exponential sums with multiplicative coefficientsandrew/pdfpre/circlemethod.pdf · exponential sums...

EXPONENTIAL SUMS WITH MULTIPLICATIVE COEFFICIENTS

Andrew Granville and K. Soundararajan

Abstract. We show that if an exponential sum with multiplicative coefficients is large then

the associated multiplicative function is “pretentious”. This leads to applications in the circlemethod.

1. Introduction

Throughout f is a totally multiplicative function with |f(n)| ≤ 1 for all n. Define

S(α,N) =∑

n≤Nf(n)e(nα)

where e(t) = e2iπt.Montgomery and Vaughan showed that

(1.1) S(α, x) ≪ x

log x+

x√R

(logR)3/2

uniformly in f , if |α − a/q| ≤ q−2 with (a, q) = 1 and 2 ≤ R ≤ q ≤ x/R. This is, in asense, sharp, aside from the factor (logR)3/2. Bachmann conjectured [B2] that

(1.2) S(α, x) ≪ x

log x+

1√R

∑

n≤x|f(n)|,

uniformly in f . Towards this end, he showed that

(1.3) S(α, x) ≪ x

log x+ x

(logR)1/2+o(1)√R

exp

−∑

p≤x

1 − |f(p)|p

.

Along these lines we prove the following. Select a/q with (a, q) = 1 so that |α−a/q| ≤ 1/qQwhere Q = x/((logx)1/6(log log x)). The proof of Montgomery and Vaughan gives

(1.4) S(α, x) ≪ xlog log x

(log x)1/12+ x

(log q)3/2√q

Le premier auteur est partiellement soutenu par une bourse de la Conseil de recherches en sciences

naturelles et en genie du Canada. The second author is partially supported by the National ScienceFoundation and the American Institute of Mathematics (AIM).

Typeset by AMS-TEX

1

2 ANDREW GRANVILLE AND K. SOUNDARARAJAN

(see (4.3) and (4.4) below). Along the lines of Bachmann’s conjecture, (3.1) below gives

(1.5) S(a/q, x) ≪√qx

(log x)1/4+

1√q

q

φ(q)

∑

n≤x|f(n)|.

However this type of upper bound is not of much use when |f(n)| = 1 for all n. It followsnaturally from our work below to proceed as follows: Define a distance function D betweentwo multiplicative functions f and g with |f(n)|, |g(n)| ≤ 1:1

Dr(f(n), g(n); x)2 :=∑

p≤xp∤r

1 − Re f(p)g(p)

p.

Let x ≥ Q,A ≥ 1 are given. Of all primitive characters with conductor below Q, let ψ(mod r) be that character for which

(1.6) min|t|≤A

D(f(n), ψ(n)nit; x)

is a minimum (where D = D1),2 and let t = t(x,Q,A) denote a value of t that gives the

minimum value in (1.6). Then

(1.7) S(a/q, x) ≪√qx

(log x)1/4+

1√q

q

φ(q)

∑

n≤xf(n)ψ(n)

and we also have, analogous to but improving (1.3),

(1.8) S(a/q, x) ≪√qx

(log x)1/4+

x√q

(

1 + D(f(n), ψ(n)nit; x)2)

e−Dr(f(n),ψ(n)nit;x)2 .

Finally, if q ≤ x/Q then

(1.9) S(α, x) ≪ x√q· log log x

(log x)1/12+

x√q

(

1 + D(f(n), ψ(n)nit; x)2)

e−Dr(f(n),ψ(n)nit;x)2 .

Let f1, f2, f3 be multiplicative functions with |fi(n)| ≤ 1. Select ψi, ti as in (1.6). Forany given integers c1 ≥ c2 > 0 > c3 with (c1, c2, c3) = 1, we will show that

(1.10)1

N2/2

∑

n1,n2,n3≤Nc1n1+c2n2+c3n3=0

f1(n1)f2(n2)f3(n3) = o(1)

1This distance function is particularly useful since it satisfies the triangle inequality D(f1(n), g1(n); x)+D(f2(n), g2(n); x) ≥ D((f1f2)(n), (g1g2)(n); x), (see Lemma 3.1 of [7], or the general discussion in [8]).

2If there are several possibilities for ψ, simply pick one of those choices.

EXPONENTIAL SUMS WITH MULTIPLICATIVE COEFFICIENTS 3

unless ψ1ψ2ψ3 is principal (i.e., f1f2f3 is 1-pretentious), and the cj and rj are bounded.In this case (1.10) equals o(1) plus

(1.11)1

N

∑

n≤N

f1(n)ψ1(n)

nit1· 1

N

∑

n≤N

f2(n)ψ2(n)

nit2· 1

N

∑

n≤N

f3(n)ψ3(n)

nit3

times

(1.12) g(ψ1)g(ψ2)g(ψ3) ·N i(t1+t2+t3) · I(

t1, t2, t3c1, c2, c3

)

·∏

p

E(p),

where g(ψ) is the Gauss sum, and

I

(

t1, t2, t3c1, c2, c3

)

:=2

|c1c2c3|

∫

Bvit11 vit22 vit33 d(c1v1)d(c2v2)

with B = B(c1, c2, c3) := {0 ≤ v1, v2, v3 ≤ 1 : c1v1 + c2v2 + c3v3 = 0}. The E(p) dependonly on the cj , ψj and fj(p)/p

itj . If p ∤ c1c2c3r1r2r3 then

E(p) = 1 +1

(p− 1)2

3∏

ℓ=1

(

fℓ(p)

ψℓ(p)pitℓ− 1

)

(

1 − 1

p2

3∏

ℓ=1

fℓ(p)

pitℓ

)−1

.

If p|c1c2c3r1r2r3 then E(p) is more complicated.Notice that the quantity in (1.11) tends to a limit as N → ∞ by Halasz’s theorem.In [10] we showed that if f is a totally multiplicative function taking on only the values

1 and −1 then there are at least (1 − δ0)N/2 solutions to f(n) = 1 with n ≤ N , and thatthis is best possible, where

δ0 = −1 + 2 log(1 +√e) − 4

∫

√e

1

log t

t+ 1dt = 0.656999 . . . .

If these solutions are “randomly-distributed with respect to addition” then we would expectat least (1 − δ0)

3N2/16 solutions to a + b = c with 1 ≤ a, b, c ≤ N and f(a) = f(b) =f(c) = 1.

Theorem 1. Let f be a totally multiplicative function taking on only the values 1 and

−1. There are at most κN2/2 solutions to a + b = c with 1 ≤ a, b, c ≤ N and f(a) =f(b) = f(c) = −1, and at least κ′N2/2 solutions to a + b = c with 1 ≤ a, b, c ≤ N and

f(a) = f(b) = f(c) = 1. Here κ = (1 + δ0)3/8 = .56869 . . . and κ′ = (1 − δ0)

3/16 =.005044 . . . . Finally if f, g, h are totally multiplicative functions taking on only the values

1 and −1 then there are at most κN2/2 solutions to a + b = c with 1 ≤ a, b, c ≤ N and

f(a) = g(b) = h(c) = −1.

Similarly

(1.13)1

N2/2

∑

n1,n2,n3≥1n1+n2+n3=N

f1(n1)f2(n2)f3(n3)


equals o(1) plus (1.11) times

(ψ1ψ2ψ3)(N)g(ψ1ψ2ψ3)g(ψ1)g(ψ2)g(ψ3)Ni(t1+t2+t3)I(t1, t2, t3) ·

∏

p: (p,N)=1

E ′(p)

where I(t1, t2, t3) := 2∫

v1,v2,v1+v2≤1vit11 vit22 (1−v1 −v2)it3dv1dv2 and the E ′(p) are compli-

cated. In particular there exists a constant c such that for N with only large prime factorsthe sum in (1.13) equals

c ψ1(N)N it1 · ψ2(N)N it2 · ψ3(N)N it3 + o(1).

2. Collecting technical results

Theorem 2.1 (Halasz). For any T ≥ 1 we have

(2.2)1

x

∑

n≤xf(n) ≪ max

|t|≤T

(

1 + D(f(n), nit; x)2)

e−D(f(n),nit;x)2 +1√T.

From the proof of Corollary 2.2 in [3] we have, for d, r ≤ √x and |t| ≤ T (≤ (log x)1/2),

(2.3)∑

n≤x/d(n,r)=1

f(n) =1

d1+it

∏

p|r

(

1 − f(p)

p1+it

)

∑

n≤xf(n) +O

(

r

φ(r)· xd

log log x

(log x)2−√

3

)

.

(Note that 2−√

3 > 1/4. Remark: It would be nice to improve (log x)1/2 to (log x)2/3 forthe application below.)

In the second-to-last displayed equation in the proof of Corollary 3 in [6] we showedthat

(2.4)∑

n≤xf(n) =

xit

1 + it

∑

n≤xf(n)n−it +O

(

log log x

(log x)2−√

3

)

.

In [3] we proved that f cannot pretend to be two different functions of the formψ(n)nit:

Corollary to Lemmas 3.1 and 3.3 of [3]. Fix T ≤ log2 x. For each primitive character

ψ with conductor below log x select tψ, |tψ| ≤ T for which D(f(n), ψ(n)nitψ ; x) is minimal,

and then label these pairs so that (ψj, tj) is that pair which gives the j-th smallest distance

D(f(n), ψj(n)nitj ; x). We have

D(f(n), ψ2(n)nit2 ; x)2 ≥{

1

3+ o(1)

}

log log x;

and

D(f(n), ψj(n)nitj ; x)2 ≥ log log x+O(√

log log x);

for all j ≥ log log x.


3. The proof

We assume that q ≤ logN .We begin with the identity, if (b, d) = 1 then

e

(

b

d

)

=

d−1∑

j=0

e

(

j

d

)

· 1

φ(d)

∑

χ (mod d)

χ(b)χ(j).

Therefore if (a, q) = 1 and if f is totally multiplicative then, writing n = mq/d when(n, q) = q/d,

S(a/q,N) =∑

d|q

∑

m≤N/(q/d)(m,d)=1

f(mq/d)e(am

d

)

=∑

d|q

f(q/d)

φ(d)

d−1∑

j=0

e

(

j

d

)

∑

m≤dN/q(m,d)=1

f(m)∑

χ (mod d)

χ(am)χ(j)

=∑

d|q

f(q/d)

φ(d)

∑

χ (mod d)

χ(a)g(χ)∑

m≤dN/q(m,d)=1

f(m)χ(m)

which g(χ) =∑d−1j=0 χ(j)e

(

jd

)

is the Gauss sum. Now suppose χ (mod d) was induced

from ψ (mod r) (where r|d) so that g(χ) = µ(d/r)ψ(d/r)g(ψ) (see Lemma 4.1 of [7]), sothat the last line equals

=∑

d|q

f(q/d)

φ(d)

∑

r|d

∑

ψ (mod r)ψ primitive

ψ(a)µ(d/r)ψ(d/r)g(ψ)∑

m≤dN/q(m,d)=1

f(m)ψ(m)

=∑

r|q

1

φ(r)

∑


ψ(a)g(ψ)∑

k|(q/r)(k,r)=1

f((q/r)/k)

φ(k)µ(k)ψ(k)

∑

m≤krN/q(m,kr)=1

f(m)ψ(m)

writing d = kr and noting that if ψ(k) 6= 0 then (k, r) = 1. We now apply (2.3), whereeach |tψ| ≤ T (≤ (logN)1/2), in the form

∑

n≤krN/q(n,k)=1

(fψ)(n) =1

(q/kr)1+itψ

∏

p|k

(

1 − (fψ)(p)

p1+itψ

)

∑

n≤N(fψ)(n)+O

(

k

φ(k)· krN

q

log logN

(logN)2−√

3

)

,

to obtain a main term of

∑

r|q

1

φ(r)

∑


ψ(a)g(ψ)∑

k|(q/r)(k,r)=1

f((q/r)/k)

(q/kr)1+itψµ(k)ψ(k)

φ(k)

∏

p|k

(

1 − (fψ)(p)

p1+itψ

)

∑

n≤N(fψ)(n)


and an error term which is

≪∑

r|q

1

φ(r)

∑


√r∑

k|(q/r)(k,r)=1

1

q/kr

µ(k)2k

φ(k)2N log logN

(logN)2−√

3≪ N ·

√q log logN

(logN)2−√

3.

We need to bound the contribution of the part of the main term that comes from theprimitive characters χj , j ≥ 2, so we take T = (logN)1/2 in the Corollary above and thenin (2.2), to get that

∑

n≤N(fψ)(n) ≪ N√

T.

Thus, in total, these characters contribute

≪∑

r|q

1

φ(r)

∑


√r∑

k|(q/r)(k,r)=1

1

(q/kr)

µ(k)2k

φ(k)2N√T

≪√q N√T

.

Finally we must consider the terms with ψ = ψ1. We have

ψ(a)g(ψ)∑

n≤N(fψ)(n)

times1

φ(r)

∑

k|(q/r)(k,r)=1

f((q/r)/k)

(q/kr)1+itψµ(k)ψ(k)

φ(k)

∏

p|k

(

1 − (fψ)(p)

p1+itψ

)

.

Write q = RS where (R, S) = 1 and R has exactly the same prime factors as r. Then thisbecomes, with t = tψ and s =

∏

p|S p,

1

r· r

φ(r)

f(R/r)

(R/r)1+it

∑

k|S

f(S/k)

(S/k)1+itµ(k)ψ(k)

φ(k)

∏

p|k

(

1 − (fψ)(p)

p1+it

)

=1

r· R

φ(R)

f(R/r)

(R/r)1+itS

φ(S)

1

S1+it

∏

pb‖Sf(p)b−1(f(p) − pitψ(p))

=1

φ(q)

f(q/rs)

(q/rs)it

∏

p|s

(

f(p)

pit− ψ(p)

)

=1

φ(q)

∏

pb‖(q/r)

(

(

f(p)

pit

)b

− ψ(p)

(

f(p)

pit

)b−1)

.

Collecting all of this together we have proved: If ψ, t are chosen as in (1.6) then

(3.1) S(a/q,N) = ψ(a)κ(q)∑

n≤N(fψ)(n) +O

( √qN

(logN)1/4

)


where κ(q) = κ(f(n)/nit, ψ, q) with κ(h, ψ, q) = 0 if r does not divide q, and

κ(h, ψ, q) =g(ψ)

φ(q)

∏

pb‖(q/r)(h(p)b − ψ(p)h(p)b−1)

if r does divide q. By (2.4) we can rewrite (3.1) as

(3.2) S(a/q,N) = ψ(a)κ(q)N it

1 + it

∑

n≤N

f(n)

ψ(n)nit+O

( √qN

(logN)1/4

)

.

3.1 Bounding S(a/q,N). We immediately have

|κ(q)| ≤√r

φ(q)2ω(q)−ω(r) ≪ 1√

q

r

φ(r),

which implies (1.5); and by Theorem 2.1

1

N

∑

n≤N(fψ)(n) ≪

(

1 + D(f(n), ψ(n)nit; x)2)

e−D(f(n),ψ(n)nit;x)2 +1

(logN)1/4.

Note then that

|κ(q)|e−D(f(n),ψ(n)nit;x)2 ≪ 1√qe−Dr(f(n),ψ(n)nit;x)2 .

4. Evaluating the exponential sum at irrationals

Theorem 1 of [9] states that if q ≤ N and (a, q) = 1 then

S(a/q,N) ≪ N

logN+

N

φ(q)1/2+ (qN)1/2(log(2N/q))3/2.

Fix Q. For any α there exists a/q with q ≤ Q and |α− a/q| ≤ 1/qQ. From the identity

(4.1) S(α,N) = e((α− a/q)N)S(a/q,N)− 2πi(α− a/q)

∫ N

1

e((α− a/q)v)S(a/q, v)dv

we immediately deduce that

(4.2) S(α,N) ≪ (1 + |α− a/q|N)

(

N

logN+

N

φ(q)1/2+ (qN)1/2(log(2N/q))3/2

)

.

since the contribution to the integral from values of v ≤ q is ≪ |α − a/q|∫ q

1vdv ≪

(1/qQ)q2 = q/Q ≤ 1.


Assume Q > 2√N . If q > 2N/Q then |α− a/q|N < N/qQ < 1/2, so (4.2) yields

(4.3) S(α,N) ≪ N

logN+N

(log(N/Q))3/2

(N/Q)1/2

If q ≤ 2N/Q <√N . Select r < 2N/q so that |α − b/r| ≤ q/(2rN). Either r = q in

which case

(4.4a) S(α,N) ≪ N

logN+

N

φ(q)1/2≪ N

logN+N

(log q)3/2

q1/2;

or 1/qr ≤ 1/qQ+ q/(2rN) ≤ 1/qQ+ 1/(2qr), so that r ≥ Q/2 ≥ q/2 and by (4.2) we get

(4.4b) S(α,N) ≪ N

logN+N

(log q)3/2

q1/2.

Now suppose that (3.1) holds in a range. By (4.1), S(α,N) equals ψ(a)κ(q) times

(4.5) e((α− a/q)N)∑

n≤N(fψ)(n) − 2πi(α− a/q)

∫ N

1

e((α− a/q)v)∑

n≤v(fψ)(n)dv

plus an error term which is

≪√qN

(logN)1/4+ |α− a/q|

∫ N

1

√qv

(log v)1/4dv ≪

√qN

(logN)1/4(1 +N |α− a/q|).

Using the bound |(fψ)(n)| ≤ 1 for v ≤√N , and (2.3) otherwise, we find that (4.5) equals

{

e((α− a/q)N) − 2πi(α− a/q)

∫ N

√N

e((α− a/q)v)

(N/v)1+itdv

}

∑

n≤N(fψ)(n)

+O

(

|α− a/q|(

∫

√N

1

vdv +

∫ N

√N

vlog logN

(logN)2−√

3dv

))

.

This error term is certainly smaller than the previous one. Extending the remainingintegral down to 0 gives more acceptable error term, and then integrating by parts andchanging variable we get, with β = α− a/q,

e(βN) − 2πiβ

∫ N

0

e(βv)

(N/v)1+itdv =

(1 + it)

N1+itI(N, β, t) where I(N, β, t) :=

∫ N

0

e(βv)vitdv.

Combining this with (2.4) we deduce that(4.6)

S(α,N) = ψ(a)κ(q)I(N,α− a/q, t)1

N

∑

n≤N

f(n)ψ(n)

nit+O

( √qN

(logN)1/4(1 +N |α− a/q|)

)

In section 4.2 we write this as S(α) = M(α) + E(α).


4.1. A tricky integral.

Let J(x, t) := (1 + it)∫ 1

0e(wx)witdw so that

J(βN, t) =(1 + it)

N1+it

∫ 1

0

e(βwN)(wN)itd(wN) =(1 + it)

N1+itI(N, β, t).

If |t| ≤ 2 then |J(x, t)| ≤ 3∫ 1

01dw = 3. Integrating by parts we have

J(x, t) = (1 + it)

∫ 1

0

e(wx)witdw = [e(wx)w1+it]10 − (2iπx)

∫ 1

0

e(wx)w1+itdw

= e(x) − 2iπx

2 + it[e(wx)w2+it]10 +

(2iπx)2

2 + it

∫ 1

0

e(wx)w2+itdw

= e(x)

1 +

k−1∑

j=1

(−2iπx)j∏j+1ℓ=2(ℓ+ it)

+(−2iπx)k∏kℓ=2(ℓ+ it)

∫ 1

0

e(wx)wk+itdw.

Therefore if x ≤ |t|/7 we can take k ≫ log(2 + x) to obtain J(x, t) = e(x)(1 +O(x/|t|).If x ≥ 1 then J(x, t)/(1 + it) equals

∫ 1/x

0

e(wx)witdw +1

2iπx

(

[

e(wx)wit]1

1/x− it

∫ 1

1/x

e(wx)wit−1dw

)

which, in absolute value, is

≤∫ 1/x

0

1dw +1

2πx

(

2 + |t|∫ 1

1/x

dw

w

)

≪ 1 + |t| logx

x.

Therefore |J(x, t)| ≤ (1 + |t|)(1 + |t| logx)/x.To obtain (1.9) we need to show that |J(x, t)| ≪ 1 when 3 ≤ |t| ≤ x ≤ |t|2 log |t|.4.2. Exponential sum estimates.

We write S(α) = M(α) + E(α) in (4.6). Now select Q = N/((logN)1/6(log logN)).Using (4.3) for q ≥ N/Q, and (4.6) for q ≤ N/Q so that its error term is

≪√qN

(logN)1/4(1 +

N

qQ) ≪

√qN

(logN)1/4+

N2

√qQ(logN)1/4

≪ N√

N/Q

(logN)1/4+

N(N/Q)

(logN)1/4,

we find that

(4.7) |E(α)| ≪ Nlog logN

(logN)1/12

for all α.


Note that∫ 1

0|S(α)|2 =

∑

n≤N |f(n)|2 ≤ N . Now, if M(α) 6= 0 then q ≤ N/Q, so by

(4.6),

|E(α)|2 ≪ qN2

(logN)1/2· N2

q2Q2=N2(N/Q)2

q(logN)1/2.

Hence

∫

α: M(α)6=0

|E(α)|2dα≪∑

q≤N/Q

∑

(a,q)=1

∫ aq+ 1

qQ

aq− 1

qQ

N2(N/Q)2

q(logN)1/2dα

≪∑

q≤N/Q

∑

(a,q)=1

N(N/Q)3

q2(logN)1/2≪ N(N/Q)3

(logN)1/2log(N/Q)

≪ N(log logN)4.(4.8)

From this we deduce

(4.9)

∫

α

|M(α)|2dα ≤ 2

∫

α: M(α)6=0

(|S(α)|2 + |E(α)|2)dα≪ N(log logN)4,

and

(4.10)

∫

α

|E(α)|2dα ≤ 2

∫

α

(|S(α)|2 + |M(α)|2)dα≪ N(log logN)4.

5. Circle method

∑

n1,n2,n3≤Nc1n1+c2n2+c3n3=0

f1(n1)f2(n2)f3(n3)

=∑

n1,n2,n3≤Nf1(n1)f2(n2)f3(n3)

∫ 1

0

e(t(c1n1 + c2n2 + c3n3))dt

=

∫ 1

0

3∏

ℓ=1

∑

nℓ≤Nfℓ(nℓ)e(cℓnℓt)dt

=

∫ 1

0

S1(c1t)S2(c2t)S3(c3t)dt(5.1)

Now∣

∣

∣

∣

∫ 1

0

S1(c1t)S2(c2t)E3(c3t)dt

∣

∣

∣

∣

≤ maxt

|E3(c3t)|∫ 1

0

|S1(c1t)S2(c2t)|dt

≪ Nlog logN

(logN)1/12

(∫ 1

0

|S1(c1t)|2dt∫ 1

0

|S2(c2t)|2dt)1/2

≪ N2 log logN

(logN)1/12.


Similarly by (4.9)

∣

∣

∣

∣

∫ 1

0

S1(c1t)E2(c2t)M3(c3t)dt

∣

∣

∣

∣

≤ maxt

|E2(c2t)|(∫ 1

0

|S1(c1t)|2dt∫ 1

0

|M3(c3t)|2dt)1/2

≪ N2 (log logN)3

(logN)1/12,

and

∣

∣

∣

∣

∫ 1

0

E1(c1t)M2(c2t)M3(c3t)dt

∣

∣

∣

∣

≤ maxt

|E1(c1t)|(∫ 1

0

|M2(c1t)|2dt∫ 1

0

|M3(c3t)|2dt)1/2

≪ N2 (log logN)5

(logN)1/12.

Therefore (5.1) equals

(5.2)

∫ 1

0

M1(c1t)M2(c2t)M3(c3t)dt+O

(

N2 (log logN)5

(logN)1/12

)

.

Suppose that this integrand is non-zero for some t; then there exist rationals, say with leastcommon denominator q such that |c1t− a/q|, |c2t− b/q|, |c3t− c/q| are all very small. Butsince (c1, c2, c3) = 1 there exist integers u, v, w with uc1 +vc2 +wc3 = 1 and so γ := t−d/qis small where d = ua+vb+wc. We may assume (d, q) = 1 else q was not the least commondenominator. Now, we need to have |c1γ| = |c1t − d(c1/(c1, q))

/

q1)| ≤ 1/(Qq1) so that|γ| ≤ 1/qQ (where qj = q/(cj , q)).

Now for the main term in (5.2) to be non-zero on an interval of t-values around a/q,we must have that rj |qj for each j, whence (4.6) gives that the integrand equals (1.11)times

3∏

ℓ=1

{

ψℓ(aℓ)κℓ(qℓ) · I(N, cℓ(α− a/q), tℓ)}

where aj = acj/(cj, q). Integrating over our interval and summing over all a with (a, q) = 1gives (1.11) times

(5.5)

∫ 1

qQ

− 1

qQ

I(N, c1β, t1)I(N, c2β, t2)I(N, c3β, t3)dβ

times

(5.6)

3∏

ℓ=1

g(ψℓ)

φ(qℓ)

∏

pb‖(qℓ/rℓ)

(

(

fℓ(p)

pitℓ

)b

− ψℓ(p)

(

fℓ(p)

pitℓ

)b−1)

times

(5.6b)∑

(a,q)=1

3∏

ℓ=1

ψℓ(aℓ) =

{

φ(q)∏3ℓ=1 ψℓ(cℓ/(cℓ, q)) if ψ1ψ2ψ3 = 1

0 otherwise.


If ψ1ψ2ψ3 = 1 then (5.6) times (5.6b) equals g(ψ1)g(ψ2)g(ψ3) times

(5.7) φ(q)

3∏

ℓ=1

ψℓ(cℓ/(cℓ, q))

φ(qℓ)

∏

pb‖(qℓ/rℓ)

(

(

fℓ(p)

pitℓ

)b

− ψℓ(p)

(

fℓ(p)

pitℓ

)b−1)

.

Therefore we have proved that the integral in (5.2) equals 0 unless ψ1ψ2ψ3 = 1 inwhich case it equals g(ψ1)g(ψ2)g(ψ3) times

(5.8) (1.11) ·∑

q≤N/Qrj |qj for each j

(5.5) · (5.7).

5.1. Estimates for the integral (5.5). We have that (5.5) equalsN3+i(t1+t2+t3)/∏

ℓ(1+itℓ) times

1

N

∫ NqQ

− NqQ

J(c1v, t1)J(c2v, t2)J(c3v, t3)dv

From the estimates in section 4.1 we have

∫

|v|>V

3∏

ℓ=1

|J(cℓv, tℓ)|dv ≪∫

|v|>V

3∏

ℓ=1

(1 + |tℓ|)(1 + |tℓ| log(c1c2c3v))dv

ijkv3

≪∏3ℓ=1(1 + |tℓ|)(1 + |tℓ| log(ijkV ))

c1c2c3V 2

We deduce that if each |tℓ| ≤ T with T ≥ 1 then the integral above is

∫ ∞

−∞

3∏

ℓ=1

J(cℓv, tℓ) dv +O

(

T 6(log logN)3

c1c2c3(N/qQ)2

)

This contributes an error term to (5.8) of

≪ N2T6(log logN)3

c1c2c3(N/Q)2(r1r2r3)

1/2∑

q≤N/Qrj |q for each j

q2

φ(q)2

3∏

ℓ=1

2ω(q)−ω(rℓ).

Let R = [r1, r2, r3] and write each q = Rm. If a ≤ b ≤ c are the powers of prime p dividingr1, r2, r3 then b = c since (ψ1ψ2ψ3) = 1. Thus this sum becomes

R2

φ(R)22ω(R)

∑

m≤N/QR

∏

p|mp∤R

8

(

p

p− 1

)2

≪ R2

φ(R)22ω(R)N/Q

Rlog(N/QR)7.


Now pa/2 is the power of prime p dividing (r1r2r3)1/2/R and so (r1r2r3)

1/2/R ≤ (r1r2r3)1/6 ≤

(N/Q)1/2. So this term is

≪ N2T6(log logN)10

(N/Q)1/2· 3ω(r1r2r3)

c1c2c3≪ N2 T 6

(logN)1/12+o(1).

Then the main term of (5.8) is

(1.11) ·3∏

ℓ=1

g(ψℓ) ·N2+i(t1+t2+t3)

∏

ℓ(1 + itℓ)

∫ ∞

−∞J(c1v, t1)J(c2v, t2)J(c3v, t3)dv·

·∑

q≤N/Qrj |qj for each j

φ(q)3∏

ℓ=1


φ(qℓ)

∏

pb‖(qℓ/rℓ)

(

(

fℓ(p)

pitℓ

)b

− ψℓ(p)

(

fℓ(p)

pitℓ

)b−1)

.

To determine the integral here consider the case fj(n) = nitj so that each ψj = 1, and(1.11) equals 1. Hence if the final product in the line above is non-zero then each b = 0;that is each qℓ = rℓ = 1. Therefore q divides (c1, c2, c3) = 1. Therefore the second lineabove equals 1. From this we deduce that

N2+i(t1+t2+t3)

∏

ℓ(1 + itℓ)

∫ ∞

−∞J(c1v, t1)J(c2v, t2)J(c3v, t3)dv =

∑

n1,n2,n3≤Nc1n1+c2n2+c3n3=0

nit11 nit22 nit33

The idea now is to replace this by an integral. If we restrict our attention to n1, n2

each in an interval of length m = |c3| then their residues e1, e2 (mod m) must satisfyc1e1 + c2e2 ≡ 0 (mod m): one can easily show that there are exactly m solutions; that isone in every m pairs e1, e2. Hence the above sum is very well approximated by

1

m

∫

0≤u1,u2,u3≤Nc1u1+c2u2+c3u3=0

uit11 uit22 uit33 du1du2 = N2+i(t1+t2+t3)1

2I

(

t1, t2, t3c1, c2, c3

)

.

We wish to complete this sum, so what one adds is

≪ N2(r1r2r3)1/22ω(R)

∑

q>N/QR|q

8ω(q)−ω(R)

φ(q)2≪ N2 (r1r2r3)

1/2

R

1

(N/Q)1+o(1)≪ N2

(logN)1/12+o(1).

Therefore the main term becomes

(1.11) ·3∏

ℓ=1

g(ψℓ) ·N2

2·N i(t1+t2+t3)I

(

t1, t2, t3c1, c2, c3

)

times

∑

q≥1rj |qj for each j

φ(q)

3∏

ℓ=1


φ(qℓ)

∏

pb‖(qℓ/rℓ)

(

(

fℓ(p)

pitℓ

)b

− ψℓ(p)

(

fℓ(p)

pitℓ

)b−1)

=∏

p

E(p).


We can determine each E(p) as follows: Let γℓ be the exact power of p dividing cℓ, and ρℓbe the exact power of p dividing rℓ. Consider those p which do not divide r1r2r3. In theterm corresponding to pa let aℓ := max{a − γℓ, 0} be the exact power of p dividing qℓ sowe have

E(p) =∑

a≥0

φ(pa)

3∏

ℓ=1

ψℓ(pγℓ+aℓ−a)

φ(paℓ)

(

(

fℓ(p)

pitℓ

)aℓ

− ψℓ(p)

(

fℓ(p)

pitℓ

)aℓ−1)

=3∏

ℓ=1

ψℓ(pγℓ)∑

a≥0

φ(pa)3∏

ℓ=1

1

φ(paℓ)

(

(

fℓ(p)

ψℓ(p)pitℓ

)aℓ

−(

fℓ(p)

ψℓ(p)pitℓ

)aℓ−1)

=

3∏

ℓ=1

ψℓ(pγℓ)

1 +∑

a≥1

pa−1(p− 1)∏

ℓ: a>γℓ

1

(p− 1)

(

fℓ(p)

ψℓ(p)p1+itℓ

)a−γℓ−1(fℓ(p)

ψℓ(p)pitℓ− 1

)

since ψ1ψ2ψ3 is principal. In particular note that if p ∤ c1c2c3 then

E(p) = 1 +1

(p− 1)2

3∏

ℓ=1

(

fℓ(p)

ψℓ(p)pitℓ− 1

)

(

1 − 1

p2

3∏

ℓ=1

fℓ(p)

pitℓ

)−1

.

If p|r1r2r3 let α = maxj:ρj≥1(γj + ρj) so that a ≥ α. Therefore cℓ/(cℓ, q) is notdivisible by p whenever p|rℓ. So if p divides each of r1, r2, r3 then

E(p) =∑

a≥αφ(pa)

3∏

ℓ=1

1

φ(pa−γℓ)

(

fℓ(p)

pitℓ

)a−γℓ−ρℓ

=1

p2α−2−P

ℓ γℓ(p− 1)2

3∏

ℓ=1

(

fℓ(p)

pitℓ

)α−γℓ−ρℓ(

1 − 1

p2

3∏

ℓ=1

fℓ(p)

pitℓ

)−1

Finally note that if p|r1r2r3 then p divides at least two ri’s since ψ1ψ2ψ3 is principal.We therefore are only left to deal with the case where exactly two of the ri’s are divisibleby p, say r1 and r2.

E(p) =∑

γ3≥a≥α

ψ3(pa−γ3)

pa−γ1−γ2−1(p− 1)

2∏

ℓ=1

(

fℓ(p)

pitℓ

)a−γℓ−ρℓ

+∑

a≥max{α,γ3+1}

ψ3(pa−γ3)

p2a−2−P

ℓγℓ(p− 1)2

2∏

ℓ=1

(

fℓ(p)

pitℓ

)a−γℓ−ρℓ·(

f3(p)

ψ3(p)pit3

)a−γ3−1 (f3(p)

ψ3(p)pit3− 1

)

Remarks. 1) It is unclear how much more these series can be usefully simplified. Onething to notice is that if a > maxℓ(ρℓ + γℓ) then the (a + 1)st term of the series is(f1f2f3)(p)/p

2+i(t1+t2+t3) times the ath term, therefore we can write E(p) as a finite lengthDirichlet series divided by 1 − (f1f2f3)(p)/p

2+i(t1+t2+t3).


2) All of these Euler products are functions of the following variables only: cj , ψj, fj(p)/pitj .

3) The number of terms in our original sum is ≪ N2/(|c1| + |c2| + |c3|) so if the sumif ≫ N2 then |c1|, |c2|, |c3| ≫ 1.

4) Since (c1, c2, c3) = 1, at least one γℓ equals 0. We then deduce that |∏

p E(p)| ≪1/[r1, r2, r3]. Therefore if the main term in (5.2) is ≫ N2 then r1, r2, r3 ≪ 1.

5) The absolute values of the powers of p (and p−1) in the leading term of the Eulerproducts are ≫ 1/[c1r1, c2r2, c3r3]

2+o(1). Therefore for this part to be small one of theE(p) must be so.

6) If E(p) = 0 then one can show that either p ≤ 5 or one has fℓ(p) = 0 withγℓ + ρℓ < α, or when p|r1, r2 but not r3, if f3(p) = ψ3(p)p

it3 or f3(p) = 0 with γℓ + 1 < α

7) If f3 = 1 then ψ3 = 1 and t3 = 0. Hence ψ2 = ψ1 and let r = r1 =

r2. Also E(p) =∏3ℓ=1 ψℓ(p

γℓ) if p ∤ r, = 0 if p|r and α > γ3, and otherwise equals∑

γ3≥a≥α1

pa−γ1−γ2−1(p−1)

∏2ℓ=1

(

fℓ(p)pitℓ

)a−γℓ−ρℓ. Now if c3 = −1 then for E(p) to be non-

zero we must have α = 0 and so p could not have divided r. In other words r1 = r2 = 1, and

therefore∏

p E(p) = 1. Therefore the main term becomes (1.11) ·N i(t1+t2+t3)I(

t1,t2,t3c1,c2,−1

)

.

(Analogous remarks apply with f1 or f2 in place of f3.)

6. Special case

We take c1 = c2 = 1, c3 = −1 with each fj real-valued, from which we can deducethat each tj = 0 and each ψj is real-valued, if our sum is to have a significant size. Henceour main term becomes, after dividing through by N2/2,

(1.11) ·3∏

ℓ=1

g(ψℓ) ·1

φ([r1, r2, r3])2·

∏

p∤r1r2r3

{

1 +1

(p− 1)2

3∏

ℓ=1

(

fℓ(p)

ψℓ(p)− 1

)(

1 − (f1f2f3)(p)

p2

)−1}

·

(6.1)∏

p|r1r2r3

3∏

ℓ=1

fℓ(p)α−ρℓ−1(fℓ(p) − ψℓ(p)) ·

(

1 − (f1f2f3)(p)

p2

)−1

since∫

0≤v1,v2,v1+v2≤11dv1dv2 = 1

2 , where α = maxj ρj.

Proof of Theorem 1. We now only consider fj which take values 1 and −1. The numberof solutions to f1(a) = f2(b) = f3(c) = −1 with a+ b = c ≤ N is

(6.2)1

N2/2

∑

a,b,c≤Na+b=c

1

8(1 − f1(a))(1 − f2(b))(1 − f3(c)).


By remark 7 of the previous section this equals o(1) plus 1/8 times

(6.3)∑

I({1,2,3}ψi=1∀i∈I

∏

i∈Iδi + (6.1),

where∑

n≤N fj(n) = −δjN , which we seek to maximize. Each term in this sum has

absolute value ≤ 1, so for (6.3) to be > 8κ > 4 we must have at least five non-zero terms.Hence at least two of the ψi = 1 and so the third one equals 1, since their product is 1.Therefore (6.3) equals

(6.4)∑

I({1,2,3}

∏

i∈Iδi + δ1δ2δ3 ·

∏

pfℓ(p)=−1∀ℓ

{

1 − 8

(p− 1)2

(

1 +1

p2

)−1}

.

Therefore every δi > 0 else there are at most four positive terms in (6.4) and so the valueis ≤ 4 < 8κ. Now if, for a given prime p, we have fi(p) = −1 then let Fi(p) = 1, andFi(q) = fi(q) for all primes q 6= p, so that δFi = (1 + 2/(p − 1))δi. But then each termin (6.4) is increased, or at worst stays the same. So (6.4) is maximized by letting eachfj(p) = 1 for all small primes p, so that (6.4) becomes (1 + δ1)(1 + δ2)(1 + δ3) + o(1). Themaximum value that this can take is when each δi = δ0, as in [5].

To prove the second part of Theorem 1, we must minimize

1

8

(

1 − 3δ + 3δ2 − Cfδ3)

where Cf :=∏

pf(p)=−1

{

1 − 8

(p− 1)2

(

1 +1

p2

)−1}

.

Now suppose f(q) = −1 and let g(p) = f(p) except when p = q. By [5, Proposition 4.4]

δf = (1 − ρ)δg where ρ := 2q+1 , and Cf = Cg

(

1 − 8q2

(q−1)2(q2+1)

)

. If f gives the minimal

quantity then 1 − 3(1 − ρ)δg + 3(1 − ρ)2δ2g − Cf (1 − ρ)3δ3g ≤ 1 − 3δg + 3δ2g − Cgδ3g so that

3ρδg+(Cg−Cf (1−ρ)3)δ3g ≤ 3(2ρ−ρ2)δ2g and therefore 1+(Cg−Cf (1−ρ)3)δ2g/3ρ ≤ (2−ρ)δg.Therefore 1 ≤ (2 − ρ)δg, so that δg ≥ 1/(2 − ρ) = 1

2· q+1

q. Thus q ≥ 5 and therefore

(2−ρ)δg ≥ 1+Cg1−(1−ρ)3

3ρ δ2g ≥ 1+Cg(1−ρ)δ2g . This then implies that 2δg ≥ 1+Cgδ2g (as

Cgδ2g < 1) which implies δg > 0.656999 . . . (which is impossible) provided C ≥ .7274404803

This is certainly the case if f(p) = 1 for all p ≤ 7. This can be established in a case-by-caseanalysis.

Remark. We could do f(a) = f(b) = 1, f(c) = −1 for variety!

7. Heuristics

We show that if

∑

c1n1+c2n2+c3n3≡0 (mod m)n1,n2,n3 (mod m)

ψ1(n1)ψ2(n2)ψ3(n3)


is non-zero, where [r1c1, r2c2, r3c3] divides m then ψ1ψ2ψ3 is principal and r1c1 = r2c2 =−r3c3, and the value we get, taking m = r1c1 is

φ(m)

m(−ψ3(−1))

3∏

i=1

cig(ψi).

Proof.

Multiplying n1, n2, n3 through by r (mod m) for any (r,m) = 1, we obtain the samesum. Hence if the sum is non-zero we must have ψ1ψ2ψ3 principal.

By the Chinese Remainder theorem we can determine our sum as a product of thevalues over the prime power factors of m, decomposing the characters ψi into prime powercomponents. Note that a component cannot be principal since the ψi are primitive –instead we must use 1. So now assume m = pe and that each ψi is a primitive charactermod pρi , with ρi ≥ 1. Suppose that pγi is the exact power of p dividing ci, so thatci = Cip

γi — let e = maxi ρi + γi and suppose ρi + γi = e. Then the above equals

1

pe

pe∑

a=0

∑

n1,n2,n3 (mod pe)

ψ1(n1)ψ2(n2)ψ3(n3) e

(

a(c1n1 + c2n2 + c3n3)

pe

)

,

=1

pe

pe∑

a=0

3∏

i=1

∑

ni (mod pe)

ψi(ni) e

(

apγiCinipe

)

.

Suppose that each ρi ≥ 1. If a term in the sum is non-zero then the lemma below impliesthat pe−ρi−γi‖a, and so p ∤ a and e = ρi + γi for every i. In that case we obtain

=1

pe

∑

0≤a≤pe−1(a,p)=1

3∏

i=1

pe−ρiψi(aCi)g(ψi).

This is non-zero only if ψ1ψ2ψ3 is a principal character, in which case we obtain

(1 − 1/p)

3∏

i=1

pγiψi(Ci)g(ψi).

If ψ3 = 1 then ψ2 = ψ1 which is not 1. We must have γ3 > 0 else each pair n1, n2

determine n3 and the sum is 0. Therefore we restrict our attention to pairs n1, n2 satisfyingc1n1 + c2n2 ≡ 0 (mod pγ3), and then n3 is determined mod pe−γ3 so there are pγ3 choicesfor n3. Since γ3 > 0 we know that at least one of γ1 and γ2 is 0. But we can only havenon-zero terms ψ1(n1)ψ2(n2) if γ1 and γ2 are equal, so we have γ1 = γ2 = 0. Thereforeour sum equals pγ3ψ1(c1)ψ2(c2) times

∑

c1n1+c2n2≡0 (mod pγ3 )n1,n2 (mod m)

ψ1(c1n1)ψ2(c2n2) =∑

r1+r2≡0 (mod pγ3 )r1,r2 (mod m)

ψ1(r1)ψ1(r2)

=∑

r (mod m)(r,p)=1

∑

s≡1 (mod pγ3 )s (mod m)

ψ1(−s)


writing r2 = r and r1 ≡ −sr (mod m). This last sum is 0 (as in the final step of the proofof the lemma) unless γ3 ≥ e = ρ1; thus e = γ3. Therefore our sum equals

peφ(pe)ψ1(c1)ψ2(c2)ψ1(−1) = φ(pe)ψ1(c1)ψ2(c2)g(ψ1)g(ψ2) = (1−1/p)

3∏

i=1

pγiψi(Ci)g(ψi),

as we again have e = ρi + γi for every i.Now we wish to combine the contributions of each prime, for fixed i. If ψi has con-

ductor ri =∏

j pρi,ji,j and ψi =

∏

j ψi,j then g(ψi) =∏

j ψi,j(ri/pρi,jj )g(ψi,j). In conclusion

if our sum is non-zero then r1c1 = r2c2 = −r3c3 (not quite true: only at primes

dividing r1r2r3 for now), a number that we will denote by m, and ψ1ψ2ψ3 is principal.

The overall product then is φ(m)m

|c1c2c3| times

3∏

i=1

∏

j

ψi,j(ci/pγi,jj )g(ψi,j) =

3∏

i=1

∏

j

ψi,j(ciri/pγi,j+ρi,jj )ψi,j(ri/p

ρi,jj )g(ψi,j)

=3∏

i=1

g(ψi)∏

j

3∏

i=1

ψi,j(m/pejj )ψ3,j(−1)

= ψ3(−1).

3∏

i=1

g(ψi)

as∏3i=1 ψi,j is principal for each j

Lemma. If ψ is a primitive character mod pe, for some e ≥ 1 and E ≥ e+ γ then

∑

n (mod pE)

ψ(n) e

(

apγn

pE

)

= 0

unless pE−e−γ‖a, whence our sum equals pE−eψ(A)g(ψ) with a = pE−e−γA.

Proof. Write n = bpe + r so the sum becomes

∑

0≤b≤pE−e−1

e

(

apγb

pE−e

)

∑

r (mod pe)

ψ(r) e

(

apγr

pE

)

.

The first sum equals 0 unless pE−e−γ divides a, whence it equals pE−e. Let us writea = pE−e−γA, so the above becomes

pE−e∑

r (mod pe)

ψ(r) e

(

Ar

pe

)

.

If p does not divide A then this sum equals ψ(A)g(ψ). If p does divide A, then the abovesum becomes, writing r = kpe−1 + ℓ

∑

0≤k≤p−1

∑

0≤ℓ≤pe−1−1(ℓ,p)=1

ψ(kpe−1 + ℓ) e

(

Aℓ

pe

)

=∑

0≤ℓ≤pe−1−1(ℓ,p)=1

ψ(ℓ) e

(

Aℓ

pe

)

∑

0≤j≤p−1

ψ(1+ jpe−1)


taking k ≡ jℓ (mod p). But this last sum is 0 since the residues {1+jpe−1 : 0 ≤ j ≤ p−1}can be written as {(1 + pe−1)i : 0 ≤ i ≤ p− 1} and ψ(1 + pe−1) 6= 1 else ψ is a charactermod pe−1 contradicting the fact that ψ is primitive mod pe.

8. Summing to N .

∑

n1,n2,n3≥1n1+n2+n3=N

f1(n1)f2(n2)f3(n3)

=∑

n1,n2,n3≥1

f1(n1)f2(n2)f3(n3)

∫ 1

0

e(t(n1 + n2 + n3 −N))dt

=

∫ 1

0

e(−Nt)3∏

ℓ=1

∑

nℓ≤Nfℓ(nℓ)e(nℓt)dt

=

∫ 1

0

e(−Nt)S1(t)S2(t)S3(t)dt(8.1)

As before (8.1) equals

(8.2)

∫ 1

0

e(−Nt)M1(t)M2(t)M3(t)dt+O

(

N2 (log logN)5

(logN)1/12

)

.

For this integrand to be large for some t there exists a rational a/q with |t−a/q| very smallthat is ≤ 1/qQ. We must have that rj|q for each j, whence (4.6) gives that the integrandequals (1.11) times

e(−Na/q) · e(−N(α− a/q))3∏

ℓ=1

ψℓ(a)κℓ(q)I(N,α− a/q, tℓ).

Integrating over our interval, and extending the integral as before, and summing over alla with (a, q) = 1 gives (1.11) times

(8.5)

∫ ∞

−∞e(−βN)I(N, β, t1)I(N, β, t2)I(N, β, t3)dβ

times∏3ℓ=1 κℓ(q) times

∑

(a,q)=1

e(−Na/q)(ψ1ψ2ψ3)(a) =∑

a (mod q)

e(−Na/q)χq(a) = χq(N)g(χq)

= χq(N)µ(q/R)χR(q/R)g(χR),


where χq is the character (mod q) induced by ψ1ψ2ψ3, and χR is principal with R|r =[r1, r2, r3]. Summing over q we have χR(N)g(χR)g(ψ1)g(ψ2)g(ψ3) times

(8.8)∑

q≥1r|q

(q,N)=1

µ(q/R)χR(q/R)

φ(q)3

3∏

ℓ=1

∏

pb‖(q/rℓ)

(

(

fℓ(p)

pitℓ

)b

− ψℓ(p)

(

fℓ(p)

pitℓ

)b−1)

.

This is 0 if (r,N) > 1 or if µ(r/R) = 0, so suppose that (r,N) = 1 and µ(r/R) 6= 0. Letpρj‖rj and ρ = maxj ρj, so the above becomes µ(r/R)χR(r/R) times

1

φ(r)3

∏

p∤Np∤r

(

1 − χR(p)

(p− 1)3

3∏

ℓ=1

(

fℓ(p)

pitℓ− ψℓ(p)

)

)

∏

p∤Np‖(r/R)

3∏

ℓ=1

(

(

fℓ(p)

pitℓ

)ρ−ρℓ− ψℓ(p)

(

fℓ(p)

pitℓ

)ρ−ρℓ−1)

∏

p∤Np∤(r/R), p|R

3∏

ℓ=1

(

(

fℓ(p)

pitℓ

)ρ−ρℓ− ψℓ(p)

(

fℓ(p)

pitℓ

)ρ−ρℓ−1)

(8.9) −χR(p)

p3

3∏

ℓ=1

(

(

fℓ(p)

pitℓ

)ρ−ρℓ (fℓ(p)

pitℓ− ψℓ(p)

)

)

.

Therefore we get (8.9) times

µ(r/R)χR(r/R)χR(N)g(χR)g(ψ1)g(ψ2)g(ψ3) = (ψ1ψ2ψ3)(N)g(ψ1ψ2ψ3)g(ψ1)g(ψ2)g(ψ3).

If we select each fj(n) = nitj then the above equals (8.5). Therefore (8.5) equals

∑

n1,n2,n3≤Nn1+n2+n3=N

nit11 nit22 nit33 ≈∫

u1,u2,u1+u2≤Nuit11 uit22 (N − u1 − u2)

it3du1du2

= N2+i(t1+t2+t3)

∫

v1,v2,v1+v2≤1

vit11 vit22 (1 − v1 − v2)it3dv1dv2(8.10)

as desired.


References

[1] G. Bachman, On a Brun-Titchmarsh inequality for multiplicative functions, Acta Arith 106 (2003),

1–25.

[2] G. Bachman, On exponential sums with multiplicative coefficients. II, Acta Arith 106 (2003),41–57.

[3] A. Balog, A. Granville and K. Soundararajan, Multiplicative functions in arithmetic progressions

(to appear).

[4] H. Davenport, Multiplicative number theory, Springer Verlag, New York, 1980.

[5] A. Granville and K. Soundararajan, The Spectrum of Multiplicative Functions, Ann. of Math 153

(2001), 407–470.

[6] , Decay of mean-values of multiplicative functions, Can. J. Math 55 (2003), 1191-1230.

[7] , Large character sums: Pretentious characters and the Polya-Vinogradov theorem, J. Amer.Math. Soc 20 (2007), 357-384.

[8] , Pretentious multiplicative functions and an inequality for the zeta-function, Proceedings,Anatomy of Integers workshop, (Montreal 2006).

[9] H.L. Montgomery and R.C. Vaughan, Exponential sums with multiplicative coefficients, Invent.

Math 43 (1977), 69-82.

Department de Mathematiques et Statistique, Universite de Montreal, CP 6128 succ

Centre-Ville, Montreal, QC H3C 3J7, Canada

E-mail address: [email protected]

Department of Mathematics, 450 Serra Mall, Bldg. 380, Stanford University, Palo

Alto CA 94305, USA

E-mail address: [email protected]

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