Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 2, Article 26, 2004
PRICE AND HAAR TYPE FUNCTIONS AND UNIFORM DISTRIBUTION OF
SEQUENCES
V. GROZDANOV AND S. STOILOVA
D EPARTMENT OF M ATHEMATICS ,
S OUTH W EST U NIVERSITY
66 I VAN M IHAILOV STR .
2700 B LAGOEVGRAD , B ULGARIA .
vassgrozdanov@yahoo.com
stanislavast@yahoo.com
Received 08 August, 2003; accepted 10 December, 2003
Communicated by N. Elezović
A BSTRACT. The Weyl criterion is shown in the terms of Price functions and Haar type functions. We define the so-called modified integrals of Price and Haar type functions and obtain the
analogues of the criterion of Weyl, the inequalities of LeVeque and Erdös-Turan and the formula
of Koksma in the terms of the modified integrals of Price and Haar type functions.
Key words and phrases: Uniform distribution of sequences; Price and Haar type functions; Weyl criterion; Inequalities of
LeVeque and Erdös-Turan; Formula of Koksma.
2000 Mathematics Subject Classification. 65C05; 11K38; 11K06.
1. I NTRODUCTION
Let ξ = (xi )i≥0 be a sequence in the unit interval [0, 1). We define A(ξ; J; N ) = {i : 0 ≤
i ≤ N − 1, xi ∈ J} for an arbitrary integer N ≥ 1 and an arbitrary subinterval J ⊆ [0, 1). The
sequence ξ is called uniformly distributed (abbreviated u. d.) if for every subinterval J of [0, 1)
)
the equality limN →∞ A(ξ;J;N
= µ(J) holds, and where µ(J) is the length of J.
N
Let ξN = {x0 , . . . , xN −1 } be an arbitrary net of real numbers in [0, 1). The extreme and
quadratical discrepancies D(ξN ) and T (ξN ) of the net ξN are defined respectively as
D(ξN ) = sup |N −1 A(ξN ; J; N ) − µ(J)|,
J⊆[0,1)
Z
|N
T (ξN ) =
0
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
138-03
1
−1
2
A(ξN ; [0, x); N ) − x| dx
12
.
2
V. G ROZDANOV AND S. S TOILOVA
The discrepancy DN (ξ) of the sequence ξ is defined as DN (ξ) = D(ξN ), for each integer
N ≥ 1, and ξN is the net, composed of the first N elements of the sequence ξ. It is well-known
that the sequence ξ is u. d. if and only if limN →∞ DN (ξ) = 0.
According to Kuipers and Niederreiter [8, Corollaries 1.1 and 1.2], the sequence ξ is u. d. if
and only if for each complex-valued and integrable in the sense of Riemann function f, defined
on R and periodical with period 1, the following equality
Z 1
N −1
1 X
(1.1)
lim
f (xi ) =
f (x)dx
N →∞ N
0
i=0
holds.
The theory of uniformly distributed sequences is divided into quantitative and qualitative
parts. Quantitative theory considers measures, showing the deviation of the distribution of a
concrete sequence from an ideal distribution. Qualitative theory main idea of uniformly distributed sequences is to find necessary and sufficient conditions for uniformity of the distribution of sequences.
Weyl [18] obtains such a condition (the so-called Weyl criterion) which is based on the use
of the trigonometric functional system T = {ek (x) = exp(2πikx), k ∈ Z, x ∈ R}. The
criterion of Weyl is: The sequence ξ = (xi )i≥0 is uniformly distributed if and only if the
P −1
equality limN →∞ N1 N
i=0 ek (xi ) = 0 holds for each integer k 6= 0.
The Walsh functional system has been recently used as an appropriate means of studying the
uniformity of the distribution of sequences. Sloss and Blyth [13] use this system to obtain future
necessary and sufficient conditions for a sequence to be u. d.
The link, which is realized for studying sequences in [0, 1), constructed in a generalized
number system and some orthonormal functional systems on [0, 1), constructed in the same
system, is quite natural. The purpose of our paper is to reveal the possibility some other classes
of orthonormal functional system, as the Price functional system and two systems of Haar type
functions to be used as a means of obtaining new necessary and sufficient conditions for uniform
distribution of sequences.
In Section 2 we obtain new necessary and sufficient conditions for uniform distribution of
sequences, which are analogues of the classical criterion of Weyl. These conditions are based
on the functions of Price and Haar type functions.
In Section 3 we introduce the so-called modified integrals of the Price functions and Haar
type functions. Integral analogues of the Weyl criterion are obtained in terms of these integrals. Analogues of the classical inequalities of LeVeque [9] and Erdös-Turan (see Kuipers and
Niederreiter [8]), and the formula of Koksma, (see Kuipers [7]) are obtained.
In Section 4 we prove some preliminary statements, which are used to prove the main results.
The proofs of the main results are given in Section 5. In Section 6 we give a conclusion, where
we announce some open problems, having to do with the problems, solved in our paper.
The results of this paper were announced in Grozdanov and Stoilova [3] and [4]. Here we
explain the full proofs of them. The results which are based on the Price functions generalize
the ones of Sloss and Blyth [13]. The results which are based on the Haar type functions are
new.
2. P RICE F UNCTIONAL S YSTEM , H AAR T YPE F UNCTIONAL S YSTEM AND
A NALOGUES OF THE C RITERION OF W EYL
Let B = {b1 , b2 , . . . , bj , . . . : bj ≥ 2, j ≥ 1} be an arbitrary fixed sequence of integer
numbers. We define ωj = exp 2πi
for each integer j ≥ 1. We define the set of the generalized
bj
Q
∞
powers {Bj }j=0 as: B0 = 1 and for each integer j ≥ 1, Bj = js=1 bs .
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
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P RICE AND H AAR T YPE F UNCTIONS AND U NIFORM D ISTRIBUTION
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3
Definition 2.1.
P∞
−1
(i) For real x ∈ [0, 1) in the B−adic form x =
i=1 xi Bi , where for i ≥ 1 xi ∈
{0, 1, . . . , bi − 1} and each integer j ≥ 0, Price [10] defines the functions χBj (x) =
xj+1
ωj+1
.
P
(ii) For each integer k ≥ 0 in the B−adic form k = nj=0 kj+1 Bj , where for 1 ≤ j ≤ n + 1,
kj ∈ {0, 1, . . . , bj −Q
1}, kn+1 6= 0 and real x ∈ [0, 1), the k-th function of Price χk (x) is
defined as χk (x) = nj=0 (χBj (x))kj+1 .
The system χ(B) = {χk }∞
k=0 is called the Price functional system. This system is a complete
orthonormal system in L2 [0, 1).
Let bj = b in the sequence B for each j ≥ 1. Then, the system {χ0 } ∪ {χbk }∞
k=0 is the
(b) ∞
(b) ∞
Rademacher [11] system {φk }k=0 of order b. The system of Chrestenson [2] {ψk }k=0 of
order b is obtained from the system χ(B). If for each j ≥ 1 bj = 2, then the original system of
Walsh [17] is obtained.
In 1947 Vilenkin [15] introduced the system χ(B) and Price [10] defined it independently
of him in 1957. Some names are used about the system χ(B) in special literature: both Price
system (see Agaev, Vilenkin, Dzafarly, Rubinstein [1]) and Vilenkin system (see Schipp, Wade,
Simon [12]). We use the name Price functional system in this paper.
We will consider two kinds of the so-called Haar type functions. Starting from the original
Haar [5] system, Vilenkin [16] proposes a new system of functions, which is called a Haar type
system, (see Schipp, Wide, Simon [12]). This definition is:
Definition 2.2. For x ∈ [0, 1) the k th Haar type function h0k (x), k ≥ 0 to the base B is defined
as follows: If k = 0, then h00 (x) = 1, ∀x ∈ [0, 1). If k ≥ 1 is an arbitrary integer and
(2.1)
k = Bn + p(bn+1 − 1) + s − 1,
where for some integer n ≥ 0, 0 ≤ p ≤ Bn − 1 and s ∈ {1, . . . , bn+1 − 1}, then
 √
+a+1
+a
sa
, if pbBn+1
≤ x < pbn+1
and a = 0, 1, . . . , bn+1 − 1,
 Bn ωn+1
Bn+1
n+1
0
hk (x) =

0,
otherwise.
We will consider another one:
Definition 2.3. For x ∈ [0, 1) the k th Haar type function h00k (x), k ≥ 0 to the base B is defined
as follows: If k = 0, then h000 (x) = 1, ∀x ∈ [0, 1). If k ≥ 1 is an arbitrary integer and
(2.2)
k = kn Bn + p,
where for some integer n ≥ 0, 0 ≤ p ≤ Bn − 1 and kn ∈ {1, . . . , bn+1 − 1}, then
 √
+a
+a+1
kn a
≤ x < pbn+1
and a = 0, 1, . . . , bn+1 − 1,
, if pbBn+1
 Bn ωn+1
Bn+1
n+1
h00k (x) =

0,
otherwise.
00 ∞
It can be easily seen that the systems {h0k }∞
k=0 and {hk }k=0 are complete orthonormal systems
in L2 [0, 1). In the case when for each j ≥ 1 bj = 2 the original system of Haar is obtained from
00 ∞
the systems {h0k }∞
k=0 and {hk }k=0 .
Theorem 2.1 (Analogues of the criterion of Weyl). The sequence (xi )i≥0 of [0, 1) is u. d. if and
only if:
N −1
1 X
lim
χk (xi ) = 0, for each k ≥ 1,
N →∞ N
i=0
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
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V. G ROZDANOV AND S. S TOILOVA
N −1
1 X 0
lim
hk (xi ) = 0, for each k ≥ 1,
N →∞ N
i=0
and
N −1
1 X 00
hk (xi ) = 0, for each k ≥ 1.
N →∞ N
i=0
lim
The proof of this theorem is based on the equality (1.1) and the properties of Price and Haar
type functional systems.
3. P RICE AND H AAR T YPE I NTEGRALS AND U. D .
S EQUENCES
Rx
We
consider
the
integrals
of
Price
and
Haar
type
functions
J
(x)
=
χk (t)dt, Ψ0k (x) =
k
0
Rx 0
R
x
hk (t)dt and Ψ00k (x) = 0 h00k (t)dt for each integer k ≥ 1 and x ∈ [0, 1).
0
For an arbitrary integer k ≥ 1 we define the integer n ≥ 0 by the condition Bn ≤ k < Bn+1 .
We define the modified integrals of Price function as
1
1
· q
δq.Bn ,k ,
(3.1)
Jn,q,k (x) = Jk (x) +
Bn+1 ωn+1 − 1
OF
for all x ∈ [0, 1) and each q = 1, 2, . . . , bn+1 − 1, and for arbitrary integers i, j ≥ 0, δi,j is the
Kronecker’s symbol.
If k is an integer of the kind (2.1), we define
1
1
−3
(3.2)
Ψ0n,s,k (x) = Ψ0k (x) +
Bn 2 s
,
bn+1
ωn+1 − 1
for all x ∈ [0, 1) and each s = 1, 2, . . . , bn+1 − 1.
If k is an integer of the kind (2.2), we define
(3.3)
Ψ00n,kn ,k (x) = Ψ00k (x) +
1
bn+1
1
−3
Bn 2
kn
ωn+1
−1
,
for all x ∈ [0, 1) and each kn = 1, 2, . . . , bn+1 − 1.
We will call the integrals Ψ0n,s,k (x) and Ψ00n,kn ,k (x) modified integrals of Haar type functions.
The next theorems hold:
Theorem 3.1 (Analogues of the inequality of LeVeque). Let ξN = {x0 , x1 . . . , xN −1 } be an
arbitrary net, composed of N ≥ 1 points of [0, 1). The discrepancy D(ξN ) of the net ξN
satisfies the inequalities:

2  31
−1 (q+1)Bn −1 N −1
∞ bn+1
X
X
X
X
12
Jn,q,k (xm )  ,
D(ξN ) ≤  2
N
n=0

D(ξN ) ≤ 
12
N2
q=1
k=qBn
m=0
2  13

0
Ψ
(x
)
,
m m=0 n,s,k
∞ bn+1
−1
X
X−1 (s+1)B
Xn −1 N
X
n=0
s=1
k=sBn
2  31
−1 (kn +1)Bn −1 N −1
∞ bn+1
X
X
X
X
12
D(ξN ) ≤  2
Ψ00n,kn ,k (xm )  .
N n=0 k =1 k=k B m=0

n
n
n
Theorem 3.2 (Integral analogues of the criterion of Weyl). Let an absolute constant B exist,
such as for each j ≥ 1 bj ≤ B. Let k ≥ 1 be an arbitrary integer and Bn ≤ k < Bn+1 . The
sequence ξ = (xi )i≥0 of [0, 1) is u. d. if and only if:
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
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P RICE AND H AAR T YPE F UNCTIONS AND U NIFORM D ISTRIBUTION
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5
(i)
N −1
1 X
lim
Jn,q,k (xi ) = 0 for each q = 1, 2, . . . , bn+1 − 1,
N →∞ N
i=0
(ii) If k ≥ 1 is of the kind (2.1) then
N −1
1 X 0
Ψn,s,k (xi ) = 0 for each s = 1, 2, . . . , bn+1 − 1,
N →∞ N
i=0
lim
(iii) If k ≥ 1 is of the kind (2.2) then
N −1
1 X 00
Ψn,kn ,k (xi ) = 0 for each kn = 1, 2, . . . , bn+1 − 1.
N →∞ N
i=0
lim
Theorem 3.3 (An analogue of the inequality of Erdös-Turan). Let an absolute constant B exist,
such that bj ≤ B for each j ≥ 1 and we signify b = min{bj : j ≥ 1}. Let ξN = {x0 , . . . , xN −1 }
be an arbitrary net, composed of N ≥ 1 points of [0, 1). For an arbitrary integer H > 0 we
define the integers M ≥ 0 and q ∈ {1, 2, . . . , bM +1 − 1} as qBM ≤ H < (q + 1)BM . Then the
following inequality holds

2
M
−1 bn+1
−1
X
X−1 (s+1)B
Xn −1 1 N
X
D(ξN ) ≤ 12
Jn,s,k (xm )
N
n=0 s=1
m=0
k=sBn
2
q−1 (s+1)BM −1 −1
X
X 1 N
X
JM,s,k (xm )
+ 12
N
s=1 k=sBM
m=0
 13
2
N −1
H
1 X
π 2
X
3B(1 + 2b sin B ) 1
 .
JM,q,k (xm ) +
+12
2 π
N
B
(b
−
1)b
sin
M
B
m=0
k=qB
M
Theorem 3.4 (Analogues of the formula of Koksma). Let ξN = {x0 , x1 . . . , xN −1 } be an arbitrary net, composed of N ≥ 1 points of [0, 1). The quadratical discrepancy T (ξN ) of the net ξN
satisfies the equalities
2
!2 X
N
−1 ∞ bn+1
−1
X
X−1 (q+1).B
Xn −1 N
X
1
(N T (ξN ))2 =
xm −
+
Jn,q,k (xm ) ,
m=0
2
m=0
n=0 q=1
k=q.Bn
(N T (ξN ))2 =
N
−1 X
m=0
1
xm −
2
!2
2
∞ bn+1
−1
X
X−1 (s+1)B
Xn −1 N
X
0
+
Ψn,s,k (xm ) ,
m=0
n=0 s=1
k=sBn
and
(N T (ξN ))2 =
N
−1 X
m=0
1
xm −
2
!2
2
∞ bn+1
−1
X
X−1 (kn +1)B
Xn −1 N
X
00
+
Ψ
(x
)
n,kn ,k m .
n=0
m=0
kn =1
k=kn Bn
4. P RELIMINARY S TATEMENTS
P∞
−1
Let x ∈ [0, 1) have the B−adic representation x =
j=0 xj+1 Bj+1 , where for j ≥ 0,
b
xj+1 ∈ {0, 1, . . . , bj+1 − 1}. For each integer j ≥ 0 we have that xj+1 = j+1
arg χBj (x).
2π
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
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6
V. G ROZDANOV AND S. S TOILOVA
Hence, we obtain the representation
∞
1 X 1
x=
arg χBj (x).
2π j=0 Bj
(4.1)
Lemma 4.1. Let k ≥ 1 be an arbitrary integer and k = βn+1 Bn + k 0 , where βn+1 ∈ {1, . . . ,
bn+1 − 1} and 0 ≤ k 0 < Bn . For x ∈ [0, 1) the k th Price integral satisfies the following equality
βn+1 bn+1
arg χBn (x)
∞
1 X −r
(4.2) Jk (x) =
χk0 (x) +
bn+1 arg χBn brn+1 x χk (x).
β
n+1
Bn+1
2πBn r=1
1 − ωn+1
. For an arbitrary integer β, 1 ≤ β ≤ b−1
Proof. Let b ≥ 2 be a fixed integer and ω = exp 2πi
b
and real x ∈ [0, 1) let
Z
1
1 − ωn+12π
x
(b)
Jβ (x) =
(b)
ψβ (t)dt.
0
We will prove the following equality
(b)
βb
(4.3)
(b)
Jβ (x)
1 1 − ω 2π arg φ0
=
b
1 − ωβ
(x)
∞
1 X −r
(b)
(b)
+
b arg ψbr (x)ψβ (x).
2π r=1
We put s = [bx], where [bx] denotes the integer part of bx and we have that
Z x
(b)
(b)
(4.4)
Jβ (x) =
[φ0 (t)]β dt
=
0
s−1
X Z (h+1)/b
Z
hβ
h/b
h=0
x
ω dt +
(b)
[φ0 (t)]β dt
s/b
β[bx]
11−ω
[bx]
(b)
=
+
ψ
(x)
x
−
β
b 1 − ωβ
b
From (4.1) and (4.4) we obtain (4.3).
If
Z x
(bn+1 )
Jβn+1 ·Bn (x) =
χβn+1 Bn (t)dt,
.
0
then
(b
)
n+1
Jk (x) = χk0 (x)Jβn+1
Bn (x).
(4.5)
We have the equalities
(bn+1 )
Jβn+1
·Bn (x)
Z
x
β
χBn+1
(t)dt
n
=
Z0 x h
iβn+1
(b
)
φ0 n+1 (Bn t)
dt
0
Z Bn x
1
1 (bn+1 )
(bn+1 )
ψβn+1
(t)dt =
J
(Bn x),
=
Bn 0
Bn βn+1
=
so that
(b
)
n+1
Jβn+1
·Bn (x) =
1 (bn+1 )
J
(Bn x).
Bn βn+1
From the last equality and (4.5) we obtain that
1
(bn+1 )
χk0 (x) · Jβn+1
(Bn x).
(4.6)
Jk (x) =
Bn
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P RICE AND H AAR T YPE F UNCTIONS AND U NIFORM D ISTRIBUTION
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S EQUENCES
7
From (4.3) we obtain
(4.7)
(bn+1 )
(Bn x)
Jβn+1
=
1 1−ω
βn+1 bn+1
2π
arg χBn (x)
β
bn+1
n+1
1 − ωn+1
+
∞
1 X −r
b
arg χBn (brn+1 x)χβn+1 Bn (x).
2π r=1 n+1
From (4.6) and (4.7) we obtain (4.2).
For every integer n ≥ 1 we consider the set B(n) = {b1 , b2 , . . . , bn }. We define the “reverse”
e
e1 = bn , B
e2 = bn bn−1 , . . . , B
en = bn · · · b1 . For an
set B(n)
= {bn , bn−1 , . . . , b1 }, so that B
, Bpn
and
arbitrary integer p, 0 ≤ p < Bn and for a B-adic rational Bpn let (p)B(n) , (p)B(n)
e
B(n)
p
e
be the corresponding representations of p and Bpn to the systems B(n) and B(n).
Bn
e
B(n)
Lemma 4.2. (Relationships between the Price and the Haar type functions)
(i) Let k ≥ 1 be an arbitrary integer of the kind (2.1). Then for all x ∈ [0, 1)
!
Bn+1 −1 bn+1 −1
X X
α
1
a.s
√
h0k (x) =
ωn+1
χ(pbn+1 +a)B(n+1)
χα (x);
e
Bn+1 B(n+1)
bn+1 Bn α=0 a=0
e
Ψ0n,s,k (x) =
1
√
bn+1 Bn
×
n bj+1
X
X−1 (t+1)B
Xj −1 bn+1
X−1
j=0
t=0
α=tBj
a.s
ωn+1
χ(pbn+1 +a)B(n+1)
e
Jj,t,α (x).
Bn+1
a=0
!
α
e
B(n+1)
(ii) Let k ≥ 1 be an arbitrary integer of the kind (2.2). Then for all x ∈ [0, 1)
!
Bn+1 −1 bn+1 −1
X X
1
α
√
(4.8)
h00k (x) =
ω a.kn χ(pbn+1 +a)B(n+1)
χα (x);
e
Bn+1 B(n+1)
bn+1 Bn α=0 a=0 n+1
e
(4.9) Ψ00n,kn ,k (x) =
×
1
√
bn+1 Bn
n bj+1
X
X−1 (t+1)B
Xj −1 bn+1
X−1
j=0
t=0
α=tBj
a.kn
ωn+1
χ(pbn+1 +a)B(n+1)
e
a=0
!
α
Bn+1
Jj,t,α (x).
e
B(n+1)
Proof. For an arbitrary integer p, 0 ≤ p < Bn and x ∈ [0, 1), following Kremer [6] we define
the function


p+1
p

, Bn
1, if x ∈


Bn

B(n)
B(n)
;
x)
=
q(Bn ; (p)B(n)
e



p
p+1

6
, Bn
.
 0 if x ∈
Bn
B(n)
B(n)
The equality
Bn −1
1 X
q(Bn ; (p)B(n)
; x) =
χ e
e
Bn α=0 (p)B(n)
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
α
Bn
!
χα (x)
e
B(n)
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8
V. G ROZDANOV AND S. S TOILOVA
holds. Let us use the significations: For an arbitrary integer p, 0 ≤ p < Bn , (p)B(n)
=
e
(p̃1 p̃2 . . . p̃n )B(n)
, for an arbitrary integer α, 0 ≤ α < Bn , Bαn
= (0 · αn αn−1 . . . α1 )B(n)
,
e
e
e
B(n)
for real x ∈ [0, 1), x = (0 · x1 . . . xn+1 . . .)B . Then, we obtain the equalities
!
α
αn−1 p̃n−1
= ωnαn p̃n ωn−1
. . . ω1α1 p̃1
(4.10)
χ(p)B(n)
e
Bn B(n)
e
and
χα (x) = ω1α1 x1 ω2α2 x2 . . . ωnαn xn .
(4.11)
If x ∈
p
Bn
,
B(n)
p+1
Bn
, then for all j = 1, . . . , n, xj = p̃j . From (4.10) and (4.11) we
B(n)
obtain
χ(p)B(n)
e
If x 6∈
p
Bn
,
B(n)
p+1
Bn
α
Bn
!
χα (x) = 1.
e
B(n)
, then, some δ, 1 ≤ δ ≤ n exists, so that xδ 6= p̃δ . Then, we
B(n)
have that
bX
δ −1
α (xδ −p̃δ )
ωδ δ
= 0.
αδ =0
From (4.10) and (4.11), we obtain
BX
n −1
χ(p)B(n)
e
α=0
α
Bn
!
χα (x) =
e
B(n)
j −1
n bX
Y
α (xj −p̃j )
ωj j
= 0.
j=1 αj =0
Now let k ≥ 1 be an integer of the kind (2.2). In order to prove (4.8) we note that for all
x ∈ [0, 1)
h00k (x)
X−1
p bn+1
a.kn
= Bn
ωn+1
q Bn+1 ; (pbn+1 + a)B(n+1)
;x .
e
a=0
We will prove (4.9). Using the proved formula for h00k (x), we have that
Z x
00
(4.12) Ψk (x) =
h00k (t)dt
0
=
=
1
√
bn+1 Bn
1
√
bn+1 Bn
Bn+1 −1 bn+1 −1
X
X
α=0
a=0
a.kn
ωn+1
χ(pbn+1 +a)B(n+1)
e
n bj+1
X
X−1 (t+1)B
Xj −1 bn+1
X−1
j=0
t=0
α=tBj
α
Bn+1
!Z
× Jj,t,α (x) −
χα (t)dt
a.kn
ωn+1
χ(pbn+1 +a)B(n+1)
e
1
1
t
Bj+1 ωj+1
−1
x
0
e
B(n+1)
a=0
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
α
Bn+1
!
e
B(n+1)
δtBj ,α .
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P RICE AND H AAR T YPE F UNCTIONS AND U NIFORM D ISTRIBUTION
OF
S EQUENCES
9
It is not difficult to prove that
(4.13)
n
X
j=0
1
Bj+1
bj+1 −1
X
t=0
(t+1)Bj −1 bn+1 −1
1
t
ωj+1
−1
X
X
α=tBj
a=0
a.kn
ωn+1
× χ(pbn+1 +a)B(n+1)
e
α
Bn+1
!
δtBj ,α =
e
B(n+1)
Bn−1
.
kn
ωn+1
−1
From (4.12) and (4.13), we obtain (4.9).
In the following lemma the relationships in the opposite direction are proved.
Lemma 4.3. Let k ≥ 1 be an arbitrary integer and k = sBn + p, where s ∈ {1, . . . , bn+1 − 1}
and 0 ≤ p ≤ Bn − 1. Then, for all x ∈ [0, 1) the equalities
(4.14)
Bn −1
1 X
(n)
√
χk (x) =
αp,j̃ h00sBn +j̃ (x);
Bn j=0
(4.15)
Bn −1
1 X
(n)
Jn,s,k (x) = √
αp,j̃ Ψ00n,s,sBn +j̃ (x);
Bn j=0
Bn −1
1 X
(n)
χk (x) = √
αp,j̃ h0Bn +j̃(bn+1 −1)+s−1 (x)
Bn j=0
and
Bn −1
1 X
(n)
Jn,s,k (x) = √
αp,j̃ Ψ0n,s,Bn +j̃(bn+1 −1)+s−1 (x),
Bn j=0
P n −1 (n)
(n)
hold, where αp,j̃ are complex mumbers, so that B
j=0 αp,j̃ = Bn · δsBn ,k .
≤ x <
Proof. Let x ∈ [0, 1) be fixed. We define t̃ = (t)B(n)
, 0 ≤ t̃ < Bn as Bt̃n
e
B(n)
h
S n+1 −1 (n+1)
(n)
(n)
t̃+1
. We denote ∆t̃ = Bt̃n , t̃+1
. It is obvious that ∆t̃ = ba=0
∆t̃bn+1 +a . There is
Bn
Bn
B(n)
(n+1)
some a, 0 ≤ a ≤ bn+1 − 1, so that x ∈ ∆t̃bn+1 +a . We have the equalities
!
p
t̃
a.s
a.s
χk (x) = χp
ωn+1
and h00s·Bn +t̃ (x) = Bn ωn+1
.
Bn B(n)
Hence, we obtain
1
χk (x) = √ χp
Bn
t̃
Bn
!
h00s·Bn +t̃ (x).
B(n)
(n)
Let t̃0 be an arbitrary integer, so that 0 ≤ t̃0 < Bn , and t̃0 6= t̃. For x ∈ ∆t̃
h00s·Bn +t̃0 (x) = 0. Hence, we obtain the equality
!
BX
n −1
1
j̃
χk (x) = √
χp
h00s·Bn +j̃ (x).
Bn B(n)
Bn j=0
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
we have that
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10
V. G ROZDANOV AND S. S TOILOVA
(n)
αp,j̃
j̃
Bn
Let for integers 0 ≤ p < Bn and 0 ≤ j < Bn we signify
= χp
. We use
B(n)
the representations p = (pn pn−1 . . . p1 )B(n) and Bj̃n
= (0 · j1 j2 . . . jn )B(n) , where for
B(n)
1 ≤ τ ≤ n pτ , jτ ∈ {0, 1, . . . , bτ − 1}. Then, we obtain the equality
(4.16)
BX
n −1
(n)
αp,j̃
=
n bX
τ −1
Y
ωτpτ jτ .
τ =1 jτ =0
j=0
If p = 0, then, for 1 ≤ τ ≤ n, pτ = 0 and from (4.16) we obtain
PBn −1
j=0
then, some δ, 1 ≤ δ ≤ n exists, so that pδ 6= 0. From (4.16), we obtain
We will prove (4.15). From (4.14) for all x ∈ [0, 1) we have
(n)
αp,j̃ = Bn . If p 6= 0
PBn −1
j=0
(n)
αp,j̃ = 0.
Bn −1
1 X
1
1
− 32
(n)
00
Jk (x) = √
α
ΨsBn +j̃ (x) +
Bn s
bn+1
ωn+1 − 1
Bn j=0 p,j̃
−
1
bn+1
Bn−2
BX
n −1
1
s
ωn+1
−1
(n)
αp,j̃ .
j=0
From the last equality and (3.3) we obtain (4.15).
Sobol [14] proved a similar result, giving the relationship between the original Haar and the
Walsh functions.
For an arbitrary net ξN = {x0 , . . . , xN −1 }, composed of N ≥ 1 points of [0, 1) and x ∈ [0, 1)
we signify R(ξN ; x) = A(ξN ; [0, x); N ) − N x. Then, the next lemma holds:
Lemma 4.4.
(i) The Fourier-Price coefficiens of R(ξN ; x) satisfy the equalities:
N
−1 X
1
(χ)
xm −
;
a0 = −
2
m=0
and for each integer k ≥ 1, k = kn Bn + p, kn ∈ {1, 2, . . . , bn+1 − 1}, 0 ≤ p < Bn
(χ)
ak = −
N
−1
X
Jn,kn ,k (xm ).
m=0
(ii) The Fourier-Haar type coefficiens of R(ξN ; x) satisfy the equalities:
N
−1 N
−1 X
X
1
1
(h0 )
(h00 )
a0 = −
xm −
, a0 = −
xm −
;
2
2
m=0
m=0
Let k ≥ 1 be an arbitrary integer of kind (2.1). Then,
(h0 )
ak
=−
N
−1
X
Ψ0n,s,k (xm ).
m=0
Let k ≥ 1 be an arbitrary integer of kind (2.2). Then,
(4.17)
(h00 )
ak
=−
N
−1
X
Ψ00n,kn ,k (xm ).
m=0
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
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P RICE AND H AAR T YPE F UNCTIONS AND U NIFORM D ISTRIBUTION OF S EQUENCES
11
Proof. Let for 0 ≤ m ≤ N − 1, cm (x) be the characteristic function of the interval (xm , 1).
Then,
N
−1
X
R(ξN ; x) =
cm (x) − N x.
m=0
We will prove only (4.17). The proof of the remaining equalities of the lemma is similar. For
an arbitrary integer k ≥ 1 of the kind (2.2) we have:
(4.18)
(h00 )
ak
Z
=
1
R(ξN ; x)h00k (x)dx
=
0
N
−1 Z 1
X
m=0
cm (x)h00k (x)dx
1
Z
xh00k (x)dx.
−N
0
0
The equalities
(4.19)
N
−1 Z 1
X
m=0
cm (x)h00k (x)dx
=
0
N
−1 Z 1
X
m=0
h00k (x)dx
Z
xm
−
0
h00k (x)dx
=−
0
N
−1
X
Ψ00k (xm )
m=0
hold. From (4.1) and (4.8) we obtain
Z 1
(4.20)
xh00k (x)dx
0
=
1
√
X
X
α=0
a=0
2πbn+1 Bn
×
=
Bn+1 −1 bn+1 −1
∞
X
1
Br
r=0
1
√
2πbn+1 Bn
×
∞
X
r=0
1
Br
Z
a·kn
ωn+1
χ(pbn+1 +a)B(n+1)
e
α
Bn+1
e
B(n+1)
1
arg χBr (x)χα (x)dx
0
BX
n −1
χ(pbn+1 )B(n+1)
e
α=0
Z
α
Bn+1
! "bn+1 −1
X
e
B(n+1)
#
a·kn
ωn+1
a=0
1
arg χBr (x)χα (x)dx
0
1
bn+1 −1 (t+1)Bn −1
X
X
α
√
χ(pbn+1 )B(n+1)
e
Bn+1
2πbn+1 Bn t=1 α=tB
n
"bn+1 −1
# ∞
Z 1
X (t−k )a X
1
n
×
ωn+1
arg χBr (x)χα (x)dx
Br 0
a=0
r=0
!
(kn +1)Bn −1
X
1
α
= √
χ(pbn+1 )B(n+1)
e
Bn+1 B(n+1)
2π Bn α=k B
e
n n
Z 1
∞
X
1
×
arg χBr (x)χα (x)dx.
Br 0
r=0
+
!
!
e
B(n+1)
The following equality can be proved: Let n ≥ 0 and q ∈ {1, 2, . . . , bn+1 − 1} be fixed integers.
Then, for each integer α ≥ 0
Z 1
1
1
1
(4.21)
arg χBn (x)χα (x)dx =
· q
δq·Bn ,α .
2π 0
bn+1 ωn+1 − 1
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
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12
V. G ROZDANOV AND S. S TOILOVA
From (4.20) and (4.21) we obtain
Z 1
(4.22)
xh00k (x)dx =
0
1
1
−3
bn+1
Bn 2
kn
ωn+1
−1
.
From (4.18), (4.19), (4.22) and (3.3) we obtain (4.17).
5. P ROOFS OF THE M AIN R ESULTS
Proof of Theorem 3.1. For an arbitrary net ξN = {x0 , . . . , xN −1 }, composed of N ≥1 points
P −1
1
of [0, 1), following Kuipers and Niederreiter [8] we denote S(ξN ) = N
m=0 xm − 2 and for
x ∈ [0, 1), Q(ξN ; x) = N1 (R(ξN ; x) + S(ξN )). The inequality
Z 1
3
D (ξN ) ≤ 12
Q2 (ξN ; x)dx.
0
is proved. Hence, we obtain
12
D (ξN ) ≤ 2
N
3
(5.1)
1
Z
R (ξN ; x)dx − S (ξN ) .
2
2
0
We obtain
2  31
−1 (kn +1)Bn −1 N −1
∞ bn+1
X
X
X
X
12
D(ξN ) ≤  2
Jn,kn ,k (xm ) 
N n=0 k =1 k=k B m=0

n
n
n
from Lemma 4.4 (i) and Parseval’s equality.
The other inequalities of Theorem 3.1 follow from (5.1) and Lemma 4.4 (ii).
Proof of Theorem 3.2.
Necessity of (i): We assume that the sequence is u. d. We will prove the equality
lim
(5.2)
N
−1
X
N →∞
Jn,q,k (xi ) = 0.
i=0
We use the representation k = q·Bn +p, where n ≥ 0, q ∈ {1, 2, . . . , bn+1 −1} and 0 ≤ p < Bn .
From (3.1) and Lemma 4.1, we obtain
(5.3) Jn,q,k (x) = −
1
χp (x)
q
Bn+1 ωn+1 − 1
·
qbn+1
+
1
Bn+1
+
1
Bn+1
arg χ
Bn
ω 2π
· n+1q
ωn+1 − 1
·
1
q
ωn+1
(x)
χp (x) +
−1
δq.Bn ,k
∞
1 X −r
bn+1 arg χBn brn+1 x χk (x).
2πBn r=1
Firstly, we assume that k = q · Bn , hence, p = 0 and δq·Bn ,k = 1. From (5.3) for Jn,q,k (x), we
have
qbn+1
Jn,q,k (x) =
1
Bn+1
arg χ
Bn
ω 2π
· n+1q
ωn+1 − 1
(x)
∞
1 X −r
+
bn+1 arg χBn brn+1 x χk (x).
2πBn r=1
We will prove the equalities
N −1
n+1
1 X qb2π
arg χBn (xi )
lim
ωn+1
= 0 and
N →∞ N
i=0
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
N −1 ∞
1 X X −r
lim
bn+1 arg χBn (brn+1 xi )χk (xi ) = 0.
N →∞ N
i=0 r=1
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P RICE AND H AAR T YPE F UNCTIONS AND U NIFORM D ISTRIBUTION OF S EQUENCES
13
From (1.1), it is sufficient to prove that
Z 1 qbn+1
arg χBn (x)
2π
dx = 0
ωn+1
0
and
Z
∞
1X
0
r
b−r
n+1 arg χBn bn+1 x χk (x)dx = 0.
r=1
We have the following equalities
Z
1
qbn+1
2π
ωn+1
arg χBn (x)
dx =
0
s=0
j=0
=
jbn+1 +s+1
Bn+1
BX
n −1 bn+1
X−1 Z
jbn+1 +s
Bn+1
qbn+1
2π
ωn+1
arg χBn (x)
dx
bn+1 −1
1
X
bn+1
sq
ωn+1
= 0.
s=0
Since k = q·Bn from (4.21), we have the equalities
#
Z 1 "X
∞
r
b−r
n+1 arg χBn bn+1 x χk (x) dx
0
r=1
=
=
=
∞
X
r=1
∞
X
r=1
∞
X
b−r
n+1
Z
b−r
n+1
Z
b−r
n+1
Z
arg χBn brn+1 x χk (x)dx
0
1
h
iq
(b
)
(b
)
arg φ0 n+1 (brn+1 Bn x) φ0 n+1 (Bn x) dx
0
Bn
h
iq
(bn+1 )
n+1 )
dt
arg φ(b
(t)
φ
(t)
0
r
0
r=1
= Bn
1
∞
X
b−r
n+1
Z
0
r=1
1
(b
)
n+1
arg ψbrn+1
(t)ψq(bn+1 ) (t)dt = 0.
If k 6= q·Bn , then, from (5.3), we will obtain a useful formula for Jn,q,k (x) and by analogy, we
can prove the equality (5.2).
Sufficiency of (i): We assume that the sequence ξ = (xi )i≥0 is not u. d. Then, limN →∞ DN (ξ) =
D > 0.
Let M > 0 be a fixed integer. From Theorem 3.1 we have
3
(5.4) DN
(ξ) ≤ 12
N −1
2
X
1
Jn,q,k (xm )
N
m=0
M
−1 bn+1
X
X−1 (q+1)B
Xn −1 n=0
q=1
k=qBn
2
∞ bn+1
−1
X
X−1 (q+1)B
Xn −1 1 N
X
+ 12
J
(x
)
n,q,k m .
N m=0
n=M q=1
k=qB
n
For arbitrary integers n ≥ 0, q ∈ {1, 2, . . . , bn+1 − 1} and qBn ≤ k < (q + 1)Bn we can prove
−1
1 N
1
1
X
(5.5)
Jn,q,k (xm ) ≤ B +
.
π
N
2 sin B Bn+1
m=0
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
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14
V. G ROZDANOV AND S. S TOILOVA
Let b = min{bn : n ≥ 1}. Then, using the inequality (5.5), we have
2 2 X
∞ bn+1
−1
∞
X
X−1 (q+1)B
Xn −1 1 N
X
1
1
(5.6)
J
(x
)
<
B
+
n,q,k m π
N
2 sin B
Bn+1
m=0
n=M q=1
n=M
k=q.Bn
2 X
∞
1
1
≤ B+
π
n+1
b
2 sin B
n=M
2
1
1
1
.
=
B+
π
bM
b−1
2 sin B
Now we choose M , so that
12
b−1
(5.7)
1
B+
2 sin Bπ
2
1
1
< D3 .
M
b
2
3
Let ε > 0 be arbitrary. Then, an integer N0 exists, so that for each N ≥ N0 , D3 − ε < DN
(ξ).
From (5.4), (5.6), (5.7) and the last inequality, we obtain
2
M
−1 bn+1
−1
X
X−1 (q+1)B
Xn −1 1 N
X
1 3
D − ε < 12
Jn,q,k (xm ) .
N
2
n=0 q=1
m=0
k=qB
n
We choose an integer ν > 0, so that 21 D3 − ε > ν1 D3 and we obtain
2
M
−1 bn+1
−1
X
X−1 (q+1)B
Xn −1 1 N
X
1 3
0 < D < 12
Jn,q,k (xm ) .
N
ν
n=0 q=1
m=0
k=qB
n
Finally, (n, q, k) exists, such that
N −1
1 X
Jn,q,k (xm ) > 0.
lim
N →∞ N
m=0
The demonstration of (ii) and (iii) of the theorem is a consequence of the formulae obtained in
Lemma 4.2, Lemma 4.3 and (i) of Theorem 3.2.
Proof of Theorem 3.3.
(5.8)
2
N
−1
X
1
D3 (ξN ) ≤ 12
J
(x
)
n,s,k m N
n=0 s=1
m=0
k=sBn
2
q−1 (s+1)BM −1 −1
X
X 1 N
X
+ 12
J
(x
)
M,s,k m N
s=1 k=sBM
m=0


2
(q+1)BM −1 H
N −1
 X
X
1 X
+ 12
+
JM,q,k (xm )

 N
m=0
k=H+1
k=qBM
2
bM +1 −1 (s+1)BM −1 −1
X
X 1 N
X
+ 12
JM,s,k (xm )
N
s=q+1
m=0
k=sB
M
−1 bn+1
X
X−1 (s+1)B
Xn −1 M
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P RICE AND H AAR T YPE F UNCTIONS AND U NIFORM D ISTRIBUTION OF S EQUENCES
+ 12
15
N −1
2  31
X
1
Jn,s,k (xm )  .
N
m=0
bn+1 −1 (s+1)Bn −1 ∞
X
n=M +1
X
X
s=1
k=sBn
We have the following inequality
2 b −1 (s+1)B −1 2
(q+1)BM −1 −1
−1
M +1
M
1 N
X
X
X
X
X 1 N
(5.9)
Σ=
J
(x
)
+
J
(x
)
M,q,k m M,s,k m N
N
m=0
s=q+1
m=0
k=H+1
k=sBM
2
bn+1 −1 (s+1)Bn −1 −1
∞
X
X
X
X 1 N
+
Jn,s,k (xm )
N
m=0
n=M +1 s=1
k=sBn
2
−1
∞ bn+1
X
X−1 (s+1)B
Xn −1 1 N
X
≤
Jn,s,k (xm ) .
N
n=M
s=1
k=sBn
m=0
The following inequalities
2
−1
1 N
X
(5.10)
Jn,s,k (xm ) ≤
N
m=0
!2
1
|Jn,s,k (xm )|
N
m=0
N
−1
X
and
(5.11)
|Jn,s,k (x)| ≤ |Jk (x)| +
1
1
, ∀x ∈ [0, 1)
π ·
2b sin B Bn
hold. It is not difficult to prove that for each k, Bn ≤ k < Bn+1
1
(5.12)
|Jk (x)| <
, ∀x ∈ [0, 1).
Bn
From (5.11) and (5.12), we obtain
1
1
(5.13)
|Jn,s,k (x)| ≤ 1 +
, ∀x ∈ [0, 1).
π
2b sin B Bn
From (5.9), (5.10) and (5.13) the following inequalities
2
∞ bn+1
X
X−1 (s+1)B
Xn −1 B(1 + 2b sin Bπ )2 1
1
1
Σ≤
1+
<
2b sin Bπ
Bn2
4(b − 1)b sin2 Bπ BM
n=M s=1
h=sB
n
hold. The statement of the theorem holds from the last inequality and (5.8).
Theorem 3.4 is a direct consequence of Lemma 4.4 and Parseval’s equality.
6. C ONCLUSION
In conclusion, the authors will present possible variants to extend this study.
The obtained results show that Price functions and Haar type functions can be used as a
means of examining uniformly distributed sequences. The proved results raise the issue of their
generalization. The most natural generalization is to obtain results in the s−dimensional cube
[0, 1)s , s ≥ 2.
This is not difficult to do when Price functions and Haar type functions are used (see Kuipers
and Niederreiter [8, Corollaries 1.1 and 1.2]).
The obtaining of results, in which the modified integrals from the corresponding s−dimensional
functions are in use, is connected with great technical difficulties. In the one-dimensional case
J. Inequal. Pure and Appl. Math., 5(2) Art. 26, 2004
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16
V. G ROZDANOV AND S. S TOILOVA
the proof of sufficiency of Theorem 3.2 is based on the analog of the LeVeque inequality, exposed in Theorem 3.1. A multidimensional variant of the inequality of Le Veque is not known in
this form to the authors. In this case, the proof of sufficiency of the multidimensional variant of
Theorem 3.2 is connected with proving the multidimensional variant of LeVeque’s inequality.
Another direction to generalize the obtained results is the possibility to use arbitrary orthonormal bases in L2 [0, 1) as a means of examining uniformity distributed sequences. The obtained
results in such a study would have more general nature. The definition and the use of the modified integrals of the functions of an arbitrary system will be difficult in practice because these
integrals depend on the concrete values of the corresponding functions.
Regarding the applications of uniformly distributed sequences, the inequality of KoksmaHlawka gives an estimation of the error of s−dimensional quadrature formula in the terms of
discrepancy of the used net. In this sense, the problem of obtaining quantitative estimations of
discrepancy is interesting, as the functions presented in this paper can be used as a means of
solving the above problem.
The shown generalizations are a part of the problems to solve in connection to the study of
uniformity distributed sequences by orthonormal bases in L2 [0, 1).
The methods to prove the theorems in this paper, by suitable adaptation may be used to solve
some of the exposed problems.
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