# Sharp maximal function inequalities and boundedness for Toeplitz type operator associated to singular integral operator with non-smooth kernel

## Abstract

In this paper, we establish the sharp maximal function inequalities for the Toeplitz type operator associated to some singular integral operator with non-smooth kernel. As an application, we obtain the boundedness of the operator on Morrey and Triebel-Lizorkin spaces.

MSC:42B20, 42B25.

## 1 Introduction and preliminaries

As the development of the singular integral operators, their commutators and multilinear operators have been well studied (see ). In [1, 2], the authors prove that the commutators generated by the singular integral operators and BMO functions are bounded on $L p ( R n )$ for $1. Chanillo (see ) proves a similar result when singular integral operators are replaced by the fractional integral operators. In [5, 6], the boundedness for the commutators generated by the singular integral operators and Lipschitz functions on Triebel-Lizorkin and $L p ( R n )$ ($1) spaces are obtained. In , some Toeplitz type operators associated to the singular integral operators and strongly singular integral operators are introduced, and the boundedness for the operators generated by BMO and Lipschitz functions are obtained. In [10, 11], some singular integral operators with non-smooth kernel are introduced, and the boundedness for the operators and their commutators are obtained (see [9, 1216]). The main purpose of this paper is to study the Toeplitz type operator generated by the singular integral operator with non-smooth kernel and the Lipschitz and BMO functions.

Definition 1 A family of operators $D t$, $t>0$ is said to be an ‘approximation to the identity’ if, for every $t>0$, $D t$ can be represented by a kernel $a t (x,y)$ in the following sense:

$D t (f)(x)= ∫ R n a t (x,y)f(y)dy$

for every $f∈ L p ( R n )$ with $p≥1$, and $a t (x,y)$ satisfies

$| a t ( x , y ) | ≤ h t (x,y)=C t − n / 2 ρ ( | x − y | 2 / t ) ,$

where ρ is a positive, bounded, and decreasing function satisfying

$lim r → ∞ r n + ϵ ρ ( r 2 ) =0$

for some $ϵ>0$.

Definition 2 A linear operator T is called a singular integral operator with non-smooth kernel if T is bounded on $L 2 ( R n )$ and associated with a kernel $K(x,y)$ such that

$T(f)(x)= ∫ R n K(x,y)f(y)dy$

for every continuous function f with compact support, and for almost all x not in the support of f.

1. (1)

There exists an ‘approximation to the identity’ ${ B t ,t>0}$ such that $T B t$ has the associated kernel $k t (x,y)$ and there exist $c 1 , c 2 >0$ so that

2. (2)

There exists an ‘approximation to the identity’ ${ A t ,t>0}$ such that $A t T$ has the associated kernel $K t (x,y)$ which satisfies

and

for some $δ>0$, $c 3 , c 4 >0$.

Let b be a locally integrable function on $R n$ and T be the singular integral operator with non-smooth kernel. The Toeplitz type operator associated to T is defined by

$T b = ∑ k = 1 m ( T k , 1 M b I α T k , 2 + T k , 3 I α M b T k , 4 ) ,$

where $T k , 1$ are the singular integral operator with non-smooth kernel T or ±I (the identity operator), $T k , 2$ and $T k , 4$ are the linear operators, $T k , 3 =±I$, $k=1,…,m$, $M b (f)=bf$ and $I α$ is the fractional integral operator ($0<α) (see ).

Note that the commutator $[b,T](f)=bT(f)−T(bf)$ is a particular operator of the Toeplitz type operator $T b$. The Toeplitz type operator $T b$ is the non-trivial generalizations of the commutator. It is well known that commutators are of great interest in harmonic analysis and have been widely studied by many authors (see ). In [10, 11], the boundedness of the singular integral operator with non-smooth kernel are obtained. In , the boundedness of the commutator associated to the singular integral operator with non-smooth kernel are obtained. Our works is motivated by these papers. In this paper, we will prove the sharp maximal inequalities for the Toeplitz type operator $T b$. As the application, we obtain the Morrey and Triebel-Lizorkin spaces boundedness for the Toeplitz type operator $T b$.

Definition 3 Let $0<β<1$ and $1≤p<∞$. The Triebel-Lizorkin space associated with the ‘approximations to the identity’ ${ A t ,t>0}$ is defined by

$F ˙ p , A β , ∞ ( R n ) = { f ∈ L loc 1 ( R n ) : ∥ f ∥ F ˙ p , A β , ∞ < ∞ } ,$

where

$∥ f ∥ F ˙ p , A β , ∞ = ∥ sup Q ∋ ⋅ 1 | Q | 1 + β / n ∫ Q | f ( x ) − A t Q ( f ) ( x ) | d x ∥ L p ,$

and the supremum is taken over all cubes Q of $R n$ with sides parallel to the axes, $t Q =l ( Q ) 2$ and $l(Q)$ denotes the side length of Q.

Now, let us introduce some notations. Throughout this paper, $Q=Q(x,r)$ will denote a cube of $R n$ with sides parallel to the axes and center at x and edge is r. For any locally integrable function f, the sharp function of f is defined by

$f # (x)= sup Q ∋ x 1 | Q | ∫ Q | f ( y ) − f Q | dy,$

where, and in what follows $f Q = | Q | − 1 ∫ Q f(x)dx$. It is well known that (see )

$f # (x)≈ sup Q ∋ x inf c ∈ C 1 | Q | ∫ Q | f ( y ) − c | dy$

and

We say that f belongs to $BMO( R n )$ if $f #$ belongs to $L ∞ ( R n )$ and $∥ f ∥ BMO = ∥ f # ∥ L ∞$.

Let M be the Hardy-Littlewood maximal operator defined by

$M(f)(x)= sup Q ∋ x | Q | − 1 ∫ Q | f ( y ) | dy.$

For $η>0$, let $M η (f)(x)=M ( | f | η ) 1 / η (x)$.

For $0≤η and $1≤r<∞$, set

$M η , r (f)(x)= sup Q ∋ x ( 1 | Q | 1 − r η / n ∫ Q | f ( y ) | r d y ) 1 / r .$

The $A 1$ weight is defined by (see )

$A 1 = { w ∈ L loc p ( R n ) : M ( w ) ( x ) ≤ C w ( x ) , a.e. } .$

The sharp maximal function $M A (f)$ associated with the ‘approximation to the identity’ ${ A t ,t>0}$ is defined by

$M A # (f)(x)= sup x ∈ Q 1 | Q | ∫ Q | f ( y ) − A t Q ( f ) ( y ) | dy,$

where $t Q =l ( Q ) 2$ and $l(Q)$ denotes the side length of Q.

For $β>0$, the Lipschitz space $Lip β ( R n )$ is the space of functions f such that

$∥ f ∥ Lip β = sup x , y ∈ R n x ≠ y | f ( x ) − f ( y ) | | x − y | β <∞.$

Throughout this paper, φ will denote a positive, increasing function on $R +$ for which there exists a constant $D>0$ such that

Let f be a locally integrable function on $R n$. Set, for $0≤η and $1≤p,

$∥ f ∥ L p , η , φ = sup x ∈ R n , d > 0 ( 1 φ ( d ) 1 − p η / n ∫ Q ( x , d ) | f ( y ) | p d y ) 1 / p .$

The generalized fractional Morrey spaces are defined by

$L p , η , φ ( R n ) = { f ∈ L loc 1 ( R n ) : ∥ f ∥ L p , η , φ < ∞ } .$

We write $L p , η , φ ( R n )= L p , φ ( R n )$ if $η=0$, which is the generalized Morrey space. If $φ(d)= d δ$, $δ>0$, then $L p , φ ( R n )= L p , δ ( R n )$, which is the classical Morrey space (see [18, 19]). As the Morrey space may be considered as an extension of the Lebesgue space (the Morrey space $L p , λ$ becomes the Lebesgue space $L p$ when $λ=0$), it is natural and important to study the boundedness of the operator on the Morrey spaces $L p , λ$ with $λ>0$ (see ). The purpose of this paper is twofold. First, we establish some sharp inequalities for the Toeplitz type operator $T b$, and, second, we prove the boundedness for the Toeplitz type operator by using the sharp inequalities.

## 2 Theorems and lemmas

We shall prove the following theorems.

Theorem 1 Let T be the singular integral operator with non-smooth kernel as Definition  2, $0<β<1$, $1 and $b∈ Lip β ( R n )$. If $T 1 (g)=0$ for any $g∈ L u ( R n )$ ($1), then there exists a constant $C>0$ such that, for any $f∈ C 0 ∞ ( R n )$ and $x ˜ ∈ R n$,

$M A # ( T b ( f ) ) ( x ˜ )≤C ∥ b ∥ Lip β ∑ k = 1 m ( M β , s ( I α T k , 2 ( f ) ) ( x ˜ ) + M β + α , s ( T k , 4 ( f ) ) ( x ˜ ) ) .$

Theorem 2 Let T be the singular integral operator with non-smooth kernel as Definition  2, $0<β, $1 and $b∈ Lip β ( R n )$. If $T 1 (g)=0$ for any $g∈ L u ( R n )$ ($1), then there exists a constant $C>0$ such that, for any $f∈ C 0 ∞ ( R n )$ and $x ˜ ∈ R n$,

$sup Q ∋ x ˜ 1 | Q | 1 + β / n ∫ Q | T b ( f ) ( x ) − A t Q ( T b ( f ) ) ( x ) | d x ≤ C ∥ b ∥ Lip β ∑ k = 1 m ( M s ( I α T k , 2 ( f ) ) ( x ˜ ) + M α , s ( T k , 4 ( f ) ) ( x ˜ ) ) .$

Theorem 3 Let T be the singular integral operator with non-smooth kernel as Definition  2, $1 and $b∈BMO( R n )$. If $T 1 (g)=0$ for any $g∈ L u ( R n )$ ($1), then there exists a constant $C>0$ such that, for any $f∈ C 0 ∞ ( R n )$ and $x ˜ ∈ R n$,

$M A # ( T b ( f ) ) ( x ˜ )≤C ∥ b ∥ BMO ∑ k = 1 m ( M s ( I α T k , 2 ( f ) ) ( x ˜ ) + M α , s ( T k , 4 ( f ) ) ( x ˜ ) ) .$

Theorem 4 Let T be the singular integral operator with non-smooth kernel as Definition  2, $0<β<1$, $1, $1/q=1/p−(α+β)/n$, $0 and $b∈ Lip β ( R n )$. If $T 1 (g)=0$ for any $g∈ L u ( R n )$ ($1) and $T k , 2$ and $T k , 4$ are the bounded operators on $L p , φ ( R n )$ for $1, $k=1,…,m$, then $T b$ is bounded from $L p , α + β , φ ( R n )$ to $L q , φ ( R n )$.

Theorem 5 Let T be the singular integral operator with non-smooth kernel as Definition  2, $0<β, $1, $1/q=1/p−α/n$ and $b∈ Lip β ( R n )$. If $T 1 (g)=0$ for any $g∈ L u ( R n )$ ($1) and $T k , 2$ and $T k , 4$ are the bounded operators on $L p ( R n )$ for $1, $k=1,…,m$, then $T b$ is bounded from $L p ( R n )$ to $F ˙ q , A β , ∞ ( R n )$.

Theorem 6 Let T be the singular integral operator with non-smooth kernel as Definition  2, $0, $1, $1/q=1/p−α/n$ and $b∈BMO( R n )$. If $T 1 (g)=0$ for any $g∈ L u ( R n )$ ($1) and $T k , 2$ and $T k , 4$ are the bounded operators on $L p , φ ( R n )$ for $1, $k=1,…,m$, then $T b$ is bounded from $L p , α , φ ( R n )$ to $L q , φ ( R n )$.

Corollary 1 Let $[b,T](f)=bT(f)−T(bf)$ be the commutator generated by the singular integral operator T with non-smooth kernel and b. Then Theorems  1-6 hold for $[b,T]$.

To prove the theorems, we need the following lemmas.

Lemma 1 ([10, 11])

Let T be the singular integral operator with non-smooth kernel as Definition  2. Then, for every $f∈ L p ( R n )$, $1,

$∥ T ( f ) ∥ L p ≤C ∥ f ∥ L p .$

Lemma 2 ([10, 11])

Let ${ A t ,t>0}$ be an ‘approximation to the identity’. For any $γ>0$, there exists a constant $C>0$ independent of γ such that

$| { x ∈ R n : M ( f ) ( x ) > D λ , M A # ( f ) ( x ) ≤ γ λ } | ≤Cγ | { x ∈ R n : M ( f ) ( x ) > λ } |$

for $λ>0$, where D is a fixed constant which only depends on n. Thus, for $f∈ L p ( R n )$, $1 and $w∈ A 1$,

$∥ M ( f ) ∥ L p ( w ) ≤C ∥ M A # ( f ) ∥ L p ( w ) .$

Lemma 3 (See )

Let ${ A t ,t>0}$ be an ‘approximation to the identity’ and $K ˜ α , t (x,y)$ be the kernel of difference operator $I α − A t I α$. Then

$| K ˜ α , t ( x , y ) | ≤C t | x − y | n + 2 − α .$

Lemma 4 (See [4, 17])

Suppose that $0≤α, $1≤s, $1/q=1/p−α/n$ and $w∈ A 1$. Then

$∥ I α ( f ) ∥ L q ( w ) ≤C ∥ f ∥ L p ( w )$

and

$∥ M α , s ( f ) ∥ L q ( w ) ≤C ∥ f ∥ L p ( w ) .$

Lemma 5 Let ${ A t ,t>0}$ be an ‘approximation to the identity’ and $0. Then

1. (a)

$∥ M ( f ) ∥ L p , φ ≤C ∥ M A # ( f ) ∥ L p , φ$ for $1;

2. (b)

$∥ I α ( f ) ∥ L q , φ ≤C ∥ f ∥ L p , α , φ$ for $0<α, $1 and $1/q=1/p−α/n$;

3. (c)

$∥ M α , s ( f ) ∥ L q , φ ≤C ∥ f ∥ L p , α , φ$ for $0≤α, $1≤s and $1/q=1/p−α/n$.

Proof (a) For any cube $Q=Q( x 0 ,d)$ in $R n$, we know $M( χ Q )∈ A 1$ for any cube $Q=Q(x,d)$ by . Noticing that $M( χ Q )≤1$ and $M( χ Q )(x)≤ d n / ( | x − x 0 | − d ) n$ if $x∈ Q c$, by Lemma 2, we have

$∫ Q M ( f ) ( x ) p d x = ∫ R n M ( f ) ( x ) p χ Q ( x ) d x ≤ ∫ R n M ( f ) ( x ) p M ( χ Q ) ( x ) d x ≤ C ∫ R n M A # ( f ) ( x ) p M ( χ Q ) ( x ) d x = C ( ∫ Q M A # ( f ) ( x ) p M ( χ Q ) ( x ) d x + ∑ k = 0 ∞ ∫ 2 k + 1 Q ∖ 2 k Q M A # ( f ) ( x ) p M ( χ Q ) ( x ) d x ) ≤ C ( ∫ Q M A # ( f ) ( x ) p d x + ∑ k = 0 ∞ ∫ 2 k + 1 Q ∖ 2 k Q M A # ( f ) ( x ) p | Q | | 2 k + 1 Q | d x ) ≤ C ( ∫ Q M A # ( f ) ( x ) p d x + ∑ k = 0 ∞ ∫ 2 k + 1 Q M A # ( f ) ( x ) p 2 − k n d y ) ≤ C ∥ M A # ( f ) ∥ L p , φ p ∑ k = 0 ∞ 2 − k n φ ( 2 k + 1 d ) ≤ C ∥ M A # ( f ) ∥ L p , φ p ∑ k = 0 ∞ ( 2 − n D ) k φ ( d ) ≤ C ∥ M A # ( f ) ∥ L p , φ p φ ( d ) ,$

thus

$∥ M ( f ) ∥ L p , φ ≤C ∥ M A # ( f ) ∥ L p , φ .$

The proofs of (b) and (c) are similar to that of (a) by Lemma 4, we omit the details. □

## 3 Proofs of theorems

Proof of Theorem 1 It suffices to prove for $f∈ C 0 ∞ ( R n )$, the following inequality holds:

$1 | Q | ∫ Q | T b ( f ) ( x ) − A t Q ( T b ( f ) ) ( x ) | d x ≤ C ∥ b ∥ Lip β ∑ k = 1 m ( M β , s ( I α T k , 2 ( f ) ) ( x ˜ ) + M β + α , s ( T k , 4 ( f ) ) ( x ˜ ) ) ,$

where $t Q = ( l ( Q ) ) 2$ and $l(Q)$ denotes the side length of Q. Without loss of generality, we may assume $T k , 1$ are T ($k=1,…,m$). Fix a cube $Q=Q( x 0 ,d)$ and $x ˜ ∈Q$. We write, by $T 1 (g)=0$,

$T b ( f ) ( x ) = ∑ k = 1 m T k , 1 M b I α T k , 2 ( f ) ( x ) + ∑ k = 1 m T k , 3 I α M b T k , 4 ( f ) ( x ) = U b ( x ) + V b ( x ) = U b − b 2 Q ( x ) + V b − b 2 Q ( x ) ,$

where

$U b − b 2 Q ( x ) = ∑ k = 1 m T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( x ) + ∑ k = 1 m T k , 1 M ( b − b 2 Q ) χ ( 2 Q ) c I α T k , 2 ( f ) ( x ) = U 1 ( x ) + U 2 ( x )$

and

$V b − b 2 Q ( x ) = ∑ k = 1 m T k , 3 I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( x ) + ∑ k = 1 m T k , 3 I α M ( b − b 2 Q ) χ ( 2 Q ) c T k , 4 ( f ) ( x ) = V 1 ( x ) + V 2 ( x ) .$

Then

$1 | Q | ∫ Q | T b ( f ) ( x ) − A t Q ( T b ( f ) ) ( x ) | d x ≤ 1 | Q | ∫ Q | U 1 ( x ) | d x + 1 | Q | ∫ Q | V 1 ( x ) | d x + 1 | Q | ∫ Q | A t Q ( U 1 ) ( x ) | d x + 1 | Q | ∫ Q | A t Q ( V 1 ) ( x ) | d x + 1 | Q | ∫ Q | U 2 ( x ) − A t Q ( U 2 ) ( x ) | d x + 1 | Q | ∫ Q | V 2 ( x ) − A t Q ( V 2 ) ( x ) | d x = I 1 + I 2 + I 3 + I 4 + I 5 + I 6 .$

Now, let us estimate $I 1$, $I 2$, $I 3$, $I 4$, $I 5$, and $I 6$, respectively. For $I 1$, by Hölder’s inequality and Lemma 1, we obtain

$1 | Q | ∫ Q | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( x ) | d x ≤ ( 1 | Q | ∫ R n | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( x ) | s d x ) 1 / s ≤ C | Q | − 1 / s ( ∫ R n | M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( x ) | s d x ) 1 / s ≤ C | Q | − 1 / s ( ∫ 2 Q ( | b ( x ) − b 2 Q | | I α T k , 2 ( f ) ( x ) | ) s d x ) 1 / s ≤ C | Q | − 1 / s ∥ b ∥ Lip β | 2 Q | β / n | 2 Q | 1 / s − β / n ( 1 | 2 Q | 1 − s β / n ∫ 2 Q | I α T k , 2 ( f ) ( x ) | s d x ) 1 / s ≤ C ∥ b ∥ Lip β M β , s ( I α T k , 2 ( f ) ) ( x ˜ ) ,$

thus

$I 1 ≤ ∑ k = 1 m 1 | Q | ∫ R n | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( x ) | d x ≤ C ∥ b ∥ Lip β ∑ k = 1 m M β , s ( I α T k , 2 ( f ) ) ( x ˜ ) .$

For $I 2$, by Lemma 4, we obtain, for $1/r=1/s−α/n$,

$I 2 ≤ ∑ k = 1 m ( 1 | Q | ∫ R n | I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( x ) | r d x ) 1 / r ≤ C ∑ k = 1 m | Q | − 1 / r ( ∫ 2 Q ( | b ( x ) − b 2 Q | | T k , 4 ( f ) ( x ) | ) s d x ) 1 / s ≤ C ∥ b ∥ Lip β ∑ k = 1 m | Q | − 1 / r | 2 Q | β / n | 2 Q | 1 / s − ( β + α ) / n ( 1 | 2 Q | 1 − s ( β + α ) / n ∫ 2 Q | T k , 4 ( f ) ( x ) | s d x ) 1 / s ≤ C ∥ b ∥ Lip β ∑ k = 1 m M β + α , s ( T k , 4 ( f ) ) ( x ˜ ) .$

For $I 3$, by the condition on $h t Q$ and notice for $x∈Q$, $y∈ 2 j + 1 Q∖ 2 j Q$, then $h t Q (x,y)≤C t Q − n / 2 ρ( 2 2 ( j − 1 ) )$, we obtain, similar to the proof of $I 1$,

$1 | Q | ∫ Q | A t Q ( T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ) ( x ) | d x ≤ C | Q | ∫ Q ∫ R n h t Q ( x , y ) | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y d x ≤ C | Q | ∫ Q ∫ 2 Q h t Q ( x , y ) | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y d x + C | Q | ∫ Q ∫ ( 2 Q ) c h t Q ( x , y ) | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y d x ≤ C | Q | ∫ Q ∫ 2 Q t Q − n / 2 | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y d x + C ∑ j = 1 ∞ t Q − n / 2 ρ ( 2 2 ( j − 1 ) ) ( 2 j l ( Q ) ) n 1 | 2 j + 1 Q | ∫ 2 j + 1 Q ∖ 2 j Q | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y ≤ C | Q | ∫ 2 Q | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y + C ∑ j = 1 ∞ 2 j n ρ ( 2 2 ( j − 1 ) ) ( 1 | 2 j + 1 Q | ∫ R n | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | s d y ) 1 / s ≤ C ∥ b ∥ Lip β M β , s ( I α T k , 2 ( f ) ) ( x ˜ ) + C ∑ j = 1 ∞ 2 j n ρ ( 2 2 ( j − 1 ) ) ( 1 | 2 j + 1 Q | ∫ R n | M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | s d y ) 1 / s ≤ C ∥ b ∥ Lip β M β , s ( I α T k , 2 ( f ) ) ( x ˜ ) + C ∑ j = 1 ∞ 2 ( j − 1 ) ( n + ϵ ) ρ ( 2 2 ( j − 1 ) ) 2 − j ( ϵ + n / s ) ( 1 | 2 Q | 1 / s − β / n ∫ 2 Q | I α T k , 2 ( f ) ( y ) | s d y ) 1 / s ≤ C ∥ b ∥ Lip β M β , s ( I α T k , 2 ( f ) ) ( x ˜ ) ,$

thus

$I 3 ≤ ∑ k = 1 m 1 | Q | ∫ R n | A t Q ( T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ) ( x ) | d x ≤ C ∥ b ∥ Lip β ∑ k = 1 m M β , s ( I α T k , 2 ( f ) ) ( x ˜ ) .$

Similarly, by Lemma 4, for $1/r=1/s−α/n$,

$I 4 ≤ ∑ k = 1 m 1 | Q | ∫ Q | A t Q ( T k , 3 I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ) ( x ) | d x ≤ ∑ k = 1 m C | Q | ∫ Q ∫ 2 Q h t Q ( x , y ) | T k , 3 I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( y ) | d y d x + ∑ k = 1 m C | Q | ∫ Q ∫ ( 2 Q ) c h t Q ( x , y ) | T k , 3 I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( y ) | d y d x ≤ C ∑ k = 1 m t Q − n / 2 | Q | 1 − 1 / r ( ∫ R n | I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( y ) | r d y ) 1 / r + C ∑ k = 1 m ∑ j = 1 ∞ t Q − n / 2 ρ ( 2 2 ( j − 1 ) ) | Q | 1 − 1 / r ( ∫ R n | I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( y ) | r d y ) 1 / r ≤ C ∑ k = 1 m | Q | − 1 / r ( ∫ 2 Q | ( b ( y ) − b 2 Q ) T k , 4 ( f ) ( y ) | s d y ) 1 / s + C ∑ k = 1 m ∑ j = 1 ∞ ρ ( 2 2 ( j − 1 ) ) | Q | − 1 / r ( ∫ 2 Q | ( b ( y ) − b 2 Q ) T k , 4 ( f ) ( y ) | s d y ) 1 / s ≤ C ∥ b ∥ Lip β ∑ k = 1 m | Q | − 1 / r | 2 Q | β / n | 2 Q | 1 / s − ( β + α ) / n ( 1 | 2 Q | 1 − s ( β + α ) / n ∫ 2 Q | T k , 4 ( f ) ( y ) | s d y ) 1 / s + C ∑ k = 1 m ∑ j = 1 ∞ 2 ( j − 1 ) ( n + ϵ ) ρ ( 2 2 ( j − 1 ) ) 2 − j ( n + ϵ ) ( 1 | 2 Q | 1 − s ( β + α ) / n ∫ 2 Q | T k , 4 ( f ) ( y ) | s d y ) 1 / s ≤ C ∥ b ∥ Lip β ∑ k = 1 m M β + α , s ( T k , 4 ( f ) ) ( x ˜ ) .$

For $I 5$, we get, for $x∈Q$,

$| T k , 1 M ( b − b 2 Q ) χ ( 2 Q ) c I α T k , 2 ( f ) ( x ) − A t Q ( T k , 1 M ( b − b 2 Q ) χ ( 2 Q ) c I α T k , 2 ( f ) ) ( x ) | ≤ ∫ ( 2 Q ) c | b ( y ) − b 2 Q | | K ( x − y ) − K t Q ( x − y ) | | I α T k , 2 ( f ) ( y ) | d y ≤ C ∑ j = 1 ∞ ∫ 2 j d ≤ | y − x 0 | < 2 j + 1 d ∥ b ∥ Lip β | 2 j + 1 Q | β / n l ( Q ) δ | x 0 − y | n + δ | I α T k , 2 ( f ) ( y ) | d y ≤ C ∥ b ∥ Lip β ∑ j = 1 ∞ 2 − j δ ( 1 | 2 j + 1 Q | 1 − s β / n ∫ 2 j + 1 Q | I α T k , 2 ( f ) ( y ) | s d y ) 1 / s ≤ C ∥ b ∥ Lip β M β , s ( I α T k , 2 ( f ) ) ( x ˜ ) ,$

thus

$I 5 ≤ 1 | Q | ∫ Q ∑ k = 1 m | T k , 1 M ( b − b 2 Q ) χ ( 2 Q ) c I α T k , 2 ( f ) ( x ) − A t Q ( T k , 1 M ( b − b 2 Q ) χ ( 2 Q ) c I α T k , 2 ( f ) ) ( x ) | d x ≤ C ∥ b ∥ Lip β ∑ k = 1 m M β , s ( I α T k , 2 ( f ) ) ( x ˜ ) .$

Similarly, by Lemma 3, we get

$I 6 ≤ 1 | Q | ∫ Q ∑ k = 1 m | T k , 3 I α M ( b − b 2 Q ) χ ( 2 Q ) c T k , 4 ( f ) ( x ) − A t Q ( T k , 3 I α M ( b − b 2 Q ) χ ( 2 Q ) c T k , 4 ( f ) ) ( x ) | d x ≤ C ∑ k = 1 m 1 | Q | ∫ Q ∫ ( 2 Q ) c | b ( y ) − b 2 Q | | K ˜ t Q ( x − y ) | | T k , 4 ( f ) ( y ) | d y d x ≤ C ∑ k = 1 m 1 | Q | ∑ j = 1 ∞ ∫ 2 j d ≤ | y − x 0 | < 2 j + 1 d ∥ b ∥ Lip β ( 2 j + 1 d ) β t Q | x − y | n + 2 − α | T k , 4 ( f ) ( y ) | d y ≤ C ∥ b ∥ Lip β ∑ k = 1 m ∑ j = 1 ∞ 2 − 2 j ( 1 | 2 j + 1 Q | 1 − s ( β + α ) / n ∫ 2 j + 1 Q | T k , 4 ( f ) ( y ) | s d y ) 1 / s ≤ C ∥ b ∥ Lip β ∑ k = 1 m M β + α , s ( T k , 4 ( f ) ) ( x ˜ ) .$

These complete the proof of Theorem 1. □

Proof of Theorem 2 It suffices to prove for $f∈ C 0 ∞ ( R n )$, the following inequality holds:

$1 | Q | 1 + β / n ∫ Q | T b ( f ) ( x ) − A t Q ( T b ( f ) ) ( x ) | d x ≤ C ∥ b ∥ Lip β ∑ k = 1 m ( M s ( I α T k , 2 ( f ) ) ( x ˜ ) + M α , s ( T k , 4 ( f ) ) ( x ˜ ) ) ,$

where $t Q = ( l ( Q ) ) 2$ and $l(Q)$ denotes the side length of Q. Without loss of generality, we may assume $T k , 1$ are T ($k=1,…,m$). Fix a cube $Q=Q( x 0 ,d)$ and $x ˜ ∈Q$. Similar to the proof of Theorem 1, we have

$1 | Q | 1 + β / n ∫ Q | T b ( f ) ( x ) − A t Q ( T b ( f ) ) ( x ) | d x ≤ 1 | Q | 1 + β / n ∫ Q | U 1 ( x ) | d x + 1 | Q | 1 + β / n ∫ Q | V 1 ( x ) | d x + 1 | Q | 1 + β / n ∫ Q | A t Q ( U 1 ) ( x ) | d x + 1 | Q | 1 + β / n ∫ Q | A t Q ( V 1 ) ( x ) | d x + 1 | Q | 1 + β / n ∫ Q | U 2 ( x ) − A t Q ( U 2 ) ( x ) | d x + 1 | Q | 1 + β / n ∫ Q | V 2 ( x ) − A t Q ( V 2 ) ( x ) | d x = J 1 + J 2 + J 3 + J 4 + J 5 + J 6 .$

By using the same argument as in the proof of Theorem 1, we get

$J 1 ≤ ∑ k = 1 m | Q | − β / n ( 1 | Q | ∫ R n | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( x ) | s d x ) 1 / s J 1 ≤ C ∑ k = 1 m | Q | − β / n − 1 / s ( ∫ R n | M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( x ) | s d x ) 1 / s J 1 ≤ C ∑ k = 1 m | Q | − β / n − 1 / s ∥ b ∥ Lip β | 2 Q | β / n | 2 Q | 1 / s ( 1 | 2 Q | ∫ 2 Q | I α T k , 2 ( f ) ( x ) | s d x ) 1 / s J 1 ≤ C ∥ b ∥ Lip β ∑ k = 1 m M s ( I α T k , 2 ( f ) ) ( x ˜ ) , J 2 ≤ ∑ k = 1 m | Q | − β / n ( 1 | Q | ∫ R n | I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( x ) | r d x ) 1 / r J 2 ≤ C ∑ k = 1 m | Q | − β / n − 1 / r ( ∫ 2 Q ( | b ( x ) − b 2 Q | | T k , 4 ( f ) ( x ) | ) s d x ) 1 / s J 2 ≤ C ∥ b ∥ Lip β ∑ k = 1 m | Q | − β / n − 1 / r | 2 Q | β / n | 2 Q | 1 / s − α / n ( 1 | 2 Q | 1 − s α / n ∫ 2 Q | T k , 4 ( f ) ( x ) | s d x ) 1 / s J 2 ≤ C ∥ b ∥ Lip β ∑ k = 1 m M α , s ( T k , 4 ( f ) ) ( x ˜ ) , J 3 ≤ C ∑ k = 1 m | Q | − 1 − β / n ∫ Q ∫ 2 Q t Q − n / 2 | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y d x J 3 ≤ + C ∑ k = 1 m ∑ j = 1 ∞ | Q | − β / n 2 j n ρ ( 2 2 ( j − 1 ) ) J 3 ≤ × ( 1 | 2 j + 1 Q | ∫ R n | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | s d y ) 1 / s J 3 ≤ ∑ k = 1 m C | Q | 1 + β / n ∫ 2 Q | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y J 3 ≤ + C ∑ k = 1 m ∑ j = 1 ∞ | Q | − β / n 2 j n ρ ( 2 2 ( j − 1 ) ) ( 1 | 2 j + 1 Q | ∫ R n | M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | s d y ) 1 / s J 3 ≤ C ∥ b ∥ Lip β ∑ k = 1 m M s ( I α T k , 2 ( f ) ) ( x ˜ ) J 3 ≤ + C ∑ k = 1 m ∑ j = 1 ∞ 2 ( j − 1 ) ( n + ϵ ) ρ ( 2 2 ( j − 1 ) ) 2 − j ( ϵ + n / s ) ( 1 | 2 Q | ∫ 2 Q | I α T k , 2 ( f ) ( y ) | s d y ) 1 / s J 3 ≤ C ∥ b ∥ Lip β ∑ k = 1 m M s ( I α T k , 2 ( f ) ) ( x ˜ ) , J 4 ≤ C ∑ k = 1 m | Q | − β / n − 1 / r ( ∫ R n | I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( y ) | r d y ) 1 / r J 4 ≤ + C ∑ k = 1 m ∑ j = 1 ∞ ρ ( 2 2 ( j − 1 ) ) | Q | − 1 / r ( ∫ R n | I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( y ) | r d y ) 1 / r J 4 ≤ C ∑ k = 1 m | Q | − β / n − 1 / r ( ∫ 2 Q | ( b ( y ) − b 2 Q ) T k , 4 ( f ) ( y ) | s d y ) 1 / s J 4 ≤ + C ∑ k = 1 m ∑ j = 1 ∞ ρ ( 2 2 ( j − 1 ) ) | Q | − 1 / r ( ∫ 2 Q | ( b ( y ) − b 2 Q ) T k , 4 ( f ) ( y ) | s d y ) 1 / s J 4 ≤ C ∥ b ∥ Lip β ∑ k = 1 m | Q | − β / n − 1 / r | 2 Q | β / n | 2 Q | 1 / s − α / n ( 1 | 2 Q | 1 − s α / n ∫ 2 Q | T k , 4 ( f ) ( y ) | s d y ) 1 / s J 4 ≤ + C ∑ k = 1 m ∑ j = 1 ∞ 2 ( j − 1 ) ( n + ϵ ) ρ ( 2 2 ( j − 1 ) ) 2 − j ( n + ϵ ) ( 1 | 2 Q | 1 − s α / n ∫ 2 Q | T k , 4 ( f ) ( y ) | s d y ) 1 / s J 4 ≤ C ∥ b ∥ Lip β ∑ k = 1 m M α , s ( T k , 4 ( f ) ) ( x ˜ ) , J 5 ≤ 1 | Q | 1 + β / n ∫ Q ∑ k = 1 m ∫ ( 2 Q ) c | b ( y ) − b 2 Q | | K ( x − y ) − K t Q ( x − y ) | | I α T k , 2 ( f ) ( y ) | d y J 5 ≤ C ∑ k = 1 m | Q | − β / n ∑ j = 1 ∞ ∫ 2 j d ≤ | y − x 0 | < 2 j + 1 d ∥ b ∥ Lip β | 2 j + 1 Q | β / n l ( Q ) δ | x 0 − y | n + δ | I α T k , 2 ( f ) ( y ) | d y J 5 ≤ C ∥ b ∥ Lip β ∑ k = 1 m ∑ j = 1 ∞ 2 j ( β − δ ) ( 1 | 2 j + 1 Q | ∫ 2 j + 1 Q | I α T k , 2 ( f ) ( y ) | s d y ) 1 / s J 5 ≤ C ∥ b ∥ Lip β ∑ k = 1 m M s ( I α T k , 2 ( f ) ) ( x ˜ ) , J 6 ≤ ∑ k = 1 m 1 | Q | 1 + β / n ∫ Q ∫ ( 2 Q ) c | b ( y ) − b 2 Q | | K ˜ t Q ( x − y ) | | T k , 4 ( f ) ( y ) | d y d x J 6 ≤ C ∑ k = 1 m 1 | Q | β / n ∑ j = 1 ∞ ∫ 2 j d ≤ | y − x 0 | < 2 j + 1 d ∥ b ∥ Lip β ( 2 j + 1 d ) β t Q | x − y | n + 2 − α | T k , 4 ( f ) ( y ) | d y J 6 ≤ C ∥ b ∥ Lip β ∑ k = 1 m ∑ j = 1 ∞ 2 j ( β − 2 ) ( 1 | 2 j + 1 Q | 1 − s α / n ∫ 2 j + 1 Q | T k , 4 ( f ) ( y ) | s d y ) 1 / s J 6 ≤ C ∥ b ∥ Lip β ∑ k = 1 m M α , s ( T k , 4 ( f ) ) ( x ˜ ) .$

These complete the proof of Theorem 2. □

Proof of Theorem 3 It suffices to prove for $f∈ C 0 ∞ ( R n )$, the following inequality holds:

$1 | Q | ∫ Q | T b ( f ) ( x ) − A t Q ( T b ( f ) ) ( x ) | d x ≤ C ∥ b ∥ BMO ∑ k = 1 m ( M s ( I α T k , 2 ( f ) ) ( x ˜ ) + M α , s ( T k , 4 ( f ) ) ( x ˜ ) ) ,$

where $t Q = ( l ( Q ) ) 2$ and $l(Q)$ denotes the side length of Q. Without loss of generality, we may assume $T k , 1$ are T ($k=1,…,m$). Fix a cube $Q=Q( x 0 ,d)$ and $x ˜ ∈Q$. Similar to the proof of Theorem 1, we have

$1 | Q | ∫ Q | T b ( f ) ( x ) − A t Q ( T b ( f ) ) ( x ) | d x ≤ 1 | Q | ∫ Q | U 1 ( x ) | d x + 1 | Q | ∫ Q | V 1 ( x ) | d x + 1 | Q | ∫ Q | A t Q ( U 1 ) ( x ) | d x + 1 | Q | ∫ Q | A t Q ( V 1 ) ( x ) | d x + 1 | Q | ∫ Q | U 2 ( x ) − A t Q ( U 2 ) ( x ) | d x + 1 | Q | ∫ Q | V 2 ( x ) − A t Q ( V 2 ) ( x ) | d x = L 1 + L 2 + L 3 + L 4 + L 5 + L 6 .$

Now, let us estimate $L 1$, $L 2$, $L 3$, $L 4$, $L 5$ and $L 6$, respectively. For $L 1$, we obtain, for $1,

$L 1 ≤ ∑ k = 1 m ( 1 | Q | ∫ R n | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( x ) | r d x ) 1 / r ≤ C ∑ k = 1 m | Q | − 1 / r ( ∫ 2 Q ( | b ( x ) − b 2 Q | | I α T k , 2 ( f ) ( x ) | ) r d x ) 1 / r ≤ C ∑ k = 1 m ( 1 | 2 Q | ∫ 2 Q | I α T k , 2 ( f ) ( x ) | s d x ) 1 / s ( 1 | 2 Q | ∫ 2 Q | b ( x ) − b 2 Q | s r / ( s − r ) d x ) ( s − r ) / s r ≤ C ∥ b ∥ BMO ∑ k = 1 m M s ( I α T k , 2 ( f ) ) ( x ˜ ) .$

For $L 2$, we obtain, for $1 and $1/v=1/u−α/n$,

$L 2 ≤ ∑ k = 1 m ( 1 | Q | ∫ R n | I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( x ) | v d x ) 1 / v ≤ C ∑ k = 1 m | Q | − 1 / v ( ∫ 2 Q ( | b ( x ) − b 2 Q | | T k , 4 ( f ) ( x ) | ) u d x ) 1 / u ≤ C ∑ k = 1 m ( 1 | 2 Q | 1 − s α / n ∫ 2 Q | T k , 4 ( f ) ( x ) | s d x ) 1 / s ( 1 | 2 Q | ∫ 2 Q | b ( x ) − b 2 Q | s u / ( s − u ) d x ) ( s − u ) / s u ≤ C ∥ b ∥ BMO ∑ k = 1 m M α , s ( T k , 4 ( f ) ) ( x ˜ ) .$

For $L 3$, by $h t Q (x,y)≤C t Q − n / 2 ρ( 2 2 ( j − 1 ) )$ for $x∈Q$, $y∈ 2 j + 1 Q∖ 2 j Q$, we obtain, for $1,

$L 3 ≤ ∑ k = 1 m C | Q | ∫ Q ∫ 2 Q h t Q ( x , y ) | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y d x + ∑ k = 1 m C | Q | ∫ Q ∫ ( 2 Q ) c h t Q ( x , y ) | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y d x ≤ ∑ k = 1 m C | Q | ∫ Q ∫ 2 Q t Q − n / 2 | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y d x + C ∑ k = 1 m ∑ j = 1 ∞ t Q − n / 2 ρ ( 2 2 ( j − 1 ) ) ( 2 j l ( Q ) ) n × ( 1 | 2 j + 1 Q | ∫ R n | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | r d y ) 1 / r ≤ ∑ k = 1 m C | Q | ∫ 2 Q | T k , 1 M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | d y + C ∑ k = 1 m ∑ j = 1 ∞ 2 j n ρ ( 2 2 ( j − 1 ) ) ( 1 | 2 j + 1 Q | ∫ R n | M ( b − b 2 Q ) χ 2 Q I α T k , 2 ( f ) ( y ) | r d y ) 1 / r ≤ C ∥ b ∥ BMO ∑ k = 1 m M s ( I α T k , 2 ( f ) ) ( x ˜ ) + C ∑ k = 1 m ∑ j = 1 ∞ 2 ( j − 1 ) ( n + ϵ ) ρ ( 2 2 ( j − 1 ) ) 2 − j ( ϵ + n / r ) ( 1 | 2 Q | ∫ 2 Q | I α T k , 2 ( f ) ( y ) | s d y ) 1 / s × ( 1 | 2 Q | ∫ 2 Q | b ( y ) − b 2 Q | s r / ( s − r ) d y ) ( s − r ) / s r ≤ C ∥ b ∥ BMO ∑ k = 1 m M s ( I α T k , 2 ( f ) ) ( x ˜ ) .$

Similarly, for $1/s+1/ s ′ =1$, $1 and $1/v=1/u−α/n$, we get

$L 4 ≤ C ∑ k = 1 m t Q − n / 2 | Q | 1 − 1 / v ( ∫ R n | I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( y ) | v d y ) 1 / v L 4 ≤ + C ∑ k = 1 m ∑ j = 1 ∞ t Q − n / 2 ρ ( 2 2 ( j − 1 ) ) | Q | 1 − 1 / v ( ∫ R n | I α M ( b − b 2 Q ) χ 2 Q T k , 4 ( f ) ( y ) | v d y ) 1 / v L 4 ≤ C ∑ k = 1 m | Q | − 1 / v ( ∫ 2 Q | ( b ( y ) − b 2 Q ) T k , 4 ( f ) ( y ) | u d y ) 1 / u L 4 ≤ + C ∑ k = 1 m ∑ j = 1 ∞ ρ ( 2 2 ( j − 1 ) ) | Q | − 1 / v ( ∫ 2 Q | ( b ( y ) − b 2 Q ) T k , 4 ( f ) ( y ) | u d y ) 1 / u L 4 ≤ C ∑ k = 1 m ( 1 | 2 Q | 1 − s α / n ∫ 2 Q | T k , 4 ( f ) ( y ) | s d y ) 1 / s L 4 ≤ × ( 1 | 2 Q | ∫ 2 Q | b ( y ) − b 2 Q | s u / ( s − u ) d y ) ( s − u ) / s u L 4 ≤ + C ∑ k = 1 m ∑ j = 1 ∞ 2 ( j − 1 ) ( n + ϵ ) ρ ( 2 2 ( j − 1 ) ) 2 − j ( n + ϵ ) ( 1 | 2 Q | 1 − s α / n ∫ 2 Q | T k , 4 ( f ) ( y ) | s d y ) 1 / s L 4 ≤ × ( 1 | 2 Q | ∫ 2 Q | b ( y ) − b 2 Q | s u / ( s − u ) d y ) ( s − u ) / s u L 4 ≤ C ∥ b ∥ BMO ∑ k = 1 m M α , s ( T k , 4 ( f ) ) ( x ˜ ) , L 5 ≤ 1 | Q | ∫ Q ∑ k = 1 m ∫ ( 2 Q ) c | b ( y ) − b 2 Q | | K ( x − y ) − K t Q ( x − y ) | | I α T k , 2 ( f ) ( y ) | d y d x L 5 ≤ C ∑ k = 1 m ∑ j = 1 ∞ ∫ 2 j d ≤ | y − x 0 | < 2 j + 1 d | b ( y ) − b 2 Q | l ( Q ) δ | x 0 − y | n + δ | I α T k , 2 ( f ) ( y ) | d y L 5 ≤ C ∑ k = 1 m ∑ j = 1 ∞ 2 − j δ ( 1 | 2 j + 1 Q | ∫ 2 j + 1 Q | b ( y ) − b 2 Q | s ′ d y ) 1 / s ′ L 5 ≤ × ( 1 | 2 j + 1 Q | ∫ 2 j + 1 Q | I α T k , 2 ( f ) ( y ) | s d y ) 1 / s L 5 ≤ C ∑ k = 1 m ∑ j = 1 ∞ j 2 − j δ ∥ b ∥ BMO M s ( I α T k , 2 ( f ) ) ( x ˜ ) L 5 ≤ C ∥ b ∥ BMO ∑ k = 1 m M s ( I α T k , 2 ( f ) ) ( x ˜ ) , L 6 ≤ ∑ k = 1 m 1 | Q | ∫ Q ∫ ( 2 Q ) c | b ( y ) − b 2 Q | | K ˜ t Q ( x − y ) | | T k , 4 ( f ) ( y ) | d y d x L 5 ≤ C ∑ k = 1 m ∑ j = 1 ∞ ∫ 2 j d ≤ | y − x 0 | < 2 j + 1 d | b ( y ) − b 2 Q | t Q | x − y | n + 2 − α | T k , 4 ( f ) ( y ) | d y L 5 ≤ C ∑ k = 1 m ∑ j = 1 ∞ j 2 − 2 j ( 1 | 2 j + 1 Q | 1 − s α / n ∫ 2 j + 1 Q | T k , 4 ( f ) ( y ) | s d y ) 1 / s L 5 ≤ × ( 1 | 2 j + 1 Q | ∫ 2 j + 1 Q | b ( y ) − b 2 Q | s ′ d y ) 1 / s ′ L 5 ≤ C ∥ b ∥ BMO ∑ k = 1 m M α , s ( T k , 4 ( f ) ) ( x ˜ ) .$

These complete the proof of Theorem 3. □

Proof of Theorem 4 Choose $1 in Theorem 1 and set $1/r=1/p−α/n$. We have, by Lemma 5,

$∥ T b ( f ) ∥ L q , φ ≤ ∥ M ( T b ( f ) ) ∥ L q , φ ≤ C ∥ M A # ( T b ( f ) ) ∥ L q , φ ≤ C ∥ b ∥ Lip β ∑ k = 1 m ( ∥ M β , s ( I α T k , 2 ( f ) ) ∥ L q , φ + ∥ M β + α , s ( T k , 4 ( f ) ) ∥ L q , φ ) ≤ C ∥ b ∥ Lip β ∑ k = 1 m ( ∥ I α T k , 2 ( f ) ∥ L r , β , φ + ∥ T k , 4 ( f ) ∥ L p , α + β , φ ) ≤ C ∥ b ∥ Lip β ∑ k = 1 m ( ∥ T k , 2 ( f ) ∥ L p , α + β , φ + ∥ f ∥ L p , α + β , φ ) ≤ C ∥ b ∥ Lip β ∥ f ∥ L p , α + β , φ .$

This completes the proof of the theorem. □

Proof of Theorem 5 Choose $1 in Theorem 2. We have, by Lemma 4,

$∥ T b ( f ) ∥ F ˙ q , A β , ∞ ≤ C ∥ sup Q ∋ ⋅ 1 | Q | 1 + β / n ∫ Q | T b ( f ) ( x ) − A t Q ( T b ( f ) ) ( x ) | d x ∥ L q ≤ C ∥ b ∥ Lip β ∑ k = 1 m ( ∥ M s ( I α T k , 2 ( f )$