By the standard argument, it suffices to verify that for any atomic block
b as in Definition
1.3 with
\(\rho=4\),
\(p=\infty\) and
\(\gamma=2\),
Tb is in
\(H^{1}(\mu)\) with norm
\(C|{b}|_{H^{1,\infty}_{atb,2}}\). By Definition
1.3, it follows
$$\begin{aligned} \|{a_{j}}\|_{L^{\infty}(\mu)}\leq \bigl( \mu(4Q_{j})K^{2}_{Q_{j},R} \bigr)^{-1}, \end{aligned}$$
(2.1)
where
\(j=1,2\). The assumption that
\(T^{*}1=0\) tells us that
\(\int_{R^{d}}Tb\,d(\mu)=0\). Recalling that
T is bounded from
\(H^{1}(\mu)\) into
\(L^{1}(\mu)\) (see [
6]), we obtain
$$\begin{aligned} \|{Tb}\|_{L^{1}(\mu)}\leq C|{b}|_{H^{1,\infty}_{atb}(\mu)}. \end{aligned}$$
By this and Theorem
1.1, we deduce that the proof of Theorem
1.2 can be reduced to proving that
$$\begin{aligned} \bigl\| {M_{\Phi}(Tb)}\bigr\| _{L^{1}(\mu)}\leq C|{b}|_{H^{1,\infty}_{atb}(\mu)}. \end{aligned}$$
(2.2)
We can write
$$\begin{aligned} \int_{R^{d}}M_{\Phi}(Tb) (x)\,d\mu(x)=\int _{R^{d}\backslash4R}M_{\Phi}(Tb) (x)\,d\mu (x)+\int _{4R}M_{\Phi}(Tb) (x)\,d\mu(x)=I_{1}+I_{2}. \end{aligned}$$
Let us now estimate
\(I_{1}\). Let
\(x_{R}\) be the center of the cube
R. From the fact
\(T^{*}1=0\), we obtain
$$\begin{aligned} I_{1}={}&\int_{R^{d}\backslash4R}\sup_{\varphi\sim x} \biggl\vert {\int_{R^{d}}Tb(y)\bigl[\varphi(y)- \varphi(x_{R})\bigr]\,d\mu(y)}\biggr\vert \,d\mu(x) \\ \leq{}&\int_{R^{d}\backslash4R}\sup_{\varphi\sim x}\biggl\vert { \int_{2R}Tb(y)\bigl[\varphi(y)-\varphi(x_{R}) \bigr]\,d\mu(y)}\biggr\vert \,d\mu(x) \\ &{}+\int_{R^{d}\backslash4R}\sup_{\varphi\sim x}\biggl\vert {\int _{R^{d}\backslash 2R}Tb(y)\bigl[\varphi(y)-\varphi(x_{R})\bigr]\,d\mu(y)}\biggr\vert \,d\mu(x) \\ ={}&I_{11}+I_{12}. \end{aligned}$$
Note that for any
\(z\in2R\),
\(x\in2^{k+1}R\backslash2^{k}R\), and
\(k\geq 2\), we have
\(|{x-z}|\geq l(2^{k-2}R)\).
This together with Definition
1.2 and the mean value theorem leads to
$$\begin{aligned} \bigl|{\varphi(y)-\varphi(x_{R})}\bigr|\leq C\frac{l(R)}{l(2^{k-2}R)^{n+1}}. \end{aligned}$$
(2.3)
For
\(j=1,2\), denote
\(N_{Q_{j},2R}\) simply by
\(N_{j}\) for
\(y\in2R\). By (
2.3), (
1.4), Hölder’s inequality, the boundedness of
T in
\(L^{2}(\mu)\) and (
2.1), we have
$$\begin{aligned} I_{11}={}&\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{\infty}\int _{2^{k+1}R\backslash2^{k}R}\sup_{\varphi\sim x} \biggl[\int _{2R\backslash 2Q_{j}}\bigl|{Ta_{j}(y)}\bigr|\bigl|{\varphi(y)- \varphi(x_{R})}\bigr|\,d\mu(y) \biggr]\,d\mu(x) \\ &{}+\sum_{j=1}^{2}|{\lambda_{j}}| \sum_{k=2}^{\infty}\int_{2^{k+1}R\backslash 2^{k}R} \sup_{\varphi\sim x} \biggl[\int_{2Q_{j}}\bigl|{Ta_{j}(y)}\bigr|\bigl|{ \varphi (y)-\varphi(x_{R})}\bigr|\,d\mu(y) \biggr]\,d\mu(x) \\ \leq{}& C\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{\infty}\int _{2^{k+1}R\backslash2^{k}R}\frac{l(R)}{l(2^{k-2}R)^{n+1}}\sum_{l=1}^{N_{j}-1} \int_{2^{l+1}Q_{j}\backslash2^{l}Q_{j}}\int_{Q_{j}}\frac {|{a_{j}(z)}|}{|{y-z}|^{n}}\,d\mu(z)\,d\mu(y)\,d\mu(x) \\ &{}+C\sum_{j=1}^{2}|{\lambda_{j}}| \sum_{k=2}^{\infty}\int_{2^{k+1}R\backslash 2^{k}R} \frac{l(R)}{l(2^{k-2}R)^{n+1}}\bigl\| {(Ta_{j})\chi_{2Q_{j}}}\bigr\| _{L^{1}(\mu )}\,d\mu(x) \\ \leq{}& C\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{\infty}2^{-k} \sum_{l=1}^{N_{j}-1}\frac{\mu(2^{l+1}Q_{j})}{l(2^{l+1}Q_{j})^{n}} \|{a_{j}}\|_{L^{\infty}(\mu)}\mu(Q_{j}) \\ &{}+C\sum_{j=1}^{2}|{\lambda_{j}}| \sum_{k=2}^{\infty}2^{-k} \bigl\| {(Ta_{j})\chi _{2Q_{j}}}\bigr\| _{L^{2}(\mu)}\mu(2Q_{j})^{1/2} \\ \leq{}& C\sum_{j=1}^{2}|{ \lambda_{j}}|K_{Q_{j},R}\|{a_{j}}\|_{L^{\infty}(\mu )}\mu (Q_{j})+C\sum_{j=1}^{2}|{ \lambda_{j}}| \|{a_{j}}\|_{L^{2}(\mu)} \mu(2Q_{j})^{1/2} \\ \leq{}& C\sum_{j=1}^{2}|{ \lambda_{j}}|, \end{aligned}$$
where we have used the fact that
$$\begin{aligned} K_{Q_{j},2R}\leq CK_{Q_{j},R}. \end{aligned}$$
For
\(I_{12}\), we get
$$\begin{aligned} I_{12}={}&\sum_{k=2}^{\infty}\int _{2^{k+1}R\backslash2^{k}R}\sup_{\varphi \sim x}\biggl\vert {\int _{R^{d}\backslash2R}Tb(y)\bigl[\varphi(y)-\varphi (x_{R})\bigr]\,d\mu (y)}\biggr\vert \,d\mu(x) \\ \leq{}&\sum_{k=2}^{\infty}\int _{2^{k+1}R\backslash2^{k}R}M_{\Phi}\bigl[|{Tb}|\chi_{2^{k+2}R\backslash2^{k-1}R} \bigr](x)\,d\mu(x) \\ &{}+\sum_{k=2}^{\infty}\int _{2^{k+1}R\backslash2^{k}R}\sup_{\varphi\sim x} \biggl[\int _{2^{k+2}R\backslash2^{k-1}R}\bigl|{Tb(y)}\bigr|\varphi(x_{R})\,d\mu (y) \biggr]\,d\mu(x) \\ &{}+\sum_{k=2}^{\infty}\int _{2^{k+1}R\backslash2^{k}R}\sup_{\varphi\sim x} \biggl[\int _{R^{d}\backslash2^{k+2}R}\bigl|{Tb(y)}\bigr|\bigl(\varphi(y)+\varphi (x_{R}) \bigr)\,d\mu(y) \biggr]\,d\mu(x) \\ &{}+\sum_{k=2}^{\infty}\int _{2^{k+1}R\backslash2^{k}R}\sup_{\varphi\sim x} \biggl[\int _{2^{k-1}R\backslash2R}\bigl|{Tb(y)}\bigr|\bigl(\varphi(y)+\varphi (x_{R}) \bigr)\,d\mu(y) \biggr]\,d\mu(x) \\ ={}&I_{121}+I_{122}+I_{123}+I_{124}. \end{aligned}$$
From Lemma
2.1, the fact that
\(\int_{R^{d}}b\,d(\mu)=0\) and (
1.5), we can deduce that
$$\begin{aligned} I_{121}\leq{}&\sum_{k=2}^{\infty}\mu \bigl(2^{k+1}R\bigr)^{1/2}\bigl\Vert {M_{\Phi}\bigl[|{Tb}|\chi_{2^{k+2}R\backslash2^{k-1}R} \bigr]}\bigr\Vert _{L^{2}(\mu)} \\ \leq{}& C\sum_{k=2}^{\infty}\mu \bigl(2^{k+1}R\bigr)^{1/2} \biggl(\int_{2^{k+2}R\backslash2^{k-1}R} \biggl\vert {\int_{R}\bigl(K(y,z)-K(y,x_{R}) \bigr)b(z)\,d\mu (z)}\biggr\vert ^{2}\,d\mu(y) \biggr)^{1/2} \\ \leq{}& C\sum_{k=2}^{\infty}\mu \bigl(2^{k+1}R\bigr)^{1/2} \\ &{}\times\biggl(\int_{2^{k+2}R\backslash2^{k-1}R} \biggl[\int_{R}\theta \biggl(\frac {|{z-x_{R}}|}{|{y-x_{R}}|} \biggr) |{y-x_{R}}|^{-n}\bigl|{b(z)}\bigr|\,d\mu(z) \biggr]^{2}\,d\mu(y) \biggr)^{1/2} \\ \leq{}& C\sum_{k=2}^{\infty}\frac{\mu(2^{k+1}R)}{l(2^{k}R)^{n}} \theta \bigl(2^{-k}\bigr)\|{b}\|_{L^{1}(\mu)} \leq C\int _{0}^{1}\frac{\theta(t)}{t}\,dt\|{b}\|_{L^{1}(\mu)} \leq C\sum_{j=1}^{2}|{\lambda_{j}}|, \end{aligned}$$
where we have used the following inequality:
$$\begin{aligned} \int_{0}^{1}\frac{\theta(t)}{t}\geq\sum \int_{2^{k}}^{2^{1-k}}\frac {\theta(2^{-k})}{2^{1-k}}\geq C \sum _{k=1}^{\infty}\theta\bigl(2^{-k}\bigr), \end{aligned}$$
and the fact
$$\begin{aligned} \|{b}\|_{L^{1}(\mu)}\leq\sum_{j=1}^{2}|{ \lambda_{j}}|\|{a_{j}}\|_{L^{1}(\mu )}\leq C\sum _{j=1}^{2}|{\lambda_{j}}|. \end{aligned}$$
An argument similar to the estimate for
\(I_{121}\) tells us that
$$\begin{aligned} I_{122}\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|. \end{aligned}$$
Finally, we estimate
\(I_{123}\). By the fact that
\(\int_{R^{d}}b\,d\mu=0\), Definition
1.2 and (
1.5), we obtain
$$\begin{aligned} I_{123}\leq{}&\sum_{k=2}^{\infty}\int _{2^{k+1}R\backslash2^{k}R}\sum_{l=k+2}^{\infty}\int _{2^{l+1}R\backslash2^{l}R}\int_{R}\bigl|{K(y,z)-K(y,x_{R})}\bigr|\bigl|{b(z)}\bigr|\,d\mu(z) \\ &{}\times \biggl[\frac{1}{|{y-x}|^{n}}+\frac{1}{|{x_{R}-x}|^{n}} \biggr]\,d\mu (y)\,d\mu(x) \\ \leq{}& C\sum_{k=2}^{\infty}\int _{2^{k+1}R\backslash2^{k}R}\sum_{l=k+2}^{\infty}\int _{2^{l+1}R\backslash2^{l}R} \int_{R}\theta \biggl( \frac{|{z-x_{R}}|}{|{y-x_{R}}|} \biggr) |{y-x_{R}}|^{-n}\bigl|{b(z)}\bigr|\,d\mu(z) \\ &{}\times \biggl[\frac{1}{|{y-x}|^{n}}+\frac{1}{|{x_{R}-x}|^{n}} \biggr]\,d\mu (y)\,d\mu(x) \\ \leq{}& C\sum_{k=2}^{\infty}\sum _{l=k+2}^{\infty}\theta\bigl(2^{-l}\bigr) \frac {\mu (2^{l+1}R)}{l(2^{l+1}R)^{n}} \frac{\mu(2^{k+1}R)}{l(2^{k+1}R)^{n}}\|{b}\|_{L^{1}(\mu)} \\ \leq{}& C\sum_{j=1}^{2}|{ \lambda_{j}}|. \end{aligned}$$
An argument similar to the estimate for
\(I_{123}\) indicates that
$$\begin{aligned} I_{124}\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|. \end{aligned}$$
Combining the estimate for
\(I_{121}\),
\(I_{122}\),
\(I_{123}\) and
\(I_{124}\), we obtain the desired estimate for
\(I_{12}\). The estimates for
\(I_{11}\) and
\(I_{12}\) tell us that
$$\begin{aligned} I_{1}=\int_{R^{d}\backslash4R}M_{\Phi}(Tb) (x)\,d\mu(x)\leq C|{b}|_{H^{1,\infty}_{atb,2}}(\mu). \end{aligned}$$
(2.4)
For
\(I_{2}\), by the sublinearity of
\(M_{\Phi}\), it follows
$$\begin{aligned} I_{2}\leq\int_{4R}M_{\Phi}\bigl[(Tb) \chi_{8R} \bigr](x)\,d\mu(x)+ \int_{4R}M_{\Phi}\bigl[(Tb)\chi_{R^{d}\backslash8R} \bigr](x)\,d\mu (x)=I_{21}+I_{22}. \end{aligned}$$
From
\(Q_{j}\subset R\), Definition
1.2 and (
2.1), we obtain
$$\begin{aligned} I_{22}&\leq\int_{4R}\sup_{\varphi\sim x} \biggl[\int_{R^{d}\backslash 8R}\bigl|{Tb(y)}\bigr|\varphi(y)\,d\mu(y) \biggr]\,d\mu(x) \\ &\leq\sum_{j=1}^{2}|{\lambda_{j}}| \int_{4R}\sum_{k=2}^{\infty}\int_{2^{k+1}R\backslash2^{k}R}\biggl\vert {\int_{Q_{j}}K(y,z)a_{j}(z)\,d\mu (z)}\biggr\vert \frac {1}{|{x-y}|^{n}}\,d\mu(y)\,d\mu(x) \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=3}^{\infty}\|{a_{j}}\| _{L^{\infty}(\mu)}\mu(Q_{j}) \frac{\mu(2^{k+1}R)}{l(2^{k-2}R)^{n}} \frac{\mu(4R)}{l(2^{k-2}R)^{n}} \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|. \end{aligned}$$
In order to estimate
\(I_{21}\), we write
$$\begin{aligned} I_{21}\leq{}&\sum_{j=1}^{2}|{ \lambda_{j}}|\int_{4Q_{j}}M_{\Phi}\bigl[(Ta_{j})\chi _{8R} \bigr](x)\,d\mu(x) +\sum_{j=1}^{2}|{\lambda_{j}}| \int_{4R\backslash4Q_{j}}M_{\Phi}\bigl[(Ta_{j}) \chi_{2Q_{j}} \bigr](x)\,d\mu(x) \\ &{}+\sum_{j=1}^{2}|{\lambda_{j}}| \int_{4R\backslash4Q_{j}}M_{\Phi}\bigl[(Ta_{j}) \chi_{8R\backslash2Q_{j}} \bigr](x)\,d\mu(x) \\ &=I_{211}+I_{212}+I_{213}. \end{aligned}$$
Hölder’s inequality, Lemma
2.1, the boundedness of
T in
\(L^{2}(\mu )\) and (
2.1) lead to
$$\begin{aligned} I_{211}&\leq\sum_{j=1}^{2}|{ \lambda_{j}}|\mu(4Q_{j})^{1/2}\bigl\| {M_{\Phi}\bigl[(Ta_{j})\chi_{8R}\bigr]}\bigr\| _{L^{2}(\mu)} \\ &\leq C\sum _{j=1}^{2}|{\lambda_{j}}| \mu(4Q_{j})^{1/2}\|{Ta_{j}}\|_{L^{2}(\mu )} \leq C\sum_{j=1}^{2}|{ \lambda_{j}}|\mu(4Q_{j})^{1/2}\|{a_{j}} \|_{L^{2}(\mu)} \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|\mu(4Q_{j})\|{a_{j}} \|_{L^{\infty}(\mu)} \leq C\sum_{j=1}^{2}|{ \lambda_{j}}|. \end{aligned}$$
By Definition
1.2, Hölder’s inequality, the boundedness of
T in
\(L^{2}(\mu)\) and (
2.1), we get
$$\begin{aligned} I_{212}&\leq\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{N_{Q_{j},4R}}\int _{2^{k+1}Q_{j}\backslash2^{k}Q_{j}}\sup_{\varphi\sim x} \biggl\vert {\int _{2Q_{j}}Ta_{j}(y)\varphi(y)\,d\mu(y)}\biggr\vert d \mu(x) \\ &\leq\sum_{j=1}^{2}|{\lambda_{j}}| \sum_{k=2}^{N_{Q_{j},4R}}\int_{2^{k+1}Q_{j}\backslash2^{k}Q_{j}} \frac{1}{l(2^{k-2}Q_{j})^{n}}\,d\mu(x) \int_{2Q_{j}}\bigl|{Ta_{j}(y)}\bigr|\,d\mu(y) \\ &\leq\sum_{j=1}^{2}|{\lambda_{j}}| \sum_{k=2}^{N_{Q_{j},4R}}\frac{\mu (2^{k+1}Q_{j})}{l(2^{k-2}Q_{j})^{n}} \|{Ta_{j}}\| _{L^{2}(\mu)}\mu(2Q_{j})^{1/2} \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|K_{Q_{j},R}\mu(2Q_{j})^{1/2} \|{a_{j}}\| _{L^{2}(\mu)} \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|, \end{aligned}$$
where we have used the fact that
$$\begin{aligned} K_{Q_{j},4R}\leq CK_{Q_{j},R}. \end{aligned}$$
(2.5)
For
\(I_{213}\), we can write
$$\begin{aligned} I_{213}={}&\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{N_{Q_{j},4R}}\int _{2^{k+1}Q_{j}\backslash2^{k}Q_{j}}M_{\Phi}\bigl[ (Ta_{j}) \chi_{8R\backslash2Q_{j}} \bigr](x)\,d\mu(x) \\ \leq{}&\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{N_{Q_{j},4R}}\int _{2^{k+1}Q_{j}\backslash2^{k}Q_{j}}M_{\Phi}\bigl[|{Ta_{j}}|\chi _{2^{k+2}Q_{j}\backslash2^{k-1}Q_{j}} \bigr](x)\,d\mu(x) \\ &{}+\sum_{j=1}^{2}|{\lambda_{j}}| \sum_{k=2}^{N_{Q_{j},4R}}\int_{2^{k+1}Q_{j}\backslash2^{k}Q_{j}}M_{\Phi}\bigl[|{Ta_{j}}|\chi_{\max\{ 2^{k+2}Q_{j},8R\}\backslash2^{k+2}Q_{j}} \bigr](x)\,d\mu (x) \\ &{}+\sum_{j=1}^{2}|{\lambda_{j}}| \sum_{k=2}^{N_{Q_{j},4R}}\int_{2^{k+1}Q_{j}\backslash2^{k}Q_{j}}M_{\Phi}\bigl[|{Ta_{j}}|\chi _{2^{k-1}Q_{j}\backslash2Q_{j}} \bigr](x)\,d\mu(x) \\ ={}&J_{1}+J_{2}+J_{3}. \end{aligned}$$
Lemma
2.1, (
1.4) and (
2.1) imply that
$$\begin{aligned} J_{1}&=\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{N_{Q_{j},4R}}\mu \bigl(2^{k+1}Q_{j}\bigr)^{1/2}\bigl\Vert {M_{\Phi}\bigl[f |{Ta_{j}}| \chi _{2^{k+2}Q_{j}\backslash2^{k-1}Q_{j}} \bigr]}\bigr\Vert _{L^{2}(\mu)} \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{N_{Q_{j},4R}}\mu \bigl(2^{k+1}Q_{j}\bigr)^{1/2} \times\biggl(\int _{2^{k+2}Q_{j}\backslash 2^{k-1}Q_{j}}\biggl\vert {\int_{Q_{j}}K(y,z)a_{j}(z)\,d\mu(z)}\biggr\vert ^{2}\,d\mu (y) \biggr)^{1/2} \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{N_{Q_{j},4R}} \frac{\mu (2^{k+2}Q_{j})}{l(2^{k-3}Q_{j})^{n}} \|{a_{j}}\|_{L^{\infty}(\mu)}\mu(Q_{j}) \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|. \end{aligned}$$
By (ii) of Definition
1.2, (
1.4), (
2.5) and (
2.1), we have
$$\begin{aligned} J_{2}&=\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{N_{Q_{j},4R}}\int _{2^{k+1}Q_{j}\backslash2^{k}Q_{j}}\sup_{\varphi\sim x} \biggl[ \int _{2^{k-1}Q_{j}\backslash2Q_{j}}\bigl|{Ta_{j}(y)}\bigr|\varphi(y)\,d\mu(y) \biggr]\,d\mu(x) \\ &\leq\sum_{j=1}^{2}|{\lambda_{j}}| \sum_{k=2}^{N_{Q_{j},4R}}\int_{2^{k+1}Q_{j}\backslash2^{k}Q_{j}} \sum_{l=1}^{k-2}\int_{2^{l+1}Q_{j}\backslash2^{l}Q_{j}} \biggl\vert {\int_{Q_{j}}K(y,z)a_{j}(z)\,d\mu (z)} \biggr\vert \frac {1}{|{y-x}|^{n}}\,d\mu(y)\,d\mu(x) \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{N_{Q_{j},4R}} \frac{\mu (2^{k+1}Q)}{l(2^{k+1}Q_{j})^{n}} \sum_{l=1}^{k-2} \frac{\mu(2^{l+1}Q)}{l(2^{l+1}Q_{j})^{n}}\|{a_{j}}\| _{L^{\infty}(\mu)}\mu(Q_{j}) \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|(K_{Q_{j},R})^{2}\|{a_{j}} \|_{L^{\infty}(\mu )}\mu(Q_{j}) \\ &\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|. \end{aligned}$$
With the argument similar to the estimate for
\(J_{2}\) it follows that
$$\begin{aligned} J_{3}=\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{N_{Q_{j},4R}}\int _{2^{k+1}Q_{j}\backslash2^{k}Q_{j}}M_{\Phi}\bigl[|{Ta_{j}}|\chi _{2^{k-1}Q_{j}\backslash2Q_{j}} \bigr](x)\,d\mu(x)\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|. \end{aligned}$$
Thus
$$\begin{aligned} I_{213}=\sum_{j=1}^{2}|{ \lambda_{j}}|\sum_{k=2}^{N_{Q_{j},4R}}\int _{2^{k+1}Q_{j}\backslash2^{k}Q_{j}}M_{\Phi}\bigl[ (Ta_{j}) \chi_{8R\backslash2Q_{j}} \bigr](x)\,d\mu(x)\leq C\sum_{j=1}^{2}|{ \lambda_{j}}|. \end{aligned}$$
From the estimation of
\(I_{21}\) and
\(I_{22}\), we obtain
$$\begin{aligned} I_{2}=\int_{4R}M_{\Phi}(Tb) (x)\,d\mu(x)\leq C\sum_{j=1}^{2}|{\lambda _{j}}|=C|{b}|_{H^{1,\infty}_{atb,2}}. \end{aligned}$$
(2.6)
The estimates (
2.4) and (
2.6) lead to (
2.2), and this completes the proof of our theorem. □