Proof. Denote by $dE(\lambda )$ the spectral resolution of the operator L. Since $L^\alpha $ is a nonnegative, densely defined and self-adjoint operator on $\mathbb {R}^n$ , by spectral analysis,

$$ \begin{align*} e^{- {tL^\alpha}}=\int_0^\infty e^{- {t\lambda^\alpha}}\,dE(\lambda), \end{align*} $$

which is expressed as

$$ \begin{align*}\langle e^{-t L^\alpha}f,\ g\rangle_{L^2(\mathbb{R}^n)}=\int_0^\infty e^{-t\lambda^\alpha}\,dE_{f,g}(\lambda) \quad \mbox{for all } f, g\in L^2(\mathbb{R}^n),\end{align*} $$

where $dE_{f,g}(\lambda )$ is a regular Borel complex measure of bounded variation concentrated on the spectrum of ${\mathcal L}$ with $d\vert E_{f,g}\vert (0,\infty )\leq \Vert f\Vert _{L^2({\mathbb R}^n)}\Vert g\Vert _{L^2({\mathbb R}^n)}$ . Therefore, we write

$$ \begin{align*} T^{L^{\alpha}}_N =\sum_{j=N_1}^{N_2} v_j\int_0^\infty (e^{-a_{j+1}\lambda^\alpha} -e^{-a_{j}\lambda^\alpha}) \,dE(\lambda) \end{align*} $$

and we obtain that, for any $f, g\in L^2(\mathbb {R}^n)$ ,

$$ \begin{align*} \vert\langle T^{L^{\alpha}}_N f,g\rangle\vert &\le \bigg| \!\sum_{j=N_1}^{N_2} v_j\int_0^\infty \int_{a_j}^{a_{j+1}} \partial_s e^{-s\lambda^\alpha} \,ds \,dE_{f,g}(\lambda)\bigg| \\ &= \bigg|\! \int_0^\infty\sum_{j=N_1}^{N_2} v_j \int_{a_j}^{a_{j+1}} \partial_s e^{-s\lambda^\alpha} \,ds \,dE_{f,g}(\lambda)\bigg| \\ &\le \|v\|_{\ell^\infty(\mathbb Z)} \int_0^\infty \int_{0}^{\infty} \vert\partial_s e^{-s\lambda^\alpha}\vert \,ds ~\vert \,dE_{f,g}\vert(\lambda)\\ &\lesssim \Vert f\Vert_{L^2({\mathbb R}^{n})} \Vert \Vert g\Vert_{L^2({\mathbb R}^{n})}. \end{align*} $$

This completes the proof of Proposition 3.1.

Proof. (i) For the size condition for $K^{\alpha }_{N}(\cdot ,\cdot )$ , by Lemma 3.2,

$$ \begin{align*} | K^\alpha_N (x,y)| &\leq \sum_{j=N_1}^{N_2}\vert v_j\vert \vert K^{L}_{\alpha,a_{j+1}}(x,y)-K^{L}_{\alpha,a_{j}} (x,y)\vert\\[5pt] &\le \|v\|_{l^\infty(\mathbb{Z})} \sum_{j=-\infty}^{\infty} \bigg|\!\int_{a_j}^{a_{j+1}}\partial_t K^{L}_{\alpha,t}(x,y) \,dt\bigg|\\[5pt] &\le \|v\|_{l^\infty(\mathbb{Z})} \sum_{j=-\infty}^{\infty} \int_{a_j}^{a_{j+1}}|\partial_t K^{L}_{\alpha,t}(x,y)| \,dt\\[5pt] &\lesssim \|v\|_{l^\infty(\mathbb{Z})} \sum_{j=-\infty}^{\infty} \int_{a_j}^{a_{j+1}}\frac 1{(t^{1/{(2\alpha)}}+|x-y|)^{n+2\alpha}}\bigg(1+\frac{t^{1/(2\alpha)}}{\rho(x)}+\frac{t^{1/(2\alpha)}}{\rho(y)}\bigg)^{-N} \,dt\\[5pt] &\lesssim \|v\|_{l^\infty(\mathbb{Z})} \sum_{j=-\infty}^{\infty} \int_{a_j}^{a_{j+1}}\frac 1{(t^{1/{(2\alpha)}}+|x-y|)^{n+2\alpha}}\,dt. \end{align*} $$

Since $\{a_{j}\}_{j\in \mathbb {Z}}$ is an increasing sequence of positive numbers, we see that

$$ \begin{align*} | K^\alpha_N (x,y)| &\lesssim \|v\|_{l^\infty(\mathbb{Z})}\int_{0}^{\infty}\frac 1{(t^{1/{(2\alpha)}}+|x-y|)^{n+2\alpha}} \,dt \\[5pt] & \lesssim {1\over {|x-y|^{n}}}. \end{align*} $$

Furthermore, we also have

$$ \begin{align*} &| K^\alpha_N (x, y)|& \\[5pt] &\quad\lesssim \int_0^{|x-y|^{2\alpha}}\frac{t^{N/\alpha}}{|x-y|^{n+2\alpha+2N}}\bigg(\frac{t^{1/2\alpha}}{\rho(x)}\bigg)^{-N}\,dt +\int_{|x-y|^{2\alpha}}^{+\infty}\frac{1}{t^{n/2\alpha+1}}\bigg(\frac{t^{1/2\alpha}}{\rho(x)}\bigg)^{-N}\,dt\\[5pt] &\quad \lesssim \frac{1}{|x-y|^{n}}\bigg({|x-y|\over \rho(x)}\bigg)^{-N}. \end{align*} $$

Then, using the symmetry of $K^{L}_{\alpha ,t}(\cdot , \cdot )$ , we get

$$ \begin{align*}|{K^\alpha_N(x, y)}|\lesssim \frac{1}{|x-y|^{n}}\bigg(1+\frac{|x-y|}{\rho(x)}+\frac{|x-y|}{\rho(y)}\bigg)^{-N}.\end{align*} $$

(ii) Note that if $|x-y|>2|y-z|$ , then $\vert x-y\vert \sim \vert x-z\vert .$ For any $0<\delta '\leq \delta =\min \{2\alpha ,\delta _{0}\}$ , if $\vert y-z\vert \le {t}^{1/2\alpha }$ , using a similar argument as in (i) and by Lemma 3.3, we get

$$ \begin{align*} &\vert K^\alpha_N(x,y)-K^\alpha_N(x,z)\vert\\ \nonumber &\quad\lesssim\Vert v\Vert_{\ell^\infty(\mathbb Z)}\int_0^{+\infty}\vert\partial_t K^{L}_{t}(x,y)-\partial_t K^{L}_{t}(x,z)\vert \,dt\\ \nonumber &\quad\lesssim\Vert v\Vert_{\ell^\infty(\mathbb Z)}\int_0^{|x-y|^{2\alpha}}\frac{1}{|x-y|^{n+2\alpha}}\bigg(\frac{|y-z|}{t^{1/2\alpha}}\bigg)^{\delta'}\,dt +\int_{|x-y|^{2\alpha}}^{+\infty}\bigg(\frac{|y-z|}{t^{1/2\alpha}}\bigg)^{\delta'}\frac{1}{t^{n/2\alpha+1}}\,dt\\ &\quad\lesssim\Vert v\Vert_{\ell^\infty(\mathbb Z)}\frac{\vert y-z\vert^{\delta'}}{\vert x-y\vert^{n+\delta'}}. \end{align*} $$

If $\vert y-z\vert>{t}^{1/2\alpha }$ , Lemma 3.2 gives

$$ \begin{align*} \vert K^\alpha_N(x,y)\vert &\lesssim \int_{0}^{+\infty} |\partial_t K^{L}_{t}(x,y)| \,dt \\ & \lesssim\int_{0}^{+\infty} \bigg(\frac{\vert y-z\vert}{{t}^{1/2\alpha}}\bigg)^{\delta'} \frac{1}{(t^{1/2\alpha}+|x-y|)^{n+2\alpha}}\,dt\\ &\lesssim\frac{\vert y-z\vert^{\delta'}}{\vert x-y\vert^{n+\delta'}}. \end{align*} $$

The same bound is valid for $K^\alpha _N(x,z)$ because $\vert x-z\vert \sim \vert x-y\vert $ . Then the estimate can be obtained directly owing to the inequality

$$ \begin{align*}\vert K^\alpha_N(x,y)-K^\alpha_N (x,z)\vert\le\vert K^\alpha_N(x,y)\vert+\vert K^\alpha_N(x,z)\vert.\end{align*} $$

The symmetry of the kernel $K^\alpha _N(x,y)=K^\alpha _N(y,x)$ gives the conclusion of (ii). This completes the proof of Proposition 3.5.

Proof. Since $L^\infty ({\mathbb R}^n) \subset BMO_{\mathcal L}({\mathbb R}^n)$ and $\|f\|_{BMO_{\mathcal L}({\mathbb R}^n)} \le \|f\|_{L^\infty ({\mathbb R}^n)}$ , it is enough to prove that (ii) of Theorem 3.6 holds. We first show that $T_N^{L^{\alpha }} f<\infty $ if $f\in BMO_L({\mathbb R}^n)$ . Let ${B=B(x_0, \rho (x_0))}$ and $B^*=B(x_0, 2C_0\rho (x_0))$ with some $x_0\in {\mathbb R}^{n}$ and $C_0>1$ . We decompose f as

$$ \begin{align*}f=f\chi_{B^*}+f\chi_{(B^*)^c}=:f_1+f_2.\end{align*} $$

The function $f_1$ is integrable, and hence $T^{L^{\alpha }}_N f_1< \infty\ \text{for almost every}\ x\in {\mathbb R}^{n}.$ When ${x\in B}$ and $y\in (B^{*})^{c}$ , we have $|y-x_0|/2\le |x-y|\le 2|y-x_0|$ and $\rho (x)\sim \rho (x_0).$ It can be deduced from Proposition 3.5 and Hölder’s inequality that

$$ \begin{align*} \vert T^{L^{\alpha}}_N f_2(x)\vert&=\bigg\vert\!\int_{{\mathbb R}^n} K^\alpha_N(x, y)f_2(y)\,dy\bigg\vert\nonumber \\ &\lesssim {\int_{(B^{*})^{c}} {1\over |x-y|^n}{\rho^2(x)\over |x-y|^2 } \vert f_2(y)\vert \,dy}\nonumber \\ &\lesssim {\int_{(B^{*})^{c}} {\rho^2(x_0)\over |x_0-y|^{n+2} }\vert f(y)\vert \,dy}\nonumber\\ &\lesssim\bigg(\int_{(B^{*})^{c}} {\rho^2(x_0)\over |x_0-y|^{n+2} } \,dy\bigg)^{1/2}\bigg(\int_{(B^{*})^{c}} {\rho^2(x_0)\over |x_0-y|^{n+2} }\vert f(y)\vert^2 \,dy\bigg)^{1/2}\nonumber\\ &\lesssim\bigg(\sum_{k=1}^\infty\int_{2^{k+1}B\setminus 2^kB} {\rho^2(x_0)\,dy\over |x_0-y|^{n+2} } \bigg)^{1/2}\bigg(\sum_{k=1}^\infty\int_{2^{k+1}B\setminus 2^kB} {\rho^2(x_0)\vert f(y)\vert^2\over |x_0-y|^{n+2} } \,dy\bigg)^{1/2}\nonumber\\ &\lesssim\bigg(\sum_{k=1}^\infty{\rho^2(x_0)\over (2^kC_0\rho(x_0))^2}{1\over \vert2^{k+1}B\vert}\int_{2^{k+1}B} \vert f(y)\vert^2 \,dy\bigg)^{1/2}\nonumber\\ &\lesssim \bigg(\sum_{k=1}^\infty2^{-2k}\Vert f\Vert^2_{BMO_L({\mathbb R}^n)}\bigg)^{1/2}\nonumber\\ &\lesssim\Vert f\Vert_{BMO_L({\mathbb R}^n)}<\infty.\nonumber \end{align*} $$

Hence, we have proved that $T^{L^{\alpha }}_N f(x)<\infty\ \text{for almost every}\ x\in {\mathbb R}^n.$ Then the proof of

$$ \begin{align*}\Vert T^{L^{\alpha}}_N f\Vert_{BMO_L({\mathbb R}^n)}\lesssim\Vert f\Vert_{BMO_L({\mathbb R}^n)}\end{align*} $$

is similar to that for the maximal operator $T^*_{L^{\alpha }}$ , which can be seen from Theorem 4.6. So we omit the details here.

Proof. We have previously remarked that the constants in the $L^{p}$ -boundedness only depend on the initial constant in $L^{p_{0}}(\mathbb {R}^{n})$ (in our case, $p_{0}=2$ ), the size constant and the smoothness constant of the kernel. Hence, the uniform $L^{p}$ -boundedness of the operators $T^{L^{\alpha }}_N $ is a direct consequence of the vector-valued Calderón–Zygmund theory.

(i) For any given $f\in L^{\rho ,\infty }_{p,\kappa }(\mathbb {R}^{n})$ with $1\leq p<\infty $ and $0<\kappa <1$ , suppose that $f\in L^{\rho ,\theta ^{\ast }}_{p,\kappa }(\mathbb {R}^{n})$ for some $\theta ^{\ast }>0$ , where

$$ \begin{align*}\theta^{\ast}=\inf\{\theta>0:f\in L^{\rho,\theta}_{p,\kappa}(\mathbb{R}^{n})\}\quad \text{and}\quad\|f\|_{L^{\rho,\infty}_{p,\kappa}(\mathbb{R}^{n})} =\|f\|_{L^{\rho,\theta^{\ast}}_{p,\kappa}(\mathbb{R}^{n})}.\end{align*} $$

Then we only need to prove that, for each ball $B(x_{0},r)\subset \mathbb {R}^{n}$ , there is some $\nu>0$ such that

(3-1) $$ \begin{align} \bigg(\frac{1}{|B(x_{0},r)|^{\kappa}}\int_{B(x_{0},r)}|T^{L^{\alpha}}_{N}(f)(x)|^{p}\,dx\bigg)^{1/p}\lesssim \bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{\nu} \end{align} $$

holds for all functions $f\in L^{\rho ,\theta ^{\ast }}_{p,\kappa }(\mathbb {R}^{n})$ with $(p,\kappa )\in (1,\infty )\times (0,1)$ .

We first decompose the function f as

$$ \begin{align*} \begin{cases} f=f_{1}+f_{2}\in L^{\rho,\theta^{\ast}}_{p,\kappa}(\mathbb{R}^{n}),\\ f_{1}=f\cdot\chi_{2B},\\ f_{2}=f\cdot\chi_{(2B)^{c}}, \end{cases} \end{align*} $$

where $2B$ is the ball centered at $x_{0}$ of radius $2r$ , $(2B)^{c}=\mathbb {R}^{n}\setminus (2B)$ and $\chi _{E}$ denotes the characteristic function of the set E. Then we can write

$$ \begin{align*}\bigg(\frac{1}{|B(x_{0},r)|^{\kappa}}\int_{B(x_{0},r)}|T^{L^{\alpha}}_{N}(f)(x)|^{p}\,dx\bigg)^{1/p}\leq I_{1}+I_{2},\end{align*} $$

where

$$ \begin{align*} \left\{ \begin{aligned} I_{1}:=\bigg(\frac{1}{|B(x_{0},r)|^{\kappa}}\int_{B(x_{0},r)}|T^{L^{\alpha}}_{N}(f_{1})(x)|^{p}\,dx\bigg)^{1/p},\\ I_{2}:=\bigg(\frac{1}{|B(x_{0},r)|^{\kappa}}\int_{B(x_{0},r)}|T^{L^{\alpha}}_{N}(f_{2})(x)|^{p}\,dx\bigg)^{1/p}. \end{aligned} \right.\end{align*} $$

In what follows, we estimate $I_{1}$ and $I_{2}$ separately. For the term $I_{1}$ ,

$$ \begin{align*} I_{1} &\lesssim \frac{1}{|B|^{\kappa/p}}\bigg(\int_{\mathbb{R}^{n}}|f_{1}(x)|^{p}\,dx\bigg)^{1/p} \\ &= \frac{1}{|B|^{\kappa/p}}\bigg(\int_{2B}|f(x)|^{p}\,dx\bigg)^{1/p} \\ &\lesssim \|f\|_{L^{\rho,\theta^{\ast}}_{p,\kappa}(\mathbb{R}^{n})}\frac{|2B|^{\kappa/p}}{|B|^{\kappa/p}}\bigg(1+\frac{2r}{\rho(x_{0})}\bigg)^{\theta^{\ast}}. \end{align*} $$

Moreover, for any fixed $\theta ^{\ast }>0$ , the inequalities

(3-2) $$ \begin{align} 1\leq \bigg(1+\frac{2r}{\rho(x_{0})}\bigg)^{\theta^{\ast}}\leq 2^{\theta^{\ast}}\bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{\theta^{\ast}} \end{align} $$

further indicate that

$$ \begin{align*}I_{1}\lesssim \|f\|_{L^{\rho,\infty}_{p,\kappa}(\mathbb{R}^{n})}\bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{\theta^{\ast}}.\end{align*} $$

Next, we estimate $I_{2}$ . From Proposition 3.5, we deduce that

(3-3) $$ \begin{align} T^{L^{\alpha}}_{N}(f_{2})(x)\lesssim \int_{(2B)^{c}}\frac{1}{|x-y|^{n}}\bigg(1+\frac{|x-y|}{\rho(x)}\bigg)^{-N}|f(y)|\,dy. \end{align} $$

Since $x\in B(x_{0},r)$ and $y\in (2B)^{c}$ , we have $|x-y|\sim |x_{0}-y|$ . This fact, along with (3-3), implies that, for any $x\in B(x_{0},r)$ ,

$$ \begin{align*} |T^{L^{\alpha}}_{N}(f_{2})(x)| &\lesssim \sum^{\infty}_{k=1}\int_{2^{k}r\leq |x_{0}-y|<2^{k+1}r}\bigg(1+\frac{|x_{0}-y|}{\rho(x)}\bigg)^{-N}\frac{1}{|x_{0}-y|^{n}}|f(y)|\,dy \\ \notag &\lesssim \sum^{\infty}_{k=1}\frac{1}{|B(x_{0},2^{k+1}r)|}\int_{|x_{0}-y|<2^{k+1}r}\bigg(1+\frac{2^{k}r}{\rho(x)}\bigg)^{-N}|f(y)|\,dy. \end{align*} $$

In view of (2-3) and (3-2), we deduce that

(3-4) $$ \begin{align} & |T^{L^{\alpha}}_{N}(f_{2})(x)|\\ \notag &\quad\lesssim \sum^{\infty}_{k=1}\frac{1}{|B(x_{0},2^{k+1}r)|}\int_{B(x_{0},2^{k+1}r)}\bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{N\cdot({N_{0}}/{N_{0}+1})} \bigg(1+\frac{2^{k+1}r}{\rho(x_{0})}\bigg)^{-N}|f(y)|\,dy.\notag \end{align} $$

We consider each term in the sum of (3-4) separately. Using Hölder’s inequality, we deduce that, for each fixed $k\in \mathbb {N}$ ,

$$ \begin{align*} \frac{1}{|B(x_{0},2^{k+1}r)|}\int_{B(x_{0},2^{k+1}r)}|f(y)|\,dy \lesssim \|f\|_{L^{\rho,\theta^{\ast}}_{p,\kappa}(\mathbb{R}^{n})}\frac{|B(x_{0},2^{k+1}r)|^{\kappa/p}}{|B(x_{0},2^{k+1}r)|^{1/p}} \bigg(1+\frac{2^{k+1}r}{\rho(x_{0})}\bigg)^{\theta^{\ast}}. \end{align*} $$

This allows us to obtain

$$ \begin{align*} I_{2} \lesssim \|f\|_{L^{\rho,\theta^{\ast}}_{p,\kappa}(\mathbb{R}^{n})} \bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{N\cdot({N_{0}}/{N_{0}+1})}\sum_{k=1}^{\infty}\frac{|B(x_{0},r)|^{(1-\kappa)/p}}{|B(x_{0},2^{k+1}r)|^{(1-k)/p}} \bigg(1+\frac{2^{k+1}r}{\rho(x_{0})}\bigg)^{-N+\theta^{\ast}}. \end{align*} $$

We can choose N large enough such that $N\geq \theta ^{\ast }$ and, consequently, the last series is convergent. Then we have

$$ \begin{align*} I_{2} &\lesssim \|f\|_{L^{\rho,\infty}_{p,\kappa}(\mathbb{R}^{n})}\bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{N\cdot({N_{0}}/{N_{0}+1})}\sum_{k=1}^{\infty} \bigg(\frac{|B(x_{0},r)|}{|B(x_{0},2^{k+1}r)|}\bigg)^{(1-\kappa)/p} \\ &\lesssim \|f\|_{L^{\rho,\infty}_{p,\kappa}(\mathbb{R}^{n})}\bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{N\cdot({N_{0}}/{N_{0}+1})}, \end{align*} $$

where the last inequality follows from the fact that $1-\kappa>0$ . Summing the above estimates for $I_{1}$ and $I_{2}$ , and letting $\nu =\max \{\theta ^{\ast },N\cdot ({N_{0}}/{N_{0}+1})\}$ with $N\geq \theta ^{\ast }$ , we obtain the desired inequality (3-1).

(ii) According to Definition 2.4, it suffices to prove that, for each given ball ${B=B(x_{0},r)\subset \mathbb {R}^{n}}$ , there exists some $\nu>0$ such that

(3-5) $$ \begin{align} \frac{1}{|B(x_{0},r)|^{\kappa}}\sup_{\lambda>0}\lambda\cdot|\{x\in B(x_{0},r):\ |T^{L^{\alpha}}_{N}(f)(x)|>\lambda\}|\lesssim \bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{\nu} \end{align} $$

holds for any function $f\in L^{\rho ,\theta ^{\ast }}_{1,\kappa }(\mathbb {R}^{n})$ with some $\theta ^{\ast }>0$ and $0<\kappa <1$ . We decompose the function f as

$$ \begin{align*} \begin{cases} f=f_{1}+f_{2}\in L^{\rho,\theta^{\ast}}_{1,\kappa}(\mathbb{R}^{n}),\\ f_{1}=f\cdot\chi_{2B},\\ f_{2}=f\cdot\chi_{(2B)^{c}}. \end{cases} \end{align*} $$

Then, for any given $\lambda>0$ , we can write

$$ \begin{align*}\frac{1}{|B(x_{0},r)|^{\kappa}}\lambda\cdot|\{x\in B(x_{0},r):|T^{L^{\alpha}}_{N}(f)(x)|>\lambda\}|\leq J_{1}+J_{2},\end{align*} $$

where

$$ \begin{align*} \left\{ \begin{aligned} J_{1}:= \frac{1}{|B(x_{0},r)|^{\kappa}}\lambda\cdot|\{x\in B(x_{0},r):|T^{L^{\alpha}}_{N}(f_{1})(x)|>\lambda/2\}|,\\ J_{2}:= \frac{1}{|B(x_{0},r)|^{\kappa}}\lambda\cdot|\{x\in B(x_{0},r):|T^{L^{\alpha}}_{N}(f_{2})(x)|>\lambda/2\}|. \end{aligned} \right.\end{align*} $$

With the aid of (3-2), we first estimate the term $J_{1}$ as

$$ \begin{align*} J_{1} &\lesssim \frac{1}{|B|^{\kappa}}\bigg(\int_{2B}|f(x)|\,dx\bigg) \\ &\lesssim \|f\|_{L^{\rho,\theta^{\ast}}_{1,\kappa}(\mathbb{R}^{n})}\frac{|2B|^{\kappa}}{|B|^{\kappa}}\bigg(1+\frac{2r}{\rho(x_{0})}\bigg)^{\theta^{\ast}}\\ &\lesssim \|f\|_{L^{\rho,\infty}_{1,\kappa}(\mathbb{R}^{n})}\bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{\theta^{\ast}}. \end{align*} $$

For the second term $J_{2}$ , using (3-4) and Chebyshev’s inequality, we have

(3-6) $$ \begin{align}\notag \notag J_{2} &\leq \frac{2}{|B(x_{0},r)|^{\kappa}}\bigg(\int_{B(x_{0},r)}|T^{L^{\alpha}}_{N}(f_{2})(x)|\,dx\bigg) \\ \notag &\lesssim \frac{|B(x_{0},r)|}{|B(x_{0},r)|^{\kappa}}\sum^{\infty}_{k=1}\bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{N\cdot({N_{0}}/{N_{0}+1})}\bigg(1+\frac{2^{k+1}r}{\rho(x_{0})}\bigg)^{-N}\\ &\quad \times\frac{1}{|B(x_{0},2^{k+1}r)|}\int_{B(x_{0},2^{k+1}r)} |f(y)|\,dy. \end{align} $$

Note that, in (3-6), for each fixed $k\in \mathbb {N}$ ,

$$ \begin{align*} \frac{1}{|B(x_{0},2^{k+1}r)|}\int_{B(x_{0},2^{k+1}r)}|f(y)|\,dy \lesssim C\|f\|_{L^{\rho,\theta^{\ast}}_{1,\kappa}(\mathbb{R}^{n})}\frac{|B(x_{0},2^{k+1}r)|^{\kappa}}{|B(x_{0},2^{k+1}r)|} \bigg(1+\frac{2^{k+1}r}{\rho(x_{0})}\bigg)^{\theta^{\ast}}. \end{align*} $$

Consequently,

$$ \begin{align*} J_{2} \lesssim \|f\|_{L^{\rho,\theta^{\ast}}_{1,\kappa}(\mathbb{R}^{n})} \bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{N\cdot({N_{0}}/{N_{0}+1})}\sum_{k=1}^{\infty}\frac{|B(x_{0},r)|}{|B(x_{0},r)|^{k}} \bigg(1+\frac{2^{k+1}r}{\rho(x_{0})}\bigg)^{-N+\theta^{\ast}}\frac{|B(x_{0},2^{k+1}r)|^{\kappa}}{|B(x_{0},2^{k+1}r)|}. \end{align*} $$

Therefore, by choosing N large enough such that $N\geq \theta ^{\ast }$ , we obtain

$$ \begin{align*} J_{2} &\lesssim \|f\|_{L^{\rho,\infty}_{1,\kappa}(\mathbb{R}^{n})}\bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{N\cdot({N_{0}}/{N_{0}+1})}\sum_{k=1}^{\infty} \bigg(\frac{|B(x_{0},r)|}{|B(x_{0},2^{k+1}r)|}\bigg)^{1-\kappa} \\[4pt] &\lesssim \|f\|_{L^{\rho,\infty}_{1,\kappa}(\mathbb{R}^{n})}\bigg(1+\frac{r}{\rho(x_{0})}\bigg)^{N\cdot({N_{0}}/{N_{0}+1})}, \end{align*} $$

since $0<\kappa <1$ . Now we choose $\nu =\max \{\theta ^{\ast },N\cdot ({N_{0}}/{N_{0}+1})\}$ with $N\geq \theta ^{\ast }$ . Summing the above estimates for $J_{1}$ and $J_{2}$ and taking the supremum over all $\lambda>0$ , we obtain the desired inequality (3-5).

Proof. For (i), by the mean value theorem and Lemma 3.2, there exists $\xi _j\in [a_j, a_{j+1}]$ such that

$$ \begin{align*}&\bigg\vert\!\sum_{j=m}^{M}v_j (K^{L}_{\alpha, a_{j+1}}(x,y)- K^{L}_{\alpha, a_{j}}(x,y) ) \bigg\vert\\ &\quad\le C\Vert v\Vert_{l^\infty(\mathbb Z)} \sum_{j=m}^{M} (a_{j+1}-a_j)\vert\partial_t K^{L}_{\alpha, t}(x,y)|_{t=\xi_j}\vert\\ &\quad\le C_{v} \sum_{j=m}^{M} (a_{j+1}-a_j) {\frac 1{(\xi_j^{{1}/{(2\alpha)}}+|x-y|)^{n+2\alpha}}} \\ &\quad\le C_{v} \sum_{j=m}^{M} (\lambda^2-1)a_j ^{-{n/(2\alpha)}}\\ &\quad \le C_{v, \lambda} \frac1{a_m^{n/(2\alpha)}} \sum_{j=m}^{M} \frac1{\lambda^{n(j-m)/(2\alpha)}}\\ &\quad \le C_{v, \lambda,\alpha, n} {1\over a_m^{n/(2\alpha)}}, \end{align*} $$

where we have used the fact that $\displaystyle \lambda \le {a_{j+1}/ a_j}\le \lambda ^2.$

Now we prove (ii). Using the mean value theorem again, we can see that there exists $a_j\le \xi _j\le a_{j+1}$ such that

$$ \begin{align*} & \bigg\vert\!\sum_{j=-M}^{m-1}v_j (K^{L}_{\alpha, a_{j+1}}(z-y)-K^{L}_{\alpha, a_{j}}(z-y) ) \bigg\vert\\ &\quad\le C\Vert v\Vert_{l^\infty(\mathbb Z)} \sum_{j=m}^{M} (a_{j+1}-a_j)\vert\partial_t K^{L}_{\alpha, t}(x,y)|_{t=\xi_j}\vert\\&\quad\le C_{v,\alpha, n}\sum_{j=-M}^{m-1}(\lambda^2-1)a_j \frac 1{(\xi_j^{ 1/{(2\alpha)}}+|x-y|)^{n+2\alpha}} \\&\quad\le C_{\lambda, v,\alpha, n}\sum_{j=-M}^{m-1}{a_j \over (\xi_j^{ 1/{(2\alpha)}}+|x-y|)^{n+2\alpha}} \\ &\quad\le C_{\lambda, v,\alpha, n} \sum_{j=-M}^{m-1} {\frac{ a_j }{a_k}}\bigg({1\over a_k^{n/(2\alpha)}}\bigg) \le C_{\lambda, v,\alpha, n} {1\over a_k^{n/(2\alpha)}}\lambda^{-(k-m+1)}, \end{align*} $$

since $k \ge m.$

Proof. Observe that, for any $x_0\in {\mathbb R}^n$ and $N=(N_1, N_2),$

$$ \begin{align*}T^{L^{\alpha}}_N f(x_0)=T^{L^{\alpha}}_{(N_1, M)} f(x_0)-T^{L^{\alpha}}_{(N_2+1, M)} f(x_0)\end{align*} $$

with $-M\le N_1<N_2\le M.$ Then it suffices to estimate $|{T^{L^{\alpha }}_{(m, M)} f(x_0)}|$ for $\vert m\vert \le M$ with constants independent of m and $M.$ Denote $B_k=B(x_0, a_k^{1/(2\alpha )})$ for each $k\in \mathbb N$ . We split f as $f=f\chi _{B_m}+f\chi _{B_m^c}:= f_1+f_2$ . Then $|T^{L^{\alpha }}_{(m,M)} f(x_0)|\le I+II,$ where

$$ \begin{align*} \left\{\begin{aligned} I&:=\vert T^{L^{\alpha}}_{(m,M)} f_1(x_0)\vert,\\ II&:=\vert T^{L^{\alpha}}_{(m,M)} f_2(x_0)\vert. \end{aligned}\right.\end{align*} $$

By Lemma 4.2(i),

$$ \begin{align*} I&=\bigg\vert\!\int_{{\mathbb R}^n} \sum_{j=m}^{M}v_j (K^{L}_{\alpha, a_{j+1}}(x_0,y)- K^{L}_{\alpha, a_{j}}(x_0,y)) f_1(y)\,dy\bigg\vert\\ &\le {C_{n, v, \lambda, \alpha}\over |{a^{n/(2\alpha)}_m}|} \int_{{\mathbb R}^n} \vert f_1(y)\vert \,dy \le C_{v, \lambda, n, \alpha}{\mathcal M} f(x_0). \end{align*} $$

Also,

$$ \begin{align*} II&=\frac{2^{n/2}}{a_{m-1}^{n/(2\alpha)}}\int_{B(x_0, {1/ 2}a_{m-1}^{1/(2\alpha)})} \vert T^{L^{\alpha}}_{(m,M)} f_2(x_0)\vert \,dz\leq \sum^{4}_{i=1}A_i, \end{align*} $$

where

$$ \begin{align*}\left\{\begin{aligned} A_{1}&:=\frac{2^{n/2}}{a_{m-1}^{n/(2\alpha)}}\int_{B(x_0, {1/ 2}a_{m-1}^{1/(2\alpha) })}\vert T^{L^{\alpha}}_{(-M,M)} f(z)\vert \,dz;\\ A_{2}&:=\frac{2^{n/2}}{a_{m-1}^{n/(2\alpha)}}\int_{B(x_0, {1/2}a_{m-1}^{1/(2\alpha)})}\vert T^{L^{\alpha}}_{(-M,M)} f_1(z)\vert \,dz;\\ A_{3}&:=\frac{2^{n/2}}{a_{m-1}^{n/(2\alpha)}}\int_{B(x_0, {1/2}a_{m-1}^{1/(2\alpha)})}\vert T^{L^{\alpha}}_{(m,M)} f_2(z)-T^{L^{\alpha}}_{(m,M)} f_2(x_0)\vert \,dz;\\ A_{4}&:=\frac{2^{n/2}}{a_{m-1}^{n/(2\alpha)}}\int_{B(x_0, {1/2}a_{m-1}^{1/(2\alpha)})}\vert T^{L^{\alpha}}_{(-M,m-1)} f_2(z)\vert \,dz. \end{aligned}\right.\end{align*} $$

It should be noted that $A_4=0$ if $m+1=-M$ . It is obvious that

$$ \begin{align*} A_1\le {\mathcal M} (T^{L^{\alpha}}_{(-M,M)} f)(x_0). \end{align*} $$

It follows from the uniform boundedness of $T^{L^{\alpha }}_{N}$ that

$$ \begin{align*} A_2&\le \bigg(\frac{2^{n/2}}{a_{m-1}^{n/(2\alpha)}}\int_{B_{m-1}}\vert T^{L^{\alpha}}_{(-M,M)} f_1(z)\vert^q\,dz\bigg)^{1/q}\\ &\lesssim \bigg(\frac{1}{a_{m-1}^{n/(2\alpha)}}\int_{\mathbb{R}^n}\vert f_1(z)\vert^q\,dz\bigg)^{1/q}\\ &\thicksim\bigg(\frac{1}{a_{m-1}^{n/(2\alpha)}}\int_{B_{m}}\vert f(z)\vert^q\,dz\bigg)^{1/q}\\ &\lesssim \bigg(\frac{\lambda^{n/(2\alpha)}}{a_{m}^{n/(2\alpha)}}\int_{B_{m}}\vert f(z)\vert^q\,dz\bigg)^{1/q}\\ &\lesssim {\mathcal M}_q f(x_0). \end{align*} $$

For the third term $A_3$ , we can see that $ A_3\lesssim {\mathcal M} f(x_0)$ , since, for $z\in B(x_0, {1\over 2}a_{m-1}^{1/(2\alpha )})$ ,

$$ \begin{align*} &\vert T^{L^{\alpha}}_{(m,M)} f_2(z)-T^{L^{\alpha}}_{(m,M)} f_2(x_0)\vert\\ &\quad\le \int_{B_m^c} \vert K^\alpha_{(m,M)}(z,y)-K^\alpha_{(m,M)}(x_0,y)\vert\vert f(y)\vert \,dy\\ &\quad\le \sum_{j=1}^{+\infty}\int_{B_{2^jm} \setminus B_{2^{j-1}m}}\frac{\vert z-x_0\vert^{\delta'}}{\vert x_0-y\vert^{n+\delta'}}\vert f(y)\vert \,dy\\ &\quad\lesssim\sum_{j=1}^{+\infty}\frac 1{2^{j\delta'}}\frac{a_{m-1}^{\delta'/(2\alpha)}}{a_m^{\delta'/(2\alpha)}}{1\over |2^ja_{m}|^{n/(2\alpha)}}\int_{B_{2^jm}} \vert f(y)\vert \,dy\lesssim {\mathcal M} f(x_0), \end{align*} $$

where $B_{2^jm}=B(x_0, 2^ja_m^{1/ (2\alpha )})$ for any $j\ge 1$ .

For $A_4,$ we have

(4-2) $$ \begin{align} A_4\le \frac{2^{n/2}}{a_{m-1}^{n/(2\alpha)}}\int_{B(x_0, {1/ 2}a_{m-1}^{1/(2\alpha)})}\int_{B_m^c}\vert K^\alpha_{(-M,m-1)}(z,y) f(y)\vert \,dy\,dz. \end{align} $$

For the inner integral in (4-2), since $z\in B(x_0, {1\over 2}a_{m-1}^{1/(2\alpha )})$ , $y\in B_m^c$ and the sequence $\{a_j\}_{j\in \mathbb Z}$ is a $\lambda $ -lacunary sequence, we have $\vert z-y\vert \sim \vert y-x_0\vert .$ From the previous arguments and by Lemma 4.2(ii), we get

$$ \begin{align*} &\int_{B_m^c}\vert K^\alpha_{(-M,m-1)}(z, y) f(y)\vert \,dy\\ &\quad\le C_{\lambda, v,\alpha, n} \sum_{k=m}^{+\infty}\lambda^{-(k-m+1)}\bigg({1\over a_k^{n/(2\alpha)}}\int_{B_{k+1}\setminus B_{k}} \vert f(y)\vert \,dy\bigg)\\ &\quad\le C_{\lambda, v,\alpha, n} {\mathcal M} f(x_0)\sum_{k=m}^{+\infty}\lambda^{-(k-m+1)}\\ &\quad\le C_{\lambda, v,\alpha, n}{\mathcal M} f(x_0), \end{align*} $$

which implies $A_4\lesssim {\mathcal M} f(x_0)$ . Combining the above estimates for $A_i, i=1,2,3,4$ , we get

$$ \begin{align*}II\lesssim {\mathcal M} ( T^{L^{\alpha}}_{(-M,M)} f)(x_0)+ {\mathcal M}_q f(x_0) \end{align*} $$

and

$$ \begin{align*} \vert T^{L^{\alpha}}_{(m,M)} f(x_0)\vert\lesssim( {\mathcal M} (T^{L^{\alpha}}_{(-M,M)} f)(x_0)+{\mathcal M}_q f(x_0)). \end{align*} $$

Since the constants C appearing above only depend on $\Vert v\Vert _{l^\infty (\mathbb Z)}$ , $\lambda $ , $\alpha $ and n,

$$ \begin{align*} T^{*}_{{L^{\alpha}},M} f(x_0)\lesssim\{{\mathcal M}( T^{L^{\alpha}}_{(-M, M)} f)(x_0)+{\mathcal M}_q f(x_0)\}.\end{align*} $$

This completes the proof of Theorem 4.3.

Proof. First, it follows from [Reference Duoandikoetxea4, Theorem 2.5] that the maximal operators ${\mathcal M}$ and $ {\mathcal M}_q$ are bounded on $L^p({\mathbb R}^n) $ . On the other hand, since the operators $T^{L^{\alpha }}_N$ are uniformly bounded on $L^p({\mathbb R}^n)$ ,

$$ \begin{align*} \Vert T_{L^{\alpha},M}^*f\Vert_{L^p({\mathbb R}^n)}&\leq C \Vert{\mathcal M} (T^{L^{\alpha}}_{(-M, M)} f)\Vert_{L^p({\mathbb R}^n)}+\Vert{\mathcal M}_q f\Vert_{L^p({\mathbb R}^n)}\\&\leq C \Vert T^{L^{\alpha}}_{(-M, M)} f\Vert_{L^p({\mathbb R}^n)}+\Vert f\Vert_{L^p({\mathbb R}^n)}\\ &\leq C\Vert f\Vert_{L^p({\mathbb R}^n)}. \end{align*} $$

Note that the constants C appearing above are independent of M. Consequently, letting M increase to infinity yields to the $L^p$ -boundedness of $T^*_{L^{\alpha }}$ . This completes the proof of part (i) of this theorem.

In order to prove (ii), we consider the $\ell ^\infty (\mathbb Z^2)$ -valued operator

$$ \begin{align*}\mathcal{T}f(x) = \{ T^{L^{\alpha}}_N f(x) \}_{N\in \mathbb Z^2}.\end{align*} $$

Since

(4-3) $$ \begin{align}\|\mathcal{T}f(x) \|_{\ell^\infty(\mathbb Z^2)}= T^*f(x),\end{align} $$

using (i), we conclude that the operator $\mathcal {T}$ is bounded from $L^p({\mathbb R}^n) $ into $L^p_{\ell ^\infty (\mathbb Z^2)}(\mathbb {R}^n)$ for every $1<p<\infty $ . The kernel of $\mathcal {T}$ is given by $\mathcal {K}(x) = \{ K^\alpha _N(x)\} _{N\in \mathbb Z^2}$ . Therefore, by the vector-valued Calderón–Zygmund theory, the operator $\mathcal {T}$ is bounded from $L^1(\mathbb {R}^n)$ into weak- $L^1_{\ell ^\infty (\mathbb Z^2)}(\mathbb {R}^n)$ . Hence, using (4-3) again, we get the proof of (ii).

Proof. By an argument similar to that used in Theorem 3.6, we know that $ T^*_{L^{\alpha }} f<\infty \text{for almost every}\ x \in \mathbb {R}^{n}$ . Since (i) can be deduced from (ii), we only need to prove (ii). By Theorem 4.4(i) and Proposition 3.5, $T^{L^{\alpha }}_N$ and ${T}^{*} =\| \, \{ T^{L^{\alpha }}_N\}_{N=(N_1, N_2)}\, \|_{\ell ^\infty (\mathbb Z^2)}$ are $\gamma $ -Schrödinger–Calderón–Zygmund operators with $\gamma =0$ . Then it is enough to verify that both of the operators satisfy (4-4). More precisely,

$$ \begin{align*} | {T}^{*}1(x)-({T}^{*}1)_B | &\le \bigg| {T}^{*}1(x) - \frac1{|B|} \int_B {T}^{*}1(y) \,dy \bigg| \\ & \le \frac1{|B|} \int_B | {T}^{*}1(x) - {T}^{*}1(y) | \,dy \\ &= \frac1{|B|} \int_B | \| \, \{ T^{L^{\alpha}}_N 1(x)\}_N\, \|_{\ell^\infty(\mathbb Z^2)} - \| \, \{ T^{L^{\alpha}}_N 1(y) \}_N\, \|_{\ell^\infty(\mathbb Z^2)} | \, dy \\ & \le \frac1{|B|} \int_B \| \, T^{L^{\alpha}}_N 1(x) - T^{L^{\alpha}}_N 1(y)\, \|_{\ell^\infty(\mathbb Z^2)}\, dy\\ &= \frac1{|B|} \int_B \sup_N | T^{L^{\alpha}}_N 1(x) - T^{L^{\alpha}}_N 1(y)\, | \, dy. \end{align*} $$

Hence, to prove (4-4), it is sufficient to prove that, for $0<\delta <\min \{1, 2\alpha , 2-n/q\} $ and for any $x, y \in B=B(x_0,t),$ if $t\le \rho (x_0)/2$ , then

(4-5) $$ \begin{align} \vert T^{L^{\alpha}}_N 1(x)-T^{L^{\alpha}}_N1(y)\vert\lesssim(t/ {\rho(x_0)})^\delta. \end{align} $$

Note that $\rho (x)\sim \rho (y)\sim \rho (x_0)$ when $x, y \in B$ . Since $e^{- {t(-\Delta )^\alpha }}f$ is a convolution operator, it follows from (1-1) that $(\mathcal {F}K^{-\Delta }_{\alpha , s}(\cdot ))(\xi )=e^{-t|\xi |^{2\alpha }}$ and

(4-6) $$ \begin{align} \int_{{\mathbb R}^{n}} K^{-\Delta}_{\alpha, s}(x-z)\,dz=\int_{{\mathbb R}^{n}} K^{-\Delta}_{\alpha, s}(y-z)\,dz=e^{-t|\xi|^{2\alpha}}|_{\xi=0}=1 \end{align} $$

for any $s>0.$ Then we can write

$$ \begin{align*} &\vert T^{L^{\alpha}}_N 1(x)-T^{L^{\alpha}}_N 1(y)\vert\\ &\quad= \bigg|\! \int_{{\mathbb R}^{n}} \sum_{j=N_1}^{N_2}v_j[( (K^{L}_{\alpha, a_{j+1}}(x,z)- K^{-\Delta}_{\alpha, a_{j+1}}(x-z)) -(K^{L}_{\alpha, a_{j}}(x,z) - K^{-\Delta}_{\alpha, a_{j}}(x-z) ) ) \\ &\qquad - ((K^{L}_{\alpha, a_{j+1}}(y,z)- K^{-\Delta}_{\alpha, a_{j+1}}(y-z))- (K^{L}_{\alpha, a_{j}}(y,z) - K^{-\Delta}_{\alpha, a_{j}}(y-z))) ] \,dz\bigg| \\ &\quad\lesssim\Vert v\Vert_{\ell^\infty(\mathbb Z)} \int_{{\mathbb R}^{n}}\int_0^\infty|\partial_s((K^{L}_{\alpha, s}(x,z)-K^{-\Delta}_{\alpha, s}(x-z)) -(K^{L}_{\alpha, s}(y,z)- K^{-\Delta}_{\alpha, s}(y-z) )\, | \,ds \,dz\\ &\quad\lesssim I+II+III, \end{align*} $$

where

$$ \begin{align*}\left\{\begin{aligned} I&:=\int_{{\mathbb R}^{n}}\int_0^{(2t)^{2\alpha}} |\partial_s((K^{L}_{\alpha, s}(x,z)-K^{-\Delta}_{\alpha, s}(x-z))\\ &\quad -(K^{L}_{\alpha, s}(y,z)- K^{-\Delta}_{\alpha, s}(y-z) )\, | \,ds \,dz;\\ II&:=\int_{{\mathbb R}^{n}}\int_{(2t)^{2\alpha}}^{ \rho^{2\alpha}(x_0)}|\partial_s((K^{L}_{\alpha, s}(x,z)-K^{-\Delta}_{\alpha, s}(x-z))\\ &\quad -(K^{L}_{\alpha, s}(y,z)- K^{-\Delta}_{\alpha, s}(y-z) )\, | \,ds \,dz;\\ III&:=\int_{{\mathbb R}^{n}}\int_{\rho^{2\alpha}(x_0)}^\infty |\partial_s((K^{L}_{\alpha, s}(x,z)-K^{-\Delta}_{\alpha, s}(x-z))\\ &\quad -(K^{L}_{\alpha, s}(y,z)- K^{-\Delta}_{\alpha, s}(y-z) )\, | \,ds \,dz. \end{aligned}\right.\end{align*} $$

For $I,$ by Lemma 3.4, we get $I\leq C(I_{1}+I_{2})$ , where

$$ \begin{align*}\left\{\begin{aligned} I_{1}&:=\int_{{\mathbb R}^{n}} \int_0^{(2t)^{2\alpha}} |\partial_s(K^{L}_{\alpha, s}(x,z)-K^{-\Delta}_{\alpha, s}(x-z) ) |\,ds \,dz;\\ I_{2}&:=\int_{{\mathbb R}^{n}} \int_0^{(2t)^{2\alpha}}| \partial_s(K^{L}_{\alpha, s}(y,z)-K^{-\Delta}_{\alpha, s}(y-z)) |\,ds \,dz. \end{aligned}\right.\end{align*} $$

For $I_{1}$ , when $\sqrt {s^{1/\alpha }}\leq |x-z|$ , by choosing $\delta _{0}<2\alpha $ , we get

$$ \begin{align*} I_{1} &\lesssim \int^{(2t)^{2\alpha}}_{0}\int_{\mathbb{R}^{n}}\bigg(\frac{|x-z|}{\rho(x)}\bigg)^{\delta_{0}}\frac{1}{(|x-z|+s^{1/(2\alpha)})^{n+2\alpha}}\,dz\,ds \\ &\lesssim \int^{(2t)^{2\alpha}}_{0}\frac{1}{\rho(x_{0})^{\delta_{0}}}\int_{\mathbb{R}^{n}}\frac{|x-z|^{\delta_{0}}}{(|x-z|+s^{1/(2\alpha)})^{n+2\alpha}}\,dz\,ds \\ &\lesssim \int^{(2t)^{2\alpha}}_{0}\bigg(\frac{s^{1/(2\alpha)}}{\rho(x_{0})}\bigg)^{\delta_{0}}\frac{1}{s}\,ds \\ &\lesssim \bigg(\frac{t}{\rho(x_{0})}\bigg)^{\delta_{0}}. \end{align*} $$

When $\sqrt {s^{1/\alpha }}>|x-z|$ , we obtain

$$ \begin{align*} I_{1} &\lesssim \int^{(2t)^{2\alpha}}_{0}\int_{\mathbb{R}^{n}}\bigg(\frac{\sqrt{s^{1/\alpha}}}{\rho(x)}\bigg)^{\delta_{0}}\frac{1}{(|x-z|+s^{1/(2\alpha)})^{n+2\alpha}}\,dz\,ds \\ &\lesssim \int^{(2t)^{2\alpha}}_{0}\bigg(\frac{s^{1/(2\alpha)}}{\rho(x_{0})}\bigg)^{\delta_{0}}\frac{1}{s}\,ds \\ &\lesssim \bigg(\frac{t}{\rho(x_{0})}\bigg)^{\delta_{0}}. \end{align*} $$

The proof for $I_{2}$ is similar to that for $I_{1}$ , so we omit the details.

We have

$$ \begin{align*} II &= \int_{{\mathbb R}^{n}} \int_{(2t)^{2\alpha}}^{ \rho^{2\alpha}(x_0)} |\partial_s(K^{L}_{\alpha, s}(x,z)-K^{L}_{\alpha, s}(y,z) )\\ &\quad-\partial_s(K^{-\Delta}_{\alpha, s}(x-z)-K^{-\Delta}_{\alpha, s}(y-z)) | \,ds \,dz\\ &\lesssim II_1+II_2+II_3, \end{align*} $$

where

$$ \begin{align*}\left\{\begin{aligned} II_{1}&:=\int_{\vert x-z\vert>c\rho(x)}\int_{(2t)^{2\alpha}}^{ \rho^{2\alpha}(x_0)}\vert\partial_s(K^{L}_{\alpha, s}(x,z)-K^{L}_{\alpha, s}(y,z)) \vert\\ &\quad +\vert\partial_s(K^{-\Delta}_{\alpha, s}(x-z)-K^{-\Delta}_{\alpha, s}(y-z))\vert\,ds \,dz; \\ II_{2}&:= \int_{4|x-y|<\vert x-z\vert\le c\rho(x)}\int_{(2t)^{2\alpha}}^{ \rho^{2\alpha}(x_0)}|\partial_s K^{L}_{\alpha, s}(x,z)-K^{-\Delta}_{\alpha, s}(x-z))\\ &\quad -\partial_s(K^{L}_{\alpha, s}(y,z)-K^{-\Delta}_{\alpha, s}(y-z)) |\,ds \,dz;\\ II_{3}&:=\int_{\vert x-z\vert\le 4|x-y|}\int_{(2t)^{2\alpha}}^{ \rho^{2\alpha}(x_0)}|\partial_s(K^{L}_{\alpha, s}(x,z)-K^{-\Delta}_{\alpha, s}(x-z))\\ &\quad -\partial_s(K^{L}_{\alpha, s}(y,z)-K^{-\Delta}_{\alpha, s}(y-z))| \,ds \,dz. \end{aligned}\right.\end{align*} $$

For $II_1$ , noting that $\vert x-y\vert \le 2t\le s^{1/(2\alpha )},$ by Lemma 3.3 and a similar result for the Laplacian as in Lemma 3.3, we have

$$ \begin{align*} II_{11} &\lesssim\int_{\vert x-z\vert>c\rho(x)}{\vert x-y\vert^\delta\over \vert x-z\vert^{2\alpha+n}} \int_{(2t)^{2\alpha}}^ { \rho^{2\alpha}(x_0)}~\frac{1}{s^{\delta/(2\alpha)}}\,ds\,dz\\ &\lesssim\bigg(\frac{t}{\rho(x_{0})}\bigg)^{\delta}\int_{\vert x-z\vert>c\rho(x)}\frac{1}{|x-z|^{n+2\alpha}}\rho^{2\alpha}(x_0)\,dz\\ &\lesssim\bigg(\frac{t}{\rho(x_{0})}\bigg)^{\delta}. \end{align*} $$

For the term $II_2$ , in this case, $|x-y|< c\rho (x)$ and $\displaystyle |x-y|<{|x-z|/4}$ . By Lemma 3.4, we conclude that, for some $0<\varepsilon <\min \{\alpha ,\delta \}$ small enough,

$$ \begin{align*} II_2&\lesssim\int_{4|x-y|<\vert x-z\vert\le c\rho(x)}\int_{(2t)^{2\alpha}}^{\rho^{2\alpha}(x_0)} \bigg({\vert x-y\vert\over \rho(z)}\bigg)^\delta \frac{1}{(s^{1/(2\alpha)}+|x-z|)^{n+2\alpha}}\,ds \,dz\\ &\lesssim\bigg({\vert x-y\vert\over \rho(x_0)}\bigg)^\delta \int_{4|x-y|<\vert x-z\vert\le c\rho(x)}\int_{(2t)^{2\alpha}}^{ \rho^{2\alpha}(x_0)} \frac{1}{s^{n/(2\alpha)+1}} \frac{|x-z|^{\epsilon}\,ds\,dz}{(1+|x-z|/s^{1/(2\alpha)})^{n+2\epsilon}|x-z|^{\epsilon}}\\ &\lesssim\bigg({\vert x-y\vert\over \rho(x_0)}\bigg)^\delta\bigg({\rho(x_0) \over\vert x-y\vert}\bigg)^{\varepsilon}\\ &\lesssim\bigg({\vert x-y\vert\over \rho(x_0)}\bigg)^{\delta'}. \end{align*} $$

For the term $II_3,$ since $|x-y|\leq 2t\leq s^{1/(2\alpha )}$ , we have $|x-z|\leq s^{1/(2\alpha )}$ . By Lemma 3.4,

$$ \begin{align*} II_3 &\le\int_{\vert x-z\vert\le 4|x-y|}\int_{(2t)^{2\alpha}}^{ \rho^{2\alpha}(x_0)}\\&\quad \times \vert{\partial_s(K^{L}_{\alpha, s}(x,z)-K^{-\Delta}_{\alpha, s}(x-z))- \partial_s(K^{L}_{\alpha, s}(y,z)-K^{-\Delta}_{\alpha, s}(y-z)) }\vert\,ds \,dz\\&\lesssim\int_{\vert x-z\vert\le 4|x-y|}\int_{(2t)^{2\alpha}}^{\rho^{2\alpha}(x_0)}\bigg\{\bigg(\frac{\sqrt{s^{1/\alpha}}}{\rho(x)}\bigg)^\delta \frac{1}{(s^{1/(2\alpha)}+|x-z|)^{n+2\alpha}}\\&\quad +\bigg(\frac{\sqrt{s^{1/\alpha}}}{\rho(y)}\bigg)^\delta \frac{1}{(s^{1/(2\alpha)}+|y-z|)^{n+2\alpha}}\bigg\}\,ds \,dz\\&\lesssim\bigg\{\!\int_{\vert x-z\vert\le 4|x-y|}\int_{(2t)^{2\alpha}}^{ \rho^{2\alpha}(x_0)}\bigg(\frac{\sqrt{s^{1/\alpha}}}{\rho(x)}\bigg)^\delta \frac{1}{(s^{1/(2\alpha)}+|x-z|)^{n+2\alpha}}\,ds\,dz\\&\quad +\int_{\vert y-z\vert\le 5|x-y|}\int_{(2t)^{2\alpha}}^{\rho^{2\alpha}(x_0)}\bigg(\frac{\sqrt{s^{1/\alpha}}}{\rho(y)}\bigg)^\delta \frac{1}{(s^{1/(2\alpha)}+|y-z|)^{n+2\alpha}}\,ds \,dz\bigg\}\\&\lesssim \int_{(2t)^{2\alpha}}^{\rho^{2\alpha}(x_0)} \int_{\vert\xi\vert\le 5{|x-y|/ c}}\bigg(\frac{\sqrt{s^{1/\alpha}}}{\rho(x)}\bigg)^\delta \frac{1}{(s^{1/(2\alpha)}+|\xi|)^{n+2\alpha}}\, d\xi \,ds\\&\lesssim \int_{(2t)^{2\alpha}}^{+\infty} \bigg(\frac{\sqrt{s^{1/\alpha}}}{\rho(x_0)}\bigg)^\delta {1\over s^{n/(2\alpha)+1} } |x-y|^n \,ds \\&\lesssim\bigg({t\over \rho(x_0)}\bigg)^{\delta}, \end{align*} $$

which gives $II\lesssim ({t/ \rho (x_0)})^{\delta }$ with $0<\delta <2-n/q.$

We next consider the last term $III$ . Now

$$ \begin{align*} III&\lesssim \int_{{\mathbb R}^{n}} \int_{ \rho^{2\alpha}(x_0)}^{+\infty}\vert\partial_s(K^{L}_{\alpha, s}(x,z)-K^{L}_{\alpha, s}(y,z))\vert\,ds \,dz\\ &\quad +\int_{{\mathbb R}^{n}} \int_{ \rho^{2\alpha}(x_0)}^{+\infty} \vert\partial_s(K^{-\Delta}_{\alpha, s}(x-z)-K^{-\Delta}_{\alpha, s}(y-z)) \vert \,ds \,dz. \end{align*} $$

In this case, since $s\ge \rho ^{2\alpha }(x_0)>(2t)^{2\alpha }$ , we have $s^{1/(2\alpha )}>2t>|x-y|$ . Regarding the first term in the summand above, we can apply Lemma 3.3 to deduce that

$$ \begin{align*} &{ \int_{{\mathbb R}^{n}}\int_{ \rho^{2\alpha}(x_0)}^{+\infty}{\vert\partial_s(K^{L}_{\alpha, s}(x,z)-K^{L}_{\alpha, s}(y,z))\vert}\,ds \,dz}\\& \quad \lesssim \int_{\rho^{2\alpha}(x_0)}^{+\infty}{1\over s} \bigg({|x-y|\over s^{1/(2\alpha)}}\bigg)^\delta\bigg(\int_{{\mathbb R}^{n}}\frac{s}{(s^{1/(2\alpha)}+|x-z|)^{n+2\alpha}} \,dz\bigg)\,ds \\& \quad \lesssim \int_{\rho^{2\alpha}(x_0)}^{+\infty}{1\over s} \bigg({|x-y|\over s^{1/(2\alpha)}}\bigg)^\delta\,ds \\& \quad \lesssim\bigg({|x-y|\over \rho(x_0)}\bigg)^\delta \le C\bigg({t\over \rho(x_0)}\bigg)^\delta, \quad 0<\delta<2-n/q. \end{align*} $$

For the second part of the summand, by Lemma 3.3, which is also valid for the Laplacian, when $4|x-y| \le |x-z|$ , we obtain the same estimate as before but with $0<\delta <\min \{1, 2-n/q\}$ . If $|x-z| \le 4|x-y|$ , then $|y-z|\le |x-y|+|x-z|\le 5|x-y|$ . Hence,

$$ \begin{align*} &\int_{{\mathbb R}^{n}} \int_{ \rho^{2\alpha}(x_0)}^{+\infty} |\partial_s K^{-\Delta}_{\alpha, s}(x-z)-\partial_s K^{-\Delta}_{\alpha, s}(y-z)| \,ds \,dz\\ &\quad\lesssim\int_{{\mathbb R}^{n}} \int_{ \rho^{2\alpha}(x_0)}^{+\infty} \frac{1}{(s^{1/(2\alpha)}+|x-z|)^{n+2\alpha}} \,ds \,dz\\ &\quad\lesssim \int_{ \rho^{2\alpha}(x_0)}^{+\infty} \frac{|x-y|^n}{s^{n/(2\alpha)+1}} \,ds \le C \frac{|x-y|^n}{\rho^n(x_0)}\\ &\quad\lesssim \bigg({t\over \rho(x_0)}\bigg)^\delta. \end{align*} $$

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 $T^{L^{\alpha }}_N$ and $T^{*}_{L^{\alpha }}.$

For the case $f\in BMO_L({\mathbb R}^n),$ by (4-5),

$$ \begin{align*} \bigg(\frac{\rho(x_0)}{t}\bigg)^{\gamma}\frac{1}{|B|}\int_B| T_N^{L^{\alpha}} 1(x)-( T_N^{L^{\alpha}} 1)_B|\,dx\lesssim \bigg(\frac{\rho(x_0)}{t}\bigg)^{\delta-\gamma}\le C \end{align*} $$

when $\delta -\gamma \ge 0,$ which can be guaranteed by $\gamma <\min \{1, 2\alpha , 2-n/q\}$ and choosing $\delta \ge \gamma .$ This completes the proof of Theorem 4.6.

Proof. Using Theorem 4.4 and Proposition 3.5, we deduce the desired results. Since the proofs are similar to those of Theorem 3.6, we omit the details.

Proof. By (4-6), we know that

$$ \begin{align*}K_{\alpha,t^{2\alpha}}^{-\Delta}=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-t^{2\alpha}|\xi|^{2\alpha}}e^{ix\xi}d\xi =\frac{1}{t}K_{\alpha,1}^{-\Delta}(x/t)\end{align*} $$

and

$$ \begin{align*}\int_{\mathbb{R}}K_{\alpha,1}^{-\Delta}(x)\,dx=1.\end{align*} $$

Moreover,

$$ \begin{align*} K_{\alpha,1}^{-\Delta}(x) &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-|\xi|^{2\alpha}}e^{ix\xi}d\xi \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-|\eta|^{2\alpha}}e^{i(-x)\eta}d\eta= K_{\alpha,1}^{-\Delta}(-x), \end{align*} $$

where $\eta =-\xi $ . Therefore,

$$ \begin{align*}\int^{\infty}_{0}K_{\alpha,1}^{-\Delta}(x)\,dx=\frac{1}{2}\int_{\mathbb{R}}K_{\alpha,1}^{-\Delta}(x)\,dx=\frac{1}{2}.\end{align*} $$

Then, there exists $A>0$ such that $\int ^{A}_{0}K_{\alpha ,1}^{-\Delta }(w)dw=1/4+\gamma /2.$ Let $w=Az$ . Then,

$$ \begin{align*} K_{\alpha,1}^{-\Delta}(Az) &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{iAz\xi}e^{-|\xi|^{2\alpha}}d\xi \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{iz\eta}e^{-A^{-2\alpha}|\eta|^{2\alpha}}A^{-1}d\eta=A^{-1}K_{\alpha,A^{-2\alpha}}^{-\Delta}(z), \end{align*} $$

where $\eta =A\xi $ . So,

$$ \begin{align*}\int^{1}_{0}K_{\alpha,A^{-2\alpha}}^{-\Delta}(z)\,dz=1/4+\gamma/2.\end{align*} $$

Assuming that $A=t_{0}^{-1}$ ,

$$ \begin{align*}\int^{1}_{0}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\,dz=1/4+\gamma/2.\end{align*} $$

This completes the proof of (i). The proofs of (ii) and (iii) follow easily.

Proof. Let $t_{0}$ and a be as in Lemma 4.8 and let f be the function defined as

$$ \begin{align*}f(x)=\sum_{k\in\mathbb{Z}}(-1)^{k+1}\chi_{(a^{k},a^{k+1}]}(x).\end{align*} $$

It is easy to see that

(4-7) $$ \begin{align} f(a^{j}x)=(-1)^{j}f(x). \end{align} $$

Considering the sequence $\{a_{j}\}$ with $a_{j}=(t_{0}a^{j})^{2\alpha }$ , we have

$$ \begin{align*} e^{-a_{j}(-\Delta)^{\alpha}}f(x) = \int_{\mathbb{R}}K_{\alpha,(t_{0}a^{j})^{2\alpha}}^{-\Delta}(y)f(x-y)\,dy = \int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(x-a^{j}z)\,dz. \end{align*} $$

By (4-7),

$$ \begin{align*} e^{-a_{j}(-\Delta)^{\alpha}}f(0) =\int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(-a^{j}z)\,dz =(-1)^{\hspace{1.2pt}j}\int^{\infty}_{0}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(z)\,dz. \end{align*} $$

On the other hand, using Lemma 4.8, we get

$$ \begin{align*} & \int^{\infty}_{0}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(z)\,dz \\ &\quad= \int^{1/a}_{0}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(z)\,dz+\int^{1}_{1/a}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(z)\,dz + \int^{\infty}_{1}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(z)\,dz \\ &\quad\geq \int^{1}_{1/a}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\,dz-\bigg(\int^{1/a}_{0}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\,dz+ \int^{\infty}_{1}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\,dz\bigg) \\ &\quad> \frac{1+\gamma}{4}-\bigg(\frac{\gamma}{4}+\frac{1}{4}-\frac{\gamma}{2}\bigg)=\frac{\gamma}{2}. \end{align*} $$

By (4-7), we get

$$ \begin{align*} &e^{-a_{j+1}(-\Delta)^{\alpha}}f(x)-e^{-a_{j}(-\Delta)^{\alpha}}f(x) \\ &\quad= (-1)^{j+1}\bigg\{\int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(x/a^{j+1}-z)\,dz+ \int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(x/a^{j}-z)\,dz\bigg\}. \end{align*} $$

By the dominated convergence theorem, we obtain

$$ \begin{align*} \lim_{h\rightarrow 0}\int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(h-z)\,dz &=\int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(-z)\,dz \\ &= \int^{\infty}_{0}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(z)\,dz>\frac{\gamma}{2}. \end{align*} $$

So, there exists $0<\eta _{0}<1$ such that, for $|h|<\eta _{0}$ ,

$$ \begin{align*}\int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(h-z)\,dz\geq \gamma/4.\end{align*} $$

Then, for each $x\neq 0$ , we can choose $j\in \mathbb {Z}$ such that $|x/a^{j}|<\eta _{0}$ ; at this time, there are infinitely many j satisfying this condition, and we get

$$ \begin{align*}\int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(x/a^{j+1}-z)\,dz+\int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z) f(x/a^{j}-z)\,dz\geq\gamma/2.\end{align*} $$

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

$$ \begin{align*} T^{\ast}_{L^{\alpha},N}f(x) &\geq \sum_{|t/a^{j}|<\eta_{0}}(-1)^{j+1}(e^{-a_{j+1}(-\Delta)^{\alpha}}f(x)-e^{-a_{j}(-\Delta)^{\alpha}}f(x)) \\ &\geq \sum_{|t/a^{j}|<\eta_{0}}(-1)^{j+1}\bigg(\int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(x/a^{j+1}-z)\,dz \\ &\quad+\int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(x/a^{j}-z)\,dz\bigg)=\infty.\\[-42pt] \end{align*} $$

Proof. For (i), we prove this theorem only for the case $1<p<\infty $ and in the Schrödinger setting. For the cases $p=1$ and $p=\infty $ , the proof follows by introducing the obvious changes. Since $2r<1$ , we know that $B\setminus B_{2r}\neq \varnothing $ . Let $f=f_{1}+f_{2}$ , where $f_{1}=f\chi _{B_{2r}}$ and $f_{2}=f\chi _{B\setminus B_{2r}}$ . Then,

$$ \begin{align*}|T^*_{L^{\alpha}} f(x)|\leq | T^*_{L^{\alpha}} f_{1}(x)|+|T^*_{L^{\alpha}} f_{2}(x)|.\end{align*} $$

By Theorem 4.4,

$$ \begin{align*} \frac{1}{|B_{r}|}\int_{B_{r}}| T^*_{L^{\alpha}} f_{1}(x)|\,dx &\leq\bigg(\frac{1}{|B_{r}|}\int_{B_{r}} | T^*_{L^{\alpha}} f_{1}(x)|^{2}\,dx\bigg)^{1/2} \\ &\lesssim \bigg(\frac{1}{|B_{r}|}\int_{\mathbb{R}^{n}}|f_{1}(x)|^{2}\,dx\bigg)^{1/2}\lesssim\|f\|_{L^{\infty}(\mathbb{R}^{n})}. \end{align*} $$

We should note that, for any $j\in \mathbb {Z}$ , by (1-2),

(4-8) $$ \begin{align} &\int_{\mathbb{R}^{n}}|K^{L}_{\alpha, a_{j+1}}(x,y)-K^{L}_{\alpha, a_{j}}(x,y)|\,dy \notag \\\notag &\quad\leq \int_{\mathbb{R}^{n}}|K^{L}_{\alpha, a_{j+1}}(x,y)|\,dy+\int_{\mathbb{R}^{n}}|K^{L}_{\alpha, a_{j}}(x,y)|\,dy \\\notag &\quad\lesssim\int_{\mathbb{R}^{n}}\frac{a_{j+1}}{(a_{j+1}^{1/(2\alpha)}+|x-y|)^{n+2\alpha}}\,dy+ \int_{\mathbb{R}^{n}}\frac{a_{j}}{(a_{j}^{1/(2\alpha)}+|x-y|)^{n+2\alpha}}\,dy \\&\quad= C. \end{align} $$

For $1<p<\infty $ and any $N=(N_{1},N_{2})$ , by Hölder’s inequality, (4-8) and Fubini’s theorem, we can use Proposition 3.5 to obtain

$$ \begin{align*} &\bigg|\hspace{-1.4pt}\sum^{N_{2}}_{j=N_{1}}v_{j}(e^{-a_{j+1}L^{\alpha}}f_{2}(x)-e^{-a_{j}L^{\alpha}}f_{2}(x))\bigg| \\&\quad\lesssim \sum^{N_{2}}_{j=N_{1}}\bigg|v_{j}\int_{\mathbb{R}^{n}}(K^{L}_{\alpha, a_{j+1}}(x,y)-K^{L}_{\alpha, a_{j}}(x,y))f_{2}(y)\,dy\bigg| \\&\quad\lesssim \|v\|_{l^{p}(\mathbb{Z})}\bigg(\sum^{N_{2}}_{j=N_{1}}\bigg(\int_{\mathbb{R}^{n}} |K^{L}_{\alpha, a_{j+1}}(x,y)-K^{L}_{\alpha, a_{j}}(x,y)||f_{2}(y)|\,dy\bigg)^{p'}\bigg)^{1/p'} \\&\quad\lesssim \|v\|_{l^{p}(\mathbb{Z})}\bigg(\sum^{N_{2}}_{j=N_{1}}\bigg(\int_{\mathbb{R}^{n}} |K^{L}_{\alpha, a_{j+1}}(x,y)-K^{L}_{\alpha, a_{j}}(x,y)||f_{2}(y)|^{p'}\,dy\bigg) \\&\qquad\times \bigg(\int_{\mathbb{R}^{n}} |K^{L}_{\alpha, a_{j+1}}(x,y)-K^{L}_{\alpha, a_{j}}(x,y)|\,dy\bigg)^{p'/p}\bigg)^{1/p'} \\&\quad\lesssim \|v\|_{l^{p}(\mathbb{Z})}\bigg(\sum^{N_{2}}_{j=N_{1}}\int_{\mathbb{R}^{n}} |K^{L}_{\alpha, a_{j+1}}(x,y)-K^{L}_{\alpha, a_{j}}(x,y)||f_{2}(y)|\bigg)\,dy^{1/p'}\\&\quad\lesssim \|v\|_{l^{p}(\mathbb{Z})}\bigg(\int_{\mathbb{R}^{n}}\sum^{+\infty}_{j=-\infty}|K^{L}_{\alpha, a_{j+1}}(x,y)- K^{L}_{\alpha, a_{j}}(x,y)||f_{2}(y)|\,dy\bigg)^{1/p'} \\&\quad\lesssim\|v\|_{l^{p}(\mathbb{Z})}\bigg(\int_{\mathbb{R}^{n}}\frac{1}{|x-y|^{n}}|f_{2}(y)|^{p'}\,dy\bigg)^{1/p'}. \end{align*} $$

For $y\in B\setminus B_{2r}$ and $x\in B_{r}$ , we have $r\leq |x-y|\leq 2$ . Then,

$$ \begin{align*} \frac{1}{|B_{r}|}\int_{B_{r}}| T^*_{L^{\alpha}} f_{2}(x)|\,dx &\lesssim\frac{1}{|B_{r}|}\int_{B_{r}}\bigg(\int_{\mathbb{R}^{n}}\frac{1}{|x-y|^{n}}|f_{2}(y)|^{p'}\,dy\bigg)^{1/p'}\,dx \\ &\lesssim \frac{\|f\|_{L^{\infty}(\mathbb{R}^{n})}}{|B_{r}|}\int_{B_{r}}\bigg(\int_{r\leq |x-y|\leq 2}\frac{1}{|x-y|^{n}}\,dy\bigg)^{1/p'}\,dx \\ &\sim \bigg(\log\frac{2}{r}\bigg)^{1/p'} \|f\|_{L^{\infty}(\mathbb{R}^{n})}. \end{align*} $$

Hence,

$$ \begin{align*} \frac{1}{|B_{r}|}\int_{B_{r}}| T^*_{L^{\alpha}} f(x)|\,dx\ {\lesssim}\ \, \bigg(1+\bigg(\log\frac{2}{r}\bigg)^{1/p'}\bigg)\|f\|_{L^{\infty}(\mathbb{R}^{n})} \lesssim\bigg(\log\frac{2}{r}\bigg)^{1/p'}\|f\|_{L^{\infty}(\mathbb{R}^{n})}. \end{align*} $$

For (ii), we only need to consider the case $n=1$ as the multidimensional cases are similar. In what follows, we assume that $1<p<\infty $ and we choose $\epsilon $ such that $0<\epsilon <p-1$ . Let

$$ \begin{align*}f(x)=\sum^{0}_{k=-\infty}(-1)^{k}\chi_{(-a^{k},-a^{k-1}]}(x),\end{align*} $$

where $a>1$ is the constant defined in Lemma 4.8. We check that

$$ \begin{align*}f(a^{j}x)=(-1)^{j}f(x)+(-1)^{j}\sum^{-j}_{\mathrm{l}=1}(-1)^{\mathrm{l}}\chi_{(-a^{\mathrm{l}},-a^{\mathrm{l}-1}]}(x)\end{align*} $$

when $j\leq 0$ . We always assume that $j\leq 0$ in what follows.

Choose a lacunary sequence $\{a_{j}\}_{j\in \mathbb {Z}}$ with $a_{j}=(t_{0}a^{\,j})^{2\alpha }$ , $j\in \mathbb {Z}$ . Then, similarly to the proof of Proposition 4.9, we have

$$ \begin{align*} &e^{-a_{j+1}(-\Delta)^{\alpha}}f(x)-e^{-a_{j}(-\Delta)^{\alpha}}f(x) \\&\quad= C(-1)^{j+1}\bigg(\int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(x/a^{j+1}-z)\,dz+ \int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(x/a^{j}-z)\,dz \\&\qquad+ \int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\sum^{-j-1}_{\mathrm{l}=1}(-1)^{\mathrm{l}} \chi_{(-a^{\mathrm{l}},-a^{\mathrm{l}-1}]}\bigg(\frac{x}{a^{j+1}}-z\bigg)\,dz \\&\qquad+ \int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\sum^{-j}_{\mathrm{l}=1}(-1)^{\mathrm{l}} \chi_{(-a^{\mathrm{l}},-a^{\mathrm{l}-1}]}\bigg(\frac{x}{a^{j}}-z\bigg)\,dz\bigg). \end{align*} $$

Using Lemma 4.8, we obtain

$$ \begin{align*} & \int^{\infty}_{0}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(-z)\,dz \\ &\quad= \bigg\{\int^{1/a}_{0}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(-z)\,dz+\int^{1}_{1/a}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(-z)\,dz\bigg\}\\ &\quad\geq \int^{1}_{1/a}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\,dz-\int^{1/a}_{0}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\,dz\\ & \quad>{(1+\gamma)}/{4}-{\gamma}/{4}={1}/{4}. \end{align*} $$

Therefore, there exists $0<\eta _{0}<1$ small enough such that, for $|h|<\eta _{0}$ ,

(4-9) $$ \begin{align} \int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)f(h-z)\,dz>1/4. \end{align} $$

Moreover, using Lemma 4.8 again, we can choose the last $\eta _{0}$ small enough such that

(4-10) $$ \begin{align} \int^{\infty}_{1-\eta_{0}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\,dz<{(1-\gamma)}/{4}. \end{align} $$

For the given $\eta _{0}$ in (4-10), let $2r<1$ be such that $r<\eta _{0}^{2}$ and $r\sim a^{\mathrm {J}_{0}}\eta _{0}$ for a certain negative integer $\mathrm {J}_{0}$ . If $\mathrm {J}_{0}\leq j\leq 0$ , we have $r/a^{j}<\eta _{0}$ . And, for any $-r\leq x\leq r$ , we get

$$ \begin{align*}-1\cdot\chi_{[1-\eta_{0},+\infty)}(z)\leq\sum^{-j-1}_{\mathrm{l}=1}(-1)^{\mathrm{l}}\chi_{(-a^{\mathrm{l}},-a^{\mathrm{l}-1}]} \bigg(\frac{x}{a^{j+1}}-z\bigg)\leq \chi_{[a-\eta_{0},+\infty)}(z)\end{align*} $$

and

$$ \begin{align*}-1\cdot\chi_{[1-\eta_{0},+\infty)}(u)\leq\sum^{-j}_{\mathrm{l}=1}(-1)^{\mathrm{l}}\chi_{(-a^{\mathrm{l}},-a^{\mathrm{l}-1}]} \bigg(\frac{x}{a^{j}}-z\bigg)\leq \chi_{[a-\eta_{0},+\infty)}(z).\end{align*} $$

Then, by (4-9),

$$ \begin{align*} & \int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\sum^{-j-1}_{\mathrm{l}=1}(-1)^{\mathrm{l}} \chi_{(-a^{\mathrm{l}},-a^{\mathrm{l}-1}]} \bigg(\frac{x}{a^{j+1}}-z\bigg)\,dz \\[5pt] &\qquad+ \int_{\mathbb{R}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\sum^{-j}_{\mathrm{l}=1}(-1)^{\mathrm{l}} \chi_{(-a^{\mathrm{l}},-a^{\mathrm{l}-1}]} \bigg(\frac{x}{a^{j}}-z\bigg)\,dz \\[5pt] &\quad\geq -2\int^{\infty}_{1-\eta_{0}}K_{\alpha,t_{0}^{2\alpha}}^{-\Delta}(z)\,dz\geq \frac{\gamma-1}{2}. \end{align*} $$

So, for any $t\in [-r,r]$ and $\mathrm {J}_{0}\leq j\leq 0$ , combining (4-9) with (4-10) yields

$$ \begin{align*} |e^{-a_{j+1}(-\Delta)^{\alpha}}f(x)-e^{-a_{j}(-\Delta)^{\alpha}}f(x)| \geq C\frac{\gamma}{2}>0. \end{align*} $$

Choose a sequence $\{v_{j}\}_{j\in \mathbb {Z}}\in l^{p}(\mathbb {Z})$ , where $v_{j}=(-1)^{j+1}(-j)^{-{1}/{(p-\epsilon )}}$ . Then, for $N=(\mathrm {J}_{0},0)$ ,

$$ \begin{align*} \frac{1}{2r}\int_{[-r,r]}|T^{\ast}_{(-\Delta)^{\alpha}}f(x)|\,dx &\geq \frac{1}{2r}\int_{[-r,r]}|T^{(-\Delta)^{\alpha}}_{N}f(x)|\,dx \\[5pt] &\geq\frac{C\gamma}{4r}\int_{[-r,r]}\sum^{-1}_{j=\mathrm{J}_{0}}((-j)^{-{1}{(p-\epsilon)}})\,dx \\[5pt] &\geq C\cdot(-\mathrm{J}_{0})^{{1}/{(p-\epsilon)^{'}}}\sim \bigg(\log\frac{2}{r}\bigg)^{{1}/{(p-\epsilon)^{'}}}. \end{align*} $$

For (iii), let $v_{j}=(-1)^{j+1}, a_{j}=(t_{0}a^{j})^{2}$ and $0<\eta _{0}<1$ as fixed in the proof of (ii). Consider the same function f as in (ii). Then $\|v\|_{l^{\infty }(\mathbb {Z})}=1$ and $\|f\|_{L^{\infty }(\mathbb {R})}=1$ . By the same arguments as in (ii), with $N=(\mathrm {J}_{0},0)$ ,

$$ \begin{align*} \frac{1}{2r}\int_{[-r,r]}|T^{\ast}_{(-\Delta)^{\alpha}}f(x)|\,dx &\geq \frac{1}{2r}\int_{[-r.r]}|T^{(-\Delta)^{\alpha}}_{N}f(x)|\,dx\\&\geq \frac{C}{2r}\int_{[-r,r]}\sum^{0}_{j=\mathrm{J}_{0}}C_{1}\,dx \\[-2pt]&\geq C\cdot C_{1}\cdot(-\mathrm{J}_{0})\sim \log\frac{2}{r}.\\[-38.5pt] \end{align*} $$