Gal-type GCD sums beyond the critical line

GÁL-TYPE GCD SUMS BEYOND THE CRITICAL LINE

ANDRIY BONDARENKO, TITUS HILBERDINK, AND KRISTIAN SEIP

ABSTRACT. We prove that

N

X

k,`=1

(n_k,n_`)^2α

(n_kn_`)^α ¿N^2−2α(logN)^b(α)

holds for arbitrary integers 1≤n₁< · · · <n_Nand 0<α<1/2 and show by an example that this bound is optimal, up to the precise value of the exponentb(α). This estimate complements recent results for 1/2≤α≤1 and shows that there is no “trace” of the functional equation for the Riemann zeta function in estimates for such GCD sums when 0<α<1/2.

1. INTRODUCTION

The study of greatest common divisor (GCD) sums of the form (1)

N

X

k,`=1

(n_k,n_{`</sub>)<sup>2α</sup> (n<sub>k</sub>n<sub>`})^α

begins with Gál’s theorem [6] which asserts that when α=1, C N(log logN)² is an optimal upper bound for (1), withC an absolute constant independent ofNand the distinct positive integers n₁, ...,n_N (the best possible value forC is 6e^2γ/π², where γ is Euler’s constant, as shown recently by Lewko and Radziwiłł [8]). Dyer and Harman [5], motivated by applications in the metric theory of diophantine approximation, obtained the first estimates for the range 1/2≤α<1. Recent work of Aistleitner, Berkes, and Seip [2] for 1/2<α<1 and Bondarenko and Seip [3, 4] forα=1/2 has led to the bounds

(2)

XN k,`=1

(n_k,n_{`</sub>)<sup>2α</sup> (n<sub>k</sub>n<sub>`})^α ¿







Nexp³

c(α)_{(log log}^(logN)1−α_N)α

´

, 1/2<α<1 Nexp

³ A

qlogNlog log logN log logN

´

, α=1/2,

2010Mathematics Subject Classification. 11C20.

Bondarenko and Seip were supported by Grant 227768 of the Research Council of Norway.

1

which are optimal, up to the precise values of the constantsc(α) and A; the asymptotic be- havior ofc(α) has been clarified both whenα&1/2 andα%1.

Bounds for the sum in (1) have a long history, and they have had a number of different applications; see the recent papers [2, 3, 8] and the references found there. In recent years, an additional interesting application has surfaced: Lower bounds for specific sums of the form (1) or corresponding quadratic forms have turned out to be useful for detecting large values of the Riemann zeta functionζ(s). This line of research was initiated in work of Soundararajan [11] and Hilberdink [7] and later pursued by Aistleitner [1] who was the first to make the link to Gál-type estimates. Recently, using Soundararajan’s resonance method [11] and a certain large Gál-type sum forα=1/2, Bondarenko and Seip showed that for everyc, 0<c <1/p

2, there exists aβ, 0<β<1, such that the maximum of|ζ(1/2+i t)|on the intervalT^β≤t≤T exceeds exp¡

cp

logTlog log logT/ log logT¢

for allT large enough.

These developments have led us to look more closely at the “phase transition” atα=1/2 by seeking estimates for (1) also in the range 0<α<1/2, which could possibly correspond to large values ofζ(σ+i t) beyond the critical lineσ=1/2. The present paper shows, however, that there is no symmetry in the estimates for (1) whenαis replaced by 1−α, as one might have expected from the functional equation forζ(s).

To state our main result, we let M denote an arbitray finite set of positive integers and introduce the quantity

Γ_α(N) := max

|M|=N

1 N

X

m,n∈M

(m,n)^2α (mn)^α .

Theorem 1. For everyα,0<α<1/2, there exist positive constants a(α)and b(α)such that (3) N^1−2α(logN)^a(α)≤Γα(N)≤N^1−2α(logN)^b(α)

for sufficiently large N .

Before giving the proof of this theorem, we will in the next section set the stage by consid- ering the simpler but closely related question of finding the largest eigenvalue of the positive definite matrix (m,n)^2α/(mn)^α, 1≤m,n≤N:

Theorem 2. For everyα,0<α<1/2, there exists a constant C_αsuch that

(4) max

|a1|²+···|aN|²=1

X

m,n≤N

a_ma¯_n(m,n)^2α

(mn)^α ≤C_αN^1−2α.

We refer to [7] for further information and for the precise asymptotics of the maximum in (4) in the range 1/2≤α≤1.

We notice that there is no logarithmic power in (4). Nevertheless, we will see that the idea for the proof of Theorem 2 (to be given in the next section) is used again as the starting point for the proof of the bound from above in Theorem 1. Resorting to some further ideas and estimates from [3], we will prove the latter bound in Section 3. In Section 4, we construct an example giving the inequality from below in (3). As expected, this example involves a large number of primes (a positive power ofN). One may notice that it would not be possible to construct a similar example if we required the set to consist only of square-free numbers.

Hence it remains an open problem to prove the analogue of Theorem 1 in the square-free case. More specifically, we may ask whether the logarithmic power can be discarded in this case as well.

2. PROOF OFTHEOREM2 We begin by noticing that

S(a) :=

N

X

d=1

X

m,n≤N (m,n)=d

|a_ma_n|d^2α (mn)^α ≤

N

X

d=1

à X

m,n≤N/d

|a_md| m^α

!₂ .

We introduce the multiplicative function g(m) :=P

d|md^−1/2+α. By the Cauchy–Schwarz in- equality,

(5) S(a)≤

XN d=1

à X

m≤N/d

|amd| m^α

!2

≤ XN d=1

X

m≤N/d

|amd|² g(m)

X

m≤N/d

g(m) m^2α. To estimate the sumP

m≤N/d g(m)

m^2α, we notice that X

n≤x

g(n) n^2α = X

n≤x

1 n^2α

X

d|n

1

d^12−α= X

d≤x

1 d^12+α

X

n≤x/d

1

n^2α¿ X

d≤x

1 d^12+α

³x d

´_1−2α

¿x^1−2α.

Hence by (5) we have

S(a)¿N^1−2α

N

X

d=1

X

m≤N/d

|amd|²

d^1−2αg(m)=N^1−2α

N

X

n=1

|a_n|²X

d|n

d^2α−1 g(n/d). So to finish the proof of Theorem 2, it is sufficient to show that

(6) h(n) :=X

d|n

d^2α−1 g(n/d)

is a bounded arithmetic function. We observe thath(n) is a multiplicative function, which means that it suffices to consider

h(p^m)=

m

X

`=0

p^−2α`

Pk≤m−`p^{−k(1/2−α)} = 1 P

k≤mp^{−k(1/2−α)} +p^−2αh(p^m−1).

This recursive formula implies, by induction, thath(p^m)≤1 for psufficiently large and that lim_m→∞h(p^m)=0 for every prime p. We infer from this thath(n) is a bounded arithmetic function.

As far as the numerical value of the constantC_αin Theorem 2 is concerned, we have con- fined ourselves to the following special case which seems to be of independent interest:

(7) 1

N

N

X

m,n=1

(m,n)^2α

(mn)^α = ζ(2−2α)

ζ(2)(1−α)²N^1−2α+O(1).

Proof of (7). WriteF_α(N) for the sum on the left and putS_α(x) :=P

m,n≤x (m,n)=1

1

(mn)^α. Then F_α(N)= X

d≤N

X

m,n≤N (m,n)=d

(m,n)^2α (mn)^α = X

d≤N

S_α(N/d).

Also letT_α(x)=P

n≤x 1

n^α =_1−α¹ x^1−α+O(1). Then

T_α(x)²= X

m,n≤x

1

(mn)^α= X

d≤x

1 d^2α

X

m,n≤x/d (m,n)=1

1

(mn)^α= X

d≤x

1 d^2αS_α³x

d

´ .

By Möbius inversion,S_α(x)=P

d≤x µ(d)

d^2αT_α(_d^x)²and so F_α(N)= X

d≤N

β(n)T_α³N n

´2

,

whereβ(n)=P

d|nµ(d)

d^2α. We note that 0<β(n)≤1 for alln. Thus F_α(N)= 1

(1−α)² X

n≤N

β(n)³³N n

´_2−2α

+O³N n

´_1−α´

= N^2−2α (1−α)²

X

n≤N

β(n) n^2−2α+O³

N^1−α X

n≤N

1 n^1−α

´ . The final term isO(N), whileP

n>N β(n) n^2−2α≤P

n>N 1

n^2−2α¿N^2α−1. Finally X∞

n=1

β(n)

n^2−2α =ζ(2−2α) ζ(2) ,

giving the result. ■

3. PROOF OF THE BOUND FROM ABOVE INTHEOREM1

In what follows, ω(n) denotes the number of distinct prime factors inn andd(n) is the divisor function.

We begin by stating the main auxiliary result used to prove the upper bound in (3).

Lemma 1. For every finite setM of positive integers there exists a divisor closed setM⁰of posi- tive integers with|M⁰| = |M|such that

X

m,n∈M

(m,n)^2α

(mn)^α ≤ X

m,n∈M⁰

(m,n)^2α

(mn)^α 2^{ω(mn/(m,n)2)}.

Proof. Following the proof of [2, Lemma 2], we transformM intoM⁰by means of the follow- ing algorithm. Fix a primepsuch thatpdivides some number inM. Then there exist distinct numbersm_j, j=1, ...,`with`≤ |M|such that we may write

M = [` j=1

Mj,

whereMjconsists of thoseminMsuch thatm/m_jis a power ofp. We then replace the num- bers inMjby the numbersm_j,m_jp, ...,m_jp^|Mj|−1. This transformation is then performed for every prime dividing some number inM. A close inspection of the largest possible change in the GCD sum in each step of this series of transformations (carried out in detail in the proof

of [2, Lemma 2]) gives the desired estimate. ■

We will also need the following two lemmas.

Lemma 2. Suppose that0<α<1/2and thatβis a real number. Then for everyβ⁰>β/(2α) there exists a positive constant C with the following property. IfK is a set of positive integers with|K| =K , then

(8) X

m∈K

d(m)^β

m^2α ≤C K^1−2α[logK]^2β

0−1.

Proof. We begin by observing that X

m∈K

d(m)^β m^2α ≤

K

X

m=1

d(m)^β m^2α +

X∞

`=0

X

2^{`</sup>K<m≤2<sup>`+1}K d(m)^β>2^2α`

d(m)^β m^2α . Now

X

2^{`</sup>K<m≤2<sup>`+1}K d(m)^β>2^2α`

d(m)^β

m^2α ≤2^{−(β0/β−1)2α`} X

2^{`</sup>K<m≤2<sup>`+1}K

d(m)^β0 m^2α

¿2^{−(β0/β−1)2α`</sup>·2<sup>(1−2α)`}K^1−2α(log 2^`K)^2β

0−1,

where we used the classical formula X

n≤x

d(n)^β0=B x(logx)^2β

0−1¡

1+O((logx)⁻¹)¢

which holds withB an absolute constant [12, 10]. It follows that the sum over`is dominated

by a convergent geometric series ifβ⁰>β/(2α). ■

We mention without proof that a more careful analysis shows that the exponent 2^β0−1 on the right-hand side of (8) can be replaced by 2α(2^β0−1) with the same requirement thatβ⁰>

β/(2α). Using results on the distribution of ‘large’ values ofd(n) (see [9]), we can show that this is optimal in the sense that the inequality fails with any exponent less than 2α(2^β/(2α)−1).

Lemma 3. If M is a divisor closed set of square-free numbers, then |_p¹M| ≤ ¹₂|M|for every prime p inM.

Proof. Suppose that|¹_pM| =`and write <sub>p</sub><sup>1</sup>M ={m<sub>1</sub>, . . . ,m<sub>`</sub>}. ThenM containspm₁, . . . ,pm_` and hence alsom₁, . . . ,m_M, since it is divisor closed. AsM is square-free, these numbers are

all distinct, and so it follows that|M| ≥2`. ■

We are now prepared to prove the bound from above in (3). To begin with, we define eΓα(N) := max

Mdivisor closed,|M|=N

1 N

X

m,n∈M

(m,n)^2α

(mn)^α 2^{ω(mn/(m,n)2)}.

By Lemma 1, we haveΓα(N)≤Γeα(N), which means that it suffices to estimateeΓα(N). Hence we assume that the setM is divisor closed and estimate instead the sum

Se:= X

m,n∈M

(m,n)^2α

(mn)^α 2^{ω(mn/(m,n)2)}.

In what follows,M^∗will denote the subset ofM consisting of the square-free numbers inM. In addition, givenminM^∗, we letM(m) denote the subset ofM consisting of those numbers ninM such thatp|nif and only ifp|m. Hence

M = [

m∈M^∗M(m) and X

m∈M^∗

|M(m)| =N.

Now suppose thatkand`are inM<sup>∗</sup>and that|M(k)| ≥ |M(`)|. We then find that X

m∈M(k),n∈M(`)

(m,n)^2α

(mn)^α 2^{ω(mn/(m,n)2)}≤(k,`)^2α

(k`)<sup>α</sup> 2<sup>ω(k`/(k,`)2)</sup> X

n∈M(`)

Y

p|k

¡1+4 X∞ ν=1

p^−ν¢

¿(k,`)^2α

(k`)<sup>α</sup> 2<sup>ω(k`/(k,`)2)</sup>|M(`)|d(k)^ε,

where the implicit constant in the latter relation only depends onαandε. Hereεcan be any positive number, but in what follows we will require that 0<ε<1−2α. We infer from the latter relation that

Se¿ X

m,n∈M^∗|M(m)|^1/2d(m)^ε|M(n)|^1/2d(n)^ε(m,n)^2α

(mn)^α 2^{ω(mn/(m,n)2)}. This leads to the bound

Se¿ X

k∈M^∗



 X

m∈¹_kM^∗

|M(mk)|^1/2d(mk)^εd(m)^1+ε m^α





2

. By the Cauchy–Schwarz inequality, we obtain from this that

Se¿ X

k∈M^∗

d(k)^2ε X

n∈_k¹M^∗

|M(nk)|

d(n)^β X

m∈¹_kM^∗

d(m)^β+2+4ε m^2α ,

whereβis a positive parameter to be chosen later. Using Lemma 2 and the estimate

|1

kM| ≤N2^−ω(k), which we get from Lemma 3, we therefore get

Se¿N^1−2α(logN)^2β

0−1 X

k∈M^∗

d(k)^2α+2ε−1 X

n∈_k¹M^∗

|M(nk)|

d(n)^β

=N^1−2α(logN)^2β

0−1 X

m∈M^∗|M(m)|X

k|m

1

2^{(1−2(α+ε))ω(k)+βω(n/k)} withβ⁰>(β+2+4ε)/(2α). Since

n7→X

d|n

1

2^{(1−2(α+ε))ω(d)+βω(n/d)}

is a multiplicative function, andnis squarefree, it suffices to make sure that 1

2^1−2(α+ε)+ 1 2^β≤1.

This means that we need

β≥log₁ ¹

−2^2(α+ε)−1

log 2 to obtain the uniform bound

X

k|m

1

2(1−2(α+ε))ω(k)+βω(n/k) ≤1.

We then find that

Se¿N^1−2α(logN)^2β

0−1 X

m∈M^∗

|M(m)| =N^2−2α(logN)^2β

0−1

, which in turn leads to the desired conclusion.

4. PROOF OF THE BOUND FROM BELOW INTHEOREM1

This section will make extensive use of the Euler totient functionφ(n). We will also need an additional multiplicative function, namely

f(n) :=Y

p|n

µ

p^2α−1³

1−_p¹´_2α +³

1−_p¹´¶ µ

1 p

³ 1−_p¹

´ +

³ 1−_p¹

´_2−2α¶ .

We now fix a positive numberM and setk:=Q

p≤Mp. We will need the following lemma.

Lemma 4. For every c such that c|k, we have

(9) 1

k² X

d|k

(c,d)^2α

³φ¡k c

¢φ¡k d

¢´_α

¡φ(c)φ(d)¢1−α

=Y

p|k

µ1 p µ

1−1 p

¶ +

µ 1−1

p

¶_2−2α¶ c kf¡k

c

¢.

Moreover, we have (10)

1 k²

X

c,d|k

(c,d)^2α³ φ¡k

c

¢φ¡k d

¢´_α

¡φ(c)φ(d)¢1−α

=Y

p|k

µ p^2α−2

µ 1−1

p

¶2α

+2 p µ

1−1 p

¶ +

µ 1−1

p

¶2−2α¶ . Proof. Since every divisord ofk has a unique representationd =d₁d₂, whered₁|c andd₂|^k_c we find that the left-hand sideLH Sof (9) is

LH S= 1

k²φ(k/c)^αφ(c)^1−αX

d1|c

X

d2|^k_c

d₁^2αφ¡ c d₁

¢_α φ¡k/c

d₂

¢_α

φ(d₁)^1−αφ(d₂)^1−α

= 1

k²φ(k/c)^αφ(c)^1−α Ã

X

d1|c

d₁^2αφ¡ c d₁

¢_α

φ(d₁)^1−α

!

 X

d2|^k_c

φ¡k/c d₂

¢_α

φ(d₂)^1−α



. (11)

Using the formulaφ(d)=dQ

p|d

³ 1−_p¹

´

and the fact that the respective sums represent mul- tiplicative functions, we find that

X

d1|c

d₁^2αφ¡ c d1

¢_α

φ(d₁)^1−α=Y

p|c

µ p^α

µ 1−1

p

¶_α

+p^1+α µ

1−1 p

¶1−α¶ , and

X

d2|^k_c

φ¡k/c d₂

¢_α

φ(d₂)^1−α=Y

p|^k_c

µ p^α

µ 1−1

p

¶_α

+p^1−α µ

1− 1 p

¶_1−α¶ . Returning to (11) and using again thatφ(d)=dQ

p|d

³ 1−¹_p

´

, we therefore obtain LH S= 1

k² X

d|k

(c,d)^2α³ φ¡k

c

¢φ¡k d

¢´_α

¡φ(c)φ(d)¢_1−α

=c k

Y

p|c

µ1 p µ

1−1 p

¶ +

µ 1−1

p

¶_2−2α¶ Y

p|^k_c

µ p^2α−1

µ 1−1

p

¶_2α +

µ 1−1

p

¶¶

=Y

p|k

µ1 p µ

1−1 p

¶ +

µ 1−1

p

¶_2−2α¶ c kf¡k

c

¢,

where the last expression is the right-hand side of (9). Finally, we get (10) from (9) by using thatn7→P

d|n 1

df(d) is a multiplicative function. ■

In addition to the identities of the preceding lemma, we need following quantitative esti- mate.

Lemma 5. For everyα,0<α<1/2, there exists a positive constant c_αsuch that Y

p≤M

µ p^2α−2

µ 1−1

p

¶2α

+2 p µ

1−1 p

¶ +

µ 1−1

p

¶2−2α¶

≥c_α(logM)^2α.

Proof. The result follows from the fact that p^2α−2

µ 1−1

p

¶_2α +2

p µ

1−1 p

¶ +

µ 1−1

p

¶_2−2α

=1+2α

p +O(p^2α−2), p→ ∞, along with Mertens’s third theorem, i.e., the fact thatQ

p≤M(1−1/p)∼e^γ/ logMwhenM→ ∞.

■

The following theorem yields the bound from below in Theorem 1.

Theorem 3. For everyα, 0<α<1/2, there exists a positive constant c_α such that if N is a positive integer, then there exists a set of integersM of cardinality N such that

X

m,n∈M

(m,n)^2α

(mn)^α ≥c_αN^2−2α(logN)^2α.

Proof. Fixα, 0<α<1/2, and letN be positive integer. SetM =N^δ, whereδ, 0<δ<1, is a constant depending only onαto be chosen later,k =Q

p≤Mp. LetA be the set of the first [N^1/3]M-smooth square-free numbers andDbe the set of integers of the formk/awithain A. For every numberd inDdenote byS_d the set of the first [Nφ(d)/k] integerss such that (s,k/d)=1, and by s_d the maximal number inS_d. Also, letd S_d be the set of integers of the formd s, wheres∈S_d andd∈D. Finally, setM :=S

d∈Dd S_d.

It is clear that all numbersd inD are square-free and also that the setsd S_d are pairwise disjoint. Moreover, sinceP

d|kφ(d)=kwe have that|M| <N. Also,

(12) |Sd| ≥ N^2/3

2 logN

for sufficiently largeN and everyd inD; this follows from the formulaφ(d)=nQ

p|d

³ 1−¹_p

´ and an application Mertens’s third theorem. We can use this to get an upper bound fors_d. Indeed, each setS_d is just a set of numbers of the forma mod (k/d), whereais from a set of cardinalityφ(k/d). Therefore ifdis inD, then

s_d≤ 2Nφ(d) dφ(k/d).

Here we used that|S_d| ≥k/dwhich follows from (12). Now we have, using (12) in the last step, X

m,n∈M

(m,n)^2α (mn)^α ≥ X

c,d∈D

(c,d)^2α (cd)^α

X

m∈Sc

1 m^α

X

n∈Sd

1

n^α≥ X

c,d∈D

(c,d)^2α (cd)^α

|S_c||S_d| s^α_cs^α_d

≥c_αN^2−2α 1 k²

X

c,d∈D

(c,d)^2α(φ(k/c)φ(k/d))^α(φ(c)φ(d))^1−α

for some positive constantc_αdepending only onα. In view of (10) of Lemma 4 and Lemma 5, the proof will be complete if we can prove that

X

c,d∈D

(c,d)^2α(φ(k/c)φ(k/d))^α(φ(c)φ(d))^1−α≥1 3

X

c,d|k

(c,d)^2α(φ(k/c)φ(k/d))^α(φ(c)φ(d))^1−α.

The latter inequality will follow from the bound X

c6∈D,d|k

(c,d)^2α(φ(k/c)φ(k/d))^α(φ(c)φ(d))^1−α≤1 3

X

c,d|k

(c,d)^2α(φ(k/c)φ(k/d))^α(φ(c)φ(d))^1−α. By (9), this is equivalent to

(13) X

n∈F

f(n) n ≤1

3 Y

p≤M

µ

1+f(p) p

¶ ,

whereF is the set of allM-smooth square-free numbers larger than [N^1/3]. By Rankin’s trick, we have

X

n∈F

f(n)

n ≤N^−3δlogN1 Y

p≤M

µ

1+f(p) p p^δlogN1

¶

≤e^−31δ Y

p≤M

µ

1+ f(p) p

¶ Y

p≤M

µ

1+f(p) p

³

p^δlog1N −1

´¶ .

Now we note that the second product is bounded by a constant depending only onα(not on δ). Indeed,

X

p≤M

f(p) p

³

p^δlog1N −1

´

≤2 X

p≤M

f(p)−1

p +2 X

p≤M

logp plogM.

The first sum is bounded becausef(p)=1+p^2α−1+o(p^2α−1) asp→ ∞, and the second sum is bounded sinceP

p≤M(logp)/p−logM ≤2 by Mertens’s first theorem. Therefore, choosing δsufficiently small (depending only onα), we get (13). Theorem 3 is proved. ■

REFERENCES

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Ann., to appear; arXiv:1409.6035.

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