Boundedness of the differential transforms for the generalized Poisson operators generated by Laplacian

Abstract

In this paper, we consider the boundedness of the differential transforms for the generalized Poisson operators associated with the Laplace operator $\Delta$. The related results of the differential transforms for the heat semigroup are proved previously. By using the subordination formula method, we prove the boundedness of the maximal operator related to the differential transforms in weighted Lebesgue spaces. Moreover, we get some $L^\infty$-behavior results and the local growth of the maximal operator related to the differential transforms. Also, we get some similar results of the differential transforms related to the generalized Poisson operators generated by Schrödinger operator $-\Delta +V$, where the nonnegative potential V belongs to the reverse Hölder class $B_q$ with $q\ge n/2.$

1 Introduction

Let $\displaystyle \Delta = \sum _{j=1}^n\frac{\partial ^2}{\partial x^2_j}$ be the Laplace operator in $\mathbb {R}^n$. Consider its heat semigroup

$$\begin{aligned} e^{t\Delta }\varphi (x)=\int _{{\mathbb R}^n} W_t(x-y)\varphi (y){\textrm{d}}y,~x\in {\mathbb R}^{n},\ \ t>0, \end{aligned}$$

where W is the Gauss–Weierstrass kernel

$$\begin{aligned}W_t(x)=\frac{1}{(4\pi t)^{n/2}} e^{-\frac{|x|^2}{4t}}.\end{aligned}$$

For more information related with this semigroup, see [17].

For $0<\alpha <1$, the generalized Poisson formula of f is given by

$$\begin{aligned} {\mathcal P}_t^\alpha f(x)&=\frac{t^{2\alpha }}{4^\alpha \Gamma (\alpha )} \int _0^{+\infty } e^{-t^2/(4s)}e^{s\Delta } f(x)\,\frac{{\textrm{d}}s}{s^{1+\alpha }}\nonumber \\&=\frac{t^{2\alpha }}{4^\alpha \Gamma (\alpha )} \int _0^{+\infty } \int _{{\mathbb R}^n} \frac{e^{-(t^2+|y|^2)/(4 s)}}{(4\pi s)^{n/2}} f (x-y){\textrm{d}}y \frac{{\textrm{d}}s}{s^{1+\alpha }}, \quad x\in {\mathbb R}^n, \ t>0. \end{aligned}$$

(1.1)

It means that the generalized Poisson formula can be obtained via the heat semigroup $\displaystyle \{e^{t\Delta }\}_{t>0}$. In [6], Carffarelli and Silvestre studied the generalized Poisson formula to solve an extension problem. Stinga and Torrea defined this kind of Poisson formula for Hermite operator $L=-\Delta +|x|^2$ in [19]. In the case $\alpha =1/2$, ${\mathcal P}_t^{1/2}$ is the Bochner subordinated Poisson semigroup of $e^{t\Delta }$; see [17].

Let $\{a_j\}_{j\in {\mathbb Z}}$ be an increasing sequence of positive real numbers, and $\{v_j\}_{j\in {\mathbb Z}}$ be a bounded sequence of real or complex numbers. Let $\{T_t\}_{t>0}$ be an operator sequence. We consider the differential transform series

$$\begin{aligned} \sum _{j\in {\mathbb Z}} v_j(T_{a_{j+1}} f(x)-T_{a_j} f(x)). \end{aligned}$$

(1.2)

In [12], Jones and Rosenblatt studied the behavior of the series of the differences of ergodic averages and the differences of differentiation operators along lacunary sequences in the context of the $L^p$ spaces. In [2], the authors solved these problems with a different approach, which relied heavily on the method of Calderón–Zygmund singular integrals (see [15]). In [3], the authors considered the series (1.2) with the Poisson operator related with translation semigroups $f(t-s)$.

In order to analyze the series

$$\begin{aligned} \sum _{j\in {\mathbb Z}} v_j({\mathcal P}^\alpha _{a_{j+1}} f(x)-{\mathcal P}^\alpha _{a_j} f(x)), \end{aligned}$$

where ${\mathcal P}_t^\alpha$ is the generalized Poisson operator defined in (1.1), we can consider the convergence of its partial sums. For each $N\in {\mathbb Z}^2,~N=(N_1,N_2)$ with $N_1<N_2,$ we define the sum

$$\begin{aligned} T_N^\alpha f(x)&=\sum _{j=N_1}^{N_2} v_j({\mathcal P}_{a_{j+1}}^\alpha f(x)-{\mathcal P}_{a_j}^\alpha f(x)) \nonumber \\&= \frac{1}{4^\alpha \Gamma (\alpha )} \sum _{j=N_1}^{N_2}v_j \int _{{\mathbb R}^n}\int _0^\infty \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }} \frac{e^{-|y|^2/(4s)}}{(4\pi s)^{n/2}} f(x-y)~{\textrm{d}}s {\textrm{d}}y \nonumber \\&= \frac{1}{4^\alpha \Gamma (\alpha )} \int _{{\mathbb R}^{n}} \int _0^{+\infty }\sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }} \frac{e^{-|y|^2/(4s)}}{(4\pi s)^{n/2}}~{\textrm{d}}s f(x-y) {\textrm{d}}y. \end{aligned}$$

(1.3)

We denote the kernel of $T_N^\alpha$ by

$$\begin{aligned}K_N^\alpha (y)=\frac{1}{4^\alpha \Gamma (\alpha )}\int _0^{+\infty } \sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} \frac{e^{-|y|^2/(4s)}}{(4\pi s)^{n/2}}{\textrm{d}}s.\end{aligned}$$

We shall also consider the maximal operators

$$\begin{aligned} T^*f(x)=\sup _N \left| T^\alpha _N f(x)\right| , \quad x\in {\mathbb R}^n, \end{aligned}$$

where the supremum are taken over all $N=(N_1,N_2)\in {\mathbb Z}^2$ with $N_1< N_2$. We shall consider the boundedness problem related to these operators. In [3], the authors proved the boundedness of the above operators related with the one-sided generalized Poisson type operator sequence.

Some of our results will be valid only when the sequence $\{a_j\}_{j\in \mathbb Z}$ is lacunary. It means that there exists a $\rho >1$ such that $\displaystyle \frac{a_{j+1}}{a_j} \ge \rho , \, j \in \mathbb {Z}$. In particular, we shall prove the boundedness of the operators $T^*$ in the weighted spaces $L^p(\mathbb R^{n}, \omega ),$ where $\omega$ is the usual Muckenhoupt weights on $\mathbb R^{n}$. We refer the reader to the book by J. Duoandikoetxea [7, Chapter 7] for definitions and properties of the $A_p$ classes. We have the following results:

Theorem 1

• (a) For any $1<p<\infty$ and $\omega \in A_p$, there exists a constant C depending on $n, p, \rho , \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned}\left\| T^*f\right\| _{L^p(\mathbb R^n, \omega )}\le C\left\| f\right\| _{L^p(\mathbb R^n, \omega )},\end{aligned}$$

  for all functions $f\in L^p(\mathbb R^n, \omega ).$

• (b) For any $\omega \in A_1$, there exists a constant C depending on $n, \rho , \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned}\omega \left( {\{x\in {\mathbb R}^n:\left| T^*f(x)\right|>\lambda \}}\right) \le C\frac{1}{\lambda }\left\| f\right\| _{L^1(\mathbb R^n, \omega )}, \quad \lambda >0,\end{aligned}$$

  for all functions $f\in L^1(\mathbb R^n, \omega ).$

• (c) Given $f\in L^\infty ({\mathbb R}^n),$ then either $T^* f(x) =\infty$ for all $x\in \mathbb R^n$, or $T^* f(x) < \infty$ for a.e. $x\in \mathbb R^n$. And in this latter case, there exists a constant C depending on $n, \rho$, $\alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned}\left\| T^*f\right\| _{BMO(\mathbb R^n)}\le C\left\| f\right\| _{L^\infty (\mathbb R^n)}.\end{aligned}$$

• (d) Given $f\in BMO({\mathbb R}^n),$ then either $T^* f(x) =\infty$ for all $x\in \mathbb R^n$, or $T^* f(x) < \infty$ for a.e. $x\in \mathbb R^n$. And in this latter case, there exists a constant C depending on $n, \rho$, $\alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned} \left\| T^*f\right\| _{ BMO(\mathbb R^n)}\le C\left\| f\right\| _{BMO(\mathbb R^n)}. \end{aligned}$$

  (1.4)

Remark 1

From the conclusions we got in Theorem 1, for $f\in L^p({\mathbb R}^{n}, \omega )$ with $\omega \in A_p$, in Theorem 8 we shall see that we can define Tf by the limit of $T_N f$ in $L^p$-norm

$$\begin{aligned}T f(x)=\lim _{(N_1,N_2)\rightarrow (-\infty , +\infty )} T_N f(x),\quad \quad ~x\in {\mathbb R}^{n}.\end{aligned}$$

In classical harmonic analysis, if $f= \chi _{(0,1)}$ and $\mathcal {H}$ is the Hilbert transform, it is easy to see that $\displaystyle \frac{1}{r} \int _{-r}^0 \mathcal {H}(f)(x){\textrm{d}}x\ \sim \log \frac{e}{r}$ as $r \rightarrow 0^+$. In general, this is the growth of a singular integral applied to a bounded function at the origin. The following theorem shows that the growth of the function $T^*f$ for bounded function f at the origin is of the same order of a singular integral operator. Some related results about the local behavior of variation operators can be found in [1]. One-dimensional results about the variation of some convolutions operators can be found in [14]. And the one-dimensional results about the differential transforms of one-sided fractional Poisson type operator sequence is proved in [3]. In [4], the authors got local growth of the differential transforms of heat semigroup generated by Laplacian.

Theorem 2

• (a) Let $\{v_j\}_{j\in \mathbb Z}\in l^p(\mathbb Z)$ for some $1 \le p\le \infty .$ For every $f\in L^\infty (\mathbb {R}^n)$ with support in the unit ball $B=B(0, 1)$, for any ball $B_r\subset B$ with $2r<1$, we have

  $$\begin{aligned}\frac{1}{|B_r|} \int _{B_r} \left| T^* f (x)\right| {\textrm{d}}x\le C\left( \log \frac{2}{r}\right) ^{1/p'}\left\| v\right\| _{l^p(\mathbb Z)}\Vert f\Vert _{L^\infty (\mathbb R^n)}.\end{aligned}$$

• (b) When $1< p<\infty$, for any $0<\varepsilon <p-1$, there exist a $\rho$-lacunary sequence $\{a_j\}_{j\in \mathbb Z}$, a sequence $\{v_j\}_{j\in \mathbb Z}\in \ell ^p(\mathbb Z)$ and a function $f\in L^\infty (\mathbb {R}^n)$ with support in the unit ball $B=B(0, 1),$ satisfying the following statement: for any ball $B_r\subset B$ with $2r<1$, we have

  $$\begin{aligned}\frac{1}{|B_r|} \int _{B_r} \left| T^* f (x)\right| {\textrm{d}}x\ge C\left( \log \frac{2}{r}\right) ^{1/(p-\varepsilon )'}\left\| v\right\| _{l^p(\mathbb Z)}\Vert f\Vert _{L^\infty (\mathbb R^n)}.\end{aligned}$$

• (c) When $p=\infty ,$ there exist a $\rho$-lacunary sequence $\{a_j\}_{j\in \mathbb Z}$, a sequence $\{v_j\}_{j\in \mathbb Z}\in l^\infty (\mathbb Z)$ and $f\in L^\infty (\mathbb {R}^n)$ with support in the unit ball $B=B(0, 1)$, satisfying the following statements: for any ball $B_r\subset B$ with $2r<1$,

  $$\begin{aligned}\frac{1}{|B_r|} \int _{B_r} \left| T^* f (x)\right| {\textrm{d}}x\ge C\left( \log \frac{2}{r}\right) \left\| v\right\| _{l^\infty (\mathbb Z)}\Vert f\Vert _{L^\infty (\mathbb R^n)}.\end{aligned}$$

In the statements above, $\displaystyle p' = \frac{p}{p-1},$ and if $p=1$, $\displaystyle p'=\infty .$

The statements in Theorem 2 shows that, when $1< p<\infty ,$ the growth of $T^*$ is between the growth of the standard singular integral and the growth of the Hardy–Littlewood maximal operator. And when $p=\infty ,$ the growth of $T^*$ is the same with the standard singular integral operator.

The organization of the paper is as follows: Sect. 2 is devoted to prove the boundedness of the maximal operators $T^*$. And we will give the proof of the local growth of $T^*$, i.e. Theorem 2, in Sect. 3. In Sect. 4, we will get some similar results in the Schrödinger setting.

Throughout this paper, the symbol C in an inequality always denotes a constant which may depend on some indices, but never on the functions f under consideration.

2 Proof of Theorem 1

In this section, we will prove Theorem 1. In order to prove Theorem 1, we need to prove the uniform boundedness of $T_N^\alpha$ first. By the Fourier transform, we can prove that the operators $T_N^\alpha$ are uniform bounded in $L^2({\mathbb R}^{n})$ for all $N\in {\mathbb Z}^2,~N_1< N_2$. Since the kernel $K_N^\alpha (y,s)$ satisfies the size and smoothness conditions (see Theorem 5), we can deduce the $L^p$-boundedness results by using the Calderón–Zygmund theorem. Thus, we have the following results:

Theorem 3

For the operator $T_N^\alpha$ defined in (1.3), we have the following statements.

• (a) For any $1<p<\infty$ and $\omega \in A_p$, there exists a constant C depending on $n, p, \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned}\left\| T_N^\alpha f\right\| _{L^p(\mathbb R^{n}, \omega )}\le C\left\| f\right\| _{L^p(\mathbb R^{n}, \omega )},\end{aligned}$$

  for all functions $f\in L^p({\mathbb R}^{n}, \omega ).$

• (b) For any $\omega \in A_1$, there exists a constant C depending on $n, \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned}\omega \left( {\{x\in {\mathbb R}^{n}:\left| T_N^\alpha f(x)\right|>\lambda \}}\right) \le C\frac{1}{\lambda }\left\| f\right\| _{L^1(\mathbb R^{n}, \omega )},\quad \lambda >0,\end{aligned}$$

  for all functions $f\in L^1({\mathbb R}^{n}, \omega ).$

The constants C appeared above all are independent of N.

We shall use the Calderón–Zygmund theory in proving the $L^p$-boundedness of the differential transforms $T_N^\alpha$ associated with the generalized Poisson operators. We will prove the $L^2$-estimates first. And then, it remains to give the estimates about the kernels of the differential transforms. By a standard argument, the results in Theorem 3 will be obtained.

First, we present a lemma which will be used later.

Lemma 1

( [3, Lemma 2.1]) Let $0<\alpha <1$. Then for any complex number $z_0$ with $Re z_0 > 0$ and $\displaystyle |\arg z_0 |\le {\pi }/{4}$, we have

$$\begin{aligned} \int _0^{+\infty } e^{-z_0 u} e^{-\frac{z_0}{u} }\,\frac{{\textrm{d}}u}{u^{\alpha }}= z_0^{1-\alpha }\int _{0}^{+\infty } \frac{e^{-r}e^{- z_0^2/r}}{r^{2-\alpha }} {\textrm{d}}r.\end{aligned}$$

Now we present the uniform $L^2$-boundedness of the operator $T^\alpha _N$ in the following theorem:

Theorem 4

There exists a constant $C>0$, depending on $n, \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$, such that

$$\begin{aligned}\sup _N \Vert T_N^\alpha f \Vert _{L^2({\mathbb R}^{n})}\le C \Vert f \Vert _{L^2({\mathbb R}^{n})}.\end{aligned}$$

Proof

Let $f\in L^2({\mathbb R}^{n})$. Using the Plancherel theorem, we have

$$\begin{aligned} \left\| T_N^\alpha f \right\| _{L^2({\mathbb R}^{n})}&= \left\| \sum _{j=N_1}^{N_2} v_j({\mathcal P}_{a_{j+1}}^\alpha f -{\mathcal P}_{a_j}^\alpha f)\right\| _{L^2({\mathbb R}^{n})} \le C\left\| v\right\| _{l^\infty (\mathbb Z)} \left\| \sum _{j=-\infty }^{\infty } \int _{a_j}^{a_{j+1}} \left| \partial _t \widehat{{\mathcal P}_{t}^\alpha f }\right| {\textrm{d}}t\right\| _{L^2({\mathbb R}^{n})}. \end{aligned}$$

By using the second identity in (1.1), we have

$$\begin{aligned} \partial _t \widehat{{\mathcal P}_{t}^\alpha f }(\xi )&= C \partial _t \int _0^\infty e^{-r} \widehat{ e^{-\frac{t^2}{4r} \Delta } f}(\xi )\,\frac{{\textrm{d}}r}{r^{1-\alpha }}\\&= C \partial _t \int _0^\infty e^{-r} e^{-\frac{t^2}{4r}|\xi |^2}{\widehat{f}} (\xi ) \,\frac{{\textrm{d}}r}{r^{1-\alpha }}\\&= C \int _0^\infty e^{-r}t |\xi |^2 e^{-\frac{t^2}{4r}|\xi |^2}{\widehat{f}} (\xi ) \,\frac{{\textrm{d}}r}{r^{2-\alpha }}. \end{aligned}$$

Note that the Fourier transform above can be well defined. Then we deduce that

$$\begin{aligned} \left\| T_N^\alpha f \right\| _{L^2({\mathbb R}^{n})}&\le C \left\| {\widehat{f}} (\xi ) \int _{0}^\infty \left| \int _0^\infty e^{-r}t |\xi |^2 e^{-\frac{t^2}{4r}|\xi |^2}\,\frac{{\textrm{d}}r}{r^{2-\alpha }} \right| {\textrm{d}}t\right\| _{L^2({\mathbb R}^{n})}. \end{aligned}$$

Thus again by the Plancherel theorem, the remainder is devoted to prove the uniform boundedness of the multiplier

$$\begin{aligned}\left| \widehat{K_N^\alpha }(\xi )\right| \le C \left| \int _{0}^\infty \left| \int _0^\infty e^{-r}t |\xi |^2 e^{-\frac{t^2}{4r}|\xi |^2}\,\frac{{\textrm{d}}r}{r^{2-\alpha }}\right| {\textrm{d}}t\right| \le C, \quad \xi \in {\mathbb R}^{n}.\end{aligned}$$

Taking $z_0=t|\xi |$, we rewrite the above inequality in

$$\begin{aligned}\left| \widehat{K_N^\alpha }(\xi )\right| \le C\int _{0}^\infty \left| \int _0^\infty e^{-r}z_0 e^{-\frac{z_0^2}{4r} } \,\frac{{\textrm{d}}r}{r^{2-\alpha }} \right| {\textrm{d}}z_0, \quad \xi \in {\mathbb R}^{n}.\end{aligned}$$

By Lemma 1, for any $\xi \in {\mathbb R}^{n}$, we have

$$\begin{aligned} \int _{0}^\infty \left| \int _0^\infty e^{-r}z_0 e^{-\frac{z_0^2}{4r} }\,\frac{{\textrm{d}}r}{r^{2-\alpha }} \right| {\textrm{d}}z_0 = 2^{1-\alpha }\int _{0}^\infty \left| z_0^\alpha \int _0^\infty e^{-\frac{z_0}{2u}}e^{-\frac{z_0}{2}u} \frac{{\textrm{d}}u}{u^\alpha } \right| {\textrm{d}}z_0.\end{aligned}$$

Since $\left| \arg z_0\right| \le {\pi }/{4}$, we have $|e^{-z_0/(2u)}| \le e^{-c|z_0|/u}$ and $|e^{-z_0 u/2}| \le e^{-c|z_0| u}$, where $c={ \sqrt{2}/ {4}}$. Then

$$\begin{aligned} \left| \int _{0}^\infty z_0^\alpha \int _0^\infty e^{-z_0/u}e^{-z_0 u} \frac{{\textrm{d}}u}{u^\alpha } {\textrm{d}}z_0\right|&\le \int _{0}^\infty \, |z_0|^\alpha \int _0^\infty e^{-c|z_0|/u}e^{-c|z_0| u} \frac{{\textrm{d}}u}{u^\alpha } {\textrm{d}}z_0\\&\le \int _{0}^\infty \,|z_0|^{2\alpha -1}\int _0^\infty e^{-c|z_0|^2/v}e^{-c v} \frac{{\textrm{d}}v}{v^\alpha } {\textrm{d}}z_0. \end{aligned}$$

Recall that $z_0=t |\xi |$. Then, we have

$$\begin{aligned}&\int _{0}^\infty \,|z_0|^{2\alpha -1}\int _0^\infty e^{-c|z_0|^2/v}e^{-c v} \frac{{\textrm{d}}v}{v^\alpha } {\textrm{d}}z_0\\&= \int _{0}^\infty |\xi |^{2\alpha }\,t^{2\alpha -1}\int _0^\infty e^{-c(|\xi |t)^2/v}e^{-c v} \frac{{\textrm{d}}v}{v^\alpha } {\textrm{d}}t\\&= \int _{0}^\infty \int _0^\infty (|\xi |t)^{2\alpha -1} e^{-c(|\xi |t)^2/v} d(|\xi |t) e^{-c v} \frac{{\textrm{d}}v}{v^\alpha }\\&= \int _{0}^\infty \int _0^\infty t^{2\alpha -1} e^{-ct^2/v} {\textrm{d}}t e^{-c v} \frac{d v}{v^\alpha } \le C \int _{0}^\infty e^{-c v} {\textrm{d}}v \le C, \end{aligned}$$

where the constants C appeared above all are independent of N. Then the proof of the theorem is complete. $\square$

Also, we can get the kernel estimates in the following:

Theorem 5

There exists constant $C>0$ depending on $n, \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$(not on N) such that, for any $y\ne 0,$

• (i) $\displaystyle |K_N^\alpha (y)|\le \frac{C}{|y|^{n}}$,

• (ii) $\displaystyle |\nabla _y K_N^\alpha (y)| \le \frac{C}{|y|^{n+1}}$.

Proof

i) This is the size condition of the kernel. We have

$$\begin{aligned}{} & {} |K_N^\alpha (y)| \le C\int _0^{+\infty }\sum _{j=-\infty }^{\infty } \left| \frac{a_{j+1}^{2\alpha } e^{-a_{j+1}^2/(4s)}-a_j^{2\alpha } e^{-a_j^2/(4s)}}{s^{1+\alpha }} \frac{e^{-|y|^2/(4s)}}{s^{n/2}} \right| {\textrm{d}}s\\{} & {} \quad = C\int _0^{+\infty }\sum _{j=-\infty }^{\infty } \left| a_{j+1}^{2\alpha } e^{-a_{j+1}^2/(4s)}-a_j^{2\alpha } e^{-a_j^2/(4s)}\right| \frac{e^{-|y|^2/(4s)}}{s^{1+\alpha +n/2}}{\textrm{d}}s. \end{aligned}$$

Observe that

$$\begin{aligned}&\sum _{j=-\infty }^{\infty } \left| a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4s)}-a_j^{2\alpha }e^{-a_j^2/(4s)}\right| = \sum _{j=-\infty }^{\infty } \left| \int _{a_j}^{a_{j+1}}\partial _u \left( u^{2\alpha } e^{-u^2/(4s)}\right) {\textrm{d}}u\right| \nonumber \\&\quad \le \int _{0}^{+\infty }\left| ( 2\alpha u^{2\alpha -1}-\frac{u^{2\alpha +1}}{2s})e^{-u^2/(4s)} \right| {\textrm{d}}u \le C\int _{0}^{+\infty }\left| (u^{2\alpha -1}+\frac{u^{2\alpha +1}}{2s})e^{-u^2/(4s)} \right| {\textrm{d}}u\nonumber \\&\quad \le C\sqrt{s}\Big (\int _{0}^{+\infty } (\sqrt{s} )^{2\alpha -1} \left( \frac{u}{\sqrt{s}} \right) ^{2\alpha -1} e^{-\frac{1}{4} \left( u/\sqrt{s}\right) ^2} d\frac{u}{\sqrt{s}}\nonumber \\&\qquad \,+ s^{\alpha -1/2}\int _{0}^{+\infty }\left( \frac{u}{\sqrt{s}}\right) ^{2\alpha +1} e^{-\frac{1}{4}\left( u/\sqrt{s} \right) ^2} d\frac{u}{\sqrt{s}}\Big )\nonumber \\&\quad \le C s^{\alpha }. \end{aligned}$$

(2.1)

Then, we have

$$\begin{aligned} |K_N^\alpha (y)| \le C \int _0^{+\infty }\frac{e^{-|y|^2/(4s)}}{s^{n/2+1}}{\textrm{d}}s=C \left( \int _0^{|y|^2}+\int _{|y|^2}^{+\infty }\right) \frac{e^{-|y|^2/(4s)}}{s^{n/2+1}}{\textrm{d}}s\\ \le C \left( \int _0^{|y|^2}\frac{1}{|y|^{n+2}}{\textrm{d}}s+\int _{|y|^2}^{+\infty }\frac{1}{s^{n/2+1}}{\textrm{d}}s\right) \le \frac{C}{|y|^{n}}. \end{aligned}$$

ii) It suffices to prove that for the first variable $y_1\in {\mathbb R}$, we have

$$\begin{aligned} |\partial _{y_1} K_N^\alpha (y) | \le \frac{C}{|y|^{n+1}},\end{aligned}$$

where

$$\begin{aligned} \partial _{y_1} K_N^{\alpha }(y)&=-\int _0^{+\infty }\sum _{j=N_1}^{N_2} v_j \frac{a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4s)}-a_j^{2\alpha }e^{-a_j^2/(4s)}}{s^{1+\alpha }} \frac{y_1 e^{-|y|^2/(4s)}}{2 s^{n/2+1}}{\textrm{d}}s. \end{aligned}$$

Then, by (2.1) we conclude that

$$\begin{aligned}&|\partial _{y_1} K_N^\alpha (y)| \le C \int _0^{+\infty } s^{\alpha }~\frac{y_1 e^{-|y|^2/(4s)}}{ s^{2+\alpha +n/2} }{\textrm{d}}s\le C \int _0^{+\infty }\frac{\frac{|y|}{\sqrt{s}} e^{-|y|^2/(4s)}}{ s^{n/2+3/2}}{\textrm{d}}s\\&\quad \le C\int _0^{+\infty } \frac{ e^{-|y|^2/(8s)}}{ s^{n/2+3/2}}{\textrm{d}}s=C\left( \int _0^{|y|^2}+\int _{|y|^2}^{+\infty }\right) \frac{ e^{-|y|^2/(8s)}}{ s^{n/2+3/2}}{\textrm{d}}s \le \frac{C}{|y|^{n+1}}. \end{aligned}$$

The proof of the theorem is complete. $\square$

Remark 2

If we consider an $l^\infty (\mathbb Z^2)$-valued operator $Q: f\mapsto \left\{ T_N^\alpha f(x)\right\} _{N\in \mathbb Z^2}$ on the homogeneous space $({\mathbb R}^{n}, d, {\textrm{d}}x)$, then $T^*_\alpha f(x)=\left\| Qf(x)\right\| _{l^\infty ({\mathbb R}^{n})}$, and by Theorem 5, we know that the kernel of the operator Q is an $l^\infty (\mathbb Z^2)$-valued Calderón–Zygmund kernel.

In the next result, we will take care of the behavior of $T_N^\alpha$ on $BMO({\mathbb R}^n)$.

Theorem 6

Let $\{a_j\}_{j\in \mathbb Z}$ be an increasing sequence. There exists a constant C depending on $n, \alpha$ and $\left\| v\right\| _{\ell ^\infty (\mathbb Z)}$(not on N) such that

$$\begin{aligned}\left\| T_N^\alpha f\right\| _{BMO({\mathbb R}^n)}\le C\left\| f\right\| _{L^\infty ({\mathbb R}^n)},\end{aligned}$$

and

$$\begin{aligned}\left\| T_N^\alpha f\right\| _{BMO({\mathbb R}^n)}\le C\left\| f\right\| _{BMO({\mathbb R}^n)}.\end{aligned}$$

Proof

The finiteness of $T_N^\alpha$ for functions in $L^\infty ({\mathbb R}^n)$ is obvious, since for each N, $K_N^\alpha$ is an integrable function. On the other hand, assume that $f\in BMO({\mathbb R}^n)$. Let $B=B(x_0, r_0)$ and $B^*=B(x_0, 2r_0)$ with $x_0\in {\mathbb R}^n$ and $r_0>0$. We decompose f to be

$$\begin{aligned}f=(f-f_B)\chi _{B^*}+(f-f_B)\chi _{(B^*)^c}+f_B=f_1+f_2+f_3.\end{aligned}$$

By Theorem 4, we have

$$\begin{aligned} \int _{{\mathbb R}^n}\left| T_N^\alpha f_1\right| ^2{\textrm{d}}x\le C \left\| f_1\right\| ^2_{L^2({\mathbb R}^n)}\le C\int _{B^*}\left| f(x)-f_B\right| ^2{\textrm{d}}x\le C|B|\left\| f\right\| _{BMO({\mathbb R}^n)}^2.\nonumber \\ \end{aligned}$$

(2.2)

This means that $T_N^\alpha f_1(x)<\infty ,$ $a.e.\ x\in {\mathbb R}^n.$ And we should note that $T_N^\alpha f_3(x)\equiv 0$, since ${\mathcal P}_{a_j}^\alpha f_3\equiv f_B$ for any $j\in \mathbb Z.$ For $T_N^\alpha f_2$, we note that, for any $x\in B$ and $t>0$,

$$\begin{aligned} {\mathcal P}_t^\alpha f_2(x)&=\frac{t^{2\alpha }}{4^\alpha \Gamma (\alpha )} \int _0^{+\infty } \int _{{\mathbb R}^n} \frac{e^{-(t^2+|x-y|^2)/(4 s)}}{(4\pi s)^{n/2}} f_2 (y){\textrm{d}}y \frac{{\textrm{d}}s}{s^{1+\alpha }}\\&\le C\frac{t^{2\alpha }}{4^\alpha \Gamma (\alpha )} \int _0^{+\infty } \sum _{k=1}^{+\infty }\int _{2^kr_0<|x_0-y|\le 2^{k+1}r_0} {1\over |x-y|^{n+2\alpha '}}\left| f(y)-f_B\right| {\textrm{d}}y\ {e^{-{t^2\over 4 s}}}\frac{{\textrm{d}}s}{s^{1+\alpha -\alpha '}}\\&\le C\frac{t^{2\alpha '}\Gamma (\alpha -\alpha ')}{4^{\alpha '}\Gamma (\alpha )} \sum _{k=1}^{+\infty }{(2^kr_0)}^{-2\alpha '}{1\over {(2^kr_0)}^{n}}\int _{|x_0-y|\le 2^{k+1}r_0} \left| f(y)-f_B\right| {\textrm{d}}y\\&\le C\frac{t^{2\alpha '}\Gamma (\alpha -\alpha ')}{4^{\alpha '}\Gamma (\alpha )} \sum _{k=1}^{+\infty }{(2^kr_0)}^{-2\alpha '}(1+2k)\left\| f\right\| _{BMO({\mathbb R}^n)}<\infty , \end{aligned}$$

where $0<\alpha '<\alpha .$ So, ${\mathcal P}_t^\alpha f_2(x)$ is well defined for $x\in B$ and $t>0.$ Since $T_N^\alpha f_2(x)$ is a finite summation and $x_0, r_0$ is arbitrary, $T_N^\alpha f_2(x)<\infty$ a.e. $x\in {\mathbb R}^n.$ Hence, $T_N^\alpha f(x)<\infty$ a.e. $x\in {\mathbb R}^n.$

Now, let us prove the two inequalities. Since $L^\infty ({\mathbb R}^n)\subset BMO({\mathbb R}^n),$ we only need to prove the inequality in the case of $f\in BMO({\mathbb R}^n).$ Choose some $x_1\in B(x_0, r_0)$ such that $T_N^\alpha f_2(x_1)<\infty .$ Now, taking a constant $c_B=T_N^\alpha f_2(x_1)$, we can write

$$\begin{aligned}&{1\over |B|}\int _B\left| T_N^\alpha f(x)-c_B\right| {\textrm{d}}x \\&= {1\over |B|}\int _B\left| T_N^\alpha (f_1+f_2+f_3)(x)-T_N^\alpha f_2(x_1)\right| {\textrm{d}}x\\&\le {1\over |B|}\int _B\left| T_N^\alpha f_1(x)\right| {\textrm{d}}x+{1\over |B|}\int _B\left| T_N^\alpha f_2(x)-T_N^\alpha f_2(x_1)\right| {\textrm{d}}x\\&=:I_1+I_2. \end{aligned}$$

For the first term $I_1$, by Hölder’s inequality and (2.2) we have

$$\begin{aligned} I_1={1\over |B|}\int _B\left| T_N^\alpha f_1(x)\right| {\textrm{d}}x\le \left( {1\over |B|}\int _B\left| T_N^\alpha f_1(x)\right| ^2{\textrm{d}}x\right) ^{1/2}\le C\left\| f\right\| _{BMO({\mathbb R}^n)}. \end{aligned}$$

For the term $I_2$, by Theorem 5ii) we have

$$\begin{aligned} I_2&={1\over |B|}\int _B\left| T_N^\alpha f_2(x)-T_N^\alpha f_2(x_1)\right| {\textrm{d}}x\\&={1\over |B|}\int _B\left| \int _{{\mathbb R}^n}\left( K_N^\alpha (x-y)-K_N^\alpha (x_1-y) \right) f_2(y){\textrm{d}}y\right| {\textrm{d}}x\\&\le {1\over |B|}\int _B\int _{{\mathbb R}^n}\left| K_N^\alpha (x-y)-K_N^\alpha (x_1-y)\right| \left| f_2(y)\right| {\textrm{d}}y{\textrm{d}}x\\&\le C{1\over |B|}\int _B\int _{(B^*)^c}{\left| x-x_1\right| \over \left| x-y\right| ^{n+1}}\left| f(y)-f_B\right| {\textrm{d}}y{\textrm{d}}x\\&\le C{r_0\over |B|}\int _B\int _{(B^*)^c}{1\over (\left| y-x_0\right| -\left| x_0-x\right| )^{n+1}}\left| f(y)-f_B\right| {\textrm{d}}y{\textrm{d}}x\\&= C{r_0\over |B|} \int _B \sum _{k=1}^{+\infty }\int _{2^kr_0\le \left| y-x_0\right|<2^{k+1}r_0}{1\over (\left| y-x_0\right| -\left| x_0-x\right| )^{n+1}}\left| f(y)-f_B\right| {\textrm{d}}y{\textrm{d}}x\\&\le C\sum _{k=1}^{+\infty }{1\over 2^k}{1\over (2^kr_0)^n} \int _{ \left| y-x_0\right| <2^{k+1}r_0}\left| f(y)-f_B\right| {\textrm{d}}y\\&\le C\left\| f\right\| _{BMO({\mathbb R}^n)}. \end{aligned}$$

Hence, we deduce

$$\begin{aligned} {1\over |B|}\int _B\left| T_N^\alpha f(x)-c_B\right| {\textrm{d}}x\le C\left\| f\right\| _{BMO({\mathbb R}^n)}. \end{aligned}$$

Thus, we proved that $T_N^\alpha f\in BMO({\mathbb R}^n)$ and

$$\begin{aligned}\left\| T_N^\alpha f\right\| _{BMO({\mathbb R}^n)}\le C\left\| f\right\| _{BMO({\mathbb R}^n)}.\end{aligned}$$

$\square$

In the following, we aim to prove Theorem 1. The next proposition, parallel to Proposition 3.2 in [2](also Proposition 3.1 in [3]), shows that, without lost of generality, we may assume that

$$\begin{aligned} 1<\rho \le {a_{j+1} \over a_j}\le \rho ^2, \quad j\in \mathbb Z. \end{aligned}$$

Proposition 1

Given a $\rho$-lacunary sequence $\{a_j\}_{j\in \mathbb Z}$ and a multiplying sequence $\{v_j\}_{j\in \mathbb Z}\in \ell ^\infty (\mathbb Z)$, we can define a $\rho$-lacunary sequence $\{\eta _j\}_{j\in \mathbb Z}$ and $\{\omega _j\}_{j\in \mathbb Z}\in \ell ^\infty (\mathbb Z)$ verifying the following properties:

• (i) $1<\rho \le \eta _{j+1}/\eta _j\le \rho ^2,\quad \quad \left\| \{\omega _j\}\right\| _{\ell ^\infty (\mathbb Z)}=\left\| \{v_j\}\right\| _{\ell ^\infty (\mathbb Z)}$.

• (ii) For all $N=(N_1, N_2)$ there exists $N'=(N_1', N_2')$ with $T^\alpha _N={\hat{T}}^\alpha _{N'},$ where ${\hat{T}}^\alpha _{N'}$ is the operator defined in (1.3) with the new sequences $\{\eta _j\}_{j\in {\mathbb Z}}$ and $\{\omega _j\}_{j\in {\mathbb Z}}.$

In order to prove Theorem 1, we need a Cotlar’s type inequality. For any $M\in \mathbb Z^+,$ let

$$\begin{aligned}T_M^*f(x)=\sup _{-M\le N_1<N_2\le M}\left| T_N f(x)\right| ,\quad x\in {\mathbb R}^n,\end{aligned}$$

where $T_N$ denotes the differential transform operator related with the heat-diffusion semigroup generated by $-\Delta$. By a similar(in fact, easier) argument as in the proof of Theorem 4, we can prove that $T_N$ is uniform bounded on $L^2({\mathbb R}^n)$. Also, we can prove that $T_N$ is uniform bounded in $L^p({\mathbb R}^n, \omega )$ for $1<p<\infty$, uniform weak-(1, 1) bounded and uniform BMO-bounded, because it is a Calderón–Zygmund operator. For these results, see [4].

Theorem 7

(See [4, Theorem 2.4]) For each $q\in (1, +\infty ),$ there exists a constant C depending on $n, \left\| v\right\| _{l^\infty (\mathbb Z)}$ and $\rho$ such that for every $x\in {\mathbb R}^n$ and every $M\in \mathbb Z^+$,

$$\begin{aligned} T_M^*f(x)\le C\left\{ {\mathcal M}(T_{-M, M} f)(x)+{\mathcal M}_q f(x)\right\} , \end{aligned}$$

where

$$\begin{aligned}T_{-M, M} f(x)={\sum _{j=-M}^{M} v_j(e^{a_{j+1}\Delta } f(x)-e^{a_j\Delta } f(x))}\end{aligned}$$

and

$$\begin{aligned}{\mathcal M}_qf(x)=\sup _{\varepsilon >0} \left( \frac{1}{|B(x, \varepsilon )|}\int _{B(x, \varepsilon )}\left| f(y)\right| ^q{\textrm{d}}y\right) ^{1\over q},\quad 1< q<\infty .\end{aligned}$$

Now, we are in a position to prove Theorem 1.

Proof of Theorem 1

For each $\omega \in A_p,$ choose $1<q<p<\infty$ such that $\omega \in A_{p/q}.$ Then, it is well known that the maximal operators ${\mathcal M}$ and ${\mathcal M}_q$ are bounded on $L^p({\mathbb R}^n, \omega )$. On the other hand, since the operators $T_N$ are uniformly bounded in $L^p({\mathbb R}^n, \omega )$ with $\omega \in A_p$. Hence, by Theorem 7, we have

$$\begin{aligned} \left\| T_M^*f\right\| _{L^p({\mathbb R}^n, \omega )}&\le C\left( \left\| {\mathcal M}(T_{-M, M} f)\right\| _{L^p({\mathbb R}^n, \omega )}+\left\| {\mathcal M}_q f\right\| _{L^p({\mathbb R}^n, \omega )}\right) \\ {}&\le C\left( \left\| T_{-M, M} f\right\| _{L^p({\mathbb R}^n, \omega )}+\left\| f\right\| _{L^p({\mathbb R}^n, \omega )}\right) \le C\left\| f\right\| _{L^p({\mathbb R}^n, \omega )}. \end{aligned}$$

Note that the constants C appeared above do not depend on M. Consequently, letting M increase to infinity, we get the proof of the $L^p$ boundedness of the maximal operator $T_\Delta ^*,$ where $\displaystyle T_\Delta ^*f(x)=\sup _N \left| T_Nf(x)\right| .$

We should note that

$$\begin{aligned} T^* f(x)&=\sup _{N\in \mathbb Z^2}\left| T_{N}^\alpha f(x)\right| =\sup _{N\in \mathbb Z^2}\left| \sum _{j=N_1}^{N_2} v_j({\mathcal P}_{a_{j+1}}^\alpha f(x)-{\mathcal P}_{a_j}^\alpha f(x))\right| \\&=C_\alpha \sup _{N\in \mathbb Z^2}\left| \sum _{j=N_1}^{N_2} v_j\left( \int _0^{+\infty } e^{-s} \left( e^{-\frac{a_{j+1}^2}{4s} \Delta } f(x)- e^{-\frac{a_j^2}{4s} \Delta } f(x)\right) \,\frac{d s}{s^{1-\alpha }}\right) \right| \\&\le C_\alpha \int _0^{+\infty } e^{-s} \sup _{N\in \mathbb Z^2}\left| \sum _{j=N_1}^{N_2} v_j\left( e^{-\frac{a_{j+1}^2}{4s} \Delta } f(x)- e^{-\frac{a_j^2}{4s} \Delta } f(x)\right) \right| \,\frac{d s}{s^{1-\alpha }}\\&=C_{\alpha , \rho , v, n} \int _0^{+\infty } e^{-s} {\bar{T}}_\Delta ^*f(x)\,\frac{d s}{s^{1-\alpha }}, \end{aligned}$$

where the operator ${\bar{T}}_{\Delta }^*$(which is bounded on $L^p(\mathbb R^n, \omega )$ and the boundedness constant is not depending on s) denotes the maximal differential transform related with $\{v_j\}_{j\in \mathbb Z}$ and $\rho ^2$-lacunary sequence $\left\{ {a_j^2/ 4s}\right\} _{j\in \mathbb Z}$. Then,

$$\begin{aligned} \left\| T^* f\right\| _{L^p({\mathbb R}^n, \omega )}\le C_{\alpha , \rho , v, n} \int _0^{+\infty } e^{-s} \left\| {\bar{T}}_\Delta ^*f\right\| _{L^p({\mathbb R}^n, \omega )}\,\frac{d s}{s^{1-\alpha }}\le C_{\alpha , \rho , v, n} \left\| f\right\| _{L^p({\mathbb R}^n, \omega )}. \end{aligned}$$

This completes the proof of part (a) of the theorem.

In order to prove (b), we consider the $\ell ^\infty (\mathbb Z^2)$-valued operator $\mathcal {T}f(x) = \{ T_N^\alpha f(x) \}_{N\in \mathbb Z^2}$. Since $\Vert \mathcal {T}f(x) \Vert _{\ell ^\infty (\mathbb Z^2)}= T^*f(x)$, by using (a) we know that the operator $\mathcal {T}$ is bounded from $L^p({\mathbb R}^n, \omega )$ into $L^p_{\ell ^\infty (\mathbb Z^2)}(\mathbb {R}^n, \omega )$, for every $1<p<\infty$ and $\omega \in A_p$. The kernel of the operator $\mathcal {T}$ is given by $\mathcal {K}^\alpha (x) = \{ K^\alpha _N(x)\} _{N\in \mathbb Z^2}$. By Theorem 5 and the vector valued version of Theorem 7.12 in [7], we get that the operator $\mathcal {T}$ is bounded from $L^1(\mathbb {R}^n, \omega )$ into weak- $L^1_{\ell ^\infty (\mathbb Z^2)}(\mathbb {R}^n, \omega )$ for $\omega \in A_1$. Hence, as $\Vert \mathcal {T}f(x) \Vert _{\ell ^\infty (\mathbb Z^2)}= T^*f(x)$, we get the proof of (b).

For (c),  we shall prove that if $f\in L^\infty (\mathbb R^n)$ and there exists $x_0\in \mathbb R^n$ such that $T^*f(x_0)<\infty ,$ then $T^*f(x)<\infty$ for a.e. $x\in {\mathbb R}^n.$ Given $x\ne x_0.$ Set $f_1=f\chi _{B{(x_0, 4|x-x_0|)}}$ and $f_2 = f-f_1$. Note that $T^*$ is $L^p$-bounded for any $1<p<\infty .$ Then $T^*f_1(x)<\infty$, because $f_1\in L^p(\mathbb R^n)$ for any $1<p<\infty .$ On the other hand, by Theorem 5 we have

$$\begin{aligned}&\Big |T_N^\alpha f_2(x)-T_N^\alpha f_2(x_0)\Big |\\ {}&=\Big |\int _{{\mathbb R}^n} K_N^\alpha (x, y)f_2(y){\textrm{d}}y-\int _{{\mathbb R}^n} K_N^\alpha (x_0, y)f_2(y){\textrm{d}}y \Big |\\&=\Big | \int _{B^c(x_0, 4|x-x_0|)}\left( K_N^\alpha (x, y) - K_N^\alpha (x_0, y)\right) f(y){\textrm{d}}y\Big |\\&\le C \int _{B^c(x_0, 4|x-x_0|)} \frac{\left| x-x_0\right| }{\left| y-x_0\right| ^{n+1}}\left| f(y)\right| {\textrm{d}}y \\&\le C\left\| f\right\| _{L^\infty (\mathbb R)}< +\infty . \end{aligned}$$

Hence

$$\begin{aligned} \left\| T_N^\alpha f_2(x)-T_N^\alpha f_2(x_0)\right\| _{l^\infty (\mathbb Z^2)} \le C\left\| f\right\| _{L^\infty (\mathbb R^n)} \end{aligned}$$

and therefore $T^*f(x) = \left\| T_N^\alpha f(x)\right\| _{l^\infty (\mathbb Z^2)} \le C < \infty .$ For the $L^\infty -BMO$ boundedness, we will prove it later.

(d) Let $x_0\in {\mathbb R}^n$ be one point such that $T^*f(x_0)<\infty .$ Set $B=B(x_0, 4\left| x-x_0\right| )$ with $x\ne x_0$. And we decompose f to be

$$\begin{aligned} f=(f-f_B)\chi _B+(f-f_B)\chi _{B^c}+f_B=:f_1+f_2+f_3. \end{aligned}$$

Note that $T^*$ is $L^p$-bounded for any $1<p<\infty .$ Then $T^*f_1(x)<\infty$, because $f_1\in L^p(\mathbb R^n)$, for any $1<p<\infty .$ And $T^* f_3=0$, since ${\mathcal P}_{a_j}^\alpha f_3=f_3$ for any $j\in \mathbb Z.$ On the other hand, by Theorem 5 we have

$$\begin{aligned}&\Big |T_N^\alpha f_2(x)-T_N^\alpha f_2(x_0)\Big |\\ {}&=\Big |\int _{{\mathbb R}^n} K_N^\alpha (x, y)f_2(y){\textrm{d}}y-\int _{{\mathbb R}^n} K_N^\alpha (x_0, y)f_2(y){\textrm{d}}y \Big |\\&=\Big | \int _{B^c}\left( K_N^\alpha (x, y) - K_N^\alpha (x_0, y)\right) f_2(y){\textrm{d}}y\Big |\\&\le C \int _{B^c} \frac{\left| x-x_0\right| }{\left| y-x_0\right| ^{n+1}}\left| f(y)-f_B\right| {\textrm{d}}y \\&\le C \sum _{k=1}^{+\infty }{ |x-x_0|} \int _{2^{k} B\setminus 2^{k-1}B} {\left| f(y)-f_B\right| \over |y-x_0|^{n+1}}{\textrm{d}}y\\&\le C \sum _{k=1}^{+\infty }{ |x-x_0|\over (2^{k+1}|x-x_0|)^{n+1}} \int _{2^kB} {\left| f(y)-f_B\right| }{\textrm{d}}y\\&\le C \sum _{k=1}^{+\infty }2^{-(k+1)}{ 1\over \left| 2^{k}B\right| } \int _{2^{k}B} \left( {\left| f(y)-f_{2^kB}\right| }+\sum _{l=1}^{k}\left| f_{2^lB}-f_{2^{l-1}B}\right| \right) {\textrm{d}}y\\&\le C\sum _{k=1}^{+\infty }2^{-(k+1)}{ 1\over \left| 2^{k}B\right| } \int _{2^{k}B} \left( {\left| f(y)-f_{2^kB}\right| }+2k\left\| f\right\| _{BMO({\mathbb R}^n)}\right) {\textrm{d}}y\\&\le C\sum _{k=1}^{+\infty }2^{-(k+1)}{(1+2k)\left\| f\right\| _{BMO({\mathbb R}^n)}}\\&\le C\left\| f\right\| _{BMO({\mathbb R}^n)}, \end{aligned}$$

where $2^kB=B(x_0, 2^{k}\cdot 4|x-x_0|)$ for any $k\in \mathbb N.$ Hence

$$\begin{aligned} \left\| T_N^\alpha f_2(x)-T_N^\alpha f_2(x_0)\right\| _{l^\infty (\mathbb Z^2)} \le C\left\| f\right\| _{BMO(\mathbb R^n)} \end{aligned}$$

and therefore $T^*f(x) = \left\| T_N^\alpha f(x)\right\| _{l^\infty (\mathbb Z^2)} \le C < \infty .$

Now, we shall prove the estimate (1.4) for functions such that $T^*f(x) < \infty \, \, a.e.$ For any $r>0$ and $x_0$ such that $T^*f(x_0) < \infty$, consider the ball $B=B(x_0, r)$ and $\displaystyle f_B={1\over |B|}\int _B f(x){\textrm{d}}x.$ Let

$$\begin{aligned}f=(f-f_B)\chi _{2B}+(f-f_B)\chi _{(2B)^c}+f_B=:f_1+f_2+f_3.\end{aligned}$$

We have $T^*f_3(x)=0.$ Then,

$$\begin{aligned}&{1\over |B|}\int _{B}\left| T^*f(x)-(T^*f)_B\right| {\textrm{d}}x={1\over |B|}\int _{B}\left| {1\over |B|}\int _{B}\left( T^*f(x)-T^*f(y)\right) {\textrm{d}}y\right| {\textrm{d}}x\\&\le {1\over |B|^2}\int _{B}\int _{B}\left| T^*f(x)-T^*f(y)\right| {\textrm{d}}y{\textrm{d}}x\\&={1\over |B|^2}\int _{B}\int _{B}\left| \left\| T^\alpha _Nf(x)\right\| _{l^\infty (\mathbb Z^2)}-\left\| T^\alpha _Nf(y)\right\| _{l^\infty (\mathbb Z^2)}\right| {\textrm{d}}y{\textrm{d}}x\\&\le {1\over |B|^2}\int _{B}\int _B{\left\| T^\alpha _Nf(x)-T^\alpha _Nf(y)\right\| _{l^\infty (\mathbb Z^2)}}{\textrm{d}}y{\textrm{d}}x\\&\le {1\over |B|^2}\int _{B}\int _{B}{\left\| T^\alpha _Nf_1(x)-T^\alpha _N f_1(y)\right\| _{l^\infty (\mathbb Z^2)}}{\textrm{d}}y{\textrm{d}}x\\&\quad + {1\over |B|^2}\int _{B}\int _{B}{\left\| T^\alpha _Nf_2(x)-T^\alpha _Nf_2(y)\right\| _{l^\infty (\mathbb Z^2)}}{\textrm{d}}y{\textrm{d}}x\\&=:I+II. \end{aligned}$$

The Hölder inequality and $L^2$-boundedness of $T^*$ imply that

$$\begin{aligned} I&\le {1\over |B|}\int _{B}{\left\| T^\alpha _Nf_1(x)\right\| _{l^\infty (\mathbb Z^2)}}{\textrm{d}}x +{1\over |B|}\int _{B}{\left\| T^\alpha _Nf_1(y)\right\| _{l^\infty (\mathbb Z^2)}}{\textrm{d}}y\\&\le \left( {1\over |B|}\int _{B}{\left\| T^\alpha _Nf_1(x)\right\| ^2_{l^\infty (\mathbb Z^2)}}{\textrm{d}}x\right) ^{1/2} +\left( {1\over |B|}\int _{B}{\left\| T^\alpha _Nf_1(y)\right\| ^2_{l^\infty (\mathbb Z^2)}}{\textrm{d}}y\right) ^{1/2}\\&\le C{1\over |B|^{1/2}}\left\| f_1\right\| _{L^2(\mathbb R^n)}\le C\left\| f\right\| _{BMO(\mathbb R^n)}. \end{aligned}$$

For II, since $x, y \in B$ and the support of $f_2$ is $(2B)^c$, by Theorem 5 we have

$$\begin{aligned}&\Big |T_N^\alpha f_2(x)-T_N^\alpha f_2(y)\Big |\\&=\Big |\int _{{\mathbb R}^n} K_N^\alpha (x, z)f_2(z){\textrm{d}}z-\int _{{\mathbb R}^n} K_N^\alpha (y, z)f_2(z){\textrm{d}}z \Big |\\&=\Big | \int _{(2B)^c}\left( K_N^\alpha (x, z) - K_N^\alpha (y, z)\right) f_2(z){\textrm{d}}z\Big |\\&\le C \sum _{k=2}^{+\infty }{ r} \int _{2^{k} B\setminus 2^{k-1}B} {\left| f(z)-f_B\right| \over |z-x_0|^{n+1}}{\textrm{d}}z\\&\le C \sum _{k=2}^{+\infty }{ r\over (2^{k}r)^{n+1}} \int _{2^kB} {\left| f(z)-f_B\right| }{\textrm{d}}z\\&\le C \sum _{k=2}^{+\infty }2^{-k}{ 1\over \left| 2^{k}B\right| } \int _{2^{k}B} \left( {\left| f(z)-f_{2^kB}\right| }+\sum _{l=1}^{k}\left| f_{2^lB}-f_{2^{l-1}B}\right| \right) {\textrm{d}}z\\&\le C\sum _{k=2}^{+\infty }2^{-k}{ 1\over \left| 2^{k}B\right| } \int _{2^{k}B} \left( {\left| f(z)-f_{2^kB}\right| }+2k\left\| f\right\| _{BMO({\mathbb R}^n)}\right) {\textrm{d}}z\\&\le C\sum _{k=2}^{+\infty }2^{-k}{(1+2k)\left\| f\right\| _{BMO({\mathbb R}^n)}}\\&\le C\left\| f\right\| _{BMO({\mathbb R}^n)}, \end{aligned}$$

where $2^kB=B(x_0, 2^kr)$. Hence, we have $II\le C \left\| f\right\| _{BMO({\mathbb R}^n)}.$ Then by the arbitrary of $x_0$ and $r>0$, we proved

$$\begin{aligned}\left\| T^*f\right\| _{BMO(\mathbb R^n)}\le C\left\| f\right\| _{BMO(\mathbb R^n)}.\end{aligned}$$

For the second part of (c), we can deduce it from the BMO-boundedness of $T^*$ and the inclusion $L^\infty ({\mathbb R}^n)\subset BMO({\mathbb R}^n).$ This completes the proof of Theorem 1. $\square$

From the conclusions we got in Theorem 1, we have the following result:

Theorem 8

• (a) If $1<p<\infty$ and $\omega \in A_p$, then $T^\alpha _N f$ converges a.e. and in $L^p(\mathbb R^n, \omega )$ norms for all $f\in L^p(\mathbb R^n, \omega )$ as $N=(N_1,N_2)$ tends to $(-\infty , +\infty ).$

• (b) If $p=1$ and $\omega \in A_1$, then $T^\alpha _N f$ converges a.e. and in measure for all $f\in L^1(\mathbb R^n, \omega )$ as $N=(N_1,N_2)$ tends to $(-\infty , +\infty ).$

Proof

First, we shall see that if $\varphi$ is a test function, then $T_N^\alpha \varphi (x)$ converges for all $x\in {\mathbb R}^n$. In order to prove this, it is enough to see that for any $N=(L,M)$ with $0<L<M$, the series

$$\begin{aligned} A= \sum _{j=L}^M v_j ( {\mathcal P}^\alpha _{a_{j+1}} \varphi (x) - {\mathcal P}_{a_j}^\alpha \varphi (x)) \hbox { and } B= \sum _{j=-M}^{-L} v_j ( {\mathcal P}^\alpha _{a_{j+1}} \varphi (x) - {\mathcal P}_{a_j}^\alpha \varphi (x)) \end{aligned}$$

converge to zero, when $L, M\rightarrow +\infty$. For A, by the mean value theorem and the $\rho$-lacunarity of the sequence $\{a_j\}_{j\in \mathbb Z}$ we have

$$\begin{aligned} |A|&=C_\alpha \left| \sum _{j=L}^{M} v_j\left( \int _0^{+\infty } e^{-s} \left( e^{\frac{a_{j+1}^2}{4s} \Delta } \varphi (x)- e^{\frac{a_j^2}{4s} \Delta } \varphi (x)\right) \,\frac{d s}{s^{1-\alpha }}\right) \right| \\&\le C_\alpha \int _0^{+\infty } e^{-s}\left| \sum _{j=L}^{M} v_j \left( e^{\frac{a_{j+1}^2}{4s} \Delta } \varphi (x)- e^{\frac{a_j^2}{4s} \Delta } \varphi (x)\right) \right| \,\frac{d s}{s^{1-\alpha }}\\&= C_{n,\alpha } \int _0^{+\infty } e^{-s}\left| \sum _{j=L}^{M} v_j \left( \int _{{\mathbb R}^n}\frac{s^{n/2}}{a_{j+1}^n} e^{-\frac{s|y|^2}{a_{j+1}^2} } \varphi (x-y){\textrm{d}}y-\int _{{\mathbb R}^n}\frac{s^{n/2}}{a_{j}^n} e^{-\frac{s|y|^2}{a_{j}^2} } \varphi (x-y){\textrm{d}}y \right) \right| \,\frac{d s}{s^{1-\alpha }}\\&\le C_{n, \alpha ,v,\rho } \int _0^{+\infty } e^{-s}\int _{{\mathbb R}^n}\sum _{j=L}^{M} \frac{s^{n/2}}{a_j^n}|\varphi (x-y|{\textrm{d}}y \,\frac{d s}{s^{1-\alpha }}\\&\le C_{n, \alpha ,v,\rho }\left( \frac{1}{a_L^{n}}\sum _{j=L}^{M} \frac{a_L^{n}}{a_j^n}\right) \int _0^{+\infty } e^{-s}\int _{{\mathbb R}^n}|\varphi (x-y|{\textrm{d}}y \,\frac{d s}{s^{1-\alpha -n/2}}\\&\le C_{n, \alpha ,v,\rho }\left\| \varphi \right\| _{L^1({\mathbb R}^n)}\frac{1}{a_L^n}\rightarrow 0, \quad \text {as}\, \, \, L,M\rightarrow +\infty . \end{aligned}$$

For B,  as the integral of the kernels are zero, we can write

$$\begin{aligned} B&=C_{n,\alpha } \int _{{\mathbb R}^{n+1}} \sum _{j=-M}^{-L}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} {e^{-{|y|^n/(4s)}}\over (4\pi s)^{n/2}}( \varphi (x-y)- \varphi (x)){\textrm{d}}y{\textrm{d}}s\\&= C_{n,\alpha } \int _0^1 \int _{{\mathbb R}^n}\sum _{j=-M}^{-L}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} {e^{-{|y|^n/(4s)}}\over (4\pi s)^{n/2}}( \varphi (x-y)- \varphi (x)){\textrm{d}}y{\textrm{d}}s\\&\quad +C_{n,\alpha } \int _1^\infty \int _{{\mathbb R}^n}\sum _{j=-M}^{-L}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} {e^{-{|y|^n/(4s)}}\over (4\pi s)^{n/2}}( \varphi (x-y)- \varphi (x)){\textrm{d}}y{\textrm{d}}s\\&=: B_1+B_2. \end{aligned}$$

Proceeding as in the case A, and by using the fact that $\varphi$ is a test function, we have

$$\begin{aligned} |B_1|&= C_{n,\alpha } \Big | \int _0^1 \int _{{\mathbb R}^n} \sum _{j=-M}^{-L}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} {e^{-{|y|^2/(4s)}}\over (4\pi s)^{n/2}}( \varphi (x-y)- \varphi (x)){\textrm{d}}y{\textrm{d}}s\Big | \\&\le C_{n,\alpha } \Big | \int _0^1 \int _{{\mathbb R}^n} \sum _{j=-M}^{-L}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} {e^{-{|y|^2/(4s)}}\over (4\pi s)^{n/2}}\left\| \nabla \varphi \right\| _{L^\infty ({\mathbb R}^{n+1})}|y|{\textrm{d}}y{\textrm{d}}s\Big | \\&\le C_{n,\alpha } \left\| \nabla \varphi \right\| _{L^\infty ({\mathbb R}^{n})} \int _0^{1}\int _{{\mathbb R}^n} \sum _{j=-M}^{-L}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{\alpha +1/2}}{e^{-{|y|^2/(4s)}}\over (4\pi s)^{n/2}} {\textrm{d}}y{\textrm{d}}s \\&\le C_{n,\alpha ,v,\rho } \left\| \nabla \varphi \right\| _{L^\infty ({\mathbb R}^{n})} \int _0^1 \sum _{j=-M}^{-L} \frac{a_{j}^{2\alpha } e^{-a_{j}^2/(4 s)}}{s^{\alpha +1/2}} {\textrm{d}}s. \end{aligned}$$

If $\displaystyle 0<\alpha \le {1\over 2},$ then, for any $0<\varepsilon <2\alpha ,$ we have

$$\begin{aligned} |B_1|&\le C_{n,\alpha , v,\rho } \left\| \nabla \varphi \right\| _{L^\infty ({\mathbb R}^{n})} \sum _{j=-M}^{-L} {a_{j}^{2\alpha -\varepsilon }}\int _0^1{s^{\varepsilon /2-\alpha -1/2}} {\textrm{d}}s\\&\le C_{n,\alpha ,v,\rho ,\varepsilon }\left\| \nabla \varphi \right\| _{L^\infty ({\mathbb R}^{n})} a_{-L}^{2\alpha -\varepsilon } \sum _{j=-M}^{-L} \frac{a_{j}^{2\alpha -\varepsilon }}{a_{-L}^{2\alpha -\varepsilon }} \\&\le C_{n,\alpha ,v,\rho ,\varepsilon } \left\| \nabla \varphi \right\| _{L^\infty ({\mathbb R}^{n})}a_{-L}^{2\alpha -\varepsilon }\longrightarrow 0,\quad \hbox {as}\ { L,M \rightarrow +\infty }. \end{aligned}$$

If $\displaystyle {1\over 2}<\alpha <1,$ then

$$\begin{aligned} |B_1|&\le C_{n,\alpha , v,\rho } \left\| \nabla \varphi \right\| _{L^\infty ({\mathbb R}^{n})} \int _0^1 \sum _{j=-M}^{-L} \frac{a_{j}^{2\alpha } e^{-a_{j}^2/(4 s)}}{s^{\alpha +1/2}} {\textrm{d}}s\\&\le C_{n,\alpha ,v,\rho }\left\| \nabla \varphi \right\| _{L^\infty ({\mathbb R}^{n})} a_{-L}^{2\alpha -1} \sum _{j=-M}^{-L} \frac{a_{j}^{2\alpha -1}}{a_{-L}^{2\alpha -1}} \int _0^1 \frac{1}{s^{\alpha }} {\textrm{d}}s\\&\le C_{n,\alpha ,v,\rho } \left\| \nabla \varphi \right\| _{L^\infty ({\mathbb R}^{n})}a_{-L}^{2\alpha -1}\longrightarrow 0,\quad \hbox {as}\ { L,M \rightarrow +\infty }. \end{aligned}$$

Therefore, we get

$$\begin{aligned}|B_1|\longrightarrow 0,\quad \hbox {as}\ { L,M \rightarrow +\infty }.\end{aligned}$$

On the other hand,

$$\begin{aligned} |B_2|&\le C_{n,\alpha ,\rho } \left\| v\right\| _{l^\infty (\mathbb Z)} \Vert \varphi \Vert _{L^\infty ({\mathbb R}^{n}) }\int _1^\infty \sum _{j=-M}^{-L}\frac{ a_{j}^{2\alpha }}{s^{1+\alpha }} {\textrm{d}}s\\&\le C_{n, \alpha , v }\Vert \varphi \Vert _{L^\infty ({\mathbb R}^{n}) } \sum _{j=-M}^{-L}{ a_j^{2\alpha }}\int _1^\infty {1\over s^{1+\alpha }} {\textrm{d}}s \\&\le C_{n,\alpha , v, \rho } \Vert \varphi \Vert _{L^\infty ({\mathbb R}^{n}) } a_{-L}^{2\alpha }\sum _{j=-M}^{-L} {a_j^{2\alpha }\over a_{-L}^{2\alpha }} \\&\le C_{n, \alpha , v, \rho }\Vert \varphi \Vert _{L^\infty ({\mathbb R}^{n}) } {\rho ^{2\alpha }\over \rho ^{2\alpha }-1}a_{-L}^{2\alpha } \longrightarrow 0,\quad \hbox {as}\ { L,M \rightarrow +\infty }. \end{aligned}$$

As the set of test functions is dense in $L^p(\mathbb {R}^n)$, by Theorem 1 we get the a.e. convergence for any function in $L^p(\mathbb {R}^n)$. Analogously, since $L^p(\mathbb {R}^n) \cap L^p(\mathbb {R}^n, \omega )$ is dense in $L^p(\mathbb {R}^n, \omega )$, we get the a.e. convergence for functions in $L^p(\mathbb {R}^n, \omega )$ with $1\le p<\infty$. By using the dominated convergence theorem, we can prove the convergence in $L^p(\mathbb {R}^n, \omega )$-norm for $1<p<\infty$, and also in measure. $\square$

3 Proof of Theorem 2

The dichotomy results announced in Theorem 1, parts (c) and (d), about $L^\infty (\mathbb R^{n})$ and $BMO(\mathbb R^{n})$ are motivated, in part, by the existence of a bounded function f such that $T^*f(x)=\infty$ as the following theorem shows. In [5], we can find some related results for the variation operators.

Theorem 9

There exist bounded sequence $\{v_j\}_{j\in \mathbb Z}$, $\rho$-lacunary sequence $\{a_j\}_{j\in \mathbb Z}$ and $f\in L^\infty (\mathbb R^n)$ such that $T^* f(x) =\infty$ for all $x\in \mathbb R^n$.

Proof

We will only consider the case $n=1$. For the multi-dimensional case, it is similar just with minor modifications. Let f be the function defined by

$$\begin{aligned} f(x) = \sum _{k\in \mathbb {Z}} (-1)^{k} \chi _{(-a^{2k+1}, -a^{2k}]}(x), \quad x\in {\mathbb R}, \end{aligned}$$

where $a>1$ is a real number that we shall fix it later. It is easy to see that

$$\begin{aligned} f(a^{2j}x) = (-1)^j f(x). \end{aligned}$$

(3.1)

By changing variable, we have

$$\begin{aligned} \mathcal {P}_{a_j}^\alpha f(x)&= \frac{1}{4^\alpha \Gamma (\alpha ) } \int _0^{+\infty } \int _{\mathbb R}\frac{a_j^{2\alpha } e^{-a_j^{2}/(4s)}}{s^{1+\alpha }}{e^{-|y|^2/4s} \over (4\pi s)^{1/2}}f(x-y){\textrm{d}}y {\textrm{d}}s \\&=\frac{1}{\sqrt{\pi }\Gamma (\alpha ) } \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f(x-a_j y){\textrm{d}}y \frac{{\textrm{d}}s}{s}. \end{aligned}$$

Let $a_j= a^{2j}.$ Then,

$$\begin{aligned} \mathcal {P}_{a_j}^\alpha f(0)&=\frac{1}{\sqrt{\pi }\Gamma (\alpha ) } \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f(-a^{2j} y){\textrm{d}}y \frac{{\textrm{d}}s}{s} \\&= \frac{(-1)^j}{\sqrt{\pi }\Gamma (\alpha ) } \int _0^{+\infty } s^\alpha e^{-s}\int _0^{+\infty } s^{1/2} e^{-s|y|^2} f(- y){\textrm{d}}y \frac{{\textrm{d}}s}{s}. \end{aligned}$$

We observe that

$$\begin{aligned} \int _0^{+\infty } s^{1/2} e^{-s|y|^2} \left| f(- y)\right| {\textrm{d}}y\le \int _0^{+\infty } s^{1/2} e^{-s|y|^2} {\textrm{d}}y=\int _0^{+\infty } e^{-t^2} {\textrm{d}}t={\sqrt{\pi }\over 2}. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _0^{+\infty } s^\alpha e^{-s}\int _0^{+\infty } s^{1/2} e^{-s|y|^2}\left| f(- y)\right| {\textrm{d}}y \frac{{\textrm{d}}s}{s}\le {\sqrt{\pi }\over 2}\int _0^{+\infty } s^\alpha e^{-s}\frac{{\textrm{d}}s}{s}={\sqrt{\pi }\over 2}\Gamma (\alpha )< \infty . \end{aligned}$$

Also, we have

$$\begin{aligned}\lim _{R\rightarrow {+\infty }} \int _0^{+\infty }s^\alpha e^{-s}\int _R^{+\infty } s^{1/2} e^{-s|y|^2} f(- y){\textrm{d}}y \frac{{\textrm{d}}s}{s}=0\end{aligned}$$

and

$$\begin{aligned}\lim _{\varepsilon \rightarrow 0^+} \int _0^{+\infty }s^\alpha e^{-s}\int _0^\varepsilon s^{1/2} e^{-s|y|^2} f(- y){\textrm{d}}y\frac{{\textrm{d}}s}{s}=0. \end{aligned}$$

On the other hand, there exists a constant $C>0$ such that

$$\begin{aligned}{} & {} \displaystyle \lim _{a\rightarrow {+\infty }}\int _0^{+\infty }s^\alpha e^{-s} \int _1^a s^{1/2} e^{-s|y|^2} f(- y){\textrm{d}}y\frac{{\textrm{d}}s}{s}\\ {}{} & {} \quad = \displaystyle \lim _{a\rightarrow {+\infty }}\int _0^{+\infty }s^\alpha e^{-s} \int _1^a s^{1/2} e^{-s|y|^2} {\textrm{d}}y\frac{{\textrm{d}}s}{s}=C>0.\end{aligned}$$

Hence we can choose $a>1$ big enough such that

$$\begin{aligned}&\int _0^{+\infty }s^\alpha e^{-s}\int _1^a s^{1/2} e^{-s|y|^2} f(- y){\textrm{d}}y\frac{{\textrm{d}}s}{s} = \int _0^{+\infty }s^\alpha e^{-s}\int _1^a s^{1/2} e^{-s|y|^2} {\textrm{d}}y\frac{{\textrm{d}}s}{s} \\&> \int _0^{+\infty }s^\alpha e^{-s}{\int _0^{1/a} s^{1/2} e^{-s|y|^2} {\textrm{d}}y}\frac{{\textrm{d}}s}{s} + \int _0^{+\infty }s^\alpha e^{-s}{\int _{a^2}^{+\infty } s^{1/2} e^{-s|y|^2} {\textrm{d}}y}\frac{{\textrm{d}}s}{s}\\&> \int _0^{+\infty }s^\alpha e^{-s} \left| \int _0^{1/a} s^{1/2} e^{-s|y|^2} f(- y){\textrm{d}}y\right| \frac{{\textrm{d}}s}{s}\\ {}&\quad +\int _0^{+\infty }s^\alpha e^{-s} \left| \int _{a^2}^{+\infty } s^{1/2} e^{-s|y|^2} f(- y){\textrm{d}}y\right| \frac{{\textrm{d}}s}{s}. \end{aligned}$$

In other words, with the $a>1$ fixed above, there exists constant $C_1>0$ such that

$$\begin{aligned} \int _0^{+\infty }s^\alpha e^{-s}\int _0^{+\infty } s^{1/2} e^{-s|y|^2} f(- y){\textrm{d}}y\frac{{\textrm{d}}s}{s}=C_1. \end{aligned}$$

(3.2)

Hence

$$\begin{aligned} \Big | \mathcal {P}_{a_{j+1}}^\alpha f(0)-\mathcal {P}_{a_j}^\alpha f(0)\Big |= {{2 C_1}\over {\sqrt{\pi }\Gamma (\alpha )}}>0. \end{aligned}$$

Therefore, we have

$$\begin{aligned}\sum _{j\in \mathbb Z} \Big |\mathcal {P}_{a_{j+1}}^\alpha f(0)- \mathcal {P}_{a_{j}}^\alpha f(0)\Big | = \infty .\end{aligned}$$

By using (3.1) and changing variable we get

$$\begin{aligned} \mathcal {P}_{a_j}^\alpha f(x)&= \frac{1}{\sqrt{\pi }\Gamma (\alpha ) } \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f(x-a^{2j} y){\textrm{d}}y \frac{{\textrm{d}}s}{s}\\&=\frac{(-1)^j}{\sqrt{\pi }\Gamma (\alpha ) } \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( {x\over a^{2j}}- y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}. \end{aligned}$$

Then

$$\begin{aligned}&{\mathcal {P}_{a_{j+1}}^\alpha f(t)-\mathcal {P}_{a_j}^\alpha f(t)}\nonumber \\&= \frac{ (-1)^{j+1}}{\sqrt{\pi }\Gamma (\alpha ) } \Big \{\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( {x\over a^{2(j+1)}}- y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}\nonumber \\&\quad +\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( {x\over a^{2j}}- y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}\Big \}. \end{aligned}$$

(3.3)

By the dominated convergence theorem, we know that

$$\begin{aligned}&\lim _{h\rightarrow 0}\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( h- y\right) {\textrm{d}}y\frac{{\textrm{d}}s}{s}\\&=\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( - y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s} \\&=\int _0^{+\infty } s^\alpha e^{-s}\int _0^{+\infty }s^{1/2} e^{-s|y|^2} f\left( - y\right) {\textrm{d}}y\frac{{\textrm{d}}s}{s} \\&= C_1>0, \end{aligned}$$

where $C_1$ is the constant appeared in (3.2). So, there exists $0<\eta _0<1,$ such that, for $|h|<\eta _0,$

$$\begin{aligned} \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( h- y\right) {\textrm{d}}y\frac{{\textrm{d}}s}{s}\ge \frac{C_1}{2}. \end{aligned}$$

Then, for each $x\in {\mathbb R}$, we can choose $j\in \mathbb Z$ such that $\displaystyle {|x|\over {a^j}}<\eta _0$ (there are infinite j satisfying this condition), and we have

$$\begin{aligned}{} & {} \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( {x\over a^{2(j+1)}}- y\right) {\textrm{d}}y\frac{{\textrm{d}}s}{s} \\{} & {} \quad +\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( {x\over a^{2j}}- y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s} \ge C_1>0. \end{aligned}$$

Choosing $v_j=(-1)^{j+1},\ j\in \mathbb Z$, by (3.3) we have, for any $x\in \mathbb R,$

$$\begin{aligned} T^* f(x)&\ge \sum _{\left| x\over {a^j}\right|<\eta _0} (-1)^{j+1}\big ({\mathcal P}_{a_{j+1}}^\alpha f(x)-{\mathcal P}_{a_j}^\alpha f(x)\big )\\&=\frac{ 1}{\sqrt{\pi }\Gamma (\alpha ) }\sum _{\left| x\over {a^j}\right|<\eta _0} \Big \{\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( {x\over a^{2(j+1)}}- y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}\\&\quad +\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( {x\over a^{2j}}- y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}\Big \}\\&\ge \frac{ 1}{\sqrt{\pi }\Gamma (\alpha ) }\sum _{\left| x\over {a^j}\right| <\eta _0}C_1=\infty . \end{aligned}$$

We complete the proof of Theorem 9. $\square$

At the end of this section, we will give the proof of Theorem 2.

Proof of Theorem 2

First, we prove the theorem in the case $1<p<\infty .$ Since $2r<1,$ we know that $B\backslash B_{2r}\ne \emptyset .$ Let $f(x)=f_1(x)+f_2(x)$, where $f_1(x)=f(x)\chi _{B_{2r}}(x)$ and $f_2(x)=f(x)\chi _{B\backslash B_{2r}}(x)$. Then

$$\begin{aligned}\left| T^* f(x)\right| \le \left| T^* f_1(x)\right| +\left| T^* f_2(x)\right| .\end{aligned}$$

By Theorem 1,

$$\begin{aligned} \frac{1}{|B_r|} \int _{B_r} \left| T^* f_1 (x)\right| {\textrm{d}}x\le & {} \left( \frac{1}{|B_r|} \int _{B_r} \left| T^* f_1 (x)\right| ^2 {\textrm{d}}x\right) ^{1/2}\\\le & {} C \left( \frac{1}{|B_r|} \int _{\mathbb {R}}\left| f_1 (x)\right| ^2 {\textrm{d}}x\right) ^{1/2}\le C\left\| f\right\| _{L^\infty (\mathbb {R}^n)}. \end{aligned}$$

We also know that, for any $j\in \mathbb Z,$

$$\begin{aligned}{} & {} \int _0^{+\infty }\int _{{\mathbb R}^n}\left| \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} \right| {e^{-|x-y|^2/4s} \over (4\pi s)^{n/2}}~ {\textrm{d}}y{\textrm{d}}s\nonumber \\{} & {} \quad \le \int _0^{+\infty }\int _{{\mathbb R}^n}{\frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}+a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} }{e^{-|x-y|^2/4s} \over (4\pi s)^{n/2}}~ {\textrm{d}}y{\textrm{d}}s =2\cdot 4^\alpha \Gamma (\alpha ).\nonumber \\ \end{aligned}$$

(3.4)

Then, by Hölder’s inequality, (3.4) and Fubini’s Theorem, for $1< p < \infty$ and any $N=(N_1, N_2)$, we have

$$\begin{aligned}&\left| \sum _{j=N_1}^{N_2}v_j\left( {\mathcal P}_{a_{j+1}}^\alpha f_2(x)-{\mathcal P}_{a_j}^\alpha f_2(x)\right) \right| \\&\le C\sum _{j=N_1}^{N_2} \left| v_j \int _0^{+\infty }\frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s))}}{s^{1+\alpha }}\int _{{\mathbb R}^n}{e^{-|x-y|^2/4s} \over (4\pi s)^{n/2}} f_2(y)~{\textrm{d}}y {\textrm{d}}s\right| \\&\le C\left\| v\right\| _{l^p(\mathbb Z)}\left( \sum _{j=N_1}^{N_2}\left( \int _0^{+\infty }\left| \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} \right| \int _{{\mathbb R}^n}{e^{-|x-y|^2/4s} \over (4\pi s)^{n/2}}\left| f_2(y)\right| ~{\textrm{d}}y {\textrm{d}}s\right) ^{p'}\right) ^{1/p'} \\&\le C\left\| v\right\| _{l^p(\mathbb Z)}\Big (\sum _{j=N_1}^{N_2} \Big \{\int _0^{+\infty }\int _{{\mathbb R}^n}\left| \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} \right| {e^{-|x-y|^2/4s} \over (4\pi s)^{n/2}}\left| f_2(y)\right| ^{p'}~{\textrm{d}}y {\textrm{d}}s\Big \} \\&\quad \quad \times \Big \{\int _0^{+\infty }\int _{{\mathbb R}^n}\left| \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} \right| {e^{-|x-y|^2/4s} \over (4\pi s)^{n/2}}~ {\textrm{d}}y{\textrm{d}}s \Big \}^{p'/p} \Big )^{1/p'} \\&\le C \left\| v\right\| _{l^p(\mathbb Z)}\left( \sum _{j=N_1}^{N_2} \int _0^{+\infty }\int _{{\mathbb R}^n}\left| \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} \right| {e^{-|x-y|^2/4s} \over (4\pi s)^{n/2}}\left| f_2(y)\right| ^{p'}~{\textrm{d}}y {\textrm{d}}s \right) ^{1/p'} \\&\le C \left\| v\right\| _{l^p(\mathbb Z)}\left( \int _0^{+\infty }\int _{{\mathbb R}^n}\sum _{j=-\infty }^{+\infty } \left| \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} \right| {e^{-|x-y|^2/4s} \over (4\pi s)^{n/2}}\left| f_2(y)\right| ^{p'}~{\textrm{d}}y {\textrm{d}}s \right) ^{1/p'}\\&\le C \left\| v\right\| _{l^p(\mathbb Z)}\left( \int _0^{+\infty } \int _{{\mathbb R}^n}{1\over s} {e^{-|x-y|^2/4s} \over (4\pi s)^{n/2}}\left| f_2(y)\right| ^{p'}~{\textrm{d}}y {\textrm{d}}s \right) ^{1/p'}. \end{aligned}$$

For $x\in B\backslash B_{2r}$ and $y\in B_r$, we have $r\le |x-y|\le 2$. Then, by integration with polar coordinates we get

$$\begin{aligned} \frac{1}{|B_r|} \int _{B_r} \left| T^* f_2(x)\right| {\textrm{d}}x&\le C\frac{1}{|B_r|} \int _{B_r} \left( \int _0^{+\infty } \int _{{\mathbb R}^n}{1\over s} {e^{-|x-y|^2/4s} \over (4\pi s)^{n/2}}\left| f_2(y)\right| ^{p'}~{\textrm{d}}y {\textrm{d}}s \right) ^{1/p'}{\textrm{d}}t \\&\le C\frac{\left\| f\right\| _{L^\infty ({\mathbb R}^n)}}{|B_r|} \int _{B_r} \left( \int _0^{+\infty }\int _{S^{n-1}}\int _{r\le |t| \le 2} \frac{1}{s} {e^{-t^2/4s} \over (4\pi s)^{n/2}}~{\textrm{d}}td_{\omega _{n-1}} {\textrm{d}}s \right) ^{1/p'}{\textrm{d}}t \\&\sim \Big (\log \frac{2}{r}\Big )^{1/p'}\left\| f\right\| _{L^\infty ({\mathbb R}^n)}. \end{aligned}$$

Hence,

$$\begin{aligned}\frac{1}{|B_r|} \int _{B_r} \left| T^* f(x)\right| {\textrm{d}}x\le & {} C\left( 1+\Big (\log \frac{2}{r}\Big )^{1/p'}\right) \left\| f\right\| _{L^\infty ({\mathbb R}^n)}\\\le & {} C\Big (\log \frac{2}{r}\Big )^{1/p'}\left\| f\right\| _{L^\infty ({\mathbb R}^n)}.\end{aligned}$$

For the case $p=1$ and $p=\infty$, the proof is similar and easier. Then we get the proof of (a).

For (b),  we only consider the case $n=1$. It is similar in the multi-dimensional case. When $1< p<\infty ,$ for any $0<\varepsilon <p-1$, let

$$\begin{aligned}\displaystyle f(x) = \sum _{k=-\infty }^{0} (-1)^{k} \chi _{(-a^{2k}, -a^{2k-1}]}(x) \,\, \hbox {and }\, \, a_j=a^{2j}, \end{aligned}$$

with $a>1$ being fixed later. Then, the support of f is contained in $[-1, 0),$ and $\{a_j\}_{j\in \mathbb Z}$ is a $\rho$-lacunary sequence with $\rho =a^2>1.$ We observe that

$$\begin{aligned}{} & {} \left| \int _0^{+\infty } s^\alpha e^{-s}\int _0^{+\infty } s^{1/2} e^{-s|y|^2} f(-y){\textrm{d}}y\ \frac{{\textrm{d}}s}{s}\right| \le \int _0^{+\infty } s^\alpha e^{-s}\int _0^{+\infty } s^{1/2} e^{-s|y|^2} {\textrm{d}}y\frac{{\textrm{d}}s}{s}\\ {}{} & {} \quad ={\sqrt{\pi }\over 2}\Gamma (\alpha )< \infty . \end{aligned}$$

Hence

$$\begin{aligned}\lim _{R\rightarrow {+\infty }} \int _0^{+\infty } s^\alpha e^{-s}\int _R^{+\infty } s^{1/2} e^{-s|y|^2} f(-y){\textrm{d}}y =0\end{aligned}$$

and

$$\begin{aligned}\lim _{\varepsilon \rightarrow 0^+} \int _0^{+\infty } s^\alpha e^{-s}\int _0^\varepsilon s^{1/2} e^{-s|y|^2} f(-y){\textrm{d}}y\frac{{\textrm{d}}s}{s}=0. \end{aligned}$$

Also there exists a constant $C>0$ such that

$$\begin{aligned}{} & {} \displaystyle \lim _{a\rightarrow {+\infty }} \int _0^{+\infty } s^\alpha e^{-s}\int _{a^{-1}}^1 s^{1/2} e^{-s|y|^2} f(-y){\textrm{d}}y\frac{{\textrm{d}}s}{s} \\ {}{} & {} \quad = \displaystyle \lim _{a\rightarrow {+\infty }}\int _0^{+\infty } s^\alpha e^{-s} \int _{a^{-1}}^1 s^{1/2} e^{-s|y|^2} {\textrm{d}}y\frac{{\textrm{d}}s}{s} =C.\end{aligned}$$

So, we can choose $a>1$ big enough such that

$$\begin{aligned}&\int _0^{+\infty } s^\alpha e^{-s}\int _{a^{-1}}^1 s^{1/2} e^{-s|y|^2} f(-y){\textrm{d}}y\frac{{\textrm{d}}s}{s} = \int _0^{+\infty } s^\alpha e^{-s}\int _{a^{-1}}^1 s^{1/2} e^{-s|y|^2} {\textrm{d}}y\frac{{\textrm{d}}s}{s}\\&\ge 10 \left( \int _0^{+\infty } s^\alpha e^{-s}\int _0^{1/a^2} s^{1/2} e^{-s|y|^2} {\textrm{d}}y+\int _0^{+\infty } s^\alpha e^{-s}\int _{a-1}^{+\infty } s^{1/2} e^{-s|y|^2} {\textrm{d}}y\frac{{\textrm{d}}s}{s}\right) \nonumber \\&> 10 \left( \left| \int _0^{+\infty } s^\alpha e^{-s}\int _0^{1/a^2} s^{1/2} e^{-s|y|^2} f(-y){\textrm{d}}y\frac{{\textrm{d}}s}{s}\right| \right. \\ {}&\quad \left. +\left| \int _0^{+\infty } s^\alpha e^{-s}\int _{a-1}^{+\infty } s^{1/2} e^{-s|y|^2} f(-y){\textrm{d}}y\frac{{\textrm{d}}s}{s}\right| \right) . \end{aligned}$$

Therefore, there exists a constant $C_1>0$ such that

$$\begin{aligned} \int _0^{+\infty } s^\alpha e^{-s}\int _0^{+\infty }s^{1/2} e^{-s|y|^2} f(-y){\textrm{d}}y\frac{{\textrm{d}}s}{s}=C_1>0 \end{aligned}$$

(3.5)

and

$$\begin{aligned} 0< & {} \int _0^{+\infty } s^\alpha e^{-s}\int _0^{1/ a^2} s^{1/2} e^{-s|y|^2} {\textrm{d}}y\frac{{\textrm{d}}s}{s}\nonumber \\ {}{} & {} +\int _0^{+\infty } s^\alpha e^{-s}\int _{a-1}^{+\infty } s^{1/2} e^{-s|y|^2} {\textrm{d}}y\frac{{\textrm{d}}s}{s}\le {C_1\over 9}. \end{aligned}$$

(3.6)

On the other hand, by the dominated convergence theorem, we have

$$\begin{aligned}{} & {} \lim _{h\rightarrow 0}\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f(h-y){\textrm{d}}y\frac{{\textrm{d}}s}{s}\\ {}{} & {} \quad =\int _0^{+\infty } s^\alpha e^{-s}\int _0^{+\infty } s^{1/2} e^{-s|y|^2} f(-y){\textrm{d}}y\frac{{\textrm{d}}s}{s} = C_1>0, \end{aligned}$$

where $C_1$ is the constant appeared in (3.5). So, there exists $0<\eta _0<1,$ such that, for $|h|<\eta _0,$

$$\begin{aligned}{} & {} \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f(h-y){\textrm{d}}y\frac{{\textrm{d}}s}{s}\nonumber \\ {}{} & {} \quad \ge {1\over 2}\int _0^{+\infty } s^\alpha e^{-s}\int _0^{+\infty }s^{1/2} e^{-s|y|^2} f(-y){\textrm{d}}y\frac{{\textrm{d}}s}{s} = \frac{C_1}{2}. \end{aligned}$$

(3.7)

It can be checked that

$$\begin{aligned}f(a^{2j}x) = (-1)^j f(x)+(-1)^j\sum _{k=1}^{-j}(-1)^k \chi _{(-a^{2k}, -a^{2k-1}]}(x)\end{aligned}$$

when $j\le 0.$ We will always assume $j\le 0$ in the following. By changing variable,

$$\begin{aligned}{} & {} \mathcal {P}_{a_j}^\alpha f(x)= \frac{1}{\sqrt{\pi }\Gamma (\alpha ) } \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( x-a^{2j} y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}\\{} & {} \quad = \frac{(-1)^j }{\sqrt{\pi }\Gamma (\alpha ) } \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2}\\ {}{} & {} \quad \left\{ f\left( {x\over a^{2j}}- y\right) +\sum _{k=1}^{-j}(-1)^k \chi _{(-a^{2k}, -a^{2k-1}]}\left( {x\over a^{2j}}-y\right) \right\} {\textrm{d}}y \frac{{\textrm{d}}s}{s}. \end{aligned}$$

Then

$$\begin{aligned}&{\mathcal {P}_{a_{j+1}}^\alpha f(x)-\mathcal {P}_{a_j}^\alpha f(x)}\nonumber \\&=\frac{ (-1)^{j+1}}{\sqrt{\pi }\Gamma (\alpha ) } \Big \{\, \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( {x\over a^{2j+2}}- y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}\nonumber \\&\quad +\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2} f\left( {x\over a^{2j}}- y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s} \nonumber \\&\quad +\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2}\sum _{k=1}^{-j-1}(-1)^k \chi _{(-a^{2k}, -a^{2k-1}]}\left( {x\over a^{2j+2}}-y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}\nonumber \\&\quad +\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2}\sum _{k=1}^{-j}(-1)^k \chi _{(-a^{2k}, -a^{2k-1}]}\left( {x\over a^{2j}}-y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}\, \Big \}. \end{aligned}$$

(3.8)

For given $\eta _0$ as above, let $2r<1$ such that $r< \eta _0^2$ and $r \sim a^{2J_0}\eta _0$ for a certain negative integer $J_0$. If $J_0\le j\le 0$, we have $\displaystyle {r\over a^{2j}} <\eta _0$. And, for any $-r\le x\le r$ we have

$$\begin{aligned}\displaystyle -1\cdot \chi _{[a-1,+\infty )}(y)\le \sum _{k=1}^{-j-1}(-1)^k\chi _{(-a^{2k}, -a^{2k-1}]}\left( {x\over a^{2j+2}}-y\right) \le \chi _{[a-1,+\infty )}(y)\end{aligned}$$

and

$$\begin{aligned}\displaystyle -1\cdot \chi _{[a-1,+\infty )}(y) \le \sum _{k=1}^{-j} (-1)^k\chi _{(-a^{2k}, -a^{2k-1}]}\left( {x\over a^{2j}}-y\right) \le \chi _{[a-1,+\infty )}(y). \end{aligned}$$

Hence, for the third and fourth integrals in (3.8), by (3.6) we have

$$\begin{aligned}&\int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2}\sum _{k=1}^{-j-1}(-1)^k \chi _{(-a^{2k}, -a^{2k-1}]}\left( {x\over a^{2j+2}}-y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}\nonumber \\&\quad \quad \quad \quad + \int _0^{+\infty } s^\alpha e^{-s}\int _{\mathbb R}s^{1/2} e^{-s|y|^2}\sum _{k=1}^{-j}(-1)^k \chi _{(-a^{2k}, -a^{2k-1}]}\left( {x\over a^{2j}}-y\right) {\textrm{d}}y \frac{{\textrm{d}}s}{s}\nonumber \\&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ge (-2)\int _0^{+\infty } s^\alpha e^{-s}\int _{a-1}^{+\infty } s^{1/2} e^{-s|y|^2}{\textrm{d}}y \frac{{\textrm{d}}s}{s}\ge -{2C_1\over 9}. \end{aligned}$$

(3.9)

So, for any $x\in [-r, r]$ and $J_0\le j\le 0$, combining (3.8), (3.7) and (3.9), we have

$$\begin{aligned} \left| \mathcal {P}_{a_{j+1}}^\alpha f(x)-\mathcal {P}_{a_j}^\alpha f(x)\right| \ge C_\alpha \cdot \left( C_1-{2C_1\over 9}\right) =C\cdot C_1>0. \end{aligned}$$

We choose the sequence $\{v_j\}_{j\in \mathbb Z} \in \ell ^p(\mathbb Z)$ given by $\displaystyle v_j=(-1)^{j+1}(-j)^{-{1\over p-\varepsilon }}$, then for $N=(J_0, 0),$ we have

$$\begin{aligned}{} & {} \frac{1}{2r}\int _{[-r,r]} \left| T^* f(x)\right| {\textrm{d}}x\\ {}{} & {} \quad \ge \frac{1}{2r}\int _{[-r,r]} \left| T_N^\alpha f(x)\right| {\textrm{d}}x \ge \frac{1}{\sqrt{\pi }\Gamma (\alpha )} \frac{1}{2r}\int _{[-r,r]} \sum _{j =J_0}^{0} \left( C\cdot C_1\cdot (-j)^{-{1\over p-\varepsilon }}\right) {\textrm{d}}x\\{} & {} \quad \ge C_{p,\varepsilon ,\alpha }\cdot C_1\cdot (-J_0)^{1\over {(p-\varepsilon )'}} \sim \left( \log \frac{2}{r}\right) ^{1\over {(p-\varepsilon )'}}. \end{aligned}$$

For (c), let $v_j=(-1)^{j+1}$, $a_j=a^{2j}$ with $a>1$ and $0<\eta _0<1$ fixed in the proof of (b). Consider the same function f as in (b). Then, $\left\| v\right\| _{l^\infty (\mathbb Z)}=1$ and $\left\| f\right\| _{L^\infty (\mathbb R)}=1.$ By the same argument as in (b), with $N=(J_0, 0)$ and $0<\alpha <1$, we have

$$\begin{aligned}{} & {} \frac{1}{2r}\int _{[-r,r]} \left| T^* f(x)\right| {\textrm{d}}x\ge \frac{1}{2r}\int _{[-r,r]} \left| T_N^\alpha f(x)\right| {\textrm{d}}x\\{} & {} \quad \ge \frac{1}{\sqrt{\pi }\Gamma (\alpha )} \frac{1}{2r}\int _{[-r,r]} \sum _{j = J_0}^{0} C_1 {\textrm{d}}t \ge \frac{ C_1}{\sqrt{\pi }\Gamma (\alpha )}\cdot (-J_0) \sim \log \frac{2}{r}. \end{aligned}$$

$\square$

4 Boundedness of the differential transforms related to Schrödinger operator $-\Delta +V$

In this section, we would consider the differential transforms related with the generalized Poisson operators generated by the Schrödinger operator ${\mathcal L}=-\Delta +V$ in ${\mathbb R}^n$ with $n\ge 3$, where the nonnegative potential V belongs to the reverse Hölder class $RH_q$ with $q\ge n/2$, that is, there exists $C>0$, such that

$$\begin{aligned} \left( \frac{1}{|B|}\int _B V(x)^q{\textrm{d}}x\right) ^{\frac{1}{q}}\le \frac{C}{|B|}\int _B V(x){\textrm{d}}x, \end{aligned}$$

for every ball B in ${\mathbb R}^n$. Associated with this potential, Z. Shen defines the critical radii function in [16] as

$$\begin{aligned} \rho (x):=\sup \Big \{r>0:\frac{1}{r^{n-2}}\int _{B(x,r)}V(y)~{\textrm{d}}y\le 1\Big \},\quad x\in {\mathbb R}^n. \end{aligned}$$

(4.1)

We will abuse $\rho$ in this article, and it should be easy to distinct the $\rho$-lacunary with $\rho (x)$ for the reader. For more information related with Schödinger operators, see [8, 16].

Lemma 2

(See [16, Lemma 1.4]) There exist $c>0$ and $k_0\ge 1$ such that for all $x,y\in {\mathbb R}^n$

$$\begin{aligned} c^{-1}\rho (x)\left( 1+\frac{\left| x-y\right| }{\rho (x)}\right) ^{-k_0}\le \rho (y)\le c\rho (x) \left( 1+\frac{\left| x-y\right| }{\rho (x)}\right) ^{\frac{k_0}{k_0+1}}. \end{aligned}$$

In particular, there exists a positive constant $C_1<1$ such that

$$\begin{aligned}\hbox {if}\quad \left| x-y\right| \le \rho (x)\quad \hbox {then}\quad C_1\rho (x)<\rho (y)<C_1^{-1}\rho (x).\end{aligned}$$

Let $\left\{ {\mathcal T}_t\right\} _{t>0}$ be the heat–diffusion semigroup associated with ${\mathcal L}$:

$$\begin{aligned} {\mathcal T}_tf(x)\equiv e^{-t{\mathcal L}}f(x)=\int _{{\mathbb R}^n}e^{-t{\mathcal L}}(x,y)f(y)~{\textrm{d}}y,\qquad f\in L^2({\mathbb R}^n),~x\in {\mathbb R}^n,~t>0. \end{aligned}$$

Lemma 3

(See [9, 11]) For every $M>0$ there exists a constant $C_M$ such that

$$\begin{aligned} 0\le e^{-t{\mathcal L}}(x,y)\le C_M t^{-n/2}e^{-\frac{\left| x-y\right| ^2}{5t}}\left( 1+\frac{\sqrt{t}}{\rho (x)} +\frac{\sqrt{t}}{\rho (y)}\right) ^{-M},\quad x,y\in {\mathbb R}^n,~t>0. \end{aligned}$$

Lemma 4

(See [9, Proposition 2.16]) There exists a nonnegative Schwartz class function $\omega$ on ${\mathbb R}^n$ such that

$$\begin{aligned} \left| e^{-t{\mathcal L}}(x,y)-W_t(x-y)\right| \le \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{\delta _0}\omega _t(x-y),\quad x,y\in {\mathbb R}^n,~t>0, \end{aligned}$$

where $W_t$ is the Gauss–Weierstrass kernel, $\omega _t(x-y):=t^{-n/2}\omega \left( (x-y)/\sqrt{t}\right)$ and

$$\begin{aligned} \delta _0:=2-\frac{n}{q}>0. \end{aligned}$$

(4.2)

Lemma 5

(See [10, Proposition 4.11]) For every $0<\delta <\delta _0,$ there exists a constant $c>0$ such that for every $M>0$ there exists a constant $C>0$ such that for $\left| x-y\right| <\sqrt{t}$ we have

$$\begin{aligned}\left| e^{-t{\mathcal L}}(x,z)-e^{-t{\mathcal L}}(y, z)\right| \le C\left( \frac{\left| x-y\right| }{\sqrt{t}}\right) ^\delta t^{-n/2}~e^{-c\left| x-z\right| ^2/t}\left( 1+\frac{\sqrt{t}}{\rho (x)}+\frac{\sqrt{t}}{\rho (z)}\right) ^{-M}.\end{aligned}$$

Lemma 6

(See [9, Proposition 2.17]) For every $0<\delta <\min \{1,\delta _0\}$,

$$\begin{aligned}\left| \left( e^{-t{\mathcal L}}(x,z)-e^{t\Delta }(x-z)\right) -\left( e^{-t{\mathcal L}}(y,z)-e^{t\Delta }(y-z)\right) \right| \le C\left( \frac{\left| x-y\right| }{\rho (z)}\right) ^\delta \omega _t(x-z),\end{aligned}$$

for all $x,z\in {\mathbb R}^n$ and $t>0$, with $\left| x-y\right| <C\rho (x)$ and $\left| x-y\right| <\tfrac{1}{4}\left| x-z\right|$.

In fact, going through the proof of [9] we see that $\omega (x)=e^{-\left| x\right| ^2}$.

Then by (1.1), we can define the generalized Poisson operators associated with the Schrödinger operator ${\mathcal L}$ as follows:

$$\begin{aligned} {\tilde{{\mathcal P}}}_t^\alpha f(x) =\frac{t^{2\alpha }}{4^\alpha \Gamma (\alpha )} \int _0^{+\infty } \int _{{\mathbb R}^n} {e^{-\frac{t^2}{4 s}}}e^{-s{\mathcal L}}(x,y) f (y){\textrm{d}}y \frac{{\textrm{d}}s}{s^{1+\alpha }}. \end{aligned}$$

Let $\{a_j\}_{j\in {\mathbb Z}}$ be an increasing sequence of positive real numbers, and $\{v_j\}_{j\in {\mathbb Z}}$ be a bounded sequence of real or complex numbers. We shall consider the differential transform series

$$\begin{aligned} \sum _{j\in {\mathbb Z}} v_j({\tilde{{\mathcal P}}}_{a_{j+1}}^\alpha f(x)-{\tilde{{\mathcal P}}}_{a_j}^\alpha f(x)). \end{aligned}$$

For each $N\in {\mathbb Z}^2,~N=(N_1,N_2)$ with $N_1<N_2,$ we define the sum

$$\begin{aligned} {{\tilde{T}}}_N^\alpha f(x)=\sum _{j=N_1}^{N_2} v_j({\tilde{{\mathcal P}}}_{a_{j+1}}^\alpha f(x)-{\tilde{{\mathcal P}}}_{a_j}^\alpha f(x)),\quad x\in \mathbb R^n. \end{aligned}$$

(4.3)

Then, we have the following formula:

$$\begin{aligned}&{{\tilde{T}}}_N^{\alpha } f(x) =\sum _{j=N_1}^{N_2} v_j({\tilde{{\mathcal P}}}_{a_{j+1}}^\alpha f(x)-{\tilde{{\mathcal P}}}_{a_j}^\alpha f(x)) \\&= \frac{1}{4^\alpha \Gamma (\alpha )} \sum _{j=N_1}^{N_2}v_j \int _{{\mathbb R}^n}\int _0^\infty \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }} e^{-s{\mathcal L}}(x,y) f(y)~{\textrm{d}}s {\textrm{d}}y \\&= \frac{1}{4^\alpha \Gamma (\alpha )} \int _{{\mathbb R}^{n}} \int _0^{+\infty }\sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }} e^{-s{\mathcal L}}(x,y)~{\textrm{d}}s f(y) {\textrm{d}}y. \end{aligned}$$

We denote the kernel of ${{\tilde{T}}}_N^\alpha$ by

$$\begin{aligned}{{\tilde{K}}}_N^\alpha (x, y)=\frac{1}{4^\alpha \Gamma (\alpha )}\int _0^{+\infty } \sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} e^{-s{\mathcal L}}(x,y){\textrm{d}}s.\end{aligned}$$

By Lemma 3, we can prove the following theorem as in the proof of Theorem 5, which indicate that the kernel ${{\tilde{K}}}_N^\alpha$ is an $\ell ^\infty (\mathbb Z^2)$-valued Calderón–Zygmund kernel.

Theorem 10

For any $x, y\in {\mathbb R}^n,$ $x\ne y$, and $M>0,$ there exists constants C depending on $n, M, \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

• (i) $\displaystyle \left| {{\tilde{K}}}_N^\alpha (x, y)\right| \le \frac{C}{|x-y|^{n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-M}$,

• (ii) $\displaystyle | {{\tilde{K}}}_N^\alpha (x,y)- {{\tilde{K}}}_N^\alpha (x,z)|+|{{\tilde{K}}}_N^\alpha (y,x)-{{\tilde{K}}}_N^\alpha (z,x)| \le C\frac{\left| y-z\right| ^\delta }{|x-y|^{n+\delta }}$, whenever $\left| x-y\right| >{2}\left| y-z\right|$, for any $\displaystyle 0<\delta <2-{n\over q}.$

Proof

(i) For any $M>0,$ we have

$$\begin{aligned}{} & {} |{{\tilde{K}}}_N^\alpha (x, y)| \le C\left\| v\right\| _{l^\infty (\mathbb Z)}\int _0^{+\infty }\sum _{j=-\infty }^{\infty } \left| \frac{a_{j+1}^{2\alpha } e^{-a_{j+1}^2/(4s)}-a_j^{2\alpha } e^{-a_j^2/(4s)}}{s^{1+\alpha }} e^{-s{\mathcal L}}(x, y)\right| {\textrm{d}}s\\{} & {} \quad = C\int _0^{+\infty }\sum _{j=-\infty }^{\infty } \left| a_{j+1}^{2\alpha } e^{-a_{j+1}^2/(4s)}-a_j^{2\alpha } e^{-a_j^2/(4s)}\right| \frac{e^{-s{\mathcal L}}(x, y)}{s^{1+\alpha }}{\textrm{d}}s. \end{aligned}$$

Then, by (2.1) and Lemma 3 we have

$$\begin{aligned} |{{\tilde{K}}}_N^\alpha (x, y)|\le & {} C \int _0^{+\infty }\frac{e^{-s{\mathcal L}}(x, y)}{s}{\textrm{d}}s=C \left( \int _0^{|x-y|^2}+\int _{|x-y|^2}^{+\infty }\right) \frac{e^{-|x-y|^2/(cs)}}{s^{n/2+1}}{\textrm{d}}s\\\le & {} C \left( \int _0^{|x-y|^2}\frac{1}{|x-y|^{n+2}}{\textrm{d}}s+\int _{|x-y|^2}^{+\infty }\frac{1}{s^{n/2+1}}{\textrm{d}}s\right) \le \frac{C}{|x-y|^{n}}, \end{aligned}$$

and

$$\begin{aligned} |{{\tilde{K}}}_N^\alpha (x, y)|&\le C \int _0^{+\infty }\frac{e^{-s{\mathcal L}}(x, y)}{s}{\textrm{d}}s\\ {}&=C \left( \int _0^{|x-y|^2}+\int _{|x-y|^2}^{+\infty }\right) s^{-n/2-1}{e^{-|x-y|^2/(cs)}}\left( {\sqrt{s}\over \rho (x)}\right) ^{-M}{\textrm{d}}s\\&\le C \left( \int _0^{|x-y|^2}\frac{\rho ^M(x)}{|x-y|^{n+M+2}}{\textrm{d}}s+\int _{|x-y|^2}^{+\infty }\frac{\rho ^M(x)}{s^{n/2+M/2+1}}{\textrm{d}}s\right) \\&\le \frac{C}{|x-y|^{n}}\left( {|x-y|\over \rho (x)}\right) ^{-M}. \end{aligned}$$

Then, together with the symmetry of $e^{-s{\mathcal L}}(x, y)$, we have

$$\begin{aligned}{{{\tilde{K}}}_N^\alpha (x, y)}\le \frac{C}{|x-y|^{n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-M}.\end{aligned}$$

(ii) It can be proved with the same method as in i) with Proposition 3.10 in [4]. $\square$

Theorem 11

For the operator ${{\tilde{T}}}_N^\alpha$ defined in (4.3), we have the following statements:

• (a) For any $1<p<\infty$ and $\omega \in A_p$, there exists a constant $C>0$ depending on $n, p, \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned}\left\| {{\tilde{T}}}_N^\alpha f\right\| _{L^p(\mathbb R^{n}, \omega )}\le C\left\| f\right\| _{L^p(\mathbb R^{n}, \omega )},\end{aligned}$$

  for all functions $f\in L^p({\mathbb R}^{n}, \omega ).$

• (b) For any $\omega \in A_1$, there exists a constant $C>0$ depending on $n, \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned}\omega \left( {\{x\in {\mathbb R}^{n}:\left| {{\tilde{T}}}_N^\alpha f(x,t)\right|>\lambda \}}\right) \le C\frac{1}{\lambda }\left\| f\right\| _{L^1(\mathbb R^{n}, \omega )},\quad \lambda >0,\end{aligned}$$

  for all functions $f\in L^1({\mathbb R}^{n}, \omega ).$

The constants C appeared above all are independent with N.

Proof

For any $f\in L^p({\mathbb R}^n, \omega )(1\le p<+\infty ),$ we have

$$\begin{aligned} \left| {{\tilde{T}}}_N^\alpha f(x)\right|&= \left| \int _{{\mathbb R}^n}{{\tilde{K}}}_N^\alpha (x, y)f(y){\textrm{d}}y\right| \\ {}&\le \left| \int _{\left| x-y\right| \le \rho (x)}{{\tilde{K}}}_N^\alpha (x, y)f(y){\textrm{d}}y\right| +\left| \int _{\left| x-y\right|> \rho (x)}{{\tilde{K}}}_N^\alpha (x, y)f(y){\textrm{d}}y\right| \\&\le \left| \int _{\left| x-y\right| \le \rho (x)}\left( {{\tilde{K}}}_N^\alpha (x, y)-K_N^\alpha (x-y)\right) f(y){\textrm{d}}y\right| \\ {}&+\left| \int _{\left| x-y\right| \le \rho (x)} K_N^\alpha (x, y)f(y){\textrm{d}}y\right| \\&\quad +\left| \int _{\left| x-y\right| > \rho (x)}{{\tilde{K}}}_N^\alpha (x, y)f(y){\textrm{d}}y\right| \\&=:I_1+I_2+I_3. \end{aligned}$$

For $I_1,$ by (2.1) and Lemma  4, we get

$$\begin{aligned}&I_1=C\left| \int _{\left| x-y\right| \le \rho (x)}\int _0^{+\infty } \sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} \left( e^{-s{\mathcal L}}(x,y)-W_s(x-y)\right) {\textrm{d}}sf(y){\textrm{d}}y\right| \\&\le C\left\| v\right\| _{l^\infty (\mathbb Z)}\int _{\left| x-y\right| \le \rho (x)}\int _0^{+\infty } \sum _{j=-\infty }^{+\infty }\left| \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }}\right| \left( \frac{\sqrt{s}}{\rho (x)}\right) ^{\delta _0} \frac{ e^{-c\frac{\left| x-y\right| ^2}{4s}}}{s^{n/2}}{\textrm{d}}s|f(y)|{\textrm{d}}y\\&\le C\int _{\left| x-y\right| \le \rho (x)}\int _0^{+\infty } \frac{1}{s} \left( \frac{\sqrt{s}}{\rho (x)}\right) ^{\delta _0} \frac{ e^{-c\frac{\left| x-y\right| ^2}{4s}}}{s^{n/2}}{\textrm{d}}s|f(y)|{\textrm{d}}y\\&=C\int _{\left| x-y\right| \le \rho (x)}\int _0^{\rho ^2(x)}\frac{1}{s} \left( \frac{\sqrt{s}}{\rho (x)}\right) ^{\delta _0} \frac{ e^{-c\frac{\left| x-y\right| ^2}{4s}}}{s^{n/2}}{\textrm{d}}s|f(y)|{\textrm{d}}y\\&\quad +C\int _{\left| x-y\right| \le \rho (x)}\int _{\rho ^2(x)}^{+\infty } \frac{1}{s} \left( \frac{\sqrt{s}}{\rho (x)}\right) ^{\delta _0} \frac{ e^{-c\frac{\left| x-y\right| ^2}{4s}}}{s^{n/2}}{\textrm{d}}s|f(y)|{\textrm{d}}y\\&=:I_{11}+I_{12}. \end{aligned}$$

For the term $I_{11}$, we have

$$\begin{aligned} I_{11}&=C\int _{\left| x-y\right| \le \rho (x)}\int _0^{\rho ^2(x)}{s^{\frac{\delta _0-n}{2}-1}}{\rho (x)^{-\delta _0}} { e^{-c\frac{\left| x-y\right| ^2}{4s}}}{\textrm{d}}s|f(y)|{\textrm{d}}y\\&\le C{\rho (x)^{-\delta _0}}\int _0^{\rho ^2(x)}{s^{\frac{\delta _0}{2}-1}}\cdot \frac{1}{s^{{n}/{2}}}\int _{{\mathbb R}^n} { e^{-c\frac{\left| x-y\right| ^2}{4s}}}|f(y)|{\textrm{d}}y {\textrm{d}}s\\&\le C \sup _{s>0} \frac{1}{s^{{n}/{2}}}\int _{{\mathbb R}^n} { e^{-c\frac{\left| x-y\right| ^2}{4s}}}|f(y)|{\textrm{d}}y\\&\le C {\mathcal M}(f)(x), \quad x\in {\mathbb R}^n, \end{aligned}$$

where ${\mathcal M}$ denotes the classical Hardy–Littlewood maximal function.

For the term $I_{12}$, since $\displaystyle \delta _0=2-\frac{n}{q}$(see (4.2)) and $n\ge 3,$ we have

$$\begin{aligned} I_{12}&\le C\int _{\left| x-y\right| \le \rho (x)}\int _{\rho ^2(x)}^{+\infty }{s^{\frac{\delta _0-n}{2}-1}}{\rho (x)^{-\delta _0}} {\textrm{d}}s|f(y)|{\textrm{d}}y\\&\le C{\rho (x)^{-n}}\int _{\left| x-y\right| \le \rho (x)}|f(y)|{\textrm{d}}y \\&\le C {\mathcal M}(f)(x), \quad x\in {\mathbb R}^n. \end{aligned}$$

Hence, we have

$$\begin{aligned}I_1\le C{\mathcal M}(f)(x), \quad x\in {\mathbb R}^n.\end{aligned}$$

For $I_2,$ we can write

$$\begin{aligned} I_2&=\left| \int _{\left| x-y\right| \le \rho (x)} K_N^\alpha (x, y)f(y){\textrm{d}}y\right| \le \left| \int _{{\mathbb R}^n} K_N^\alpha (x, y)f(y){\textrm{d}}y\right| \\ {}&\quad +\sup _{\varepsilon>0}\left| \int _{\left| x-y\right| >\varepsilon } K_N^\alpha (x, y)f(y){\textrm{d}}y\right| . \end{aligned}$$

Now, let us consider the operator defined by

$$\begin{aligned} T:L^2({\mathbb R}^n)&\longrightarrow L^2({\mathbb R}^n)\\ f&\mapsto Tf(x)=\int _{{\mathbb R}^n}K_N^\alpha (x,y) f(y){\textrm{d}}y. \end{aligned}$$

From Theorem 4, we know that T is bounded on $L^2({\mathbb R}^n)$. And T is a Calderón–Zygmund operator associated with the kernel $K_N^\alpha (x,y)$(see Theorem 5). Then, by proving a Cotlar’s inequality as in [18, p. 34, Proposition 2] and the argument in [18, p. 36, Corollary 2], we can prove that the maximal operator $T^{\alpha ,*}_N$ defined by

$$\begin{aligned}T^{\alpha ,*}_Nf(x)=\sup _{\varepsilon>0}\left| \int _{\left| x-y\right| >\varepsilon }K_N^\alpha (x,y)f(y){\textrm{d}}y\right| \end{aligned}$$

is bounded on $L^p({\mathbb R}^n, \omega )$, for every $1<p<\infty$, and from $L^1({\mathbb R}^n, \omega )$ in to $L^{1,\infty }({\mathbb R}^n, \omega ).$ Combining this fact with Theorem 3, we conclude that

$$\begin{aligned}\left\| I_2\right\| _{L^p({\mathbb R}^n, \omega )}\le C\left\| f\right\| _{L^p({\mathbb R}^n, \omega )},\quad 1<p<\infty \end{aligned}$$

and

$$\begin{aligned}\left\| I_2\right\| _{L^{1,\infty }({\mathbb R}^n, \omega )}\le C\left\| f\right\| _{L^1({\mathbb R}^n, \omega )}.\end{aligned}$$

For the last term $I_3,$ by (2.1) and Lemma 3 with $M>1$,

$$\begin{aligned} I_3&=\left| \int _{\left| x-y\right|> \rho (x)}{{\tilde{K}}}_N^\alpha (x, y)f(y){\textrm{d}}y\right| \\&=C\left| \int _{\left| x-y\right|> \rho (x)}\int _0^{+\infty } \sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }} e^{-s{\mathcal L}}(x,y){\textrm{d}}sf(y){\textrm{d}}y\right| \\&\le C\left\| v\right\| _{l^\infty (\mathbb Z)}\int _{\left| x-y\right|> \rho (x)}\int _0^{+\infty } \sum _{j=-\infty }^{+\infty }\left| \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)}}{s^{1+\alpha }}\right| e^{-s{\mathcal L}}(x,y){\textrm{d}}s|f(y)|{\textrm{d}}y\\&\le C \int _{\left| x-y\right|> \rho (x)}\int _0^{+\infty }\frac{1}{ s^{\frac{n}{2}+1}}e^{-\frac{\left| x-y\right| ^2}{5s}} \left( 1+\frac{\sqrt{s}}{\rho (x)} +\frac{\sqrt{s}}{\rho (y)}\right) ^{-M}{\textrm{d}}s|f(y)|{\textrm{d}}y\\&= C \int _{\left| x-y\right|> \rho (x)}\int _0^{\rho ^2(x)}\frac{1}{ s^{\frac{n}{2}+1}}e^{-\frac{\left| x-y\right| ^2}{5s}} \left( 1+\frac{\sqrt{s}}{\rho (x)} +\frac{\sqrt{s}}{\rho (y)}\right) ^{-M}{\textrm{d}}s|f(y)|{\textrm{d}}y\\&\quad +C\int _{\left| x-y\right| > \rho (x)}\int _{\rho ^2(x)}^{+\infty }\frac{1}{ s^{\frac{n}{2}+1}} e^{-\frac{\left| x-y\right| ^2}{5s}}\left( 1+\frac{\sqrt{s}}{\rho (x)} +\frac{\sqrt{s}}{\rho (y)}\right) ^{-M}{\textrm{d}}s|f(y)|{\textrm{d}}y\\&=:I_{31}+I_{32}. \end{aligned}$$

For the term $I_{31}$,

$$\begin{aligned} I_{31}&\le C\int _{\left| x-y\right|> \rho (x)}\int _0^{\rho ^2(x)}\frac{1}{ s^{\frac{n}{2}+1}}e^{-\frac{\left| x-y\right| ^2}{5s}} {\textrm{d}}s|f(y)|{\textrm{d}}y\\&\le C\int _0^{\rho ^2(x)}\frac{1}{s} e^{-c_1\frac{\rho ^2(x)}{s}}\left( \frac{1}{ s^{n /2}}\int _{{\mathbb R}^n}e^{-c_2\frac{\left| x-y\right| ^2}{s}}|f(y)|{\textrm{d}}y\right) {\textrm{d}}s\\&\le C\sup _{s>0}\frac{1}{ s^{n /2}}\int _{{\mathbb R}^n}e^{-c_2\frac{\left| x-y\right| ^2}{s}}|f(y)|{\textrm{d}}y\\&\le C{\mathcal M}(f)(x),\quad x\in {\mathbb R}^n. \end{aligned}$$

For the other term $I_{32}$, by changing variable we have

$$\begin{aligned} I_{32}&\le C\int _{\left| x-y\right|> \rho (x)}\int _{\rho ^2(x)}^{+\infty }\frac{1}{ s^{\frac{n}{2}+1}} e^{-\frac{\left| x-y\right| ^2}{5s}}\left( 1+\frac{\sqrt{s}}{\rho (x)} \right) ^{-M}{\textrm{d}}s|f(y)|{\textrm{d}}y\\&=C\int _{\left| x-y\right|> \rho (x)}|f(y)|\int _{1}^{+\infty }\frac{1}{ (u\rho (x))^n} (1+u)^{-M}e^{-c\frac{\left| x-y\right| ^2}{u^2\rho ^2(x)}}\frac{{\textrm{d}}u}{u}{\textrm{d}}y\\&\le C\frac{1}{\rho ^n(x)}\int _{\left| x-y\right|> \rho (x)}|f(y)|\int _{1}^{+\infty }(1+u)^{-M}\left( \frac{u\rho (x)}{\left| x-y\right| }\right) ^{n+1}\frac{{\textrm{d}}u}{u^{n+1}}{\textrm{d}}y\\&\le C\frac{1}{\rho ^n(x)}\int _{\left| x-y\right| > \rho (x)}|f(y)|\left( \frac{\rho (x)}{\left| x-y\right| }\right) ^{n+1}{\textrm{d}}y\\&=C\frac{1}{\rho ^n(x)}\sum _{k=0}^{+\infty }\int _{2^k\rho (x)<\left| x-y\right| \le 2^{k+1} \rho (x)}|f(y)|\left( \frac{\rho (x)}{\left| x-y\right| }\right) ^{n+1}{\textrm{d}}y\\&\le C\sum _{k=0}^{+\infty }\frac{1}{2^k}\frac{1}{(2^k\rho (x))^n}\int _{\left| x-y\right| \le 2^{k+1} \rho (x)}|f(y)|{\textrm{d}}y\\&\le C{\mathcal M}(f)(x), \quad x\in {\mathbb R}^n. \end{aligned}$$

Hence, we get

$$\begin{aligned}I_3\le C{\mathcal M}(f)(x), \quad x\in {\mathbb R}^n.\end{aligned}$$

Then, combining the above estimates of $I_1, I_2, I_3$ and the $L^p$-boundedness of ${\mathcal M}$, we conclude that ${{\tilde{T}}}_N^\alpha$ is a bounded operator on $L^p({\mathbb R}^n, \omega )$ for every $1<p<\infty$, and from $L^1({\mathbb R}^n, \omega )$ into $L^{1,\infty }({\mathbb R}^n, \omega )$. We should note that the constants C appeared in the above estimates all are independent with $N=(N_1, N_2)$. Thus, the proof of the theorem is complete. $\square$

We shall also analyze the behavior in $L^\infty$ and $BMO_{{\mathcal L}}$. The space $BMO_{{\mathcal L}}({\mathbb R}^n)$, introduced in [8], is defined as the set of functions f such that

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{1}{|B|} \int _B\Big |f(z)- \frac{1}{|B|} \int _B f \Big | {\textrm{d}}z\le C_1, \hbox { for all } B= B_R(x): R \le \rho (x), \\ \displaystyle \frac{1}{|B|} \int _B |f| \le C_2, \hbox { for all } B= B_R(x): R >\rho (x), \end{array}\right. \end{aligned}$$

where $\rho (x)$ is the critical radii associated with ${\mathcal L}$, see (4.1). The norm $\Vert f\Vert _{BMO_{\mathcal L}(\mathbb {R}^n)}$ is defined as $\min \{C_1,C_2\}$.

Theorem 12

Given $f\in BMO_{\mathcal L}({\mathbb R}^n)$, then there exists a constant C depending only on n,  $\alpha$, $\rho$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

$$\begin{aligned}\left\| {{\tilde{T}}}_N^\alpha f\right\| _{BMO_{\mathcal L}(\mathbb R^n)}\le C\left\| f\right\| _{BMO_{\mathcal L}(\mathbb R^n)},\end{aligned}$$

for all functions $f\in BMO_{\mathcal L}(\mathbb R^n).$

Proof

We first show that, if $f\in BMO_{\mathcal L}({\mathbb R}^n)$, then $\tilde{T}_N^\alpha f$ is finite almost everywhere. This information is contained in the lemma as follows:

Lemma 7

Given $f\in BMO_{\mathcal L}({\mathbb R}^n)$ and any $x_0\in {\mathbb R}^n, C_0\ge 1,$ then ${{\tilde{T}}}_N^\alpha f(x)<\infty$ at almost every $x\in B=B(x_0, C_0\rho (x_0))$.

Proof

Let us split the function f to be

$$\begin{aligned}f=(f-f_B)\chi _{B^*}+(f-f_B)\chi _{{B^*}^c}+f_B=:f_1+f_2+f_3,\end{aligned}$$

where $B^*=B(x_0, 2C_0\rho (x_0)).$ Since $f\in BMO_{\mathcal L}({\mathbb R}^n)$, $f_1\in L^1({\mathbb R}^n)$. By Theorem 11 (b), we know that ${{\tilde{T}}}_N^\alpha f_1(x)<\infty$ a.e. $x\in B.$ For $\tilde{T}_N^\alpha f_2$, we note that, for any $x\in B$ and $t>0$,

$$\begin{aligned}&{{\tilde{{\mathcal P}}}}_t^\alpha f_2(x)\le \frac{t^{2\alpha }}{4^\alpha \Gamma (\alpha )} \int _0^{+\infty } \int _{{\mathbb R}^n} \frac{e^{-(t^2+|x-y|^2)/(4 s)}}{(4\pi s)^{n/2}} f_2 (y){\textrm{d}}y \frac{{\textrm{d}}s}{s^{1+\alpha }}\nonumber \\&\le C\frac{t^{2\alpha }}{4^\alpha \Gamma (\alpha )} \int _0^{+\infty }\sum _{k=1}^{+\infty }\int _{2^kC_0\rho (x_0)<|x_0-y|\le 2^{k+1}C_0\rho (x_0)} {1\over |x-y|^{n+2\alpha '}}\left| f(y)-f_B\right| {\textrm{d}}y\ {e^{-{t^2\over 4 s}}}\frac{{\textrm{d}}s}{s^{1+\alpha -\alpha '}}\nonumber \\&\le C\frac{t^{2\alpha '}\Gamma (\alpha -\alpha ')}{4^{\alpha '}\Gamma (\alpha )} \sum _{k=1}^{+\infty }{(2^kC_0\rho (x_0))}^{-2\alpha '}{1\over {(2^kC_0\rho (x_0))}^{n}}\int _{|x_0-y|\le 2^{k+1}C_0\rho (x_0)} \left| f(y)-f_B\right| {\textrm{d}}y\\&\le C\frac{t^{2\alpha '}\Gamma (\alpha -\alpha ')}{4^{\alpha '}\Gamma (\alpha )} \sum _{k=1}^{+\infty }{(2^kC_0\rho (x_0))}^{-2\alpha '}(1+2k)\left\| f\right\| _{BMO_{\mathcal L}({\mathbb R}^n)}<\infty ,\nonumber \end{aligned}$$

where $0<\alpha '<\alpha .$ So, ${{\tilde{{\mathcal P}}}}_t^\alpha f_2(x)$ is well defined for $x\in B$ and $t>0.$ Since ${{\tilde{T}}}_N^\alpha f_2(x)$ is a finite summation and $x_0, r_0$ is arbitrary, ${{\tilde{T}}}_N^\alpha f_2(x)<\infty$ a.e. $x\in {\mathbb R}^n.$ And, we should note that

$$\begin{aligned} \left| {{\tilde{{\mathcal P}}}}_t^\alpha f_3(x)\right|&\le \frac{t^{2\alpha }}{4^\alpha \Gamma (\alpha )} \int _0^{+\infty } \int _{{\mathbb R}^n} \frac{e^{-(t^2+|x-y|^2)/(4 s)}}{(4\pi s)^{n/2}} f_B {\textrm{d}}y \frac{{\textrm{d}}s}{s^{1+\alpha }}\\ {}&\le C \cdot f_B\le C\left\| f\right\| _{BMO_{\mathcal L}({\mathbb R}^n)}. \end{aligned}$$

So, ${{\tilde{T}}}_N^\alpha f_3(x)<\infty .$ Hence, ${{\tilde{T}}}_N^\alpha f(x)<\infty$ at almost every $x\in B=B(x_0, C_0\rho (x_0))$. This completes the proof of the lemma. $\square$

Assume that $f\in BMO_{\mathcal L}({\mathbb R}^n)$. Our goal is to show that $\tilde{T}_N^\alpha f\in BMO_{\mathcal L}({\mathbb R}^n).$ By Theorem 10, we know that the operator ${\tilde{T}}_N^\alpha$ is a $\gamma$-Schrödinger-Calderón–Zygmund operator with $\gamma =0$ appeared in [13]. By Theorem 1.2 in [13], we can prove the BMO-boundedness of ${{\tilde{T}}}_N^\alpha$ by checking a condition related with ${{\tilde{T}}}_N^\alpha 1:$

$$\begin{aligned} \log \left( \frac{\rho (x_0)}{t}\right) \frac{1}{|B|}\int _B|\tilde{T}_N^\alpha 1(x)-({{\tilde{T}}}_N^\alpha 1)_B|~{\textrm{d}}x\le C \end{aligned}$$

(4.4)

for every ball $B=B(x_0,t)$, $x_0\in {\mathbb R}^n$ and $0<t\le \tfrac{1}{2}\rho (x_0)$. In fact, since

$$\begin{aligned} |{{\tilde{T}}}_N^\alpha 1(x)-({{\tilde{T}}}_N^\alpha 1)_B|\le \frac{1}{\left| B\right| }\int _B\left| {{\tilde{T}}}_N^\alpha 1(x)-\tilde{T}_N^\alpha 1(y)\right| ~{\textrm{d}}y, \end{aligned}$$

we only need to prove that, with some $0<\delta <\delta _0$,

$$\begin{aligned} \left| {{\tilde{T}}}_N^\alpha 1(x)-{{\tilde{T}}}_N^\alpha 1(y)\right| \le C\left( \frac{t}{\rho (x_0)}\right) ^\delta , \quad x, y \in B, \end{aligned}$$

(4.5)

and then, (4.4) follows.

We shall note that, when $x, y \in B,$ $\rho (x)\sim \rho (y)\sim \rho (x_0).$ First, we have

$$\begin{aligned}&\left| {{\tilde{T}}}_N^\alpha 1(x)-{{\tilde{T}}}_N^\alpha 1(y)\right| \\&=C \left| \int _{{\mathbb R}^{n}} \int _0^{+\infty }\sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }} \left( e^{-s{\mathcal L}}(x,z)-e^{-s{\mathcal L}}(y,z)\right) ~{\textrm{d}}s {\textrm{d}}z\right| \\&\le C \left| \int _{{\mathbb R}^{n}} \int _0^{4t^2}\sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }} \left( e^{-s{\mathcal L}}(x,z)-e^{-s{\mathcal L}}(y,z)\right) ~{\textrm{d}}s {\textrm{d}}z\right| \\&\quad +C \left| \int _{{\mathbb R}^{n}} \int _{4t^2}^{\rho ^2(x_0)}\sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }} \left( e^{-s{\mathcal L}}(x,z)-e^{-s{\mathcal L}}(y,z)\right) ~{\textrm{d}}s {\textrm{d}}z\right| \\&\quad +C \left| \int _{{\mathbb R}^{n}} \int _{\rho ^2(x_0)}^{+\infty }\sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }} \left( e^{-s{\mathcal L}}(x,z)-e^{-s{\mathcal L}}(y,z)\right) ~{\textrm{d}}s {\textrm{d}}z\right| \\&=:I+II+III. \end{aligned}$$

For the term I,  since $\displaystyle \int _{{\mathbb R}^{n}}W_s(x, z){\textrm{d}}z=\int _{{\mathbb R}^{n}}W_s(y, z){\textrm{d}}z=1$ and by Lemma 4, we get

$$\begin{aligned} I&\le C \left| \int _{{\mathbb R}^{n}} \int _0^{4t^2}\sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }} \left( e^{-s{\mathcal L}}(x,z)-W_s(x,z)\right) ~{\textrm{d}}s {\textrm{d}}z\right| \\&\quad +C \left| \int _{{\mathbb R}^{n}} \int _0^{4t^2}\sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }} \left( e^{-s{\mathcal L}}(y,z)-W_s(y,z)\right) ~{\textrm{d}}s {\textrm{d}}z\right| \\&\le C\int _0^{4t^2}\sum _{j=N_1}^{N_2} \left| v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }}\right| \left( \sqrt{s} \over \rho (x)\right) ^{\delta _0}\int _{{\mathbb R}^{n}} \omega _s(x-z)~{\textrm{d}}z {\textrm{d}}s\\&\quad +C\int _0^{4t^2}\sum _{j=N_1}^{N_2} \left| v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }}\right| \left( \sqrt{s} \over \rho (y)\right) ^{\delta _0}\int _{{\mathbb R}^{n}} \omega _s(y-z)~{\textrm{d}}z {\textrm{d}}s\\&\le C\int _0^{4t^2} {1\over s} \left( \sqrt{s} \over \rho (x)\right) ^{\delta _0} {\textrm{d}}s\le C\left( t \over \rho (x_0)\right) ^{\delta _0}. \end{aligned}$$

For II,  we have

$$\begin{aligned} II&= C \Big | \int _{{\mathbb R}^{n}} \int _{4t^2}^{\rho ^2(x_0)}\sum _{j=N_1}^{N_2}v_j \frac{ a_{j+1}^{2\alpha }e^{-a_{j+1}^2/(4 s)}-a_j^{2\alpha } e^{-a_j^2/(4 s)} }{s^{1+\alpha }}\left( e^{-s{\mathcal L}}(x,z)-e^{-s{\mathcal L}}(y,z)\right) ~{\textrm{d}}s {\textrm{d}}z\Big |\\&=C\Big (\int _{\left| x-z\right| >c\rho (x)}\int _{4t^2}^{\rho ^2(x_0)}{1\over s} \left| e^{-s{\mathcal L}}(x,z)-e^{-s{\mathcal L}}(y,z)\right| ~{\textrm{d}}s {\textrm{d}}z\\&\quad +\int _{4|x-y|<\left| x-z\right| \le c\rho (x)}\int _{4t^2}^{\rho ^2(x_0)}{1\over s} \left| e^{-s{\mathcal L}}(x,z)-W_s(x, z)+W_s(y, z)-e^{-s{\mathcal L}}(y,z)\right| ~{\textrm{d}}s {\textrm{d}}z\\&\quad +\int _{\left| x-z\right| \le 4|x-y|}\int _{4t^2}^{\rho ^2(x_0)}{1\over s} \left| e^{-s{\mathcal L}}(x,z)-W_s(x, z)+W_s(y, z)-e^{-s{\mathcal L}}(y,z)\right| ~{\textrm{d}}s {\textrm{d}}z\Big )\\&=:C\left( II_1+II_2+II_3\right) . \end{aligned}$$

Since $\left| x-y\right| \le 2t\le \sqrt{s},$ by Lemma 5 we have

$$\begin{aligned} II_1&\le C\int _{\left| x-z\right|>c\rho (x)}\int _{4t^2}^{\rho ^2(x_0)}{1\over s} \left( \left| x-y\right| \over \sqrt{s}\right) ^\delta \omega _s(x-z)~{\textrm{d}}s{\textrm{d}}z\\&\le C\int _{\left| x-z\right|>c\rho (x)}{\left| x-y\right| ^\delta \over \left| x-z\right| ^{2+\delta }} \int _{4t^2}^{\rho ^2(x_0)}\omega _s(x-z)~{\textrm{d}}s{\textrm{d}}z\\&\le C\int _{\left| x-z\right|>c\rho (x)}{\left| x-y\right| ^\delta \over \left| x-z\right| ^{n+2+\delta }} \int _{4t^2}^{\rho ^2(x_0)}1~{\textrm{d}}s{\textrm{d}}z\\&\le C\int _{\left| x-z\right| >c\rho (x)}{\left| x-y\right| ^\delta \over \left| x-z\right| ^{n+2+\delta }} \rho ^2(x_0){\textrm{d}}z\\&\le C\left( t\over \rho (x_0)\right) ^\delta . \end{aligned}$$

For the term $II_2$, in this case, $|x-y|< c\rho (x)$ and $\displaystyle |x-y|<{|x-z|\over 4}$. By Lemma 6 we get

$$\begin{aligned} II_2&\le C\int _{4|x-y|<\left| x-z\right| \le c\rho (x)}\int _{4t^2}^{\rho ^2(x_0)}{1\over s} \left( {\left| x-y\right| \over \rho (z)}\right) ^\delta \omega _s(x-z)~{\textrm{d}}s {\textrm{d}}z\\&\le C\int _{4|x-y|<\left| x-z\right| \le c\rho (x)}\left( {\left| x-y\right| \over \rho (z)}\right) ^\delta {1\over |x-z|^{n+2}}\int _{4t^2}^{\rho ^2(x_0)}1~{\textrm{d}}s {\textrm{d}}z\\&\le C\left( {\left| x-y\right| \over \rho (x_0)}\right) ^\delta \int _{4|x-y|<\left| x-z\right| \le c\rho (x)} {\rho ^2(x_0)\over |x-z|^{n+2}} {\textrm{d}}z\\&\le C\left( {\left| x-y\right| \over \rho (x_0)}\right) ^\delta \le C \left( {t\over \rho (x_0)}\right) ^{\delta }. \end{aligned}$$

For the term $II_3,$ by Lemma 4, we have

$$\begin{aligned} II_3&\le C\int _{\left| x-z\right| \le 4|x-y|}\int _{4t^2}^{\rho ^2(x_0)}{1\over s} \left[ {\left( {\sqrt{s}\over \rho (x)}\right) }^{\delta _0}\omega _s(x-z)+{\left( {\sqrt{s}\over \rho (y)}\right) }^{\delta _0}\omega _s(y-z)\right] ~{\textrm{d}}s {\textrm{d}}z\\&\le C\Big (\int _{\left| x-z\right| \le 4|x-y|}\int _{4t^2}^{\rho ^2(x_0)}{1\over s}{\left( {\sqrt{s}\over \rho (x)}\right) }^{\delta _0}\omega _s(x-z)~{\textrm{d}}s{\textrm{d}}z\\&\quad +\int _{\left| y-z\right| \le 5|x-y|}\int _{4t^2}^{\rho ^2(x_0)}{1\over s}{\left( {\sqrt{s}\over \rho (y)}\right) }^{\delta _0}\omega _s(y-z)~{\textrm{d}}s {\textrm{d}}z\Big )\\&\le C \int _{4t^2}^{\rho ^2(x_0)} \int _{\left| \xi \right| \le 5{|x-y|\over \sqrt{s}}}{1\over s} {\left( {\sqrt{s}\over \rho (x)}\right) }^{\delta _0}\omega _s(\xi )~ d\xi {\textrm{d}}s\\&\le C \int _{4t^2}^{+\infty } {1\over s} {\left( {\sqrt{s}\over \rho (x_0)}\right) }^{\delta _0}\left( {|x-y|\over \sqrt{s}}\right) ^n {\textrm{d}}s\\&\le C {{|x-y|^n \over \rho (x_0)^{\delta _0}}}~t^{\delta _0-n}\le C \left( {t\over \rho (x_0)}\right) ^{\delta _0}. \end{aligned}$$

Then, we get

$$\begin{aligned} II\le C\left( {t\over \rho (x_0)}\right) ^{\delta } \end{aligned}$$

with some $0<\delta <\delta _0.$

We shall treat the latest term III. In this case, since $s\ge \rho ^2(x_0)>4t^2$, then $\sqrt{s}>2t>|x-y|$. By Lemma 5, we have, for $0<\delta <\delta _0,$

$$\begin{aligned}{} & {} III\le C \int _{\rho ^2(x_0)}^{+\infty }{1\over s} \left( {|x-y|\over \sqrt{s}}\right) ^\delta \int _{{\mathbb R}^{n}}\omega _s(x-z) {\textrm{d}}z~{\textrm{d}}s \\{} & {} \quad \le C \int _{\rho ^2(x_0)}^{+\infty }{1\over s} \left( {|x-y|\over \sqrt{s}}\right) ^\delta ~{\textrm{d}}s \le C\left( {|x-y|\over \rho (x_0)}\right) ^\delta \le C\left( {t\over \rho (x_0)}\right) ^\delta . \end{aligned}$$

Combining the above estimates for I, II and III,  we have proved (4.5). Hence, we get the estimation (4.4) and the BMO-boundedness of ${{\tilde{T}}}_N^\alpha .$ This completes the proof of Theorem 12. $\square$

As in the Laplacian case, we also can consider the maximal operator

$$\begin{aligned} {{\tilde{T}}}^*f(x)=\sup _N \left| {{\tilde{T}}}_N^\alpha f(x)\right| , \quad x\in {\mathbb R}^n, \end{aligned}$$

where the supremum are taken over all $N=(N_1,N_2)\in {\mathbb Z}^2$ with $N_1< N_2$.

Now we present our results as follows:

Theorem 13

• (a) For any $1<p<\infty$ and $\omega \in A_p$, there exists a constant C depending on $n, p, \rho , \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned}\left\| {{\tilde{T}}}^*f\right\| _{L^p(\mathbb R^n, \omega )}\le C\left\| f\right\| _{L^p(\mathbb R^n, \omega )},\end{aligned}$$

  for all functions $f\in L^p(\mathbb R^n, \omega ).$

• (b) For any $\lambda >0$ and $\omega \in A_1$, there exists a constant C depending on $n, \rho , \alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned}\omega \left( {\{x\in {\mathbb R}^n:\left| {{\tilde{T}}}^*f(x)\right| >\lambda \}}\right) \le C\frac{1}{\lambda }\left\| f\right\| _{L^1(\mathbb R^n, \omega )},\end{aligned}$$

  for all functions $f\in L^1(\mathbb R^n, \omega ).$

• (c) There exists a constant C depending on $n, \rho$, $\alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned}\left\| {{\tilde{T}}}^*f\right\| _{BMO_{\mathcal L}(\mathbb R^n)}\le C\left\| f\right\| _{L^\infty (\mathbb R^n)},\end{aligned}$$

  for any $f\in L^\infty ({\mathbb R}^n)$.

• (d) There exists a constant C depending on $n, \rho$, $\alpha$ and $\left\| v\right\| _{l^\infty (\mathbb Z)}$ such that

  $$\begin{aligned} \left\| {{\tilde{T}}}^*f\right\| _{ BMO_{\mathcal L}(\mathbb R^n)}\le C\left\| f\right\| _{BMO_{\mathcal L}(\mathbb R^n)}, \end{aligned}$$

  for all functions $f\in BMO_{\mathcal L}({\mathbb R}^n).$

With a similar argument as in the proof of Theorem 1, we can prove a Cotlar’s type inequality in the Schrödinger setting. And then, all the statements in Theorem 13 can be gotten just with minor changes. We omit the proof at here.