Subconvexity of L-Functions

V Vinay Kumaraswamy, Chennai Mathematical InstituteAdvisor: Prof R Balasubramanian, Institute of Mathematical Sciences

AbstractMany problems in number theory involve getting bounds for L-functions inside thecritical strip 0 ≤ σ ≤ 1. On the line Re(s) = 1/2 one can prove quite easily, using thePhragmen-Lindelof convexity principle, that L(s, f) �ε Q(s, f)1/4+ε - this bound iscalled the convexity bound. The Lindelof Hypothesis states that L(s, f)�ε Q(s, f)ε,where Q(s, f) is the conductor of L(s, f). For instance, for the Riemann zeta functionthis translates to the bound ζ(1/2+ it)�ε (|t|+3)ε. The Lindelof Hypothesis followsfrom the Riemann Hypothesis and although the Lindelof Hypothesis is much weaker,no proof is known. For many applications, however, we do not need the full strength ofthe Lindelof Hypothesis while estimating L-functions in the critical strip; breaking theconvexity bound suffices, i.e., proving that L(s, f) � Q(s, f)1/4−δ for some δ > 0.In this thesis, we study the subconvexity problem for various GL1 and GL2 automor-phic L-functions over Q and briefly indicate some of their applications, especially toproblems in equidistribution.

1

AcknowledgementsI would first of all like to thank my advisor, Prof R Balasubramanian for his constantencouragement and guidance, not only during the duration of this thesis, but also for allthe number theory he has taught me over the last three years; it has been an immenselyrewarding experience learning from him. It also gives me great pleasure to thank theCMI and the IMSc for having taught me all the mathematics I know. I would also like tothank the AQUA summer school organised by Harald Helfgott and Alexander Gorodnikfor getting me interested in the subject of automorphic forms and the subconvexityproblem for L-functions. This non-original thesis has been greatly influenced by ProfPhilippe Michel’s notes [10], I would like to thank him for making them availableonline. Thanks also to my friend Krishnan Rajkumar for all his help.

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Contents

1 Theory of L-functions 41.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 The Convexity Bound . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Approximate Functional Equation . . . . . . . . . . . . . . . . . . . 10

2 Review of Automorphic Forms 132.1 Spectral Expansion Formula . . . . . . . . . . . . . . . . . . . . . . 132.2 Kloosterman Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3 Trace Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 Estimates for Fourier Coefficients of Modular Forms . . . . . . . . . 14

3 Tools Used to Prove Subconvexity 193.1 Exponential Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.2 Character Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.3 The δ-Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.4 Poisson Type Formulae . . . . . . . . . . . . . . . . . . . . . . . . . 293.5 Method of Families and Amplification . . . . . . . . . . . . . . . . . 323.6 Shifted Convolution Problem . . . . . . . . . . . . . . . . . . . . . . 353.7 Quadratic Divisor Problem . . . . . . . . . . . . . . . . . . . . . . . 35

4 Subconvexity for GL(1) L-functions 414.1 Subconvexity for ζ(s) . . . . . . . . . . . . . . . . . . . . . . . . . . 414.2 Subconvexity for L(s, χ) . . . . . . . . . . . . . . . . . . . . . . . . 42

5 Subconvexity for GL(2) L-functions 445.1 Subconvexity for GL(2)×GL(1) L-functions . . . . . . . . . . . . 445.2 Subconvexity for L-functions Attached to Newforms . . . . . . . . . 495.3 Other Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

6 Weak Subconvexity 716.1 Mean Values of Multiplicative Functions . . . . . . . . . . . . . . . . 726.2 Some preliminary lemmas . . . . . . . . . . . . . . . . . . . . . . . 746.3 Deduction of the Main Theorem from Theorem 6.2.1 . . . . . . . . . 786.4 Successive Maxima . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

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6.5 Proof of the main theorem . . . . . . . . . . . . . . . . . . . . . . . 85

7 Applications of Subconvexity 947.1 Least Quadratic Non-residue . . . . . . . . . . . . . . . . . . . . . . 947.2 Equidistribution of Lattice Points on the Sphere . . . . . . . . . . . . 957.3 Mass Equidistribution of Hecke Eigenforms . . . . . . . . . . . . . . 97

A Complex Analysis 100A.1 Phragmen-Lindelof Principle . . . . . . . . . . . . . . . . . . . . . . 100A.2 Stirling’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100A.3 Bounds for the Analytic Conductor . . . . . . . . . . . . . . . . . . . 101A.4 Functions of Finite Order . . . . . . . . . . . . . . . . . . . . . . . . 101

B Bessel Functions 102B.1 Recurrence Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . 102B.2 Bounds for Bessel Functions . . . . . . . . . . . . . . . . . . . . . . 102B.3 Definite Integrals Involving Bessel Functions . . . . . . . . . . . . . 102

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Chapter 1

Theory of L-functions

In this chapter, we collect all the basic facts about L-functions that we will need later.We start off with the

1.1 Basic DefinitionsLet m ≥ 1. Let L(s, f) be given by the Dirichlet series and Euler product

L(s, f) =

∞∑n=1

λf (n)

ns=∏p

m∏j=1

(1− αj(p)

ps

)−1

(1.1)

with the sum and product converging absolutely for Re(s) > 1. We write

L∞(f, s) = q(f)s/2m∏j=1

ΓR(s+ µj) (1.2)

where ΓR(s+µj) = π−s/2Γ(s/2), the µj1 are complex numbers and q(f) ≥ 1, calledthe conductor of L(s, f), is an integer such that αi(p) 6= 0 for p - q(f). The localfactors αi(p) satisfy2

|αi(p)| < p.

The completed L-function Λ(s, f) = L(s, f)L∞(f, s) has analytic continuation to theentire complex plane(except for possible poles at 0 and 1), is of finite order and satisfiesthe functional equation,

Λ(s, f) = κ(f)Λ(1− s, f) (1.3)

1The Selberg conjecture predicts that Re(µj) > 0. The best known result is that Re(µj) ≥ −1/2 +

1/(m2 + 1)2The Ramanujan-Petersson conjecture predicts that |αi(p)| = 1 for all p - q(f) and |αi(p)| ≤ 1

otherwise.

5

where κ(f) is a complex number with modulus one, called the root number, and

L(s, f) =

∞∑n=1

λf (n)

ns(1.4)

L∞(s, f) = q(f)s/2m∏j=1

ΓR(s+ µj) (1.5)

and f is an object associated with f .Any L(s, f) that satisfy conditions (1) to (5) is called an L-function. We define the

analytic conductor

Q(s, f) = q(f)

m∏j=1

(|s+ µj |+ 3) (1.6)

We also denote

Q(f) = Q(0, f) = q(f)

m∏j=1

(|µj |+ 3). (1.7)

We will use Q(s, f) to estimate the growth of L(s, f) inside the critical strip. SinceQ(f) ≥ 3mq(f) we have m < logQ(f). And so,

Q(s, f) ≤ Q(f)(|s|+ 3)m. (1.8)

Also define

γ(s, f) =

m∏j=1

ΓR(s+ µj). (1.9)

1.2 ExamplesIn this section, we define all the L-functions that we will later encounter in the thesisand we also compute their respective analytic conductors. The prototype for all L-functions is of course the well known Riemann zeta function,

1.2.1 Riemann zeta function

ζ(s) =

∞∑n=1

1

ns=∏p

(1− 1

ps

)−1

(1.10)

defined for Re(s) > 1. Let

ξ(s) = s(s− 1)π−s/2Γ(s/2)ζ(s).

Then ξ(s) extends to an entire function and satisfies the functional equation

ξ(s) = ξ(1− s). (1.11)

Therefore, the analytic conductor for ζ(s) is (|s|+ 3).

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1.2.2 Dirichlet L-functionLet χ be a Dirichlet character (mod q). Let q∗ be the conductor of χ.To χ we mayattach the following Dirichlet series,

L(s, χ) =

∞∑n=1

χ(n)

ns=∏p

(1− χ(p)

ps

)−1

(1.12)

for Re(s) > 1. When χ is a primitive character, i.e., q = q∗, let a =χ(1)− χ(−1)

2.

Let

ξ(s, χ) = π−s/2Γ(s+ a

2)L(s, χ).

ξ(s) then extends to an entire function and satisfies the following functional equation

ξ(s) = κ(χ)ξ(1− s, χ). (1.13)

where κ(χ) =τ(χ)

wq1/2and w = 1 if χ(−1) = 1 and w = i if χ(−1) = −1. Since

χ(p) = 0 whenever p|q, the conductor of L(s, χ) is q and the analytic conductor is

q(|s + a| + 3). Here τ(χ) =

q∑n=1

χ(n)e2πin/q is the Gauss sum and we know that

|τ(χ)| = q1/2.

1.2.3 L-functions attached to holomorphic cusp formsLet f be a primitive cusp form of weight k, level q with nebentypus χ then f has thefourier expansion

f(z) =

∞∑n=1

nk−12 λf (n)e(nz).

To f we attach the L-function,

L(s, f) =

∞∑n=1

λf (n)

ns=∏p

(1− λf (p)

ps+χ(p)

p2s

)−1

. (1.14)

where λf (n) is the nth Hecke eigenvalue of f . Let

L∞(f, s) = qs/2π−sΓ(s+ (k − 1)/2

2)Γ(

s+ (k + 1)/2

2).

Λ(f, s) has analytic continuation to the entire complex plane, and satisfies the func-tional equation

Λ(s, f) = κ(χ)2i−kΛ(1− s, f) (1.15)

7

with κ(χ) as before. Then

Q(s, f) = q(|s+k − 1

2|)(|s+

k + 1

2) ≤ q(|s|+ k + 3)2, (1.16)

Q(f) = q(k − 1

2)(k + 1

2) � qk2 (1.17)

In this thesis, we will prove subconvexity results for all the above defined L-functions.

1.2.4 L-functions attached to Maass cusp formsLet f be a primitive Maass cusp form with nebentypus χ which is an eigenfunction ofall the Hecke operators with eigenvalue λ = 1/4 + r2, where 0 ≤ r < 1/2. f has thefourier expansion

f(z) =√y∑n 6=0

ρ(n)Kir(2π|n|y)e(nx). (1.18)

To it we associate the L-function of conductor q,

L(s, f) =∑n≥1

ρ(n)

ns=∏p

(1− ρ(p)

ps+χ(p)

p2s)−1. (1.19)

L∞(f, s) = qs/2π−sΓ(s+ δ − ir

2)Γ(

s+ δ + ir

2)

where δ is 0 if f is even and 1 otherwise. Once again Γ(f, s) extends to an analyticfunction on the entire complex plane. The analytic conductor Q(f) � λq.

1.2.5 Rankin-Selberg L-functionLet χD and χq be Dirichlet characters mod D and q and let χ = χqχD be a charactermod qD. Let f be a normalised newform of weight k and level q and let g be anormalised newform of weight k′ and level D(or a Maass form with eigenvalue λ =1/4 + r2 ≥ 1/4). The Rankin-Selberg convolution L-function is

L(f × g, s) = L(χ, 2s)∑n≥1

λf (n)λg(n)

ns=∏p

2∏i=1

2∏j=1

(1− αf,i(p)αg,j(p)

ps)−1

(1.20)

The αf,i and αg,j are the Euler factors of f and g respectively. L(f × g) has analyticcontinuation to the entire complex plane except when f = g in which case there aresimple poles at s = 0, 1. Set

L∞(f × g, s) = Γ(s+|k − k′|

2)Γ(s+

k + k′

2− 1) (1.21)

if g is a holomorphic form. And

L∞(f × g, s) = Γ(s+k + 2ir − 1

2)Γ(s+

k − 2ir − 1

2− 1) (1.22)

otherwise.

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1.2.6 Symmetric square L-functionLet f be an eigenform of weight k for the full modular group. Then we form the Eulerproduct

L(s, sym2f) =∏p

(1− αf (p)2

ps

)−1(1− αf (p)βf (p)

ps

)−1(1− βf (p)2

ps

)−1

.

(1.23)

Shimura proved that the completed L-function

Λ(s, sym2f) = ΓR(s+ 1)ΓR(s+ k − 1)ΓR(s+ k)L(s, sym2f)

is entire and that Λ(s, sym2f) = Λ(1− s, sym2f). The analytic conductor Q(s, f) �k2(|s|+ 3)3.

We then have the identity

−L′

L(s, sym2f) =

∑n≥2

λsym2f (n)Λ(n)ns

so that λsym2f (pk) = αf (p)2k+(αf (p)βf (p))k+βf (p)2k, and since αf (p)βf (p) = 1,this is nothing but αf (p)2k+1+βf (p)2k. And hence λf (pk) ≤ 3 by the Deligne bound.Hence L(s, sym2f) satisfies the weak-Ramanujan hypothesis(see Chapter 6).

1.2.7 Triple product L-functionLet f be a holomorphic form as above, and let φ be a Maass eigenform. We writetheir corresponding pth Hecke eigenvalues as αf (p) + βf (p) and αφ(p) + βφ(p). Letλ = 1/4 + t2 ≥ 1/4 be the Laplace eigenvalue of φ. Define the triple product L-function, L(s, f × f × φ) by the Euler product

∏p

(1− αf (p)2αφ(p)

ps)−1(1− αφ(p)

ps)−2(1− βf (p)2αφ(p)

ps)−1

(1− αf (p)2βφ(p)

ps)−1(1− βφ(p)

ps)−2(1− βf (p)2βφ(p)

ps)−1.

(1.24)

Consider the product of Γ factors

L∞(f ×f ×φ, s) =∏±

ΓR(s+k−1± it)ΓR(s+k± it)ΓR(s+±it)ΓR(s+1+±it)

Garrett proved that the completed L-function Γ(s, f×f×φ) is entire and that Γ(s, f×f×φ) = Γ(1−s, f×f×φ). And we easily have the estimateQ(s, f)� k4(1+ |s|)8.

We then have the identity

−L′

L(s, sym2f) =

∑n≥2

λf (n)Λ(n)ns

9

where λf×f×φ = λf (n)2λφ(n). We need to verify that∑

x<n≤ex

|λf (n)2λφ(n)|2Λ(n)

ns≤

A2 +A0

log exfor some A,A0 ≥ 1. But since |λf (n)| = |αf (p)k + βf (p)k| ≤ 2 by the

Deligne bound, it suffices to check that∑

x<n≤ex

|λφ(n)|2Λ(n)

ns≤ A2 +

A0

log exwhich

follows from the prime number theorem applied to the Rankin-Selberg convolutionL(s, φ× φ).

1.3 The Convexity BoundWe now estimate the growth of L(s, f) in the critical strip. Let 0 ≤ σ ≤ 1 ands = σ + it. Assume that L(s, f) has no singularity at s = 0. Let ε > 0. Then sinceL(s, f) is given by a convergent sum to the right of 1,

L(f, 1 + ε+ it)�ε 1.

Hence by the functional equation and Appendix A.3 we obtain

L(f,−ε− it)� q(f)1/2+ε γ(f, 1 + ε+ it)

γ(f ,−ε− it)

� q(f)1/2+εQ∞(−ε− it)1/2+ε∏j

|1 + ε+ it+ κj || − ε− it+ κj |

� q(f)1/2+εQ∞(−ε− it)1/2+ε.

Hence by the Phragmen-Lindelof principle(see Appendix A.1), we get

L(s, f)� Q(s, f)(1−σ)

2 +ε. (1.25)

Now if L(s, f) has a zero or pole of order r, we kill it by multiplying by

pr(s) =

(s− 1

s+ 1

)rto obtain

pr(s)L(s, f)� Q(s, f)(1−σ)

2 +ε. (1.26)

This yields on the critical line the bound

L(s, f)�ε Q(s, f)1/4+ε (1.27)

This is called the convexity bound, and in what follows we will break the convexitybound for certain GL(1) and GL(2) L-functions, i.e., prove that there exists δ > 0 suchthat

L(s, f)�ε Q(s, f)1/4−δ+ε

on the line Re(s) = 1/2. And the Generalised Lindelof Hypothesis states that δ canbe taken to be 1/4.

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1.4 Approximate Functional EquationIn this section, we prove a tool that will be of great use while proving subconvexbounds.

Theorem 1.4.1. Let L(s, f) be an L-function. Let G(u) be any function which isholomorphic and bounded in the strip −4 < Re(u) < 4, even, and normalised byG(0) = 1. Let X > 0. Then for s in the strip 0 ≤ σ ≤ 1 we have

L(s, f) =

∞∑n=1

λf (n)

nsVs

(n

X√q

)+ κ(f, s)

∞∑n=1

λf (n)

n1−s V1−s

(nX√q

)+R (1.28)

where Vs(y) is a smooth function defined by

Vs(y) =1

2πi

∫(3)

y−uG(u)γ(f, s+ u)

γ(f, s)

du

u(1.29)

and

κ(f, s) = κ(f)q(f)1/2−s γ(f, 1− s)γ(f, s)

. (1.30)

The last term R = 0 if Λ(f, s) is entire, otherwise

R = (resu=1−s + resu=−s)Λ(f, s+ u)

qs/2γ(f, s)

G(u)

uXu. (1.31)

Proof. Let

I(X, f, s) =1

2πi

∫(3)

XuΛ(f, s+ u)G(u)du

u.

The integral exists because Λ(f, s) decays rapidly as t → ∞ for fixed σ by Stirling’sformula. Because L(s, f) is at most of polynomial growth, we may move the line ofintegration to Re(s) = −3. Applying the functional equation there yields

Λ(f, s) = I(X, f, s) + κ(f)I(X−1, f , 1− s) +Rqs/2γ(f, s) (1.32)

where Λ(f, s) comes from the simple pole of u−1G(u) at u = 0 and the last term Rcomes from possible residues at u = 1− s and u = −s. By expanding into absolutelyconverging Dirichlet series we have

I(X, f, s) = qs/2∞∑n=1

λf (n)

ns1

2πi

∫(3)

γ(f, s+ u)

(X√q

n

)uG(u)

du

u

= qs/2γ(f, s)

∞∑n=1

λf (n)

nsVs

(n

X√q

).

I(X−1, f , 1− s) = q1−s/2∞∑n=1

λf (n)

n1−s1

2πi

∫(3)

γ(f , 1− s+ u)

(X−1√q

n

)uG(u)

du

u

= q1−s/2γ(f , 1− s)∞∑n=1

λf (n)

n1−s V1−s

(n

X−1√q

).

11

Plug into the relation above, divide both sides by qs/2γ(f, s) to get the theorem.

Such a theorem was well known for the zeta function, and it was generalised by RNarasimhan and K Chandrasekharan in the paper The Approximate Functional Equa-tion For a Class of Zeta Functions, Math. Ann. 152 (1963).

For suitable test functions both the sums in the theorem are effectively limited tothe terms n�

√Q(s, f). We take

G(u) =(

cosπu

4A

)−4mA

where A is a positive integer.

Proposition 1.4.2. SupposeRe(s+κj) ≥ 3α > 0 for 1 ≤ j ≤ m. Then the derivativesof Vs(y) satisfy

yaV (a)s (y) �

(1 +

y√Q∞(s)

)−A, (1.33)

yaV (a)s (y) = δa=0 +O

((y√

Q∞(s)

)α)(1.34)

and the implied constants depend only on α, a,A and m.

Proof. For s and u with Re(s) = σ > 0 and Re(u) = β > −σ we derive by Stirling’sformula(see Appendix 2)

Γ(s+ u)

Γ(s)� |s+ u|σ+β− 1

2

|s|σ− 12

exp(π

2(|s| − |s+ u|)

)� (|s|+ 3)β exp(

πd

2|u|).

Differentiating under the integral sign we have

yaV (a)s (y) =

1

2πi

∫(3)

y−uG(u)(−u)aγ(f, s+ u)

γ(f, s)

du

u.

Moving the line of integration to the line Re(u) = β = −α we derive the first es-timate while moving to the line Re(u) = A we derive the bound O((

√Q∞)A).This

combined with the first estimate gives the second.

The upshot of the above approximate functional equation is this: choosing X = 1,and using the rapid decay of Vs(z), we are left with estimating the sums,

∑n≤N

λf (n)

nsand

∑n≤N

λf (n)

ns

12

where N � q1/2. Therefore to estimate L(s, f), it suffices to estimate sums of theabove type. And by breaking into smooth dyadic partitions, it suffices to study sums oftype ∑

H≤n≤2H

λf (n)g(n)

ns

where g(x) is a smooth function supported on [H, 2H] with sufficient decay properties.We end this section with the remark that the approximate functional equation roughlyyields the convexity bound. By the Rankin-Selberg method(see Chapter 2), we get that∑

n≤t

λf (n)� (Q(f)t)εt.

And the approximate functional equation is essentially the fact that

L(1/2, f) ∼∑

n≤√Q(f)

λf (n)n−1/2.

Which on partial summation yields the convexity bound. This just illustrates the factthat although the approximate functional equation is a convenient tool to estimate L-functions, we still haven’t gained anything so far as subconvexity is concerned; it justreflects the truth of the convexity principle.

13

Chapter 2

Review of Automorphic Forms

In this chapter, we collect all the facts about automorphic forms for GL(2) that we willneed.

2.1 Spectral Expansion FormulaWe have the formula,

Theorem 2.1.1. Let f(z) ∈ L2(SL2(Z)\H), then we have

f(z) =

∞∑j=1

< f, uj > uj(z) +1

4π

∫ ∞−∞

< f,E(∗, 1/2 + it) > E(∗, 1/2 + it) dt,

(2.1)

where {uj(z)} denotes an orthonormal basis for Maass cusp forms and E(z, 1/2+ it)denotes the unitary Eisenstein series for SL2(Z).

This formula is a close analogue of Fourier theory on the real line. The aboveformula states that L2 functions on SL2(Z)\H can be expanded in terms of eigen-

functions of the hyperbolic Laplacian ∆ = − 1

y2

(∂2

∂x2+

∂2

∂y2

)with the Maass cusp

forms spanning the discrete spectrum and the unitary eisenstein series spanning thecontinuous spectrum.

2.2 Kloosterman SumsWe define the sum

S(m,n; c) =∑∗

x≡1(c)

e

(mx+ nx

c

). (2.2)

14

When n = 0, this is the Ramanujan sum, and we have the following bounds,

S(m,n; c)� τ(c)c1/2(m,n, c)1/2, (2.3)

due to Weil, and

S(0, n; c)� (n, c). (2.4)

We remark that the bound S(m,n; c)� c1/2(m,n, c)1/2, although optimal, is notreally needed to solve the shifted convolution problem(see chapter 3). For instance, ifwe only had the bound

S(m,n; c)� (m,n, c)1/2cθ

for some θ < 1, then Theorem 5.1.1 can be proved with the exponentθ + 2

θ + 5which

still breaks subconvexity. Hence, Kloosterman’s bound, which can be obtained viaelementary methods, with θ = 3/4 will suffice.

2.3 Trace FormulaeThe following inequality is an example of a large sieve inequality, and using it we willprove another large sieve inequality in the next section.

Theorem 2.3.1. For N,Q ≥ 1 and any an ∈ C, we have∑a(q)

|∑n≤N

ane

(an

q

)|2 � (q +N)

∑n≤N

|an|2 (2.5)

The following theorem, called the Petersson trace formula is a very useful ’orthog-onality relation’.

Theorem 2.3.2. Let k ≥ 2, and Fq be an orthonormal basis for Sk(q). Let ψf (n) beas in equation (2.7). Then for any m,n ≥ 1 we have∑f∈Fq

ψf (m)ψf (n) = q(k − 1)δm=n + q(k − 1)2πik∑c=0(q)

1

cS(m,n; c)Jk−1

(4π√mn

c

)(2.6)

2.4 Estimates for Fourier Coefficients of Modular FormsLet f be a holomorphic cusp form of weight k, level q. Then f has the Fourier expan-sion

f(z) =∑n≥1

αf (n)e(nz).

In light of the Petersson trace formula, we renormalise αf (n) by setting

ψf (n) =

(q(k − 1)!

(4πn)k−1

)1/2

αf (n). (2.7)

15

And a similar formula fourier expansion (see equation 1.18) holds for Maass cuspforms. In this section we will gather some results on the Fourier coefficients of modularforms for use later. Since f is a cusp form, we obtain by Parseval,

∑n≥1

|αf (n)|2e−4πny =

∫ 1

0

|f(z)|2 dx� y−k

for any y > 0, so that for any N ≥ 1,∑n≤N

|αf (n)|2 � y−ke4πNy

By choosing y = N−1 we get the bound∑n≤N

|αf (n)|2 � Nk (2.8)

which by positivity yields αf (n) � nk/2. The best possible bound was obtained byDeligne, who proved

|αf (n)| ≤ τ(n)nk−12 (2.9)

and henceψf (n)� (qkn)ε. (2.10)

Another useful bound that we will use time and again is the bound that comes from theRankin-Selberg method, ∑

n≤N

|αf (n)|2 ∼ cfNk (2.11)

where cf > 0 is a constant depending only on f . Whereas Deligne’s proof is verydifficult, the proof of the above equation isn’t. The above equation is to be interpretedas the truth of the Deligne bound ’holding on average’.

For ρf (n), coefficients of a Maass cusp form f , the analogue of equation (2.9) isthat

ρf (n)� nε. (2.12)

But this is far from being proved. Serre proved that

ρf (n)� n1/5+ε. (2.13)

2.4.1 Estimates for bilinear formsWe will estimate general linear forms

Lf (a) =∑n≤N

anψf (n) (2.14)

on average over the basis F . We assume that k ≥ 2.

16

Theorem 2.4.1. For any sequence of complex numbers an we have∑f∈F

|Lf (a)|2 = (k − 1){q +O(N logN)}||a||2 (2.15)

where ||a|| denotes the l2-norm and the implied constant is absolute.

Proof. Let Sq(a) denote the left-hand side of equation (2.9), so we have

Sq(a) =∑m≤M

∑n≤N

aman∆(m,n)

For ∆(m,n) we have the identity

∆(m,n) = (k − 1)qδmn + 2πik(k − 1)q∑

c = 0 (mod q)

c−1S(m,n; c)Jk−1

(4π√mn

c

)(2.16)

The diagonal contributes the main term and we explain below how to compute the errorterm. For any complex numbers bn,∑

m

∑n

bmbnS(m,n; c) =∑m,n

∑∗

x(c)

bmbne

(mx+ nx

c

)=∑m,n

∑∗

x(c)

e(mxc

)e(nxc

)=∑∗

x(c)

|∑n

bne(nxc

)|2

� (c+N)||b||2

(2.17)

For large c,∑m

∑n

bmbnS(m,n; c)�∑m

∑n

bmbn√cτ(c)(m,n, c)1/2

�√cτ(c)N ||b||2

(2.18)

using the Weil bound. Therefore,∑m

∑n

bmbnS(m,n; c)� min{c+N,√cτ(c)N}||b||2. (2.19)

We now expand Jk−1 into its power series, to get

Jk−1(x) =

∞∑l=0

(−1)l

l!(l + k − 1)!

(x2

)2l+k−1

(2.20)

17

To apply the above formula, we first assume that q ≥ N . The case q < N willfollow from this. We have∑m,n

amanS(m,n; c)Jk−1

(4π√mn

c

)

=

∞∑l=0

(−1)l

l!(l + k − 1)!

∑m,n

(am

2√πm√c

)(an

2√πn√c

)S(m,n; c)

� 1

(k − 1)!min{c+N,

√cτ(c)N}

∞∑l=0

1

l!

∑m≤N

|am|2(

2πm

c

)2l

� 1

(k − 1)!min{c+N,

√cτ(c)N}||a||2

(2.21)

because ∀ l ∑m≤N

|am|2(

2πm

c

)2l

� ||a||2

Hence ∑c≡0(q)

1

c

∑amanS(m,n; c)Jk−1

(4π√mn

c

)

�∑c≡0(q)

1

c

(N

c

)k−1

min{c+N,√cτ(c)N}N ||a||2

� N

qlogN ||a||2

where the logN appears only in the case k = 2(for k ≥ 3, we get the boundO(N

q||a||2).This

completes the proof when q ≥ N . For the case q < N we use the following trick.Let p be a prime such that N ≤ pq ≤ 2N . Let Fq and Fpq be orthonormal bases

for Sk(q) and Sk(pq) respectively. Then we may embed1

[Γ0(q) : Γ0(pq)]1/2Fq inside

Fpq . Since [Γ0(q) : Γ0(pq)] ≤ p+ 1

1

p+ 1Sq(a) ≤ 1

pSpq(a).

∴ Sq(a) ≤(

1 +1

p

)Spq(a)

(2.22)

The 1/p factor appears on the right hand side because the fourier coefficients of cuspforms for Γ0(pq) have different normalisation, see equation (2.7). Since pq ≥ N , wemay appeal to the previous case and hence we are done.

Now that we have a bound for the bilinear form

B =∑m,n

aman∆(m.n)

18

we very easily obtain the following

Corollary 2.4.2. Let r and s be positive integers such that (q, rs) = 1 and F (x, y) areal-valued smooth function supported on [M, 2M ]× [N, 2N ] satisfying

F (i,j) �M−iN−j (2.23)

and let am and bm be complex numbers. Then

B(r, s) =∑m,n

ambn∆(rm, sn)F (m,n)� (rsMN)ε(q +M)1/2(q +N)1/2||a||||b||

(2.24)

Proof. Clearly it is sufficient to consider the form

B0(r, s) =∑m≤M

∑n≤N

ambn∆(rm, sn)

By equation (5.57) we have can convert sums involving ψf (rm)ψf (sn) into sums ofsums involving on ψf (m′)ψf (n′).

B0(r, s) =∑

uvw2=r

∑xyz2=s

µ(vw)µ(w)τ(v)µ(yz)µ(z)τ(y)

×∑d

∑e

∑f∈F

λf (x)λf (u)

(∑m

amvw2d2τ(m)ψf (m)

)(∑n

bnyz2e2τ(n)ψf (n)

).

Hence by Cauchy-Schwarz applied to the f summation and the bound |λf (n)| ≤ τ(n)the corollary follows from the theorem.

Set am = bm = τ(m). Then Corollary 2.4.2 gives,

B(r, s)� (rs)εq1+ε(MN)1/2

when M,N � q1+ε. By equation (5.102), this is exactly the convexity bound. Tobreak convexity, we need better bounds for this bilinear form(see Chapter 5).

19

Chapter 3

Tools Used to ProveSubconvexity

To prove subconvexity, we need various tools and we develop them systematically inthis chapter for use later.

3.1 Exponential Sums

Since we are reduced to studying sums of the type∑n

λf (n)

n1/2+it, we are essentially

dealing with a sum of the type∑n

a(n)e(f(n)). This is how the theory of exponential

sums enters the picture. This method is particularly useful in proving subconvexity

for the zeta function, where a(n) = n−1/2 and f(t) = − log t

2π. We develop van der

Corput’s method below.

3.1.1 van der Corput’s methodLemma 3.1.1. Let f(x) be a smooth function such that f ′(x) is monotonic and f ′(x) ≥m > 0 or f ′(x) ≤ −m < 0 for x ∈ [a, b]. Then

|∫ b

a

eif(x) dx |≤ 4m−1. (3.1)

Proof. Since the conjugate of∫ b

a

eiF (x) dx is∫ b

a

e−if(x) dx we may suppose that

f ′(x) > 0. Now eif(x) = cos f(x)+i sin f(x), and by the second mean value theorem

20

for integrals,

∫ b

a

f(x)g(x) dx =

f(b)

∫ b

c

g(x) dx if f(x) ≥ 0, f ′(x) ≥ 0 in [a, b];

f(a)

∫ c

a

g(x) dx if f(x) ≥ 0, f ′(x) ≤ 0 in [a, b].

(3.2)

where c is some number in (a, b). Hence riting∫ b

a

cos(f(x)) dx =

∫ b

a

(f ′(x))−1d(sin f(x))

we see that

|∫ b

a

cos(f(x)) dx |≤ 2m−1

because ((f ′(x))−1 is monotonic in [a, b]. The same bound holds for the integral in-volving sin, and hence the lemma follows.

Lemma 3.1.2. Let f be a smooth function on [a, b] such that f ′′(x) ≥ m > 0 orf ′′(x) ≤ −m < 0. Then

|∫ b

a

eiF (x) dx |≤ 8m−1/2 (3.3)

Proof. Assume that f ′′(x) > 0, so that f ′ is monotonic increasing and has at most onezero c ∈ (a, b). We write,∫ b

a

eiF (x) dx =

∫ c−u

a

+

∫ c+u

c−u+

∫ b

c+u

= I1 + I2 + I3,

where we will choose u later. Trivially, I2 � 2u, and for u < c−a and a ≤ x ≤ c−u

we have |F ′(x)| = |∫ c

x

f ′′(t) dt| ≥ um, so that the previous lemma gives |I1| ≤

4(um)−1. A similar estimate holds for I3 if u < b− c. If u ≥ c− a, or u ≥ b− c, orif f ′ has no zero in [a, b], the analysis is similar, and all the cases lead to

|∫ b

a

eiF (x) dx |≤ 8(um)−1 + 2u = 8m−1/2

taking u = 2m−1/2.

Lemma 3.1.3. Let f(x) be a smooth real valued function on [a, b] such that f ′′(x) < 0in [a, b], and let f ′(b) = α, f ′(a) = β. Then for any η with 0 < η < 1 we have

∑a≤n≤b

e(f(n)) =∑

α−η<m<β+η

∫ b

a

e(f(x)−mx) dx+O (log(β − α+ 2)) . (3.4)

21

Proof. By the Euler-Maclaurin formula we get

∑a≤n≤b

e(f(n)) =

∫ b

a

e(f(x)) dx+ 2πi

∫ b

a

ψ(x)f ′(x)e(f(x)) dx+O(1),

where ψ(x) = x− 1/2. Without loss of generality we may suppose that η − 1 < α ≤η(so that m ≥ 0), for if k is an integer such that η − 1 < α − k ≤ η, then equation(3.4) becomes with h(x) = f(x)− kx,∑a<n≤b

e(f(n)) =∑

a<n≤b

e(h(n))

=∑

α′−η<m−k<β′+η

∫ b

a

e(h(x)− (m− k)x) dx+O (log(β′ − α′ + 2)) ,

(3.5)

where α′ = α − k and β′ = β − k. Thus equation (3.5) implies equation (3.3) andm ≥ k by the choice of k. Using the Fourier expansion for ψ(x), the integral is equalto

−2i

∞∑m=1

∫ b

a

1

msin(2πmx)e(f(x))f ′(x) dx =

∞∑m=1

1

m

∫ b

a

(e(−mx)− e(mx))e(f(x))f ′(x) dx

=

∞∑m=1

1

2πm

∫ b

a

f ′(x)

f ′(x)−md(e(f(x)−mx))

−∞∑m=1

1

2πm

∫ b

a

f ′(x)

f ′(x) +md(e(f(x)) +mx).

We have assumed that f ′(x) is monotonically decreasing, so thatf ′(x)

f ′(x) +malso

decreases monotonically. We see easily that the second integral is� β

β +muniformly

in m, so that the second sum is� 1 + log(β + 2)Similarly, it is seen that

∑m≥β+η

1

m

∫ b

a

f ′(x)

f ′(x)−md(e(f(x)−mx))�

∑m≥β+η

1

m

β

m− β

�∑

β+η≤m≤2β

1

m− β+∑m≥2β

β

m2� 1 + log(β + 2).

22

We now estimate the remaining sum,

∑1≤m<β+η

1

2πm

∫ b

a

f ′(x)

f ′(x)−md(e(f(x)−mx)) =

∑1≤m<β+η

1

m

∫ b

a

f ′(x)e(f(x)−mx) dx

=1

2πi

∑1≤mβ+η

1

m

∫ b

a

d(e(f(x)−mx))

+∑

1≤m<β+η

∫ b

a

e(f(x)−mx) dx

= O (log(β + η)) +∑

1≤m<β+η

∫ b

a

e(f(x)−mx) dx.

Hence the lemma is proved.

We now prove van der Corput’s lemma

Lemma 3.1.4. Let f(x) be a real function on [a, b] and let H > 0. Then

∑a<n≤b

e(f(n))� (b− a)H−1/2 +

(b− a)H−1H−1∑h=1

|∑

a<n≤b−h

e(f(n+ h)− f(n)) |

1/2

.

(3.6)

Proof. We may suppose that a and b are integers and 2 ≤ H < b − a, since trivially∑a<n≤b

e(f(n)) � b − a + 1 and b − a � (b − a)H−1/2 for H < 2. If H ≥ b − a

the LHS of equation (3.6) is trivially O(H). We may also suppose that H is an integer,since the RHS of equation (3.6) remains unchanged in magnitude if H is replaced bythe integer nearest to it. Hence we are reduced to showing that

∑a<n≤b

e(f(n))� (b−a)H−1/2+

(b− a)H−1H−1∑h=1

|∑

a<n≤b−h

e(f(n+ h)− f(n)) |

1/2

.

Observe that

H∑

a<n≤b

e(f(n)) =

H∑m=1

b−m∑n=a−m+1

e(f(m+ n)), (3.7)

23

and define f(k) = 0 if k is an integer such that k ≤ a or k > b. Then writingS =

∑a<n≤b

e(f(n)), and inverting the order or summation we obtain,

HS =

b−1∑n=a−H+1

H∑m=1

e(f(m+ n)), (3.8)

so that n takes at most b− a+H ≤ 2(b− a) values. By Cauchy-Schwarz,

H2S2 ≤ 2(b− a)

b−1∑n=a−H+1

|H∑m=1

e(f(m+ n)) |2 .

≤ 2(b− a)H + 2 |b−1∑

n=a−H+1

∑1≤r<s≤H

e(f(n+ s)− f(n+ r)) | .

(3.9)

In the last sum for a fixed k,H such that 1 ≤ h ≤ H − 1, a < k ≤ b − h, wehave f(n + s) − f(n + r) = f(k + h) − f(k) exactly H − h times, that is, forr = 1, 2, . . . ,H − h, s = r + h, n = k − r. Hence modulus of the double sum doesnot exceed

|H−1∑h=1

(H − h)∑

a<k≤b−h

e(f(k + h)− f(k)) |� H

H−1∑h=1

|∑

a<n≤b−h

e(f(n+ h)− f(n)) |,

(3.10)

and hence the proof follows.

Proposition 3.1.5. Let k ≥ 2 be a fixed integer and let f be a 3 times differentiablefunction on [a, b]. If b ≥ a+ 1, and for a ≤ x ≤ b

0 < λ3 ≤ f (3)(x) ≤ Aλ3

0 < λ2 ≤ f (2)(x) ≤ Aλ2, A > 1

or

λ3 ≤ −f (3)(x) ≤ Aλ3

λ2 ≤ −f (2)(x) ≤ Aλ2, A > 1,

then ∑a<n≤b

e(f(n))� A1/2(b− a)λ1/63 + (b− a)1/2λ

−1/63 . (3.11)

Proof. We may suppose that λ3 < 1, for otherwise the result is trivial. Lemma 3.1.2and Lemma 3.1.3 give∑

a<n≤b

e(f(n))� (β − α+ 1)λ−1/22 + log(β − α+ 2)

� (β − α+ 1)λ−1/22 � |f ′(a)− f ′(b)|λ−1/2

2 + λ−1/22

� λ2A(b− a)λ−1/22 + λ

−1/22 ,

24

We now apply van der Corput’s lemma in the following way. Set g(x) = f(x +h) − f(x). Then g′′(x) = hf ′′′(ξ) for some x < ξ < x + h. By assumption on f ,hλ3 ≤ g′′(x) ≤ Ahλ3. Hence by equation (3.11), we get∑

a<n≤b−h

e(g(n))� A(b− a)(hλ3)1/2 + (hλ3)−1/2

Hence by equation (3.6) we get∑a<n≤b

e(f(n))� (b− a)H−1/2 + (b− a)H1/2λ1/43 A1/2 + (b− a)1/2λ

−1/43 H1/4

and we the lemma is proved on setting H = [λ−1/33 ], since we have assumed that

λ3 ≤ 1.

3.2 Character SumsLet χ be a character to the modulus q. We are interested in sums of the type

S(M,N) =∑

M<n≤M+N

χ(n). (3.12)

We have the trivial bound, |S(M,N)| ≤ min(N, q). Our intention now is to improveon this bound for shorter intervals. Polya and Vinogradov independently proved the

Theorem 3.2.1.

S(M,N)� q1/2 log q. (3.13)

We see that the above bound is trivial for N � q1/2, and from the point of view ofsubconvexity, this is precisely the convexity bound L(1/2, χ)� q1/4 log q. Thereforeto break subconvexity, we need cancellation from short character sums. We have the

Theorem 3.2.2. If χ is a primitive character modulo q. Then

S(M,N)� N1−1/rqr+1

4r2+ε (3.14)

for r = 2, 3, and for any r ≥ 1 when q is cubefree.

Proof. We only give the proof when q = p is a prime. Notice that the Burgess boundis trivial unless p1/4+1/4r+ε ≤ N ≤ p1/2+1/4r+ε, which we now assume. Apply theshift n→ n+ h with 1 ≤ h ≤ H < N/10 to obtain

S(M,N) =∑

M<n≤M+N

χ(n+ h) + S(M,M + h)− S(M +N,M +N + h).

(3.15)

25

By induction, assume that S(M,M +h)−S(M +N,M +N +h) = 2θE(H), whereE(H) = cH1−1/rp

r+1

4r2+ε and |θ| ≤ 1. Let H = AB, where A,B are positive integers.

We use the shifts of type h = ab with 1 ≤ a ≤ A and 1 ≤ b ≤ B. Averaging, we get

S(M,N) =1

H

∑a,b

∑M<n≤M+N

χ(n+ ab) + 2θE(H). (3.16)

Since χ(n+ ab) = χ(a)χ(an+ b), we get

|S(M,N)| ≤ V

H+ 2E(H) (3.17)

where

V =∑x(p)

ν(x) |∑

1≤b≤B

χ(n+ b) | (3.18)

and ν(x) is the number of representations of x as an (mod p) with 1 ≤ a ≤ A andM < N ≤ M + N . By Holder applied to the functions ν1−1/r(x), ν1/r(x), w(x) =|∑1≤b≤B

χ(x+ b) | to the exponentsr

r − 1, 2r, 2r respectively, we get

V ≤ V 1−1/r1 V

1/2r2 W 1/2r (3.19)

where

V1 =∑x(p)

ν(x), V2 =∑

x(modp)

ν2(x),W =∑x(p)

|∑

1≤b≤B

χ(x+ b) |2r .

It is clear that V1 = AN , and that V2 �ε′ ANpε′ . To see the latter, observe that∑

x(p)

ν2(x) =∑

a1n1=a2n2

1

for ai ≤ A and ni ∈ [M + 1,M + N ]. Hence the bound follows. The key to theBurgess bound is evaluating W , and for that we need the following important

Lemma 3.2.3.

W ≤ (2rB)rp+ 2rB2rp1/2. (3.20)

Proof. By the Riemann Hypothesis for curves over finite fields, we have

|∑x(p)

χ((x+ b1) . . . (x+ br))χ((x+ br+1) . . . (x+ b2r)) |≤ 2rp1/2 (3.21)

if one of the classes bν(mod p), is different from any other. We have

W =∑

. . .∑

1≤b1,...,b2r≤B

∑x(p)

χ((x+ b1) . . . (x+ br))χ((x+ br+1) . . . (x+ b2r)).

(3.22)

26

The complete sum above satisfies (3.22) except possibly for the (b1, . . . , b2r) whichcan be arranged in r equal pairs. Hence the number of exceptions is bounded by

r

(2r

r

)Br ≤ (2rB)r. And if there is no perfect matching among the brs, then us-

ing the Weil bound we get 2rB2rp1/2.

Choosing A = [N

9rp1/2r] and B = [rp1/2r] we get W ≤ (2r)2rp3/2 and so

V ≤ 2r(AN)1−1/2r(4p3/4pε)1/r ≤ N2−1/r(pr+14r +ε)1/r. (3.23)

on choosing ε′ = ε/2 and since H ≥ N/10, we get

|S(M,N)| � N1−1/rpr+1

4r2+ε (3.24)

3.3 The δ-SymbolDefine

δ(n) =

{1 if n = 0;0 if n 6= 0.

(3.25)

Let ω(t) be a smooth compactly supported function on R such that

∞∑n=1

ω(n) = 1.

Set

δk(n) = ω(k)− ω(n

k). (3.26)

We then have

δ(n) =∑k|n

δk(n) (3.27)

Define

∆c(n) =

∞∑r=1

δcr(n)

r(3.28)

27

Using additive characters,

δ(n) =∑k≥1

δk(n)

k

∑h(k)

e(hn

k)

=∑k≥1

δk(n)

k

∑ar(cr)

e(hn

k)

=∑c,r≥1

δcr(n)

cr

∑∗

a(c)

e(an

c)

=∑c≥1

c−1∑∗

a(c)

e(an

c)∆c(n)

(3.29)

where r = (h, k) in the third step. In practice, we apply the δ-symbol to integers |n| <N

2, with test function ω(t) supported on

K

2< |t| < K and whose derivatives satisfy

ω(j)(t) � K−j−1. Then δk(n) vanishes save for 1 ≤ k < max (K,N

K) = K by

choosing K = N1/2. Hence ∆c(n) vanishes outside 1 ≤ c < K and ∆c(n)� K−1.We now get slightly more precise estimates for ∆c(n) and examine its properties.

Lemma 3.3.1. Let f ∈ C∞0 (R). Then∫ ∞−∞

f(u)∆q(u) du = f(0)

∫ ∞0

ω(r) dr

− qj∫ ∞

0

ψj

(r

q

)∫ ∞−∞

(f(u)

(ω(r)

r

)(j)

− w(u)ujf (j)(ru)

)du dr

(3.30)

where j ≥ 1 and

ψj(z) = −∞∑m=1

(2πim)−j(e(mz) + (−1)je(−mz)).

Proof. We split into two parts as follows∫ ∞−∞

f(u)∆q(u) du =

∫ ∞−∞

f(u)

( ∞∑r=1

(qr)−1ω(qr)

)du,−

∫ ∞−∞

f(u)

( ∞∑r=1

(qr)−1ω

(u

qr

))du.

Call the integrals I1 and I2. In I2, set u = qrt. So that,∫ ∞−∞

f(u)∆q(u) du = I1 −∫ ∞−∞

ω(u)

( ∞∑r=1

f(qru)

)du.

Euler-Maclaurin gives∞∑r=1

(qr)−1ω(qr) =

∫ ∞0

(qr)−1ω(qr) dr +

∫ ∞0

ψj(r)∂j

∂rj

(ω(qr)

qr

)dr.

28

Since ω is even,

I2 = f(0)

∫ ∞0

ω(u) du−∫ ∞

0

ω(u)

( ∞∑r=−∞

f(qru)

)du.

Again by Euler-Maclaurin,

∞∑r=1

f(qru) =

∫ ∞−∞

f(qru) du+

∫ ∞−∞

ψj(r)∂j

∂rjf(qru) dr.

∴ I2 = f(0)

∫ ∞0

ω(u) du−∫ ∞

0

ω(u)

∫ ∞−∞

f(qru) dr du−∫ ∞

0

ω(u)

∫ ∞−∞

ψj(r)∂j

∂rjf(qru) du dr.

= f(0)

∫ ∞0

ω(u) du−∫ ∞−∞

f(u)

∫ ∞0

(qu)−1ω(tu) du dt− qj(∫ ∞

0

ψj(t/q)

∫ ∞−∞

f (j)(tu)ω(u)uj du dt

).

on setting qru = t in the first integral, and qr = t in the second integral. We nowevaluate I1.

I1 =

∫ ∞−∞

f(u)

∫ ∞0

(qr)−1ω(qr) dr du+

∫ ∞−∞

f(u)

∫ ∞0

ψj(r)∂j

∂rj

(ω(qr)

qr

)dr du.

=

∫ ∞−∞

f(u)

∫ ∞0

ω(tu)

qudt du+ qj

(∫ ∞0

ψj(t/q)

∫ ∞−∞

f(u)

(ω(t)

t

)(j)

)

)dt du.

on setting qru = t in the first integral and qr = t in the second. Adding I1 and I2, weget the lemma.

Recall that ω(u) is supported in Q ≤ |u| ≤ 2Q and its derivatives have nice decay.Since |ψj(z)| ≤ 1 the terms on the right side of (3.6) are bounded by

f(0)(1 +O(Q−j−1)),

qjQ−j−1|∫f(u) du|,

qjQj − 1

∫|f (j)(u)| du,

respectively. We also have

f(0)�∫

(|f(u)|+ |f (j)(u)|) du.

Adding the above four, we get

Corollary 3.3.2. Let j ≥ 1. We have∫ ∞−∞

f(u)∆q(u) du = f(0) +O

(Q−1qj

∫(Q−j |f(u)|+Qj |f (j)(u)|) du

).

(3.31)

29

If q < Q1−ε this shows that ∆q(u) approximates to the Dirac distribution very wellon test functions such that f (j) � (qQ1+ε)−j .

Lemma 3.3.3. We have

∆q(u)� (q +Q)−1 + (qQ+ |u|)−1. (3.32)

Proof. By partial summation we get

∆q(u) =

∫ ∞1

{rq} d

ω(r)− ω(u/r)

r,

Hence, we get

|∆q(u)| ≤∫ ∞

1

min

(1,r

q

)| dω(r)− ω(u/r)

r|

≤∫ ∞

1

min

(1,

2Q

q

)| dω(r)

r| +∫ ∞

1

min

(1

|u|,

1

qQ

)| drω(r) |

� min

(1

q,

1

Q

)+min

(1

|u|,

1

qQ

).

Where we have made a change of variables u/r = t to get the second integral in thesecond step; the last step follows from estimates for the decay of ω. Finally, the prooffollows from the identity min(1/x, 1/y) ≤ (x+ y)−1.

Proceeding in a similar manner to Lemma 3.3.3, for j ≥ 1, we also get

∆(j)q (x)�j

1

(qQ+ |x|)(qQ)j+1(3.33)

for the j-th derivative.

3.4 Poisson Type FormulaeWe will need generalisations of the Poisson summation formula for GL(2) automor-phic forms which are crucial to the study of subconvexity(see Section 3.6 and Chapter5). We record some of these formulas below. Theorems 3.4.1 and 3.4.3 are the so-called Voronoi summation formulas for GL(2).

Theorem 3.4.1. Let f =∑

ann(k−1)/2e(nz) ∈ Mk(SL2(Z)). Let a, c be positive

integers such that (a, c) = 1 and F a smooth function supported on R+. Then∑m

ame(amc

)F (m) =

∑r

are(− arc

)F (r), (3.34)

and F is the Hankel-type transform

F (y) = 2πikc−1

∫ ∞0

F (x)Jk−1

(4π√xy

c

)dx. (3.35)

30

Proof. Let bn = ann(k−1)/2. Let z = −d

c+

i

cζfor some complex number ζ with

Re(ζ) > 0. Then we have by the modularity of f at z,∑n

bne(anc

)exp

(−2πζ

n

c

)=

(i

ζ

)k∑n

bne

(−dnc

)exp

(−2π

n

ζc

)(3.36)

Let p(x) =F (x)

x(k−1)/2. Consider the Mellin transform,

p(x) =1

2πi

∫(σ)

exp(−xζ)q(ζ) dζ

q(ζ) =

∫ ∞0

exp(xζ)p(x) dx

Then ∑n

bne(anc

)p(

2πn

c) = ik

∑n

bne

(−dnc

)r(

2πn

c)

where

r(y) =1

2πi

∫(σ)

exp(−xζ)ζ−kq(ζ) dζ

=

∫ ∞0

p(x)

2πi

∫(σ)

exp(xζ − yζ−1)ζ−k dζ

=

∫ ∞0

p(x)x(k−1)/2

y(k−1)/2Jk−1(2

√xy) dx

Set d = a. Then switching back to F (x) = p(x)x(k−1)/2 to go from bn to an, we get∑m

ame(amc

)F (m) =

∑r

are(− arc

)F (r)

Renormalising the Hankel-type transform, we get our result.

Corollary 3.4.2. For a primitive character χ to modulus q we have∑m

amχ(m)F (m) = ε2χ∑r

arχ(r)F (r) (3.37)

where εχ = τ(χ)q−1/2.

Proof. This follows from the identities

χ(m) =1

τ(χ)

∑a(modq)

χ(a)e

(am

q

),

χ(−r) =1

τ(χ)

∑a(modq)

χ(a)e

(− arq

).

31

Theorem 3.4.3. Let g(x) be a smooth, compactly supported function on R+ and let(c, d) = 1. Then we have

c

∞∑m=1

τ(m)e

(dm

c

)g(m) =

∫(log x+ 2γ − 2 log c)g(x) dx

− 2π

∞∑m=1

τ(m)e

(−dmc

)∫ ∞0

Y0

(4π√mx

c

)g(x) dx

+ 4

∞∑m=1

τ(m)e

(dm

c

)∫ ∞0

K0

(4π√mx

c

)g(x) dx

(3.38)

Theorem 3.4.3 depends on the crucial fact that the series

L(s, h/k) =

∞∑n=1

τ(n) cos(2πnh/k)n−s

satisfies a functional equation relating its value at s to its value at 1−s. Notice that τ(n)

are the coefficients of the Maass form g =∂

∂sE(z, s) |s=1/2, the level one Eisenstein

series(see Section 5.2). We may analogously define L(s, h/k) for any Maass cusp formand it is known that it also satisfies a functional equation relating its value at s to itsvalue at 1− s. Hence Theorem 3.4.3 may be extended to any Maass cusp form.

Theorem 3.4.4. Let L(s, f) be entire. Let F be a smooth, compactly supported func-tion. Then, ∑

n≥1

λf (n)F (n) =ε(f)√q

∑m≥1

λf (n)H(n) (3.39)

where

H(y) =1

2πi

∫(3)

F (1− s) γ(f, s)

γ(f, 1− s)y−s ds (3.40)

Proof. The RHS is nothing but,

ε(f)√q

1

2πi

∫(3)

F (1− s) γ(f, s)

γ(f, 1− s)L(f , s) ds

where we have interchanged the sum and the integral by appealing to absolute conver-gence. By the functional equation this is∫

(3)

F (1− s)L(f, 1− s) ds =

∫(3)

F (s)L(s, f) ds

by shifting the line of integration to Re(s) = −2, which we can do because L(s, f) isentire and F has compact support. We are once again in a domain of absolute conver-gence, and interchanging the order of integration and summation once more, we get by

32

the Mellin inversion formula,∫(3)

F (s)L(s, f) ds =∑n≥1

λf (n)F (n).

This proves our theorem.

Corollary 3.4.5. Let f be a newform. Set

ψf (1)L(s, f)2 =

∞∑n=1

ρf (n)

ns.

Let g be a nice smooth, compactly supported function. Then we have

∞∑n=1

ρf (n)√ng(n) =

∞∑m=1

ρf (m)√m

h(16π4q−2m) (3.41)

where

h(y) =1

2πi

∫(0)

g(s)Γ2(k2 − s)Γ2(k2 + s)

ys ds

and g is the Mellin transform of g(x),

g(s) =

∫ ∞0

g(x)xsdx

x.

Proof. This follows from the functional equation

L2(1

2− s, f) =

( q

4π2

)2s Γ2(k2 + s)

Γ2(k2 − s)L2(

1

2+ s, f).

The factor 1/2 that appears in L2(1

2± s, f) contributes the n−1/2 term and renormal-

ising, we get the result.

3.5 Method of Families and AmplificationIn this section we explain the idea of amplification, due to Iwaniec, that can provesubconvexity for all GL(2) L-functions. Let L(s, f0) be a fixed L-function for whichwe would like to prove subconvexity. This section is essentially Section 4.2 of the notes[10] by Philippe Michel.

By the approximate functional equation, we are lead to studying sums of the form

L(f0, X) =∑n

anλf0(n) (3.42)

33

where an =1

n1/2V( nX

), for a rapidly decaying function V (x). By the Rankin-

Selberg method and Cauchy-Schwarz, we have

L(f0, X)� Q(f0)εX1/2

∑n≤X

|an|21/2

. (3.43)

This is the bound that we want to beat. To do this, we first choose a ’nice’ family Fof automorphic forms that contains f0 and average the moments of L(f,X) over thisfamily. To do this, form the sum

L′(f,X) = Lk(f,X) =∑n≤Xk

bnλf (n)

where bn =τf,k(n)

n1/2, by the multiplicativity of λf (n) and the the function τf,k(n)

behaves much like the usual k-divisor function and we have the bound τf,k(n)�ε nε.

We now average the square of L′(f,X) over F , to get

1

|F|∑f∈F

|∑n≤Xk

bnλf (n) |2=∑n,m

bnbm1

|F|∑f∈F

λf (n)λf (m). (3.44)

The moment method works when we can prove quasi-orthogonality in the last sumabove in the range n ≤ Xk, in which case we would get,

1

|F|∑f∈F

| L′(f,X) |2 �∑n,m

bnbmδm=n

� Q(f0)ε

∑n≤Xk

|bn|2 .

(3.45)

Hence, by positivity, we get

1

|F|L2k(f0, X)� Q(f0)ε

∑n≤Xk

|bn|2� (Q(f0)X)ε

∑n≤X

|an|2k

. (3.46)

and so,

L(f0, X)� (Q(f0)X)ε|F|1/2k∑n≤X

|an|21/2

. (3.47)

Hence to get an improvement on equation (3.43), we should be able to prove quasi-

orthogonality in the range n ≤ Xk for k strictly bigger than k0 =log |F|logX

. And doing

this is the non-trivial step in the moments method. There are two clear ways to choosea nice k,

34

1. Keep F as it is, and try and choose k bigger than k0. The difficulty one encoun-ters then is that obtaining quasi-orthogonality becomes difficult when one takeshigher moments, even if one can get non-trivial results for lower moments.

2. Make k0 smaller by choosing a smaller family, but this method also doesn’t workvery well in general.

When both these methods fail, one can use the amplification method, which we nowexplain.

Keep k = k0. Let L be an integer. Choose an amplifier,

A(f, L) =∑l≤L

cnλf (n). (3.48)

We now average over F , the modified linear form,

1

|F|∑f∈F

| A(f, L) |2|∑n≤Xk

bnλf (n) |2 . (3.49)

We hope to be able to prove quasi-orthogonality for this modified linear form, gettinga bound of the form

� (Q(f0)N)ε(∑n≤X

|an|2)k∑l≤L

|cl|2

The key step is to be able to choose an amplifier such that∑l≤L

|cl|2 � Q(f0)εLα+ε, and |A(f0, L)| � Q(f0)−εLα−ε (3.50)

for some α > 0. Then by positivity we get

L(f0, X)� (Q(f0)X)εX1/2L−α(∑n≤X

|an|2)1/2. (3.51)

Choosing L to be a small power of X , we improve on the trivial bound. The role ofthe amplifier, therefore, is to amplify the contribution coming from f while averagingover F .

If the family F consists of Dirichlet characters, then choosing cl = χ0(l), fora fixed character χ0 ∈ F , we get an amplifier with α = 1. When averaging overmodular forms, the obvious choice of cl = λf0(l) does not work because we do nothave a bound of the form ∑

l≤L

|cl|2 � Q(f0)εL1−ε.

To get around this problem, Iwaniec observed that since λf (p)2 − λf (p2) = 1, bothλf (p) and λf (p2) can’t be small simultaneously. Hence choosing

cl =

λf (l) if l ≤ L1/2 is a prime;

−1 if l is square of a prime that is less than L1/2;0 otherwise.

(3.52)

we get an amplifier with α = 1/2.

35

3.6 Shifted Convolution ProblemConsider g a primitive form, h 6= 0, l1, l2 coprime integers and F (x, y) a smoothfunction supported in [X, 2X]× [Y, 2Y ]. Suppose that F satisfies

F (i,j) �i,j X−iY −j . (3.53)

Set

S(g, l1, l2, h) =∑

l1m−l2n=h

λg(m)λg(n)F (l1m, l2n). (3.54)

By the Rankin-Selberg method, we obtain the trivial bound

S(g, l1, l2, h)�ε (Q(g)XY )εMax(X/l1, Y/l2)

Many problems in subconvexity boil down to getting a non-trivial bound for the aboveshifted convolution sum. The main idea to solve the problem is to ”smooth out” thecondition l1m − l2n = h in order to separate the variables m and n. There are twoways to do this, one may use the spectral theory of automorphic forms or one may usethe δ-symbol. In this thesis, we will explain how to use the latter.

Applying the δ-symbol to get rid of the condition l1m− l2n = h, we get

S(g, l1, l2, h) =∑

1≤q≤Q

∑a=1(r)

e

(−arh

)∑m,n

λg(m)λg(n)e

(a(l1m− l2n)

q

)Eq(m,n, h)

whereEq(m,n, h) = F (l1x, l2y)∆q(l1x− l2y − h)

Now, the functionEq has very nice decay and applying the Voronoi summation formulato above, we get

S(g, l1, l2, h) =∑q≤Q

(l1l2, q)

q2

∑m,n

λg(m)λg(n)S(−l′1m+ l′2n,−h; q)Iq(m,n, h)

where Iq is the Hanke-type transform obtained from applying Voronoi to the m and nsummations, We can get bounds for Iq using bounds for Eq and the Bessel functions,and applying the Weil bound for Kloosterman sums, we will get an improvement onthe trivial bound. For more details see Section 4.4 of [10] from which this section wasborrowed.

3.7 Quadratic Divisor ProblemIn this section we consider an instance of the shifted convolution problem related toproving GL(2) subconvexity, the problem of estimating the sum

Df (a, b;h) =∑

am±bn=h

τ(m)τ(n)f(am, bn), (3.55)

36

where f is a nice smooth function on R+ × R+. We also assume that f satisfies,

xiyjf (i,j)(x, y)�(

1 +x

X

)−1 (1 +

y

Y

)−1

P i+j , (3.56)

with some P,X, Y ≥ 1 for all i, j ≥ 0, the implied constant depending only on i, j.The main term is expressed in terms of the series

Λabh(x, y) =1

ab

∞∑q=1

q−2(ab, q)cq(h)(log x− λaq)(log y − λbq), (3.57)

where

λaq = 2γ + logaq2

(a, q)2. (3.58)

Theorem 3.7.1. Suppose a, b ≥ 1, (a, b) = 1, h 6= 0 and f satisfies equation (3.56).Then we have

Df (a, b;h) =

∫ ∞0

g(x,±x∓ h) +O(P 5/4 (X + Y )

1/4(XY )

1/4+ε), (3.59)

where g(x, y) = f(x, y)Λabh(x, y) and the implied constant depends on ε only.

Proof. The proof proceeds in three steps. First we apply the δ-symbol to identify therelation am−bn = h. Then we apply Voronoi summation formula to both them and nsummations to get complete Kloosterman sums. We finally estimate the resulting sumusing the Weil bound for Kloosterman sums.

3.7.1 Applying the δ-symbolWe only deal with the case am − bn = h, the other case being similar. We alsoassume that f(x, y) is supported in the box [X, 2X]× [Y, 2Y ]. To f(x, y) we attach aredundant factor ϕ(x− y − h) where ϕ(u) is a smooth function supported on |u| < Usuch that ϕ(0) = 1 and ϕ(i) � U−i(so that ϕ(am − bn − h) = 1). Let F (x, y) =f(x, y)ϕ(x− y − h). Then

F (i,j)(x, y) =∑l≤i

∑m≤j

(i

l

)(j

m

)f (l,m)ϕ(i+j−l−m)

�∑l≤i

∑m≤j

(i

l

)(j

m

)(1 +

x

X

)−1 (1 +

y

Y

)−1

P l+mX−lY −mU l+m−i−j

�(

1

U+P

X

)i(1

U+P

Y

)j

Choosing U ≤ P−1min (X,Y ), we get

F (i,j) � U−i−j . (3.60)

37

For the test function ω(u) choose U = Q1/2, so ∆q(u) vanishes if q ≥ 2Q. Thereforewe get

Df (a, b;h) = DF (a, b;h)

=∑

1≤q<2Q

∑∗

d(q)

e

(−dhq

)∑m

∑n

τ(m)τ(n)e

(dam− dbn

q

)E(m,n)

(3.61)

where E(x, y) = F (ax, by)∆q(ax− by − h).

3.7.2 Applying Voronoi Summation FormulaApplying the formula to each variable, we get

q2

(ab, q)

∑m

∑n

= I +∑m

τ(m)e

(m−adq

)Ia(m) +

∑n

τ(n)e

(nbd

q

)Ib(n)

+∑m

∑n

τ(m)τ(n)

(−m−ad

q+ n

bd

q

)Iab(m,n) + ?

(3.62)

where a/q and b/q have to be but in reduced forms because we need (a, q) = 1 toapply Voronoi, which is why we get the extra factor (ab, q) in the denominator of LHSabove. Where

I =

∫ ∫(log ax− λaq)(log by − λbq)E(x, y) dx dy

Ia(m) = −2π

∫ ∫Y0

(4π(a, q)

√mx

q

)(log by − λbq)E(x, y) dx dy

Ib(n) = −2π

∫ ∫(log ax− λaq)Y0

(4π(a, q)

√ny

q

)E(x, y) dx dy

Ia,b(m,n) = 4π2

∫ ∫Y0

(4π(a, q)

√mx

q

)Y0

(4π(b, q)

√ny

q

)E(x, y) dx dy,

? denotes similar summation terms involving integrals with the K0 Bessel function.Inserting into equation (3.61) we get full Kloosterman sums:

Df (a, b;h) =∑q<2Q

q−2(ab, q){S(h, 0; q)I +∑m

τ(m)S(h, am; q)Ia(m)

+∑n

τ(n)S(h,−bn; q)Ib(n)

+∑m

∑n

τ(m)τ(n)S(h, am− bn; q)Iab(m,n) + ?

(3.63)

38

3.7.3 The main termSince the τ(n) may be viewed as coefficients of an Eisenstein series, we get a mainterm. We have

abI =

∫ ∫C(x, y)∆q(x− y − h) dx dy

=

∫ ∫C(x, x− h+ u)∆q(u) du, dx

where C(x, y) = (log x− λaq)(log y − λbq)F (x, y). By the lemma,∫C(x, x− h+ u)∆q(u) = C(x, x− h) +O

((q

Q

)j).

Assuming q < Q1−ε we make the error term very small by taking j very big(j ofcourse depends on X). Hence we obtain

abI =

∫C(x, x− h) dx+O

(Q−A

).

Moreover,∫ ∫|F (ax, by)∆q(ax− by − h)| dx dy = (ab)−1

∫ ∫|F (x, x− h− u)∆q(u)| dx du

� (ab)−1min (X,Y )

∫ U

−U|∆q(u)| du

� (ab)−1(X + Y )−1XY logQ

(3.64)

Hence for all q we have

I � (ab)−1(X + Y )−1(XY )1+ε logQ. (3.65)

Therefore, the first part yields,∑q<2Q

q−2(ab, q)S(h, 0; q)I = (ab)−1∑q<2Q

q−2(ab, q)cq(h)

∫C(x, x− h) dx

= (ab)−1∞∑q=1

q−2(ab, q)cq(h)

∫C(x, x− h) dx

+O

(ab)−1∑q>2Q

q−2(ab, q)cq(h)(XY )1+ε(X + Y )−1 logQ

= (ab)−1

∞∑q=1

q−2(ab, q)cq(h)

∫C(x, x− h) dx

+O((ab)−1(XY )1+ε(X + Y )−1Q−1+ε

)(3.66)

39

3.7.4 The error termIt is a straightforward check using the chain rule that

E(i,j) � 1

qQ

(ab

qQ

)i+j. (3.67)

Ia(m) = −2π

∫ ∫Y0

(4π(a, q)

√mx

q

)(log by − λbq)E(x, y) dx dy

Set4π(a, q)

√mx

q= t. Then

Ia(m) = 2π

(q2

8π2(a, q)2m

)∫ ∫tY0(t)(log by − λbq)E(

q2t2

16π2(a, q)m, y) dt dy

Using the fact that tY0(t) = (tY1(t))′

we get,

Ia(m) = 2π

(q2

8π2(a, q)2m

)2 ∫ ∫t2Y1(t)(log by−λbq)E(1,0)(

q2t2

16π2(a, q)m, y) dt dy

Repeating this process j − 1 times we get,

Ia(m) = 2π

(q2

8π2(a, q)2m

)j ∫ ∫tjYj−1(t)(log by − λbq)E(j,0)(

q2t2

16π2(a, q)m, y) dt dy

�(Xa

mQ2

)j(3.68)

Hence we may choose j so that Ia(m)� m−2 if m�(aX

Q2−ε

).

For m�(aX

Q2−ε

), we estimate Ia(m) trivially using the bound Y0(z)� z−1/2:

Ia(m)�(aq2

mX

)1/4

(logQ)

∫ ∫And in a similar way we estimate Ib(n) and Iab(m,n) also,

Ib(n)�(bq2

nY

)1/4

(logQ)

∫ ∫Iab(m,n)�

(abq4

mnXY

)1/4

(logQ)

∫ ∫

40

where∫ ∫

is the integral in equation (3.64). By summing over m,n in the range

m�(aX

Q2−ε

)and n�

(bY

Q2−ε

)we obtain

∑m

τ(m)|Ia(m)| �∑

m�(

aX

Q2−ε

)�q1/2

b

X3/2Y

X + YQ−3/2+ε,

∑n

τ(n)|Ib(n)| �∑

n�(

bY

Q2−ε

)�q1/2

a

XY 3/2

X + YQ−3/2+ε,

∑m

τ(m)τ(n)|Iab(m,n)| �∑

m�(

aX

Q2−ε

)∑

n�(

bY

Q2−ε

)� q(XY )3/2

X + YQ−3+ε.

Finally, summing over q < 2Q, we get the error term

(ab)−1 (XY )1+ε

X + YQ−1+ε +

(XY )3/2

X + YQ−5/2+ε.

And this becomes the claimed error term on setting U = Q2 = P−1(X + Y )−1XY .

Remark: It is remarked in [3] that the exponent 5/4 may be replaced by 3/4 inequation (3.61) by refining the argument and it is expected that the exponent 1/2 musthold.

We end this section by stating the

Corollary 3.7.2. For a, h,M ≥ 1 we have∑n≤x

τ(m)τ(am+ h) =

∫ x

0

λ(y, ay + h) dy +O(a1/9(ax+ h)2/9x2/3+ε) (3.69)

where

λ(x, y) =

∞∑q=1

q−2(a, q)cq(h)

(log x− 2γ − 2 log

q

(a, q)

)(log y − 2γ − 2 log q).

(3.70)Notice that the result is non-trivial only if h < a−1/2x3/2−ε.

Applying Corollary 3.7.2 with a = h = 1, we get∑n≤x

τ(n)τ(n+ 1) = xP2(log x) +O(x8/9+ε). (3.71)

This is an improvement on Estermann’s result whose error term has exponent 11/12.Here P2(x) is a quadratic polynomial. The quadratic divisor problem has many ap-plications. For instance, Heath-Brown has used it to study the fourth moment of theRiemann zeta function in his paper, The Fourth Power Moment of the Riemann ZetaFunction, Proc. London Math. Soc. s3-38(3) (1979).

41

Chapter 4

Subconvexity for GL(1)L-functions

Let L(s, f) be an automorphic L-function for GL(1). Then we know that such an L-function is either the Riemann zeta function or a Dirichlet L-function with character χ.Recall that the overall goal is to obtain a bound of the form L(s, f) � Q(s, f)1/4−δ .In our case,Q(s, f) � q|s+3|, where χ is a character to the modulus q. Hence we mayprove subconvexity either in the s-aspect or the q aspect. We then have the followingtheorem

Theorem 4.0.3. Let δ > 0 be a constant. Let χ be a primitive Dirichlet character tothe modulus q. Then

1. s aspect: as |s| → ∞, we have L(s, χ)� |s+ 3|1/4−δ

2. q aspect: as |q| → ∞, we have L(s, χ)� |q|1/4−δ

4.1 Subconvexity for ζ(s)For the ζ(s), q = 1, hence we only have subconvexity in the s-aspect. The first proofwas given by Weyl(who proved the exponent 1/6) whose proof we give below. Therehave been several advances in improving the exponent 1/6 for ζ(s), and the currentrecord is held by M N Huxley, who proved the exponent 32/205.

4.1.1 Proof using van der Corput’s methodBy the approximate functional equation, to estimate |ζ(1/2 + it)| it suffices to studythe sum

S =∑n≤t

1

n1/2+it. (4.1)

42

To do this we will first study the sum ∑n≤t1/2

n−it (4.2)

and then use partial summation to get a bound for S. Notice that x−it = e(f(x)) where

f(x) =−t log x

2π. We then have f ′′′(x) = − t

πx3. Applying Proposition 3.1.5 with

A = 1, λ3 = − t

a3, we get∑

a<n≤b

n−it � a(t/a3)1/6 + a1/2(t/a3)−1/6 � a1/2t1/6 + at−1/6 (4.3)

under the assumption that b ≤ 2a. By partial summation for a ≤ t2/3, we then get∑a<n≤b

1

n1/2+it� t1/6 (4.4)

By the proof of Proposition 3.1.5, we get∑a<n<b

n−it � t1/2 + at−1/2. (4.5)

And hence by partial summation,∑a<n<b

1

n1/2+it� (t/a)1/2 + (a/t)1/2 (4.6)

for t2/3 < a < t. Hence we get ∑a<n<b

1

n1/2+it� t1/6 (4.7)

for 1 < a < t. Splitting [1, t] into dyadic regions, and applying the above result log ttimes, we get

ζ(1/2 + it)� t1/6 log t (4.8)

4.2 Subconvexity for L(s, χ)By the approximate functional equation we only need to estimate the sum∑

n

χ(n)

nsVs

(n√q

). From equation (1.33), we infer that

d

dx

(x−sVs(x/

√q))� |s|

x3/2

(1 +

x

q|s|

)−1

. (4.9)

43

Hence applying Theorem 3.2.1 with r = 2, we get

S(x) =∑n≤x

χ(n)� min(q, x1/2q3/16+ε). (4.10)

By the partial summation formula, we get

Theorem 4.2.1. Let χ be a primitive character to the modulus q and let ε > 0. Then

L(s, χ)� |s|q3/16+ε (4.11)

Note that we have only proved subconvexity in the q-aspect. The proof in the s-aspect is similar to Weyl’s proof. The current record, in the q-aspect, for real charactersis held by Friedlander-Iwaniec who proved that L(s, χ)� q1/6+ε, which matches theexponent given by Weyl for the zeta function. But this result is quite hard and requiresthe spectral theory of automorphic forms for GL(2).

44

Chapter 5

Subconvexity for GL(2)L-functions

Let L(s, f) be a GL(2) automorphic L-function. The analytic conductor, Q(s, f) �qk2|s+3|2, where q is the level, k the weight for holomorphic forms. For Maass forms,Q(s, f) � qλ|s+ 3|2, where λ is the eigenvalue of the form. We can study the size ofL(s, f) inside the critical strip in terms of the level aspect qf , the s-aspect and also thespectral aspect(which is essentially (kf −1)/2 if f is holomorphic). We may prove the

Theorem 5.0.2. Let f and g any primitive modular forms. For any s such thatRe(s) =1/2, we have

1. s-aspect: as |s| → ∞,

L(s, f)�Q(f),Q(g) (|s+ 3|2)1/4−δ, L(f × g, s)�Q(f)Q(g) (|s+ 3|4)1/4−δ

2. q-aspect: as qf →∞,

L(s, f)� q1/4−δ, L(f × g, s)� (q2f )1/4−δ

for some δ > 0.

A similar result may be proved in the spectral aspect also, but we omit stating ithere. In this thesis, we prove certain cases of the above theorem, namely, the caseGL(2) × GL(1) and subconvexity for newforms of level q, weight k ≥ 2 and trivialnebentypus. We will also prove the theorem for certain GL(2) × GL(2) automorphicL-functions also. All the results will be in the level aspect.

5.1 Subconvexity for GL(2)×GL(1) L-functionsIn this section we prove theorem due to Duke, Friedlander and Iwaniec[2]

45

Theorem 5.1.1. Let f be a cusp form of weight k for the full modular group and χ bea primitive character modulo q. Then

L(s, f × χ)� |s|2q5/11+ετ(q)2 (5.1)

with the implied constant depending only on f .

Here f × χ is the twisted cusp form

(f × χ)(z) =∑n≥1

afχ(n)e(nz)

It is easy to show that f × χ ∈ Sk(q2, χ2). Hence the conductor of L(s, f × χ) is q2.

5.1.1 Reduction to the shifted convolution problem

Let af (n) = bf (n)nk−12 . By the approximate functional equation, we are interested in

sums of the typeSχ =

∑n

bf (n)χ(n)g(n)

where g(n) is a rapidly decreasing function supported on [X, 2X] for X � q. Weintend to prove this theorem using the amplification method, so we must estimate sumsof type

∑∗

χ(q)

| A(χ,L) |2| Sχ |2 =∑∗

χ(q)

∑l1,l2,m1,m2

cl1χ(l1)bf (m1)g(m1)cl2χ(l2)bf (m2)g(m2)

=∑∗

χ(q)

|∑n

anχ(n) |2= S

where an is the convolution

an =∑lm=n

clbf (m)g(m),

and A(χ,L) is the amplifier

A(χ,L) =∑l≤L

clχ(l).

Note that the family we are averaging over is the set of primitive characters to themodulus q.

Using the fact that χ(n) = τ(χ)−1∑

k(modq)

χ(k)e

(nk

q

)and extending the sum to

46

all characters we get

S =∑∗

χ(q)

1

| τ(χ) |2|∑n

an∑k(q)

χ(k)e

(nk

q

)|2

=∑∗

χ(q)

1

q|∑n

an∑k(q)

χ(k)e

(nk

q

)|2

≤ 1

q

∑n,m

anam∑k,l(q)

e

(nk −ml

q

)∑χ(q)

χ(k)χ(l)

=ϕ(q)

q

∑n,m

anam∑k(q)

e

((n−m)k

q

)

=ϕ(q)

q

∑k(q)

|∑

ane

(nk

q

)|2

≤ ϕ(q)

q

∑k(q)

|∑

ane

(nk

q

)|2

≤ ϕ(q)∑

h≡0(q)

Sh

whereSh =

∑n−m=h

anam =∑l1,l2

cl1cl2S(f, l1, l2, h)

and S(f, l1, l2, h) is the shifted convolution sum

S(f, l1, l2, h) =∑

l1m−l2n=h

bf (m)bf (n)g(m)g(n)

Having reduced to the shifted convolution problem, our first task is to evaluate thediagonal term.

5.1.2 Evaluating the term S0

|an|2 =∑

l1m1=l2m2=n

cl1g(m1)bf (m1)cl2g(m2)bf (m2) (5.2)

By Cauchy-Schwarz,

|an|2 ≤ (∑lm=n

1)(∑lm=n

|cl|2|g(m)bf (m)|2). (5.3)

Since τ(n) ≤ τ(l)τ(m), we get

|an|2 � (∑lm=n

|cl|2τ(l)|g(m)bf (m)|2τ(m)) (5.4)

47

Hence separating the sums,∑n

|an|2 � (∑l≤L

|cl|2τ(l))(∑

M≤m≤2M

|bf (m)|2τ(m)) = (M logM)(∑l≤L

|cl|2τ(l))

(5.5)

5.1.3 Solving the shifted convolution problemDenote the remaining h 6= 0 sum by S∗. Given h 6= 0 we apply the δ-symbol to split

Sh =∑c

c−1Shc,

where

Shc =∑∗

a(c)

e

(−ahc

) ∑n1,n2

an1an2

e

(a(n1 − n2)

c

)∆c(n1 − n2 − h).

Sincean =

∑lm=n

clbf (m)g(m)

with cl complex numbers for 1 ≤ l ≤ L and g a smooth function supported in [M, 2M ]satisfying | g(j)(m) |≤M−j .

Shc =∑l1,l2

cl1cl2Tl1l2(c), (5.6)

where

Tl1l2(c) =∑∗

a(c)

e

(−ahc

) ∑m1,m2

bf (m1)bf (m2)e

(a(l1m1 − l2m2)

c

)F (m1,m2)

(5.7)

and F (m1,m2) = g(m1)g(m2)∆c(l1m1− l2m2− h) with K = N1/2 = 2(LM)1/2.We then see that

F (i,j) � K−1(cM/K)−i−j (5.8)

We now transform Tl1l2 using the Voronoi formula applied to both variables to get

Tl1l2(c) =∑∗

a(c)

e

(−ahc

)∑r1,r2

br1bf (r2)e

(al′

c′1r1 −

al′′

c′2r2

)F (r1, r2) (5.9)

where l′ = l1/(l1, c), l′′ = l2/(l2, c), c

′1 = c/(l1, c), c

′2 = c/(l2, c) so that we may

apply the Voronoi formula. And

F (r1, r2) =4π2

c′1c′2

∫ ∞0

∫ ∞0

F (x1, x2)Jk−1

(4π

c′1

√x1r1

)Jk−1

(4π

c′2

√x2r2

)dx1 dx2.

(5.10)

48

Integrating by parts twice, we see that F (r1, r2) < (r1r2q)−100, if either r1 > c′21 K

2qε/c2Mor r2 > c′22 K

2qε/c2M . In the remaining range we get, without integrating by parts,

F (r1, r2)� (c′1c′2)−1

(1 +

(Mr1)1/2

c′1

)−1/2(1 +

(Mr2)1/2

c′2

)−1/2

||F ||1 (5.11)

Using the bounds on ∆c(u)(see equation (3.33)) one gets,

||F ||1 ≤∫ 2M

M

∫ 2M

M

|∆c(l1x1 − l2x2 − h)| dx1 dx2 � cK−2M2 log(2K/c)

(5.12)

Hence, using the bound∑r≤x

| br |2� x we infer

∑r1,r2

br1br2 | F (r1, r2) |� Kqε. (5.13)

For the sum over a we apply the Weil bound to get

Tl1l2(c) =∑r1,r2

br1¯br2 | F (r1, r2) | S(h, ∗; c)

� (h, c)1/2c1/2τ(c)(LM)1/2qε(5.14)

HenceShc � (h, c)1/2c1/2τ(c)(LM)1/2qε

(∑| cl |

)2

.

Next summing over c < 2(LM)1/2 we obtain

Sh � τ(h)(LM)3/4(logLM)qε(∑

| cl |)2

.

Summing over h = 0(modq), 0 < |h| ≤ LM we obtain

S∗ � τ(q)(LM)7/4qε(logLM)2(∑

| cl |)2

(5.15)

S0 � M(logM)∑l

| cl |2 τ(l). (5.16)

Theorem 5.1.2. Let

S =∑∗

χ(modq)

|∑m

bf (m)χ(m)g(m) |2|∑l

clχ(l) |2 .

Then

S � ϕ(q)M(logM)∑l

| cl |2 τ(l) + τ(q)qε(LM)7/4(logLM)2(∑

| cl |)2

(5.17)

49

Corollary 5.1.3. Let χ be a primitive character mod q. We have

Bχ =∑m

bf (m)χ(m)g(m)� (q7/22M7/11 +M7/8)qετ(q)2 logM. (5.18)

Proof. Choose cl = χ(l). The contribution to S from χ is

|l ≤ L : (l, q) = 1|2|Bχ|2.

Notice that |l ≤ L : (l, q) = 1| � ϕ(q)

qL, (

∑l≤L

|cl|)2 � ϕ2(q)

q2L2 and

∑l≤L

|cl|2τ(l)�

L logL

q. Hence by positivity and choosing L = q4/11+εM−3/11 + 2qτ(q)ϕ(q)−1 the

corollary follows.

5.1.4 Proof of Theorem 5.1.1Proof. Let f have coefficients bf (m)m(k−1)/2, then it suffices to estimate sums of type

H =∑m

bf (m)χ(m)m−sh(m).

Applying equation (5.18) with g(m) = m−sh(m) we get

H � |s|2(q7/22+εM3/22 +M3/8)τ(q)2.

Hence the contribution from M � q is O(|s|2q5/11+ετ(q)2). If M � q, we applyCorollary 3.4.2 to replace g by g. By partial integration g satisfies

g(r)� |s|2M1/2q−1(1 + q−2Mr)−5/4.

Making a dyadic partition, it follows from the corollary above that

H � |s|2(q7/22R7/11 +R7/8)M1/2q−1(1 + q−2MR)−5/4τ(q)2qε

for some R ≥ 1. This is greatest when R = q2M−1 and the resulting bound is greatestwhen M = q. Hence the theorem is proved.

5.2 Subconvexity forL-functions Attached to NewformsIn this section we prove the following theorem due to Duke, Friedlander and Iwaniec[3,4]

Theorem 5.2.1. Let f be a newform for Γ0(q) with trivial nebentypus of weight k ≥ 2.Then

L(s, f)� q1/4−1/192+ε (5.19)

50

Like before, we prove subconvexity using the amplification method. The proofof this theorem is very similar to that of Theorem 5.1.1 because like the L-functionL(s, f × χ), we look at

L(s, f × g) =

∞∑n=1

λf (n)τ(n)

ns.

Since τ(n) are the coefficients of the Maass form

g =∂

∂sE(z, s) |s=1/2= y1/2 log y + 4y1/2

∞∑n=1

τ(n)K0(2πny) cos(2πnx)

we view the current situation as the newform f being twisted by the Maass form g.Hence an analogue of the theorem stated below can be proved for any modular formwhich is orthogonal to Sk(Γ0(q)), which includes the space of Maass cusp forms, andwe may therefore obtain subconvexity for such GL(2)×GL(2) L-functions also. Thisis possible because of the generality of the Voronoi formula.

We have already noted that the bound (2.9) is equivalent to the convexity bound.Hence to obtain a subconvex bound, we need to improve on (2.9) specialised to an =bn = τ(n). To this effect, we have the following

Theorem 5.2.2. For M,N � q1+ε

B(r, s)� min{q1+ε(MN)1/2, qε[q(r, s)τ(r)(rs)−1/2 + q3/4(rs)15/8](MN)1/2}(5.20)

Proof. The proof has the following key steps,

1. Evaluate B(s) = B(1, s).

2. To prove the bound forB(s), use the Petersson trace formula to get Kloostermansums.

3. Transform the Kloosterman sums using the Voronoi formula for τ(n) to getshifted convolution sums.

4. Use the Hecke multiplication rules to go from B(s) to B(r, s).

For the proof we need a truncated version of the Petersson formula(see [1])

Proposition 5.2.3. Let ∆η(m,n) denote a truncated version of ∆(m,n)

∆η(m,n) = 2πik(k − 1)q∑

c = 0 (mod) q

S(m,n; c)Jk−1(

√mn

c)η(c)

where η(c) is supported in [C, 2C]; that its derivatives satisfy η(i)(c) � C−i forC >

√MN . Then

∑Bη =

∑m≤M

∑n≤N

ambn∆η(m,n)� Cε

(√MN

C

)k−3/2

(q +M)1/2(q +N)1/2||a||||b||

(5.21)

51

5.2.1 Transforming B(s) into Kloosterman sumsBy the Petersson trace formula,

B(s) = q(k − 1)∑m,n

δm=snF (m,n) + 2πikq(k − 1)∑

c = 0 (mod q)

1

c

∑m,n

τ(m)τ(n)S(m, sn; c)Jk−1(4π√smn

c)

= (k − 1)qT (0) + 2πikq(k − 1)∑

c = 0 (mod q)

1

c2T (c)

(5.22)

where

T (0) =∑n

τ(sn)τ(n)F (sn, n) (5.23)

T (c) = c∑m

∑n

τ(m)τ(n)S(m, sn; c)Jk−1(4π√smn

c) (5.24)

5.2.2 Evaluating T(0)We have the trivial estimate using the fact that F(.,y) is supported on N ≤ y ≤ 2N .

T (0)� min(s−1M,N

)qε ≤

(MN

s

)1/2

qε (5.25)

and the bound τ(n) ≤ nε.

5.2.3 Transforming T(c) into Ramanujan sumsBy Theorem 3.4.3 applied to the m summation alone we get,

T (c) = T ∗(c) + T−(c) + T+(c) (5.26)

where

T ∗(c) =∑n

τ(n)S(0, sn; c)G∗(n) (5.27)

which we get from the integral in the Voronoi formula and summing over n, and

T±(c) =∑m

∑n

τ(m)τ(n)S(0, sn∓m; c)G∓(m,n). (5.28)

G∗(y) =

∫(log x+ 2γ − 2 log c)Jk−1

(4π√sxy

c

)F (x, y) dx (5.29)

G−(z, y) = −2π

∫Y0

(4π√xz

c

)Jk−1

(4π√sxy

c

)F (x, y) dx (5.30)

G+(z, y) = 4

∫Y0

(4π√xz

c

)Jk−1

(4π√sxy

c

)F (x, y) dx (5.31)

52

To reduce to the case of the divisor problem we write

T± =∑h

S(0, h; c)T±h (c), (5.32)

where

T±h (c) =∑

m∓sn=h

τ(m)τ(n)G∓(m,n). (5.33)

5.2.4 Estimating T ∗(c)We use the bound S(0, sn; c)� (c, sn) and G∗(n)�M logM log c getting

T ∗(c)� (c, s)MN logMcε (5.34)

5.2.5 Estimating T−h (c)As mentioned earlier, we use the solution to the quadratic divisor problem to sumT−h (c) when h 6= 0. When h = 0, we have to deal separately with the sum

T−0 (c) = −2π∑n

τ(sn)τ(n)

∫ ∞0

Y0

(4π

c

√snx

)Jk−1

(4π

c

√snx

)F (x, n) dx.

(5.35)

To apply Theorem 3.7.1 we must first check that

f(z, y) = G−(z,y

s

)= 2π

∫Y0

(4π√xz

c

)Jk−1

(4π√xy

c

)F (x,

y

s) dx

satisfies the condition (3.56). Set Y = sN and Z = c2P 2M−1 where P = 1 +c−1(sMN)1/2. We integrate by parts if z > Z several times in x and don’t integrateby parts otherwise. Then differentiate i times in x and j times in y and use the boundsfor the Bessel functions to get,

ziyjf (i,j)(z, y)�(

1 +z

Z

)−1 (1 +

y

Y

)−1√sMN

cMP i+j−2 log c (5.36)

Hence

(Y +Z)(Y Z)1/4+ε =

(sN +

c2P 2

M

)1/4

,

(sNc2P 2

M

)1/4+ε

� P 3/4 c√M

(sNP )1/4

sinceSN

P 2� c2

M. Hence by Theorem 3.7.1

53

T−h (c) =

∫ ∞0

g(h+ sy, y) dy +O

(√sMN

cMP−2 log c× P 3/4 c√

M(sNP )1/4

)

=

∫ ∞0

g(h+ sy, y) dy +O(

(sN)3/4P 1/4Mcε)

(5.37)

where g(x, y) = G−(z, y)Λh(z, y) and

Λh(z, y) =

∞∑w=1

(s, w)

sw2S(0, h;w)[log z − λw][log y − λsw] (5.38)

where λsw = 2γ + logsw

(s, w)2. We truncate the series to w < q with error

∞∑w=c+1

(s, w)

w2S(0, h;w)[log z − λw][log y − λsw]

�∞∑w=q

(s, w)(h,w)

w2log2 q

� τ(|h|)τ(s)q−1 log2 q

so that the resulting change in T−h (c) is O(τ(|h|)τ(s)MNq−1 log2 q), which is ab-sorbed into the error term in equation (5.36).

Since G−(z, y) is very small if |h| � Zcε, it follows that T−h (c) and the integralin equation (5.36) are both very small if |h| � Zcε. Therefore the error term inequation (5.36) is very small for such h. Which means that in evaluating T (c) themain contribution for the error term in equation (5.33) comes from |h| � Zcε. So wemay freely introduce a factor of

(1 + |h|Z−1

)−2into the error term of equation (5.36).

Therefore

T−h (c) =

∫ ∞0

g(h+ sy, y) dy +O((

1 + |h|Z−1)−2

(sNP )3/4Mcε)

= −2π∑

1≤w<c

(s, w)

sw2S(0, h;w)V (h) +O

((1 + |h|Z−1

)−2P 1/4(sN)3/4Mcε

)

(5.39)

where

Y (h) =

∫ ∞0

∫ ∞0

[log(h+ sy)− λw][log y − λsw]

Y0

(4π√

(h+ sy)x

c

)Jk−1

(4π√sxy

c

)F (x, y) dx dy

(5.40)

54

Here the range of integration is restricted by the support of F (x, y) and by the conditionz = h+ sy > 0. Similarly we show that

T+h (c) =

∫ ∞0

g(h+ sy, y) dy +O((

1 + |h|Z−1)−2

(sNP )3/4Mcε)

= −2π∑

1≤w<c

(s, w)

sw2S(0, h;w)K(h) +O

((1 + |h|Z−1

)−2P 1/4(sN)3/4Mcε

)(5.41)

where

K(h) =

∫ ∞0

∫ ∞0

[log(h− sy)− λw][log y − λsw]

K0

(4π√

(h− sy)x

c

)Jk−1

(4π√sxy

c

)F (x, y) dx dy

(5.42)

5.2.6 Estimating T (c)We get by summing (5.33) over h and equation (5.38) that

T−(c) =ϕ(c)T−0 (c)− 2π∑

1≤w<c

(s, w)

sw2

∑h6=0

S(0, h; c)S(0, h;w)V (h)

+O(P 9/4(sN)3/4Mc2+ε

) (5.43)

using Z = c2P 2M−1. We will exploit the fact that Y (h) depends on the summationvariables c and w to get some cancellation after the h sum is executed.

We shall treat the sum over h in equation (5.40) by the following lemma on theorthogonality property of Ramanujan sums.

Lemma 5.2.4. Let f be a twice differentiable function such that (1 + x2)f l(x) � 1for l = 0, 1, 2. Then we have∑h 6=0

S(0, h; c)S(0, h;w)f(h) = δc=wϕ(w)f(0) +∑u6=0

vcw(u)f

((c, w)

cwu

)(5.44)

for some 0 ≤ vcw(u) ≤ (c, w).

Proof. Split the sum over h into progressions modulo [c, w] followed by Poisson sum-mation formula,∑h 6=0

S(0, h; c)S(0, h;w)f(h) =∑

a (mod [c,w])

∞∑t=0

S(0, a+ t[c, w]; c)S(0, a+ t[c, w];w)f(a+ t[c, w])

=1

[c, w]

∑a (mod [c,w])

S(0, a; c)S(0, a;w)

∞∑t=0

e

(−at[c, w]

)f

(t

[c, w]

)

55

because f (a+ t[c, w]) = e

(−at[c, w]

)f

(t

[c, w]

). Now fix t and open the Ramanujan

sums:

∑h 6=0

S(0, h; c)S(0, h;w)f(h) =

∞∑t=0

f

(t

[c, w]

)1

[c, w]

∑x∗(c),y∗(w)

∑a([c,w])

e

(ax

c+ay

w− at

[c, w]

)

=

∞∑t=0

f

(t

[c, w]

)1

[c, w]

∑x∗(c),y∗(w))

∑a([c,w])

e

(a

[c, w]

(wx

(c, w)+

cy

(c, w)− t))

=

∞∑t=0

f

(t

[c, w]

) ∑d∗(c),δ∗(w)

(#solutions to

dw

(c, w)+

δc

(c, w)= t(

cw

(c, w)2)

)

If t = 0 and c 6= w,

cw

(c, w)2| dw

(c, w)+

δc

(c, w)=⇒ cwλ

(c, w)2

dw

(c, w)+

δc

(c, w)

=⇒ λ =d(c, w)

c+δ(c, w)

w

which is absurd since d, δ belong to coprime residue classes mod c and w respectively.Similarly we see that if c = w, then t = 0. And this gives us the diagonal term inequation (5.41). And it is obvious that if t 6= 0 the above congruence has exactlyϕ((c, w)) solutions if (

u,cw

(c, w)2

)= 1

. And this completes the proof of the lemma.

Lemma 5.2.4 therefore shows an orthogonality property of Ramanujan sums whichis particularly strong if f(y) has small support. In our applications the diagonal termdoesn’t appear because w < q ≤ c.

Applying Lemma 5.2.4 to Y (h) we get∑h 6=0

S(0, h; c)S(0, h;w)Y (h) = −ϕ(c)ϕ(w)Y (0) +∑u≥0

vcw(u)Y (u(c, w)

cw) (5.45)

where

Y (v) = −2π

∫ ∫ ∫ ∞0

[log z − λw]Y0

(4π

c

√zx

)cos 2πv(z − sy) dz

[log y − λsw]Jk−1

(4π

c

√sxy

)F (x, y) dx dy.

(5.46)

56

by combining the u and −u terms, which we can do because u 6= 0. Hence we get

T−(c) =ϕ(c)T−0 (c)− ϕ(c)∑

1≤w<q

ϕ(w)(s, w)w−2Y (0)

+∑

1≤w<q

(s, w)w−2∑u>0

vcw(u)Y (u(c, w)/cw)

+O(P 9/4(sN)3/4c2+ε).

(5.47)

Next we evaluate T+(C). The only difference is that T+0 (c) = 0 because there

are no solutions to the equation m + sn = 0 has no solutions in positive integers.Therefore,

T+(c) =∑

1≤w<q

(s, w)w−2∑h

S(0, h; c)S(0, h;w)K(h) +O(P 9/4(sN)3/4c2+ε).

(5.48)

Applying Lemma 5.2.4 for K(h) and collecting the u and −u terms(recall u 6= 0)we get

T+(c) =∑

1≤w<q

(s, w)w−2∑u>0

vcw(u)K(u(c, w)/cw) +O(P 9/4(sN)3/4c2+ε)

(5.49)

To evaluate the sums involving K and Y , we use the following functions

C(z) = 4K0(2z)− 2πY0(2z), (5.50)S(z) = 4K0(2z) + 2πY0(2z). (5.51)

We then have

T (c) = T ∗(c) + ϕ(c)T0(c)− ϕ(c)∑

1≤w<q

ϕ(w)(s, w)w−2Y (0)

+∑

1≤wq

(s, w)w−2∑u≥1

vcw(u)(C − S)(u(c, w)/cw)

+O(P 9/4(sN)3/4c2+ε),

(5.52)

where

C(v) =

∫ ∫ (∫ ∞0

[log z − λw]C

(2π

c

√zx

)cos(2πzv) dz

)[log y − λsw]Jk−1

(4π

c

√sxy

)cos(2πsyv)F (x, y) dx dy,

(5.53)

S(v) =

∫ ∫ (∫ ∞0

[log z − λw]S

(2π

c

√zx

)sin(2πzv) dz

)[log y − λsw]Jk−1

(4π

c

√sxy

)sin(2πsyv)F (x, y) dx dy.

(5.54)

57

The innermost integrals in the z-variable are evaluated in B3. Changing z intoz/2πv, we get from formulas in Appendix B that

C(v) =1

v

∫ ∫[log

x

c2v2− λw][log y − λsw]F (x, y)

Jk−1

(4π

c

√sxy

)cos(2πsyv) cos

(2πx

c2v

)dx dy,

(5.55)

S(v) =1

v

∫ ∫[log

x

c2v2− λw][log y − λsw]F (x, y)

Jk−1

(4π

c

√sxy

)sin(2πsyv) sin

(2πx

c2v

)dx dy.

(5.56)

Subtracting we arrive at

C(v)− S(v) =1

v

∫ ∫[log

x

c2v2− λw][log y − λsw]F (x, y)

Jk−1

(4π

c

√sxy

)cos(2πsyv) cos 2π

(syv +

2πx

c2v

)dx dy,

(5.57)

Integrating by parts we get

C(v)− S(v)� v−1(1 + vsNP−1)−1MN(logMN)2 (5.58)

which is good enough to truncate the u summation to 1 ≤ u < q, the tail beingabsorbed by the error already present.

5.2.7 Estimating B(s)

We need to estimate

B(s) = (k − 1)qT (0) + 2πikq(k − 1)∑

c = 0 (mod q)

1

c2T (c)

58

We split the sum into two parts, c ≤ 2C and c > 2C for some C ≥√sMN . Using

equation (5.51) for the part c ≤ 2C and Proposition 5.2.3 for c > 2C, we get

B(s)� (MN/s)1/2q1+ε

+MNq−1+ε

+ q∑

N<n<2N

τ(sn)τ(n)|Q(n)|

+ q∑

1≤w<q

(s, w)

w|R(w)|

+ q∑

1≤u,w<q

∑ (s, w)

uw|R(u,w)|

+ (sMN)1/2MNC−1+ε

+(C + q−5/4(sMN)9/8

)(sN)3/4Cε

+ q(MN)3/4C−1/2+ε

(5.59)

The first term comes from the estimation of T (0) given by equation (5.25), the secondterm is given by equation (5.34), and the third term comes from summing up T−0 (c)for c < 2C

Q(n) =∑

c mod 0(mod)q

ϕ(c)c−2

∫ ∞0

Y0

(4π

c

√snx

)Jk−1

(4π

c

√snx

)F (x, n) dx.

(5.60)

by equation (5.25). In the fifth term we have The fourth term comes by estimatingϕ(w) ≤ w in equation (5.52) where

R(w) =∑

c mod 0(q)

ϕ(c)c−2

∫ ∫[log sy − λw][log y − λsw]F (x, y)

Y0

(4π

c

√sxy

)Jk−1

(4π

c

√sxy

)dx dy

(5.61)

by equation (5.40). In the fifth term we have

R(u,w) =∑

c≡0 mod q

ϕ((c, w))

c(c, w)

∫ ∫[log

xw2

u2(c, w)2− λw][log y − λsw]

F (x, y)Jk−1

(4π

c

√sxy

)cos

2π

c

(syu

(c, w)

w+

xw

u(c, w)

)dx dy,

(5.62)

The restriction c ≤ 2C has been removed and the tails contribute the sixth and sev-enth terms in equation (5.59). Finally the eighth term comes from the application ofProposition 5.2.3 for c > 2C, and using M,N � q1+ε,

59

∑c=0(q),c>2C

c−2T (c) =

∞∑l=1

∑c=0(q),2lC≤c≤2l+1

c−2T (c)

�∞∑l=1

(sMN

2lC

)k−3/2

(q +M)1/2(q +N)1/2(MN)1/2

�(sMN

C

)k−3/2

(q +M)1/2(q +N)1/2(MN)1/2

� (q +M)1/2(q +N)1/2(MN)3/4C−1/2+ε

� q(MN)3/4C−1/2+ε

on using the fact that C2−k2 ≤ (sMN)1−k because C >

√sMN . Choosing C =

s√MN we make the second, sixth and eighth terms negligible by comparison to the

first. All we have to do now is to estimate Q(n), R(w), R(u,w). The idea is to elimi-nate the c from the oscillating Bessel sums while estimating all these three quantities.For Q(n), we change the variable x into c2t2 to get

Q(n) =

∫ ∞0

Y0(4πt√sn)Jk−1(4πt

√sn)

∑c≡0(q)

ϕ(c)F (c2t2, n)

2t dt (5.63)

For the last sum we use the following

Lemma 5.2.5. Let f be a smooth function compactly supported on R+. Then∑c≡0(q)

ϕ(c)

cf(c) =

1

ζ(2)ν(q)

∫f(x) dx+O

(ϕ(q)

q

∫|f ′(x)| log(1 +

x

q) dx

)(5.64)

where

ν(q) = q∏p|q

(1 +

1

p

)(5.65)

Proof. Our sum is equal to

S =∑d

µ(d)

d

∑n

f(n[d, q]).

Hence we derive by the Euler-Maclaurin formula∑l

F (l) =

∫(F (t) + tF ′(t)) dt

60

that

S =

(∑d

µ(d)

d[d, q]

)∫f(x) dx+

∫ξq(x)f ′(x) dx.

Here the factor in front of∫f(x) dx equals 1/ζ(2)ν(q) and

ξq(x) =∑d

µ(d)

d{ x

[d, q]}.

Another expression for this is

ξq(x) =ϕ(q)

q

∑(d,q)=1

µ(d)

d{ xdq}.

Hence it follows that

ξq(x)� ϕ(q)

q

∑d

1

dmin

(1,x

dq

)� ϕ(q)

qlog

(1 +

x

q

).

This completes the proof.

Applying Lemma 5.2.5 to equation (5.63) we get

(2ζ(2)ν(q)t2)−1

∫F (x, n) dx+O

(√Mt−1 log(1 +Mt−1)

). Hence

Q(n) =1

ζ(2)ν(q)

(∫ ∞0

Y0(t)Jk−1(t)dt

t

)∫F (x, n) dx

+O

(√M

∫ ∞0

|Y0(4πt√sn)Jk−1(4πt

√sn| log(1 +

M

t) dt

).

The first integral vanishes and by applying the usual bounds for the Bessel functionswe get

Q(n)�√M

∫ ∞0

1 + | log t√sn|

1 + t√sn

log(1 +M

t) dt�

(M

sN

)1/2

.(log snM)2

(5.66)

Summing over n we derive that

∑N<n<2N

τ(sn)τ(n)|Q(n)| � τ(s)

(MN

s

)1/2

(log sMN)5. (5.67)

Similarly we get

R(w)�(MN

s

)1/2

(log sMN)5 (5.68)

61

We now estimate R(u,w). Once again we eliminate c from the cosine and the Besselfunction by changing the variables. We write d = (c, w), c = l[d, q] so the conditionsof summation are now over d | w such that(

w

d,

q

(d, q)

)= 1,

(wq

d(d, q), u

)= 1 (5.69)

and (l, uw/d) = 1. We change the variables of integration x, y into lxy, lx/sy getting

R(u,w) =2

s

∑d|w

ϕ(d)

d[d, q]

∫ ∞0

∫ ∞0

Jk−1

(4πx

[d, q]

)cos

(2πx

[d, q]

(du

wy+wy

du

))Φd(x, y)

dxdy

y,

(5.70)

where

Φd(x, y) =∑

(l,uw/d)=1

[loglxyw2

u2d2− λw][log

lx

sy− λsw]F (lxy,

lx

sy)lx (5.71)

and∑

is restricted by the condition (5.69).

Lemma 5.2.6. Let F (t) be a smooth function compactly supported on R+. Then∑(l,k)=1

F (l) =ϕ(k)

k

∫F (t) dt+

∫ηk(t)F ′(t) dt (5.72)

where

ηk(t) =∑m|k

µ(m){ tm}. (5.73)

Proof. The proof follows from Euler-Maclaurin.

Note that ηk(t) = tϕ(k)/k if 0 ≤ t < 1 and that |ηk(t)| ≤ τ(k). Hence

|ηk(t)| ≤ τ(k)min(1, t). (5.74)

In particular equation (5.72) for k = uw/d yields

Φd(x, y) =ϕ(k)

k

I(y)

x+

∫ηk

(t

x

)H(t, y) dt (5.75)

where

I(y) =

∫[log

tyw2

u2d2− λw][log

t

sy− λsw]F (ty,

t

sy)t dt (5.76)

and

H(t, y) =∂

∂t[log

tyw2

u2d2− λw][log

t

sy− λsw]F (ty,

t

sy)t. (5.77)

62

Now we introduce equation (5.75) into equation (5.70). From the leading term weobtain the following integral in the x-variable∫ ∞

0

Jk−1

(4πx

[d, q]

)cos

(2πx

[d, q]

(du

wy+wy

du

))dx

x= 0

by formula (B.4) becausedu

wy+wy

du≥ 2. Therefore we are left with

R(u,w) =2

s

∑d|w

ϕ(d)

d[d, q]

∫Jk−1

(4πx

[d, q]

)∫ηk

(t

x

)I(t, x) dt dx (5.78)

where

I(t, x) =

∫H(t, y) cos

(2πx

[d, q]

(du

wy+wy

du

))dy

y. (5.79)

We estimate I(t, x) by means of the following

Lemma 5.2.7. Suppose H(y) is a smooth function supported on [Y, 2Y ] such that|H(y)| ≤ 1 and |H ′(y)| ≤ Y −1, where Y > 0. Then for any a, b > 0 we have∫

H(y) cos

(ay +

b

y

)dy

y� (1 + ab)−1/4 (5.80)

Proof. By Lemma 4.5 of [] applied twice, together with the trivial bound, we deducethat ∫ y

Y

cos

(ax+

b

x

)dx

x� min(1, a−1/2, b−1/2)

for Y ≤ y ≤ 2Y . Hence∫H(y) cos

(ay +

b

y

)dy

y= −

∫ 2Y

Y

(∫ y

Y

cos

(ax+

b

x

)dx

x

)H ′(y) dy

which is� min(1, a−1/2, b−1/2)

and this proves the lemma.

Before applying the lemma toH(t, y), from the support of F (x, y) we observe that√sMN ≤ t ≤ 2

√sMN (5.81)

and

H(t, y)� (log q)2. (5.82)

Therefore Lemma yields

I(t, x)�(

1 +x

[d, q]

)−1/2

(log q)2. (5.83)

63

Inserting this into equation (5.70) and using Jk−1(z)� (1 + z)−1/2 we obtain

R(u,w)�(MN

s

)1/2

(log q)2∑d|w

τ(uwd

)∫ ∞0

(x+[d, q])−1min(1,√sMN/x) dx.

Hence we finally get

R(u,w)� τ(u)τ3(w)

(MN

s

)1/2

(log q)3. (5.84)

Collecting all the results together, we get

B(s)�(MN

s

)1/2

q1+ε +(s√MN + q−5/4(sMN)9/8

)(sN)3/4qε

� qε(qs−1/2 + q3/4s15/8)(MN)1/2.

(5.85)

And the theorem is proved.

To prove the general case B(r, s), we need to use Hecke relations which we nowstudy.

5.2.8 Hecke multiplication rulesAssume that f is an eigenfunction of the Hecke operators Tn for all (n, q) = 1. Wehave the well-known relations

λf (m)λf (n) =∑

d|(m,n)

λf (mn

d2) (5.86)

if (mn, q) = 1 and

λf (n)ψf (m) =∑

d|(m,n)

ψf (mn

d2) (5.87)

if (n, q) = 1 where

λf (n) =ψf (n)

ψf (1)(5.88)

is the eigenvalue of Tn. These relations give rise to the factorisations∞∑n=1

ψf (n)n−s = Gf (s)Hf (s)

where

Gf (s) =∑n|q∞

ψf (n)n−s

Hf (s) =∑

(n,q)=1

λf (n)n−s =∏p-q

(1− λf (p)

ps+

1

p2s

)−1

.

64

Here n | q∞ if n has the same prime divisors as q.Notice that the divisor function also has the same multiplicativity properties as

λf (n) being the coefficients of the rank one Eisenstein series∂

∂sE(z, s) |s=1/2. Hence,

we get the same factorisation as above

∞∑n=1

τ(n)ψf (n)n−s = Gτf (s)Hτf (s)

where

Gτf (s) =∑n|q∞

τ(n)ψf (n)n−s

Hτf (s) =∑

(n,q)=1

τ(n)λf (n)n−s

=∏p-q

(1− λf (p)

ps+

1

p2s

)−2 (1− p−2s

)−1= ζq(2s)H

2f (s).

The last expression coming from Rankin-Selberg theory.We now work out expressions writing λf (an) and τ(n)λf (an) in terms of sums of

products of λf (n). For (a, q) = 1,

∑(n,q)=1

λf (an)n−s = Hf (s)∏

pα||a,α≥1

(λf (pα)− λf (pα−1)

ps

)(5.89)

∑(n,q)=1

τ(n)λf (an)n−s = Hτf (s)∏

pα||a,α≥1

(1− p−s

)−1

(λf (pα)− 2

λf (pα−1)

ps+λf (pα−2)

p2s

)(5.90)

By multiplicativity, it suffices to prove the above two in the case when a is a primepower. To illustrate the method of proof, we will prove the first relation in the casea = p2, where p is a prime not dividing q. The general case is the same. And themethod extends to prove the second formula also. Observe that

λf (n)λf (p2) =

λf (p2n) if (n, p2) = 1;λf (p2n) + λf (n) if p || n;λf (p2n) + λf (n) + λf (n/p2) if p2 | n.

(5.91)

65

∑ λf (an)

ns=

∑(n,p2)=1

+∑p||n

+∑p2||n

= λf (p2){∑

(n,p2)=1

+∑p||n

+∑p2||n

(λf (n)

ns

)}

−∑p||n

λf (n)

ns−∑p2||n

λf (n)

ps−∑p2|n

λf (n/p2)

ns

= λf (p2)Hf (s)− 1

ps

∑(n,p)=1

λf (pn)

ns−∑p|n

λf (pn)

ns−∑n

λf (n)

p2sns

= λf (p2)Hf (s)− λf (p)

psHf (s)

Comparing coefficients, we get

λf (an) =∑

a1n′=n

a0a1=a

µ(a1)λf (a0)λf (n′) (5.92)

τ(n)λf (an) =∑

a1a22d

2n′=n

a0a1a22=a

µ(a1a2)µ(a2)τ(a1)λf (a0)τ(n′)λf (n′) (5.93)

Furthermore, we get the following rule by applying equation (5.86) to (5.93)

τ(n)λf (n)λf (a) =∑

a1a22n′=n

a0a1a2=a

µ(a1)τ(a2)τ(n′)λf (a0a1n′) (5.94)

Note that all the above rules are valid for all a,n with (a, q) = 1. We may relax thecondition (n, q) = 1 by changing λf to ψf in all places in which the argument containsn or n′ by appealing to equation (5.87). We record two such formulas below

τ(m)ψf (rm) =∑

bc2d2m′=mabc2=r

µ(bc)µ(c)τ(b)λf (a)τ(m′)ψf (m′) (5.95)

τ(m)τ(n)ψf (rm)ψf (sn) =∑

αβγabc2=rγδ=s

(αβ,δ)=1

µ(β)µ(bc)µ(c)τ(α)τ(b)

.∑

bc2d2m′=m

∑αβ2n′=n

τ(m′)τ(n′)ψf (m′)ψf (βδan′).

(5.96)

Making use of the above rules we may complete the proof of Theorem 5.2.2 for thegeneral case

66

Proof.

B(r, s) =∑f,m,n

τ(m)τ(n)ψf (rm)ψf (sn)F (m,n)

�∑f,m,n

∑αβγabc2=r

γδ=s(αβ,δ)=1

τ(α)τ(β) |∑

bc2d2m′=mαβ2n′=n

τ(m′)τ(n′)ψ(m′)ψ(βδan′)F (m,n) |

�∑

αβγabc2=rγδ=s,(αβ,δ)=1

τ(α)τ(β) | B(1, βδa) |

� qε∑

αβγabc2=rγδ=s,(αβ,δ)=1

τ(α)τ(β)[q(βδa)−1/2 + q3/4(βδa)15/8]

(MN

bc2d2αβ2

)1/2

� qε[∑

αβγabc2=rγδ=s,(αβ,δ)=1

q

(γ

β

)(rs)−1/2 + q3/4(rs)15/8

∑ 1

α2β2](MN)1/2

� qε[q(r, s)τ(r)r(rs)−1/2 + q3/4(rs)15/8](MN)1/2

But B(r, s)� q1+ε(MN)1/2 also. This proves our theorem.

5.2.9 The amplification stepCorollary 5.2.8. Let (l, q) = 1, Then∑f∈F

λf (l) |∑m

τ(m)ψf (m)g(m) |2� qεmin(q, ql−1/2τ(l) + q3/4l15/8)M (5.97)

Proof. By equation (5.57)

λf (l)λf (m) =∑

a0a1a2=la1a

22n′=n

µ(a1)τ(a2)τ(n′)λf (a0a1n′)

∑f∈F

λf (l) |∑m

τ(m)ψf (m)g(m) =∑f.m,n

λf (l)τ(m)τ(n)ψf (m)ψf (n)g(m)g(n)

=∑f,m,n

∑a0a1a2=la1a

22m′=m

µ(a1)µ(a2)τ(m′)ψf (a0a1m′)ψf (n)g(m)g(n)τ(n)

=∑

a0a1a2=l

µ(a1)µ(a2)∑m,n

τ(m)τ(n)∆(m, a0a1n)g(m)g(a1a22n)

�∑

a0a1a2=l

τ(a2) | B(1, a0a1) |

� qε∑

a0a1a2=l

[q(a0a1)−1/2 + q3/4(a0a1)15/8]M

67

which completes the proof.

Corollary 5.2.9. Let cl be a sequence of complex numbers supported on (l, q) = 1.Set Λf (c) =

∑l≤L

clλf (l). Then

∑f

| Λf (c) |2|∑m

τ(m)ψf (m)g(m) |2� qε(q || c ||22 Lε + q3/4L15/8 || c ||21)M.

(5.98)where || c ||2 and || c ||1 represent the l2 and l1 norms of cl respectively.

Proof.∑f

| Λf (c) |2∑m

τ(m)ψf (m)g(m) |2=∑

l1,l2≤L

cl1cl2∑

f,v|(l1,l2)

λf (l1l2v2

) |∑m

τ(m)ψf (m)g(m) |2

Hence we are done by the above corollary applied to each l =l1l2v2

.

Corollary 5.2.10.

|∑m

τ(m)ψf (m)g(m) |2� q18/19+εM (5.99)

Proof. By positivity, the above corollary yields

| Λf (c) |2|∑m

τ(m)ψf (m)g(m) |2� qε(q || c ||22 Lε + q3/4L15/8 || c ||21)M.

We now choose a sequence cl that will maximise the value of the amplifier. Choose

cl =

λf (l) if l ≤ L1/2 is a prime;

−1 if l is square of a prime that is less than L1/2;0 otherwise.

(5.100)

Then

Λf (c) =∑

clλf (l) =∑

l∈Primesl≤L1/2

λf (l)2 +∑

l∈Primes2l≤L

(−1)λf (l)

=∑

l∈primesl≤L1/2

λ2f (l)− λf (l2)

� L1/2

logL.

Because λ2f (l)− λf (l2) = 1 for primes (l, q) = 1.

68

And

|| c ||22 =∑

l∈Primesl≤L1/2

| λf (l) |2 +∑

l∈primesl≤L1/2

1

≤∑

l∈Primesl≤L1/2

(4 + 1)

≤ 5Λf (c)

|| c ||1 =∑

l∈Primesl≤L1/2

| λf (l) | +∑

l∈Primesl≤L1/2

1

≤∑

l∈Primesl≤L1/2

(2 + 1)

≤ 3Λf (c)

where we have used the Deligne bound for λf (n). Now choosing L = q2/19 we getour result.

We are now ready to prove the main theorem.

5.2.10 Proof of theorem 5.2.1Since f is a newform we know that for all m,n

λf (m)λf (n) =∑

d|(m,n)

εq(d)λf (mn

d2). (5.101)

Define

L(s, f) =

∞∑n=1

λf (n)

ns.

Then it follows immediately by equation (5.65) that

L(s, f)2 = ζq(2s)L(s, f × g) (5.102)

where

L(s, f × g) =

∞∑n=1

τ(n)λf (n)

ns

is the Rankin-Selberg convolution L-function. Set

ψf (1)L2f (s) =

∞∑n=1

ρf (n)

ns

69

whereρf (n) =

∑d2m=n(d,q)=1

τ(m)ψf (m)

Set

S(g) =

∞∑m=1

ρf (m)g(m)m−1/2

where g is a nice function supported in [M, 2M ]. Then by equation the approximatefunctional equation, it suffices to estimate S(g) to get a bound for L(s, f) for M � Qwhere Q � q. By Corollay 5.2.8 we get

S(g)� q1/4−1/76+ε (5.103)

Let g∗(t) = h(16π4q−2t), where

h(y) =1

2πi

∫(0)

g(s)Γ2(k2 − s)Γ2(k2 + s)

ys ds

and g is the Mellin transform of g(x),

g(s) =

∫ ∞0

g(x)xsdx

x.

We now remove the restriction M � q1+ε. To do this we appeal to the Poisson typeformula (3.41) to get S(g) = S(g∗). By partial integration g(s)�A (1+ |s|)−A. Thenby partial integration one again shows that

yjg(j)(y)� (1 + yM)−A

for all j ≥ 0 and any A > 0. Although h(y) is not compactly supported it has a rapiddecay in the range y � Y = M−1. Breaking into dyadic intervals, we get the samebound as equation (5.103) for S(g∗) providedM � q. Hence, theorem 5.2.1 is proved.

5.3 Other CasesWe mention some results in the q-aspect due to Duke-Friedlander-Iwaniec which takecare of some of the remaining cases of subconvexity for GL(2). In Bounds for Auto-morphic L-functions III, Invent. Math. 143 (2001) it is proved that

Theorem 5.3.1. Let k ≥ 3 and q be squarefree and χ a primitve character modulo q.Let f ∈ Sk(q, χ). Then, on the line Re(s) = 1/2, we have

L(s, f)� |s|2q1/4−1/218+ε (5.104)

In Subconvexity for Artin L-functions, Invent. Math. 149 (2002) they prove thefollowing

70

Theorem 5.3.2. Let k ≥ 0 be an integer and χ(modq) a primitive character withχ(−1) = (−1)k. Let f be a Hecke-Maass cusp form of weight k, level q, character χand eigenvalue λj = 1/4 + t2j . Then

L(s, f)� (|tj |+ |s|)10q1/4−1/23041+ε (5.105)

Building on the work of Duke-Friedlander-Iwaniec, many authors, Valentin Blomer,Gergely Harcos, Emmanuel Kowalski, Philippe Michel, Peter Sarnak, to name a few,solved the subconvexity problem for mostGL(2)L-functions. Finally in 2010, PhilippeMichel and Akshay Venkatesh in The Subconvexity Problem for GL(2), Publ. Math.de l’ IHES, 111 (2010) completely solved the subconvexity problem for GL(2) L-functions over an arbitrary number field. The subconvex bound they obtain is of ahybrid nature, i.e., they prove subconvexity in the q, s and spectral aspects simultane-ously with the same subconvexity exponent.

71

Chapter 6

Weak Subconvexity

Let L(s, f) be a Dirichlet series∞∑n=1

af (n)

nsconverging to the right of the line Re(s) =

1. We assume that L(s, f) satisfies conditions (1.1) to (1.3) of Chapter 1. Furthermore,we assume the following properties on αj(p) and µj . Let

−L′

L(s, f) =

∑n≥1

Λf (n)

ns=∑n≥1

λf (n)Λ(n)

ns

where Λ(n) is the von-Mangoldt function and λf (pk) =

m∑j=1

αj(p)k. We then assume

that ∑x≤n≤ex

| λf (n) |2 Λ(n)

n≤ A2 +

A0

log ex; with A1, A0 ≥ 1. (6.1)

Equation (6.1) is called the weak-Ramanujan hypothesis. We also assume that

Re(µj) ≥ 1− δm (6.2)

for some δm > 0,∀ 1 ≤ j ≤ m. The main theorem that will be proved in this section,due to K Soundararajan[11], is

Theorem 6.0.3. Let L(s, f) be as above, then

L(1/2, f)�εQ(f)1/4

(logQ(f))1−ε (6.3)

The convexity principle would give us,∑n≤x

af (n)� x

log x.

for x ≥ Q(f)1/2(logQ(f))B . And as usual, to break convexity we need similar can-cellation for shorter sums. To that effect we have the

72

Theorem 6.0.4. Let L(s, f) be an L-function as above. For any ε > 0 and B > 0,and all x ≥ Q(f)1/2(logQ(f))−B ,∑

n≤x

af (n)� x

(log x)1−ε . (6.4)

Notice that the above theorem is not as strong as the subconvexity theorems wehave seen so far. We are only saving a power of log and hence Soundararajan calls itweak subconvexity. It is also remarked in [8] that the limit of the method would be

obtain a boundQ(f)

logQ(f)in Theorem 6.0.9 and

x

log xin Theorem 6.0.10.

6.1 Mean Values of Multiplicative FunctionsThe main idea behind the proof of Theorem is that mean values of multiplicative func-tions vary slowly. Let f(n) be a multiplicative function(i.e., f(mn) = f(m)f(n)

whenever (m,n) = 1). Denote by S(x) = S(x; f) =∑n≤x

f(x). Knowing equa-

tion (6.4) in the range x ≥ Q(f)1/2(logQ(f))B , this fact about the mean valuesof multiplicative functions allows us to extrapolate equation (6.4) to the range x ≥Q(f)1/2(logQ(f))−B .

This was first accomplished by Hildebrand for for multiplicative functions such that−1 ≤ f(n) ≤ 1. If 1 ≤ w ≤

√x, he proved that

1

x

∑n≤x

f(n) =w

x

∑n≤x

f(n) +O

((log

log x

log 2w

)−1/2). (6.5)

In other words, the mean value of f at x does not change very much from the meanvalue at x/w. Using this idea, Hildebrand was able to extrapolate Burgess’s bound(seeequation (3.15) or the discussion at the end of Section 7.1) for x ≥ q1/4−ε. The directextension for complex f is wrong(take f(n) = nit for instance), and Elliott gavethe following generalisation: if f(n) is a complex valued multiplicative function suchthat |f(n)| ≤ 1, there exists a real number τ = τ(x) with |τ | ≤ log x such that for1 ≤ w ≤

√x

S(x) = w1+iτS(x/w) +O

(x

(log 2w

log x

)1/19). (6.6)

To establish Theorem, we need to consider complex valued functions no longerconstrained in the disc. Instead of showing that a suitable combination of S(x/w)/(x/w)is small, we will need to show that linear combinations involving S(x/wj)/(x/wj)with j = 0, 1, . . . , J . is small. We give two examples to motivate the approach.

1. Let k be a natural number. Take f(n) = τk(n), the k-divisor function. It canthen be shown that S(x) = xPk(log x)+O(x1−1/k+ε) where Pk is a polynomial

73

of degree k − 1. If k ≥ 2, it follows that S(x)/x − S(x/w)/(x/w) is of size(logw)(log x)k−2, which is not o(1). However, if 1 ≤ w ≤ x1/2k

k∑j=0

(−1)j(k

j

)S(x/wj)

x/wj=

k∑j=0

(−1)j(k

j

)Pk(log x/wj) +O(x−1/2k) = O(x−1/2k).

(6.7)

2. Let τ1, . . . , τR be distinct real numbers and let k1, . . . , kR be natural numbers.

Let f be the multiplicative function F (s) =

∞∑n=1

f(n)n−s =

R∏j=1

ζ(s − iτj)kj .

Consider the linear combination

1

x

k1∑j1=0

. . .

kR∑jR=0

(−1)j+...+jR(k1

j1

). . .

(kRjR

)wj1(1+iτ1)+...+jR(1+iτR)S

( x

wj1+...+jR

).

(6.8)

By Perron’s formula we may express this as, for c > 1,

1

2πi

∫(c)

R∏j=1

ζ(s− iτj)kj (1− w1+iτj−s)kjxs−1 ds

s. (6.9)

Notice that the poles of the zeta-functions at 1 + iτj have been cancelled by thefactors (1 − w1+iτj−s)kj . Thus the integrand only has a pole at s = 0. By theestimates ζ(s) � 1 for Re(s) > 1 and ζ(s) � |t|

1−σ2 we see that we may

move the line of integration slightly to the left of 1 to get the integral is� x−δ

for some δ > 0. With these examples as motivation, we make the followingdefinitions,

Let f(n) be a multiplicative function. Let

S(x) = S(x; f) =∑n≤x

f(n) (6.10)

F (s) =∑n≥1

f(n)

nsconvergent for Re(s) > 1. (6.11)

Let

−F ′

F(s) =

∑n≥1

λf (n)Λ(n)

ns(6.12)

and assume the weak-Ramanujan hypothesis∑x≤n<ex

| λf (n) |2 Λ(n)

ns≤ A2 +

A0

log ex; with A1, A0 ≥ 1 (6.13)

74

Let R be a natural number, and let τ1, . . . , τR denote R real numbers. Let l =(l1, . . . , lR) and j = (j1, . . . , jR) denote vectors of non-negative integers, withthe notation j ≤ l indicating that 0 ≤ ji ≤ li for all 1 ≤ i ≤ R. Define(

l

j

)=

(l1j1

). . .

(lRjR

). (6.14)

Finally, we define a measure of the oscillation of the mean-values of f by setting

Ol(x,w) = Ol(x,w; τ1, . . . , τR)

=∑j≤l

(−1)j1+...+jR

(l

j

)wj1(1+iτ1)+...+jR(1+iτR)S

( x

wj1+...+jR

).

(6.15)

The theorem that we want to prove is the following extrapolation theorem,

Theorem 6.1.1. For f satisfying (6.10) to (6.13), let ε > 0 be given. Set R =[10A2/ε2] + 1 and put L = [10AR] and L = (L, . . . , L). Let w be such that 0 ≤logw ≤ (logX)1/3R. There exist real numbers τ1, . . . , τR with |τj | ≤ exp((log logX)2)such that 2 ≤ x ≤ X we have

|OL(x,w; τ1, . . . , τR)| � x

log x(logX)ε. (6.16)

6.2 Some preliminary lemmasLemma 6.2.1. Let b(n) ∈ C. Define the sequence a(n) by setting a(0) = 0, a(n) by

exp

(∑l

b(k)xk

k

)=∑n

a(n)xn. (6.17)

For j = 1, 2 define the sequences Aj(0) = 1, Aj(1), Aj(2) by means of the powerseries identity

exp

(∑k

| b(k) |j xk

k

)=∑n

Aj(n)xn. (6.18)

Then Aj(n) ≥| a(n) |j for all n.

Proof. Expanding the LHS of equation (6.17),

exp

(∑k

b(k)xk

k

)=∏k

exp(

((b(k)xk

k

)=∏k

∑m

b(k)mxkm

m!km

∴ a(n) =∑

λ,λ1+λ2+...+λr=nλ1≥λ2≥...≥λr≥1

b(λ1) . . . b(λr)W (λ)

75

where the sums runs over all partitions of n. We may calculate W (λ) explicitly, but itsuffices to observe that it’s strictly positive and that it is independent of b(k). Hence

setting b(k) = 1, we see that exp

(∑ xk

k

)= exp

(log

(1

1− x

))=

1

1− x. This

implies that a(n) = 1. Hence we obtain the identity∑λ

W (λ) = 1. (6.19)

To prove the lemma when j = 1, we use the triangle inequality,

| a(n) |≤∑| b(λ1) | . . . | b(λr) ||W (λ) |,

but the RHS is precisely A1(n).

∴ A1(n) ≥| a(n) | .

When j = 2,

| a(n) |2 ≤(| b(λ1) |2 . . . | b(λr) |2|W (λ) |

)∑λ

W (λ)

= A2(n)

and this proves the lemma.

Lemma 6.2.2. Let f be a multiplicative function satisfying the weak-Ramanujan hy-pothesis, then for all x ≥ 2 we have∑

n≤x

| f(n) |n

� (log x)A, (6.20)

∑n≤x

| f(n) |2

n� (log x)A

2

. (6.21)

Moreover, for all 2 ≥ σ > 1 we have

|F (σ + it)| �(

1

σ − 1

)A, (6.22)

|F (σ + it)| �(

1

σ − 1

)A+1

(6.23)

The implied constants may depend on A,A0.

Proof. ∑n=2

|λf (n)|2Λ(n)

nσ log n=

∞∑k=0

∑ek<n≤ek+1

|λf (n)|2Λ(n)

nσ log n

76

To evaluate the inner sum we need equation (6.13). By partial summation,

∑ek<n≤ek+1

|λf (n)|2Λ(n)

n.nσ−1 log n=

1

(ek+1)σ−1k + 1

∑ek<n≤ek+1

|λf (n)|2Λ(n)

n+

∫ ek+1

ek

Sk(t)

tσ log t((log t)−1+σ−1) dt

where Sk(t) =∑

ek<n≤t

|λf (n)|2Λ(n)

n� A2 log t/ek + A0 log log t/ek. Therefore as

σ > 1 we have∞∑n=0

|λf (n)|2Λ(n)

nσ log n≤∞∑k=0

A2 +A0/(k + 1)

(eσ−1)k+1(k + 1)+O(1)

= A2∞∑k=0

1

(eσ−1)k+1(k + 1)+O(1)

= A2 log

(1

1− e1−σ

)+O(1).

(6.24)

Hence ∑n=2

|λf (n)|2Λ(n)

nσ log n≤ A2 log

(1

1− σ

)+O(1) (6.25)

as 1 < σ ≤ 2. By Cauchy-Schwarz,

∞∑n=2

|λf (n)|Λ(n)

nσ log n≤

(∑n=2

|λf (n)|2Λ(n)

nσ log n

)1/2( ∞∑n=2

Λ(n)

nσ log n

)1/2

≤(A2 log

(1

1− σ

)+O(1)

)1/2(log

(1

σ − 1

))1/2

as ζ(σ)� (σ − 1)−1. Hence we obtain

∞∑n=2

|λf (n)|Λ(n)

nσ log n≤ A log

(1

σ − 1

)+O(1). (6.26)

Now,

∑n≤x

| f(n) |n

=∑n≤x

| f(n) |n1+1/ log ex

n1/ log ex ≤

∑n≤x

| f(n) |n1+1/ log ex

1/2∑n≤x

| f(n) |n

n1/ log ex

1/2

.

Since x1/ log x = 1 and n ≤ x, n1/ log ex ≤ 1. Hence extending the first sum to∞ weobtain

∑n≤x

| f(n) |n

�

( ∞∑n=1

| f(n) |n1+1/ log ex

n1/ log ex

)1/2∑n≤x

|f(n)|n

1/2

77

which gives ∑n≤x

| f(n) |n

�

( ∞∑n=1

| f(n) |n1+1/ log ex

n1/ log ex

)By the multiplicativity of f(n),∑

n

f(pn)

pns= exp

(λf (p)

kpks

).

Therefore by the previous lemma if we set

exp

(∑k

|λf (p)|kpks

)= exp

(∑k

|λf (p)|Λ(pk)

log pkpks

)=∑n

f (1)(pn)

pns,

then f (1)(n) ≥ |f(n)|. Note that this is the only place where multiplicativity of f(n)and f (1)(n) are used. Hence

∞∑n=1

f (1)(n)

ns≥∞∑n=1

|f(n)|ns

,

i.e.,∞∑n=1

| f(n) |n1+1/ log ex

≤ exp

(∑n

|λf (n)|Λ(n)

log nn1+1/ log ex

)By equation (6.26) above, the last term is� (log x)A. Hence we get equation (6.20).

To estimate∑n≤x

| f(n) |2

n� (log x)A

2

, we use equation (6.25) and proceed in the

same way to get

∑n≤x

| f(n) |2

n� (log x)A

2

≤∞∑n=1

|f(n)|2

n1+1/ log ex≤ exp

(∑n

|λf (n)|2Λ(n)

n1+1/ log ex log n

)� (log x)A

2

.

To prove the third inequality,

|F (σ+it)| =|∞∑n=1

f(n)

nσ+it|≤

∞∑n=1

|f(n)|nσ

� exp

(∑n

|λf (n)|Λ(n)

nσ log n

)�(

1

σ − 1

)A.

The last inequality follows similarly,

| F′

F(σ + it) |≤

∞∑n=1

|λf (n)Λ(n)

nσ� 1

σ − 1

by Cauchy-Schwarz. Hence we are done.

78

Lemma 6.2.3. Let f be a multiplicative function as in the above lemma. Then for allx ≥ 2,

∑n≤x

|f(n)| � x(log x)A. Moreover, for 1 ≤ y ≤ x we have∑

x<n≤x+y

|f(n)| ≤

(yx)1/2(log x)A2/2.

Proof. Since∑n≤x

|f(n)| ≤ x∑n≤x

|f(n)|n

, the first assertion follows from the previous

lemma. For the next, we use Cauchy-Schwarz.

|∑

x<n≤x+y

|f(n)|2 |≤ y∑n

|f(n)|2 � yx∑n≤2x

|f(n)|2

n� yx(log x)A

2

.

6.3 Deduction of the Main Theorem from Theorem 6.2.1Lemma 6.3.1. LetL(s, f) be anL-function satisfying the weak-Ramanujan hypothesisand the assumption that Re(µj) ≥ 1 − δm for some δm > 0. Then for all t ∈ R, wehave

|L(1/2 + it, f)| � Q(f)1/4(1 + |t|)m/4+1(log(Q(f))A (6.27)

Proof. Define Λ(s) = L(s, f)L∞(s, f)e(s−1/2−it)2 . By the Phragmen-Lindelof prin-ciple we may bound | Λ(1/2+ it) | by the maximum value taken by |Λ(s)| on the lines

Re(s) = 1 +1

logQand Re(s) =

−1

logQ. But by the functional equation the maximum

on Re(s) = 1 +1

logQand Re(s) =

−1

logQare the same.

|L(1/2 + it, f)| ≤ maxy∈R|Λ(1 + 1/ logQ+ it+ iy)||L∞(1/2 + it, f)|

≤ maxy∈Re−y2 |L(1 + 1/ logQ+ it+ iy, f)||L∞(1 + 1/ logQ+ it+ iy, f)|

|L∞(1/2 + it, f)|

� maxy∈Re−y2(logQ)A

|L∞(1 + 1/ logC + it+ iy, f)||L∞(1/2 + it, f)|

by Lemma (6.2.2). Now by Stirling’s formula,

|ΛR(1 + 1/ logQ+ it+ iy + µj)||ΛR(1/2 + it+ µj)|

� Γ(1/2 + 1/ logQ2 + it/2 + iy/2 + µj/2)

Γ(1/4 + µj/2 + it/2)

� e2|y| (1 + |t|+ |y|+ |µj |)1

2 logQ

(1 + |t|+ |µj |)−1/4

� e2|y|(1 + |t|)1

2 logQ+1/4(1 + |µj |)1

2 logQ+1/4.

From which the result follows. Note that in the above lemma, we have used the as-sumption Re(µj) ≥ 1− δm to ensure that the gamma factors have no poles.

79

Lemma 6.3.2. For x ≥ x0 = Q(f)1/2 (logQ(f))50mA2

we have∑n≤x

af (n)� x

log x. (6.28)

Proof. Observe that for any c > 0, y > 0, λ > 0 and any k ∈ N, since∫ λ

0

exis dxi =

eλs − 1

s,

1

2πi

∫ c+i∞

c−i∞

ys

s

(eλs − 1

λs

)kds =

1

λk

∫ λ

0

. . .

∫ λ

0

1

2πi

∫ c+i∞

c−i∞(yex1+x2+...+xk)s

ds

sdx1 . . . dxk.

By Perron’s formula,

1

2πi

∫ c+i∞

c−i∞

ys

s

(eλs − 1

λs

)kds =

1 if y ≥ 1;

∈ [0, 1] if e−λk ≤ y < 1;

0 if y < e−λk.

For any c > 1,

1

2πi

∫ c+i∞

c−i∞L(s, f)

ys

s

(eλs − 1

λs

)kds =

∑n≤x

af (n) +O

∑x<n≤ekλx

| af (n) |

(6.29)

Choosing k, λ appropriately for x large(λ = λ(x) we ensure that ekλ ≤ 2. Hence∑x<n≤ekλx

| λf (n) | � (ekλ − 1)1/2x(log x)A2/2

=

(ekλ−1

1(log x)2+A2

)1/2x

log x

on choosing λ = (log x)−2−A2

. Hence we see that∑x<n≤ekλx

| af (n) |� x

log x(6.30)

Now move the line of integration to the line Re(c) = 1/2. Using Lemma 6.2.2 we seethat the integral is

� Q(f)1/4x1/2λ−k (logQ)A∫ ∞−∞

(1 + |t|)m/4+1 dt

(1 + |t|)k+1� Q(f)1/4x1/2 (logQx)

12mA

(6.31)

where we have chosen k = [m

4] + 3 to make the integral converge. Hence we are

done.

80

We now deduce Theorem 6.0.10 from Theorem 6.2.1.

Proof. Let R = [10A2

ε2] + 1, L = [10AR]. Let x0 = Q(f)1/2 (logQ(f))

50mA2

and

x0 ≥ x ≥ Q(f)1/2

(logQ(f))B. Take w =

x0

x, X = xwLR. Applying Theorem 6.2.1 to the

multiplicative function af (n), we find for an appropriate choice of τ1, τ2, . . . , τR that

|OL(x,w)| � X

(logX)1−ε . (6.32)

The LHS is by definition

wLR|∑

n≤X/wLRaf (n)|+O

∑j<L

(L

j

)|wj ||

∑n≤X/wj1+...+jR

|af (n)|

= wLR|

∑n≤X/wLR

af (n)|+O

LR−1∑j=0

|wj ||∑

n≤X/wj|af (n)|

.

(6.33)

The constants depend only on A and ε in the above equation. Now X/wLR = x, andfor 0 ≤ j ≤ LR − 1 we have X/wj ≥ xw = x0 so that the bound of Lemma 6.3.2

holds. Hence the above equation equals wLR|∑n≤x

af (n)| + O

(X

logX

). Hence by

equation (6.32) we are done.

We can now prove weak-subconvexity from the previous theorem.

Proof. Let c > 1/2.

1

2πi

∫(c)

L(s+1/2, f)L∞(s+ 1/2, f)

L∞(1/2, f)es

2 ds

s= L(1/2, f)+

1

2πi

∫(−c)

L(s+1/2, f)L∞(s+ 1/2, f)

L∞(1/2, f)es

2 ds

s

Changing variables s to −s and using the functional equation,

L(1/2, f) =1

2πi

∫(c)

L(s+1/2, f)L∞(s+ 1/2, f)

L∞(1/2, f)es

2 ds

s+

κ

2πi

∫(c)

L(s+1/2, f)L∞(s+ 1/2, f)

L∞(1/2, f)es

2 ds

s

We estimate the first integral, the same bounds hold for the second. By partial summa-tion,

L(s+ 1/2, f) = (s+ 1/2)

∫ ∞1

∑n≤x

af (n)dx

xs+3/2.

Hence the first integral equals∫ ∞1

∑n≤x

af (n)1

2πi

(∫(c)

(s+ 1/2)L(s+ 1/2, f)L∞(s+ 1/2, f)

L∞(1/2, f)es

2

x−sds

s

)dx

x3/2.

(6.34)

81

We estimate the integral over s by first moving the line of integration to Re(s) = 2.Then∫

(c)

(s+ 1/2)L(s+ 1/2, f)L∞(s+ 1/2, f)

L∞(1/2, f)es

2

x−sds

s=

∫(2)

(s+ 1/2)L(s+ 1/2, f)L∞(s+ 1/2, f)

L∞(1/2, f)es

2

x−sds

s.

(6.35)

On Re(s) = 2,

|ΓR(s+ 1/2 + µj)| � |Γ(1 + it+ 1/4 + µj/2)| = |Γ(5/4 + µj/2 + it/2)|� (1 + |t|+ |µj |)5/4−12

= (1 + |t|)3/4(1 + |µj |)3/4

(6.36)

and|ΓR(1/2 + µj)| = |Γ(1/4 + µj/2)| � (1 + µj)

−1/4.

Hence the integral is

� x−2

∫ ∞−∞

Q(f)3/4

Q(f)−1/4(1 + |t|)3m/4e−t

2

dt�

(√Qf

x

)2

.

Now move the line of integration to Re(s) = 1/2 − δm/2. Then by Stirling’sformula we get on Re(s) = 1/2− δm/2,

|ΓR(s+ 1/2 + µj)| � |Γ(1/2− δm/4 + µj/2 + it/2)|� (1 + |t|)−δm/4+1/2−1/2(1 + µj)

−δm/4+1/2−1/2(6.37)

Hence the integral is

� xδm/2−1/2

∫ ∞−∞

Q(f)−δm/4

Q(f)−1/4(1 + |t|)−δmm/4e−t

2

dt�(√

Q

x

)1/2−δm/2

.

Therefore, the integral is

� min

((√Q

x

)1/2−δm/2

,

(√Q

x

)2).

We now break the inner integral into two parts,

∫ √Q1

dx

x2−δm/2=

∫ √Q/(logQ)4A/δm

1

+

∫ √Q√Q/(logQ)4A/δm

�∫ √Q

1

x(log x)Adx

x2−δm/2+

∫ √Q√Q/(logQ)4A/δm

x

(log x)1−εdx

x2−δm/2

82

Now ∫ √Q1

x(log x)Adx

x2−δm/2� Cδm/4

(logQ)A,∫ √Q

√Q/(logQ)4A/δm

x

(log x)1−εdx

x2−δm/2� Qδm/4

(logQ)1−ε ,

Q

∫ ∞√Q

|∑n≤x

| dxx7/2

� 1.

Combining the above three estimates the theorem is proved.

6.4 Successive MaximaLet T = exp((log logX)2).Define τ1 to be that point t in [−T, T ] where the maximumof |F (1+1/ logX+ it)| is attained. Now remove the interval (τ1−(logX)−1/R, τ1 +(logX)−1/R) from C1 = [−T, T ], and let C2 denote the remaining compact set. Wedefine τ2 to be that point inC2 where the maximum of |F (1+1/ logX+it)| is attained.Next remove the interval (τ2−(logX)−1/R, τ2+(logX)−1/R) fromC2 leaving behindthe compact set C3. Proceed in a similar manner, defining the successive maximaτ1, . . . , τR and the nested compact sets C1 ⊃ C2 ⊃ . . . ⊃ CR. Notice that all thepoints τ1, . . . , τR lie in the interval [−T, T ] and moreover the points are well-spaced,i.e., |τj − τk| ≥ (logX)−1/R for j 6= k. By Lemma 6.2.2 |F (1 + 1/ logX + it)| �(logX)A; we will improve on this bound, so long as t is not too close to the τjs.

Lemma 6.4.1. Let 1 ≤ j ≤ R, t ∈ Cj . Then

|F (1 + 1/ logX + it)| � (logX)A√

1j+ j−1

jR (6.38)

In particular if t ∈ CR we have |F (1 + 1/ logX + it)| � (logX)ε/2

Proof. If t ∈ Cj then for all 1 ≤ r ≤ j

|F (1 + 1/ logX + it)| ≤ |F (1 + 1/ logX + iτj)| ≤ |F (1 + 1/ logX + iτr)|.

Hence

|F (1 + 1/ logX + iτj)| ≤

(j∏r=1

|F (1 + 1/ logX + iτr)|

)1/j

≤ exp

Re1

j

∑n≥2

λf (n)Λ(n)

n1+1/ logX log n(n−iτ1 + . . .+ n−iτj )

By Cauchy-Schwarz,

∑n≥2

|λf (n)|Λ(n)

n1+1/ logX log n|j∑r=1

n−iτr | ≤

∑n≥2

|λf (n)|2Λ(n)

n1+1/ logX log n

1/2∑n≥2

λf (n)Λ(n)

n1+1/ logX log n|j∑r=1

n−iτr |21/2

83

By equation (6.25) the first factor is ≤ (A2 log logX +O(1))1/2. For the second,

∑n≥2

λf (n)Λ(n)

n1+1/ logX log n|j∑r=1

n−iτr |2 = j∑n≥2

Λ(n)

n1+1/ logX log n+ 2Re

∑1≤r≤s≤j

∑n≥2

Λ(n)

n1+1/ logX+i(τr−τs) log n

= j(log logX +O(1)) + 2∑

1≤r<s≤j

log |ζ(1 + 1/ logX + i(τr − τs))|

≤ j(log logX) +(j)(j − 1)

Rlog logX +O(1)

using the bound log |ζ(1 + 1/ logX + it| ≤ (logX)1/R + O(1) for (logX)−1/R ≤|τr−τs| ≤ 2T . To see this, if (logX)−1/R ≤ |τr−τs| ≤ 1, we use the fact that ζ(s) �| 1

s− 1| ≤ (logX)1/R + O(1). For the remaining range use the first approximation

formula. Hence we have the result.

Note that the definition R = [10A2

ε2] + 1 comes from this lemma.

Let l = (l1, . . . , lR) be a vector of non-negative integers. We study the function

Fl(s) = F (s)

R∏j=1

(1− w1+iτj−s)lj (6.39)

because by Perron’s formula,

1

2πi

∫(c)

F (s)

R∏j=1

(1− w1+iτj−s)ljxs

sds = Ol(x,w).

We now bound this function.

Lemma 6.4.2. Let σ ≥ 1 + 1/ logX. Then

max|t|≤T/2|Fl(σ + it)| ≤ max|t|≤TFl(1 + 1/ logX + it)|+O((logX)−1).(6.40)

Proof. Let σ = 1 + 1/ logX + α for α > 0. The Fourier transform of k(z) = e−α|z|

is

k(ξ) =

∫ ∞−∞

e−α|z|−iξz dz =2α

α2 + ξ2

By Fourier inversion, for any z ≥ 1

z−α = k(log z) = k(− log z) =1

2π

∫ ∞−∞

k(ξ)x−iξ dξ

=1

π

∫ ∞−∞

k(ξ)x−iξ dξ +O(α

T)

84

Hence

Fl(σ+it) =1

π

∫ ∞−∞

2α

α2 + ξ2Fl(1+1/ logX+it+iξ) dξ+O

α

T

∑n≥1

|f(n)|n1+1/ logX

By lemma 6.2.2∑

n

|f(n)|n1+1/ logX

=1

logX

∫ ∞1

(log t)A

t1+1/ logX� (logX)A

Hence the error term is O(αT

(logX)A)

= O((logX)−1

)by the definition of T .

Now if |t| ≤ T/2 then |t+ ξ| ≤ T . Hence we are done.

Lemma 6.4.3. Suppose lj ≥ L − 1 for all 1 ≤ j ≤ R. Then provided 0 ≤ logw ≤(logX)1/3R we have

max|t|≤T |Fl(1 + 1/ logX + it)| � (logX)ε/2. (6.41)

Proof. Suppose first that |t| ≤ T but |t − τj | ≥ (logX)−1/R for all 1 ≤ j ≤R. Then Lemma 6.4.1 gives that |F (1 + 1/ logX + it)| � (logX)ε/2, and hencemax|t|≤T |Fl(1+1/ logX+it)| � (logX)ε/2. Now suppose that |t−τj | ≤ (logX)−1/R

for some 1 ≤ j ≤ R. By Lemma 6.2.2 we have that |F (1 + 1/ logX + it)| �(logX)A. Moreover,

|1− w−1/ logX−it+iτj |lj = |1− e−logwlogX−i logw(t−τj)|lj

≤(|1− e−

logwlogX |+ |1− e−i logw(t−τj)|

)lj� sin (logw(t− τj))lj �

(logw

(logX)1/R

)lj� (logX)−

2(L−1)3R

� (logX)−A.

The lemma follows.

Proposition 6.4.4. Let lj ≥ L − 1 for all 1 ≤ j ≤ R and that 0 ≤ logw ≤(logX)1/3R. For x ≤ X we have Ol(x,w)� x(logX)2ε/3

Proof. Since S(x)� x(log x)A by lemma 6.2.3 we may assume that log x ≥ (logX)ε/2A,and in particular that x ≥ w2RL is large(otherwise (logX)ε/2A ≤ log x ≤ 2RL(logX)1/3R

i.e., log x � (logX)ε). By Perron’s formula we have for c > 1,

1

2πi

∫(c)

F (s)zs

(es/√T − 1

s/√T

)ds

s=∑n≤z

f(n) +O

∑z<n≤ze1/

√T

|f(n)|

.

85

By Lemma 6.2.3, the error term above is� z(e1/√T − 1)1/2(log z)A

2/2 � z

log z.

Ol(x,w) =1

2πi

∫(c)

F (s)xs∏j

(1− w1+iτj−s)lj

(es/√T − 1

s/√T

)ds

s+O

(x

log x

).

Choose c = 1 + 1/ logX and split the integral into two parts: when |Im(s)| ≤ T andwhen |Im(s)| > T . For the first, we use Lemma 6.4.3 to get

� x(logX)ε/2∫|Im(s)≤T |

|ds||s|� x(logX)2ε/3

using the fact that T = exp((log logX)2). For the second region useF (s)� (logX)A

to get

� x(logX)A∫|Im(s)|>T

√T

|s|2|ds| � x.

6.5 Proof of the main theoremLet τ1, τ2, . . . , τR be the successive maxima of the previous section. Recall that R =

[10A2

ε2] + 1, that L = [10AR] L denotes the vector (L,L, . . . , L) and 0 ≤ logw ≤

(logX)1/R, x ≤ X .

Lemma 6.5.1.

(log x)OL(x,w) =∑d≤x

Λf (d)OL(x/d,w) +O (x(logX)ε) (6.42)

Proof. Write log x = log(x/wj1+...+jR) + (j1 + . . . + jR) logw. Hence we mayexpress (log x)OL(x,w) as

∑j≤L

(−1)j1+...+jR

(L

j

)log(x/wj1+...+jR)S(x/wj1+...+jR)wj1(1+iτ1)+...+jR(1+iτR)

+ logw∑j≤L

(−1)j1+...+jR

(L

j

)log(x/wj1+...+jR)wj1(1+iτ1)+...+jR(1+iτR)

Using the fact that(L

k

)=L

k

(L− 1

k − 1

)we see that the second term is nothing but

−LR∑k=1

w1+iτk logwOL−ek(x/w,w), where ek = (0, . . . , 1, 0, . . . , 0 with the 1 at the

86

kth place. Since the coordinates of L − ek are all at least L − 1, by Proposition 6.4.4we have that the second term is� x(logX)ε. For the first term,

(log x)S(x) =∑n≤x

f(n) log n+∑n≤x

f(n) log x/n.

Since∑d|n

Λf (d)f(n/d) = Λf ? f , L(Λ ? f) = L(Λf )L(f) =F ′

F.F = F ′ =

∑ f(n) log n

ns. Hence,∑n≤x

f(n) log n =∑n≤x

∑d|n

Λf (d)f(n/d).

(log x)S(x) =∑n≤x

∑d|n

Λf (d)f(n/d) +

∫ x

1

S(t)

tdt

=∑d≤x

Λf (d)S(x/d) +

∫ x

1

S(t)

tdt.

∴ the first term is ∑d≤x

OL(x/d,w) +

∫ x

1

OL(t, w)dt

t

Again using Propositon 6.4.4 we see that the integral is� x(logX)ε. Hence thelemma is proved.

Now the idea is to estimate Λf (d)OL(x/d,w) and show that it is � x(logX)ε,and Theorem would follow.

Lemma 6.5.2. For 1 ≤ z ≤ y with y + z ≤ X we have

| |OL(y, w)|2 − |OL(y + z, w)|2 |� y(logX)ε∑j≤LR

wj∑

y

wj≤n≤ y+z

wj

|f(n)| (6.43)

Proof. The quantity we wish to estimate is

�(|OL(y, w)|+ |OL(y + z, w)|

) (OL(y + z, w)−OL(y, w)

).

By Proposition 6.4.4 the first factor is� y(logX)ε, since z ≤ y. The second factor is

�∑j≤LR

wj |S(y + z

wj)− S(

y

wj)| �

∑j≤LR

wj∑

y

wj<n≤ y+z

wj

|f(n)|

and the proof of the lemma is complete.

Proposition 6.5.3.

log x|OL(x,w)| � x(log log x)1/2

(∫ x

1

log(ey)|OL(y, w)|2 dyy3

)1/2

+ x(logX)ε.

(6.44)

87

Proof. By Lemma 6.5.1 we must estimate∑d≤x

OL(x/d,w) +

∫ x

1

OL(t, w)dt

t. We

split the sum into two parts, d ≤ D = [exp(log logX)6] and d > D. For the firstrange we use Proposition 6.4.4 and obtain that the contribution is∑

d≤D

|Λf (d)|xd

(logX)2ε/3 � x(logX)2ε/3∑d≤D

|Λf (d)|d

.

But∑d≤D

|Λf (d)|d

≤

∑d≤D

|λf (d)|2Λ(d)

d

1/2∑d≤D

Λ(d)

d

1/2

� logD

by the weak-Ramanujan hypothesis. Hence∑d≤D

|Λf (d)|xd

(logX)2ε/3 � x(logX)ε

Define the function g(t) = t log(ex/t) for 1 ≤ t ≤ x. By Cauchy-Schwarz,

∑D<d≤x

|Λf (d)||OL(x/d,w)| ≤

∑D<d≤x

|λf (d)|2Λ(d)

g(d)

1/2 ∑D<d≤x

g(d)|Λ(d)||OL(x/d,w)|

1/2

� (log log x)1/2

∑D<d≤x

g(d)Λ(d)|OL(x/d,w)|21/2

.

Hence we must show that the last sum is(∫ x

1

log(ey)|OL(y, w)|2 dyy3

). Define the functionψ0(x) =

∑n≤x

(Λ(n)−1) = ψ(x)−

x, so that ψ0(x)� xe−c√

log x. Then∑D<d≤x

g(d)Λ(d)|OL(x/d,w)|2 =∑

D<d≤x

g(d)|OL(y, w)|2

+∑

D<d≤x

(ψ0(d)− ψ0(d− 1))g(d)|OL(x/d,w)|2

(6.45)

We rewrite the second term above as∑D<d≤x

ψ0(d)(g(d)|OL(x/d,w)|2 − g(d+ 1)|OL(x/d+ 1, w)|2

)−ψ0(D)g(D + 1)|OL(x/D + 1, w)|2.

(6.46)

88

Since d > D, ψ0(d) = O(d exp(−(log log x)2) = O

(d

(logX)A+2

). Therefore,

ψ0(D)g(D + 1)|OL(x/D + 1, w)|2 � x2

(D + 1)2(logX)ε(de−c

√log d)d log(ex/d)

� x2

On adding and subtracting g(d)|OL(x/d+ 1, w)|2 the first term in equation (6.46)is

�∑

D<d≤x

d

(logX)A+2

(g(d)(|OL(x/d,w)|2 − |OL(x/d+ 1, w)|2) + |g(d+ 1)− g(d)||OL(x/d+ 1, w)|2

)

∑D<d≤x

d

(logX)A+2|g(d+ 1)− g(d)||OL(x/d+ 1, w)|2

�∑

D<d≤x

d

(logX)A+2|g(d+ 1)− g(d)| x2

(d+ 1)2(logX)

ε

�∑

D<d≤x

d2

(d+ 1)2log(ex/d)

(logX)ε

(logX)A+2

� x2.

By Lemma 6.5.2 the first term is

�∑

D<d≤x

d2

(logX)Ax

d

LR∑j=0

wj∑

x

(d+1)wj<n≤ x

dwj

|f(n)|

Interchanging the d and n sums this is,

� x

(logX)A

LR∑j=0

wj∑

n≤x/wj|f(n)|d� x

(logX)A

LR∑j=0

wj∑

n≤x/wj|f(n)| x

nwj

� x2

(logX)A

LR∑j=0

∑n≤x/wj

|f(n)|n� x2

(logX)A(logX)A � x2,

as d ≤ x/nwj .We now estimate the first term in equation (6.45). For D ≤ d ≤ x, and d − 1 ≤

t ≤ d, g(d) = g(t) +O(log x), and by Lemma 6.5.2

|OL(x/d,w)|2 = |OL(x/t, w)|2 +O

xd

(logX)εLR∑j=0

wj∑

x

dwj<n≤ x

(d−1)wj

|f(n)|

.

89

By Proposition 6.4.4 and the fact that∫ d

d−1

dt = 1 we therefore get

g(d)|OL(x/d,w)|2 =

∫ d

d−1

g(t)|OL(x/t, w)|2 +O

(x2

d2(logX)1+2ε

)

+O

x(logX)1+εLR∑j=0

wj∑

x

dwj<n≤ x

(d−1)wj

|f(n)|

.

By summing over D < d ≤ x and changing variables we get∫ x

D

t log(ex/t)|OL(x/t, w)|2 dt ≤ x2

∫ x

1

|OL(y, w)|2 log(ey)dy

y3.

Now

x2∑

D<d≤x

(logX)1+2ε

d2� x2

D(logX)1+2ε � x2.

x(logX)1+ε∑

D<d≤x

LR∑j=0

wj∑

x

dwj<n≤ x

(d−1)wj

|f(n)| � x(logX)1+εLR∑j=0

wj∑

n≤ x

Dwj

|f(n)|.

� x2

D(logX)1+ε

LR∑j=0

∑n≤ x

Dwj

|f(n)|n� x2.

by Lemma 6.2.2. And this completes the proof of the proposition.

We must analyse the integral in the previous proposition. We therefore study thefunctions

S =∑n≤x

f(n) log n

Ol(x,w) =∑j≤l

(−1)j1+...+jR

(l

j

)wj1(1+iτ1)+...+jR(1+iτR)S(x/wj1+...+jR)

Lemma 6.5.4.(∫ x

1

|Ol(t, w)|2 log(et)dt

t3

)1/2

�(∫ x

1

|Ol(t, w)|2 dt

t3 log(et)

)1/2

+ (logX)7ε/8

(6.47)

Proof. Starting as in the proof of Lemma 6.5.1,

log t(Ol(t, w)

)= −

R∑k=1

Lw1+iτk(logw)OL−ek(t/w,w)

+∑j≤L

(−1)j1+...+jR

(L

j

)log(t/wj1+...+jR

)S(t/wj1+...+jR

)wj1(1+iτ1)+...+jR(1+iτR)

90

Now,

(log z)S(z) = S(z) +∑n≤z

f(n) log(z/n) = S(z) +

∫ z

1

S(y)

ydy.

The last term above is

OL(t, w) +

∫ t

1

OL(t, w)

ydy

Hence,

(log t)OL(t, w) = OL(t, w) +

∫ t

1

OL(y, w)

ydy +O

((logX)2ε/3

)= OL(t, w) +O

(t(1 + logw)(logX)2ε/3

)= OL(t, w) +O

(t(logX)5ε/6

).

∴ (log(et))2|OL(t, w)|2 � |OL(t, w)|2 +O(t2(logX)5ε/3

).And so(∫ x

1

|OL(t, w)| log(et)

t3dt

)1/2

�(∫ x

1

|OL(t, w)|2 dt

(log et)t3

)1/2

+O(

(logX)7ε/8).

By Proposition 6.5.3 and Lemma 6.5.4,

|OL(x,w)| � x

log x(log log x)1/2

(∫ x

1

|OL(t, w)|2 dt

t3 log(et)

)1/2

+x

log x(logX)ε.

Theorem 6.2.1 will follow if we can prove that the integral is� (logX)ε.

Proposition 6.5.5. ∫ x

1

|OL(t, w)|2 dt

t3 log(et)� (logX)2ε/3 (6.48)

Proof. Set t = ey .∫ x

1

|OL(t, w)|2 dt

t3 log(et)=

∫ log x

0

|OL(ey, w)|2e−2y dy

1 + y

=

∫ 1/ log x

0

+

∫ ∞1/ log x

e−2α

(∫ log x

0

|OL(ey, w)|2e−2y(1+α) dy

)dα

Since∫ log x

1

e−2α(1+y) � 1

1 + y, we get∫ x

1

|OL(t, w)|2 dt

t3 log(et)�∫ ∞

1/ log x

e−2α

(∫ ∞0

|OL(t, w)|2e−2y(1+α) dy

)dα.

(6.49)

91

We now use Plancherel’s formula to evaluate the integral over y in equation (6.49).Note that the Fourier transform of OL(ey, w)e−y(1+α) is

∫ ∞−∞

OL(ey, w)e−y(1+α+it) dy =∑j≤L

(−1)j1+...+jR

(l

j

)wj1(1+iτ1)+...+jR(1+iτR)S(x/wj1+...+jR)

×∫ ∞−∞

∑n≤ey/wj1+...+jR

f(n) log ne−y(1+α+it) dy.

(6.50)

Now,∫ ∞−∞

∑n≤ey/wj1+...+jR

f(n) log ne−y(1+α+it) =

∫ ∞0

∞∑n=1

f(n) log ne−y(1+α+it)

=

∞∑n=1

f(n) log n

∫ ∞logn+(j1+...+jR) logw

e−y(1+α+it) dy

= − 1

1 + α+ it

∞∑n=1

f(n) log n

n1+α+it.

Hence the integral in equation (6.50) is

=∑j≤L

(−1)j1+...+jR

(L

j

)wj1(1+iτ1)+...+jR(1+iτR)S(x/wj1+...+jR)

×− 1

1 + α+ it

∞∑n=1

f(n) log n

n1+α+it

− 1

1 + α+ it

R∏k=1

(1− w−α−it+iτk)F ′(1 + α+ it).

By Plancherel’s formula,∫ ∞0

|OL(ey, w)|2e−y(2+2α) �∫ ∞−∞|F ′(1 + α+ it)|2

R∏k=1

|1− w−α−it+iτk |2L dt

|1 + α+ it|2.

(6.51)

We split the integral in equation (6.50) into two parts, |t| ≤ T/2 and |t| > T/2.For |t| > T/2,∫|t|>T/2

|F ′(1 + α+ it)|2R∏k=1

|1− w−α−it+iτk |2L dt

|1 + α+ it|2

� (logX)2A+2

∫|t|>T/2

dt

|1 + α+ it|2� 1

(6.52)

92

using Lemma 6.2.2 to bound F ′(1 + α+ it) since α ≥ 1/ logX .For the region |t| ≤ T/2, using Lemmas 6.2.2 and 6.4.3 we obtain that

|F ′(1 + α+ it)|R∏k=1

|1− w−α−it+iτk |2 �| F′

F(1 + α+ it) || FL(1 + α+ it) |

�| F′

F(1 + α+ it) | (logX)ε/2

Hence,∫|t|≤T/2

� (logX)ε∫|t|≤T/2

| F′

F(1 + α+ it) |2 dt

|1 + α+ it|2

� (logX)ε∫ ∞−∞| F′

F(1 + α+ it) |2 dt

|1 + α+ it|2

(6.53)

THe Fourier transform of the function e−y(1+α)∑n≤ey

Λf (n) is

∫ ∞−∞

∑n≤ey

Λf (n)e−y(1+α+it) dy =∑n

Λf (n)

∫ ∞logn

e−y(1+α+it) dt

=1

1 + α+ it

∑n

Λf (n)

n1+α+it= − 1

1 + α+ it

F ′

F(1 + α+ it)

Therefore by Plancherel, the integral in equation (6.53) is

� (logX)ε∫ ∞

0

|∑n

Λf (n)|2e−(2+2α)y dy

We conclude that∫ ∞0

|OL(ey, w)|2e−y(2+2α) dy � (1 + (logX)ε)

∫ ∞0

|∑n

Λf (n)|2e−2y(1+α) dy.

(6.54)

Substituting equation (6.54) in equation (6.49),∫ ∞0

|OL(ey, w)|2 dt

t3 log(et)� (1+(logX)ε)

∫ ∞1/ logX

e−2α

∫ ∞0

|∑n≤ey

Λf (n)|2e−2y(1+α) dy

93

Expanding,∫ ∞1/ logX

e−2α

∫ ∞0

|∑n≤ey

Λf (n)|2e−2y(1+α) dy

=∑ni≤ey

Λf (n1)Λf (n2)

∫ ∞1/ logX

∫ ∞0

e−2y(2+α) dy dα

�∑

n1≤n2≤ey|Λf (n1)Λf (n2)|

∫ ∞1/ logX

∫ ∞0

e−2y(2+α) dy dα; But then y ≥ log n2. So,

�∑n1≤n2

|Λf (n1)Λf (n2)|∫

1/ logX

∫ ∞0

e−2α(1+y+logn2)e−2y dy dα

�∑n1≤n2

|Λf (n1)Λf (n2)|n

2+2/ logX2

∫ ∞0

e−2y/ logX−2y

1 + y + log n2dy

�∑n1≤n2

|Λf (n1)Λf (n2)|n

2+2/ logX2 log n2

=∑n1≤n2

|λf (n1)λf (n2)Λ(n1)Λ(n2)

n2+2/ logX2 log n2

�∑n1≤n2

(|λf (n1)|2 + |λf (n2)|2

) Λ(n1)Λ(n2)

n2+2/ logX2 log n2

.

By the prime number theorem, this is

�∑n≥2

|λf (n)|2λ(n)

n1+2/ logX log n� log logX

and the last step is by Lemma 6.2.2. This complete the proof of Theorem 6.2.1.

94

Chapter 7

Applications of Subconvexity

In this chapter we illustrate some applications of subconvexity. We do not give com-plete proofs; the idea is to demonstrate the power of getting subconvex bounds, espe-cially in proving equidistribution results.

7.1 Least Quadratic Non-residueIn this section we study the very basic problem of studying smoothed sums of the form

S(X, f) =∑n

λf (n)V (n/X),

where V (x) is a smooth, rapidly decreasing compactly supported function. Our aim isto improve on the trivial estimate

S(X, f)� (Q(f)X)εX,

which follows from Rankin’s trick. We have by the inverse Mellin transform,

S(X, f) =

∫(3)

V (s)L(s, f)Xs ds

s(7.1)

where V (s) is the Mellin transform of V . Shifting the contour to Re(s) = 1/2(assumethat there are no poles at s = 1), and integrating V by parts several times to gainconvergence in the integral in equation (7.1), we get

S(X, f)�V,A X1/2supRe(s)=1/2

|L(s, f)||s|A

�V,A,ε X1/2supRe(s)=1/2

Q(f, t)1/4−δ+ε

|s|A�V,A,ε X

1/2Q(f)1/4−δ+ε

(7.2)

95

This is non-trivial for X � Q(f)1/2−δ and even for δ = 0 this is an improvement onthe trivial bound. The Lindelof hypothesis would of course give

S(X, f)�V,A,ε Q(f)εX1/2,

giving cancellation for sums with X � Q(f)ε.For instance, when λf (n) = χ(n), where χ is a real quadratic character to the

modulus q(assume χ is primitive and that q is squarefree for convenience) such sumshelp in estimating the size of the least quadratic non-residue, denoted nχ. Notice thatif ∑

1≤n≤x

χ(n) = o(x) (7.3)

then clearly nχ � x. By Polya-Vinogradov we have for all M,N

S(M,N) =∑

M≤n≤M+N

χ(n)� q1/2 log q, (7.4)

which immediately gives us the bound that nχ �ε p1/2+ε. By Burgess’s subcon-

vexity bound(see Section 4.2) we have, S(M,N) � N1−1/rqr+1

4r2+ε, which gives

nχ �ε q1/4+ε. To see this, observe by taking r very large,

r + 1

4r2∼ 1/4r. So we get

non-trivial bounds for S(M,N) whenever N � q1/4+1/2r. And using Vinogradov’ssieving trick, we can improve this to nχ �ε q

14√e

+ε.

7.2 Equidistribution of Lattice Points on the SphereThe material in this section is from Section 5.3.1 of [10]. Given n ≥ 1, Gauss provedthat n is of the form x2 + y2 + z2 if and only if n is not of the form 4k(8l− 7). Denoteby R3(n) = {x = (x, y, z) ∈ Z3 | x2 + y2 + z2 = n} (resp. R∗3(n) = {x =(x, y, z) ∈ Z3 | x2 + y2 + z2 = n, (x, y, z) = 1}) the (resp. primitive) representationsof n. Denote by r3(n) and r∗3(n) the number of representations in either case. Thenr3(n) =

∑d2|n

r∗3(n/d2). We have a formula for r∗3(n) in terms of h(−n), namely

r∗3(n) =

12h(−n) if n ≡ 1, 2(mod 4);8h(−n) if n ≡ 3(mod 4);0 otherwise.

By the class number formula, it follows that

r∗3(n)� n1/2−ε

so long as r∗3(n) > 0. Hence, for the unrestricted representations one has

r3(n)� n1/2−ε if n 6≡ 0, 4, 7 (mod 8) (7.5)

96

Thus if n 6≡ 0, 4, 7(mod 8), there are many vectors in R3(n); one may then lookat the distribution of their projections on the unit sphere as n → ∞. Linnik, using er-godic methods proved equidistribution under the condition that−n is quadratic residuemodulo some fixed odd prime. We will now give a rough outline of the proof of thisstatement using subconvexity.

Theorem 7.2.1. As n→∞, through the integers n 6≡ 0, 4, 7(mod 8), the setR3(n)/n1/2

becomes equidistributed on the unit sphere with respect to the standard Lebesgue mea-sure. We have for any continuous function on V on S2,

1

r3(n)

∑x∈R3(n)

V( x

n1/2

)→∫S2

V (u) dµ (7.6)

Proof. By Weyl’s equidistribution criterion, it is sufficient to show that for any har-monic polynomial P on R3 of degree k ≥ 1(the statement is obvious when k = 0), theWeyl sum

W (n, P ) =1

r3(n)

∑x∈R3(n)

P( x

n1/2

)→∫S2

P (u) dµ = 0

Observe that W (n, P ) = 0 if k is odd. If k is even, one has

W (n, P ) =n−k/2

r3(n)

∑x∈R3(n)

P (x) =n−k/2

r3(n)rP (n) (7.7)

In view of equation (7.7), it is sufficient to show that

rP (n) = O(nk+12 −δ)

for some δ > 0 fixed. The theta series

ΘP (z) =∑n≥0

rP (n)e(nz)

is a holomorphic modular form of weight l = k + 3/2, level 4 and a cusp form whenk ≥ 1(see [7]). For a cusp form f of half-integral weight l of some level q = 0(mod 4)with Fourier expansion

f(z) =∑n≥1

ρf (n)nl/2e(nz)

one has the following bound

ρf (n)�f,ε n−1/4+ε.

This bound yields rP (n)�P,ε nk+12 +ε which is barely insufficient to prove equidistri-

bution. This is achieved through the theory work of Shimura and Waldspurger. SinceΘP (z) can be written as a sum of primitive forms, we may assume without loss ofgenerality that f is a primitive form of weight l ≥ 5/2. Shimura proved that to f we

97

may associate a holomorphic cusp form of weight 2l − 1, g, say. Then for squarefreen and χ(m) =

( nm

)1, Waldspurger’s theorem states that

|ρf (n)|2nl = C(f, g, n)nl−1L(1/2, g × χ). (7.8)

for some proportionality constant C(f, g, n) which is bounded independently of n. ByTheorem 5.1.1, we get for squarefree n2

|ρf (n)| � n−1/4−1/44+ε. (7.9)

If n is not squarefree, we have by the Shimura lift that,

ρf (nt2) = ρf (n)∑b|t

µ(b)

b1/2ψ(b)λg(t/b), (7.10)

where

ψ(p) =

(−1

p

)l−1/2(n

p

).

Therefore, by Deligne’s bound for λg(n), we get (7.9) for non-squarefree n also. Ap-plying this bound to ΘP (z), we get

rP (n)�P,ε nk+12 −

144 ,

and this solves the equidistribution problem.

As an aside, notice that by the Waldspurger formula, the Lindelof Hypothesis forL(s, g × χ) becomes equivalent to the Ramanujan-Petersson conjecture

ρf (n)�ε n−1/2+ε.

7.3 Mass Equidistribution of Hecke EigenformsTheorem 7.3.1. Consider a holomorphic eigencuspform f ∈ Sk(SL2(Z)). Considerthe measure µf

µf = yk|f(z)|2 dxdyy2

and suppose that f is a normalised. That is,∫Γ\H

yk|f(z)|2 dxdyy2

= 1

1If n 6= 0, the map m → (n

m) is a character modulo a divisor of 4n,whose conductor is the conductor

of Q(√n) over Q

2In Thoerem 5.1.1 we assumed that g to be a cusp form for SL2(Z), but the proof is essentially the samefor any primitive cusp form in Sk(q

′, χ′) See [10].

98

Then for any h ∈ L2(Γ\H)∫Γ\H

h(z) dµf →∫

Γ\Hh(z)

dxdy

y2(7.11)

as k →∞.

A nice application of the above theorem is the following result due to Z Rudnick,On the Asymptotic Distribution of Zeros of Modular Forms, IMRN 34 (2005),

Theorem 7.3.2. Let fk be a sequence of Hecke cuspforms for SL2(Z), then as k →∞their zeros are equidistributed in SL2(Z)\H with respect to the normalised hyperbolic

volume measuredxdy

y2.

A key ingredient for proving the above theorem is weak subconvexity forL-functions.We will now sketch that part of the proof of the QUE conjecture, due to Soundararajan.

Let Fk(z) = yk/2f(z). Let X = Γ\H. Denote by <,> the inner product offunctions g1, g2 on X

< g1, g2 >=

∫X

f(z)dxdy

y2

By Theorem 2.1.1, it suffices to prove the above theorem for the unitary Eisensteinseries and Maass cusp forms, i.e.,

< φFk, Fk >→ 0,

< E(1/2 + it)Fk, Fk >→ 0,

as k →∞, where φ is a fixed Maass cusp form and t is also fixed. It would then followthat the only contribution to the integral on the LHS of equation (7.11) comes fromthe constant term in the spectral expansion for L2 proving Theorem 7.3.1. These innerproducts are nothing but,

| < φFk, Fk >2 | =

1

8

L∞(1/2, f × f × φ)L(1/2, f × f × φ)

Λ(1, sym2f)2Λ(1, sym2φ)

< E(1/2 + it)Fk, Fk > = | π3/2 ζ(1/2 + it)L(1/2 + it, sym2f)

ζ(1 + 2it)L(1, sym2f)

Γ(k − 1/2 + it)

Γ(k)| .

Since |Γ(k−1/2+ it)| ≤ Γ(k−1/2), |ζ(1/2+ it)� (1+ |t|)1/4, and |ζ(1+2it)| �1/log(1 + |t|), L(1, sym2f)� (log k)−1 we get

< E(1/2 + it)Fk, Fk > � | L(1/2 + it, sym2f) | log kΓ(k − 1/2)

Γ(k)log(1 + |t|)(1 + |t|)1/4,

|< φFk, Fk >|2 � (log k)2L(1/2, f × f × φ).

Hence it is clear that getting subconvex bounds for L(s, f × f × φ) and L(s, sym2f)would prove Theorem 7.3.1, and we would get

|< E(1/2 + it)Fk, Fk >| � k−δ1(log k) log(1 + |t|)(1 + |t|)2, (7.12)|< φFk, Fk >| � k−δ2(log k) (7.13)

99

for some positive numbers δ1 and δ2.However, we do not have subconvex bounds for L(s, f × f ×φ) and L(s, sym2f).

We only have weakly subconvex bounds by Theorem 6.0.9 and sections 1.2.6 and 1.2.7. We would then obtain,

| < E(1/2 + it)Fk, Fk > | �ε(1 + |t|)2

(log k)1−εL(1, sym2f)(7.14)

| < φFk, Fk > | �ε1

(log k)1/2−εL(1, sym2f)(7.15)

Given δ > 0, equation (7.15) shows that if f ranges over those Hecke eigencusp-forms such that L(1, sym2f) ≥ (log k)−1/2+δ , then as k → ∞, the measure µf

converges to3

π

dxdy

y2. This condition holds for all Hecke eigencuspforms under the

Riemann hypothesis for L(s, sym2f). And using sieve estimates one can show thatthe number of exceptional eigenforms with weight k ≤ K for which the criterion failsis � Kε. Hence, barring sequences that contain such exceptional eigencuspforms,theorem 7.3.1 holds. The criterion L(1, sym2f) ≥ (log k)−1/2+δ is however comple-mented by the work of Holowinsky who proves Theorem 7.3.1 for these exceptionaleigencuspforms, thus proving QUE.

100

Appendix A

Complex Analysis

A.1 Phragmen-Lindelof PrincipleTheorem A.1.1. Let f be a function holomorphic on an open neighbourhood of astrip a ≤ σ ≤ b, for some real numbers a < b, such that |f(s)| � exp(|s|A) for someA ≥ 0 and a ≤ σ ≤ b.

1. Assume that |f(s)| ≤M for all s on the boundary of the strip, i.e., for σ = a orσ = b. Then we have |f(s)| ≤M for all s in the strip.

2. Assume that

|f(a+ it)| ≤ Ma(1 + |t|)α,|f(b+ it)| ≤ Mb(1 + |t|)β

for t ∈ R. Then

|f(σ + it)| ≤M l(σ)a M

1−l(σ)b (1 + |t|)αl(σ)+β(1−l(σ)) (A.1)

for all s in the strip, where l is the linear function such that l(a) = 1, l(b) = 0.

A.2 Stirling’s FormulaThe Stirling asymptotic formula

Γ(s) =

(2π

s

)1/2 (se

)s(1 +O

(1

|s|

))(A.2)

is valid in the angle |args| ≤ π − ε with the implied constant depending on ε. Hencefor s = σ + it, t 6= 0, σ fixed we have

Γ(σ + it) =√

2π(it)σ−1/2e−π2 |t|(|t|e

)it(1 +O

(1

|t|

)). (A.3)

101

A.3 Bounds for the Analytic ConductorLet

Q∞(f, s) =

m∏j=1

(|t+ µj |+ 3) (A.4)

let −1/2 ≤ Re(s) ≤ 2. From A.3 we deduce

|γf (s)|∏j

|s+ µj | = Q∞(f, s)1/2(k+σ+1)exp

−π4

∑j

|t+ Im(µj)|+O(d)

(A.5)

where k = Re∑

µj . Hence we have

γf (1− s)γf (s)

� Q∞(f, s)(1/2−σ)∏j

s+ µj1− s+ µj

. (A.6)

A.4 Functions of Finite OrderA entire function f is said to be of finite order if there exists β > 0 such that

|f(s)| � exp(|s|β).

f is said to be of order ≤ 1 if we can take any β < 1. If no β < 1 is possible, f is saidto be of order 1.

102

Appendix B

Bessel Functions

Let Jν(z), Yν(z) and K0(z) be the standard Bessel functions.

B.1 Recurrence Formulae

(zνJν(z))′ = zνJν−1(z)

(zνYν(z))′ = zνYν−1(z)(B.1)

B.2 Bounds for Bessel FunctionsWe also record some of the estimates for the Bessel functions that are used in the thesis.For ν > 0,

Jν(z)� min{z−1/2, zν}Yν(z)� min{z−1/2, z−ν + zν | log z|}K0(z)� min{e−zz−1/2, 1 + | log z|}.

(B.2)

B.3 Definite Integrals Involving Bessel FunctionsWe now record the formulas that are used in the thesis here.

First are the orthogonal formulas∫ ∞0

Y0(z)Jk−1(z)z−1 dz = 0 (B.3)

if k is even, and ∫ ∞0

Jk−1(za) cos(bz)z−1 dz = 0 (B.4)

103

if k is even and b ≥ a > 0. We have the following formulas for any a > 0 involvingthe functions C(z) and S(z) defined in Chapter 5.∫ ∞

0

C(√az) cos z dz = 2π cos a, (B.5)∫ ∞

0

S(√az) sin z dz = 2π sin a, (B.6)∫ ∞

0

C(√az)(log z) cos z dz = 2π(log a) cos a, (B.7)∫ ∞

0

S(√az)(log z) sin z dz = 2π(log a) sin a. (B.8)

104

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