We will only need to prove (
5), since (
6) can be obtained by a similar argument. To prove (
5), it suffices to prove that for any fixed
\(x_{0}\in\mathbb{R}^{n}\) and some
\({F_{\rho}}\)-valued constant
\(c_{1}\), such that for every ball
\(B=B(x_{0},l)\) with radius
l, centered at
\(x_{0}\), the following inequality
$$\begin{aligned} \frac{1}{ \vert B \vert } \int_{B} \bigl\Vert V\bigl(T^{b} \bigr)f(x)-c_{1} \bigr\Vert _{E}\,dx \lesssim\sum _{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}{b} \bigr\Vert _{*} M_{s}(f) (x_{0}) \end{aligned}$$
holds.
We write
\(f=f_{1}+f_{2}=f\chi_{10B}+f\chi_{\mathbb{R}^{n}\setminus10B}\). Then
$$\begin{aligned} V\bigl(T^{b}\bigr)f(x) ={}&\biggl\{ \int_{\{\epsilon_{i+1}< \vert x-y \vert < \epsilon_{i}\}}\frac {R_{m+1}({b};x,y)}{ \vert x-y \vert ^{m}}K(x-y)f(y)\,dy\biggr\} _{\beta=\{\epsilon_{i}\} \in\Theta} \\ ={}&V(T) \biggl(\frac{R_{m+1}({b};x,\cdot)}{ \vert x-\cdot \vert ^{m}}f_{1} \biggr) (x)+V \bigl(T^{b}\bigr) (f_{2}) (x). \end{aligned}$$
Let
\(x_{1}\) be a point at the boundary of 2
B, and let
$$\begin{aligned} c_{1}=\biggl\{ \int_{\{\epsilon_{i+1}< \vert x_{1}-y \vert < \epsilon_{i}\}}\frac {R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}}K(x_{1}-y)f_{2}(y)\,dy \biggr\} _{\beta=\{ \epsilon_{i}\}\in\Theta}=V(T) (f_{2}) (x_{1}). \end{aligned}$$
Then
$$\begin{aligned} &\frac{1}{ \vert B \vert } \int_{B} \bigl\Vert V\bigl(T^{b} \bigr)f(x)-c_{1} \bigr\Vert _{F_{\rho}}\,dx \\ &\quad\leq\frac{1}{ \vert B \vert } \int_{B} \biggl\Vert V(T) \biggl(\frac{R_{m+1}({b};x,\cdot )}{ \vert x-\cdot \vert ^{m}}f_{1} \biggr) (x) \biggr\Vert _{F_{\rho}}\,dx \\ &\qquad{}+\frac{1}{ \vert B \vert } \int_{B} \bigl\Vert V\bigl(T^{b}\bigr) (f_{2}) (x)-V\bigl(T^{b}\bigr) (f_{2}) (x_{1}) \bigr\Vert _{F_{\rho}}\,dx \\ &\quad=M_{1}+M_{2}. \end{aligned}$$
For any
\(x\in B, k\in\mathbb{Z}\), let
\(E_{k}=\{y:2^{k}\cdot3l\leq \vert y-x \vert <2^{k+1}\cdot3l\}\), let
\(F_{k}=\{y: \vert y-x \vert <2^{k+1}\cdot3l\}\), and let
$$\begin{aligned} {b}_{k}(z)=b(z)-\sum_{ \vert \alpha \vert =m} \frac{1}{\alpha!}\bigl(D^{\alpha}b\bigr)_{F_{k}}z^{m}. \end{aligned}$$
By [
11] we have
\(R_{m+1}({b};x,y)=R_{m+1}(b_{k};x,y)\) for any
\(y\in E_{k}\). From Lemma
3, we know
\(V_{\rho}(T)\) is bounded on
\(L^{u}(\mathbb{R}^{n})\) for
\(\max\{r',p_{0}\}< u< s\). Then, using Hölder’s inequality, we deduce
$$\begin{aligned} M_{1}\lesssim{}& \biggl(\frac{1}{ \vert B \vert } \int_{B} \biggl\Vert V(T) \biggl(\frac{R_{m+1}({b};x,\cdot )}{ \vert x-\cdot \vert ^{m}}f_{1} \biggr) \biggr\Vert ^{u}_{F_{\rho}}\,dx\biggr)^{1/u} \\ \lesssim{}& \biggl(\frac{1}{ \vert B \vert } \int_{\{y: \vert y-x \vert < 11l\}}\biggl\vert \frac{R_{m+1}({b};\cdot ,y)}{ \vert y-\cdot \vert ^{m}}f(y) \biggr\vert ^{u}\,dy \biggr)^{1/u} \\ ={}& \Biggl(\frac{1}{ \vert B \vert } \int_{\{y: \vert y-x \vert < 11l\}} \Biggl\vert \sum_{k=-\infty }^{1} \biggl(\frac{R_{m+1}({b}_{k};\cdot,y)}{ \vert y-\cdot \vert ^{m}}\chi_{E_{k}}(y) \biggr)f(y) \Biggr\vert ^{u}\,dy\Biggr)^{1/u} \\ \lesssim{}& \Biggl(\frac{1}{ \vert B \vert } \int_{\{y: \vert y-x \vert < 11l\}} \Biggl\vert \Biggl(\sum_{k=-\infty }^{1} \frac{R_{m}({b_{k}};\cdot,y)}{ \vert y-\cdot \vert ^{m}}\chi_{E_{k}}(y) \\ &{}-\frac{1}{ \vert y-\cdot \vert ^{m}}\sum_{ \vert \alpha \vert =m}\frac{1}{\alpha!} \sum_{k=-\infty}^{1}D^{\alpha}b_{k}(y)\chi_{E_{k}}(y) (y-\cdot)^{\alpha}\Biggr)f(y) \Biggr\vert ^{u}\,dy\Biggr)^{1/u} \\ \lesssim{}& \Biggl(\frac{1}{ \vert B \vert }\sum_{k=-\infty}^{1} \int_{E_{k}}\biggl\vert \frac {R_{m}({b_{k}};\cdot,y)}{ \vert y-\cdot \vert ^{m}}f(y) \biggr\vert ^{u}\,dy \Biggr)^{1/u} \\ &{}+\sum_{ \vert \alpha \vert =m}\Biggl(\frac{1}{ \vert B \vert } \int_{\{y: \vert y-x \vert < 11l\}} \Biggl(\Biggl(\sum_{k=-\infty}^{1} \bigl\vert D^{\alpha}b_{k}(y) \bigr\vert \chi_{E_{k}}(y)\Biggr) \bigl\vert f(y)\bigr\vert \Biggr)^{u}\,dy\Biggr)^{1/u} \\ = {}&M_{11}+M_{12}. \end{aligned}$$
For any
\(y\in E_{k}\), by Lemma
2 and Lemma
1,
$$\begin{aligned} \bigl\vert R_{m}({b}_{k};x,y) \bigr\vert & \lesssim \vert x-y \vert ^{m}\sum_{ \vert \alpha \vert =m}\biggl( \frac {1}{ \vert Q(x,y) \vert } \int_{Q(x,y)} \bigl\vert D^{\alpha}b(z) \bigr\vert ^{q}\,dz\biggr)^{1/q} \\ &\lesssim \vert x-y \vert ^{m}\sum_{ \vert \alpha \vert =m} \biggl(\frac{1}{(\sqrt{n}2^{k}\cdot 30l)^{n}} \int_{ \vert z-x \vert < \sqrt{n}2^{k}\cdot30l} \bigl\vert D^{\alpha}b(z)- \bigl(D^{\alpha}b\bigr)_{F_{k}} \bigr\vert ^{q}\,dz \biggr)^{1/q} \\ &\lesssim \vert x-y \vert ^{m}\sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}{b} \bigr\Vert _{*}. \end{aligned}$$
Then we have
$$\begin{aligned} M_{11} \lesssim{}&\sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}{b} \bigr\Vert _{*}\Biggl(\frac {1}{ \vert B \vert }\sum _{k=-\infty}^{1} \int_{E_{k}} \bigl\vert f(y) \bigr\vert ^{u}\,dy \Biggr)^{1/u} \\ = {}&\sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}{b} \bigr\Vert _{*}\biggl(\frac{1}{ \vert B \vert } \int _{12B} \bigl\vert f(y) \bigr\vert ^{u}\,dy \biggr)^{1/u} \\ \lesssim{}& \sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}{b} \bigr\Vert _{*} M_{s}(f) (x_{0}). \end{aligned}$$
By
\(D^{\alpha}b_{k}(y)=D^{\alpha}b(y)-(D^{\alpha}b)_{F_{k}}\), applying Hölder’s inequality and Lemma
2, we get
$$\begin{aligned} M_{12} \lesssim{}&\Biggl(\frac{1}{ \vert B \vert }\biggl( \int_{12B} \bigl\vert f(y) \bigr\vert ^{s}\,dy \biggr)^{u/s}\Biggl(\sum_{ \vert \alpha \vert =m}\sum _{k=-\infty}^{1} \int_{F_{k}} \bigl\vert D^{\alpha}b(y)- \bigl(D^{\alpha}b\bigr)_{F_{k}} \bigr\vert ^{\frac{us}{s-u}}\,dy \Biggr)^{1-u/s} \Biggr)^{1/u} \\ \lesssim{}&\sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}{b} \bigr\Vert _{*} \Biggl(\frac{1}{ \vert B \vert } \biggl( \int_{12B} \bigl\vert f(y) \bigr\vert ^{s}\,dy \biggr)^{u/s} \sum_{k=-\infty}^{1} \vert F_{k} \vert ^{1-u/s}\Biggr)^{1/u} \\ \lesssim{}&\sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}{b} \bigr\Vert _{*}\biggl(\frac{1}{ \vert B \vert } \int _{12B} \bigl\vert f(y) \bigr\vert ^{s}\,dy \biggr)^{1/s} \lesssim\sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}{b} \bigr\Vert _{*} M_{s}(f) (x_{0}). \end{aligned}$$
By Minkowski’s inequality and
\(\Vert \{\epsilon _{i+1}< \vert x-y \vert <\epsilon_{i}\}_{\beta=\{\epsilon_{i}\}\in\Theta} \Vert _{F_{\rho}}\leq1\), we obtain
$$\begin{aligned} N_{1}\leq{}&\int_{\mathbb{R}^{n}} \bigl\Vert \{\chi_{\{t_{i+1}< \vert x-y \vert < s\}} \}_{\beta=\{\epsilon_{i}\}\in\Theta} \bigr\Vert _{F_{\rho}} \\ &{}\times \biggl\vert \frac{R_{m+1}({b};x,y)}{ \vert x-y \vert ^{m}}K(x-y) -\frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}}K(x_{1}-y) \biggr\vert \bigl\vert f_{2}(y) \bigr\vert \,dy \\ \leq{}&\int_{\mathbb{R}^{n}} \biggl\vert \frac{R_{m+1}({b};x,y)}{ \vert x-y \vert ^{m}}K(x-y) - \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}}K(x_{1}-y) \biggr\vert \bigl\vert f_{2}(y) \bigr\vert \,dy \\ \leq{}&\sum_{k=1}^{\infty}\int_{E_{k}} \biggl\vert \biggl(\frac {R_{m+1}({b}_{k};x,y)}{ \vert x-y \vert ^{m}}- \frac {R_{m+1}({b}_{k};x_{1},y)}{ \vert x_{1}-y \vert ^{m}}\biggr)K(x-y) \biggr\vert \bigl\vert f_{2}(y) \bigr\vert \,dy \\ &{}+ \int_{\mathbb{R}^{n}} \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert \bigl\vert K(x-y)-K(x_{1}-y) \bigr\vert \bigl\vert f_{2}(y) \bigr\vert \,dy \\ \leq{}& \sum_{k=1}^{\infty}\int_{E_{k}}\frac {1}{ \vert x-y \vert ^{m}}\vert R_{m}({b}_{k};x,y)-R_{m}({b}_{k};x_{1},y) \bigl\vert K(x-y) \bigr\vert \bigl\vert f(y) \bigr\vert \,dy \\ &{}+\sum _{k=1}^{\infty}\int_{E_{k}} \bigl\vert R_{m}({b}_{k};x_{1},y) \bigr\vert \biggl\vert \frac {1}{ \vert x-y \vert ^{m}}-\frac{1}{ \vert x_{1}-y \vert ^{m}} \biggl\vert \bigl\vert K(x-y) \bigr\vert \bigl\vert f(y) \bigr\vert \,dy\\ &{} +\sum _{k=1}^{\infty}\int_{E_{k}}\sum_{ \vert \alpha \vert =m} \frac{1}{\alpha !} \bigr\vert D^{\alpha}{b}_{k}(y) \bigr\vert \biggl\vert \frac{(x-y)^{\alpha}}{ \vert x-y \vert ^{m}}- \frac{(x_{1}-y)^{\alpha}}{ \vert x_{1}-y \vert ^{m}} \biggr\vert \bigl\vert K(x-y) \bigr\vert \bigl\vert f(y) \bigr\vert \,dy\\ &{} + \int_{(10B)^{c}} \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert \bigl\vert K(x-y)-K(x_{1}-y) \bigr\vert \bigl\vert f(y) \bigr\vert \,dy\\ ={}& N_{11}+N_{12}+N_{13}+N_{14}. \end{aligned}$$
Applying the formula ([
11] p. 448), we have
$$\begin{aligned} R_{m}({b}_{k};x,y)-R_{m}({b}_{k};x_{1},y) =R_{m}({b}_{k};x,x_{1})+\sum _{0< \vert \beta \vert < m}\frac{(x-x_{1})^{\beta}}{\beta !}R_{m-\beta}\bigl(D^{\beta}{b}_{k};x_{1},y \bigr). \end{aligned}$$
Then, by Lemmas
1 and
2, we have
$$\begin{aligned} \bigl\vert R_{m}({b}_{k};x,x_{1}) \bigr\vert \lesssim{}&\vert x-x_{1} \vert ^{m}\sum _{ \vert \alpha \vert =m}\biggl(\frac {1}{Q(x,x_{1})} \int_{Q(x,x_{1})} \bigl\vert D^{\alpha}b_{k}- \bigl(D^{\alpha}b_{k}\bigr)_{F_{k}} \bigr\vert ^{q} \biggr)^{1/q} \\ \lesssim{}&kl^{m}\sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*} \end{aligned}$$
and
$$\begin{aligned} \bigl\vert R_{m-\beta}\bigl(D^{\beta}{b}_{k};x_{1},y \bigr) \bigr\vert \lesssim \vert x-y \vert ^{m-\beta}\sum _{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}. \end{aligned}$$
So
$$\begin{aligned} \bigl\vert R_{m}({b}_{k};x,y)-R_{m}({b}_{k};x_{1},y) \bigr\vert \lesssim{}& \sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}\biggl(kl^{m}+\sum _{0< \vert \beta \vert < m}l^{\beta} \vert x-y \vert ^{m-\beta} \biggr) \\ \lesssim{}&kl \vert x-y \vert ^{m-1}\sum _{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}. \end{aligned}$$
By (
1) we have
$$\begin{aligned} \bigl\vert K(x-y) \bigr\vert \lesssim \vert x-y \vert ^{-n}. \end{aligned}$$
Then
$$\begin{aligned} N_{11}={}& \sum_{k=1}^{\infty}\int_{E_{k}}\frac {1}{ \vert x-y \vert ^{m}} \bigl\vert R_{m}({b}_{k};x,y)-R_{m}({b}_{k};x_{1},y) \bigr\vert K(x-y)\vert \bigl\vert f(y) \bigr\vert \,dy \\ \lesssim{}& \sum _{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*} \sum_{k=1}^{\infty}\int _{E_{k}}\frac{kl}{ \vert x-y \vert ^{n+1}} \bigl\vert f(y) \bigr\vert \,dy\\ \lesssim{}& \sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}\sum_{k=1}^{\infty}\frac {k}{2^{k}}\frac{1}{ \vert F_{k} \vert } \int_{F_{k}} \bigl\vert f(y) \bigr\vert \,dy\\ \lesssim{}& \sum _{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}M_{s}(f) (x_{0}). \end{aligned}$$
Let us estimate
\(N_{14}\) now. For
\(y\in(10B)^{c}\), we have
\(\vert y-x_{1} \vert \geq \vert y-x_{0} \vert - \vert x_{0}-x_{1} \vert >8l\). For
\(k=1,2,\ldots\) , let
\(\widetilde{E}_{k}=\{y:2^{k}\cdot 3l\leq \vert y-x_{1} \vert <2^{k+1}\cdot3l\}\), let
\(\widetilde{F}_{k}=\{y: \vert y-x_{1} \vert <2^{k+1}\cdot3l\}\), and let
$$\begin{aligned} {\widetilde{b}}_{k}(z)=b(z)-\sum_{ \vert \alpha \vert =m} \frac{1}{\alpha!}\bigl(D^{\alpha}b\bigr)_{\widetilde{F}_{k}}z^{m}. \end{aligned}$$
Note that
\(Q(x,y)\subset2\sqrt{n}Q(x_{1},y)\), then for
\(x\in B, y\in \widetilde{E}_{k}\) we have
$$\begin{aligned} \bigl\vert R_{m}({\widetilde{b}}_{k};x,y) \bigr\vert \lesssim{}& \vert x-y \vert ^{m}\sum_{ \vert \alpha \vert =m} \biggl(\frac{1}{ \vert 2\sqrt{n}Q(x_{1},y) \vert } \int_{2\sqrt{n}Q(x_{1},y)} \bigl\vert D^{\alpha}{\widetilde{b}}_{k}(z) \bigr\vert ^{q}\,dz\biggr)^{1/q} \\ \lesssim{}& \vert x_{1}-y \vert ^{m}\sum _{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}{b} \bigr\Vert _{*}. \end{aligned}$$
So
$$\begin{aligned} \bigl\vert R_{m+1}({\widetilde{b}}_{k};x,y) \bigr\vert \lesssim \vert x_{1}-y \vert ^{m}\biggl(\sum _{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}+\sum _{ \vert \alpha \vert =m} \bigl\vert D^{\alpha}b(y)- \bigl(D^{\alpha}b\bigr)_{\widetilde{F}_{k}} \bigr\vert \biggr). \end{aligned}$$
Then
$$\begin{aligned} N_{14} \lesssim{}& \sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}\sum_{k=1}^{\infty}\int _{\widetilde{E}_{k}} \bigl\vert K(x-y)-K(x_{1}-y) \bigr\vert \bigl\vert f(y) \bigr\vert \,dy \\ & {}+\sum_{k=1}^{\infty}\sum _{ \vert \alpha \vert =m} \int_{\widetilde{E}_{k}} \bigl\vert D^{\alpha}b(y)- \bigl(D^{\alpha}b\bigr)_{\widetilde{F}_{k}} \bigr\vert \bigl\vert K(x-y)-K(x_{1}-y) \bigr\vert \bigl\vert f(y) \bigr\vert \,dy \\ ={}& N_{141}+N_{142}. \end{aligned}$$
Taking
\(R=3l\), then
\(\vert x-x_{1} \vert < R\). By Hölder’s inequality and
\(K\in H_{r,1}\subset H_{r,0} \), we have
$$\begin{aligned} N_{141}\lesssim{}& \sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}\sum_{k=1}^{\infty}\biggl( \int_{\widetilde{E}_{k}} \bigl\vert K(x-y)-K(x_{1}-y) \bigr\vert ^{r}\,dy \biggr)^{1/r}\biggl( \int_{\widetilde{E}_{k}} \bigl\vert f(y) \bigr\vert ^{r'}\,dy \biggr)^{1/r'} \\ \lesssim{}& \sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}\sum_{k=1}^{\infty}\bigl(2^{k} \cdot3l\bigr)^{n}\biggl(\frac{1}{(2^{k}\cdot3l)^{n}} \int_{\widetilde{E}_{k}} \bigl\vert K(x-y)-K(x_{1}-y) \bigr\vert ^{r}\,dy\biggr)^{1/r} \\ &{}\times \biggl(\frac{1}{(2^{k} \cdot3l)^{n}} \int_{\widetilde {E}_{k}} \bigl\vert f(y) \bigr\vert ^{r'}\,dy \biggr)^{1/r'} \\ \lesssim{}& \sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*} M_{r'}(f) (x)\lesssim \sum _{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}M_{s}(f) (x_{0}) \end{aligned}$$
and
$$\begin{aligned} N_{142}\lesssim{}& \sum_{k=1}^{\infty}\bigl(2^{k} \cdot3l\bigr)^{n}\biggl(\frac {1}{(2^{k}\cdot3l)^{n}} \int_{\widetilde{E}_{k}} \bigl\vert K(x-y)-K(x_{1}-y) \bigr\vert ^{r}\,dy\biggr)^{1/r} \\ &{}\times\sum_{ \vert \alpha \vert =m}\biggl(\frac{1}{(2^{k}\cdot3l)^{n}} \int_{\widetilde {E}_{k}} \bigl\vert \bigl(D^{\alpha}b(y)- \bigl(D^{\alpha}b\bigr)_{\widetilde{F}_{k}}\bigr)f(y) \bigr\vert ^{r'}\,dy \biggr)^{1/r'} \\ \lesssim{}& \sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}M_{s}(f) (x_{0}). \end{aligned}$$
In case (i) we observe that
\(\epsilon_{i+1}< \vert x-y \vert \leq \vert x_{1}-x \vert + \vert x_{1}-y \vert <3l+\epsilon_{i+1}\) as
\(\vert x-x_{0} \vert < l\). Similarly, in case (iii) we have
\(\epsilon_{i+1}< \vert x_{1}-y \vert <3l+\epsilon _{i+1}\) as
\(\vert x-x_{0} \vert < l\). In case (ii) we have
\(\epsilon_{i}< \vert x_{1}-y \vert <3l+\epsilon_{i}\), and in case (iv) we have
\(\epsilon_{i}< \vert x-y \vert <3l+\epsilon_{i}\). By (
1) we have
$$\begin{aligned} & \biggl\vert \int_{\mathbb{R}^{n}}\bigl(\chi_{\{\epsilon_{i+1}< \vert x-y \vert < \epsilon _{i}\}}(y)-\chi_{\{\epsilon_{i+1}< \vert x_{1}-y \vert < \epsilon_{i}\}}(y) \bigr) \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}}K(x_{1}-y)f_{2}(y)\,dy \biggr\vert \\ &\quad\lesssim \int_{\mathbb{R}^{n}}\chi_{\{\epsilon_{i+1}< \vert x-y \vert < \epsilon_{i}\}}(y) \chi_{\{\epsilon_{i+1}< \vert x-y \vert < 3l+\epsilon_{i+1}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert \frac { \vert f_{2}(y) \vert }{ \vert x_{1}-y \vert ^{n}}\,dy \\ &\qquad{}+ \int_{\mathbb{R}^{n}}\chi_{\{\epsilon_{i+1}< \vert x-y \vert < \epsilon_{i}\}}(y) \chi_{\{\epsilon_{i}< \vert x_{1}-y \vert < 3l+\epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert \frac { \vert f_{2}(y) \vert }{ \vert x_{1}-y \vert ^{n}}\,dy \\ &\qquad{}+ \int_{\mathbb{R}^{n}}\chi_{\{\epsilon_{i+1}< \vert x_{1}-y \vert < \epsilon_{i}\}}(y) \chi_{\{\epsilon_{i+1}< \vert x_{1}-y \vert < 3l+\epsilon_{i+1}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert \frac { \vert f_{2}(y) \vert }{ \vert x_{1}-y \vert ^{n}}\,dy \\ &\qquad{}+ \int_{\mathbb{R}^{n}}\chi_{\{\epsilon_{i+1}< \vert x_{1}-y \vert < \epsilon_{i}\}}(y) \chi_{\{\epsilon_{i}< \vert x-y \vert < 3l+\epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert \frac { \vert f_{2}(y) \vert }{ \vert x_{1}-y \vert ^{n}}\,dy \\ &\quad= P_{1}+P_{2}+P_{3}+P_{4}. \end{aligned}$$
It is easy to check that
\(\vert x-y \vert \geq9l\),
\(\vert x_{1}-y \vert \geq8l\), and
\(\frac{8}{11} \vert x-y \vert \leq \vert x_{1}-y \vert \leq\frac{4}{3} \vert x-y \vert \) for
\(x\in B,y\in (10B)^{c}\); moreover, if
\(3l\geq\epsilon_{i+1},i\in\mathbb{N}\), we have
$$\begin{aligned} \bigl\{ y\in(10B)^{c}: \epsilon_{i+1}< \vert x-y \vert < 3l+\epsilon_{i+1}\bigr\} \subset\bigl\{ y\in (10B)^{c}: \vert x-y \vert < 6l\bigr\} =\phi \end{aligned}$$
and
$$\begin{aligned} \bigl\{ y\in(10B)^{c}: \epsilon_{i+1}< \vert x_{1}-y \vert < 3l+\epsilon_{i+1}\bigr\} \subset\bigl\{ y \in(10B)^{c}: \vert x-y \vert < 6l\bigr\} =\phi; \end{aligned}$$
this means
\(P_{1}=P_{3}=0\). Similarly,
\(P_{2}=P_{4}=0\) for
\(3l\geq\epsilon _{i}, i\in\mathbb{N}\). By Hölder’s inequality with
t satisfying
\(1< t<\sqrt{\min(r',\rho)}\), we get
$$\begin{aligned} &P_{1}\lesssim\biggl( \int_{\mathbb{R}^{n}}\chi_{\{\epsilon _{i+1}< \vert x-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t} \frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{nt}}\,dy \biggr)^{1/t}\bigl[(3l+\epsilon _{i+1})^{n}-( \epsilon_{i+1})^{n}\bigr]^{1/t'}, \\ &P_{2}\lesssim\biggl( \int_{\mathbb{R}^{n}}\chi_{\{\epsilon _{i+1}< \vert x-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac { \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{nt}}\,dy \biggr)^{1/t}\bigl[(3l+\epsilon _{i})^{n}-( \epsilon_{i})^{n}\bigr]^{1/t'}, \\ &P_{3} \lesssim\biggl( \int_{\mathbb{R}^{n}}\chi_{\{\epsilon _{i+1}< \vert x_{1}-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac { \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{nt}}\,dy \biggr)^{1/t}\bigl[(3l+\epsilon _{i+1})^{n}-( \epsilon_{i+1})^{n}\bigr]^{1/t'} \end{aligned}$$
and
$$\begin{aligned} P_{4}\lesssim\biggl( \int_{\mathbb{R}^{n}}\chi_{\{\epsilon _{i+1}< \vert x_{1}-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac { \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{nt}}\,dy \biggr)^{1/t}\bigl[(3l+\epsilon _{i})^{n}-( \epsilon_{i})^{n}\bigr]^{1/t'}. \end{aligned}$$
Note that for
\(3l<\epsilon_{i+1}\), we have
\((3l+\epsilon _{i+1})^{n}-(\epsilon_{i+1})^{n}\lesssim(\epsilon_{i+1})^{n-1}l\). Then
$$\begin{aligned} P_{1}\lesssim{}& \frac{(3l+\epsilon_{i+1})^{n}-(\epsilon_{i+1})^{n}}{(\epsilon _{i+1})^{(n-1)/t'}}\biggl( \int_{\mathbb{R}^{n}}\chi_{\{\epsilon _{i+1}< \vert x-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac { \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{nt}}\,dy \biggr)^{1/t} \\ \lesssim{}& l^{1/t'}\biggl( \int_{\mathbb{R}^{n}}\chi_{\{\epsilon _{i+1}< \vert x-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy \biggr)^{1/t}. \end{aligned}$$
Similarly, we have
$$\begin{aligned} &P_{2}\lesssim l^{1/t'}\biggl( \int_{\mathbb{R}^{n}}\chi_{\{\max\{\epsilon _{i+1},\frac{3}{4}\epsilon_{i}\}< \vert x-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac {R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy \biggr)^{1/t}, \\ &P_{3}\lesssim l^{1/t'}\biggl( \int_{\mathbb{R}^{n}}\chi_{\{\epsilon _{i+1}< \vert x_{1}-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy \biggr)^{1/t} \end{aligned}$$
and
$$\begin{aligned} P_{4}\lesssim l^{1/t'}\biggl( \int_{\mathbb{R}^{n}}\chi_{\{\max\{\epsilon _{i+1},\frac{8}{11}\epsilon_{i}\}< \vert x_{1}-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac {R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy \biggr)^{1/t}. \end{aligned}$$
Then
$$\begin{aligned} N_{2}= {}&\biggl\Vert \biggl\{ \int_{\mathbb{R}^{n}}\bigl(\chi_{\{\epsilon _{i+1}< \vert x-y \vert < \epsilon_{i}\}}(y)-\chi_{\{\epsilon_{i+1}< \vert x_{1}-y \vert < \epsilon _{i}\}}(y) \bigr) \\ &{} \times\frac {R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}}K(x_{1}-y)f_{2}(y)\,dy \biggr\} _{\beta=\{\epsilon _{i}\}\in\Theta} \biggr\Vert _{F_{\rho}}\\ \lesssim {}& l^{1/t'} \biggl\Vert \biggl\{ \biggl( \int_{\mathbb{R}^{n}}\chi_{\{ \epsilon_{i+1}< \vert x-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy \biggr)^{1/t} \biggr\} _{\beta =\{\epsilon_{i}\}\in\Theta} \biggr\Vert _{F_{\rho}} \\ &{}+ l^{1/t'} \biggl\Vert \biggl\{ \biggl( \int_{\mathbb{R}^{n}}\chi_{\{\epsilon _{i+1}< \vert x_{1}-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy \biggr)^{1/t} \biggr\} _{\beta =\{\epsilon_{i}\}\in\Theta} \biggr\Vert _{F_{\rho}}\\ ={}& N_{21} +N_{22}. \end{aligned}$$
A straight computation deduces that
$$\begin{aligned} N_{21}\lesssim {}& l^{1/t'}\sup_{\epsilon_{i}\searrow0 }\biggl( \sum_{i} \biggl( \int_{\mathbb{R}^{n}}\chi_{\{\epsilon_{i+1}< \vert x-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy \biggr)^{\rho/t} \biggr)^{1/\rho}\\ \lesssim{}& l^{1/t'}\biggl( \int_{(10B)^{c}} \biggl\vert \frac {R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy\biggr)^{1/t} \\ \lesssim{}& l^{1/t'}\Biggl(\sum_{k=1}^{\infty}\int_{\widetilde{E}_{k}} \biggl\vert \frac{R_{m+1}(\widetilde{b}_{k};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy\Biggr)^{1/t}. \end{aligned}$$
Similar to the estimates for
\(N_{13}\), for
\(x\in B, y\in\widetilde {E}_{k}\), we have
$$\begin{aligned} \bigl\vert R_{m+1}({\widetilde{b}}_{k};x_{1},y) \bigr\vert \lesssim \vert x_{1}-y \vert ^{m} \sum _{ \vert \alpha \vert =m}\bigl( \bigl\Vert D^{\alpha}b \bigr\Vert _{*}+ \bigl\vert D^{\alpha}b(y)-\bigl(D^{\alpha}b \bigr)_{\widetilde {F}_{k}} \bigr\vert \bigr). \end{aligned}$$
Then
$$\begin{aligned} N_{21}\lesssim{}& l^{1/t'}\sup_{\epsilon_{i}\searrow0 }\biggl( \sum_{i} \biggl( \int_{\mathbb{R}^{n}}\chi_{\{\epsilon_{i+1}< \vert x-y \vert < \epsilon_{i}\}}(y) \biggl\vert \frac{R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f_{2}(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy \biggr)^{\rho/t} \biggr)^{1/\rho} \\ \lesssim{}& l^{1/t'}\biggl( \int_{(10B)^{c}} \biggl\vert \frac {R_{m+1}({b};x_{1},y)}{ \vert x_{1}-y \vert ^{m}} \biggr\vert ^{t}\frac{ \vert f(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy\biggr)^{1/t} \\ \lesssim{}& l^{1/t'}\Biggl(\sum_{k=1}^{\infty}\int_{\widetilde{E}_{k}}\sum_{ \vert \alpha \vert =m}\bigl( \bigl\Vert D^{\alpha}b \bigr\Vert ^{t}_{*}+ \bigl\vert D^{\alpha}b(y)-\bigl(D^{\alpha}b\bigr)_{\widetilde{F}_{k}} \bigr\vert ^{t}\bigr)\frac{ \vert f(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy \Biggr)^{1/t} \\ \lesssim{}& l^{1/t'}\sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*}\sum_{k=1}^{\infty}\biggl( \int_{\widetilde{E}_{k}}\frac{ \vert f(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy \biggr)^{1/t} \\ &{} +l^{1/t'}\sum_{k=1}^{\infty}\biggl( \int_{\widetilde{E}_{k}}\sum_{ \vert \alpha \vert =m} \bigl\vert D^{\alpha}b(y)-\bigl(D^{\alpha}b\bigr)_{\widetilde{F}_{k}} \bigr\vert ^{t}\frac { \vert f(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy\biggr)^{1/t}. \end{aligned}$$
However,
$$\begin{aligned} &\sum_{k=1}^{\infty}\biggl( \int_{\widetilde{E}_{k}}\frac { \vert f(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy\biggr)^{1/t} \\ &\quad \lesssim \sum_{k=1}^{\infty}\frac{1}{(2^{k}\cdot3l)^{n/t+1/t'}}\biggl( \int _{ \vert x_{0}-y \vert < 2^{k+1}\cdot3l} \bigl\vert f(y) \bigr\vert ^{t}dt \biggr)^{1/t} \\ &\quad \lesssim M_{t}f(x_{0})\sum_{k=1}^{\infty}\frac{(2^{k}\cdot 3l)^{n/t}}{(2^{k}\cdot3l)^{n/t+1/t'}} \\ &\quad \lesssim l^{-1/t'}M_{r'}f(x_{0})\lesssim l^{-1/t'}M_{s}f(x_{0}) \end{aligned}$$
and
$$\begin{aligned} &\sum_{k=1}^{\infty}\biggl( \int_{\widetilde{E}_{k}}\sum_{ \vert \alpha \vert =m} \bigl\vert D^{\alpha}b(y)-\bigl(D^{\alpha}b\bigr)_{\widetilde{F}_{k}} \bigr\vert ^{t}\frac { \vert f(y) \vert ^{t}}{ \vert x_{1}-y \vert ^{n+t-1}}\,dy\biggr)^{1/t} \\ &\quad \lesssim\sum_{k=1}^{\infty}\frac{1}{(2^{k}\cdot3l)^{n/t+1/t'}}\biggl( \int _{\widetilde{E}_{k}}\sum_{ \vert \alpha \vert =m} \bigl\vert D^{\alpha}b(y)-\bigl(D^{\alpha}b\bigr)_{\widetilde{F}_{k}} \bigr\vert ^{t} \bigl\vert f(y) \bigr\vert ^{t} \,dy \biggr)^{1/t} \\ &\quad\lesssim \sum_{k=1}^{\infty}\frac{1}{(2^{k}\cdot3l)^{n/t+1/t'}}\biggl( \int _{\widetilde{F}_{k}} \bigl\vert f(y) \bigr\vert ^{t^{2}} \,dy \biggr)^{1/t^{2}} \biggl(\sum_{ \vert \alpha \vert =m} \int_{\widetilde{F}_{k}} \bigl\vert D^{\alpha}b(y)- \bigl(D^{\alpha}b\bigr)_{\widetilde{F}_{k}} \bigr\vert ^{t't} \,dy \biggr)^{1/(tt')} \\ &\quad \lesssim \sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*} M_{r'}f(x_{0}) \sum _{k=1}^{\infty}\frac{(2^{k}\cdot3l)^{n/t^{2}+1/(tt')}}{(2^{k}\cdot 3l)^{n/t+1/t'}} \\ &\quad = \sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*} M_{r'}f(x_{0}) \sum_{k=1}^{\infty}\frac {(2^{k}\cdot3l)^{n/t}}{(2^{k}\cdot3l)^{n/t+1/t'}} \\ &\quad \lesssim l^{-1/t'}\sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*} M_{s}f(x_{0}). \end{aligned}$$
Then
$$\begin{aligned} N_{21}\lesssim\sum_{ \vert \alpha \vert =m} \bigl\Vert D^{\alpha}b \bigr\Vert _{*} M_{s}f(x_{0}). \end{aligned}$$