Acta Mathematica Academiae Paedagogicae Nyı́regyháziensis
22 (2006), 33–39
www.emis.de/journals
ISSN 1786-0091
ABSOLUTE CONVERGENCE OF THE DOUBLE SERIES OF
FOURIER – HAAR COEFFICIENTS
ALEXANDER APLAKOV
Abstract. In this paper we study the absolute convergence of the double
series of Fourier-Haar coefficients of the class PBV p .
1. Introduction
The problems related to the behaviour of single series of Fourier-Haar are well
studied [9]. Namely, P. Ulianov [14] and B. Golubov [8] received the results related to the problems of absolute convergence of the series of Fourier–Haar coefficients. Some generalization of these results related were received by Z. Chanturia
[3], T. Akhobadze [1], U. Goginava [7] and by the author [2]. In the term of modulus of smoothness the problem of absolute convergence of the series of Fourier-Haar
coefficients was studied by V. Krotov [10]. Multidimensional analogies corresponding to the results of V. Krotov were formulated in the works of V. Tsagareishvili
[13] and G. Tabatadze [12].
The estimates of Fourier coefficients of functions of bounded fluctuation with
respect to Walsh system were studied in [11] and with respect to Vilenkin system
were studied by G. Gát and R. Toledo [4].
We consider the double Haar system {χn (x) × χm (y) : n, m = 0, 1, 2, . . .} on the
unit square I 2 = [0, 1] × [0, 1]. As usual, Lp I 2 (p > 1) denotes the set of all
measurable functions defined on I 2 , for which

kf kp = 
Z1 Z1
0
0
p
 p1
|f (x, y)| dxdy  < ∞
and C I 2 is the space of continuous functions on I 2 equipped with maximum
norm
kf kc = max |f (x, y)| .
x,y∈I
2
If f ∈ L I , then
Cn,m (f ) =
Z1 Z1
0
f (x, y) χn (x) χm (y) dxdy
0
is the (n, m)th Fourier-Haar coefficient of f .
2000 Mathematics Subject Classification. 42C10.
Key words and phrases. Fourier-Haar coefficients, bounded variation, absolute convergence.
33
34
ALEXANDER APLAKOV
2
We say that f ∈ Lip α on [0, 1] , if
kf (· + h, · + η) − f (·, ·)kc = O
We have the following theorem.
h2 + η 2
α2 , α ∈ (0, 1] .
2
Theorem A ([12]). a) Let f ∈ Lip α on [0, 1] , α ∈ (0, 1]. If β > 0 and γ + 1 <
β (α+1)
2 , then
∞ X
∞
X
γ
β
(nm) |Cn,m (f )| < ∞.
b) Let γ + 1 =
Lip α for which
n=1 m=1
(α+1)
β 2 , for some
∞ X
∞
X
α ∈ (0, 1). Then there exists a function fα ∈
(nm)γ |Cn,m (fα )|β = ∞.
n=1 m=1
The case for γ = 0 was considered earlier by V. Tsagareishvili [13].
Let f ∈ Lp I 2 . The partial integrated modulus of continuity are defined by
o
n
ω1 (δ1 , f )p = sup kf (x + u, y) − f (x, y)kp : |u| 6 δ1 ,
n
o
ω2 (δ2 , f )p = sup kf (x, y + v) − f (x, y)kp : |v| 6 δ2 .
We also use the notion of the mixed integrated modulus of continuity. It is
defined as follows
ω1,2 (δ1 , δ2 , f )p
= sup kf (x + u, y + v) − f (x + u, y) − f (x, y + v) + f (x, y)kp :
|u| 6 δ1 , |v| 6 δ2 , f ∈ Lp I 2 .
It is not difficult to show that
(1)
ω1,2 (δ1 , δ2 , f )p 6 2
q
q
ω1 (δ1 , f )p ω2 (δ2 , f )p .
We study the problem of absolute convergence of the series of Fourier-Haar
coefficients for the classes of functions with bounded partial p-variations, which
were first considered by U. Goginava (see [5] for p = 1 and [6] for p > 1).
Definition. A
function f : I 2 → R is said to be of bounded partial p-variation
f ∈ PBVp I 2 if there exists a constant K such that for any partition
∆1 : 0 6 x0 < x1 < x2 < . . . < xn 6 1,
∆2 : 0 6 y0 < y1 < y2 < . . . < ym 6 1,
we have
V1 (f )p
= sup sup
V2 (f )p
= sup sup
y
x
n−1
X
p
|f (xi , y) − f (xi+1, y)| 6 K,
∆1 i=0
m−1
X
p
|f (x, yj ) − f (x, yj+1 )| 6 K.
∆2 j=0
Given a function f (x, y), periodic in both variables with period 1. Denote by
∆h1 f (x, y)1 = f (x + h1 , y) − f (x, y) ,
∆h2 f (x, y)2 = f (x, y + h2 ) − f (x, y) ,
∆h1 ,h2 f (x, y) = ∆h1 (∆h2 f (x, y)2 )1 = ∆h2 (∆h1 f (x, y)1 )2
= f (x, y) − f (x + h1 , y) − f (x, y + h2 ) + f (x + h1 , y + h2 ) .
ABSOLUTE CONVERGENCE OF SERIES. . .
35
2. Main Results
The main results of this paper are presented in the following propositions.
Theorem 1. Let f ∈ PBVp I 2 , p > 1 and β >
∞ X
∞
X
2p
1+p .
Then
β
|Cn,m (f )| < ∞.
n=0 m=0
Theorem 2. Let f ∈ PBVp I 2 , p > 1 and α <
∞ X
∞
X
1
2p
− 21 . Then
[(n + 1) (m + 1)]α |Cn,m (f )| < ∞.
n=0 m=0
1
+ 12 . Then
Theorem 3. Let f ∈ PBVp I 2 , p > 1 and β > 0, α + 1 < β 2p
∞ X
∞
X
α
β
[(n + 1) (m + 1)] |Cn,m (f )| < ∞.
n=0 m=0
Since Lip p1 ⊂ PBVp in case p > 1 the sharpness of Theorems 1-3 follows from
the works [12, 13].
Lemma 1. Let f ∈ PBVp
3. Auxiliary results
I 2 , p > 1. Then
1
1
ωi (δ, f )p 6 3 p δ p Vi (f )p (i = 1, 2), 0 < δ < 1,
where Vi (f )p is a partial p-variation of function.
Using the method of [8], we can easily obtain the validity of Lemma 1.
4. Proof of main results
Proof of Theorem 1. We write
∞ X
∞
X
n=0 m=0
β
|Cn,m (f )| =
∞
X
n=0
β
|Cn,0 (f )| +
∞
X
m=1
β
|C0,m (f )| +
∞ X
∞
X
n=1 m=1
β
|Cn,m (f )| .
36
ALEXANDER APLAKOV
Let n = 2n1 + i, m = 2m1 + j, n1 = 0, 1, . . . i = 1, . . . , 2n1 , m1 = 0, 1, . . . , j =
1, . . . , 2m1 . Then using Hölder inequality, from the Lemma 1 we get
n
2 1
X
|C2n1 +i,0 (f )|p
i=1
=2
62
pn1
2
pn1
2
˛ 0 2i−1
1 ˛p
˛
˛ 1
n1 +1
2
˛
˛
»
„
«–
Z
Z
2
X
C ˛
˛ B
1
C dy ˛
˛ B
f
(x,
y)
−
f
x
+
,
y
dx
˛ @
A ˛
2n1 +1
˛
i=1 ˛ 0
2i−2
˛
˛
2n1 +1
2 0 2i−1
1 3p
2n
˛
Z1 +1 ˛
Z1
2n1
X
˛ C 7
6 B
˛
6 B
˛∆ 1 f (x, y) ˛ dxC dy 7
1˛
˛
4 @
A 5
n1
2n1 +1
i=1
(2)
62
pn1
2
n1
0
20
2
6B
X
6B
6@
4
i=1
62
pn1
2
2i−2
2n1 +1
Z1
1
0
0
B
B
@
2i−1
2n1 +1
Z
2i−2
2n1 +1
1 1 p1
˛p
˛
˛
C C
˛
˛∆ 1 f (x, y) ˛ dxC dy C
1˛
˛ 2n1 +1
A A
0
B
B
@
2i−1
Z1 2nZ1 +1
0
2i−2
2n1 +1
˛p
Z1 Z1 ˛
˛
˛
˛∆ 1 f (x, y) ˛ dxdy
1
˛ 2n1 +1
˛
11− p1 3p
7
C
7
1dxdy C
7
A
5
2n1 (p−1)
0 0
«
„
n1 p
p
1
1
p
n1 (1− p
2 )ω
,
f
6 2n1 (1− 2 ) 3 n V1p (f )p 6 c2− 2 V1p (f )p .
62
1
1
2n1 +1
2
p
Let
2p
1+p
< β < p. Using Hölder inequality, from (2) we get
2 n1 β
X
(i)
Cn1 ,0 (f ) 6
i=1
(3)
2 n1 p
X
(i)
Cn1 ,0 (f )
i=1
! βp
2n1 (1− p )
β
βp
n1 p
β
6 2n1 (1− p ) c2− 2 V1p (f )p
6 c2n1 (1− p ) 2−
β
n1β
2
6 c2n1 [1− p − 2 ] .
β
β
By (3) and from the condition of the Theorem 1 we obtain
∞
X
n=2
n
β
|Cn,0 (f )| =
∞ X
2 1
X
|C2n1 +i,0 (f )|β 6
∞
X
2n1 [1− p − 2 ] < ∞.
n1 =0
n1 =0 i=1
Analogously, we obtain that
∞
X
m=1
β
|C0,m (f )| < ∞, for β >
2p
.
1+p
β
β
ABSOLUTE CONVERGENCE OF SERIES. . .
37
Using Hölder inequality, by (1) and from Lemma 1 we get
i+1 j+1
p
Z2n1 Z
2m1
f (x, y) χ2n1 +i (x) χ2m1 +j (y) dxdy j=0 2i
2j
n
m
1
2
2 1
 2i+1 2j+1
p
n1 +1 2m1 +1
n1
m1
2
Z
Z
2X
−1 2 X
−1


n1 +m1

1
1
6 2p 2
f
(x,
y)
∆
dxdy 
,


n +1
m +1
1 −1 2m1 −1
2n
X X
i=0
i=0
n1
6 2p
n1 +m1
2


2X
−1 2 X
−1 

×

Z
Z
62
Z
Z
∆
2i
2j
2n1 +1 2m1 +1
Z1 Z1 ∆
1
n1 +m1
2
2j+1
2i+1
2n1 +1 2m1 +1
1− p1 p



1dxdy 



2i
2j
2n1 +1 2m1 +1
p



j=0
2j+1
2i+1
2n1 +1 2m1 +1
1
2
2i
2j
2n1 +1 2m1 +1
m1
i=0
(4)
2 1
j=0
1
2n1 +1
,
1
2n1 +1
1
2m1 +1
,
 p1
p

f (x, y) dxdy 

1
2m1 +1
p
f (x, y) dxdy
2(n1 +m1 )(p−1)
0 0
p
1
1
p
6 2(n1 +m1 )(1− 2 ) ω1,2
,
,
f
2n1 +1 2m1 +1
p
p
p
p
1
1
(n1 +m1 )(1− 2 ) p 2
2
, f ω2
,f
2 ω1
62
2n1 +1
2m1 +1
p
p
6 c2(n1 +m1 )(1− 2 )
2p
p
Let
(5)
2p
1+p
2
n1 +m1
2
6 c2
(n1 +m1 )
( 21 − p2 )
.
< β < p. Using Hölder inequality, by (4) we write
1 −1 2m1 −1
2n
X
X
i=0
|C2n1 +i,2m1 +j (f )|β 6 c2(n1 +m1 )(1− 2 ) p 2(n1 +m1 )(1− p )
p
β
β
j=0
= c2(n1 +m1 )[ 2p − 2 +1− p ]
β
β
β
= c2n1 [1− 2 − 2p ] 2m1 [1− 2 − 2p ] .
β
β
β
β
By (5) and from the condition of the Theorem 1 we get
∞ X
∞
X
n=1 m=1
n
β
|Cn,m (f )| =
m
1 −1 2 1 −1
∞
∞ 2X
X
X
X
n1 =0 m1 =0 i=0
∞
β
X
2
6c
n1 =0
The proof of Theorem 1 is complete.
[
n1 1−
2
−
β
2p
|C2n1 +i,2m1 +j (f )|
j=0
∞
X
]
β
2m1 [1− 2 − 2p ] < ∞.
β
β
m1 =0
38
ALEXANDER APLAKOV
Proof of Theorem 2. We write
∞ X
∞
X
[(n + 1) (m + 1)]α |Cn,m (f )| =
n=0 m=0
+
∞
X
∞
X
(n + 1)α |Cn,0 (f )|
n=0
∞ X
∞
X
α
(m + 1) |C0,m (f )| +
m=1
n=1 m=1
Let β = 1. Then from (3) we get
n
n
(6)
2 1
X
α
[(n + 1) (m + 1)] |Cn,m (f )| .
(2
n1
α
+ i + 1) |C2n1 +i,0 (f )| 6 c2
n1 α
2 1
X
|C2n1 +i,0 (f )|
i=1
i=1
6 c2n1 α 2n1 ( 2 − p ) = c2n1 (α+ 2 − p ) .
1
1
1
1
By (6) and from the condition of the Theorem 2 we obtain
∞
X
n
α
(n + 1) |Cn,0 (f )| =
∞ X
2 1
X
α
(2n1 + i + 1) |C2n1 +i,0 (f )|
n1 =0 i=1
n=0
6c
∞
X
2
n1 =0
2 1
X
|C
2n1 +i,0
(f )| 6 c
∞
X
2n1 (α+ 2 − p ) < ∞.
i=1
n1 =0
(m + 1)α |C0,m (f )| < ∞, for α <
1
1
− .
2p 2
Analogously, we obtain that
∞
X
n
n1 α
m=1
1
1
Let β = 1. Then by (5) and from the condition of the Theorem 2 we get
∞ X
∞
X
α
[(n + 1) (m + 1)] |Cn,m (f )|
n=1 m=1
6
∞ X
∞
X
n
2(n1 +m1 )α
n1 =0 m1 =0
∞
X
1
n1 (α+ 12 − 2p
)
2
6c
n1 =0
The proof of Theorem 2 is complete.
m
2 1 2
X
X1
|C2n1 +i,2m1 +j
i=1 j=1
∞
X
1
m1 (α+ 12 − 2p
)
2
(f )|
< ∞.
m1 =0
Combining the methods of Theorems 1-2 we can prove validity of Theorem 3.
Observe that the result of this paper can be proved in the same way for dimension
more than 2.
References
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ABSOLUTE CONVERGENCE OF SERIES. . .
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Received May 7, 2005.
Department of Agro-Business Engineers,
Georgian State University of Subtropical Agriculture,
I. Chavchavadze str. 21, Kutaisi,
4616 Georgia
E-mail address: a aplakov@wanex.ge