# Smooth Decompositions of Triebel-Lizorkin and Besov Spaces on Product Spaces of Homogeneous Type

Authors: Fanghui Liao; Zongguang Liu; Xiaojin Zhang
Journal: Journal of Function Spaces
 (2014)

---

## Abstract

We introduce Triebel-Lizorkin and Besov spaces by Calderón’s reproducing formula on product spaces of homogeneous type. We also obtain smooth atomic and molecular decompositions for these spaces.


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## Body

## 1. Introduction and Main Results

Atomic and molecular decompositions are significant tools in studying function spaces and operators in harmonic analysis. The atomic decomposition for Hardy spaces onR was first introduced by Coifman in [1] and was extended to R
n by Latter in [2]. Molecules in Hardy spaces were also introduced by Coifman in [3]. Coifman and Weiss in [4] extended molecules to more general setting in Hardy spaces.The atomic decomposition of product Hardy spacesH
p
(
R
×
R
) was established by Chang and Fefferman in [5, 6]. Recently, general atomic decomposition on product Hardy spaces was constructed by Han et al. in [7] and Han et al. in [8].The smooth atomic decomposition for the Triebel-Lizorkin spaces onR
n first is considered by Frazier and Jawerth in [9] and later extended to spaces of homogeneous type by Han and Sawyer in [10]; for similar results, see [11]. Smooth molecules of the Triebel-Lizorkin spaces on bidisc were given by Wang in [12].In [13], Hardy spaces associated with different homogeneities were constructed. Wu in [14] introduced related Triebel-Lizorkin and Besov spaces similarly. In [7, 15], the authors studied the Hardy spaces on product spaces of homogeneous type and showed some properties of these spaces. A natural question arises: whether we introduce the Triebel-Lizorkin and Besov spaces on product spaces of homogeneous type and study some properties of these spaces.The purpose of this paper is to answer the question. Namely, we introduce the Triebel-Lizorkin and Besov spaces on product spaces of homogeneous type. A theory of atomic and molecular decompositions for these spaces is also presented.We begin with some necessary definitions and notation on spaces of homogeneous type.A quasimetricρ on a set X is a function ρ: X
×
X
→
[
0
,
∞
) satisfying that (i) ρ
(
x
,
y
)
=
0 if and only if x
=
y, (ii) ρ
(
x
,
y
)
=
ρ
(
y
,
x
) for all x
,
y
∈
X, and (iii) there exists a constant A
∈
[
1
,
∞
) such that for all x, y, and z
∈
X
(1)
ρ
(
x
,
y
)
≤
A
[
ρ
(
x
,
z
)
+
ρ
(
z
,
y
)
]
.
Any quasimetric defines a topology, for which the balls B
(
x
,
r
)
=
{
y
∈
X
:
ρ
(
y
,
x
)
<
r
} with all x
∈
X and all r
>
0 form a basis.We now state definitions of spaces of homogeneous type.Definition 1.
A space of homogeneous type(
X
,
ρ
,
μ
) is a set X with a quasimetric ρ and a nonnegative Borel regular measure μ on X such that 0
<
μ
(
B
(
x
,
r
)
)
<
∞, and there exists a positive constant C
<
∞ such that
(2)
μ
(
B
(
x
,
2
r
)
)
≤
C
μ
(
B
(
x
,
r
)
)
for all x
∈
X and all r
>
0, where μ is assumed to be defined on a σ-algebra which contains all Borel sets and all balls B
(
x
,
r
).We suppose thatμ
(
X
)
=
∞ and μ
(
{
x
}
)
=
0 for all x
∈
X. Further, suppose there exist constant C
>
0 and regularity exponent 0
<
θ
<
1 such that, for all 0
<
r
<
∞ and all x
,
x
′
,
y
∈
X, μ
(
B
(
x
,
r
)
)
~
r and |
ρ
(
x
,
y
)
-
ρ
(
x
′
,
y
)
|
≤
C
ρ
(
x
,
x
′
)
θ
[
ρ
(
x
,
y
)
+
ρ
(
x
′
,
y
)
]
1
-
θ.Let us recall the definition of an approximation to the identity on spaces of homogeneous type.Definition 2 (see [7]).
LetX be a space of homogeneous type as in Definition 1 and constant A satisfying (1). A sequence {
S
k
}
k
∈
Z of linear operators is said to be an approximation to the identity of order ϵ
∈
(
0
,
θ
] if there exists C
>
0 such that, for all x, x
′, y, and y
′
∈
X, S
k
(
x
,
y
), the kernel of S
k, is a function from X
×
X into C satisfying(i)|
S
k
(
x
,
y
)
|
≤
C
(
2
-
k
ϵ
/
(
2
-
k
+
ρ
(
x
,
y
)
)
1
+
ϵ
);
(ii)|
S
k
(
x
,
y
)
-
S
k
(
x
′
,
y
)
|
≤
C
(
ρ
(
x
,
x
′
)
/
(
2
-
k
+
ρ
(
x
,
y
)
)
)
ϵ
(
2
-
k
ϵ
/
(
2
-
k
+
ρ
(
x
,
y
)
)
1
+
ϵ
)
forρ
(
x
,
x
′
)
≤
(
1
/
2
A
)
(
2
-
k
+
ρ
(
x
,
y
)
);
(iii)|
S
k
(
x
,
y
)
-
S
k
(
x
,
y
′
)
|
≤
C
(
ρ
(
y
,
y
′
)
/
(
2
-
k
+
ρ
(
x
,
y
)
)
)
ϵ
(
2
-
k
ϵ
/
(
2
-
k
+
ρ
(
x
,
y
)
)
1
+
ϵ
)
forρ
(
y
,
y
′
)
≤
(
1
/
2
A
)
(
2
-
k
+
ρ
(
x
,
y
)
);
(iv)|
[
S
k
(
x
,
y
)
-
S
k
(
x
,
y
′
)
]
-
[
S
k
(
x
′
,
y
)
-
S
k
(
x
′
,
y
′
)
]
|
≤
C
(
ρ
(
x
,
x
′
)
/
(
2
-
k
+
ρ
(
x
,
y
)
)
)
ϵ
(
ρ
(
y
,
y
′
)
/
2
-
k + ρ
(
x
,
y
)
)
ϵ
(
2
-
k
ϵ
/
(
2
-
k
+
ρ
(
x
,
y
)
)
1
+
ϵ
)
forρ
(
x
,
x
′
)
≤
(
1
/
2
A
)
(
2
-
k
+
ρ
(
x
,
y
)
) and ρ
(
y
,
y
′
)
≤
(
1
/
2
A
)
(
2
-
k
+
ρ
(
x
,
y
)
);
(v)∫
X
‍
S
k
(
x
,
y
)
d
μ
(
y
)
=
∫
X
‍
S
k
(
x
,
y
)
d
μ
(
x
)
=
1.Moreover, a sequence {
S
k
}
k
∈
Z of linear operators is said to be an approximation to the identity of older ϵ
∈
(
0
,
θ
] having compact support if there exists a constant C
>
0 such that, for all k
∈
Z and all x
,
x
′
,
y, and y
′
∈
X, S
k
(
x
,
y
), the kernel of S
k is a function from X
×
X into C satisfying (
v
) and(vi)
S
k
(
x
,
y
)
=
0 if ρ
(
x
,
y
)
≥
C
2
-
k and ∥
S
k
∥
L
∞
(
X
×
X
)
≤
C
2
k;
(vii)
|
S
k
(
x
,
y
)
-
S
k
(
x
′
,
y
)
|
≤
C
2
k
(
1
+
ϵ
)
ρ
(
x
,
x
′
)
ϵ;
(viii)
|
S
k
(
x
,
y
)
-
S
k
(
x
,
y
′
)
|
≤
C
2
k
(
1
+
ϵ
)
ρ
(
y
,
y
′
)
ϵ;
(ix)
|
[
S
k
(
x
,
y
)
-
S
k
(
x
,
y
′
)
]
-
[
S
k
(
x
′
,
y
)
-
S
k
(
x
′
,
y
′
)
]
|
≤
C
2
k
(
1
+
2
ϵ
)
ρ
(
x
,
x
′
)
ϵ
ρ
(
y
,
y
′
)
ϵ.We present test functions on spaces of homogeneous typeX before we give test functions on product spaces of homogeneous type X
1
×
X
2.Definition 3 (see [15]).
For fixedx
0
∈
X, let 0
<
β, γ
<
θ, and r
>
0, where θ is the regularity exponent on X. A function f
(
x
,
y
) defined on X
×
X is said to be a test function of type (
x
0
,
r
,
β
,
γ
), if, for all x
,
y
∈
X, f satisfies the following conditions:(i)
|
f
(
x
)
|
≤
C
(
r
/
(
r
+
ρ
(
x
,
x
0
)
)
)
1
+
γ;
(ii)
|
f
(
x
)
-
f
(
y
)
|
≤
C
(
ρ
(
x
,
y
)
/
(
r
+
ρ
(
x
,
x
0
)
)
)
β
(
r
/
(
r
+
ρ
(
x
,
x
0
)
)
)
1
+
γ for ρ
(
x
,
y
)
≤
(
1
/
2
A
)
(
r
+
ρ
(
x
,
x
0
)
).
Iff is a test function of type (
x
0
,
r
,
β
,
γ
), we write f
∈
G
(
x
0
,
r
,
β
,
γ
) and the norm of f is defined by
(3)
∥
f
∥
G
(
x
0
,
r
,
β
,
γ
)
=
inf
⁡
{
C
:
(
i
)

-

(
ii
)
hold
}
.We denote byG
(
β
,
γ
) the class of G
(
x
0
,
r
,
β
,
γ
) with r
=
1 for fixed x
0
∈
X. Set f
(
x
)
∈
G
0
(
β
,
γ
) if f
(
x
)
∈
G
(
β
,
γ
) with ∫
X
‍
f
(
x
)
d
μ
(
x
)
=
0. It is easy to see that G
(
x
1
,
r
,
β
,
γ
)
=
G
(
β
,
γ
) with an equivalent norm for all x
1
∈
X and r
>
0. Furthermore, we can check that G
(
β
,
γ
) is a Banach space with respect to the norm in G
(
β
,
γ
). Let G
°
(
β
,
γ
) be the completion of the space G
0
(
θ
,
θ
) in G
(
β
,
γ
) with 0
<
β
,
γ
<
θ. If f
∈
G
(
β
,
γ
), we then define ∥
f
∥
G
°
(
β
,
γ
)
=
∥
f
∥
G
(
β
,
γ
).We define the distribution space(
G
°
(
β
,
γ
)
)
′ by all linear functionals L from G
°
(
β
,
γ
) to C with the property that there exists C
≥
0 such that |
L
(
f
)
|
≤
C
∥
f
∥
G
°
(
β
,
γ
) for all f
∈
G
°
(
β
,
γ
).Let(
X
i
,
ρ
i
,
μ
i
) for i
=
1,2 be two spaces of homogeneous type as in Definition 1 and ρ
i satisfies (1) with A replaced by A
i for i
=
1,2. We define spaces of test functions on product space X
1
×
X
2 of spaces of homogeneous type.Definition 4 (see [15]).
Let(
x
0
,
y
0
)
∈
X
1
×
X
2
,
0
<
β
i
,
γ
i
<
θ
i, and r
i
>
0, where θ
i is the regularity exponent on X
i for i
=
1,2. A function f
(
x
,
y
) defined on X
1
×
X
2 is said to be a test function of type (
x
0
,
y
0
;
r
1
,
r
2
;
β
1
,
β
2
;
γ
1
,
γ
2
) if, for any fixed y
,
y
′
∈
X
2, f
(
x
,
y
), as a function of variable of x, is a test function of G
(
x
0
,
r
1
,
β
1
,
γ
1
) on X
1. Similarly, for any fixed x
,
x
′
∈
X
1, f
(
x
,
y
), as a function of variable of y, is a test function of G
(
y
0
,
r
2
,
β
2
,
γ
2
) on X
2. Moreover the following conditions are satisfied:(i)
∥
f
(
·
,
y
)
∥
G
(
x
0
,
r
1
,
β
1
,
γ
1
)
≤
C
(
r
2
/
(
r
2
+
ρ
2
(
y
,
y
0
)
)
)
1
+
γ
2;
(ii)
∥
f
(
·
,
y
)
-
f
(
·
,
y
′
)
∥
G
(
x
0
,
r
1
,
β
1
,
γ
1
)
≤
C
(
ρ
2
(
y
′
,
y
)
/
(
r
2
+
ρ
2
(
y
,
y
0
)
)
)
β
2
(
r
2
/
(
r
2
+
ρ
2
(
y
,
y
0
)
)
)
1
+
γ
2  for all y
,
y
′
∈
X
2 with ρ
2
(
y
,
y
′
)
≤
(
1
/
2
A
2
)
(
r
2
+
ρ
2
(
y
,
y
0
)
);
(iii)
properties (i)-(ii) also hold withx and y interchanged.
Iff is a test function of type (
x
0
,
y
0
;
r
1
,
r
2
;
β
1
,
β
2
;
γ
1
,
γ
2
), we write
(4)
f
∈
G
(
x
0
,
y
0
;
r
1
,
r
2
;
β
1
,
β
2
;
γ
1
,
γ
2
)
,
and the norm of f is defined by
(5)
∥
f
∥
G
(
x
0
,
y
0
;
r
1
,
r
2
;
β
1
,
β
2
;
γ
1
,
γ
2
)
=
inf
⁡
{
C
:
(
i
)
–
(
iii
)
hold
}
.Similarly, we denote byG
(
β
1
,
β
2
;
γ
1
,
γ
2
) the class of G
(
x
0
,
y
0
;
r
1
,
r
2
;
β
1
,
β
2
;
γ
1
,
γ
2
) with r
1
=
r
2
=
1 for fixed (
x
0
,
y
0
)
∈
X
1
×
X
2. Set f
(
x
,
y
)
∈
G
0
(
β
1
,
β
2
;
γ
1
,
γ
2
) if f
(
x
,
y
)
∈
G
(
β
1
,
β
2
;
γ
1
,
γ
2
) with ∫
X
1
‍
f
(
x
,
y
)
d
μ
1
(
x
)
=
∫
X
2
‍
f
(
x
,
y
)
d
μ
2
(
y
)
=
0. It is easy to see that
(6)
G
(
x
0
,
y
0
;
r
1
,
r
2
;
β
1
,
β
2
;
γ
1
,
γ
2
)
=
G
(
β
1
,
β
2
;
γ
1
,
γ
2
)
with an equivalent norm for all (
x
0
,
y
0
)
∈
X
1
×
X
2 and r
1
,
r
2
>
0. Furthermore, we can check that G
(
β
1
,
β
2
;
γ
1
,
γ
2
) is a Banach space with respect to the norm in G
(
β
1
,
β
2
;
γ
1
,
γ
2
).LetG
°
(
β
1
,
β
2
,
γ
1
,
γ
2
) be the completion of the space G
0
(
θ
1
,
θ
2
;
θ
1
,
θ
2
) in G
(
β
1
,
β
2
,
γ
1
,
γ
2
) with 0
<
β
i
,
γ
i
<
θ
i for i
=
1,2. If f
∈
G
(
β
1
,
β
2
;
γ
1
,
γ
2
), we then define ∥
f
∥
G
∘
(
β
1
,
β
2
,
γ
1
,
γ
2
)
=
∥
f
∥
G
(
β
1
,
β
2
,
γ
1
,
γ
2
).Also, we define the distribution space(
G
°
(
β
1
,
β
2
;
γ
1
,
γ
2
)
)
′ by all linear functionals L from G
°
(
β
1
,
β
2
;
γ
1
,
γ
2
) to C with the property that there exists a constant C
>
0 such that |
L
(
f
)
|
≤
C
∥
f
∥
G
∘
(
β
1
,
β
2
;
γ
1
,
γ
2
) for all f
∈
G
∘
(
β
1
,
β
2
;
γ
1
,
γ
2
).We give Calderón’s reproducing formulas on product spaces of homogeneous type.Lemma 5 (see [7, 15]).
Suppose thati
=
1,2. Let ϵ
i
∈
(
0
,
θ
i
], let {
S
k
i
}
k
i
∈
Z be an approximation to the identity of order ϵ
i on spaces of homogeneous type X
i, and let D
k
i
=
S
k
i
-
S
k
i
-
1 for k
i
∈
Z. Then there are families of linear operators {
D
~
k
i
}
k
i
∈
Z and {
D
¯
k
i
}
k
i
∈
Z on X
i such that, for all f
∈
(
G
°
(
β
1
,
β
2
;
γ
1
,
γ
2
)
)
′ with β
i
,
γ
i
∈
(
0
,
ϵ
i
),
(7)
f
=
∑
k
1
,
k
2
∈
Z
‍
D
~
k
1
D
~
k
2
D
k
1
D
k
2
(
f
)
=
∑
k
1
,
k
2
∈
Z
‍
D
k
1
D
k
2
D
¯
k
1
D
¯
k
2
(
f
)
,
where the series converges in the norm of both the spaces (
G
°
(
β
1
′
,
β
2
′
;
γ
1
′
,
γ
2
′
)
)
′ with β
i
′
∈
(
0
,
β
i
)
,
γ
i
′
∈
(
0
,
γ
i
). Moreover, for x
i
,
y
i
∈
X
i and all k
i
∈
Z, D
~
k
i
(
x
i
,
y
i
), the kernel of D
~
k
i, satisfies conditions (i) and (ii) in Definition 2 with ϵ
i replaced by any ϵ
i
′
∈
(
0
,
ϵ
i
) and
(8)
∫
X
i
‍
D
~
k
i
(
x
i
,
y
i
)
d
μ
i
(
y
i
)
=
∫
X
i
‍
D
~
k
i
(
x
i
,
y
i
)
d
μ
i
(
x
i
)
=
0
.
D
¯
k
(
x
,
y
), the kernel of D
¯
k, satisfies conditions (
i
) and (
i
i
i
) in Definition 2 with ϵ replaced by ϵ
′ for 0
<
ϵ
′
<
ϵ and (8).We also need the following result, which gives an analogue of the grid Euclidean dyadic cubes on a space of homogeneous type.Lemma 6 (see [15]).
LetX be a space of homogeneous type as in Definition 1. Then there exists a collection {
Q
α
k
⊂
X
:
k
∈
Z
,
α
∈
I
k
} of open subsets, where I
k is some index set, and constants C
1
,
C
2
>
0 such that(i)
μ
(
X
∖
∪
α
Q
α
k
)
=
0 for each fixed k and Q
α
k
∩
Q
β
k
=
∅, if α
≠
β;
(ii)
for anyα
,
β
,
k
,
l with l
>
k, either Q
β
l
⊂
Q
α
k or Q
β
l
∩
Q
α
k
=
∅;
(iii)
for each(
k
,
α
) and each l
<
k, there is a unique β such that Q
α
k
⊂
Q
β
l;
(iv)
diam
⁡
(
Q
α
k
)
≤
C
1
2
-
k;
(v)
eachQ
α
k contains some ball B
(
y
α
k
,
C
2
2
-
k
), where y
α
k
∈
X.We think ofQ
α
k as being a dyadic cube with diameter 2
-
k and center y
α
k. When k
∈
Z and τ
∈
I
k, we denote by Q
τ
k
,
v, where v
=
1,2
,
…
,
N
(
τ
,
k
), the set of all cubes Q
τ
k
+
j
⊂
Q
τ
k, where j is a fixed large positive integer, and y
τ
k
,
v a point in Q
τ
k
,
v.Throughout the paper, we useC to denote positive constants, whose value may change from one occurrence to the next. We denote by f
~
g that there exists a constant C
>
0 independent of the main parameters such that C
-
1
g
<
f
<
C
g. χ
A is the characteristic function of A. Let a
+
=
max
⁡
(
0
,
a
) for any a
∈
R. We also denote M
1, M
2 are the Hardy-Littlewood maximal function on spaces of homogeneous type X
1, X
2, respectively.We introduce Triebel-Lizorkin spaceF
˙
p
s
,
q and Besov space B
˙
p
s
,
q on product spaces of homogeneous type.Definition 7.
Suppose thati
=
1,2, s
=
(
s
1
,
s
2
), s
i
∈
(
-
ϵ
i
,
ϵ
i
) and let {
D
k
i
}
k
i
∈
Z be the same as in Lemma 5. Let 1
<
p
,
q
<
∞ and 0
<
β
i
,
γ
i
<
ϵ
i
<
θ
i, f
∈
(
G
∘
(
β
1
,
β
2
;
γ
1
,
γ
2
)
)
′. Then Triebel-Lizorkin space F
˙
p
s
,
q
(
X
1
×
X
2
) is defined by
(9)
∥
f
∥
F
˙
p
s
,
q
:
=
∥
(
∑
k
1
,
k
2
∈
Z
‍
2
k
1
s
1
q
2
k
2
s
2
q
|
D
k
1
D
k
2
(
f
)
|
q
)
1
/
q
∥
p
<
∞
.
The Besov space B
˙
p
s
,
q
(
X
1
×
X
2
) is defined by
(10)
∥
f
∥
B
˙
p
s
,
q
:
=
(
∑
k
1
,
k
2
∈
Z
‍
2
k
1
s
1
q
2
k
2
s
2
q
∥
D
k
1
D
k
2
(
f
)
∥
p
q
)
1
/
q
<
∞
.In order to check that the definitions forF
˙
p
s
,
q and B
˙
p
s
,
q are independent of the choice of approximations to the identity, we recall almost orthogonality estimate.Lemma 8 (see [15]).
Leti
=
1,2. Suppose that {
S
k
i
}
k
i
∈
Z and {
P
k
i
}
k
i
∈
Z are two approximations to the identity on spaces of homogeneous type, and D
k
i
=
S
k
i
-
S
K
i
-
1, E
k
i
=
P
k
i
-
P
k
i
-
1. Then for ϵ
i
∈
(
0
,
θ
i
), there exists a positive constant C depending only on ϵ
i such that D
l
1
D
l
2
E
k
1
E
k
2
(
x
1
,
y
1
;
x
2
,
y
2
), the kernel of D
l
1
D
l
2
E
k
1
E
k
2, satisfies the following estimate:
(11)
|
D
l
1
D
l
2
E
k
1
E
k
2
(
x
1
,
y
1
;
x
2
,
y
2
)
|
≤
C
2
-
|
k
1
-
l
1
|
ϵ
1
2
-
|
k
2
-
l
2
|
ϵ
2
2
-
(
k
1
∧
l
1
)
ϵ
1
(
2
-
(
k
1
∧
l
1
)
+
ρ
1
(
x
1
,
y
1
)
)
1
+
ϵ
1
×
2
-
(
k
2
∧
l
2
)
ϵ
2
(
2
-
(
k
2
∧
l
2
)
+
ρ
2
(
x
2
,
y
2
)
)
1
+
ϵ
2
.By Calderón’s reproducing formula and almost orthogonality estimate, we can get the following product-type Plancherel-Pôlya inequalities. The proofs of the following theorems are similar to Proposition  4.1 in [10]. Here we omit the details.Theorem 9.
Suppose thatD
k and E
k are the same as in Lemma 8 for k
∈
Z. Let 1
<
p
,
q
<
∞, and |
s
i
|
<
ϵ
i for i
=
1,2. Then there exists a constant C
>
0 such that for all f
∈
(
G
°
(
β
1
,
β
2
;
γ
1
,
γ
2
)
)
′ with 0
<
β
i, γ
i
<
ϵ
i
<
θ
i for i
=
1,2, and then
(12)
∥
{
∑
k
1
∈
Z
‍
∑
k
2
∈
Z
‍
(
2
k
1
s
1
2
k
2
s
2
|
D
k
1
D
k
2
(
f
)
|
)
q
}
1
/
q
∥
L
p
~
∥
{
∑
l
1
∈
Z
‍
∑
l
2
∈
Z
‍
(
2
l
1
s
1
2
l
2
s
2
|
E
l
1
E
l
2
(
f
)
|
)
q
}
1
/
q
∥
L
p
,
{
∑
k
1
∈
Z
‍
∑
k
2
∈
Z
‍
(
2
k
1
s
1
2
k
2
s
2
∥
D
k
1
D
k
2
(
f
)
∥
L
p
)
q
}
1
/
q
~
{
∑
l
1
∈
Z
‍
∑
l
2
∈
Z
‍
(
2
l
1
s
1
2
l
2
s
2
∥
E
l
1
E
l
2
(
f
)
∥
L
p
)
q
}
1
/
q
.Remark 10.
LetD
l
1, D
l
2, D
¯
k
1, and D
¯
k
2 be as in Lemma 5. Then the kernel of D
l
1
D
l
2 has compact support, but not of D
¯
k
1
D
¯
k
2. Since (11) holds for D
l
1 and D
l
2 satisfying only the smoothness condition for the second variable, we conclude that Theorem 9 holds with D
l
1
D
l
2 being replaced by D
¯
k
1
D
¯
k
2.


## 2. Smooth Atomic Decomposition

In this section, smooth atomic decomposition is presented. We first give definitions of the smooth atoms forB
˙
p
s
,
q and F
˙
p
s
,
q.Definition 11.
Suppose collections of open subset{
Q
τ
1
k
1
,
v
1
⊂
X
1
:
k
1
∈
Z
,
τ
1
∈
I
k
1
,
v
1
=
1
,
…
,
N
(
k
1
,
τ
1
)
} and {
Q
τ
2
k
2
,
v
2
⊂
X
2
:
k
2
∈
Z
,
τ
2
∈
I
k
2
,
v
2
=
1
,
…
,
N
(
k
2
,
τ
2
)
} satisfying the conditions in Lemma 6. Let ϵ
=
(
ϵ
1
,
ϵ
2
) and R
=
Q
τ
1
k
1
,
v
1
×
Q
τ
2
k
2
,
v
2, and a function a
R defined on X
1
×
X
2 is said to be a ϵ-smooth atom for R if(i)
(13)
supp
⁡
a
R
⊂
B
1
(
z
τ
1
k
1
,
v
1
,
A
1
C
2
-
k
1
)
×
B
2
(
z
τ
2
k
2
,
v
2
,
A
2
C
2
-
k
2
)
,
where z
τ
i
k
i
,
v
i is the center of Q
τ
i
k
i
,
v
i for i
=
1,2;
(ii)
(14)
∫
X
2
‍
a
R
(
x
1
,
x
2
)
d
μ
2
(
x
2
)
=
∫
X
1
‍
a
R
(
x
1
,
x
2
)
d
μ
1
(
x
1
)
=
0
;
(iii)
(15)
|
a
R
(
x
,
y
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
,
(16)
|
a
R
(
x
,
y
)
-
a
R
(
x
′
,
y
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
-
ϵ
1
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
ρ
1
(
x
,
x
′
)
ϵ
1
,
(17)
|
a
R
(
x
,
y
)
-
a
R
(
x
,
y
′
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
-
ϵ
2
ρ
2
(
y
,
y
′
)
ϵ
2
,
(18)
|
a
R
(
x
,
y
)
-
a
R
(
x
′
,
y
)
-
a
R
(
x
,
y
′
)
-
a
R
(
x
′
,
y
′
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
-
ϵ
1
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
-
ϵ
2
×
ρ
1
(
x
,
x
′
)
ϵ
1
ρ
2
(
y
,
y
′
)
ϵ
2
.
Now, we also define certain spaces of sequences of indexed by dyadic rectangles{
Q
τ
1
k
1
,
v
1
×
Q
τ
2
k
2
,
v
2
} in X
1
×
X
2 which will character the coefficients in our decomposition of F
˙
p
s
,
q and B
˙
p
s
,
q. Suppose that s
=
(
s
1
,
s
2
), ϵ
=
(
ϵ
1
,
ϵ
2
) with -
ϵ
i
<
s
i
<
ϵ
i for i
=
1,2, 1
<
p, q
<
∞. Let R be as in Definition 11. Let f
˙
p
s
,
q and b
˙
p
s
,
q be the collection of sequences S
=
{
s
Q
1
×
Q
2
}
Q
1
×
Q
2
∈
{
R
} such that
(19)
∥
S
∥
f
˙
p
s
,
q
=
∥
{
∑
k
1
,
k
2
‍
∑
τ
1
,
τ
2
‍
∑
v
1
,
v
2
‍
(
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
s
1
-
1
/
2
iiiiiiiiiiiiiiiiiiiiiiiiiiiii
×
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
s
2
-
1
/
2
|
s
R
|
χ
R
)
q
∑
v
1
,
v
2
‍
(
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
s
1
-
1
/
2
}
1
/
q
∥
L
p
<
∞
,
∥
S
∥
b
˙
p
s
,
q
=
{
∑
k
1
,
k
2
‍

(
∑
τ
1
,
τ
2
‍
∑
v
1
,
v
2
‍
(
×
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
s
2
)
-
1
/
2
+
1
/
p
|
s
R
|
(
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
s
1
iiiiiiiiiiiiiiiiiiiiiiiiiiii
×
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
s
2
)
-
1
/
2
+
1
/
p
|
s
R
|
)
p
∑
τ
1
,
τ
2
‍
∑
v
1
,
v
2
‍
(
(
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
s
1
)
q
/
p
∑
k
1
,
k
2
‍

(
∑
τ
1
,
τ
2
‍
∑
v
1
,
v
2
‍
(
(
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
s
1
}
1
/
q
<
∞
.Smooth atomic decompositions forB
˙
p
s
,
q and F
˙
p
s
,
q can be stated as follows.Theorem 12.
LetR be as in Definition 11. Suppose that 1
<
p, q
<
∞, s
=
(
s
1
,
s
2
), ϵ
=
(
ϵ
1
,
ϵ
2
) with -
ϵ
i
<
s
i
<
ϵ
i for i
=
1,2.(i)
Then there exist sequenceS
=
{
s
R
}
∈
b
˙
p
s
,
q and ε-smooth atoms {
a
R
} such that
(20)
f
=
∑
k
1
,
k
2
∈
Z
‍
∑
τ
1
∈
I
k
1
‍
∑
v
1
=
1
N
(
k
1
,
τ
1
)
‍
∑
τ
2
∈
I
k
2
‍
∑
v
2
=
1
N
(
k
2
,
τ
2
)
‍
s
R
a
R
with convergence inB
˙
p
s
,
q, and
(21)
∥
S
∥
b
˙
p
s
,
q
≤
C
∥
f
∥
B
˙
p
s
,
q
.
Similarly, there exist sequenceS
=
{
s
R
}
∈
f
˙
p
s
,
q and ϵ-smooth atoms {
a
R
} such that
(22)
f
=
∑
k
1
,
k
2
∈
Z
‍
∑
τ
1
∈
I
k
1
‍
∑
v
1
=
1
N
(
k
1
,
τ
1
)
‍
∑
τ
2
∈
I
k
2
‍
∑
v
2
=
1
N
(
k
2
,
τ
2
)
‍
s
R
a
R
with convergence inF
˙
p
s
,
q and
(23)
∥
S
∥
f
˙
p
s
,
q
≤
C
∥
f
∥
F
˙
p
s
,
q
.
(ii)
Conversely, ifa
R are ϵ-smooth atoms and
(24)
f
=
∑
k
1
,
k
2
∈
Z
‍
∑
τ
1
∈
I
k
1
‍
∑
v
1
=
1
N
(
k
1
,
τ
1
)
‍
∑
τ
2
∈
I
k
2
‍
∑
v
2
=
1
N
(
k
2
,
τ
2
)
‍
s
R
a
R
,
then(25)
∥
f
∥
B
˙
p
s
,
q
≤
C
∥
S
∥
b
˙
p
s
,
q
,
∥
f
∥
F
˙
p
s
,
q
≤
C
∥
S
∥
f
˙
p
s
,
q
.Proof.
We first prove part (i). LetR be as in Definition 11. Suppose f
∈
B
˙
p
s
,
q; by Lemma 5, we have
(26)
f
(
x
,
y
)
=
∑
k
1
,
k
2
∈
Z
‍
∑
τ
1
∈
I
k
1
‍
∑
v
1
=
1
N
(
k
1
,
τ
1
)
‍
∑
τ
2
∈
I
k
2
‍
∑
v
2
=
1
N
(
k
2
,
τ
2
)
∫
R
‍
D
k
1
(
x
,
s
)
‍
×
D
k
2
(
t
,
y
)
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
d
μ
1
(
s
)
d
μ
2
(
t
)
∶
=
∑
k
1
,
k
2
∈
Z
‍
∑
τ
1
∈
I
k
1
‍
∑
v
1
=
1
N
(
k
1
,
τ
1
)
‍
∑
τ
2
∈
I
k
2
‍
∑
v
2
=
1
N
(
k
2
,
τ
2
)
‍
s
R
a
R
,
where
(27)
s
R
=
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
×
∫
R
‍
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
,
(28)
a
R
=
C
-
1
μ
1
(
Q
τ
1
k
1
,
v
1
)
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
1
/
2
×
(
∫
R
‍
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
)
-
1
×
∫
R
‍
D
k
1
(
x
,
s
)
D
k
2
(
t
,
y
)
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
×
d
μ
1
(
s
)
d
μ
2
(
t
)
.
Similar to [7, 10], we can prove the convergence of the above series converges in B
˙
p
s
,
q and (
G
∘
(
β
1
′
,
β
2
′
,
γ
1
′
,
γ
2
′
)
)
′ with β
i
<
β
i
′
<
ϵ
i and γ
i
<
γ
i
′
<
ϵ
i for i
=
1,2. We now show that a
R satisfies conditions in Definition 11. Here we only show that a
R satisfies condition (iii) and we first prove that a
R satisfies (15). By the size condition of D
k
i and μ
i
(
Q
τ
i
k
i
,
v
i
)
~
2
-
k
i for i
=
1,2, we have
(29)
|
a
R
(
x
,
y
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
1
/
2
×
(
∫
R
‍
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
)
-
1
×
∫
R
‍
2
-
k
1
ϵ
1
(
2
-
k
1
+
ρ
1
(
x
,
s
)
)
1
+
ϵ
1
2
-
k
2
ϵ
2
(
2
-
k
2
+
ρ
2
(
y
,
t
)
)
1
+
ϵ
2
iiii
i
i
×
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
1
/
2
2
k
1
2
k
2
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
.
We now verify (16); here, we only consider the case ρ
1
(
x
,
x
′
)
≤
(
1
/
2
A
)
(
2
-
k
1
+
ρ
1
(
x
,
s
)
) and similarly we estimate the case ρ
1
(
x
,
x
′
)
>
(
1
/
2
A
)
(
2
-
k
1
+
ρ
1
(
x
,
s
)
); we get
(30)
|
a
R
(
x
,
y
)
-
a
R
(
x
′
,
y
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
1
/
2
×
(
∫
R
‍
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
)
-
1
×
∫
R
‍
|
D
k
1
(
x
,
s
)
-
D
k
1
(
x
′
,
s
)
|
|
D
k
2
(
t
,
y
)
|
×
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
1
/
2
×
(
∫
R
‍
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
)
-
1
×
∫
R
‍
(
ρ
1
(
x
,
x
′
)
2
-
k
1
+
ρ
1
(
x
,
s
)
)
ϵ
1
×
2
-
k
1
ϵ
1
(
2
-
k
1
+
ρ
1
(
x
,
s
)
)
1
+
ϵ
1
×
2
-
k
2
ϵ
2
(
2
-
k
2
+
ρ
2
(
y
,
t
)
)
1
+
ϵ
2
×
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
-
ϵ
1
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
ρ
1
(
x
,
x
′
)
ϵ
1
,
which is a desired estimate. The proof of (17) is similar to that of (16) by symmetry. Here we omit the details.
We now estimate (18). If ρ
1
(
x
,
x
′
)
≤
(
1
/
2
A
)
(
2
-
k
1
+
ρ
1
(
x
,
s
)
) and ρ
2
(
y
,
y
′
)
≤
(
1
/
2
A
)
(
2
-
k
2
+
ρ
2
(
y
,
t
)
), then
(31)
|
a
R
(
x
,
y
)
-
a
R
(
x
′
,
y
)
-
a
R
(
x
,
y
′
)
-
a
R
(
x
′
,
y
′
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
1
/
2
×
(
∫
R
‍
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
)
-
1
×
∫
R
‍
|
D
k
1
(
x
,
s
)
-
D
k
1
(
x
′
,
s
)
|
iiiiiiii
×
|
D
k
2
(
t
,
y
)
-
D
k
2
(
t
,
y
′
)
|
iiiiiiii
×
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
1
/
2
×
(
∫
R
‍
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
)
-
1
×
∫
R
‍
(
ρ
1
(
x
,
x
′
)
2
-
k
1
+
ρ
1
(
x
,
s
)
)
ϵ
1
(
ρ
2
(
y
,
y
′
)
2
-
k
2
+
ρ
2
(
y
,
t
)
)
ϵ
2
iiiiiiii
×
2
-
k
1
ϵ
1
(
2
-
k
1
+
ρ
1
(
x
,
s
)
)
1
+
ϵ
1
iiiiiiii
×
2
-
k
2
ϵ
2
(
2
-
k
2
+
ρ
2
(
y
,
t
)
)
1
+
ϵ
2
iiiiiiii
×
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
-
ϵ
1
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
-
ϵ
2
×
ρ
1
(
x
,
x
′
)
ϵ
1
ρ
2
(
y
,
y
′
)
ϵ
2
,
which is the desired result. Similarly we can consider another three cases.
Then, by Hölder’s inequality and Remark10, we obtain
(32)
∥
S
∥
b
˙
p
s
,
q
=
{
∑
k
1
,
k
2
∈
Z
‍
(
∑
τ
1
∈
I
k
1
‍
∑
v
1
=
1
N
(
k
1
,
τ
1
)
‍
∑
τ
2
∈
I
k
2
‍
∑
v
2
=
1
N
(
k
2
,
τ
2
)
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
s
1
-
1
/
2
+
1
/
p
‍
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
(
∫
R
‍
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
×
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
s
2
-
1
/
2
+
1
/
p
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∫
R
‍
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
)
p
∑
k
1
,
k
2
∈
Z
‍
)
q
/
p
∑
k
1
,
k
2
∈
Z
‍
}
1
/
q
≤
C
{
∑
k
1
,
k
2
∈
Z
‍
(
∑
τ
1
∈
I
k
1
‍
∑
v
1
=
1
N
(
k
1
,
τ
1
)
‍
∑
τ
2
∈
I
k
2
‍
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
∑
v
2
=
1
N
(
k
2
,
τ
2
)
‍
2
k
1
s
1
p
2
k
2
s
2
p
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
p
/
p
′
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
p
/
p
′
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
(
∫
R
‍
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
d
μ
1
(
s
)
d
μ
2
(
t
)
∫
R
‍
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
)
p
∑
k
1
,
k
2
∈
Z
‍
)
q
/
p
}
1
/
q
≤
C
{
∑
k
1
,
k
2
∈
Z
‍
(
2
k
1
s
1
2
k
2
s
2
∥
D
¯
k
1
D
¯
k
2
(
f
)
∥
p
)
q
}
1
/
q
≤
C
∥
f
∥
B
˙
p
s
,
q
.
The proof of the casef
∈
F
˙
p
s
,
q goes by the analogous argument to the case that f
∈
B
˙
p
s
,
q, but using the vector-valued maximal inequality. Indeed, suppose that f
∈
F
˙
p
s
,
q; as in the proof of B
˙
p
s
,
q, we have
(33)
f
=
∑
k
1
,
k
2
∈
Z
‍
∑
τ
1
∈
I
k
1
‍
∑
v
1
=
1
N
(
k
1
,
τ
1
)
‍
∑
τ
2
∈
I
k
2
‍
∑
v
2
=
1
N
(
k
2
,
τ
2
)
‍
s
R
a
R
,
where s
R and a
R are as in (27) and (28), respectively. From [7, 10], we can also similarly obtain the series converges in the norm of F
˙
p
s
,
q and in (
G
°
(
β
1
′
,
β
2
′
,
γ
1
′
,
γ
2
′
)
)
′ with β
i
<
β
i
′
<
ϵ
i and γ
i
<
γ
i
′
<
ϵ
i for i
=
1,2. To show the conclusion, using Fefferman-Stein vector-valued maximal inequality and Remark 10, we have
(34)
∥
S
∥
f
˙
p
s
,
q
=
∥
{
∑
k
1
,
k
2
∈
Z
‍
∑
τ
1
∈
I
k
1
‍
∑
v
1
=
1
N
(
k
1
,
τ
1
)
‍
∑
τ
2
∈
I
k
2
‍
∑
v
2
=
1
N
(
k
2
,
τ
2
)
(
∫
R
‍
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
s
1
-
1
/
2
‍
×
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
s
2
-
1
/
2
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
×
∫
R
‍
|
D
¯
k
1
D
¯
k
2
(
f
)
(
s
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
χ
Q
τ
1
k
1
,
v
1
χ
Q
τ
2
k
2
,
v
2
)
q
∑
k
1
,
k
2
∈
Z
‍
}
1
/
q
∥
p
≤
C
∥
{
∑
k
1
,
k
2
∈
Z
‍
(
2
k
1
s
1
2
k
2
s
2
M
2
(
M
1
(
D
¯
k
1
D
¯
k
2
(
f
)
)
)
)
q
}
1
/
q
∥
p
≤
C
∥
{
∑
k
1
,
k
2
∈
Z
‍
(
2
k
1
s
1
2
k
2
s
2
D
¯
k
1
D
¯
k
2
(
f
)
)
q
}
1
/
q
∥
p
≤
C
∥
f
∥
F
˙
p
s
,
q
.
To prove part (ii), we need the discrete version of the Hardy-Littlewood maximal function estimate on one single factorX, which is an analogue to a result of [9]. Also see [10].
Lemma  13.Suppose that
X
is a space of homogeneous type as in Definition1 Let
0
<
a
≤
r
<
∞
and
λ
>
r
/
a
. Fix
μ
,
η
with
μ
,
η
∈
Z
and for any
x
∈
X
(35)
(
∑
τ
∈
I
μ
‍
∑
v
=
1
N
(
μ
,
τ
)
‍
|
s
Q
τ
μ
,
v
|
r
(
1
+
min
⁡
{
2
η
,
2
μ
}
ρ
(
x
,
z
τ
μ
,
v
)
)
-
λ
)
1
/
r
≤
C
2
(
μ
-
η
)
+
/
a
(
M
(
∑
τ
∈
I
μ
‍
∑
v
=
1
N
(
μ
,
τ
)
‍
|
s
Q
τ
μ
,
v
|
a
χ
Q
τ
μ
,
v
)
(
x
)
)
1
/
a
,
where
z
τ
μ
,
v
is the center of dyadic cube
Q
τ
μ
,
v
and
C
depends only on
λ
,
r
,
a
.
We now turn to proof of part (ii). In this part, we denoteR
=
Q
τ
1
μ
1
,
v
1
×
Q
τ
2
μ
2
,
v
2. Suppose that {
D
k
i
}
k
i
∈
Z for i
=
1,2 is as in Lemma 5 and a
R is an ε-smooth atom. For D
k
1
D
k
2
(
a
R
)
(
x
,
y
), we consider it by four cases. We first estimate the case k
1
>
μ
1
,
k
2
>
μ
2. In this case, applying the cancellation conditions of D
k
1 and D
k
2,
(36)
|
D
k
1
D
k
2
(
a
R
)
(
x
,
y
)
|
=
|
∫
X
1
×
X
2
‍
D
k
1
(
x
,
s
)
D
k
2
(
y
,
t
)
m
m
m
m
m
m
×
[
a
R
(
s
,
t
)
-
a
R
(
x
,
t
)
-
a
R
(
s
,
y
)
+
a
R
(
x
,
y
)
]
iiiiiiii
iii
iiiiiiii
×
d
μ
1
(
s
)
d
μ
2
(
t
)
∫
X
1
×
X
2
‍
|
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
-
ϵ
1
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
-
ϵ
2
×
∫
ρ
1
(
s
,
x
)
≤
2
C
2
-
k
1
ρ
2
(
t
,
y
)
≤
2
C
2
-
k
2
2
-
k
1
ϵ
1
(
2
-
k
1
+
ρ
1
(
x
,
s
)
)
1
+
ϵ
1
iiiiiiii
×
2
-
k
2
ϵ
2
(
2
-
k
2
+
ρ
2
(
t
,
y
)
)
1
+
ϵ
2
ρ
1
(
x
,
s
)
ϵ
1
mmii
×
ρ
2
(
t
,
y
)
ϵ
2
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
-
ϵ
1
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
-
ϵ
2
×
∫
ρ
1
(
s
,
x
)
≤
2
C
2
-
k
1
ρ
2
(
t
,
y
)
≤
2
C
2
-
k
2
d
μ
1
(
s
)
d
μ
2
(
t
)
iiiiiiii
×
2
k
1
2
-
k
1
ϵ
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
ϵ
1
iiiiiiii
×
2
k
2
2
-
k
2
ϵ
2
(
1
+
2
μ
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
ϵ
2
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
k
1
-
μ
1
)
ϵ
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
ϵ
1
×
2
-
(
k
2
-
μ
2
)
ϵ
2
(
1
+
2
μ
2
ρ
2
(
x
,
z
τ
2
μ
2
,
v
2
)
)
1
+
ϵ
2
.
To estimate the casek
1
>
μ
1, k
2
≤
μ
2, we only estimate ρ
2
(
t
,
z
τ
2
u
2
,
v
2
)
≤
(
1
/
2
A
)
(
2
-
k
2
+
ρ
2
(
y
,
t
)
), and afterwards we can consider the case ρ
2
(
t
,
z
τ
2
u
2
,
v
2
)
>
(
1
/
2
A
)
(
2
-
k
2
+
ρ
2
(
y
,
t
)
) similarly. In this case, we have
(37)
|
D
k
1
D
k
2
(
a
R
)
(
x
,
y
)
|
=
|
∫
X
1
×
X
2
‍
D
k
1
(
x
,
s
)
[
D
k
2
(
y
,
t
)
-
D
k
2
(
y
,
z
τ
2
u
2
,
v
2
)
]
i
i
i
i
×
[
a
R
(
s
,
t
)
-
a
R
(
x
,
t
)
]
d
μ
1
(
s
)
d
μ
2
(
t
)
∫
X
1
×
X
2
‍
|
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
-
ϵ
1
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
∫
ρ
1
(
s
,
x
)
≤
2
C
2
-
k
1
ρ
2
(
t
,
z
τ
2
u
2
,
v
2
)
≤
2
C
2
-
μ
2
2
-
k
1
ϵ
1
(
2
-
k
1
+
ρ
1
(
x
,
s
)
)
1
+
ϵ
1
i
i
i
i
i
i
i
i
i
×
(
ρ
2
(
t
,
z
τ
2
u
2
,
v
2
)
2
-
k
2
+
ρ
2
(
t
,
y
)
)
ϵ
2
2
-
k
2
ϵ
2
(
2
-
k
2
+
ρ
2
(
t
,
y
)
)
1
+
ϵ
2
i
i
i
i
i
i
i
i
i
×
ρ
1
(
x
,
s
)
ϵ
1
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
-
ϵ
1
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
∫
ρ
1
(
s
,
x
)
≤
2
C
2
-
k
1
ρ
2
(
t
,
z
τ
2
u
2
,
v
2
)
≤
2
C
2
-
μ
2
2
k
1
2
-
k
1
ϵ
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
ϵ
1
i
i
i
i
i
i
i
i
i
×
2
-
μ
2
ϵ
2
2
k
2
2
k
2
ϵ
2
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
ϵ
2
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
-
ϵ
1
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
∫
ρ
1
(
s
,
x
)
≤
2
C
2
-
k
1
ρ
2
(
t
,
z
τ
2
u
2
,
v
2
)
≤
2
C
2
-
μ
2
d
μ
1
(
s
)
d
μ
2
(
t
)
×
2
k
1
2
-
k
1
ϵ
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
ϵ
1
i
i
i
i
i
i
×
2
-
μ
2
ϵ
2
2
k
2
2
-
k
2
ϵ
2
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
ϵ
2
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
i
i
i
i
i
i
×
2
-
(
k
1
-
μ
1
)
ϵ
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
ϵ
1
i
i
i
i
i
i
×
2
-
(
μ
2
-
k
2
)
(
1
+
ϵ
2
)
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
ϵ
2
.
The estimate of the casek
1
≤
μ
1, k
2
>
μ
2 is similar to the case k
1
>
μ
1, k
2
≤
μ
2 by symmetry and we obtain
(38)
|
D
k
1
D
k
2
(
a
R
)
(
x
,
y
)
|
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
μ
1
-
k
1
)
(
1
+
ϵ
1
)
(
1
+
2
k
1
ρ
1
(
y
,
z
τ
1
μ
1
,
v
1
)
)
1
+
ϵ
1
2
-
(
k
2
-
μ
2
)
ϵ
2
(
1
+
2
μ
2
ρ
2
(
x
,
z
τ
2
μ
2
,
v
2
)
)
1
+
ϵ
2
.
Fork
1
≤
μ
1
,
k
2
≤
μ
2, when ρ
1
(
s
,
z
τ
1
u
1
,
v
1
)
≤
(
1
/
2
A
)
(
2
-
k
1
+
ρ
1
(
s
,
x
)
) and ρ
2
(
t
,
z
τ
2
u
2
,
v
2
)
≤
(
1
/
2
A
)
(
2
-
k
2
+
ρ
2
(
y
,
t
)
), then
(39)
|
D
k
1
D
k
2
(
a
R
)
(
x
,
y
)
|
=
|
∫
X
1
×
X
2
‍
[
D
k
1
(
x
,
s
)
-
D
k
1
(
x
,
z
τ
1
u
1
,
v
1
)
]
m
m
m
m
m
m
i
×
[
D
k
2
(
y
,
t
)
-
D
k
2
(
y
,
z
τ
2
u
2
,
v
2
)
]
m
m
m
m
m
m
i
×
a
R
(
s
,
t
)
d
μ
1
(
s
)
d
μ
2
(
t
)
∫
X
1
×
X
2
‍
|
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
∫
ρ
1
(
s
,
z
τ
1
u
1
,
v
1
)
≤
2
C
2
-
μ
1
ρ
2
(
t
,
z
τ
2
u
2
,
v
2
)
≤
2
C
2
-
μ
2
(
ρ
1
(
s
,
z
τ
1
u
1
,
v
1
)
2
-
k
1
+
ρ
1
(
x
,
s
)
)
ϵ
1
iiiiiiiiii
×
2
-
k
1
ϵ
1
(
2
-
k
1
+
ρ
1
(
x
,
s
)
)
1
+
ϵ
1
(
ρ
2
(
t
,
z
τ
2
u
2
,
v
2
)
2
-
k
2
+
ρ
2
(
t
,
y
)
)
ϵ
2
iiiiiiiiii
×
2
-
k
2
ϵ
2
(
2
-
k
2
+
ρ
2
(
t
,
y
)
)
1
+
ϵ
2
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
∫
ρ
1
(
s
,
z
τ
1
u
1
,
v
1
)
≤
2
C
2
-
μ
1
ρ
2
(
t
,
z
τ
2
u
2
,
v
2
)
≤
2
C
2
-
μ
2
2
-
μ
1
ϵ
1
(
2
-
k
1
+
ρ
1
(
x
,
s
)
)
ϵ
1
iiiiiiiiii
×
2
k
1
(
1
+
2
k
1
ρ
1
(
x
,
s
)
)
1
+
ϵ
1
iiiiiiiiii
×
2
-
μ
2
ϵ
2
(
2
-
k
2
+
ρ
2
(
t
,
y
)
)
ϵ
2
iiiiiiiiii
×
2
k
2
(
1
+
2
k
2
ρ
2
(
t
,
y
)
)
1
+
ϵ
2
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
μ
1
-
k
1
)
(
1
+
ϵ
1
)
(
1
+
2
k
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
ϵ
1
×
2
-
(
μ
2
-
k
2
)
(
1
+
ϵ
2
)
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
ϵ
2
.
We can similarly get the same result for another three cases; here we omit the details.
Thus, iff
=
∑
μ
1
,
μ
2
∈
Z

‍
∑
τ
1
∈
I
μ
1

‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍

∑
τ
2
∈
I
μ
2
‍

∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
s
R
a
R, by Lemma 13 and above estimates, then
(40)
{
∑
k
1
,
k
2
∈
Z
‍
(
2
k
1
s
1
2
k
2
s
2
∥
D
k
1
D
k
2
(
f
)
∥
p
)
q
}
1
/
q
=
{
∑
k
1
,
k
2
∈
Z
‍
(
2
k
1
s
1
2
k
2
s
2
∥
D
k
1
D
k
2
(
∑
μ
1
,
μ
2
‍
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
s
R
a
R
)
∥
p
)
q
}
1
/
q
≤
C
{
∥
M
2
(
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
)
∥
p
)
q
∑
k
1
,
k
2
∈
Z
‍
(
∥
M
2
(
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
)
∥
p
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
-
∞
k
2
‍
2
k
1
s
1
2
k
2
s
2
2
μ
1
/
2
2
μ
2
/
2
2
(
μ
1
-
k
1
)
ϵ
1
2
(
μ
2
-
k
2
)
ϵ
2
mmm
×
∥
M
2
(
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
)
∥
p
)
q
}
1
/
q
+
C
{
×
∥
2
(
μ
2
-
k
2
)
M
2
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
∥
p
)
q
∑
k
1
,
k
2
∈
Z
‍
(
∥
2
(
μ
2
-
k
2
)
M
2
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
∥
p
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
k
2
+
1
∞
‍
2
k
1
s
1
2
k
2
s
2
2
μ
1
/
2
2
μ
2
/
2
2
(
μ
1
-
k
1
)
ϵ
1
2
-
(
μ
2
-
k
2
)
2
(
1
+
ϵ
2
)
mmm
i
×
∥
2
(
μ
2
-
k
2
)
M
2
(
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
)
∥
p
)
q
}
1
/
q
+
C
{
∑
k
1
,
k
2
∈
Z
‍
(
∥
2
(
μ
1
-
k
1
)
M
2
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
∥
p
∥
2
(
μ
1
-
k
1
)
M
2
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
∥
p
∑
μ
1
=
k
1
+
1
∞
‍
∑
μ
2
=
-
∞
k
2
‍
2
k
1
s
1
2
k
2
s
2
2
μ
1
/
2
2
μ
2
/
2
2
-
(
μ
1
-
k
1
)
(
1
+
ϵ
1
)
2
(
μ
2
-
k
2
)
ϵ
2
mmm
i
×
∥
2
(
μ
1
-
k
1
)
M
2
(
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
)
∥
p
)
q
}
1
/
q
+
C
{
∑
k
1
,
k
2
∈
Z
‍
(
∥
2
(
μ
1
-
k
1
)
2
(
μ
2
-
k
2
)
M
2
(
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
)
∥
p
∑
μ
1
=
k
1
+
1
∞
‍
∑
μ
2
=
k
2
+
1
∞
‍
2
k
1
s
1
2
k
2
s
2
2
μ
1
/
2
2
μ
2
/
2
2
-
(
μ
1
-
k
1
)
(
1
+
ϵ
1
)
2
-
(
μ
2
-
k
2
)
(
1
+
ϵ
2
)
mmmmm
×
∥
2
(
μ
1
-
k
1
)
2
(
μ
2
-
k
2
)
M
2
(
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
χ
Q
τ
2
μ
2
,
v
2
)
∥
p
)
q
}
1
/
q
∶
=
I
1
+
I
2
+
I
3
+
I
4
.
Let A
μ
=
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
|
s
R
|
p
χ
R
)
1
/
p. We now consider I
1. By Hölder’s inequality and noting that k
i
≥
μ
i, ϵ
i
>
s
i for i
=
1,2, we have(41)
I
1
≤
C
{
∑
k
1
,
k
2
∈
Z
‍
(
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
-
∞
k
2
‍
2
k
1
s
1
2
k
2
s
2
2
μ
1
/
2
2
μ
2
/
2
2
(
μ
1
-
k
1
)
ϵ
1
2
(
μ
2
-
k
2
)
ϵ
2
A
μ
2
-
μ
1
/
p
2
-
μ
2
/
p
)
q
}
1
/
q
≤
C
{
∑
k
1
,
k
2
∈
Z
‍
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
-
∞
k
2
‍
2
-
(
k
1
-
μ
1
)
(
ϵ
1
-
s
1
)
2
-
(
k
2
-
μ
2
)
(
ϵ
2
-
s
2
)
(
2
μ
1
(
1
/
2
+
s
1
-
1
/
p
)
2
μ
2
(
1
/
2
+
s
2
-
1
/
p
)
A
μ
)
q
}
1
/
q
≤
C
{
∑
μ
1
,
μ
2
∈
Z
‍
(
2
μ
1
(
1
/
2
+
s
1
-
1
/
p
)
2
μ
2
(
1
/
2
+
s
2
-
1
/
p
)
A
μ
)
q
}
1
/
q
≤
C
∥
S
∥
b
˙
p
s
,
q
.Similar to the proof of I
1, we get desired results for I
2, I
3, and I
4.
To show the second inequality of (25), let a
(
x
,
y
)
=
{
a
μ
1
,
μ
2
(
x
,
y
)
}
μ
1
,
μ
2
∈
Z with
(42)
a
μ
1
,
μ
2
(
x
,
y
)
=
M
2
(
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
-
s
2
k
k
k
k
k
k
×
M
1
(
∑
τ
1
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
-
s
1
|
s
R
|
χ
Q
τ
1
μ
1
,
v
1
)
k
k
k
k
k
k
∑
τ
2
∈
I
μ
2
‍
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
×
(
x
)
χ
Q
τ
2
μ
2
,
v
2
)
(
y
)
.
Applying Lemma 13 and Hölder’s inequality, we get
(43)
{
∑
k
1
,
k
2
∈
Z
‍
(
2
k
1
s
1
2
k
2
s
2
|
D
k
1
D
k
2
(
f
)
|
)
q
}
1
/
q
≤
C
{
∑
k
1
,
k
2
∈
Z
‍
(
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
-
∞
k
2
‍
2
-
(
k
1
-
μ
1
)
(
ϵ
1
-
s
1
)
2
-
(
k
2
-
μ
2
)
(
ϵ
2
-
s
2
)
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
×
a
μ
1
,
μ
2
(
x
,
y
)
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
-
∞
k
2
‍
)
q
}
1
/
q
+
C
{
∑
k
1
,
k
2
∈
Z
‍
(
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
k
2
+
1
∞
‍
2
-
(
k
1
-
μ
1
)
(
ϵ
1
-
s
1
)
2
(
k
2
-
μ
2
)
(
ϵ
2
+
s
2
)
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
×
a
μ
1
,
μ
2
(
x
,
y
)
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
-
∞
k
2
‍
)
q
}
1
/
q
+
C
{
∑
k
1
,
k
2
∈
Z
‍
(
∑
μ
1
=
k
1
+
1
∞
‍
∑
μ
2
=
-
∞
k
2
‍
2
(
k
1
-
μ
1
)
(
ϵ
1
+
s
1
)
2
-
(
k
2
-
μ
2
)
(
ϵ
2
-
s
2
)
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
×
a
μ
1
,
μ
2
(
x
,
y
)
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
-
∞
k
2
‍
)
q
}
1
/
q
+
C
{
∑
k
1
,
k
2
∈
Z
‍
(
∑
μ
1
=
k
1
+
1
∞
‍
∑
μ
2
=
k
2
+
1
∞
‍
2
(
k
1
-
μ
1
)
(
ϵ
1
+
s
1
)
2
(
k
2
-
μ
2
)
(
ϵ
2
+
s
2
)
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
m
m
×
a
μ
1
,
μ
2
(
x
,
y
)
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
-
∞
k
2
‍
)
q
}
1
/
q
≤
C
{
∑
μ
1
,
μ
2
∈
Z
‍
∑
k
1
=
μ
1
∞
‍
∑
k
2
=
μ
2
∞
‍
2
-
(
k
1
-
μ
1
)
(
ϵ
1
-
s
1
)
2
-
(
k
2
-
μ
2
)
(
ϵ
2
-
s
2
)
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
×
a
μ
1
,
μ
2
(
x
,
y
)
q
∑
μ
1
,
μ
2
∈
Z
‍
∑
μ
1
=
-
∞
k
1
‍
∑
μ
2
=
-
∞
k
2
‍
}
1
/
q
+
C
{
∑
μ
1
,
μ
2
∈
Z
‍
∑
k
1
=
μ
1
∞
‍
∑
k
2
=
-
∞
μ
2
‍
2
-
(
k
1
-
μ
1
)
(
ϵ
1
-
s
1
)
2
(
k
2
-
μ
2
)
(
ϵ
2
+
s
2
)
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
×
a
μ
1
,
μ
2
(
x
,
y
)
q
∑
μ
1
,
μ
2
∈
Z
‍
∑
k
1
=
μ
1
∞
‍
}
1
/
q
+
C
{
∑
μ
1
,
μ
2
∈
Z
‍
∑
k
1
=
-
∞
μ
1
‍
∑
k
2
=
μ
2
∞
‍
2
(
k
1
-
μ
1
)
(
ϵ
1
+
s
1
)
2
-
(
k
2
-
μ
2
)
(
ϵ
2
-
s
2
)
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
×
a
μ
1
,
μ
2
(
x
,
y
)
q
∑
μ
1
,
μ
2
∈
Z
‍
}
1
/
q
+
C
{
∑
μ
1
,
μ
2
∈
Z
‍
∑
k
1
=
-
∞
μ
1
‍
∑
k
2
=
-
∞
μ
2
‍
2
(
k
1
-
μ
1
)
(
ϵ
1
+
s
1
)
2
(
k
2
-
μ
2
)
(
ϵ
2
+
s
2
)
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
×
a
μ
1
,
μ
2
(
x
,
y
)
q
∑
μ
1
,
μ
2
∈
Z
‍
}
1
/
q
≤
C
{
∑
μ
1
,
μ
2
∈
Z
‍
a
μ
1
,
μ
2
(
x
,
y
)
q
}
1
/
q
.
Hence, by Fefferman-Stein vector-valued maximal inequality, we obtain
(44)
∥
f
∥
F
˙
p
s
,
q
=
∥
{
∑
k
1
,
k
2
∈
Z
‍
(
2
k
1
s
1
2
k
2
s
2
|
D
k
1
D
k
2
(
f
)
|
)
q
}
1
/
q
∥
p
≤
C
∥
{
∑
μ
1
,
μ
2
∈
Z
‍
(
∑
τ
2
∈
I
μ
1
‍
∑
v
1
=
1
N
(
μ
1
,
τ
1
)
‍
∑
τ
2
∈
I
μ
2
∑
v
2
=
1
N
(
μ
2
,
τ
2
)
‍
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
-
s
1
‍
i
i
i
i
i
i
i
i
×
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
-
s
2
|
s
R
|
χ
R
)
q
}
1
/
q
∥
p
≤
C
∥
S
∥
f
˙
p
s
,
q
,
which completes the proof of Theorem 12.


## 3. Smooth Molecular Decomposition

In this section, we show smooth molecular decomposition of Triebel-Lizorkin and Besov spaces. We first give definitions of the smooth molecules as follows.Definition 14.
Suppose that collections of open subset{
Q
τ
1
k
1
,
v
1
⊂
X
1
:
k
1
∈
Z
,
τ
1
∈
I
k
1
,
v
1
=
1
,
…
,
N
(
k
1
,
τ
1
)
} and {
Q
τ
2
k
2
,
v
2
⊂
X
2
:
k
2
∈
Z
,
τ
2
∈
I
k
2
,
v
2
=
1
,
…
,
N
(
k
2
,
τ
2
)
} satisfy the conditions in Lemma 5. Let β
=
(
β
1
,
β
2
), γ
=
(
γ
1
,
γ
2
) and R
=
Q
τ
1
k
1
,
v
1
×
Q
τ
2
k
2
,
v
2, and z
τ
1
k
1
,
v
1
,
z
τ
1
k
1
,
v
1 are the centers of Q
τ
1
k
1
,
v
1
,
Q
τ
2
k
2
,
v
2, respectively. A function m
R defined on X
1
×
X
2 is said to be a (
β
,
γ
)-smooth molecule for R if
(i)
(45)
∫
X
2
‍
m
R
(
x
1
,
x
2
)
d
μ
2
(
x
2
)
=
∫
X
1
‍
m
R
(
x
1
,
x
2
)
d
μ
1
(
x
1
)
=
0
;
(ii)
(46)
|
m
R
(
x
,
y
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
×
(
1
+
2
-
k
1
ρ
1
(
x
,
z
τ
1
k
1
,
v
1
)
)
-
(
1
+
γ
1
)
×
(
1
+
2
-
k
2
ρ
2
(
y
,
z
τ
2
k
2
,
v
2
)
)
-
(
1
+
γ
2
)
,
|
m
R
(
x
,
y
)
-
m
R
(
x
′
,
y
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
-
β
1
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
×
ρ
1
(
x
,
x
′
)
β
1
(
1
+
2
-
k
2
ρ
2
(
y
,
z
τ
2
k
2
,
v
2
)
)
(
-
1
+
γ
2
)
×
{
(
1
+
2
-
k
1
ρ
1
(
x
,
z
τ
1
k
1
,
v
1
)
)
-
(
1
+
γ
1
)
i
i
i
i
i
i
+
(
1
+
2
-
k
1
ρ
1
(
x
′
,
z
τ
1
k
1
,
v
1
)
)
-
(
1
+
γ
1
)
}
,
|
m
R
(
x
,
y
)
-
m
R
(
x
,
y
′
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
-
β
2
×
ρ
2
(
y
,
y
′
)
β
2
(
1
+
2
-
k
1
ρ
1
(
x
,
z
τ
1
k
1
,
v
1
)
)
-
(
1
+
γ
1
)
×
{
(
1
+
2
-
k
2
ρ
2
(
y
,
z
τ
2
k
2
,
v
2
)
)
-
(
1
+
γ
2
)
i
i
i
i
+
(
1
+
2
-
k
2
ρ
2
(
y
′
,
z
τ
2
k
2
,
v
2
)
)
-
(
1
+
γ
2
)
}
,
|
m
R
(
x
,
y
)
-
m
R
(
x
′
,
y
)
-
m
R
(
x
,
y
′
)
-
m
R
(
x
′
,
y
′
)
|
≤
C
μ
1
(
Q
τ
1
k
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
k
2
,
v
2
)
-
1
/
2
×
ρ
1
(
x
,
x
′
)
β
1
ρ
2
(
y
,
y
′
)
β
2
×
{
(
1
+
2
-
k
1
ρ
1
(
x
,
z
τ
1
k
1
,
v
1
)
)
-
(
1
+
γ
1
)
i
i
i
i
i
+
(
1
+
2
-
k
1
ρ
1
(
x
′
,
z
τ
1
k
1
,
v
1
)
)
-
(
1
+
γ
1
)
}
×
{
(
1
+
2
-
k
2
ρ
2
(
y
,
z
τ
2
k
2
,
v
2
)
)
(
-
1
+
γ
2
)
i
i
i
i
i
+
(
1
+
2
-
k
2
ρ
2
(
y
′
,
z
τ
2
k
2
,
v
2
)
)
-
(
1
+
γ
2
)
}
.
Now we can state the smooth molecular decomposition for Triebel-Lizorkin and Besov spaces.Theorem 15.
LetR be as in Definition 11. Suppose that for i
=
1,2, s
=
(
s
1
,
s
2
), ϵ
=
(
ϵ
1
,
ϵ
2
), β
=
(
β
1
,
β
2
), γ
=
(
γ
1
,
γ
2
), -
ϵ
i
<
s
i
<
ϵ
i, 1
<
p, q
<
∞, m
R is a (
β
,
γ
)-smooth molecule with max
⁡
⁡
(
0
,
s
i
)
<
β
i
<
ϵ
i and max
⁡
⁡
(
0
,
-
s
i
)
<
γ
i
<
ϵ
i for i
=
1,2. If
(47)
f
=
∑
k
1
,
k
2
∈
Z
‍
∑
τ
1
∈
I
k
1
‍
∑
v
1
=
1
N
(
k
1
,
τ
1
)
‍
∑
τ
2
∈
I
k
2
‍
∑
v
2
=
1
N
(
k
2
,
τ
2
)
‍
s
R
m
R
,
then ∥
f
∥
B
˙
p
s
,
q
≤
C
∥
S
∥
b
˙
p
s
,
q and ∥
f
∥
F
˙
p
s
,
q
≤
C
∥
S
∥
f
˙
p
s
,
q.Proof.
Suppose thatR
=
Q
τ
1
μ
1
,
v
1
×
Q
τ
2
μ
2
,
v
2. From the proof of Theorem 12, we only need to claim the following estimates. If k
1
≤
μ
1, k
1
≤
μ
2, then
(48)
|
D
k
1
D
k
2
(
m
R
)
(
x
,
y
)
|
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
μ
1
-
k
1
)
(
1
+
γ
1
)
(
1
+
2
k
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
2
-
(
μ
2
-
k
2
)
(
1
+
γ
2
)
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
.
If k
1
>
μ
1, k
1
≤
μ
2, then
(49)
|
D
k
1
D
k
2
(
m
R
)
(
x
,
y
)
|
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
k
1
-
μ
1
)
β
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
2
-
(
μ
2
-
k
2
)
(
1
+
γ
2
)
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
.
If k
1
≤
μ
1, k
1
>
μ
2, then
(50)
|
D
k
1
D
k
2
(
m
R
)
(
x
,
y
)
|
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
μ
1
-
k
1
)
(
1
+
γ
1
)
(
1
+
2
k
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
2
-
(
k
2
-
μ
2
)
β
2
(
1
+
2
μ
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
.
If k
1
>
μ
1, k
1
>
μ
2, then
(51)
|
D
k
1
D
k
2
(
m
R
)
(
x
,
y
)
|
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
k
1
-
μ
1
)
β
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
2
-
(
k
2
-
μ
2
)
β
2
(
1
+
2
μ
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
.
We now prove the above estimates. LetW
1
=
{
x
:
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
≤
4
C
1
2
-
k
1
}, W
2
=
{
y
:
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
≤
4
C
2
2
-
k
2
}, W
3
=
{
x
:
ρ
1
(
x
,
s
)
≤
4
C
1
2
-
k
1
}, and W
4
=
{
y
:
ρ
2
(
y
,
t
)
≤
4
C
2
2
-
k
2
}, where 2
C
i is the smallest constant which satisfies supp D
k
i
(
x
,
y
)
⊂
{
(
x
,
y
)
∈
X
i
×
X
i
:
ρ
i
(
x
,
y
)
≤
2
C
i
2
-
k
i
} for i
=
1,2, and W
1
C, W
2
C, W
3
C, and W
4
C are their complements, respectively. We first give the proof of (48). Note that k
1
≤
μ
1, k
2
≤
μ
2, and then
(52)
|
D
k
1
D
k
2
(
m
R
)
(
x
,
y
)
|
≤
∫
X
1
×
X
2
‍
|
D
k
1
(
x
,
s
)
-
D
k
1
(
x
,
z
τ
1
μ
1
,
v
1
)
|
m
×
|
D
k
2
(
y
,
t
)
-
D
k
2
(
y
,
z
τ
2
μ
2
,
v
2
)
|
m
×
|
m
R
(
s
,
t
)
|
d
μ
1
(
s
)
μ
2
(
t
)
≤
(
χ
W
1
χ
W
2
∫
W
3
×
W
4
‍
+
χ
W
1
C
χ
W
2
∫
W
3
×
W
4
‍
m
+
χ
W
1
χ
W
2
C
∫
W
3
×
W
4
‍
+
χ
W
1
χ
W
2
∫
W
3
C
×
W
4
‍
m
+
χ
W
1
χ
W
2
∫
W
3
×
W
4
C
‍
+
χ
W
1
C
χ
W
2
C
∫
W
3
×
W
4
‍
m
+
χ
W
1
C
χ
W
2
∫
W
3
C
×
W
4
‍
+
χ
W
1
C
χ
W
2
∫
W
3
×
W
4
C
‍
m
+
χ
W
1
χ
W
2
C
∫
W
3
C
×
W
4
‍
+
χ
W
1
χ
W
2
C
∫
W
3
×
W
4
C
‍
m
+
χ
W
1
χ
W
2
∫
W
3
C
×
W
4
C
‍
+
χ
W
1
C
χ
W
2
C
∫
W
3
C
×
W
4
‍
m
+
χ
W
1
C
χ
W
2
C
∫
W
3
×
W
4
C
‍
+
χ
W
1
C
χ
W
2
∫
W
3
C
×
W
4
C
‍
m
+
χ
W
1
χ
W
2
C
∫
W
3
C
×
W
4
C
‍
+
χ
W
1
C
χ
W
2
C
∫
W
3
C
×
W
4
C
‍
)
×
|
D
k
1
(
x
,
s
)
-
D
k
1
(
x
,
z
τ
1
μ
1
,
v
1
)
|
×
|
D
k
2
(
y
,
t
)
-
D
k
2
(
y
,
z
τ
2
μ
2
,
v
2
)
|
×
|
m
R
(
s
,
t
)
|
d
μ
1
(
s
)
μ
2
(
t
)
:
=
(
48
)
1
+
(
48
)
2
+
⋯
+
(
48
)
16
.
By the support ofD
k
i and the definition of W
j for i
=
1,2 and j
=
1,2
,
3,4, we can easily get (48)7 = (48)10 = (48)12 = (48)13 = (48)14 = (48)15 = (48)16 = 0. Here we only estimate the first five terms, and we can similarly estimate other terms. Now, we first prove (48)1:
(53)
|
(
48
)
1
|
≤
C
2
k
1
(
1
+
ϵ
1
)
2
k
2
(
1
+
ϵ
2
)
×
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
χ
W
1
χ
W
2
×
∫
W
3
×
W
4
‍
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
ϵ
1
(
1
+
2
μ
1
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
)
(
1
+
γ
1
)
iiiiii
×
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
ϵ
2
(
1
+
2
μ
2
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
)
(
1
+
γ
2
)
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
k
1
(
1
+
ϵ
1
)
2
k
2
(
1
+
ϵ
2
)
2
-
μ
1
(
1
+
γ
1
)
2
μ
2
(
1
+
γ
2
)
χ
W
1
χ
W
2
×
∫
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
≤
8
C
1
2
-
k
1
‍
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
ϵ
1
-
γ
1
-
1
d
μ
1
(
s
)
×
∫
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
≤
8
C
2
2
-
k
2
‍
ρ
1
(
t
,
z
τ
2
μ
2
,
v
2
)
ϵ
2
-
γ
2
-
1
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
μ
1
-
k
1
)
(
1
+
γ
1
)
2
-
(
μ
2
-
k
2
)
(
1
+
γ
2
)
χ
W
1
χ
W
2
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
μ
1
-
k
1
)
(
1
+
γ
1
)
(
1
+
2
k
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
×
2
-
(
μ
2
-
k
2
)
(
1
+
γ
2
)
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
,
which is the desired estimate.
We now estimate (48)2 by(54)
|
(
48
)
2
|
≤
C
2
k
1
2
k
2
(
1
+
ϵ
2
)
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
χ
W
1
C
χ
W
2
×
∫
ρ
1
(
x
,
s
)
≤
2
C
1
2
-
k
1
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
≥
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
/
2


(
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
)
(
1
+
γ
1
)
)
-
1
1
×
(
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
)
(
1
+
γ
1
)
)
(
(
s
,
z
τ
1
μ
1
,
v
1
)
1
+
2
μ
1
m
m
m
m
m
m
m
m
i
m
m
m
m
m
i
×
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
)
(
1
+
γ
1
)
)
-
1
)
d
μ
1
(
s
)
×
∫
W
4
‍
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
ϵ
2
(
1
+
2
μ
2
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
)
(
1
+
γ
2
)
d
μ
2
(
t
)
≤
C
2
k
1
2
k
2
(
1
+
ϵ
2
)
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
×
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
χ
W
1
C
χ
W
2
×
2
-
k
1
2
μ
1
(
1
+
γ
1
)
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
(
1
+
γ
1
)
2
-
μ
2
(
1
+
γ
2
)
×
∫
W
4
‍
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
ϵ
2
-
1
-
γ
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
μ
1
-
k
1
)
(
1
+
γ
1
)
(
1
+
2
k
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
×
2
-
(
μ
2
-
k
2
)
(
1
+
γ
2
)
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
,
which verifies (48)2.
The verification of (48)3 is similar to that of (48)2 by symmetry.
Similarly we have(55)
|
(
48
)
4
|
≤
C
2
k
1
2
k
2
(
1
+
ϵ
2
)
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
χ
W
1
χ
W
2
×
∫
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
≥
2
C
1
2
-
k
1
‍
1
(
2
μ
1
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
)
(
1
+
γ
1
)
d
μ
1
(
s
)
×
∫
W
4
‍
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
ϵ
2
(
1
+
2
μ
2
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
)
(
1
+
γ
2
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
χ
W
1
χ
W
2
×
2
-
(
μ
1
-
k
1
)
(
1
+
γ
1
)
2
-
(
μ
2
-
k
2
)
(
1
+
γ
2
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
μ
1
-
k
1
)
(
1
+
γ
1
)
(
1
+
2
k
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
×
2
-
(
μ
2
-
k
2
)
(
1
+
γ
2
)
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
,
which is also a desired estimate.
By an analogou argument to (48)4, we can verify (48)5.
We now prove (49)1
(56)
|
D
k
1
D
k
2
(
m
R
)
(
x
,
y
)
|
=
|
∫
X
1
×
X
2
D
k
1
(
x
,
s
)
D
k
2
(
y
,
t
)
m
R
(
s
,
t
)
d
μ
1
(
s
)
d
μ
2
(
t
)
|
≤
∫
X
1
×
X
2
|
D
k
1
(
x
,
s
)
|
|
D
k
2
(
y
,
t
)
-
D
k
2
(
y
,
z
τ
2
(
μ
2
,
v
2
)
)
|
×
|
m
R
(
s
,
t
)
-
m
R
(
x
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
=
(
χ
w
2
∫
X
3
×
W
4
+
χ
w
2
∫
X
3
×
W
4
C
+
χ
W
2
C
∫
X
3
×
W
4
)
×
|
D
k
1
(
x
,
s
)
|
|
D
k
2
(
y
,
t
)
-
D
k
2
(
y
,
z
τ
2
(
μ
2
,
v
2
)
)
|
|
m
R
(
s
,
t
)
-
m
R
(
x
,
t
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
∶
=
(
49
)
1
+
(
49
)
2
+
(
49
)
3
.
For (49)1, we have
(57)
|
(
49
)
1
|
≤
χ
W
2
2
k
1
2
k
2
(
1
+
ϵ
2
)
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
-
β
1
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
∫
W
3
×
W
4
‍
ρ
1
(
x
,
s
)
β
1
(
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
+
1
(
1
+
2
μ
1
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
)
mm
×
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
ϵ
2
(
1
+
2
μ
2
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
χ
W
2
2
k
1
2
k
2
(
1
+
ϵ
2
)
×
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
-
β
1
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
∫
W
3
‍
(
ρ
1
(
x
,
s
)
β
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
iiiiiiiiiiiiiii
+
ρ
1
(
x
,
s
)
β
1
(
1
+
2
μ
1
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
)
d
μ
1
(
s
)
×
∫
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
≤
8
C
2
2
-
k
2
‍
2
-
μ
2
(
1
+
γ
2
)
mmm
×
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
ϵ
2
-
1
-
γ
2
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
μ
2
-
k
2
)
(
1
+
γ
2
)
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
×
(
+
2
-
(
k
1
-
μ
1
)
β
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
2
μ
1
β
1
+
k
1
×
∫
ρ
1
(
x
,
s
)
≤
4
C
1
2
-
k
1
‍
ρ
1
(
x
,
s
)
β
1
(
1
+
2
μ
1
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
d
μ
1
(
s
)
m
m
m
m
m
m
m
+
2
-
(
k
1
-
μ
1
)
β
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
k
1
-
μ
1
)
β
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
×
2
-
(
μ
2
-
k
2
)
(
1
+
γ
2
)
(
1
+
2
k
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
,
since 1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
≤
2
(
2
C
1
+
1
)
(
1
+
2
μ
1
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
) by k
1
>
μ
1, which is the proof of (48)1.
The verification of terms (49)2 and (49)3 is similar to that of (49)1 and proof of (50) is similar to that of (49) by symmetry.
We now prove (51). In this case, we have
(58)
|
D
k
1
D
k
2
(
m
R
)
(
x
,
y
)
|
≤
∫
X
1
×
X
2
‍
|
D
k
1
(
x
,
s
)
|
|
D
k
2
(
y
,
t
)
|
×
|
m
R
(
s
,
t
)
-
m
R
(
x
,
t
)
-
m
R
(
s
,
y
)
-
m
R
(
x
,
y
)
|
d
μ
1
(
s
)
d
μ
2
(
t
)
≤
2
k
1
2
k
2
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
-
β
1
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
-
β
2
×
∫
W
3
‍
ρ
1
(
x
,
s
)
β
1
(
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
iiiiiiiiiiiiiiiiiiiiiiiiiii
+
1
(
1
+
2
μ
1
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
)
d
μ
1
(
s
)
×
∫
W
4
‍
ρ
2
(
y
,
t
)
β
2
(
1
(
1
+
2
μ
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
iiiiiiiiiiiiiiiiiiiiiiiiiii
+
1
(
1
+
2
μ
2
ρ
2
(
t
,
z
τ
2
μ
1
,
v
2
)
)
1
+
γ
2
)
d
μ
2
(
t
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
(
2
-
(
k
1
-
μ
1
)
β
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
iiiii
+
2
μ
1
β
1
+
k
1
∫
W
3
‍
ρ
1
(
x
,
s
)
β
1
(
1
+
2
μ
1
ρ
1
(
s
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
d
μ
1
(
s
)
)
×
(
2
-
(
k
2
-
μ
2
)
β
2
(
1
+
2
μ
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
iiiiii
+
2
μ
2
β
2
+
k
2
∫
W
4
‍
ρ
2
(
y
,
t
)
β
2
(
1
+
2
μ
2
ρ
2
(
t
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
d
μ
2
(
t
)
)
≤
C
μ
1
(
Q
τ
1
μ
1
,
v
1
)
-
1
/
2
μ
2
(
Q
τ
2
μ
2
,
v
2
)
-
1
/
2
×
2
-
(
k
1
-
μ
1
)
β
1
(
1
+
2
μ
1
ρ
1
(
x
,
z
τ
1
μ
1
,
v
1
)
)
1
+
γ
1
×
2
-
(
k
2
-
μ
2
)
β
2
(
1
+
2
μ
2
ρ
2
(
y
,
z
τ
2
μ
2
,
v
2
)
)
1
+
γ
2
.



---

*Source: 396418-2014-07-22.xml*