# Graphs with Sudoku number $n-1$

Alexey Pokrovskiy 111Email<EMAIL_ADDRESS>Address:
Department of Mathematics, University College London, UK.

###### Abstract

Recently Lau-Jeyaseeli-Shiu-Arumugam introduced the concept of the “Sudoku
colourings” of graphs — partial $\chi(G)$-colourings of $G$ that have a unique
extension to a proper $\chi(G)$-colouring of all the vertices. They introduced
the Sudoku number of a graph as the minimal number of coloured vertices in a
Sudoku colouring. They conjectured that a connected graph has Sudoku number
$n-1$ if, and only if, it is complete. In this note we prove that this is
true.

##  1 Introduction

All colourings in this note are vertex-colourings. A vertex colouring of a
graph is _proper_ if adjacent vertices receive different colours. The
chromatic number, $\chi(G)$ is the minimal number of colours in a proper
colouring of $G$. A _partial_ proper colouring of a graph is a colouring of
some subset of $V(G)$ which doesn’t give adjacent vertices the same colour. We
say that a colouring $\phi$ extends a partial colouring $\psi$ if all vertices
coloured in $\psi$ receive the same colour in $\phi$. A variety of tasks can
be encoded as taking a partial proper colouring in a graph and then extending
it to a full proper colouring of all the vertices.

For example the well known “Sudoku puzzle” can be encoded in this form.
Consider a graph $G_{\mathrm{Sudoku}}$ on $27$ vertices that are identified
with the cells in a $9\times 9$ grid. Add an edge between any two vertices in
the same column, between any two vertices in the same row, and between any two
vertices in the same $3\times 3$ box. It is easy to see that a proper
$9$-colouring of $G_{\mathrm{Sudoku}}$ exactly corresponds to filling in
$9\times 9$ array according to the rules of Sudoku puzzles. Thus a Sudoku
puzzle can be summarized as “you are given a partial colouring of
$G_{\mathrm{Sudoku}}$ and need to complete it to a proper colouring of all the
vertices of $G_{\mathrm{Sudoku}}$”.

One of the conventions for designing Sudoku puzzles is that there should
always be precisely one way of filling in the $9\times 9$ array i.e. there
should always exist one solution, and there shouldn’t exist multiple
solutions. This motivates the definition of a Sudoku colouring of a graph.
Lau-Jeyaseeli-Shiu-Arumugam defined it as follows in [1].

######  Definition 1.

A Sudoku colouring of a graph $G$ is a partial proper $\chi(G)$-colouring of
$G$ which has precisely one extension to a $\chi(G)$-colouring of $G$.

Sudoku colourings of $G_{\mathrm{Sudoku}}$ are thus in one-to-one
correspondence with uncompleted Sudoku puzzles (that have unique completion).
But we can also investigate Sudoku colourings of general graphs.

Lau-Jeyaseeli-Shiu-Arumugam [1] defined the Sudoku number of $G$, $sn(G)$ as
the smallest number of coloured vertices in a Sudoku colouring of $G$. The
motivation for this is that $sn(G_{\mathrm{Sudoku}})$ now asks for the minimum
number of clues (i.e. non-blank entries) in a Sudoku puzzle with unique
solution. This number has been determined as $sn(G_{\mathrm{Sudoku}})=17$ by
McGuire, Tugemann, and Civario using a computer-assisted proof [2].

For general graphs, Lau-Jeyaseeli-Shiu-Arumugam [1] determined $sn(G)$ for
various classes of graphs $G$ and obtained bounds for other classes. For
example, they showed that $sn(G)=1$ if, and only if, $G$ is connected and
bipartite. On the other extreme, they showed that $sn(G)\leq|G|-1$ for all
graphs and conjectured that for connected graphs, equality holds if, and only
if, $G$ is complete. Here we show that this is the case.

######  Theorem 2.

A connected graph has $sn(G)=|G|-1$ if, and only if, $G$ is complete.

The backwards direction already appears in [1] (see Corollary 3.4), so we
focus on proving the statement “let $G$ be a connected graph with
$sn(G)=|G|-1$. Then $G$ is complete”. This amounts to showing that non-
complete graphs have partial proper $\chi(G)$-colourings with $\geq 2$
uncoloured vertices which have a unique extension to a full proper
$\chi(G)$-colouring of $G$.

##  2 Proofs

In a $k$-coloured graph our set of colours will always be
$[k]=\\{1,\dots,k\\}$. For a colouring $\phi$ and vertices
$u_{1},\dots,u_{k}$, we use $\phi-u_{1}-\dots-u_{k}$ to mean the partial
colouring formed by uncolouring the vertices $u_{1},\dots,u_{k}$. For a
partial colouring $\phi$ and a set of vertices $S$, we define
$\phi(S):=\\{\phi(s):s\in S\\}$ to mean the set of colours appearing in $S$.
We use $N(v)$ to denote the set of vertices connected to $v$ by an edge (our
graphs are simple, so this never includes $v$ itself).

The following is the tool we use for constructing Sudoku colourings in this
note. It gives two kinds of Sudoku colourings with two uncoloured vertices.

######  Lemma 3.

Let $\psi$ be a partial proper $\chi(G)$-colouring with exactly two uncoloured
vertices $u,v$. Suppose that either of the following holds:

  1. (i)

$uv$ is a nonedge and $|\phi(N(u))|=|\phi(N(v))|=\chi(G)-1$.

  2. (ii)

$uv$ is an edge, $|\phi(N(u))|=\chi(G)-1$, $|\phi(N(v))|=\chi(G)-2$, and
$\phi(N(v))\subset\phi(N(u))$.

Then $\psi$ is a Sudoku colouring.

###### Proof.

In case (i), there is precisely one colour missing from $N(u)$ and precisely
one colour missing from $N(v)$. In a proper $\chi(G)$-colouring, $u$ and $v$
must receive exactly these colours, and so the extension is unique.

In case (ii), there are two colours $c,d$ missing from $N(v)$ and one of these
colours (say $c$), is missing from $N(u)$. To complete the colouring $u$ must
receive colour $c$, and $v$ must receive colour $d$, so the extension is
unique. ∎

The following definition is crucial for us.

######  Definition 4.

Let $\phi$ be a proper $\chi(G)$-colouring of $G$. A vertex is full if it is
adjacent to vertices of all colours (aside from its own) i.e. if
$\phi(N(v))=[\chi(G)]\setminus\phi(v)$ (or equivalently if
$|\phi(N(v))|=\chi(G)-1$).

In graphs with Sudoku number $|G|-1$, it turns out that the full vertices form
a complete subgraph.

######  Lemma 5.

Let $sn(G)=|G|-1$ and let $\phi$ be a proper $\chi(G)$-colouring of $G$. Then
any two full vertices are connected by an edge.

###### Proof.

Let $u,v$ be full and suppose for contradiction that $uv$ is not an edge in
$G$. Consider the partial colouring $\psi:=\phi-u-v$. Since $u,v$ are full, we
have $|\phi(N(u))|=|\phi(N(v)))|=\chi(G)-1$. Since $uv$ is a non-edge, the
neighbours of $u,v$ all remain coloured in $\psi$ and so
$|\psi(N(u))|=|\psi(N(v)))|=\chi(G)-1$. Thus, by Lemma 3 (i), $\psi$ is a
Sudoku colouring. It has two uncoloured vertices, contradicting $sn(G)=|G|-1$.
∎

We say that a proper $k$-colouring of $G$ is $c$-minimal if it has as few
colour $c$ vertices as possible (for a $k$-colouring of $G$). The following
lemma shows that there are a lot of full vertices around all $c$-coloured
vertices in a $c$-minimal colouring.

######  Lemma 6.

Let $\phi$ be a $1$-minimal proper $\chi(G)$-colouring of $G$. Let $v$ be a
vertex with $\phi(v)=1$. Then for every colour $c=2,\dots,\chi(G)$, the vertex
$v$ has at least one colour $c$ neighbour $u$ with $u$ full. In particular,
$v$ is full.

###### Proof.

First notice that is impossible that $v$ has no colour $c$ neighbours — indeed
otherwise, we could recolour $v$ with colour $c$ to get a proper colouring
with one fewer colour 1 vertex (contradicting $1$-minimality). This proves the
“in particular $v$ is full” part — since we’ve shown that every colour other
than $\phi(v)=1$ appears on $N(v)$.

Now let the set of colour $c$ neighbours of $v$ be $\\{u_{1},\dots,u_{k}\\}$.
Suppose for contradiction that none of these are full — equivalently there are
colours $c_{1},\dots,c_{k}\in[\chi(G)]\setminus c$ with $c_{i}$ missing from
$N(u_{i})$. Note that $c_{i}\neq 1$ for all $i$, since $v\in N(u_{i})$ and
$\phi(v)=1$. Note that $\\{u_{1},\dots,u_{k}\\}$ is an independent set since
all these vertices have colour $c$ and the colouring is proper.

Now recolour $u_{i}$ by $c_{i}$ for each $i$ and recolour $v$ by $c$. Notice
that this colouring is proper. To show this, we need to check that the
recoloured vertices $v,u_{1},\dots,u_{k}$ have different colours to all their
neighbours (everywhere else the colouring remains proper just because $\phi$
was proper). Indeed $v$ has no colour $c$ neighbours since
$\\{u_{1},\dots,u_{k}\\}$ was the set of all colour $c$ neighbours of $v$ and
these have all been recoloured by colours $c_{i}\neq c$. Vertex $u_{i}$ has no
colour $c_{i}$ neighbour since it initially had no colour $c_{i}$ neighbours
and the only neighbour of $u_{i}$ that was recoloured was $v$ (which received
colour $c\neq c_{i}$).

But the new colouring we have has one fewer colour $1$ vertex, contradicting
$1$-minimality. ∎

Applying the above lemma to a graph with Sudoku number $|G|-1$ gives even more
structure in a minimal colouring.

######  Lemma 7.

Let $sn(G)=|G|-1$ and let $\phi$ be a $1$-minimal proper $\chi(G)$-colouring
of $G$. Then there is precisely one colour $1$ vertex. Additionally, this
vertex $v$ is full and has $|N(v)|=\chi(G)-1$.

###### Proof.

Let $v$ be a colour $1$ vertex. By Lemma 6 $v$ is full. There can’t be another
colour $1$ vertex $z$ — because otherwise $z$ would also be full, which would
give two disconnected full vertices (contradicting Lemma 5).

Suppose that $|N(v)|>\chi(G)-1$. Then, by the pigeonhole principle there must
be some colour $c$ which occurs more than once on $N(v)$. By Lemma 6 $v$ has
some full colour $c$ neighbour $u$. Let $w$ be some other colour $c$ neighbour
of $v$. Now let $\psi:=\phi-u-v$. Note that
$\psi(N(v))=\phi(N(v))=[\chi(G)]\setminus 1$ (the first equality holds because
precisely one neighbour $u$ of $v$ was uncoloured, but that neighbour $u$ had
colour $c$ which is still present at $w$. The second equality holds because
$v$ is full and has colour $1$). Also $\psi(N(u))=\phi(N(u))\setminus
1=[\chi(G)]\setminus\\{1,c\\}$ (the first equality holds because precisely one
neighbour $v$ or $u$ was uncoloured, and that neighbour had colour $1$ which
isn’t present anywhere else in the graph. The second equality holds because
$\phi(N(u))=[\chi(G)]\setminus c$ since $u$ is full and has colour $c$). Thus
by Lemma 3 (ii), $\psi$ is a Sudoku colouring with two uncoloured vertices,
contradicting $sn(G)=|G|-1$. ∎

We are ready to prove our main theorem.

###### Proof of Theorem 2.

The backwards direction already appears in [1] (see Corollary 3.4), so it
remains to prove that every connected $G$ with $sn(G)=|G|-1$ is complete. To
that end, let $G$ be connected with $sn(G)=|G|-1$.

Consider a 1-minimal proper $\chi(G)$-colouring of $G$. By Lemma 7, there is
precisely one colour 1 vertex, call it $v$. Also by Lemma 7, $v$ is full and
$|N(v)|=\chi(G)-1$ — which, using the definition of “full”, implies that all
neighbours of $v$ have different colours. By Lemma 6, we have that all
neighbours of $v$ are full. Now we have that all vertices in $v\cup N(v)$ are
full, and so Lemma 5 tells us that $v\cup N(v)$ is complete.

Unless $G$ is complete, then, by connectedness, there is some vertex $w$
outside $v\cup N(v)$ with a neighbour $u$ in $N(v)$. Now construct a colouring
$\psi$ by recolouring $w$ by colour $1$ and uncolouring $u,v$. First notice
that this is a proper (partial) colouring — indeed $w$ has no colour $1$
neighbours since $v$ is the unique colour $1$ vertex and $w\not\in N(v)$. We
have that
$\psi(N(v))=\phi(N(v))\setminus\phi(u)=[\chi(G)]\setminus\\{1,\phi(u)\\}$ (the
first equality holds because the neighbour $u$ of $v$ was uncoloured and
$\phi(u)$ doesn’t appear anywhere else on $N(v)$ since the neighbours of $v$
have different colours. The second equality holds because
$\phi(N(v))=[\chi(G)]\setminus 1$ since $v$ is full and has colour $1$). Also
$\psi(N(u))=[\chi(G)]\setminus\\{\phi(u)\\}$ (colour $1$ is present on $N(u)$
in $\psi$ because $\psi(w)=1$ and $w\in N(u)$. All colours in
$[\chi(G)]\setminus\\{1,\phi(u)\\}$ are present on $N(u)$ in $\psi$ because
the vertices of $N(v)\setminus u$ have exactly these colours in $\phi$, $u$ is
connected to all of them, and their colours don’t change going from $\phi$ to
$\psi$). Thus by Lemma 3 (ii), $\psi$ is a Sudoku colouring with two
uncoloured vertices, contradicting $sn(G)=|G|-1$. ∎

## References

  * [1] Gee-Choon Lau, J. Maria Jeyaseeli, Wai-Chee Shiu, and S. Arumugam.  Sudoku number of graphs.  arXiv:2206.08106, 2022.
  * [2] Gary McGuire, Bastian Tugemann, and Gilles Civario.  There is no 16-clue sudoku: Solving the sudoku minimum number of clues problem via hitting set enumeration.  Experimental Mathematics, 23(2):190–217, 2014.


#  Hardy spaces, Regularized BMO spaces and the boundedness of Calderón-
Zygmund operators on non-homogeneous spaces

The Anh Bui The Anh Bui was supported by a Macquarie University scholarship
Xuan Thinh Duong Xuan Thinh Duong was supported by a research grant from
Macquarie University
2010 Mathematics Subject Classification: 42B20, 42B35.
Key words: non-homogeneous spaces, Hardy spaces, BMO, Calderón - Zygmund
operator.

###### Abstract

One defines a non-homogeneous space $(X,\mu)$ as a metric space equipped with
a non-doubling measure $\mu$ so that the volume of the ball with center $x$,
radius $r$ has an upper bound of the form $r^{n}$ for some $n>0$. The aim of
this paper is to study the boundedness of Calderón-Zygmund singular integral
operators $T$ on various function spaces on $(X,\mu)$ such as the Hardy
spaces, the $L^{p}$ spaces and the regularized BMO spaces. This article thus
extends the work of X. Tolsa [T1] on the non-homogeneous space
$(\mathbb{R}^{n},\mu)$ to the setting of a general non-homogeneous space
$(X,\mu)$. Our framework of the non-homogeneous space $(X,\mu)$ is similar to
that of [Hy] and we are able to obtain quite a few properties similar to those
of Calderón-Zygmund operators on doubling spaces such as the weak type $(1,1)$
estimate, boundedness from Hardy space into $L^{1}$, boundedness from
$L^{\infty}$ into the regularized BMO and an interpolation theorem.
Furthermore, we prove that the dual space of the Hardy space is the
regularized BMO space, obtain a Calderón-Zygmund decomposition on the non-
homogeneous space $(X,\mu)$ and use this decomposition to show the boundedness
of the maximal operators in the form of Cotlar inequality as well as the
boundedness of commutators of Calderón-Zygmund operators and BMO functions.

###### Contents

  1. 1 Introduction
  2. 2 Non-homogeneous spaces, families of doubling balls and singular integrals
    1. 2.1 Non-homogeneous spaces and families of doubling balls
    2. 2.2 Calderón-Zygmund operators
  3. 3 The RBMO spaces
    1. 3.1 Definition of RBMO$(\mu)$
    2. 3.2 Some characterizations of RBMO$(\mu)$
  4. 4 Interpolation results
    1. 4.1 The sharp maximal operator
    2. 4.2 An Interpolation Theorem for linear operators
  5. 5 Atomic Hardy spaces and their dual spaces
    1. 5.1 The space $H^{1,\infty}_{at}(\mu)$
    2. 5.2 The space $H^{1,p}_{at}(\mu)$
  6. 6 Calderón-Zygmund decomposition
    1. 6.1 Calderón-Zygmund decomposition
    2. 6.2 The weak $(1,1)$ boundedness of Calderón-Zygmund operators
    3. 6.3 Cotlar inequality
  7. 7 The boundedness of Calderón-Zygmund operators
    1. 7.1 The boundedness of Calderón-Zygmund operators from $L^{\infty}$ to RBMO space
    2. 7.2 The boundedness of Calderón-Zygmund operators on Hardy spaces
    3. 7.3 Commutators of Calderón-Zygmund operators with RBMO functions

##  1 Introduction

In the last few decades, Calderón-Zygmund theory of singular integrals has
played a central part of modern harmonic analysis with lots of extensive
applications to other fields of mathematics. This theory has established
criteria for singular integral operators to be bounded on various function
spaces including $L^{p}$ spaces, $1<p<\infty$, Hardy spaces, BMO spaces and
Besov spaces.

One of the main features of the standard Calderón-Zygmund singular integral
theory is the requirement that the underlying spaces or domains to possess the
doubling (volume) property. Recall that a space $X$ equipped with a distance
$d$ and a measure $\mu$ is said to have the doubling property if there exists
a constant $C$ such that for all $x\in X$ and all $r>0$,

$\mu(B(x,2r))\leq C\mu(B(x,r))$

where $B(x,r)$ denotes the ball with center $x$ and radius $r>0$.

In the last ten years or so, there has been substantial progress in obtaining
boundedness of singular integrals acting on spaces without the doubling
property. Many features of the standard Calderón-Zygmund singular integral
theory was extended to spaces with a mild volume growth condition in place of
doubling property through the works of Nazarov, Treil, Volberg, Tolsa and
others. See, for example [NTV1], [NTV2], [NTV3], [T1] and [T2]. These
breakthroughs disproved the long held belief of the decades of 70’s and 80’s
that the doubling property is indispensable in the theory of Calderón-Zygmund
singular integrals and lead to more powerful techniques and estimates in
harmonic analysis.

Let $X$ be a metric space equipped with a measure $\mu$, possibly non-
doubling, satisfying

$\mu(B(x,r))\leq Cr^{n}$

for some positive constants $C$, $n$ and all $r>0$. We will call such a space
$(X,\mu)$ a non-homogeneous space. Let $T$ be a Calderón-Zygmund operator
acting on a non-homogeneous space $X$, i.e. the associated kernel of $T$
satisfies appropriate bounds and has Hölder continuity (for the precise
definition, see Section 2.1). Assume that $T$ is bounded on $L^{2}(X)$, then
it is shown in [NTV2] that the Calderón-Zygmund operator $T$ is of weak type
$(1,1)$, hence by interpolation, is bounded on $L^{p}(X),1<p<\infty$. See also
[T2].

Hardy spaces and BMO spaces on a non-homogeneous space $X$ were studied by a
number of authors, for example [T1], [MMNO], [Hy]. In [MMNO], the authors
studied the spaces BMO$(\mu)$ and $H^{1}_{at}(\mu)$ on $\mathbb{R}^{n}$ (with
BMO$(\mu)$ space being defined via the standard bounded oscillations and the
Hardy space $H^{1}_{at}(\mu)$ being defined by an atomic decomposition) for a
non doubling measure $\mu$ and showed some standard properties of these spaces
such as the John-Nirenberg inequality, an interpolation theorem between
BMO$(\mu)$ and $H^{1}_{at}(\mu)$, and BMO$(\mu)$ being the dual space of
$H^{1}_{at}(\mu)$. However, it is shown by Verdera [V] that an $L^{2}$ bounded
Calderón-Zygmund operator may be unbounded from $L^{\infty}(\mu)$ into
BMO$(\mu)$ as well as from $H^{1}_{at}(\mu)$ into $L^{1}(\mu)$. This shows the
need to introduce variants of the BMO spaces characterized by bounded
oscillation estimates so that the Calderón-Zygmund operators are bounded from
$L^{\infty}(\mu)$ into these variants of BMO spaces.

In [T1], the author introduced the RBMO space, a variant of the space BMO, on
the non-homogeneous space $(\mathbb{R}^{n},\mu)$ which retains some of the
properties of the standard BMO such as the John-Nirenberg inequality. See
Section 3 for the definition of RBMO spaces. While Calderón-Zygmund operators
might not be bounded from $L^{\infty}(\mathbb{R}^{n},\mu)$ into
BMO$(\mathbb{R}^{n},\mu)$, they are bounded from
$L^{\infty}(\mathbb{R}^{n},\mu)$ into RBMO$(\mathbb{R}^{n},\mu)$, [T1].

Recently, Hytönen studied the RBMO spaces on non-homogeneous spaces $(X,\mu)$
(instead of $(\mathbb{R}^{n},\mu)$) [Hy]. He proved that the space RBMO$(\mu)$
on $X$ still satisfies John-Nirenberg inequality. However, the boundedness of
Calderón-Zygmund operators from $L^{\infty}(\mu)$ into RBMO$(\mu)$ and a
number of other properties are still open questions for the setting of general
non-homogeneous spaces $(X,\mu)$.

In this article, our aim is to conduct an extensive study on the RBMO spaces
on general non-homogeneous spaces. More specifically, for a non-homogeneous
space $(X,\mu)$ equipped with a measure $\mu$ which is dominated by some
doubling measure (the same setting as in [Hy]), we are able to prove the
following new results:

  1. (i)

An $L^{2}$ bounded Calderón-Zygmund operator is bounded from $L^{\infty}(\mu)$
into the RBMO space, see Theorem 7.1.

  2. (ii)

The dual space of the atomic Hardy spaces is shown to be the RBMO space. We
also show that an $L^{2}$ bounded Calderón-Zygmund operator is bounded from
the atomic Hardy space $H^{1}_{at}(\mu)$ into $L^{1}(\mu)$, see Theorems 5.6
and 7.3.

  3. (iii)

An interpolation theorem between the RBMO space and the Hardy space
$H^{1}_{at}(\mu)$: if an operator is bounded from $H^{1}_{at}(\mu)$ into
$L^{1}(\mu)$ and from $L^{\infty}(\mu)$ into the RBMO space, then the operator
is bounded on $L^{p}(\mu)$ for all $1<p<\infty$, see Theorem 6.4.

  4. (iv)

A Cotlar type inequality for Calderón-Zygmund operators which gives the
boundedness of several maximal operators associated with $T$, see Theorem 6.6.

  5. (v)

The boundedness of commutators of Calderón-Zygmund operators and RBMO
functions on $L^{p}$ spaces, see Theorem 7.6.

  6. (vi)

A Calderón-Zygmund decomposition using a variant of Vitali covering lemma, see
Theorem 6.3, and the weak type $(1,1)$ of an $L^{2}$ bounded Calderón-Zygmund
operator, see Theorem 6.5.

We remark that, while this manuscript was in finishing touch, we learned that
similar results concerning the Hardy spaces as in (ii) have been obtained
independently in [HyYY].

We now give a brief comment about some techniques used in this paper. In
addition to using some ideas and techniques in [T1], we obtain certain key
estimates through careful investigation of the family of doubling balls in a
non-homogeneous space $(X,\mu)$. Let us recall that the main techniques used
in [T1] rely on the Besicovitch covering lemma and the construction of the
$(\alpha,\beta)$-doubling balls in $\mathbb{R}^{n}$. However, the Besicovitch
covering lemma is only applicable to $\mathbb{R}^{n}$ and it is not applicable
in the setting of general non-homogeneous spaces. In the general setting, one
can construction the $(\alpha,\beta)$-doubling balls by using a covering lemma
in [He] in place of the Besicovitch covering lemma. In [Hy], the author used
this substitution to obtain the John-Nirenberg inequality for the RBMO spaces.
However, it seems that to obtain further results similar to the standard
theory as in the case of doubling spaces, more refined techniques are needed.

An important technical detail in this paper is our construction of the three
consecutive $(\alpha,\beta)$-doubling balls (see, Proposition 2.4) which we
employ successfully to obtain the important characterizations (9) and (10),
similar to those in [T1, Lemma 2.10]. By using these three consecutive
$(\alpha,\beta)$-doubling balls, we show the boundedness of Calderón-Zygmund
operators from $L^{\infty}(X,\mu)$ into the space RBMO (see Theorem 7.1) as
well as an interpolation theorem of RBMO spaces (see Theorem 4.3).

Acknowledgement:  The second named author would like to thank El Maati Ouhabaz
for helpful discussion.

##  2 Non-homogeneous spaces, families of doubling balls and singular
integrals

###  2.1 Non-homogeneous spaces and families of doubling balls

In this paper, for the sake of simplicity we always assume that $(X,d)$ is a
metric space. With minor modifications, similar results hold when $X$ is a
quasi-metric space.

Geometrically doubling regular metric spaces. We adopt the definition that the
space $(X,d)$ is geometrically doubling if there exists a number
$N\in\mathbb{N}$ such that every open ball $B(x,r)=\\{y\in X:d(y,x)<r\\}$ can
be covered by at most $N$ balls of radius $r/2$. Our using of this somewhat
non-standard name is to differentiate this property from other types of
doubling properties. If there is no specification, the ball $B$ means the ball
center $x_{B}$ with radius $r_{B}$. Also, we set $n=\log_{2}N$, which can be
viewed as (an upper bound for) a geometric dimension of the space. Let us
recall the following well-known lemma. See, for example [Hy].

###### Lemma 2.1

In a geometrically doubling regular metric space, a ball $B(x,r)$ can contain
the centers $x_{i}$ of at most $N\alpha^{-n}$ disjoint balls $B(x_{i},\alpha
r)$ for any $\alpha\in(0,1]$.

Upper doubling measures. A measure $\mu$ in the metric space $(X,\mu)$ is said
to be an upper doubling measure if there exists a dominating function
$\lambda$ with the following properties:

  1. (i)

$\lambda:X\times(0,\infty)\mapsto(0,\infty)$;

  2. (ii)

for any fixed $x\in X$, $r\mapsto\lambda(x,r)$ is increasing;

  3. (iii)

there exists a constant $C_{\lambda}>0$ such that $\lambda(x,2r)\leq
C_{\lambda}\lambda(x,r)$ for all $x\in X$, $r>0$;

  4. (iv)

the inequality $\mu(x,r):=\mu(B(x,r))\leq\lambda(x,r)$ holds for all $x\in X$,
$r>0$;

  5. (v)

and $\lambda(x,r)\approx\lambda(y,r)$ for all $r>0,\ x,y\in X$ and $d(x,y)\leq
r$.

We note that in [Hy], the condition (v) is not assumed.

###### Lemma 2.2

Every family of balls $\\{B_{i}\\}_{i\in F}$ of uniformly bounded diameter in
a metric space $X$ contains a disjoint sub-family $\\{B_{i}\\}_{i\in E}$ with
$E\subset F$ such that

$\cup_{i\in F}B_{i}\subset\cup_{i\in E}5B_{i}.$

For a proof of Lemma 2.2, see [He].

Assumptions: Throughout the paper, we always assume that $(X,\mu)$ is a
geometrically doubling regular metric space and the measure $\mu$ is an upper
doubling measure.

We adopt the following definition as in [T1]. For $\alpha,\beta>1$, a ball
$B\subset X$ is called $(\alpha,\beta)$-doubling if
$\mu(\alpha)\leq\beta\mu(B)$. The following result states the existence of
plenty of doubling balls with small radii and with large radii.

######  Lemma 2.3 ([Hy])

The following statements hold:

  1. (i)

If $\beta>C_{\lambda}^{\log_{2}\alpha}$, then for any ball $B\subset X$ there
exists $j\in\mathbb{N}$ such that $\alpha^{j}B$ is $(\alpha,\beta)$-doubling.

  2. (ii)

If $\beta>\alpha^{n}$ where $n$ is the doubling order of $\lambda$, then for
any ball $B\subset X$ there exists $j\in\mathbb{N}$ such that $\alpha^{-j}B$
is $(\alpha,\beta)$-doubling.

Our following result which shows the existence of three consecutive
$(\alpha,\beta)$ doubling balls will play an important role in this paper.

###### Proposition 2.4

If $B$ is a $(\alpha^{3},\beta)$ doubling ball ($\alpha>1$), then $B,\alpha B$
and $\alpha^{2}B$ are three consecutive $(\alpha,\beta)$ doubling balls.

_Proof:_ The proof of Proposition 2.4 is simple, hence we omit the details
here.

For any two balls $B\subset Q$, we defined

$K_{B,Q}=1+\int_{r_{B}\leq d(x,x_{B})\leq
r_{Q}}\frac{1}{\lambda(x_{B},d(x,x_{B}))}d\mu(x).$ (1)

This definition is a variant of the definition in [T1, pp.94-95]. Similarly to
the results [T1, Lemma 2.1] we have the following properties:

###### Lemma 2.5

  1. (i)

If $Q\subset R\subset S$ are balls in $X$, then

$\max\\{K_{Q,R},K_{R,S}\\}\leq K_{Q,S}\leq C(K_{Q,R}+K_{R,S}).$

  2. (ii)

If $Q\subset R$ are compatible size, then $K_{Q,R}\leq C.$

  3. (iii)

If $\alpha Q,\ldots\alpha^{N-1}Q$ are non $(\alpha,\beta)$-doubling balls
$(\beta>C_{\lambda}^{\log_{2}\alpha})$ then $K_{Q,\alpha^{N}Q}\leq C.$

The proof of Lemma 2.5 is not difficult, hence we omit the details here.

As in [T1], for two balls $B\subset Q$ we can define the coefficient
$K^{\prime}_{B,Q}$ as follows: let $N_{B,Q}$ be the smallest integer
satisfying $6^{N_{B,Q}}r_{B}\geq r_{Q}$, then we set

$K^{\prime}_{B,Q}:=1+\sum_{k=1}^{N_{B,Q}}\frac{\mu(6^{k}B)}{\lambda(x_{B},6^{k}r_{B})}.$

In the case that $\lambda(x,ar)=a^{m}\lambda(x,r)$ for all $x\in X$ and
$a,r>0$, it is not difficult to show that $K_{B,Q}\approx K^{\prime}_{B,Q}$.
However, in general, we only have $K_{B,Q}\leq CK^{\prime}_{B,Q}$.

###  2.2 Calderón-Zygmund operators

A kernel $K(\cdot,\cdot)\in L^{1}_{{\rm loc}}(X\times
X\backslash\\{(x,y):x=y\\})$ is called a Calderón-Zygmund kernel if

  1. (i)

$|K(x,y)|\leq
C\min\Big{\\{}\frac{1}{\lambda(x,d(x,y))},\frac{1}{\lambda(y,d(x,y))}\Big{\\}}.$
(2)

  2. (ii)

There exists $0<\delta\leq 1$ such that

$|K(x,y)-K(x^{\prime},y)|+|K(y,x)-K(y,x^{\prime})|\leq
C\frac{d(x,x^{\prime})^{\delta}}{d(x,y)^{\delta}\lambda(x,d(x,y))}$ (3)

if $d(x,x^{\prime})\leq Cd(x,y).$

A linear operator $T$ is called a Calderón-Zygmund operator with kernel
$K(\cdot,\cdot)$ satisfying (2) and (3) if for all $f\in L^{\infty}(\mu)$ with
bounded support and $x\notin{supp}f$,

$Tf(x)=\int_{X}K(x,y)f(y)d\mu(y).$

The maximal operator $T_{*}$ associated with the Calderón-Zygmund operator $T$
is defined by

$T_{*}f(x)=\sup_{\epsilon>0}|T_{\epsilon}f(x)|,$

where $T_{\epsilon}f(x)=\int_{d(x,y)\geq\epsilon}K(x,y)f(y)d\mu(y)$.

We would like to give an example for the operator whose the associated kernel
satisfies the conditions (2) and (3). As in [Hy], we consider Bergman-type
operators which are studied by Volberg and Wick. In [VW], the authors obtained
a characterization of measures $\mu$ in the unit ball $\mathbb{B}_{2n}$ of
$\mathbb{C}^{n}$ for which the analytic Besov-Sobolev space
$B^{\sigma}_{2}(\mathbb{B}^{2n})$ embeds continuously into $L^{2}(\mu)$. Their
proof goes through a new $T1$ theorem for what they call Bergman-type
operators. Let us describe the situation of this application. The measures
$\mu$ in [VW] satisfy the upper power bound $\mu(B(x,r))\leq r^{m}$, except
possibly when $B(x,r)\subset H$, where $H$ is a fixed open set. However, in
the exceptional case there holds $r\leq\delta(x):=d(x,H^{c})$, and hence

$\mu(B(x,r))\leq\lim_{\epsilon\rightarrow
0}B(x,\delta(x)+\epsilon)\leq\lim_{\epsilon\rightarrow
0}(\delta(x)+\epsilon)^{m}=\delta^{m}.$

Thus we find that their measures are actually upper doubling with

$\mu(B(x,r))\leq\max\\{\delta(x)^{m},r^{m})=:\lambda(x,r).$

It is not difficult to show that $\lambda(\cdot,\cdot)$ satisfies the
conditions (i)-(v) in definition of upper doubling measures.

In [VW], as a main application concerning the Besov-Sobolev spaces, the
authors introduced the operator associated to the kernel

$K(x,y)=(1-\overline{x}\cdot y)^{-m},$ (4)

for $x,y\in\overline{\mathbb{B}}_{2n}\subset\mathbb{C}^{n}$. Here
$\overline{x}$ stands for the componentwise complex conjugation, and
$\overline{x}\cdot y$ designates the usual dot product of $n$-vectors
$\overline{x}$ and $y$. Moreover, one equips $\overline{\mathbb{B}}_{2n}$ with
the regular quasi-distance, see [Tch, Lemma 2.6],

$d(x,y):=\Big{|}|x|-|y|\Big{|}+\Big{|}1-\frac{\overline{x}\cdot
y}{|x||y|}\Big{|}.$

Finally, the set $H$ related to the exceptional balls is now the open unit
ball $\overline{\mathbb{B}}_{2n}$. It was proved in [HyM] that the kernel
$K(x,y)$ defined by (4) satisfies (2) and (3).

##  3 The RBMO spaces

###  3.1 Definition of RBMO$(\mu)$

The RBMO (Regularized BMO) space was introduced by Tolsa for
$(\mathbb{R}^{n},\mu)$ in [T1] and it was adopted by T. Hytönen for general
non-homogeneous space $(X,\mu)$ in [Hy].

###### Definition 3.1

Fix a parameter $\rho>1$. A function $f\in L^{1}_{{\rm loc}}(\mu)$ is said to
be in the space RBMO$(\mu)$ if there exists a number $C$, and for every ball
$B$, a number $f_{B}$ such that

$\frac{1}{\mu(\rho B)}\int_{B}|f(x)-f_{B}|d\mu(x)\leq A$ (5)

and, for any two balls $B$ and $B_{1}$ such that $B\subset B_{1}$,

$|f_{B}-f_{B_{1}}|\leq CK_{B,B_{1}}.$ (6)

The infimum of the values $C$ in (6) is taken to be the RBMO norm of $f$ and
denoted by $\|f\|_{{\rm RBMO}(\mu)}$.

The RBMO norm $\|\cdot\|_{{\rm RBMO}(\mu)}$ is independent of $\rho>1$.
Moreover the John-Nirenberg inequality holds for RBMO$(X)$. More precisely, we
have the following result (see Corollary 6.3 in [Hy]).

###### Proposition 3.2

For any $\rho>1$ and $p\in[1,\infty)$, there exists a constant $C$ so that for
every $f\in{\rm RBMO}(\mu)$ and every ball $B_{0}$,

$\Big{(}\frac{1}{\mu(\rho
B_{0})}\int_{B_{0}}|f(x)-f_{B_{0}}|^{p}d\mu(x)\Big{)}^{1/p}\leq C\|f\|_{{\rm
RBMO}(\mu)}.$

###  3.2 Some characterizations of RBMO$(\mu)$

In the rest of paper, unless $\alpha$ and $\beta$ are specified otherwise, by
an $(\alpha,\beta)$ doubling ball we mean a $(6,\beta_{0})$-doubling with a
fixed number $\beta_{0}>\max\\{C_{\lambda}^{3\log_{2}6},6^{3n}\\}$.

Given a ball $B\subset X$, let $N$ be the smallest non-negative integer such
that $\widetilde{B}=6^{N}B$ is doubling. Such a ball $\widetilde{B}$ exists
due to Lemma 2.5.

Let $\rho>1$ be some fixed constant. We say that $f\in L^{1}_{{\rm loc}}(\mu)$
is in RBMO$(\mu)$ if there exists some constant $C>0$ such that for any ball
$Q$

$\frac{1}{\mu(\rho B)}\int_{B}|f(x)-m_{\widetilde{B}}f|d\mu(x)\leq C$ (7)

and

$|m_{Q}f-m_{R}f|\leq CK_{Q,R},\ \ \text{for any two doubling balls $Q\subset
R$},$ (8)

here $m_{B}f$ is the mean value of $f$ over the ball $B$. Then we take

$\|f\|_{*}:=\inf\\{C:\text{(\ref{cond1-RBMOdefn2}) and (\ref{cond2-RBMOdefn2})
hold}\\}.$

By the same proof as in Lemma 2.8 of [T1], we have the following result.

###### Proposition 3.3

For a fixed $\rho>1$, the norms $\|\cdot\|_{*}$ and $\|\cdot\|_{{\rm
RBMO}(\mu)}$ are equivalent.

We now extend certain characterizations of ${\rm RBMO}(\mu)$ in [T1] in the
case of $(\mathbb{R}^{n},\mu)$ to the case of non-homogeneous spaces
$(X,\mu)$. In the case of $\mathbb{R}^{n}$, Besicovitch covering lemma was
used but this lemma is not applicable in our setting. We overcome this problem
by using the three consecutive doubling balls in Proposition 2.4.

###### Proposition 3.4

For $f\in L^{1}_{{\rm loc}}(\mu)$, the following are equivalent:

  1. (a)

$f\in{\rm RBMO}(\mu)$.

  2. (b)

There exists some constant $C_{b}$ such that for any ball $B$

$\frac{1}{\mu(6B)}\int_{B}|f(x)-m_{B}f|d\mu(x)\leq C_{b}$ (9)

and

$|m_{Q}f-m_{R}f|\leq
C_{b}K_{Q,R}\Big{(}\frac{\mu(6Q)}{\mu(Q)}+\frac{\mu(6R)}{\mu(R)}\Big{)},\ \
\text{for any two balls $Q\subset R$}.$ (10)

  3. (c)

There exists some constant $C_{c}$ such that for any doubling ball $B$

$\frac{1}{\mu(B)}\int_{B}|f(x)-m_{B}f|d\mu(x)\leq C_{c}$ (11)

and

$|m_{Q}f-m_{R}f|\leq C_{c}K_{Q,R},\ \ \text{for any two doubling balls
$Q\subset R$}.$ (12)

Moreover, the best constants $C_{b}$ and $C_{c}$ are comparable to the
RBMO$(\mu)$ norm of $f$.

_Proof:_ $(a)\rightarrow(b):$ If $f\in{\rm RBMO}(\mu)$, then (9) and (10) hold
for $C_{b}=C\|f\|_{*}$ for some constant $C$. Indeed, for any ball $B$ we have

$|m_{B}f-m_{\widetilde{B}}f|\leq
m_{Q}(|f-m_{\widetilde{B}}f|)\leq\|f\|_{*}\frac{\mu(6Q)}{Q}.$

Therefore,

$\frac{1}{\mu(6B)}\int_{B}|f(x)-m_{B}f|d\mu(x)\leq\frac{1}{\mu(6B)}\int_{B}(|f-m_{\widetilde{B}}f|+|m_{B}f-m_{\widetilde{B}}f|)\leq
2\|f\|_{*}.$ (13)

On the other hand, for any two balls $Q\subset R$, one has

$|m_{Q}f-m_{R}f|\leq|m_{Q}f-m_{\widetilde{Q}}f|+|m_{\widetilde{Q}}f-m_{\widetilde{R}}f|+|m_{R}f-m_{\widetilde{R}}f|.$

Applying (13) for the first and the third terms, we have

$|m_{Q}f-m_{\widetilde{Q}}f|+|m_{R}f-m_{\widetilde{R}}f|\leq\|f\|_{*}\Big{(}\frac{\mu(6Q)}{Q}+\frac{\mu(6R)}{R}\Big{)}.$

We can follow the argument in [T1] to obtain the estimate for the second term.
Let us remark that for any two balls $Q\subset R$ such that
$\widetilde{Q}\subset\widetilde{R}$, it follows from (8) that

$|m_{\widetilde{Q}}f-m_{\widetilde{R}}f|\leq\|f\|_{*}K_{\widetilde{Q},\widetilde{R}}.$

By Lemma 2.5, we have

$K_{\widetilde{Q},\widetilde{R}}\leq
C(K_{Q,\widetilde{Q}}+K_{Q,R}+K_{R,\widetilde{R}})\leq
C(C_{1}+K_{Q,R}+C_{2})\leq CK_{Q,R}.$

In general, $Q\subset R$ does not imply $\widetilde{Q}\subset\widetilde{R}$.
We consider two cases:
Case 1: If $r_{\widetilde{Q}}\geq r_{\widetilde{R}}$, then
$\widetilde{Q}\subset 3\widetilde{R}$. Setting
$R_{0}=\widetilde{3\widetilde{R}}$, then it follows from Lemma 2.5 and (8)
that

$\displaystyle|m_{\widetilde{Q}}f-m_{\widetilde{R}}f|$
$\displaystyle\leq|m_{\widetilde{Q}}f-m_{R_{0}}f|+m_{R_{0}}f-m_{\widetilde{R}}f$
$\displaystyle\leq(K_{\widetilde{Q},R_{0}}+K_{\widetilde{R},R_{0}})\|f\|_{*}.$

For the term $K_{\widetilde{Q},R_{0}}$ we have

$\displaystyle K_{\widetilde{Q},R_{0}}$ $\displaystyle\leq CK_{Q,R_{0}}$
$\displaystyle\leq C(K_{Q,R}+K_{R,R_{0}})$ $\displaystyle\leq
C(K_{Q,R}+K_{R,\widetilde{R}}+K_{\widetilde{R},3\widetilde{R}}+K_{3\widetilde{R},R_{0}})$
$\displaystyle\leq CK_{Q,R}.$

The remaining term $K_{\widetilde{R},R_{0}}$ is dominated by

$C(K_{\widetilde{R},3\widetilde{R}}+K_{3\widetilde{R},R_{0}})\leq CK_{Q,R}.$

So in this case, we obtain $|m_{\widetilde{Q}}f-m_{\widetilde{R}}f|\leq
CK_{Q,R}\|f\|_{*}$.

Case 2: If $r_{\widetilde{R}}<r_{\widetilde{Q}}$, then $\widetilde{R}\subset
6^{2}\widetilde{Q}$. Obviously, we can find some $m\geq 1$ such that
$r_{\widetilde{R}}\geq\frac{r_{5^{m}Q}}{25}$ and $\widetilde{R}\subset
6^{m}Q\subset 6^{2}\widetilde{Q}$. Therefore, $\widetilde{R}$ and $5^{m}Q$ are
comparable sizes. This implies $K_{\widetilde{R},5^{m}Q}\leq C$. Setting
$Q_{0}=\widetilde{6^{2}\widetilde{Q}}$ we have

$\displaystyle|m_{\widetilde{Q}}f-m_{\widetilde{R}}f|$
$\displaystyle\leq|m_{\widetilde{Q}}f-m_{Q_{0}}f|+|m_{Q_{0}}f-m_{\widetilde{R}}f|$
$\displaystyle\leq(K_{\widetilde{Q},Q_{0}}+K_{\widetilde{R},Q_{0}})\|f\|_{*}.$

Let us estimate $K_{\widetilde{Q},Q_{0}}$. We have

$K_{\widetilde{Q},Q_{0}}\leq
C(K_{\widetilde{Q},6^{2}\widetilde{Q}}+K_{6^{2}\widetilde{Q},Q_{0}})\leq
CK_{Q,R}.$

For the term $K_{\widetilde{R},Q_{0}}$, one has

$\displaystyle K_{\widetilde{R},Q_{0}}$ $\displaystyle\leq
C(K_{\widetilde{R},5^{m}Q}+K_{5^{m}Q,6^{2}\widetilde{Q}}+K_{6^{2}\widetilde{Q},Q_{0}})$
$\displaystyle\leq
C(K_{\widetilde{R},5^{m}Q}+K_{Q,6^{2}\widetilde{Q}}+K_{6^{2}\widetilde{Q},Q_{0}})$
$\displaystyle\leq CK_{Q,R}.$

Therefore, in this case we also obtain
$|m_{\widetilde{Q}}f-m_{\widetilde{R}}f|\leq CK_{Q,R}\|f\|_{*}$.

$(b)\rightarrow(c)$: the proof of this implication it easy and hence we omit
the detail here.

$(c)\rightarrow(a)$: Let $B$ be some ball. We need to show that (7) holds for
$\rho=6$. For any $x\in B$, there exists some $(6^{3},\beta_{0})$-doubling
ball centered $x$ with radius $r_{6^{-2j}B}$ for some $j\in\mathbb{N}$. We
denote by $B_{x}$ the biggest ball satisfying these properties. Let us recall
that by Proposition 2.4, the balls $B_{x},6B_{x}$ and $6^{2}B_{x}$ are three
$(6,\beta_{0})$-doubling balls. Moreover, by Lemma 2.5 we have

$|m_{6B_{x}}f-m_{\widetilde{B}}f|\leq|m_{6B_{x}}f-m_{B_{x}}f|+|m_{B_{x}}f-m_{\widetilde{B}}f|\leq
CC_{c}.$

By Lemma 2.2, we can pick a disjoint subcollection $B_{x_{i}}$, $i\in I$, such
that $B\subset\cup_{i\in I}5B_{x_{i}}\subset\cup_{i\in I}6B_{x_{i}}$. Thus, we
have

$\displaystyle\int_{B}|f-m_{\widetilde{B}}f|d\mu$ $\displaystyle\leq\sum_{i\in
I}\int_{B_{x_{i}}}|f-m_{\widetilde{B}}f|d\mu$ $\displaystyle\leq\sum_{i\in
I}\int_{B_{x_{i}}}(|f-m_{6B_{x_{i}}}f|+|m_{6B_{x_{i}}}f-m_{\widetilde{B}}f|)d\mu$
$\displaystyle\leq\sum_{i\in I}CC_{c}\mu(6B_{x_{i}})$
$\displaystyle\leq\sum_{i\in I}C\beta_{0}C_{c}\mu(B_{x_{i}})$
$\displaystyle\leq C\beta_{0}C_{c}\mu(6B).$

This completes our proof.

##  4 Interpolation results

###  4.1 The sharp maximal operator

Adapting an idea in [T1], we define the sharp maximal operator as follows:

$M^{\sharp}f(x)=\sup_{B\ni
x}\frac{1}{\mu(6B)}\int_{B}|f-m_{\widetilde{B}}f|d\mu+\sup\limits_{(Q,R)\in\Delta_{x}}\frac{|m_{Q}f-m_{R}f|}{K_{Q,R}},$
(14)

here $\Delta_{x}:=\\{(Q,R):x\in Q\subset R\ \text{and}\ Q,R:{\rm
doubling}\\}$.

Note that in our sharp maximal operator, the term $\mu(6B)$ was chosen with
the fixed constant $6$ throughout the paper. It is clear that

$f\in{\rm RBMO}(\mu)\Leftrightarrow M^{\sharp}f\in L^{\infty}(\mu).$

We define, for $\rho\geq 1$, the non-centered maximal operator $M_{(\rho)}$ by
setting

$M_{(\rho)}f(x)=\sup_{x\in Q}\frac{1}{\mu(\rho Q)}\int_{Q}|f|d\mu.$

It was proved that $M_{(\rho)}$ is of weak type $(1,1)$ for $\rho\geq 5$ and
hence $M_{(\rho)}$ is bounded on $L^{p}(\mu)$ for all $p\in(1,\infty]$, see
[Hy, Proposition 3.5]. When $\rho=1$, we write $Mf$ instead of $M_{(1)}f$.
From the boundedness of $M_{(\rho)}$ for $\rho\geq 5$, the non-centered
doubling maximal operator is defined by

$Nf(x)=\sup_{x\in Q:\ {\rm doubling}}\frac{1}{\mu(Q)}\int_{Q}|f|d\mu$

where the supremum is taken over all $(6,\beta_{0})$ doubling balls, is of
weak type $(1,1)$ and hence bounded on $L^{p}(\mu)$ for all $p\in(1,\infty]$.
Note that it is not difficult to show that

$M^{\sharp}f(x)\leq M_{(6)}f(x)+3Nf(x)$

for all $x\in X$. Therefore the operator $M^{\sharp}$ is of type weak $(1,1)$
and bounded on $L^{p}(\mu)$ for all $1<p<\infty$.

###### Lemma 4.1

For $f\in L^{1}_{{\rm loc}}(\mu)$, we have

$M^{\sharp}|f|(x)\leq 5\beta_{0}M^{\sharp}f(x).$

The proof is similar to that of Remark 6.1 in [T1].

We now show that the non-centered doubling maximal operator is dominated by
the sharp maximal operator in the following theorem. Although, some estimates
are inspired from [T1, Theorem 6.2], there are some main differences in our
proof. More specifically, the three consecutive doubling balls argument will
be used to replace the Besicovitch covering lemma.

###### Theorem 4.2

Let $f\in L^{1}_{\rm loc}(\mu)$ with the extra condition $\int fd\mu=0$ if
$\|\mu\|:=\mu(X)<\infty$. Assume that for some $p$, $1<p<\infty$,
$\inf\\{1,Nf\\}\in L^{p}(\mu)$. Then we have

$\|Nf\|_{L^{p}(\mu)}\leq C\|M^{\sharp}f\|_{L^{p}(\mu)}.$

_Proof:_ We assume that $\|\mu\|=\infty$. The proof for $\|\mu\|<\infty$ is
similar. By standard argument, it suffices to prove the following
$\lambda$-good inequality: for some fixed $\nu<1$ and all $\epsilon>0$ there
exists some $\delta>0$ such that for any $\lambda>0$ we have

$\mu\\{x:Nf(x)>(1+\epsilon)\lambda,M^{\sharp}f(x)\leq\delta\lambda\\}\leq\nu\mu\\{x:Nf(x)>\lambda\\}.$
(15)

Setting
$E_{\lambda}=\\{x:Nf(x)>(1+\epsilon)\lambda,M^{\sharp}f(x)\leq\delta\lambda\\}$
and $\Omega_{\lambda}=\\{x:Nf(x)>\lambda\\}$, for $f\in L^{p}(\mu)$. For each
$x\in E_{\lambda}$, we can choose the doubling ball $Q_{x}$ containing $x$
satisfying that $m_{Q_{x}}|f|>(1+\epsilon/2)\lambda$ and if $Q$ is any
doubling ball containing $x$ with $r_{Q}>2r_{Q_{x}}$ then
$m_{Q}|f|\leq(1+\epsilon/2)\lambda$. Such a ball $Q_{x}$ exists due to $f\in
L^{p}(\mu)$.

Let $R_{x}$ be the ball centered $x$ with radius $6r_{Q_{x}}$ and $S_{x}$ be
the smallest $(6^{3},\beta_{0})$-doubling ball in the form $6^{3j}R_{x}$.
Then, by Proposition 2.4, $S_{x},6S_{x}$ and $6^{2}S_{x}$ are three
$(6,\beta_{0})$-doubling balls. Moreover, one has

$K_{Q_{x},6S_{x}}\leq C(K_{Q_{x},R_{x}}+K_{R_{x},S_{x}}+K_{S_{x},6S_{x}})\leq
C.$

Therefore, it follows from Lemma 4.1 that

$|m_{Q_{x}}|f|-m_{6S_{x}}|f||\leq K_{Q_{x},6S_{x}}M^{\sharp}|f|(x)\leq
C\beta_{0}M^{\sharp}f(x)\leq C\beta_{0}\delta\lambda.$

This implies that for sufficiently small $\delta$ we have

$m_{6S_{x}}|f|>\lambda$

and hence $6S_{x}\subset\Omega_{\lambda}$.

Note that by Lemma 2.2, we can pick a disjoint collection
$\\{S_{x_{i}}\\}_{i\in I}$ with $x_{i}\in E_{\lambda}$ and
$E_{\lambda}\subset\cup_{i\in I}5S_{x_{i}}\subset\cup_{i\in I}6S_{x_{i}}$.
Setting $W_{x_{i}}=6S_{x_{i}}$, we will show that

$\mu(6S_{x_{i}}\cap E_{\lambda})\leq C\frac{\nu}{\beta_{0}}\mu(6S_{x_{i}})$
(16)

for all $i\in I$.

Once (16) is proved, (15) follows readily. Indeed, from (16) we have

$\mu(E_{\lambda})\leq\sum_{i\in I}\mu(6S_{x_{i}}\cap
E_{\lambda})\leq\sum_{i\in I}\frac{\nu}{\beta_{0}}\mu(6S_{x_{i}})\leq
C\sum_{i\in I}\nu\mu(S_{x_{i}})\leq C\nu\mu(\Omega_{\lambda}).$

Now we show the proof of (16). Let $y\in W_{x_{i}}\cap E_{\lambda}$. For any
doubling ball $Q\ni y$ satisfying $m_{Q}|f|>(1+\epsilon)\lambda$, it follows
that $r_{Q}\leq r_{W_{x_{i}}}/8$. Indeed, if $r_{Q}>r_{W_{x_{i}}}/8$ then we
have $Q_{x_{i}}\subset W_{x_{i}}\subset\widetilde{16Q}$ and

$|m_{Q}|f|-m_{\widetilde{16Q}}|f||\leq
K_{Q,\widetilde{16Q}}M^{\sharp}|f|(y)\leq
C\delta\lambda\leq\frac{\epsilon}{2}$

for sufficiently small $\delta$. This implies
$m_{\widetilde{16Q}}|f|>(1+\epsilon/2)\lambda$ which is a contradiction to the
choice of $Q_{x_{i}}$. So, $r_{Q}\leq r_{W_{x_{i}}}/8$. This, together with
$m_{Q}|f|>(1+\epsilon)\lambda$, imply

$N(f\chi_{\frac{5}{4}W_{x_{i}}})(y)>(1+\epsilon)\lambda$

and

$m_{\widetilde{\frac{5}{4}W_{x_{i}}}}|f|\leq(1+\epsilon/2)\lambda\
\text{(since $r_{\widetilde{\frac{5}{4}W_{x_{i}}}}>2r_{Q_{x_{i}}}$}).$

This yields,

$N(\chi_{\frac{5}{4}W_{x_{i}}}|f|-m_{\widetilde{\frac{5}{4}W_{x_{i}}}}|f|)(y)>\frac{\epsilon}{2}\lambda.$

Therefore, by using the weak $(1,1)$ boundedness of $N$, we have

$\displaystyle\mu(W_{x_{i}}\cap E_{\lambda})$
$\displaystyle\leq\mu\\{y:N(\chi_{\frac{5}{4}W_{x_{i}}}|f|-m_{\widetilde{\frac{5}{4}W_{x_{i}}}}|f|)(y)>\frac{\epsilon}{2}\lambda\\}$
$\displaystyle\leq\frac{C}{\epsilon\lambda}\int_{\frac{5}{4}W_{x_{i}}}(|f|-m_{\widetilde{\frac{5}{4}W_{x_{i}}}}|f|)d\mu$
$\displaystyle\leq\frac{C}{\epsilon\lambda}\mu(\frac{15}{2}W_{x_{i}})M^{\sharp}|f|(x_{i})$
$\displaystyle\leq\frac{C\delta}{\epsilon}\beta_{0}\mu(6^{3}S_{x_{i}})$
$\displaystyle\leq\frac{C\delta}{\epsilon}\beta_{0}\mu(S_{x_{i}}).$

Thus, (16) holds provided $\delta<\epsilon/C\nu\beta_{0}$.
For the case $f\notin L^{p}(\mu)$, we define the sequence of functions
$\\{f_{k}\\},\ k=1,2,\cdots$ by setting

$f_{k}(x)=\begin{cases}f(x),&|f(x)|\leq k,\\\ k\frac{f(x)}{|f(x)|},\
&|f(x)|>k.\end{cases}$

Then we have $M^{\sharp}f_{k}(x)\leq CM^{\sharp}f(x)$. On the other hand,
$|f_{k}(x)|\leq k\inf\\{1,|f|(x)\\}\leq k\inf(1,Nf)(x)$ and so $f_{k}\in
L^{p}(\mu)$. Hence,

$\|Nf_{k}\|_{L^{p}(\mu)}\leq C\|M^{\sharp}f_{k}\|_{L^{p}(\mu)}\leq
C\|M^{\sharp}f\|_{L^{p}(\mu)}.$

Taking the limit as $k\rightarrow\infty$, we obtain the required result and
the proof is completed.

###  4.2 An Interpolation Theorem for linear operators

###### Theorem 4.3

Let $1<p<\infty$ and let $T$ be a linear operator bounded on $L^{p}(\mu)$ and
from $L^{\infty}(\mu)$ into ${\rm RBMO(\mu)}$. Then $T$ extends to a bounded
operator on $L^{r}(\mu)$ for $p<r<\infty$.

_Proof:_ We consider 2 cases:
Case 1: $\|\mu\|=\infty$: Since $T$ is bounded on $L^{p}(\mu)$, $M^{\sharp}T$
is sublinear bounded on $L^{p}(\mu)$ and on $L^{\infty}(\mu)$. Therefore, by
interpolation, $M^{\sharp}T$ is bounded on $L^{r}(\mu)$ for $p<r<\infty$,

$\|M^{\sharp}Tf\|_{L^{r}(\mu)}\leq C\|f\|_{L^{r}(\mu)}.$

Assume that $f\in L^{r}(\mu)$ is supported in compact set. Then $f\in
L^{p}(\mu)$ and so $Tf\in L^{p}(\mu)$. Hence $Nf\in L^{p}(\mu)$ and
$\inf\\{1,Nf\\}\in L^{r}(\mu)$. By invoking Theorem 4.2,

$\|Tf\|_{L^{r}(\mu)}\leq\|M^{\sharp}Tf\|_{L^{r}(\mu)}\leq
C\|f\|_{L^{r}(\mu)}.$

Case 2: Assume that $\|\mu\|<\infty$. For $f\in L^{r}(\mu)$, set $f=(f-\int
fd\mu)+\int fd\mu=f_{1}+f_{2}$. Since $\int f_{1}d\mu=0$, we can apply the
same argument as for $\|\mu\|=\infty$. It is not difficult to show that
$\|T1\|_{L^{r}(\mu)}\leq C\|1\|_{L^{r}(\mu)}$. This completes the proof.

##  5 Atomic Hardy spaces and their dual spaces

###  5.1 The space $H^{1,\infty}_{at}(\mu)$

For a fixed $\rho>1$, a function $b\in L^{1}_{loc}(\mu)$ is called an atomic
block if

  1. (i)

there exists some ball $B$ such that ${\rm supp}b\subset B$;

  2. (ii)

$\int bd\mu=0;$

  3. (iii)

there are functions $a_{j}$ supported on cubes $B_{j}\subset B$ and numbers
$\lambda_{j}\in\mathbb{R}$ such that

$b=\sum_{j=1}^{\infty}\lambda_{j}a_{j},$ (17)

where the sum converges in $L^{1}(\mu)$, and
$\|a_{j}\|_{L^{\infty}(\mu)}\leq(\mu(\rho B_{j})K_{B_{j},B})^{-1}$ and the
constant $K_{B_{j},B}$ being given in the paragraph before Lemma 2.5.

We denote $|b|_{H^{1,\infty}_{at}(\mu)}=\sum_{j=1}^{\infty}|\lambda_{j}|$. We
say that $f\in H^{1,\infty}_{at}(\mu)$ if there are atomic blocks $b_{i}$ such
that

$f=\sum_{i=1}^{\infty}b_{i}$ (18)

with $\sum_{i=1}^{\infty}|b_{i}|_{H^{1,\infty}_{at}(\mu)}<\infty$. The
$H^{1,\infty}_{at}(\mu)$ norm of $f$ is defined by

$\|f\|_{H^{1,\infty}_{at}(\mu)}:=\inf\sum_{i=1}^{\infty}|b_{i}|_{H^{1,\infty}_{at}(\mu)}$

where the infimum is taken over all the possible decompositions of $f$ in
atomic blocks.

We have the following basic properties of $H_{at}^{1,\infty}(\mu)$.

###### Proposition 5.1

  1. (a)

$H_{at}^{1,\infty}(\mu)$ is a Banach space.

  2. (b)

$H_{at}^{1,\infty}(\mu)\subset L^{1}(\mu)$ and
$\|f\|_{L^{1}(\mu)}\leq\|f\|_{H_{at}^{1,\infty}(\mu)}$.

  3. (c)

The space $H_{at}^{1,\infty}(\mu)$ is independent of the constant $\rho$ when
$\rho>1$.

_Proof:_ The proofs of (a) and (b) are standard and we omit the details here.

The proof of (c): Given $\rho_{1}>\rho_{2}>0$, it is clear that
$H^{1,\infty}_{at,\rho_{1}}\subset H^{1,\infty}_{at,\rho_{2}}$ with
$\|f\|_{H^{1,\infty}_{at,\rho_{2}}}\leq\|f\|_{H^{1,\infty}_{at,\rho_{1}}}$.
Conversely, if $b=\sum_{i=1}^{\infty}\lambda_{i}a_{i}$ is an atomic block with
supp $a_{i}\subset B_{i}\subset B$ in $H^{1,\infty}_{at,\rho_{1}}$, then by
Lemma 2.1 we can cover each $B_{i}$ by
$N\Big{[}\frac{\rho_{1}}{\rho_{2}}\Big{]}^{n}$ balls, says $\\{B_{ik}\\}$,
with the same radius $\frac{\rho_{2}}{\rho_{1}}r_{B}$. Therefore, we can
decompose $a_{i}:=\sum_{k}a_{ik}$ where
$a_{ik}:=a_{i}\frac{\chi_{B_{ik}}}{\sum_{j}\chi_{B_{ij}}}$. It is not
difficult to verify that $b$ is also an atomic block in
$H^{1,\infty}_{at,\rho_{2}}$. This completes our proof.

We now show that the space ${\rm RBMO}(\mu)$ is embedded in the dual space of
$H_{at}^{1,\infty}(\mu).$

###### Lemma 5.2

We have

${\rm RBMO}(\mu)\subset H_{at}^{1,\infty}(\mu)^{*}.$

That is, for $g\in{\rm RBMO}(\mu)$, the linear functional

$L_{g}(f)=\int_{X}fgd\mu$

defines a continuous linear functional $L_{g}$ over $H_{at}^{1,\infty}(\mu)$
with $\|L_{g}\|_{H_{at}^{1,\infty}(\mu)^{*}}\leq C\|g\|_{{\rm RBMO}(\mu)}.$

_Proof:_ Following standard argument, see for example [CW2, p.64], we only
need to check that for an atomic block $b$ and $g\in$RBMO$(\mu)$, we have

$\Big{|}\int bgd\mu\Big{|}\leq C|b|_{H_{at}^{1,\infty}(\mu)}\|g\|_{{\rm
RBMO}(\mu)}.$

Assume that supp$b\subset B$ and $b=\sum_{j}^{\infty}\lambda_{j}a_{j}$, where
$a_{j}$’s are functions satisfying (a) and (b) in the definition of atomic
blocks. If $g\in L^{\infty}$, by using $\int bd\mu=0$, we have

$\Big{|}\int bgd\mu\Big{|}=\Big{|}\int
b(g-g_{B})d\mu\Big{|}\leq\sum_{j}^{\infty}|\lambda_{j}|\|a_{j}\|_{L^{\infty}(\mu)}\int_{B_{i}}|g-g_{B}|d\mu.$
(19)

Since $g\in L^{\infty}(\mu)\subset{\rm RBMO}(\mu)$, we have

$\displaystyle\int_{B_{i}}|g-g_{B}|d\mu$
$\displaystyle\leq\int_{B_{i}}|g-g_{B_{i}}|d\mu+\int_{B_{i}}|g_{B}-g_{B_{i}}|d\mu$
$\displaystyle\leq CK_{B_{i},B}\|g\|_{{\rm RBMO}(\mu)}\mu(\rho B_{j}).$

From (19), we obtain

$\Big{|}\int bgd\mu\Big{|}\leq C|b|_{H_{at}^{1,\infty}(\mu)}\|g\|_{{\rm
RBMO}(\mu)}.$

In general case, if $g\in{\rm RBMO}(\mu)$, define

$g_{N}(x):=\begin{cases}f(x),&\ \ |f(x)|<N,\\\ N\frac{f(x)}{|f(x)|},&\ \
|f(x)|\geq N.\end{cases}$

It can be verified that $\|g_{N}\|_{{\rm RBMO}(\mu)}\leq C\|g\|_{{\rm
RBMO}(\mu)}$. As above, since $g_{N}\in L^{\infty}(\mu)$, we have

$\Big{|}\int fg_{N}d\mu\Big{|}\leq
C\|f\|_{H_{at}^{1,\infty}(\mu)}\|g_{N}\|_{{\rm RBMO}(\mu)}\leq
C\|f\|_{H_{at}^{1,\infty}(\mu)}\|g\|_{{\rm RBMO}(\mu)}.$

Let us denote $L^{\infty}_{0}:=\\{f:f\ \text{in $L^{\infty}(\mu)$ with bounded
support}\\}$ and $D=H_{at}^{1,\infty}(\mu)\cap L^{\infty}_{0}$. So, the
functional $L_{g}:f\mapsto\int gf$ is well-defined on $D$ whenever $g\in{\rm
RBMO}(\mu)$ (since $g\in L^{1}_{loc}(\mu)$). By the dominated convergence
theorem

$\lim_{N\rightarrow\infty}\int fg_{N}d\mu=\int fgd\mu$

for all $f\in D$. We claim that $D$ is dense in $H_{at}^{1,\infty}(\mu)$. To
verify this claim, denote by $H^{1,\infty}_{at,fin}(\mu)$ the set of all
elements in $H_{at}^{1,\infty}(\mu)$ where the sums (17) and (18) are taken
over finite elements. Obviously, $H^{1,\infty}_{at,fin}(\mu)$ is dense in
$H_{at}^{1,\infty}(\mu)$ and each functional $f\in H^{1,\infty}_{at,fin}(\mu)$
is also in $L^{\infty}_{0}$. Therefore , $L_{b}$ is a unique extension on
$H_{at}^{1,\infty}(\mu)$ and hence

$\Big{|}\int fgd\mu\Big{|}\leq C\|f\|_{H_{at}^{1,\infty}(\mu)}\|g\|_{{\rm
RBMO}(\mu)}.$

This completes our proof.

The following lemma can be obtained by the same argument as in [T1, Lemma
4.4].

###### Lemma 5.3

If $g\in{\rm RBMO}(\mu)$, we have

$\|L_{g}\|_{H_{at}^{1,\infty}(\mu)}\approx\|g\|_{{\rm RBMO}(\mu)}.$

###  5.2 The space $H^{1,p}_{at}(\mu)$

For a fixed $\rho>1$, a function $b\in L^{1}_{loc}(\mu)$ is called a
$p$-atomic block, $1<p<\infty$, if

  1. (i)

there exists some ball $B$ such that ${\rm supp}b\subset B$;

  2. (ii)

$\int bd\mu=0;$

  3. (iii)

there are functions $a_{j}$ supported on cubes $B_{j}\subset B$ and numbers
$\lambda_{j}\in\mathbb{R}$ such that

$b=\sum_{j=1}^{\infty}\lambda_{j}a_{j},$ (20)

where the sum converges in $L^{1}(\mu)$, and

$\|a_{j}\|_{L^{p}(\mu)}\leq(\mu(\rho B_{j}))^{1/p-1}K_{B_{j},B}^{-1}.$

We denote $|b|_{H^{1,p}_{at}(\mu)}=\sum_{j=1}^{\infty}|\lambda_{j}|$. We say
that $f\in H^{1,p}_{at}(\mu)$ if there are $p$-atomic blocks $b_{i}$ such that

$f=\sum_{i=1}^{\infty}b_{i}$ (21)

with $\sum_{i=1}^{\infty}|b_{i}|_{H^{1,p}_{at}(\mu)}<\infty$. The
$H^{1,p}_{at}(\mu)$ norm of $f$ is defined by

$\|f\|_{H^{1,p}_{at}(\mu)}:=\inf\sum_{i=1}^{\infty}|b_{i}|_{H^{1,p}_{at}(\mu)}$

where the infimum is taken over all the possible decompositions of $f$ in
$p$-atomic blocks.

Similarly to $H_{at}^{1,\infty}(\mu)$, we have the following basic properties
of $H_{at}^{1,p}(\mu)$

###### Proposition 5.4

  1. (a)

$H_{at}^{1,p}(\mu)$ is a Banach space.

  2. (b)

$H_{at}^{1,p}(\mu)\subset L^{1}(\mu)$ and
$\|f\|_{L^{1}(\mu)}\leq\|f\|_{H_{at}^{1,p}(\mu)}$.

  3. (c)

The space $H_{at}^{1,p}(\mu)$ is independent of the constant $\rho$ when
$\rho>1$.

The proofs of this proposition is in line with Proposition 5.1, so we omit the
details here.

###### Lemma 5.5

We have

${\rm RBMO}(\mu)\subset H_{at}^{1,p}(\mu)^{*}.$

That is, for $g\in{\rm RBMO}(\mu)$, the linear functional

$L_{g}(f)=\int_{X}fgd\mu$

defines a continuous linear functional $L_{g}$ over $H_{at}^{1,p}(\mu)$ with

$\|L_{g}\|_{H_{at}^{1,p}(\mu)^{*}}\leq C\|g\|_{{\rm RBMO}(\mu)}.$

_Proof:_ The proof of this lemma is analogous to that of Lemma 5.2 with minor
modifications. We leave the details to the interested reader.

We remark that a main difference between the Hardy space in Tolsa’s setting
[T1] and our Hardy space in this article is the sense of convergence in the
atomic decomposition. This leads to different approaches in proving the
inclusions RBMO$(\mu)\subset H_{at}^{1,\infty}(\mu)^{*}$ and RBMO$(\mu)\subset
H_{at}^{1,p}(\mu)^{*}$. However, for the inverse inclusion
$H_{at}^{1,p}(\mu)^{*}\subset$ RBMO$(\mu)$, by a careful investigation, Tolsa
[T1] showed that one only needs to consider the sums in (20) and (21) over
finite $p$-atoms and $p$-atomic blocks, hence the sense of convergence in (20)
and (21) does not matter in both settings. This is the reason why we can use
the arguments in [T1] for our setting with minor modifications to obtain the
duality result of $H_{at}^{1,\infty}(\mu)$ and $H_{at}^{1,p}(\mu)$ as in the
next Theorem.

###### Theorem 5.6

For $1<p<\infty,H_{at}^{1,p}(\mu)=H_{at}^{1,\infty}(\mu)$. Also
$H_{at}^{1,p}(\mu)^{*}=H_{at}^{1,\infty}(\mu)^{*}={\rm RBMO}(\mu)$.

As explained above, we omit the details of the proof.

##  6 Calderón-Zygmund decomposition

###  6.1 Calderón-Zygmund decomposition

The following two technical lemmas will be useful for the construction of a
Calderón-Zygmund decomposition on non-homogeneous spaces.

###### Lemma 6.1

Assume that $Q$, $S$ are two concentric balls, $Q\subset R$, such that there
are no $(\alpha,\beta)$-doubling balls with
$\beta>C_{\lambda}^{\log_{2}\alpha}$ in the form $\alpha^{k}Q,k\in\mathbb{N}$
such that $Q\subset\alpha^{k}Q\subset R$. Then we have

$\int_{R\backslash Q}\frac{1}{\lambda(x_{Q},d(x_{Q},x))}d\mu(x)\leq C.$

_Proof:_ Let $N$ be the smallest integer such that $R\subset\alpha^{N}Q$.
Then, $\mu(\alpha^{k}Q)\geq\beta\mu(\alpha^{k-1}Q)$ for all $k=1,\ldots,N$.
Therefore, we have,

$\displaystyle\int_{R\backslash Q}$
$\displaystyle\frac{1}{\lambda(x_{Q},d(x_{Q},x))}d\mu(x)$
$\displaystyle\leq\sum_{k=1}^{N}\int_{\alpha^{k-1}r_{Q}\leq
d(x,y)\leq\alpha^{k}r_{Q}}\frac{1}{\lambda(x_{Q},d(x_{Q},x))}d\mu(x)$
$\displaystyle\leq\sum_{k=1}^{N}\frac{\mu(\alpha^{k}Q)}{\lambda(x_{Q},\alpha^{k-1}r_{Q})}$
$\displaystyle\leq\sum_{k=1}^{N}\frac{\beta^{N-k}\mu(\alpha^{N}Q)}{(C_{\lambda})^{(N-k)\log_{2}\alpha}\lambda(x_{Q},\alpha^{N}r_{Q})}$
$\displaystyle\leq\sum_{k=1}^{N}\Big{[}\frac{\beta}{(C_{\lambda})^{\log_{2}\alpha}}\Big{]}^{N-k}$
$\displaystyle\leq\sum_{j=1}^{\infty}\Big{[}\frac{\beta}{(C_{\lambda})^{\log_{2}\alpha}}\Big{]}^{j}$
$\displaystyle\leq C\ \ \text{(since $\beta>\log_{2}\alpha$)}.$

This completes the proof.

While the Covering Lemma 2.2 for $(X,\mu)$ can be used to replace the
Besicovich covering lemma for $(\mathbb{R}^{n},\mu)$ in certain estimates, the
Calderón-Zygmund decomposition in $(X,\mu)$ will need a covering lemma which
gives the finite overlapping property at all points $x\in X$. This is given in
the next lemma.

###### Lemma 6.2

Every family of balls $\\{B_{i}\\}_{i\in F}$ of uniformly bounded diameter in
a metric space $X$ contains a disjoint sub-family $\\{B_{i}\\}_{i\in E}$ with
$E\subset F$ such that

  1. (i)

$\cup_{i\in F}B_{i}\subset\cup_{i\in E}6B_{i},$

  2. (ii)

For each $x\in X$, $\sum_{i\in E}\chi_{6B_{i}}<\infty$.

We remark that in (ii), the sum $\sum_{i\in E}\chi_{6B_{i}}<\infty$ at each
$x$ but these sums are not necessarily uniformly bounded on $X$.

_Proof:_ By Lemma 2.2 we can pick a disjoint subfamily
$\\{B_{i}:B_{i}=B(x_{B_{i}},r_{B_{i}})\\}_{i\in E}$ with $E\subset F$
satisfying (i). Moreover, we can assume that for $i,j\in E$, neither
$6B_{i}\subset 6B_{j}$ nor $6B_{j}\subset 6B_{i}$.

To prove (ii), we assume in contradiction that there exists some $x\in X$ such
that there exists an infinite family of balls $\\{B_{i}:i\in I_{x}\subset
E\\}$ such that $x\in B_{i}$ for all $i\in I_{x}$. We will show that
$\lim\inf_{i\in I_{x}}r_{B_{i}}>0$. Otherwise, for any $\epsilon>0$ there
exists $i_{\epsilon}\in I_{x}$ such that $r_{B_{i_{\epsilon}}}<\epsilon$.
Therefore, if $B_{0}$ is any ball in the family $\\{B_{i}:i\in I_{x}\\}$,
there exists $r>0$ such that $B(x,r)\subset 6B_{0}$. For
$\epsilon=\frac{r}{30}$, we have $x\in 6B_{i_{\epsilon}}$ and
$r_{6B_{i_{\epsilon}}}<\frac{r}{4}$. This implies $6B_{i_{\epsilon}}\subset
6B_{0}$ which is a contradiction.

Thus $\lim\inf_{i\in I_{x}}r_{B_{i}}>0$. This together with the uniform
boundedness of diameter of the family of balls shows that there exist $m$ and
$M>0$ such that $m<r_{B_{i}}<M$ for all $i\in I_{x}$. Obviously, $\cup_{i\in
I_{x}}B(x_{B_{i}},m)\subset B(x,2M)$ and the balls $\\{B(x_{B_{i}},m):i\in
I_{x}\\}$ are pairwise disjoint. By Lemma 2.1, there exists a finite family of
balls with radius $\frac{m}{30}$ such that
$B(x,2M)\subset\cup_{i=1}^{K}B(x_{i},\frac{m}{30})$. Therefore, there exist a
ball, says $B_{k}\in\\{B(x_{i},\frac{m}{30}):i\in 1,\ldots K\\}$, and at least
two balls $B_{1}$ and $B_{2}$ in $\\{B_{i}:i\in I_{x}\\}$ such that
$B_{k}\cap\frac{1}{6}B_{1}\notin\emptyset$ and
$B_{k}\cap\frac{1}{6}B_{1}\notin\emptyset$. Since
$\min\\{r_{\frac{1}{6}B_{1}},r_{\frac{1}{6}B_{1}}\\}>\frac{1}{6}m=5r_{B_{k}}$,
we have $B_{k}\subset B_{1}\cap B_{2}$. This is a contradiction, because the
family of balls $\\{B(x_{B_{i}},m):i\in I_{x}\\}$ is pairwise disjoint. Our
proof is completed.

We now give a Calderón-Zygmund decomposition on a non-homogenous space
$(X,\mu)$ which is an extension of a Calderón-Zygmund decomposition on the
non-homogeneous space $(\mathbb{R}^{n},\mu)$ in [T1].

###### Theorem 6.3

(Calderón-Zygmund decomposition) Assume $1\leq p<\infty$. For any $f\in
L^{p}(\mu)$ and any $\lambda>0$ (with $\lambda>\beta_{0}||f||_{p}/||\mu||$ if
$||\mu||<\infty$), the following statements hold.

(a) There exists a family of finite overlapping balls $\\{6Q_{i}\\}_{i}$ such
that $\\{Q_{i}\\}_{i}$ is a pairwise disjoint family and

$\frac{1}{\mu(6^{2}Q_{i})}\int_{Q_{i}}|f|^{p}d\mu>\frac{\lambda^{p}}{\beta_{0}},$
(22) $\frac{1}{\mu(6^{2}\eta Q_{i})}\int_{\eta
Q_{i}}|f|^{p}d\mu\leq\frac{\lambda^{p}}{\beta_{0}},\ \text{for all $\eta>1$},$
(23) $|f|\leq\lambda\ \text{a.e. $(\mu)$ on $X\backslash\bigcup_{i}6Q_{i}$}.$
(24)

(b) For each $i$, let $R_{i}$ be a $(3\times
6^{2},C_{\lambda}^{\log_{2}3\times 6^{2}+1})$-doubling ball concentric with
$Q_{i}$, with $l(R_{i})>6^{2}l(Q_{i})$ and denote
$\omega_{i}=\frac{\chi_{6Q_{i}}}{\sum_{k}\chi_{6Q_{k}}}$. Then there exists a
family of functions $\varphi_{i}$ with constant signs and supp
$(\varphi_{i})\subset R_{i}$ satisfying

$\int\varphi_{i}d\mu=\int_{6Q_{i}}f\omega_{i}d\mu,$ (25)
$\sum_{i}|\varphi_{i}|\leq\kappa\lambda,$ (26)

(where $\kappa$ is some constant which depends only on $(X,\mu)$), and

  1. (i)

$||\varphi_{i}||_{\infty}\mu(R_{i})\leq C\int_{X}|w_{i}f|d\mu\ \text{if
$p=1$};$ (27)

  2. (ii)

$||\varphi_{i}||_{L^{p}(\mu)}\mu(R_{i})^{1/p^{\prime}}\leq\frac{C}{\lambda^{p-1}}\int_{X}|w_{i}f|^{p}d\mu\
\text{if $1<p<\infty$}.$ (28)

(c) For $1<p<\infty$, if $R_{i}$ is the smallest $(3\times
6^{2},C_{\lambda}^{\log_{2}3\times 6^{2}+1})$-doubling ball of the family
$\\{3\times 6^{2}Q_{i}\\}_{k\geq 1}$, then

$\|b\|_{H_{at}^{1,p}(\mu)}\leq\frac{C}{\lambda^{p-1}}\|f\|_{L^{p}(\mu)}^{p}$
(29)

where $b=\sum_{i}(w_{i}f-\varphi_{i}))$.

_Proof:_ For the sake of simplicity, we only give the proof for the case $p=1$
for (a) and (b). When $p>1$, by setting $g=f^{p}\in L^{1}(\mu)$, we can reduce
to the problem $p=1$. Then, with a simple modification, we will obtain (28)
instead of (27).

(a) Set $E:=\\{x:|f(x)|>\lambda\\}$. For each $x\in E$, there exists some ball
$Q_{x}$ such that

$\frac{1}{\mu(6^{2}Q_{x})}\int_{Q_{x}}|f|d\mu>\frac{\lambda}{\beta_{0}}$ (30)

and such that if $Q^{\prime}_{x}$ is centered at $x$ with
$l(Q^{\prime}_{x})>l(Q_{x})$, then

$\frac{1}{\mu(6^{2}Q^{\prime}_{x})}\int_{Q^{\prime}_{x}}|f|d\mu\leq\frac{\lambda}{\beta_{0}}$

Now we can apply Lemma 6.2 to get a family of balls
$\\{Q_{i}\\}_{i}\subset\\{Q_{x}\\}_{x}$ such that
$\sum_{j}\chi_{6Q_{j}}(x)<\infty$ for all $x\in X$ and (22), (23) and (24) are
satisfied.

(b) Assume first that the family of balls $\\{Q_{i}\\}$ is finite. Without
loss of generality, suppose that $l(R_{i})\leq l(R_{i+1})$. The functions
$\varphi$ will be constructed of the form
$\varphi_{i}=\alpha_{i}\chi_{A_{i}},A_{i}\subset R_{i}$.

First, set $A_{1}=R_{1}$ and $\varphi_{1}=\alpha_{1}\chi_{R_{1}}$ such that
$\int\varphi_{1}=\int_{6Q_{i}}f\omega_{1}$. Assume that
$\varphi_{1},\ldots,\varphi_{k-1}$ have been constructed satisfying (25) and

$\sum_{i=1}^{K-1}\varphi_{i}\leq\kappa\lambda,$

where $\kappa$ is some constant which will be fixed later. There are two
cases:

Case 1: There exists some $i\in\\{1,\ldots,k-1\\}$ such that $R_{i}\cap
R_{k}\neq\emptyset$. Let $R_{s_{1}},\ldots,R_{s_{m}}$ be the family of
$R_{1},\ldots,R_{k-1}$ such that $R_{s_{j}}\cap R_{k}\neq\emptyset$. Since
$l(R_{s_{j}})\leq l(R_{k})$, $R_{s_{j}}\subset 3R_{k}$. By using $R_{k}$ is
$(3\times 6^{2},C_{\lambda}^{\log_{2}3\times 6^{2}+1})$-doubling and (23), we
get

$\displaystyle\sum_{j}|\varphi_{s_{j}}|$
$\displaystyle\leq\sum_{j}\int_{X}|f\omega_{s_{j}}|d\mu$ $\displaystyle\leq
C\sum_{j}\int_{X}\omega_{s_{j}}|f|d\mu\leq C\sum_{j}\int_{3R_{k}}|f|d\mu\leq
C\lambda\mu(3.6^{2}R_{k})\leq C_{1}\lambda\mu(R_{k}).$

Therefore,

$\mu\\{\sum_{j}\|\varphi_{s_{j}}|>2C_{1}\lambda\\}\leq\frac{\mu(R_{k})}{2}.$

Thus,

$\mu(A_{k})\geq\frac{\mu(R_{k})}{2},A_{k}=R_{k}\cap\\{\sum_{j}|\varphi_{s_{j}}|\leq
2C_{1}\lambda\\}.$

The constant $\alpha_{k}$ will be chosen such that
$\int\varphi_{k}=\int_{Q_{k}}f\omega_{k}d\mu$ where
$\varphi_{k}=\alpha_{k}\chi_{A_{k}}$. Then we obtain

$\alpha_{k}\leq\frac{C}{\mu(A_{k})}\int_{X}w_{i}|f|d\mu\leq
C\frac{2}{\mu(R_{k})}\int_{\frac{1}{6^{2}}R_{k}}|f|d\mu\leq C_{2}\lambda\ \
(\text{by using (\ref{cz2})}).$

If we choose $\kappa=2C_{1}+C_{2}$, (26) follows.

Case 2: $R_{i}\cap R_{k}=\emptyset$ for all $i=1,\ldots,k-1$. Set
$A_{k}=R_{k}$ and $\varphi_{k}=\alpha_{k}\chi_{R_{k}}$ such that
$\int\varphi_{k}=\int_{Q_{k}}f\omega_{k}d\mu$. We also get (26).

By the construction of the functions $\varphi_{i}$, it is easy to see that
$\mu(R_{i})\leq 2\mu(A_{k})$. Hence,

$||\varphi_{i}||_{\infty}\mu(R_{i})\leq C\alpha_{i}\mu(A_{k})\leq
C\int_{X}|f\omega_{i}|d\mu.$

When the collection of balls $\\{Q_{i}\\}$ is not finite, we can argue as in
[T1, p.134]. This completes the proofs of (a) and (b).

(c) Since $R_{i}$ is the smallest $(3\times 6^{2},C_{\lambda}^{\log_{2}3\times
6^{2}+1})$-doubling ball of the family $\\{3\times 6^{2}Q_{i}\\}_{k\geq 1}$,
one has $K_{Q_{i},R_{i}}\leq C$. For each $i$, we consider the atomic block
$b_{i}=fw_{i}-\varphi_{i}$ supported in ball $R_{i}$. By (22) and (28) we have

$|b_{i}|_{H_{at}^{1,p}(\mu)}\leq\frac{C}{\lambda^{p-1}}\int_{X}|fw_{i}|^{p}d\mu$

which implies

$|b|_{H_{at}^{1,p}(\mu)}\leq\frac{C}{\lambda^{p-1}}\int_{X}\sum_{i}|fw_{i}|^{p}d\mu\leq\frac{C}{\lambda^{p-1}}\int_{X}(\sum_{i}w_{i})^{p}|f|^{p}d\mu=\frac{C}{\lambda^{p-1}}\int_{X}|f|^{p}d\mu.$

Our proof is completed.

Using the Calderón-Zygmund decomposition and a standard argument, see for
example [J, pp.43-44] (also [T1, p.135]), we obtain the following
interpolation result for a linear operator. For clarity and completeness, we
sketch the proof below.

###### Theorem 6.4

Let $T$ be a linear operator which is bounded from $H_{at}^{1,\infty}(\mu)$
into $L^{1}(\mu)$ and from $L^{\infty}(\mu)$ into ${\rm RBMO}(\mu)$. Then $T$
can be extended to a bounded operator on $L^{p}(\mu)$ for all $1<p<\infty$.

_Proof:_ For simplicity we may assume that $\|\mu\|=\infty$. Let $f$ be a
function in $L^{\infty}(\mu)$ with bounded support satisfying $\int fd\mu=0$.
Let us recall that the set of all such functions is dense in $L^{p}(\mu)$ for
all $1<p<\infty$. For such functions $f$, we need only to show that

$\|M^{\sharp}Tf\|_{L^{p}(\mu)}\leq C\|f\|_{L^{p}(\mu)},\ \ 1<p<\infty.$ (31)

Once (31) is proved, Theorem 6.4 follows from Theorem 4.2.
For such a function $f$ and $\lambda>0$, we can decompose the function $f$ as
in Theorem 6.3

$f:=b+g=\sum_{i}(w_{i}f-\varphi_{i})+g.$

By (24) and (26), we have $\|g\|_{L^{\infty}(\mu)}\leq C\lambda$, and by (29)

$\|b\|_{H^{1,p}_{at}(\mu)}\leq\frac{C}{\lambda^{p-1}}\|f\|^{p}_{L^{p}(\mu)}.$

Since $T$ is bounded from $L^{\infty}(\mu)$ into RBMO$(\mu)$, we have

$\|M^{\sharp}Tg\|_{L^{\infty}(\mu)}\leq C_{0}\lambda.$

Therefore,

$\\{M^{\sharp}Tf>(C_{0}+1)\lambda\\}\subset\\{M^{\sharp}Tb>\lambda\\}.$

The fact that $M^{\sharp}$ is of weak type $(1,1)$ gives

$\mu\\{M^{\sharp}Tb>\lambda\\}\leq C\frac{\|Tb\|_{L^{1}(\mu)}}{\lambda}.$

Moreover, since $T$ is bounded from $H^{1,\infty}_{at}(\mu)$ into
$L^{1}(\mu)$, we have

$\|Tb\|_{L^{1}(\mu)}\leq
C\|b\|_{H^{1,\infty}_{at}(\mu)}\leq\frac{C}{\lambda^{p-1}}\|f\|^{p}_{L^{p}(\mu)}.$

This implies

$\mu\\{M^{\sharp}Tf>(C_{0}+1)\lambda\\}\leq
C\frac{\|f\|^{p}_{L^{p}(\mu)}}{\lambda^{p}}.$

So the sublinear operator $M^{\sharp}T$ is of weak type $(p,p)$ for all
$1<p<\infty$. By Marcinkiewicz interpolation theorem the operator
$M^{\sharp}T$ is bounded for all $1<p<\infty$. This completes our proof.

###  6.2 The weak $(1,1)$ boundedness of Calderón-Zygmund operators

###### Theorem 6.5

If a Calderón-Zygmund operator $T$ is bounded on $L^{2}(\mu)$, then $T$ is of
weak type $(1,1)$.

_Proof:_ Let $f\in L^{1}(\mu)$ and $\lambda>0$. We can assume that
$\lambda>\beta_{0}\|f\|_{L^{1}(\mu)}/\|\mu\|$. Otherwise, there is nothing to
prove. Using the same notations as in Theorem 6.3 with $R_{i}$ which is chosen
as the smallest $(3\times 6^{2},C_{\lambda}^{\log_{2}3\times
6^{2}+1})$-doubling ball of the family $\\{3\times 6^{2}Q_{i}\\}_{k\geq 1}$,
we can write $f=g+b$, with

$g=f\chi_{{}_{X\backslash\cup_{i}6Q_{i}}}+\sum_{i}\varphi_{i}$

and

$b:=\sum_{i}b_{i}=\sum_{i}(w_{i}f-\varphi_{i}).$

Taking into account (22), one has

$\mu(\cup_{i}6^{2}Q_{i})\leq\frac{C}{\lambda}\int_{Q_{i}}|f|d\mu\leq\frac{C}{\lambda}\int_{X}|f|d\mu$

where in the last inequality we use the pairwise disjoint property of family
$\\{Q_{i}\\}_{i}$.
We need only to show that

$\mu\\{x\in
X\backslash\cup_{i}6^{2}Q_{i}:|Tf(x)|>\lambda\\}\leq\frac{C}{\lambda}\int_{X}|f|d\mu.$

We have

$\displaystyle\mu\\{x\in X\backslash\cup_{i}6^{2}Q_{i}:|Tf(x)|>\lambda\\}$
$\displaystyle\leq\mu\\{x\in
X\backslash\cup_{i}6^{2}Q_{i}:|Tg(x)|>\lambda/2\\}$
$\displaystyle~{}~{}~{}+\mu\\{x\in
X\backslash\cup_{i}6^{2}Q_{i}:|Tb(x)|>\lambda/2\\}:=I_{1}+I_{2}.$

Let us estimate the term $I_{1}$ related to the “good part” first. Since
$|g|\leq C\lambda$ then

$\displaystyle\mu\\{x\in
X\backslash\cup_{i}6^{2}Q_{i}:|Tg(x)|>\lambda/2\\}\leq\frac{C}{\lambda^{2}}\int|g|^{2}d\mu\leq\frac{C}{\lambda}\int|g|d\mu.$

Furthermore, we have

$\displaystyle\int|g|d\mu$
$\displaystyle\leq\int_{X\backslash\cup_{i}6Q_{i}}|f|d\mu+\sum_{i}\int_{R_{i}}|\varphi_{i}|$
$\displaystyle\leq\int_{X}|f|d\mu+\sum_{i}\mu(R_{i})\|\varphi_{i}\|_{L^{\infty}(\mu)}$
$\displaystyle\leq\int_{X}|f|d\mu+C\sum_{i}\int_{X}|fw_{i}|d\mu$
$\displaystyle\leq C\int_{X}|f|d\mu\ .$

Therefore,

$\mu\\{x\in
X\backslash\cup_{i}6^{2}Q_{i}:|Tg(x)|>\lambda/2\\}\leq\frac{C}{\lambda}\int|f|d\mu.$

For the term $I_{2}$, we have

$\displaystyle I_{2}$
$\displaystyle\leq\frac{C}{\lambda}\sum_{i}\Big{(}\int_{X\backslash
2R_{i}}|Tb_{i}|d\mu+\int_{2R_{i}}|T\varphi_{i}|d\mu+\int_{2R_{i}\backslash
6^{2}Q_{i}}|Tw_{i}f|d\mu\Big{)}$
$\displaystyle\leq\frac{C}{\lambda}\sum_{i}\Big{(}K_{i1}+K_{i2}+K_{i3}\Big{)}$

Note that $\int b_{i}d\mu=0$ for all $i$. We have, by (3),

$\displaystyle K_{i1}=\int_{X\backslash 2R_{i}}|Tb_{i}|d\mu$
$\displaystyle\leq C\int|b_{i}|d\mu$
$\displaystyle\leq\int_{X}|fw_{i}|d\mu+\int_{R_{i}}|\varphi_{i}|d\mu$
$\displaystyle\leq\int_{X}|fw_{i}|d\mu+\mu(R_{i})\|\varphi_{i}\|_{L^{\infty}(\mu)}$
$\displaystyle\leq C\sum_{i}\int_{X}|fw_{i}|d\mu$ $\displaystyle\leq
C\sum_{i}\int_{X}|f|d\mu.$

On the other hand, by the $L^{2}$ boundedness of $T$ and $R_{i}$ is a
$(3\times 6^{2},C_{\lambda}^{\log_{2}3\times 6^{2}+1})$-doubling ball, we get

$\displaystyle K_{i2}$
$\displaystyle\leq\Big{(}\int_{2R_{i}}|T\varphi_{i}|^{2}\Big{)}^{1/2}(\mu(2R_{i}))^{1/2}$
$\displaystyle\leq\Big{(}\int_{2R_{i}}|\varphi_{i}|^{2}\Big{)}^{1/2}(\mu(2R_{i}))^{1/2}$
$\displaystyle\leq C\|\varphi_{i}\|_{L^{\infty}(\mu)}\mu(2R_{i})$
$\displaystyle\leq C\int|w_{i}f|d\mu.$

Moreover, taking into account the fact that supp$w_{i}f\subset 6Q_{i}$, for
$x\in 2R_{i}\backslash 6^{2}Q_{i}$ we have, by Lemma 6.1,

$\displaystyle K_{i3}$ $\displaystyle\leq C\int_{2R_{i}\backslash
6^{2}Q_{i}}\frac{1}{\lambda(x_{Q_{i}},d(x,x_{Q_{i}}))}d\mu(x)\times\int_{X}|w_{i}f|d\mu.$

Hence we obtain

$I_{2}\leq\frac{C}{\lambda}\sum_{i}\int_{X}|w_{i}f|d\mu\leq\frac{C}{\lambda}\sum_{i}\int_{X}|f|d\mu$

and the proof is completed.

###  6.3 Cotlar inequality

We note that from the weak type $(1,1)$ estimate of $T$, we can obtain a
Cotlar inequality on $T$. More precisely, we have the following result.

###### Theorem 6.6

Assume that $T$ is a Calderón-Zygmund operator and that $T$ is bounded on
$L^{2}(X,\mu)$. Then there exists a constant $C>0$ such that for any bounded
function $f$ with compact support and $x\in X$ we have

$T_{*}f(x)\leq C\Big{(}M_{6,\eta}(Tf)(x)+M_{(5)}f(x)\Big{)}$

where

$M_{p,\rho}f(x)=\sup_{Q\ni x}\Big{(}\frac{1}{\mu(\rho
Q)}\int_{Q}|f|^{p}\Big{)}.$

_Proof:_ For any $\epsilon>0$ and $x\in X$, let $Q_{x}$ be the biggest
$(6,\beta)$-doubling ball centered $x$ of the form $6^{-k}\epsilon,k\geq 1$
and $\beta>6^{n}$. Assume that $Q_{x}=B(x,6^{-k_{0}}\epsilon)$. Then, we can
break $f=f_{1}+f_{2}$, where $f_{1}=f\chi_{\frac{6}{5}Q_{x}}$. Obviously
$T_{*}f_{1}(x)=0$. This follows that

$T_{*}f(x)\leq|Tf_{2}(x)|+\Big{|}\int_{d(x,y)\leq\epsilon}K(x,y)f_{2}(y)d\mu(y)\Big{|}=I_{1}+I_{2}.$

Let us estimate $I_{1}$ first. For any $z\in Q_{x}$, we have

$|Tf_{2}(x)|\leq|Tf_{2}(x)-Tf_{2}(z)|+|Tf(z)|+|Tf_{1}(z)|.$ (32)

On the other hand, it follows from (3) that

$\displaystyle|Tf_{2}(x)-Tf_{2}(z)|$
$\displaystyle\leq\int_{X\backslash\frac{6}{5}Q_{x}}|K(x,y)-K(z,y)||f(y)|d\mu(y)$
$\displaystyle\leq
C\int_{X\backslash\frac{6}{5}Q_{x}}\frac{d(x,z)^{\delta}}{d(x,y)^{\delta}\lambda(x,d(x,y))}|f(y)|d\mu(y)$
$\displaystyle\leq C\int_{X\backslash
Q_{x}}\frac{r_{Q_{x}}^{n+\delta}}{d(x,y)^{\delta}\lambda(x,d(x,y))}|f(y)|d\mu(y)$
$\displaystyle\leq C\sum_{k=0}^{\infty}\int_{6^{k+1}Q_{x}\backslash
6^{k}Q_{x}}\frac{r_{Q_{x}}^{\delta}}{d(x,y)^{\delta}\lambda(x,d(x,y))}|f(y)|d\mu(y)$
$\displaystyle\leq C\sum_{k=0}^{\infty}\int_{6^{k+1}Q_{x}\backslash
6^{k}Q_{x}}\frac{r_{Q_{x}}^{\delta}}{(6^{k}r_{Q_{x}})^{\delta}\lambda(x,6^{k}r_{Q_{x}})}|f(y)|d\mu(y)$
$\displaystyle\leq C\sum_{k=0}^{\infty}6^{-k}\frac{\mu(6\times
6^{k+1}Q_{x})}{\lambda(x,6^{k}r_{Q_{x}})}\frac{1}{\mu(6\times
6^{k+1}Q_{x})}\int_{6^{k+1}Q_{x}}|f(y)|d\mu(y)$ $\displaystyle\leq
C\sum_{k=0}^{\infty}6^{-k}M_{(6)}f(x)=CM_{(6)}f(x).$

This together with (33) implies

$|Tf_{2}(x)|\leq CM_{(6)}f(x)+|Tf_{1}(z)|+|Tf(z)|$ (33)

for all $z\in Q_{x}$.
At this stage, taking the $L^{\eta}(Q_{x},\frac{d\mu(x)}{\mu(Q)})$-norm with
respect to $z$, we have

$|Tf_{2}(x)|\leq
CM_{(6)}f(x)+\Big{(}\frac{1}{\mu(Q_{x})}\int_{Q_{x}}|Tf_{1}(z)|^{\eta}d\mu(z)\Big{)}^{1/\eta}+\Big{(}\frac{1}{\mu(Q_{x})}\int_{Q_{x}}|Tf(z)|^{\eta}d\mu(z)\Big{)}^{1/\eta}.$

By the Kolmogorov inequality and the weak type $(1,1)$ boundedness of $T$, we
have, for $\eta<1$,

$\displaystyle\Big{(}\frac{1}{\mu(Q_{x})}\int_{Q_{x}}|Tf_{1}(z)|^{\eta}d\mu(z)\Big{)}^{1/\eta}$
$\displaystyle\leq\frac{1}{\mu(Q_{x})}\int_{\frac{6}{5}Q_{x}}|f_{1}(z)|d\mu(z)$
$\displaystyle\leq\frac{C}{\mu(15Q_{x})}\int_{\frac{6}{5}Q_{x}}|f_{1}(z)|d\mu(z)\
\ \ \text{(since $Q_{x}$ is $(6,\beta)$- doubling)}$ $\displaystyle\leq
CM_{(5)}f(x).$

Furthermore, since $Q_{x}$ is $(6,\beta)$-doubling,

$\Big{(}\frac{1}{\mu(Q_{x})}\int_{Q_{x}}|Tf(z)|^{\eta}d\mu(z)\Big{)}^{1/\eta}\leq
CM_{\eta,6}M(Tf)(x).$

Therefore, $I_{1}\leq CM_{(6)}f(x)+CM_{\eta,6}M(Tf)(x)$.

For the term $I_{2}$ we have

$\displaystyle I_{2}$
$\displaystyle\leq\int_{d(x,y)\leq\epsilon}|K(x,y)||f_{2}(y)|d\mu(y)$
$\displaystyle\leq C\int_{B(x,\epsilon)\backslash
B(x,6^{-k_{0}}\epsilon)}\frac{1}{\lambda(x,d(x,y))}|f_{2}(y)|d\mu(y)$
$\displaystyle\leq
C\sum_{k=0}^{k_{0}-1}\int_{B(x,6^{k+1-k_{0}}\epsilon)\backslash
B(x,6^{k-k_{0}}\epsilon)}\frac{1}{\lambda(x,d(x,y))}|f_{2}(y)|d\mu(y)$
$\displaystyle\leq
C\sum_{k=0}^{k_{0}-1}\int_{B(x,6^{k+1-k_{0}}\epsilon)}\frac{1}{\lambda(x,6^{k-k_{0}}\epsilon)}|f_{2}(y)|d\mu(y)$
$\displaystyle\leq C\sum_{k=0}^{k_{0}-1}\frac{\mu(x,6\times
6^{k+1-k_{0}}\epsilon)}{\lambda(x,6\times
6^{k+1-k_{0}}\epsilon)}\frac{1}{\mu(x,6\times
6^{k+1-k_{0}}\epsilon)}\int_{B(x,6^{k+1-k_{0}}\epsilon)}|f_{2}(y)|d\mu(y)$
$\displaystyle\leq C\sum_{k=0}^{k_{0}-1}\frac{\mu(x,6\times
6^{k+1-k_{0}}\epsilon)}{\lambda(x,6\times 6^{k+1-k_{0}}\epsilon)}M_{(6)}f(x).$

At this stage, by repeating the argument in the proof of Lemma 6.1, we have

$\sum_{k=0}^{k_{0}-1}\frac{\mu(x,6\times
6^{k+1-k_{0}}\epsilon)}{\lambda(x,6\times 6^{k+1-k_{0}}\epsilon)}\leq C.$

Therefore,

$I_{2}\leq CM_{(6)}f(x).$

This completes our proof.

###### Remark 6.7

From the boundedness of $M_{6,\eta}(\cdot)$ and $M_{(5)}(\cdot)$, the Cotlar
inequality tells us that if $T$ is bounded on $L^{2}(X,\mu)$ then the maximal
operator $T_{*}$ is bounded on $L^{p}(X,\mu)$ for $1<p<\infty$. Note that the
endpoint estimate of $T_{*}$ will be investigated in [AD].

The Calderón-Zygmund decomposition Theorem 6.3 does not require the property
(v) of $\lambda(\cdot,\cdot)$.

##  7 The boundedness of Calderón-Zygmund operators

The main results of this section are Theorems 7.1, 7.3 and 7.6.

###  7.1 The boundedness of Calderón-Zygmund operators from $L^{\infty}$ to
RBMO space

The following result shows that on a non-homogeneous space $(X,\mu)$, a
Calderón-Zygmund operator which is bounded on $L^{2}$ is also bounded from
$L^{\infty}(\mu)$ into the regularized BMO space RBMO$(\mu)$.

###### Theorem 7.1

Assume that $T$ is a Calderón-Zygmund operator and $T$ is bounded on
$L^{2}(\mu)$, then $T$ is bounded from $L^{\infty}(\mu)$ into RBMO$(\mu)$.
Therefore, by interpolation and duality, $T$ is bounded on $L^{p}(\mu)$ for
all $1<p<\infty$.

_Proof:_ We use the RBMO characterizations (9) and (10). The condition (9) can
be obtained by the standard method used in the case of doubling measure. We
omit the details here.

We will check (10). To do this, we have to show that

$|m_{Q}(Tf)-m_{R}(Tf)|\leq
CK_{Q,R}\Big{(}\frac{\mu(6Q)}{\mu(Q)}+\frac{\mu(6R)}{\mu(R)}\Big{)}\|f\|_{L^{\infty}(\mu)}$

for all $Q\subset R$.
Let $N$ be the first integer $k$ such that $R\subset 6^{k}Q$. We denote
$Q_{R}=6^{N+1}Q$. Then for $x\in Q$ and $y\in R$, we set

$\displaystyle Tf(x)-Tf(y)$
$\displaystyle=Tf\chi_{6Q}(x)+Tf\chi_{6^{N}Q\backslash
6Q}(x)+Tf\chi_{X\backslash Q_{R}}(x)$
$\displaystyle~{}~{}~{}-(Tf\chi_{Q_{R}}(y)+Tf\chi_{X\backslash Q_{R}}(y))$
$\displaystyle\leq|Tf\chi_{6Q}(x)|+|Tf\chi_{6^{N}Q\backslash 6Q}(x)|$
$\displaystyle~{}~{}~{}+|Tf\chi_{X\backslash Q_{R}}(x)-Tf\chi_{X\backslash
Q_{R}}(y)|+|Tf\chi_{Q_{R}}(y)|$ $\displaystyle\leq I_{1}+I_{2}+I_{3}+I_{4}.$

Let us estimate $I_{3}$ first. We have

$\displaystyle I_{3}$ $\displaystyle\leq\int\limits_{X\backslash
Q_{R}}|K(x,z)-K(y,z)||f(z)|d\mu(z)$
$\displaystyle\leq\sum_{k=N+1}^{\infty}\int\limits_{6^{k+1}Q\backslash
6^{k}Q}\frac{d(x,y)^{\delta}}{d(x,z)^{\delta}\lambda(x,d(x,z))}|f(z)|d\mu(z)$
$\displaystyle\leq\sum_{k=N+1}^{\infty}6^{-(k-N)\delta}\frac{\mu(6^{k+1}Q)}{\lambda(x,6^{k-1}r_{Q})}\|f\|_{L^{\infty}(\mu)}$
$\displaystyle\leq
C\sum_{k=N+1}^{\infty}6^{-(k-N)\delta}\frac{\mu(6^{k+1}Q)}{\lambda(x,6^{k+1}r_{Q})}\|f\|_{L^{\infty}(\mu)}$
$\displaystyle\leq
C\sum_{k=N+1}^{\infty}6^{-(k-N)\delta}\|f\|_{L^{\infty}(\mu)}=C\|f\|_{L^{\infty}(\mu)},$

where in the last inequality we use the fact that
$\mu(6^{k+1}Q)\leq{\lambda(x,6^{k+1}r_{Q})}$, since $x\in Q\subset 2^{k+1}Q$.
As to the term $I_{2}$, we have

$\displaystyle Tf\chi_{6^{N}Q\backslash 6Q}(x)$
$\displaystyle\leq\int_{6^{N}Q\backslash 6Q}|K(x,y)||f(y)|d\mu(y)$ (34)
$\displaystyle\leq\int_{6^{N}Q\backslash
6Q}\frac{C}{\lambda(x_{Q},d(x_{Q},y))}|f(y)|d\mu(y)$ $\displaystyle\leq
K_{6Q,6^{N}Q}\|f\|_{L^{\infty}(\mu)}.$

Therefore, $I_{2}\leq CK_{Q,R}\|f\|_{\infty}$. So, we get

$\displaystyle Tf(x)-Tf(y)$
$\displaystyle=Tf\chi_{6Q}(x)+CK_{Q,R}\|f\|_{L^{\infty}(\mu)}+|Tf\chi_{Q_{R}}(y)|+C\|f\|_{L^{\infty}(\mu)}.$

Taking the mean over $Q$ and $R$ for $x$ and $y$, respectively, we have

$\displaystyle|m_{Q}(Tf)-m_{R}(Tf)|$ $\displaystyle\leq
m_{Q}(|Tf\chi_{6Q}|)+CK_{Q,R}\|f\|_{L^{\infty}(\mu)}+|Tf\chi_{Q_{R}}(y)|$
$\displaystyle+C\|f\|_{L^{\infty}(\mu)}+m_{R}(Tf\chi_{Q_{R}}).$

For the boundedness on $L^{2}(\mu)$ of $T$, we have

$\displaystyle m_{Q}(|Tf\chi_{6Q}|)$
$\displaystyle\leq\Big{(}\frac{1}{\mu(Q)}\int_{Q}|Tf\chi_{6Q}|^{2}\Big{)}^{1/2}$
$\displaystyle\leq
C\Big{(}\frac{\mu(6Q)}{\mu(Q)}\Big{)}^{1/2}\|f\|_{L^{\infty}(\mu)}$
$\displaystyle\leq
C\Big{(}\frac{\mu(6Q)}{\mu(Q)}\Big{)}\|f\|_{L^{\infty}(\mu)}.$

Next, we write

$m_{R}(Tf\chi_{Q_{R}})\leq m_{R}(Tf\chi_{Q_{R}\cap
6R})+m_{R}(Tf\chi_{Q_{R}\backslash 6R}).$

By similar argument in estimate of $m_{Q}(|Tf\chi_{6Q}|)$, the term
$m_{R}(Tf\chi_{Q_{R}\cap 6R})$ is dominated by

$C\Big{(}\frac{\mu(6R)}{\mu(R)}\Big{)}\|f\|_{L^{\infty}(\mu)}.$

The second term $m_{R}(Tf\chi_{Q_{R}\backslash 6R})$ can be treated as in
(34). Since $r_{Q_{R}}\approx r_{R}$, we have

$m_{R}(Tf\chi_{Q_{R}\backslash 6R})\leq C\|f\|_{L^{\infty}(\mu)}$

To sum up, we have

$\displaystyle|m_{Q}(Tf)-m_{R}(Tf)|$ $\displaystyle\leq
CK_{Q,R}\|f\|_{L^{\infty}(\mu)}+\Big{(}\frac{\mu(6Q)}{\mu(Q)}+\frac{\mu(6R)}{\mu(R)}\Big{)}\|f\|_{L^{\infty}(\mu)}$
$\displaystyle\leq
CK_{Q,R}\Big{(}\frac{\mu(6Q)}{\mu(Q)}+\frac{\mu(6R)}{\mu(R)}\Big{)}\|f\|_{L^{\infty}(\mu)}.$

###### Remark 7.2

By similar argument in [T1, Theorem 2.11], we can replace the assumption of
$L^{2}(\mu)$ boundedness by the weaker assumption: for any ball $B$ and any
function $a$ supported on $B$,

$\int_{B}|Ta|d\mu\leq C\|a\|_{L^{\infty}}\mu(6B)$

uniformly on $\epsilon>0$.

###  7.2 The boundedness of Calderón-Zygmund operators on Hardy spaces

We now show that an $L^{2}$ bounded Calderón-Zygmund operator maps the atomic
Hardy space boundedly into $L^{1}$.

###### Theorem 7.3

Assume that $T$ is a Calderón-Zygmund operator and $T$ is bounded on
$L^{2}(X,\mu)$, then $T$ is bounded from $H_{at}^{1,\infty}(\mu)$ into
$L^{1}(X,\mu)$. Therefore, by interpolation and duality, $T$ is bounded on
$L^{p}(\mu)$ for all $1<p<\infty$.

_Proof:_ By [HoM, Lemm 4.1], it is enough to show that

$\|Tb\|_{L^{1}(\mu)}\leq C|b|_{H_{at}^{1,\infty}(\mu)}$ (35)

for any atomic block $b$ with supp$b\subset B$ and $=\sum_{j}\lambda_{j}a_{j}$
where the $a_{j}$’s are functions satisfying (a) and (b) in definition of
atomic blocks. At this stage we can use the same argument as in [T1, Theorem
4.2] with minor modifications as in Theorem 7.1 to obtain the estimate (35).
We omit the details here.

###  7.3 Commutators of Calderón-Zygmund operators with RBMO functions

In this section we assume that the dominating function $\lambda$ satisfies
$\lambda(x,ar)=a^{m}\lambda(x,r)$ for all $x\in X$ and $a,r>0$. Then, for two
balls $B$, $Q$ such that $B\subset Q$ we can define the coefficient
$K^{\prime}_{B,Q}$ as follows: let $N_{B,Q}$ be the smallest integer
satisfying $6^{N_{B,Q}}r_{B}\geq r_{Q}$, we set

$K^{\prime}_{B,Q}:=1+\sum_{k=1}^{N_{B,Q}}\frac{\mu(6^{k}B)}{\lambda(x_{B},6^{k}r_{B})}.$
(36)

It is not difficult to show that the coefficient $K_{B,Q}\approx
K^{\prime}_{B,Q}$. Note that in the definition of $K^{\prime}_{B,Q}$ we can
replace $6$ by any number $\eta>1$.

To establish the boundedness of commutators of Calderón-Zygmund operators with
RBMO functions, we need the following two lemmas. Note that these lemmas are
similar to those in [T1]. However, due to the difference of choices of
coefficient $K_{Q,R}$, we would like to provide the proof for the first one.
Meanwhile, the proof of Lemma 7.5 is completely analogous to that of Lemma 9.3
in [T1], hence we omit the details.

###### Lemma 7.4

If $B_{i}=B(x_{0},r_{i}),i=1,\ldots,m$ are concentric balls $B_{1}\subset
B_{2}\subset\ldots\subset B_{m}$ with $K_{B_{i},B_{i+1}}>2$ for
$i=1,\ldots,m-1$ then

$\sum_{i=1}^{m-1}K_{B_{i},B_{i+1}}\leq 2K_{B_{1},B_{m}}.$ (37)

_Proof:_ By definition,

$K_{B_{i},B_{i+1}}=1+\int\limits_{r_{i}\leq d(x,x_{0})\leq
r_{i+1}}\frac{1}{\lambda(x_{0},d(x,x_{0}))}d\mu(x).$

Since $K_{B_{i},B_{i+1}}>2$, we have

$K_{B_{i},B_{i+1}}<2\int\limits_{r_{i}\leq d(x,x_{0})\leq
r_{i+1}}\frac{1}{\lambda(x_{0},d(x,x_{0}))}d\mu(x)$

for all $i=1,\ldots,m-1$.
This implies

$\displaystyle\sum_{i=1}^{m-1}K_{B_{i},B_{i+1}}$
$\displaystyle<2\sum_{i=1}^{m-1}\int\limits_{r_{i}\leq d(x,x_{0})\leq
r_{i+1}}\frac{1}{\lambda(x_{0},d(x,x_{0}))}d\mu(x)$ $\displaystyle\leq
2\int\limits_{r_{1}\leq d(x,x_{0})\leq
r_{m}}\frac{1}{\lambda(x_{0},d(x,x_{0}))}d\mu(x)$ $\displaystyle\leq
2K_{B_{1},B_{m}}.$

###### Lemma 7.5

There exists some constant $P_{0}$ such that if $x\in X$ is some fixed point
and $\\{f_{B}\\}_{B\ni x}$ is collection of numbers such that
$|f_{Q}-f_{R}|\leq C_{x}$ for all doubling balls $Q\subset R$ with $x\in Q$
and $K_{Q,R}\leq P_{0}$, then

$|f_{Q}-f_{R}|\leq CK_{Q,R}C_{x}\ \ \text{for all doubling balls $Q\subset R$
with $x\in Q$}.$

###### Theorem 7.6

If $b\in{\rm RBMO}(\mu)$ and $T$ is a Calderón-Zygmund bounded on
$L^{2}(\mu)$, then the commutator $[b,T]$ defined by

$[b,T](f)=bT(f)-T(bf)$

is bounded on $L^{p}(\mu)$ for $1<p<\infty$.

_Proof:_ For $1<p<\infty$ we will show that

$M^{\sharp}([b,T]f)(x)\leq C\|b\|_{{\rm
RBMO}(\mu)}\Big{(}M_{p,5}f(x)+M_{p,6}Tf(x)+T_{*}f(x)\Big{)},$ (38)

where

$M_{p,\rho}f(x)=\sup_{Q\ni x}\Big{(}\frac{1}{\mu(\rho
Q)}\int_{Q}|f|^{p}\Big{)}.$

Once (38) is proved, it follows from the boundedness of $T_{*}$ on
$L^{p}(\mu)$ and $M_{p,\rho}$ on $L^{r}(\mu),r>p$ and $\rho\geq 5$, and from a
standard argument that we can obtain the boundedness of $[b,T]$ on
$L^{p}(\mu)$.

Let $\\{b_{B}\\}$ be a family of numbers satisfying

$\int_{B}|b-b_{B}|d\mu\leq 2\mu(6B)\|b\|_{{\rm RBMO}}$

for balls $B$, and

$|b_{Q}-b_{R}|\leq 2K_{Q,R}\|b\|_{{\rm RBMO}}$

for balls $Q\subset R$. Denote

$h_{Q}:=m_{Q}(T((b-b_{Q})f\chi_{X\backslash\frac{6}{5}Q}).$

We will show that

$\frac{1}{\mu(6B)}\int_{B}|[b,T]f-h_{Q}|d\mu\leq C\|b\|_{{\rm
RBMO}}(M_{p,5}f(x)+M_{p,6}Tf(x))$ (39)

for all $x$ and $B$ with $x\in B$, and

$|h_{Q}-h_{R}|\leq C\|b\|_{{\rm RBMO}}(M_{p,5}f(x)+T_{*}f(x))K^{2}_{Q,R}$ (40)

for all $x\in Q\subset R$.
The proof of (39) is similar to that in Theorem 9.1 in [T1] with minor
modifications and we omit it here.
It remains to check (40). For two balls $Q\subset R$, let $N$ be an integer
such that $(N-1)$ is the smallest number satisfying $r_{R}\leq 6^{N-1}r_{Q}$.
Then, we break the term $|h_{Q}-h_{R}|$ into five terms:

$\displaystyle|m_{Q}(T((b-b_{Q})$ $\displaystyle
f\chi_{X\backslash\frac{6}{5}Q})-m_{R}(T((b-b_{R})f\chi_{X\backslash\frac{6}{5}R})|$
$\displaystyle\leq|m_{Q}(T((b-b_{Q})f\chi_{6Q\backslash\frac{6}{5}Q})|+|m_{Q}(T((b_{Q}-b_{R})f\chi_{X\backslash
6Q})|$ $\displaystyle~{}~{}~{}+|m_{Q}(T((b-b_{R})f\chi_{6^{N}Q\backslash
6Q})|$ $\displaystyle~{}~{}~{}+|m_{Q}(T((b-b_{R})f\chi_{X\backslash
6^{N}Q})-m_{R}(T((b-b_{R})f\chi_{X\backslash 6^{N}Q})|$
$\displaystyle~{}~{}~{}+|m_{R}(T((b-b_{R})f\chi_{6^{N}Q\backslash\frac{6}{5}R})$
$\displaystyle=M_{1}+M_{2}+M_{3}+M_{4}+M_{5}.$

Let us estimate $M_{1}$ first. For $y\in Q$ we have, by Proposition 3.2

$\displaystyle|T((b-b_{Q})$ $\displaystyle
f\chi_{6Q\backslash\frac{6}{5}Q})(x)|$
$\displaystyle\leq\frac{C}{\lambda(x,r_{Q})}\int_{6Q}|b-b_{Q}||f|d\mu$
$\displaystyle\leq\frac{\mu(30Q)}{\lambda(x,30r_{Q})}\Big{(}\frac{1}{\mu(5\times
6Q)}\int_{6Q}|b-b_{Q}|^{p^{\prime}}d\mu\Big{)}^{1/p^{\prime}}\Big{(}\frac{1}{\mu(5\times
6Q)}\int_{6Q}|f|^{p}d\mu\Big{)}^{1/p}$ $\displaystyle C\|b\|_{{\rm
RBMO}}M_{p,5}f(x).$

The term $M_{5}$ can be treated by similar way. So, we have $M_{1}+M_{5}\leq
C\|b\|_{{\rm RBMO}}M_{p,5}f(x).$ For the term $M_{2}$, we have for $x,y\in Q$

$\displaystyle|Tf\chi_{X\backslash 6Q}(y)|$
$\displaystyle=\Big{|}\int_{X\backslash 6Q}K(y,z)f(z)d\mu(z)\Big{|}$
$\displaystyle\leq\Big{|}\int_{X\backslash
6Q}K(x,z)f(z)d\mu(z)\Big{|}+\int_{X\backslash 6Q}|K(y,z)-K(x,z)||f(z)|d\mu(z)$
$\displaystyle\leq T_{*}f(x)+CM_{p,5}f(x).$

This implies

$|m_{Q}(T((b_{Q}-b_{R})f\chi_{X\backslash 6Q})|\leq
CK_{Q,R}(T_{*}f(x)+M_{p,6}f(x)).$

For the term $M_{4}$, we have, for $y,z\in R$

$\displaystyle|T((b-b_{R})$ $\displaystyle f\chi_{X\backslash
6^{N}Q}(y)-T((b-b_{R})f\chi_{X\backslash 6^{N}Q}(z)|$
$\displaystyle\leq\int_{X\backslash
2R}|K(y,w)-K(z,w)||(b(w)-b_{R})||f(x)|d\mu(w)$
$\displaystyle\leq\int_{X\backslash
2R}\frac{d(y,z)^{\delta}}{d(w,y)^{\delta}\lambda(y,d(w,y))}|(b(w)-b_{R})||f(x)|d\mu(w)$
$\displaystyle\leq\sum_{k=1}^{\infty}\int_{2^{k+1}R\backslash
2^{k}R}\frac{d(y,z)^{\delta}}{d(w,y)^{\delta}\lambda(y,d(w,y))}|(b(w)-b_{R})||f(x)|d\mu(w)$
$\displaystyle\leq\sum_{k=1}^{\infty}2^{-k\delta}\frac{1}{\lambda(y,2^{k-1}r_{R})}\int_{2^{k+1}R}|(b(w)-b_{R})||f(x)|d\mu(x)$

By Hölder inequality we have

$\displaystyle|T((b-b_{R})$ $\displaystyle f\chi_{X\backslash
6^{N}Q}(y)-T((b-b_{R})f\chi_{X\backslash 6^{N}Q}(z)|$ $\displaystyle\leq
C\sum_{k=1}^{\infty}2^{-k\delta}\frac{\mu(5\times 2^{k+1}R)}{\lambda(y,5\times
2^{k+1}r_{R})}\Big{(}\frac{1}{\mu(5\times
2^{k+1}R)}\int_{2^{k+1}R}|b-b_{R}|^{p^{\prime}}d\mu\Big{)}^{1/p^{\prime}}$
$\displaystyle~{}~{}~{}~{}\Big{(}\frac{1}{\mu(5\times
2^{k+1}R)}\int_{2^{k+1}R}|f|^{p}d\mu\Big{)}^{1/p}$ $\displaystyle\leq
C\sum_{k=1}^{\infty}2^{-k\delta}\Big{[}\Big{(}\frac{1}{\mu(5\times
2^{k+1}R)}\int_{2^{k+1}R}|b-b_{2^{k+1}R}|^{p^{\prime}}d\mu\Big{)}^{1/p^{\prime}}$
$\displaystyle~{}~{}~{}~{}+\Big{(}\frac{1}{\mu(5\times
2^{k+1}R)}\int_{2^{k+1}R}|b_{R}-b_{2^{k+1}R}|^{p^{\prime}}d\mu\Big{)}^{1/p^{\prime}}\Big{]}\Big{(}\frac{1}{\mu(5\times
2^{k+1}R)}\int_{2^{k+1}R}|f|^{p}d\mu\Big{)}^{1/p}$ $\displaystyle\leq
C\sum_{k=1}^{\infty}C(k+1)2^{-k\delta}\|b\|_{{\rm RBMO}}M_{p,5}f(x)$
$\displaystyle\leq C\|b\|_{{\rm RBMO}}M_{p,5}f(x).$

Taking the mean over $Q$ and $R$ for $y$ and $z$ respectively, we obtain

$M_{4}\leq C\|b\|_{{\rm RBMO}}M_{p,5}f(x).$

Concerning the last estimate for $M_{3}$, we have for $y\in Q$

$\displaystyle|T((b-b_{R})$ $\displaystyle f\chi_{6^{N}Q\backslash 6Q}(y)|$
(41) $\displaystyle\leq
C\sum_{k=1}^{N-1}\frac{1}{\lambda(6^{k}Q)}\int_{6^{k+1Q}\backslash
6^{k}Q}|b-b_{R}||f|d\mu$ $\displaystyle\leq
C\sum_{k=1}^{N-1}\frac{1}{\lambda(y,6^{k}Q)}\Big{[}\int\limits_{6^{k+1}Q\backslash
6^{k}Q}|b-b_{5^{k+1}Q}||f|d\mu+\int\limits_{6^{k+1Q}\backslash
6^{k}Q}|b_{R}-b_{6^{k+1}Q}||f|d\mu\Big{]}$ $\displaystyle\leq
C\sum_{k=1}^{N-1}\frac{\mu(5\times
6^{k+1}Q)}{\lambda(x_{Q},6^{k}Q)}\Big{[}\frac{1}{\mu(6^{k+2}Q)}\int\limits_{6^{k+1}Q\backslash
6^{k}Q}|b-b_{6^{k+1}Q}||f|d\mu$
$\displaystyle~{}~{}~{}~{}+\frac{1}{\mu(5\times
6^{k+1}Q)}\int\limits_{6^{k+1Q}\backslash
6^{k}Q}|b_{R}-b_{6^{k+1}Q}||f|d\mu\Big{]}.$

By Hölder inequality and a similar argument to the estimate of the term
$M_{4}$, we have

$\frac{1}{\mu(5\times 6^{k+2}Q)}\int\limits_{6^{k+1}Q\backslash
6^{k}Q}|b-b_{6^{k+1}Q}||f|d\mu\leq\|b\|_{{\rm RBMO}}M_{p,5}f(x)$

and

$\frac{1}{\mu(5\times 6^{k+1}Q)}\int\limits_{6^{k+1Q}\backslash
6^{k}Q}|b_{R}-b_{6^{k+1}Q}||f|d\mu\leq CK_{Q,R}\|b\|_{{\rm RBMO}}M_{p,5}f(x).$

These two estimates together with (41) give

$|T((b-b_{R})f\chi_{6^{N}Q\backslash 6Q}(y)|\leq CK^{2}_{Q,R}\|b\|_{{\rm
RBMO}}M_{p,5}f(x).$

This implies $M_{3}\leq CK^{2}_{Q,R}\|b\|_{{\rm RBMO}}M_{p,5}f(x).$ From the
estimates $M_{1},M_{2},M_{3},M_{4},M_{5}$, we obtain (40).
To obtain (38) from (39) and (40), we use a trick of [T1]. From (39), if $Q$
is a doubling ball and $x\in Q$, we have

$\displaystyle|m_{Q}([b,T]f)-h_{Q}|$
$\displaystyle\leq\frac{1}{\mu(Q)}\int_{Q}|[b,T]f-h_{Q}|d\mu$ (42)
$\displaystyle\leq C\|b\|_{{\rm RBMO}}(M_{p,5}f(x)+M_{p,6}Tf(x)).$

Also, for any ball $Q\ni x$ (non doubling, in general),
$K_{Q,\widetilde{Q}}\leq C$, and then by (39) and (40) we have

$\displaystyle\frac{1}{\mu(6Q)}$
$\displaystyle\int_{Q}|[b,T]f-m_{\widetilde{Q}}[b,T]f|d\mu$ (43)
$\displaystyle\leq\frac{1}{\mu(6Q)}\int_{Q}|[b,T]f-h_{Q}|d\mu+|h_{Q}-h_{\widetilde{Q}}|+|h_{\widetilde{Q}}-m_{\widetilde{Q}}[b,T]f|$
$\displaystyle\leq C\|b\|_{{\rm
RBMO}(\mu)}\Big{(}M_{p,5}f(x)+M_{p,6}Tf(x)+T_{*}f(x)\Big{)}.$

In addition, for all doubling balls $Q\subset R$ with $x\in Q$ such that
$K_{Q,R}\leq P_{0}$ where $P_{0}$ is a constant in Lemma 7.5, by (40) we have

$|h_{Q}-h_{R}|\leq C\|b\|_{{\rm
RBMO}(\mu)}\Big{(}M_{p,5}f(x)+T_{*}f(x)\Big{)}P_{0}^{2}.$

Due to Lemma 7.5 we get

$|h_{Q}-h_{R}|\leq C\|b\|_{{\rm
RBMO}(\mu)}\Big{(}M_{p,5}f(x)+T_{*}f(x)\Big{)}K_{Q,R},$

for all doubling balls $Q\subset R$ with $x\in Q$. At this stage, applying
(41), we obtain

$\displaystyle m_{Q}([b,T]f)$ $\displaystyle-m_{R}([b,T]f)$ $\displaystyle\leq
C\|b\|_{{\rm
RBMO}(\mu)}\Big{(}M_{p,5}f(x)+M_{p,6}Tf(x)+T_{*}f(x)\Big{)}K_{Q,R}.$

This completes our proof.

###### Remark 7.7

As mentioned earlier in this paper, the results of this article still hold
when $X$ is a quasi-metric space. Indeed, one can see that the main problem in
quasi-metric space setting is that the covering lemma, Lemma 2.1, may not be
true. However, instead of using this covering property, we can adapt the
covering lemma in [FGL, Lemma 3.1] to our situation. This problem is not
difficult and we leave it to the interested reader.

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Department of Mathematics, Macquarie University, NSW 2109, Australia and
Department of Mathematics, University of Pedagogy, Ho chi Minh city, Vietnam
Email<EMAIL_ADDRESS>
Department of Mathematics, Macquarie University, NSW 2109, Australia
Email<EMAIL_ADDRESS>