Characterization of asymptotic distribution functions of ratio block sequences

Abstract

In this paper we give necessary and sufficient conditions for the block sequence of the set X = {x
1 < x
2 < … < x

n
< …} ⊂ ℕ to have an asymptotic distribution function in the form x
λ.

Key words and phrasesasymptotic distribution function-uniform distribution-block sequence
Mathematics subject classification number11B05

Full text

Periodica Mathematica Hungarica Vol. 60 (2), 2010, pp. 115–126
DOI: 10.1007/s10998-010-2115-2
CHARACTERIZATION
OF ASYMPTOTIC DISTRIBUTION FUNCTIONS
OF RATIO BLOCK SEQUENCES
Ferdin´
and Filip1and J´
anos T. T´
oth2
1Department of Mathematics, J. Selye University
Bratislavsk´a cesta 3322, 945 01 Kom´arno, Slovakia
E-mail: filip.ferdinand@selyeuni.sk
2Department of Mathematics, J. Selye University
Bratislavsk´a cesta 3322, 945 01 Kom´arno, Slovakia
E-mail: toth.janos@selyeuni.sk
(Received November 15, 2009; Accepted March 5, 2010)
[Communicated by Attila Peth˝o]
Abstract
In this paper we give necessary and sufficient conditions for the block
sequence of the set X={x1<x
2<··· <x
n<···} ⊂ Nto have an
asymptotic distribution function in the form xλ.
1. Introduction
Denote by Nand R+the set of all positive integers and positive real numbers,
respectively. In the whole paper we will assume that Xis an infinite set of positive
integers. Denote by R(X)={x
y;x∈X, y ∈X}the ratio set of Xand say that a
set Xis (R)-dense if R(X) is (topologically) dense in the set R+. Let us note that
the concept of (R)-density was defined and first studied in the papers [12] and [13].
Now let X={x1,x
2,...}where xn<x
n+1 are positive integers. The (R)-
density of the set Xis equivalent to the everywhere density in [0,1] of the sequence
x1
x1
,x1
x2
,x2
x2
,x1
x3
,x2
x3
,x3
x3
,..., x1
xn
,x2
xn
,...,xn
xn
,... (1)
It is also called the ratio block sequence of the set Xand we see that it is composed
by blocks
Xn=x1
xn
,x2
xn
,...,xn
xn,n=1,2,... (2)
Mathematics subject classification number: 11B05.
Key words and phrases: asymptotic distribution function, uniform distribution, block se-
quence.
Supported by grant APVV SK-HU-0009-08 and VEGA 1/0753/10.
0031-5303/2010/$20.00 Akad´emiai Kiad´o, Budapest
c
Akad´emiai Kiad´o, Budapest Springer, Dordrecht

116 F. FILIP and J. T. T ´
OTH
which can be studied individually. The density of the sequence of induvidual blocks
(2) implies the (R)-density of the set X. Also, if the distribution functions of (1)
or (2) are increasing, then again the set Xis (R)-dense. This is a motivation for
the study of sets G(xm/xn)andG(Xn) of distribution functions of (1) and (2),
respectively (defined in Par. 2), cf. [17], [18] and [7].
The second motivation for the study of block sequence (2) is that also other
kinds of block sequences were studied by several autors, see [3], [8],[9], [14], [20],
etc.
2. Definitions and basic results
In the following we use standard notations and definitions from [2], [9] and
[16].
By a distribution function we mean any function f:0,1→0,1such that
f(0) = 0,f(1) = 1 and fis nondecreasing in 0,1.
For the block sequence (1), for n∈Nand x∈0,1denote
A(Xn,x)=#
i:i≤n, xi
xn≤xand A(Xn,x)=
n

j=1
A(Xj,x).
Then we can attach to the sequence of blocks (Xn) and to the block sequence (1)
the following distribution functions:
F(Xn,x)=A(Xn,x)
n,
FN(xm/xn,x)=#{[i, j ]:1≤i≤j≤k, xi
xj≤x}+#{i:i≤l, xi
xk+1 ≤x}
N
=A(Xk,x)+O(k)
N=A(Xk,x)
N+O1
√N,
where x∈0,1and N=k(k+1)
2+lwith 0 ≤l<k+ 1. Consequently
lim
N→∞ FNxm
xn
,x
−A(Xk,x)
k(k+1)/2=0.
Denote by G(Xn) the set of all distribution functions g(x)forwhichthere
exists an increasing sequence of indices {nk}∞
k=1 such that
lim
k→∞ F(Xnk,x)=g(x)
almost everywhere (abbreviated as a.e.) in 0,1.

DISTRIBUTION FUNCTION OF BLOCK SEQUENCES 117
Similarly G(xm/xn) denotes the set of all distribution functions g(x)ofthe
block sequence (1) for which there exists an increasing sequence of indices {Nk}∞
k=1
such that
lim
k→∞ FNkxm
xn
,x
=g(x)
a.e. in 0,1.
If the set G(Xn) is a singleton: G(Xn)={g(x)}, then we say that the sequence
Xnhas the asymptotic distribution function g(x) (abbreviated as a.d.f.). Similarly
if G(xm/xn)={g(x)}, then we say that the block sequence (1) of the set Xhas
a.d.f. g(x). In these cases
lim
n→∞ F(Xn,x)=g(x) and lim
N→∞ FNxm
xn
,x
=g(x)
holds for almost all x∈0,1.
Especially, if G(Xn)={g(x)=x}, resp. G(xm/xn)={g(x)=x},thenwe
say that the sequence Xnis uniformly distributed (abbreviated as u.d.), resp. the
block sequence (1) of the set Xis uniformly distributed.
Distribution functions of the sequence Xnand the block sequence (1) of the
set Xwere first investigated in the paper [17], where the next statement is proved
([17], Theorem 8.1, Theorem 8.2, Theorem 8.4).
(A1) If G(Xn)={g(x)},thenG(xm/xn)={g(x)}.
(A2) Let G(Xn)={g(x)}. Then one of the following equalities holds:
(i) g(x)=c0(x)=0,if x=0
1,if x∈(0,1,or
(ii) g(x)=xλfor some 0 <λ≤1.
(A3) G(Xn)={c0(x)}iff one of the following equalities holds:
(i)
lim
n→∞
1
nxn
n

i=1
xi=0,
(ii)
lim
m,n→∞
1
mn
m

i=1
n

j=1 


xi
xm−xj
xn

=0.
We will use the following notation.
Let X={x1<x
2<···}⊂R,andE⊂0,1. We define
A(E,Xn)=#
i:i≤n, xi
xn∈E
and
A(E,Xn)=#
[i, j]:1≤i≤j≤n, xi
xj∈E=
n

j=1
A(E,Xj).

118 F. FILIP and J. T. T ´
OTH
Suppose that G(xm/xn)={g(x)}and that gis continuous on 0,1.Let
0≤a<b≤1berealnumbers. IfE=(a, b),a, b),(a, bor a, b, then it is obvious
that
lim
n→∞
A(E,Xn)
n(n+1)/2=g(b)−g(a).
3. Results
In this paper we give necessary and sufficient conditions for the block sequence
of the set Xto have an a.d.f. of the form xλ. The main result of this paper is the
following theorem.
Theorem 1. Let λ>0be a real number and X={x1<x
2<···} ⊂ N.
Then G(xm/xn)={xλ}iff for every k∈N
lim
n→∞
xkn
xn
=k1
λ.(3)
We give a sufficient condition for block sequences not to have an asymptotic
distribution function in the form xλ.
Lemma 1. Let λ>0be a real number and X={x1<x
2<···} ⊂ N.
Suppose that there exist numbers αand βwith 1<α<βsuch that for infinitely
many n, m ∈N,m<n
xn
xm
>β and n
m<α
λ,(4)
or xn
xm
<α and n
m>β
λ.(5)
Then G(xm/xn)={xλ}.
Proof. We prove the first part of Lemma 1 by contradiction.
Let X={x1<x
2<···}⊂ Nbe such set that (4) and G(xm/xn)={xλ}
hold.
By (4) there exists a sequence {[mi,n
i]}∞
i=1, such that for every i∈Nmi<n
i,
ni<n
i+1 and xni
xmi
>β and ni
mi
<α
λ.(6)
Choose d∈Rso that
1<d<2β
αλ−1.(7)

DISTRIBUTION FUNCTION OF BLOCK SEQUENCES 119
Wenowinvestigatethevalueof
A1
β,1,X
[dni].
It is obvious that
A1
β,1,X
[dni]=A1
β,1,X
ni+
[dni]

j=ni+1
A1
β,1,X
j.(8)
From (6) we obtain
A1
β,1,X
ni<n
i1−1
αλ.(9)
Then
A1
β,1,X
ni+j≤A1
β,1,X
ni+j, (10)
since if xk
xni+j>1
β,then xk
xni
>1
βor k>n
i. Inequalities (8) and (10) imply that
A1
β,1,X
[dni]≤A1
β,1,X
ni+([dni]−ni)ni1−1
αλ+
+([dni]−ni)([dni]−ni+1)
2.
(11)
Then G(xm/xn)={xλ}yields
lim
i→∞
A(( 1
β,1,X
[dni])
([dni])([dni]+1)/2=1−1
βλ(12)
and
lim
i→∞
A(( 1
β,1,X
ni)
ni(ni+1)/2=1−1
βλ.(13)
On the other hand, using (13) we obtain
lim
i→∞
A(( 1
β,1,X
ni)+([dni]−ni)ni(1 −1
αλ)+([dni]−ni)([dni]−ni+1)
2
[dni]([dni]+1)/2
= lim
i→∞
A(( 1
β,1,Xni)
ni(ni+1)/2
ni(ni+1)
2+(d−1)n2
i(1 −1
αλ)+n2
i(d−1)2
2
n2
id2/2
=(1 −1
βλ)+2(d−1)(1 −1
αλ)+(d−1)2
d2.
From this, (11) and (12) we derive that
1−1
βλ≤(1 −1
βλ)+2(d−1)(1 −1
αλ)+(d−1)2
d2.

120 F. FILIP and J. T. T ´
OTH
Elementary calculations give
1−1
βλ(d2−1) ≤2(d−1)1−1
αλ+(d−1)2,
1−1
βλ(d+1)≤21−1
αλ+(d−1) ,
d−d
βλ+1−1
βλ≤1−2
αλ+d,
d≥2β
αλ−1,
contradicting (7).
Now we prove by a contradiction the second part of Lemma 1. Let X=
{x1<x
2<···}be such that (5) and G(xm/xn)={xλ}hold. By (5) there exists
a sequence {[mi,n
i]}∞
i=1 such that mi<n
i<n
i+1 and
xni
xmi
<α and ni
mi
>β
λ(14)
for every i∈N.
Let γ∈(α, β) be arbitrary.
First we show that for every sufficiently large ni
x[(γ/α)λni]
xni≤β
α.(15)
For a contradiction assume that for infinitely many ni
x[(γ/α)λni]
xni
>β
α.(16)
Let γ<γ
<β. It is obvious that
lim
i→∞
[(γ/α)λni]
ni
=γ
αλ<γ
αλ.
Thus there exists i0∈N, such that for every i≥i0
[(γ/α)λni]
ni
<γ
αλ.
From this, (16) and the fact that γ
α<β
α, using the first part of the proof one
deduces that G(xm/xn)={xλ}. Hence (15) holds for every sufficiently large ni.
Let nibe sufficiently large. We will make estimations for
A1
β,1,X
[(γ/α)λni].

DISTRIBUTION FUNCTION OF BLOCK SEQUENCES 121
It is obvious that
A1
β,1,X
[( γ
α)λni]=A1
β,1,X
ni+
[( γ
α)λni]−ni

j=1
A1
β,1,X
ni+j.(17)
Moreover, for every ni<k≤[(γ/α)λni] inequalities (14) and (15) imply that
xmi
xk≥xmi
x[(γ/α)λni]
=xmi
xni
xni
x[(γ/α)λni]
>1
α
α
β=1
β.
From this using (14) we obtain for every 1 ≤j≤[(γ/α)λni]−ni
A1
β,1,X
ni+j≥ni−mi+j≥ni1−1
βλ+j. (18)
Putting (18) into (17) we obtain
A1
β,1,X
[( γ
α)λni]≥A1
β,1,X
ni+
[(γ/α)λni]−ni

j=1
ni1−1
βλ+j
≥
A1
β,1,X
ni
ni(ni+1)/2
ni(ni+1)
2+
+γ
αλni−nini1−1
βλ+
+([( γ
α)λni]−ni)([(γ/α)λni]−ni+1)
2.
(19)
On the other hand, the assumption G(xm/xn)={xλ}implies that
lim
i→∞
A(( 1
β,1,X
[(γ/α)λni])
[(γ/α)λni]([(γ/α)λni]+1)/2=1−1
βλ(20)
and
lim
i→∞
A(( 1
β,1,X
ni)
ni(ni+1)/2=1−1
βλ.(21)
From (19), (20) and (21) we can derive
1−1
βλ≥(1 −1
βλ) + 2((γ/α)λ−1)(1 −1
βλ)+((γ/α)λ−1)2
(γ/α)2λ.
Using elementary computations we obtain
1−1
βλγ
α2λ−2γ
αλ+1
≥γ
αλ−12,
hence
1−1
βλ≥1,
a contradiction.
The following lemma is required for the proof of Theorem 1.

122 F. FILIP and J. T. T ´
OTH
Lemma 2. Let ε>0and X={x1<x
2<···}⊂N. Suppose that for every
k∈N
lim
n→∞
xkn
xn
=kε.(22)
Then for every real number α>0
lim
n→∞
x[αn]
xn
=αε.
Proof. We will prove the lemma in two steps. First we prove the following
claim. If lim
n→∞
xkn
xn=kεfor some ε>0 and every natural number k,then
lim
n→∞
xn+1
xn
=1.(23)
Let kbe a fixed positive integer and na sufficiently large positive integer.
Then there exists a positive integer msuch that
km ≤n<(k+1)m.
Then obviously km ≤n+1≤(k+1)m. From this we obtain
1≤xn+1
xn≤x(k+1)m
xkm
=x(k+1)m/xm
xkm/xm
=V(m).
If n→∞,thenm→∞and obviously V(m)→(k+1
k)ε. Hence
1≤lim sup
n→∞
xn+1
xn≤k+1
kε.
This inequality holds for every k∈N.Fork→∞we obtain from this inequality
that
lim
n→∞
xn+1
xn
=1.
Now we prove the lemma for the case α∈Q+.
Let α=p/q where p, q ∈N.Letn≥qbe an integer. Let lbe the integer
given by
ql ≤n<q(l+1).
Then obviously
xpl/xl
xq(l+1)/xl+1
xl
xl+1
=xpl
xq(l+1) ≤x[αn]
xn≤xp(l+1)
xql
=xp(l+1)/xl+1
xql/xl
xl+1
xl
.
From this, (22), (23) we obtain
lim
n→∞
x[αn]
xn
=pε
qε=αε.

DISTRIBUTION FUNCTION OF BLOCK SEQUENCES 123
It remains to prove that the lemma is valid also for irrational α>0.
Let δ>0. From the facts that the set Q+is dense in R+and that the function
f(x)=xεis increasing and continuous it follows that there are numbers α1,α
2∈Q
such that 0 <α
1<α<α
2and
αε−αε
1<δ and αε
2−αε<δ.
Then from the inequalities
αε−δ<α
ε
1= lim
n→∞
x[α1n]
xn≤lim inf
n→∞
x[αn]
xn
≤lim sup
n→∞
x[αn]
xn≤lim
n→∞
x[α2n]
xn
=αε
2<α
ε+δ
it follows that
lim
n→∞
x[αn]
xn
=αε.
Using Lemma 1 and Lemma 2 we can prove Theorem 1.
Proof of Theorem 1. We will prove the theorem in two steps.
I) The condition (3) is necessary for block sequences to have asymptotic dis-
tribution function in the form xλ.
Suppose that there exists a set X={x1<x
2<···} ⊂ Nand k∈Nsuch
that G(xm/xn)={xλ}, but
lim
n→∞
xkn
xn=k1/λ .
(It is obvious that k≥2.) Then there are two possibilities:
1) lim inf n→∞(xkn/xn)<k
1/λ,
2) lim supn→∞ (xkn/xn)>k
1/λ.
Inthefirstcaseput
α= lim inf
n→∞
xkn
xn
and β=k1/λ .
Since 1 ≤α<β
, we can take numbers α, β such that α<α<β<β
.Thenwe
have for infinitely many n∈N
xkn
xn
<α and kn
n=k=(β)λ>β
λ.
Using Lemma 1 we obtain G(xm/xn)={xλ}, a contradiction.
In the second case put
β= lim sup
n→∞
xkn
xn
and α=k1/λ .

124 F. FILIP and J. T. T ´
OTH
(Itispossiblethatβ=+∞.) Since 1 ≤α<β
, we can take numbers α, β,such
that α<α<β<β
. Then we have for infinitely many n∈N
xkn
xn
>β and kn
n=k=(α)λ<α
λ.
Using Lemma 1 we obtain G(xm/xn)={xλ}.
II) The condition (3) is sufficient for block sequences to have asymptotic
distribution function in the form xλ.
From (A1) it follows that it is sufficient to prove that G(Xn)={xλ}.Let
α∈(0,1). We will show that
lim
n→∞ F(Xn,α) = lim
n→∞
#{i:i≤n, xi
xn<α}
n=αλ.
Let 0 <α
1<α. Lemma 2 implies that
lim
n→∞
x[αλ
1n]
xn
=(αλ
1)1/λ =α1<α.
Then there exists n0∈Nsuch that for every n≥n0
x[αλ
1n]
xn
<α.
Hence, if i≤[αλ
1n], then xi/xn<α.Thismeansthat
#i:i≤n, xi
xn
<α
≥[αλ
1n].
Then
lim inf
n→∞
#{i:i≤n, xi/xn<α}
n≥lim
n→∞
[αλ
1n]
n=αλ
1.(24)
Since (24) holds for every 0 <α
1<α,wehave
lim inf
n→∞
#{i:i≤n, xi/xn<α}
n≥αλ.(25)
Similarly
lim sup
n→∞
#{i:i≤n, xi/xn<α}
n≤αλ.(26)
Inequalities (25) and (26) imply that
lim
n→∞
#{i:i≤n, xi/xn<α}
n=αλ.
This completes the proof of Theorem 1.

DISTRIBUTION FUNCTION OF BLOCK SEQUENCES 125
Corollary 1. The block sequence (1) is uniformly distributed iff
lim
n→∞
xkn
xn
=k
for all k∈N.
Proof. The statement is an immediate consequence of Theorem 1.
Acknowledgement
The authors would like to thank Prof. Attila Peth˝o and the anonymous referee
for a careful reading of the manuscript and suggestions that improved the quality
of this paper.
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