Sharp boundedness for vector-valued multilinear integral operators

Abstract

In this paper, the sharp inequalities for some vector-valued multilinear integral operators are obtained. As applications, we get the weighted $L^p$ ($p>1$) norm inequalities and an $LlogL$-type estimate for the vector-valued multilinear operators.

MSC:42B20, 42B25.

1 Preliminaries and theorems

As the development of singular integral operators and their commutators, multilinear singular integral operators have been well studied (see [1–5]). In this paper, we study some vector-valued multilinear integral operators as follows.

Suppose that $m_j$ are positive integers ($j=1,\dots ,l$), $m_1+\cdots +m_l=m$ and $A_j$ are functions on $R^n$ ($j=1,\dots ,l$). Let $F_t(x,y)$ be defined on $R^n\times R^n\times [0,+\mathrm{\infty })$. Set

$$F_t(f)(x)={\int }_{R^n}{F}_t(x,y)f(y)dy$$

and

$$F_t^A(f)(x)={\int }_{R^n}\frac{{\prod }_{j=1}^l{R}_{m_j+1}(A_j;x,y)}{{|x-y|}^m}{F}_t(x,y)f(y)dy$$

for every bounded and compactly supported function f, where

$$R_{m_j+1}(A_j;x,y)=A_j(x)-\sum _{|\alpha |\le m_j}\frac{1}{\alpha !}{D}^{\alpha }{A}_j(y){(x-y)}^{\alpha }.$$

Let H be the Banach space $H=\{h:\parallel h\parallel <\mathrm{\infty }\}$ such that, for each fixed $x\in R^n$, $F_t(f)(x)$ and $F_t^A(f)(x)$ may be viewed as a mapping from $[0,+\mathrm{\infty })$ to H. For $1<s<\mathrm{\infty }$, the vector-valued multilinear operator related to $F_t$ is defined by

$${|T^A(f)(x)|}_s={\left(\sum _{i=1}^{\mathrm{\infty }}{(T^A(f_i)(x))}^s\right)}^{1/s},$$

where

$$T^A(f_i)(x)=\parallel F_t^A(f_i)(x)\parallel ,$$

and $F_t$ satisfies: for fixed $\epsilon >0$,

$$\parallel F_t(x,y)\parallel \le C{|x-y|}^{-n}$$

and

$$\parallel F_t(y,x)-F_t(z,x)\parallel \le C{|y-z|}^{\epsilon }{|x-z|}^{-n-\epsilon }$$

if $2|y-z|\le |x-z|$. Set

$${|T(f)(x)|}_s={\left(\sum _{i=1}^{\mathrm{\infty }}{|T(f_i)(x)|}^s\right)}^{1/s}\text{and}{|f|}_s={\left(\sum _{i=1}^{\mathrm{\infty }}{|f_i(x)|}^s\right)}^{1/s}.$$

Suppose that ${|T|}_s$ is bounded on $L^p(R^n)$ for $1<p<\mathrm{\infty }$ and weak $(L^1,L^1)$-bounded.

Note that when $m=0$, $T^A$ is just a vector-valued multilinear commutator of T and A (see [6]). While when $m>0$, $T^A$ is a non-trivial generalization of the commutator. It is well known that multilinear operators are of great interest in harmonic analysis and have been studied by many authors (see [1–5]). In [7], Hu and Yang proved a variant sharp estimate for the multilinear singular integral operators. In [6], Pérez and Trujillo-Gonzalez proved a sharp estimate for some multilinear commutator. The main purpose of this paper is to prove a sharp inequality for the vector-valued multilinear integral operators. As applications, we obtain the weighted $L^p$ ($p>1$) norm inequalities and an $LlogL$-type estimate for the vector-valued multilinear operators.

First, let us introduce some notations. Throughout this paper, Q will denote a cube of $R^n$ with sides parallel to the axes. For any locally integrable function f, the sharp function of f is defined by

$$f^{\mathrm{\#}}(x)=\underset{x\in Q}{sup}\frac{1}{|Q|}{\int }_Q|f(y)-f_Q|dy,$$

where, and in what follows, $f_Q={|Q|}^{-1}{\int }_{Q}f(x)dx$. It is well known that (see [8])

$$f^{\mathrm{\#}}(x)\approx \underset{x\in Q}{sup}\underset{c\in C}{inf}\frac{1}{|Q|}{\int }_Q|f(y)-c|dy.$$

We say that f belongs to $BMO(R^n)$ if $f^{\mathrm{\#}}$ belongs to $L^{\mathrm{\infty }}(R^n)$ and ${\parallel f\parallel }_{BMO}={\parallel f^{\mathrm{\#}}\parallel }_{L^{\mathrm{\infty }}}$. For $0<r<\mathrm{\infty }$, we denote $f_r^{\mathrm{\#}}$ by

$$f_r^{\mathrm{\#}}(x)={[{\left({|f|}^r\right)}^{\mathrm{\#}}(x)]}^{1/r}.$$

Let M be the Hardy-Littlewood maximal operator, that is, $M(f)(x)={sup}_{x\in Q}{|Q|}^{-1}{\int }_Q|f(y)|dy$. For $k\in N$, we denote by $M^k$ the operator M iterated k times, i.e., $M^1(f)(x)=M(f)(x)$ and $M^k(f)(x)=M(M^{k-1}(f))(x)$ for $k\ge 2$.

Let Φ be a Young function and $\tilde{\mathrm{\Phi }}$ be the complementary associated to Φ, we denote the Φ-average by, for a function f,

$${\parallel f\parallel }_{\mathrm{\Phi },Q}=inf\{\lambda >0:\frac{1}{|Q|}{\int }_Q\mathrm{\Phi }\left(\frac{|f(y)|}{\lambda }\right)dy\le 1\}$$

and the maximal function associated to Φ by

$$M_{\mathrm{\Phi }}(f)(x)=\underset{x\in Q}{sup}{\parallel f\parallel }_{\mathrm{\Phi },Q}.$$

The Young functions to be used in this paper are $\mathrm{\Phi }(t)=exp(t^r)-1$ and $\mathrm{\Psi }(t)=t{log}^r(t+e)$, the corresponding Φ-average and maximal functions are denoted by ${\parallel \cdot \parallel }_{exp{L}^r,Q}$, $M_{exp{L}^r}$ and ${\parallel \cdot \parallel }_{L{(logL)}^r,Q}$, $M_{L{(logL)}^r}$. We have the following inequality, for any $r>0$ and $m\in N$ (see [6])

$$M(f)\le M_{L{(logL)}^r}(f),M_{L{(logL)}^m}(f)\approx M^{m+1}(f).$$

For $r\ge 1$, we denote that

$${\parallel b\parallel }_{{osc}_{exp{L}^r}}=\underset{Q}{sup}{\parallel b-b_Q\parallel }_{exp{L}^r,Q},$$

the space ${Osc}_{exp{L}^r}$ is defined by

$${Osc}_{exp{L}^r}=\{b\in L_{log}^1\left(R^n\right):{\parallel b\parallel }_{{osc}_{exp{L}^r}}<\mathrm{\infty }\}.$$

It has been known that (see [6])

$${\parallel b-b_{2Q}\parallel }_{exp{L}^r,2^{k}Q}\le Ck{\parallel b\parallel }_{{Osc}_{exp{L}^r}}.$$

It is obvious that ${Osc}_{exp{L}^r}$ coincides with the $BMO$ space if $r=1$, and ${Osc}_{exp{L}^r}\subset BMO$ if $r>1$. We denote the Muckenhoupt weights by $A_p$ for $1\le p<\mathrm{\infty }$ (see [8]).

Now we state our main results as follows.

Theorem 1 Let $1<s<\mathrm{\infty }$, $r_j\ge 1$ and $D^{\alpha }{A}_j\in {Osc}_{exp{L}^{r_j}}$ for all α with $|\alpha |=m_j$ and $j=1,\dots ,l$. Define $1/r=1/r_1+\cdots +1/r_l$. Then, for any $0<p<1$, there exists a constant $C>0$ such that for any $f=\{f_i\}\in C_0^{\mathrm{\infty }}(R^n)$ and $x\in R^n$,

$${\left({|T^A(f)|}_s\right)}_p^{\mathrm{\#}}(x)\le C\prod _{j=1}^l\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(x).$$

Theorem 2 Let $1<s<\mathrm{\infty }$, $r_j\ge 1$ and $D^{\alpha }{A}_j\in {Osc}_{exp{L}^{r_j}}$ for all α with $|\alpha |=m_j$ and $j=1,\dots ,l$.

• (1) If $1<p<\mathrm{\infty }$ and $w\in A_p$, then

  $${\parallel {|T^A(f)|}_s\parallel }_{L^p(w)}\le C\prod _{j=1}^l\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right){\parallel {|f|}_s\parallel }_{L^p(w)};$$

• (2) If $w\in A_1$. Define $1/r=1/r_1+\cdots +1/r_l$ and $\mathrm{\Phi }(t)=t{log}^{1/r}(t+e)$. Then there exists a constant $C>0$ such that for all $\lambda >0$,

  $$\begin{array}{r}w\left(\{x\in R^n:{|T^A(f)(x)|}_s>\lambda \}\right)\\ \le C{\int }_{R^n}\mathrm{\Phi }\left[{\lambda }^{-1}\prod _{j=1}^l\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right){|f(x)|}_s\right]w(x)dx.\end{array}$$

Remark The conditions in Theorems 1 and 2 are satisfied by many operators.

Now we give some examples.

Example 1 Littlewood-Paley operators.

Fix $\epsilon >0$ and $\mu >(3n+2)/n$. Let ψ be a fixed function which satisfies the following properties:

• (1) ${\int }_{R^n}\psi (x)dx=0$,

• (2) $|\psi (x)|\le C{(1+|x|)}^{-(n+1)}$,

• (3) $|\psi (x+y)-\psi (x)|\le C{|y|}^{\epsilon }{(1+|x|)}^{-(n+1+\epsilon )}$ when $2|y|<|x|$.

We denote that $\mathrm{\Gamma }(x)=\{(y,t)\in R_{+}^{n+1}:|x-y|<t\}$ and the characteristic function of $\mathrm{\Gamma }(x)$ by ${\chi }_{\mathrm{\Gamma }(x)}$. The Littlewood-Paley multilinear operators are defined by

$$\begin{array}{c}{g}_{\psi }^A(f)(x)={\left({\int }_0^{\mathrm{\infty }}{|F_t^A(f)(x)|}^2\frac{dt}{t}\right)}^{1/2},\hfill \\ S_{\psi }^A(f)(x)={[\int {\int }_{\mathrm{\Gamma }(x)}{|F_t^A(f)(x,y)|}^2\frac{dydt}{t^{n+1}}]}^{1/2}\hfill \end{array}$$

and

$$g_{\mu }^A(f)(x)={[\int {\int }_{R_{+}^{n+1}}{\left(\frac{t}{t+|x-y|}\right)}^{n\mu }{|F_t^A(f)(x,y)|}^2\frac{dydt}{t^{n+1}}]}^{1/2},$$

where

$$\begin{array}{c}{F}_t^A(f)(x)={\int }_{R^n}\frac{{\prod }_{j=1}^l{R}_{m_j+1}(A_j;x,y)}{{|x-y|}^m}{\psi }_t(x-y)f(y)dy,\hfill \\ F_t^A(f)(x,y)={\int }_{R^n}\frac{{\prod }_{j=1}^l{R}_{m_j+1}(A_j;x,z)}{{|x-z|}^m}f(z){\psi }_t(y-z)dz\hfill \end{array}$$

and ${\psi }_t(x)=t^{-n}\psi (x/t)$ for $t>0$. Set $F_t(f)(y)=f\ast {\psi }_t(y)$. We also define that

$$\begin{array}{c}{g}_{\psi }(f)(x)={\left({\int }_0^{\mathrm{\infty }}{|F_t(f)(x)|}^2\frac{dt}{t}\right)}^{1/2},\hfill \\ S_{\psi }(f)(x)={(\int {\int }_{\mathrm{\Gamma }(x)}{|F_t(f)(y)|}^2\frac{dydt}{t^{n+1}})}^{1/2}\hfill \end{array}$$

and

$$g_{\mu }(f)(x)={(\int {\int }_{R_{+}^{n+1}}{\left(\frac{t}{t+|x-y|}\right)}^{n\mu }{|F_t(f)(y)|}^2\frac{dydt}{t^{n+1}})}^{1/2},$$

which are the Littlewood-Paley operators (see [9]). Let H be the space

$$H=\{h:\parallel h\parallel ={({\int }_0^{\mathrm{\infty }}{|h(t)|}^{2}dt/t)}^{1/2}<\mathrm{\infty }\}$$

or

$$H=\{h:\parallel h\parallel ={(\int {\int }_{R_{+}^{n+1}}{|h(y,t)|}^{2}dydt/t^{n+1})}^{1/2}<\mathrm{\infty }\}.$$

Then, for each fixed $x\in R^n$, $F_t^A(f)(x)$ and $F_t^A(f)(x,y)$ may be viewed as the mapping from $[0,+\mathrm{\infty })$ to H, and it is clear that

$$\begin{array}{c}{g}_{\psi }^A(f)(x)=\parallel F_t^A(f)(x)\parallel ,g_{\psi }(f)(x)=\parallel F_t(f)(x)\parallel ,\hfill \\ S_{\psi }^A(f)(x)=\parallel {\chi }_{\mathrm{\Gamma }(x)}{F}_t^A(f)(x,y)\parallel ,S_{\psi }(f)(x)=\parallel {\chi }_{\mathrm{\Gamma }(x)}{F}_t(f)(y)\parallel \hfill \end{array}$$

and

$$g_{\mu }^A(f)(x)=\parallel {\left(\frac{t}{t+|x-y|}\right)}^{n\mu /2}{F}_t^A(f)(x,y)\parallel ,g_{\mu }(f)(x)=\parallel {\left(\frac{t}{t+|x-y|}\right)}^{n\mu /2}{F}_t(f)(y)\parallel .$$

It is easy to see that $g_{\psi }$, $S_{\psi }$ and $g_{\mu }$ satisfy the conditions of Theorems 1 and 2 (see [10–12]), thus Theorems 1 and 2 hold for $g_{\psi }^A$, $S_{\psi }^A$ and $g_{\mu }^A$.

Example 2 Marcinkiewicz operators.

Fix $\lambda >max(1,2n/(n+2))$ and $0<\gamma \le 1$. Let Ω be homogeneous of degree zero on $R^n$ with ${\int }_{S^{n-1}}\mathrm{\Omega }(x^{\prime })d\sigma (x^{\prime })=0$. Assume that $\mathrm{\Omega }\in {Lip}_{\gamma }(S^{n-1})$. The Marcinkiewicz multilinear operators are defined by

$$\begin{array}{c}{\mu }_{\mathrm{\Omega }}^A(f)(x)={\left({\int }_0^{\mathrm{\infty }}{|F_t^A(f)(x)|}^2\frac{dt}{t^3}\right)}^{1/2},\hfill \\ {\mu }_S^A(f)(x)={[\int {\int }_{\mathrm{\Gamma }(x)}{|F_t^A(f)(x,y)|}^2\frac{dydt}{t^{n+3}}]}^{1/2}\hfill \end{array}$$

and

$${\mu }_{\lambda }^A(f)(x)={[\int {\int }_{R_{+}^{n+1}}{\left(\frac{t}{t+|x-y|}\right)}^{n\lambda }{|F_t^A(f)(x,y)|}^2\frac{dydt}{t^{n+3}}]}^{1/2},$$

where

$$F_t^A(f)(x)={\int }_{|x-y|\le t}\frac{{\prod }_{j=1}^l{R}_{m_j+1}(A_j;x,y)}{{|x-y|}^m}\frac{\mathrm{\Omega }(x-y)}{{|x-y|}^{n-1}}f(y)dy$$

and

$$F_t^A(f)(x,y)={\int }_{|y-z|\le t}\frac{{\prod }_{j=1}^l{R}_{m_j+1}(A_j;y,z)}{{|y-z|}^m}\frac{\mathrm{\Omega }(y-z)}{{|y-z|}^{n-1}}f(z)dz.$$

Set

$$F_t(f)(x)={\int }_{|x-y|\le t}\frac{\mathrm{\Omega }(x-y)}{{|x-y|}^{n-1}}f(y)dy.$$

We also define that

$$\begin{array}{c}{\mu }_{\mathrm{\Omega }}(f)(x)={\left({\int }_0^{\mathrm{\infty }}{|F_t(f)(x)|}^2\frac{dt}{t^3}\right)}^{1/2},\hfill \\ {\mu }_S(f)(x)={(\int {\int }_{\mathrm{\Gamma }(x)}{|F_t(f)(y)|}^2\frac{dydt}{t^{n+3}})}^{1/2}\hfill \end{array}$$

and

$${\mu }_{\lambda }(f)(x)={(\int {\int }_{R_{+}^{n+1}}{\left(\frac{t}{t+|x-y|}\right)}^{n\lambda }{|F_t(f)(y)|}^2\frac{dydt}{t^{n+3}})}^{1/2},$$

which are the Marcinkiewicz operators (see [13]). Let H be the space

$$H=\{h:\parallel h\parallel ={({\int }_0^{\mathrm{\infty }}{|h(t)|}^{2}dt/t^3)}^{1/2}<\mathrm{\infty }\}$$

or

$$H=\{h:\parallel h\parallel ={(\int {\int }_{R_{+}^{n+1}}{|h(y,t)|}^{2}dydt/t^{n+3})}^{1/2}<\mathrm{\infty }\}.$$

Then it is clear that

$$\begin{array}{c}{\mu }_{\mathrm{\Omega }}^A(f)(x)=\parallel F_t^A(f)(x)\parallel ,{\mu }_{\mathrm{\Omega }}(f)(x)=\parallel F_t(f)(x)\parallel ,\hfill \\ {\mu }_S^A(f)(x)=\parallel {\chi }_{\mathrm{\Gamma }(x)}{F}_t^A(f)(x,y)\parallel ,{\mu }_S(f)(x)=\parallel {\chi }_{\mathrm{\Gamma }(x)}{F}_t(f)(y)\parallel \hfill \end{array}$$

and

$${\mu }_{\lambda }^A(f)(x)=\parallel {\left(\frac{t}{t+|x-y|}\right)}^{n\lambda /2}{F}_t^A(f)(x,y)\parallel ,{\mu }_{\lambda }(f)(x)=\parallel {\left(\frac{t}{t+|x-y|}\right)}^{n\lambda /2}{F}_t(f)(y)\parallel .$$

It is easy to see that ${\mu }_{\mathrm{\Omega }}$, ${\mu }_S$ and ${\mu }_{\lambda }$ satisfy the conditions of Theorems 1 and 2 (see [13, 14]), thus Theorems 1 and 2 hold for ${\mu }_{\mathrm{\Omega }}^A$, ${\mu }_S^A$ and ${\mu }_{\lambda }^A$.

Example 3 Bochner-Riesz operators.

Let $\delta >(n-1)/2$, $B_t^{\delta }(\hat{f})(\xi )={(1-t^2{|\xi |}^2)}_{+}^{\delta }\hat{f}(\xi )$ and $B_t^{\delta }(z)=t^{-n}{B}^{\delta }(z/t)$ for $t>0$. Set

$$F_{\delta ,t}^A(f)(x)={\int }_{R^n}\frac{{\prod }_{j=1}^l{R}_{m_j+1}(A_j;x,y)}{{|x-y|}^m}{B}_t^{\delta }(x-y)f(y)dy.$$

The maximal Bochner-Riesz multilinear operators are defined by

$$B_{\delta ,\ast }^A(f)(x)=\underset{t>0}{sup}|B_{\delta ,t}^A(f)(x)|.$$

We also define that

$$B_{\delta ,\ast }(f)(x)=\underset{t>0}{sup}|B_t^{\delta }(f)(x)|,$$

which is the maximal Bochner-Riesz operator (see [15]). Let H be the space $H=\{h:\parallel h\parallel ={sup}_{t>0}|h(t)|<\mathrm{\infty }\}$, then

$$B_{\delta ,\ast }^A(f)(x)=\parallel B_{\delta ,t}^A(f)(x)\parallel ,B_{\ast }^{\delta }(f)(x)=\parallel B_t^{\delta }(f)(x)\parallel .$$

It is easy to see that $B_{\delta ,\ast }^A$ satisfies the conditions of Theorems 1 and 2 (see [16]), thus Theorems 1 and 2 hold for $B_{\delta ,\ast }^A$.

2 Some lemmas

We give some preliminary lemmas.

Lemma 1 ([3])

Let A be a function on $R^n$ and $D^{\alpha }A\in L^q(R^n)$ for all α with $|\alpha |=m$ and some $q>n$. Then

$$|R_m(A;x,y)|\le C{|x-y|}^m\sum _{|\alpha |=m}{\left(\frac{1}{|\tilde{Q}(x,y)|}{\int }_{\tilde{Q}(x,y)}{|D^{\alpha }A(z)|}^{q}dz\right)}^{1/q},$$

where $\tilde{Q}$ is the cube centered at x and having side length $5\sqrt{n}|x-y|$.

Lemma 2 ([[8], p.485])

Let $0<p<q<\mathrm{\infty }$ and for any function $f\ge 0$, we define that, for $1/r=1/p-1/q$,

$${\parallel f\parallel }_{W{L}^q}=\underset{\lambda >0}{sup}\lambda {\left|\{x\in R^n:f(x)>\lambda \}\right|}^{1/q},N_{p,q}(f)=\underset{E}{sup}{\parallel f{\chi }_E\parallel }_{L^p}/{\parallel {\chi }_E\parallel }_{L^r},$$

where the sup is taken for all measurable sets E with $0<|E|<\mathrm{\infty }$. Then

$${\parallel f\parallel }_{W{L}^q}\le N_{p,q}(f)\le {(q/(q-p))}^{1/p}{\parallel f\parallel }_{W{L}^q}.$$

Lemma 3 ([6])

Let $r_j\ge 1$ for $j=1,\dots ,m$, we denote that $1/r=1/r_1+\cdots +1/r_m$. Then

$$\frac{1}{|Q|}{\int }_Q|f_1(x)\cdots f_m(x)g(x)|dx\le {\parallel f\parallel }_{exp{L}^{r_1},Q}\cdots {\parallel f\parallel }_{exp{L}^{r_m},Q}{\parallel g\parallel }_{L{(logL)}^{1/r},Q}.$$

3 Proof of the theorem

It is only to prove Theorem 1.

Proof of Theorem 1 It suffices to prove for $f\in C_0^{\mathrm{\infty }}(R^n)$ and some constant $C_0$ that the following inequality holds:

$${\left(\frac{1}{|Q|}{\int }_Q{|{|T^A(f)(x)|}_s-C_0|}^{p}dx\right)}^{1/p}\le C\prod _{j=1}^l\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(x).$$

Without loss of generality, we may assume $l=2$. Fix a cube $Q=Q(x_0,d)$ and $\tilde{x}\in Q$. Let $\tilde{Q}=5\sqrt{n}Q$ and ${\tilde{A}}_j(x)=A_j(x)-{\sum }_{|\alpha |=m}\frac{1}{\alpha !}{(D^{\alpha }{A}_j)}_{\tilde{Q}}{x}^{\alpha }$, then $R_m(A_j;x,y)=R_m({\tilde{A}}_j;x,y)$ and $D^{\alpha }{\tilde{A}}_j=D^{\alpha }{A}_j-{(D^{\alpha }{A}_j)}_{\tilde{Q}}$ for $|\alpha |=m_j$. We split $f=g+h=\{g_i\}+\{h_i\}$ for $g_i=f_i{\chi }_{\tilde{Q}}$ and $h_i=f_i{\chi }_{R^n\setminus \tilde{Q}}$. Write

$$\begin{array}{rcl}{F}_t^A(f_i)(x)& =& {\int }_{R^n}\frac{{\prod }_{j=1}^2{R}_{m_j+1}({\tilde{A}}_j;x,y)}{{|x-y|}^m}{F}_t(x,y)f_i(y)dy\\ =& {\int }_{R^n}\frac{{\prod }_{j=1}^2{R}_{m_j+1}({\tilde{A}}_j;x,y)}{{|x-y|}^m}{F}_t(x,y)h_i(y)dy+{\int }_{R^n}\frac{{\prod }_{j=1}^2{R}_{m_j}({\tilde{A}}_j;x,y)}{{|x-y|}^m}{F}_t(x,y)g_i(y)dy\\ -\sum _{|{\alpha }_1|=m_1}\frac{1}{{\alpha }_1!}{\int }_{R^n}\frac{R_{m_2}({\tilde{A}}_2;x,y){(x-y)}^{{\alpha }_1}}{{|x-y|}^m}{D}^{{\alpha }_1}{\tilde{A}}_1(y)F_t(x,y)g_i(y)dy\\ -\sum _{|{\alpha }_2|=m_2}\frac{1}{{\alpha }_2!}{\int }_{R^n}\frac{R_{m_1}({\tilde{A}}_1;x,y){(x-y)}^{{\alpha }_2}}{{|x-y|}^m}{D}^{{\alpha }_2}{\tilde{A}}_2(y)F_t(x,y)g_i(y)dy\\ +\sum _{|{\alpha }_1|=m_1,|{\alpha }_2|=m_2}\frac{1}{{\alpha }_1!{\alpha }_2!}{\int }_{R^n}\frac{{(x-y)}^{{\alpha }_1+{\alpha }_2}{D}^{{\alpha }_1}{\tilde{A}}_1(y)D^{{\alpha }_2}{\tilde{A}}_2(y)}{{|x-y|}^m}{F}_t(x,y)g_i(y)dy.\end{array}$$

Then, by Minkowski’s inequality, we have

$$\begin{array}{r}{\left[\frac{1}{|Q|}{\int }_Q\right|{|T^A(f)(x)|}_s-{|T^{\tilde{A}}(h)(x_0)|}_s{|}^{p}dx]}^{1/p}\\ \le {\left[\frac{1}{|Q|}{\int }_Q\right|{\parallel F_t^A(f)(x)\parallel }_s-{\parallel F_t^{\tilde{A}}(h)(x_0)\parallel }_s{|}^{p}dx]}^{1/p}\\ \le {\left[\frac{1}{|Q|}{\int }_Q{\left(\sum _{i=1}^{\mathrm{\infty }}{\parallel F_t^A(f_i)(x)-F_t^{\tilde{A}}(h_i)(x_0)\parallel }^s\right)}^{p/s}dx\right]}^{1/p}\\ \le {\left[\frac{C}{|Q|}{\int }_Q{\left(\sum _{i=1}^{\mathrm{\infty }}{\parallel {\int }_{R^n}\frac{{\prod }_{j=1}^2{R}_{m_j}({\tilde{A}}_j;x,y)}{{|x-y|}^m}{F}_t(x,y)g_i(y)dy\parallel }^s\right)}^{p/s}dx\right]}^{1/p}\\ +[\frac{C}{|Q|}{\int }_Q(\sum _{i=1}^{\mathrm{\infty }}\parallel \sum _{|{\alpha }_1|=m_1}\frac{1}{{\alpha }_1!}\\ {{\times {\int }_{R^n}\frac{R_{m_2}({\tilde{A}}_2;x,y){(x-y)}^{{\alpha }_1}}{{|x-y|}^m}{D}^{{\alpha }_1}{\tilde{A}}_1(y)F_t(x,y)g_i(y)dy{\parallel }^s)}^{p/s}dx]}^{1/p}\\ +[\frac{C}{|Q|}{\int }_Q(\sum _{i=1}^{\mathrm{\infty }}\parallel \sum _{|{\alpha }_2|=m_2}\frac{1}{{\alpha }_2!}\\ {{\times {\int }_{R^n}\frac{R_{m_1}({\tilde{A}}_1;x,y){(x-y)}^{{\alpha }_2}}{{|x-y|}^m}{D}^{{\alpha }_2}{\tilde{A}}_2(y)F_t(x,y)g_i(y)dy{\parallel }^s)}^{p/s}dx]}^{1/p}\\ +[\frac{C}{|Q|}{\int }_Q(\sum _{i=1}^{\mathrm{\infty }}\parallel \sum _{|{\alpha }_1|=m_1,|{\alpha }_2|=m_2}\frac{1}{{\alpha }_1!{\alpha }_2!}\\ {{\times {\int }_{R^n}\frac{{(x-y)}^{{\alpha }_1+{\alpha }_2}{D}^{{\alpha }_1}{\tilde{A}}_1(y)D^{{\alpha }_2}{\tilde{A}}_2(y)}{{|x-y|}^m}{F}_t(x,y)g_i(y)dy{\parallel }^s)}^{p/s}dx]}^{1/p}\\ +{\left[\frac{C}{|Q|}{\int }_Q{\left(\sum _{i=1}^{\mathrm{\infty }}{\parallel F_t^{\tilde{A}}(h_i)(x)-F_t^{\tilde{A}}(h_i)(x_0)\parallel }^s\right)}^{p/s}dx\right]}^{1/p}\\ :=I_1+I_2+I_3+I_4+I_5.\end{array}$$

Now, let us estimate $I_1$, $I_2$, $I_3$, $I_4$ and $I_5$, respectively. First, for $x\in Q$ and $y\in \tilde{Q}$, by Lemma 1, we get

$$R_{m_j}({\tilde{A}}_j;x,y)\le C{|x-y|}^{m_j}\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}.$$

Thus, by Lemma 2 and the weak type $(1,1)$ of ${|T|}_s$, we obtain

$$\begin{array}{rcl}{I}_1& \le & C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right){\left(\frac{1}{|Q|}{\int }_Q{|T(g)(x)|}_s^{p}dx\right)}^{1/p}\\ =& C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right){|Q|}^{-1}\frac{{\parallel {|T(g)|}_s{\chi }_Q\parallel }_{L^p}}{{|Q|}^{1/p-1}}\\ \le & C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right){|Q|}^{-1}{\parallel {|T(g)|}_s\parallel }_{W{L}^1}\\ \le & C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right){|Q|}^{-1}{\parallel {|g|}_s\parallel }_{L^1}\\ \le & C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M\left({|f|}_s\right)(\tilde{x})\\ \le & C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).\end{array}$$

For $I_2$, note that ${\parallel {\chi }_Q\parallel }_{exp{L}^{r_2},Q}\le C$, similar to the proof of $I_1$ and by using Lemma 3, we get

$$\begin{array}{rcl}{I}_2& \le & C\sum _{|{\alpha }_2|=m_2}{\parallel D^{{\alpha }_2}{A}_2\parallel }_{{Osc}_{exp{L}^{r_2}}}\sum _{|{\alpha }_1|=m_1}{\left(\frac{1}{|Q|}{\int }_{R^n}{|T\left(D^{{\alpha }_1}{\tilde{A}}_{1}g\right)(x)|}_s^{p}dx\right)}^{1/p}\\ \le & C\sum _{|{\alpha }_2|=m_2}{\parallel D^{{\alpha }_2}{A}_2\parallel }_{{Osc}_{exp{L}^{r_2}}}\sum _{|{\alpha }_1|=m_1}{|Q|}^{-1}{\parallel {|T\left(D^{{\alpha }_1}{\tilde{A}}_{1}g\right)(x)|}_s{\chi }_Q\parallel }_{W{L}^1}\\ \le & C\sum _{|{\alpha }_2|=m_2}{\parallel D^{{\alpha }_2}{A}_2\parallel }_{{Osc}_{exp{L}^{r_2}}}\sum _{|{\alpha }_1|=m_1}\frac{1}{|Q|}{\int }_{R^n}|D^{{\alpha }_1}{\tilde{A}}_1(x)|{|g(x)|}_{s}dx\\ \le & C\sum _{|{\alpha }_2|=m_2}{\parallel D^{{\alpha }_2}{A}_2\parallel }_{{Osc}_{exp{L}^{r_2}}}{\parallel {\chi }_Q\parallel }_{exp{L}^{r_2},Q}\\ \times \sum _{|{\alpha }_1|=m_1}{\parallel D^{{\alpha }_1}{A}_1-{\left(D^{{\alpha }_1}{A}_1\right)}_{\tilde{Q}}\parallel }_{exp{L}^{r_1},\tilde{Q}}{\parallel {|f|}_s\parallel }_{L{(logL)}^{1/r},\tilde{Q}}\\ \le & C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).\end{array}$$

For $I_3$, similar to the proof of $I_2$, we get

$$I_3\le C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).$$

Similarly, for $I_4$, by using Lemma 3, we get

$$\begin{array}{rcl}{I}_4& \le & C\sum _{|{\alpha }_1|=m_1,|{\alpha }_2|=m_2}{\left(\frac{1}{|Q|}{\int }_{R^n}{|T\left(D^{{\alpha }_1}{\tilde{A}}_1{D}^{{\alpha }_2}{\tilde{A}}_2{f}_1\right)(x)|}_s^{p}dx\right)}^{1/p}\\ \le & C\sum _{|{\alpha }_1|=m_1,|{\alpha }_2|=m_2}{|Q|}^{-1}{\parallel {\left|T\left(D^{{\alpha }_1}{\tilde{A}}_1{D}^{{\alpha }_2}{\tilde{A}}_{2}g\right)\right|}_s{\chi }_Q\parallel }_{W{L}^1}\\ \le & C\sum _{|{\alpha }_1|=m_1,|{\alpha }_2|=m_2}\frac{1}{|Q|}{\int }_{R^n}|D^{{\alpha }_1}{\tilde{A}}_1(x)D^{{\alpha }_2}{\tilde{A}}_2(x)|{|g(x)|}_{s}dx\\ \le & C\sum _{|{\alpha }_1|=m_1}{\parallel D^{{\alpha }_1}{A}_1-{\left(D^{{\alpha }_1}{A}_1\right)}_{\tilde{Q}}\parallel }_{exp{L}^{r_1},\tilde{Q}}\\ \times \sum _{|{\alpha }_2|=m_2}{\parallel D^{{\alpha }_2}{A}_2-{\left(D^{{\alpha }_2}{A}_2\right)}_{\tilde{Q}}\parallel }_{exp{L}^{r_2},\tilde{Q}}{\parallel {|f|}_s\parallel }_{L{(logL)}^{1/r},\tilde{Q}}\\ \le & C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).\end{array}$$

For $I_5$, we write

$$\begin{array}{r}{F}_t^{\tilde{A}}(h_i)(x)-F_t^{\tilde{A}}(h_i)(x_0)\\ ={\int }_{R^n}(\frac{F_t(x,y)}{{|x-y|}^m}-\frac{F_t(x_0,y)}{{|x_0-y|}^m})\prod _{j=1}^2{R}_{m_j}({\tilde{A}}_j;x,y)h_i(y)dy\\ +{\int }_{R^n}(R_{m_1}({\tilde{A}}_1;x,y)-R_{m_1}({\tilde{A}}_1;x_0,y))\frac{R_{m_2}({\tilde{A}}_2;x,y)}{{|x_0-y|}^m}{F}_t(x_0,y)h_i(y)dy\\ +{\int }_{R^n}(R_{m_2}({\tilde{A}}_2;x,y)-R_{m_2}({\tilde{A}}_2;x_0,y))\frac{R_{m_1}({\tilde{A}}_1;x_0,y)}{{|x_0-y|}^m}{F}_t(x_0,y)h_i(y)dy\\ -\sum _{|{\alpha }_1|=m_1}\frac{1}{{\alpha }_1!}{\int }_{R^n}[\frac{R_{m_2}({\tilde{A}}_2;x,y){(x-y)}^{{\alpha }_1}}{{|x-y|}^m}{F}_t(x,y)-\frac{R_{m_2}({\tilde{A}}_2;x_0,y){(x_0-y)}^{{\alpha }_1}}{{|x_0-y|}^m}{F}_t(x_0,y)]\\ \times D^{{\alpha }_1}{\tilde{A}}_1(y)h_i(y)dy\\ -\sum _{|{\alpha }_2|=m_2}\frac{1}{{\alpha }_2!}{\int }_{R^n}[\frac{R_{m_1}({\tilde{A}}_1;x,y){(x-y)}^{{\alpha }_2}}{{|x-y|}^m}{F}_t(x,y)-\frac{R_{m_1}({\tilde{A}}_1;x_0,y){(x_0-y)}^{{\alpha }_2}}{{|x_0-y|}^m}{F}_t(x_0,y)]\\ \times D^{{\alpha }_2}{\tilde{A}}_2(y)h_i(y)dy\\ +\sum _{|{\alpha }_1|=m_1,|{\alpha }_2|=m_2}\frac{1}{{\alpha }_1!{\alpha }_2!}{\int }_{R^n}[\frac{{(x-y)}^{{\alpha }_1+{\alpha }_2}}{{|x-y|}^m}{F}_t(x,y)-\frac{{(x_0-y)}^{{\alpha }_1+{\alpha }_2}}{{|x_0-y|}^m}{F}_t(x_0,y)]\\ \times D^{{\alpha }_1}{\tilde{A}}_1(y)D^{{\alpha }_2}{\tilde{A}}_2(y)h_i(y)dy\\ =I_5^{(1)}+I_5^{(2)}+I_5^{(3)}+I_5^{(4)}+I_5^{(5)}+I_5^{(6)}.\end{array}$$

By Lemma 1, we know that, for $x\in Q$ and $y\in 2^{k+1}\tilde{Q}\setminus 2^k\tilde{Q}$,

$$\begin{array}{rcl}|R_{m_j}({\tilde{A}}_j;x,y)|& \le & C{|x-y|}^{m_j}\sum _{|{\alpha }_j|=m_j}({\parallel D^{{\alpha }_j}A\parallel }_{{Osc}_{exp{L}^{r_j}}}+|{\left(D^{{\alpha }_j}A\right)}_{\tilde{Q}(x,y)}-{\left(D^{{\alpha }_j}A\right)}_{\tilde{Q}}|)\\ \le & Ck{|x-y|}^{m_j}\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}A\parallel }_{{Osc}_{exp{L}^{r_j}}}.\end{array}$$

Note that $|x-y|\sim |x_0-y|$ for $x\in Q$ and $y\in R^n\setminus \tilde{Q}$, we obtain, by the condition of $F_t$,

$$\begin{array}{rcl}\parallel I_5^{(1)}\parallel & \le & C{\int }_{R^n}(\frac{|x-x_0|}{{|x_0-y|}^{m+n+1}}+\frac{{|x-x_0|}^{\epsilon }}{{|x_0-y|}^{m+n+\epsilon }})\prod _{j=1}^2{R}_{m_j}({\tilde{A}}_j;x,y)|h_i(y)|dy\\ \le & C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)\\ \times \sum _{k=0}^{\mathrm{\infty }}{\int }_{2^{k+1}\tilde{Q}\setminus 2^k\tilde{Q}}{k}^2(\frac{|x-x_0|}{{|x_0-y|}^{n+1}}+\frac{{|x-x_0|}^{\epsilon }}{{|x_0-y|}^{n+\epsilon }})|f_i(y)|dy\\ \le & C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)\sum _{k=1}^{\mathrm{\infty }}{k}^2(2^{-k}+2^{-\epsilon k})\frac{1}{|2^k\tilde{Q}|}{\int }_{2^k\tilde{Q}}|f_i(y)|dy.\end{array}$$

Thus, by Minkowski’s inequality, we get

$$\begin{array}{rcl}{\left(\sum _{i=1}^{\mathrm{\infty }}{\parallel I_5^{(1)}\parallel }^s\right)}^{1/s}& \le & C\prod _{j=1}^2\left(\sum _{|\alpha |=m_j}{\parallel D^{\alpha }{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)\\ \times \sum _{k=1}^{\mathrm{\infty }}{k}^2(2^{-k}+2^{-\epsilon k})\frac{1}{|2^k\tilde{Q}|}{\int }_{2^k\tilde{Q}}{|f(y)|}_{s}dy\\ \le & C\prod _{j=1}^2\left(\sum _{|\alpha |=m_j}{\parallel D^{\alpha }{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M\left({|f|}_s\right)(\tilde{x})\\ \le & C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).\end{array}$$

For $I_5^{(2)}$, by the formula (see [3])

$$R_{m_j}(\tilde{A};x,y)-R_{m_j}(\tilde{A};x_0,y)=\sum _{|\beta |<m_j}\frac{1}{\beta !}{R}_{m_j-|\beta |}(D^{\beta }\tilde{A};x,x_0){(x-y)}^{\beta }$$

and Lemma 1, we have

$$|R_{m_j}(\tilde{A};x,y)-R_{m_j}(\tilde{A};x_0,y)|\le C\sum _{|\beta |<m_j}\sum _{|\alpha |=m_j}{|x-x_0|}^{m_j-|\beta |}{|x-y|}^{|\beta |}{\parallel D^{\alpha }A\parallel }_{{Osc}_{exp{L}^{r_j}}},$$

thus

$$\begin{array}{rl}{\left(\sum _{i=1}^{\mathrm{\infty }}{\parallel I_5^{(2)}\parallel }^s\right)}^{1/s}& \le C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)\sum _{k=0}^{\mathrm{\infty }}{\int }_{2^{k+1}\tilde{Q}\setminus 2^k\tilde{Q}}k\frac{|x-x_0|}{{|x_0-y|}^{n+1}}{|f(y)|}_{s}dy\\ \le C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).\end{array}$$

Similarly,

$${\left(\sum _{i=1}^{\mathrm{\infty }}{\parallel I_5^{(3)}\parallel }^s\right)}^{1/s}\le C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).$$

For $I_5^{(4)}$, similar to the proof of $I_5^{(1)}$, $I_5^{(2)}$ and $I_2$, we get

$$\begin{array}{r}{\left(\sum _{i=1}^{\mathrm{\infty }}{\parallel I_5^{(4)}\parallel }^s\right)}^{1/s}\\ \le C\sum _{|{\alpha }_1|=m_1}{\int }_{R^n\setminus \tilde{Q}}\parallel \frac{{(x-y)}^{{\alpha }_1}{F}_t(x,y)}{{|x-y|}^m}-\frac{{(x_0-y)}^{{\alpha }_1}{F}_t(x_0,y)}{{|x_0-y|}^m}\parallel \\ \times |R_{m_2}({\tilde{A}}_2;x,y)||D^{{\alpha }_1}{\tilde{A}}_1(y)|{|f(y)|}_{s}dy\\ +C\sum _{|{\alpha }_1|=m_1}{\int }_{R^n\setminus \tilde{Q}}|R_{m_2}({\tilde{A}}_2;x,y)-R_{m_2}({\tilde{A}}_2;x_0,y)|\\ \times \frac{\parallel {(x_0-y)}^{{\alpha }_1}{F}_t(x_0,y)\parallel }{{|x_0-y|}^m}|D^{{\alpha }_1}{\tilde{A}}_1(y)|{|f(y)|}_{s}dy\\ \le C\sum _{|{\alpha }_2|=m_2}{\parallel D^{{\alpha }_2}{A}_2\parallel }_{{Osc}_{exp{L}^{r_2}}}\\ \times \sum _{|{\alpha }_1|=m_1}\sum _{k=1}^{\mathrm{\infty }}k(2^{-k}+2^{-\epsilon k})\frac{1}{|2^k\tilde{Q}|}{\int }_{2^k\tilde{Q}}|D^{{\alpha }_1}{\tilde{A}}_1(y)|{|f(y)|}_{s}dy\\ \le C\sum _{|{\alpha }_2|=m_2}{\parallel D^{{\alpha }_2}{A}_2\parallel }_{{Osc}_{exp{L}^{r_2}}}\sum _{|{\alpha }_1|=m_1}\sum _{k=1}^{\mathrm{\infty }}k(2^{-k}+2^{-\epsilon k})\\ \times {\parallel D^{{\alpha }_1}{A}_1-{\left(D^{{\alpha }_1}{A}_1\right)}_{\tilde{Q}}\parallel }_{exp{L}^{r_1},2^k\tilde{Q}}{\parallel {|f|}_s\parallel }_{L{(logL)}^{1/r},2^k\tilde{Q}}\\ \le C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).\end{array}$$

Similarly,

$${\left(\sum _{i=1}^{\mathrm{\infty }}{\parallel I_5^{(5)}\parallel }^s\right)}^{1/s}\le C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).$$

For $I_5^{(6)}$, by using Lemma 3, we obtain

$$\begin{array}{r}{\left(\sum _{i=1}^{\mathrm{\infty }}{\parallel I_5^{(6)}\parallel }^s\right)}^{1/s}\\ \le C\sum _{|{\alpha }_1|=m_1,|{\alpha }_2|=m_2}{\int }_{R^n\setminus \tilde{Q}}\parallel \frac{{(x-y)}^{{\alpha }_1+{\alpha }_2}{F}_t(x,y)}{{|x-y|}^m}-\frac{{(x_0-y)}^{{\alpha }_1+{\alpha }_2}{F}_t(x_0,y)}{{|x_0-y|}^m}\parallel \\ \times |D^{{\alpha }_1}{\tilde{A}}_1(y)||D^{{\alpha }_2}{\tilde{A}}_2(y)|{|f(y)|}_{s}dy\\ \le C\sum _{|{\alpha }_1|=m_1,|{\alpha }_2|=m_2}\sum _{k=1}^{\mathrm{\infty }}(2^{-k}+2^{-\epsilon k})\frac{1}{|2^k\tilde{Q}|}{\int }_{2^k\tilde{Q}}|D^{{\alpha }_1}{\tilde{A}}_1(y)||D^{{\alpha }_2}{\tilde{A}}_2(y)|{|f(y)|}_{s}dy\\ \le C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).\end{array}$$

Thus

$$I_5\le C\prod _{j=1}^2\left(\sum _{|{\alpha }_j|=m_j}{\parallel D^{{\alpha }_j}{A}_j\parallel }_{{Osc}_{exp{L}^{r_j}}}\right)M_{L{(logL)}^{1/r}}\left({|f|}_s\right)(\tilde{x}).$$

This completes the proof of Theorem 1. □

By Theorem 1 and the $L^p$-boundedness of $M_{L{(logL)}^{1/r}}$, we may obtain the conclusions (1), (2) of Theorem 2.