# Approximation byT-Transformation of Double Walsh-Fourier Series to Multivariable Functions

Authors: Yi Zhao; Dansheng Yu
Journal: ISRN Mathematical Analysis
 (2014)

---

## Abstract

We study the Walsh series expansion of multivariate functions inL
p
(
1
≤
p
≤
∞
) and, in particular, in L
i
p
(
α
,
p
). The rate of uniform approximation by T-transformation of rectangular partial sums of double Walsh to these functions is investigated. By extending the concepts of rest (head) bounded variation series, which was introduced by Leindler (2004), we generalize the related results of Móricz and Rhoades (1996), Nagy (2012). Our results can be applied to many summability methods, including the Nörlund summability and weighted summability.


---

## Body

## 1. Introduction

Letr
0
(
x
) be the function defined on I
≔
I
0
=
[
0,1
) by
(1)
r
0
(
x
)
=
{
1
,
x
∈
[
0
,
1
2
)
,
-
1
,
x
∈
[
1
2
,
1
)
,
r
0
(
x
+
1
)
=
r
0
(
x
)
.The Rademacher system is defined by(2)
r
n
(
x
)
=
r
0
(
2
n
x
)
.Let(3)
w
0
(
x
)
=
1
,
w
n
(
x
)
=
∏
k
=
0
∞
‍
(
r
k
(
x
)
)
n
k
be the Walsh functions, where x
∈
[
0,1
),n
∈
ℕ with the dyadic expansion x
=
∑
i
=
0
∞
‍
x
i
2
-
i
-
1 and n
=
∑
j
=
0
∞
‍
n
j
2
j, respectively; here x
i
,
n
j
∈
{
0,1
} for i
,
j
=
0,1
,
…. We also write ⌈
n
⌉
≔
{
max
⁡
j
,
n
j
≠
0
}. The idea of using products of Rademacher’s functions to construct the Walsh system originated from Paley [1].As an important orthonormal bases, Walsh functions have most of the properties of Fourier series but are more suited to nonlinear studies. If a zero-memory nonlinear transformation is applied to a Walsh series, the output series can be derived by simple algebraic processes. Corrington [2] proved that nonlinear differential and integral equations can be solved by Walsh series. Meanwhile, the Walsh functions are of great practical interest. They have many applications in signal processing [3], dynamic systems, identification, control [4, 5], and so on.In the above-mentioned issues, Walsh series expansion of certain function and its convergence to that function play very important roles. In this paper, we are interested in the Walsh expansion of multivariate functions and discuss the convergence of itsT-transformation to these functions (we mention here that, in order to avoid notational difficulties, we restrict ourselves to the case of bivariate functions). Furthermore, the results are applied to some summability methods.The Walsh-Dirichlet kernels and Walsh-Fejér kernels are defined by(4)
D
n
(
x
)
=
∑
k
=
0
n
-
1
‍
w
k
(
x
)
,
D
0
(
x
)
≔
0
,
(5)
K
n
(
x
)
=
1
n
∑
k
=
1
n
D
k
(
x
)
,
‍
(
n
∈
ℕ
)
,
K
0
(
x
)
=
0
.It is known that [6]
(6)
D
2
n
(
x
)
=
{
2
n
,
x
∈
I
n
0
,
x
∈
I
∖
I
n
,
where I
n denotes the dyadic interval in [
0,1
) defined by I
n
=
[
(
i
/
2
n
)
,
(
(
i
+
1
)
/
2
n
)
], and i
=
0,1
,
…
,
2
n
-
1, n
∈
ℕ.In addition, we point out that the standard representations for the Walsh-Dirichlet kernel [7, 8] are
(7)
D
2
n
+
j
(
x
)
=
D
2
n
(
x
)
+
r
n
(
x
)
D
j
(
x
)
,
(8)
D
2
j
+
i
(
x
)
-
D
2
j
+
1
(
x
)
=
-
w
2
j
+
1
-
1
(
x
)
D
2
j
-
i
(
x
)
.For the Walsh-Fejér kernelK
i, let i
≥
1; Yano [9] proved that
(9)
∥
K
i
∥
1
≤
2
.LetT
≔
{
t
m
n
j
k
} be a doubly infinite matrix. It is said to be doubly triangular if t
m
n
j
k
=
0 for j
>
m or k
>
n. In the recent research [10], the authors established necessary conditions for a general inclusion theorem involving a pair of doubly triangular matrices.Given a double sequence{
s
j
k
:
j
,
k
=
0,1
,
…
} of complex numbers, the m
nth term of the T-transformation of {
s
j
k
} is defined by
(10)
t
m
n
≔
∑
μ
=
0
m
‍
∑
ν
=
0
n
t
m
n
μ
ν
s
μ
ν
‍
.If∑
μ
=
0
m
‍
∑
ν
=
0
n
‍
t
m
n
μ
ν
=
1, then we say that T is normal.LetP
=
{
p
j
k
:
j
,
k
=
0,1
,
…
} be a double sequence of nonnegative numbers; p
00
>
0. Taking
(11)
t
m
n
j
k
=
p
m
-
j
,
n
-
k
P
m
n
,
0
≤
j
≤
m
,
0
≤
k
≤
n
,
t
m
n
j
k
=
0
,
if
j
>
m
or
k
>
n
,
where P
m
n
≔
∑
j
=
0
m
‍
∑
k
=
0
n
‍
p
j
k. Then the corresponding t
m
n is known as the Nörlund means. The Cesàro summability of orders γ
,
δ
>
-
1, denoted by (
C
,
γ
,
δ
), is a special case of the Nörlund summability with
(12)
p
j
k
=
A
j
γ
-
1
A
k
δ
-
1
,
j
,
k
=
0,1
,
2
,
…
,
A
l
γ
=
(
γ
+
l
l
)
=
(
γ
+
1
)
(
γ
+
2
)
⋯
(
γ
+
l
)
l
!
for l
=
1,2
,
…, and A
0
γ
=
1. In this case,
(13)
P
m
n
=
A
m
γ
A
n
δ
,
m
,
n
=
0,1
,
2
,
…
.If we take(14)
t
m
n
j
k
=
p
j
k
P
m
n
,
0
≤
j
≤
m
,
0
≤
k
≤
n
,
t
m
n
j
k
=
0
,
if
j
>
m
or
k
>
n
,
the corresponding t
m
n is the well-known Riesz means of {
s
j
k
}.LetL
p (1
≤
p
≤
∞) denote the Lebesgue function spaces on the torus I
2; that is, f
∈
L
p
(
I
2
). The double Walsh (Walsh-Fourier) series of such function is defined by
(15)
f
(
x
,
y
)
~
S
(
x
,
y
)
=
∑
i
=
0
∞
∑
j
=
0
∞
f
^
(
i
,
j
)
w
i
(
x
)
w
j
(
y
)
‍
‍
,
and the u
vth rectangular partial sum of S is
(16)
S
u
v
(
x
,
y
)
=
∑
i
=
0
u
-
1
∑
j
=
0
v
-
1
f
^
(
i
,
j
)
w
i
(
x
)
w
j
(
y
)
‍
,
‍
where
(17)
f
^
(
i
,
j
)
=
∬
0
1
f
(
s
,
t
)
w
i
(
s
)
w
j
(
t
)
d
s
d
t
.Them
nthT-transformation of S
u
v is defined by
(18)
T
m
n
≔
T
m
n
(
f
,
x
,
y
)
=
∑
u
=
0
m
∑
v
=
0
n
t
m
n
u
v
S
u
v
(
x
,
y
)
‍
‍
.By (16), we have
(19)
T
m
n
(
f
,
x
,
y
)
=
∬
0
1
f
(
x
+
s
,
y
+
t
)
K
m
n
(
s
,
t
)
d
s
d
t
,
where
(20)
K
m
n
(
s
,
t
)
≔
∑
u
=
0
m
∑
v
=
0
n
t
m
n
u
v
D
u
(
s
)
D
v
(
t
)
‍
‍
,
m
,
n
=
0,1
,
…
,
D
u
(
s
) and D
v
(
t
) are the Walsh-Dirichlet kernels, in terms of u and v respectively.For any functionf
∈
L
p
(
I
2
), when T is normal,
(21)
T
m
n
(
x
,
y
)
-
f
(
x
,
y
)
=
∬
0
1
∑
u
=
0
m
∑
v
=
0
n
t
m
n
u
v
D
u
(
s
)
D
v
(
t
)
‍
‍
h
h
h
h
h
h
h
h
h
i
h
h
h
h
l
×
(
f
(
x
+
s
,
y
+
t
)
-
f
(
x
,
y
)
)
d
s
d
t
=
∬
0
1
K
m
n
(
s
,
t
)
(
f
(
x
+
s
,
y
+
t
)
-
f
(
x
,
y
)
)
d
s
d
t
.Recall that the modulus of continuity of the functionf
(
x
,
y
)
∈
L
p
(
I
2
)
,
1-periodic in each variable, is defined by
(22)
ω
p
(
f
,
δ
)
≔
sup
⁡
|
t
|
≤
δ
{
∥
f
(
x
+
u
,
y
+
v
)
-
f
(
x
,
y
)
∥
p
,
u
2
+
v
2
h
h
h
l
h
|
t
|
=
u
2
+
v
2
,
δ
≥
0
}
.For eachα
>
0, the Lipschitz classes in L
p are defined by
(23)
L
i
p
(
α
,
p
)
≔
{
f
∈
L
p
,
ω
p
(
f
,
δ
)
=
𝒪
(
δ
α
)
}
.The (total) modulus of continuity of functionf
(
x
,
y
)
∈
L
p
(
I
2
), 1-periodic in each variable, is defined by
(24)
ω
1,2
p
(
f
,
δ
1
,
δ
2
)
≔
sup
⁡
{
∥
f
(
x
+
u
,
y
+
v
)
-
f
(
x
+
u
,
y
)
h
h
h
h
h
h
h
l
-
f
(
x
,
y
+
v
)
+
f
(
x
,
y
)
∥
p
,
h
h
h
h
l
h
h
|
u
|
≤
δ
1
,
|
v
|
≤
δ
2
}
.It is easy to verify that there is a constant𝒞
>
0 such that
(25)
ω
1,2
p
(
f
,
δ
1
,
δ
2
)
≤
𝒞
(
ω
p
(
f
,
δ
1
)
+
ω
p
(
f
,
δ
2
)
)
.Móricz and Siddiqi [11] studied the rate of uniform approximation by Nörlund means of Walsh (Walsh-Fourier) series of f
∈
L
p
[
0,1
). Later, Móricz and Rhoades [12] studied the corresponding approximation problem by weighted means of Walsh-Fourier series. Their main results in [12] can be read as follows.Theorem A.
Let
f
∈
L
p
,
1
≤
p
≤
∞
,
n
=
2
m
+
k
,
1
≤
k
≤
2
m
,
m
≥
1.
(i) If{
p
k
} is nondecreasing and satisfies the condition
(26)
n
p
n
P
n
=
𝒪
(
1
)
,
then
(27)
∥
t
n
(
f
)
-
f
∥
p
≤
3
P
n
∑
j
=
0
m
-
1
2
j
p
2
j
+
1
-
1
ω
p
(
f
,
2
-
j
)
+
𝒪
(
ω
p
(
f
,
2
-
m
)
)
‍
.
(ii) If{
p
k
} is nonincreasing, then
(28)
∥
t
n
(
f
)
-
f
∥
p
≤
3
P
n
∑
j
=
0
m
-
1
2
j
p
2
j
ω
p
(
f
,
2
-
j
)
+
𝒪
(
ω
p
(
f
,
2
-
m
)
)
‍
.Theorem B.
Letf
∈
L
i
p
(
α
,
p
) for α
>
0 and 1
≤
p
≤
∞. If  {
p
k
} is nondecreasing, then one has the following estimates:
(29)
∥
t
n
(
f
)
-
f
∥
L
p
=
{
𝒪
(
n
-
α
)
,
0
<
α
<
1
,
𝒪
(
n
-
1
log
⁡
n
)
,
α
=
1
,
𝒪
(
n
-
1
)
,
α
>
1
.For any fourfold sequence{
a
m
n
j
k
}, write
(30)
Δ
11
a
m
n
j
k
≔
a
m
n
j
k
-
a
m
,
n
,
j
+
1
,
k
-
a
m
,
n
,
j
,
k
+
1
+
a
m
,
n
,
j
+
1
,
k
+
1
,
Δ
01
a
m
n
j
k
≔
a
m
n
j
k
-
a
m
,
n
,
j
,
k
+
1
,
Δ
10
a
m
n
j
k
≔
a
m
n
j
k
-
a
m
,
n
,
j
+
1
,
k
.The sequence{
a
m
n
j
k
} is called nondecreasing if it is nondecreasing in both j and k; that is, Δ
01
a
m
n
j
k
≤
0 and Δ
10
a
m
n
j
k
≤
0 for every j
,
k
=
0,1
,
…. The nonincreasing case is defined analogously.Recently, Nagy [13] did some research on the approximation by Nörlund means of double Walsh-Fourier series for Lipschitz functions and generalized Theorems A and B to the functions of two variables. We present one of the main results in [13] here.Theorem C.
Letf
∈
L
i
p
(
α
,
p
) for some α
>
0 and 1
≤
p
≤
∞; let {
q
j
k
} be a double sequence of nonnegative numbers such that it is nondecreasing; Δ
11
q
j
k is of fixed sign and satisfies the regularity condition:
(31)
(
m
+
1
)
(
n
+
1
)
q
m
n
Q
m
n
=
𝒪
(
1
)
.
Q
m
n
≔
∑
i
=
0
m
∑
j
=
0
m
q
i
j
‍
‍
(
m
,
n
=
0,1
,
…
)
,
then,
(32)
∥
W
m
n
(
f
)
-
f
∥
L
p
≤
{
𝒪
(
m
-
α
+
n
-
α
)
,
0
<
α
<
1
,
𝒪
(
m
-
1
log
⁡
m
+
n
-
1
log
⁡
n
)
,
α
=
1
,
𝒪
(
m
-
1
+
n
-
1
)
,
α
>
1
,
where
(33)
W
m
n
(
f
,
x
,
y
)
≔
1
Q
m
n
∑
i
=
0
m
∑
j
=
0
m
q
m
n
,
m
-
i
,
n
-
j
S
i
j
(
x
,
y
)
‍
‍
,
and  S
i
j
(
x
,
y
) is defined as in (16).We know that in the theory of Fourier series it is of main interest how to approximate the function from the partial sums of its Fourier series. The purpose of the present paper is to get the rate of uniform approximation byT-transformation with doubly triangular. We give the outline of the paper. In Section 2, we state the main results. Some auxiliary lemmas are given in Section 3, and the proofs of the main theorems are presented in Section 4. Our new results can be applied to many classical summability methods such as Nörlund summability and Riesz summability. As an important application, we will apply them to the Nörlund summability and weighted means in Section 5. We will see that not only Theorems A, B, and C are corollaries of our results but also some other new types of estimates are presented in this paper.


## 2. The Main Results

For a fixedn, α
n
≔
{
a
n
k
} of nonnegative numbers tending to zero is calledrest bounded variation, or briefly α
n
∈
RBVS, if there is a constant K
(
α
n
), only depending on α
n, such that
(34)
∑
k
=
m
∞
|
a
n
k
-
a
n
,
k
+
1
|
‍
≤
K
(
α
n
)
a
n
m
holds for all natural numbers m.For a fixedn
,
α
n
≔
{
a
n
k
} of nonnegative numbers tending to zero is calledhead bounded variation, or briefly α
n
∈
HBVS, if there is a constant K
(
α
n
) only depending on α
n such that
(35)
∑
k
=
0
m
|
a
n
k
-
a
n
,
k
+
1
|
‍
≤
K
(
α
n
)
a
n
m
for all natural numbers m, or only for all m
≤
N if the sequence α
n has only finite nonzero terms, and the last nonzero term is a
n
N.Remark A. The definitions of RBVS and HBVS are introduced by Leindler [14] to generalize the monotonicity conditions on sequences. In fact, RBVS and HBVS generalized monotone nonincreasing sequences and monotone nondecreasing sequences, respectively.Remark B. Since it involves a sequence K
(
α
n
), there should be a constant K such that 0
<
K
(
α
n
)
≤
K.Now, we extend the concepts of RBVS and HBVS to the double sequences as follows.Definition 1.
A double sequenceα
m
n
=
{
a
m
n
j
k
} is called DRBVS, if there is a constant K
(
α
m
n
) such that for j
,
k
=
0,1
,
2
,
…,
(36)
∑
μ
=
j
∞
|
Δ
10
a
m
n
μ
k
|
‍
∑
ν
=
k
∞
|
Δ
01
a
m
n
j
ν
|
‍
∑
μ
=
j
∞
∑
ν
=
k
∞
|
Δ
11
a
m
n
j
ν
|
‍
‍
}
≤
K
(
α
m
n
)
a
m
n
j
k
.Definition 2.
A double sequenceα
m
n
=
{
a
m
n
j
k
} is called DHBVS, if there is a constant K
(
α
m
n
) such that for j
,
k
=
0,1
,
2
,
…,
(37)
∑
μ
=
0
j
|
Δ
10
a
m
n
μ
k
|
‍
∑
ν
=
0
k
|
Δ
01
a
m
n
j
ν
|
‍
∑
μ
=
0
j
∑
ν
=
0
k
|
Δ
11
a
m
n
j
ν
|
‍
‍
}
≤
K
(
α
m
n
)
a
m
n
j
k
.It is also required that the sequenceK
(
α
m
n
) is bounded; that is, there is a positive constant K such that 0
<
K
(
α
m
n
)
≤
K.We state our main theorems as follows.Theorem 3.
Letf
∈
L
p, {
t
m
n
u
v
} be a a doubly normal triangular matrix of nonnegative numbers, {
t
m
n
u
v
}
∈ DRBVS, and let E
≔
⌈
m
⌉
,
F
≔
⌈
n
⌉. Then, one has
(38)
∥
T
m
n
-
f
∥
p
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
2
k
2
l
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
‍
‍
.
Also, iff
∈
L
p, {
t
m
n
u
v
} is doubly normal triangular of nonnegative numbers; {
t
m
n
u
v
}
∈ DHBVS; then one has the conclusion as follows:
(39)
∥
T
m
n
-
f
∥
p
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
(
2
k
+
1
-
1
)
(
2
l
+
1
-
1
)
‍
‍
h
h
h
h
h
h
h
i
h
h
h
h
h
×
(
ω
p
(
f
,
2
-
k
)
+
𝒞
ω
p
(
f
,
2
-
l
)
)
+
𝒞
∑
k
=
0
E
-
1
2
k
(
n
-
2
F
)
t
m
n
(
2
k
+
1
-
1
)
n
‍
h
h
h
h
h
h
h
l
h
h
h
×
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
F
)
)
+
𝒞
∑
l
=
0
F
-
1
2
l
(
m
-
2
E
)
t
m
n
m
(
2
l
+
1
-
1
)
‍
h
h
h
h
h
h
h
l
h
h
h
×
(
ω
p
(
f
,
2
-
E
)
+
ω
p
(
f
,
2
-
l
)
)
+
𝒞
(
m
-
2
E
)
(
n
-
2
F
)
t
m
n
m
n
+
×
(
ω
p
(
f
,
2
-
E
)
+
ω
p
(
f
,
2
-
F
)
)
.Theorem 4.
Suppose thatf
∈
L
i
p
(
p
,
α
) for α
>
0
,
1
≤
p
≤
+
∞. Let {
t
m
n
u
v
} be nonnegative, doubly normal triangular, and {
t
m
n
u
v
}
∈ DHBVS, satisfying  m
n
t
m
n
m
n
=
𝒪
(
1
). Then, one has
(40)
∥
T
m
n
-
f
∥
p
=
{
𝒪
(
m
-
α
+
n
-
α
)
,
0
<
α
<
1
,
𝒪
(
m
-
1
log
⁡
m
+
n
-
1
log
⁡
n
)
,
α
=
1
,
𝒪
(
m
-
1
+
n
-
1
)
,
α
>
1
.
While for f
∈
L
i
p
(
p
,
α
) and for α
>
0
,
1
≤
p
≤
+
∞, let {
t
m
n
u
v
} be nonnegative, normal doubly triangular, and {
t
m
n
u
v
}
∈ DRBVS; one has
(41)
∥
T
m
n
-
f
∥
p
=
𝒪
(
1
)
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
(
2
k
(
1
-
α
)
+
l
+
2
k
+
l
(
1
-
α
)
)
t
m
n
2
k
2
l
‍
‍
.


## 3. Lemmas and Proofs

Lemma 1.
Suppose that{
t
m
n
u
v
} is doubly normal triangular and D
u is defined as (4); write E
≔
⌈
m
⌉
,
F
≔
⌈
n
⌉; then one has
(42)
A
1
≔
∑
u
=
0
2
E
-
1
∑
v
=
0
2
F
-
1
t
m
n
u
v
D
u
(
s
)
D
v
(
t
)
‍
‍
=
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
w
2
k
+
1
-
1
(
s
)
w
2
l
+
1
-
1
(
t
)
‍
‍
h
h
h
h
h
h
h
×
(
2
k
+
l
t
m
n
2
k
2
l
K
2
k
(
s
)
K
2
l
(
t
)
∑
j
=
1
2
l
-
1
‍
hhhhhhhhhhh
+
∑
i
=
1
2
k
-
1
i
2
l
Δ
10
t
m
n
(
2
k
+
1
-
i
)
2
l
K
i
(
s
)
K
2
l
(
t
)
‍
hhhhhhhhhhh
+
∑
j
=
1
2
l
-
1
j
2
k
Δ
01
t
m
n
2
k
(
2
l
+
1
-
j
)
K
j
(
t
)
K
2
k
(
s
)
‍
hhhhhhhhhhh
+
∑
i
=
1
2
k
-
1
∑
j
=
1
2
l
-
1
Δ
11
t
m
n
(
2
k
+
1
-
i
)
(
2
l
+
1
-
j
)
‍
‍
hhhhhhhhhhhhhhhhhhh
∑
i
=
1
2
k
-
1
∑
j
=
1
2
l
-
1
‍
‍
×
i
j
K
i
(
s
)
K
j
(
t
)
)
-
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
w
2
k
+
1
-
1
(
s
)
D
2
l
+
1
(
t
)
‍
‍
hhhhhhhhh
×
(
2
k
K
2
k
(
s
)
∑
j
=
0
2
l
-
1
t
m
n
2
k
(
2
l
+
j
)
‍
hhhhhhhhhhhhhh
+
∑
i
=
1
2
k
-
1
∑
j
=
0
2
l
-
1
Δ
10
t
m
n
(
2
k
+
1
-
i
)
(
2
l
+
j
)
i
K
i
(
s
)
‍
‍
)
-
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
D
2
k
+
1
(
s
)
w
2
l
+
1
-
1
(
t
)
‍
‍
hhhhhhhhh
×
(
2
l
K
2
l
(
t
)
∑
i
=
0
2
k
-
1
t
m
n
(
2
k
+
i
)
2
l
‍
hhhhhhhhhhhhh
+
∑
i
=
0
2
k
-
1
∑
j
=
1
2
l
-
1
Δ
01
t
m
n
(
2
k
+
i
)
(
2
l
+
1
-
j
)
j
K
j
(
s
)
‍
‍
)
+
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
2
l
-
1
t
m
n
(
2
k
+
i
)
(
2
l
+
j
)
D
2
k
+
1
(
s
)
D
2
l
+
1
(
t
)
‍
‍
‍
‍
.Proof.
We can rewriteA
1 as
(43)
A
1
=
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
2
l
-
1
t
m
n
(
2
k
+
i
)
(
2
l
+
j
)
D
2
k
+
i
(
s
)
D
2
l
+
j
(
t
)
‍
‍
‍
‍
,
and keep dividing the above into 4 parts:
(44)
A
1
=
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
2
l
-
1
t
m
n
(
2
k
+
i
)
(
2
l
+
j
)
(
D
2
k
+
i
(
s
)
-
D
2
k
+
1
(
s
)
)
‍
‍
‍
‍
h
h
h
h
h
h
h
h
h
h
h
i
h
h
h
×
(
D
2
l
+
j
(
t
)
-
D
2
l
+
1
(
t
)
)
+
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
2
l
-
1
t
m
n
(
2
k
+
i
)
(
2
l
+
j
)
‍
‍
‍
‍
h
h
h
h
h
h
h
h
h
i
h
h
h
h
h
×
(
D
2
k
+
i
(
s
)
-
D
2
k
+
1
(
s
)
)
D
2
l
+
1
(
t
)
+
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
2
l
-
1
t
m
n
(
2
k
+
i
)
(
2
l
+
j
)
‍
‍
‍
‍
h
h
h
h
h
h
h
h
h
h
h
i
h
h
h
h
×
(
D
2
l
+
j
(
t
)
-
D
2
l
+
1
(
t
)
)
D
2
k
+
1
(
s
)
+
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
2
l
-
1
t
m
n
(
2
k
+
i
)
(
2
l
+
j
)
D
2
k
+
1
(
s
)
D
2
l
+
1
(
t
)
‍
‍
‍
‍
≔
A
11
+
A
12
+
A
13
+
A
14
.
By using (8), we have
(45)
A
11
=
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
2
l
-
1
t
m
n
(
2
k
+
i
)
(
2
l
+
j
)
w
2
k
+
1
-
1
(
s
)
‍
‍
‍
‍
h
h
h
h
h
h
h
h
h
h
h
h
×
D
2
k
-
i
(
s
)
w
2
l
+
1
-
1
(
t
)
D
2
l
-
j
(
t
)
=
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
i
=
1
2
k
∑
j
=
1
2
l
t
m
n
(
2
k
+
1
-
i
)
(
2
l
+
1
-
j
)
w
2
k
+
1
-
1
(
s
)
‍
‍
‍
‍
h
h
h
h
h
h
h
h
h
h
h
×
D
i
(
s
)
w
2
l
+
1
-
1
(
t
)
D
j
(
t
)
.
Applying double Abel’s transformation,(46)
A
11
=
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
w
2
k
+
1
-
1
(
s
)
w
2
l
+
1
-
1
(
t
)
‍
‍
h
h
h
h
h
×
(
∑
i
1
=
1
2
k
∑
j
1
=
1
2
l
D
i
1
(
s
)
D
j
1
(
t
)
t
m
n
2
k
2
l
‍
‍
h
h
h
h
h
h
h
h
h
+
∑
j
1
=
1
2
l
D
j
1
(
t
)
∑
i
=
1
2
k
-
1
Δ
10
t
m
n
(
2
k
+
1
-
i
)
2
l
‍
‍
h
h
h
h
h
h
h
h
h
×
∑
i
1
=
1
i
D
i
1
(
s
)
h
h
h
h
h
h
h
h
h
+
∑
i
1
=
1
2
k
D
i
1
(
s
)
∑
j
=
1
2
l
-
1
Δ
01
t
m
n
2
k
(
2
l
+
1
-
j
)
‍
‍
h
h
h
h
h
h
h
h
h
×
∑
j
1
=
1
j
D
j
1
(
t
)
‍
h
h
h
h
h
h
h
h
h
+
∑
i
=
1
2
k
-
1
∑
j
=
1
2
l
-
1
Δ
11
t
m
n
(
2
k
+
1
-
i
)
(
2
l
+
1
-
j
)
‍
‍
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
×
∑
i
1
=
1
i
D
i
1
(
s
)
∑
j
1
=
1
j
D
j
1
(
t
)
‍
‍
)
.
Next, (5) leads to
(47)
A
11
=
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
w
2
k
+
1
-
1
(
s
)
w
2
l
+
1
-
1
(
t
)
‍
‍
h
h
h
h
h
×
(
2
k
+
l
t
m
n
2
k
2
l
K
2
k
(
s
)
K
2
l
(
t
)
∑
j
1
=
1
j
D
j
1
(
t
)
‍
h
h
h
h
h
h
h
h
h
+
∑
i
=
1
2
k
-
1
i
2
l
Δ
10
t
m
n
(
2
k
+
1
-
i
)
2
l
K
i
(
s
)
K
2
l
(
t
)
‍
h
h
h
h
h
h
h
h
h
+
∑
j
=
1
2
l
-
1
j
2
k
Δ
01
t
m
n
2
k
(
2
l
+
1
-
j
)
K
j
(
t
)
K
2
k
(
s
)
‍
h
h
h
h
h
h
h
h
h
+
∑
i
=
1
2
k
-
1
∑
j
=
1
2
l
-
1
Δ
11
t
m
n
(
2
k
+
1
-
i
)
(
2
l
+
1
-
j
)
‍
‍
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
∑
j
1
=
1
j
D
j
1
(
t
)
‍
×
i
j
K
i
(
s
)
K
j
(
t
)
)
.
Using (8), (5), and Abel’s transformation, similar to the estimate of A
11 (more easily actually),
(48)
A
12
=
-
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
w
2
k
+
1
-
1
(
s
)
D
2
l
+
1
(
t
)
2
k
K
2
k
(
s
)
‍
‍
h
h
h
h
i
h
h
h
h
×
∑
j
=
0
2
l
-
1
t
m
n
2
k
(
2
l
+
j
)
‍
-
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
w
2
k
+
1
-
1
(
s
)
D
2
l
+
1
‍
‍
h
h
h
i
h
h
h
h
×
∑
i
=
1
2
k
-
1
∑
j
=
0
2
l
-
1
Δ
10
t
m
n
(
2
k
+
1
-
i
)
(
2
l
+
j
)
i
K
i
(
s
)
‍
‍
.
Analogously,(49)
A
13
=
-
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
w
2
l
+
1
-
1
(
t
)
D
2
k
+
1
(
s
)
2
l
K
2
l
(
t
)
‍
‍
h
h
h
i
h
h
h
h
h
×
∑
i
=
0
2
k
-
1
t
m
n
(
2
k
+
i
)
2
l
‍
-
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
D
2
k
+
1
w
2
k
+
1
-
1
(
s
)
‍
‍
h
h
h
h
i
h
h
h
×
∑
i
=
0
2
k
-
1
∑
j
=
1
2
l
-
1
Δ
10
t
m
n
(
2
k
+
i
)
(
2
l
+
1
-
j
)
j
K
j
(
s
)
‍
‍
.
Combining the above estimates ofA
1
i
(
i
=
1,2
,
3,4
), we obtain the conclusion of Lemma 1.Lemma 2.
Let{
t
m
n
u
v
} be the same matrix as in Lemma 1; then one has
(50)
A
2
≔
∑
u
=
0
2
E
-
1
∑
v
=
2
F
n
t
m
n
u
v
D
u
(
s
)
D
v
(
t
)
‍
‍
=
-
∑
k
=
0
E
-
1
∑
j
=
0
n
-
2
F
w
2
k
+
1
-
1
(
s
)
D
2
F
(
t
)
‍
‍
h
h
h
h
l
h
h
h
h
×
(
2
k
K
2
k
(
s
)
t
m
n
2
k
(
2
F
+
j
)
∑
i
=
1
2
k
-
1
‍
h
h
h
h
h
h
l
h
h
h
h
h
+
∑
i
=
1
2
k
-
1
Δ
10
t
m
n
(
2
k
+
1
-
i
)
(
2
F
+
j
)
i
K
i
(
s
)
‍
)
-
∑
k
=
0
E
-
1
w
2
k
+
1
-
1
r
F
(
t
)
‍
h
h
h
h
×
(
t
m
n
2
k
n
2
k
(
n
-
2
F
)
K
2
k
(
s
)
K
n
-
2
F
(
t
)
∑
i
=
1
2
k
-
1
‍
hhhhhhhh
+
∑
i
=
0
2
k
-
1
Δ
10
t
m
n
(
2
k
+
1
-
i
)
n
i
(
n
-
2
F
)
‍
hhhhhhhhhhhhh
×
K
i
(
s
)
K
n
-
2
F
(
t
)
hhhhhhhh
+
∑
j
=
0
n
-
2
F
-
1
Δ
01
t
m
n
2
k
(
2
F
+
j
)
2
k
j
K
2
k
(
s
)
K
j
(
t
)
‍
hhhhhhhh
+
∑
i
=
1
2
k
-
1
∑
j
=
0
n
-
2
F
-
1
Δ
11
t
m
n
(
2
k
+
1
-
i
)
(
2
F
+
j
)
‍
‍
hhhhhhhhhhhhhhhhh
×
i
j
K
i
(
s
)
K
j
(
t
)
∑
i
=
1
2
k
-
1
‍
)
+
∑
k
=
0
E
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
n
-
2
F
‍
t
m
n
(
2
k
+
i
)
(
2
F
+
j
)
D
2
k
+
1
(
s
)
D
2
F
(
t
)
‍
‍
-
∑
k
=
0
E
-
1
‍
D
2
k
+
1
(
s
)
r
F
(
t
)
hhhhh
×
(
∑
i
=
0
2
k
-
1
t
m
n
(
2
k
+
i
)
n
(
n
-
2
F
)
K
n
-
2
F
(
t
)
‍
hhhhhhhhh
+
∑
i
=
0
2
k
-
1
∑
j
=
0
n
-
2
F
-
1
Δ
01
t
m
n
(
2
k
+
i
)
(
2
F
+
j
)
j
K
j
(
t
)
‍
‍
)
.
Meanwhile,
(51)
A
3
≔
∑
u
=
2
E
m
∑
v
=
0
2
F
-
1
t
m
n
u
v
D
u
(
s
)
D
v
(
t
)
‍
‍
=
-
∑
l
=
0
F
-
1
∑
i
=
0
m
-
2
E
w
2
l
+
1
-
1
(
t
)
D
2
E
(
s
)
‍
‍
h
h
h
h
h
h
h
h
×
(
2
l
K
2
l
(
t
)
t
m
n
(
2
E
+
i
)
2
l
+
∑
j
=
1
2
l
-
1
‍
h
h
h
h
h
h
h
h
h
h
h
h
+
∑
j
=
1
2
l
-
1
Δ
01
t
m
n
(
2
E
+
i
)
(
2
l
+
1
-
j
)
j
K
j
(
t
)
‍
)
-
∑
l
=
0
F
-
1
w
2
l
+
1
-
1
(
t
)
r
E
(
s
)
‍
h
h
h
l
h
×
(
t
m
n
m
2
l
2
l
(
m
-
2
E
)
K
2
l
(
t
)
K
m
-
2
E
(
s
)
+
∑
i
=
0
m
-
2
E
-
1
‍
h
h
h
h
h
h
h
h
h
+
∑
i
=
0
m
-
2
E
-
1
Δ
10
t
m
n
(
2
E
+
i
)
2
l
i
2
l
K
i
(
s
)
K
2
l
(
t
)
‍
h
h
h
h
h
h
h
h
h
+
∑
j
=
1
2
l
-
1
Δ
01
t
m
n
m
(
2
l
+
1
-
j
)
(
m
-
2
E
)
‍
h
h
h
h
h
h
h
h
h
h
h
h
h
h
×
j
K
m
-
2
E
(
s
)
K
j
(
t
)
h
h
h
h
h
h
h
h
h
+
∑
i
=
0
m
-
2
E
-
1
∑
j
=
1
2
l
-
1
Δ
11
t
m
n
(
2
E
+
i
)
(
2
l
+
1
-
j
)
‍
‍
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
∑
i
=
0
m
-
2
E
-
1
‍
×
i
j
K
i
(
s
)
K
j
(
t
)
)
-
∑
l
=
0
F
-
1
∑
j
=
0
2
l
-
1
∑
i
=
0
m
-
2
E
t
m
n
(
2
E
+
i
)
(
2
l
+
j
)
D
2
E
(
s
)
D
2
l
+
1
(
t
)
‍
‍
‍
-
∑
l
=
0
F
-
1
r
E
(
s
)
D
2
l
+
1
(
t
)
‍
h
h
h
h
×
(
∑
j
=
0
2
l
-
1
t
m
n
m
(
2
l
+
j
)
(
m
-
2
E
)
K
m
-
2
E
(
s
)
‍
h
h
h
h
h
h
h
h
+
∑
j
=
0
2
l
-
1
∑
i
=
0
m
-
2
A
-
1
Δ
10
t
m
n
(
2
E
+
i
)
(
2
l
+
j
)
i
K
i
(
s
)
‍
‍
)
.Proof.
RewriteA
2 by some different decomposition method; combining the decomposition we used in Lemma 1, it follows that
(52)
A
2
=
∑
k
=
0
E
-
1
∑
i
=
0
2
k
-
1
∑
v
=
2
F
n
t
m
n
(
2
k
+
i
)
v
D
2
k
+
i
(
s
)
D
v
(
t
)
‍
‍
‍
=
∑
k
=
0
E
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
n
-
2
F
t
m
n
(
2
k
+
i
)
(
2
F
+
j
)
D
2
k
+
i
(
s
)
D
2
F
+
j
(
t
)
‍
‍
‍
=
∑
k
=
0
E
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
n
-
2
F
t
m
n
(
2
k
+
i
)
(
2
F
+
j
)
‍
‍
‍
h
h
h
h
h
h
h
h
h
l
×
(
D
2
k
+
i
(
s
)
-
D
2
k
+
1
(
s
)
)
D
2
F
+
j
(
t
)
+
∑
k
=
0
E
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
n
-
2
F
t
m
n
(
2
k
+
i
)
(
2
F
+
j
)
D
2
k
+
1
(
s
)
D
2
F
+
j
(
t
)
‍
‍
‍
.
Furthermore, by properties (7) and (8) of D
n, we have
(53)
A
2
=
∑
k
=
0
E
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
n
-
2
F
t
m
n
(
2
k
+
i
)
(
2
F
+
j
)
(
-
w
2
k
+
1
-
1
(
s
)
D
2
k
-
i
(
s
)
)
‍
‍
‍
h
h
h
h
h
h
h
h
h
×
(
D
2
F
(
t
)
+
r
F
(
t
)
D
j
(
t
)
)
+
∑
k
=
0
E
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
n
-
2
F
t
m
n
(
2
k
+
i
)
(
2
F
+
j
)
D
2
k
+
1
(
s
)
‍
‍
‍
h
h
h
h
h
h
h
h
h
h
h
×
(
D
2
F
(
t
)
+
r
F
(
t
)
D
j
(
t
)
)
=
∑
k
=
0
E
-
1
∑
i
=
0
2
k
-
1
∑
j
=
0
n
-
2
F
t
m
n
(
2
k
+
i
)
(
2
F
+
j
)
‍
‍
‍
h
h
h
h
h
h
h
h
×
(
-
D
2
F
(
t
)
w
2
k
+
1
-
1
(
s
)
D
2
k
-
i
(
s
)
D
j
(
t
)
h
h
h
h
h
h
h
h
h
h
h
-
w
2
k
+
1
-
1
(
s
)
D
2
k
-
i
(
s
)
r
F
(
t
)
D
j
(
t
)
h
h
h
h
h
h
h
h
h
h
h
+
D
2
k
+
1
(
s
)
D
2
F
(
t
)
h
h
h
h
h
h
h
h
h
h
h
+
D
2
k
+
1
(
s
)
r
F
(
t
)
D
j
(
t
)
)
.
≔
A
21
+
A
22
+
A
23
+
A
24
.
By changingD
2
f
-
i to D
i and applying Abel’s transformation and (5),
(54)
A
21
=
-
∑
k
=
0
E
-
1
∑
i
=
1
2
k
∑
j
=
0
n
-
2
F
t
m
n
(
2
k
+
1
-
i
)
(
2
F
+
j
)
D
2
F
(
t
)
w
2
k
+
1
-
1
(
s
)
D
i
(
s
)
‍
‍
‍
=
-
∑
k
=
0
E
-
1
∑
j
=
0
n
-
2
F
w
2
k
+
1
-
1
(
s
)
D
2
F
(
t
)
‍
‍
h
h
h
h
h
h
h
h
×
(
2
k
K
2
k
(
s
)
t
m
n
2
k
(
2
F
+
j
)
∑
i
=
1
2
k
-
1
Δ
10
t
m
n
(
2
k
+
1
-
i
)
(
2
F
+
j
)
i
K
i
(
s
)
‍
h
h
h
h
h
h
h
h
h
h
h
h
h
+
∑
i
=
1
2
k
-
1
Δ
10
t
m
n
(
2
k
+
1
-
i
)
(
2
F
+
j
)
i
K
i
(
s
)
‍
)
.
Similarly,(55)
A
22
=
-
∑
k
=
0
E
-
1
w
2
k
+
1
-
1
r
F
(
t
)
‍
h
h
h
h
h
×
(
t
m
n
2
k
n
2
k
(
n
-
2
F
)
K
2
k
(
s
)
K
n
-
2
F
(
t
)
∑
i
=
0
2
k
-
1
‍
h
h
h
h
h
h
l
h
h
+
∑
i
=
0
2
k
-
1
Δ
10
t
m
n
(
2
k
+
1
-
i
)
n
i
(
n
-
2
F
)
‍
h
h
h
h
h
h
h
h
l
h
h
h
h
h
h
×
K
i
(
s
)
K
n
-
2
F
(
t
)
h
h
h
h
h
l
h
h
h
+
∑
j
=
0
n
-
2
F
-
1
Δ
01
t
m
n
2
k
(
2
F
+
j
)
2
k
j
K
2
k
(
s
)
K
j
(
t
)
‍
h
h
h
h
l
h
h
h
h
+
∑
i
=
1
2
k
-
1
∑
j
=
0
n
-
2
F
-
1
Δ
11
t
m
n
(
2
k
+
1
-
i
)
(
2
F
+
j
)
‍
‍
h
h
h
h
h
h
h
h
h
h
l
h
h
h
h
h
h
h
+
∑
i
=
1
2
k
-
1
‍
∑
j
=
0
n
-
2
F
-
1
‍
×
i
j
K
i
(
s
)
K
j
(
t
)
)
.
We also obtain the estimate for(56)
A
24
=
-
∑
k
=
0
E
-
1
D
2
k
+
1
(
s
)
r
F
(
t
)
‍
h
h
h
h
h
h
×
(
∑
i
=
0
2
k
-
1
t
m
n
(
2
k
+
i
)
n
(
n
-
2
F
)
K
n
-
2
F
(
t
)
‍
∑
i
=
0
2
k
-
1
‍
h
h
h
h
h
l
h
h
h
h
+
∑
i
=
0
2
k
-
1
∑
j
=
0
n
-
2
F
-
1
Δ
01
t
m
n
(
2
k
+
i
)
(
2
F
+
j
)
j
K
j
(
t
)
‍
‍
)
.
Combining the estimates ofA
2
i
(
i
=
1,2
,
3,4
), we have the conclusion for A
2 in Lemma 2, and the discussion for A
3 is similar (as for the decomposition of A
2, we denote A
3
=
∑
i
=
1
4
‍
A
3
i). This completes the proof of Lemma 2.Lemma 3.
Let{
t
m
n
u
v
} be the same matrix as in Lemma 1. Then, one has
(57)
A
4
≔
∑
u
=
2
E
m
∑
v
=
2
F
n
t
m
n
u
v
D
u
(
s
)
D
v
(
t
)
‍
‍
=
∑
i
=
0
m
-
2
E
∑
j
=
0
n
-
2
F
t
m
n
(
2
E
+
i
)
(
2
F
+
j
)
D
2
E
(
s
)
D
2
F
(
t
)
‍
‍
+
∑
i
=
0
m
-
2
E
D
2
E
(
s
)
r
F
(
t
)
‍
h
h
h
h
h
×
(
t
m
n
(
2
E
+
i
)
n
(
n
-
2
F
)
K
n
-
2
F
(
t
)
+
∑
j
=
0
n
-
2
F
-
1
‍
h
h
h
h
h
h
h
h
h
+
∑
j
=
0
n
-
2
F
-
1
Δ
01
t
m
n
(
2
E
+
i
)
(
2
F
+
j
)
j
K
j
(
t
)
‍
)
+
∑
j
=
0
n
-
2
F
r
E
(
s
)
D
2
F
(
t
)
‍
h
h
h
h
h
×
(
t
m
n
m
(
2
F
+
j
)
(
m
-
2
E
)
K
m
-
2
E
(
s
)
+
∑
i
=
0
m
-
2
E
-
1
‍
h
h
h
h
h
h
h
h
h
+
∑
i
=
0
m
-
2
E
-
1
Δ
10
t
m
n
(
2
E
+
i
)
(
2
F
+
j
)
i
K
i
(
s
)
)
+
r
E
(
s
)
r
F
(
t
)
×
(
t
m
n
m
n
(
m
-
2
E
)
(
n
-
2
F
)
K
m
-
2
E
(
s
)
K
n
-
2
F
(
t
)
+
∑
i
=
0
m
-
2
E
-
1
‍
hhhh
+
∑
i
=
0
m
-
2
E
-
1
Δ
10
t
m
n
(
2
E
+
i
)
n
i
(
n
-
2
F
)
K
i
(
s
)
K
n
-
2
F
(
t
)
‍
hhhh
+
∑
j
=
0
n
-
2
F
-
1
Δ
01
t
m
n
m
(
2
F
+
j
)
j
(
m
-
2
E
)
K
m
-
2
E
(
s
)
K
j
(
t
)
‍
hhhh
+
∑
i
=
0
m
-
2
E
-
1
Δ
11
t
m
n
(
2
E
+
i
)
(
2
F
+
j
)
i
j
K
i
(
s
)
K
j
(
t
)
‍
)
.Proof.
DecomposeA
4 into 4 parts:
(58)
A
4
=
∑
i
=
0
m
-
2
E
∑
j
=
0
n
-
2
F
t
m
n
(
2
E
+
i
)
(
2
F
+
j
)
D
2
E
+
i
(
s
)
D
2
F
+
j
(
t
)
‍
‍
=
∑
i
=
0
m
-
2
E
∑
j
=
0
n
-
2
F
t
m
n
(
2
E
+
i
)
(
2
F
+
j
)
(
D
2
E
(
s
)
+
r
E
(
s
)
D
i
(
s
)
)
‍
‍
h
h
h
h
h
h
h
×
(
D
2
F
(
t
)
+
r
F
(
t
)
D
j
(
t
)
)
=
∑
i
=
0
m
-
2
E
∑
j
=
0
n
-
2
F
t
m
n
(
2
E
+
i
)
(
2
F
+
j
)
D
2
E
(
s
)
D
2
F
(
t
)
‍
‍
+
∑
i
=
0
m
-
2
E
∑
j
=
0
n
-
2
F
t
m
n
(
2
E
+
i
)
(
2
F
+
j
)
D
2
E
(
s
)
r
F
(
t
)
D
j
(
t
)
‍
‍
+
∑
i
=
0
m
-
2
E
∑
j
=
0
n
-
2
F
t
m
n
(
2
E
+
i
)
(
2
F
+
j
)
r
E
(
s
)
D
i
(
s
)
D
2
F
(
t
)
‍
‍
+
∑
i
=
0
m
-
2
E
∑
j
=
0
n
-
2
F
t
m
n
(
2
E
+
i
)
(
2
F
+
j
)
r
E
(
s
)
r
F
(
t
)
D
i
(
s
)
D
j
(
t
)
‍
‍
≔
A
41
+
A
42
+
A
43
+
A
44
.
Using the techniques as in Lemmas1 and 2, we can easily get the result of Lemma 3.
By Lemmas1, 2, and 3 and (20), K
m
n
(
s
,
t
) can be written as
(59)
K
m
n
(
s
,
t
)
=
∑
u
=
0
m
∑
v
=
0
n
t
m
n
u
v
D
u
(
s
)
D
v
(
t
)
‍
‍
=
∑
i
=
1
4
A
i
‍
=
∑
i
=
1
4
∑
j
=
1
4
A
i
j
‍
‍
.
Therefore, by (21), we have
(60)
T
m
n
(
x
,
y
)
-
f
(
x
,
y
)
=
∬
0
1
∑
i
=
1
4
∑
j
=
1
4
A
i
j
(
f
(
x
+
s
,
y
+
t
)
-
f
(
x
,
y
)
)
d
s
d
t
‍
‍
‍
.
We denote by𝒫
n the set of Walsh polynomials of order less than n; that is,
(61)
𝒫
n
≔
{
P
(
x
)
,
P
(
x
)
≔
∑
i
=
0
n
-
1
‍
c
i
w
i
(
x
)
}
,
where n
≥
1 and c
i denotes the real or complex numbers. On the torus I
2, we define the two-dimensional Walsh polynomials of order less than (
m
,
n
) as
(62)
𝒫
m
,
n
≔
{
P
1
(
s
)
×
P
2
(
t
)
,
P
1
∈
𝒫
m
,
P
2
∈
𝒫
n
,
s
,
t
∈
I
}
.Lemma 4.
Consider(63)
Θ
≔
∥
∫
I
2
D
2
k
(
s
)
D
2
l
(
t
)
(
f
(
·
+
s
,
·
+
t
)
-
f
(
·
,
·
)
)
d
s
d
t
‍
∥
p
≤
𝒞
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
.Proof.
By (6) and Hölder inequality, we have
(64)
Θ
≤
∥
∫
I
k
×
I
l
D
2
k
(
s
)
D
2
l
(
t
)
(
f
(
·
+
s
,
·
+
t
)
-
f
(
·
,
·
)
)
d
s
d
t
‍
∥
p
≤
[
∫
I
2
(
∫
I
k
×
I
l
D
2
k
(
s
)
D
2
l
(
t
)
‍
(
f
(
·
+
s
,
·
+
t
)
-
f
(
·
,
·
)
)
p
d
s
d
t
‍
h
h
h
h
h
h
h
h
h
h
∫
I
k
×
I
l
D
2
k
(
s
)
D
2
l
(
t
)
‍
×
(
f
(
·
+
s
,
·
+
t
)
-
f
(
·
,
·
)
)
p
d
s
d
t
)
d
x
d
y
]
1
/
p
.
Furthermore, using the generalized Minkowski inequality, we have(65)
Θ
≤
∫
I
k
×
I
l
D
2
k
(
s
)
D
2
l
(
t
)
‍
h
h
h
×
(
∫
I
2
(
f
(
·
+
s
,
·
+
t
)
-
f
(
·
,
·
)
)
p
d
x
d
y
‍
)
1
/
p
d
s
d
t
≤
ω
p
(
f
,
2
-
k
,
2
-
l
)
≤
𝒞
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
.
This completes the proof of Lemma4.Lemma 5.
LetP
∈
𝒫
2
k
,
2
l
,
Q
∈
𝒫
2
k, 1
≤
p
≤
∞
,
k
,
l
∈
ℕ; then one has(i)
(66)
∥
∬
0
1
(
f
(
·
+
s
,
·
+
t
)
-
f
(
·
,
·
)
)
r
k
(
s
)
r
l
(
t
)
P
(
s
,
t
)
d
s
d
t
∥
p
≤
C
∥
P
∥
1
ω
1,2
p
(
f
,
2
-
k
,
2
-
l
)
,
(ii)
(67)
∥
∬
0
1
(
f
(
·
+
s
,
·
+
t
)
-
f
(
·
,
·
)
)
r
k
(
s
)
D
2
l
(
t
)
Q
(
s
)
d
s
d
t
‍
∥
p
≤
C
∥
Q
∥
1
ω
1,2
p
(
f
,
2
-
k
,
2
-
l
)
.Proof.
Note thatP
∈
𝒫
2
k
,
2
l and Q
∈
𝒫
2
k are constants on the sets I
k
×
I
l and I
k, respectively. By Lemma 4, it is not difficult to prove Lemma 5. We can also find the conclusions in [13].Lemma 6.
Suppose that(i)
{
t
m
n
u
v
}
∈
D
R
B
V
S, and it is doubly triangular; then for any i
1
≥
i
2 or j
1
≥
j
2, one has t
m
n
i
1
j
=
𝒪
(
t
m
n
i
2
j
) or t
m
n
i
j
1
=
𝒪
(
t
m
n
i
j
2
).
(ii)
{
t
m
n
u
v
}
∈
D
H
B
V
S, and it is doubly triangular; then for any i
1
≤
i
2 or j
1
≤
j
2, one has t
m
n
i
1
j
=
𝒪
(
t
m
n
i
2
j
) or t
m
n
i
j
1
=
𝒪
(
t
m
n
i
j
2
).Proof.
(i) Since{
t
m
n
u
v
} is triangular,
(68)
t
m
n
i
1
j
≤
∑
u
=
i
1
m
|
Δ
10
t
m
n
u
j
|
‍
≤
∑
u
=
i
2
m
|
Δ
10
t
m
n
u
j
|
‍
=
𝒪
(
t
m
n
i
2
j
)
.
Similarly, we have
(69)
t
m
n
i
j
1
≤
∑
v
=
j
1
m
|
Δ
10
t
m
n
i
v
|
‍
≤
∑
v
=
j
2
m
|
Δ
10
t
m
n
i
v
|
‍
=
𝒪
(
t
m
n
i
j
2
)
.
(ii) Since(70)
|
t
m
n
i
1
j
-
t
m
n
i
2
j
|
≤
∑
i
=
i
1
i
2
|
Δ
10
t
m
n
i
j
|
‍
=
𝒪
(
t
m
n
i
2
j
)
,
it implies that
(71)
t
m
n
i
1
j
=
𝒪
(
t
m
n
i
2
j
)
.
Similarly, we have(72)
t
m
n
i
j
1
=
𝒪
(
t
m
n
i
j
2
)
.
This completes the proof of Lemma6.


## 4. Proofs of Theorems3 and 4

We denote the norm of∬
0
1
A
i
j
(
f
(
x
+
s
,
y
+
t
)
-
f
(
x
,
y
)
)
d
s
d
t by
(73)
Φ
i
j
≔
∥
∬
0
1
A
i
j
(
f
(
·
+
s
,
·
+
t
)
-
f
(
·
,
·
)
)
d
s
d
t
‍
∥
p
.
Then by (60) and Minkowski inequality, we have
(74)
∥
T
m
n
-
f
∥
p
≤
∑
i
=
1
4
∑
j
=
1
4
Φ
i
j
‍
‍
.Proof of Theorem3.
(I) When{
t
m
n
i
j
}
∈
DRBVS, we deduce Φ
i
j separately, by (9) and Lemma 5(i),
(75)
Φ
11
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
2
k
2
l
ω
1,2
p
(
f
,
2
-
k
,
2
-
l
)
‍
‍
+
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
i
=
1
2
k
-
1
i
2
l
|
Δ
10
t
m
n
(
2
k
+
1
-
i
)
2
l
|
ω
1,2
p
(
f
,
2
-
k
,
2
-
l
)
‍
‍
‍
+
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
j
=
1
2
l
-
1
j
2
k
|
Δ
01
t
m
n
2
k
(
2
l
+
1
-
j
)
|
ω
1,2
p
(
f
,
2
-
k
,
2
-
l
)
‍
‍
‍
+
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
∑
i
=
1
2
k
-
1
∑
j
=
1
2
l
-
1
|
Δ
11
t
m
n
(
2
k
+
1
-
i
)
(
2
l
+
1
-
j
)
|
‍
‍
‍
‍
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
×
i
j
ω
1,2
p
(
f
,
2
-
k
,
2
-
l
)
.
Since {
t
m
n
i
j
}
∈
DRBVS, by the definition of DRBVS, we have
(76)
∑
i
=
1
2
k
-
1
i
2
l
|
Δ
10
t
m
n
(
2
k
+
1
-
i
)
2
l
|
‍
≤
2
k
+
l
t
m
n
2
k
2
l
,
∑
j
=
1
2
l
-
1
j
2
k
|
Δ
01
t
m
n
2
k
(
2
l
+
1
-
j
)
|
‍
≤
2
k
+
l
t
m
n
2
k
2
l
,
∑
i
=
1
2
k
-
1
∑
j
=
1
2
l
-
1
|
Δ
11
t
m
n
(
2
k
+
1
-
i
)
(
2
l
+
1
-
j
)
|
i
j
‍
‍
≤
2
k
+
l
t
m
n
2
k
2
l
.
Substituting the above into Φ
11 yields
(77)
Φ
11
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
2
k
2
l
ω
1,2
p
(
f
,
2
-
k
,
2
-
l
)
‍
‍
.
Now by (9) and Lemma 5(ii),
(78)
Φ
12
≤
𝒞
∑
k
=
0
E
-
1
(
∑
l
=
0
F
-
1
∑
j
=
1
2
l
-
1
2
k
t
m
n
2
k
(
2
l
+
j
)
‍
‍
‍
h
h
h
h
l
h
h
+
∑
l
=
0
F
-
1
∑
i
=
1
2
k
-
1
∑
j
=
1
2
l
-
1
i
|
Δ
10
t
m
n
(
2
k
+
1
-
i
)
(
2
l
+
j
)
|
‍
‍
‍
)
×
ω
p
(
f
,
2
-
k
)
.
Applying Lemma 6(i) and still the definition of DRBVS, we have
(79)
Φ
12
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
ω
p
(
f
,
2
-
k
)
‍
‍
h
h
h
h
h
h
h
h
h
×
(
∑
j
=
1
2
l
-
1
t
m
n
2
k
(
2
l
+
j
)
‍
+
t
m
n
2
k
2
l
)
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
ω
p
(
f
,
2
-
k
)
t
m
n
2
k
2
l
‍
‍
.
Analogously,
(80)
Φ
13
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
ω
p
(
f
,
2
-
l
)
t
m
n
2
k
2
l
‍
‍
.
By Lemmas 4 and 6(i), it is easy to verify that
(81)
Φ
14
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
2
k
2
l
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
‍
‍
.
Similar discussion deduces that (by using Lemmas 4 and 5 (i) and (ii) separately)
(82)
Φ
21
≤
𝒞
∑
k
=
0
E
-
1
2
k
(
n
-
2
F
)
t
m
n
2
k
2
F
ω
p
(
f
,
2
-
k
)
‍
,
Φ
22
≤
𝒞
(
∑
k
=
0
E
-
1
‍
2
k
(
n
-
2
F
)
t
m
n
2
k
n
h
h
h
h
h
+
∑
k
=
0
E
-
1
‍
2
k
(
n
-
2
F
)
t
m
n
2
k
2
F
)
×
ω
p
(
f
,
2
-
k
,
2
-
F
)
,
Φ
23
≤
𝒞
∑
k
=
0
E
-
1
2
k
(
n
-
2
F
)
t
m
n
2
k
2
F
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
F
)
)
‍
,
Φ
24
≤
𝒞
∑
k
=
0
E
-
1
2
k
(
n
-
2
F
)
t
m
n
2
k
n
ω
p
(
f
,
2
-
F
)
‍
+
𝒞
∑
k
=
0
E
-
1
(
n
-
2
F
)
t
m
n
2
k
2
F
ω
p
(
f
,
2
-
F
)
‍
.
Note the definition ofE and F; clearly, m
+
1
≤
2
E
+
1 and n
+
1
≤
2
F
+
1; thus we have
(83)
m
-
2
E
≤
2
E
-
1
=
∑
k
=
0
E
-
1
2
k
‍
,
n
-
2
F
≤
2
F
-
1
=
∑
l
=
0
F
-
1
2
l
‍
.
Applying inequality (83) and Lemma 6 to (82), also using the monotonicity of continuous modulus, it is not difficult to deduce that
(84)
∑
i
=
1
4
Φ
2
i
‍
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
2
k
2
l
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
‍
‍
.
Φ
3
i
(
i
=
1,2
,
3,4
) goes analogously as Φ
2
i
(
i
=
1,2
,
3,4
); thus we have
(85)
∑
i
=
1
4
Φ
3
i
‍
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
2
k
2
l
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
‍
‍
.
Next, we estimateΦ
4
i
,
(
i
=
1,2
,
3,4
). Applying Lemmas 4 and 6, we easily get
(86)
Φ
41
≤
𝒞
(
m
-
2
E
)
(
n
-
2
F
)
t
m
n
2
E
2
F
Φ
41
≤
×
(
ω
p
(
f
,
2
-
E
)
+
ω
p
(
f
,
2
-
F
)
)
,
Φ
42
≤
𝒞
ω
p
(
f
,
2
-
F
)
Φ
41
≤
×
(
(
m
-
2
E
)
(
n
-
2
F
)
t
m
n
2
E
n
+
(
n
-
2
F
)
t
m
n
2
E
2
F
)
,
Φ
43
≤
𝒞
ω
p
(
f
,
2
-
E
)
Φ
41
≤
×
(
(
m
-
2
E
)
(
n
-
2
F
)
t
m
n
m
2
F
+
(
m
-
2
E
)
t
m
n
2
E
2
F
)
,
and the last term
(87)
Φ
44
≤
𝒞
ω
p
(
f
,
2
-
E
,
2
-
F
)
(
m
-
2
E
)
(
n
-
2
F
)
×
(
t
m
n
m
n
+
t
m
n
2
E
n
+
t
m
n
m
2
F
+
t
m
n
2
E
2
F
)
.
Applying Lemma6(i) and (83) to (86) and (87), we have
(88)
∑
i
=
1
4
Φ
4
i
‍
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
2
k
2
l
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
‍
‍
.
Combining all the estimates, (77)–(81), (84), (85), and (88) of Φ
i
j, yields that
(89)
∥
T
m
n
-
f
∥
L
p
≤
∑
i
=
1
4
∑
j
=
1
4
Φ
i
j
‍
‍
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
2
k
2
l
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
‍
‍
.
(II) When{
t
m
n
i
j
}
∈
DHBVS, the proof goes similarly. We mainly point out the differences in this case and prove some terms of Φ
i
j as examples.
Since{
t
m
n
i
j
}
∈
DHBVS, by the definition of DHBVS, we have
(90)
∑
i
=
1
2
k
-
1
i
2
l
|
Δ
10
t
m
n
(
2
k
+
1
-
i
)
2
l
|
‍
≤
2
k
+
l
t
m
n
(
2
k
+
1
-
1
)
2
l
,
∑
j
=
1
2
l
-
1
j
2
k
|
Δ
01
t
m
n
2
k
(
2
l
+
1
-
j
)
|
‍
≤
2
k
+
l
t
m
n
2
k
(
2
l
+
1
-
1
)
,
∑
i
=
1
2
k
-
1
∑
j
=
1
2
l
-
1
|
Δ
11
t
m
n
(
2
k
+
1
-
i
)
(
2
l
+
1
-
j
)
|
i
j
‍
‍
≤
2
k
+
l
t
m
n
(
2
k
+
1
-
1
)
(
2
l
+
1
-
1
)
.
Thus by Lemma5(i), (9), (71), and (72), we have
(91)
Φ
11
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
(
2
k
+
1
-
1
)
(
2
l
+
1
-
1
)
ω
1,2
p
(
f
,
2
-
k
,
2
-
l
)
‍
‍
.
Lemma5(ii), (9), (71), (72), and (83) yield that
(92)
Φ
1
i
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
t
m
n
(
2
k
+
1
-
1
)
(
2
l
+
1
-
1
)
ω
1,2
p
(
f
,
2
-
k
,
2
-
l
)
‍
‍
.
Now the definition ofDHBVS, Lemmas 5 and 6, and (9) deduce that
(93)
Φ
21
≤
𝒞
∑
k
=
0
E
-
1
2
k
(
(
n
-
2
F
)
t
m
n
2
k
n
+
t
m
n
(
2
k
+
1
-
1
)
n
)
ω
p
(
f
,
2
-
k
)
‍
Φ
21
≤
𝒞
∑
k
=
0
E
-
1
2
k
(
n
-
2
F
)
t
m
n
(
2
k
+
1
-
1
)
n
ω
p
(
f
,
2
-
k
)
‍
,
Φ
22
≤
𝒞
∑
k
=
0
E
-
1
2
k
(
n
-
2
F
)
t
m
n
(
2
k
+
1
-
1
)
n
ω
p
(
f
,
2
-
k
,
2
-
F
)
‍
,
Φ
23
≤
𝒞
∑
k
=
0
E
-
1
2
k
(
n
-
2
F
)
t
m
n
(
2
k
+
1
-
1
)
n
‍
h
h
h
h
h
h
h
h
h
h
×
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
F
)
)
,
Φ
24
≤
𝒞
∑
k
=
0
E
-
1
2
k
(
n
-
2
F
)
t
m
n
(
2
k
+
1
-
1
)
n
ω
p
(
f
,
2
-
F
)
‍
.
While in this case we keept
m
n
(
2
k
+
1
-
1
)
n in the sum, so we have
(94)
∑
i
=
1
4
Φ
2
i
‍
≤
𝒞
∑
k
=
0
E
-
1
2
k
(
n
-
2
F
)
t
m
n
(
2
k
+
1
-
1
)
n
‍
h
h
i
h
h
h
×
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
F
)
)
.
Analogously,(95)
∑
i
=
1
4
Φ
3
i
‍
≤
𝒞
∑
l
=
0
F
-
1
(
m
-
2
E
)
2
l
t
m
n
m
(
2
l
+
1
-
1
)
‍
h
i
h
h
h
h
×
(
ω
p
(
f
,
2
-
E
)
+
ω
p
(
f
,
2
-
l
)
)
,
(96)
∑
i
=
1
4
Φ
4
i
‍
≤
𝒞
(
m
-
2
E
)
(
n
-
2
F
)
t
m
n
m
n
×
(
ω
p
(
f
,
2
-
E
)
+
ω
p
(
f
,
2
-
F
)
)
.
Combining all the estimates of (92)–(96), we obtain the conclusion of the case of {
t
m
n
i
j
}
∈
DHBVS in Theorem 3.Remark C. Actually, in the case of {
t
m
n
i
j
}
∈
DHBVS, (96) can be represented in a more complicated form if we calculate the term with the same manner. For instance,
(97)
∑
i
=
1
4
Φ
4
i
‍
≤
𝒞
∑
i
=
0
m
-
2
E
t
m
n
(
2
E
+
i
)
n
(
n
-
2
F
)
ω
p
(
f
,
2
-
F
)
‍
+
𝒞
∑
i
=
0
m
-
2
E
t
m
n
(
2
E
+
i
)
n
(
n
-
2
F
)
ω
p
(
f
,
2
-
F
)
‍
+
𝒞
(
m
-
2
E
)
(
n
-
2
F
)
t
m
n
m
n
×
(
ω
p
(
f
,
2
-
E
)
+
ω
p
(
f
,
2
-
F
)
)
,
while we give a simpler form in Theorem 3 of the present paper.Proof of Theorem4.
Since{
t
m
n
u
v
}
∈
DHBVS, and m
n
t
m
n
m
n
=
𝒪
(
1
), in the case of DHBVS, it is easy to calculate that
(98)
∥
T
m
n
-
f
∥
p
=
{
𝒪
(
2
-
E
α
+
2
-
F
α
)
,
0
<
α
<
1
,
𝒪
(
E
2
-
E
+
F
2
-
F
)
,
α
=
1
,
𝒪
(
2
-
E
+
2
-
F
)
,
α
>
1
.
Note thatm
~
2
E and n
~
2
F; the above estimate is equivalent to
(99)
∥
T
m
n
-
f
∥
p
=
{
𝒪
(
m
-
α
+
n
-
α
)
,
0
<
α
<
1
,
𝒪
(
m
-
1
log
⁡
m
+
n
-
1
log
⁡
n
)
,
α
=
1
,
𝒪
(
m
-
1
+
n
-
1
)
,
α
>
1
.
While for{
t
m
n
u
v
}
∈
DRBVS, the conclusion is obvious from Theorem 3 and the supposed f
∈
L
i
p
(
α
,
p
).Comparing Theorems3 and 4 with Theorems A, B, and C, we find that the former are the generalizations of the latter, from the sense of monotonicity on one hand. On the other hand, T-transformation is the generalization of some means of series, such as Nörlund means, Cesàro means, and Riesz means. In the next section, we give applications of our results to some summability methods.


## 5. Applications to Nörlund Means and Weighted Means

Corollary 5.
LetT be the matrix defined by (11). Suppose that {
p
j
k
}
∈
D
H
B
V
S, for f
∈
L
p
(
I
2
). Then, one has
(100)
P
m
n
∥
T
m
n
-
f
∥
p
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
p
m
-
2
k
,
n
-
2
l
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
‍
‍
.
For{
p
j
k
}
∈
D
R
B
V
S, one has
(101)
P
m
n
∥
T
m
n
-
f
∥
p
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
p
m
-
2
k
+
1
+
1
,
n
-
2
l
+
1
+
1
‍
‍
h
h
h
h
h
i
h
h
h
l
h
h
h
×
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
+
𝒞
∑
k
=
0
E
-
1
2
k
(
n
-
2
F
)
p
m
-
2
k
+
1
+
1,0
‍
h
h
h
h
h
h
h
l
h
h
h
×
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
F
)
)
+
𝒞
∑
l
=
0
F
-
1
2
l
(
m
-
2
E
)
p
0
,
n
-
2
l
+
1
+
1
‍
h
h
h
h
h
h
l
h
h
h
h
×
(
ω
p
(
f
,
2
-
E
)
+
ω
p
(
f
,
2
-
l
)
)
+
𝒞
(
m
-
2
E
)
(
n
-
2
F
)
p
00
+
×
(
ω
p
(
f
,
2
-
E
)
+
ω
p
(
f
,
2
-
F
)
)
.Proof.
Note that in the case when{
p
j
k
}
∈
DHBVS, we have {
t
m
n
j
k
}
∈
DRBVS. By the first part of Theorem 3, it is easy to deduce that
(102)
P
m
n
∥
T
m
n
-
f
∥
p
≤
𝒞
∑
k
=
0
E
-
1
∑
l
=
0
F
-
1
2
k
+
l
p
m
-
2
k
,
n
-
2
l
(
ω
p
(
f
,
2
-
k
)
+
ω
p
(
f
,
2
-
l
)
)
‍
‍
.
When{
p
j
k
}
∈
DRBVS (corresponding to the case {
t
m
n
j
k
}
∈
DHBV), we get the conclusion from the second part of Theorem 3 immediately.Remark D. When {
p
j
k
} is nondecreasing and Δ
11
p
j
k is of fixed sign, it is obvious that {
p
j
k
}
∈
DHBVS. Therefore, Theorem 3 of [13] can be deduced directly from the first inequality of Corollary 5. Furthermore, we exclude some preconditions in this case. For the nonincreasing case of Theorem in [13], as we figure out in Remark C, the second inequality of Corollary 5 can be seen as the generalization of this case as long as it takes the more complicated form of Φ
4
i, i
=
1,2
,
3,4.The following is a Corollary of Theorem4, for the case of Nörlund means. Since the nondecreasing of {
p
j
k
} and fixed sign of Δ
11
p
j
k imply that {
p
j
k
}
∈
DHBVS, it is the generalization of Theorem C.Corollary 6.
LetT be the matrix defined by (11). Suppose that {
p
j
k
}
∈
D
H
B
V
S satisfies
(103)
(
m
+
1
)
(
n
+
1
)
p
m
n
P
m
n
=
𝒪
(
1
)
.
Then forf
∈
L
i
p
(
α
,
p
), one has
(104)
∥
T
m
n
-
f
∥
p
=
{
𝒪
(
m
-
α
+
n
-
α
)
,
0
<
α
<
1
,
𝒪
(
m
-
1
log
⁡
m
+
n
-
1
log
⁡
n
)
,
α
=
1
,
𝒪
(
m
-
1
+
n
-
1
)
,
α
>
1
.As far as the weighted means is concerned, we have the following corollary which generalizes Theorems A and B.Corollary 7.
LetT be the matrix defined by (14) and f
∈
L
p
(
I
2
). Suppose that {
p
j
k
}
∈
D
H
B
V
S. Then, one has (101). For {
p
j
k
}
∈
D
H
B
V
S, one has (100).
Suppose thatf
∈
L
i
p
(
α
,
p
), {
p
j
k
}
∈
D
H
B
V
S satisfies
(105)
(
m
+
1
)
(
n
+
1
)
p
m
n
P
m
n
=
𝒪
(
1
)
.
Then, one has(106)
∥
T
m
n
-
f
∥
p
=
{
𝒪
(
m
-
α
+
n
-
α
)
,
0
<
α
<
1
,
𝒪
(
m
-
1
log
⁡
m
+
n
-
1
log
⁡
n
)
,
α
=
1
,
𝒪
(
m
-
1
+
n
-
1
)
,
α
>
1
.



---

*Source: 713175-2014-04-22.xml*