# Boundedness of Marcinkiewicz Integrals on RBMO Spaces over Nonhomogeneous Metric Measure Spaces

Authors: Ji Cheng; Guanghui Lu
Journal: Journal of Mathematics
 (2015)

---

## Abstract

LetX
,
d
,
μ be a metric measures space satisfying the upper doubling conditions and the geometrically doubling conditions in the sense
of Hytönen. Under the assumption that the dominating function satisfies the weak reverse doubling condition, the authors prove that Marcinkiewicz
integral with kernel satisfying certain stronger Hörmander-type condition is bounded on RBMOμ space.


---

## Body

## 1. Introduction

We all know that the Littlewood-Paleyg
-function has been playing an important role in harmonic analysis and the Marcinkiewicz integral is an essential Littlewood-Paley g
-function. As an analogy to Littlewood-Paley g
-function without going into the interior of the unit disk, in 1938, Marcinkiewicz introduced the integral on one dimensional Euclidean space R, which now is called the Marcinkiewicz integral, and conjectured that it is bounded on L
p
(
[
0,2
π
]
) for p
∈
(
1
,
∞
) (see [1]). In 1944, Zygmund proved the Marcinkiewicz conjecture by using the complex variable method in [2]. Particularly, in 1958, Stein introduced the higher dimensional Marcinkiewicz integral (see [3]). Let Ω be homogeneous of degree zero in R
n for n
≥
2 and integrable and have mean value zero on the unit sphere S
n
-
1. The higher dimensional Marcinkiewicz integral M
Ω is defined by (1)
M
Ω
f
x
≔
∫
0
∞
‍
∫
x
-
y
≤
t
‍
Ω
x
-
y
x
-
y
n
-
1
f
y
d
y
2
d
t
t
3
1
/
2
,
x
∈
R
n
.Recently, many papers focus on the boundedness of this operator on various function spaces. We refer the reader to see [4, 5].Many results from real analysis and harmonic analysis on the classical Euclidean spaces have been extended to the space of homogeneous type by Coifman and Weiss in [6]. Recall that a metric space (
X
,
d
) equipped with a Borel measure μ is called a space of homogeneous type, if (
X
,
d
,
μ
) satisfies the following doubling measure condition that there exists a positive constant C
μ such that, for all balls B
(
x
,
r
)
≔
{
y
∈
X
:
d
(
x
,
y
)
<
r
} with x
∈
X and r
∈
(
0
,
∞
),(2)
μ
B
x
,
2
r
≤
C
μ
μ
B
x
,
r
.Meanwhile, many classical results concerning the theory of Calderón-Zygmund operators and function spaces have been proved still valid for nondoubling measures. In particular, letμ be a nonnegative Radon measure on R
n which only satisfies the polynomial growth condition that there exists a positive constant C
0 and d
∈
(
0
,
n
] such that, for all x
∈
R
n and r
∈
(
0
,
∞
), (3)
μ
B
x
,
2
r
≤
C
0
r
d
,where B
(
x
,
r
)
≔
{
y
∈
R
n
:
|
x
-
y
|
<
r
}. Such a measure μ need not satisfy the doubling condition (2). The analysis with nondoubling measures plays an important role in solving the longstanding open Painlevé problem by Tolsa in [7]. In 2007, Hu et al. introduced the Marcinkiewicz integral with nondoubling measures and got some excellent results in [8]. In 2012, Wang et al. established the boundedness of Marcinkiewicz integrals on RBMO
(
μ
) in [9].However, in 2010, Hytönen pointed out that the measures satisfying the polynomial growth condition are different from the doubling measures in [10]. Hytönen introduced a new class of metric measure spaces which satisfy the upper doubling condition and the geometrically doubling condition (resp., see Definitions 1 and 3 below). This new class of metric measure space is called the nonhomogeneous metric measure space, which includes both spaces of homogeneous type and metric spaces with the measures satisfying (3) as special cases. We refer the reader to the monograph [11] for several recent developments on harmonic analysis in this setting.In this paper, we mainly discuss the boundedness of Marcinkiewicz integralsM on RBMO
(
μ
) with nonhomogeneous metric measure spaces. Now we recall some notations and definitions as follows.Definition 1.
A metric measure space(
X
,
d
,
μ
) is said to be upper doubling, if μ is Borel measure on X and there exists 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
,
∞
), (4)
μ
B
x
,
r
≤
λ
x
,
r
≤
C
λ
λ
x
,
r
2
.Remark 2.
(
1
) Obviously, a space of homogeneous type is a special case of upper doubling spaces, where one can take the dominating function λ
(
x
,
r
)
=
μ
(
B
(
x
,
r
)
). Moreover, let μ be a nonnegative Radon measure on R
n which only satisfies the Polynomial growth condition. By taking λ
(
x
,
r
)
=
C
r
n, we see that (
R
n
,
|
·
|
,
μ
) is also an upper doubling measure space.
(
2
) It was proved that there exists a dominating function λ
~ related to λ satisfying the property that there exists a positive constant C
λ
~ such that λ
~
≤
λ, C
λ
~
≤
C
λ and, for all x
,
y
∈
χ with d
(
x
,
y
)
≤
λ, (5)
λ
~
x
,
r
≤
C
λ
~
λ
~
y
,
r
.Based on this, in this paper, we always assume that the dominating function λ also satisfies (5).Definition 3.
A metric space(
X
,
d
) is said to be geometrically doubling, if there exist some N
0
∈
N
=
{
1,2
,
…
} 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 4.
Let(
X
,
d
) be a metric space. Hytönen showed that the following statements are mutually equivalent (see [10]):(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
ϵ
-
n. Here and in what follows, N
0 is as in Definition 3 and n
=
log
⁡
2
N
0.
(3)
For everyϵ
∈
(
0,1
), any ball B
(
x
,
r
)
⊂
X can contain at most N
ϵ
-
n centers {
x
i
}
i of disjoint balls with radius ϵ
r.
(4)
There existsM
∈
N such that any ball B
(
x
,
r
)
⊂
X can contain at most N
ϵ
-
n centers {
x
i
}
i of disjoint balls {
B
(
x
i
,
r
/
4
)
}
i
=
1
M.Definition 5.
For any two ballsB
⊂
S, define (6)
K
B
,
S
=
1
+
∫
2
S
∖
B
‍
1
λ
c
B
,
d
x
,
c
B
d
μ
x
,where above and in that follows, for a ball B
=
B
(
c
B
,
r
B
) and ρ
>
0, ρ
B
=
B
(
c
B
,
ρ
r
B
)
.
c
B is the center of ball B.Remark 6.
The following discrete version,K
~
B
,
S, of K
B
,
S defined in Definition 5 was first introduced by Bui and Duong in nonhomogeneous metric measure spaces (see [12]), which is more close to the quantity K
Q
,
R introduced by Tolsa [13] in the setting of nondoubling measures. For any two balls B
⊂
S, let K
~
B
,
S be defined by(7)
K
~
B
,
S
≔
1
+
∑
k
=
1
N
B
,
S
‍
μ
6
k
B
λ
c
B
,
6
k
r
B
,where r
B and r
S, respectively, denote the radius of the balls B and S and N
B
,
S is the smallest integer satisfying 6
N
B
,
S
r
B
≥
r
S. Obviously, K
B
,
S
≤
C
K
~
B
,
S. That was pointed by Bui and Duong in [12]; in general, it is not true that K
B
,
S
~
C
K
~
B
,
S.Definition 7.
Letα
,
β
∈
(
0
,
∞
). A ball B
⊂
X is called (
α
,
β
)-doubling if μ
(
α
B
)
≤
β
μ
(
B
).It was proved in [10] that if a metric measure space (
X
,
d
,
μ
) is upper doubling and β
>
C
λ
log
⁡
2
α
=
α
ν, then, for every ball B
⊂
X, there exists some j
∈
Z
+ such that α
j
β is (
α
,
β
)
-doubling. Moreover, let (
X
,
d
) be geometrically doubling, β
>
α
n with n
=
log
⁡
2
N
0, and μ is Borel measure on X which is finite on bounded sets. In [10] Hytönen also showed that, for μ
-almost every x
∈
X, there exist arbitrarily small (
α
,
β
)
-doubling balls centered at x. Furthermore, the radius of these balls may be chosen to be form α
-
j
r for j
∈
N and any preassigned number r
∈
(
0
,
∞
). Throughout this paper, for any α
∈
(
1
,
∞
) and ball B, B
~ denotes the smallest (
α
,
β
α
)
-doubling ball of the form α
j
B with j
∈
Z
+, where (8)
β
α
=
max
⁡
α
3
n
,
α
3
ν
+
3
0
n
+
3
0
ν
=
α
3
(
max
⁡
(
n
,
ν
)
)
+
3
0
n
+
3
0
ν
.Ifα
=
6, we denote the ball B
~
α simply by B
~.LetK
(
x
,
y
) be a μ-locally integrable function on X
×
X
∖
{
(
x
,
x
)
:
x
∈
X
}. Assume that there exists a positive constant C such that, for any x
,
y
∈
X with x
≠
y, (9)
K
x
,
y
≤
C
d
x
,
y
λ
x
,
d
x
,
y
,and, for any x
,
y
,
y
′
∈
X, (10)
∫
d
x
,
y
≥
2
d
y
,
y
′
‍
K
x
,
y
-
K
x
,
y
′
hhhhhhhhhhhh
+
K
y
,
x
-
K
y
′
,
x
1
d
x
,
y
d
μ
x
≤
C
.

The Marcinkiewicz integralM
(
f
) associated with the above kernel K
(
x
,
y
) is defined by (11)
M
f
x
=
∫
0
∞
‍
∫
d
x
,
y
≤
t
‍
K
x
,
y
f
y
d
μ
y
2
d
t
t
3
1
/
2
,
x
∈
X
.

Obviously, by takingλ
(
x
,
r
)
=
C
r
n, we see that, in the classical Euclidean space R
n, if (12)
K
x
,
y
=
Ω
x
-
y
x
-
y
n
-
1
,with Ω homogeneous of degree zero and Ω
∈
L
i
p
α
(
S
d
-
1
) for some α
∈
(
0,1
], then K
(
x
,
y
) satisfies (5) and (8). In 2014, Lin and Yang [14] established the equivalent boundedness of Marcinkiewicz integral M with kernel K
(
x
,
y
) satisfying (9) and (10). In this note, we make some modification for the kernel. Besides satisfying the regular condition (9), K
(
x
,
y
) also satisfies that, for any x
,
x
′ and y
∈
X with d
(
x
,
y
)
≥
2
d
(
x
,
x
′
), there exists 0
<
δ
≤
1 such that(13)
K
x
,
y
-
K
x
′
,
y
+
K
y
,
x
-
K
y
,
x
′
≤
d
x
,
x
′
δ
d
x
,
y
1
-
δ
λ
x
,
d
x
,
y
,and, for any x
∈
X and any two positive real numbers ɛ
,
N with ɛ
<
N,(14)
∫
ɛ
<
d
x
,
y
<
N
‍
K
x
,
y
d
μ
y
=
0
.When the kernel satisfies condition (13), it also satisfies (10).Definition 8.
Letη
∈
(
1
,
∞
). A function f
∈
L
loc
⁡
1
(
μ
) is said to be the space RBMO
(
μ
), if there exists a positive constant C and a number f
B for any ball B such that, for all balls B, (15)
1
μ
η
B
∫
B
‍
f
y
-
f
B
d
μ
y
≤
Cand, for balls B
⊂
S, (16)
f
B
-
f
S
≤
C
K
B
,
S
.
Moreover, the norm off in RBMO
(
μ
) is defined to be the minimal constant C as above and denoted by f
RBMO
(
μ
).Definition 9.
A dominating functionλ is said to satisfy the weak reverse doubling condition if, for all r
∈
(
0,2
diam
⁡
(
X
)
) and a
∈
(
1,2
diam
⁡
(
X
)
/
r
), there exists a number C
(
a
)
∈
[
1
,
∞
), depending only on a and X, such that, for all x
∈
X, (17)
λ
x
,
a
r
≥
C
a
λ
x
,
rand, moreover, (18)
∑
‍
k
=
1
∞
1
C
a
k
<
∞
.Remark 10.
If the dominating functionλ satisfies the weak reverse doubling condition, then K
B
,
S
~
K
~
B
,
S for any two balls; see [15].Throughout this paper, we denote byC a positive constant which is independent of the main parameters involved, but it may be different from line to line. For any p
∈
[
1
,
∞
], we denote by p
′ its conjugate index; namely, 1
/
p
+
1
/
p
′
=
1.


## 2. Main Result and Its Proof

We give the main result as follows.Theorem 11.
Let(
X
,
d
,
μ
) be a nonhomogeneous metric measure space with the dominating function satisfying the weak reverse doubling condition, let K
(
x
,
y
) satisfy conditions (9), (13), and (14), and let M be defined as (11). If M is bounded on L
2
(
μ
), then, for any f
∈
R
B
M
O
(
μ
), (19)
M
f
R
B
M
O
(
μ
)
≤
C
f
R
B
M
O
(
μ
)
,where C is a positive constant independent of f, only when M
(
f
) is finite μ
-almost everywhere; otherwise, M
(
f
)
∉
R
B
M
O
(
μ
).In order to prove the theorem, we need the following two lemmas. The following useful properties ofK
B
,
S were proved in [10, 16]; see also [17].Lemma 12.
(i) For all ballsB
⊂
R
⊂
S, K
B
,
R
≤
K
R
,
S.
(ii) For anyρ
∈
[
1
,
∞
), there exists a positive constant C, depending on ρ, such that, for all balls B
⊂
S with r
S
≤
ρ
r
B, K
B
,
S
≤
C.
(iii) For anyα
∈
[
1
,
∞
), there exists a positive constant C
~, depending on α, such that, for all balls B, K
B
,
B
~
α
≤
C
~.
(iv) 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.
(v) There exists a positive constantc
~ such that, for all balls B
⊂
R
⊂
S, K
R
,
S
≤
c
~
K
B
,
S; moreover, if B and R are concentric, then K
R
,
S
≤
K
B
,
S.We also need the following characterization ofRBMO
(
μ
); see [16, Proposition 2.10] and [18, Lemma 3.2].Lemma 13.
Letα
∈
(
1
,
∞
), and let β
α be as in (8). For f
∈
L
l
o
c
1
(
μ
), the following statements are equivalent:(i)
f
∈
R
B
M
O
(
μ
);
(ii)
there exists a nonnegative constantC such that, for all (
α
,
β
α
)
-doubling balls B,(20)
1
μ
B
∫
B
‍
f
y
-
m
B
f
d
μ
y
≤
C
,and that, for all (
α
,
β
α
)
-doubling balls B
⊂
S,(21)
m
B
f
-
m
S
f
≤
C
K
B
,
S
,where and in what follows m
B
(
f
)
=
(
1
/
μ
(
B
)
)
∫
B
‍
f
(
y
)
d
μ
(
y
).Moreover, the minimal constantC as above is equivalent to f
RBMO
(
μ
).Proof of Theorem11.
Forf
∈
RBMO
(
μ
), we have that M
(
f
) is finite μ
-almost everywhere and decompose(22)
f
x
=
f
x
-
f
6
S
χ
6
S
x
+
f
x
-
f
6
S
χ
X
∖
6
S
x
+
f
6
S
≔
f
1
x
+
f
2
x
+
f
3for any two balls B
⊂
S.
By the vanishing (14), we have (23)
∫
d
x
,
y
≤
t
‍
K
x
,
y
f
3
d
μ
y
=
lim
ɛ
→
0
∫
ɛ
<
d
x
,
y
≤
t
‍
K
x
,
y
f
3
d
μ
y
=
0
.
Furthermore,(24)
M
f
3
x
=
∫
0
∞
‍
∫
d
x
,
y
≤
t
‍
K
x
,
y
f
3
d
μ
y
2
d
t
t
3
1
/
2
=
0
.
By applying the Minkowski inequality and vanishing condition (14), we have (25)
M
f
x
≤
M
f
1
x
+
M
f
2
x
,
M
f
2
x
≤
M
f
1
x
+
M
f
x
.
Then it is easy to get that, for anyx
,
y
∈
X, (26)
M
f
x
-
M
f
y
≤
M
f
1
x
+
M
f
2
x
-
M
f
2
y
+
M
f
1
y
,
M
f
y
-
M
(
f
)
(
x
)
≤
M
f
1
y
+
M
f
2
y
-
M
f
2
x
+
M
f
1
x
,which follows that (27)
M
f
x
-
M
f
y
≤
M
f
1
x
+
M
f
2
x
-
M
f
2
y
+
M
f
1
y
.
So, we deduce(28)
∫
B
‍
∫
S
‍
M
f
x
-
M
f
y
d
μ
y
d
μ
x
≤
∫
B
‍
∫
S
‍
M
f
2
x
-
M
f
2
y
d
μ
y
d
μ
x
+
μ
B
∫
S
‍
M
f
1
y
d
μ
y
+
μ
S
∫
B
‍
M
f
1
y
d
μ
y
.Applying Hölder’s inequality, [10, Corollary 6.3], and L
2
(
μ
)
-boundedness of M, we deduce that(29)
μ
B
∫
S
‍
M
f
1
y
d
μ
y
≤
μ
B
μ
S
1
/
2
∫
X
‍
M
f
1
x
2
d
μ
y
1
/
2
≤
C
μ
B
μ
S
1
/
2
∫
X
‍
f
1
y
2
d
μ
y
1
/
2
=
C
μ
B
μ
S
1
/
2
∫
6
S
‍
f
y
-
f
6
S
2
d
μ
y
1
/
2
≤
C
μ
B
μ
S
1
/
2
μ
36
S
1
/
2
f
RBMO
(
μ
)
≤
C
μ
B
μ
36
S
f
RBMO
(
μ
)
.
There exists a nonnegative integerm such that (30)
6
m
B
⊂
6
S
⊂
6
m
+
2
,
36
S
⊂
6
m
+
3
B
.
Setf
4
≔
(
f
-
f
6
m
B
)
χ
6
m
B; then we get (31)
f
1
-
f
4
=
f
-
f
6
S
χ
6
S
∖
6
m
B
+
f
6
m
B
-
f
6
S
χ
6
m
B
.
For anyx
∈
B, by applying the Minkowski inequality and a similar argument to (29) and (9), we have that(32)
M
f
-
f
6
S
χ
6
S
∖
6
m
B
x
≤
∫
6
S
∖
6
m
B
‍
K
x
,
y
f
y
-
f
6
S
∫
d
x
,
y
∞
‍
d
t
t
3
1
/
2
d
μ
y
≤
C
∫
6
S
∖
6
m
B
‍
1
λ
x
,
d
x
,
y
f
y
-
f
6
S
d
μ
y
≤
C
μ
6
m
+
3
B
λ
c
B
,
6
m
r
B
f
RBMO
μ
≤
C
f
RBMO
(
μ
)
.
By using Definition8 and Lemma 12, we can get (33)
f
6
m
B
-
f
6
S
≤
C
K
6
m
B
,
6
m
+
3
B
f
RBMO
(
μ
)
≤
C
f
RBMO
(
μ
)
.
Denotea
≔
f
6
m
B
-
f
6
S. Let us estimate M
(
a
χ
6
m
B
)
(
x
): (34)
M
a
χ
6
m
B
x
=
∫
0
∞
‍
∫
d
x
,
y
≤
t
‍
K
x
,
y
a
χ
6
m
B
y
d
μ
y
2
d
t
t
3
1
/
2
.
There are three cases about the relation of6
m
B and {
y
∈
X
:
d
(
x
,
y
)
≤
t
}.
Case I. Consider  {
y
∈
X
:
d
(
x
,
y
)
≤
t
}
⊂
6
m
B; then by using the vanishing condition (14), we have (35)
∫
d
x
,
y
≤
t
‍
K
x
,
y
a
χ
6
m
B
y
d
μ
y
=
0
.
Case II. Consider  6
m
B
⊂
{
y
∈
X
:
d
(
x
,
y
)
≤
t
}; then by using the vanishing condition (14), we have (36)
-
∫
d
x
,
y
≤
t
‍
K
x
,
y
a
χ
6
m
B
y
d
μ
y
=
-
∫
6
m
B
‍
K
x
,
y
a
d
μ
y
=
∫
y
∈
X
:
d
x
B
,
y
≤
6
m
·
3
r
B
∖
6
m
B
‍
K
x
,
y
a
d
μ
y
.
Case III. Others. By applying the vanishing condition (14), we also get (37)
∫
d
x
,
y
≤
t
‍
K
x
,
y
a
χ
6
m
B
y
d
μ
y
=
-
∫
y
∈
X
:
d
x
,
y
≤
t
∖
6
m
B
‍
K
x
,
y
a
d
μ
y
.
And, in this case, it is easy to get that{
y
∈
X
:
d
(
x
,
y
)
≤
t
}
⊂
{
y
∈
X
:
d
(
x
B
,
y
)
≤
6
m
6
r
B
}.
From this and the Minkowski inequality, it follows that(38)
M
a
χ
6
m
B
x
=
∫
0
∞
‍
∫
d
x
,
y
≤
t
‍
K
x
,
y
a
χ
6
m
B
y
d
μ
y
2
d
t
t
3
1
/
2
≤
C
f
RBMO
μ
∫
6
m
+
1
B
∖
6
m
B
‍
1
λ
x
,
d
x
,
y
d
μ
y
≤
C
f
RBMO
(
μ
)
.
From (31), (32), and (38), we deduce that (39)
∫
B
‍
M
f
1
-
f
4
y
d
μ
y
≤
C
μ
B
f
RBMO
μ
.
On the other hand,(40)
f
4
=
f
-
f
6
m
B
χ
6
m
B
=
f
-
f
6
B
χ
6
B
+
∑
‍
j
=
1
m
-
1
f
-
f
6
j
+
1
B
χ
6
j
+
1
B
∖
6
j
B
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
H
h
h
-
f
6
j
B
-
f
6
j
+
1
B
χ
6
j
B
f
-
f
6
j
+
1
B
χ
6
j
+
1
B
∖
6
j
B
.
By applying a similar argument of (29), (32), and (38), we have (41)
∫
B
‍
M
f
-
f
6
B
χ
6
B
y
d
μ
y
≤
C
μ
36
B
f
RBMO
(
μ
)
,
M
(
f
-
f
6
j
+
2
B
χ
6
j
+
2
B
∖
6
j
B
)
≤
C
μ
6
j
+
2
B
λ
c
B
,
6
j
+
2
r
B
f
RBMO
(
μ
)
,
M
f
6
j
B
-
f
6
j
+
2
B
χ
6
j
B
≤
C
μ
6
j
+
2
B
λ
c
B
,
6
j
+
2
r
B
f
RBMO
(
μ
)
.
From the above inequalities, it follows that(42)
∫
B
‍
M
f
4
x
d
μ
x
≤
C
μ
36
B
K
B
,
S
f
RBMO
(
μ
)
,and by (39), we deduce that (43)
μ
S
∫
S
‍
M
f
1
x
d
μ
x
≤
C
μ
S
μ
36
B
K
B
,
S
f
RBMO
μ
.
Now let us estimate(44)
M
f
2
x
-
M
f
2
y
≤
C
f
RBMO
μ
.
To prove (44), suppose (45)
D
1
≔
∫
0
∞
‍
∫
d
y
,
z
≤
t
≤
d
x
,
z
‍
K
x
,
z
f
2
z
d
μ
z
2
d
t
t
3
1
/
2
,



D
2
≔
∫
0
∞
‍
∫
d
x
,
z
≤
t
≤
d
y
,
z
‍
K
y
,
z
f
2
z
d
μ
z
2
d
t
t
3
1
/
2
,



D
3
≔
·
f
2
z
d
μ
z
∫
max
⁡
(
d
(
x
,
z
)
,
d
(
y
,
z
)
)
<
t
‍
K
x
,
z
-
K
y
,
z
2
d
t
t
3
∫
0
∞
‍
∫
max
⁡
(
d
(
x
,
z
)
,
d
(
y
,
z
)
)
<
t
‍
K
(
x
,
z
)
-
K
(
y
,
z
)
h
h
h
h
h
h
h
h
h
h
h
h
h
·
f
2
z
d
μ
z
∫
max
⁡
(
d
(
x
,
z
)
,
d
(
y
,
z
)
)
<
t
‍
K
x
,
z
-
K
y
,
z
2
d
t
t
3
1
/
2
.
Now we estimate|
M
(
f
2
)
(
x
)
-
M
(
f
2
)
(
y
)
|,(46)
M
f
2
x
-
M
f
2
y
=
∫
0
∞
‍
∫
d
x
,
z
≤
t
‍
K
x
,
z
f
2
z
∣
d
μ
z
2
d
t
t
3
1
/
2
h
h
h
h
h
-
∫
0
∞
‍
∫
d
y
,
z
≤
t
‍
K
y
,
z
f
2
z
d
μ
z
2
d
t
t
3
1
/
2
≤
∫
0
∞
‍
-
∫
d
y
,
z
≤
t
‍
K
y
,
z
f
2
z
d
μ
z
2
∫
d
x
,
z
≤
t
‍
K
x
,
z
f
2
z
d
μ
z
hhhhhhhhhh
-
∫
d
y
,
z
≤
t
‍
K
y
,
z
f
2
z
d
μ
z
2
d
t
t
3
∫
0
∞
‍
-
∫
d
y
,
z
≤
t
‍
K
y
,
z
f
2
z
d
μ
z
2
∫
d
x
,
z
≤
t
‍
K
x
,
z
f
2
z
d
μ
z
1
/
2
≤
∑
‍
i
=
1
3
D
i
.
Forx
,
y
∈
S, applying the Minkowski inequality and (9), we can deduce that(47)
D
1
≤
∫
d
y
,
z
≤
d
x
,
z
‍
d
x
,
z
λ
x
,
d
x
,
z
∫
d
y
,
z
d
x
,
z
‍
d
t
t
3
1
/
2
d
μ
z
≤
C
∫
d
y
,
z
≤
d
x
,
z
‍
f
2
z
λ
x
,
d
x
,
z
d
x
,
y
d
x
,
z
1
/
2
d
μ
z
≤
C
r
S
1
/
2
∫
X
∖
6
S
‍
f
z
-
f
6
S
λ
x
,
d
x
,
z
d
μ
z
d
x
,
z
1
/
2
≤
C
r
S
1
/
2
∑
‍
k
=
1
∞
∫
6
k
+
1
S
∖
6
k
S
‍
f
z
-
f
6
S
λ
x
,
d
x
,
z
d
μ
z
d
x
,
z
1
/
2
≤
C
r
S
1
/
2
∑
‍
k
=
1
∞
μ
6
k
+
1
S
λ
c
S
,
6
k
+
1
S
1
6
k
r
S
1
/
2
1
μ
6
k
+
1
S
·
∫
6
k
+
1
S
‍
f
z
-
f
6
S
d
μ
z
≤
C
∑
‍
k
=
1
∞
6
-
k
/
2
1
μ
6
k
+
1
S
·
∫
6
k
+
1
S
‍
f
z
-
f
6
k
+
1
S
+
f
6
k
+
1
S
-
f
6
S
d
μ
z
≤
C
∑
‍
k
=
1
∞
6
-
k
/
2
k
+
1
f
RBMO
(
μ
)
≤
C
f
RBMO
(
μ
)
.
ForD
2, by using a similar method of D
1, we have (48)
D
2
≤
C
f
RBMO
(
μ
)
.
Now let us estimateD
3; by applying (13), we obtain(49)
D
3
≤
∫
X
‍
K
x
,
z
-
K
y
,
z
f
2
z
·
∫
max
⁡
⁡
d
x
,
z
,
d
y
,
z
≤
t
‍
d
t
t
3
1
/
2
d
μ
z
≤
C
∫
X
∖
6
S
‍
d
x
,
y
δ
d
x
,
z
1
-
δ
λ
x
,
d
x
,
z
·
f
z
-
f
6
S
1
d
x
,
z
d
μ
z
≤
C
∫
X
∖
6
S
‍
d
x
,
y
δ
d
x
,
z
-
δ
λ
x
,
d
x
,
z
f
z
-
f
6
S
d
μ
z
≤
C
r
S
δ
∑
‍
k
=
1
∞
∫
6
k
+
1
S
∖
6
k
S
‍
f
z
-
f
6
S
λ
x
,
d
x
,
z
1
d
x
,
z
δ
d
μ
z
≤
C
r
S
δ
∑
‍
k
=
1
∞
μ
6
k
+
1
S
λ
x
S
,
r
6
k
S
1
6
k
r
S
δ
1
μ
6
k
+
1
S
·
∫
6
k
+
1
S
‍
f
z
-
f
6
S
d
μ
z
≤
C
∑
‍
k
=
1
∞
6
-
k
δ
1
μ
6
k
+
1
S
·
∫
6
k
+
1
S
‍
f
z
-
f
6
k
+
1
S
+
f
6
k
+
1
S
-
f
6
S
d
μ
z
≤
C
f
RBMO
μ
.
Combining theD
1, D
2, and D
3, we know that (44) holds. And then we have (50)
∫
B
‍
∫
S
‍
M
f
2
x
-
M
f
2
y
d
μ
y
d
μ
x
≤
C
μ
B
μ
S
f
RBMO
(
μ
)
.
Finally, we obtain that, for any two ballsB
⊂
S,(51)
∫
B
‍
∫
S
‍
M
f
x
-
M
f
y
d
μ
y
d
μ
x
≤
C
K
B
,
S
f
RBMO
μ
μ
B
μ
36
S
+
μ
36
B
μ
S
.
Suppose thatB
=
S is a (
36
,
β
36
)
-doubling ball; by (51) we have(52)
1
μ
B
∫
B
‍
M
f
x
-
m
B
M
f
d
μ
x
≤
1
μ
B
μ
B
∫
B
‍
∫
B
‍
M
f
x
-
M
f
y
d
μ
y
d
μ
x


≤
C
f
RBMO
(
μ
)
.
And whenB
⊂
S are two (
36
,
β
36
)
-doubling balls, it can be deduced that(53)
m
B
M
f
-
m
S
M
f
≤
1
μ
B
μ
S
∫
B
‍
∫
S
‍
M
f
x
-
M
f
y
d
μ
y
d
μ
x
≤
C
f
RBMO
(
μ
)
K
B
,
S
.
Applying Lemma13 and the above two inequalities, we have (54)
M
f
RBMO
(
μ
)
≤
C
f
RBMO
(
μ
)
.
Iff
∈
RBMO
(
μ
) such that M
(
f
) is infinite on a set of positive measures, it is easy to prove that we take a ball B such that (55)
∫
B
‍
M
f
x
=
∞
;namely, M
(
f
)
(
x
)
∉
L
loc
⁡
(
μ
). So, M
(
f
)
(
x
)
∉
RBMO
(
μ
).
We complete the proof of Theorem11.



---

*Source: 318797-2015-04-19.xml*