# Estimates for Parameter Littlewood-Paleyg

Authors: Guanghui Lu; Shuangping Tao
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

Let(
X
,
d
,
μ
) be a metric measure space which satisfies the geometrically doubling measure and the upper doubling measure conditions. In this paper, the authors prove that, under the assumption that the kernel of M
κ
⁎ satisfies a certain Hörmander-type condition, M
κ
⁎
,
ρ is bounded from Lebesgue spaces L
p
(
μ
) to Lebesgue spaces L
p
(
μ
) for p
≥
2 and is bounded from L
1
(
μ
) into L
1
,
∞
(
μ
). As a corollary, M
κ
⁎
,
ρ is bounded on L
p
(
μ
) for 1
<
p
<
2. In addition, the authors also obtain that M
κ
⁎
,
ρ is bounded from the atomic Hardy space H
1
(
μ
) into the Lebesgue space L
1
(
μ
).


---

## Body

## 1. Introduction

In 1958, Stein in [1] firstly introduced the Littlewood-Paley operators of the higher-dimensional case; meanwhile, the author also obtained the boundedness of the Marcinkiewicz integrals and area integrals. In 1970, Fefferman in [2] proved that the Littlewood-Paley g
κ
∗ function is weak type (
p
,
p
) for p
∈
(
1,2
) and κ
=
2
/
p. With further research about Littlewood-Paley operators, some authors turn their attentions to study the parameter Littlewood-Paley operators. For example, in 1999, Sakamoto and Yabuta in [3] considered the parameter g
κ
∗ function. Since then, many papers focus on the behaviours of the operators; among them we refer readers to see [4–6].In the past ten years or so, most authors mainly study the classical theory of harmonic analysis onR
n under nondoubling measures which only satisfy the polynomial growth condition; see [7–12]. Exactly, we assume that μ which is a positive Radon measure on R
n satisfies the following growth conditions; namely, for all x
∈
R
n and r
∈
(
0
,
∞
), there exist constant C and 0
<
d
≤
n such that (1)
μ
B
x
,
r
≤
C
r
d
,where B
(
x
,
r
)
≔
{
y
∈
R
n
:
x
-
y
<
r
}. The analysis associated with nondoubling measures μ as in (1) has important applications in solving long-standing open Painlevé’s problem and Vitushkin’s conjecture (see [13, 14]). Besides, Coifman and Weiss have showed that the measure μ is a key assumption in harmonic analysis on homogeneous-type spaces (see [15, 16]).However, Hytönen in [17] pointed that the measure μ as in (1) may not contain the doubling measure as special cases. To solve the problem, in 2010, Hytönen in [17] introduced a new class of metric measure spaces satisfying the so-called upper doubling conditions and the geometrically doubling (resp., see Definitions 1 and 2 below), which are now claimed nonhomogeneous metric measure spaces. Therefore, if we replace the underlying spaces with nonhomogeneous metric measure spaces, many known-consequences have been proved still true; for example, see [18–22].In this paper, we always assume that(
X
,
d
,
μ
) is a nonhomogeneous metric measure space. In this setting, we will establish the boundedness of the parameter Littlewood-Paley g
κ
∗ functions on (
X
,
d
,
μ
).In order to state our main results, we firstly recall some necessary notions and notation. Hytönen in [17] gave out the definition of upper doubling metric spaces as follows.Definition 1 (see [17]).
A metric measure space(
X
,
d
,
μ
) is said to be upper doubling, if μ is Borel measure on X and there exist a dominating function λ
:
X
×
(
0
,
∞
)
→
(
0
,
∞
) and a positive constant C
λ such that for each x
∈
X, r
→
λ
(
x
,
r
) is nondecreasing and, for all x
∈
X and r
∈
(
0
,
∞
), (2)
μ
B
x
,
r
≤
λ
x
,
r
≤
C
λ
λ
x
,
r
2
.Htyönen et al. in [18] proved that there exists another dominating function λ
~ such that λ
~
≤
λ, C
λ
~
≤
C
λ and (3)
λ
~
x
,
y
≤
C
λ
~
λ
~
y
,
r
,where x
,
y
∈
X and d
(
x
,
y
)
≤
r. Based on this, from now on, let the dominating function in (2) also satisfy (3).Now we recall the notion of geometrically doubling conditions given in [17].Definition 2 (see [17]).
A metric space(
X
,
d
) is said to be geometrically doubling, if there exists some N
0
∈
N such that, for any ball B
(
x
,
r
)
⊂
X, there exists a finite ball covering {
B
(
x
i
,
r
/
2
)
}
i of B
(
x
,
r
) such that the cardinality of this covering is at most N
0.Remark 3 (see [17]).
Let(
X
,
d
) be a metric space. Hytönen in [17] showed that the following statements are mutually equivalent:(1)
(
X
,
d
) is geometrically doubling.
(2)
For anyϵ
∈
(
0,1
) and ball B
(
x
,
r
)
⊂
X, there exists a finite ball covering {
B
(
x
i
,
ϵ
r
)
}
i of B
(
x
,
r
) such that the cardinality of this covering is at most N
0
ϵ
-
n. Here and in what follows, N
0 is as Definition 2 and n
=
log
2
⁡
N
0
.
(3)
For everyϵ
∈
(
0,1
), any ball B
(
x
,
r
)
⊂
X can contain at most N
0
ϵ
-
n centers of disjoint balls {
B
(
x
i
,
ϵ
r
)
}
i.
(4)
There existsM
∈
N such that any ball B
(
x
,
r
)
⊂
X can contain at most M centers {
x
i
}
i of disjoint balls {
B
(
x
i
,
r
/
4
)
}
i
=
1
M.Hytönen in [17] introduced the following coefficients K
B
,
S analogous to Tolsa’s number K
Q
,
R in [7].Given any two ballsB
⊂
S, set (4)
K
B
,
S
≔
1
+
∫
2
S
∖
B
1
λ
c
B
,
d
x
,
c
B
d
μ
x
,where c
B represents the center of the ball B.Remark 4.
Bui and Duong in [21] firstly introduced the following discrete version K
~
B
,
S of K
B
,
S as in (4) on (
X
,
d
,
μ
), which is very similar to the number K
Q
,
R introduced in [7] by Tolsa. For any two balls B
⊂
S, K
~
B
,
S is defined by (5)
K
~
B
,
S
=
1
+
∑
i
=
1
N
B
,
S
μ
6
i
B
λ
c
B
,
6
i
r
B
,where the radii of the balls B and S are denoted by r
B and r
S, respectively, and N
B
,
S is the smallest integer satisfying 6
N
B
,
S
r
B
≥
r
s. It is easy to obtain K
~
B
,
S
≤
C
K
B
,
S. Bui and Duong in [21] also pointed out that it is incorrect that K
B
,
S
~
K
~
B
,
S.Now we recall the following notion of(
α
,
β
)-doubling property (see [17]).Definition 5 (see [17]).
Letα
,
β
∈
(
1
,
∞
). A ball B
⊂
X is claimed to be (
α
,
β
)-doubling if μ
(
α
B
)
≤
β
μ
(
B
).It was stated in [17] that, there exist many balls which have the above (
α
,
β
)-doubling property. In the latter part of the paper, if α and β
α are not specified, (
α
,
β
α
)-doubling ball always stands for (
6
,
β
6
)-doubling ball with a fixed number β
6
>
max
⁡
{
C
λ
3
log
2
⁡
6
,
6
n
}, where n
≔
log
2
⁡
N
0 is considered as a geometric dimension of the space. Moreover, the smallest (
6
,
β
6
)-doubling ball of the form 6
j
B with j
∈
N is denoted by B
~
6, and sometimes B
~
6 can be simply denoted by B
~.Now we give the definition of the parameter Littlewood-Paleyg
κ
∗ functions on (
X
,
d
,
μ
).Definition 6 (see [22]).
LetK
(
x
,
y
) be a locally integrable function on (
X
×
X
)
∖
{
(
x
,
y
)
:
x
=
y
}. Assume that there exists a positive constant C such that, for all x
,
y
∈
X with x
≠
y, (6)
K
x
,
y
≤
C
d
x
,
y
λ
x
,
d
x
,
yand, for all x
,
y
,
y
′
∈
X, (7)
∫
d
x
,
y
≥
2
d
y
,
y
′
K
x
,
y
-
K
x
,
y
′
+
K
y
,
x
-
K
y
′
,
x
1
d
x
,
y
d
μ
x
≤
C
.The parameter Marcinkiewicz integralM
ρ associated with the above K
(
x
,
y
) which satisfies (6) and (7) is defined by (8)
M
ρ
f
x
=
∫
0
∞
1
t
ρ
∫
d
x
,
y
≤
t
K
x
,
y
d
x
,
y
1
-
ρ
f
y
d
μ
y
2
d
t
t
1
/
2
,
x
∈
X
,where ρ
∈
(
0
,
∞
). The parameter g
κ
∗ function M
κ
∗
,
ρ is defined by (9)
M
κ
∗
,
ρ
f
x
=
∬
X
×
0
,
∞
t
t
+
d
x
,
y
κ
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
f
z
d
μ
z
2
d
μ
y
λ
y
,
t
d
t
t
1
/
2
,where x
∈
X, X
×
(
0
,
∞
)
≔
{
(
y
,
t
)
:
y
∈
X
,
t
>
0
}, ρ
>
0 and κ
∈
(
1
,
∞
).Remark 7.
(1) Whenρ
=
1, the operator M
ρ as in (8) is just the Marcinkiewicz integral on (
X
,
d
,
μ
) (see [22]).
(2) If we take(
X
,
d
,
μ
)
=
(
R
n
,
·
,
μ
) and λ
(
y
,
t
)
≔
t
n, then the parameter g
κ
∗ function M
κ
∗
,
ρ as in (9) is just a parameter Littlewood-Paley operator with nondoubling measures in [8].The following definition of the atomic Hardy space was introduced by Htyönen et al. (see [18]).Definition 8 (see [18]).
Letζ
∈
(
1
,
∞
) and p
∈
(
1
,
∞
]. A function b
∈
L
l
o
c
1
(
μ
) is called a (
p
,
1
)
υ-atomic block if(a)
there exists a ballB such that supp
⁡
b
⊂
B,
(b)
∫
X
b
(
x
)
d
μ
(
x
)
=
0,
(c)
for anyi
∈
{
1,2
} there exist a function a
i supported on ball B
i
⊂
B and a number υ
i
∈
C such that (10)
b
=
υ
1
a
1
+
υ
2
a
2
,
a
i
L
p
μ
≤
μ
ζ
B
i
1
/
p
-
1
K
B
i
,
B
-
1
.Moreover, let b
H
a
t
b
1
,
p
μ
≔
υ
1
+
υ
2.We say a functionf
∈
L
1
(
μ
) belongs to theatomic Hardy space H
a
t
b
1
,
p
(
μ
) if there are atomic blocks {
b
i
}
i
=
1
∞ such that f
=
∑
i
=
1
∞
b
i with ∑
i
=
1
∞
b
i
H
a
t
b
1
,
p
(
μ
)
<
∞. The H
a
t
b
1
,
p
(
μ
) norm of f is denoted by f
H
a
t
b
1
,
p
(
μ
)
=
inf
⁡
{
∑
i
=
1
∞
b
i
H
a
t
b
1
,
p
(
μ
)
}, where the infimum is taken over all the possible decompositions of f as above.It was proved by Htyönen et al. in [18] that the definition of H
a
t
b
1
,
p
(
μ
) is not related to the choice of ζ and the spaces H
a
t
b
1
,
p
(
μ
) and H
a
t
b
1
,
∞
(
μ
) have the same norms for p
∈
(
1
,
∞
]. Thus, for convenience, we always denote H
a
t
b
1
,
p
(
μ
) by H
1
(
μ
).Now we give the Hörmander-type condition on(
X
,
d
,
μ
); that is, there exists a positive C, such that (11)
sup
r
>
0
d
y
,
y
′
≤
r
⁡
∑
i
=
1
∞
i
∫
6
i
r
<
d
x
,
y
≤
6
i
+
1
r
K
x
,
y
-
K
x
,
y
′
+
K
y
,
x
-
K
y
′
,
x
d
μ
x
d
x
,
y
≤
C
.Notice this condition is slightly stronger than (7).Now let us state the main theorems which generalize and improve the corresponding results in [8].Theorem 9.
LetK
(
x
,
y
) satisfy (6) and (7), and let M
κ
∗
,
ρ be as in (9) with ρ
∈
(
0
,
∞
) and κ
∈
(
1
,
∞
). Then M
κ
∗
,
ρ is bounded on L
p
(
μ
) for any p
∈
[
2
,
∞
).Theorem 10.
LetK
(
x
,
y
) satisfy (6) and (11), and let M
κ
∗
,
ρ be as in (9) with ρ
∈
(
1
/
2
,
∞
) and κ
∈
(
1
,
∞
). Then M
κ
∗
,
ρ is bounded from L
1
(
μ
) into weak L
1
(
μ
); namely, there exists a positive constant C such that, for any τ
>
0 and f
∈
L
1
(
μ
), (12)
μ
x
∈
X
:
M
κ
∗
,
ρ
f
x
>
τ
≤
C
f
L
1
μ
τ
.Theorem 11.
LetK
(
x
,
y
) satisfy (6) and (11), and let M
κ
∗
,
ρ be as in (9) with ρ
>
1
/
2 and κ
>
1. Suppose that M
κ
∗
,
ρ is bounded on L
2
(
μ
). Then, M
κ
∗
,
ρ is bounded from H
1
(
μ
) into L
1
(
μ
).Applying the Marcinkiewicz interpolation theorem and Theorems9 and 10, it is easy to get the following result.Corollary 12.
Under the assumption of Theorem10, M
κ
∗
,
ρ is bounded on L
p
(
μ
) for p
∈
(
1,2
).The organization of this paper is as follows. In Section2, we will give some preliminary lemmas. The proofs of the main theorems will be given in Section 3. Throughout this paper, C stands for a positive constant which is independent of the main parameters, but it may be different from line to line. For any E
⊂
X, we use χ
E to denote its characteristic function.


## 2. Preliminary Lemmas

In this section, we make some preliminary lemmas which are used in the proof of the main results. Firstly, we recall some properties ofK
B
,
S as in (4) (see [17]).Lemma 13 (see [17]).
(1) For all ballsB
⊂
R
⊂
S, it holds true that K
B
,
R
≤
K
B
,
S.
(2) For anyξ
∈
[
1
,
∞
), there exists a positive constant C
ξ, such that, for all balls B
⊂
S with r
S
≤
ξ
r
B
,
K
B
,
S
≤
C
ξ.
(3) For anyϱ
∈
(
1
,
∞
), there exists a positive constant C
ϱ, depending on ϱ, such that, for all balls B
,
K
B
,
B
~
ϱ
≤
C
ϱ.
(4) There exists a positive constantc such that, for all balls B
⊂
R
⊂
S, K
B
,
S
≤
K
B
,
R
+
c
K
R
,
S
. In particular, if B and R are concentric, then c
=
1.
(5) There exists a positive constantc
~ such that, for all balls B
⊂
R
⊂
S, K
B
,
R
≤
c
~
K
B
,
S; moreover, if B and R are concentric, then K
R
,
S
≤
K
B
,
S.To state the following lemmas, let us give a known-result (see [19]). For η
∈
(
0
,
∞
), the maximal operator is defined, by setting that, for all f
∈
L
l
o
c
1
(
μ
) and x
∈
X, (13)
M
η
f
x
≔
sup
Q
∋
x
,
Q
doubling
⁡
1
μ
η
Q
∫
Q
f
y
d
μ
yis bounded on L
p
(
μ
) provided that p
∈
(
1
,
∞
) and also bounded from L
1
(
μ
) into L
1
,
∞
(
μ
).The following lemma is slightly changed from [8].Lemma 14.
LetK
(
x
,
y
) satisfy (6) and (7), and η
∈
(
0
,
∞
). Assume that M
ρ is as in (8) and M
κ
∗
,
ρ is as in (9) with ρ
∈
(
0
,
∞
) and κ
∈
(
1
,
∞
). Then for any nonnegative function ϕ, there exists a positive constant C such that, for all f
∈
L
p
(
μ
) with p
∈
(
1
,
∞
), (14)
∫
X
M
κ
∗
,
ρ
f
x
2
ϕ
x
d
μ
x
≤
C
∫
X
M
ρ
f
x
2
M
η
ϕ
x
d
μ
x
.Proof.
By the definition ofM
κ
∗
,
ρ
(
f
), we have (15)
∫
X
M
κ
∗
,
ρ
f
x
2
ϕ
x
d
μ
x
=
∫
X
∬
X
×
0
,
∞
t
t
+
d
x
,
y
β
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
f
y
d
μ
z
2
d
μ
y
λ
y
,
t
d
t
t
ϕ
x
d
μ
x
≤
∫
X
∫
0
∞
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
f
y
d
μ
z
2
d
t
t
sup
t
>
0
⁡
∫
X
t
t
+
d
x
,
y
β
ϕ
x
λ
y
,
t
d
μ
x
d
μ
y
=
∫
X
M
ρ
f
y
2
sup
t
>
0
⁡
∫
X
t
t
+
d
x
,
y
β
ϕ
x
λ
y
,
t
d
μ
x
d
μ
y
.Thus, to prove Lemma 14, we only need to estimate that (16)
sup
t
>
0
⁡
∫
X
t
t
+
d
x
,
y
β
ϕ
x
λ
y
,
t
d
μ
x
≤
C
M
η
ϕ
y
.For any y
∈
X and t
>
0, write (17)
∫
X
t
t
+
d
x
,
y
β
ϕ
x
λ
y
,
t
d
μ
x
=
∫
B
y
,
t
t
t
+
d
x
,
y
β
ϕ
x
λ
y
,
t
d
μ
x
+
∫
X
∖
B
y
,
t
t
t
+
d
x
,
y
β
ϕ
x
λ
y
,
t
d
μ
x
≕
D
1
+
D
2
.For D
1, it is not difficult to obtain that (18)
D
1
≤
∫
B
y
,
t
ϕ
x
λ
y
,
t
d
μ
x
=
μ
η
B
y
,
t
λ
y
,
t
1
μ
η
B
y
,
t
∫
B
y
,
t
ϕ
x
d
μ
x
≤
C
M
η
ϕ
y
.
Now we turn to estimateD
2, by (2) and (13); we have (19)
D
2
≤
∑
k
=
1
∞
∫
B
y
,
6
k
t
∖
B
y
,
6
k
-
1
t
t
t
+
d
x
,
y
β
ϕ
x
λ
y
,
t
d
μ
x
≤
C
∑
k
=
1
∞
6
-
k
-
1
β
∫
B
y
,
6
k
t
ϕ
x
λ
y
,
t
d
μ
x
≤
C
∑
k
=
1
∞
6
-
k
-
1
β
μ
B
y
,
6
k
t
λ
y
,
t
1
μ
B
y
,
6
k
t
∫
B
y
,
6
k
t
ϕ
x
d
μ
x
≤
C
∑
k
=
1
∞
6
-
k
-
1
β
μ
B
y
,
6
k
t
λ
y
,
t
M
η
ϕ
y
≤
C
λ
y
,
6
k
t
λ
y
,
t
M
η
ϕ
y
∑
k
=
1
∞
6
-
k
-
1
β
λ
y
,
6
k
t
λ
y
,
t
≤
C
λ
y
,
6
k
t
λ
y
,
t
M
η
ϕ
y
∑
k
=
1
∞
6
-
k
-
1
β
≤
C
M
η
ϕ
y
.Combining the estimates for D
1 and D
2, we obtain (16) and hence complete the proof of Lemma 14.Finally, we recall the Calderón-Zygmund decomposition theorem (see [21]). Suppose that γ
0 is a fixed positive constant satisfying that γ
0
>
max
⁡
{
C
λ
3
log
2
⁡
6
,
6
3
n
}, where C
λ is as in (2) and n as in Remark 3.Lemma 15 (see [21]).
Letp
∈
[
1
,
∞
), f
∈
L
p
(
μ
), and t
∈
(
0
,
∞
)
(
t
>
γ
0
f
L
p
(
μ
)
/
μ
(
X
)
w
h
e
n
μ
(
X
)
<
∞
). Then(1)
there exists a family of finite overlapping balls{
6
B
i
}
i such that {
B
i
}
i is pairwise disjoint: (20)
1
μ
6
2
B
i
∫
B
i
f
x
p
d
μ
x
>
t
p
γ
0
∀
i
,
(21)
1
μ
6
2
τ
B
i
∫
τ
B
i
f
x
p
d
μ
x
≤
t
p
γ
0
∀
i
,
∀
τ
∈
2
,
∞
,
(22)
f
x
≤
t
f
o
r
μ
-
a
l
m
o
s
t
e
v
e
r
y
x
∈
X
∖
⋃
i
6
B
i
;
(2)
for eachi, let S
i be a (
3
×
6
2
,
C
λ
log
2
⁡
(
3
×
6
2
)
+
1
)-doubling ball of the family {
(
3
×
6
2
)
k
B
i
}
k
∈
N, and ω
i
=
χ
6
B
i
/
(
∑
k
‍
χ
6
B
k
). Then there exists a family {
φ
i
}
i of functions that, for each i, supp
⁡
(
φ
i
)
⊂
S
i, φ
i has a constant sign on S
i and (23)
∫
X
φ
i
x
d
μ
x
=
∫
6
B
i
f
x
ω
i
x
d
μ
x
,
∑
i
φ
i
x
≤
γ
t
f
o
r
μ
-
a
l
m
o
s
t
e
v
e
r
y
x
∈
X
,where γ is some positive constant depending only on (
X
,
μ
), and there exists a positive constant C, independent of f, t, and i, such that if p
=
1, then (24)
φ
i
L
∞
μ
μ
S
i
≤
C
∫
X
f
x
ω
i
x
d
μ
x
,and if p
∈
(
1
,
∞
), (25)
∫
S
i
φ
i
x
p
d
μ
x
1
/
p
μ
S
i
1
/
p
′
≤
C
t
p
-
1
∫
X
f
x
ω
i
x
p
d
μ
x
.


## 3. Proofs of Theorems

Proof of Theorem9.
For the case ofp
=
2, assume ϕ
(
x
)
=
1 in Lemma 14; then it is easy to get that (26)
∫
X
M
κ
∗
,
ρ
f
x
2
d
μ
x
≤
C
∫
X
M
ρ
f
x
2
d
μ
x
,which, along with L
2
(
μ
)-boundedness of M
ρ, easily yields that Theorem 9 holds.
For the case ofp
>
2, let q be the index conjugate to p
/
2. By applying Hölder inequality and Lemma 14, we can conclude (27)
M
κ
∗
,
ρ
f
L
p
μ
2
=
sup
ϕ
≥
0
ϕ
L
q
μ
≤
1
⁡
∫
X
M
κ
∗
,
ρ
f
x
2
ϕ
x
d
μ
x
≤
C
sup
ϕ
≥
0
ϕ
L
q
μ
≤
1
⁡
∫
X
M
ρ
f
x
2
M
η
ϕ
x
d
μ
x
≤
C
M
ρ
f
L
p
μ
2
sup
ϕ
≥
0
ϕ
L
q
μ
≤
1
⁡
M
η
ϕ
L
q
μ
≤
C
f
L
p
μ
2
sup
ϕ
≥
0
ϕ
L
q
μ
≤
1
⁡
ϕ
L
q
μ
≤
C
f
L
p
μ
2
,which is desired. Thus, we complete the proof of Theorem 9.Proof of Theorem10.
Without loss of generality, we may assume thatf
L
1
(
μ
)
=
1. It is easy to see that the conclusion of Theorem 10 naturally holds if τ
≤
β
6
(
f
L
1
μ
/
μ
X
) when μ
(
X
)
<
∞. Thus, we only need to discuss the case that τ
>
β
6
(
f
L
1
μ
/
μ
X
). Applying Lemma 15 to f at the level τ and letting ω
i, φ
i, B
i, and S
i be the same as in Lemma 15, we see that f
(
x
)
=
b
(
x
)
+
h
(
x
), where b
(
x
)
≔
f
χ
X
∖
⋃
i
6
B
i
(
x
)
+
∑
i
‍
φ
i
(
x
) and h
(
x
)
≔
∑
i
‍
[
ω
i
(
x
)
f
(
x
)
-
φ
i
(
x
)
]
≕
∑
i
‍
h
i
(
x
). It is easy to obtain that b
L
∞
(
μ
)
≤
C
τ and b
L
1
(
μ
)
≤
C. By L
2
(
μ
)-boundedness of M
κ
∗
,
ρ, we have (28)
μ
x
∈
X
:
M
κ
∗
,
ρ
b
x
>
τ
≤
M
κ
∗
,
ρ
b
L
2
μ
2
τ
2
≤
C
b
L
2
μ
2
τ
2
≤
C
τ
-
1
.On the other hand, by (20) with p
=
1 and the fact that the sequence of balls, {
B
i
}
i, is pairwise disjoint, we see that (29)
μ
⋃
i
6
2
B
i
≤
C
τ
-
1
∫
X
f
x
d
μ
x
≤
C
τ
-
1
,and thus the proof of the Theorem 10 can be reduced to prove that (30)
μ
x
∈
X
∖
⋃
i
6
2
B
i
:
M
κ
∗
,
ρ
h
x
>
τ
≤
C
τ
-
1
.
For each fixedi, denote the center of B
i by x
i, and let N
1 be the positive integer satisfying S
i
=
(
3
×
6
2
)
N
1
B
i. We have (31)
μ
x
∈
X
∖
⋃
i
6
2
B
i
:
M
κ
∗
,
ρ
h
x
>
τ
≤
τ
-
1
∑
i
∫
X
∖
⋃
i
6
2
B
i
M
κ
∗
,
ρ
h
i
x
d
μ
x
≤
τ
-
1
∑
i
∫
X
∖
6
S
i
M
κ
∗
,
ρ
h
i
x
d
μ
x
+
τ
-
1
∑
i
∫
6
S
i
∖
6
2
B
i
M
κ
∗
,
ρ
h
i
x
d
μ
x
≕
τ
-
1
∑
i
E
1
+
E
2
.
Firstly, let us estimateE
2 and write it as (32)
E
2
≤
∫
6
S
i
∖
6
2
B
i
M
κ
∗
,
ρ
f
ω
i
x
d
μ
x
+
∫
6
S
i
∖
6
2
B
i
M
κ
∗
,
ρ
φ
i
x
d
μ
x
≕
E
21
+
E
22
,where h
i
≔
ω
i
f
-
φ
i. By Hölder inequality, (24), and L
2
(
μ
)-boundedness of M
κ
∗
,
ρ, we have (33)
E
22
≤
∫
6
S
i
M
κ
∗
,
ρ
φ
i
x
d
μ
x
≤
∫
6
S
i
M
κ
∗
,
ρ
φ
i
x
2
d
μ
x
1
/
2
μ
6
S
i
1
/
2
≤
C
∫
6
S
i
φ
i
x
2
d
μ
x
1
/
2
μ
6
S
i
1
/
2
≤
C
∫
X
f
x
ω
i
x
d
μ
x
.For E
21, by Minkowski inequality and (6), write (34)
E
21
=
∫
6
S
i
∖
6
2
B
i
∬
X
×
0
,
∞
t
t
+
d
x
,
y
κ
/
2
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
f
z
ω
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
/
2
d
μ
x
≤
C
∫
6
S
i
∖
6
2
B
i
∫
X
f
z
ω
i
z
∬
d
y
,
z
≤
t
t
t
+
d
x
,
y
κ
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
z
d
μ
x
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∬
d
y
,
z
≤
t
t
t
+
d
x
,
y
κ
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∬
d
y
,
z
≤
t
2
d
y
,
z
>
d
x
,
z
t
t
+
d
x
,
y
κ
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
d
μ
z
+
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∬
d
y
,
z
≤
t
,
d
x
,
y
<
t
2
d
y
,
z
≤
d
x
,
z
t
t
+
d
x
,
y
κ
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
d
μ
z
+
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∬
d
y
,
z
≤
t
,
d
x
,
y
≥
t
2
d
y
,
z
≤
d
x
,
z
t
t
+
d
x
,
y
κ
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
d
μ
z
≕
F
1
+
F
2
+
F
3
.To this end, let B
i be as in Lemma 15 with c
B
i and r
B
i being, respectively, its center and radius. For any x
∈
6
S
i
∖
6
2
B
i and z
∈
6
B
i, by (2) and (3), we have (35)
F
1
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∫
2
d
y
,
z
>
d
x
,
z
∫
d
y
,
z
∞
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∫
2
d
y
,
z
>
d
x
,
z
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
1
λ
y
,
d
y
,
z
∫
d
y
,
z
∞
d
t
t
1
+
2
ρ
d
μ
y
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∫
2
d
y
,
z
>
d
x
,
z
1
λ
y
,
d
y
,
z
3
d
μ
y
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∫
2
d
y
,
z
>
d
x
,
z
1
λ
y
,
d
x
,
z
d
μ
y
λ
y
,
1
/
2
d
x
,
z
2
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
1
λ
z
,
1
/
2
d
x
,
z
2
∫
2
d
y
,
z
>
d
x
,
z
d
μ
y
λ
y
,
d
y
,
z
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∑
k
=
1
∞
∫
B
z
,
2
k
-
1
d
x
,
z
∖
B
z
,
2
k
-
2
d
x
,
z
d
μ
y
λ
y
,
d
y
,
z
1
/
2
1
λ
z
,
d
x
,
z
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∑
k
=
1
∞
∫
B
z
,
2
k
-
1
d
x
,
z
d
μ
y
λ
y
,
2
k
-
2
d
x
,
z
1
/
2
1
λ
c
B
i
,
d
x
,
c
B
i
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
1
λ
c
B
i
,
d
x
,
c
B
i
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
d
μ
z
,where we use the fact that (36)
∫
6
S
i
∖
6
2
B
i
1
λ
c
B
i
,
d
x
,
c
B
i
d
μ
x
≤
C
K
B
i
,
S
i
.
Next we estimateF
2. For any x
∈
6
S
i
∖
6
2
B
i, y
∈
X, and z
∈
6
B
i satisfying d
(
y
,
x
)
<
t, 2
d
(
y
,
z
)
≤
d
(
x
,
z
), and (
1
/
2
)
d
(
x
,
z
)
<
t, we have (37)
F
2
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∫
2
d
y
,
z
≤
d
x
,
z
∫
1
/
2
d
x
,
z
∞
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∫
2
d
y
,
z
≤
d
x
,
z
1
λ
y
,
d
x
,
z
2
d
μ
y
μ
B
y
,
d
x
,
z
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
1
λ
c
B
i
,
d
x
,
c
B
i
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
d
μ
z
.
Finally, for anyx
∈
6
S
i
∖
6
2
B
i, y
∈
X, and z
∈
6
B
i satisfying 2
d
(
y
,
z
)
≤
d
(
x
,
z
), 2
d
(
y
,
z
)
≥
d
(
x
,
z
), and d
(
x
,
y
)
<
(
3
/
2
)
d
(
x
,
z
), by applying (2), we have (38)
F
3
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∬
d
y
,
z
≤
t
,
d
x
,
y
≥
t
2
d
y
,
z
≤
d
x
,
z
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
∫
2
d
y
,
z
≤
d
x
,
z
1
λ
y
,
d
x
,
z
2
1
λ
y
,
d
x
,
z
d
μ
y
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
1
λ
z
,
d
x
,
z
2
μ
B
z
,
1
/
2
d
x
,
z
λ
z
,
d
x
,
z
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
∫
6
S
i
∖
6
2
B
i
1
λ
c
B
i
,
d
x
,
c
B
i
d
μ
x
d
μ
z
≤
C
∫
6
B
i
f
z
d
μ
z
.Combining the estimates for F
1, F
2, and F
3, we obtain that E
21
≤
C
∫
6
B
i
f
z
d
μ
z, where, together with the fact that E
22
≤
C
∫
6
B
i
f
z
d
μ
(
z
), we have(39)
E
2
≤
C
∫
6
B
i
f
x
d
μ
x
.
Now we turn to estimate forE
1. Let Q
i
=
B
(
c
B
i
,
r
S
i
), and write (40)
E
1
≤
∫
X
∖
6
S
i
∬
d
x
,
y
<
t
t
t
+
d
x
,
y
κ
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
/
2
d
μ
x
+
∫
X
∖
6
S
i
∬
d
x
,
y
≥
t
y
∈
Q
i
t
t
+
d
x
,
y
κ
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
/
2
d
μ
x
+
∫
X
∖
6
S
i
∬
d
x
,
y
≥
t
y
∈
X
∖
Q
i
t
t
+
d
x
,
y
κ
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
/
2
d
μ
x
≕
E
11
+
E
12
+
E
13
.For each fixed i, decompose E
11 as (41)
E
11
≤
∫
X
∖
6
S
i
∬
d
x
,
y
<
t
y
∈
2
S
i
t
t
+
d
x
,
y
κ
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
/
2
d
μ
x
+
∫
X
∖
6
S
i
∬
d
x
,
y
<
t
y
∈
X
∖
2
S
i
t
t
+
d
x
,
y
κ
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
/
2
d
μ
x
≕
I
1
+
I
2
.For any x
∈
X
∖
6
S
i, y
∈
2
S
i with d
(
y
,
x
)
<
t, and z
∈
S
i, d
(
x
,
c
B
i
)
-
2
r
S
i
≤
d
(
x
,
y
)
<
t and d
(
y
,
z
)
<
3
r
S
i, together with Minkowski inequality and (6), we can conclude (42)
I
1
≤
C
∫
X
∖
6
S
i
∫
6
S
i
h
i
z
∬
d
x
,
y
<
t
,
d
y
,
z
≤
t
y
∈
2
S
i
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
z
d
μ
x
≤
C
∫
6
S
i
h
i
z
∫
X
∖
6
S
i
∫
d
y
,
z
≤
3
r
S
i
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
∫
d
x
,
c
B
i
-
2
r
S
i
∞
d
t
λ
y
,
t
t
1
+
2
ρ
d
μ
y
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
S
i
h
i
z
∫
X
∖
6
S
i
∫
d
y
,
z
≤
3
r
S
i
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
1
μ
B
y
,
d
x
,
c
B
i
1
d
x
,
c
B
i
-
2
r
S
i
2
ρ
d
μ
y
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
S
i
h
i
z
∫
X
∖
6
S
i
∫
d
y
,
z
≤
3
r
S
i
1
λ
y
,
d
y
,
z
2
1
μ
B
y
,
d
x
,
c
B
i
d
μ
y
1
/
2
d
μ
x
d
μ
z
≤
C
∫
6
S
i
h
i
z
∫
X
∖
6
S
i
1
λ
c
B
i
,
d
x
,
c
B
i
d
μ
x
d
μ
z
≤
C
h
i
L
1
μ
.For I
2, write (43)
I
2
≤
∫
X
∖
6
S
i
∬
d
x
,
y
<
t
,
y
∈
X
∖
2
S
i
t
≤
d
y
,
c
B
i
+
r
S
i
t
t
+
d
x
,
y
κ
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
/
2
d
μ
x
+
∫
X
∖
6
S
i
∬
d
x
,
y
<
t
,
y
∈
X
∖
2
S
i
t
>
d
y
,
c
B
i
+
r
S
i
t
t
+
d
x
,
y
κ
1
t
ρ
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
/
2
d
μ
x
≕
I
21
+
I
22
.For I
21, by Minkowski inequality and (6), we deduce (44)
I
21
≤
C
∫
X
∖
6
S
i
∫
S
i
h
i
z
∫
y
∈
X
∖
2
S
i
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
1
λ
y
,
d
y
,
c
B
i
+
r
S
i
∫
d
y
,
z
d
y
,
c
B
i
+
r
S
i
d
t
t
1
+
2
ρ
d
μ
y
1
/
2
d
μ
z
d
μ
x
≤
C
∫
X
∖
6
S
i
∫
S
i
h
i
z
∫
y
∈
X
∖
2
S
i
1
λ
y
,
d
y
,
z
2
1
λ
y
,
d
y
,
c
B
i
+
r
S
i
d
μ
y
1
/
2
d
μ
z
d
μ
x
≤
C
∫
S
i
h
i
z
∫
X
∖
6
S
i
1
λ
c
B
i
,
d
x
,
c
B
i
∑
k
=
1
∞
∫
2
k
+
1
6
S
i
∖
2
k
6
S
i
1
λ
y
,
d
y
,
c
B
i
+
r
S
i
d
μ
y
1
/
2
d
μ
x
d
μ
z
≤
C
h
i
L
1
μ
.Now we estimate I
22. Applying Minkowski inequality and the vanishing moment, we have (45)
I
22
≤
C
∫
X
∖
6
S
i
∬
d
x
,
y
<
t
,
y
∈
X
∖
2
S
i
t
>
d
y
,
c
B
i
+
r
S
i
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
-
K
y
,
c
B
i
d
y
,
c
B
i
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
≤
C
∫
X
∖
6
S
i
∬
d
x
,
y
<
t
,
y
∈
X
∖
2
S
i
t
>
d
y
,
c
B
i
+
r
S
i
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
-
K
y
,
z
d
y
,
c
B
i
1
-
ρ
+
K
y
,
z
d
y
,
c
B
i
1
-
ρ
-
K
y
,
c
B
i
d
y
,
c
B
i
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
≤
C
∫
X
∖
6
S
i
∬
d
x
,
y
<
t
,
y
∈
X
∖
2
S
i
t
>
d
y
,
c
B
i
+
r
S
i
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
z
1
-
ρ
-
K
y
,
z
d
y
,
c
B
i
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
+
C
∫
X
∖
6
S
i
∬
d
x
,
y
<
t
,
y
∈
X
∖
2
S
i
t
>
d
y
,
c
B
i
+
r
S
i
∫
d
y
,
z
≤
t
K
y
,
z
d
y
,
c
B
i
1
-
ρ
-
K
y
,
c
B
i
d
y
,
c
B
i
1
-
ρ
h
i
z
d
μ
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
x
≕
J
1
+
J
2
.With a way similar to that used in the proof of I
1, we have J
1
≤
C
h
i
L
1
(
μ
). Thus, we only need to estimate J
2; by Minkowski inequality and (11), it follows that (46)
J
2
≤
C
∫
X
∖
6
S
i
∫
S
i
h
i
z
∫
X
∖
2
S
i
K
y
,
z
-
K
y
,
c
B
i
2
1
d
y
,
c
B
i
2
-
2
ρ
1
λ
y
,
d
y
,
c
B
i
+
r
S
i
∫
d
y
,
c
B
i
+
r
S
i
∞
d
t
t
1
+
2
ρ
d
μ
y
1
/
2
d
μ
z
d
μ
x
≤
C
∫
S
i
h
i
z
∫
X
∖
6
S
i
1
λ
c
B
i
,
d
x
,
c
B
i
∫
X
∖
2
S
i
K
y
,
z
-
K
y
,
c
B
i
2
1
d
y
,
c
B
i
2
d
μ
y
1
/
2
d
μ
x
d
μ
z
≤
C
∫
S
i
h
i
z
∫
X
∖
6
S
i
1
λ
c
B
i
,
d
x
,
c
B
i
∑
k
=
1
∞
∫
2
k
r
S
i
<
d
y
,
c
B
1
≤
2
k
+
1
r
S
i
K
y
,
z
-
K
y
,
c
B
i
2
1
d
y
,
c
B
i
2
d
μ
y
1
/
2
d
μ
x
d
μ
z
≤
C
∫
S
i
h
i
z
∫
X
∖
6
S
i
1
λ
c
B
i
,
d
x
,
c
B
i
d
μ
x
d
μ
z
≤
C
h
i
L
1
μ
.Combining the estimates for J
1, J
2, I
21, and I
1, we obtain that (47)
E
11
≤
C
h
i
L
1
μ
.
Next we estimateE
12. For any y
∈
B
i, x
∈
X
∖
6
S
i, and z
∈
S
i, we have d
(
x
,
y
)
≥
(
1
/
2
)
d
(
x
,
c
B
i
), d
(
y
,
z
)
≤
2
r
S
i, and d
(
x
,
y
)
~
d
(
x
,
c
B
i
), and together with this fact, Minkowski inequality, and (6), we get (48)
E
12
≤
C
∫
X
∖
6
S
i
∫
S
i
h
i
z
∬
d
x
,
y
≥
t
d
y
,
z
≤
t
y
∈
Q
i
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
+
2
ρ
1
/
2
d
μ
z
d
μ
x
≤
C
∫
X
∖
6
S
i
∫
S
i
h
i
z
∫
d
y
,
z
≤
2
r
S
i
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
1
λ
y
,
d
x
,
c
B
i
∫
d
y
,
z
d
x
,
y
d
t
t
1
+
2
ρ
d
μ
y
1
/
2
d
μ
z
d
μ
x
≤
C
∫
X
∖
6
S
i
∫
S
i
h
i
z
∫
d
y
,
z
≤
2
r
S
i
1
λ
y
,
d
y
,
z
2
1
λ
y
,
d
x
,
c
B
i
d
μ
y
1
/
2
d
μ
z
d
μ
x
≤
C
∫
S
i
h
i
z
∫
X
∖
6
S
i
1
λ
c
B
i
,
d
x
,
c
B
i
d
μ
x
d
μ
z
≤
C
h
i
L
1
μ
.It remains to estimate E
13. Applying Minkowski inequality and (6), we have (49)
E
13
≤
C
∫
X
∖
6
S
i
∫
S
i
h
i
x
∬
d
y
,
z
≤
t
≤
d
x
,
y
,
y
∈
X
∖
Q
i
d
x
,
c
B
i
≤
2
d
y
,
c
B
i
t
t
+
d
x
,
y
κ
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
/
2
d
μ
z
d
μ
x
+
C
∫
X
∖
6
S
i
∫
S
i
h
i
x
∬
d
y
,
z
≤
t
≤
d
x
,
y
,
y
∈
X
∖
Q
i
d
x
,
c
B
i
>
2
d
y
,
c
B
i
t
t
+
d
x
,
y
κ
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
d
μ
y
d
t
λ
y
,
t
t
1
/
2
d
μ
z
d
μ
x
≕
U
1
+
U
2
.Now we estimate U
1. For any y
∈
X
∖
Q
i, z
∈
S
i, and d
(
y
,
z
)
≤
t
≤
d
(
x
,
z
), it is easy to see d
(
y
,
z
)
~
d
(
y
,
c
B
i
). So we have(50)
U
1
≤
C
∫
X
∖
6
S
i
∫
S
i
h
i
z
∫
y
∈
X
∖
Q
i
d
x
,
c
B
i
≤
2
d
y
,
c
B
i
d
y
,
z
2
ρ
λ
y
,
d
y
,
z
2
1
λ
y
,
d
x
,
y
∫
d
y
,
z
d
x
,
y
d
t
t
1
+
2
ρ
d
μ
y
1
/
2
d
μ
z
d
μ
x
≤
C
∫
S
i
h
i
z
∫
X
∖
6
S
i
∫
y
∈
X
∖
Q
i
d
x
,
c
B
i
≤
2
d
y
,
c
B
i
1
λ
y
,
d
y
,
z
2
1
λ
y
,
d
x
,
y
d
μ
y
1
/
2
d
μ
x
d
μ
z
≤
C
∫
S
i
h
i
z
∫
X
∖
6
S
i
1
λ
c
B
i
,
d
x
,
c
B
i
∫
y
∈
X
∖
Q
i
d
x
,
c
B
i
≤
2
d
y
,
c
B
i
d
μ
y
λ
y
,
d
y
,
c
B
i
1
/
2
d
μ
x
d
μ
z
≤
C
∫
S
i
h
i
z
∫
X
∖
6
S
i
1
λ
c
B
i
,
d
x
,
c
B
i
d
μ
x
d
μ
z
≤
C
h
i
L
1
μ
.
On the other hand, by a method similar to that used in the proof ofU
1, we have (51)
U
2
≤
C
h
i
L
1
μ
.Combining the estimates U
1, U
2, E
11, E
12, and the fact that h
i
L
1
(
μ
)
≤
C
∫
6
B
i
‍
f
x
d
μ
(
x
), we conclude that (52)
E
1
≤
C
∫
6
B
i
f
x
d
μ
x
,which, together with E
2, implies (30) and the proof of Theorem 10 is finished.Proof of Theorem11.
Without loss of generality, we assumeζ
=
2. By a standard argument, it suffices to show that, for any (
∞
,
1
)-atomic block b, (53)
M
κ
∗
,
ρ
b
L
1
μ
≤
C
b
H
a
t
b
1
,
∞
μ
.Assume that supp
⁡
b
⊂
R and b
=
∑
i
=
1
2
υ
i
a
i, where, for i
∈
{
1,2
}, a
i is a function supported in B
i
⊂
R such that a
i
L
∞
(
μ
)
≤
[
μ
(
4
B
i
)
]
-
1
K
B
i
,
R
-
1 and υ
1
+
υ
2
~
b
H
a
t
b
1
,
∞
(
μ
). Write (54)
∫
X
M
κ
∗
,
ρ
b
x
d
μ
x
=
∫
2
R
M
κ
∗
,
ρ
b
x
d
μ
x
+
∫
X
∖
2
R
M
κ
∗
,
ρ
b
x
d
μ
x
≕
V
1
+
V
2
.For V
1, we see that (55)
V
1
≤
∑
i
=
1
2
υ
i
∫
2
B
i
M
κ
∗
,
ρ
a
i
x
d
μ
x
+
∑
i
=
1
2
υ
i
∫
2
R
∖
2
B
i
M
κ
∗
,
ρ
a
i
x
d
μ
x
≕
V
11
+
V
12
.Applying the Hölder inequality, L
2
(
μ
)-boundedness of M
κ
∗
,
ρ, and the fact that a
i
L
∞
(
μ
)
≤
C
[
μ
(
4
B
i
)
]
-
1
K
B
i
,
R
-
1 for i
∈
{
1,2
}, we have(56)
V
11
≤
∑
i
=
1
2
υ
i
∫
2
B
i
M
κ
∗
,
ρ
a
i
x
2
d
μ
x
1
/
2
μ
2
B
i
1
/
2
≤
C
∑
i
=
1
2
υ
i
a
i
L
2
μ
μ
2
B
i
1
/
2
≤
C
b
H
a
t
b
1
,
∞
μ
.Now we estimate V
12, with a method similar to that used in the proof of F
1 and a
i
L
∞
(
μ
)
≤
C
[
μ
(
4
B
i
)
]
-
1
K
B
i
,
R
-
1, and we see that (57)
V
12
≤
C
∑
i
=
1
2
υ
i
a
i
L
1
μ
∫
2
R
∖
2
B
i
1
λ
c
B
i
,
d
x
,
c
B
i
d
μ
x
≤
C
∑
i
=
1
2
υ
i
K
B
i
,
R
a
i
L
∞
μ
μ
B
i
≤
C
b
H
a
t
b
1
,
∞
μ
.Therefore, V
1
≤
C
b
H
a
t
b
1
,
∞
(
μ
).
On the other hand, based on the proof ofE
1 and Definition 8, it is easy to obtain that(58)
V
2
≤
C
b
L
1
μ
≤
C
b
H
a
t
b
1
,
∞
μ
.Combining the estimates for V
1 and V
2, (53) holds. Thus, Theorem 11 is completed.



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*Source: 9091478-2016-04-11.xml*