η3LKc
2π2t√
Y n
))2H(t2)tj+k−µ
dt . (4.12)
Recall that the integral in (4.12) is actually taken over 1 ≤ t ≤ √2, and
that Y ≤ n ≤ 2Y . Take r = n + h. Then |η1√
n + η2
√n + h| ≥ √
Y . Thephase is
φ(t) =4πt
√Y (η1
√n + η2
√r)
c− η3K
2c
2πt√
Y n
with derivative
φ′(t) =4π
√Y (η1
√n + η2
√r)
c+
η3K2c
2πt2√
Y n.
Note that
|φ′(t)| =4πY
c+ O
(K2c
Y
)≥ 4πLK1−ε + O
(K1+ε
L
)for 1 ≤ t ≤ √
2. Now we do integration by parts in (4.12) many timesby integrating the exponential function multiplied by φ′(t) and differen-tiating the rest of the integrand divided by φ′(t). The differentiation of
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1313
a function like 2H(t2)/tj+k−µ yields a constant bound. The differentia-tion of the sum of the two Fourier transforms in (4.12) produces a fac-tor −η3LKc/(2π2t2
√Y n) which is of size Kε. The factor 1/φ′(t) yields
O(L−1Kε−1), while (1/φ′(t))′ gives us a smaller O(L−3Kε−1). Conse-quently, integration by parts N times produces a negligible O(L−NKε−N)for arbitrary N > 0. �
4.8 The case of opposite signs. Now we consider that case of η1 �= η2
and rewrite P (c, h, Y ):
P (c, h, Y ) =∑
n≥max(1,1−h)
λg(n)λg(n + h)(√
n(n + h)2n + h
)l−1
G(2n + h) ,
where
G(2n + h) =(
2n + h√n(n + h)
)l−1 1n(k−µ)/2+1/4(n + h)j/2+1/4
× H( n
Y
)B
(µ,ν,j,k)η,Y,c (n, n + h) .
That is,
G(x) =(
2x√x2 − h2
)l−1( 2x − h
)(k−µ)/2+1/4 ( 2x + h
)j/2+1/4
× H
(x − h
2Y
)B
(µ,ν,j,k)η,Y,c
(x − h
2,x + h
2
).
The Mellin transform of G(x) is by definition
G(s) =∫ ∞
max(1,1−h)G(x)xs−1dx = (2Y )s
∫ ∞
0G(2Y z − h)
(z +
h
2Y
)s−1
dz
= (2Y )sY (µ−j−k−1)/2
∫ ∞
0G0(z)
(z +
h
2Y
)s−1
dz , (4.13)
where z = (x − h)/(2Y ) and
G0(z) =(
2z + h/Y√z(z + h/Y )
)l−1
z(µ−k)/2−1/4
(z +
h
Y
)−j/2−1/4
× H(z)B(µ,ν,j,k)η,Y,c
(zY, zY + h
). (4.14)
Using the Mellin inversion formula, we get
P (c, h, Y ) =1
2πi
∫Re s=σ
Dg(s, 1, 1, h)G(s)ds , (4.15)
1314 JIANYA LIU AND YANGBO YE GAFA
where
Dg(s, ν1, ν2, h) =∑
m,n>0,ν1m−ν2n=h
λg(n)λg(m)( √
ν1ν2mn
ν1m + ν2n
)l−1
(ν1m + ν2n)−s
for ν1, ν2 > 0, if g is a holomorphic cusp form. If g is a Maass cusp form,then define
Dg(s, ν1, ν2, h) =∑
m,n �=0,ν1m−ν2n=h
λg(n)λg(m)( √
ν1ν2|mn|ν1|m| + ν2|n|
)2il
(ν1|m|+ ν2|n|)−s
for ν1, ν2 > 0.
4.9 Bounds for D�(s, ν1, ν2, h). The following two theorems wereproved by Sarnak [S2] using bounds toward the Ramanujan conjecturein (1.3). These theorems play a crucial role in our proof.
Theorem 4.3. Let g be a holomorphic cusp form of even weight l. ForRe(s) > 1, ν1, ν2 > 0, and h ∈ Z, define Dg(s, ν1, ν2, h) as above. Thenassuming (1.3) for θ we have that Dg(s) extends to a holomorphic functionfor Re(s) ≥ 1/2 + θ + ε, for any ε > 0. Moreover, in this region it satisfies
Dg(s, ν1, ν2, h) �g,ε (ν1ν2)−1/2+ε|h|1/2+θ+ε−Re s(1 + |t|)3.
where Im(s) = t.
Theorem 4.4. Let g be a Maass cusp form of Laplace eigenvalue 1/4+ l2.For Re(s) > 1, ν1, ν2 > 0, and h ∈ Z define Dg(s, ν1, ν2, h) as above. Thenassuming (1.3) for θ we have that Dg(s) extends to a holomorphic functionfor Re(s) ≥ 1/2 + θ + ε, for any ε > 0. Moreover, in this region it satisfies
Dg(s, ν1, ν2, h) �g,ε (ν1ν2)−1/2+ε|h|1/2+θ+ε−Re s(1 + |t|)3 + |h|1−Re s ,
where Im(s) = t.
Here we can take θ = 7/64 of Kim and Sarnak [KS], and set σ =1/2 + θ + ε in (4.15).
4.10 The Mellin transform of G(x). To show that the integral in(4.15) converges absolutely, we do integration by parts five times to (4.13)by integrating the power of z + h/(2Y ) and differentiating G0(z):
G(s) =(2Y )sY (µ−j−k−1)/2
s(s + 1) · · · (s + 4)
∫ ∞
0G
(5)0 (z)
(z +
h
2Y
)s+4
dz . (4.16)
As |h| < Y , differentiation of the (· · · )l−1, H(z), and the powers of z andz + h/Y in (4.14) is all bounded.
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1315
4.11 The fifth derivative. Now we look at the fifth direvative of
B(µ,ν,j,k)η,Y,c (zY, zY + h) =
∫ ∞
0e
(2tY η1(
√z −√z + h/Y )
c− η3K
2c
4π2tY√
z
)×(
L
(u2k d2ν
du2ν(uh(u))
)∧(η3LKc
2π2tY√
z
)+ K
(u2kh(2ν)(u)
)∧( η3LKc
2π2tY√
z
))2H(t2)tj+k−µ
dt
= z(j+k−µ−1)/2
∫ ∞
0e
(2thη2
cz(1 +√
1 + h/(Y z))− η3K
2c
4π2tY
)(4.17)
×(
L
(u2k d2ν
du2ν(uh(u))
)∧(η3LKc
2π2tY
)+ K(u2kh(2ν)(u))∧
(η3LKc
2π2tY
))2H(t2/z)tj+k−µ
dt ,
where we changed the variables from t to t/√
z. A differentiation withrespect to z of B
(µ,ν,j,k)η,Y,c (zY, zY + h) as on the right side of (4.17) yields a
leading term
z(j+k−µ−1)/2
∫ ∞
0e
(2thη2
cz(1 +√
1 + h/(Y z))− η3K
2c
4π2tY
)× 4πithη1
cz2(1 +√
1 + h/(Y z))
(L
(u2k d2ν
du2ν(uh(u))
)∧(η3LKc
2π2tY
)+ K(u2kh(2ν)(u))∧
(η3LKc
2π2tY
))2H(t2/z)tj+k−µ
dt .
Therefore
G(5)0 (z) =
(2z + h/Y√z(z + h/Y )
)l−1
zj/2−3/4
(z +
h
Y
)−j/2−1/4
H(z)
×∫ ∞
0e
(2thη2
cz(1 +√
1 + h/(Y z))− η3K
2c
4π2tY
)×(
4πithη1
cz2(1 +√
1 + h/(Y z))
)5
×(
L
(u2k d2ν
du2ν(uh(u))
)∧(η3LKc
2π2tY
).
+ K(u2kh(2ν)(u))∧(
η3LKc
2π2tY
))2H(t2/z)tj+k−µ
dt
+ other terms . (4.18)
1316 JIANYA LIU AND YANGBO YE GAFA
4.12 The range of h. When h �= 0, we apply integration by partsmany times to the integrals on the right side of (4.18) by integrating
e
(2thη2
cz(1 +√
1 + h/(Y z))
)(4.19)
and differentiating the rest of the integrand. Each integration of (4.19)yields cz(1 +
√1 + h/(Y z))/(4πihη2) which is of size c/|h|. Each
differentiation of the rest of the integrand gives us a leading termiη3K
2c/(2πt2Y ) plus O(LKc/Y ). Consequently each integration by partsproduces O(K2c2/(|h|Y ). If |h| ≥ K2+εc2/Y , each integration by partsthen gives us a saving of O(K−ε). Doing this repeatedly, we know thatG
(5)0 (z) � K−N for arbitrary N > 0, and hence is negligible, when |h| ≥
K2+εc2/Y . Therefore, the innermost sum in (4.11) can be taken over|h| ≤ K2+εc2/Y .
4.13 Bounds for P (c, h, Y ). We may use the obvious boundO(K|H(z)|(|h|/c)5 + · · · + 1) for G
(5)0 (z) to get a bound for G(s):
G(s) � KY (µ−j−k−1)/2+Re s
(( |h|c
)5
+ · · · + 1)(
1 + |Ims|)−5. (4.20)
Back to (4.15), we can use (4.20) and Theorem 4.3 to get
P (c, h, Y ) � KY (µ−j−k−1)/2+σ
(( |h|c
)5
+ · · · + 1)|h|1/2+θ+ε−σ , (4.21)
when g is holomorphic, where we can choose σ = 1/2 + θ + ε. When gis a Maass cusp form, we can split Dg(s, 1, 1, h) into two terms as in theproof of Theorem 4.4 in Sarnak [S2], such that the first term is bounded by|h|1/2+θ+ε−Re s(1 + |Ims|)3, while the second term is bounded by |h|1−Re s.For this second term, we only need to do integration by parts to the rightside of (4.13) twice. Thus by Theorem 4.4 we get
P (c, h, Y ) � KY (µ−j−k−1)/2+σ
(( |h|c
)5
+ · · · + 1)|h|1/2+θ+ε−σ
+ KY (µ−j−k−1)/2+σ
(( |h|c
)2
+ · · · + 1)|h|1−σ , (4.22)
when g is Maass.
4.14 Integration term by term. If we apply the bound in (4.21) or(4.22) to (4.11) with |h| ≤ K2+εc2/Y , we will get a subconvexity bound forour L-functions. We can actually get a better bound using another trick.We want to integrate the right side of (4.15) term by term with G(s) as
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1317
given in (4.16), and to interchange the order of integration with respect tos and z. As we pointed out before, the integral (4.15) converges absolutelywhen σ ≥ 1/2+θ+ε, while the factor H(z) in (4.18) provides us the uniformconvergence of the integral in (4.16). If we take σ sufficiently large, usingthe bounds in (1.3), the series defining Dg(s, 1, 1, h) converges uniformly.This allows us to write
P (c, h, Y ) = Y (µ−j−k−1)/2∑n
λg(n)λg(n+h)(√
n(n+h)2n + h
)l−1∫ ∞
0G
(5)0 (z)dz
× 12πi
∫Re s=σ
(2n + h)−s(2Y )s
s(s + 1) · · · (s + 4)
(z +
h
2Y
)s+4
ds
= Y (µ−j−k−1)/2∑n
λg(n)λg(n + h)(√
n(n + h)2n + h
)l−1
(4.23)
×(
2Y2n + h
)σ ∫ ∞
0G
(5)0 (z)
(z +
h
2Y
)σ+4
dz
× 12π
∫�
(2Y
2n+h
(z + h
2Y
))it(σ + it)(σ + 1 + it) · · · (σ + 4 + it)
dt .
The inner integral on the right side of (4.23) can be computed. Usingpartial fractions
1(σ + it)(σ + 1 + it) · · · (σ + 4 + it)
=1
24(σ + it)− 1
6(σ + 1 + it)+
14(σ + 2 + it)
− 16(σ + 3 + it)
+1
24(σ + 4 + it)we get
12π
∫�
(2Y
2n+h
(z + h
2Y
))it(σ + it)(σ + 1 + it) · · · (σ + 4 + it)
dt
= − 124
(2Y
2n + h
(z +
h
2Y
))−σ
+16
(2Y
2n + h
(z +
h
2Y
))−σ
− · · · − 124
(2Y
2n + h
(z +
h
2Y
))−σ−4
, (4.24)
when2Y
2n + h
(z +
h
2Y
)> 1 ,
1318 JIANYA LIU AND YANGBO YE GAFA
i.e. when 2n + h < 2Y (z + h/(2Y )). When 2n + h ≥ 2Y (z + h/(2Y )),however, the integral on the left side of (4.24) vanishes. Therefore, the sumin (4.23) is indeed a finite sum over max(1, 1−h) ≤ n < 2Y . Consequently,
P (c, h, Y ) = Y (µ−j−k−1)/2∑
max(1,1−h)≤n<2Y
λg(n)λg(n + h)
×(√
n(n + h)2n + h
)l−1 ∫ ∞
0G
(5)0 (z)
(− 1
24
(z +
h
2Y
)4
+16
(2Y
2n + h
)−1(z +
h
2Y
)3
− 14
(2Y
2n + h
)−2(z +
h
2Y
)2
+16
(2Y
2n + h
)−3(z +
h
2Y
)− 1
24
(2Y
2n + h
)−4)
dz . (4.25)
4.15 A smaller range for h. Using the expression of G(5)0 (z) in (4.18),
we switch the order of integration and rewrite the integral on the right sideof (4.25) as∫ ∞
0e
(−η3K
2c
4π2tY
)(L
(u2k d2ν
du2ν(uh(u))
)∧(η3LKc
2π2tY
)+ K
(u2kh(2ν)(u)
)∧(η3LKc
2π2tY
))2dt
tj+k−µ
×∫ ∞
0
(2z + h/Y√z(z + h/Y )
)l−1
zj/2−3/4
(z +
h
Y
)−j/2−1/4
H(z)H(
t2
z
)× e
(2thη2
cz(1 +√
1 + h/(Y z))
)(4πithη1
cz2(1 +√
1 + h/(Y z))
)5
×(− 1
24
(z +
h
2Y
)4
+16
(2Y
2n + h
)−1(z +
h
2Y
)3
− 14
(2Y
2n + h
)−2(z +
h
2Y
)2
+16
(2Y
2n + h
)−3(z +
h
2Y
)− 1
24
(2Y
2n + h
)−4)dz
+ other terms . (4.26)
Recall that the integrals are indeed both taken over [1, 2]. The phase is
φ(z) =4πthη2
cz(1 +
√1 + h/(Y z)
)
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1319
with
φ′(z) = − 4πthη2
cz(1 +√
1 + h/(Y z))
(1 +
√1 +
h
Y z− h
2Y z√
1 + h/(Y z)
).
Note that for t, z ∈ [1, 2], |φ′(z)| ≥ π|h|/c + O(h2/(cY )). Now we applyintegration by parts many times to the integral with respect to z in (4.26)by integrating exp(iφ(z)) and differentiating the rest of the integrand. Thedifferentiation will always yield a constant bound, while each integrationgives us O(c/|h|). If |h| ≥ cKε, we can do this repeatedly so that (4.26)and hence (4.25) are � K−N and negligible. Therefore, the innermost sumin (4.11) may be taken over |h| ≤ cKε.
4.16 A bound for the holomorphic case. Back to the bound forP (c, h, Y ) in (4.21) for the holomorphic case with σ = 1/2 + θ + ε, its sumover |h| ≤ cKε is∑
|h|≤cKε
P (c, h, Y ) � KY (µ−j−k)/2+θ+ε∑
|h|≤cKε
(( |h|c
)5
+ · · · + 1)
� cK1+εY (µ−j−k)/2+θ+ε.
Taking the sums over c and k as in (4.11), we get
T(η)µ,ν,j(Y ) � K1+εY µ−j+1/2+θ+ε
L2ν−1
∑0≤k≤N
1(2π)kk!
L2k
Y k
∑c≤Y/(LK1−ε)
cj+k−µ+1
� K1+εY 5/2+θ+ε
L2ν−1(LK1−ε)µ−j−2
∑0≤k≤N
1(2π)kk!
Lk
K(1−ε)k
� Y 5/2+θ+ε
LK1−ε
(LK1−ε)µ
L2ν(LK1−ε)−j
for 0 ≤ 2µ ≤ ν < N and 0 ≤ j < 2N . Consequently
TK,L(Y ) � Y 5/2+θ+ε
LK1−ε
for L ≥ K1/3. Recall that we need a bound of the form LKY 1+ε. SettingY 5/2+θ+ε/(LK1−ε) ≤ LKY 1+ε, we can see that
TK,L(Y ) � LKY 1+ε, for L ≥ K1/2+θ+ε. (4.27)
With θ = 7/64, we have this bound for L ≥ K39/64+ε.
4.17 A bound for the Maass case. Finally let us turn to the case ofMaass forms. Using the bounds in (4.22) with σ = 1/2 + θ + ε, we get
1320 JIANYA LIU AND YANGBO YE GAFA
∑|h|≤cKε
P (c, h, Y ) � KY (µ−j−k)/2+θ+ε∑
|h|≤cKε
(( |h|c
)5
+ · · · + 1)
+ KY (µ−j−k)/2+θ+ε∑
|h|≤cKε
( |h|5/2−θ−ε
c2+
|h|3/2−θ−ε
c+ |h|1/2−θ−ε
)� (c + c3/2−θ)K1+εY (µ−j−k)/2+θ+ε.
Back to (4.11) we have
T(η)µ,ν,j(Y ) � K1+εY µ−j+1/2+θ+ε
L2ν−1
∑0≤k≤N
1(2π)kk!
L2k
Y k
×∑
c≤Y/(LK1−ε)
(cj+k−µ+1 + cj+k−µ+3/2−theta)
�(
Y 5/2+θ+ε
LK1−ε
Y 3+ε
(LK1−ε)3/2−θ
)(LK1−ε)µ
L2ν(LK1−ε)−j
for 0 ≤ 2µ ≤ ν < N and 0 ≤ j < 2N . Consequently, if L ≥ K1/3, then
TK,L(Y ) � KεY 5/2+θ+ε
LK+
KεY 3+ε
(LK)3/2−θ
� KεY 3+ε
(LK)3/2−θ
with θ < 1/2. Therefore
TK,L(Y ) � LKY 1+ε, for L ≥ K(3+2θ)/(5−2θ)+ε, (4.28)
when g is a Maass form. With θ = 7/64, we have this bound for L ≥K103/153+ε.
5 The Proof of the Theorems
5.1 Proof of Theorem 1.1. Now we want to use our estimate for thediagonal terms in §3 and bounds in (4.27) and (4.28). With these boundswe conclude that∑
K,L
=∑fj
(h
(kj − K
L
)+ h
(−kj + K
L
)) ∣∣SY (fj)∣∣2
� LKY 1+ε,
or simply ∑K−L≤kj≤K+L
∣∣SY (fj)∣∣2 � LKY 1+ε, (5.1)
Vol. 12, 2002 SUBCONVEXITY FOR L-FUNCTIONS 1321
when Y � K2+ε and
K1/2+θ+ε ≤ L ≤ K2−δ for g being holomorphic , (5.2)
andK(3+2θ)/(5−2θ)+ε ≤ L ≤ K2−δ for g being Maass, (5.3)
for arbitrarily small δ > 0 and ε > 0. According to the approximationformula of the central value of the L-function in (2.3), we have∑
K−L≤kj≤K+L
∣∣L (12 + it, fj ⊗ g
)∣∣2�
∑K−L≤kj≤K+L
∣∣∣∣ ∑1≤b≤K1+ε
1b
∑a≥1
λj(a)λg(a)√a
V
(ab2
K2
) ∣∣∣∣2
� 1K2
∑K−L≤kj≤K+L
∣∣∣∣∑a≥1
λj(a)λg(a)∑
1≤b≤K1+ε
V (ab2/K2)√ab2/K2
∣∣∣∣2.Applying smooth dyadic subdivisions to the function∑
1≤b≤K1+ε
V (ab2/K2)√ab2/K2
,
we get ∑K−L≤kj≤K+L
∣∣L (12 + it, fj ⊗ g
)∣∣2� log K
K2
∑K−L≤kj≤K+L
max1≤B≤K2+ε
∣∣∣∣∑a≥1
λj(a)λg(a)H(
a
K2/B
) ∣∣∣∣2,where H is essentially a fixed smooth function of compact support in [1, 2].Using the bound in (5.1) with Y = K2/B, we see that the maximum isfrom B = 1 and hence∑
K−L≤kj≤K+L
∣∣∣∣L(12
+ it, fj ⊗ g
)∣∣∣∣2 � log K
K2
∑K−L≤kj≤K+L
∣∣SK2(fj)∣∣2
� log K
K2LK(K2)1+ε � (LK)1+ε
(5.4)
for L as in (5.2) or (5.3). After scaling back to the normalization of theMaass form f by multiplying Kε, we prove Theorem 1.1. �
1322 JIANYA LIU AND YANGBO YE GAFA
5.2 Proof of Theorem 1.2. From (5.2) we see that∣∣L (12 + it, fj ⊗ g
)∣∣2 � (LK)1+ε,
i.e.L(
12 + it, fj ⊗ g
)� (LK)1/2+ε
for any j with K − L ≤ kj ≤ K + L and any L as in (5.2) and (5.3).Taking L = K1/2+θ+ε in the holomorphic case or L = K(3+2θ)/(5−2θ)+ε inthe Maass case and scaling back to the normalization of f by multiplyingkε, we prove Theorem 1.2. �
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Jianya Liu, Department of Mathematics, Shandong University, Jinan, Shandong250100, China [email protected]
Yangbo Ye, Department of Mathematics, The University of Iowa, Iowa City,Iowa 52242-1419, USA [email protected]
Submitted: January 2002
