On weighted densities

Abstract

The continuity of densities given by the weight functions n

α
, α ∈ [−1, ∞[, with respect to the parameter α is investigated.

Full text

Czechoslovak Mathematical Journal
Rita Giuliano-Antonini; Georges Grekos; Ladislav Mišík
On weighted densities
Czechoslovak Mathematical Journal, Vol. 57 (2007), No. 3, 947--962
Persistent URL: http://dml.cz/dmlcz/128218
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Czechoslovak Mathematical Journal, 57 (132) (2007), 947–962
ON WEIGHTED DENSITIES
Rita Giuliano-Antonini, Pisa, Georges Grekos, Saint-Etienne, and
Ladislav Mišík, Ostrava
(Received August 8, 2005)
Abstract. The continuity of densities given by the weight functions n
α
, α ∈ [−1, ∞[, with
respect to the parameter α is investigated.
Keywords: asymptotic density, logarithmic density, f-density, continuity
MSC 2000 : 11B05, 11K99
1. Introduction
Let f : → [0, ∞[ be a nonzero function, let A ⊂ and n ∈ . We denote
A
f
(n) =
X
a∈A,a6n
f(a)
and define
d
f
(A) = lim inf
n→∞
A
f
(n)
f
(n)
and
¯
d
f
(A) = lim sup
n→∞
A
f
(n)
f
(n)
,
i.e. the lower and the upper f -densities o f the s e t A, respectively. Put moreover
D
f
(A) = (
¯
d
f
(A), d
f
(A)) ∈ {(x, y); x ∈ [0, 1], y ∈ [0, x]}.
We call D
f
(A) the f-density point of the set A. Two important cases of densities a re
those of asymptotic densities (denoted by d
,
¯
d) with f(n) = 1, n ∈ , and logarithmic
This work is supported by MIUR Italy, Program Barrande n. 2003-009-2, MSM6198898701
and GA ČR no. 201/04/0381.
947

densities (denoted by δ,
¯
δ) with f(n) = 1/n, n ∈ . There is a well known relation
between these four values (see, for insta nce , [2], p. 241–242)
(I) 0 6 d
(A) 6 δ(A) 6
¯
δ(A) 6
¯
d(A) 6 1
which holds for every A ⊂
. Also, examples of sets are known for which the values
of the asymptotic densities differ from the corresponding ones of the logarithmic
densities. There exist even sets with arbitrary prescribed values of all four densities
respecting the relation (I) (see [4]). In this paper we will deal with the class of
densities determined by weight functions f
α
(n) = n
α
, α ∈ [−1, ∞[. Notice that the
asymptotic densities c orrespond to α = 0 and the logarithmic densities correspond
to α = −1.
In the sequel we shall write A
α
in place of A
f
α
and d
α
,
¯
d
α
, D
α
in place of d
f
α
,
¯
d
f
α
and D
f
α
, resp e c tively, and moreover we shall use the term α-density point instead
of the f
α
-density point.
In [6] it is proved that both the upper and lower α-densities vary monotonously
with respect to the parameter α. This provides an extension of inequalities (I): Let
−1 6 α < β < ∞. Then the inequalities
(R) d
β
(A) 6 d
α
(A) and
¯
d
α
(A) 6
¯
d
β
(A)
hold for every A ⊂
.
A natural question arises whether the coordinates of the α-density point of a set A
depend on the parameter α continuously. The aim of the present paper is to discuss
this problem. As there are no well known examples of sets with different α-density
points for α ∈ [−1, ∞[, we will start with the following example. It shows that there
are sets A ⊂
for which both the functions α 7→ d
α
(A) and α 7→
¯
d
α
(A) are injective
on [−1, ∞[.
Example 1. Let a > 1 be a real number. Denote A =
∞
S
k=0
][a
2k
], [a
2k+1
]] ∩
,
where [r] means the integer part of the real number r, i.e. the largest integer less
than or equal to r. Then for every α ∈ [−1, ∞[
d
α
(A) =
1
a
α+1
+ 1
and
¯
d
α
(A) =
a
α+1
a
α+1
+ 1
.
First, let α > −1. Then both densities can be calculated using the technique
in [5], integra ting the function x
α
in the corresponding intervals and cancelling the
948

constant multipliers 1/(α + 1):
¯
d
α
(A) = lim sup
n→∞
n
P
k=0
[a
2k+1
]
P
i=[a
2k
]+1
i
α
a
2n+1
P
j=1
j
α
= lim sup
n→∞
n
P
k=0
(a
2k+1
)
α+1
− (a
2k
)
α+1
(a
2n+1
)
α+1
= lim
n→∞
(a
α+1
− 1)
n
P
k=0
(a
2α+2
)
k
(a
2n+1
)
α+1
= (a
α+1
− 1) lim
n→∞
(a
2α+2
)
n+1
−1
a
2α+2
−1
(a
2n+1
)
α+1
=
1
a
α+1
+ 1
lim
n→∞
a
2αn+2n+2α+2
a
2αn+2n+α+1
=
a
α+1
a
α+1
+ 1
and, similarly, or using the fact that in this case d
α
(A) =
¯
d
α
(A)/a
α+1
, we get
d
α
(A) =
1
a
α+1
+ 1
.
Calculation of d
−1
(A) and
¯
d
−1
(A) can be performed using the same technique to get
d
−1
(A) =
1
2
=
¯
d
−1
(A).
Notice that the same result can be obtained using Theorem 2 below on continuity
at α = −1, as the set A fulfils its ass umptions and
lim
α→−1
+
a
α+1
a
α+1
+ 1
= lim
α→−1
+
1
a
α+1
+ 1
=
1
2
.
2. Continuity on ]−1, ∞[
Now we are going to answer the question about the continuity o f the dependence
of α-density points on the parameter α. First we will consider the case α ∈ ]−1, ∞[.
Theorem 1. Let α ∈ ]−1, ∞[ and δ > 0. Then for every set A ⊂
|d
α
(A) − d
α+δ
(A)| 6
2δ
α + 1
and |
¯
d
α
(A) −
¯
d
α+δ
(A)| 6
2δ
α + 1
.
P r o o f. We calculate
∆
n
(A) =



A
α
(n)
α
(n)
−
A
α+δ
(n)
α+δ
(n)



=




X
a∈A,a6n
a
α
α
(n)
−
X
a∈A,a6n
a
α+δ
α+δ
(n)




=




(α + 1)
X
a∈A,a6n
a
α
n
α+1
n
α+1
α+1
α
(n)
− (α + δ + 1)
X
a∈A,a6n
a
α+δ
n
α+δ+1
n
α+δ+1
α+δ+1
α+δ
(n)




.
949

Denote
n
α+1
α+1
α
(n)
= 1 + ε
1
(n) and, similarly,
n
α+δ+1
α+δ+1
α+δ
(n)
= 1 + ε
2
(n)
and notice that both
ε
1
(n) → 0, ε
2
(n) → 0 as n → ∞.
Let us continue the calculation:
∆
n
(A) =




(α + 1)
X
a∈A,a6n
a
α
n
α+1
+ ε
1
(n)(α + 1)
X
a∈A,a6n
a
α
n
α+1
− (α + δ + 1)
X
a∈A,a6n
a
α+δ
n
α+δ+1
− ε
2
(n)(α + δ + 1)
X
a∈A,a6n
a
α+δ
n
α+δ+1




=




α + 1
n
α+1
X
a∈A,a6n
a
α

1 −

a
n

δ

−
δ
n
α+δ+1
X
a∈A,a6n
a
α+δ
+ ε
1
(n)
α + 1
n
α+1
X
a∈A,a6n
a
α
− ε
2
(n)
α + δ + 1
n
α+δ+1
X
a∈A,a6n
a
α+δ




6 S
1
(n) + S
2
(n) + |ε
1
(n)|S
3
(n) + |ε
2
(n)|S
4
(n)
where
S
1
(n) =
α + 1
n
α+1
X
a6n
a
α

1 −

a
n

δ

,
S
2
(n) =
δ
n
α+δ+1
X
a6n
a
α+δ
,
S
3
(n) =
α + 1
n
α+1
X
a6n
a
α
and
S
4
(n) =
α + δ + 1
n
α+δ+1
X
a6n
a
α+δ
.
It is clear that
lim
n→∞
S
2
(n) =
δ
α + δ + 1
6
δ
α + 1
and also
lim
n→∞
S
3
(n) = lim
n→∞
S
4
(n) = 1.
We have
S
1
(n) = (α + 1)
n
X
k=1

k
n

α

1 −

k
n

δ

1
n
.
950

The last sum is an integra l sum of the (integrable) function ϕ(x) = x
α
(1 − x
δ
) in the
interval [0, 1]; hence
lim
n→∞
n
X
k=1

k
n

α

1 −

k
n

δ

1
n
=
1
Z
0
x
α
(1 − x
δ
) dx
=
1
α + 1
−
1
α + δ + 1
=
δ
(α + 1)(α + δ + 1)
6
δ
(α + 1)
2
.
Thus we have
|d
α
(A) − d
α+δ
(A)| 6 lim sup
n→∞
∆
n
(A)
6 lim
n→∞
(S
1
(n) + S
2
(n) + |ε
1
(n)|S
3
(n) + |ε
2
(n)|S
4
(n))
6
δ
α + 1
+
δ
α + 1
+ 0 + 0 =
2δ
α + 1
.
The corresponding inequality for the upper densities can be derived by simple ob-
servation that
|
¯
d
α
(A) −
¯
d
α+δ
(A)| = |(1 − d
α
(
− A)) − (1 − d
α+δ
(
− A))|
= |d
α
(
− A) − d
α+δ
(
− A)| 6
2δ
α + 1
as the last inequality has already be en proved for all subsets of
. 
Remark 1. Since for a ll α > −1 and all δ such that 0 < δ < α + 1 we have
α − δ > −1, the statement of the theorem can be applied to the pair a − δ > −1 and
α = (α − δ) + δ to get
|d
α
(A) − d
α−δ
(A)| 6
2δ
α − δ + 1
and |
¯
d
α
(A) −
¯
d
α−δ
(A)| 6
2δ
α − δ + 1
for all A ⊂ .
Thus we have direct consequences of the above theorem.
Corollary 1. Given a set A ⊂ , the function α 7→ D
α
(A) is Lipschitzian on
each closed half-line [a
0
, ∞[, with a
0
> −1 fixe d.
951

Corollary 2. Given a set A ⊂ , the function α 7→ D
α
(A) is continuous on
]−1, ∞[.
3. The continuity at −1
Let A be a fixed subset of
. In this section we shall study continuity of the
α-density points as α → −1
+
. We assume that the set A ⊆ is neither finite nor
cofinite, so that it can be written in the form
A =
∩

∞
[
n=1
]a
n
, b
n
]

for two suitable sequences of integers (a
n
)
n>1
and (b
n
)
n>1
such tha t a
n
< b
n
< a
n+1
for every n. We recall that
α
(n) =
n
X
k=1
k
α
, n ∈
.
By an applicatio n o f Theorem 8.2 of [1], we are able to calculate the upper and lower
α-densities of A as follows:
Theorem A. The following relations hold:
d
α
(A) = lim inf
n→∞
n−1
P
k=1
(
α
(b
k
) −
α
(a
k
))
α
(a
n
)
,(1)
¯
d
α
(A) = lim sup
n→∞
n
P
k=1
(
α
(b
k
) −
α
(a
k
))
α
(b
n
)
.
The following result is also easy to prove:
Lemma B. For α > −1 the values
¯
d
α
(A) and d
α
(A) c an be also calculated as
(2)
¯
d
α
(A) = lim sup
n→∞
n
P
k=1
(b
1+α
k
− a
1+α
k
)
b
1+α
n
952

and
(3) d
α
(A) = lim inf
n→∞
n−1
P
k=1
(b
1+α
k
− a
1+α
k
)
a
1+α
n
,
while for α = −1 we have
¯
d
−1
(A) = lim sup
n→∞
n
P
k=1
(log b
k
− log a
k
)
log b
n
,(4)
d
−1
(A) = lim inf
n→∞
n−1
P
k=1
(log b
k
− log a
k
)
log a
n
(5)
respectively.
Lemma B follows from the equivalence relations, as n → ∞,
n
α
∼



1
1 + α
((n + 1)
α+1
− n
α+1
) for α > −1,
log(n + 1) − log n for α = −1
using the same arguments as in Theorem 3.2 of [1] or in Lemma 1 of [5].
In the sequel we set, for each n,
C
n
= log b
n
− log a
n
; B
n
= log b
n
− log b
n−1
; A
n
= log a
n
− log a
n−1
.
Also we will suppose that the sequence (B
n
)
n>1
is bounded (assumption (H)).
This e asily implies that (A
n
)
n>1
and (C
n
)
n>1
are bounded as well.
We have the following result.
Theorem 2. In addition to assumption (H), suppose that
(6) L
.
= lim inf
n→∞
C
n
> 0.
Then we have
lim
α→−1
+
¯
d
α
(A) =
¯
d
−1
(A),(7)
lim
α→−1
+
d
α
(A) = d
−1
(A).(8)
The following example shows that assumption (H) canno t be dropp e d.
953

Example 2. For the set A = ∩

∞
S
n=1
]a
n
, b
n
]

with
a
n
= n((n − 1)!)
2
, b
n
= (n!)
2
,
assumption (H) is not satisfied.
In fact, we have
C
n
= log(n!)
2
− log n((n − 1)!)
2
= log n,
which is not bounded. Now, by means of Theorem A and relatio ns (2), (3), (4) and
(5) it is easy to verify that
d
−1
(A) =
¯
d
−1
(A) = d
−1
(A) =
1
2
,
while for every α > −1 we have
d
α
(A) = 0;
¯
d
α
(A) = 1,
hence neither of the functions α 7→ d
α
(A), α 7→
¯
d
α
(A) is continuous at −1.
Theorem 2 covers evidently the rather relevant case of sets such as the set E
r
of
numbers beginning by a fixed digit r (r ∈ {1, 2, . . . , 9}), i.e.
E
r
=
∩

∞
[
n=1
]r10
n
− 1, (r + 1)10
n
− 1]

,
but it is not use ful for instance for the set of even numbe rs (or the set of multiples
of any other integer, of course). In fact, her e we have a
n
= 2n − 1, b
n
= 2n and
lim inf
n→∞
(log b
n
− log a
n
) = 0.
Observe that in this case the limit
lim
n→∞
log b
n
− log a
n
log b
n
− log b
n−1
= lim
n→∞
C
n
B
n
exists (= 1/2).
In fact, for a general set A =
∩

∞
S
n=1
]a
n
, b
n
]

the following result holds (we keep
the nota tion used for Theorem 2):
954

Theorem 3. Let assumption (H) hold and suppose that the limit
lim
n→∞
C
n
B
n
exists and is equal to L. Put b
0
= 1. Then
(i) A possesses logarithmic density d
−1
(A) = L;
(ii) there exists α
0
and a p ositive constant c such that for −1 < α < α
0
we have
(9) lim sup
n→∞








n
P
k=1
(b
1+α
k
− a
1+α
k
)
n
P
k=1
(b
1+α
k
− b
1+α
k−1
)
−
n
P
k=1
C
k
n
P
k=1
B
k








6 c(1 + α).
As a consequence we get
lim
α→−1
+
d
α
(A) = lim
α→−1
+
¯
d
α
(A) = d
−1
(A).
The rest of this section is devoted to the proofs o f Theorems 2 and 3.
P r o o f o f T h e o r e m 2. We shall prove relation (7) only, since (8) has an
identical proof (simply r e place (b
n
) with (a
n
) and use the part of T heorem A that
concerns lower density, along with relations (3) and (5)). We sta rt with a remark.
Remark 2. Since C
n
6 B
n
and C
n
6 A
n+1
, assumption (6) implies that also
M
.
= lim inf
n→∞
B
n
> 0 and N
.
= lim inf
n→∞
A
n
> 0.
We need a famous result:
Lemma (Abel) [3]. Let (r
n
)
n
and (s
n
)
n
be any two s e quences of real numbers.
Then
n
X
k=1
r
k
s
k
=

n
X
k=1
r
k

s
n
−
n−1
X
k=1

k
X
h=1
r
h

(s
k+1
− s
k
).
In (2), the fraction can be replaced by
(10)
n
P
k=1
(b
1+α
k
− a
1+α
k
)
b
1+α
n
− 1
=
n
P
k=1
(b
1+α
k
− a
1+α
k
)
n
P
k=1
(b
1+α
k
− b
1+α
k−1
)
(b
0
= 1).
Concerning its numerator, Abel’s lemma will be applied with
r
k
= C
k
, s
k
=
b
1+α
k
− a
1+α
k
C
k
,
955

so we obtain
n
X
k=1
(b
1+α
k
− a
1+α
k
) =

n
X
k=1
C
k

b
1+α
n
− a
1+α
n
C
n
(11)
−
n−1
X
k=1

k
X
h=1
C
h


b
1+α
k+1
− a
1+α
k+1
C
k+1
−
b
1+α
k
− a
1+α
k
C
k

.
As to the denominator of (10), another application of Ab e l’s lemma with
r
k
= B
k
, s
k
=
b
1+α
k
− b
1+α
k−1
B
k
,
gives
n
X
k=1
(b
1+α
k
− b
1+α
k−1
) =

n
X
k=1
B
k

b
1+α
n
− b
1+α
n−1
B
n
(12)
−
n−1
X
k=1

k
X
h=1
B
h


b
1+α
k+1
− b
1+α
k
B
k+1
−
b
1+α
k
− b
1+α
k−1
B
k

.
In view o f the above formulas (11) and (12 ), in o rder to get the statement of Theo-
rem 2 it will be enough to show that
lim
α→−1
+
sup
n



b
1+α
n
− a
1+α
n
(1 + α)C
n
b
1+α
n
− 1



= 0,(13)
lim
α→−1
+
sup
n
n−1
P
k=1

k
P
h=1
C
h

b
1+α
k+1
−a
1+α
k+1
C
k+1
−
b
1+α
k
−a
1+α
k
C
k

(1 + α)b
1+α
n

n
P
k=1
C
k

= 0,(14)
and two analogous relations concerning (1 2) (with B
k
replacing C
k
and b
k−1
replac-
ing a
k
).
We shall prove only (13) a nd (14 ).
In order to get (13), put, for x > 0,
H(x) =
1 − e
−x
x
,
and recall the inequality
|H(x) − 1| 6
x
2
.
956

Hence
sup
n




b
1+α
n
− a
1+α
n
(1 + α)b
1+α
n
C
n
− 1




= sup
n


H

(1 + α)C
n

− 1


6
1
2
(1 + α) sup
n
C
n
,
which concludes the proof of (13).
The proof of (14) is longer. We remark that, by assumption (H) and relation (6),
we have
lim inf
n→∞
n
P
k=1
C
k
log b
n
> 0.
This allows us to replace the term
n
P
k=1
C
k
in the denominator of (14) by lo g b
n
; so,
we s hall prove that
(15) lim
α→−1
+
sup
n
n−1
P
k=1

k
P
h=1
C
h

b
1+α
k+1
−a
1+α
k+1
C
k+1
−
b
1+α
k
−a
1+α
k
C
k

(1 + α)b
1+α
n
log b
n
= 0.
We now need some lemmas.
Lemma 1. The sequence (n/ lo g b
n
)
n
is bo unded.
P r o o f. Recall that M = lim inf
n→∞
B
n
> 0; fix ε, with 0 < ε < M . There exists
an integer n
0
such that, for n > n
0
, we have
B
n
> M − ε,
hence, for n > k > n
0
, we get
(16) log b
n
− log b
k
=
n
X
h=k+1
B
h
> (M − ε)(n − k).
In pa rticular, for n > k = n
0
we find
log b
n
> log b
n
0
+ (M − ε)(n − n
0
),
which completes the proof. 
957

Lemma 2. Let (D
k
)
k
and (E
k
)
k
be any two positive bounded sequences and le t
m be a nonnegative integer. Then
n−1
X
k=1

k
X
h=1
D
h

b
1+α
k+m
E
k
= O(b
1+α
n
log b
n
).
P r o o f. Since (D
k
)
k
and (E
k
)
k
are bounded, it is enough to prove the statement
for D
k
= E
k
= 1 for e very k, i.e. for the sequence
n−1
X
k=1
kb
1+α
k+m
.
Let ε, n
0
be as in Lemma 1, and let n > k > n
0
. Relation (16) can be written in the
equivalent form
b
k+m
b
n
6 e
−(M−ε)(n−k−m)
,
hence ther e exist positive constants c
1
and c
2
such that for n > n
0
− 1 we have
n−1
P
k=1
kb
1+α
k+m
b
1+α
n
log b
n
=
n
0
P
k=1
kb
1+α
k+m
b
1+α
n
log b
n
+
n−1
P
k=n
0
+1
k(b
k+m
/b
n
)
1+α
log b
n
6
c
1
b
1+α
n
log b
n
+ n
n−1
P
k=1
e
−(M−ε)(1+α)(n−k−m)
log b
n
6
c
1
b
1+α
n
log b
n
+ c
2
n
log b
n
,
as
n−1
X
k=1
e
−(M−ε)(1+α)(n−k−m)
6
∞
X
i=1−m
e
−(M−ε)(1+α)i
=
e
−(M−ε)(1+α)(1−m)
(1 − e
−(M−ε)(1+α)
)
= c
2
.
An application of Lemma 1 completes the proof of Lemma 2. 
Lemma 3. For every integer k we have



b
1+α
k+1
− a
1+α
k+1
C
k+1
−
b
1+α
k
− a
1+α
k
C
k



6 (1 + α)
2
b
1+α
k+1
B
k+1
+ (1 + α)G(α)b
1+α
k
,
where G is a function such that
lim
α→−1
+
G(α) = 0.
958

P r o o f. The reader can verify the equality
b
1+α
k+1
− a
1+α
k+1
C
k+1
−
b
1+α
k
− a
1+α
k
C
k
= (1 + α)
2
b
1+α
k+1
H((1 + α)B
k+1
)H((1 + α)C
k+1
)B
k+1
+ (1 + α)

H((1 + α)C
k+1
) − H((1 + α)C
k
)

b
1+α
k
by substituting H(x) = (1 − e
−x
)/x (with co rresponding arguments) into its right
hand side.
It is now enough to recall that 0 6 H(x) 6 1 and to put
G(α) = sup
k
|H((1 + α)C
k+1
) − H((1 + α)C
k
)|.
The fac t that lim
α→−1
+
G(α) = 0 follows from the Lagrange theorem:
|H((1 + α)C
k+1
) − H((1 + α)C
k
)| 6 2(1 + α)(sup
k
C
k
) sup
x
|H
′
(x)|,
and it is easily verified that
sup
x
|H
′
(x)| = sup
x



xe
−x
− 1 + e
−x
x
2



= sup
x



1 − e
−x
x
+
1 − e
−x
− x
x
2



6
3
2
.

Relation (15) now follows by applying Lemma 3 and Lemma 2 with m = 0 and
m = 1. This concludes the proof of Theore m 2.
P r o o f o f T h e o r e m 3. (i) is immediate, since
d
−1
(A) = lim
n→∞
n
P
k=1
C
k
n
P
k=1
B
k
= lim
n→∞
C
n
B
n
= L
by Cesaro’s theorem.
(ii) We need some algebra in order to wr ite the first member of (9) in a suitable
manner. By reducing to the common denominator we get that it is e qual to
(17) A =
n
P
h,k=1
b
1+α
k
(1 − e
−(1+α)C
k
)B
h
−
n
P
h,k=1
b
1+α
k
(1 − e
−(1+α)B
k
)C
h
n
P
h,k=1
b
1+α
k
(1 − e
−(1+α)B
k
)B
h
.
959

Recall the definition o f the function H (see Lemma 3). Then each summand in the
denominator of the fraction A in (17) is equal to
(18) (1 + α)b
1+α
k
H((1 + α)B
k
)B
h
B
k
.
Moreover, in the first (second) parenthesis of the numerator of A we subtract and
add the term (1 + α)C
k
((1 + α)B
k
, respectively) and separate the sums in order to
split the fraction A of (17) into tree summands
A = R − S + T,
where (recall the expression (18))
R =
n
P
h,k=1
b
1+α
k
(1 − e
−(1+α)C
k
− (1 + α)C
k
)B
h
(1 + α)
n
P
h,k=1
b
1+α
k
H((1 + α)B
k
)B
h
B
k
,
S =
n
P
h,k=1
b
1+α
k
(1 − e
−(1+α)B
k
− (1 + α)B
k
)C
h
(1 + α)
n
P
h,k=1
b
1+α
k
H((1 + α)B
k
)B
h
B
k
,
T =
n
P
h,k=1
b
1+α
k
C
k
B
h
−
n
P
h,k=1
b
1+α
k
C
h
B
k
n
P
h,k=1
b
1+α
k
H((1 + α)B
k
)B
h
B
k
.
We recall the inequality (x > 0)
0 6 e
−x
− 1 + x 6
x
2
2
and remark that, s ince (B
n
) is positive and bounded, we have
lim
α→−1
+
sup
k
H((1 + α)B
k
) = 1,
hence ther e exists α
0
> −1 such that for −1 < α < α
0
we have
sup
k
H((1 + α)B
k
) >
1
2
,
960

so that
R 6
(1 + α)
2
n
P
h,k=1
b
1+α
k
C
2
k
B
h
(1 + α)
n
P
h,k=1
b
1+α
k
B
h
B
k
6 (1 + α)(sup
k
C
k
)
(since C
k
6 B
k
).
In an analogous way we also find
S 6 (1 + α)(sup
k
B
k
).
Last, for −1 < α < α
0
we have
|T | 6
n
P
h,k=1
b
1+α
k
|C
k
B
h
− C
h
B
k
|
n
P
h,k=1
b
1+α
k
H((1 + α)B
k
)B
h
B
k
(19)
6
n
P
h,k=1
b
1+α
k
B
h
B
k
|(C
k
/B
k
) − (C
h
/B
h
)|
(1/2)

n
P
k=1
b
1+α
k
B
k

log b
n
.
Fix ε > 0 and n
0
such that, for n > n
0
, we have
L − ε <
C
n
B
n
< L + ε.
As the sequence (C
n
/B
n
) is bounded, the last term of (19) is not greater than
n
0
P
h,k=1
b
1+α
k
B
h
B
k
|(C
k
/B
k
) − (C
h
/B
h
)|
(1/2)

n
P
k=1
b
1+α
k
B
k

log b
n
+ c
3
n
0
P
h=1
B
h
log b
n
+ c
4
n
0
P
h=1
b
1+α
h
B
h
n
P
h=1
b
1+α
h
B
h
+ 4ε,
where c
3
and c
4
are suitable positive constants. Now the statement follows since
(log b
n
) a nd

n
P
h=1
b
1+α
h
B
h

go to ∞. 
961

4. An open problem
Problem. We have seen that to any given set A ⊂
we can attach a pair of
functions
d
A
: [−1, ∞[→ [0, 1] and
¯
d
A
: [−1, ∞[→ [0, 1],
both continuous in the interval ]−1, ∞[ a nd such that d
A
is nonincreasing,
¯
d
A
is
nondecreasing and d
A
(α) 6
¯
d
A
(α) for all α ∈ [−1, ∞[.
A natural question that arises is:
For which pairs o f functions d
,
¯
d with properties listed above there exists a s e t
A ⊂
such that
d
A
= d
and
¯
d
A
=
¯
d?
References
[1] A. Fuchs and R. Antonini Giuliano: Théorie générale des densités. Rend. Acc. Naz. delle
Scienze detta dei XL, Mem. di Matematica, 108 (XIV, 14) (1990), 253–294. zbl
[2] H. Halberstam and K. F. Roth: Sequences. Oxford Univ. Press, 1966. zbl
[3] G. H. Hardy and M. Riesz: The General Theory of Dirichlet Series. Cambridge Univ.
Press, 1952.
zbl
[4] L. Mišík: Sets of positive integers with prescribed values of densities. Mathematica Slo-
vaca 52 (2002), 289–296.
zbl
[5] L. Mišík and J. T. Tóth: Logarithmic density of a sequence of integers and density of its
ratio set. Journal de Théorie des Nombres de Bordeaux 15 (2003), 309–318.
zbl
[6] C. T. Rajagopal: Some limit theorems. Amer. J. Math. 70 (1948), 157–166. zbl
Authors’ addresses: R i t a G i u l i a n o - A n t o n i n i, Dipartimento di Matematica,
Università di Pisa, Via Buonarroti 2, 56100 Pisa, Italy, e-mail: giuliano@dm.unipi.it;
G e o r g e s G r e k o s, Université Jean Monnet (Saint-Etienne), Mathématiques, 23 rue du
Dr. Paul Michelon, 42023 Saint-Etienne cedex 2, France, e-mail: grekos@univ-st-etienne.
fr; L a d i s l av M i š í k, Dept. of Mathematics, University of Ostrava, 30. dubna 22, 701 03
Ostrava, Czech Republic, e-mail: ladislav.misik@osu.cz.
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