volume 5, issue 3, article 81, 2004.
Received 27 August, 2003;
accepted 28 April, 2004.
Communicated by:S.S. Dragomir
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Journal of Inequalities in Pure and Applied Mathematics
CORRIGENDUM TO “ON SHORT SUMS OF CERTAIN MULTIPLICATIVE FUNCTIONS”
OLIVIER BORDELLÈS
2 Allée de la Combe 43000 AIGUILHE FRANCE.
EMail:borde43@wanadoo.fr
c
2000Victoria University ISSN (electronic): 1443-5756 118-03
Corrigendum to “On Short Sums of Certain Multiplicative
Functions”
Olivier Bordellès
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J. Ineq. Pure and Appl. Math. 5(3) Art. 81, 2004
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Abstract
This note is a corrigendum of the main result of the paper ”On short sums of certain multiplicative functions” (J. Ineq. Pure & Appl. Math., 3(5), Art. 70 (2002)).
2000 Mathematics Subject Classification:11N37, 11P21 Key words: Corrigendum, Short sums, Integer points
The author is indebted to the referee for insightful comments.
The purpose of this note is both to give a corrected statement of the main result in [1] and to provide the necessary changes to the arguments in that paper to justify this corrected statement. The result we now assert is the following : Theorem 1. Let ε, c0 > 0 and2 6 y 6 c0x1/2 be real numbers. Let f be a multiplicative function satisfying0 6 f(n)6 1for any positive integern and f(p) = 1for any prime numberp.We have asx→+∞:
X
x<n6x+y
f(n) =yP(f) +Oε x1/15+εy2/3 ,
where
P(f) :=Y
p
1− 1
p
1 +
∞
X
l=1
f pl pl
! .
Corrigendum to “On Short Sums of Certain Multiplicative
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Olivier Bordellès
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Before giving the proof, we note that ify < x1/5,thenx1/15 > y1/3 so that the expression x1/15+εy2/3 in the error term exceeds the main term (as well as the trivial bound ofy+ 1on the sum).
Proof. On page 5 of [1], the sum
S1 := X
y<d6x+y dsquarefull
|g(d)|
x+y d
−hx d
i
has been bounded by the sum
S2 := X
b6(x+y)1/3
X (by3)1/2<a6(x+yb3 )1/2
(x+y)b−3 a2
− xb−3
a2
by usingd=a2b3withµ2(b) = 1,andS2 has been bounded by
εxε max
16B6(x+y)1/3
Rx
b3, B, y B3
for any (small) positive real number ε,where we definedR(f, N, δ)to be the number of integer points (n, m) verifying n ∈ ]N; 2N]and |f(n)−m| 6 δ.
Unfortunately, the partP
b6y1/3 ofS2 cannot be estimated by
max
16B6(x+y)1/3
Rx
b3, B, y B3
,
Corrigendum to “On Short Sums of Certain Multiplicative
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hence we have to proceed differently: for any positive integerr,we set
τ(r)(n) :=X
dr|n
1
and recall that
τ(r)(n)εnε/r for any positive integersn, r.
InS1,d=a2b3 > yimpliesa > y1/5 orb > y1/5.Since
X
y1/5<a6(x+y)1/2
X (ay2)1/3<b6(x+ya2 )1/3
(x+y)a−2 b3
− xa−2
b3
6 X
y1/5<a6(x+y)1/2
X
x
a2<n6x+ya2
τ(3)(n)
and we have the same ifb > y1/5,then
S1 6 X
y1/5<b6(x+y)1/3
X
x
b3<n6x+y
b3
τ(2)(n) + X
y1/5<a6(x+y)1/2
X
x
a2<n6x+y
a2
τ(3)(n)
= X
y1/5<b6(2y)1/3
X
x
b3<n6x+yb3
τ(2)(n) + X
(2y)1/3<b6(x+y)1/3
X
x
b3<n6x+yb3
τ(2)(n)
+ X
y1/5<a6(2y)1/2
X
x
a2<n6x+ya2
τ(3)(n) + X
(2y)1/2<a6(x+y)1/2
X
x
a2<n6x+ya2
τ(3)(n)
:= Σ1+ Σ2+ Σ3+ Σ4.
Corrigendum to “On Short Sums of Certain Multiplicative
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• ForΣ1andΣ3 we use the trivial bound:
Σ1+ Σ3 ε xε/2 X
y1/5<b6(2y)1/3
x+y b3
−hx b3
i
+xε/3 X
y1/5<a6(2y)1/2
x+y a2
−h x a2
i
ε xε/2y
X
b>y1/5
1
b3 + X
a>y1/5
1 a2
ε y4/5xε/2.
• ForΣ2,we use the method of [1] to get
Σ2 ε xε x1/6+y1/3 .
• ForΣ4,if we supposey 6c0x1/2(wherec0 >0is sufficiently small),we have using Lemmas 2.1 and 2.2 of [1]:
Σ4 ε xε (
max
(2y)1/2<A6c−10 y
Rx a2, A, y
A2
+ max
c−10 y<A6x1/2
Rx
a2, A, y A2
)
ε xε
(xy)1/6+x1/5+x1/15y2/3 .
Hence we finally have S1 ε xε
x1/15y2/3+y4/5+ (xy)1/6+x1/5 +x1/6+y1/3 ε xε x1/15y2/3+y4/5
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ify>x1/5.Note thaty4/5 x1/15y2/3ify6c0x1/2 and that
X
x<n6x+y
f(n)−yP(f)
y x1/15y2/3
ify < x1/5.This concludes the proof of the theorem.
Corrigendum to “On Short Sums of Certain Multiplicative
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References
[1] O. BORDELLÈS, On short sums of certain multiplicative functions, J.
Inequal. Pure and Appl. Math., 3(5) (2002), Art. 70. [ONLINE: http:
//jipam.vu.edu.au/article.php?sid=222]