Sign changes of error terms related to arithmetical functions

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de Bordeaux 19 (2007), 1–25

Sign changes of error terms related to

arithmetical functions

par Paulo J. ALMEIDA

R´esum´e. Soit H(x) = Pn≤x φ(n)

n −

6

π2x. Motivé par une con-jecture de Erdös, Lau a développé une nouvelle méthode et il a démontré que #{n ≤ T : H(n)H(n + 1) < 0} T. Nous consi-dérons des fonctions arithmétiques f (n) =P

d|n bd

d dont l’addition

peut ˆetre exprim´ee commeP

n≤xf (n) = αx + P (log(x)) + E(x).

Ici P (x) est un polynˆome, E(x) = −P

n≤y(x) bn

nψ x

n + o(1) avec

ψ(x) = x − bxc − 1/2. Nous généralisons la méthode de Lau et démontrons des résultats sur le nombre de changements de signe pour ces termes d’erreur.

Abstract. Let H(x) =Pn≤x φ(n)

n −

6

π2x. Motivated by a con-jecture of Erd¨os, Lau developed a new method and proved that #{n ≤ T : H(n)H(n + 1) < 0} T. We consider arithmetical functions f (n) =P

d|n bd

d whose summation can be expressed as

P

n≤xf (n) = αx + P (log(x)) + E(x), where P (x) is a polynomial,

E(x) = −P n≤y(x) bn nψ x n + o(1) and ψ(x) = x − bxc − 1/2. We

generalize Lau’s method and prove results about the number of sign changes for these error terms.

1. Introduction

We say that an arithmetical function f (x) has a sign change on integers at x = n, if f (n)f (n + 1) < 0. The number of sign changes on integers of f (x) on the interval [1, T ] is defined as

Nf(T ) = #{n ≤ T, n integer : f (n)f (n + 1) < 0}.

We also define zf(T ) = #{n ≤ T, n integer : f (n) = 0}. Throughout this

work, ψ(x) = x − bxc − 1/2 and f (n) will be an arithmetical function such

Manuscrit re¸cu le 8 janvier 2006.

This work was funded by Funda¸c˜ao para a Ciˆencia e a Tecnologia grant number SFRH/BD/4691/2001, support from my advisor and from the department of mathematics of University of Georgia.

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that

f (n) =X

d|n

bd

d for some sequence of real numbers bn.

The motivation for our work was a paper by Y.-K. Lau [5], where he proves that the error term, H(x), given by

X n≤x φ(n) n = 6 π2x + H(x)

has a positive proportion of sign changes on integers solving a conjecture stated by P. Erd¨os in 1967.

An important tool that Lau used to prove his theorem, was that the error term H(x) can be expressed as

(1) H(x) = − X n≤ x log5 x µ(n) n ψ x n + O 1 log20x , (S. Chowla [3])

We generalize Lau’s result in the following way

Theorem 1.1. Suppose H(x) is a function that can be expressed as

(2) H(x) = − X n≤y(x) bn nψ x n + O 1 k(x) ,

where each bn is a real number and

(i) y(x) increasing, x14 y(x) x

(log x)5+ D2

, for some D > 0, and

(3) X

n≤x

b4_n x logDx;

(ii) k(x) is an increasing function, satisfying limx→∞k(x) = ∞.

(iii) H(x) = H(bxc) − α{x} + θ(x), where α 6= 0 and θ(x) = o(1).

Let ≺ ∈ {<, =, ≤}. If #{1 ≤ n ≤ T : αH(n) ≺ 0} T then there exists a positive constant c0 and c0T disjoint subintervals of [1, T ], with each of

them having at least two integers, m and n, such that αH(m) > 0 and αH(n) ≺ 0. In particular,

(1) #{n ≤ T : αH(n) > 0} T ;

(2) if #{n ≤ T : αH(n) < 0} T , then NH(T ) T or zH(T ) T .

We consider arithmetical functions f (n) for which, the error term of the summation function satisfies the conditions of Theorem 1.1. A first class is described in the following result

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Theorem 1.2. Let f (n) be an arithmetical function and suppose the se-quence bn satisfies condition (3) and

(4) X n≤x bn= Bx + O x logAx

for some B real, D > 0 and A > 6 + D₂, respectively. Let α =P∞

n=1nbn2, γb = lim x→∞ X n≤x bn n − B log x and H(x) =X n≤x f (n) − αx +B log 2πx 2 + γb 2 . If α 6= 0, then Theorem 1.1 is valid for the error term H(x). Moreover, if f (n) is a rational function, then, except when α = 0, or B = 0 and α is rational, we have

NH(T ) T if and only if #{n ≤ T : αH(n) < 0} T.

Notice that this class of arithmetical functions is closed for addition, i.e., if f (n) and g(n) are members of the class then also is (f + g)(n). In the case considered by Lau, it was known that H(x) has a positive proportion of negative values (Y.-F. S. P´etermann [6]), so the second part of Theorem 1.2 generalizes Lau’s result. Another example is f (n) = _φ(n)n .

Using a result of U. Balakrishnan and Y.-F. S. P´etermann [2] we are able to apply Theorem 1.1 to more general arithmetical functions:

Theorem 1.3. Let f (n) be an arithmetical function and suppose the se-quence bn satisfies condition (3) and

(5) ∞ X n=1 bn ns = ζ β_(s)g(s)

for some β real, D > 0, and a function g(s) with a Dirichlet series expan-sion absolutely convergent for σ > 1−λ, for some λ > 0. Let α = ζβ(2)g(2) and H(x) = ( _P n≤xf (n) − αx, if β < 0, P n≤xf (n) − αx − Pbβc j=0Bj(log x)β−j if β > 0,

where the constants Bj are well defined. If α 6= 0, then Theorem 1.1 is

valid for the error term H(x).

Theorem 1.3 is valid for the following examples, where r 6= 0 is real: φ(n) n r , σ(n) n r , φ(n) σ(n) r .

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2. Main Lemma

The main tool used by Y.-K. Lau was his Main Lemma, where he proved that if H(x) =P n≤x φ(n) n − 6 π2x then Z 2T T Z t+h t H(u) du 2 dt T h,

for sufficiently large T and any 1 ≤ h log4T . Lau’s argument depends essentially on the formula (1). In this section, we obtain a generalization of Lau’s Main Lemma.

Main Lemma. Suppose H(x) is a function that can be expressed as (2) and satisfies conditions (i) and (ii) of Theorem 1.1. Then, for all large T and h ≤ min log T, k2(T ), we have

(6) Z 2T T Z t+h t H(u) du 2 dt T h32.

For any positive integer N , define

(7) HN(x) = − X d≤N bd dψ x d . The Main Lemma will follow from the next result.

Lemma 2.1. Assume the conditions of the Main Lemma and take D > 0 satisfying condition (i). Let E = 4 +D₂, then

(a) For any δ > 0, large T , any Y T and N ≤ y(T ), we have Z T +Y T (H(u) − HN(u))2 du Y N1−δ + Y k2_{(T )}+ y(T + Y ) (log T ) E ; (b) For all large T , N ≤ y(T ) and 1 ≤ h ≤ min log T, k2_{(T ), we have}

Z 2T T Z t+h t HN(u) du 2 dt T h32 + N3(log N )E.

Now we prove the Main Lemma:

Proof. Take N = T14 and δ > 0 small. Cauchy’s inequality gives us

Z t+h t H(u) du 2 ≤ 2 Z t+h t HN(u) du 2 + 2 Z t+h t (H(u) − HN(u)) du 2 .

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Since N = T14 then, for sufficiently large T , N3logEN T . So, using

part (b) of Lemma 2.1 we have Z 2T T Z t+h t HN(u) du 2 dt T h32 + N3logEN T h 3 2.

Using Cauchy’s inequality and interchanging the integrals, Z 2T T Z t+h t (H(u) − HN(u)) du 2 dt ≤ h Z 2T T Z t+h t (H(u) − HN(u))2 du dt ≤ h Z 2T +h T Z min(u,2T ) max(u−h,T ) (H(u) − HN(u))2dt ! du ≤ h2 Z 2T +h T (H(u) − HN(u))2 du h2 T + h N1−δ + T + h k2_{(T )} + y(2T + h) (log T ) E T + T h T h32

since y(2T + h) _{(log T )}TE+1 and h ≤ min(log T, k2(T )). Hence

Z 2T T Z t+h t H(u) du 2 dt T h32. 3. Step I

In this section, we will prove part (a) of Lemma 2.1. Using expression (2) and Cauchy’s inequality, we obtain

Z T +Y T (H(u) − HN(u))2du ≤ 2 Z T +Y T   y(u) X m=N +1 bm mψ u m   2 du+O Y k2_{(T )} . Let η(T, m, n) = max T, y−1(m), y−1(n), then

2 Z T +Y T   y(u) X m=N +1 bm mψ u m   2 du = 2 y(T +Y ) X m,n=N +1 bmbn mn Z T +Y η(T,m,n) ψ u m ψ u n du.

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The Fourier series of ψ(u) = u − buc − 1₂, when u is not an integer, is given by (8) ψ(u) = −1 π ∞ X k=1 sin(2πku) k , so we obtain 2 π2 y(T +Y ) X m,n=N +1 bmbn mn ∞ X k,l=1 1 kl Z T +Y η(T,m,n) sin 2πku m sin 2πlu n du. Now, the integral above is equal to

1 2 Z T +Y η(T,m,n) cos 2πu k m + l n − cos 2πu k m − l n du. For the first term we get

Z T +Y η(T,m,n) cos 2πu k m+ l n du _k 1 m+ l n . If _mk = _nl, then Z T +Y η(T,m,n) cos 2πu k m− l n du ≤ Y, otherwise Z T +Y η(T,m,n) cos 2πu k m− l n du 1 _mk − _nl .

Part (a) of Lemma 2.1 will now follow from the next three lemmas. Lemma 3.1. Let E = 4 +D₂ as in Lemma 2.1. Then

X m,n≤X |b_mbn| ∞ X k,l=1 kn6=lm 1 kl | kn − lm| X (log X) E_.

Lemma 3.2. If D > 0 satisfies condition (3), then X m,n≤X |bmbn| ∞ X k,l=1 1 kl ( kn + lm) X (log X) 1+D₂ . Lemma 3.3. For any δ > 0,

X N <m,n≤X |bmbn| mn ∞ X k,l=1 kn=lm 1 kl 1 N1−δ.

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In order to finish the proof of part (a) of Lemma 2.1 we just need to take X = y(T + Y ) in the previous lemmas. Hence

Z T +Y T (H(u) − HN(u))2 du Y N1−δ + y(T + Y ) (log T ) E + Y k2_{(T )}. 2 Before we prove the three lemmas above, we need the following technical result

Lemma 3.4. Let bn be a sequence satisfying condition (3). Then

X n≤N b2_n N logD2 N, X n≤N |b_n| N logD4 N, X n≤N b2_n n (log N ) 1+D₂ _, X n≤N |bn| n (log N ) 1+D 4 _, X n>N b4_n n2τ (n) 1 N1−δ, for any δ > 0.

Proof. Follows from Cauchy’s inequality, partial summations and the fact

that, for any > 0, τ (n) = O(n).

Remark. If H(x) can be expressed in the form (2), then

(9) |H(x)| ≤ X n≤y(x) |bn| n + O 1 k(x) (log x)1+D4 _.

Proof of Lemma 3.2 : Since the arithmetical mean is greater or equal to the geometrical mean, we have

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To complete the proof of Lemma 3.3 we use Cauchy’s inequality and Lemma 3.4. For any δ > 0,

X N <m,n≤X |bmbn| mn ∞ X k,l=1 kn=lm 1 kl X d≤X 1 d min 1, d 3 N3 1 2 X N <m≤X d|m b4_m m2 1 2 ! , therefore X N <m,n≤X |b_mbn| mn ∞ X k,l=1 kn=lm 1 kl !2 X d≤X 1 d2 min 1, d 3 N3 X D≤X X N <m≤X D|m b4_m m2 ! 1 N3 X d≤N d +X d>N 1 d2 _X N <m≤X b4 m m2 X D|m 1 1 N X N <m≤X b4_m m2τ (m) 1 N2−δ. Proof of Lemma 3.1 : This lemma is a generalization of Hilfssatz 6 in [9] of A. Walfisz. Notice first that

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For the first term, we use Lemma 3.4,

X m≤n≤X |bmbn| ∞ X k,l=1 lm≤kn 2 1 kl | kn − lm| ≤ 2 X m≤n≤X |bmbn| ∞ X k,l=1 lm≤kn 2 1 k2_ln = 2 X m≤n≤X |b_m||bn| n ∞ X k=1 1 k2 X l≤_2mkn 1 l X m≤n≤X |bm| |bn| n ∞ X k=1 log k k2 + log X k2 log X X n≤X |b_n| n X m≤n |b_m| X (log X)1+D2.

From the forth inequality of Lemma 3.4, we get X m≤n≤X |b_mbn| ∞ X k,l=1 kn≤lm₂ 1 kl | kn − lm| X (log X) 1+D₂ _.

The estimation of the third term is more complicated and we have to use a different approach. In this case, 1_l < 2m_kn, so that

X m≤n≤X |bmbn| ∞ X k,l=1 kn 2 <lm<kn 1 kl | kn − lm| < 2 X m≤n≤X |b_m||bn| n ∞ X k=1 1 k2 X kn 2m<l< kn m m kn − lm

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< 2 X m≤n≤X |bm| |bn| n ∞ X k=1 1 k2 X l≤kn_m−1 1 kn m − l ! + 2 X m≤n≤X |bm| |b_n| n ∞ X k=1 1 k2 X kn m−1<l< kn m m kn − lm. Now, taking L =kn_m − l, X m≤n≤X |b_m||bn| n ∞ X k=1 1 k2 X l≤kn_m−1 1 kn m − l ! ≤ X m≤n≤X |b_m||bn| n ∞ X k=1 1 k2 X L≤kn_m−1 1 L X (log X)1+D2 _,

as in the first term. If there exists an integer l with kn_m − 1 < l < kn m, then

m - kn. In this case, kn − lm = mkn_m and m < n. So, we have to estimate (12) X m<n≤X |b_m||bn| n ∞ X k=1 m- kn 1 k2_kn m .

Notice that the fractional part of kn_m is at least _m1. So, when k ≥ m, X m<n≤X |bm| |bn| n ∞ X k=m m k2 X m<n≤X |bm| |bn| n X (log X) D 2 _.

We are left with the estimation of X m<n≤X |b_m||bn| n X k<m m- kn 1 k2_kn m .

Since m - kn, given k and n, we can take ak,n, such that 1 ≤ ak,n < m and

ak,n ≡ kn mod m. Then, X m<n≤X |b_m||bn| n X k<m m- kn 1 k2_kn m ≤ X m<n≤X |b_m||bn| n X k<m m k2_a k,n ≤ X a,k≤X 1 ak2 X max(a,k)<m≤X m|bm| X m<n≤X kn≡a mod m |b_n| n .

We need to estimate the inner sums. In order to do that, we partition the interval [1, X] in intervals of the form [M, 2M ) and apply Cauchy’s

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inequality. Take 1 ≤ P ≤ Q ≤ X, then, X P ≤m<2P m|bm| X Q≤n<2Q kn≡a mod m |b_n| n P Q X P ≤m<2P |bm| X Q≤n<2Q kn≡a mod m |bn|.

Next, we apply Cauchy’s inequality twice, first to the first sum on the right and afterwards to the second sum:

X P ≤m<2P |b_m| X Q≤n<2Q kn≡a mod m |b_n| !!2 ≤ X P ≤M <2P b2_M X P ≤m<2P X Q≤n<2Q kn≡a mod m |bn| !2 P logD2 P X P ≤m<2P X Q≤n<2Q kn≡a mod m b2_n X Q≤N <2Q kN ≡a mod m 1 ! P logD2 P X P ≤m<2P 1 + Q_m (k,m) X Q≤n<2Q kn≡a mod m b2_n ! . Since m ≤ 2P ≤ 2Q, we have _mQ ≥ 1

2. Using also (k, m) ≤ k, we obtain

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By a theorem of S. Ramanujan [7],P

n≤Xτ2(n) ∼ X log

3_{X and by another}

application of Cauchy inequality and condition (3), we get

_X Q≤n<2Q b2_n τ (kn − a)2≤ X Q≤n<2Q b4_n X Q≤n<2Q τ2(kn − a) Q logDQ X kQ−a≤N <2kQ−a τ2(N ) kQ2logD+3X. Therefore, X P ≤m<2P m|bm| X Q≤n<2Q kn≡a mod m |bn| n P Q kQ(log P )D2 X Q≤n<2Q b2_nτ (kn − a) 1 2 P Q kQ(log X)D2(kQ2logD+3X) 1 2 1₂ P k34(log X)1+ D 2.

The number of pairs of intervals of the form ([P, 2P ), [Q, 2Q)) to be con-sidered is at most log2X, hence

X a,k≤X 1 ak2 X a<m≤X m|bm| X m<n≤X kn≡amod m |b_n| n X a,k≤X k34 ak2 X P,Q P (log X)1+D2 X (log X)4+D2 _.

The fourth term is treated as the third, hence X m≤n≤X |b_mbn| ∞ X k,l=1 kn<lm<2 kn 1 kl | kn − lm| X (log X) 4+D₂ _.

This completes the proof of Lemma 3.1.

4. Step II

In this section we prove part (b) of Lemma 2.1. From equation (8), we get Z b a ψ(u) du = 1 2π2 ∞ X k=1 cos(2πkb) k2 − cos(2πka) k2 .

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Using the definition of HN stated in (7), we obtain Z t+h t HN(u) du = − X m≤N bm m Z t+h t ψ(u m) du = − 1 2π2 X m≤N bm ∞ X k=1 cos 2πk(t+h)_m − cos 2πkt_m k2 .

As usual, let us write e(t) for e2πit, then

Z t+h t HN(u)du = 1 4π2 X m≤N bm ∞ X k=1 e kh_m − 1 e kt_me−k(2t+h)_m − 1 k2 . Therefore, 16π4 Z 2T T Z t+h t HN(u)du 2 dt = Z 2T T X m≤N bm ∞ X k=1 e kh_m − 1 e kt_m e −k(2t+h)_m − 1 k2 2 dt = X m,n≤N bmbn ∞ X k,l=1 e kh_m − 1 e −lh n − 1 (kl)2 × Z 2T T e kt m e −lt n e −k (2t + h) m − 1 e l (2t + h) n − 1 dt.

After multiplying the terms inside the integral above, we obtain the follow-ing four terms that we will estimate below:

Z 2T T e kt m − lt n dt + e lh n − kh m Z 2T T e lt n− kt m dt − e −kh m Z 2T T e −lt n − kt m dt − e lh n Z 2T T e lt n + kt m dt.

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Notice that, R2T T e 2πirt_dt ≤ 1

π|r|, for any r 6= 0. We begin with the last

term and use |e(t) − 1)| ≤ 2 and Lemma 3.2. Then X m,n≤N bmbn ∞ X k,l=1 e kh_m − 1 e −lh_n − 1 e lh_n (kl)2 Z 2T T e lt n + kt m dt ≤ 4 π X m,n≤N |bmbn| ∞ X k,l=1 1 (kl)2 l n+ k m X m,n≤N |bmbn| ∞ X k,l=1 m k n l 1 kl(lm + kn) N 3_{(log N )}1+D₂ _.

The third term is treated similarly to obtain X m,n≤N bmbn ∞ X k,l=1 e kh_m − 1 e −lh n − 1 e − kh m (kl)2 Z 2T T e −lt n − kt m dt N3(log N )1+D2 _. Now, if kn = lm, then Z 2T T e kt m − lt n dt + e lh n − kh m Z 2T T e lt n − kt m dt = 2T. If kn 6= lm then, Z 2T T e kt m − lt n dt + e lh n − kh m Z 2T T e lt n − kt m dt 1 _mk −_nl . Let us study first the case when kn 6= lm,

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by Lemma 3.1. Similarly, X m,n≤N bmbn ∞ X k,l=1 kn6=lm e kh_m − 1 e −lh_n − 1 e lh_n −kh m (kl)2 Z 2T T e lt n − kt m dt N3logEN. If kn = ml, we will use |e(t) − 1| min(1, |t|) instead. The expression obtained has some similarities with Lemma 3.3. We are going to use the same argument to prove:

(13) X m,n≤N |bmbn| ∞ X k,l=1 kn=lm 1 (kl)2 min 1,kh m min 1,lh n h32.

As in Lemma 3.3, take d = (m, n), α = m_d, β = n_d and γ = _αk. So l = βγ and then ∞ X k,l=1 kn=lm 1 (kl)2 min 1,kh m min 1,lh n = 1 α2_β2 ∞ X γ=1 1 γ4 min 1,hγ d 2 . If d ≤ h, we obtainP∞ γ=1 γ14 min1,hγ_d2= π₉₀4, and if h < d ≤ N , ∞ X γ=1 1 γ4 min 1,hγ d 2 = h d 2 X γ≤d_h 1 γ2 + X γ>d_h 1 γ4 h d 2 + h d 3 h d 2 . Therefore, X m,n≤N |bmbn| ∞ X k,l=1 kn=lm 1 (kl)2 min 1,kh m min 1,lh n = X m,n≤N |bmbn| (m, n)4 m2_n2 ∞ X γ=1 1 γ4 min 1, hγ (m, n) 2 ≤ X d≤N d4 X m,n≤N d=(m,n) | bmbn| m2_n2 ∞ X γ=1 1 γ4 min 1,hγ d 2 X d≤h d2 X m≤N d|m |bm| m2 2 + h2 X h<d≤N dX m≤N d|m |bm| m2 2 .

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Since d|m we have m > h and so, h2P h<d≤N dP m≤N d|m |bm| m2 2 h1+δ_{. To}

estimate the first term we begin with H¨older inequality: X d≤h d4 X m≤N d|m |b_m| m2 2 ≤X d≤h d4 X m≤N d|m |b_m|4 m2 1₂_X M ≤N d|M 1 M2 3₂ .

The third sum is O _d13 (similar to (11)). Then

X d≤h d4 X m≤N d|m |b_m|4 m2 1₂_X M ≤N d|M 1 M2 3₂ X d≤h d X d≤m≤N d|m |b_m|4 m2 1₂ X d≤h d2 1₂_{h X} D≤h _X D≤m≤N D|m |bm|4 m2 i1₂ h32 X m≤N |bm|4 m2 X D≤h D|m 1 1 2 h32 X m≤N |b_m|4 m2 τ (m) 1₂ .

The last part of Lemma 3.4 implies P

m≤N |bm|4

m2 τ (m) = O(1), where the

underlying constant doesn’t depend on N . Therefore, we obtain inequality

(13) and part (b) of Lemma 2.1, follows. 2

5. A general Theorem

In this section, we prove Theorem 1.1, from which the main Theorems 1.2 and 1.3, will be deduced.

Proof of Theorem 1.1 : From the Main Lemma, we have, for all large T and h ≤ min log T, k2(T ),

Z 2T T Z t+h t H(u) du 2 dt T h32.

Assume #{n ≤ T : αH(n) ≺ 0} T . Let c > 0 be a constant and T be sufficiently large, such that #{n ≤ 2T : αH(n) ≺ 0} > cT. Divide the interval [1, 2T ] into subintervals of length h, where h is a sufficiently large integer satisfying h ≤ log T . Then more than cT /h of those subintervals must have at least one integer n with αH(n) ≺ 0. Let C be the set of the subintervals which satisfy this property. Write C = {Jr| 1 ≤ r ≤ R}, where

the subintervals are indexed by their positions in the interval [1, 2T ] and where R > cT /h. Define Ks = J3s−2, for 1 ≤ s ≤ R/3, and let D be the

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set of these subintervals. We have #(D) > cT /3h. Notice that any two members of D are separated by a distance of at least 2h.

Let M be the number of subintervals K in D for which there exists an integer n in K such that αH(n) ≺ 0 and αH(m) ≤ 0 for every integer m ∈ (n, n + 2h), and let S be the set of the corresponding values of n. Lemma 5.1. For some absolute constant c1, we have M ≤ c1 T

h32.

Proof. Since H(x) = H(bxc) − α{x} + θ(x), then αH(x) − αH(bxc) = −α2{x} + αθ(x). So, if x is sufficiently large and not an integer then

(14) −5

4α

2_{{x} < αH(x) − αH(bxc) < −}3

4α

2_{x}.

Let n1 be the smallest integer such that any non integer x > n1 satisfies

condition (14). If #{n ∈ S : n ≥ n1} = 0 then M ≤ n1, so, the lemma is

clearly true for sufficiently large T . Otherwise,

#{n ∈ S : n ≥ n1} ≥ M − n1 M.

Take n ∈ S with n ≥ n1 and t ∈ [n, n + h]. For any integer m ∈ [t, t + h],

αH(m) ≤ 0. Now, for any 1 ≤ j < h, Z [t]+j+1 [t]+j H(u) du = Z [t]+j+1 [t]+j (H(u) − H([t] + j)) du + Z [t]+j+1 [t]+j H([t] + j) du. Therefore, by (14), Z [t]+j+1 [t]+j αH(u) du < Z 1 0 −3 4α 2_x dx + αH([t] + j) < −3 8α 2_,

because αH([t] + j) ≤ 0. Since [t] ≥ n, we also have Z [t]+1 t αH(u) du < Z 1 {t} −3 4α 2_x dx + αH([t]) (1 − {t}) < 0 and Z t+h [t]+h αH(u) du < Z {t} 0 −3 4α 2_x dx + αH([t] + h){t} ≤ 0. Hence, Z t+h t H(u) du ≥ 3 8|α|(h − 1).

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Take an integer r = r(T ) such that 2r> (log T )3+D2_{. Using (6) and (9), we} obtain Z 2T 0 Z t+h t H(u) du 2 dt = Z T 2r 0 Z t+h t H(u) du 2 dt + r X j=0 Z T 2j−1 T 2j Z t+h t H(u) du 2 dt T 2rh 2_{(log T )}2+D₂ _{+ h}3 2 r X j=0 T 2j T h32,

due to h ≤ log T . On the other hand, Z 2T 0 Z t+h t H(u) du 2 dt ≥ X n∈S Z n+h n Z t+h t H(u) du 2 dt ≥ X n∈S n≥n1 Z n+h n 3 8|α|(h − 1) 2 dt M h3_. Hence M ≤ c1 T h32

for some absolute constant c1.

Take c0 = c/6h. If h is a suitably large integer such that c1T < c0T h

3 2,

then there are at least c0T intervals K in D such that αH(n) ≺ 0 for some

integer n ∈ K and αH(m) > 0 for some integer m lying in (n, n + 2h). Now, suppose #{n ≤ T : αH(n) ≤ 0} T . Take T sufficiently large and take the order relation ‘ ≺’ to be ‘ ≤’. Therefore, we have c0T integers m

in the interval [1, 2T ], for which αH(m) is positive. In this case, #{n ≤ T : αH(n) > 0} T.

If we don’t have #{n ≤ T : αH(n) ≤ 0} T , then #{n ≤ T : αH(n) > 0} = T (1 + o(1)).

Hence, part 1 of Theorem 1.1 is proved. Next, we prove part 2. Take ‘ ≺’ to be ‘ <’. Then, there exists a positive constant c0 and c0T disjoint

subintervals of [1, T ], with each of them having at least two integers, m and n, such that H(m) > 0 and H(n) < 0. Therefore, in each of those intervals we have at least one l with either H(l) = 0 or H(l)H(l + 1) < 0. Whence, zH(T ) > c₂0T or NH(T ) > c₂0T .

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6. A class of arithmetical functions

In this section, we consider arithmetical functions f (n), such that the sequence bnsatisfies conditions (3) and (4). We begin with some elementary

results about this class of arithmetical functions. Using condition (4), we immediately obtain the following lemma:

Lemma 6.1. Let bn be a sequence of real numbers satisfying (4), for some

constants B and A > 1, then there exist constants γb and α such that

X n≤x bn n = B log x + γb+ O 1 logA−1x , (15) ∞ X n=1 bn n2 = α, (16) X n>x bn n2 = B x + O 1 x logA−1x . (17)

Next, we calculate the sum of f (n) and describe the error term H(x). Lemma 6.2. Let bn be a sequence of real numbers as in Lemma 6.1, then

(18) X n≤x f (n) =X n≤x X d|n bd d = αx − B log 2πx 2 − γb 2 + H(x), where, (19) H(x) = − X n≤ x logC x bn nψ x n + O 1 logCx + O 1 logA−C−1x , for any 0 < C < A − 1. Proof. We have X n≤x f (n) =X n≤x X d|n bd d = X d≤x bd d X n≤x d|n 1 = X m≤x X d≤_mx bd d.

We separate the double sum above in two parts. Let 0 < C < A − 1 and y = logCx. Then (20) X m≤x X d≤_mx bd d = X n≤y X d≤x_n bd d + X d≤x_y bd d X y<n≤x_d 1.

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In order to evaluate the first term on the right, we start with an application of formula (15): X n≤y X d≤_nx bd d = X n≤byc B log x − B log n + γb+ O 1 logA−1(x_n) !! = Bbyc log x − B X n≤byc log n + γbbyc + O y logA−1(x_y) ! .

Recall that by Stirling formula, P

n≤byclog n = byc logbyc − byc + logbyc 2 + log 2π 2 + O 1 y . Hence, X n≤y X d≤_nx bd

d = Bbyc (log x − logbyc + 1) + γbbyc −

B(log 2πy) 2 + O y logA−1x + O 1 y . For the second term, we get

X d≤x_y bd d X y<n≤x_d 1 = X d≤x_y bd d jx d k − byc = x ∞ X d=1 bd d2 − x X d>x_y bd d2 − X d≤x_y bd d ψ x d − 1 2+ byc X d≤x_y bd d = αx − By − X d≤x_y bd d ψ x d −γb+ B log x 2 − Bbyc(log x − log y) +B log y

2 − γbbyc + O y logA−1x . Notice also that Bbyc(log y − logbyc) = Bbyc− log1 −{y}_y = B{y} + O1_y. Joining everything together, we obtain

H(x) = X n≤x f (n) − αx − B log 2πx 2 − γb 2 = − X d≤ x logC x bd dψ x d + O 1 logA−C−1x + O 1 logCx .

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Proof of Theorem 1.2 : We just have to show that H(x) satisfies the con-ditions of Theorem 1.1. From Lemma 6.2, for any x

H(x) − H(bxc) = −α{x} −B 2 (log 2πbxc − log 2πx) = −α{x} +B 2 {x} x + O 1 x2 . In Lemma 6.2, we also obtained

H(x) = − X n≤ x logC x bn nψ x n + O 1 logCx + O 1 logA−C−1x ,

for any 0 < C < A − 1. Take C = 5 +D₂, y(x) = _logxC_x and k(x) =

min logCx, logA−C−1x. Since A > 6 + D/2, then C < A − 1 and A − C − 1 > 0. The first part of Theorem 1.2 now follows from Theo-rem 1.1.

Suppose that f (n) takes only rational values. In order to prove the second part of Theorem 1.2, we use the following result of A. Baker [1]. Proposition. Let α1, . . . , αnand β0, . . . , βndenote nonzero algebraic

num-bers. Then β0+ β1log α1+ · · · + βnlog αn6= 0.

Using the result above, we obtain the next lemma.

Lemma 6.3. Let f (n) be a rational valued arithmetical function and sup-pose the sequence bnsatisfies condition (4) for some real B and A > 1. Let

r be a real number and suppose H(x) is given by (18). Then

(1) If B = 0 and α is irrational then #{n integer : H(n) = r} ≤ 1; (2) If B is a nonzero algebraic number then #{n integer : H(n) = r} ≤ 2; (3) If B is transcendental then there exists a constant C that depends on

r and on the function f (n), such that

#{n ≤ T, n integer : H(n) = r} < (log T )C.

Proof. Suppose that B = 0 and α is irrational. Suppose also that there are two integers, say M 6= N , such that H(M ) = H(N ). Then

X n≤M f (n) − αM +γb 2 = X n≤N f (n) − αN + γb 2. But this implies that α is rational, a contradiction.

Next, suppose B 6= 0 is algebraic number and that there are M > N > Q integers, satisfying H(M ) = H(N ) = H(Q). We have

X n≤M f (n) − αM + B log 2πM 2 + γb 2 = X n≤N f (n) − αN +B log 2πN 2 + γb 2

(22)

which implies α = B M − N log M N + 1 M − N X N <n≤M f (n). Consequently B log    M N _{M −N}1 M Q _{M −Q}1   = 1 M − Q X Q<n≤M f (n) − 1 M − N X N <n≤M f (n).

We are going to prove that

(21) M N _{M −N}1 6= M Q _{M −Q}1 .

Since B is a nonzero algebraic number and the values of f (n) are rational, for any integer n, the proposition implies

B log    M N _{M −N}1 M Q _{M −Q}1   6= 1 M − Q X Q<n≤M f (n) − 1 M − N X N <n≤M f (n),

and so we get a contradiction, which implies #{n integer : H(n) = r} ≤ 2, for any real number r. In fact, instead of proving (21), we are going to prove that

(22) MN −QQM −N < NM −Q,

for any positive integers M > N > Q. Clearly, this implies (21). The inequality (22) is just a particular case of the geometric mean-analytic mean inequality (23) n Y i=1 ui !_n1 ≤ 1 n n X i=1 ui,

where equality happens only if u1 = u2 = · · · = un. In fact, if we take

n = M − Q, ui = M for 1 ≤ i ≤ N − Q and ui = Q for N − Q < i ≤ M − Q,

we derive

MN −QQM −N

1

M −Q _< 1

M − Q((N − Q)M + (M − N )Q) = N. Hence, we obtain (22) and part 2 of the Lemma.

Finally we prove part 3. Suppose r is a real number such that #{n ≤ T : H(n) = r} ≥ 4.

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Let Q < N < M be the three smallest positive integers in the above set, then 0 6= B log    M N _{M −N}1 M Q _{M −Q}1   = 1 M − Q X Q<n≤M f (n) − 1 M − N X N <n≤M f (n).

Suppose L is such that H(L) = r. Then L > N > Q, and as in part 2:

0 6= B log    L N _L−N1 L Q _L−Q1   = 1 L − Q X Q<n≤L f (n) − 1 L − N X N <n≤L f (n).

After we cross multiply the two expressions above, we obtain

log    M N _{M −N}1 M Q 1 M −Q   = r1log    L N _L−N1 L Q 1 L−Q   ,

for some rational r1. Therefore, there are four rational numbers r2, r3, r4

and r5, such that

Lr2 _{= M}r3_Nr4_Qr5_.

Now, any prime dividing L must divide M N Q. Notice that, if p is a prime, k is an integer and pk≤ x then k ≤ log x_{log p}. Therefore, the number of integers smaller than x, which have all prime divisors smaller than M is smaller than (log x)π(M ). This finishes the proof. Except when α = 0, or B = 0 and α is rational, we cannot have zH(T )

T . Hence, we obtain the second part of Theorem 1.2. Example. We finish this section by proving that Theorem 1.2 is valid for the arithmetical function _φ(n)n .

Notice that n φ(n) = Y p|n 1 + 1 p − 1 =X d|n µ2_(d) φ(d). Let bn= µ 2_(n)n φ(n) , then f (n) = P d|n bd d. In [8], R. Sitaramachandrarao proved that X n≤x µ2(n)n φ(n) = x + O x12 ,

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so condition (4) is satisfied for any A and with B = 1. By Merten’s Theorem Q p|n 1 −1_p−1 ≤Q p≤n 1 −1_p−1 ∼ eγ_{log n, hence} X n≤x b4_n=X n≤x µ2(n) n 4 φ4_(n) = X n≤x O log4n = O x log4x ,

and condition (3) is satisfied for D ≥ 4. In this case, α = ζ(2)ζ(3) ζ(6) , γb = γ + X p log p p(p − 1) and H(x) =X n≤x n φ(n) − ζ(2)ζ(3) ζ(6) x + log x 2 + log 2π + γ +P p log p p(p−1) 2 .

Since B = 1 we can apply Theorem 6.3, and so z(T ) ≤ 2. Therefore, if #{n ≤ T : αH(n) < 0} T, then NH(T ) T .

7. Second class of arithmetical functions

Given a sequence of real numbers bn, and a complex number s, we define

the Dirichlet series B(s) = P∞

n=1nbns. In this section, we consider

arith-metical functions f (n), such that the sequence bn satisfies conditions (3)

and (5) for some D > 0, β real and a function g(s) with a Dirichlet series expansion absolutely convergent for σ > 1 − λ, for some λ > 0.

U. Balakrishnan and Y.-F. S. P´etermann [2] proved that:

Proposition. Let f (n) be a complex valued arithmetical function satisfying

∞ X n=1 f (n) ns = ζ(s)ζ β_{(s + 1)g(s + 1),}

for a complex number β, and g(s) having a Dirichlet series expansion g(s) = ∞ X n=1 cn ns,

which is absolutely convergent in the half plane σ > 1 − λ for some λ > 0. Let β0 be the real part of β. If

ζβ(s)g(s) = ∞ X n=1 bn ns

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then there is a real number b, 0 < b < 1/2, and constants Bj, such that,

taking y(x) = x exp −(log x)b_{and α = ζ}β_(2)g(2),

X n≤x f (n) = ( αx −P n≤y(x) bn nψ x n + o(1) if β0< 0, αx +P[β0] j=0Bj(log x)β−j−Pn≤y(x) bnnψ x n + o(1) if β0≥ 0,

The real version of the previous proposition allows us to prove Theo-rem 1.3:

Proof. Notice that, for any c > 0, logcbxc = logcx − c{x}_x logc−1x + O 1_x . So, H(x) = H(bxc) − α{x} + o(1). From the previous proposition, there is an increasing function k(x), with limx→∞k(x) = ∞, such that

H(x) = − X n≤y(x) bn nψ x n + O 1 k(x) ,

where y(x) = x exp −(log x)b, for some 0 < b < 1/2. Hence, the result

follows from Theorem 1.1.

Acknowledgements: I would like to thank my advisor Andrew Granville for his guidance and encouragement in this research and to the Referee for his useful comments.

References

[1] A. Baker, Linear forms in the logarithms of algebraic numbers (III). Mathematika 14 (1967), 220–228.

[2] U. Balakrishnan, Y.-F. S. P´etermann, The Dirichlet series of ζ(s)ζα(s + 1)f (s + 1): On an error term associated with its coefficients. Acta Arithmetica 75 (1996), 39–69. [3] S. Chowla, Contributions to the analytic theory of numbers. Math. Zeit. 35 (1932), 279–

299.

[4] G. H. Hardy, E. M. Wright, An Introduction to the Theory of Numbers. 5th ed. Oxford 1979.

[5] Y.-K. Lau, Sign changes of error terms related to the Euler function. Mathematika 46 (1999), 391–395.

[6] Y.-F. S. P´etermann, On the distribution of values of an error term related to the Euler function. Th´eorie des Nombres, (Quebec, PQ, 1987), 785–797.

[7] S. Ramanujan, Collected papers. Cambridge, 1927, 133–135.

[8] R. Sitaramachandrarao, On an error term of Landau - II. Rocky Mount. J. of Math. 15 2 (1985), 579–588.

[9] A. Walfisz, Teilerprobleme II. Math. Zeit. 34 (1931), 448–472. Paulo J. Almeida

Departamento de Matem´atica Universidade de Aveiro

Campus Universit´ario de Santiago 3810-193 Aveiro, Portugal E-mail : paulo@mat.ua.pt