Journal of Inequalities in Pure and
Applied Mathematics
A NOTE ON THE TRACE INEQUALITY FOR PRODUCTS
OF HERMITIAN MATRIX POWER
volume 3, issue 5, article 70,
2002.
OLIVIER BORDELLÈS
22 rue Jean Barthélemy
43000 Le Puy-En-Velay,
FRANCE.
EMail: borde43@wanadoo.fr
Received 30 August, 2002;
accepted 16 October, 2002.
Communicated by: L. Toth
Abstract
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c
2000
Victoria University
ISSN (electronic): 1443-5756
092-02
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Abstract
We use the recent theory of integer points close to a smooth curve developed
by Huxley-Sargos and Filaseta-Trifonov to get an asymptotic formula for short
sums of a class of multiplicative functions.
2000 Mathematics Subject Classification: 11N37, 11P21
Key words: Multiplicative Functions, Short Sums, Integer points close to a curve
On Short Sums of Certain
Multiplicative Functions
Contents
1
Introduction and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
Olivier Bordellès
3
6
9
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1.
Introduction and Notation
Let k > 2 be an integer. A positive integer n is said to be k−free (resp. k−full)
if, for any prime p | n, the p−adic valuation vp (n) of n satisfies vp (n) < k
(resp. vp (n) > k), and we use the terms squarefree or squarefull when k = 2.
We denote by µk the multiplicative function defined by

 1, if n is k − free
µk (n) :=

0, otherwise.
Obtaining gap results for k−free (or k−full) numbers is a very famous problem in analytic number theory (see [1] and the references). The best estimation
in this direction has been obtained by Filaseta and Trifonov
([1]) who showed
that, for x sufficiently large, any interval of the type x; x + cx1/(2k+1) log x
(c := c (k) > 0) contains a k−free number.
A dual problem is to get an asymptotic formula for µk . This requires estimations for short sums of multiplicative functions, but such results are still
relatively rare in the literature (see [2, 6]). In this paper, we are motivated by
finding asymptotic results for short sums of the following class of arithmetical functions: define M to be the set of multiplicative functions f verifying
0 6 f (n) 6 1 for any positive integer n and f (p) = 1 for any prime number
p. If f ∈ M, we set
!
∞
Y
X
f pl
1
P (f ) :=
1−
1+
.
p
pl
p
l=1
On Short Sums of Certain
Multiplicative Functions
Olivier Bordellès
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We prove:
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Theorem 1.1. Let ε > 0 and x, y > 2 be real numbers such that x6ε 6 y 6 x.
If f ∈ M, we have as x → +∞ :
X
f (n) = yP (f ) + O x1/6+ε + y 1/2 .
x<n6x+y
Moreover, if y 6 x1/3 , then:
X
f (n) = yP (f ) + O x1/7+ε + x1/21+ε y 2/3 .
x<n6x+y
Olivier Bordellès
For example, using this, we get:
X
x<n6x+y
µ2 (n) =
On Short Sums of Certain
Multiplicative Functions
6y
1/6+ε
1/2
+
y
+
O
x
π2
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6ε
with x 6 y 6 x. The proof of this theorem uses a convolution argument,
and the new theory of integer points close to a curve (see [1, 3, 4]) arises as the
crucial point to estimate the difference of integer parts.
In what follows, a, b, d, k, l, m, n, q, B, N will always denote positive integers, p will be a prime number and [x] is the integral part of x. For any X, Y > 0,
the notation X Y means there exists C > 0 such that X 6 CY. the notation
X Y means X Y and Y X simultaneously.
If f, g are two arithmetical functions, the Dirichlet convolution product f ∗ g
is defined by
X
(f ∗ g) (n) :=
f (d) g (n/d) .
d|n
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P
µ (n) is the Möbius function, τ (n) :=
d|n 1 the classical divisor function,
and, more generally, τ(k) (n) is the arithmetical function defined by
τ(k) (n) :=
X
1.
dk |n
If ϕ is any smooth function and if δ > 0 is any real number, we set
R (ϕ, N, δ) := |{n ∈ ]N ; 2N ] ∩ Z, ∃ m ∈ Z, |ϕ (n) − m| 6 δ}| .
Thus, if δ is sufficiently small, R (ϕ, N, δ) counts the number of integer points
close to the curve y = ϕ (x) when N < x 6 2N, and then
X
([ϕ (n) + δ] − [ϕ (n) − δ]) .
R (ϕ, N, δ) =
N <n62N
Olivier Bordellès
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We will therefore use the following principle: let M be a large real number,
ϕ (n) and δ (n) two functions such that δ := maxn6M δ (n) satisfies 0 < δ 6
1/4. Then we have
X
(1.1)
([ϕ (n) + δ (n)] − [ϕ (n)]) max R (ϕ, N, δ) log M.
n6M
On Short Sums of Certain
Multiplicative Functions
N 6M
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2.
Auxiliary Results
We collect here the lemmas we will use to show the theorem. The first was
proved by Huxley and Sargos in [3, 4] using divided differences and the reduction principle, and the second is due to a remarkable work by Filaseta and
Trifonov ([1, Theorem 7]) who used divided differences and a very useful polynomial identity:
Lemma 2.1. Let δ > 0 be a real number and ϕ ∈ C k (]N ; 2N ] 7→ R) such that
there exists a real number λk > 0 such that
(k)
ϕ (x) λk
On Short Sums of Certain
Multiplicative Functions
Olivier Bordellès
for any real number x ∈ ]N ; 2N ] . Then:
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(i) If k > 2,
2
k(k+1)
R (ϕ, N, δ) N λk
Contents
+ Nδ
2
k(k−1)
+
1/k
δλ−1
k
+ 1.
(ii) Moreover, if k > 5 and ϕ(k−1) (x) λk−1 = N λk for any real number
x ∈ ]N ; 2N ] , then:
2
k(k+1)
R (ϕ, N, δ) N λk
2
+ N δ (k−1)(k−2) + δλ−1
k−1
1
k−1
+ 1.
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Lemma 2.2. Let k > 2 be an integer and x, c0 > 0, δ > 0 be real numbers (c0
sufficiently small) satisfying N k−1 δ 6 c0 and N 6 x1/k . Then we have:
x
6k2 +k−1
1
1
R k , N, δ x 2k+1 + x 6k+3 δN 6k+3 .
n
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Lemma 2.3. Let f ∈ M and z > 1 any real number. Then:
(i)
X
1 < 3z 1/2
and
X
n>z
n squarefull
n6z
n squarefull
1
< 8z −1/2 .
n
(ii)
X
|(f ∗ µ) (n)| < 4z 1/2
and
n6z
X |(f ∗ µ) (n)|
< 8z −1/2 .
n
n>z
Proof. (i) Since every squarefull number n can be written in a unique way as
n = a2 b3 with b squarefree, we have:
∞
X
X
X
X
3
1/2
−3/2
1<z
b
=ζ
z 1/2
16
2
√
1/3
n6z
b=1
−3
b6z
a6 zb
On Short Sums of Certain
Multiplicative Functions
Olivier Bordellès
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n squarefull
and the well-known inequality ζ (σ) 6 σ/ (σ − 1) gives the first part of
the result. In the same way, let Z > z be any real number. We have:
X 1
X 1
X
X
1
1 X 1
6
+
n
b3 √ −3
a2
b3 √ −3 a2
√
1/3
1/3
1/3
z<n6Z
−3
b6z
zb
z
<a6 Zb
<b6Z
a6 Zb
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n squarefull
< 2z −1/2
X
b6z 1/3
2
6 6z −1/2 +
1
b3/2
+
π2 X 1
6
b3
1/3
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b>z
π −2/3
z
< 8z −1/2 .
6
J. Ineq. Pure and Appl. Math. 3(5) Art. 70, 2002
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(ii) We set g := f ∗ µ. The hypothesis f (p) = 1 implies |g (p)| = 0 and,
using multiplicativity, |g (n1 )| = 0 for any positive squarefree integer n1 >
1. Since any positive integer n can be written in a unique way as n =
n1 n2 with n1 squarefree, n2 squarefull and (n1 , n2 ) = 1, we deduce that
|g (n)| =
6 0 if either n = 1 or n > 1 is squarefull. The result follows by
using (i) and the fact that |g (n)| 6 1 for any positive integer n.
Lemma 2.4. Let k > 1 be an integer and ε > 0 be a fixed real number. Then,
for any positive integer d, we have:
τ(k) (d) 6
2
eε log 2
21/ε
On Short Sums of Certain
Multiplicative Functions
Olivier Bordellès
ε/k
d
.
Title Page
21/ε
−21/ε
Proof. We set c (ε) := 2
(eε log 2)
. The bound τ (d) 6 c (ε) dε is wellknown (see [5]). Since any positive integer n can be represented in a unique
way as n = qmk with q a positive k−free integer, we have
τ(k) (d) = τ (m) 6 c (ε) mε 6 c (ε) dε/k .
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3.
Proof of the Theorem
Let ε > 0 be a fixed real number. We take g := f ∗ µ again, and we have:
X
X X
f (n) =
g (d)
x<n6x+y
x<n6x+y d|n
X
=
g (d)
d6x+y
=
X
d6y
X
+
h i
x+y
x
−
d
d
On Short Sums of Certain
Multiplicative Functions
:= Σ1 + Σ2 .
y<d6x+y
Olivier Bordellès
1. Using Lemma 2.3, we have:
Σ1 = y
X g (d)
d6y
=y
d
∞
X
g (d)
d=1
d
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!
+O
X
|g (d)|
Contents
d6y
+O y
X |g (d)|
d>y
d
!
+O y
1/2
!
f p − f pl−1
1/2
=y
1+
+
O
y
pl
p
l=1
!
∞
Y
1 X f pl
1
=y
1+ 1−
−
+ O y 1/2
l
p l=1 p
p
p
= yP (f ) + O y 1/2 .
Y
∞
X
l
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2. Writing again d = a2 b3 with µ2 (b) = 1, we have:
h i
X
x+y
x
|Σ2 | 6
|g (d)|
−
d
d
y<d6x+y
d squarefull
6
X
b6(x+y)1/3
=
X
√y
<a6
b3
X
q
x+y
b3
−3 xb
(x + y) b−3
−
2
a
a2
X
X
−3
−3
b6(x+y)1/3 xb <d6(x+y)b
a2 |d
q
x+y
3
3 <a6
√y
b
6
X
b6(x+y)
X
1/3
1
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b
τ(2) (d)
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xb−3 <d6(x+y)b−3
h i
x+y
x
6 c (ε) (2x)
− 3
3
b
b
b6(x+y)1/3
x
y ε
x
max
R 3 , B, 3 ,
b
B
16B6(x+y)1/3
ε/2
X
where we used (1.1) , Lemma 2.4 (with k = 2) and the inequality log x 6
2 (eε)−1 xε/2 . Now Lemma 2.1(i) with k = 3 and λ3 = xB −6 gives
|Σ2 | xε x1/6 + y 1/3 x1/6+ε + y 1/2
since y > x6ε .
On Short Sums of Certain
Multiplicative Functions
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3. We suppose now that y 6 x1/3 . One can improve the former estimation
by using Lemma 2.2 instead of Lemma 2.1. The hypothesis N k−1 δ 6 c0
compels us to be more careful:
X x + y h x i
− 3
b3
b
1/3
b6(x+y)
h i
X x + y h x i
X
x+y
x
=
− 3
+
− 3
3
3
b
b
b
b
1/3
b6c−1
c−1
0 y
0 y<b6(x+y)
)
(
x
x
y y max
R 3 , B, 3 +
max
R 3 , B, 3
log x.
1/3
b
B
b
B
B6c−1
c−1
0 y
0 y<B6(x+y)
Lemma 2.1(ii) with k = 6 for the first sum and Lemma 2.2 with k = 3 for
the second yield:
X
b6(x+y)1/3
h i
x
x+y
− 3
3
b
b
x−1/5 y 6/5 + y 4/5 + x1/7 + x1/21 y 2/3 log x
On Short Sums of Certain
Multiplicative Functions
Olivier Bordellès
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and one easily checks that
x−1/5 y 6/5 + y 4/5 x1/21 y 2/3
if y 6 x1/3 . The proof of the theorem is complete.
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References
[1] M. FILASETA AND O. TRIFONOV, The distribution of fractional parts
with applications to gap results in Number Theory, Proc. London Math.
Soc., 73 (1996), 241–278.
[2] A. HILDEBRAND, Multiplicative functions in short intervals, Can. J.
Math., 3 (1987), 646–672.
[3] M.N. HUXLEY AND P. SARGOS, Points entiers au voisinage d’une courbe
plane de classe C n , Acta Arith., 69 (1995), 359–366.
On Short Sums of Certain
Multiplicative Functions
[4] M.N. HUXLEY AND P. SARGOS, Points entiers au voisinage d’une courbe
plane de classe C n II, les prépublications de l’Institut Elie Cartan de Nancy
(1997).
Olivier Bordellès
[5] D.S. MITRINOVIĆ AND J. SÁNDOR (in cooperation with B. Crstici),
Handbook of Number Theory, Kluwer Academic Publisher (1996).
Contents
[6] P. SHIU, A Brun-Titchmarsh theorem for multiplicative functions, J. Reine
Angew. Math., 313 (1980), 161–170.
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