Proof We first show the sufficiency. Assume that $(u,v)$ satisfies (2.1). By this and Definition 2.4, we find that, for any $R\in \mathcal {R}_p^{n+1}$ and for almost every $(x,t)\in R^+(\gamma )$ ,

Taking the essential infimum over all $(x,t)\in R^+(\gamma )$ , we then obtain

and hence $(u,v)\in TA_{1,q}^+(\gamma )$ with $[u,v]_{TA_{1,q}^+(\gamma )}\leq C$ . This finishes the proof of the sufficiency.

Now, we prove the necessity. Assume that $(u,v)\in TA_{1,q}^+(\gamma )$ and let

$$ \begin{align*} \mathcal{N}:=\left\{(x,t)\in\mathbb{R}^{n+1}:\ M^{\gamma-}_0\left(u^q\right)(x,t)>[u,v]_{TA_{1,q}^+(\gamma)} [v(x,t)]^q\right\}. \end{align*} $$

To show that (2.1) holds almost everywhere for some $C\in (0,\infty )$ , it suffices to prove that $|\mathcal {N}|=0$ . Indeed, from Definition 2.4, we infer that, for any given $(x,t)\in \mathcal {N}$ , there exist $R_{(x,t)}\in \mathcal {R}_p^{n+1}$ and $\epsilon \in (0,\infty )$ such that $(x,t)\in R_{(x,t)}^+(\gamma )$ and

(2.2)

Since $\text {pr}_t(R_{(x,t)}^+(\gamma ))$ is an open interval and $(x,t)\in R_{(x,t)}^+(\gamma )$ , it follows that there exists $\widetilde {R}_{(x,t)}:=Q(x_0,L_0)\times (t_0-L_0^p,t_0+L_0^p)\in \mathcal {R}_p^{n+1}$ with $(x_0,t_0)\in \mathbb {R}^{n+1}$ and $L_0\in (0,\infty )$ such that the following statements hold:

1. (i) $(x,t)\in \widetilde {R}_{(x,t)}^+(\gamma )$ ;

2. (ii) $R_{(x,t)}^-(\gamma )\subset \widetilde {R}_{(x,t)}^-(\gamma )$ and $|\widetilde {R}_{(x,t)}^-(\gamma )\setminus R_{(x,t)}^-(\gamma )|<\epsilon $ ;

3. (iii) all the vertices of $Q(x_0,L_0)$ belong to $\mathbb {Q}^n$ and $t_0-L_0^p\in \mathbb {Q}$ .

Combining (2.2), (ii), and $(u,v)\in TA_{1,q}^+(\gamma )$ , we conclude that

which further implies that

(2.3) $$ \begin{align} v(x,t)<\mathop{\mathrm{ess\,inf}}_{(y,s) \in\widetilde{R}_{(x,t)}^+(\gamma)}v(y,s). \end{align} $$

Let $\{R_k\}_{k\in \mathbb {N}}$ be the sequence of all $R:=Q(z,L)\times (r-L^p,r+L^p)\in \mathcal {R}_p^{n+1}$ with $(z,r)\in \mathbb {R}^{n+1}$ and $L\in (0,\infty )$ such that all the vertices of $Q(z,L)$ belong to $\mathbb {Q}^n$ and $r-L^p\in \mathbb {Q}$ and, for any $k\in \mathbb {N}$ , let

$$ \begin{align*} \mathcal{N}_k:=\left\{(x,t)\in R_k^+(\gamma):\ v(x,t)<\mathop{\mathrm{ess\,inf}}_{(y,s) \in R_k^+(\gamma)}v(y,s)\right\}. \end{align*} $$

Then $|\mathcal {N}_k|=0$ for any $k\in \mathbb {N}$ . Moreover, from (i), (iii), and (2.3), it follows that $\mathcal {N}\subset \bigcup _{k\in \mathbb {N}} \mathcal {N}_k$ , which further implies that $|\mathcal {N}|\leq \sum _{k\in \mathbb {N}}|\mathcal {N}_k|=0$ . This finishes the proof of the necessity and hence Proposition 2.5.

Proof Let $(u,v),(\widetilde {u},\widetilde {v})\in TA_{r,q}^+(\gamma )$ . We first show (i). Let $W_1:=\max \{u,\,\widetilde {u}\}$ and $W_2:=\max \{v,\,\widetilde {v}\}$ . We consider the following two cases for r.

Case 1: $r=1$ . In this case, by Definition 2.1(ii), we find that, for any given $R\in \mathcal {R}_p^{n+1}$ ,

Taking the supremum over all $R\in \mathcal {R}_p^{n+1}$ , we conclude that $(W_1,W_2)\in TA_{1,q}^+(\gamma )$ and

$$ \begin{align*} [W_1,W_2]_{TA_{1,q}^+(\gamma)}\leq[u,v] _{TA_{1,q}^+(\gamma)}+[\widetilde{u},\widetilde{v}] _{TA_{1,q}^+(\gamma)}. \end{align*} $$

Case 2: $r\in (1,\infty )$ . In this case, from Definition 2.1(i), we deduce that, for any given $R\in \mathcal {R}_p^{n+1}$ ,

Taking the supremum over all $R\in \mathcal {R}_p^{n+1}$ , we obtain $(W_1,W_2)\in TA_{r,q}^+(\gamma )$ and

$$ \begin{align*} [W_1,W_2]_{TA_{r,q}^+(\gamma)}\leq[u,v] _{TA_{r,q}^+(\gamma)}+[\widetilde{u},\widetilde{v}] _{TA_{r,q}^+(\gamma)}. \end{align*} $$

Combining the above two cases, we then finish the proof of (i).

Next, we prove (ii). Let $w_1:=\min \{u,\,\widetilde {u}\}$ and $w_2:=\min \{v,\,\widetilde {v}\}$ . Similarly to the proof of (i), we consider the following two cases for r.

Case 1: $r=1$ . In this case, using Definition 2.1(ii), we conclude that, for any $R\in \mathcal {R}_p^{n+1}$ ,

Taking the supremum over all $R\in \mathcal {R}_p^{n+1}$ , we find that $(w_1,w_2)\in TA_{1,q}^+(\gamma )$ and

$$ \begin{align*} [w_1,w_2]_{TA_{1,q}^+(\gamma)}\leq\max\left\{[u,v] _{TA_{1,q}^+(\gamma)},\,[\widetilde{u},\widetilde{v}] _{TA_{1,q}^+(\gamma)}\right\}. \end{align*} $$

Case 2: $r\in (1,\infty )$ . In this case, from Definition 2.1(i), we infer that, for any $R\in \mathcal {R}_p^{n+1}$ ,

Taking the supremum over all $R\in \mathcal {R}_p^{n+1}$ , we obtain $(w_1,w_2)\in TA_{r,q}^+(\gamma )$ and

$$ \begin{align*} [w_1,w_2]_{TA_{r,q}^+(\gamma)}\leq[u,v] _{TA_{r,q}^+(\gamma)}+[\widetilde{u},\widetilde{v}] _{TA_{r,q}^+(\gamma)}. \end{align*} $$

Combining the above two cases, we then complete the proof of (ii) and hence Proposition 2.6.

Proof We first show $(\mathrm {i})\Longrightarrow (\mathrm {ii})$ . Assume that $(u,v)\in TA_{q,q}^+(\gamma )$ for some $q\in (1,\infty )$ . Then, from the Hölder inequality, we deduce that, for any $R\in \mathcal {R}_p^{n+1}$ and any measurable set $E\subset R^+(\gamma )$ ,

which further implies that (2.4) with $r:=q$ , $\delta \in (0,\frac {1}{q}]$ , and $C:=[u,v]_{TA_q^+(\gamma )}^{\delta }$ holds. This finishes the proof of $(\mathrm {i})\Longrightarrow (\mathrm {ii})$ .

Now, we prove $(\mathrm {ii})\Longrightarrow (\mathrm {i})$ . Assume that (ii) holds. According to (2.4), we conclude that, for any $R\in \mathcal {R}_p^{n+1}$ and any measurable set $E\subset R^+(\gamma )$ ,

(2.5) $$ \begin{align} C^{-\frac{1}{\delta}}\left(u^r\right)(R^-(\gamma)) \left[\frac{|E|}{|R^+(\gamma)|}\right] ^{\frac{1}{\delta}}\leq\left(v^r\right)(E). \end{align} $$

Fix $R\in \mathcal {R}_p^{n+1}$ and, for any given $\lambda \in (0,\infty )$ , let $E_\lambda :=R^+(\gamma )\cap \{v^{-r}>\lambda \}$ . Then $(v^r)(E_\lambda )\leq |E_\lambda |/\lambda $ . Combining this and (2.5), we find that, for any $\lambda \in (0,\infty )$ ,

$$ \begin{align*} C^{-\frac{1}{\delta}}\left(u^r\right)(R^-(\gamma)) \left[\frac{|E_\lambda|}{|R^+(\gamma)|} \right]^{\frac{1}{\delta}}\leq\left(v^r\right)(E_\lambda) \leq\frac{|E_\lambda|}{\lambda}, \end{align*} $$

which further implies that

(2.6) $$ \begin{align} |E_\lambda|\leq\frac{C^{\frac{1}{1-\delta}}|R^-(\gamma)| ^{\frac{1}{1-\delta}}}{\lambda^{\frac{\delta}{1-\delta}} [(u^r)(R^-(\gamma))]^{\frac{\delta}{1-\delta}}}. \end{align} $$

Since $r\in (\frac {1}{\delta },\infty )$ , it follows that $\frac {r'}{r}\in (0,\frac {\delta }{1-\delta })$ . From the Cavalieri principle (see, for example, [Reference Grafakos34, Proposition 1.1.4]), the obvious fact that $E_\lambda \subset R^+(\gamma )$ for any $\lambda \in (0,\infty )$ , and (2.6), we infer that

where the implicit positive constant depends only on C, r, and $\delta $ . Taking the supremum over all $R\in \mathcal {R}_p^{n+1}$ , we conclude that $(u,v)\in TA_{r,r}^+(\gamma )$ , which completes the proof of the sufficiency and hence Proposition 2.8.

Proof We first show the necessity of (i) and (ii). Assume that $(u,v)\in TA_{r,q}^+(\gamma )$ . Fix $R:=R(x,t,L)\in \mathcal {R}_p^{n+1}$ with $(x,t)\in \mathbb {R}^{n+1}$ and $L\in (0,\infty )$ . Let

$$ \begin{align*} P:=R\left(x,t-\frac{(\tau-1)(1+\alpha)L^p}{2}, \left[1+\frac{(\tau-1)(1+\alpha)}{2}\right]^{\frac{1}{p}}L\right). \end{align*} $$

Then

(3.3) $$ \begin{align} R^+(\alpha)\subset P^+(\widetilde{\alpha}) \ \ \mbox{and}\ \ R^+(\alpha)-(\mathbf{0},\tau(1+\alpha)L^p)\subset P^-(\widetilde{\alpha}), \end{align} $$

where $\widetilde {\alpha }:=\frac {2\alpha +(\tau -1)(1+\alpha )} {2+(\tau -1)(1+\alpha )}\in (\gamma ,1)$ . Moreover, we have

(3.4) $$ \begin{align} \left|P^\pm(\widetilde{\alpha})\right|=2^n \left[1+\frac{(\tau-1)(1+\alpha)}{2}\right]^{\frac{n}{p}} \left|R^\pm(\alpha)\right|. \end{align} $$

From Remark 2.2(ii), we deduce that $(u,v)\in TA_{r,q}^+(\widetilde {\alpha })$ , which, combined with (3.3) and (3.4), further implies that

This finishes the proof of the necessity of (i). Letting $r\to 1^+$ in the above argument, we find that the necessity of (ii) holds. Here and thereafter, $r\to 1^+$ means that $r\in (1,\infty )$ and $r\to 1$ .

Then we prove the sufficiency of (i) by the chaining argument. Suppose that (3.1) holds. Fix $R\in \mathcal {R}_p^{n+1}$ . Without loss of generality, we may suppose that R is centered at the origin $(\mathbf {0},0)$ . Let $m\in \mathbb {N}$ be such that there exists $\epsilon \in [0,1)$ satisfying

(3.5) $$ \begin{align} m=\log_2\left[\frac{\tau(1+\alpha)}{1-\alpha}\right]+ \frac{1}{p-1}\left\{1+\log_2\left[\frac{\tau(1+\alpha)} {\gamma}\right]\right\}+2+\epsilon. \end{align} $$

Partition each spatial edge of $R^+(\gamma )$ into $2^m$ equally long intervals and divide the temporal edge of $R^+(\gamma )$ into $J:=\lceil (1-\gamma )2^{pm}/(1-\alpha )\rceil $ equally long intervals. Then we obtain $2^{nm}J$ subrectangles of $R^+(\gamma )$ and denote them by $\{V_{i,j}^+\}_{i\in \mathbb {N}\cap [1,2^{nm}], j\in \mathbb {N}\cap [1,J]}$ . Fix $i\in \mathbb {N}\cap [1,2^{nm}]$ and $j\in \mathbb {N}\cap [1,J]$ . Notice that there exists $R_{i,j}\in \mathcal {R}_p^{n+1}$ such that the tops of both $R_{i,j}$ and $V_{i,j}^+$ coincide, $V_{i,j}^+\subset R_{i,j}^+(\alpha )$ , and

(3.6) $$ \begin{align} \frac{|V_{i,j}^+|}{|R_{i,j}^+(\alpha)|}= \frac{(1-\gamma)2^{pm}}{(1-\alpha)J}. \end{align} $$

Divide $R^-(\gamma )$ in the same manner as we partition $R^+(\gamma )$ . Then we obtain $2^{nm}J$ subrectangles of $R^-(\gamma )$ and denote them by $\{U_{k,\iota }^-\} _{k\in \mathbb {N}\cap [1,2^{nm}],\iota \in \mathbb {N}\cap [1,J]}$ . Fix $k\in \mathbb {N}\cap [1,2^{nm}]$ and $\iota \in \mathbb {N}\cap [1,J]$ . Observe that there exists $\widetilde {R}_{k,\iota }\in \mathcal {R}_p^{n+1}$ such that the bottoms of both $\widetilde {R}_{k,\iota }^+(\alpha )-(\mathbf {0}, \tau (1+\alpha )[l(\widetilde {R}_{k,\iota })]^p)$ and $U_{k,\iota }^-$ coincide, $U_{k,\iota }^-\subset \widetilde {R}_{k,\iota }^+(\alpha )-(\mathbf {0}, \tau (1+\alpha )[l(\widetilde {R}_{k,\iota })]^p)$ , and

(3.7) $$ \begin{align} \frac{|U_{k,\iota}^-|}{|\widetilde{R}_{k,\iota}^+(\alpha) -(\mathbf{0},\tau(1+\alpha)[l(\widetilde{R}_{k,\iota})]^p)|} =\frac{(1-\gamma)2^{pm}}{(1-\alpha)J}. \end{align} $$

We claim that there exists $\mathfrak {R}\in \mathcal {R}_p^{n+1}$ satisfying that, for any $i,k\in \mathbb {N}\cap [1,2^{nm}]$ and $j,\iota \in \mathbb {N}\cap [1,J]$ , there exist $N_j,\widetilde {N}_\iota \in \mathbb {N}$ and chains $\{P_d^{i,j}\}_{d=0}^{N_j}$ and $\{\widetilde {P}_d^{k,\iota }\}_{d=0}^{\widetilde {N}_\iota }$ consisting of congruent parabolic rectangles such that the following statements hold.

1. (i) $\mathfrak {R}\subset R^+(0)$ and $\mathfrak {R}^+(\alpha )-(\mathbf {0},\tau (1+\alpha ) [l(\mathfrak {R})]^p)\subset R^+(0)$ .

2. (ii) For any $i\in \mathbb {N}\cap [1,2^{nm}]$ and $j\in \mathbb {N}\cap [1,J]$ , $P_0^{i,j}=R_{i,j}$ and $P_{N_j}^{i,j}=\mathfrak {R}$ . For any $k\in \mathbb {N}\cap [1,2^{nm}]$ and $\iota \in \mathbb {N}\cap [1,J]$ , $\widetilde {P}_0^{k,\iota }=\widetilde {R}_{k,\iota }$ and $\widetilde {P}_{\widetilde {N}_\iota }^{k,\iota }=\mathfrak {R}$ .

3. (iii) For any $j\in \mathbb {N}\cap [1,J]$ ,

   $$ \begin{align*} N_j\leq2^{\frac{p}{p-1}+3p+1}\left[\frac{\tau(1+\alpha)} {1-\alpha}\right]^p\left[\frac{\tau(1+\alpha)}{\gamma}\right] ^{\frac{p}{p-1}}=:C_1. \end{align*} $$
   For any $\iota \in \mathbb {N}\cap [1,J]$ , $\widetilde {N}_\iota \leq 2C_1$ .

4. (iv) For any $i\in \mathbb {N}\cap [1,2^{nm}]$ , $j\in \mathbb {N}\cap [1,J]$ , and $d\in \mathbb {N}\cap [1,N_j]$ ,

   $$ \begin{align*} \frac{|(P_d^{i,j})^+(\alpha)\cap(S_{d-1}^{i,j})^-(\alpha)|} {|(P_d^{i,j})^+(\alpha)|}\in \left[\frac{1}{2^{n+1}},1\right], \end{align*} $$
   where $(S_h^{i,j})^-(\alpha ):= (P_h^{i,j})^+(\alpha )-(\mathbf {0}, \tau (1+\alpha )[l(P_h^{i,j})]^p)$ for any $h\in \mathbb {Z}_+ \cap [0,N_j]$ . Moreover, for any $k\in \mathbb {N}\cap [1,2^{nm}]$ , $\iota \in \mathbb {N}\cap [1,J]$ , and $d\in \mathbb {N}\cap [1,\widetilde {N}_\iota ]$ ,
   $$ \begin{align*} \frac{|(\widetilde{S}_d^{k,\iota}) ^-(\alpha)\cap(\widetilde{P}_{d-1}^{k,\iota})^+(\alpha)|} {|(\widetilde{S}_d^{k,\iota})^-(\alpha)|}\in \left[\frac{1}{2^{n+1}},1\right], \end{align*} $$
   where $(\widetilde {S}_h^{k,\iota })^-(\alpha ):= (\widetilde {P}_h^{k,\iota })^+(\alpha )-(\mathbf {0}, \tau (1+\alpha )[l(\widetilde {P}_h^{k,\iota })]^p)$ for any $h\in \mathbb {Z}_+\cap [0,\widetilde {N}_\iota ]$ .

Indeed, fix $i\in \mathbb {N}\cap [1,2^{nm}]$ and $j\in \mathbb {N}\cap [1,J]$ . We will specify $\mathfrak {R}$ and $N_j$ and construct the chain $\{P_d^{i,j}\}_{d=0}^{N_j}$ in the following two steps.

Step 1. In this step, we construct the chain corresponding to spatial variables. Assume that $R_{i,j}=Q(x_i,l)\times (t_j-2l^p,t_j)$ with $(x_i,t_j)\in \mathbb {R}^{n+1}$ and $l:=l(R)/2^m$ . Let $Q_0^i:=Q(x_i,l)$ . For any $d\in \mathbb {N}\cap [1,M_i]$ with $M_i\in \mathbb {N}$ determined later, let

(3.8) $$ \begin{align} Q_d^i:=Q_{d-1}^i-\frac{x_i} {\vert x_i\vert}\frac{\theta_il}{2}, \end{align} $$

where $\theta _i\in [1,\sqrt {n}]$ , depending only on the angle between $x_i$ and the spatial axes, is such that the center of $Q_d^i$ belongs to the boundary of $Q_{d-1}^i$ . Notice that there exists $b_i\in \mathbb {N}\cap [1,2^m-1]$ , depending only on $\vert x_i\vert $ , such that

(3.9) $$ \begin{align} \vert x_{i}\vert=\frac{\theta_i}{2}[l(R)-b_il]. \end{align} $$

From this and (3.8), it follows that, to ensure that $Q_{M_{i}}^i=Q(\mathbf {0},l)$ , we need to choose

(3.10) $$ \begin{align} M_{i}:=\frac{2\vert x_i\vert}{\theta_il} =\frac{l(R)}{l}-b_i=2^m-b_i. \end{align} $$

Observe that, for any $d\in \mathbb {N}\cap [1,M_i]$ ,

(3.11) $$ \begin{align} \frac{1}{2^n}\leq\frac{\vert Q_d^i\cap Q_{d-1}^i\vert} {\vert Q_d^i\vert}\leq\frac{1}{2}. \end{align} $$

Then $\{Q_d^i\}_{d=0}^{M_{i}}$ forms a chain in $\mathbb {R}^n$ starting from $Q(x_i,l)$ and ending with $Q(\mathbf {0},l)$ .

Step 2. In this step, we specify $\mathfrak {R}$ and $N_j$ and construct the chain $\{P_d^{i,j}\}_{d=0}^{N_j}$ based on the chains in Step 1. For any $d\in \mathbb {Z}_+\cap [0,M_i]$ , let

$$ \begin{align*} P_d^{i,j}:=Q_d^i\times\left(t_j-d\tau(1+\alpha)l^p-2l^p, t_j-d\tau(1+\alpha)l^p\right). \end{align*} $$

Then $P_0^{i,j}=R_{i,j}$ and $\text {pr}_x(P_{M_i}^{i,j}) =Q(\mathbf {0},l)$ . To determine $\mathfrak {R}$ , we first assume that $j=1$ and $i\in \mathbb {N}\cap [1,2^{nm}]$ such that $Q(x_i,l)$ intersects with the boundary of $\text {pr}_x(R)=Q(\mathbf {0},l(R))$ . From (3.9) and (3.10), we infer that $b_i=1$ and hence $M_i=2^m-1$ in this case. Let $N:=2^m-1$ and $\mathfrak {R}:=P_N^{i,1}$ . We show that (i) holds. Indeed, on the one hand, notice that

(3.12) $$ \begin{align} Q_N=Q(\mathbf{0},l)\subset Q(\mathbf{0},l(R))\ \ \mbox{and}\ \ t_1-N\tau(1+\alpha)l^p<t_1<[l(R)]^p. \end{align} $$

On the other hand, by (3.5), we obtain

$$ \begin{align*} m\geq\frac{1}{p-1}\left\{1+\log_2\left[\frac{\tau(1+\alpha)} {\gamma}\right]\right\}, \end{align*} $$

which further implies that $2^{(p-1)m}\geq \frac {2\tau (1+\alpha )}{\gamma }$ and hence $\gamma -\frac {2\tau (1+\alpha )}{2^{(p-1)m}}\geq 0$ . From this, the definitions of both J and l, and the fact that $\lceil s\rceil \leq 2s$ for any $s\in [1,\infty )$ , we deduce that

$$ \begin{align*} &t_1-(N+1)\tau(1+\alpha)l^p-(1-\alpha)l^p\\ &\quad=\left[\gamma+\frac{1-\gamma}{\lceil(1-\gamma)2^{pm} /(1-\alpha)\rceil}-\frac{(N+1)\tau(1+\alpha)}{2^{pm}} -\frac{(1-\alpha)}{2^{pm}}\right][l(R)]^p\\ &\quad\geq\left[\gamma+\frac{1-\gamma}{(1-\gamma)2^{pm+1} /(1-\alpha)}-\frac{(N+1)\tau(1+\alpha)}{2^{pm}} -\frac{(1-\alpha)}{2^{pm}}\right][l(R)]^p\\ &\quad=\left[\gamma-\frac{1-\alpha} {2^{pm+1}}-\frac{\tau(1+\alpha)}{2^{(p-1)m}}\right][l(R)]^p \geq\left[\gamma-\frac{2\tau(1+\alpha)}{2^{(p-1)m}}\right] [l(R)]^p\geq0. \end{align*} $$

This, together with both the fact that the bottom of $\mathfrak {R}^+(\alpha )-(\mathbf {0},\tau (1+\alpha ) [l(\mathfrak {R})]^p)$ is

$$ \begin{align*} Q(\mathbf{0},l)\times \left\{t_1-(N+1)\tau(1+\alpha)l^p-(1-\alpha)l^p\right\} \end{align*} $$

and (3.12), further implies that (i) holds.

Then we suppose that $j=1$ and $i\in \mathbb {N}\cap [1,2^{nm}]$ such that $Q(x_i,l)$ does not intersect with the boundary of $\text {pr}_x(R)=Q(\mathbf {0},l(R))$ . In this case, $b_i\neq 1$ and hence $M_{i}\in \mathbb {N}\cap [1,N)$ . For any $d\in \mathbb {N}\cap [M_i,N-1]$ , let $P_{d+1}^{i,j}:=P_d^{i,j}-(\mathbf {0},\tau (1+\alpha )l^{p})$ . Then we are easy to see that $P_{N}^{i,j}=\mathfrak {R}$ .

Now, we consider the residual situation $j\in \mathbb {N}\cap [2,J]$ and $i\in \mathbb {N}\cap [1,2^{nm}]$ . In this case, we first extend the chain $\{P_d^{i,j}\}_{d=0}^{M_i}$ to the chain $\{P_d^{i,j}\}_{d=0}^N$ in the same way as we did in the above case. Then we can easily verify that $\mathrm {pr}_x(P_N^{i,j})=\mathrm {pr}_x(P_N^{i,1})$ . However, in the temporal variable, the distance between the top of $P_N^{i,j}$ and that of $P_N^{i,1}$ is $(j-1)(1-\gamma )[l(R)]^p/J$ . We can shift every $P_d^{i,j}$ for $d\in \mathbb {N}\cap [1,2^{m-1}]$ and add $\widetilde {M}_j$ parabolic rectangles into the chain to guarantee that the eventual parabolic rectangle is $\mathfrak {R}$ . To be specific, we can choose $\beta _j\in [0,1)$ and $\widetilde {M}_j\in \mathbb {Z}_{+}$ such that

(3.13) $$ \begin{align} \frac{2^{m-1}\beta_j(1-\alpha)[l(R)]^p}{2^{pm}} +\frac{\widetilde{M}_j\tau(1+\alpha)[l(R)]^{p}}{2^{pm}} =\frac{(j-1)(1-\gamma)[l(R)]^p}{J}. \end{align} $$

Indeed, notice that there exists $\eta \in [1,2)$ such that

$$ \begin{align*} J=\left\lceil\frac{(1-\gamma)2^{pm}} {1-\alpha}\right\rceil =\frac{\eta(1-\gamma)2^{pm}}{1-\alpha} \end{align*} $$

and hence we can rewrite (3.13) as

(3.14) $$ \begin{align} 2^{m-1}\beta_j(1-\alpha)+\widetilde{M}_j\tau(1+\alpha) =\frac{(j-1)(1-\alpha)}{\eta}. \end{align} $$

Choose $\widetilde {M}_j\in \mathbb {Z}_{+}$ such that

$$ \begin{align*} \widetilde{M}_j\tau(1+\alpha)\leq\frac{(j-1)(1-\alpha)}{\eta} <(\widetilde{M}_j+1)\tau(1+\alpha). \end{align*} $$

That is, let $\widetilde {M}_j\in \mathbb {Z}_{+}$ be such that

(3.15) $$ \begin{align} \frac{(j-1)(1-\alpha)}{2\tau(1+\alpha)}-1&\leq \frac{(j-1)(1-\alpha)}{\eta\tau(1+\alpha)}-1<\widetilde{M}_j\\ &\leq\frac{(j-1)(1-\alpha)}{\tau(1+\alpha)} \leq\frac{(j-1)(1-\alpha)}{\tau(1+\alpha)}.\notag \end{align} $$

Let $\xi _j:=\frac {(j-1)(1-\alpha )}{\eta } -\widetilde {M}_j(1+\alpha )$ . Using this and the choice of $\widetilde {M}_j$ , we find that $\xi _j \in [0,\tau (1+\alpha ))$ . Select $\beta _j\in [0,\infty )$ such that $\xi _j=2^{m-1}\beta _{j}(1-\alpha )$ . From this and (3.5), we infer that

$$ \begin{align*} \beta_j=2^{\frac{-1}{p-1}-1-\epsilon}\left(\frac{\gamma} {1+\alpha}\right)^{\frac{1}{p-1}}\frac{\xi_j}{\tau(1+\alpha)}. \end{align*} $$

By this and $\xi _j\in [0,\tau (1+\alpha ))$ , we conclude that

$$ \begin{align*} 0\leq\beta_j<2^{-1}\left(\frac{\gamma}{1+\alpha}\right) ^{\frac{1}{p-1}}<\frac{1}{2}, \end{align*} $$

which, together with the choices of both $\widetilde {M}_j$ and $\beta _j$ , further implies that (3.14) holds. Finally, for any $d\in \mathbb {N}\cap [1,2^{m-1}]$ , we modify the definition of $P_d^{i,j}$ by setting

$$ \begin{align*} P_d^{i,j}:=&Q_d^i\times\left(t_j-d\left[\tau(1+\alpha)+ \beta_j(1-\alpha)\right]l^p-2l^p,\right.\\ &\qquad\qquad\left. t_j-d\left[\tau(1+\alpha)+\beta_j(1-\alpha)\right]l^p\right). \end{align*} $$

If $\widetilde {M}_j\in \mathbb {N}$ , then, for any $d\in \mathbb {N}\cap [N,N+\widetilde {M}_j-1]$ , let $P_{d+1}^{i,j}:=P_d^{i,j}-\left (\mathbf {0},\tau (1+\alpha ) l^p\right )$ . Then $P_{N+\widetilde {M}_j}^{i,j}=\mathfrak {R}$ . For convenience, let $\widetilde {M}_1:=0$ .

In conclusion, for any $i\in \mathbb {N}\cap [1,2^{nm}]$ and $j\in \mathbb {N}\cap [1,J]$ , we have constructed a chain $\{P_d^{i,j}\}_{d=0}^{N+\widetilde {M}_j}$ starting from $R_{i,j}$ and ending with $\mathfrak {R}$ . Let $N_j:=N+ \widetilde {M}_j$ . Applying the definitions of both N and J, (3.15), and (3.5), we obtain

$$ \begin{align*} N_j&=2^m-1+\widetilde{M}_j\leq2^m+ \frac{(j-1)(1-\alpha)}{\tau(1+\alpha)}\\ &\leq2^m+\frac{(J-1)(1-\alpha)}{\tau(1+\alpha)}\leq 2^m+\frac{(1-\gamma)2^{pm}}{1-\alpha}\frac{1-\alpha} {\tau(1+\alpha)}\\ &\leq2^{pm+1}\leq2^{\frac{p}{p-1}+3p+1} \left[\frac{\tau(1+\alpha)} {1-\alpha}\right]^p\left[\frac{\tau(1+\alpha)}{\gamma}\right] ^{\frac{p}{p-1}}. \end{align*} $$

From (3.11), the fact that $\beta _j\in [0,\frac {1}{2})$ , and the construction of $\{P_d^{i,j}\}_{d=0}^{N_j}$ , we deduce that, for any $i\in \mathbb {N}\cap [1,2^{nm}]$ , $j\in \mathbb {N}\cap [1,J]$ , and $d\in \mathbb {N}\cap [1,N_j]$ ,

$$ \begin{align*} \frac{|(P_d^{i,j})^+(\alpha)\cap(S_{d-1}^{i,j})^-(\alpha)|} {|(P_d^{i,j})^+(\alpha)|}\in \left[\frac{1}{2^{n+1}},1\right], \end{align*} $$

Similarly, for any $k\in \mathbb {N}\cap [1,2^{nm}]$ and $\iota \in \mathbb {N} \cap [1,J]$ , we can construct a chain $\{\widetilde {P}_d^{k,\iota }\}_{d=0}^{\widetilde {N_\iota }}$ such that $\widetilde {N}_\iota \in \mathbb {N}\cap [1,2C_1]$ , $\widetilde {P}_0^{k,\iota }=\widetilde {R}_{k,\iota }$ , $\widetilde {P}_{\widetilde {N}_\iota }^{k,\iota }=\mathfrak {R}$ , and

$$ \begin{align*} \frac{|(\widetilde{S}_d^{k,\iota}) ^-(\alpha)\cap(\widetilde{P}_{d-1}^{k,\iota})^+(\alpha)|} {|(\widetilde{S}_d^{k,\iota})^-(\alpha)|}\in \left[\frac{1}{2^{n+1}},1\right] \end{align*} $$

for any $d\in \mathbb {N}\cap [1,\widetilde {N}_\iota ]$ . This finishes the proof of (ii)–(iv) and hence the above claim.

Next, based on (i)–(iv), we can build a chain connecting $R_{i,j}$ and $\widetilde {R}_{k,\iota }$ for any $i,k\in \mathbb {N}\cap [1,2^{nm}]$ and $j,\iota \in \mathbb {N}\cap [1,J]$ . More precisely, for any $i,k\in \mathbb {N}\cap [1,2^{nm}]$ , $j,\iota \in \mathbb {N}\cap [1,J]$ , and $d\in \mathbb {Z}_+\cap [0,\widetilde {N}_\iota +N_j]$ , let

$$ \begin{align*} P_{d}^{i,j,k,\iota}:= \begin{cases} \widetilde{P}_{d}^{k,\iota},&\text{if }d\in\mathbb{Z}_+ \cap\left[0,\widetilde{N}_\iota\right],\\ P_{\widetilde{N}_\iota+N_j-d}^{i,j},&\text{if }d\in\mathbb{Z}_+ \cap\left[\widetilde{N}_\iota+1,\widetilde{N}_\iota+N_j\right], \end{cases} \end{align*} $$

and $(S_d^{i,j,k,\iota })^-(\alpha ):= (P_d^{i,j,k,\iota })^+(\alpha )-(\mathbf {0},\tau (1+\alpha ) [l(P_d^{i,j,k,\iota })]^p)$ . Then, for any $i,k\in \mathbb {N}\cap [1,2^{nm}]$ and $j,\iota \in \mathbb {N}\cap [1,J]$ , we have constructed a chain $\{P_{d}^{i,j,k,\iota }\}_{d=0}^{\widetilde {N}_\iota +N_j}$ consisting of congruent parabolic rectangles. From (i) through (iv), it follows that the following statements hold.

1. (v) For any $i,k\in \mathbb {N}\cap [1,2^{nm}]$ , $j,\iota \in \mathbb {N}\cap [1,J]$ , and $d\in \mathbb {Z}_+\cap [0,\widetilde {N}_\iota +N_j]$ , $P_d^{i,j,k,\iota }\subset R$ .

2. (vi) For any $i,k\in \mathbb {N}\cap [1,2^{nm}]$ and $j,\iota \in \mathbb {N}\cap [1,J]$ , $P_0^{i,j,k,\iota }=\widetilde {R}_{k,\iota }$ and $P_{\widetilde {N}_\iota +N_j}^{i,j,k,\iota }=R_{i,j}$ .

3. (vii) For any $i,k\in \mathbb {N}\cap [1,2^{nm}]$ , $j,\iota \in \mathbb {N}\cap [1,J]$ , and $d\in \mathbb {N}\cap [1,\widetilde {N}_\iota +N_j]$ ,

   $$ \begin{align*} \frac{|(S_d^{i,j,k,\iota})^-(\alpha) \cap(P_{d-1}^{i,j,k,\iota}) ^+(\alpha)|}{|(S_d^{i,j,k,\iota})^-(\alpha)|} \in\left[\frac{1}{2^{n+1}},1\right]. \end{align*} $$

Now, we prove that there exists a positive constant $C_2$ such that, for any given $i,k\in \mathbb {N}\cap [1,2^{nm}]$ and $j,\iota \in \mathbb {N}\cap [1,J]$ ,

(3.16)

Indeed, by [Reference Kinnunen and Saari50, Lemma 7.4], [Reference Kinnunen and Myyryläinen45, Theorems 3.1 and 4.1], and the assumption that $u\in A_\infty ^+(\gamma )$ , we find that there exist $K,\delta \in (0,\infty )$ such that, for any $R\in \mathcal {R}_p^{n+1}$ and any measurable set $E\subset R^+(\alpha )$ ,

$$ \begin{align*} \left(u^q\right)\left(R^+(\alpha)- (0,\tau(1+\alpha)l(R)^p)\right)\leq K\left[\frac{|R^+(\alpha)|}{|E|}\right]^{\delta} \left(u^q\right)(E). \end{align*} $$

From this and (vii), we infer that, for any $d\in \mathbb {N} \cap [1,\widetilde {N}_\iota +N_j]$ ,

Iterating this inequality and using (iii) and (vi), we obtain

Hence, (3.16) holds with $C_2:=[2^{(n+1)\delta }K]^{3C_1}$ .

To show $(u,v)\in TA_{r,q}^+(\gamma )$ , we still need the following estimate. From the fact that $\lceil s\rceil \leq 2s$ for any $s\in [1,\infty )$ , (3.5), and the definition of J, we deduce that

(3.17) $$ \begin{align} 2^{nm}J&=2^{nm}\left\lceil\frac{(1-\gamma)2^{pm}}{1-\alpha} \right\rceil\leq\frac{1-\gamma}{1-\alpha}2^{(n+p)m+1}\\ &\leq\frac{1-\gamma}{1-\alpha}\left(\frac{1+\alpha}{1-\alpha} \right)^{n+p}\left(\frac{1+\alpha}{\gamma}\right) ^{n+p}2^{\frac{n+p}{p-1}+3(n+p)+1}=:C_3.\notag \end{align} $$

Applying the obvious facts that

$$ \begin{align*} R^+(\gamma)=\bigcup_{i=1}^{2^{nm}}\bigcup_{j=1}^J V_{i,j}^+ \ \ \mbox{and}\ \ R^-(\gamma)=\bigcup_{k=1}^{2^{nm}} \bigcup_{\iota=1}^JU_{k,\iota}^-, \end{align*} $$

$V_{i,j}^+\subset R_{i,j}^+(\alpha )$ for any $i\in \mathbb {N}\cap [1,2^{nm}]$ and $j\in \mathbb {N}\cap [1,J]$ , (3.6), $U_{k,\iota }^-\subset \widetilde {R}_{k,\iota }^+(\alpha ) -(\mathbf {0},\tau (1+\alpha )l(\widetilde {R}_{k,\iota })^p)$ for any $k\in \mathbb {N}\cap [1,2^{nm}]$ and $\iota \in \mathbb {N}\cap [1,J]$ , (3.7), (3.16), (3.17), (3.1), the definition of J, and the fact that $\lceil s\rceil \leq 2s$ for any $s\in [1,\infty )$ , we conclude that

(3.18)

Taking the supremum over all $R\in \mathcal {R}_p^{n+1}$ , we obtain $(u,v)\in TA_{r,q}^+(\gamma )$ and

$$ \begin{align*} [u,v]_{TA_{r,q}^+(\gamma)}\leq2^{1+\frac{q}{r'}}\max\left\{1,\, C_3^{1-\frac{q}{r'}}\right\}C_2C, \end{align*} $$

which completes the proof of the sufficiency of (i).

Finally, to prove the present theorem, it remains to show the sufficiency of (ii). Indeed, using [Reference Grafakos34, Exercises 1.1.3(a)] and letting $r\to 1^+$ in (3.18) with the assumption therein replaced by (3.2), we find that the sufficiency of (ii) holds, which then completes the proof of Theorem 3.1.

Proof From the assumption that $u\in A_\infty ^+(\gamma )$ and [Reference Kinnunen and Saari50, Lemma 7.4], it follows that there exists $q_u\in (1,\infty )$ such that $u^q\in A_{q_u}^+(\gamma )$ . By this and [Reference Kinnunen and Myyryläinen45, Corollary 5.3], we conclude that there exist positive constants C and $\delta _0$ , depending only on n, p, $\gamma $ , q, and the weight constant of $u^q$ , such that, for any $R\in \mathcal {R}_p^{n+1}$ ,

which further implies that

From this and Corollary 3.3(ii), we infer that $(u,v)\in TA_{r,q+\delta _0}^+(\gamma )$ . This, together with Proposition 2.3(ii), further implies that, for any $\delta \in (0,\delta _0)$ , $(u,v)\in TA_{r,q+\delta }^+(\gamma )$ . This finishes the proof of Corollary 3.5.

Proof We first show the sufficiency. Assume that (4.1) holds for any $f\in L^r(\mathbb {R}^{n+1},v^r)$ . Fix $R\in \mathcal {R}_p^{n+1}$ . From Definition 2.4, we deduce that, for any $\lambda \in (0,|R^+(\gamma )|^\beta (v^{-r'})_{R^+(\gamma )})$ and for almost every $(x,t)\in R^-(\gamma )$ ,

which further implies that $R^-(\gamma )\subset \{M^{\gamma +}_\beta (v^{-r'} \boldsymbol {1}_{R^+(\gamma )})>\lambda \}$ up to a set of measure zero. Combining this and (4.1), we obtain

$$ \begin{align*} \int_{R^-(\gamma)}u^q&\leq \left(u^q\right)\left(\left\{M^{\gamma+}_\beta\left(v^{-r'} \boldsymbol{1}_{R^+(\gamma)}\right)>\lambda\right\}\right) \leq\frac{C}{\lambda^q}\left[\int _{R^+(\gamma)}v^{-r'}\right]^{\frac{q}{r}}. \end{align*} $$

Letting $\lambda \to |R^+(\gamma )|^\beta (v^{-r'})_{R^+(\gamma )}$ , dividing both sides by $|R^+(\gamma )|$ , and using $\frac {1}{r}-\frac {1}{q}=\beta $ , we find that

Taking the supremum over all $R\in \mathcal {R}_p^{n+1}$ , we conclude that $(u,v)\in TA_{r,q}^+(\gamma )$ and $[u,v]_{TA_{r,q}^+(\gamma )}\leq C$ , which completes the proof of the sufficiency.

Next, we prove the necessity. Let $(u,v)\in TA_{r,q}^+(\gamma )$ , $f\in L^r(\mathbb {R}^{n+1},v^r)$ , and $\lambda \in (0,\infty )$ . We need to show that (4.1) holds. To this end, we divide the proof into the following five steps.

Step 1. In this step, we make the following simplification.

1. (i) We may assume that f is bounded and has compact support. Indeed, for any $k\in \mathbb {N}$ , let

   $$ \begin{align*} f_k:=\max\left\{|f|,\,k\right\}\mathbf{1}_{R(\mathbf{0},0,k)}. \end{align*} $$
   Then, for any $k\in \mathbb {N}$ , $f_k$ is bounded and has compact support and $f_k\to f$ almost everywhere on $\mathbb {R}^{n+1}$ as $k\to \infty $ . From this and the monotone convergence theorem, it follows that
   (4.2) $$ \begin{align} \int_{\mathbb{R}^{n+1}}|f_k|^rv^r\to\int_{\mathbb{R}^{n+1}} |f|^rv^r \end{align} $$
   as $k\to \infty $ . In addition, by an argument similar to that used in the proof of [Reference Cruz-Uribe and Fiorenza18, Lemma 3.30] and the monotone convergence theorem again, we find that
   $$ \begin{align*} \left(u^q\right)\left(\left\{M^{\gamma+}_\beta(f_k)>\lambda \right\}\right)\to\left(u^q\right)\left(\left\{M^{\gamma+} _\beta(f)>\lambda\right\}\right) \end{align*} $$
   as $k\to \infty $ . From this and (4.2), we deduce that, to prove (4.1) for f, it suffices to show that (4.1) holds for any $f_k$ with $k\in \mathbb {N}$ and the positive constant C independent of k, $\lambda $ , and f.

2. (ii) We may assume that both $u^q$ and $v^r$ have a lower bound A for some $A\in (0,\infty )$ . Indeed, applying Proposition 2.6, we conclude that $(\max \{u,\, A^{\frac {1}{q}}\},\max \{v,\,A^{\frac {1}{r}}\})\in TA_{r,q}^+(\gamma )$ . If we have

   $$ \begin{align*} \left(\max\{u^q,\,A\}\right) \left(\left\{M^{\gamma+}_\beta(f)>\lambda\right\}\right) \lesssim\frac{1}{\lambda^q}\left[\int_{\mathbb{R}^{n+1}}|f|^r \max\left\{v^r,\,A\right\}\right]^{\frac{q}{r}}, \end{align*} $$
   where the implicit positive constant is independent of A, $\lambda $ , and f, then we obtain (4.1) by letting $A\to 0$ and taking the supremum over all $\lambda \in (0,\infty )$ .

3. (iii) Fix $a\in (0,1)$ . Let $M_{\beta ,a}^{\gamma +}$ denote the truncated uncentered parabolic forward in time maximal operator, that is, for any $g\in L_{\mathrm {loc}}^1(\mathbb {R}^{n+1})$ and $(x,t)\in \mathbb {R}^{n+1}$ ,

   
   To prove (4.1), we only need to show that
   (4.3) $$ \begin{align} \left(u^q\right)\left(\left\{M_{\beta,a}^{\gamma+}(f)>\lambda \right\}\right)\lesssim\frac{1}{\lambda^q} \left(\int_{\mathbb{R}^{n+1}}|f|^rv^r\right)^{\frac{q}{r}}, \end{align} $$
   where the implicit positive constant is independent of a, $\lambda $ , and f. Indeed, in (4.3), letting $a\to 0$ and taking the supremum over all $\lambda \in (0,\infty )$ , we then obtain (4.1).

4. (iv) To prove (4.3), it suffices to show that

   (4.4) $$ \begin{align} \left(u^q\right)\left(\left\{\lambda<M_{\beta,a}^{\gamma+}(f) \leq2\lambda\right\}\right) \leq\frac{1}{\lambda^q}\left(\int_{\mathbb{R}^{n+1}}|f|^rv^r \right)^{\frac{q}{r}}, \end{align} $$
   where the implicit positive constant is independent of a, $\lambda $ , and f. Indeed, from (4.4), we infer that
   $$ \begin{align*} \left(u^q\right)\left(\left\{M_{\beta,a}^{\gamma+}(f)>\lambda \right\}\right)&\leq\sum_{k\in\mathbb{Z}_+}\left(u^q\right) \left(\left\{2^k\lambda<M_{\beta,a}^{\gamma+}(f)\leq 2^{k+1}\lambda\right\}\right)\\ &\lesssim\sum_{k\in\mathbb{Z}_+}\frac{1}{2^{kq}\lambda^q} \left(\int_{\mathbb{R}^{n+1}}|f|^rv^r\right)^{\frac{q}{r}}\\ &\sim\frac{1}{\lambda^q} \left(\int_{\mathbb{R}^{n+1}}|f|^rv^r\right)^{\frac{q}{r}} \end{align*} $$
   and hence (4.3) holds.

5. (v) Fix any compact set $K\subset \{\lambda <M_{\beta ,a}^{\gamma +}(f) \leq 2\lambda \}$ . Then, to prove (4.4), it suffices to show

   (4.5) $$ \begin{align} \left(u^q\right)(K)\lesssim\frac{1}{\lambda^q} \left(\int_{\mathbb{R}^{n+1}}|f|^rv\right)^{\frac{q}{r}}, \end{align} $$
   where the implicit positive constant is independent of K, a, $\lambda $ , and f. Indeed, using the inner regularity of the Lebesgue measure (see, for example, [Reference Rudin83, Theorem 2.14(d)]) and taking the supremum over all compact subsets of $\{\lambda <M_{\beta ,a}^{\gamma +}(f)\leq 2\lambda \}$ , we then obtain (4.4).

Step 2. In this step, we aim to prove that there exist $N\in \mathbb {N}$ and a sequence $\{P_i\}_{i=1}^N$ of parabolic rectangles such that

(4.6) $$ \begin{align} \left(u^q\right)(K)\leq2\sum_{i=1}^{N}\left(u^q\right) \left(P_i^-(\alpha)\right), \end{align} $$

where $\alpha :=\frac {\gamma }{5^p}$ . To begin with, from the definitions of both K and $M_{\beta ,a}^{\gamma +}$ , we deduce that, for any given $(x,t)\in K$ , there exists $R_{(x,t)}\in \mathcal {R}_p^{n+1}$ such that $(x,t)\in R_{(x,t)}^-(\gamma )$ , $l(R_{(x,t)})\in [a,\infty )$ , and

(4.7)

Using the assumption on f in (i) of Step 1, we find that $f\in L^1(\mathbb {R}^{n+1})$ . This, together with (4.7), further implies that

$$ \begin{align*} 2^n(1-\gamma)\left[l(R_{(x,t)})\right]^{n+p}&= \left|R_{(x,t)}^+(\gamma)\right| <\left[\frac{1}{\lambda}\int_{R_{(x,t)}^+(\gamma)}|f|\right] ^{\frac{1}{1-\beta}}\\ &\leq\left[\frac{\|f\|_{L^1(\mathbb{R}^{n+1})}}{\lambda}\right] ^{\frac{1}{1-\beta}}<\infty. \end{align*} $$

Combining this and the fact that $l(R_{(x,t)})\in [a,\infty )$ , we conclude that

(4.8) $$ \begin{align} a\leq l\left(R_{(x,t)}\right) \leq\left[\frac{\|f\|_{L^1(\mathbb{R}^{n+1})}} {\lambda}\right]^{\frac{1}{(n+p)(1-\beta)}} \left[\frac{1}{2^n(1-\gamma)}\right]^{\frac{1}{n+p}}. \end{align} $$

Let $P_{(x,t)}:=5R_{(x,t)}$ . Here and thereafter, for any $R:=\prod _{i=1}^{n}[y_i-L_i,y_i+L_i]\times (s-L^p,s+L^p)\subset \mathbb {R}^{n+1}$ with $\{y_i\}_{i=1}^{n}\subset \mathbb {R}$ , $\{L_i\}_{i=1}^{n}\subset (0,\infty )$ , $s\in \mathbb {R}$ , and $L\in (0,\infty )$ and for any $\Lambda \in (0,\infty )$ , define $\Lambda R:=\prod _{i=1}^{n}[y_i-\Lambda L_i,y_i+\Lambda L_i]\times (s-(\Lambda L)^p,s+(\Lambda L)^p)$ . Then it is easy to verify that $R_{(x,t)}^+(\gamma )\subset P_{(x,t)}^+(\alpha )$ and $R_{(x,t)}^-(\gamma )\subset P_{(x,t)}^-(\alpha )$ . In addition, from the fact that K is compact and (4.8), it follows that $\bigcup _{(x,t)\in K}P_{(x,t)}$ is bounded and hence $u^q$ is integrable on $\bigcup _{(x,t)\in K}P_{(x,t)}$ . Applying this, the absolute continuity of the Lebesgue integral, (4.8), and (ii) in Step 1, we find that there exists $\epsilon \in (0,1)$ such that, for any $(x,t)\in K$ ,

$$ \begin{align*} &\left(u^q\right)\left((1+\epsilon)P_{(x,t)}^-(\alpha)\setminus P_{(x,t)}^-(\alpha)\right)\\ &\quad\leq A2^n5^{n+p}a^{n+p}(1-\alpha) \leq A\left|P_{(x,t)}^-(\alpha)\right| \leq(u^q)\left(P_{(x,t)}^-(\alpha)\right), \end{align*} $$

which further implies that

(4.9) $$ \begin{align} \left(u^q\right)\left((1+\epsilon)P_{(x,t)}^-(\alpha)\right) \leq2\left(u^q\right)\left(P_{(x,t)}^-(\alpha)\right). \end{align} $$

On the other hand, let

(4.10) $$ \begin{align} \widetilde{\epsilon}:=\min\left\{ \left(\frac{1-\alpha}{2}\right)^{\frac{1}{p}} \left[(1+\epsilon)^p-1\right]^{\frac{1}{p}},\,\epsilon\right\}. \end{align} $$

From the finite covering theorem and the fact that

$$ \begin{align*} K\subset\bigcup_{(x,t)\in K} \mathrm{int\,}(R(x,t,5a\widetilde{\epsilon})), \end{align*} $$

we deduce that there exist $N_1\in \mathbb {N}$ and $\{(x_k,t_k)\}_{k=1}^{N_1}\subset K$ such that

(4.11) $$ \begin{align} K\subset\bigcup_{k=1}^{N_1}\mathrm{int\,}(R(x_k,t_k, 5a\widetilde{\epsilon}))\subset\bigcup_{k=1}^{N_1} R(x_k,t_k,5a\widetilde{\epsilon}), \end{align} $$

where, for any $E\subset \mathbb {R}^{n+1}$ , $\mathrm {int\,}(E)$ denotes the interior of E.

Now, we select a subsequence of $\{R_{(x_k,t_k)}\} _{k=1}^{N_1}$ by two steps. Assume that, for any ${k\in \mathbb {N}\cap [1,N_1]}$ , $R_{(x_k,t_k)}:=R(y_k,s_k,L_k)$ with $(y_k,s_k)\in \mathbb {R}^{n+1}$ and $L_k\in (0,\infty )$ . Without loss of generality, we may assume that the tops of $\{R_{(x_k,t_k)}\}_{k=1}^{N_1}$ are monotonically descending, that is, for any $k,j\in \mathbb {N}\cap [1,N_1]$ with $k\leq j$ , $s_k+L_k^p\geq s_j+L_j^p$ ; otherwise, we can rearrange $\{R_{(x_k,t_k)}\}_{k=1}^{N_1}$ in terms of the t-coordinates of their tops.

Then we make the first selection inductively. Select $R_{(x_1,t_1)}$ and denote it by $R_{(x_{k_1},t_{k_1})}$ . Suppose that we have selected the subsequence $\{R_{(x_{k_i},t_{k_i})}\}_{i=1}^{m}$ of $\{R_{(x_k,t_k)}\}_{k=1}^{N_1}$ , where $m\in \mathbb {N} \cap [1,N_1)$ and $k_m<N_1$ . Let

$$ \begin{align*} j_m:=\min\left\{j\in\mathbb{N}:\ k_m+j\leq N_1\mbox{ and } \left(x_{k_m+j},t_{k_m+j}\right)\notin\bigcup_{i=1}^{m} P_{(x_{k_i},t_{k_i})}^-(\alpha)\right\} \end{align*} $$

with the convention $\inf \emptyset =\infty $ . If $j_m\in \mathbb {N}$ , then select $R_{(x_{k_m+j_m},t_{k_m+j_m})}$ and denote it by $R_{(x_{k_{m+1}},t_{k_{m+1}})}$ ; otherwise, we terminate the selection process. In this manner, we have obtained a subsequence $\{R_{(x_{k_i},t_{k_i})}\}_{i=1}^{N_2}$ of $\{R_{(x_k,t_k)}\}_{k=1}^{N_1}$ , where $N_2\in \mathbb {N} \cap [1,N_1]$ . This finishes the first step of selection process.

Without loss of generality, we may assume that the edge lengths of $\{R_{(x_{k_i},t_{k_i})}\}_{i=1}^{N_2}$ are monotonically decreasing; otherwise, we can rearrange $\{R_{(x_{k_i},t_{k_i})}\}_{i=1}^{N_2}$ in terms of their edge lengths. Then we make the second selection inductively. Select $R_{(x_{k_1},t_{k_1})}$ and denote it by $R_{(x_{k_{r_1}},t_{k_{r_1}})}$ . Assume that we have selected the subsequence $\{R_{(x_{k_{r_i}},t_{k_{r_i}})}\}_{i=1}^m$ of $\{R_{(x_{k_i},t_{k_i})}\}_{i=1}^{N_2}$ , where $m\in \mathbb {N}\cap [1,N_2)$ and $k_{r_m}<N_2$ . Let

with the convention $\inf \emptyset =\infty $ . If $j_{r_m}\in \mathbb {N}$ , then select $R_{(x_{k_{r_m+}+j_{r_m}},t_{k_{r_m}+j_{r_m}})}$ and denote it by $R_{(x_{k_{r_{m+1}}},t_{k_{r_{m+1}}})}$ ; otherwise, we stop the selection process. By this way, we have obtained a subsequence $\{R_{(x_{k_{r_i}},t_{k_{r_i}})}\}_{i=1}^N$ of $\{R_{(x_{k_i},t_{k_i})}\}_{i=1}^{N_2}$ , where $N\in \mathbb {N}\cap [1,N_2]$ , which completes the second step of selection process. For convenience, for any $i\in \mathbb {N} \cap [1,N]$ , we simply write $R_{(x_{k_{r_i}},t_{k_{r_i}})}$ as $R_i$ . In conclusion, we have selected a subsequence $\{R_i\}_{i=1}^N$ of $\{R_{(x_k,t_k)}\}_{k=1}^{N_1}$ .

According to the above selection, we conclude that the following statements hold:

1. (i) For any $i,j\in \mathbb {N}\cap [1,N]$ with $i\neq j$ , , which is a direct consequence of the first selection.

2. (ii) For any $k\in \mathbb {N}\cap [1,N_1]$ , there exists $i\in \mathbb {N}\cap [1,N]$ such that $(x_k,t_k)\in P_i^-(\alpha )$ . Indeed, if $R_{(x_k,t_k)}\notin \{R_{(x_{k_i},t_{k_i})}\} _{i=1}^{N_2}$ , then, based on the first selection, there exists $i\in \mathbb {N}\cap [1,N_2]$ such that $(x_k,t_k)\in P_{(x_{k_i},t_{k_i})}^-(\alpha )$ . If $R_{(x_{k_i},t_{k_i})}\notin \{R_i\}_{i=1}^N$ , then, in light of the second selection, we have $P_{(x_{k_i},t_{k_i})}^-(\alpha )\subset \bigcup _{i=1}^N P_i^-(\alpha )$ .

3. (iii) For any $k\in \mathbb {N}\cap [1,N_2]$ , there exists $i\in \mathbb {N}\cap [1,N]$ such that

   (4.12) $$ \begin{align} R(x_k,t_k,5a\widetilde{\epsilon})\subset(1+\epsilon) P_i^-(\alpha).\\ \end{align} $$
   Indeed, let $i\in \mathbb {N}\cap [1,N]$ be such that
   $$ \begin{align*} (x_k,t_k)&\in P_i^-(\alpha)=P_{(x_{k_{r_i}},t_{k_{r_i}})}^-(\alpha)\\ &=Q\left(y_{k_{r_i}},5L_{k_{r_i}}\right)\times \left(s_{k_{r_i}}-\left(5L_{k_{r_i}}\right)^p, s_{k_{r_i}}-\alpha\left(5L_{k_{r_i}}\right)^p\right). \end{align*} $$
   Then $\|x_k-y_{k_{r_i}}\|_{\infty }\in [0,5L_{k_{r_i}}]$ and $s_{k_{r_i}}-t_k\in (\alpha (5L_{k_{r_i}})^p, (5L_{k_{r_i}})^p)$ . From this, (4.8), and (4.10), we infer that, for any $(y,s)\in R(x_k,t_k,5a\widetilde {\epsilon })$ ,
   $$ \begin{align*} s&<t_k+(5a)^p\widetilde{\epsilon}^p <s_{k_{r_i}}-\alpha\left(5L_{k_{r_i}}\right)^p +(5a)^p\widetilde{\epsilon}^p\\ &\leq s_{k_{r_i}}+(\widetilde{\epsilon}^p-\alpha) \left(5L_{k_{r_i}}\right)^p\\ &\leq s_{k_{r_i}}-\frac{1+\alpha}{2}\left(5L_{k_{r_i}}\right)^p +\frac{1-\alpha}{2}(1+\epsilon)^p\left(5L_{k_{r_i}}\right)^p, \end{align*} $$
   $$ \begin{align*} s&>t_k-(5a)^p\widetilde{\epsilon}^p >s_{k_{r_i}}-\left(5L_{k_{r_i}}\right)^p -(5a)^p\widetilde{\epsilon}^p\\ &\geq s_{k_{r_i}}-\left(5L_{k_{r_i}}\right)^p (1+\widetilde{\epsilon}^p)\\ &\geq s_{k_{r_i}}-\frac{1+\alpha}{2}\left(5L_{k_{r_i}}\right)^p -\frac{1-\alpha}{2}(1+\epsilon)^p\left(5L_{k_{r_i}}\right)^p, \end{align*} $$
   and
   $$ \begin{align*} \left\|y-y_{k_{r_i}}\right\|_\infty&\leq\|y-x_k\|_\infty+ \left\|x_k-y_{k_{r_i}}\right\|_\infty\\ &\leq5a\widetilde{\epsilon} +5L_{k_{r_i}}\leq5L_{k_{r_i}}(1+\epsilon), \end{align*} $$
   which further implies that $(y,s)\in (1+\epsilon )P_i^-(\alpha )$ . By the arbitrariness of $(y,s)$ , we obtain $R(x_k,t_k,5a\widetilde {\epsilon })\subset (1+\epsilon ) P_i^-(\alpha )$ .

4. (iv) For any given $k\in \mathbb {Z}$ and $R_i,R_j\in \{R_i\}_{i=1}^N$ with $l(R_i),l(R_j)\in (\frac {1}{2^{k+1}},\frac {1}{2^k}]$ , we have

   (4.13) $$ \begin{align} R_i^-(\gamma)\cap R_j^-(\gamma)=\emptyset. \end{align} $$
   We show this by contradiction. Indeed, from the first selection, we deduce that or . Without loss of generality, we may assume that . Suppose that there exist $k_0\in \mathbb {Z}$ and $R_i,R_j\in \{R_i\}_{i=1}^N$ satisfying $l(R_i),l(R_j)\in (\frac {1}{2^{k_0+1}}, \frac {1}{2^{k_0}}]$ and $R_i^-(\gamma )\cap R_j^-(\gamma )\neq \emptyset $ . Fix $(y_0,s_0)\in R_i^-(\gamma )\cap R_j^-(\gamma )$ . Then, for any $(y,s)\in R_i^-(\gamma )=R_{(x_{k_{r_i}},t_{k_{r_i}})}^-(\gamma )$ ,
   $$ \begin{align*} s_{k_{r_j}}-s\geq\gamma\left(L_{k_{r_j}}\right)^p=\alpha \left(5L_{k_{r_j}}\right)^p, \end{align*} $$
   $$ \begin{align*} s-s_{k_{r_j}}&=s-s_0+s_0-s_{k_{r_j}}>s-s_0- \left(L_{k_{r_j}}\right)^p\\ &>-(1-\gamma)\left(L_{k_{r_i}}\right)^p -\left(L_{k_{r_j}}\right)^p\\ &\geq-\left[2^p(1-\gamma)+1\right]\left(L_{k_{r_j}}\right)^p >-\left(5L_{k_{r_j}}\right)^p, \end{align*} $$
   and
   $$ \begin{align*} \left\|y-y_{k_{r_j}}\right\|_\infty\leq\|y-y_0\|_\infty+ \left\|y_0-y_{k_{r_j}}\right\|_\infty\leq2L_{k_{r_i}} +L_{k_{r_j}}\leq5L_{k_{r_j}}, \end{align*} $$
   which implies that $R_i^-(\gamma )\subset P_j^-(\alpha )$ . This contradicts . Thus, (4.13) holds.

Using (4.11) and (4.12), we conclude that

$$ \begin{align*} K\subset\bigcup_{k=1}^{N_1}R(x_k,t_k,5a\widetilde{\epsilon}) \subset\bigcup_{i=1}^N(1+\epsilon)P_i^-(\alpha), \end{align*} $$

which, combined with (4.9), completes the proof of (4.6) and hence Step 2.

Step 3. In this step, we prove that there exists a positive constant C such that, for any $i\in \mathbb {N}\cap [1,N]$ ,

(4.14) $$ \begin{align} \sum_{j\in\Gamma_i}\left[\int_{R_j^+(\gamma)}|f|\right]^{\frac 1{1-\beta}}\leq C_1\left[\int_{R_i^+(\gamma)}|f|\right]^{\frac 1{1-\beta}}, \end{align} $$

where $\Gamma _i:=\{j\in \mathbb {N}\cap [1,N]:\ R_i^+(\gamma )\cap R_j^+(\gamma )\neq \emptyset ,\ l(R_j)<l(R_i)\}$ . Fix $i\in \mathbb {N}\cap [1,N]$ and let

Then $\Gamma _{i,1}\cup \Gamma _{i,2}=\Gamma _{i}$ and $\Gamma _{i,1}\cap \Gamma _{i,2}=\emptyset $ . To show (4.14), it suffices to prove that there exist positive constants $C_2$ and $C_3$ , depending only on n, p, and $\gamma $ , such that, for any $h\in \{1,2\}$ ,

(4.15) $$ \begin{align} \sum_{j\in\Gamma_{i,h}}\left|R_j^+(\gamma)\right|\leq C_{h+1}\left|R_i^+(\gamma)\right|. \end{align} $$

Indeed, from this and (4.7), we infer that

$$ \begin{align*} \sum_{j\in\Gamma_i}\left[\int_{R_j^+(\gamma)}|f|\right]^{\frac 1{1-\beta}} &=\left(\sum_{j\in\Gamma_{i,1}}+\sum_{j\in\Gamma_{i,2}}\right) \left[\int_{R_j^+(\gamma)}|f|\right]^{\frac 1{1-\beta}}\\ &\leq(2\lambda)^{\frac 1{1-\beta}}\left(\sum_{j\in\Gamma_{i,1}} +\sum_{j\in\Gamma_{i,2}}\right)\left|R_j^+(\gamma)\right|\\ &\leq(C_2+C_3)(2\lambda)^{\frac 1{1-\beta}}\left|R_i^+(\gamma)\right|\\ &\leq(C_2+C_3)2^{\frac 1{1-\beta}}\left[\int_{R_i^+(\gamma)}|f|\right]^{\frac 1{1-\beta}} \end{align*} $$

and hence (4.14) holds with $C_1:=(C_2+C_3)2^{\frac 1{1-\beta }}$ .

Next, we turn to show (4.15). Let $k_0\in \mathbb {Z}$ be such that $l(R_i)\in (\frac {1}{2^{k_0+1}}, \frac {1}{2^{k_0}}]$ . We first consider the case $h=1$ . We claim that, for any $k\in \mathbb {Z}\cap [k_0,\infty )$ , there exists a measurable set $E_k\subset \mathbb {R}^{n+1}$ such that

$$ \begin{align*} \bigcup_{j\in\Gamma_{i,1},\,l(R_j) \in(\frac{1}{2^{k+1}},\frac{1}{2^k}]}R_j\subset E_k \end{align*} $$

and

$$ \begin{align*} |E_k|\leq\frac{2\times3^n(3-\gamma)}{2^{kp}}[l(R_i)]^n +\frac{2^33^{n-1}(2-\gamma)n}{2^k}[l(R_i)]^{n-1+p}. \end{align*} $$

Indeed, fix $k\in \mathbb {Z}\cap [k_0,\infty )$ . Denote the $2n+2$ faces of $R_i^+(\gamma )$ by $\{S_m\}_{m=1}^{2n+2}$ , where $S_1$ and $S_2$ denote, respectively, the top and the bottom of $R_i^+(\gamma )$ . For any given $j\in \Gamma _{i,1}$ such that $l(R_j)\in (\frac {1}{2^{k+1}},\frac {1}{2^k}]$ , since and $R_j^+(\gamma )\cap R_i^+(\gamma )\neq \emptyset $ , it follows that $R_j^+(\gamma )$ intersects with the boundary of $R_i^+(\gamma )$ , that is,

(4.16) $$ \begin{align} R_j^+(\gamma)\cap\left(\bigcup_{m=1}^{2n+2}S_m\right) \neq\emptyset. \end{align} $$

For any $S_m$ with $m\in \mathbb {N}\cap [1,2n+2]$ , there exists a rectangle $E_{k,m}\subset \mathbb {R}^{n+1}$ such that

and

$$ \begin{align*} \left|E_{k,m}\right|= \begin{cases} (3-\gamma)\left[l(R_i)+2l(R_j)\right]^n [l(R_j)]^p&\text{if }m\in\{1,2\},\\ 2l(R_j)\left[l(R_i)+2l(R_j)\right] ^{n-1}\Big\{(1-\gamma)\\ \ \ \left.\times[l(R_i)]^p +(3-\gamma)[l(R_j)]^p\right\}&\text{if}\ m\in\mathbb{N}\cap[3,2n+2]. \end{cases} \end{align*} $$

Applying this, (4.16), $l(R_j)<l(R_i)$ , and $l(R_j)\in (\frac {1}{2^{k+1}}, \frac {1}{2^k}]$ , we find that

and

$$ \begin{align*} |E_k|&\leq\sum_{m=1}^{2n+2}\left|E_{k,m}\right|\\ &\leq 2[3l(R_i)]^n\frac{3-\gamma}{2^{kp}}+ 2n\frac{2}{2^k}[3l(R_i)]^{n-1}(4-2\gamma)[l(R_i)]^p\\ &=\frac{2\times3^n(3-\gamma)}{2^{kp}}[l(R_i)]^n +\frac{2^33^{n-1}(2-\gamma)n}{2^k}[l(R_i)]^{n-1+p}. \end{align*} $$

This finishes the proof of the above claim.

Observe that, for any $j\in \Gamma _{i,1}$ , $l(R_j)<l(R_i)$ and hence

$$ \begin{align*} \Gamma_{i,1}=\bigcup_{k=k_0}^{\infty}\left\{j\in\Gamma_{i,1}: \ l\left(R_j\right)\in\left(\frac{1}{2^{k+1}}, \frac{1}{2^k}\right]\right\}. \end{align*} $$

From this, (iv) in Step 2, the above claim, and $l(R_i)\in (\frac {1}{2^{k_0+1}},\frac {1}{2^{k_0}}]$ , we deduce that

(4.17) $$ \begin{align} \sum_{j\in\Gamma_{i,1}}\left|R_j^+(\gamma)\right|& =\sum_{k=k_0}^\infty\sum_{\{j\in\Gamma_{i,1}:\ l(R_j)\in(\frac{1}{2^{k+1}}, \frac{1}{2^k}]\}}\left|R_j^+(\gamma)\right|\\&=\sum_{k=k_0}^\infty \sum_{\{j\in\Gamma_{i,1}:\ l(R_j)\in(\frac{1}{2^{k+1}}, \frac{1}{2^k}]\}}\left|R_j^-(\gamma)\right| \leq\sum_{k=k_0}^\infty|E_k|\notag\\&\leq\sum_{k=k_0}^\infty \left\{\frac{3^n(3-\gamma)}{2^{kp-1}} +\frac{3^{n-1}(2-\gamma)n}{2^{k-3}}[l(R_i)]^{p-1} \right\}[l(R_i)]^n\notag\\&=\left\{\frac{2^{2p+1}3^n(3-\gamma)}{2^p-1} \frac{1}{2^{(k_0+1)p}}\right.\notag\\&\quad\left.+2^53^{n-1}(2-\gamma)n[l(R_i)]^{p-1} \frac{1}{2^{k_0+1}}\right\}[l(R_i)]^n\notag\\&\leq\left[\frac{2^{2p+1}3^n(3-\gamma)}{2^p-1} +2^53^{n-1}(2-\gamma)n\right][l(R_i)]^{n+p}\notag\\&=\frac{2^{2p+1}3^n(3-\gamma)+2^53^{n-1}(2-\gamma) (2^p-1)n}{2^n(1-\gamma)(2^p-1)}\left|R_i^+(\gamma)\right|\nonumber\\&=:C_2\left|R_i^+(\gamma)\right|,\notag \end{align} $$

which completes the proof of (4.15) when $h=1$ .

Now, we prove (4.15) in the case $h=2$ . Let

$$ \begin{align*} \Omega_{k_0}:=\left\{j\in\Gamma_{i,2}:\ l(R_j)\in\left(\frac{1}{2^{k_0+1}}, \frac{1}{2^{k_0}}\right]\right\}. \end{align*} $$

For any $k\in \mathbb {N}\cap [k_0+1,\infty )$ , we define $\Omega _k$ inductively by setting

$$ \begin{align*} \Omega_k:=&\left\{j\in\Gamma_{i,2}:\ l(R_j)\in\left(\frac{1}{2^{k+1}}, \frac{1}{2^k}\right]\quad\mbox{and}\quad R_j^-(\gamma)\cap\bigcup_{d=k_0}^{k-1}\bigcup_{m\in\Omega_d} R_m^-(\gamma)=\emptyset\right\}. \end{align*} $$

Let $\Omega :=\bigcup _{k=k_0}^\infty \Omega _k$ . Notice that $\{\Omega _k\}_{k=k_0}^\infty $ are pairwise disjoint. In addition, for any $j\in \Omega $ , we have $R_j^+(\gamma )\subset R_i^+(\gamma )$ and hence $R_j^-(\gamma )\subset R_i$ . From these two facts and (iv) of Step 2, it follows that

(4.18) $$ \begin{align} \sum_{j\in\Omega}\left|R_j^+(\gamma)\right| &=\sum_{k=k_0}^\infty\sum_{j\in\Omega_k} \left|R_j^+(\gamma)\right| =\sum_{k=k_0}^\infty\sum_{j\in\Omega_k} \left|R_j^-(\gamma)\right|\\ &=\sum_{k=k_0}^\infty \left|\bigcup_{j\in\Omega_k}R_j^-(\gamma)\right| =\left|\bigcup_{k=k_0}^\infty\bigcup_{j\in\Omega_k} R_j^-(\gamma)\right|\notag\\ &\leq\left|\bigcup_{j\in\Gamma_{i,2}}R_j^-(\gamma)\right|\leq \left|R_i^-(\gamma)\right|=\left|R_i^+(\gamma)\right|.\notag \end{align} $$

Next, we show that, for any $m\in \Gamma _{i,2}\setminus \Omega $ , there exists $j\in \Omega $ such that $R_{j}^-(\gamma )\cap R_m^-(\gamma )\neq \emptyset $ , , and $l(R_{j})\in (l(R_m),\infty )$ . Indeed, let $k_{\mathrm {min}}\in [k_0,\infty )$ be the smallest integer such that there exists $j\in \Omega _{k_{\mathrm {min}}}$ satisfying $R_m^-(\gamma )\cap R_j^-(\gamma )\neq \emptyset $ . From (i) and (iv) in Step 2, we deduce that $k_{\mathrm {min}}>k_0$ and . Moreover, $l(R_{j})\in (l(R_m),\infty )$ . Otherwise, by the definition of $k_{\mathrm {min}}$ and (iv) in Step 2, we conclude that there exists $k'\in \mathbb {N}\cap [k_0,k_{\mathrm {min}})$ such that $m\in \Omega _{k'}$ , which contradicts the assumption that $m\in \Gamma _{i,2}\setminus \Omega $ . Thus, $l(R_{j})\in (l(R_m),\infty )$ .

For any $j\in \Omega $ , define

Let $j\in \Omega $ . Then there exists $\widetilde {k}\in \mathbb {Z} \cap [k_0,\infty )$ such that $l(R_j)\in (\frac {1}{2^{\widetilde {k}+1}},\frac {1}{2^{\widetilde {k}}}]$ . We claim that, for any $k\in \mathbb {Z}\cap [\widetilde {k},\infty )$ , there exists a measurable set $\widetilde {E}_k\subset \mathbb {R}^{n+1}$ such that

(4.19) $$ \begin{align} \bigcup_{m\in\widetilde{\Omega}_j,\,l(R_m) \in(\frac{1}{2^{k+1}},\frac{1}{2^k}]}R_m^-(\gamma) \subset\widetilde{E}_k \end{align} $$

and

(4.20) $$ \begin{align} \left|\widetilde{E}_k\right|\leq\frac{2^23^n (1-\gamma)}{2^{kp}}[l(R_j)]^n+\frac{2^23^n (1-\gamma)n}{2^k}[l(R_j)]^{n-1+p}. \end{align} $$

Indeed, fix $k\in \mathbb {Z}\cap [\widetilde {k},\infty )$ . Denote the $2n+2$ faces of $R_j^-(\gamma )$ by $\{\widetilde {S}_d\} _{d=1}^{2n+2}$ , where $\widetilde {S}_1$ and $\widetilde {S}_2$ denote, respectively, the top and the bottom of $R_j^-(\gamma )$ . For any given $m\in \widetilde {\Omega }_j$ such that $l(R_m)\in (\frac {1}{2^{k+1}},\frac {1}{2^k}]$ , since and $R_m^-(\gamma )\cap R_j^-(\gamma )\neq \emptyset $ , it follows that $R_m^-$ intersects with the boundary of $R_j^-(\gamma )$ , that is,

(4.21) $$ \begin{align} R_m^-(\gamma)\cap\left(\bigcup_{d=1}^{2n+2} \widetilde{S}_d\right)\neq\emptyset. \end{align} $$

For any $\widetilde {S}_d$ with $d\in \mathbb {N}\cap [1,2n+2]$ , there exists a rectangle $\widetilde {E}_{k,d}\subset \mathbb {R}^{n+1}$ such that

and

$$ \begin{align*} \left|\widetilde{E}_{k,d}\right|= \begin{cases} \left[l(R_j)+2l(R_m)\right]^n 2(1-\gamma)[l(R_m)]^p&\text{if }d\in\{1,2\},\\ 2l(R_m)\left[l(R_j)+2l(R_m)\right] ^{n-1}\Big\{(1-\gamma)\\ \ \ \left.\times[l(R_j)]^p +2(1-\gamma)[l(R_m)]^p\right\}&\text{if } d\in\mathbb{N}\cap[3,2n+2]. \end{cases} \end{align*} $$

Applying this, (4.21), $l(R_m)<l(R_j)$ , and $l(R_m)\in (\frac {1}{2^{k+1}}, \frac {1}{2^k}]$ , we obtain

and

$$ \begin{align*} \left|\widetilde{E}_k\right|&\leq\sum_{d=1}^{2n+2} \left|\widetilde{E}_{k,d}\right|\\ &\leq 2\left[3l(R_j)\right]^n \frac{2(1-\gamma)}{2^{kp}}+ 2n\frac{2}{2^k}\left[3l(R_j)\right] ^{n-1}3(1-\gamma)\left[l(R_j)\right]^p\\ &=\frac{2^23^n(1-\gamma)}{2^{kp}}[l(R_j)]^n +\frac{2^23^n(1-\gamma)n}{2^k} [l(R_j)]^{n-1+p}. \end{align*} $$

This finishes the proofs of (4.19) and (4.20). From this, the fact that

$$ \begin{align*} \widetilde{\Omega}_k=\bigcup_{k=\widetilde{k}}^\infty \left\{m\in\widetilde{\Omega}_k:\ l(R_m)\in\left(\frac{1}{2^{k+1}},\frac{1}{2^k}\right]\right\}, \end{align*} $$

(iv) in Step 2, and $l(R_j)\in (\frac {1}{2^{\widetilde {k}+1}},\frac {1}{2^{\widetilde {k}}}]$ , we infer that

$$ \begin{align*} \sum_{m\in\widetilde{\Omega}_j}\left|R_m^+(\gamma)\right| &=\sum_{k=\widetilde{k}}^\infty \sum_{\{m\in\widetilde{\Omega} _j:\ l(R_m)\in(\frac{1}{2^{k+1}},\frac{1}{2^k}]\}} \left|R_m^+(\gamma)\right|\\&=\sum_{k=\widetilde{k}}^\infty \sum_{\{m\in\widetilde{\Omega} _j:\ l(R_m)\in(\frac{1}{2^{k+1}},\frac{1}{2^k}]\}} \left|R_m^-(\gamma)\right| \leq\sum_{k=\widetilde{k}}^\infty\left|\widetilde{E}_k\right|\\&\leq\sum_{k=\widetilde{k}}^\infty \left\{\frac{3^n(1-\gamma)}{2^{kp-2}}+\frac{ 3^n(1-\gamma)n}{2^{k-2}} [l(R_j)]^{p-1}\right\}[l(R_j)]^n\\&=\left\{\frac{2^{2p+2}3^n}{2^p-1} \frac{1}{2^{(\widetilde{k}+1)p}}+2^43^nn [l(R_j)]^{p-1}\frac{1}{2^{\widetilde{k}+1}}\right\}[l(R_j)]^n\\&\leq\left[\frac{2^{2p+2}3^n(1-\gamma)}{2^p-1} +2^43^n(1-\gamma)n\right][l(R_j)]^{n+p}\\&=\frac{2^{2p+2}3^n(1-\gamma) +2^43^n(1-\gamma)(2^p-1)n}{2^n(1-\gamma)(2^p-1)} \left|R_j^+(\gamma)\right|\\&=:\widetilde{C}_3\left|R_j^+(\gamma)\right|. \end{align*} $$

Combining this and (4.18), we conclude that

$$ \begin{align*} \sum_{j\in\Gamma_{i,2}}\left|R_j^+(\gamma)\right| &=\left(\sum_{j\in\Omega}+\sum_{m\in\Gamma_{i,2} \setminus\Omega}\right)\left|R_j^+(\gamma)\right|\\ &\leq\sum_{j\in\Omega}\left|R_j^+(\gamma)\right| +\sum_{j\in\Omega}\sum_{m\in\widetilde{\Omega}_j} \left|R_m^+(\gamma)\right|\\ &\leq\sum_{j\in\Omega}\left|R_j^+(\gamma)\right| +\widetilde{C}_3\sum_{j\in\Omega}\left|R_j^+(\gamma)\right| =\left(\widetilde{C}_3+1\right)\sum_{j\in\Omega} \left|R_j^+(\gamma)\right|\\ &\leq\left(\widetilde{C}_3+1\right) \left|R_i^+(\gamma)\right|=:C_3\left|R_i^+(\gamma)\right|, \end{align*} $$

which completes the proof of (4.15) when $h=2$ . This, together with (4.17), further implies that (4.15) holds and hence (4.14).

Step 4. In this step, we prove that there exists a positive constant $C_4$ , depending only on n and p, such that, for any $i\in \mathbb {N}\cap [1,N]$ , there exists a measurable set $F_i\subset R_i^+(\gamma )$ such that

(4.22) $$ \begin{align} \frac{1}{|R_i^+(\gamma)|^{1-\beta}}\int_{F_i}|f| \in\left(\frac{\lambda}{2},\infty\right) \end{align} $$

and

(4.23) $$ \begin{align} \sum_{i=1}^N\boldsymbol{1}_{F_i}\leq C_4. \end{align} $$

In what follows, for any finite subset $A\subset \mathbb {N}$ , we denote the cardinality of A by $\sharp\! A$ . For any given $i\in \mathbb {N}\cap [1,N]$ , we define $F_i$ by considering the following two cases for $\sharp \Gamma _{i}$ .

Case 1: $\sharp \Gamma _{i}\leq 2^{\frac 1{1-\beta }}\lceil C_1\rceil $ . In this case, let $F_i:=R_i^+(\gamma )$ .

Case 2: $\sharp \Gamma _{i}>2^{\frac 1{1-\beta }}\lceil C_1\rceil $ . In this case, for any $k\in \mathbb {N}$ , let

$$ \begin{align*} E_i^k:=R_i^+(\gamma)\cap\left\{\sum_{j\in\Gamma_i} \boldsymbol{1}_{R_j^+(\gamma)}\in[k,\infty)\right\}. \end{align*} $$

Notice that, for any $k,k'\in \mathbb {N}$ with $k<k'$ , we have

(4.24) $$ \begin{align} E_i^{k'}\subset E_i^k \end{align} $$

and, for any $(x,t)\in R_i^+(\gamma )$ ,

(4.25) $$ \begin{align} \sum_{k=1}^{\sharp \Gamma_{i}} \boldsymbol{1}_{E_i^k}(x,t)=\sum_{j\in\Gamma_i} \boldsymbol{1}_{R_j^+(\gamma)}(x,t). \end{align} $$

Then we define $F_i:=R_i^+(\gamma )\setminus E_i^{2\lceil C_1\rceil }$ .

We show that, for any $i\in \mathbb {N}\cap [1,N]$ , $F_i$ satisfies (4.22). We consider the following two cases for $\sharp \Gamma _i$ .

Case 1: $\sharp \Gamma _i\leq 2^{\frac 1{1-\beta }}\lceil C_1\rceil $ . In this case, from the definition of $F_i=R_i^+(\gamma )$ and (4.7), it follows that

Case 2: $\sharp \Gamma _i>2^{\frac 1{1-\beta }}\lceil C_1\rceil $ . In this case, using (4.24), $E_i^k\subset R_i^+(\gamma )$ for any $k\in \mathbb {N}\cap [1,\sharp \Gamma _i]$ , (4.25), and (4.14), we obtain

$$ \begin{align*} &(\sharp \Gamma_i)^{\frac \beta{1-\beta}}2^{\frac 1{1-\beta}} \lceil C_1\rceil\left(\int_{E_i^{2\lceil C_1\rceil}}|f|\right)^{\frac 1{1-\beta}}\\ &\quad\leq\left(\sum_{k=1}^{\sharp \Gamma_i}\int_{E_i^k}|f|\right)^{\frac 1{1-\beta}} =\left[\int_{R_i^+(\gamma)}|f|\sum_{k=1}^{\sharp \Gamma_i} \boldsymbol{1}_{E_i^k}\right]^{\frac 1{1-\beta}}\\ &\quad=\left[\int_{R_i^+(\gamma)}|f| \sum_{j\in\Gamma_i}\boldsymbol{1}_{R_j^+(\gamma)}\right]^{\frac 1{1-\beta}} \leq(\sharp \Gamma_i)^{\frac \beta{1-\beta}} \sum_{j\in\Gamma_i}\left[\int_{R_j^+(\gamma)}|f|\right]^{\frac 1{1-\beta}}\\ &\quad\leq(\sharp \Gamma_i)^{\frac \beta{1-\beta}} \lceil C_1\rceil\left[\int_{R_i^+(\gamma)}|f|\right]^{\frac 1{1-\beta}}, \end{align*} $$

which, together with (4.7) and the definition of $F_i$ , further implies that

$$ \begin{align*} \int_{F_i}|f|=\int_{R_i^+(\gamma)}|f| -\int_{E_i^{2\lceil C_1\rceil}}|f| \geq\frac{1}{2}\int_{R_i^+(\gamma)}|f|>\frac{\lambda}{2} \left|R_i^+(\gamma)\right|{}^{1-\beta}. \end{align*} $$

Combing the above two cases, we complete the proof of (4.22).

Now, we turn to prove (4.23). To this end, we first show that, for any $(x,t)\in F_i$ ,

(4.26) $$ \begin{align} \sum_{j\in\Gamma_i}\boldsymbol{1}_{R_j^+(\gamma)}(x,t)\leq 2^{\frac 1{1-\beta}}\lceil C_1\rceil. \end{align} $$

Indeed, if $\sharp \Gamma _i\leq 2^{\frac 1{1-\beta }}\lceil C_1\rceil $ , then, for any $(x,t)\in F_i$ ,

(4.27) $$ \begin{align} \sum_{j\in\Gamma_i}\boldsymbol{1}_{R_j^+(\gamma)}(x,t)\leq \sharp \Gamma_i\leq2^{\frac 1{1-\beta}}\lceil C_1\rceil. \end{align} $$

If $\sharp \Gamma _i>2^{\frac 1{1-\beta }}\lceil C_1\rceil $ , then, from (4.25), we deduce that, for any $(x,t)\in F_i$ ,

$$ \begin{align*} \sum_{j\in\Gamma_i}\boldsymbol{1}_{R_j^+}(x,t)= \sum_{k=1}^{2\lceil C_1\rceil}\boldsymbol{1}_{E_i^k}(x,t) \leq2^{\frac 1{1-\beta}}\lceil C_1\rceil, \end{align*} $$

which, together with (4.27), completes the proof of (4.26).

Then we prove that there exists a positive constant $\widetilde {C}_4$ , depending only on n, p, and $\gamma $ , such that, for any given $(x,t)\in \mathbb {R}^{n+1}$ and $k\in \mathbb {Z}$ ,

(4.28) $$ \begin{align} \sum_{i\in\mathbb{N}\cap[1,N],\, l(R_i)\in(\frac{1}{2^{k+1}}, \frac{1}{2^k}]}\boldsymbol{1}_{R_i^+(\gamma)}(x,t) \leq\widetilde{C}_4. \end{align} $$

To do this, let

$$ \begin{align*} I_{(x,t)}^k:=\left\{i\in\mathbb{N}\cap[1,N]:\ l(R_i)\in\left(\frac{1}{2^{k+1}},\frac{1}{2^k}\right] \mbox{ and }(x,t)\in R_i^+(\gamma)\right\} \end{align*} $$

and

$$ \begin{align*} S_{(x,t)}:=Q\left(x,\frac{1}{2^{k-1}}\right)\times \left(t-\frac{1}{2^{kp-1}},t+\frac{1}{2^{kp-1}}\right). \end{align*} $$

Then, for any $(x,t)\in \mathbb {R}^{n+1}$ ,

(4.29) $$ \begin{align} \bigcup_{i\in I_{(x,t)}^k}R_i^-(\gamma)\subset S_{(x,t)}. \end{align} $$

Indeed, by the definition of $I_{(x,t)}^k$ , we find that, for any $i\in I_{(x,t)}^k$ and $(y,s)\in R_i^-(\gamma )$ ,

$$ \begin{align*} \|y-x\|_\infty\leq2l(R_i)\leq\frac{1}{2^{k-1}} \ \ \mbox{and}\ \ |t-s|\leq2[l(R_i)]^p\leq\frac{1}{2^{kp-1}}. \end{align*} $$

Therefore, $(y,s)\in S_{(x,t)}$ and hence (4.29) holds. From the definitions of $I_{(x,t)}^k$ and $S_{(x,t)}$ , (iv) in Step 2, and (4.29), we infer that

$$ \begin{align*} \sum_{\genfrac{}{}{0pt}{}{i\in\mathbb{N}\cap[1,N]}{l(R_i)\in(\frac{1}{2^{k+1}}, \frac{1}{2^k}]}}\boldsymbol{1}_{R_i^+(\gamma)}(x,t) &=\sum_{i\in I_{(x,t)}^k}1=\sum_{i\in I_{(x,t)}^k} \frac{|R_i^-(\gamma)|}{2^n(1-\gamma)[l(R_i)]^{n+p}}\\&\leq\frac{2^{(k+1)(n+p)-n}}{1-\gamma}\left|\bigcup_{i\in I_{(x,t)}^k}R_i^-(\gamma)\right|\\&\leq\frac{2^{(k+1)(n+p)-n}}{1-\gamma}\left|S_{(x,t)}\right|\\&=\frac{2^{(k+1)(n+p)-n}}{1-\gamma} \left(\frac{2}{2^{k-1}}\right)^n \frac{2}{2^{kp-1}}\\&=\frac{2^{2n+p+2}}{1-\gamma} =:\widetilde{C}_4, \end{align*} $$

which completes the proof of (4.28).

Finally, we show (4.23). For any $(x,t)\in \bigcup _{i=1}^N F_i$ , there exists $i_0\in \mathbb {N}\cap [1,N]$ such that $(x,t)\in F_{i_0}$ and $R_{i_0}$ is the largest parabolic rectangle in the following sense: for any $i\in \mathbb {N}\cap [1,N]$ satisfying $(x,t)\in F_i$ , $|R_i|\leq |R_{i_0}|$ . Let $k_0\in \mathbb {Z}$ be such that $l(R_{i_0})\in (\frac {1}{2^{k_0+1}}, \frac {1}{2^{k_0}}]$ . Applying the construction of $i_0$ , $F_i\subset R_i^+(\gamma )$ for any $i\in \mathbb {N}\cap [1,N]$ , the definition of $\Gamma _{i_0}$ , (4.26), and (4.28), we obtain

$$ \begin{align*} \sum_{i=1}^N\boldsymbol{1}_{F_i} &=\left\{\sum_{\genfrac{}{}{0pt}{}{i\in\mathbb{N}\cap[1,N]}{l(R_i)\in (\frac{1}{2^{k_0+1}},\frac{1}{2^{k_0}}]}}+ \sum_{\genfrac{}{}{0pt}{}{i\in\mathbb{N}\cap[1,N]}{l(R_i)\in (\frac{1}{2^{k+1}},\frac{1}{2^k}]}}\right\} \boldsymbol{1}_{F_i}\\ &\leq\left\{\sum_{\genfrac{}{}{0pt}{}{i\in\mathbb{N}\cap[1,N]}{l(R_i)\in (\frac{1}{2^{k_0+1}},\frac{1}{2^{k_0}}]}} +\sum_{i\in\Gamma_{i_0}}\right\}\boldsymbol{1} _{R_i^+(\gamma)} \leq\widetilde{C}_4+2^{\frac 1{1-\beta}}\lceil C_1\rceil=:C_4. \end{align*} $$

This finishes the proof of (4.23) and hence Step 4.

Step 5. In this step, we prove (4.5) by considering the following two cases for r.

Case 1: $r\in (1,\infty )$ . In this case, from (4.6), (4.22), the Hölder inequality, $F_i\subset R_i^+(\gamma )$ , $\alpha =\frac {\gamma }{5^p}$ , $P_i=5R_i$ for any $i\in \mathbb {N}\cap [1,N]$ , $\frac {1}{r}-\frac {1}{q}=\beta $ , Definition 2.1(i), and (4.23), we deduce that

(4.30)

This, together with Corollary 3.3, finishes the proof of (4.5) in this case.

Case 2: $r=1$ . In this case, by (4.6), (4.22), $\alpha =\frac {\gamma }{5^p}$ , $P_i=5R_i$ for any $i\in \mathbb {N}\cap [1,N]$ , $\frac {1}{r}-\frac {1}{q}=\beta $ , Definition 2.1(ii), and (4.23), we conclude that

$$ \begin{align*} \left(u^q\right)(K)&\leq2\sum_{i=1}^{N}\left(u^q\right) \left(P_i^-(\alpha)\right)\\ &\leq \frac{2^{q+1}}{\lambda^q}\sum_{i=1}^{N}\left(u^q\right) \left(P_i^-(\alpha)\right)\left[\frac{1}{|R_i^+(\gamma)| ^{1-\beta}}\int_{F_i}|f|\right]^q\\ &\leq\frac{2^{q+1}5^{n+p}(1-\alpha)}{(1-\gamma)\lambda^q} \sum_{i=1}^{N}\left(u^q\right)_{P_i^-(\alpha)} \left(\int_{F_i}|f|\right)^q\\ &\leq\frac{2^{q+1}5^{n+p}(1-\alpha) [u,v]_{TA_{1,q}^+(\alpha)}} {(1-\gamma)\lambda^q}\sum_{i=1}^{N} \left(\int_{F_i}|f|v\right)^q\\ &\leq\frac{2^{q+1}5^{n+p}(5^p-\gamma) [u,v]_{TA_{1,q}^+(\alpha)}C_4^q} {5^p\lambda^q}\left(\int_{\mathbb{R}^{n+1}}|f|v\right)^q. \end{align*} $$

This, together with Corollary 3.3, finishes the proof of (4.5) in this case. Combining this and (4.30), we obtain (4.5). This finishes the proof of the necessity and hence Theorem 4.1.

Proof We first show the sufficiency of both (i) and (ii). Let $\gamma \in [0,1)$ . Assume $\mathcal {M}^{\gamma +}_\beta $ is bounded from $L^r(\mathbb {R}^{n+1},v^r)$ to $L^{q,\infty }(\mathbb {R}^{n+1},u^q)$ . Fix $R:=R(x,t,L)\in \mathcal {R}_p^{n+1}$ with $(x,t)\in \mathbb {R}^{n+1}$ and $L\in (0,\infty )$ . Define $S^+(\gamma ):=R^-(\gamma )+(1-\gamma )L^p+2^p\gamma L^p$ and, for any $\epsilon \in (0,\infty )$ , let $f_\epsilon :=(v+\epsilon )^{-r'}\boldsymbol {1}_{S^+(\gamma )}$ . Then, for any $(y,s)\in R^-(\gamma )$ , $S^+(\gamma )\subset R(y,s,2L)^+(\gamma )$ and $|R(y,s,2L)^+(\gamma )|=2^{n+p}|S^+(\gamma )|$ . From this and Definition 2.4, it follows that, for any $\lambda \in (0,2^{(n+p)(\beta -1)} |S^+(\gamma )|^\beta (f_\epsilon )_{S^+(\gamma )})$ and $(y,s)\in R^-(\gamma )$ ,

which further implies that $R^-(\gamma )\subset \{M^{\gamma +}_\beta (f_\epsilon )>\lambda \}$ . Combining this and the assumption that $\mathcal {M}^{\gamma +}_\beta $ is bounded from $L^r(\mathbb {R}^{n+1},v^r)$ to $L^{q,\infty }(\mathbb {R}^{n+1},u^q)$ , we obtain

$$ \begin{align*} \int_{R^-(\gamma)}u^q&=(u^q)(R^-(\gamma))\leq (u^q)\left(\left\{M^{\gamma+}_\beta(f_\epsilon)>\lambda\right\}\right)\\ &\leq\frac{C}{\lambda^q}\left(\int_{\mathbb{R}^{n+1}} f_\epsilon^rv^r\right)^{\frac{q}{r}} =\frac{C}{\lambda^q}\left[\int _{S^+(\gamma)}(v+\epsilon)^{-r'r}v^r\right]^{\frac{q}{r}}. \end{align*} $$

Letting $\lambda \to 2^{(n+p)(\beta -1)} |S^+(\gamma )|^\beta (f_\epsilon )_{S^+(\gamma )}$ and $\epsilon \to 0$ , dividing both sides of the above inequality by $|R^+(\gamma )|$ , using $\frac {1}{r}-\frac {1}{q}=\beta $ , and taking the supremum over all $R\in \mathcal {R}_p^{n+1}$ , we find that

Thus, if $\gamma =0$ , this and Definition 2.1(i) imply $(u,v)\in TA_{r,q}^+(0)$ and, if $\gamma \in (0,1)$ , then applying this and Theorem 3.1 [here, we use the assumption that $u\in A_\infty ^+(\gamma )$ ] we also conclude $(u,v)\in TA_{r,q}^+(\gamma )$ , which then completes the proof of the sufficiency of both (i) and (ii).

Next, we prove the necessity of (i). Let $f\in L_{\mathrm {loc}}^1(\mathbb {R}^{n+1})$ . From Corollary 3.3(i) and Theorem 4.1, we infer that, to show that $\mathcal {M}^{\gamma +}_\beta $ is bounded from $L^r(\mathbb {R}^{n+1},v^r)$ to $L^{q,\infty }(\mathbb {R}^{n+1},u^q)$ , it remains to prove that there exists a positive constant K, depending only on n, p, $\gamma $ , and $\beta $ , such that, for any $(x,t)\in \mathbb {R}^{n+1}$ ,

(4.31) $$ \begin{align} \mathcal{M}^{\gamma+}_\beta(f)(x,t)\leq KM^{\frac{\gamma}{4}+}_\beta(f)(x,t). \end{align} $$

Indeed, fix $L\in (0,\infty )$ and let $P:=R(x,t+\frac {\gamma L^p}{2},L)$ . Then we are easy to show that $(x,t)\in P^-(\frac {\gamma }{4})$ , $R(x,t,L)^+(\gamma )\subset P^+(\frac {\gamma }{4})$ , and $|P^+(\frac {\gamma }{4})|= \frac {1-\frac {\gamma }{4}}{1-\gamma }|R(x,t,L)^+(\gamma )|$ . This further implies that

Taking the supremum over all $L\in (0,\infty )$ , we obtain (4.31), which completes the proof of necessity of (i).

The proof of the necessity of (ii) is a slight modification of that of (i) with Theorem 4.1 replaced by Theorem 4.3; we omit the details. This finishes the proof of Theorem 4.7.

Proof Let $(x,t)\in \mathbb {R}^{n+1}$ and $L_0\in (0,\infty )$ . Simply denote $R(x,t,L_0)$ by R. Observe that, for any $(y,s)\in R^+(\gamma )$ , $d_p((x,t),(y,s))\leq L$ . From this, the fact that $|R^+(\gamma )|=2^n(1-\gamma )L^{n+p}$ , and Definition 5.1, we deduce that

Taking the supremum over all $L_0\in (0,\infty )$ , we conclude that (5.1) with $C:=2^{n(\beta -1)}(1-\gamma )^{\beta -1}$ holds, which completes the proof of Lemma 5.3.

Proof Let $f\in L_{\mathrm {loc}}^{1}(\mathbb {R}^{n+1})$ , $(x,t)\in \mathbb {R}^{n+1}$ , and $\Omega _{(x,t)}^{\gamma +}:=\bigcup _{L\in (0,\infty )}R(x,t,L) ^+(\gamma )$ . Without loss of generality, we may assume that $\mathcal {M}^{\gamma ^2+}_{\beta -\epsilon }(f)(x,t) \mathcal {M}^{\gamma ^2+}_{\beta +\epsilon }(f)(x,t)\in (0,\infty )$ ; otherwise, (5.2) holds automatically. Fix $\delta \in (0,\infty )$ (which will be determined later) and define

$$ \begin{align*} \Omega_1:= \left\{(y,s)\in\Omega_{(x,t)}^{\gamma+}:\ d_p((x,t),(y,s))\in[0,\delta)\right\} \end{align*} $$

and $\Omega _2:=\Omega _{(x,t)}^{\gamma +}\setminus \Omega _1$ . Then

(5.3) $$ \begin{align} I^{\gamma+}_\beta(|f|)(x,t)&= \int_{\Omega_1}\frac{|f(y,s)|\,dy\,ds}{[d_p((x,t),(y,s))] ^{(n+p)(1-\beta)}}+ \int_{\Omega_2}\ldots\\ &=:I(x,t)+II(x,t).\nonumber \end{align} $$

We first estimate $I(x,t)$ . Let $\eta :=(\frac {1}{\gamma })^{\frac {1}{p}}$ and, for any $i\in \mathbb {N}$ ,

$$ \begin{align*} V_i^+:=Q\left(x,\eta^{-i+2}\delta\right)\times \left(t+\gamma^i\delta^p,t+\gamma^{i-1}\delta^p\right).\\ \end{align*} $$

Then the following four statements hold obviously:

1. (i) For any $i\in \mathbb {N}$ and $(y,s)\in V_i^+$ , $d_p((x,t),(y,s))\geq (\gamma ^i\delta ^p)^{\frac {1}{p}} =\eta ^{-i}\delta $ .

2. (ii) For any $i,j\in \mathbb {N}$ with $i\neq j$ , $V_i^+\cap V_j^+=\emptyset $ .

3. (iii) For any $i\in \mathbb {N}$ , the bottom of $V_i^+$ contains the top of $V_{i+1}^+$ .

4. (iv) $\Omega _1\subset \bigcup _{i\in \mathbb {N}}V_i^+$ .

For any $i\in \mathbb {N}$ , let $R_i:=R(x,t,\eta ^{-i+2}\delta )$ . Then $V_i^+\subset R_i^+(\gamma ^2)$ and $|V_i^+|=\frac {\gamma }{1+\gamma } |R_i^+(\gamma ^2)|$ . From this, the monotone convergence theorem, (i)–(iv), and the fact that

$$ \begin{align*} \left|R_i^+\left(\gamma^2\right)\right| =2^n\left(1-\gamma^2\right)\eta^{2(n+p)} \left(\eta^{-i}\delta\right)^{n+p}, \end{align*} $$

it follows that

(5.4) $$ \begin{align} I(x,t)&\leq\sum_{i\in\mathbb{N}}\int_{V_i^+} \frac{|f(y,s)|}{[d_p((x,t),(y,s))]^{(n+p)(1-\beta)}}\,dy\,ds\\ &\leq\sum_{i\in\mathbb{N}}\left(\eta^{-i}\delta\right) ^{(n+p)(\beta-1)}\int_{V_i^+}|f|\notag\\ &\leq2^{n(1+\epsilon-\beta)}\left(1-\gamma^2\right) ^{1+\epsilon-\beta}\eta^{2(n+p)(1+\epsilon-\beta)}\notag\\ &\quad\times\sum_{i\in\mathbb{N}}\left(\eta^{-i}\delta\right) ^{(n+p)\epsilon}\left|R_i^+(\gamma^2)\right| ^{\beta-\epsilon-1}\int_{R_i^+(\gamma^2)}|f|\notag\\ &\leq\frac{2^{n(1+\epsilon-\beta)}(1-\gamma^2) ^{1+\epsilon-\beta}\eta^{2(n+p)(1+\epsilon-\beta)}} {\eta^{(n+p)\epsilon}-1}\delta^{(n+p)\epsilon} \mathcal{M}_{\beta-\epsilon}^{\gamma^2+}(f)(x,t).\notag \end{align} $$

Now, we estimate $II(x,t)$ similarly. For any $i\in \mathbb {N}$ , let

$$ \begin{align*} U_j^+:=Q\left(x,\eta^j\delta\right)\times \left(t+\gamma^{-j+2}\delta^p,t+\gamma^{-j+1}\delta^p\right). \end{align*} $$

Then the following other four statements hold obviously:

1. (v) For any $j\in \mathbb {N}$ and $(y,s)\in U_j^+$ , $d_p((x,t),(y,s))\geq (\gamma ^{-j+2}\delta ^p)^{\frac {1}{p}} =\eta ^{j-2}\delta $ .

2. (vi) For any $j,k\in \mathbb {N}$ with $j\neq k$ , $U_j^+\cap U_k^+=\emptyset $ .

3. (vii) For any $j\in \mathbb {N}$ , the bottom of $U_{j+1}^+$ contains the top of $U_j^+$ .

4. (viii) $\Omega _2\subset \bigcup _{j\in \mathbb {N}}U_j^+$ .

For any $j\in \mathbb {N}$ , let $P_i:=R(x,t,\eta ^j\delta )$ . Then $U_j^+\subset P_j^+(\gamma ^2)$ and $|U_j^+|=\frac {\gamma }{1+\gamma } |P_j^+(\gamma ^2)|$ . From this, the monotone convergence theorem, (v)–(viii), and the fact that

$$ \begin{align*} \left|P_j^+\left(\gamma^2\right)\right| =2^n\left(1-\gamma^2\right)\eta^{2(n+p)} \left(\eta^{j-2}\delta\right)^{n+p}, \end{align*} $$

we deduce that

$$ \begin{align*} II(x,t)&\leq\sum_{j\in\mathbb{N}}\int_{U_j^+} \frac{|f(y,s)|}{[d_p((x,t),(y,s))]^{(n+p)(1-\beta)}}\,dy\,ds\\&\leq\sum_{j\in\mathbb{N}}\left(\eta^{j-2}\delta\right) ^{(n+p)(\beta-1)}\int_{U_j^+}|f|\notag\\&\leq2^{n(1-\epsilon-\beta)}\left(1-\gamma^2\right) ^{1-\epsilon-\beta}\eta^{2(n+p)(1-\epsilon-\beta)}\notag\\&\quad\times\sum_{j\in\mathbb{N}}\left(\eta^{j-2}\delta\right) ^{-(n+p)\epsilon}\left|P_j^+(\gamma^2)\right| ^{\beta+\epsilon-1}\int_{P_j^+(\gamma^2)}|f|\notag\\&\leq\frac{2^{n(1-\epsilon-\beta)}(1-\gamma^2) ^{1-\epsilon-\beta}\eta^{2(n+p)(1-\epsilon-\beta)}} {\eta^{(n+p)\epsilon}-1}\delta^{-(n+p)\epsilon} \mathcal{M}_{\beta+\epsilon}^{\gamma^2+}(f)(x,t). \end{align*} $$

Combining this, (5.3), and (5.4) and choosing

$$ \begin{align*} \delta:=\left[\frac{\mathcal{M}_{\beta+\epsilon} ^{\gamma^2+}(f)(x,t)}{\mathcal{M}_{\beta-\epsilon} ^{\gamma^2+}(f)(x,t)}\right]^{\frac{1}{2(n+p)\epsilon}}, \end{align*} $$

we then obtain (5.2) and hence finish the proof of Lemma 5.4.

Proof We first prove the sufficiency. Assume that (5.5) holds. By this, Lemma 5.3, and Theorem 4.7(i), we conclude that $(u,v)\in TA_{r,q}^+(\gamma )$ , which completes the proof of the sufficiency.

Then we show the necessity. Assume that $(u,v)\in TA_{r,q}^+(\gamma )$ . Let $f\in L_{\mathrm {loc}}^{1}(\mathbb {R}^{n+1})$ and $\lambda \in (0,\infty )$ . From Corollary 3.5, we infer that there exists $\delta _0\in (0,\infty )$ such that $(u,v)\in A_{r,q+\delta }^+(\gamma )$ for any $\delta \in (0,\delta _0)$ . Choose $\epsilon \in (0,\min \{\beta ,\,1-\beta \})$ such that

(5.6) $$ \begin{align} \frac{1}{\frac{1}{r}-(\beta+\epsilon)}-q\in(0,\delta_0), \end{align} $$

and let $q_1,q_2\in (1,\infty )$ satisfy

(5.7) $$ \begin{align} \frac{1}{q_1}=\frac{1}{r}-(\beta-\epsilon)\ \ \mbox{and}\ \ \frac{1}{q_2}=\frac{1}{r}-(\beta+\epsilon). \end{align} $$

Then $1<q_1<q<q_2<q+\delta _0<\infty $ . Applying Lemma 5.4 and (4.31), we find that there exists a positive constant C, depending only on n, p, $\gamma $ , $\beta $ , and $\epsilon $ , such that

$$ \begin{align*} \left\{\left|I^{\gamma+}_\beta(f)\right|>\lambda\right\}&\subset\left\{I^{\gamma+}_\beta(|f|)>\lambda\right\}\subset\left\{C\left[M^{\frac{\gamma^2}{4}+}_{\beta-\epsilon}(f)M^{\frac{\gamma^2}{4}+}_{\beta+\epsilon}(f)\right]^\frac{1}{2}>\lambda\right\}\\&\subset\left\{M^{\frac{\gamma^2}{4}+}_{\beta-\epsilon}(f)>\frac{\lambda}{C}\right\}\cup\left\{M^{\frac{\gamma^2}{4}+}_{\beta+\epsilon}(f)>\frac{\lambda}{C}\right\}\\\end{align*} $$

and hence

(5.8) $$ \begin{align} \left(u^q\right)\left(\left\{\left|I^{\gamma+}_\beta(f)\right|>\lambda\right\}\right)&\leq \left(u^q\right)\left(\left\{M^{\frac{\gamma^2}{4}+} _{\beta-\epsilon}(f)>\frac{\lambda}{C}\right\}\right)\\ &\quad+ \left(u^q\right)\left(\left\{M^{\frac{\gamma^2}{4}+} _{\beta+\epsilon}(f)>\frac{\lambda}{C}\right\}\right).\nonumber \end{align} $$

On the one hand, according to the proven conclusion that $q_1<q$ , Proposition 2.3(ii), and Corollary 3.3(i), we obtain $(u,v)\in TA_{r,q_1}^+(\frac {\gamma ^2}{4})$ , which, together with (5.7) and Theorem 4.1, further implies that there exists a positive constant $K_1$ , depending only on n, p, r, q, $\beta $ , $\epsilon $ , and $[u,v]_{TA_{r,q_1}^+(\frac {\gamma ^2}{4})}$ , such that

(5.9) $$ \begin{align} \left(u^q\right)\left(\left\{M^{\frac{\gamma^2}{4}+} _{\beta-\epsilon}(f)>\frac{\lambda}{C}\right\}\right) \leq\frac{K_1}{\lambda^q}\left(\int_{\mathbb{R}^{n+1}} |f|^rv^r\right)^{\frac{q}{r}}. \end{align} $$

On the other hand, from the proven conclusions that $q<q_2<q+\delta _0$ and $(u,v)\in TA_{r,q+\delta }^+(\gamma )$ for any $\delta \in (0,\delta _0)$ and from Corollary 3.3(i), it follows that $(u,v)\in TA_{r,q_2}^+(\frac {\gamma ^2}{4})$ , which, combined with (5.7) and Theorem 4.1, further implies that there exists a positive constant $K_2$ , depending only on n, p, r, q, $\beta $ , $\epsilon $ , and $[u,v]_{TA_{r,q_1}^+(\frac {\gamma ^2}{4})}$ , such that

$$ \begin{align*} \left(u^q\right)\left(\left\{M^{\frac{\gamma^2}{4}+} _{\beta+\epsilon}(f)>\frac{\lambda}{C}\right\}\right) \leq\frac{K_2}{\lambda^q}\left(\int_{\mathbb{R}^{n+1}} |f|^rv^r\right)^{\frac{q}{r}}. \end{align*} $$

By this, (5.8), and (5.9), we find that

$$ \begin{align*} \left(u^q\right)\left(\left\{\left|I^{\gamma+}_\beta(f)\right|>\lambda\right\}\right)\leq\frac{K_1+K_2}{\lambda^q} \left(\int_{\mathbb{R}^{n+1}}|f|^rv^r\right)^{\frac{q}{r}}. \end{align*} $$

Taking the supremum over all $\lambda \in (0,\infty )$ , we then conclude that (5.5) holds. This finishes the proof of the necessity and hence Theorem 5.5.

Proof To show the sufficiency, assume that $I^{\gamma +}_\beta $ is bounded from $L^r(\mathbb {R}^{n+1},\omega ^r)$ to $L^q(\mathbb {R}^{n+1},\omega ^q)$ . Then, by Lemma 5.3, we find that $\mathcal {M}^{\gamma +}_\beta $ is bounded from $L^r(\mathbb {R}^{n+1},\omega ^r)$ to $L^q(\mathbb {R}^{n+1},\omega ^q)$ , which, together with Corollary 4.9, implies that $\omega \in A_{r,q}^+(\gamma )$ . This finishes the proof of the sufficiency.

Now, we prove the necessity. Assume that $\omega \in A_{r,q}^+(\gamma )$ . Using Corollary 3.5, we conclude that there exists $\delta _0\in (0,\infty )$ such that $\omega \in A_{r,q+\delta }^+(\gamma )$ for any $\delta \in (0,\delta _0)$ . Fix $\epsilon \in (0,\min \{\beta ,1-\beta \})$ such that (5.6) holds and let $q_1,q_2\in (1,\infty )$ satisfy (5.7). Then $1<q_1<q<q_2<q+\delta _0<\infty $ and

$$ \begin{align*} \frac{1}{2q_1}+\frac{1}{2q_2}=\frac{1}{q}. \end{align*} $$

From this, Lemma 5.4, and the Hölder inequality, we infer that, for any $f\in L^r(\mathbb {R}^{n+1},\omega ^r)$ ,

(5.10) $$ \begin{align} &\left[\int_{\mathbb{R}^{n+1}}\left|I^{\gamma+}_\beta(f)\right| ^q\omega^q\right]^{\frac{1}{q}}\\ &\quad\lesssim\left\{\int_{\mathbb{R}^{n+1}} \left[\mathcal{M}_{\beta-\epsilon}^{\gamma^2+}(f)\right] ^{\frac{q}{2}}\omega^{\frac{q}{2}}\left[\mathcal{M} _{\beta+\epsilon}^{\gamma^2+}(f)\right]^{\frac{q}{2}} \omega^{\frac{q}{2}}\right\}^{\frac{1}{q}}\nonumber\\ &\quad\leq\left\{\int_{\mathbb{R}^{n+1}}\left[\mathcal{M} _{\beta-\epsilon}^{\gamma^2+}(f)\right]^{q_1} \omega^{q_1}\right\}^{\frac{1}{2q_1}} \left\{\int_{\mathbb{R}^{n+1}}\left[\mathcal{M} _{\beta+\epsilon}^{\gamma^2+}(f)\right]^{q_2} \omega^{q_2}\right\}^{\frac{1}{2q_2}}\nonumber\\ &\quad=:\mathrm{I}\times\mathrm{II}.\notag \end{align} $$

To estimate $\mathrm {I}$ , by the fact that $q_1<q$ and Proposition 2.3(ii), we find that $\omega \in A_{r,q_1}^+(\gamma )$ . From this, (5.7), and Corollary 4.9, we deduce that

(5.11) $$ \begin{align} \mathrm{I}\lesssim\left(\int_{\mathbb{R}^{n+1}}|f|^r \omega^r\right)^{\frac{1}{2r}}. \end{align} $$

To estimate $\mathrm {II}$ , by the proven conclusions that $q_2-q\in (0,\delta _0)$ and $\omega \in A_{r,q+\delta }^+(\gamma )$ for any $\delta \in (0,\delta _0)$ , we obtain $\omega \in A_{r,q_2}^+(\gamma )$ . From this, (5.7), and Corollary 4.9, it follows that