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Oscillation and variation inequalities for the multilinear singular integrals related to Lipschitz functions

Journal of Inequalities and Applications20172017:292

https://doi.org/10.1186/s13660-017-1568-8

  • Received: 12 October 2017
  • Accepted: 9 November 2017
  • Published:

Abstract

The main purpose of this paper is to establish the weighted \((L^{p},L^{q})\) inequalities of the oscillation and variation operators for the multilinear Calderón-Zygmund singular integral with a Lipschitz function.

Keywords

  • oscillation operator
  • variation operator
  • multilinear operator
  • Lipschitz function

MSC

  • 42B25
  • 42B20
  • 47G10

1 Introduction and results

Let K be a kernel on \(\mathbb{R}\times\mathbb{R}\setminus\{ (x,x): x\in\mathbb{R}\}\). Suppose that there exist two constants δ and C such that
$$\begin{aligned} &\bigl\vert K(x,y) \bigr\vert \leq\frac{C}{ \vert x-y \vert }\quad \mbox{for } x\neq y; \end{aligned}$$
(1.1)
$$\begin{aligned} &\bigl\vert K(x,y)-K\bigl(x',y\bigr) \bigr\vert \leq\frac{C \vert x-x' \vert ^{\delta}}{ \vert x-y \vert ^{1+\delta}}\quad \mbox{for } \vert x-y \vert \geq2 \bigl\vert x-x' \bigr\vert ; \end{aligned}$$
(1.2)
$$\begin{aligned} &\bigl\vert K(x,y)-K\bigl(x,y'\bigr) \bigr\vert \leq\frac{C \vert y-y' \vert ^{\delta}}{ \vert x-y \vert ^{1+\delta}} \quad \mbox{for } \vert x-y \vert \geq2 \bigl\vert y-y' \bigr\vert . \end{aligned}$$
(1.3)
We consider the family of operators \(T=\{T_{\epsilon}\}_{\epsilon>0}\) given by
$$\begin{aligned} T_{\epsilon}f(x)= \int_{ \vert x-y \vert >\epsilon}K(x,y) f(y)\,dy. \end{aligned}$$
(1.4)
A common method of measuring the speed of convergence of the family \(T_{\epsilon}\) is to consider the square functions
$$\begin{aligned} \Biggl(\sum_{i=1}^{\infty} \vert T_{\epsilon_{i}}f-T_{\epsilon _{i+1}}f \vert ^{2} \Biggr)^{1/2}, \end{aligned}$$
where \(\epsilon_{i}\) is a monotonically decreasing sequence which approaches 0. For convenience, other expressions have also been considered. Let \(\{t_{i}\}\) be a fixed sequence which decreases to zero. Following [1], the oscillation operator is defined as
$$\begin{aligned} \mathcal{O}(Tf) (x)= \Biggl(\sum_{i=1}^{\infty}\sup_{t_{i+1}\leq \epsilon_{i+1}< \epsilon_{i}\leq t_{i}} \bigl\vert T_{\epsilon _{i+1}}f(x)-T_{\epsilon_{i}}f(x) \bigr\vert ^{2} \Biggr)^{1/2} \end{aligned}$$
and the ρ-variation operator is defined as
$$\begin{aligned} \mathcal{V}_{\rho}(Tf) (x)=\sup_{\epsilon_{i}\searrow0 } \Biggl(\sum _{i=1}^{\infty}\bigl\vert T_{\epsilon_{i+1}}f(x)-T_{\epsilon_{i}}f(x) \bigr\vert ^{\rho}\Biggr)^{1/\rho}, \end{aligned}$$
where the sup is taken over all sequences of real number \(\{ \epsilon_{i}\}\) decreasing to zero.

The oscillation and variation for some families of operators have been studied by many authors on probability, ergodic theory, and harmonic analysis; see [24]. Recently, some authors [58] researched the weighted estimates of the oscillation and variation operators for the commutators of singular integrals.

Let m be a positive integer, let b be a function on \(\mathbb{R}\), and let \(R_{m+1}(b;x,y)\) be the \(m+1\)th Taylor series remainder of b at x expander about y, i.e.
$$\begin{aligned} R_{m+1}(b;x,y)=b(x)-\sum_{\alpha\leq m} \frac{1}{\alpha!} b^{(\alpha )}(y) (x-y)^{\alpha}. \end{aligned}$$
We consider the family of operators \(T^{b}=\{T^{b}_{\epsilon}\}_{\epsilon >0}\), where \(T^{b}_{\epsilon}\) are the multilinear singular integral operators of \(T_{\epsilon}\),
$$\begin{aligned} T^{b}_{\epsilon}f(x)= \int_{ \vert x-y \vert >\epsilon}\frac{ R_{m+1}(b;x,y)}{ \vert x-y \vert ^{m}}K(x,y)f(y)\,dy. \end{aligned}$$
(1.5)
Note that when \(m=0\), \(T^{b}_{\epsilon}\) is just the commutator of \(T_{\epsilon}\) and b, which is denoted by \(T_{\epsilon,b}\), that is to say
$$\begin{aligned} T_{\epsilon,b} f(x)= \int_{ \vert x-y \vert >\epsilon}\bigl(b(x)-b(y)\bigr)K(x,y)f(y)\,dy. \end{aligned}$$
(1.6)
However, when \(m>0\), \(T^{b}_{\epsilon}\) is a non-trivial generation of the commutator. It is well known that multilinear operators are of great interest in harmonic analysis and have been widely studied by many authors (see [913]).
A locally integrable function b is said to be in Lipschitz space \(\mathrm{Lip}_{\beta}(\mathbb{R})\) if
$$\begin{aligned} \Vert b \Vert _{\dot{\wedge}_{\beta}}=\sup_{I}\frac{1}{ \vert I \vert ^{1+\beta}} \int _{I} \bigl\vert b(x)-b_{I} \bigr\vert \,dx< \infty, \end{aligned}$$
where
$$\begin{aligned} b_{I}=\frac{1}{ \vert I \vert } \int_{I}b(x)\,dx. \end{aligned}$$

In this paper, we will study the boundedness of oscillation and variation operators for the family of the multilinear singular integral related to a Lipschitz function defined by (1.5) in weighted Lebesgue space. Our main results are as follows.

Theorem 1.1

Suppose that \(K(x,y)\) satisfies (1.1)-(1.3), \(b^{(m)}\in\dot{\wedge}_{\beta}\), \(0<\beta\leq\delta<1\), where δ is the same as in (1.2). Let \(\rho>2\), \(T=\{T_{\epsilon}\} _{\epsilon>0}\) and \(T^{b}=\{T^{b}_{\epsilon}\}_{\epsilon>0}\) be given by (1.4) and (1.5), respectively. If \(\mathcal{O}(T)\) and \(\mathcal{V}_{\rho}(T)\) are bounded on \(L^{p_{0}}(\mathbb {R},dx)\) for some \(1< p_{0}<\infty\), then, for any \(1< p<1/\beta\) with \(1/q=1/p-\beta\), \(\omega\in A_{p,q}(\mathbb{R})\), \(\mathcal{O}(T^{b})\) and \(\mathcal{V}_{\rho}(T^{b})\) are bounded from \(L^{p}(\mathbb{R},\omega^{p} \,dx)\) into \(L^{q}(\mathbb{R},\omega^{q} \,dx)\).

Corollary 1.1

Suppose that \(K(x,y)\) satisfies (1.1)-(1.3), \(b\in\dot{\wedge}_{\beta}\), \(0<\beta\leq\delta<1\), where δ is the same as in (1.2). Let \(\rho>2\), \(T=\{T_{\epsilon}\} _{\epsilon>0}\) and \(T_{b}=\{T_{b,\epsilon}\}_{\epsilon>0}\) be given by (1.4) and (1.6), respectively. If \(\mathcal{O}(T)\) and \(\mathcal{V}_{\rho}(T)\) are bounded on \(L^{p_{0}}(\mathbb {R},dx)\) for some \(1< p_{0}<\infty\), then, for any \(1< p<1/\beta\) with \(1/q=1/p-\beta\), \(\omega\in A_{p,q}(\mathbb{R})\), \(\mathcal{O}(T_{b})\) and \(\mathcal{V}_{\rho}(T_{b})\) are bounded from \(L^{p}(\mathbb{R},\omega^{p} \,dx)\) into \(L^{q}(\mathbb{R},\omega^{q} \,dx)\).

In this paper, we shall use the symbol \(A\lesssim B\) to indicate that there exists a universal positive constant C, independent of all important parameters, such that \(A\leq CB\). \(A\thickapprox B\) means that \(A\lesssim B\) and \(B\lesssim A\).

2 Some preliminaries

2.1 Weight

A weight ω is a nonnegative, locally integrable function on \(\mathbb{R}\). The classical weight theories were introduced by Muckenhoupt and Wheeden in [14] and [15].

A weight ω is said to belong to the Muckenhoup class \(A_{p}(\mathbb{R})\) for \(1< p<\infty\), if there exists a constant C such that
$$\begin{aligned} \biggl(\frac{1}{ \vert I \vert } \int_{I}\omega(x)\,dx \biggr) \biggl(\frac {1}{ \vert I \vert } \int_{I}\omega(x)^{-\frac{1}{p-1}}\,dx \biggr)^{p-1}\leq C \end{aligned}$$
for every interval I. The class \(A_{1}(\mathbb{R})\) is defined by replacing the above inequality with
$$\begin{aligned} \frac{1}{ \vert I \vert } \int_{I}\omega(x)\,dx\lesssim \mathop{\operatorname{ess}\operatorname{inf}}_{x\in I} w(x)\quad\mbox{for every ball } I\subset\mathbb{R}. \end{aligned}$$
When \(p=\infty\), we define \(A_{\infty}(\mathbb{R})=\bigcup_{1\leq p<\infty}A_{p}(\mathbb{R})\).
A weight \(\omega(x)\) is said to belong to the class \(A_{p,q}(\mathbb{R})\), \(1< p\leq q<\infty\), if
$$\begin{aligned} \biggl(\frac{1}{ \vert I \vert } \int_{I}\omega(x)^{q}\,dx \biggr)^{1/q} \biggl(\frac {1}{ \vert I \vert } \int_{I}\omega(x)^{-p'}\,dx \biggr)^{1/p'}\leq C. \end{aligned}$$
It is well known that if \(\omega\in A_{p.q}(\mathbb{R})\), then \(\omega^{q}\in A_{\infty}(\mathbb{R})\).

2.2 Function of \(\mathrm{Lip}_{\beta}(\mathbb{R})\)

The function of \(\mathrm{Lip}_{\beta}(\mathbb{R})\) has the following important properties.

Lemma 2.1

Let \(b\in \mathrm{Lip}_{\beta}(\mathbb{R})\). Then
  1. (1)
    \(1\leq p<\infty\)
    $$\begin{aligned} \sup_{I}\frac{1}{ \vert I \vert ^{\beta}} \biggl(\frac{1}{ \vert I \vert } \int _{I} \bigl\vert b(x)-b_{I} \bigr\vert ^{p}\,dx \biggr)^{1/p}\leq C \Vert b \Vert _{\dot{\wedge}_{\beta}}; \end{aligned}$$
     
  2. (2)
    for any \(I_{1}\subset I_{2}\),
    $$\begin{aligned} \frac{1}{ \vert I_{2} \vert } \int_{I_{2}} \bigl\vert b(y)-b_{I_{1}} \bigr\vert \,dy \lesssim\frac { \vert I_{2} \vert }{ \vert I_{1} \vert } \vert I_{2} \vert ^{\beta} \Vert b \Vert _{\dot{\wedge}_{\beta}}. \end{aligned}$$
     

2.3 Maximal function

We recall the definition of Hardy-Littlewood maximal operator and fractional maximal operator. The Hardy-Littlewood maximal operator is defined by
$$\begin{aligned} M(f) (x)=\sup_{I\ni x}\frac{1}{ \vert I \vert } \int_{I} \bigl\vert f(y) \bigr\vert \,dy. \end{aligned}$$
The fractional maximal function is defined as
$$\begin{aligned} M_{\beta,r}(f) (x)=\sup_{I\ni x} \biggl(\frac{1}{ \vert I \vert ^{1-r\beta}} \int _{I} \bigl\vert f(y) \bigr\vert ^{r}\,dy \biggr)^{1/r} \end{aligned}$$
for \(1\leq r<\infty\). In order to simplify the notation, we set \(M_{\beta}(f)(x)=M_{\beta,1}(f)(x)\).

Lemma 2.2

Let \(1< p<\infty\) and \(\omega\in A_{\infty}(\mathbb{R})\). Then
$$\begin{aligned} \Vert M f \Vert _{L^{p}(\omega)}\lesssim \bigl\Vert M^{\sharp}f \bigr\Vert _{L^{p}(\omega)} \end{aligned}$$
for all f such that the left hand side is finite.

Lemma 2.3

Suppose \(0<\beta<1\), \(1\leq r< p<1/\beta\), \(1/q=1/p-\beta\). If \(\omega\in A_{p,q}(\mathbb{R})\), then
$$\begin{aligned} \Vert M_{\beta,r} f \Vert _{L^{q}(\omega^{q})}\lesssim \Vert f \Vert _{L^{p}(\omega^{p})}. \end{aligned}$$

2.4 Taylor series remainder

The following lemma gives an estimate on Taylor series remainder.

Lemma 2.4

[10] Let b be a function on \(\mathbb{R}\) and \(b^{(m)}\in L^{s}(\mathbb {R})\) for any \(s>1\). Then
$$\begin{aligned} \bigl\vert R_{m}(b;x,y) \bigr\vert \lesssim \vert x-y \vert ^{m} \biggl(\frac{1}{ \vert I_{x}^{y} \vert } \int _{I_{x}^{y}} \bigl\vert b^{(m)}(z) \bigr\vert ^{s}\,dz \biggr)^{1/s}, \end{aligned}$$
where \(I_{x}^{y}\) is the interval \((x-5 \vert x-y \vert , x+5 \vert x-y \vert )\).

2.5 Oscillation and variation operators

We consider the operator
$$\begin{aligned} \mathcal{O}'(Tf) (x)= \Biggl(\sum_{i=1}^{\infty}\sup_{t_{i+1}< \delta _{i}< t_{i}} \bigl\vert T_{t_{i+1}}f(x)-T_{\delta_{i}}f(x) \bigr\vert ^{2} \Biggr)^{1/2}. \end{aligned}$$
It is easy to check that
$$\begin{aligned} \mathcal{O}'(Tf)\thickapprox\mathcal{O}(Tf). \end{aligned}$$
Following [4], we denote by E the mixed norm Banach space of two variable function h defined on \(\mathbb{R}\times\mathbb{N}\) such that
$$\begin{aligned} \Vert h \Vert _{E}\equiv \biggl(\sum _{i} \Bigl(\sup_{s} \bigl\vert h(s,i) \bigr\vert \Bigr)^{2} \biggr)^{1/2}< \infty. \end{aligned}$$
Given \(T=\{T_{\epsilon}\}_{\epsilon>0}\), where \(T_{\epsilon}\) defined as (1.4), for a fixed decreasing sequence \(\{t_{i}\}\) with \(t_{i}\searrow0\), let \(J_{i}=(t_{i+1},t_{i}]\) and define the E-valued operator \(\mathcal{U}(T): f\rightarrow\mathcal{U}(T)f\) by
$$\begin{aligned} \mathcal{U}(T)f(x)=\bigl\{ T_{t_{i+1}}f(x)-T_{s}f(x)\bigr\} _{s\in J_{i},i\in\mathbb{N}}= \biggl\{ \int_{\{t_{i+1}< \vert x-y \vert < s\} }K(x,y)f(y)\,dy \biggr\} _{s\in J_{i},i\in\mathbb{N}}. \end{aligned}$$
Then
$$\begin{aligned} \mathcal{O}'(Tf) (x)={}& \bigl\Vert \mathcal{U}(T)f(x) \bigr\Vert _{E}= \bigl\Vert \bigl\{ T_{t_{i+1}}f(x)-T_{s}f(x) \bigr\} _{s\in J_{i},i\in\mathbb{N}} \bigr\Vert _{E} \\ ={}& \biggl\Vert \biggl\{ \int_{\{t_{i+1}< \vert x-y \vert < s\}}K(x,y)f(y)\,dy \biggr\} _{s\in J_{i},i\in\mathbb{N}} \biggr\Vert _{E}. \end{aligned}$$
On the other hand, let \(\Theta=\{\beta: \beta=\{\epsilon_{i}\} ,\epsilon_{i}\in\mathbb{R},\epsilon_{i}\searrow0\}\). We denote by \(F_{\rho}\) the mixed norm space of two variable functions \(g(i,\beta)\) such that
$$\begin{aligned} \Vert g \Vert _{F_{\rho}}\equiv\sup_{\beta}\biggl(\sum _{i} \bigl\vert g(i,\beta) \bigr\vert ^{\rho}\biggr)^{1/\rho}. \end{aligned}$$
We also consider the \(F_{\rho}\)-valued operator \(\mathcal {V}(T):f\rightarrow\mathcal{V}(T)f\) given by
$$\begin{aligned} \mathcal{V}(T)f(x)=\bigl\{ T_{t_{i+1}}f(x)-T_{t_{i}}f(x)\bigr\} _{\beta=\{ \epsilon_{i}\}\in\Theta}. \end{aligned}$$
Then
$$\begin{aligned} \mathcal{V}_{\rho}(T)f(x)= \bigl\Vert \mathcal{V}(T)f(x) \bigr\Vert _{F_{\rho}}. \end{aligned}$$
Next, let B be a Banach space and φ be a B-valued function, we define the sharp maximal operator as follows:
$$\begin{aligned} \varphi^{\sharp}(x)=\sup_{x\in I}\frac{1}{ \vert I \vert } \int_{I} \biggl\Vert \varphi (y)-\frac{1}{ \vert I \vert } \int_{I}\varphi(z)\,dz \biggr\Vert _{B}\,dy \thickapprox\sup_{ x\in I}\inf_{c} \frac{1}{ \vert I \vert } \int_{I} \bigl\Vert \varphi(y)-c \bigr\Vert _{B}\,dy. \end{aligned}$$
Then
$$\begin{aligned} M^{\sharp}\bigl(\mathcal{O}'(Tf)\bigr)\leq2\bigl( \mathcal{U}(T)f\bigr)^{\sharp}(x) \end{aligned}$$
and
$$\begin{aligned} M^{\sharp}\bigl(\mathcal{\mathcal{V}}_{\rho}(Tf)\bigr)\leq2\bigl( \mathcal {V}(T)f\bigr)^{\sharp}(x). \end{aligned}$$

Finally, let us recall some results about oscillation and variation operators.

Lemma 2.5

([5])

Suppose that \(K(x,y)\) satisfies (1.1)-(1.3), \(\rho >2\). Let \(T=\{T_{\epsilon}\}_{\epsilon>0}\) be given by (1.4). If \(O(T)\) and \(V_{\rho}(T)\) are bounded on \(L^{p_{0}}(R)\) for some \(1< p_{0}<\infty\), then, for any \(1< p<\infty\), \(\omega\in A_{p}(\mathbb{R})\),
$$\begin{aligned} \bigl\Vert \mathcal{O}'(Tf) \bigr\Vert _{L^{p}(\omega)}\leq \bigl\Vert \mathcal{O}(Tf) \bigr\Vert _{L^{p}(\omega)}\lesssim \Vert f \Vert _{L^{p}(\omega)} \end{aligned}$$
and
$$\begin{aligned} \bigl\Vert \mathcal{V}_{\rho}(Tf) \bigr\Vert _{L^{p}(\omega)}\lesssim \Vert f \Vert _{L^{p}(\omega)}. \end{aligned}$$

3 The proof of main results

Note that if \(\omega\in A_{p,q}(\mathbb{R})\), then \(\omega^{q}\in A_{\infty}(\mathbb{R})\). By Lemma 2.2 and Lemma 2.3, we only need to prove
$$\begin{aligned} M^{\sharp}\bigl(\mathcal{O}'\bigl(T^{b} \bigr)f\bigl)(x) \lesssim \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge }_{\beta}} \bigl( M_{\beta,r}(f) (x)+ M_{\beta}(f) (x) \bigr) \end{aligned}$$
(3.1)
and
$$\begin{aligned} M^{\sharp}\bigl(\mathcal{V}_{\rho}\bigl(T^{b} \bigr)f\bigl)(x) \lesssim \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot {\wedge}_{\beta}} \bigl( M_{\beta,r}(f) (x)+ M_{\beta}(f) (x) \bigr) \end{aligned}$$
(3.2)
hold for any \(1< r<\infty\).
We will prove only inequality (3.1), since (3.2) can be obtained by a similar argument. Fix f and \(x_{0}\) with an interval \(I=(x_{0}-l,x_{0}+l)\). Write \(f=f_{1}+f_{2}=f\chi_{5I}+f\chi_{\mathbb{R}\setminus5I}\), and let
$$\begin{aligned} C_{I}= \biggl\{ \int_{\{t_{i+1}< \vert x_{0}-y \vert < s\}}\frac {R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}}K(x_{0},y)f_{2}(y)\,dy \biggr\} _{s\in J_{i},i\in \mathbb{N}}=\mathcal{U}\bigl(T^{b}\bigr)f_{2}(x_{0}). \end{aligned}$$
Then
$$\begin{aligned} \mathcal{U}\bigl(T^{b}\bigr)f (x) ={}& \biggl\{ \int_{\{t_{i+1}< \vert x-y \vert < s\}}\frac {R_{m+1}({b};x,y)}{ \vert x-y \vert ^{m}}K(x,y)f(y)\,dy \biggr\} _{s\in J_{i},i\in\mathbb {N}} \\ ={}& \mathcal{U}(T) \biggl(\frac{R_{m+1}({b};x,\cdot)}{ \vert x-\cdot \vert ^{m}}f_{1} \biggr)+\mathcal{U} \bigl(T^{b}\bigr)f_{2}(x). \end{aligned}$$
Therefore
$$\begin{aligned} &\frac{1}{ \vert I \vert } \int_{I} \bigl\Vert \mathcal{U}\bigl(T^{b}\bigr)f (x)-C_{I} \bigr\Vert _{E}\,dx \\ &\quad \leq \frac{1}{ \vert I \vert } \int_{I} \biggl\Vert \mathcal{U}(T) \biggl( \frac {R_{m+1}({b};x,\cdot)}{ \vert x-\cdot \vert ^{m}}f_{1} \biggr) \biggr\Vert _{E}\,dx + \frac{1}{ \vert I \vert } \int_{I} \bigl\Vert \mathcal{U}\bigl(T^{b} \bigr)f_{2}(x)-\mathcal {U}\bigl(T^{b}\bigr)f_{2}(x_{0}) \bigr\Vert _{E}\,dx \\ &\quad = M_{1}+M_{2}. \end{aligned}$$

For \(x\in I\), \(k=0,-1,-2,\ldots\) , let \(E_{k}=\{y:2^{k-1}\cdot6l\leq \vert y-x \vert <2^{k}\cdot6l\}\), let \(I_{k}=\{y: \vert y-x \vert <2^{k}\cdot6l\}\), and let \({b}_{k}(z)=b(z)-\frac{1}{m!}(b^{(m)})_{I_{k}}z^{m}\). By [10] we have \(R_{m+1}({b};x,y)=R_{m+1}(b_{k};x,y)\) for any \(y\in E_{k}\).

By Lemma 2.5, we know \(\mathcal{O}'(T)\) is bounded on \(L^{u}(\mathbb{R})\) for \(u>1\). Then, using Hölder’s inequality, we deduce
$$\begin{aligned} M_{1} \lesssim{}& \biggl(\frac{1}{ \vert I \vert } \int_{I} \biggl\Vert \mathcal{U}(T) \biggl( \frac {R_{m+1}({b};x,\cdot)}{ \vert x-\cdot \vert ^{m}}f_{1} \biggr) \biggr\Vert ^{u}_{E}\,dx \biggr)^{1/u} \\ \lesssim& \biggl(\frac{1}{ \vert I \vert } \int_{\{y: \vert y-x \vert < 6l\}} \biggl\vert \frac {R_{m+1}({b};\cdot,y)}{ \vert y-\cdot \vert ^{m}}f(y) \biggr\vert ^{u}\,dy \biggr)^{1/u} \\ =& \Biggl(\frac{1}{ \vert I \vert }\sum_{k=-\infty}^{0} \int_{E_{k}} \biggl\vert \biggl(\frac{R_{m+1}({b}_{k};\cdot,y)}{ \vert y-\cdot \vert ^{m}}f(y) \biggr) \biggr\vert ^{r}\,dy \Biggr)^{1/r} \\ \lesssim& \Biggl(\frac{1}{ \vert I \vert }\sum_{k=-\infty}^{0} \int_{E_{k}} \biggl\vert \biggl( \biggl(\frac{R_{m}({b_{k}};\cdot,y)}{ \vert y-\cdot \vert ^{m}}- \frac{1}{m!}\frac {(y-\cdot)^{m}{b}_{k}^{(m)}(y) }{ \vert y-\cdot \vert ^{m}} \biggr)f(y) \biggr) \biggr\vert ^{u}\,dy \Biggr)^{1/u} \\ \lesssim& \Biggl(\frac{1}{ \vert I \vert }\sum_{k=-\infty}^{0} \int_{E_{k}} \biggl\vert \frac{R_{m}({b_{k}};\cdot,y)}{ \vert y-\cdot \vert ^{m}}f(y) \biggr\vert ^{u}\,dy \Biggr)^{1/u} \\ &{} + \Biggl(\frac{1}{ \vert I \vert }\sum_{k=-\infty}^{0} \int_{E_{k}} \biggl\vert \frac {1}{m!} \frac{(y-\cdot)^{m}{b}_{k}^{(m)}(y) }{ \vert y-\cdot \vert ^{m}}f(y) \biggr\vert ^{u}\,dy \Biggr)^{1/u} \\ =& M_{11}+M_{12}. \end{aligned}$$
By Lemma 2.4 and Lemma 2.1,
$$\begin{aligned} \bigl\vert R_{m}({b}_{k};x,y) \bigr\vert \lesssim& \vert x-y \vert ^{m} \biggl(\frac{1}{ \vert I_{x}^{y} \vert } \int _{I_{x}^{y}} \bigl\vert {b}_{k}^{(m)}(z) \bigr\vert ^{s}\,dz \biggr)^{1/s} \\ \lesssim& \vert x-y \vert ^{m} \biggl(\frac{1}{2^{k}\cdot30l} \int _{ \vert y-x \vert < 2^{k}\cdot30l} \bigl\vert b^{(m)}(y)- \bigl(b^{(m)}\bigr)_{I_{k}} \bigr\vert ^{s}\,dz \biggr)^{1/s} \\ \lesssim& \vert x-y \vert ^{m}\bigl(2^{k}l \bigr)^{\beta}\bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}}. \end{aligned}$$
Then
$$\begin{aligned} M_{11} \lesssim& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} l^{\beta}\Biggl(\frac{1}{ \vert I \vert }\sum _{k=-\infty}^{0}2^{k\beta u} \int _{E_{k}} \bigl\vert f(y) \bigr\vert ^{u}\,dy \Biggr)^{1/u} \\ \lesssim& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} l^{\beta}\Biggl(\frac {1}{ \vert I \vert }\sum_{k=-\infty}^{0} \int_{E_{k}} \bigl\vert f(y) \bigr\vert ^{u}\,dy \Biggr)^{1/u} \\ \lesssim& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} l^{\beta}\biggl(\frac {1}{ \vert I \vert } \int_{7I} \bigl\vert f(y) \bigr\vert ^{u}\,dy \biggr)^{1/u} \\ \lesssim& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} l^{\beta}\biggl(\frac {1}{ \vert I \vert } \int_{7I} \bigl\vert f(y) \bigr\vert ^{r}\,dy \biggr)^{1/r} \\ \lesssim& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} M_{\beta,r}(f) (x_{0}). \end{aligned}$$
Since \({b}_{k}^{(m)}(y)=b^{(m)}(y)-(b^{(m)})_{I_{k}}\), then, applying Hölder’s inequality and Lemma 2.1, we get
$$\begin{aligned} M_{12} \lesssim & \Biggl(\frac{1}{ \vert I \vert }\sum _{k=-\infty}^{0} \int _{E_{k}} \bigl\vert \bigl(b^{(m)}(y)- \bigl(b^{(m)}\bigr)_{I_{k}}\bigr)f(y) \bigr\vert ^{u}\,dy \Biggr)^{1/u} \\ \lesssim& \biggl(\frac{1}{ \vert I \vert }\sum_{k=-\infty}^{0} \biggl( \int _{I_{k}} \bigl\vert f(y) \bigr\vert ^{r}\,dy \biggr)^{u/r} \biggl( \int_{I_{k}} \bigl\vert b^{(m)}(y)- \bigl(b^{(m)}\bigr)_{I_{k}} \bigr\vert ^{\frac {ur}{r-u}} \biggr)^{1-u/r} \biggr)^{1/u} \\ \lesssim& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} \Biggl( \frac {1}{ \vert I \vert }\sum_{k=-\infty}^{0} \biggl( \int_{I_{k}} \bigl\vert f(y) \bigr\vert ^{r}\,dy \biggr)^{u/r} \vert I_{k} \vert ^{\beta u+1-u/r} \Biggr)^{1/u} \\ \lesssim& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} M_{\beta ,r}(f) (x_{0}) \Biggl(\frac{1}{ \vert I \vert }\sum _{k=-\infty}^{0} \vert I_{k} \vert \Biggr)^{1/u} \\ \lesssim& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} M_{\beta,r}(f) (x_{0}). \end{aligned}$$
We now estimate \(M_{2}\). For \(x\in I\), we have
$$\begin{aligned} & \bigl\Vert \mathcal{U}\bigl(T^{b}\bigr)f_{2}(x)- \mathcal{U}\bigl(T^{b}\bigr)f_{2}(x_{0}) \bigr\Vert _{E} \\ &\quad = \biggl\Vert \biggl\{ \int_{\{t_{i+1}< \vert x-y \vert < s\}}\frac {R_{m+1}({b};x,y)}{ \vert x-y \vert ^{m}}K(x,y)f_{2}(y)\,dy \\ &\qquad{}- \int_{\{t_{i+1}< \vert x_{0}-y \vert < s\}}\frac {R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}}K(x_{0},y)f_{2}(y)\,dy \biggr\} _{s\in J_{i}, i\in\mathbb{N}} \biggr\Vert _{E} \\ &\quad \leq \biggl\Vert \biggl\{ \int_{\{t_{i+1}< \vert x-y \vert < s\}} \biggl(\frac {R_{m+1}({b};x,y)}{ \vert x-y \vert ^{m}}K(x,y)- \frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}}K(x_{0},y) \biggr)f_{2}(y)\,dy\biggr\} _{s\in J_{i}, i\in\mathbb{N}} \biggr\Vert _{E} \\ &\qquad{} + \biggl\Vert \biggl\{ \int_{R} \bigl(\chi_{\{t_{i+1}< \vert x-y \vert < s\}}(y)-\chi _{\{t_{i+1}< \vert x_{0}-y \vert < s\}}(y) \bigr)\frac {R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}}K(x_{0},y)f_{2}(y)\,dy \biggr\} _{s\in J_{i}, i\in \mathbb{N}} \biggr\Vert _{E} \\ &\quad = N_{1}+N_{2}. \end{aligned}$$
For \(k=0,1,2,\ldots\) , let \(F_{k}=\{y:2^{k}\cdot4l\leq \vert y-x_{0} \vert <2^{k+1}\cdot4l\}\), let \(\widetilde{I}_{k}=\{y: \vert y-x_{0} \vert <2^{k}\cdot4l\}\), and let \(\widetilde{b}_{k}(z)=b(z)-\frac {1}{m!}(b^{(m)})_{\widetilde{I}_{k}}z^{m}\). Note that
$$\begin{aligned} & \frac{R_{m+1}({b};x,y)}{ \vert x-y \vert ^{m}}K(x,y) -\frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}}K(x_{0},y) \\ &\quad =\frac{R_{m+1}(\widetilde{b}_{k};x,y)}{ \vert x-y \vert ^{m}}K(x,y) -\frac{R_{m+1}(\widetilde{b}_{k};x_{0},y)}{ \vert x_{0}-y \vert ^{m}}K(x_{0},y) \\ &\quad = \frac{1}{ \vert x-y \vert ^{m}} \bigl(R_{m}(\widetilde {b}_{k};x,y)-R_{m}( \widetilde{b}_{k};x_{0},y) \bigr)K(x,y) \\ &\qquad{} +R_{m}(\widetilde{b}_{k};x_{0},y) \biggl( \frac{1}{ \vert x-y \vert ^{m}}-\frac {1}{ \vert x_{0}-y \vert ^{m}} \biggr)K(x,y) \\ &\qquad{} - \frac{1}{m!}\widetilde{b}_{k}^{(m)}(y) \biggl( \frac {(x-y)^{m}}{ \vert x-y \vert ^{m}}-\frac{(x_{0}-y)^{m}}{ \vert x_{0}-y \vert ^{m}} \biggr)K(x,y) \\ &\qquad{} + \frac{R_{m+1}(\widetilde{b}_{k};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \bigl(K(x,y)-K(x_{0},y) \bigr). \end{aligned}$$
By Minkowski’s inequalities and \(\Vert \{\chi_{\{ t_{i+1}< \vert x-y \vert <s\}}\}_{s\in J_{i}, i\in\mathbb{N}} \Vert _{E}\leq1\), we obtain
$$\begin{aligned} N_{1}\leq{}& \int_{\mathbb{R}} \bigl\Vert \{\chi_{\{t_{i+1}< \vert x-y \vert < s\} } \}_{s\in J_{i}, i\in\mathbb{N}} \bigr\Vert _{E} \\ &{}\times \biggl\vert \frac{R_{m+1}(\widetilde{b}_{k};x,y)}{ \vert x-y \vert ^{m}}K(x,y) -\frac{R_{m+1}(\widetilde{b}_{k};x_{0},y)}{ \vert x_{0}-y \vert ^{m}}K(x_{0},y) \biggr\vert \bigl\vert f_{2}(y) \bigr\vert \,dy \\ \leq{}& \sum _{k=0}^{\infty}\int_{F_{k}}\frac{1}{ \vert x-y \vert ^{m}} \bigl\vert R_{m}( \widetilde{b}_{k};x,y)-R_{m}(\widetilde{b}_{k};x_{0},y) \bigr\vert \bigl\vert K(x,y) \bigr\vert \bigl\vert f_{2}(y) \bigr\vert \,dy \\ &{}+ \sum_{k=0}^{\infty}\int_{F_{k}} \bigl\vert R_{m}(\widetilde{b}_{k};x_{0},y) \bigr\vert \biggl\vert \frac{1}{ \vert x-y \vert ^{m}}-\frac{1}{ \vert x_{0}-y \vert ^{m}} \biggr\vert \bigl\vert K(x,y) \bigr\vert \bigl\vert f_{2}(y) \bigr\vert \,dy \\ &{}+ \sum_{k=0}^{\infty}\int_{F_{k}}\frac{1}{m!} \bigl\vert \widetilde {b}_{k}^{(m)}(y) \bigr\vert \biggl\vert \frac{(x-y)^{m}}{ \vert x-y \vert ^{m}}-\frac {(x_{0}-y)^{m}}{ \vert x_{0}-y \vert ^{m}} \biggr\vert \bigl\vert K(x,y) \bigr\vert \bigl\vert f_{2}(y) \bigr\vert \,dy \\ &{}+ \sum_{k=0}^{\infty}\int_{F_{k}} \biggl\vert \frac{R_{m+1}(\widetilde {b}_{k};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert \bigl\vert K(x,y)-K(x_{0},y) \bigr\vert \bigl\vert f_{2}(y) \bigr\vert \,dy \\ ={}& N_{11}+N_{12}+N_{13}+N_{14}. \end{aligned}$$
From the mean value theorem, there exists \(\eta\in I\) such that
$$\begin{aligned} R_{m}(\widetilde{b}_{k};x,y)-R_{m}(\widetilde {b}_{k};x_{0},y)=(x-x_{0})R_{m-1}\bigl( \widetilde{b}_{k}';\eta,y\bigr). \end{aligned}$$
For \(\eta, x\in I\), \(y\in F_{k}\), we have \(\vert y-x_{0} \vert \thickapprox \vert y-x \vert \thickapprox \vert y-\eta \vert \) and \(5 \vert y-\eta \vert \approx5 \vert y-x_{0} \vert \leq 2^{k+1}\cdot20l\). By Lemma 2.4 and Lemma 2.1 we get
$$\begin{aligned} \bigl\vert R_{m-1}\bigl(\widetilde{b}'_{k}; \eta,y\bigr) \bigr\vert \lesssim{}& \vert \eta-y \vert ^{m-1} \biggl( \frac{1}{ \vert I_{\eta}^{y} \vert } \int_{I_{\eta}^{y}} \bigl\vert \widetilde {b}_{k}^{(m)}(z) \bigr\vert ^{s}\,dz \biggr)^{1/s} \\ \lesssim{}& \vert x-y \vert ^{m-1} \biggl(\frac{1}{2^{k+1}\cdot20l} \int _{ \vert z-x_{0} \vert < 2^{k+1}\cdot20l} \bigl\vert {b}^{(m)}(z)- \bigl(b^{(m)}\bigr)_{\widetilde {I}_{k}} \bigr\vert ^{s}\,dz \biggr)^{1/s} \\ \lesssim{}& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} \bigl(2^{k}l\bigr)^{\beta} \vert x-y \vert ^{m-1}. \end{aligned}$$
Then
$$\begin{aligned} \bigl\vert R_{m}(\widetilde{b}_{k};x,y)-R_{m}( \widetilde{b}_{k};x_{0},y) \bigr\vert \lesssim \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}}\bigl(2^{k}l \bigr)^{\beta} \vert x-x_{0} \vert \vert x-y \vert ^{m-1}. \end{aligned}$$
Since \(\vert K(x,y) \vert \leq C \vert x_{0}-y \vert ^{-1}\),
$$\begin{aligned} N_{11}\lesssim{}& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}}\sum_{k=0}^{\infty}\bigl(2^{k}l\bigr)^{\beta}\int_{2^{k}\cdot4l\leq \vert x_{0}-y \vert < 2^{k+1}\cdot4l}\frac{l}{(2^{k}\cdot4l)^{2}} \bigl\vert f(y) \bigr\vert \,dy \\ \lesssim{}& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}}\sum _{k=0}^{\infty}\frac{1}{2^{k}} \frac{(2^{k}l)^{\beta}}{2^{k}l} \int_{ \vert x_{0}-y \vert < 2^{k+1}\cdot4l} \bigl\vert f(y) \bigr\vert \,dy \\ \lesssim{}& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}}M_{\beta}(f) (x_{0}). \end{aligned}$$
For \(N_{12}\), since \(x\in I\), \(y\in F_{k}\),
$$\begin{aligned} \bigl\vert R_{m}(\widetilde{b}_{k};x,y) \bigr\vert \lesssim \vert x-y \vert ^{m} \biggl(\frac {1}{ \vert I_{x}^{y} \vert } \int_{I_{x}^{y}} \bigl\vert \widetilde{b}_{k}^{(m)}(z) \bigr\vert ^{s}\,dz \biggr)^{1/s}\lesssim \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}}\bigl(2^{k}l \bigr)^{\beta} \vert x-y \vert ^{m} \end{aligned}$$
and
$$\begin{aligned} \biggl\vert \frac{1}{ \vert x-y \vert ^{m}}-\frac{1}{ \vert x_{0}-y \vert ^{m}} \biggr\vert \lesssim \frac { \vert x-x_{0} \vert }{ \vert x-y \vert ^{m+1}}. \end{aligned}$$
Thus
$$\begin{aligned} N_{12}\lesssim \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}}\sum_{k=0}^{\infty}\bigl(2^{k}l\bigr)^{\beta}\int_{2^{k}\cdot4l\leq \vert x_{0}-y \vert < 2^{k+1}\cdot4l}\frac {l}{(2^{k}\cdot4l)^{2}} \bigl\vert f(y) \bigr\vert \,dy \lesssim \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge }_{\beta}}M_{\beta}(f) (x_{0}). \end{aligned}$$
As for \(N_{13}\), due to
$$\begin{aligned} \biggl\vert \frac{(x-y)^{m}}{ \vert x-y \vert ^{m}}-\frac{(x_{0}-y)^{m}}{ \vert x_{0}-y \vert ^{m}} \biggr\vert \lesssim \frac{ \vert x-x_{0} \vert }{ \vert x-y \vert }, \end{aligned}$$
and noting \(\widetilde{b}_{k}^{(m)}(y)=b^{(m)}(y)-(b^{(m)})_{\widetilde {I}_{k}}\), we have
$$\begin{aligned} N_{13}\lesssim{}& \sum_{k=0}^{\infty}\int_{F_{k}} \bigl\vert b^{(m)}(y)- \bigl(b^{(m)}\bigr)_{\widetilde{I}_{k}} \bigr\vert \frac { \vert x-x_{0} \vert }{ \vert x_{0}-y \vert ^{2}} \bigl\vert f(y) \bigr\vert \,dy \\ \lesssim{}& \sum_{k=0}^{\infty}\frac{1}{2^{k}}\frac{1}{2^{k}\cdot 4l} \int_{ \vert x_{0}-y \vert < 2^{k}\cdot4l} \bigl\vert b^{(m)}(y)- \bigl(b^{(m)}\bigr)_{\widetilde {I}_{k}} \bigr\vert \bigl\vert f(y) \bigr\vert \,dy \\ \lesssim{}& \sum_{k=0}^{\infty}\frac{1}{2^{k}} \biggl(\frac {1}{2^{k}\cdot4l} \int_{ \vert x_{0}-y \vert < 2^{k}\cdot4l} \bigl\vert f(y) \bigr\vert ^{r}\,dy \biggr)^{1/r} \\ &{} \times \biggl(\frac{1}{2^{k}\cdot4l} \int_{ \vert x_{0}-y \vert < 2^{k}\cdot 4l} \bigl\vert b^{(m)}(y)- \bigl(b^{(m)}\bigr)_{\widetilde{I}_{k}} \bigr\vert ^{r'}\,dy \biggr)^{1/r'} \\ \lesssim{}& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}}M_{r,\beta }(f) (x_{0})\sum_{k=0}^{\infty}\frac{1}{2^{k}}\lesssim \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot {\wedge}_{\beta}}M_{\beta,r}(f) (x_{0}). \end{aligned}$$
Notice
$$\begin{aligned} \bigl\vert R_{m+1}(\widetilde{b}_{k}; x_{0},y) \bigr\vert &\leq \bigl\vert R_{m}(\widetilde {b}_{k};x_{0},y) \bigr\vert +\frac{1}{m!} \bigl\vert \widetilde{b}_{k}^{(m)}(y) (x_{0}-y)^{m} \bigr\vert \\ & \lesssim \bigl\Vert b^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} \bigl(2^{k}l\bigr)^{\beta} \vert x_{0}-y \vert ^{m}+ \bigl\vert b^{(m)}(y)-\bigl(b^{(m)} \bigr)_{\widetilde{I}_{k}} \bigr\vert \vert x_{0}-y \vert ^{m} \end{aligned}$$
and by (1.2),
$$\begin{aligned} \bigl\vert K(x,y)-K(x_{0},y) \bigr\vert \lesssim \frac{ \vert x-x_{0} \vert ^{\delta}}{ \vert x_{0}-y \vert ^{1+\delta}}. \end{aligned}$$
Similar to the estimates for \(N_{11}\), we have
$$\begin{aligned} \sum_{k=0}^{\infty}\int_{F_{k}}\frac{ \vert R_{m}(\widetilde {b}_{k};x_{0},y) \vert }{ \vert x-y \vert ^{m}}\frac{ \vert x-x_{0} \vert ^{\delta}}{ \vert x_{0}-y \vert ^{1+\delta }} \bigl\vert f(y) \bigr\vert \,dy\lesssim \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}}M_{\beta}(f) (x_{0}). \end{aligned}$$
Similar to the estimates for \(N_{13}\), we have
$$\begin{aligned} \sum_{k=0}^{\infty}\int_{F_{k}}\frac{ \vert \widetilde {b}_{k}^{(m)}(y)(x_{0}-y)^{m} \vert }{ \vert x-y \vert ^{m}}\frac{ \vert x-x_{0} \vert ^{\delta}}{ \vert x_{0}-y \vert ^{1+\delta}} \bigl\vert f(y) \bigr\vert \,dy\lesssim \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge }_{\beta}}M_{\beta,r}(f) (x_{0}). \end{aligned}$$
Then
$$\begin{aligned} N_{14}\lesssim \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} \bigl(M_{\beta}(f) (x_{0})+M_{\beta,r}(f) (x_{0}) \bigr). \end{aligned}$$
Finally, let us estimate \(N_{2}\). Notice that the integral
$$\begin{aligned} \int_{R} \bigl(\chi_{\{t_{i+1}< \vert x-y \vert < s\}}(y)-\chi_{\{ t_{i+1}< \vert x_{0}-y \vert < s\}}(y) \bigr) \frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}}K(x_{0},y)f_{2}(y)\,dy \end{aligned}$$
will be non-zero in the following cases:
  1. (i)

    \(t_{i+1}< \vert x-y \vert <s\) and \(\vert x_{0}-y \vert \leq t_{i+1}\);

     
  2. (ii)

    \(t_{i+1}< \vert x-y \vert <s\) and \(\vert x_{0}-y \vert \geq s\);

     
  3. (iii)

    \(t_{i+1}< \vert x_{0}-y \vert <s\) and \(\vert x-y \vert \leq t_{i+1}\);

     
  4. (iv)

    \(t_{i+1}< \vert x_{0}-y \vert <s\) and \(\vert x-y \vert \geq s\).

     
In case (i) we have \(t_{i+1}< \vert x-y \vert \leq \vert x_{0}-x \vert + \vert x_{0}-y \vert <l+t_{i+1}\) as \(\vert x-x_{0} \vert < l\). Similarly, in case (iii) we have \(t_{i+1}< \vert x_{0}-y \vert <l+t_{i+1}\) as \(\vert x-x_{0} \vert < l\). In case (ii) we have \(s< \vert x_{0}-y \vert <l+s\) and in case (iv) we have \(s< \vert x-y \vert <l+s\). By (1.1) and taking \(1< t< r\), we have
$$\begin{aligned} &\int_{\mathbb{R}} \bigl(\chi_{\{t_{i+1}< \vert x-y \vert < s\}}(y)-\chi_{\{ t_{i+1}< \vert x_{0}-y \vert < s\}}(y) \bigr) \frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}}K(x_{0},y)f_{2}(y)\,dy \\ &\quad \lesssim \int_{\mathbb{R}}\chi_{\{t_{i+1}< \vert x-y \vert < s\}}(y)\chi_{\{ t_{i+1}< \vert x-y \vert < l+t_{i+1}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert \frac { \vert f_{2}(y) \vert }{ \vert x_{0}-y \vert }\,dy \\ &\qquad{} + \int_{\mathbb{R}}\chi_{\{t_{i+1}< \vert x-y \vert < s\}}(y)\chi_{\{ s< \vert x_{0}-y \vert < l+s\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert \frac { \vert f_{2}(y) \vert }{ \vert x_{0}-y \vert }\,dy \\ &\qquad{} + \int_{\mathbb{R}}\chi_{\{t_{i+1}< \vert x_{0}-y \vert < s\}}(y)\chi_{\{ t_{i+1}< \vert x_{0}-y \vert < l+t_{i+1}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert \frac { \vert f_{2}(y) \vert }{ \vert x_{0}-y \vert }\,dy \\ &\qquad{} + \int_{\mathbb{R}}\chi_{\{t_{i+1}< \vert x_{0}-y \vert < s\}}(y)\chi_{\{ s< \vert x-y \vert < l+s\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert \frac { \vert f_{2}(y) \vert }{ \vert x_{0}-y \vert }\,dy \\ &\quad \lesssim l^{1/t'} \biggl( \int_{\mathbb{R}}\chi_{\{ t_{i+1}< \vert x-y \vert < s\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert ^{t} \frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy \biggr)^{1/t} \\ &\qquad{} + l^{1/t'} \biggl( \int_{\mathbb{R}}\chi_{\{t_{i+1}< \vert x_{0}-y \vert < s\} }(y) \biggl\vert \frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert ^{t} \frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy \biggr)^{1/t}. \end{aligned}$$
Then
$$\begin{aligned} N_{2}\lesssim{}& l^{1/t'} \biggl\Vert \biggl\{ \biggl( \int_{\mathbb{R}}\chi _{\{t_{i+1}< \vert x-y \vert < s\}}(y) \biggl\vert \frac {R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert ^{t} \frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy \biggr)^{1/t}\biggr\} _{s\in J_{i}, i\in \mathbb{N}} \biggr\Vert _{E} \\ &{} +l^{1/t'} \biggl\Vert \biggl\{ \biggl( \int_{\mathbb{R}}\chi_{\{ t_{i+1}< \vert x_{0}-y \vert < s\}}(y) \biggl\vert \frac {R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert ^{t} \frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy \biggr)^{1/t}\biggr\} _{s\in J_{i}, i\in \mathbb{N}} \biggr\Vert _{E} \\ ={}& N_{21}+N_{22}. \end{aligned}$$
Notice
$$\begin{aligned} \bigl\vert R_{m+1}(\widetilde{b}_{k};x_{0},y) \bigr\vert \lesssim \bigl\Vert b^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} \bigl(2^{k}l\bigr)^{\beta} \vert x_{0}-y \vert ^{m}+ \bigl\vert b^{(m)}(y)-\bigl(b^{(m)} \bigr)_{\widetilde{I}_{k}} \bigr\vert \vert x_{0}-y \vert ^{m}. \end{aligned}$$
Choosing \(1< r< p\) with \(t=\sqrt{r}\), we have
$$\begin{aligned} N_{21}\lesssim{}& l^{1/t'}\biggl\{ \sum _{i\in\mathbb{N}}\sup_{s\in J_{i}} \biggl( \int_{\mathbb{R}}\chi_{\{t_{i+1}< \vert x-y \vert < s\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert ^{t} \frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy \biggr)^{2/t}\biggr\} ^{1/2} \\ \lesssim{}& l^{1/t'}\biggl\{ \sum_{i\in\mathbb{N}} \int_{\mathbb {R}}\chi_{\{t_{i+1}< \vert x-y \vert < t_{i}\}}(y) \biggl\vert \frac {R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert ^{t} \frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy\biggr\} ^{1/t} \\ \lesssim{}& l^{1/t'}\biggl\{ \int_{\mathbb{R}} \biggl\vert \frac {R_{m+1}({b};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert ^{t} \frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy\biggr\} ^{1/t} \\ \lesssim{}& l^{1/t'}\Biggl\{ \sum_{k=0}^{\infty}\int_{F_{k}} \biggl\vert \frac {R_{m+1}(\widetilde{b}_{k};x_{0},y)}{ \vert x_{0}-y \vert ^{m}} \biggr\vert ^{t} \frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy\Biggr\} ^{1/t} \\ \lesssim{}& \bigl\Vert {b}^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}}l^{1/t'} \Biggl\{ \sum_{k=0}^{\infty}\bigl(2^{k}l \bigr)^{\beta t} \int_{F_{k}}\frac { \vert f(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy\Biggr\} ^{1/t} \\ &{} +l^{1/t'}\Biggl\{ \sum_{k=0}^{\infty}\int_{F_{k}} \bigl( \bigl\vert b^{(m)}(y)- \bigl(b^{(m)}\bigr)_{\widetilde{I}_{k}} \bigr\vert \bigr)^{t} \frac{ \vert f(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy\Biggr\} ^{1/t}. \end{aligned}$$
But
$$\begin{aligned} & l^{1/t'}\Biggl\{ \sum_{k=0}^{\infty}\bigl(2^{k}l\bigr)^{\beta t} \int_{F_{k}}\frac { \vert f(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy\Biggr\} ^{1/t} \\ &\quad \lesssim l^{1/t'} \Biggl(\sum_{k=1}^{\infty}\frac{(2^{k}l)^{\beta t}}{(2^{k}\cdot4l)^{t}} \int_{ \vert x_{0}-y \vert < 2^{k+1}\cdot4l} \bigl\vert f(y) \bigr\vert ^{t}\,dy \Biggr)^{1/t} \\ &\quad \lesssim \Biggl(\sum_{k=1}^{\infty}\frac{1}{2^{k(t-1)}}\frac {(2^{k}l)^{\beta t}}{2^{k}\cdot5l} \int_{ \vert x_{0}-y \vert < 2^{k}\cdot 5l} \bigl\vert f(y) \bigr\vert ^{t}\,dy \Biggr)^{1/t} \\ &\quad \lesssim \Biggl(\sum_{k=1}^{\infty}\frac{1}{2^{k(t-1)}} \biggl(\frac {(2^{k}l)^{\beta t^{2}}}{2^{k}\cdot5l} \int_{ \vert x_{0}-y \vert < 2^{k}\cdot 5l} \bigl\vert f(y) \bigr\vert ^{t^{2}}\,dy \biggr)^{1/t} \Biggr)^{1/t} \\ &\quad \lesssim \Biggl(\sum_{k=1}^{\infty}\frac{1}{2^{k(t-1)}} \Biggr)^{1/t}M_{\beta,r}(f) (x_{0}) \lesssim M_{\beta,r}(f) (x_{0}) \end{aligned}$$
and
$$\begin{aligned} & l^{1/t'}\Biggl\{ \sum_{k=0}^{\infty}\int_{F_{k}} \bigl( \bigl\vert b^{(m)}(y)- \bigl(b^{(m)}\bigr)_{\widetilde{I}_{k}} \bigr\vert \bigr)^{t} \frac{ \vert f(y) \vert ^{t}}{ \vert x_{0}-y \vert ^{t}}\,dy\Biggr\} ^{1/t} \\ &\quad \lesssim \Biggl(\sum_{k=0}^{\infty}\frac{1}{2^{k(t-1)}}\frac {1}{2^{k}\cdot4l} \int_{ \vert x_{0}-y \vert < 2^{k}\cdot 4l} \bigl\vert b^{(m)}(y)- \bigl(b^{(m)}\bigr)_{\widetilde{I}_{k}} \bigr\vert ^{t} \bigl\vert f(y) \bigr\vert ^{t}\,dy \Biggr)^{1/t} \\ &\quad \lesssim \Biggl(\sum_{k=0}^{\infty}\frac{1}{2^{k(t-1)}} \biggl(\frac {1}{2^{k}\cdot4l} \int_{ \vert x_{0}-y \vert < 2^{k}\cdot4l} \bigl\vert f(y) \bigr\vert ^{t^{2}}\,dy \biggr)^{1/t} \\ &\qquad{}\times \biggl(\frac{1}{2^{k}\cdot4l} \int_{ \vert x_{0}-y \vert < 2^{k}\cdot 4l} \bigl\vert b^{(m)}(y)- \bigl(b^{(m)}\bigr)_{\widetilde{I}_{k}} \bigr\vert ^{tt'} \biggr)^{1/t'} \Biggr)^{1/t} \\ &\quad \lesssim \bigl\Vert b^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} \Biggl(\sum _{k=0}^{\infty}\frac{1}{2^{k(t-1)}} \biggl( \frac{(2^{k}\cdot4l)^{r\beta }}{2^{k}\cdot4l} \int_{ \vert x_{0}-y \vert < 2^{k}\cdot4l} \bigl\vert f(y) \bigr\vert ^{t^{2}}\,dy \biggr)^{1/t} \Biggr)^{1/t} \\ &\quad \lesssim \bigl\Vert b^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} M_{\beta ,r}(f) (x_{0}) \Biggl(\sum_{k=0}^{\infty}\frac{1}{2^{k(t-1)}} \Biggr)^{1/t} \\ &\quad \lesssim \bigl\Vert b^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} M_{\beta,r}(f) (x_{0}). \end{aligned}$$
Therefore
$$\begin{aligned} N_{21}\lesssim \bigl\Vert b^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} M_{\beta,r}(f) (x_{0}). \end{aligned}$$
Similarly,
$$\begin{aligned} N_{22}\lesssim \bigl\Vert b^{(m)} \bigr\Vert _{\dot{\wedge}_{\beta}} M_{\beta,r}(f) (x_{0}). \end{aligned}$$
This completes the proof of (3.1). Hence, Theorem 1.1 is proved.

Declarations

Authors’ contributions

The authors completed the paper together. They also read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
College of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo, 454003, China
(2)
Department of Mathematics, Jiaozuo University, Jiaozuo, 454003, China

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