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 Then $Q_m(A;x,y)=Q_m(\tilde A; x,y)$. We write, by the vanishing moment of $a$ and for $x \in (2Q)^c$,
\begin{eqnarray*}
 \tilde F_t^A(a)(x)&=&\int_{R^n}\frac{\psi_t(x-y)R_m(A;x,y)}{|x-y|^m}a(y)dy-\sum_{|\alpha|=m}\frac{1}{\alpha!}\int_{R^n}
 \frac{\psi_t(x-y)D^\alpha\tilde A(x)(x-y)^\alpha}{|x-y|^m}a(y)dy   \\
 &=& \int_{R^n}\left[\frac{\psi_t(x-y)}{|x-y|^m}-\frac{\psi_t(x-x_0)}{|x-x_0|^m}\right]R_m(\tilde A;x,y)a(y)dy \\
 &\;& +\int_{R^n}\frac{\psi_t(x-x_0)a(y)}{|x_0-x|^m}[R_m(\tilde A;x,y)-R_m(\tilde A;x,x_0)]dy  \\
 &\;& -\sum_{|\alpha|=m}\frac{1}{\alpha!}\int_{R^n}\left[\frac{\psi_t(x-y)(x-y)^\alpha}{|x-y|^m}-\frac{\psi_t(x-x_0)(x-x_0)^\alpha}{|x-x_0|^m}\right]D^\alpha\tilde A(x)a(y)dy  \\
&:=& JJ_1+JJ_2+JJ_3.
\end{eqnarray*}
 For $JJ_1$, by the condition of $\psi$ and Lemma 1, similar to the proof of Lemma 2 and (i), we obtain, for $x\in (2Q)^c$
\begin{eqnarray*}
 ||JJ_1|| &\le& C\int_{R^n}\left[\frac{|y-x_0|}{|x-y|^{n+m+1-\delta}}+\frac{|y-x_0|^\varepsilon}{|x-y|^{n+m+\varepsilon-\delta}}\right]
 |R_m(\tilde A;x,y)||a(y)|dy  \\
 &\le& C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}\left(|Q|^{1/n}|x-x_0|^{-n-1+\delta}+|Q|^{\varepsilon/n}|x-x_0|^{-n-\varepsilon+\delta}\right), \\
 ||JJ_2||&\le& C\int_{R^n}\frac{|R_m(\tilde A;x,y)-R_m(\tilde A;x,x_0)||a(y)|}{|x-y|^{m+n-\delta}}dy    \\
 &\le& C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}\int_{R^n}\frac{|x_0-y||a(y)|}{|x-x_0|^{n+1-\delta}}dy   \\
 &\le& C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}|Q|^{1/n}|x-x_0|^{-n-1+\delta}, \\
 ||JJ_3||&\le&C\int_{R^n}\frac{|x_0-y|}{|x-y|^{n+1-\delta}}\sum_{|\alpha|=m}|D^\alpha\tilde A(x)||a(y)|dy   \\
 &\le& C\sum_{|\alpha|=m}|D^\alpha\tilde A(x)|\left(|Q|^{1/n}|x-x_0|^{-n-1+\delta}+|Q|^{\varepsilon/n}|x-x_0|^{-n-\varepsilon+\delta}\right).
\end{eqnarray*}
Thus
\begin{eqnarray*}
 JJ &\le& \int_{(2Q)^c}(||JJ_1+JJ_2+JJ_3||)^{n/(n-\delta)}dx  \\
 &\le& C(\sum_{|\alpha|=m}||D^\alpha A||_{BMO})^{n/(n-\delta)}\sum_{k=1}^\infty k\left[2^{-kn/(n-\delta)}+2^{-kn\varepsilon/(n-\delta)}\right] \\
&\le& C.
\end{eqnarray*}
\par
 (iii).\ By the equality
$$
 R_{m+1}(A;x,y)=Q_{m+1}(A;x,y)+\sum_{|\alpha|=m}\frac{1}{\alpha!}(x-y)^\alpha(D^\alpha A(x)-D^\alpha A(y)),
$$
similar to the proof of Lemma 2, we get
$$
 g_\delta^A(f)(x)\le \tilde g_\delta^A(f)(x)+C\sum_{|\alpha|=m}\int_{R^n}\frac{|D^\alpha A(x)-D^\alpha A(y)|}{|x-y|^{n-\delta}}|f(y)|dy,
$$
by (i)(ii) and [2], we obtain
\begin{eqnarray*}
&\;&  |\{x\in R^n:g_\delta^A(f)(x)>\lambda\}|  \\
 &\le& |\{x\in R^n:\tilde g_\delta^A(f)(x)>\lambda/2\}|+\left|\left\{x\in R^n:\sum_{|\alpha|=m}\int_{R^n}
 \frac{|D^\alpha A(x)-D^\alpha A(y)|}{|x-y|^{n-\delta}}|f(y)|dy>C\lambda\right\}\right|    \\
&\le& C(||f||_{H^1}/\lambda)^{n/(n-\delta)}.
\end{eqnarray*}
\par
(iv).\ Let $a$ be $H^1(w)$-atom with supp$a\subset Q=Q(x_0, d)$. We write, by the vanishing moment of $a$ and for
$u\in 3Q\setminus2Q$,
\begin{eqnarray*}
 F_t^A(a)(x)&=& \chi_{4Q}(x)F_t^A(a)(x)+\chi_{(4Q)^c}(x)\int_{R^n}\left[\frac{R_m(\tilde A; x, y)\psi_t(x-y)}{|x-y|^m}
 -\frac{R_m(\tilde A; x, u)\psi_t(x-u)}{|x-u|^m}\right]a(y)dy  \\
 &\;& -\chi_{(4Q)^c}(x)\sum_{|\alpha|=m}\frac{1}{\alpha!}\int_{R^n}\left[\frac{\psi_t(x-y)(x-y)^\alpha}{|x-y|^m}
 -\frac{\psi_t(x-u)(x-u)^\alpha}{|x-u|^m}\right]D^\alpha\tilde A(y)a(y)dy  \\
 &\;& -\chi_{(4Q)^c}(x)\sum_{|\alpha|=m}\frac{1}{\alpha!}\int_{R^n}\frac{(x-u)^\alpha}{|x-u|^m}\psi_t(x-u)D^\alpha\tilde A(y)a(y)dy,
\end{eqnarray*}
then
\begin{eqnarray*}
 &\;& g_\delta^A(a)(x)=\left|\left|F_t^A(a)(x)\right|\right|\le \chi_{4Q}(x)\left|\left|F_t^A(a)(x)\right|\right| \\
 &\;& +\chi_{(4Q)^c}(x)\left|\left|\int_{R^n}\left[\frac{R_m(\tilde A; x, y)\psi_t(x-y)}{|x-y|^m}-\frac{R_m(\tilde A; x, u)
 \psi_t(x-u)}{|x-u|^m}\right]a(y)dy\right|\right|  \\
 &\;& +\chi_{(4Q)^c}(x)\left|\left|\sum_{|\alpha|=m}\frac{1}{\alpha!}\int_{R^n}\left[\frac{\psi_t(x-y)(x-y)^\alpha}{|x-y|^m}
 -\frac{\psi_t(x-u)(x-u)^\alpha}{|x-u|^m}\right]D^\alpha\tilde A(y)a(y)dy\right|\right| \\
 &\;& +\chi_{(4Q)^c}(x)\left|\left|\sum_{|\alpha|=m}\frac{1}{\alpha!}\int_{R^n}\frac{(x-u)^\alpha}{|x-u|^m}
 \psi_t(x-u)D^\alpha\tilde A(y)a(y)dy\right|\right| \\
 &=& I_1(x)+I_2(x,u)+I_3(x,u)+I_4(x,u).
\end{eqnarray*}
Similar to the proof of (i), we get
$$
 ||I_1(\cdot)||_{L^{n/(n-\delta)}}\le||g_\delta^A(a)||_{L^q}|4Q|^{(n-\delta)/n-1/q}\le C||a||_{L^p}|Q|^{1-1/p}\le C;
$$
\begin{eqnarray*}
 &\;& ||I_2(\cdot,u)||_{L^{n/(n-\delta)}} \\
 &\le& C\sum_{k=2}^\infty\left[\int_{2^{k+1}Q\setminus2^kQ}\left(\int_Q\left(\frac{|y-u|}{|x-y|^{m+n+1-\delta}}+\frac{|y-u|^\varepsilon}{|x-y|^{m+n+\varepsilon-\delta}}\right)
 |R_m(\tilde A; x,y)||a(y)|dy\right)^{n/(n-\delta)}dx\right]^{(n-\delta)/n} \\
 &\;& +\sum_{|\alpha|=m}||D^\alpha A||_{BMO}\sum_{k=2}^\infty\left[\int_{2^{k+1}Q\setminus2^kQ}\left(\int_Q\frac{|y-u|}
 {|x-y|^{n+1-\delta}}|a(y)|dy\right)^{n/(n-\delta)}dx\right]^{(n-\delta)/n}  \\
 &\le& C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}\sum_{k=2}^\infty\left[\int_{2^{k+1}Q\setminus2^kQ}\left(\int_Qk\left(\frac{|y-u|}
 {|x-y|^{n+1-\delta}}+\frac{|y-u|^\varepsilon}{|x-y|^{n+\varepsilon-\delta}}\right)|a(y)|dy\right)^{n/(n-\delta)}dx\right]^{(n-\delta)/n} \\
 &\le& C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}\sum_{k=2}^\infty\left[\int_{2^{k+1}Q\setminus2^kQ}k\left(\frac{d}{(2^kd)^{n+1-\delta}}
 +\frac{d^\varepsilon}{(2^kd)^{n+\varepsilon-\delta}}\right)^{n/(n-\delta)}dx\right]^{(n-\delta)/n}||a||_{L^\infty}|Q|  \\
 &\le& C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}\sum_{k=2}^\infty k(2^{-k}+2^{-\varepsilon k})\le C; \\
 &\;&||I_3(\cdot, u)||_{L^{n/(n-\delta)}} \\
 &\le& C\sum_{|\alpha|=m}\sum_{k=2}^\infty\left[\int_{2^{k+1}Q\setminus2^kQ}\left(\int_Q \left(\frac{|y-u|}{|x-y|^{n+1-\delta}}
 +\frac{|y-u|^\varepsilon}{|x-y|^{n+\varepsilon-\delta}}\right)|D^\alpha\tilde A(y)||a(y)|dy\right)^{n/(n-\delta)}dx\right]^{(n-\delta)/n}  \\
 &\le& C\sum_{|\alpha|=m}\sum_{k=2}^\infty\left(\frac{d}{(2^kd)^{n+1-\delta}}+\frac{d^\varepsilon}{(2^kd)^{n+\varepsilon-\delta}}\right)
 \left(\frac{1}{|Q|}\int_Q |D^\alpha\tilde A(y)|dy\right)||a||_{L^\infty}|Q||2^kQ|^{(n-\delta)/n}    \\
 &\le& C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}\sum_{k=2}^\infty (2^{-k}+2^{-\varepsilon k})\le C.
\end{eqnarray*}
Thus, using the condition of $I_4(x,u)$, we obtain
$$
||g_\delta^A(a)||_{L^{n/(n-\delta)}} \le C.
$$
\par
 (v).\ For any cube $Q=Q(x_0, d)$, we write, for $f=f\chi_{4Q}+f\chi_{(4Q)^c}=f_1+f_2$ and $u\in 3Q\setminus2Q$,
\begin{eqnarray*}
 \tilde F_t^A(f)(x)&=&\tilde F_t^A(f_1)(x)+\int_{R^n}\frac{R_m(\tilde A; x, y)}{|x-y|^m}\psi_t(x-y)f_2(y)dy  \\
 &\;& -\sum_{|\alpha|=m}\frac{1}{\alpha!}(D^\alpha A(x)-(D^\alpha A)_Q)\int_{R^n}\left[\frac{\psi_t(x-y)(x-y)^\alpha}{|x-y|^m}
 -\frac{\psi_t(u-y)(u-y)^\alpha}{|u-y|^m}\right]f_2(y)dy  \\
 &\;& -\sum_{|\alpha|=m}\frac{1}{\alpha!}(D^\alpha A(x)-(D^\alpha A)_Q)\int_{R^n}\frac{(u-y)^\alpha}{|u-y|^m}\psi_t(u-y)f_2(y)dy,
\end{eqnarray*}
then
\begin{eqnarray*}
 &\;& \left|\tilde g_\delta^A(f)(x)-g_\delta\left(\frac{R_m(\tilde A; x_0, \cdot)}{|x_0-\cdot|^m}f_2\right)(x_0)\right|
 =\left|\left|\left|\tilde F_t^A(f)(x)\right|\right|-\left|\left|F_t\left(\frac{R_m(\tilde A; x_0, \cdot)}
 {|x_0-\cdot|^m}f_2\right)(x_0)\right|\right|\right| \\
 &\le& \left|\left|\tilde F_t^A(f)(x)-F_t\left(\frac{R_m(\tilde A; x_0, \cdot)}{|x_0-\cdot|^m}f_2\right)(x_0)\right|\right| \\
 &\le& \left|\left|\tilde F_t^A(f_1)(x)\right|\right|+\left|\left|\int_{R^n}\left[\frac{R_m(\tilde A; x, y)}{|x-y|^m}\psi_t(x-y)
 -\frac{R_m(\tilde A; x_0, y)}{|x_0-y|^m}\psi_t(x_0-y)\right]f_2(y)dy\right|\right|  \\
 &\;& +\left|\left|\sum_{|\alpha|=m}\frac{1}{\alpha!}(D^\alpha A(x)-(D^\alpha A)_Q)\int_{R^n}\left[\frac{\psi_t(x-y)(x-y)^\alpha}
 {|x-y|^m}-\frac{\psi_t(u-y)(u-y)^\alpha}{|u-y|^m}\right]f_2(y)dy\right|\right| \\
 &\;& +\left|\left|\sum_{|\alpha|=m}\frac{1}{\alpha!}(D^\alpha A(x)-(D^\alpha A)_Q)\int_{R^n}\frac{(u-y)^\alpha}{|u-y|^m}
 \psi_t(u-y)f_2(y)dy\right|\right| \\
 &=& J_1(x)+J_2(x)+J_3(x,u)+J_4(x,u).
\end{eqnarray*}
Similar to the proof of (i) and (iv), we get
\begin{eqnarray*}
 \frac{1}{|Q|}\int_Q J_1(x)dx &\le& |Q|^{-1/q}||\tilde g_\delta^A(f_1)||_{L^q}\le C|Q|^{-1/q}||f_1||_{L^p}\le C||f||_{L^{n/\delta}}; \\
 \frac{1}{|Q|}\int_Q J_2(x)dx&\le& C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}\frac{1}{|Q|}\int_Q\sum_{k=2}^\infty\int_{2^{k+1}Q\setminus2^kQ}
 k\left(\frac{|x-x_0|}{|x_0-y|^{n+1-\delta}}+\frac{|x-x_0|^\varepsilon}{|x_0-y|^{n+\varepsilon-\delta}}\right)|f(y)|dy dx \\
 &\le& C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}||f||_{L^{n/\delta}}\sum_{k=1}^\infty k(2^{-k}+2^{-\varepsilon k})
 \le C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}||f||_{L^{n/\delta}}; \\
 \frac{1}{|Q|}\int_Q J_3(x,u)dx&\le&\sum_{|\alpha|=m}\frac{C}{|Q|}\int_Q|D^\alpha A(x)-(D^\alpha A)_Q)| \\
 &\;& \times\sum_{k=2}^\infty\int_{2^{k+1}Q\setminus2^kQ}
 \left(\frac{|x-u|}{|x-y|^{n+1-\delta}}+\frac{|x-u|^\varepsilon}{|x-y|^{n+\varepsilon-\delta}}\right)|f(y)|dy dx \\
 &\le& C\sum_{|\alpha|=m}||D^\alpha A||_{BMO}\sum_{k=2}^\infty (2^{-k}+2^{-\varepsilon k})||f||_{L^{n/\delta}}
 \le C||f||_{L^{n/\delta}}.
\end{eqnarray*}
Thus, using the condition of $J_4(x,u)$, we obtain
$$
 \frac{1}{|Q|}\int_Q \left|\tilde g_\delta^A(f)(x)-g_\delta\left(\frac{R_m(\tilde A; x_0, \cdot)}{|x_0-\cdot|^m}f_2\right)(x_0)\right|dx \le C||f||_{L^{n/\delta}}.
$$
 This completes the proof of Theorem.
\vskip4mm
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\end{document}