On ratio block sequences with extreme distribution function

Abstract

Distribution functions of ratio block sequences formed from sequences of positive integers are investigated in the paper.
We characterize the case when the set of all distribution functions of a ratio block sequence contains c
0, the greatest possible distribution function. Presented results complete some previously published results.

Full text



DOI: 10.2478/s12175-009-0123-6
Math. Slovaca 59 (2009), No. 3, 275–282
ON RATIO BLOCK SEQUENCES
WITH EXTREME DISTRIBUTION FUNCTION
Ferdin´
and Filip* — Ladislav Miˇ
s
´
ık** — J´
anos T. T ´
oth* **
(Co mmuni cated by Sta nis lav Ja kubec )
ABSTRACT. Distribution functions of ratio block sequences formed from se-
quences of positive integers are investigated in the paper. We characterize the
case when the set of all distribution functions of a ratio block sequence contains
c0, the greatest possible distribution function. Presented results complete some
previously published results.
c
2009
Mathematical Institute
Slovak Academy of Sciences
1. Introduction
In this part we recall some basic definitions. Denote by Nand R+the set
of all positive integers and positive real numbers, respectively. For X⊂Nlet
X(n)=#{x∈X:x≤n}. In the whole paper we will assume that Xis
infinite.
Now let X={x1,x
2,...}where xn<x
n+1 are positive integers. The follow-
ing sequence derived from X
x1
x1
,x1
x2
,x2
x2
,x1
x3
,x2
x3
,x3
x3
,..., x1
xn
,x2
xn
,...,xn
xn
,... (1)
is called the ratio block sequence of the set X. Thus the ratio block sequence is
formed by blocks X1,X
2,...,X
n,... where the nth block is
Xn=x1
xn
,x2
xn
,..., xn
xn.
This kind of block sequences were studied in papers [12], [14] and [3]. Also
other kinds of block sequences were studied by several authors, see [1], [6], [7],
2000 M a t h e m a t i c s Subject Classification: Primary 11B05.
K e y w o r d s: asymptotic distribution function, ratio block sequence.
Supported by grant MSM 6198898701, and VEGA no. 1/4006/07.

FERDIN´
AND FILIP — LADISLAV MIˇ
S´
IK — J´
ANOS T. T ´
OTH
[9] and [13]. Let Y=(yn) be an increasing sequence of positive integers. In [8],
extending a result in [5], it was investigated a sequence of blocks of the type
Yn=1
yn
,2
yn
,...,yn
yn.
Authors obtained a complete theory for the uniform distribution of the related
block sequence (Yn).
By 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]. We will use the following
special distribution function:
c0=c0(x)=0ifx=0
1ifx∈(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) = lim
k→∞
A(Xk,x)
k(k+1)
2
.
Denote G(Xn) the set of all distribution functions g(x) of the sequence of
single blocks (Xn) for which there exists an increasing sequence of indices (nk)
such that
lim
k→∞ F(Xnk,x)=g(x)
almost everywhere (abbrev. a.e.) in [0,1].
Similarly G(xm/xn) denotes the set of all distribution functions g(x)ofblock
sequence (1) for which there exists an increasing sequence of indices (Nk)such
that
lim
k→∞ FNk(xm/xn,x)=g(x)
a.e. in [0,1].
276

DISTRIBUTION FUNCTION OF BLOCK SEQUENCES
If the set G(Xn)isasingletonG(Xn)={g(x)}then we say that the sequence
Xnhas the asymptotic distribution function g(x) (abbrev. 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)
hold for almost all x∈[0,1].
Especially, if G(Xn)={g(x)=x},resp. G(xm/xn)={g(x)=x}then we
say that the sequence (Xn) is uniformly distributed (abbrev. 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 paper [12]. Much more information about the
mentioned concepts and their relations can be found in the monograph [11].
In this paper we will use also the following theorems.

1

([4, Theorem 5]) Let X={x1<x
2<...}⊂N.Thenc0∈G(Xn)
is equivalent to the existence of two integer sequences (mk)and (nk)such that
mk<n
k,k=1,2,... and xmk
xnk→0,mk
nk→1for k→∞.(2)

2

([2, Theorem 1]) Let G(xm/xn)={g(x)}and g(x)<1holds for
all 0≤x<1.Then lim
n→∞
xn
xn+1 =1.

3

([12, Theorem 8.4]) The set G(Xn)={c0(x)}if and only if any
of the following limit relation 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.
2. Results
In [4] the following question is asked (Question 3, see also [10]). Prove or
disprove
lim
n→∞
xn
x1+x2+···+xn
=0 ⇐⇒ c0(x)/∈G(Xn).(3)
We disprove this statement by counterexample and then we show that, after a
slight modification of the left side condition in (3), the equivalence holds.
277

FERDIN´
AND FILIP — LADISLAV MIˇ
S´
IK — J´
ANOS T. T ´
OTH
Example 1.There exists a set X={x1<x
2<...}⊂Nsuch that
lim
n→∞
xn
x1+x2+···+xn
=0 (4)
and
c0(x)∈G(Xn).(5)
We will construct the set by induction. Put a1= 9 and define inductively an+1
the closest odd integer to the number a3
2
nfor n=1,2,....Forn∈Ndefine
An=an
2,3an
2∩N
and X={x1<x
2<...}=
∞

n=1
An. Notice that the kth block of X,thesetAk,
contains exactly akconsecutive natural numbers with akas the middle term of
the block. Thus the sum of all elements belonging to the kth block is

x∈Ak
x=a2
k.(6)
A simple calculation shows, assuming xn∈Ak+1,that
lim
n→∞
xn
x1+x2+···+xn≤lim
n→∞
2ak+1

x∈Ak
x= lim
n→∞
2a3
2
k
a2
k
=0
proving (4).
For k=1,2,... denote xmk=3
2ak−1
2, the last term in the kth block Akand
put nk=mk+ 1. Then obviously xnk=1
2(ak+1 + 1) and lim
k→∞
mk
nk= 1. Finally
we have
lim
k→∞
xmk
xnk
= lim
k→∞
3
2ak−1
2
1
2(ak+1 +1) = 3 lim
k→∞
ak
ak+1
= 3 lim
k→∞
ak
a3
2
k
=0.
Thus sequences (mk)and(nk) fulfil the condition (2) of Theorem 1 proving (5).
Now notice that for every set X={x1<x
2< ...}⊂Nthe condition
lim sup
n→∞
nxn
x1+x2+···+xn<∞implies lim
n→∞
xn
x1+x2+···+xn=0. Thusonecanaskif
replacing the last condition with the above weaker one, the equivalence in (3)
would hold. The following theorem answers affirmatively this question and it
completes Theorem 3.
278

DISTRIBUTION FUNCTION OF BLOCK SEQUENCES

4

Let X={x1<x
2<...}⊂N.Then
lim sup
n→∞
nxn
x1+x2+···+xn
=∞⇐⇒c0(x)∈G(Xn).
P r o o f. First we will prove the implication =⇒.LetK>0 be arbitrary. Then
there exists a sequence (nk) of positive integers such that
L(nk)= nkxnk
x1+x2+···+xnk
>K. (7)
Then for every x∈(0,1) we have
F(Xnk,x)=
#i≤nk:xi
xnk
<x

nk
=mk
nk
.
By (7) and a simple calculation
K<L(nk)= nk
x1
xnk
+x2
xnk
+···+xnk
xnk
<nk
(nk−mk)x=1
1−mk
nkx
we obtain that the inequality
Kx < 1
1−mk
nk
holds for every K>0, proving mk
nk→1ask→∞holds for every x∈(0,1).
Consequently c0∈G(Xn).
Now we prove the reverse implication. Suppose c0∈G(Xn). Theorem 1 gives
the existence of integer sequences (mk)and(nk) such that
mk<n
k,k=1,2,... and xmk
xnk→0,mk
nk→1fork→∞.
Thus for each ε>0thereexistsak0∈Nsuch that for all k>k
0the inequality
xmk
xnk
<εholds. Thus
L(nk)= nk
x1
xnk
+x2
xnk
+···+xmk
xnk
+xmk+1
xnk
+···+xnk
xnk
>nk
mkε+(nk−mk).
Consequently
lim sup
n→∞
L(n)≥lim sup
k→∞
L(nk)≥1
ε
holds for every ε∈(0,1) proving lim sup
n→∞
L(n)=∞.
Now we are going to show that if the set G(xm/xn)={g(x)}is a singleton
and lim inf
n→∞
xn
xn+1 <1theng(x) is necessary equal to c0(x).
First we prove the useful lemma.
279

FERDIN´
AND FILIP — LADISLAV MIˇ
S´
IK — J´
ANOS T. T ´
OTH

1

Let X={x1<x
2<···<x
n<···}⊂Nand x∈(0,1).If
lim
N→∞ FN(xm/xn,x)=1,(8)
then for every ε>0there exists n0∈Nsuch that
A(Xn,x)>n(1 −ε)
holds for all n>n
0.
Proof. Let x∈(0,1) and (8) holds. Suppose that there exists an ε>0and
increasing sequence (ni) of positive integers such that
A(Xni,x)≤ni(1 −ε).(9)
Then
A(Xni,x)=AX[ni(1−ε)],x
+
ni

k=[ni(1−ε)]+1
A(Xk,x).
Using (9) and the fact that the sequence A(Xk,x) is increasing with respect
to kwe obtain
A(Xni,x)≤AX[ni(1−ε)],x
+(niε+1)ni(1 −ε).
Thus
A(Xni,x)
(ni+1)ni
2≤
A(X[ni(1−ε)],x)
([ni(1−ε)]+1)[ni(1−ε)]
2
([ni(1−ε)]+1)[ni(1−ε)]
2+(niε+1)ni(1 −ε)
(ni+1)ni
2
.(10)
The relation (8) implies
lim
n→∞
A(Xn,x)
(n+1)n
2
=1,
and applying (10) for ni→∞we obtain
1≤(1 −ε)2+2ε(1 −ε)=1−ε2,
a contradiction. 

5

Let G(xm/xn)={g(x)}and lim inf
n→∞
xn
xn+1 <1.Theng(x)=c0(x).
Proof. Let G(xm/xn)={g(x)}, lim inf
n→∞
xn
xn+1 <1andg(x)=c0(x)hold. Then
there exists a positive number a<1 and an increasing sequence (mk) of positive
integers such that for every k∈N
xmk
xmk+1
<a
holds. Theorem 2 implies the existence of x∈(0,1) such that
g(x)=1 and g(ax)<1.
280

DISTRIBUTION FUNCTION OF BLOCK SEQUENCES
As the condition xn
xmk
<x
implies xn
xmk+l≤xn
xmk+1
=xn
xmk
xmk
xmk+1
<ax
for every l∈N,weobtain
A(Xmk+l,ax)≥A(Xmk,x) (11)
for every l∈N.Nowchooseanε>0with
0<ε<1−g(ax).
As g(x) = 1, Lemma 1 and (11) give
A(Xmk+l,ax)>m
k(1 −ε)
for sufficiently large mkand every l∈N.
Let c>1 be arbitrary. Then
AX[cmk],ax
=A(Xmk,ax)+
[cmk]

i=mk+1
A(Xi,ax)
≥A(Xmk,ax)+
[cmk]

i=mk+1
mk(1 −ε)
≥A(Xmk,ax)
(mk+1)mk
2
(mk+1)mk
2+mk(c−1) −1mk(1 −ε).
Thus
AX[cmk],ax

([cmk]+1)[cmk]
2≥
A(Xmk,ax)
(mk+1)mk
2
(mk+1)mk
2+mk(c−1) −1mk(1 −ε)
([cmk]+1)[cmk]
2
.(12)
Using
lim
n→∞
A(Xn,ax)
(n+1)n
2
=g(ax),
and (12) with k→∞we obtain
g(ax)≥g(ax)+2(c−1)(1 −ε)
c2.
Consequently
g(ax)≥2(1 −ε)
c+1 ,
and for c→1+we obtain
g(ax)≥1−ε,
contradicting the choice of ε.
281

FERDIN´
AND FILIP — LADISLAV MIˇ
S´
IK — J´
ANOS T. T ´
OTH
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Received 7. 1. 2008 * Department of Mathematics
University of J. Selye
P.O. BOX 54
SK–945 01 Kom´arno
SLOVAKIA
E-mail: filip.ferdinand@selyeuni.sk
toth.janos@selyeuni.sk
** Department of Mathematics
University of Ostrava
30. dubna 22
CZ–701 03 Ostrava 1
CZECH REPUBLIC
E-mail: ladislav.misik@osu.cz
janos.toth@osu.cz
282