Let T = { T ε } ε be a family of operators such that the limit lim ε → 0 T ε f exists. It is nature to study the speed of convergence of the family { T ε } . A classic method

is to consider square function of the type ( ∑ i = 1 ∞ | T ε i f − T ε i + 1 f | 2 ) 1 / 2 , more generally, define the following oscillation operator:

O ( T f ) ( x ) = ( ∑ i = 1 ∞ sup t i + 1 ≤ ε i + 1 < ε i ≤ t i | T ε i + 1 f ( x ) − T ε i f ( x ) | 2 ) 1 / 2 ,

where { t i } is a fixed sequence decreasing to zero. Let ρ > 2 . The ρ -variation operator is defined by

V ρ ( T f ) ( x ) = sup ε i ↘ 0 ( ∑ i = 1 ∞ | T ε i + 1 f ( x ) − T ε i f ( x ) | ρ ) 1 / ρ ,

where the sup is taken over all sequence { ε i } of positive numbers decreasing to zero. The variation inequalities play important roles in probability, ergodic theory, and harmonic analysis. We refer the readers to [1] - [6] and the references therein for more background information. Gillespie and Torrea [7] show that oscillation and variation of Hilbert transforms are bounded on L p ( ω ) for 1 < p < ∞ . The weighted oscillation and variation boundedness of differential operators and Calderón-Zygmund singular integral are established in [8] [9]. This paper is devoted to studying weighted boundedness of ρ -variational operator for the families of commutator generated by one-sided Calderón-Zygmund singular integral with Lipschitz functions. Recently, Liu and Wu [10] presented a criterion on the weighted estimate of the oscillation and variation operators for the commutators of Calderón-Zygmund singular integrals with BMO functions in one dimension. Variation inequalities for the commutators one-sided singular integrals with BMO functions were established in [11]. Zhang and Wu [12] gave the oscillation and variation inequalities for the commutators of singular integrals with Lipschitz functions.

Before stating our main results, we firstly recall some notations and definitions. In [13], Aimar, Forzani and Martín-Reyes introduced the one-sided Calderón-Zygmund singular integrals defined by:

T + f ( x ) = l i m ε → 0 + T ε + f ( x ) = l i m ε → 0 + ∫ x + ε ∞     K ( x − y ) f ( y ) d y (1.1)

and

T − f ( x ) = l i m ε → 0 + T ε − f ( x ) = l i m ε → 0 + ∫ − ∞ x − ε     K ( x − y ) f ( y ) d y ,

where the kernel K is called the one-sided Calderón-Zygmund kernel (OCZK) which satisfies

| ∫ a < | x | < b     K ( x ) d x | ≤ C ,   0 < a < b , (1.2)

| K ( x ) | ≤ C / | x | ,   x ≠ 0, (1.3)

| K ( x − y ) − K ( x ) | ≤ C | y | / | x | 2 ,   | x | > 2 | y | > 0 , (1.4)

with support in ℝ − = ( − ∞ ,0 ) or ℝ + = ( 0, + ∞ ) , where (1.4) is named Hörmander’s condition. Equation (1.3) is also called the size condition for K. An interesting example is

K ( x ) = sin ( log | x | ) x log | x | χ ( − ∞ ,0 ) ( x ) ,

where χ E denotes the characteristic function of a set E, for more details one can refer to [13].

For 0 < α < 1 , a function f ∈ L i p α , if it satisfies

‖ f ‖ L i p α = sup x , h ∈ ℝ , h ≠ 0 f ( x + h ) − f ( x ) | h | α < ∞ .

Let K be one-sided Calderón-Zygmund kernel (OCZK) with support in ℝ − = ( − ∞ ,0 ) . b ∈ L i p α , we define the following one-sided operator

T b + f ( x ) = l i m ε → 0 + T b , ε + f ( x ) ,

where

T b , ε + f ( x ) = ∫ x + ε ∞ ( b ( x ) − b ( y ) ) K ( x − y ) f ( y ) d y . (1.5)

In this paper, we study variational inequalities for the commutators of one-sided singular integrals with Lipschitz functions. Our result can be formulated as follows:

Theorem 1.1. Let K be one-sided Calderón-Zygmund kernel (OCZK) with support in ℝ − = ( − ∞ ,0 ) . Let T b = { T b , ε + } ε > 0 and T = { T ε + } ε > 0 be given as in (1.5) and (1.1), respectively. The operator V ρ ( T + ) is bounded in L p 0 ( ℝ , d x ) for some p 0 ∈ ( 1, ∞ ) . Let 0 < α < 1 , b ∈ L i p α . Then, for all ω ∈ A + ( p , q ) , 1 < p < q < ∞ , 1 / p − 1 / q = α , we have

‖ V ρ ( T b + ) f ‖ L q ( w q ) ≲ ‖ b ‖ L i p a ‖ f ‖ L p ( w p ) .

The rest of this paper is organized as follows. In Section 2, we introduce and recall some basic facts and auxiliary lemmas. The proof of main theorem will be given in Sections 3.

Throughout this paper, the letter C, sometimes with additional parameters, will stand for positive constants, not necessarily the same one at each occurrence, but independent of the essential variables. We also denote f ≲ g if f ≤ C g .

In 1986, Sawyer [14] first introduced the one-sided Muckenhoupt weights A p + and A p − to treat the one-sided Hardy-Littlewood maximal operators

M + f ( x ) : = sup h > 0 1 h ∫ x x + h | f ( y ) | d y ,   M − f ( x ) : = sup h > 0 1 h ∫ x − h x | f ( y ) | d y .

A positive function ω is said to belong to A p + or A p − if it satisfies

A p + ( ω ) : = sup a < b < c 1 ( c − a ) p ∫ a b     ω ( x ) d x ( ∫ b c ω ( x ) 1 − p ′ d x ) p − 1 < ∞

or

A p − ( ω ) : = sup a < b < c 1 ( c − a ) p ∫ b c     ω ( x ) d x ( ∫ a b     ω ( x ) 1 − p ′ d x ) p − 1 < ∞ ,

when 1 < p < ∞ ; also, for p = 1 ,

A 1 + : M − ω ≤ C ω ,     A 1 − : M + ω ≤ C ω ,

for some constant C. If 1 ≤ p < ∞ , then A p ⊊ A p + and A p ⊊ A p − . Notice that the function ω ( x ) = e x mentioned above is in A p + but not in A p .

Similarly, the double weight classes A + ( p , q ) and A − ( p , q ) are denoted by

A + ( p , q ) : 1 ( c − a ) 1 − α ( ∫ a b     ω q ) 1 / q ( ∫ b c     ω − p ′ ) 1 / p ′ < C ,

A − ( p , q ) : 1 ( c − a ) 1 − α ( ∫ b c     ω q ) 1 / q ( ∫ a b     ω − p ′ ) 1 / p ′ < C

for all a < b < c ∈ ℝ , 0 < α < 1 , 1 < p < q and 1 / p − 1 / q = α ; also for p = 1 , 1 − 1 / q = α ,

A ( 1, q ) + : M − ω q ≲ ω q ,     A ( 1, q ) − : M + ω q ≲ ω q .

The one-sided A p + classes not only control the boundedness of one-sided Hardy-Littlewood maximal operators, but also serve as the right weight classes for one-sided singular integral operators. Set 1 < p < ∞ and let K be an OCZK with support in ℝ − . Then T + is bounded on L p ( ω ) if ω ∈ A p + , see [13].

Lemma 2.1. [15] Suppose that ω ∈ A 1 − , then there exists ε 1 > 0 such that, for all 1 < r ≤ 1 + ε 1 , ω r ∈ A 1 − .

Lemma 2.2. [16] Suppose ω ∈ A + ( p , q ) , then ω q ∈ A q + and ω p ∈ A p + for all 1 < p < q ≤ ∞ .

Lemma 2.3. [17] Let 1 < p 0 < ∞ and let T be sublinear operator defined in C c ∞ ( ℝ ) satisfying

‖ ω T f ‖ ∞ ≤ C ‖ f ω ‖ p 0

for every x ∈ ℝ and ω ∈ A + ( p 0 , ∞ ) ; then, for every 1 < p < p 0 , 1 / p − 1 / q = 1 / p 0 , and ω ∈ A + ( p , q ) , the inequality

‖ ω T f ‖ q ≤ C ‖ f ω ‖ p

Lemma 2.4. [18] For every p with p 0 ≤ p < ∞ , assume that ω ∈ A ∞ + , and that M + f ∈ L p 0 ( ω ) for some p 0 with 0 < p 0 < ∞ . Then

‖ M + f ‖ L p ( ω ) ≤ C ‖ M + , # f ‖ L p ( ω ) ,

where

M + , # f ( x ) = sup h > 0 1 h ∫ x x + h ( f ( y ) − 1 h ∫ x + h x + 2 h     f ( z ) d z ) + d y

and z + = max { z , 0 } .

According to [7], we denote by F ρ the mixed norm Banach space of two variable functions h defined on ℕ × Θ such that

‖ h ‖ F ρ = sup β ( ∑ i | h ( i , β ) | ρ ) 1 / ρ < ∞ ,

where Θ = { β : β = { ε i } , ε i ∈ ℝ , ε i ↘ 0 } . Given a family of operators T = { T t + } t > 0 defined on L p ( ℝ ) , we consider the F ρ -valued operator V ( T ) : f → V ( T ) f on L p ( ℝ ) defined by

V ( T ) f ( x ) = { T ε i + 1 f ( x ) − T ε i f ( x ) } β = { ε i } ∈ Θ = : { T [ ε i + 1 , ε i ] f ( x ) } β = { ε i } ∈ Θ ,

where { T [ ε i + 1 , ε i ] f ( x ) } β = { ε i } ∈ Θ is an abbreviation for the element of F ρ given by

( i , β ) = ( i , { ε i } ) → T [ ε i + 1 , ε i ] f ( x ) .

This implies

V ρ ( T f ) ( x ) = ‖ V ( T ) f ( x ) ‖ F ρ .

This section is devoted to proving Theorem 1.1. To do this, we need establish the following lemma.

Lemma 3.1. Let K be one-sided Calderón-Zygmund kernel (OCZK) with support in ℝ − = ( − ∞ ,0 ) . Let T b = { T b , ε + } ε > 0 and T = { T ε + } ε > 0 be given as in (1.5) and (1.1), respectively. The operator V ρ ( T + ) is bounded in L p 0 ( ℝ , d x ) for some p 0 ∈ ( 1, ∞ ) . Let 0 < α < 1 , b ∈ L i p α . Then, for all ω ∈ A + ( p , q ) , 1 < p < q < ∞ , 1 / p − 1 / q = α , we have

‖ M + , # ( V ρ ( T b + ) f ) ( x ) ‖ L q ( w q ) ≲ ‖ b ‖ L i p a ‖ f ‖ L p ( w p ) . (3.1)

Proof. Let x ∈ ℝ , h > 0 and I = [ x , x + 4 h ] . Suppose f = f 1 + f 2 : = f χ I + f χ I c . Notice that

M + , # ( V ρ ( T b + ) f ) ( x ) ≲ 1 h ∫ x x + 2 h | V ρ ( T b + ) f ( y ) − V ρ ( T ) ( ( b − b I ) f 2 ) ( x ) | d y .

It is easy to check

1 h ∫ x x + 2 h | V ρ ( T b + ) f ( y ) − V ρ ( T ) ( ( b − b I ) f 2 ) ( x ) | d y ≤ 1 h ∫ x x + 2 h | ‖ V ( T b + ) f ( y ) ∥ F ρ − ∥ V ( T b + ) ( ( b − b I ) f 2 ) ( x ) ‖ F ρ | d y ≤ 1 h ∫ x x + 2 h ‖ V ( T b + ) f ( y ) − V ( T b + ) ( ( b − b I ) f 2 ) ( x ) ‖ F ρ d y ≤ 1 h ∫ x x + 2 h | b ( y ) − b I | V ρ ( T + ) f ( y ) d y + 1 h ∫ x x + 2 h     V ρ ( T + ) ( ( b − b I ) f 1 ) ( y ) d y         + 1 h ∫ x x + 2 h ‖ V ( T + ) ( ( b − b I ) f 2 ) ( y ) − V ( T + ) ( ( b − b I ) f 2 ) ( x ) ‖ F ρ d y .

Consider the following three sublinear operators defined on C c ∞ ( ℝ ) :

M 1 + f ( x ) : = sup h > 0 1 h ∫ x x + 2 h | b ( y ) − b I | V ρ ( T + ) f ( y ) d y ,

M 2 + f ( x ) : = sup h > 0 1 h ∫ x x + 2 h     V ρ ( T + ) ( ( b − b I ) f 1 ) ( y ) d y ,

M 3 + f ( x ) : = sup h > 0 1 h ∫ x x + 2 h ‖ V ( T + ) ( ( b − b I ) f 2 ) ( y ) − V ( T + ) ( ( b − b I ) f 2 ) ( x ) ‖ F ρ d y .

For M 1 + f , let ω ∈ A + ( 1 / α , ∞ ) , then ω − 1 / ( 1 − α ) ∈ A 1 − . By Lemma 2.1, there exists t > 1 such that ω − t / ( 1 − α ) ∈ A 1 − . Take r = t / ( 1 − α ) and 1 / s − 1 / r = α . Using Hölder’s inequality and the L r boundedness for V ρ ( T + ) , we obtain

1 h ∫ x x + 2 h | b ( y ) − b I | V ρ ( T + ) f ( y ) d y ≲ sup y ∈ ( x , x + 2 h ) | b ( y ) − b I | ( 1 h ∫ x x + 2 h | f ( y ) | r d y ) 1 / r ≲ ‖ b ‖ L i p α h α ( 1 h ∫ x x + 2 h | f ( y ) | r ω ( y ) r ω ( y ) − r d y ) 1 / r ≲ ‖ b ‖ L i p α ( ∫ x x + 2 h | f ( y ) ω ( y ) | 1 / α d y ) α ( 1 h ∫ x x + 2 h     ω ( y ) − s d y ) 1 / s ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α ω ( x ) − 1 .

By Lemma 2.3, we have that for 1 / r − 1 / s = α and A + ( p , q )

‖ M 1 + f ‖ L q ( ω q ) ≲ ‖ b ‖ L i p α ‖ f ‖ L p ( ω p ) .

For M 2 + f , by Hölder’s inequality and the L r boundedness for V ρ ( T + ) , we get

1 h ∫ x x + 2 h     V ρ ( T + ) ( ( b − b I ) f 1 ) ( y ) d y ≤ ( 1 h ∫ x x + 2 h | V ρ ( T + ) ( ( b − b I ) f 1 ) ( y ) | r d y ) 1 / r ≲ ( 1 h ∫ x x + 2 h | ( b ( y ) − b I ) f 1 ( y ) | r d y ) 1 / r ≤ sup y ∈ ( x , x + 2 h ) | b ( y ) − b I | ( 1 h ∫ x x + 2 h | f ( y ) | r d y ) 1 / r ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α ω ( x ) − 1 ,

where ω − r = ω − t / ( 1 − α ) ∈ A 1 − for all ω ∈ A + ( 1 / α , ∞ ) . Then

‖ ω M 2 + f ‖ ∞ ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α .

For 1 / r − 1 / s = α and A + ( p , q ) , it follows from Lemma 2.3 that

‖ M 2 + f ‖ L q ( ω q ) ≲ ‖ b ‖ L i p α ‖ f ‖ L p ( ω p ) .

It remains to deal with M 3 + f . For y ∈ [ x , x + 2 h ] , we get

T [ ε i + 1 , ε i ] + ( ( b − b I ) f 2 ) ( y ) − T [ ε i + 1 , ε i ] + ( ( b − b I ) f 2 ) ( x ) = ∫ ℝ     k ( y − z ) χ ( y + ε i + 1 , y + ε i ) ( z ) ( b ( z ) − b I ) f 2 ( z ) d z         − ∫ ℝ     k ( x − z ) χ ( x + ε i + 1 , x + ε i ) ( z ) ( b ( z ) − b I ) f 2 ( z ) d z = ∫ ℝ ( k ( y − z ) − k ( x − z ) ) χ ( y + ε i + 1 , y + ε i ) ( z ) ( b ( z ) − b I ) f 2 ( z ) d z         + ∫ ℝ     k ( x − z ) ( χ ( y + ε i + 1 , y + ε i ) ( z ) − χ ( x + ε i + 1 , x + ε i ) ( z ) ) ( b ( z ) − b I ) f 2 ( z ) d z (3.2)

In view of (3.2), we have

‖ V ( T + ) ( ( b − b I ) f 2 ) ( y ) − V ( T + ) ( ( b − b I ) f 2 ) ( x ) ‖ F ρ ≤ ‖ { ∫ ℝ ( k ( y − z ) − k ( x − z ) ) χ ( y + ε i + 1 , y + ε i ) ( z ) ( b ( z ) − b I ) f 2 ( z ) d z } i ∈ ℕ , β = { ε i } ∈ Θ ‖ F ρ     + ‖ { ∫ ℝ     k ( x − z ) ( χ ( y + ε i + 1 , y + ε i ) ( z ) − χ ( x + ε i + 1 , x + ε i ) ( z ) )     × ( b ( z ) − b I ) f 2 ( z ) d z } i ∈ ℕ , β = { ε i } ∈ Θ ‖ F ρ = : J 1 + J 2 .

Since z ∈ ( x + 8 h , ∞ ) and y ∈ ( x , x + 2 h ) , we get | x − z | ≥ 2 | x − y | . Using (1.4), we have | k ( y − z ) − k ( x − z ) | ≲ h | x − z | − 2 . Note that

‖ { χ ( y + ε i + 1 , y + ε i ] ( z ) } i ∈ ℕ , γ = { ε i } ∈ Θ ‖ F ρ ≤ 1,     for   ∀   y ∈ ℝ .

Then

J 1 ≲ ∫ ℝ h | x − z | 2 ‖ { χ ( y + ε i + 1 , y + ε i ] ( z ) } i ∈ ℕ , γ = { ε i } ∈ Θ ‖ F ρ | ( b ( z ) − b I ) f 2 ( z ) | d z ≲ h ∫ ℝ | ( b ( z ) − b I ) f 2 ( z ) | | x − z | 2 d z ≲ h ∑ k = 3 ∞ ∫ x + 2 k h x + 2 k + 1 h | ( b ( z ) − b I ) f ( z ) | | x − z | 2 d z ≲ ‖ b ‖ L i p α ∑ k = 3 ∞ h α 2 k ( 1 − α ) 1 2 k + 1 h ∫ x x + 2 k + 1 h | f ( z ) | d z

≲ ‖ b ‖ L i p α ∑ k = 3 ∞ h α 2 k ( 1 − α ) ( 1 2 k + 1 h ∫ x x + 2 k + 1 h | f ( y ) | r ω ( y ) r ω ( y ) − r d y ) 1 / r ≲ ‖ b ‖ L i p α ∑ k = 3 ∞ 1 2 k ( ∫ x x + 2 k + 1 h | f ( y ) ω ( y ) | 1 / α d y ) α ( 1 2 k + 1 h ∫ x x + 2 k + 1 h     ω ( y ) − s d y ) 1 / s ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α ω ( x ) − 1 ,

where ω − r = ω − t / ( 1 − α ) ∈ A 1 − for all ω ∈ A + ( 1 / α , ∞ ) .

For { ε i } ∈ Θ , let N 1 = { i ∈ Z : ε i − ε i + 1 ≥ y − x } and N 2 = { i ∈ Z : ε i − ε i + 1 < y − x } . Then

J 2 ≤ ‖ { ∫ ℝ ( χ ( y + ε i + 1 , y + ε i ) ( z ) − χ ( x + ε i + 1 , x + ε i ) ( z ) )             × k ( x − z ) ( b ( z ) − b I ) f 2 ( z ) d z } i ∈ ℕ 1 , β = { ε i } ∈ Θ ‖ F ρ       ≤ ‖ { ∫ ℝ ( χ ( y + ε i + 1 , y + ε i ) ( z ) − χ ( x + ε i + 1 , x + ε i ) ( z ) )             × k ( x − z ) ( b ( z ) − b I ) f 2 ( z ) d z } i ∈ ℕ 2 , β = { ε i } ∈ Θ ‖ F ρ       = : J 21 + J 22 .

Now we estimate J 21 . It is easy to see

J 21 ≤ ‖ { ∫ ℝ     k ( x − z ) χ ( x + ε i + 1 , y + ε i + 1 ) ( z ) ( b ( z ) − b I ) f 2 ( z ) d z } i ∈ ℕ 1 , β = { ε i } ∈ Θ ‖ F ρ               + ‖ { ∫ ℝ     k ( x − z ) χ ( x + ε i , y + ε i ) ( z ) ( b ( z ) − b I ) f 2 ( z ) d z } i ∈ ℕ 1 , β = { ε i } ∈ Θ ‖ F ρ           = : L 1 + L 2 .

For i ∈ N 1 , using Hölder’s inequality and (1.3), we have

L 1 ≲ h 1 / r ′ ‖ { ( ∫ ℝ | b ( z ) − b I | r | f 2 ( z ) | r | x − z | r χ ( y + ε i + 1 , y + ε i ) ( z ) d z ) 1 / r } i ∈ ℕ 1 , β = { ε i } ∈ Θ ‖ F ρ = h 1 / r ′ ( sup β ∑ i ∈ ℕ 1 ( ∫ ℝ     χ ( x + ε i + 1 , x + ε i ) ( z ) | b ( z ) − b I | r | f 2 ( z ) | r | x − z | r d z ) p / r ) 1 / ρ

≲ h 1 / r ′ ( ∫ ℝ | b ( z ) − b I | r | f 2 ( z ) | r | x − z | r d z ) 1 / r ≲ ‖ b ‖ L i p α ∑ k = 3 ∞ h α 2 k ( 1 / r ′ − α ) ( 1 2 k + 1 h ∫ x x + 2 k + 1 h | f ( z ) | r d z ) 1 / r ≲ ∑ k = 3 ∞ ‖ b ‖ L i p α 2 k / r ′ ( ∫ x x + 2 k + 1 h | f ( z ) ω ( z ) | 1 / α d z ) α ( 1 2 k + 1 h ∫ x x + 2 k + 1 h     ω ( z ) − s d z ) 1 / s ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α ω ( x ) − 1 ,

where ω − r = ω − t / ( 1 − α ) ∈ A 1 − for all ω ∈ A + ( 1 / α , ∞ ) . By a similar estimate of L 1 , we have

L 2 ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α ω ( x ) − 1 .

Now we estimate J 22 . It is easy to see

J 22 ≤ ‖ { ∫ ℝ     k ( x − z ) χ ( y + ε i + 1 , y + ε i ) ( z ) ( b ( z ) − b I ) f 2 ( z ) d z } i ∈ ℕ 2 , β = { ε i } ∈ Θ ‖ F ρ           + ‖ { ∫ ℝ     k ( x − z ) χ ( x + ε i + 1 , x + ε i ) ( z ) ( b ( z ) − b I ) f 2 ( z ) d z } i ∈ ℕ 2 , β = { ε i } ∈ Θ ‖ F ρ           = : L 3 + L 4 .

By noting that N 2 = { i ∈ Z : ε i − ε i + 1 < y − x } , we have ε i − ε i + 1 < y − x ≤ 2 h with y ∈ [ x , x + 2 h ] . Using Hölder’s inequality and (1.3), we have

L 3 ≲ h 1 / r ′ ‖ { ( ∫ ℝ     χ ( y + ε i + 1 , y + ε i ) ( z ) | b ( z ) − b I | r | f 2 ( z ) | r | x − z | r d z ) 1 / r } i ∈ ℕ 2 , β = { ε i } ∈ Θ ‖ F ρ = h 1 / r ′ ( sup β ∑ i ∈ ℕ 1 ( ∫ ℝ     χ ( y + ε i + 1 , y + ε i ) ( z ) | b ( z ) − b I | r | f 2 ( z ) | r | x − z | r d z ) p / r ) 1 / ρ ≲ h 1 / r ′ ( ∫ ℝ | b ( z ) − b I | r | f 2 ( z ) | r | x − z | r d z ) 1 / r ≲ ‖ b ‖ L i p α ∑ k = 3 ∞ h α 2 k ( 1 / r ′ − α ) ( 1 2 k + 1 h ∫ x x + 2 k + 1 h | f ( z ) | r d z ) 1 / r ≲ ∑ k = 3 ∞ ‖ b ‖ L i p α 2 k / r ′ ( ∫ x x + 2 k + 1 h | f ( z ) ω ( z ) | 1 / α d z ) α ( ‖ b ‖ L i p α 2 k + 1 h ∫ x x + 2 k + 1 h     ω ( z ) − s d z ) 1 / s ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α ω ( x ) − 1 ,

where ω − r = ω − t / ( 1 − α ) ∈ A 1 − for all ω ∈ A + ( 1 / α , ∞ ) . By similar arguments, we have

L 4 ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α ω ( x ) − 1 ,

Following from the estimates of L 1 , L 3 , L 3 and L 4 , we get

J 2 ≤ J 21 + J 22 ≤ L 1 + L 2 + L 3 + L 4 ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α ω ( x ) − 1 .

This together with the estimate of J 4 implies

‖ V ( T + ) ( ( b − b I ) f 2 ) ( y ) − V ( T + ) ( ( b − b I ) f 2 ) ( x ) ‖ F ρ ≤ J 1 + J 2 ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α ω ( x ) − 1 ,

where ω − r = ω − t / ( 1 − α ) ∈ A 1 − for all ω ∈ A + ( 1 / α , ∞ ) . Then

‖ ω M 3 + f ‖ ∞ ≲ ‖ b ‖ L i p α ‖ f ω ‖ L 1 / α .

For 1 / r − 1 / s = α and A + ( p , q ) , it follows from Lemma 2.3 that

‖ M 3 + f ‖ L q ( ω q ) ≲ ‖ b ‖ L i p α ‖ f ‖ L p ( ω p ) .

This completes the proof.

Now, we turn to the proof of theorem 1.1.

Proof. For ω ∈ A + ( p , q ) , by Lemma 2.2, we have ω q ∈ A q + . Using Lemma 2.4 and Lemma 3.1, we obtain

‖ V ρ ( T b + ) f ‖ L q ( ω q ) ≲ ‖ M + ( V ρ ( T b + ) f ) ‖ L q ( ω q ) ≲ ‖ M + , # ( V ρ ( T b + ) f ) ‖ L q ( ω q ) ≲ ‖ b ‖ L i p a ‖ f ‖ L p ( ω p ) .

It remains to prove ‖ V ρ ( T b + ) f ‖ L q ( ω q ) < ∞ . By the similar arguments in the proof of Theorem 1.3 in [10], we can get ‖ V ρ ( T b + ) f ‖ L q ( ω q ) < ∞ . Then Theorem 1.1 is proved.

Supported by the Natural Science Foundation of Shandong Province (No. ZR2020QA006).

The authors declare no conflicts of interest regarding the publication of this paper.

Juan, T.T. and Hou, X.M. (2021) Variation Inequalities for the Commutators of One-Sided Calderón-Zygmund Singular Integrals with Lipschitz Functions. Advances in Pure Mathematics, 11, 978-987. https://doi.org/10.4236/apm.2021.1112063