Multiplicative Functions Supported on the $ K $-Free Integers with Small

MULTIPLICATIVE FUNCTIONS SUPPORTED ON THE k-FREE INTEGERS WITH SMALL PARTIAL SUMS

MARCO AYMONE, CAIO BUENO, KEVIN MEDEIROS

Abstract. We provide examples of multiplicative functions f supported on the k- free integers such that at primes f(p) = ±1 and such that the partial sums of f up to x are o(x1/k). Further, if we assume the Generalized Riemann Hypothesis, then we can improve the exponent 1/k: There are examples such that the partial sums up to 1/(k+ 1 )+ x are o(x 2 ), for all > 0.

1. Introduction.

We say that an integer n is k-free if n is not divisible by any prime power pk, where k ≥ 2 is an integer; We say that f : N → C is multiplicative if f(nm) = f(n)f(m) for all integers n ≥ 1 and m ≥ 1 such that gcd(n, m) = 1, and we say that f is completely multiplicative if f(nm) = f(n)f(m) for all integers n ≥ 1 and m ≥ 1. Here the Vinogradov notation f(x) g(x) means that for some constant C > 0, |f(x)| ≤ C|g(x)|, for all suﬃciently large x > 0. P In this paper we are interested in the partial sums n≤x f(n) where f : N → {−1, 0, 1} is a multiplicative function such that at primes f(p) = ±1 and supported on the k-free integers, i.e., functions f such that f(n) = 0 if n is not k-free.

When we consider the case of the squarefree integers (the case k = 2), if f is multiplicative, supported on the squarefree integers and f(p) = −1 for all primes p, then f = µ, the Möbiusfunction. A well known fact is that the Riemann hypothesis (RH) is equivalent to the square-root cancellation of the partial sums of the Möbius P 1/2+ function µ, more precisely, n≤x µ(n) x , for all > 0. Up to this date, the true arXiv:2101.00279v1 [math.NT] 1 Jan 2021 magnitude of the partial sums of the Möbiusfunction still unknown, even conditionally on RH; The unconditional upper bound obtained by the classical zero free region for P √ the Riemann zeta function is n≤x µ(n) x exp(−c log x), and the best conditional upper bound is due to Soundararajan [14] which states that under RH, P µ(n) √ √ n≤x x exp( log x(log log x)14).

These results for the M¨obiusfunction naturally raises the question of how small the partial sums of a multiplicative function supported on the squarefree integers can be. This led Wintner [18] to investigate a random model for the M¨obiusfunction which consists in randomizing the values f(p) = ±1 at primes p, and thus ending up with a 1 random multiplicative function. Wintner proved that a random multiplicative function P 1/2+ has partial sums n≤x f(n) x , for all > 0, almost surely, and also that these partial sums are not x1/2−, almost surely. Wintner’s seminal paper led to many investigations on the partial sums of a random multiplicative function. Here we refer for the interested reader: [4], [5], [7], [8], [9] and [11]. The best upper bound up to date √ P 2+ is due to Lau, Tenenbaum and Wu [11]: n≤x f(n) x(log log x) , for all > 0, almost surely; The best Ω bound is due to Harper [8]: The partial sums P f(n) are √ n≤x not x/(log log x)5/2+, for all > 0, almost surely.

By these results for a random multiplicative function we can naturally ask if there are examples of multiplicative functions f supported on the squarefree integers such that at primes f(p) = ±1, and such that its partial sums P f(n) have cancellation √ n≤x beyond the x-cancellation. In [3], the ﬁrst author exhibited examples of such functions √ such that P f(n) x exp(−c(log x)1/4), and further, by assuming RH we can go n≤x √ P 2/5+ much more beyond x-cancellation: There are examples such that n≤x f(n) x , for all > 0. Here we ask the same question over the k-free integers:

Question. For which values of exponents α > 0 there exists a multiplicative function f that is supported on the k-free integers, at primes f(p) = ±1 and such that P α n≤x f(n) x ? Here we give a reason to assume f(p) = ±1 over primes p. With this we exclude trivial solutions to our problem, because if f is supported on the squarefree integers (and hence supported on the k-free integers), and if we allow f(p) = 0 over primes, P by choosing f(p) such that p≤x |f(p)| x , for all > 0, then the partial sums P α n≤x f(n) x , for all α > 0. By assuming that f(p) = ±1 for all primes p, we guarantee that the set of the integers n ≥ 1 such that f(n) = ±1 has a positive lower density, since it contains the set of the squarefree integers which has density 6/π2. Moreover, with this condition, the event in which f has small partial sums means a non-trivial cancellation among the values f(n), as a consequence of the choice of the values f(pm), for primes p and powers m ≥ 1.

Our ﬁrst result states:

Theorem 1.1. There exists a multiplicative function f : N → {−1, 0, 1} supported on the k-free integers such that at primes p, f(p) = ±1, and such that

X x1/k f(n) , exp(c(log x)1/4) n≤x for some constant c > 0. 2 Thus, we can go beyond x1/k-cancellation when we consider k-free integers, and the exponent 1/k is critical in our results – If we assume the Generalized Riemann Hypothesis (GRH 1), then we can go beyond the exponent 1/k:

Theorem 1.2. Assume the Generalized Riemann Hypothesis. Then there exists a multiplicative function f : N → {−1, 0, 1} supported on the k-free integers such that at primes p, f(p) = ±1, and such that

X 1/(k+ 1 )+ f(n) x 2 , n≤x for any > 0.

Our examples in Theorems 1.1 and 1.2 are constructed in light of the extreme examples in the Erd¨osdiscrepancy theory for multiplicative functions.

The original formulation of the Erd¨osdiscrepancy problem (EDP) is whether there exists an arithmetic function f : N → {−1, 1} such that its partial sums over any homegenous arithmetic progression are uniformly bounded by some constant C > 0. As shown by Tao in [15], no such function exists, and moreover, we might say that the essence of the EDP is whether a completely multiplicative function f : N → {z ∈ C : |z| = 1} has bounded or unbounded partial sums, see [15].

A consequence of the solution of the EDP is that a completely multiplicative function f : N → {−1, 1} has unbounded partial sums. It is important to mention that if we allow completely multiplicative functions vanish over a ﬁnite subset of primes, then we might get bounded partial sums, for instance, any real and non-principal Dirichlet character χ. These Dirichlet characters are what we call the extreme examples in the Erd¨osdiscrepancy theory for multiplicative functions.

Following Granville and Soundararajan [6], deﬁne the “distance” up to x between two multiplicative functions |f|, |g| ≤ 1 as !1/2 X 1 − Re(f(p)g(p)) (f, g; x) := , D p p≤x where in the sum above p denotes a generic prime. Moreover, let D(f, g) := D(f, g; ∞), and we say that f pretends to be g, or that f is g-pretentious if D(f, g) < ∞. In the resolution of the EDP (see Remark 3.1 of [15] and Proposition 2.1 of [2]), it has been shown that a multiplicative function f : N → [−1, 1] with bounded partial 1We recall that ζ stands for the classical Riemann zeta function and for a Dirichlet character χ, P∞ χ(n) L(s, χ) stands for the classical L-function L(s, χ) = n=1 ns . The Riemann Hypothesis for ζ states that all zeros ζ(s0) = 0 with real part Re(s0) > 0 have real part equals 1/2. The Generalized Riemann Hypothesis is the same statement for L(s, χ). 3 sums must be χ-pretentious, for some real Dirichlet character χ : N → {−1, 0, 1}, 2 1 P√ |f(n)| provided that the logarithmic average log x x≤n≤x n 1. Thus, if we seek for multiplicative functions supported on the k-free integers with small partial sums, then we must look ﬁrstly to those that are χ-pretentious for some real and non-principal Dirichlet character χ. In our particular case, to construct examples for theorems 1.1 and 1.2 we assume much more than χ-pretentiousness. Indeed, we deduce the Theorem 1.1 from the following result:

2 Theorem 1.3. Let µk(n) ∈ {0, 1} be the indicator that n is k-free. Let g : N → {−1, 1} be a completely multiplicative function that it is χ-pretentious for some real and non- principal Dirichlet character χ, and further assume that X x1/k (1) |1 − g(p)χ(p)| √ . exp(c log x) p≤x

2 Let f = µkg be a multiplicative function supported on the k-free integers. Then there exists a constant λ > 0 such that X x1/k (2) f(n) . exp(λ(log x)1/4) n≤x

The condition (1) above essentialy says that the completely multiplicative function √ g is obtained by modifying a Dirichlet character χ in a quantity x1/k exp(−c log x) of primes p below x, and then the example for Theorem 1.1 is obtained by setting 2 f = µkg. In contrast, to get an example as in Theorem 1.2, we do not have such flexibility and we assume that our completely multiplicative function g coincides with a 2 Dirichlet character in all, but a finite subset of primes p, and then we set f = µkg just as in Theorem 1.1. Thus, these are our examples of multiplicative functions supported on the k-free integers with the smallest partial sums (under GRH), and are what we call real and non-principal Dirichlet characters restricted to the k-free integers. An interesting open question is whether these are the only examples, in the sense that a multiplicative function f needs or not to be χ-pretentious, in the situation that f is supported on the k-free integers, at primes f(p) = ±1 and such that the partial sums P 1/k− n≤x f(n) x , for some > 0. We observe that if f is within our class of examples of Theorem 1.2, then the P∞ −s 2 Dirichlet series of f, F (s) := n=1 f(n)n , is essentially the Dirichlet series of µkχ which is L(s, χ)/ζ(ks) if k is even, and it is L(s, χ)/L(ks, χ) if k is odd. Thus, under GRH, the Dirichlet series of f has analytic continuation to the half plane Re(s) > 1/2k. This suggests that the partial sums up to x of f could have cancellation x1/2k+, which would be much better than the result of Theorem 1.2. However, it is important to note that the existence of analytic continuation is not enough to ensure this cancellation. 4 Finally, if f is in the class of examples of Theorem 1.2, then f can not have partial sums up to x that are x1/2k−, for any > 0. This is because ζ(s) and L(s, χ) have an infinite number of zeros with Re(s) = 1/2. Thus, the Dirichlet series of f will have an infinite number of poles at Re(s) = 1/2k, unless that, in the case that k is even, every zero of ζ(ks) corresponds to a zero of L(s, χ) in the line Re(s) = 1/2k, or in the case that k is odd, every zero of L(ks, χ) corresponds to a zero of L(s, χ) in the line Re(s) = 1/2k. But this cannot happen due to zero density estimates, see the Grand Density Theorem (Theorem 10.4, page 260 of [10]).

2. Preliminaries

In this section we state classical results that we will need in the course of the proofs of the main results.

2.1. Partial sums of µ and of µχ. As stated in the introduction, for some constant c > 0, the partial sums of the M¨obiusfunction are (unconditionally) X x (3) µ(n) √ . exp(c log x) n≤x

This can be obtained by adapting the proof of the Prime Number Theorem as in [12], chapter 6. Indeed, this follows by combining an eﬀective Perron’s formula (Corollary 5.3 of [12]) with the classical zero free region for the Riemann zeta function, namely c those points s = σ + it such that σ ≥ 1 − log(|t|+4) , for some constant c > 0, and the estimate 1/ζ(s) log(|t| + 4), valid for s in the classical zero free region with |t| ≥ 1.

If χ is a real and non-principal Dirichlet character, then we also have that for some constant c > 0, the function µχ has (unconditionally) partial sums X x (4) µ(n)χ(n) √ . exp(c log x) n≤x

The proof of this result follows the same line of reasoning of the proof of (3) with some extra care. This extra care is related to excepetional zeros of L functions. Indeed, the Dirichlet series of µχ is 1/L(s, χ), where L(s, χ) is the Dirichlet L function associated to χ. The L function L(s, χ) has a zero free region of the same type of Riemann zeta function with at most one possible exception, a real zero β < 1 inside this region, see [12] Theorem 11.3. Further, by Theorem 11.4 of [12], whether there exists an exceptional zero or not, we have that for s = σ + it in this zero free region with suﬃciently large |t|, 1/L(s, χ) log(|t| + 4), and thus the proof of (4) is completed with an eﬀective Perron’s formula. 5 Now if we assume the Riemann Hypothesis (RH), then for any > 0 the partial sums X (5) µ(n) x1/2+. n≤x

A proof of this can be found at [17] Theorem 14.25 (C). In a few words, under

RH we can deduce that for s = σ + it with σ ≥ σ0 > 1/2 and |t| suﬃcently large, 2(1−σ)+ log ζ(s) σ0, (log(|t| + 4)) , for any > 0, where σ0, means that the implicit constant depends only in σ0 and , see Theorem 14.2 of [17]. Hence, under RH, 1/ζ(s) is analytic in Re(s) > 1/2 and 1/ζ(s) σ0, |t| , for any > 0. Then, the proof of (5) is completed with an eﬀective Perron’s formula.

Finally, under GRH, for a real and non-principal Dirichlet character χ, for any > 0 we have that X (6) µ(n)χ(n) x1/2+, n≤x and the proof follows the lines of the proof of (5).

2.2. The Dirichlet Hyperbola Method. This method allow us to evaluate the partial sums of the Dirichlet convolution between two arithmetic functions h and g. We P recall that the Dirichlet convolution is deﬁned as h ∗ g(n) = d|n h(d)g(n/d), where P the notation d|n means that d divides n. Denote Mf (x) := n≤x f(n). The Dirichlet hyperbola method states: If UV = x with U, V ≥ 1, then X X x X x (7) (h ∗ g)(n) = h(n)M + g(n)M − M (V )M (U). g n h n g h n≤x n≤U n≤V

3. Proof of the main results

3.1. Proof of Theorem 1.3. We begin with the following Lemma:

Lemma 3.1. Let h : N → [0, ∞) be a multiplicative function such that: (i) h(p) ≤ 2 and h(pr) ≤ h(p), for all primes p and all powers r ≥ 2; 1/k P x√ (ii) For some constant c > 0, p≤x h(p) exp(c log x) . Then there exists a δ > 0 such that X x1/k h(n) √ . exp(δ log x) n≤x

Proof. By partial summation, or by Kroenecker’s Lemma (see [13] page 390), we can obtain the stated conclusion if the following series converges for some δ > 0: ∞ √ X h(n) exp(δ log n) . n1/k n=1 6 Notation: By prkn we mean that r is the largest power of p that divides n. Since √ √ qP r P r log n = prkn log p ≤ prkn log p , we have that √ X h(n) exp(δ log n) X h˜(n) ≤ , n1/k n1/k n≤x n≤x √ where h˜ is the multiplicative function such that h˜(pr) = exp(δ log pr)h(pr), for all P h˜(n) primes p and all powers r ≥ 1. We will show that the series n≤x n1/k converges. For that, by the Euler product formula (see Theorem 2, page 106 of [16]), we only need to P P∞ h˜(pr) show that p∈P r=1 pr/k converges, where P stands for the set of primes. √ exp(δ log p) Let 0 < δ < c/2 be small such that p1/k < 1 for all primes p ∈ P. By the hypothesis (i)

∞ ∞ √ ∞ √ √ X h˜(pr) X h(pr) exp(δ log pr) X exp(δ r + 2 log p) r/k = r/k ≤ h(p) r+2 p p k r=2 r=2 r=0 p ∞ √ √ ∞ √ X exp(δ(r + 2) log p) exp(2δ log p) X exp(rδ log p) ≤ h(p) r+2 = h(p) 2/k r/k k p p r=0 p r=0 √ √ exp(2δ log p) 1 exp(2δ log p) = h(p) √ h(p) √ p2/k exp(δ log p) p1/k(p1/k − exp(δ log p)) 1 − p1/k √ h(p) exp(2δ log p) . δ p1/k

P Deﬁne T (x) = 0 if 0 ≤ x < 1 and T (x) = p≤x h(p) if x ≥ 1. By expressing a sum by a Riemann-Stieltjes integral and then using the integration by parts formula: √ √ X h(p) exp(2δ log p) Z x exp(2δ log t) = dT (t) p1/k t1/k p≤x 1 √ √ exp(2δ log x) Z x exp(2δ log t) T (x) + T (t) dt 1/k 1+ 1 x 1 t k 1 Z x 1 √ + √ dt exp((c − 2δ) log x) 1 t exp((c − 2δ) log t) 1, where in the penultimate inequality we used the hypothesis (ii).

Thus, ∞ √ X X h˜(pr) X h(p) exp(2δ log p) 1. pr/k δ p1/k p≤x r=1 p≤x

7 Lemma 3.2. Let g : N → {−1, 1} be a completely multiplicative function. Assume that for some real and non-principal Dirichlet character χ, g satisﬁes (1): X x1/k |1 − g(p)χ(p)| √ . exp(c log x) p≤x Then, for some δ > 0, X x1/k g(n) √ . exp(δ log x) n≤x

Proof. Let h = g ∗ χ−1, where χ−1 is the inverse of χ with respect to the Dirichlet convolution. Since χ is completely multiplicative, we have that χ−1 = µχ (Theorem 2.17 of [1]). Thus, χ−1 is multiplicative and has support on the squarefree integers. Therefore, for each prime p and any power r: r r −1 r X −1 p r r−1 −1 |h(p )| = |g ∗ χ (p )| = g(d)χ = |g(p ) + g(p )χ (p)| d d|pr = |g(pr)||1 + g(p)−1χ−1(p)| = |1 − g(p)χ(p)|.

Hence, |h| satisﬁes the hypothesis of Lemma 3.1. Since g = h ∗ χ, we have that P P P n≤x g(n) = n≤x h(n) m≤x/n χ(m), and as χ has bounded partial sums, we have P P that n≤x g(n) n≤x |h(n)|.

Before we state our next Lemma, we will introduce some notation. Given an arithmetic function f, we denote its partial sums up to x by X Mf (x) := f(n). n≤x

Lemma 3.3. Let g : N → {−1, 1} be multiplicative and let χ be a real and non-principal Dirichlet character. Assume that for some δ > 0 X x1/k |1 − g(p)χ(p)| √ , exp(δ log x) p≤x where k ≥ 2. Then, there exists a constant c > 0 such that x M (x) √ . µg exp(c log x)

Proof. Recall that χ = (µχ)−1, where the exponent −1 means that we are taking the inverse with respect to the Dirichlet convolution. Deﬁne h := µg∗(µχ)−1. Let 1/L(s, χ), Z(s) e H(s) be the Dirichlet series associated to µχ, µg and h, respectively. Recall the Euler product formula for L(s, χ): ∞ X χ(n) Y χ(p) χ(p)2 L(s, χ) = = 1 + + + ··· . ns ps p2s n=1 p∈P 8 For µg we have ∞ X µ(n)g(n) Y g(p) Z(s) = = 1 − . ns ps n=1 p∈P

By the rules of Dirichlet convolution, we can write H(s) = Z(s)L(s, χ), and hence

Y χ(p) χ(p)2 g(p) H(s) = Z(s)L(s, χ) = 1 + + + ··· 1 − ps p2s ps p∈P Y χ(p) − g(p) χ(p)n−1(χ(p) − g(p)) = 1 + + ··· + + ··· . ps pns p∈P

Since h is multiplicative, we have that h(pn) = χ(p)n−1(χ(p) − g(p)). Thus |h(p)| ≤ 2 and for all powers r, |h(pr)| ≤ |h(p)|. Moreover, since |χ(p) − g(p)| ≤ |1 − χ(p)g(p)|, we have by hypothesis that

1 X X X x k |h(p)| = |χ(p) − g(p)| ≤ |1 − χ(p)g(p)| √ . exp δ log x p≤x p≤x p≤x

Thus, by Lemma 3.1, there exists a new δ > 0 such that

1 x k |M (x)| ≤ M (x) √ x1/k. h |h| exp δ log x

P∞ |h(n)| In particular, by partial summation, the series n=1 n converges. Now we will use the Dirichlet Hyperbola method for µg = h ∗ µχ, see subsection 2.2. For all U ≥ 1 and V ≥ 1 such that UV = x, we have that

X x X x M (x) = h(n)M + µ(n)χ(n)M − M (U)M (V ) := A + B − C. µg µχ n h n µχ h n≤V n≤U

√x By (4), we have that for some constant a > 0, Mµχ(x) exp(a log x) . Now choose V = √ k−1 exp( log x), where 0 < < a. Further, let c > 0 be a parameter c < min k , a . Estimate for A.

X x X x |A| ≤ |h(n)M | |h(n)| µχ n n exp aplog x n≤V n≤V n x X |h(n)| x exp aplog x n exp aplog x V n≤V V x x √ 1 p √ 2 exp a log x − log x √ exp a log x 1 − log x x √ . exp c log x 9 Estimate for B. 1/k X x X x 1/k X 1 1 1− 1 |B| ≤ µ(n)χ(n)Mh x x k U k n n n1/k n≤U n≤U n≤U x x √ k−1 √ . exp k log x exp(c log x) Estimate for C.

1/k 1 UV x V k C = Mµχ(U)Mh(V ) √ exp(a log U) V exp(aplog x − log V ) x 1− 1 p √ V k exp a log x − log x x p √ k−1 √ exp a log x − log x + k log x x √ 1/2 √ k−1 exp log x a 1 − log x + k x √ . exp(c log x)

By combining these estimates for A, B, and C we conclude the proof.

Proof of Theorem 1.3. Let h := f ∗ g−1, where g−1 is the inverse of g with respect to the Dirichlet convolution. Let F , G and H be the Dirichlet series associated to f, g and h respectively. Since g is completely multiplicative, by the Euler product formula we have that −1 Y g(p) G(s) = 1 − . ps p∈P 2 For f = µkg, we have that Y f(p) f(p2) Y g(p) g(p)2 g(p)k−1 F (s) = 1 + + + ... = 1 + + + ··· + . ps p2s ps p2s p(k−1)s p∈P p∈P Since h = f ∗ g−1, we can write H(s) = F (s)/G(s), and hence

F (s) Y g(p) g(p)2 g(p)k−1 g(p) H(s) = = 1 + + ··· + 1 − G(s) ps p2s p(k−1)s ps p∈P Y g(p)k = 1 − . pks p∈P Since h is multiplicative, we obtain that h(pk) = −g(p)k and h(pr) = 0, for all r ≥ 1 k 1 1/k and r 6= k. Thus, h(n) = 0 unless n = m for some integer m ≥ 1. Let N(n ) be the indicator that n1/k is an integer. Now observe that if k is odd, then h(pk) = −g(p)k = 1 1/k 1/k 1/k −g(p), since g(p) = ±1, and hence h(n) = N(n )µ(n )g(n ). 10 Thus, in the case that k is odd, by Lemma 3.3, for some constant c > 0 we have that

1 X X 1 1 1 X x k Mh(x) = h(n) = 1 (n k )µ(n k )g(n k ) = µ(n)g(n) . N p 1/k n≤x n≤x 1 exp(c log x ) n≤x k

k 1 1/k 1/k If k is even, then h(p ) = −1, and hence h(n) = N(n )µ(n ). By (3), we have √x for some (new) constant c > 0, Mµ(x) exp(c log x) . Therefore,

1/k X 1 1 X x Mh(x) = 1 (n k )µ(n k ) = µ(n) . N p 1/k n≤x 1 exp(c log x ) n≤x k

Thus, for any integer k ≥ 1, for some constant c > 0,

x1/k Mh(x) . exp(cplog x1/k)

On the other hand, by Lemma 3.2, for some δ > 0, we have that

x1/k M (x) √ . g exp(δ log x)

Now we use the Dirichlet hyperbola method (subsection 2.2): For all U ≥ 1 and V ≥ 1 such that UV = x, we have that

X x X x (8) M (x) = h(n)M + g(n)M − M (V )M (U) := A + B − C. f g n h n g h n≤U n≤V √ We choose V = exp(( log x)), where 0 < < √c and U = x . Besides that, λ > 0 is a k V √ (k−1) parameter λ < min(δ , √c − ). k k Estimate for A.

X 1 X k 1 k |A| ≤ | N(n )Mg(x/n)| = |Mg(x/n )| n≤U 1 n≤U k 1 1 X x k 1 x k X 1 n n 1 p x p x 1 n≤U k exp δ log nk exp δ log U n≤U k

1 1 1 x k log U x k (log x − log V ) x k exp(log log x) √ √ 1 √ 1 exp(δ log V ) exp(δ (log x) 4 ) exp(δ (log x) 4 ) 1 x k 1 , exp(λ(log x) 4 ) √ since λ < δ . 11 Estimate for B.

1 X X X x k 1 |B| = |g(n)M (x/n)| ≤ |M (x/n)| h h q n x 1 n≤V n≤V n≤V k exp c log( n )

1 1 X 1 1 x k X 1 k x 1 1 n k n k n≤V exp √c plog x exp √c plog x n≤V k n k V 1 k−1 √ k 1 k−1 x exp k log x x k V k q c √ c 1 exp √ log x − log V exp √ log x − (log x) 2 k k

1 x k q c 1 k−1 √ exp √ log x − (log x) 2 − log x k k

1 1 x k x k 1 2 1 1 c (k−1) 2 exp (log x) 2 √ 1 − − exp λ(log x) k (log x)1/2 k

1 x k , 1 exp λ(log x) 4

(k−1) since 0 < λ < √c − . k k Estimate for C.

1 1 1 V k U k x k C = Mg(V )Mh(U) √ √ exp(δ log V ) exp(cplog U 1/k) exp(δ log V ) 1 1 x k x k p √ √ 1 exp(δ log exp( log x)) exp(δ (log x) 4 ) 1 x k 1 . exp(λ(log x) 4 )

Combining these estimates for A, B and C we complete the proof.

3.2. Proof of Theorem 1.2.

Proof. Let χ be any real and non-principal Dirichlet character mod q and g : N → {−1, 1} be the completely multiplicative function such that:

g(n) = χ(n), if gcd(n, q) = 1, g(p) = 1, for each prime p|q. 12 Then, by Lemma 2.4 of [3] :

|Mg(x)| maxy≥1 |Mχ(y)| Y 1 lim sup ω(k) ≤ . x→∞ (log x) ω(k)! log p p|k

α In particular, Mg(x) x , for each α > 0. Let h = f ∗ g−1, where g−1 is the Dirichlet inverse of g. As in the proof of theorems 1 1 1 1 k k k 1.1 and 1.3 above, if k is odd, we have that h(n) = N(n )µ(n )g(n ); If k is even, 1 1 1 k k then h(n) = N(n )µ(n ). In any case, by assuming GRH, we have (5) and (6), and hence, for any k ≥ 2, 1 + Mh(x) x 2k , for all > 0. Now we use the Dirichlet Hyperbola method following the same notation of (8), 2k 1 and setting U = x 2k+1 and V = x 2k+1 . It is noteworthy that these choices for U and V are optimal in our method.

Estimate for A. α X x X x X x α 1 (1−kα) |A| ≤ h(n)Mg = Mg x U k n nk nkα n≤U 1 1 n≤U k n≤U k

α 2 (1−kα) 2 +α 1− 2k 2 + α x x 2k+1 x 2k+1 ( 2k+1 ) x 2k+1 2k+1 .

Estimate for B. X x 1 + X 1 1 + 1− 1 − |B| ≤ g(n)Mh x 2k x 2k V 2k 1 + n 2k n≤V n≤V n

1 + 1 1− 1 − 2 + 1− 1 2 + 2k x 2k x 2k+1 ( 2k ) x 2k+1 ( 2k+1 ) x 2k+1 2k+1 .

Estimate for C.

α 1 + α 1 + 2k 1 + α + 2k C = Mg(V )Mh(U) V U 2k x 2k+1 x 2k+1 2k+1 x 2k+1 2k+1 2k+1 .

Since we have freedom to choose α and arbitrarily small, the proof is complete.

References

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Departamento de Matem´atica,Universidade Federal de Minas Gerais (UFMG), Brazil. Email address: [email protected] caiomaﬁ[email protected] [email protected]