Singular Integrals Associated with Zygmund Dilations

Abstract

The main purpose of this paper is to study multi-parameter singular integral operators which commute with Zygmund dilations. Motivated by some explicit examples of singular integral operators studied in Ricci and Stein (Ann Inst Fourier (Grenoble) 42:637–670, 1992), Fefferman and Pipher (Am J Math 11:337–369, 1997), and Nagel and Wainger (Am J Math 99:761–785, 1977), we introduce a class of singular integral operators on ${\mathbb {R}}^3$ associated with Zygmund dilations by providing suitable version of regularity conditions and cancellation conditions on convolution kernels, and then show the boundedness for this class of operators on $L^p$, $1<p<\infty$.

1 Introduction and Statement of Main Results

It is well known that Calderón and Zygmund [1] introduced certain convolution singular integral operators on ${\mathbb {R}}^n$ which generalize the Hilbert transform on ${\mathbb {R}}.$ They proved that if $T(f)=\mathcal {K}*f,$ where $\mathcal {K}$ is defined on ${\mathbb {R}}^n$ and satisfies the analogous estimates as $\frac{1}{x}$ does on ${\mathbb {R}}^1,$ namely

$$\begin{aligned} |\mathcal {K}(x)|\le \frac{C}{|x|^n},\quad \quad \quad |\nabla \mathcal {K}(x)|\le \frac{C}{|x|^{n+1}}, \end{aligned}$$

(1.1)

and

$$\begin{aligned} \int _{a<|x|<b}\mathcal {K}(x)\text {d}x=0\quad \text {for all}\ \ 0<a<b, \end{aligned}$$

(1.2)

then T is bounded on $L^p({\mathbb {R}}^n)$ for $1<p<\infty .$ The core of this theory is that the regularity and cancellation conditions are invariant with respect to the one-parameter family of dilations on ${\mathbb {R}}^n$ defined by $\delta (x_1, x_2, \ldots , x_n)= (\delta x_1, \ldots , \delta x_n)$, $\delta >0,$ in the sense that the kernel $\delta ^n{\mathcal {K}}(\delta x)$ satisfies the same conditions with the same bound as ${\mathcal {K}}(x)$.

On the other hand, the multiparameter theory of ${\mathbb {R}}^n$ began with Zygmund’s study of the strong maximal function defined by

$$\begin{aligned} {\mathcal {M}}_n(f)(x)=\sup _{R\ni x}{{1}\over {|R|}} \int _R |f(y)| dy, \end{aligned}$$

where R are the rectangles in ${\mathbb {R}}^n$ with sides parallel to the axes, and then continued with Marcinkiewicz’s proof of his multiplier theorem. If we consider the family of product dilations defined by $\delta (x_1, x_2, \ldots , x_n)=(\delta _1 x_1, \ldots , \delta _n x_n), \delta _i>0, i=1,..., n,$ then the strong maximal function and Marcinkiewicz’s multiplier are invariant under the product dilations. The multiparameter dilations are also associated with problems in the theory of differentiation of integrals. Jensen–Marcinkiewicz–Zygmund [17] proved that the strong maximal function in ${\mathbb {R}}^n$ is bounded from the Orlicz space $L(1+(\log ^+L)^{n-1})$ to weak $L^1$.

In [14], Fefferman and Stein generalized the singular integral operator theory to the product space. They took the space ${\mathbb {R}}^n\times {\mathbb {R}}^m$ along with the two-parameter family of dilations $(x, y)\mapsto (\delta _1 x, \delta _2 y), (x,y)\in {\mathbb {R}}^n\times {\mathbb {R}}^m, \delta _1,\delta _2>0.$ Those operators considered in [14] generalize the double Hilbert transform on ${\mathbb {R}}^2$ given by $H(f)=f*\frac{1}{xy}$ and are of the form $T(f)=\mathcal {K}*f,$ where the kernel $\mathcal {K}$ is characterized by the cancellation properties

$$\begin{aligned}&\int _{a<|x|<b}\mathcal {K}(x,y)\text {d}x=0\quad \text {for all}\ \ 0<a<b\ \text {and}\ y \in {\mathbb {R}}^m, \end{aligned}$$

(1.3)

$$\begin{aligned}&\int _{a<|y|<b}\mathcal {K}(x, y)dy =0\quad \text {for all}\ \ 0<a<b\ \text {and}\ x \in {\mathbb {R}}^n, \end{aligned}$$

(1.4)

and the regularity conditions

$$\begin{aligned} |\partial ^\alpha _{x} \partial ^\beta _{y} \mathcal {K}(x,y)| \le C_{\alpha , \beta }|x|^{-n-|\alpha |}|y|^{-m-|\beta |}, \quad \alpha ,\beta \ge 0. \end{aligned}$$

(1.5)

Under the conditions (1.3)–(1.5), Fefferman and Stein proved the $L^p, 1<p<\infty ,$ boundedness of the product convolution operators $T(f)={\mathcal {K}}*f.$ See [14] for more details. Note that the kernel ${\mathcal {K}}$ satisfying the conditions (1.3)–(1.5) is invariant with respect to the product dilation in the sense that the kernel $\delta _1^n\delta _2^m{\mathcal {K}}(\delta _1 x,\delta _2 y)$ satisfies conditions (1.3)–(1.5) with the same bound. For more discussions about the multiparameter product theory, see for example [2,3,4,5,6,7,8,9,10,11,12, 15, 16, 18,19,20,21, 27, 29] and in particular the survey article of Fefferman [12] for development in this area. For singular integrals with flag kernels, see [22, 24,25,26].

It has been widely considered that the next simplest multiparameter group of dilations after the product multiparameter dilations is the so-called the Zygmund dilation defined on ${\mathbb {R}}^3$ by $\rho _{s,t}(x_1,x_2,x_3) = (sx_1, tx_2, stx_3)$ for $s, t > 0$. Corresponding to this Zygmund dilation, one has the maximal function

$$\begin{aligned} {{\mathcal {M}}}_{\mathfrak {z}} f (x) = \sup _{R\ni x}\frac{1}{|R|}\int _{R}|f(u)|du, \end{aligned}$$

where supremum above is taken over all rectangles in ${\mathbb {R}}^3$ with sides parallel to the axes and side lengths of the form s, t, and st. As far as ${{\mathcal {M}}}_{\mathfrak {z}}$ is concerned, Stein was the first to link the properties of maximal operators associated with Zygmund dilations to boundary value problems for Poisson integrals on symmetric spaces, such as Siegel’s upper half space. We refer to the survey paper of Fefferman [10] on the future direction of research of multiparameter analysis on Zygmund dilations.

Besides ${{\mathcal {M}}}_{\mathfrak {z}}$, Ricci and Stein [28] introduced a class of singular integrals with more general dilations. One of those singular integrals is an explicit operator associated with Zygmund dilation of the form $T_{\mathfrak {z}}f=f* \mathcal {K}$, where

$$\begin{aligned} \mathcal {K}(x_1,x_2,x_3) =\sum _{k,j\in {{\mathbb {Z}}}}2^{-2(k+j)}\phi \Big ({{x_1}\over {2^j}},{{x_2}\over {2^k}},{{x_3}\over {2^{j+k}}}\Big ) \end{aligned}$$

(1.6)

and the function $\phi$ is supported in an unit cube in ${\mathbb {R}}^3$ with a certain amount of uniform smoothness and satisfies cancellation conditions

$$\begin{aligned} \int _{{{\mathbb {R}}}^2}\phi (x_1,x_2,x_3)\text {d}x_1 \text {d}x_2 =\int _{{{\mathbb {R}}}^2} \phi (x_1,x_2,x_3)\text {d}x_2 \text {d}x_3=\int _{{{\mathbb {R}}}^2}\phi (x_1,x_2,x_3)\text {d}x_3 \text {d}x_1=0. \end{aligned}$$

It was shown in [28] that $T_{\mathfrak {z}}$ is bounded on $L^p({\mathbb {R}}^3)$ for all $1<p< \infty .$ Particularly, as mentioned in [13], the above cancellation conditions are also necessary for the boundedness of the above-mentioned operators on $L^2({\mathbb {R}}^3).$ It is easy to see that if the dyadic Zygmund dilation is given by $(\delta _{{2^j},{2^k}}f)(x_1,x_2,x_3)=2^{2(j+k)} f(2^j x_1,2^k x_2, 2^{(j+k)} x_3),$ then we obtain that $(\delta _{{2^j},{2^k}}T_{\mathfrak {z}}(f))(x_1,x_2,x_3)= T_{\mathfrak {z}}(\delta _{{2^j},{2^k}}f)(x_1,x_2,x_3).$ This means that the operators studied by Ricci and Stein commute with Zygmund dilations of dyadic form. See [28] for more details. Fefferman and Pipher [13] further showed that $T_{\mathfrak {z}}$ is bounded in $L^p_w$ spaces for $1<p<\infty$ when the weight w satisfies an analogous condition of Muckenhoupt associated with Zygmund dilations.

Another explicit example related to the Zygmund dilation is an operator studied by Nagel–Wainger [23], which is the singular integrals along certain surfaces in ${\mathbb {R}}^3$, defined as $Tf=f* \mathcal {K}$, where

$$\begin{aligned} {\mathcal {K}}(x_1, x_2, x_3) = \text {sgn}(x_1 x_2) \bigg \{\frac{1}{|x_1|^{2} |x_2|^{2}+x_3^2}\bigg \}. \end{aligned}$$

(1.7)

Motivated by these specific operators with the convolution kernel as in (1.6) and (1.7), in the current paper we introduce a class of singular integral operators associated with Zygmund dilations by providing suitable version of regularity conditions and cancellation conditions for the convolution kernel and then show the boundedness for these operators on $L^p, 1<p<\infty$. We also provide another versions of cancellation conditions via normalized bump functions introduced by Stein [29].

To be more precise, suppose that $\mathcal {K}(x_1,x_2,x_3)$ is a function defined on ${\mathbb {R}}^3$ away from the union $\lbrace 0, x_2,x_3\rbrace \cup \lbrace x_1, 0,x_3\rbrace \cup \lbrace x_1, x_2,0\rbrace$. For integers $\alpha$, $\beta$, and $\gamma$ taking only values 0 or 1, we define

$$\begin{aligned} \Delta _{x_1,h_1}^\alpha \mathcal {K}(x_1,x_2,x_3)= & {} \alpha \mathcal {K}(x_1+h_1,x_2,x_3)- \mathcal {K}(x_1,x_2,x_3),\quad \alpha =0 \text { or } 1;\\ \Delta _{x_2,h_2}^\beta \mathcal {K}(x_1,x_2,x_3)= & {} \beta \mathcal {K}(x_1,x_2+h_2,x_3)- \mathcal {K}(x_1,x_2,x_3),\quad \beta =0 \text { or } 1; \end{aligned}$$

and

$$\begin{aligned} \Delta _{x_3,h_3}^\gamma \mathcal {K}(x_1,x_2,x_3)=\gamma \mathcal {K}(x_1,x_2,x_3+h_3)-\mathcal {K}(x_1,x_2,x_3),\quad \gamma =0 \text { or } 1. \end{aligned}$$

For simplicity, we denote $\Delta _{x_1,h_1}=\Delta _{x_1,h_1}^1,$$\Delta _{x_2,h_2}=\Delta _{x_2,h_2}^1$ and $\Delta _{x_3,h_3}=\Delta _{x_3,h_3}^1$.

The “regularity” conditions considered in this paper are characterized by

$$\begin{aligned} |\Delta _{x_1,h_1}^\alpha \Delta _{x_2,h_2}^\beta \Delta _{x_3,h_3}^\gamma \mathcal {K}(x_1,x_2,x_3)| \le \frac{C|{h_1}|^{{\alpha \theta _1}} |{h_2}|^{{\beta \theta _1}} |{h_3}|^{{\gamma \theta _1}}}{|x_1|^{{\alpha \theta _1}+1} |x_2|^{{\beta \theta _1}+1} |x_3|^{{\gamma \theta _1}+1} \big (|\frac{x_1x_2}{x_3}|+|\frac{x_3}{x_1x_2}| \big )^{{\theta _2}}} \end{aligned}$$

(R)

for all $0 \le \alpha \le 1, 0 \le \beta +\gamma \le 1$ or $0 \le \alpha +\gamma \le 1, 0 \le \beta \le 1,$ and $|x_1|\ge 2 |{h_1}|>0, |x_2|\ge 2 |{h_2}|>0, |x_3|\ge 2 |{h_3}|>0$, ${h_1},{h_2},{h_3}\in {\mathbb {R}}$ and some $0<{\theta _1}\le 1,0<{\theta _2}<1.$

Note that for any fixed non-zero two variables, say, $x_1\not =0$ and $x_2\not =0,$$\mathcal {K}(x_1,x_2,x_3)$ is an integrable function with respect to the variable $x_3$ and the resulting integral $\widetilde{K}(x_1,x_2)=\int _{{\mathbb {R}}}{\mathcal {K}}(x_1,x_2,x_3)\text {d}x_3,$ as a kernel on ${\mathbb {R}}^2,$ satisfies the regularity conditions of the classical product kernel on ${\mathbb {R}}^2$ as studied by Fefferman and Stein in [14]. These facts, as mentioned above, can also be easily checked for singular integral operator $T_{{\mathfrak {z}}}$.

In this paper, we consider three kinds of cancellation conditions. The first one is

$$\begin{aligned} \bigg |\int _{\delta _1 \le |x_1| \le r_1}\int _{\delta _2 \le |x_2| \le r_2}\int _{\delta _3 \le |x_3| \le r_3} \mathcal {K}(x_1,x_2,x_3)\text {d}x_1\text {d}x_2\text {d}x_3 \bigg | \le C \end{aligned}$$

(C1.a)

uniformly for all $\delta _1,\delta _2,\delta _3,r_1,r_2,r_3>0;$

$$\begin{aligned} \bigg |\int _{\delta _1 \le |x_1| \le r_1}\int _{\delta _2 \le |x_2| \le r_2} \Delta _{x_3,h_3}^\gamma \mathcal {K}(x_1,x_2,x_3)\text {d}x_1\text {d}x_2 \bigg | \le \frac{C|{h_3}|^{\gamma \theta _1}}{|x_3|^{\gamma {\theta _1}+1}} \end{aligned}$$

(C1.b)

uniformly for all $\delta _1,\delta _2,r_1,r_2>0,$$|x_3|\ge 2|{h_3}|>0$ and $0 \le \gamma \le 1;$

$$\begin{aligned} \bigg |\int _{\delta _2 \le |x_2| \le r_2}\int _{\delta _3 \le |x_3| \le r_3} \Delta _{x_1,h_1}^\alpha \mathcal {K}(x_1,x_2,x_3)\text {d}x_2 \text {d}x_3 \bigg | \le \frac{C|{h_1}|^{{\alpha \theta _1}}}{|x_1|^{{\alpha \theta _1}+1}} \end{aligned}$$

(C1.c)

uniformly for all $\delta _2,\delta _3,r_2,r_3>0,$$|x_1| \ge 2|{h_1}|>0$ and $0 \le \alpha \le 1;$

$$\begin{aligned} \bigg |\int _{\delta _1 \le |x_1| \le r_1}\int _{\delta _3 \le |x_3| \le r_3} \Delta _{x_2,h_2}^\beta \mathcal {K}(x_1, x_2, x_3)\text {d}x_3\text {d}x_1\bigg | \le \frac{C|{h_2}|^{\beta \theta _1}}{|x_2|^{{\beta \theta _1}+1}} \end{aligned}$$

(C1.d)

uniformly for all $\delta _1,\delta _3,r_1,r_3>0,$$|x_2| \ge 2|{h_2}|>0$ and $0 \le \beta \le 1.$

The regularity conditions (RR) and the cancellation conditions (C1.a)–(C1.d) imply the following $L^2$ boundedness.

Theorem 1.1

Suppose that $\mathcal {K}$ is a function defined on ${\mathbb {R}}^3$ and satisfies the conditions (RR) and (C1.a)–(C1.d). Set ${\mathcal {K}}_{\epsilon }^N(x_1,x_2,x_3)={\mathcal {K}}(x_1,x_2,x_3)$ if $\epsilon _1 \le |x_1| \le N_1, \epsilon _2 \le |x_2| \le N_2$ and $\epsilon _3\le |x_3| \le N_3$ and ${\mathcal {K}}_{\epsilon }^N(x_1,x_2,x_3)=0$ otherwise, where $\epsilon =(\epsilon _1,\epsilon _2,\epsilon _3)$ and $N=(N_1,N_2,N_3)$ for all $0<\epsilon _1\le N_1<\infty ,\epsilon _2\le N_2<\infty ,$ and $\epsilon _3\le N_3<\infty .$ Then, the operator ${\mathcal {K}}_{\epsilon }^N*f$ is bounded on $L^2({\mathbb {R}}^3)$ and moreover,

$$\begin{aligned} \Vert {\mathcal {K}}_{\epsilon }^N*f\Vert _{L^2({\mathbb {R}}^3)}\le A\Vert f\Vert _{L^2({\mathbb {R}}^3)} \end{aligned}$$

where the constant A depends only on the constant C but not on $\epsilon _1,\epsilon _2, \epsilon _3, N_1,N_2$ and $N_3.$

From Theorem 1.1, we will deduce the existence of the corresponding singular integrals in the $L^2$ norm as a limit of the truncated integrals.

Corollary 1.2

Suppose that ${\mathcal {K}}$ is a function defined on ${\mathbb {R}}^3$ and satisfies the conditions (RR), (C1.a)–(C1.d) and, in addition, the three integrals

$$\begin{aligned}&\int _{|x_3|\le 1}\int _{|x_2|\le 1}\int _{|x_1|\le 1} K_{\epsilon }^N(x_1,x_2,x_3)\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3,\\&\int _{|x_3|\le 1}\int _{|x_2|\le 1}{\mathcal {K}}_{\epsilon }^N(x_1,x_2,x_3)\mathrm{d}x_2\mathrm{d}x_3,\\&\int _{|x_3|\le 1}\int _{|x_1|\le 1}{\mathcal {K}}_{\epsilon }^N(x_1,x_2,x_3)\mathrm{d}x_1\mathrm{d}x_3 \end{aligned}$$

converge almost everywhere as $\epsilon _1,\epsilon _2,\epsilon _3\rightarrow 0$ and $N_1,N_2,N_3\rightarrow \infty .$ Then the limit $\lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3\rightarrow 0 \\ N_1,N_2,N_3\rightarrow \infty \end{array}} {\mathcal {K}}_{\epsilon }^N*f={\mathcal {K}}*f$ exists in the $L^2({\mathbb {R}}^3)$ norm. Moreover,

$$\begin{aligned} \Vert \mathcal {K}*f\Vert _{L^2({\mathbb {R}}^3)}\le A\Vert f\Vert _{L^2({\mathbb {R}}^3)} \end{aligned}$$

with the constant A depending only on the constant C.

We remark in advance that the proof of Corollary 1.2 indeed implies that $\lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3\rightarrow 0 \\ N_1,N_2,N_3\rightarrow \infty \end{array}} {\mathcal {K}}_{\epsilon }^N*f$ exists in the $L^p, 1<p<\infty ,$ norm for smooth functions f having compact support. This fact leads to the study of the $L^p, p\not =2,$ boundedness of the operator ${\mathcal {K}}*f.$ For this purpose, we need the second kind of the cancellation conditions which are somewhat stronger than the first ones. They are given by

$$\begin{aligned} \bigg |\int _{\delta _3 \le |x_3| \le r_3}\int _{\delta _2 \le |x_2| \le r_2}\int _{\delta _1 \le |x_1| \le r_1} \mathcal {K}(x_1,x_2,x_3)\text {d}x_1\text {d}x_2\text {d}x_3 \bigg | \le C \end{aligned}$$

(C2.a)

uniformly for all $\delta _1,\delta _2,\delta _3,r_1,r_2,r_3>0;$

$$\begin{aligned}&\bigg |\int _{\delta _1 \le |x_1| \le r_1} \Delta _{x_2,h_2}^\beta \Delta _{x_3,h_3}^\gamma \mathcal {K}(x_1,x_2,x_3)\text {d}x_1 \bigg |\\&\quad \le \frac{C|{h_2}|^{\beta \theta _1}|{h_3}|^{\gamma \theta _1}}{|x_2|^{{\beta \theta _1}+1} |x_3| ^{{\gamma \theta _1}+1}}\left( \frac{1}{\big (|\frac{r_1x_2}{x_3}|+|\frac{x_3}{r_1x_2}| \big )^{{\theta _2}}}+\frac{1}{\big (|\frac{\delta _1 x_2}{x_3}|+|\frac{x_3}{\delta _1 x_2}| \big )^{{\theta _2}}}\right) \end{aligned}$$

(C2.b)

for all $\delta _1, r_1>0, 0 \le \beta +\gamma \le 1,$$|x_2|\ge 2|{h_2}|>0, |z| \ge 2|{h_3}|>0;$

$$\begin{aligned} \bigg |\int _{\delta _3 \le |x_3| \le r_3}\int _{\delta _2 \le |x_2| \le r_2}\Delta _{x_1,h_1}^\alpha \mathcal {K}(x_1,x_2,x_3)\text {d}x_2 \text {d}x_3 \bigg | \le \frac{C|h_1|^{\alpha \theta _1}}{|x_1|^{{\alpha \theta _1}+1}} \end{aligned}$$

(C2.c)

uniformly for all $\delta _2,\delta _3,r_2,r_3>0,$$|x_1| \ge 2|{h_1}|>0$ and $0 \le \alpha \le 1.$ Or

uniformly for all $\delta _1,\delta _2,\delta _3,r_1,r_2,r_3>0;$

for all $\delta _2, r_2>0, 0 \le \alpha +\gamma \le 1$, $|x_1| \ge 2|h_1|>0$ and $|x_3| \ge 2|h_3|>0;$

uniformly for all $\delta _1,\delta _3,r_1,r_3>0,$$|x_2|\ge 2|h_2|>0$ and $0 \le \beta \le 1.$

We would like to point out that the condition (C2.b) implies (C1.b) and (C1.d) while (C$2^\prime .$b) implies (C1.b) and (C1.c), and all the above regularity and cancellation conditions are invariant with respect to the Zygmund dilation in the sense that the kernel $\delta ^2_1\delta ^2_2\mathcal {K}(\delta _1 x_1, \delta _2 x_2, \delta _1\delta _2 x_3)$ satisfies the same conditions with the exactly same bounds as $\mathcal {K}(x_1,x_2,x_3)$.

The $L^p$ estimate then is given by the following

Theorem 1.3

Suppose that ${\mathcal {K}}$ is a function defined on ${\mathbb {R}}^3$ and satisfies the conditions (RR) and (C2.a)–(C2.c) (or (RR), (C$2^\prime .$a)–(C$2^\prime$.c)) and in addition the three integrals

$$\begin{aligned}&\int _{|x_3|\le 1}\int _{|x_2|\le 1}\int _{|x_1|\le 1}K_{\epsilon }^N(x_1,x_2,x_3)\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3,\\&\int _{|x_3|\le 1}\int _{|x_2|\le 1}{\mathcal {K}}_{\epsilon }^N(x_1,x_2,x_3)\mathrm{d}x_2\mathrm{d}x_3,\\&\int _{|x_3|\le 1}\int _{|x_1|\le 1} {\mathcal {K}}_{\epsilon }^N(x_1,x_2,x_3)\mathrm{d}x_1\text {d}x_3 \end{aligned}$$

converge almost everywhere as $\epsilon _1,\epsilon _2,\epsilon _3\rightarrow 0$ and $N_1,N_2,N_3\rightarrow \infty .$ Then the operator

$$\begin{aligned} {\mathcal {K}}*f:=\lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3\rightarrow 0 \\ N_1,N_2,N_3\rightarrow \infty \end{array}} {\mathcal {K}}_{\epsilon }^N*f \end{aligned}$$

defined initially on $L^2\cap L^p,1<p<\infty ,$ extends to a bounded operator on $L^p({\mathbb {R}}^3)$ and moreover,

$$\begin{aligned} \Vert \mathcal {K}*f\Vert _{L^p({\mathbb {R}}^3)}\le A\Vert f\Vert _{L^p({\mathbb {R}}^3)} \end{aligned}$$

with the constant A depending only on the constant C.

In many applications, singular integral operators are of the form ${\mathcal {K}}*f$ where ${\mathcal {K}}$ is a distribution that equals a function $\mathcal {K}$ on ${\mathbb {R}}^3$ away from the union $\lbrace 0, x_2,x_3\rbrace \cup \lbrace x_1, 0,x_3\rbrace \cup \lbrace x_1, x_2,0\rbrace$ and satisfy certain regularity and cancellation conditions. For this purpose, we begin with recalling the bump functions introduced by Stein in [29]. A normalized bump function (n.b.f.) is a smooth function $\phi$ supported on the unit ball and is bounded by a fixed constant together with its gradient. The third kind of the cancellation conditions considered in this paper is characterized by

$$\begin{aligned} \bigg |\iiint \mathcal {K}(x_1,x_2,x_3)\phi (R_1 x_1, R_2 x_2, R_1R_2 x_3)\text {d}x_1\text {d}x_2\text {d}x_3 \bigg | \le C \end{aligned}$$

(C3.a)

for every n.b.f. $\phi$ on ${\mathbb {R}}^3$ and all $R_1,R_2>0;$

$$\begin{aligned} \bigg |\int \Delta _{x_2,h_2}^\beta \Delta _{x_3,h_3}^\gamma \mathcal {K}(x_1,x_2,x_3)\phi (R x_1)\text {d}x_1\bigg | \le \frac{C|h_2|^{\beta \theta _1}|h_3|^{\gamma \theta _1}}{|x_2|^{{\beta \theta _1}+1} |x_3|^{{\gamma \theta _1}+1} \big (\big |\frac{R x_3}{x_2}\big |+\big |\frac{x_2}{R x_3}\big | \big )^{{\theta _2}}} \end{aligned}$$

(C3.b)

for all $0 \le \beta +\gamma \le 1$, every n.b.f. $\phi$ on ${\mathbb {R}},$$|x_2| \ge 2|h_2|>0,$$|x_3| \ge 2|h_3|>0$ and all $R>0;$

$$\begin{aligned} \bigg |\iint \Delta _{x_1,h_1}^\alpha \mathcal {K}(x_1,x_2,x_3)\phi (R_1 x_2, R_2x_3)\text {d}x_2\text {d}x_3\bigg | \le \frac{C|h_1|^{\alpha \theta _1}}{|x_1|^{{\alpha \theta _1}+1}} \end{aligned}$$

(C3.c)

for all $0\le \alpha \le 1$, every n.b.f. $\phi$ on ${\mathbb {R}}^2$, $|x_1| \ge 2|h_1|>0$ and all $R_1,R_2>0.$ Or

for every n.b.f. $\phi$ on ${\mathbb {R}}^3$ and all $R_1,R_2>0;$

for all $0 \le \alpha +\gamma \le 1$, every n.b.f. $\phi$ on ${\mathbb {R}}$, $|x_1| \ge 2|h_1|>0, |x_3| \ge 2|h_3|>0$ and all $R>0;$

for all $0\le \beta \le 1$, every n.b.f. $\phi$ on ${\mathbb {R}}^2$, $|x_2| \ge 2|h_2|>0$ and all $R_1,R_2>0.$

Theorem 1.4

Let all the notation be the same as above.

• (a) Suppose that $\mathcal {K}$ is a distribution that equals a function on ${\mathbb {R}}^3$ away from the union $\lbrace 0, x_2,x_3\rbrace \cup \lbrace x_1, 0,x_3\rbrace \cup \lbrace x_1, x_2,0\rbrace$ and satisfies conditions (RR) and (C3.a)–(C3.c) (or (RR), (C$3^\prime .$a)–(C$3^\prime .$c)). Then, the operator ${\mathcal {K}}*f$ is bounded on $L^p({\mathbb {R}}^3),1<p<\infty ;$ moreover,

  $$\begin{aligned} \Vert \mathcal {K}*f\Vert _{L^p({\mathbb {R}}^3)}\le A\Vert f\Vert _{L^p({\mathbb {R}}^3)} \end{aligned}$$

  with the constant A depending only on the constant C.

• (b) Suppose that $\mathcal {K}$ is a distribution that equals a function on ${\mathbb {R}}^3$ away from the union $\lbrace 0, x_2,x_3\rbrace \cup \lbrace x_1, 0,x_3\rbrace \cup \lbrace x_1, x_2,0\rbrace$ and satisfies conditions (RR) and (C2.a)–(C2.c) (or (RR), (C$2^\prime .$a)–(C$2^\prime .$c)). Then, the operator ${\mathcal {K}}*f$ is bounded on $L^p({\mathbb {R}}^3),1<p<\infty ,$ and

  $$\begin{aligned} \Vert \mathcal {K}*f\Vert _{L^p({\mathbb {R}}^3)}\le A\Vert f\Vert _{L^p({\mathbb {R}}^3)} \end{aligned}$$

  with the constant A depending only on the constant C.

Remark 1.5

We would like to point out that all regularity and cancellation conditions given above are invariant with respect to Zygmund dilations. Moreover, The boundedness results in this paper can be extended to higher dimensions. The consideration of regularity and cancellation conditions in this paper leads naturally to the study of non-convolution singular integral operators which are associated with Zygmund dilations. We will discuss all these topics in the forthcoming works.

Theorem 1.6

Let all the notation be the same as above.

• (i) The singular integral operator $T_{\mathfrak {z}}$ given in [28] with the convolution kernel as in (1.6) can be written as $T_{{\mathfrak {z}}} =T_{{\mathfrak {z}},1}+T_{{\mathfrak {z}},2}$, where $T_{{\mathfrak {z}},i}$ is a singular integral operator with convolution kernel ${\mathcal {K}}_i$, $i=1,2$. ${\mathcal {K}}_1$ satisfies conditions (RR) and (C2.a)–(C2.c), and ${\mathcal {K}}_2$ satisfies conditions (RR), (C$2^\prime .$a)–(C$2^\prime .$c)).

• (ii) The singular integral operator T given in [23] with the convolution kernel as in (1.7) satisfy conditions (RR) and (C2.a)–(C2.c).

Remark 1.7

Note that for the operator T from [23] with the convolution kernel as in (1.7), only the $L^2$ boundedness is proved. Hence, as a consequence of Theorem 1.3, T is bounded on $L^p({\mathbb {R}}^3)$ for all $1<p<\infty .$

The organization of this paper is as follows: In the next section, we will show the $L^2$ boundedness for singular integral operators associated with Zygmund dilations, namely Theorem 1.1 and Corollary 1.2. The $L^p$ boundedness, namely Theorems 1.3 and 1.4, will be proved in Sect. 3. In the last section, we prove Theorem 1.6.

2 $L^2$ Boundedness

The main task of this section is to provide proofs of Theorem 1.1 and Corollary 1.2. Before proving Theorem 1.1, we first prove the following simple result which will be used frequently below.

Lemma 2.1

For any $f(x)\in L_{\text {loc}}^1({\mathbb {R}})$ and $N>8$, we have

$$\begin{aligned}&\Big |\int _{8 \le |x| \le N} f(x)e^{-ix}\mathrm{d}x\Big | \le {\frac{1}{2}}\int _{E_N} |f(x)|\mathrm{d}x +{\frac{1}{2}}\int _{8 \le |x| \le N} |f(x)-f(x+\pi )|\mathrm{d}x, \end{aligned}$$

where $E_N= \{x:4 \le |x| \le 12 \} \cup \{x:N-\pi \le |x| \le N+\pi \}$.

Proof

We write

$$\begin{aligned} \int _{8 \le |x| \le N} f(x)e^{-ix}\text {d}x= & {} \int _{8 \le |x+\pi | \le N} f(x+\pi )e^{-i(x+\pi )}\text {d}x\\= & {} -\int _{8 \le |x+\pi | \le N} f(x+\pi )e^{-ix}\text {d}x. \end{aligned}$$

Therefore,

$$\begin{aligned} \Big |\int _{8 \le |x| \le N} f(x)e^{-ix}\text {d}x\Big |= & {} {\frac{1}{2}}\Big |\int _{8 \le |x| \le N} f(x)e^{-ix}\text {d}x-\int _{8 \le |x+\pi | \le N} f(x+\pi )e^{-ix}\text {d}x \Big |\\\le & {} {\frac{1}{2}}\Big |\int _{8 \le |x| \le N} (f(x)-f(x+\pi ))e^{-ix}\text {d}x\Big |\\&+{\frac{1}{2}}\Big |\int _{\{x:8 \le |x| \le N\} \setminus \{x:8 \le |x+\pi | \le N\}} f(x+\pi )e^{-ix}\text {d}x \Big |\\&+ {\frac{1}{2}}\Big |\int _{\{x:8 \le |x+\pi | \le N\} \setminus \{x:8 \le |x| \le N\}} f(x+\pi )e^{-ix}\text {d}x \Big |\\= & {} {\frac{1}{2}}\Big |\int _{8 \le |x| \le N} (f(x)-f(x+\pi ))e^{-ix}\text {d}x\Big |\\&+{\frac{1}{2}}\Big |\int _{\{x:8 \le |x-\pi | \le N\} \setminus \{x:8 \le |x| \le N\}} f(x)e^{-ix}\text {d}x \Big |\\&+ {\frac{1}{2}}\Big |\int _{\{x:8 \le |x| \le N\} \setminus \{x:8 \le |x-\pi | \le N\}} f(x)e^{-ix}\text {d}x \Big |\\\le & {} {\frac{1}{2}}\int _{8 \le |x| \le N} |f(x)-f(x+\pi )|\text {d}x+{\frac{1}{2}}\int _{E_N} |f(x)|\text {d}x \end{aligned}$$

and Lemma 2.1 follows. $\square$

We now prove Theorem 1.1.

Proof of Theorem 1.1

By the Plancherel theorem, the $L^2$ boundedness of ${\mathcal {K}}_\epsilon ^N*f$ is equivalent to $|\widehat{{\mathcal {K}}_\epsilon ^N}(\chi ,\eta ,\xi )|$$\le A,$ where $\widehat{{\mathcal {K}}_\epsilon ^N}$ is the Fourier transform of ${\mathcal {K}}_\epsilon ^N$ and A is the constant depending only on the constant C but not on $\epsilon =(\epsilon _1,\epsilon _2,\epsilon _3)$ and $N=(N_1,N_2,N_3).$ To obtain such an estimate, we may assume that $\chi$ and $\eta$ are both positive. Note that,

$$\begin{aligned} \widehat{{\mathcal {K}}_\epsilon ^N}(\chi ,\eta ,\xi )= & {} \int _{\epsilon _3\le |x_3|\le N_3}\int _{\epsilon _2 \le |x_2|\le N_1}\int _{\epsilon _1 \le |x_1|\le N_1}\mathcal {K}(x_1,x_2,x_3)e^{-ix\chi }e^{-iy\eta }e^{-iz\xi }\text {d}x_1\text {d}x_2\text {d}x_3\\= & {} \int _{\frac{\epsilon _3}{\chi \eta } \le |x_3|\le \frac{N_3}{\chi \eta }} \int _{\frac{\epsilon _2}{\eta } \le |x_2|\le \frac{N_2}{\eta }}\int _{\frac{\epsilon _1}{\chi } \le |x_1|\le \frac{N_1}{\chi }} \frac{1}{\chi ^2\eta ^2}\mathcal {K}\Big (\frac{x_1}{\chi },\frac{x_2}{\eta }, \frac{x_3}{\chi \eta }\Big )\\&\quad \cdot e^{-ix_1}e^{-ix_2}e^{-ix_3\xi /(\chi \eta )}\text {d}x_1\text {d}x_2\text {d}x_3. \end{aligned}$$

As we remarked above that the assumptions on $\mathcal {K}$ are invariant in the sense that $\delta ^2_1\delta ^2_2 \mathcal {K}(\delta _1x_1,\delta _2 x_2,$$\delta _1\delta _2 x_3)$ satisfies the same assumptions as $\mathcal {K}$ with the same constant C, independent of $\delta _1, \delta _2>0.$ Thus $\frac{1}{\chi ^2\eta ^2}\mathcal {K}(\frac{x_1}{\chi },\frac{x_2}{\eta }, \frac{x_3}{\chi \eta })$ satisfies all conditions (RR) and (C1.a)–(C1.d) with the same bounds uniformly for $\chi ,\eta .$ Therefore, it suffices to show that $\widehat{{\mathcal {K}}_\epsilon ^N}(1,1,\xi )$ is a bounded function uniformly for $0<\epsilon _1,\epsilon _2, \epsilon _3, N_1,N_2,N_3<\infty .$ To do this, for simplicity, we set $\epsilon _4=\epsilon _3|\xi |$ and $N_4=N_3|\xi |$. Without loss of generality, we may assume that $\epsilon _1,\epsilon _2, \epsilon _4 \le 8 \le N_1,N_2,N_4$ since all other cases can be written as a finite linear combination of these cases and can be handled similarly.

The bound of $\widehat{{\mathcal {K}}_\epsilon ^N}(1,1,\xi )$ follows from the regularity and cancellation conditions on ${\mathcal {K}}.$ More precisely, we write

$$\begin{aligned} \widehat{{\mathcal {K}}_\epsilon ^N}(1,1,\xi )&= \int _{\epsilon _3 \le |x_3| \le N_3} \int _{\epsilon _2 \le |x_2| \le N_2}\int _{\epsilon _1 \le |x_1| \le N_1}\\&\quad \mathcal {K}(x_1,x_2,x_3) e^{-ix_1}e^{-ix_2}e^{-ix_3\xi }\text {d}x_1\text {d}x_2\text {d}x_3\\&=\int _{\epsilon _4 \le |x_3| \le N_4}\int _{\int _{\epsilon _2 \le |x_2| \le N_2}} \int _{\epsilon _1 \le |x_1| \le N_1}\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\\&\quad \times e^{-ix_1}e^{-ix_2}e^{-ix_3}\text {d}x_1\text {d}x_2\text {d}x_3 :=I+II, \end{aligned}$$

where I is the result of integrating over the set $\big \{ 8\le |x_1|\le N_1, \epsilon _2 \le |x_2|\le N_1, \epsilon _4 \le |x_3|\le N_4\big \}$ and II over the set $\big \{ \epsilon _1 \le |x_1|<8,\epsilon _2 \le |x_2|\le N_1, \epsilon _4 \le |x_3|\le N_4\big \}.$

For term I,  using Lemma 2.1 with $f(x_1)=\int _{\epsilon _2 \le |x_2| \le N_2}\int _{\epsilon _4 \le |x_3| \le N_4}\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )$$e^{-ix_2} \cdot e^{-ix_3 }\text {d}x_3\text {d}x_2$, we obtain

$$\begin{aligned} |I|&\lesssim \int _{E_{N_1}}\bigg |\int _{\epsilon _2 \le |x_2| \le N_2}\int _{\epsilon _4 \le |x_3| \le N_4}\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}e^{-ix_3 }\text {d}x_3\text {d}x_2\bigg | \text {d}x_1\\&\quad + \int _{8 \le |x_1| \le N_1} \bigg |\int _{\epsilon _2 \le |x_2| \le N_2}\int _{\epsilon _4 \le |x_3| \le N_4}\Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big )\\&\quad \times e^{-ix_2}e^{-ix_3 }\text {d}x_3\text {d}x_2\bigg |\text {d}x_1\\&:= I_1+ I_2. \end{aligned}$$

We write $I_1$ by

$$\begin{aligned} I_1&\le \int _{E_{N_1}}\bigg | \int _{8 \le |x_2| \le N_2} \int _{\epsilon _4 \le |x_3| \le N_4}\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}e^{-ix_3 }\text {d}x_3\text {d}x_2\bigg | \text {d}x_1\\&\quad +\int _{E_{N_1}}\bigg | \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _4 \le |x_3| \le N_4}\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}e^{-ix_3 }\text {d}x_3\text {d}x_2\bigg | \text {d}x_1\\&:=I_{1,1}+I_{1,2}. \end{aligned}$$

To estimate term $I_{1,1},$ using Lemma 2.1 with $f(x_2)=\int _{\epsilon _4 \le |x_3| \le N_4}\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )$$e^{-ix_3 }\text {d}x_3$ we get

$$\begin{aligned} {|I_{1,1}|}&\lesssim \int _{E_{N_1}} \int _{E_{N_2}}\bigg | \int _{\epsilon _4 \le |x_3| \le N_4}\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_3 }\text {d}x_3 \bigg | \text {d}x_2\text {d}x_1\\&\quad +\int _{E_{N_1}} \int _{8 \le |x_2| \le N_2} \bigg | \int _{\epsilon _4 \le |x_3| \le N_4}\Delta _{x_2,\pi } \Big ( \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) e^{-ix_3 }\text {d}x_3 \bigg | \text {d}x_2\text {d}x_1\\&\lesssim \int _{E_{N_1}} \int _{E_{N_2}} \int _{\epsilon _4 \le |x_3| \le N_4}\frac{1}{|x| |x_2| |x_3| \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}}\text {d}x_3 \text {d}x_2\text {d}x\\&\quad +\int _{E_{N_1}} \int _{8 \le |x_2| \le N_2} \int _{\epsilon _4 \le |x_3| \le N_4}\frac{1}{|x_1| |x_2|^{{\theta _1}+1} |x_3| \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}}\text {d}x_3 \text {d}x_2\text {d}x_1 \\&\lesssim 1, \end{aligned}$$

where we use the condition (RR) above on $\mathcal {K}$ with $\alpha =\beta =\gamma =0$ and $\alpha =0, \beta =1,\gamma =0,$ respectively.

To handle term $I_{1,2}$, we write

$$\begin{aligned} I_{1,2}&\le \int _{E_{N_1}}\bigg | \int _{\epsilon _2 \le |x_2| \le 8} \int _{8 \le |x_3| \le N_4} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}e^{-ix_3 }\text {d}x_3\text {d}x_2\bigg | \text {d}x_1\\&\quad + \int _{E_{N_1}}\bigg | \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}e^{-ix_3 }\text {d}x_3\text {d}x_2\bigg | \text {d}x_1\\&:=I_{1,2,1}+I_{1,2,2}. \end{aligned}$$

By Lemma 2.1 with $f(x_3)=\int _{\epsilon _2 \le |x_2| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2} \text {d}x_2$, we get

$$\begin{aligned} |I_{1,2,1}|&\lesssim \int _{E_{N_1}} \int _{E_{N_2}}\bigg | \int _{\epsilon _2 \le |x_2| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}\text {d}x_2\bigg | \text {d}x_3\text {d}x_1\\&\quad + \int _{E_{N_1}} \int _{8 \le |x_3| \le N_4} \bigg | \int _{\epsilon _2 \le |x_2| \le 8} \Delta _{x_3,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) e^{-ix_2}\text {d}x_2\bigg | \text {d}x_3\text {d}x_1\\&\lesssim \int _{E_{N_1}} \int _{E_{N_2}} \int _{\epsilon _2 \le |x_2| \le 8} \frac{1}{|x_1| |x_2| |x_3| \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}} \text {d}x_2 \text {d}x_3\text {d}x_1\\&\quad + \int _{E_{N_1}} \int _{8 \le |x_3| \le N_4} \int _{\epsilon _2 \le |x_2| \le 8} \frac{1}{|x_1| |x_2| |x_3|^{1+{\theta _1}} \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}} \text {d}x_2 \text {d}x_3 \text {d}x_1 \\&\lesssim 1, \end{aligned}$$

where we use the regularity condition (RR) above with $\alpha =\beta =\gamma =0$ and $\alpha =\beta =0, \gamma =1,$ respectively.

To estimate $I_{1,2,2},$ we note that

$$\begin{aligned} I_{1,2,2}&\le \int _{E_{N_1}}\bigg | \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_2}e^{-ix_3 }-1)\text {d}x_3\text {d}x_2\bigg | \text {d}x_1\\&\quad +\int _{E_{N_1}}\bigg | \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\text {d}x_3\text {d}x_2\bigg | \text {d}x_1 \\&\lesssim \int _{E_{N_1}} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \frac{|x_2|+|x_3|}{|x_1| |x_2| |x_3| \big (|\frac{x_1 x_2\xi }{x_3}|+|\frac{x_3}{x_1 x_2\xi }| \big )^{{\theta _2}}}\text {d}x_3\text {d}x_2\text {d}x_1 \\&\quad + \int _{E_{N_1}} \frac{1}{|x_1|} \text {d}x_1 \lesssim 1, \end{aligned}$$

where we use the condition (RR) with $\alpha =\beta =\gamma =0,$ and the cancellation condition (C1.c) with $\alpha =0.$

Next we consider $I_2$. Set $I_{2,1}$ and $I_{2,2}$ by

$$\begin{aligned} I_{2,1}= & {} \int _{8 \le |x_1| \le N_1} \bigg |\int _{8 \le |x_2| \le N_2} \int _{\epsilon _4 \le |x_3| \le N_4}\Delta _{x_1,\pi } \\&\times \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) e^{-ix_2}e^{-ix_3 }\text {d}x_3\text {d}x_2\bigg |\text {d}x_1 \end{aligned}$$

and

$$\begin{aligned} I_{2,2}= & {} \int _{8 \le |x_1| \le N_1} \bigg |\int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _4 \le |x_3| \le N_4}\Delta _{x_1,\pi } \\&\times \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) e^{-ix_2}e^{-ix_3 }\text {d}x_3\text {d}x_2\bigg |\text {d}x_1. \end{aligned}$$

Then $I_2\le |I_{2,1}|+|I_{2,2}|.$ Similarly, applying Lemma 2.1 with

$$\begin{aligned} f(x_2)= \int _{\epsilon _4 \le |x_3| \le N_4}\Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) e^{-ix_3 }\text {d}x_3, \end{aligned}$$

we obtain

$$\begin{aligned} |I_{2,1}|&\lesssim \int _{8 \le |x_1| \le N_1} \int _{E_{N_2}}\bigg | \int _{\epsilon _4 \le |x_3| \le N_4}\Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) e^{-ix_3 }\text {d}x_3\bigg |\text {d}x_2\text {d}x_1\\&\quad +\int _{8 \le |x_1| \le N_1} \int _{8 \le |x_2| \le N_2} \bigg | \int _{\epsilon _4 \le |x_3| \le N_4}\Delta _{x_2,\pi } \Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) \\&\quad \times e^{-ix_3 }\text {d}x_3\bigg |\text {d}x_2\text {d}x_1\\&\lesssim \int _{8 \le |x_1| \le N_1} \int _{E_{N_2}}\int _{\epsilon _4 \le |x_3| \le N_4}\frac{1}{|x_1|^{1+{\theta _1}} |x_2| |x_3| \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}}\text {d}x_3\text {d}x_2\text {d}x_1\\&\quad + \int _{8 \le |x_1| \le N_1} \int _{8 \le |x_2| \le N_2} \int _{\epsilon _4 \le |x_3| \le N_4}\frac{1}{|x_1|^{1+{\theta _1}} |x_2|^{1+{\theta _1}} |x_3| \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}}\\&\quad \times \text {d}x_3\text {d}x_2\text {d}x_1\lesssim 1, \end{aligned}$$

where we use the condition (RR) above with $\alpha =1,\beta =\gamma =0$ and $\alpha =\beta =1, \gamma =0,$ respectively.

For term $I_{2,2},$ note that

$$\begin{aligned} I_{2,2}&\le \int _{8 \le |x_1| \le N_1} \bigg |\int _{\epsilon _2 \le |x_2| \le 8} \int _{8 \le |x_3| \le N_4} \Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big )\\&\quad \times e^{-ix_2}e^{-ix_3 }\text {d}x_3\text {d}x_2\bigg |\text {d}x_1 \\&\quad + \int _{8 \le |x_1| \le N_1} \bigg |\int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big )\\&\quad \times e^{-ix_2}e^{-ix_3 }\text {d}x_3\text {d}x_2\bigg |\text {d}x_1\\&:= I_{2,2,1}+ I_{2,2,2}. \end{aligned}$$

By Lemma 2.1 with $f(x_3)=\int _{\epsilon _2 \le |x_2| \le 8} \Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) e^{-ix_2}\text {d}x_2$, we have

$$\begin{aligned} |I_{2,2,1}|&\lesssim \int _{8 \le |x_1| \le N_1} \int _{E_{N_4}} \bigg |\int _{\epsilon _2 \le |x_2| \le 8} \Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) e^{-ix_2}\text {d}x_2\bigg |\text {d}x_3\text {d}x_1\\&\quad +\int _{8 \le |x_1| \le N_1} \int _{8 \le |x_3| \le N_4} \bigg |\int _{\epsilon _2 \le |x_2| \le 8} \Delta _{x_3,\pi } \Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big )\\&\quad \times e^{-ix_2}\text {d}x_2\bigg |\text {d}x_3\text {d}x_1\\&\lesssim \int _{8 \le |x_1| \le N_1} \int _{E_{N_4}} \int _{\epsilon _2 \le |x_2| \le 8} \frac{1}{|x_1|^{1+{\theta _1}} |x_2| |x_3| \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}}\text {d}x_2\text {d}x_3\text {d}x_1\\&\quad +\int _{8 \le |x_1| \le N_1} \int _{8 \le |x_3| \le N_4} \int _{\epsilon _2 \le |x_2| \le 8} \frac{1}{|x_1|^{1+{\theta _1}} |x_2| |x_3|^{1+{\theta _1}} \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}}\\&\quad \times \text {d}x_2\text {d}x_3\text {d}x_1 \lesssim 1, \end{aligned}$$

where we use the conditions (RR) above with $\alpha =1, \beta =\gamma =0$ and $\alpha =\gamma =1, \beta =0,$ respectively.

The estimate for term $I_{2,2,2}$ follows from a similar way as term $I_{1,2,2}.$ Indeed,

$$\begin{aligned} I_{2,2,2}&= \int _{8 \le |x_1| \le N_1} \bigg |\int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) (e^{-ix_2}e^{-ix_3 }-1)\text {d}x_3\text {d}x_2\bigg |\text {d}x_1\\&\quad + \int _{8 \le |x_1| \le N_1} \bigg |\int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \Delta _{x_1,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big )\\&\quad \times \text {d}x_3\text {d}x_2\bigg |\text {d}x_1\\&\lesssim \int _{8 \le |x_1| \le N_1} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \frac{|x_2|+|x_3|}{|x_1|^{1+{\theta _1}} |x_2| |x_3| \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}}\text {d}x_2\text {d}x_3\text {d}x_1\\&\quad +\int _{8 \le |x_1| \le N_1} \frac{1}{|x_1|^{1+{\theta _1}}}\text {d}x_1 \lesssim 1, \end{aligned}$$

where we use the condition (RR) with $\alpha =1, \beta =\gamma =0$ and the condition (C1.c), respectively.

Now we turn to the estimate for term II. We first write

$$\begin{aligned} II&= \int _{\epsilon _4 \le |x_3| \le N_4}\int _{\epsilon _2 \le |x_2| \le N_2}\int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_1}-1)e^{-ix_2}\\&\quad \times e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3 \\&\quad +\int _{\epsilon _4 \le |x_3| \le N_4}\int _{\epsilon _2 \le |x_2| \le N_2}\int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\\&\quad \times e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\\&:=II_1 +II_2. \end{aligned}$$

We further write

$$\begin{aligned} |II_1|&=\int _{\epsilon _4 \le |x_3| \le N_4}\int _{8 \le |x_2| \le N_2} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_1}-1)e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\\&\quad +\int _{\epsilon _4 \le |x_3| \le N_4}\int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_1}-1)e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\\&:=II_{1,1}+ II_{1,2}. \end{aligned}$$

For term $II_{1,1},$ using Lemma 2.1 with $f(x_2)=\int _{\epsilon _4 \le |x_3| \le N_4}\int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )$$(e^{-ix_1}-1)e^{-ix_3 }\text {d}x_1\text {d}x_3$, we obtain

$$\begin{aligned} |II_{1,1}|&\lesssim \int _{E_{N_2}}\bigg |\int _{\epsilon _4 \le |x_3| \le N_4}\int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_1}-1)e^{-ix_3 }\text {d}x_1\text {d}x_3\Bigg |\text {d}x_2\\&\quad +\int _{8 \le |x_2| \le N_2} \bigg | \int _{\epsilon _4 \le |x_3| \le N_4}\int _{\epsilon _1 \le |x_1| \le 8} \Delta _{x_2,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big )\\&\quad \times (e^{-ix_1}-1)e^{-ix_3 }\text {d}x_1\text {d}x_3\Bigg |\text {d}x_2\\&\lesssim \int _{E_{N_2}}\int _{\epsilon _4 \le |x_3| \le N_4}\int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{ |x_2| |x_3| \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}}\text {d}x_1\text {d}x_3\text {d}x_2\\&\quad +\int _{8 \le |x_2| \le N_2} \int _{\epsilon _4 \le |x_3| \le N_4}\int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{ |x_2|^{1+{\theta _1}} |x_3| \big (|\frac{x_1x_2\xi }{x_3}|+|\frac{x_3}{x_1x_2\xi }| \big )^{{\theta _2}}}\text {d}x_1\text {d}x_3\text {d}x_2, \end{aligned}$$

where we use the condition (RR) above for $\alpha =\beta =\gamma =0$ and $\beta =1, \alpha =\gamma =0,$ respectively.

Similarly,

$$\begin{aligned} II_{1,2}&=\int _{8 \le |x_3| \le N_4} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\\&\quad \times (e^{-ix_1}-1)e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\\&\quad + \int _{\epsilon _4 \le |x_3| \le 8} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\\&\quad \times (e^{-ix_1}-1)e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3:=II_{1,2,1}+ II_{1,2,2}. \end{aligned}$$

The bounds of $II_{1,2,1}$ and $II_{1,2,2}$ then follow from the similar estimates as terms $I_{2,2,1}$ and $I_{2,2,2},$ respectively.

Finally, we estimate term $II_2.$ Denote $II_2=II_{2,1}+II_{2,2},$ where

$$\begin{aligned} II_{2,1}=\int _{\epsilon _4 \le |x_3| \le N_4}\int _{8 \le |x_2| \le N_2} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3 \end{aligned}$$

and

$$\begin{aligned} II_{2,2}=\int _{\epsilon _4 \le |x_3| \le N_4}\int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3. \end{aligned}$$

Note that

$$\begin{aligned} |II_{2,1}|&=\int _{8 \le |x_3| \le N_4} \int _{8 \le |x_2| \le N_2} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3 \\&\quad +\int _{\epsilon _4 \le |x_3| \le 8} \int _{8 \le |x_2| \le N_2} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3. \\&:=II_{2,1,1}+II_{2,1,2} \end{aligned}$$

Applying Lemma 2.1 with $f(x_2)=\int _{8 \le |x_3| \le N_4} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_3 }\text {d}x_1\text {d}x_3$ first, and then $f(x_3)= \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\text {d}x_1,$ combining with the condition (RR), we obtain

$$\begin{aligned} |II_{2,1,1}|&\lesssim \int _{E_{N_2}}\bigg | \int _{8 \le |x_3| \le N_4} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_3}\text {d}x_1\text {d}x_3 \bigg |\text {d}x_2 \\&\quad +\int _{8 \le |x_2| \le N_2} \bigg | \int _{8 \le |x_3| \le N_4} \int _{\epsilon _1 \le |x_1| \le 8} \Delta _{x_2,\pi } \Big ( \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big )\\&\quad \times e^{-ix_3}\text {d}x_1\text {d}x_3 \bigg |\text {d}x_2 \\&\lesssim \int _{E_{N_2}} \int _{E_{N_4}} \bigg | \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_3}\text {d}x_1 \bigg | \text {d}x_3\text {d}x_2\\&\quad + \int _{E_{N_2}} \int _{8 \le |x_3| \le N_4} \bigg | \int _{\epsilon _1 \le |x_1| \le 8} \Delta _{x_3,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) e^{-ix_3}\text {d}x_1 \bigg | \text {d}x_3\text {d}x_2\\&\quad +\int _{8 \le |x_2| \le N_2} \int _{E_{N_4}} \bigg | \int _{\epsilon _1 \le |x_1| \le 8} \Delta _{x_2,\pi } \Big ( \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) \text {d}x_1\bigg | \text {d}x_3\text {d}x_2 \\&\quad +\int _{8 \le |x_2| \le N_2} \int _{8 \le |x_3| \le N_4} \bigg | \int _{\epsilon _1 \le |x_1| \le 8} \Delta _{x_3,\pi } \Delta _{x_2,\pi } \Big ( \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big )\\&\quad \times \text {d}x_1\bigg | \text {d}x_3\text {d}x_2 \lesssim 1. \end{aligned}$$

To estimate $II_{2,1,2},$ inserting $e^{-ix_3 }=[e^{-ix_3 }-1]+1$ and then applying Lemma 2.1, we get

$$\begin{aligned} II_{2,1,2}&\lesssim \bigg |\int _{8 \le |x_2| \le N_2} \int _{\epsilon _1 \le |x_1| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2}(e^{-ix_3}-1)\\&\quad \text {d}x_3\text {d}x_1 \text {d}x_2\bigg |\\&\quad +\bigg | \int _{8 \le |x_2| \le N_2} \int _{\epsilon _1 \le |x_1| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_2} \text {d}x_3\text {d}x_1\text {d}x_2 \bigg |\\&\lesssim \int _{E_{N_2}} \int _{\epsilon _1 \le |x_1| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \Big | \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big | |x_3| \text {d}x_3\text {d}x_1\text {d}x_2 \\&\quad + \int _{8 \le |x_2| \le N_2} \int _{\epsilon _1 \le |x_1| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \Big | \Delta _{x_2,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) \Big | |x_3| \text {d}x_3\text {d}x_1\text {d}x_2 \\&\quad + \int _{E_{N_2}}\bigg | \int _{\epsilon _1 \le |x_1| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\text {d}x_3\text {d}x_1\bigg | \text {d}x_2 \\&\quad +\int _{8 \le |x_2| \le N_2} \bigg | \int _{\epsilon _1 \le |x_1| \le 8} \int _{\epsilon _4 \le |x_3| \le 8} \Delta _{x_2,\pi } \Big ( \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) \text {d}x_3\text {d}x_1\bigg | \text {d}x_2. \end{aligned}$$

The required bound then follows from the conditions (RR) for the first two integrals while the condition (C1.d) for the last two integrals.

For $II_{2,2}$, splitting the set $\{\epsilon _4 \le |x_3| \le N_4\}$ into two parts $\{\epsilon _4 \le |x_3| \le 8\}$ and $\{8 \le |x_3| \le N_4\}$, and inserting $e^{-ix_2}e^{-ix_3 }= (e^{-ix_2}-1)(e^{-ix_3}-1) +(e^{-ix_2} -1) + (e^{-ix_3}-1)+ 1$ for the integral over the first set and $e^{-ix_2}e^{-ix_3 }= (e^{-ix_2}-1)e^{-ix_3}+e^{-ix_3}$ for the integral over the second set, we obtain

$$\begin{aligned} II_{2,2}&\lesssim \bigg |\int _{\epsilon _4 \le |x_3| \le 8} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_2}-1)(e^{-ix_3 }-1)\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&\quad + \bigg |\int _{\epsilon _4 \le |x_3| \le 8} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_2}-1)\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&\quad +\bigg |\int _{\epsilon _4 \le |x_3| \le 8} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_3 }-1)\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&\quad +\bigg |\int _{\epsilon _4 \le |x_3| \le 8} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&\quad + \bigg |\int _{8 \le |x_3| \le N_4} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_2}-1)e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&\quad +\bigg |\int _{8 \le |x_3| \le N_4} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\bigg |. \end{aligned}$$

The first four items follow from the conditions (RR), (C1.d), (C1.b), and (C1.a), respectively. To estimate the fifth and sixth terms, applying Lemma 2.1, we get

$$\begin{aligned}&\bigg |\int _{8 \le |x_3| \le N_4} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_2}-1)e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&\qquad +\bigg |\int _{8 \le |x_3| \le N_4} \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&\quad \lesssim \int _{E_{N_4}} \bigg | \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )(e^{-ix_2}-1)\text {d}x_1\text {d}x_2\bigg |\text {d}x_3\\&\qquad + \int _{8 \le |x_3| \le N_4} \bigg | \int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \Delta _{x_3,\pi } \Big ( \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) (e^{-ix_2}-1)\text {d}x_1\text {d}x_2\bigg |\text {d}x_3\\&\qquad +\int _{E_{N_4}} \bigg |\int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\text {d}x_1\text {d}x_2\bigg | \text {d}x_3\\&\qquad +\int _{8 \le |x_3| \le N_4} \bigg |\int _{\epsilon _2 \le |x_2| \le 8} \int _{\epsilon _1 \le |x_1| \le 8} \Delta _{x_3,\pi } \Big (\frac{1}{\xi }\mathcal {K}\Big (x_1,x_2,\frac{x_3}{\xi }\Big )\Big ) \text {d}x_1\text {d}x_2\bigg | \text {d}x_3\lesssim 1, \end{aligned}$$

where we use the condition (RR) for the first two terms and (C1.b) for the last two terms above. Thus, these estimates yield the bound of $II_{2,2}$ and hence the required bound for term II. The $L^2$ boundedness of ${\mathcal {K}}_\epsilon ^N*f$ follows. $\square$

Proof of Corollary 1.2

It suffices to show that ${\mathcal {K}}_\epsilon ^N*f$ converges in $L^2({\mathbb {R}}^3)$, as $\epsilon _1,\epsilon _2,\epsilon _3\rightarrow 0$ and $N_1,N_2,N_3\rightarrow \infty ,$ for a dense subset of $L^2({\mathbb {R}}^3).$ For this purpose, we consider smooth functions f having compact support. We may assume that $\epsilon _1,\epsilon _2,\epsilon _3<1$ and $N_1,N_2,N_3>1.$

We write $\iiint _{{\mathbb {R}}^3}{\mathcal {K}}_\epsilon ^N(u)f(x-u))du$ as a sum of eight terms; that is, the integrals over the sets $(i) |u_1|\le 1, |u_2|\le 1, |u_3|\le 1; (ii) |u_1|\le 1, |u_2|\le 1, |u_3|> 1; (iii) |u_1|\le 1, |u_2|> 1, |u_3|\le 1; (iv) |u_1|\le 1, |u_2|> 1, |u_3|> 1; (v) |u_1|> 1,|u_2|\le 1, |u_3|\le 1; (vi) |u_1|> 1, |u_2|\le 1, |u_3|> 1; (vii) |u_1|> 1, |u_2|> 1, |u_3|\le 1; (viii) |u_1|> 1, |u_2|> 1, |u_3|> 1.$ Inserting

$$\begin{aligned} f(x-u)&=[f(x_1-u_1,x_2-u_2,x_3-u_3)-f(x_1,x_2-u_2,x_3-u_3)\\&\quad -f(x_1-u_1,x_2,x_3)\\&\quad +f(x_1,x_2,x_3)]+[f(x_1-u_1,x_2,x_3)-f(x_1,x_2,x_3)]\\&\quad +[f(x_1,x_2-u_2,x_3-u_3)-f(x_1,x_2-u_2,x_3)]\\&\quad +[f(x_1,x_2-u_2,x_3)-f(x_1,x_2,x_3)]+f(x_1,x_2,x_3). \end{aligned}$$

into the first term

$$\begin{aligned} \int _{\epsilon _1 \le |u_1|\le 1}\int _{\epsilon _2 \le |u_2|\le 1}\int _{\epsilon _3 \le |u_3|\le 1}{\mathcal {K}}(u_1,u_2,u_3)f(x_1-u_1,x_2-u_2,x_3-u_3) \text {d}u_1\text {d}u_2\text {d}u_3 \end{aligned}$$

yields five integrals. In view of the conditions of f and the condition (RR) on ${\mathcal {K}},$ the first integral is dominated by

$$\begin{aligned}&F_1(x_1)F_2(x_2)F_3(x_3)\int _{ |u_1|\le 1}\int _{|u_2|\le 1}\int _{|u_3|\le 1} \frac{1}{|u_1| |u_2| |u_3|}{|u_1|(|u_2|+|u_3|)\over \Big (\Big |\frac{u_1u_2}{u_3}\Big | +\Big |\frac{u_3}{u_1u_2}\Big | \Big )^{{\theta _2}} } \text {d}u_1\text {d}u_2\text {d}u_3, \end{aligned}$$

where $F_1(x_1)$, $F_2(x_2)$, and $F_3(x_3)$ are bounded functions with bounded supports. Thus, as $\epsilon _1, \epsilon _2,\epsilon _3\rightarrow 0,$ the limit of the first integral exists for each $x_1, x_2,$ and $x_3$ and, moreover, is dominated by a fixed bounded function with compact support. Therefore, the first integral converges in $L^2$ as $\epsilon _1, \epsilon _2,\epsilon _3\rightarrow 0.$ The third integral can be handled by the same way. To see the second integral, by the condition (C1.c) and the assumption on ${\mathcal {K}}$ we observe that the limit $\int _{\epsilon _2\le |u_2|\le 1}\int _{\epsilon _3\le |u_3|\le 1}{\mathcal {K}}(u_1,u_2,u_3)\text {d}u_2\text {d}u_3$ exists as $\epsilon _2, \epsilon _3\rightarrow 0,$ and is dominated by $C|u_1|^{-1}.$ This fact together with the smoothness condition on f implies the second integral converges in $L^2$ as $\epsilon _1, \epsilon _2,\epsilon _3\rightarrow 0$ and the limit is dominated by a bounded function with compact support. Similarly, the required results for the fourth and the last integrals follow from the conditions (C1.d) and (C1.a), respectively, together with the assumptions on ${\mathcal {K}}.$

Note that in fact ${\mathcal {K}}(u)$ is integrable over the sets $(ii) |u_1|\le 1, |u_2|\le 1, |u_3|\ge 1$ and $(vii) |u_1|\ge 1, |u_2|\ge 1, |u_3|\le 1.$ Thus we have all the required results over these two sets.

Observe that

$$\begin{aligned}&\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}\int _{|u_3|\ge 1}|{\mathcal {K}}(u)f(x-u)|du \\&\quad \lesssim \int _{{\mathbb {R}}}\int _{{\mathbb {R}}}\int _{|u_3|\ge 1} \frac{1}{|u_1| |u_2| |u_3|} \Big (\Big |\frac{u_1u_2}{u_3}\Big | +\Big |\frac{u_3}{u_1u_2}\Big |\Big )^{-{\theta _2}} \\&\qquad \times \frac{1}{(1+|x_1-u_1|)^2 (1+|x_2-u_2|)^2 (1+|x_3-u_3|)^2}du \end{aligned}$$

which belongs to $L^2({\mathbb {R}}^3).$ This implies the required results over the corresponding sets (iv), (vi), and (viii).

To handle the integral over the set $(iii) |u_1|\le 1, |u_2|\ge 1, |u_3|\le 1,$ inserting

$$\begin{aligned} f(x_1-u_1,x_2-u_2,x_3-u_3)&=[f(x_1-u_1,x_2-u_2,x_3-u_3)\\&\quad -f(x_1,x_2-u_2,x_3-u_3)]\\&\quad +[f(x_1,x_2-u_2,x_3-u_3)-f(x_1,x_2-u_2,x_3)]\\&\quad +f(x_1,x_2-u_2,x_3) \end{aligned}$$

yields three integrals over the set (iii). The first two integrals, by the condition (RR) and the smoothness of f,  are dominated by

$$\begin{aligned}&F_1(x_1)F_3(x_3)\int _{ |u_1|\le 1}\int _{|u_2|\ge 1}\int _{|u_3|\le 1} \frac{1}{|u_2| |u_3|} \Big (\Big |\frac{u_1u_2}{u_3}\Big |+\Big |\frac{u_3}{u_1u_2}\Big |\Big )^{-{\theta _2}}\\&\quad \frac{1}{(1+|x_2-u_2|)^2}\text {d}u_1\text {d}u_2\text {d}u_3, \end{aligned}$$

and

$$\begin{aligned}&F_1(x_1)F_3(x_3)\int _{ |u_1|\le 1}\int _{|u_2|\ge 1}\int _{|u_3|\le 1} \frac{1}{|u_1| |u_2|} \Big (\Big |\frac{u_1u_2}{u_3}\Big |+\Big |\frac{u_3}{u_1u_2} \Big |\Big )^{-{\theta _2}}\\&\quad \times \frac{1}{(1+|x_2-u_2|)^2}\text {d}u_1\text {d}u_2\text {d}u_3, \end{aligned}$$

where $F_1(x_1)$ and $F_3(x_3)$ are bounded functions with bounded supports. Thus, we obtain a domination, independent of $\epsilon _1,\epsilon _3$ and $N_2,$ by a function which belongs to $L^2({\mathbb {R}}^3),$ so the limits as $\epsilon _1,\epsilon _3 \rightarrow 0$ and $N_2\rightarrow \infty$ exist. Applying condition (C1.d) with $\beta =0$ yields that the last integral is bounded by

$$\begin{aligned} F_1(x_1)F_3(x_3)\int _{|u_2|\ge 1}\frac{1}{|u_2| (1+|x_2-u_2|)^2}\text {d}u_2 \end{aligned}$$

which belongs to $L^2({\mathbb {R}}^3)$ and the limit as $\epsilon _1,\epsilon _3 \rightarrow 0$ and $N_2\rightarrow \infty$ exists.

Finally, for the integral over the set $(v) |u_1|\ge 1, |u_2|\le 1,|u_3|\le 1,$ inserting

$$\begin{aligned} f(x_1-u_1,x_2-u_2,x_3-u_3)&= [f(x_1-u_1,x_2-u_2,x_3-u_3)\\&\quad -f(x_1-u_1,x_2-u_2,x_3)]\\&\quad +[f(x_1-u_1,x_2-u_2,x_3)-f(x_1-u_1,x_2,x_3)]\\&\quad +f(x_1-u_1,x_2,x_3) \end{aligned}$$

and then applying the condition (RR) for the first two integrals and (C1.c) with $\alpha =0$ on the last integral, this integral is dominated by

$$\begin{aligned} F_2(x_2)F_3(x_3)\int _{|u_1|\ge 1}\frac{1}{|u_1| (1+|x_1-u_1|)^2}\text {d}u_1, \end{aligned}$$

where $F_2(x_2)$ and $F_3(x_3)$ are bounded functions with bounded supports. The existence of the limit is concluded. The $L^2$ boundedness of ${\mathcal {K}}*f$ then follows from Theorem 1.1. $\square$

We would like to remark, as mentioned early in Sect. 1, that incidentally we have shown that ${\mathcal {K}}_\epsilon ^N*f$ converges in $L^p$ norm and almost everywhere as $\epsilon _1,\epsilon _2,\epsilon _3\rightarrow 0$ and $N_1,N_2,N_3\rightarrow \infty$ whenever f is a smooth function with compact support. We also point out that the condition (C1.b) is not used in the proof of Corollary 1.2.

3 $L^p$ Estimates for $1<p<\infty$

In this subsection, we will prove Theorems 1.3 and 1.4. The main tools to show the $L^p,1<p<\infty ,$ estimates are

• the $L^2$ boundedness of ${\mathcal {K}}*f,$

• the Littlewood–Paley theory associated with Zygmund dilation,

• the almost orthogonality argument.

We first recall the Littlewood–Paley theory. As mentioned in Sect. 1, to handle the $L^p, 1<p<\infty ,$ boundedness of operators, one only needs the continuous Littlewood–Paley square function. To do this, let ${{\mathcal {S}}} ({\mathbb {R}}^i)$ denote the Schwartz class in ${\mathbb {R}}^i, i=1, 2, 3.$ We construct a function defined on ${\mathbb {R}}^3$ given by

$$\begin{aligned} \phi (x_1,x_2,x_3)=\phi ^{(1)}(x)\phi ^{(2)}(x_2,x_3), \end{aligned}$$

(3.1)

where $\phi ^{(1)}\in {{\mathcal {S}}}({\mathbb {R}}), \phi ^{(2)}\in {{\mathcal {S}}}({\mathbb {R}}^2)$ with the supports contained in the unit ball centered at the origin in ${\mathbb {R}}^3,$ and satisfy

$$\begin{aligned} \sum _{j \in {\mathbb {Z}}}\vert \widehat{\phi ^{(1)}}(2^{j}\xi _1 )\vert ^2= & {} 1 \quad \text {for all}\ \ \xi _1\in {{\mathbb {R}}}\backslash \lbrace 0\rbrace , \end{aligned}$$

(3.2)

$$\begin{aligned} \sum _{k\in {\mathbb {Z}}}\vert \widehat{ \phi ^{(2)}}(2^{k} \xi _2, 2^{k}\xi _3 )\vert ^2= & {} 1\quad \text {for all}\ \ (\xi _2, \xi _3) \in {{\mathbb {R}}^2}\backslash \lbrace (0, 0)\rbrace , \end{aligned}$$

(3.3)

and the moment conditions

$$\begin{aligned} \int _{{\mathbb {R}}}x_1^{\alpha }\phi ^{(1)}(x_1)\text {d}x_1=\int _{{\mathbb {R}}^2} x_2^{\beta }x_3^{\gamma }\phi ^{(2)}(x_2,x_3)\text {d}x_2\text {d}x_3=0 \quad \text {for}\ \ 0\le \alpha , \beta ,\gamma \le 10. \end{aligned}$$

(3.4)

For $f\in L^p, 1<p<\infty$, $g_{\mathfrak {z}}^c(f)$, the continuous Littlewood–Paley square function of f associated with the Zygmund dilation is defined by

$$\begin{aligned} g_{\mathfrak {z}}^c(f)(x) =\bigg \lbrace \sum _{j,k\in {\mathbb {Z}}}\vert (\phi _{j,k}*f)(x)\vert ^2\bigg \rbrace ^{{{1}\over {2}}}, \end{aligned}$$

where

$$\begin{aligned} \phi _{j,k}(x_1,x_2,x_3):=2^{-2(j+k)}\phi ^{(1)}(2^{-j}x_1 ) \phi ^{(2)}(2^{-k}x_2, 2^{-(j+k)}x_3). \end{aligned}$$

(3.5)

By taking the Fourier transform, it is easy to see that the following Calderón’s identity

$$\begin{aligned} f(x) =\sum _{j,k\in {\mathbb {Z}}} (\phi _{j,k}*\phi _{j,k}*f)(x) \end{aligned}$$

(3.6)

holds on $L^2({\mathbb {R}}^3).$

Using the $L^p$ boundedness of operators for $1<p<\infty$ obtained by Ricci and Stein in [28], as mentioned in Sect. 1, we have

$$\begin{aligned} \Bigg \Vert \sum _{(j,k)\in F} \epsilon (j,k)(\phi _{j,k}*f)\Bigg \Vert _p\le C\Vert f\Vert _p \end{aligned}$$

for every sequence $\epsilon (j,k),$ taking the values 1 and $-1$, where F is any finite subset of $j, k\in {\mathbb {Z}}.$ Using Khinchin’s well-known inequality, we obtain

$$\begin{aligned} \Vert g_{\mathfrak {z}}^c(f) \Vert _p\le C_2\Vert f\Vert _p \end{aligned}$$

for $1<p<\infty .$ This estimate together with Calderón’s identity on $L^2$ allows us to obtain the $L^p$ estimates of $g_{\mathfrak {z}}^c$ for $1<p<\infty$. Namely, there exist constants $C_1$ and $C_2$ such that, for $1<p<\infty ,$

$$\begin{aligned} C_1\Vert f \Vert _p\le \Vert g_{\mathfrak {z}}^c(f) \Vert _p\le C_2\Vert f\Vert _p. \end{aligned}$$

(3.7)

Now we turn to the proof of Theorem 1.3. First note that ${\mathcal {K}}$ satisfies the conditions (C1.a)–(C1.d) since (C2.b) implies (C1.b) and (C1.d), as mentioned in Sect. 1. Therefore, by Corollary 1.2, the operator ${\mathcal {K}}*f=\lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3\rightarrow 0 \\ N_1,N_2,N_3\rightarrow \infty \end{array}} {\mathcal {K}}_\epsilon ^N*f$ is bounded on $L^2({\mathbb {R}}^3).$ To obtain the $L^p$ boundedness of ${\mathcal {K}}*f,$ it suffices to show this for all $f\in L^2\cap L^p$ since the subspace $L^2\cap L^p$ is dense in $L^p, 1<p<\infty .$ By the $L^p$ estimates of the Littlewood–Paley square function given in (3.7), the $L^p$ boundedness of ${\mathcal {K}}*f$ will follow from the estimate

$$\begin{aligned} \Vert g_{\mathfrak {z}}^c({\mathcal {K}}*f)\Vert _p \lesssim \Vert f\Vert _p. \end{aligned}$$

(3.8)

To prove (3.8) for all $f\in L^2\cap L^p,$ using the fact that ${\mathcal {K}}*f$ is bounded on $L^2,$ as mentioned above, and Calderón’s identity on $L^2$ given in (3.6), we write

$$\begin{aligned} \phi _{j,k}*({\mathcal {K}}*f)(x_1,x_2,x_3)=\sum _{j',k'\in {\mathbb {Z}}} (\phi _{j,k}*{\mathcal {K}}*\phi _{j',k'}*\phi _{j',k'}*f)(x_1,x_2,x_3). \end{aligned}$$

The proof of Theorem 1.3 now follows from the following almost orthogonality argument.

Proposition 3.1

Suppose that $\phi$ is defined as in (3.5) and ${\mathcal {K}}$ is a function on ${\mathbb {R}}^3$ satisfying the conditions (RR) and (C2.a)–(C2.c). Then, for $\lambda ={\frac{1}{2}}\min ({\theta _1},{\theta _2}),$

$$\begin{aligned}&\big | (\phi _{j,k} *{\mathcal {K}} *\phi _{j',k'})(x_1,x_2,x_3) \big |\\&\quad \le C2^{-|j-j'|}2^{-|k-k'|}\frac{2^{-(j\vee j')}}{(1+2^{-(j\vee j')}|x_1|)^{1+\lambda }}\\&\quad \times \frac{2^{-(k\vee k')}}{(1+2^{-(k\vee k')}|x_2|)^{1+\lambda }}\frac{2^{-(j\vee j')-(k\vee k')}}{(1+2^{-(j\vee j')-(k\vee k')}|x_3|)^{1+\lambda }}, \end{aligned}$$

where the constant C depends only on $\lambda$ and ${\mathcal {K}}*f$ is defined for $f\in L^2$ as in Corollary 1.2, and $j\vee j'$ means $\max (j, j')$.

Assuming Proposition 3.1 for the moment, we then observe that

$$\begin{aligned}&|[\phi _{j,k}*{\mathcal {K}}*\phi _{j',k'}*(\phi _{j',k'}*f)](x_1,x_2,x_3)| \\&\quad \le C 2^{-|j-j'|}2^{-|k-k'|}{\mathcal {M}}_s(\phi _{j',k'}*f)(x_1,x_2,x_3), \end{aligned}$$

where ${\mathcal {M}}_s$ is the strong maximal function on ${\mathbb {R}}^3.$ Hölder’s inequality implies

$$\begin{aligned} \Vert g_{\mathfrak {z}}^c({\mathcal {K}}*f)\Vert _p= & {} \bigg \Vert \bigg \lbrace \sum _{j,k}\vert \phi _{j,k}*{\mathcal {K}}*f\vert ^2\bigg \rbrace ^{{{1}\over {2}}}\bigg \Vert _p \le C \bigg \Vert \bigg \{\sum _{j',k'} |{\mathcal {M}}_s(\phi _{j',k'}*f)|^2\bigg \}^{\frac{1}{2}}\bigg \Vert _p \\\le & {} C\bigg \Vert \bigg \{\sum _{j',k'} |\phi _{j',k'}*f|^2\bigg \}^{\frac{1}{2} }\bigg \Vert _p \le C \Vert f\Vert _p, \end{aligned}$$

where we use the Fefferman and Stein’s vector-valued maximal inequality and the Littlewood–Paley square function estimate for $L^p, 1<p<\infty ,$ in the last two inequalities, respectively.

To finish the proof of Theorem 1.3, we only need to show Proposition 3.1, whose proof follows from the following lemma.

Lemma 3.2

Suppose that $\phi ^{(1)}$ and $\phi ^{(2)}$ satisfy the conditions (3.1)–(3.4) and ${\mathcal {K}}$ is a function on ${\mathbb {R}}^3$ satisfying the conditions (RR) and (C2.a)–(C2.c). Then, for $\lambda ={\frac{1}{2}}\min ({\theta _1},{\theta _2}),$ we have

$$\begin{aligned} |{\mathcal {K}}*(\phi ^{(1)}\otimes \phi ^{(2)})(x_1,x_2,x_3)|\le \frac{C_\lambda }{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}, \end{aligned}$$

where $C_\lambda$ is the constant depending only on $\lambda .$

Proof

For simplicity, let $S=\lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3 \rightarrow 0 \\ N_1,N_2,N_3 \rightarrow \infty \end{array}}\int _{\epsilon _1 \le |x-u|\le N_1}\int _{\epsilon _2 \le |x_2-u_2|\le N_2}$$\int _{\epsilon _3 \le |x_3-u_3|\le N_3} {\mathcal {K}}(x_1-u_1,x_2-u_2,x_3-u_3)\phi ^{(1)}(u_1)\phi ^{(2)}(u_2,u_3)\text {d}u_3\text {d}u_2\text {d}u_1$. We consider the following eight cases:

Case 1$|x_1|\ge 3, |x_2| \ge 3, |x_3| \ge 3$. For this case, we use the cancellation conditions in (3.4) to write

$$\begin{aligned} S&=\lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3 \rightarrow 0 \\ N_1,N_2,N_3 \rightarrow \infty \end{array}}\int _{\epsilon _1 \le |x_1-u_1|\le N_1}\int _{\epsilon _2 \le |x_2-u_2|\le N_2}\int _{\epsilon _3\le |x_3-u_3|\le N_3}\\&\quad \times \big [{\mathcal {K}}(x_1-u_1,x_2-u_2,x_3-u_3)\\&\quad - {\mathcal {K}}(x_1,x_2-u_2,x_3-u_3)-{\mathcal {K}}(x_1-u_1,x_2,x_3)+{\mathcal {K}}(x_1,x_2,x_3)\big ]\\&\quad \times \phi ^{(1)}(u_1)\phi ^{(2)}(u_2,u_3)\text {d}u_3\text {d}u_2\text {d}u_1. \end{aligned}$$

Note that ${\mathcal {K}}(x_1-u_1,x_2-u_2,x_3-u_3)-{\mathcal {K}}(x_1,x_2-u_2,x_3-u_3)- \big ({\mathcal {K}}(x_1-u_1,x_2,x_3)-{\mathcal {K}}(x_1,x_2,x_3)\big ) = \Delta _{x_2,-u_2} \Delta _{x_1,-u_1} \mathcal {K}(x_1,x_2,x_3-u_3)+\Delta _{x_3,-u_3} \Delta _{x_1,-u_1} \mathcal {K}(x_1,x_2,x_3).$ Thus, by the condition (RR) with $\alpha =\beta =1, \gamma =0$ and $\alpha =\gamma =1,\beta =0$ respectively, we have

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{|u_1|^{\theta _1}(|u_2|^{\theta _1}+|u_3|^{\theta _1})}{|x_1|^{1+ {\theta _1}} |x_2|^{1+{\theta _1}} |x_3|}\Big (\Big |\frac{x_1x_2}{x_3}\Big |+\Big | \frac{x_3}{x_1x_2}\Big |\Big )^{-\lambda }\text {d}u_3\text {d}u_2\text {d}u_1\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 2$|x_1|\ge 3, |x_2| \ge 3, |x_3| < 3$. By the cancellation condition of $\phi ^{(1)},$

$$\begin{aligned} S&=\lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3 \rightarrow 0 \\ N_1,N_2,N_3 \rightarrow \infty \end{array}}\int _{\epsilon _1 \le |x_1-u_1|\le N_1}\int _{\epsilon _2 \le |x_2-u_2|\le N_2}\int _{\epsilon _3 \le |x_3-u_3|\le N_3}\\&\quad \times \Delta _{x_1,-u_1} {\mathcal {K}}(x_1,x_2-u_2,x_3-u_3) \times \phi ^{(1)}(u_1)\phi ^{(2)}(u_2,u_3)\text {d}u_3\text {d}u_2\text {d}u_1. \end{aligned}$$

Therefore, by the condition (RR) with $\alpha =1$ and $\beta =\gamma =0,$ we obtain

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{|u_1|^{\theta _1}}{|x_1|^{1+{\theta _1}} |x_2-u_2| |x_3-u_3|}\\&\quad \times \Big ( \Big |\frac{x_1(x_2-u_2)}{x_3-u_3}\Big | + \Big |\frac{x_3-u_3}{x_1(x_2-u_2)}\Big |\Big )^{-\lambda }\text {d}u_3\text {d}u_2\text {d}u_1\\&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{1}{|x_1|^{1+{\theta _1}} |x_2| |x_3-u_3|}\Big |\frac{x_1x_2}{x_3-u_3}\Big |^{-\lambda }\text {d}u_3\text {d}u_2\text {d}u_1\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 3$|x_1|\ge 3, |x_2| < 3, |x_3| \ge 3.$ The same expression for S as in Case 2 yields

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{|u_1|^{\theta _1}}{|x_1|^{1+{\theta _1}} |x_2-u_2| |x_3|}\Big |\frac{x_3}{x_1(x_2-u_2)}\Big |^{-\lambda }\text {d}u_3\text {d}u_2\text {d}u_1\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 4$|x_1|\ge 3, |x_2|< 3, |x_3| < 3.$ Using the cancellation condition of $\phi ^{(1)}$, we write

$$\begin{aligned} S= & {} \lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3 \rightarrow 0 \\ N_1,N_2,N_3 \rightarrow \infty \end{array}}\int _{\epsilon _1 \le |x_1-u_1|\le N_1}\int _{\epsilon _2 \le |u_2|\le 4}\int _{\epsilon _3 \le |u_3|\le 4}\\&\times \big [{\mathcal {K}}(x_1-u_1,u_2,u_3)-{\mathcal {K}}(x_1,u_2,u_3)\big ]\\&\times \phi ^{(1)}(u_1)\big (\phi ^{(2)}(x_2-u_2,x_3-u_3)-\phi ^{(2)}(x_2,x_3)\big )\text {d}u_3\text {d}u_2\text {d}u_1 \\&+\lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3 \rightarrow 0 \\ N_1,N_2,N_3 \rightarrow \infty \end{array}}\int _{\epsilon _1 \le |x_1-u_1|\le N_1}\int _{\epsilon _2 \le |u_2|\le 4}\int _{\epsilon _3 \le |u_3|\le 4}\\&\times \big [{\mathcal {K}}(x_1-u_1,u_2,u_3)-{\mathcal {K}}(x_1,u_2,u_3)\big ]\phi ^{(1)}(u_1)\phi ^{(2)}(x_2,x_3)\text {d}u_3\text {d}u_2\text {d}u_1. \end{aligned}$$

By the condition (RR) with $\alpha =1,\beta =\gamma =0$ for the first integral, and the cancellation condition (C2.c) with $\alpha =1$ for the second integral, we get

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 4}\int _{|u_3| \le 4} \frac{|u_1|^{\theta _1}}{|x_1|^{1+{\theta _1}} |u_2| |u_3|}\Big (\Big |\frac{x_1u_2}{u_3}\Big |+\Big |\frac{u_3}{x_1u_2}\Big | \Big )^{-{\theta _2}}\\&\quad (|u_2|+|u_3|)\text {d}u_3\text {d}u_2\text {d}u_1\\&\quad + \int _{|u_1| \le 1} \frac{|u_1|^{\theta _1}}{|x_1|^{1+{\theta _1}}} \text {d}u_1\\&\lesssim \frac{1}{|x_1|^{1+{\theta _1}}}\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 5$|x_1|< 3, |x_2| \ge 3, |x_3| \ge 3$. Similar to Case 4, using the cancellation condition of $\phi ^{(2)},$ we write

$$\begin{aligned} S&= \lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3 \rightarrow 0 \\ N_1,N_2,N_3 \rightarrow \infty \end{array}}\int _{\epsilon _1 \le |u_1|\le 4}\int _{\epsilon _2 \le |x_2-u_2|\le N_2}\\&\quad \times \int _{\epsilon _3 \le |x_3-u_3|\le N_3} \big [{\mathcal {K}}(u_1,x_2-u_2,x_3-u_3)-{\mathcal {K}}(u_1,x_2,x_3)\big ]\\&\quad \times \big (\phi ^{(1)}(x_1-u_1)-\phi ^{(1)}(x_1)\big )\phi ^{(2)}(u_2,u_3)\text {d}u_3\text {d}u_2\text {d}u_1 \\&\quad +\lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3 \rightarrow 0 \\ N_1,N_2,N_3 \rightarrow \infty \end{array}}\int _{\epsilon _1 \le |u_1|\le 4}\int _{\epsilon _2\le |x_2-u_2|\le N_2}\\&\quad \times \int _{\epsilon _3 \le |x_3-u_3|\le N_3} \big [{\mathcal {K}}(u_1,x_2-u_2,x_3-u_3)-{\mathcal {K}}(u_1,x_2,x_3)\big ]\\&\quad \times \phi ^{(1)}(x_1)\phi ^{(2)}(u_2,u_3)\text {d}u_3\text {d}u_2\text {d}u_1. \end{aligned}$$

Note that ${\mathcal {K}}(u_1,x_2-u_2,x_3-u_3)-{\mathcal {K}}(u_1,x_2,x_3)=\Delta _{x_2,-u_2} \mathcal {K}(u_1,x_2,x_3)+ \Delta _{x_3,-u_3} \mathcal {K}(u_1,x_2,x_3)$ and ${\mathcal {K}}(u_1,x_2-u_2,x_3-u_3)-{\mathcal {K}}(u_1,x_2,x_3)=\Delta _{x_2,-u_2}$$\mathcal {K}(u_1,x_2,x_3-u_3) +\Delta _{x_3,-u_3} \mathcal {K}(u_1,x_2,x_3)$. Thus, using the condition (RR) on ${\mathcal {K}}$, the smoothness of $\phi ^{(1)}$ for the first integral, and the cancellation conditions (C2.b) with $\beta =1, \gamma =0$ and $\beta =0, \gamma =1,$ respectively, for the second integral, and applying the dominated convergence theorem, we obtain

$$\begin{aligned} |S|&\lesssim \int _{|u_1|\le 4}\int _{|u_2|\le 1}\int _{|u_3|\le 1} \bigg (\frac{|u_2|^{\theta _1}}{|u_1| |x_2|^{1+{\theta _1}} |x_3|}+\frac{|u_3|^{\theta _1}}{|u_1| |x_2| |x_3|^{1+{\theta _1}}}\Big )\\&\quad \times \Big (\Big |\frac{u_1x_2}{x_3}\Big |+\Big |\frac{x_3}{u_1x_2}\Big |\Big )^{-\lambda }|u_1|\text {d}u_3\text {d}u_2\text {d}u_1\\&\quad + \int _{|u_2| \le 1}\int _{|u_3| \le 1} \bigg (\frac{|u_2|^{\theta _1}}{|x_2|^{1+{\theta _1}} |x_3|}+\frac{|u_3|^{\theta _1}}{|x_2| |x_3|^{1+{\theta _1}}}\Big )\Big (\Big |\frac{4x_2}{x_3}\Big |+\Big |\frac{x_3}{4x_2}\Big | \Big )^{-\lambda }\text {d}u_3 \text {d}u_2\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 6$|x_1|< 3, |x_2| \ge 3, |x_3|< 3.$ Note that

$$\begin{aligned} S&= \lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3 \rightarrow 0 \\ N_1,N_2,N_3 \rightarrow \infty \end{array}}\int _{\epsilon _1 \le |u_1|\le 4}\int _{\epsilon _2 \le |x_2-u_2|\le N_2}\int _{\epsilon _3 \le |x_3-u_3|\le N_3} {\mathcal {K}}(u_1,x_2-u_2,x_3-u_3)\\&\quad \times \big (\phi ^{(1)}(x_1-u_1)-\phi ^{(1)}(x_1)\big ) \phi ^{(2)}(u_2,u_3)\text {d}u_3\text {d}u_2\text {d}u_1\\&\quad +\lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3 \rightarrow 0 \\ N_1,N_2,N_3 \rightarrow \infty \end{array}}\int _{\epsilon _1 \le |u_1|\le 4}\int _{\epsilon _2 \le |x_2-u_2|\le N_2}\int _{\epsilon _3 \le |x_3-u_3|\le N_3} {\mathcal {K}}(u_1,x_2-u_2,x_3-u_3)\\&\quad \times \phi ^{(1)}(x_1)\phi ^{(2)}(u_2,u_3)\text {d}u_3\text {d}u_2\text {d}u_1. \end{aligned}$$

By the condition (RR) with $\alpha =\beta =\gamma =0$ and the smoothness condition of $\phi ^{(1)}$ on the first integral and the condition (C2.b) with $\beta =\gamma =0$ for the second integral and the dominated convergent theorem, we have

$$\begin{aligned} |S|&\lesssim \int _{|u_1|\le 4}\int _{|u_2|\le 1}\int _{|u_3|\le 1}\frac{1}{|u_1| |x_2| |x_3-u_3|}\Big (\Big |\frac{u_1x_2}{x_3-u_3}\Big |+\Big |\frac{x_3-u_3}{u_1x_2}\Big | \Big )^{-\lambda }|u_1|\text {d}u_3\text {d}u_2\text {d}u_1\\&\quad +\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{1}{|x_2| |x_3-u_3|}\Big (\Big |\frac{4x_2}{x_3-u_3}\Big |+\Big |\frac{x_3-u_3}{4x_2}\Big | \Big )^{-\lambda }\text {d}u_3\text {d}u_2\\&\lesssim \frac{1}{|x_2|^{1+\lambda }}\\&\lesssim \frac{1}{(1+|x|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 7$|x_1|< 3, |x_2| < 3, |x_3| \ge 3$. The required estimate follows directly from the condition (RR):

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{1}{|x_1-u_1| |x_2-u_2| |x_3|}\Big |\frac{x_3}{(x_1-u_1)(x_2-u_2)}\Big |^{-\lambda }\text {d}u_3\text {d}u_2\text {d}u_1\\&\lesssim \frac{1}{|x_3|^{1+\lambda }}\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 8$|x_1|< 3, |x_2|< 3, |x_3| < 3$. Inserting

$$\begin{aligned}&\phi ^{(1)}(x_1-u_1)\phi ^{(2)}(x_2-u_2,x_3-u_3)\\&\quad = [\phi ^{(1)}(x_1-u_1)-\phi ^{(1)}(x_1)][\phi ^{(2)}(x_2-u_2,x_3-u_3)-\phi ^{(2)}(x_2,x_3)]\\&\quad \quad +\phi ^{(1)}(x_1)[\phi ^{(2)}(x_2-u_2,x_3-u_3)-\phi ^{(2)}(x_2,x_3)]\\&\quad \quad +[\phi ^{(1)}(x_1-u_1)-\phi ^{(1)}(x_1)]\phi ^{(2)}(x_2,x_3) +\phi ^{(1)}(x_1)\phi ^{(2)}(x_2,x_3), \end{aligned}$$

we write

$$\begin{aligned} S&= \lim _{\begin{array}{c} \epsilon _1,\epsilon _2,\epsilon _3 \rightarrow 0 \\ N_1,N_2,N_3 \rightarrow \infty \end{array}}\int _{\epsilon _1 \le |u_1|\le 4}\int _{\epsilon _2 \le |u_2|\le 4}\int _{\epsilon _3 \le |u_3|\le 4} {\mathcal {K}}(u_1,u_2,u_3) \\&\quad \times \Big \{[\phi ^{(1)}(x_1-u_1)-\phi ^{(1)}(x_1)][\phi ^{(2)}(x_2-u_2,x_3-u_3)-\phi ^{(2)}(x_2,x_3)]\\&\quad + \phi ^{(1)}(x_1)[\phi ^{(2)}(x_2-u_2,x_3-u_3)-\phi ^{(2)}(x_2,x_3)]\\&\quad + [\phi ^{(1)}(x_1-u_1)-\phi ^{(1)}(x_1)]\phi ^{(2)}(x_2,x_3)\\&\quad +\phi ^{(1)}(x_1)\phi ^{(2)}(x_2,x_3)\Big \}\text {d}u_3\text {d}u_2\text {d}u_1 \end{aligned}$$

as four integrals. Using the condition (RR) with $\alpha =\beta =\gamma =0$ and the smoothness condition of $\phi ^{(1)}$ for the first integral, the cancellation conditions (C2.b), (C2.c), and (C2.a) for the last three integrals, and the dominated convergent theorem, we obtain

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 4}\int _{|u_2| \le 4}\int _{|u_3| \le 4} \frac{1}{|u_1| |u_2| |u_3|}\Big (\Big |\frac{u_1u_2}{u_3}\Big |+\Big |\frac{u_3}{u_1u_2}\Big | \Big )^{-{\theta _2}}\\&\qquad |u_1|(|u_2|+|u_3|)\text {d}u_3\text {d}u_2\text {d}u_1\\&\qquad +\int _{|u_2| \le 4}\int _{|u_3| \le 4} \frac{1}{|u_2| |u_3|}\Big (\Big |\frac{4u_2}{u_3}\Big |+\Big |\frac{u_3}{4u_2}\Big | \Big )^{-{\theta _2}}(|u_2|+|u_3|)\text {d}u_3\text {d}u_2 \\&\qquad +\int _{|u_1| \le 4} \frac{1}{|u_1|} |u_1|\text {d}u_1+1 \\&\lesssim 1\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

The proof of Lemma 3.2 is complete. $\square$

Recall that $\phi _{j,k}(u_1,u_2,u_3)=2^{-2(j+k)}\phi ^{(1)}(2^{-j}u_1)\phi ^{(2)}(2^{-k}u_2,2^{-(j+k)}u_3)$, and the assumptions on ${\mathcal {K}}$ are invariant with respect to Zygmund dilation. By Lemma 3.2, we have the following estimate.

$$\begin{aligned} \big | ({\mathcal {K}} *\phi _{j,k})(x_1,x_2,x_3) \big | \le C\frac{2^{-j}}{(1+2^{-j} |x_1|)^{1+\lambda }}\frac{2^{-k}}{(1+2^{-k}|x_2|)^{1+\lambda }}\frac{2^{-(j+k)}}{(1+2^{-(j+k)}|x_3|)^{1+\lambda }}. \end{aligned}$$

Now the proof of Proposition 3.1 follows from the above estimate with replacing $\phi _{j,k}$ by $\phi _{j,k} *\phi _{j',k'}.$ Note that $\phi _{j,k} *\phi _{j',k'}$ satisfies the same properties as $\phi _{j\vee j', k\vee k'}$ but with the bound $C2^{-|j-j'|}2^{-|k-k'|}.$ Thus, the proof of Proposition 3.1 is complete and hence Theorem 1.3 is proved.

We now turn to the proof of Theorem 1.4. To prove part (a), we first show the $L^2$ boundedness of ${\mathcal {K}}*f.$ This is similar to the proof of Theorem 1.1. We only outline the proof as follows:

By the Plancherel theorem, the $L^2$ boundedness of ${\mathcal {K}}*f$ is equivalent to $|\widehat{{\mathcal {K}}}(\chi ,\eta ,\xi )|$$\le A,$ where $\widehat{{\mathcal {K}}}$ is the Fourier transform of ${\mathcal {K}}$ in the sense of distributions and A is the constant depending only on the constant C.

Let $\zeta _1(x_1)$ be a smooth function on ${\mathbb {R}}$ with $\zeta _1(x_1)=1$ if $|x_1| \le 8$ and $\zeta _1(x_1)=0$ if $|x_1| \ge 16$, and $\zeta _2=1-\zeta _1$. For simplicity, we denote by $\widetilde{\mathcal {K}}(x_1,x_2,x_3)= \frac{1}{\chi \eta \xi }\mathcal {K}(\frac{x_1}{\chi },\frac{x_2}{\eta },\frac{x_3}{\xi })$. We write

$$\begin{aligned} \widehat{\mathcal {K}}(\chi ,\eta ,\xi ) =&\iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_1)e^{-ix_1}e^{-ix_2}e^{-ix_3}\text {d}x_1\text {d}x_2\text {d}x_3 \\&+\iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1) e^{-ix_1}e^{-ix_2} e^{-ix_3}\text {d}x_1\text {d}x_2\text {d}x_3\\ :=&I+II. \end{aligned}$$

To estimate I,  we write

$$\begin{aligned} |I|&= {\frac{1}{2}}\bigg |\iiint \Delta _{x_1,\pi } \Big (\widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_1)\Big )e^{-ix_1}e^{-ix_2}e^{-ix_3}\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&\lesssim \bigg |\iiint \Delta _{x_1,\pi } \Big ( \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_1) \Big )e^{-ix_1}\zeta _2(x_2)e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\bigg | \\&\quad + \bigg |\iiint \Delta _{x_1,\pi } \Big ( \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_1) \Big ) e^{-ix_1}\zeta _1(x_2)e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&:= I_1+I_2. \end{aligned}$$

Note that

$$\begin{aligned} I_{1}&\lesssim \bigg |\iiint \Delta _{x_2,\pi } \bigg (\Delta _{x_1,\pi } \Big ( \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_1) \Big )\zeta _2(x_2)\bigg )e^{-ix_1}e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&\lesssim \int _{{\mathbb {R}}}\int _{|x_2| \ge 8}\int _{|x_1|\ge 8} \frac{1}{|x_1|^{1+{\theta _1}} |x_2|^{1+{\theta _1}} |x_3|}\Big (\Big |\frac{x_1x_2\xi }{x_3\chi \eta }\Big |+\Big |\frac{x_3\chi \eta }{x_1x_2\xi }\Big |\Big )^{-{\theta _2}}\text {d}x_1 \text {d}x_2 \text {d}x_3 \lesssim 1. \end{aligned}$$

For term $I_{2},$ note that

$$\begin{aligned} |I_{2}|&\le \big |\iiint \Delta _{x_1,\pi } \big ( \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_1) \big ) e^{-ix_1}\zeta _1(x_2)e^{-ix_2}\zeta _2(x_3)e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\big | \\&\quad + \big |\iiint \Delta _{x_1,\pi } \big ( \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_1) \big ) e^{-ix_1}\zeta _1(x_2)e^{-ix_2}\zeta _1(x_3)e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\big |\\&:= I_{2,1}+ I_{2,2}. \end{aligned}$$

and

$$\begin{aligned} I_{2,1}&= {\frac{1}{2}}\bigg | \iiint \Delta _{x_3,\pi } \Big (\Delta _{x_1,\pi } \big (\widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_1)\big )\zeta _2(x_3)\Big ) e^{-ix_1}\zeta _1(x_2)e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3 \bigg |\\&\lesssim \int _{|x_3| \ge 8}\int _{|x_2| < 16}\int _{|x_1|\ge 8} \frac{1}{|x_1|^{1+{\theta _1}} |x_2| |x_3|^{1+{\theta _1}}}\Big (\Big |\frac{x_1x_2\xi }{x_3\chi \eta }\Big |+\Big |\frac{x_3\chi \eta }{x_1x_2\xi }\Big |\Big )^{-{\theta _2}}\text {d}x_1\text {d}x_2\text {d}x_3\\&\lesssim 1. \end{aligned}$$

To estimate $I_{2,2},$ we write

$$\begin{aligned} I_{2,2}&=\iiint \Delta _{x_1,\pi } \big ( \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_1) \big ) e^{-ix_1}\zeta _1(x_2)\zeta _1(x_3)\big (e^{-ix_2}e^{-ix_3}-1\big )\text {d}x_1\text {d}x_2\text {d}x_3\\&\quad + \iiint \Delta _{x_1,\pi } \big ( \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_1) \big ) e^{-ix_1}\zeta _1(x_2)\zeta _1(x_3)\text {d}x_1\text {d}x_2\text {d}x_3. \end{aligned}$$

Inserting $|e^{-ix_2}e^{-ix_3 }-1|\le |x_2|+|x_3 |$ into the first integral together with the condition (RR) and using the cancellation condition (C3.c) for the second integral, respectively, we get

$$\begin{aligned} I_{2,2}&\lesssim \int _{|x_3 |< 16}\int _{|x_2| < 16}\int _{|x_1|\ge 8} \frac{1}{|x_1|^{1+{\theta _1}} |x_2| |x_3|}\Big (\Big |\frac{x_1x_2\xi }{x_3\chi \eta }\Big |+\Big |\frac{x_3\chi \eta }{x_1x_2\xi }\Big |\Big )^{-{\theta _2}}\\&\quad (|x_2|+|x_3|)\text {d}x_2\text {d}x_3 \\&\quad + \int _{|x_1|\ge 8}|x_1|^{-1-{\theta _1}}\text {d}x_1, \end{aligned}$$

which is dominated by the constant. Altogether, we obtain the required bound for term I.

Now we estimate term II. We first write

$$\begin{aligned} II&= \iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)(e^{-ix_1}-1)e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\\&\quad +\iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)e^{-ix_2} e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\\&:= II_1 +II_2. \end{aligned}$$

We further write

$$\begin{aligned} II_1=&\iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)(e^{-ix_1}-1)\zeta _2(x_2)e^{-ix_2} e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3 \\&+\iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)(e^{-ix_1}-1)\zeta _1(x_2) e^{-ix_2} e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\\ :=&II_{1,1}+ II_{1,2}. \end{aligned}$$

For term $II_{1,1},$ we have

$$\begin{aligned} |II_{1,1}|= {\frac{1}{2}}\bigg | \iiint \Delta _{x_2,\pi } \big (\widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_2)\big )\zeta _1(x_1)(e^{-ix_1}-1)e^{-ix_2} e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3 \bigg |. \end{aligned}$$

Then the required bound follows from the fact that $|e^{-ix_1}-1|\le |x_1|$ and the condition (RR).

Similarly, we write

$$\begin{aligned} II_{1,2}&=\iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)(e^{-ix_1}-1)\zeta _1(x_2)e^{-ix_2} \zeta _2(x_3)e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3\\&\quad + \iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)(e^{-ix_1}-1)\zeta _1(x_2)e^{-ix_2} \zeta _1(x_3)e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3 \\&:=II_{1,2,1}+ II_{1,2,2}. \end{aligned}$$

Since

$$\begin{aligned}&|II_{1,2,1}|\\&= {\frac{1}{2}}\bigg |\iiint \Delta _{x_3,\pi } \big (\widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_3)\big ) \zeta _1(x_1)(e^{-ix_1}-1)\zeta _1(x_2)e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3 \bigg |. \end{aligned}$$

The required bound for $II_{1,2,1}$ then is concluded by the fact that $|e^{-ix_1}-1|\le |x_1|$ and the condition (RR). To estimate term $II_{1,2,2},$ we write

$$\begin{aligned} II_{1,2,2}&=\iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)(e^{-ix_1}-1)\zeta _1(x_2)\zeta _1(x_3)\big (e^{-ix_2} e^{-ix_3}-1\big )\text {d}x_1\text {d}x_2\text {d}x_3\\&\quad +\iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)(e^{-ix_1}-1)\zeta _1(x_2)\zeta _1(x_3)\text {d}x_1\text {d}x_2\text {d}x_3. \end{aligned}$$

Using the facts that $|e^{-ix_1}-1|\le |x_1|$ and $|e^{-ix_2}e^{-ix_3}-1|\le |x_2|+|x_3|$ and the condition (RR) for the first integral and the condition (C3.c) for the second integral yields the desired bound for $II_{1,2,2}.$

Finally, we estimate term $II_2.$ Denote $II_2=II_{2,1}+II_{2,2},$ where $II_{2,1}$ and $II_{2,2}$ are given by $\iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)\zeta _2(x_2)e^{-ix_2}$$e^{-ix_3}\text {d}x_1\text {d}x_2\text {d}x_3$ and $\iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)\zeta _1(x_2)e^{-ix_2}$$e^{-ix_3}\text {d}x_1\text {d}x_2\text {d}x_3,$ respectively. Then

$$\begin{aligned} |II_{2,1}|= {\frac{1}{2}}\bigg |\iiint \Delta _{x_2,\pi } \big ( \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _2(x_2)\big )\zeta _1(x_1)e^{-ix_2}e^{-ix_3 }\text {d}x_1\text {d}x_2\text {d}x_3 \bigg | \lesssim 1. \end{aligned}$$

For $II_{2,2}$, inserting

$$\begin{aligned} \zeta _1(x_1)\zeta _1(x_2)e^{-ix_2} e^{-ix_3}&= \zeta _1(x_1)\zeta _1(x_2)e^{-ix_2} \zeta _2(x_3)e^{-ix_3}\\&\quad +\zeta _1(x_1)\zeta _1(x_2) \zeta _1(x_3)\big (e^{-ix_2}e^{-ix_3}-1\big ) + \zeta _1(x_1)\zeta _1(x_2) \zeta _1(x_3) \end{aligned}$$

into

$$\begin{aligned} \iiint \widetilde{\mathcal {K}}(x_1,x_2,x_3)\zeta _1(x_1)\zeta _1(x_2)e^{-ix_2} e^{-ix_3}\text {d}x_1\text {d}x_2\text {d}x_3 \end{aligned}$$

and using the condition (C1.b) and (C1.a), respectively. Thus, these estimates yield the bound of $II_{2,2}$ and hence the required bound for term II. The $L^2$ boundedness of ${\mathcal {K}}*f$ follows.

Next, to show the $L^p$ boundedness of the operator ${\mathcal {K}}*f$, similar to the proof of Theorem 1.3, it suffices to prove the following lemma.

Lemma 3.3

Suppose that $\phi ^{(1)}$ and $\phi ^{(2)}$ satisfy the conditions (3.1)–(3.4) and $\mathcal {K}$ is a distribution defined on ${\mathbb {R}}^3$ and satisfies the conditions (RR) and (C3.a)–(C3.c). Then for $\lambda ={\frac{1}{2}}\min ({\theta _1},{\theta _2}),$

$$\begin{aligned} |{\mathcal {K}}*(\phi ^{(1)}\otimes \phi ^{(2)})(x_1,x_2,x_3)|\le \frac{C_\lambda }{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}, \end{aligned}$$

where $C_\lambda$ is the constant depending only on $\lambda .$

Proof

The proof the Lemma 3.3 is similar to the proof of lemma 3.2. For simplicity, let $S={\mathcal {K}}*(\phi ^{(1)}\otimes \phi ^{(2)})(x_1,x_2,x_3)$. We consider the following eight cases:

Case 1$|x_1|\ge 3, |x_2| \ge 3, |x_3| \ge 3$. For this case, we use (3.4) to write

$$\begin{aligned} S&=\iiint \big [{\mathcal {K}}(x_1-u_1,x_2-u_2,x_3-u_3) -{\mathcal {K}}(x_1,x_2-u_2,x_3-u_3)\\&\quad -{\mathcal {K}}(x_1-u_1,x_2,x_3)+{\mathcal {K}}(x_1,x_2,x_3)\big ] \phi ^{(1)}(u_1)\phi ^{(2)}(u_2,u_3)\text {d}u_3\text {d}u_2\text {d}u_1. \end{aligned}$$

Note that ${\mathcal {K}}(x_1-u_1,x_2-u_2,x_3-u_3)-{\mathcal {K}}(x_1,x_2-u_2,x_3-u_3)- \big ({\mathcal {K}}(x_1-u_1,x_2,x_3)-{\mathcal {K}}(x_1,x_2,x_3)\big ) = \Delta _{x_2,-u_2}\Delta _{x_1,-u_1} \mathcal {K}(x_1,x_2,x_3-u_3)+\Delta _{x_3,-u_3} \Delta _{x_1,-u_1} \mathcal {K}(x_1,x_2,x_3).$ Thus, by the condition (RR) with $\alpha =\beta =1, \gamma =0$ and $\alpha =\gamma =1,\beta =0$ respectively, we have

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{|u_1|^{\theta _1}|u_2|^{\theta _1}}{|x_1|^{1+{\theta _1}} |x_2|^{1+{\theta _1}} |x_3|}\Big (\Big |\frac{x_1x_2}{x_3}\Big |+\Big | \frac{x_3}{x_1x_2}\Big |\Big )^{-\lambda }\text {d}u_3\text {d}u_2\text {d}u_1\\&\quad +\int _{|u_1| \le 1}\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{|u_1|^{\theta _1}|u_3|^{\theta _1}}{|x_1|^{1+{\theta _1}} |x_2| |x_3|^{1+{\theta _1}}}\Big (\Big |\frac{x_1x_2}{x_3}\Big |+\Big | \frac{x_3}{x_1x_2}\Big |\Big )^{-\lambda }\text {d}u_3\text {d}u_2\text {d}u_1\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 2$|x_1|\ge 3, |x_2| \ge 3, |x_3| < 3$. By the cancellation condition of $\phi ^{(1)},$

$$\begin{aligned} S&=\iiint \big [{\mathcal {K}}(x_1-u_1,x_2-u_2,x_3-u_3) -{\mathcal {K}}(x_1,x_2-u_2,x_3-u_3) \big ]\\&\quad \times \phi ^{(1)}(u_1)\phi ^{(2)}(u_2,u_3)\text {d}u_3\text {d}u_2\text {d}u_1. \end{aligned}$$

Therefore, by the condition (RR) with $\alpha =1$ and $\beta =\gamma =0,$ we obtain

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{|u_1|^{\theta _1}}{|x_1|^{1+{\theta _1}} |x_2| |x_3-u_3|}\Big |\frac{x_1x_2}{x_3-u_3} \Big |^{-\lambda }\text {d}u_3\text {d}u_2\text {d}u_1\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 3$|x_1|\ge 3, |x_2| < 3, |x_3| \ge 3.$ The same expression for S as in Case 2 yields

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{|u_1|^{\theta _1}}{|x_1|^{1+{\theta _1}} |x_2-u_2| |x_3|}\Big |\frac{x_3}{x_1(x_2-u_2)}\Big |^{-\lambda }\text {d}u_3\text {d}u_2\text {d}u_1\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Before handling the other cases, we introduce a bump function $\widetilde{\phi }$ on ${\mathbb {R}}$, with $\widetilde{\phi }(x_1)=1$ if $|x_1|\le 1/2$ and $\widetilde{\phi }(x_1)=0$ if $|x_1|\ge 1$.

Case 4$|x_1|\ge 3, |x_2|< 3, |x_3| < 3.$ Using the cancellation condition of $\phi ^{(1)}$, we write

$$\begin{aligned}&{\mathcal {K}}*(\phi ^{(1)}\otimes \phi ^{(2)})(x_1,x_2,x_3)\\&= \iiint \big ({\mathcal {K}}(x_1-u_1,u_2,u_3)-{\mathcal {K}}(x,u_2,u_3)\big ) \phi ^{(1)}(u_1)\\&\quad \times \big (\phi ^{(2)}(x_2-u_2,x_3-u_3)-\phi ^{(2)}(x_2,x_3)\big ) \widetilde{\phi }\Big (\frac{u_2}{10}\Big )\widetilde{\phi }\Big (\frac{u_3}{10}\Big ) \text {d}u_3\text {d}u_2\text {d}u_1\\&\quad +\iiint \big ({\mathcal {K}}(x_1-u_1,u_2,u_3)-{\mathcal {K}}(x_1,u_2,u_3)\big ) \phi ^{(1)}(u_1)\\&\quad \times \phi ^{(2)}(x_2,x_3) \widetilde{\phi }\Big (\frac{u_2}{10}\Big )\widetilde{\phi }\Big (\frac{u_3}{10} \Big )\text {d}u_3\text {d}u_2\text {d}u_1. \end{aligned}$$

Hence, by the condition (RR) with $\alpha =1,\beta =\gamma =0$ for the first integral, and the cancellation condition (C3.c) with $\alpha =1$ for the second integral, we get

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 10}\int _{|u_3| \le 10} \frac{|u_1|^{\theta _1}}{|x_1|^{1+{\theta _1}} |u_2| |u_3|}\Big (\Big |\frac{x_1u_2}{u_3}\Big |+\Big |\frac{u_3}{x_1u_2}\Big | \Big )^{-\lambda } (|u_2|+|u_3|)\text {d}u_3\text {d}u_2\text {d}u_1\\&\quad + \int _{|u_1| \le 1} \frac{|u_1|^{\theta _1}}{|x_1|^{1+{\theta _1}}} \text {d}u_1\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda }(1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 5$|x_1|< 3, |x_2| \ge 3, |x_3| \ge 3$. Similar to Case 4, using the cancellation condition of $\phi ^{(2)},$ we write

$$\begin{aligned} S&= \iiint \big [{\mathcal {K}}(u_1,x_2-u_2,x_3-u_3)-{\mathcal {K}}(u_1,x_2,x_3)\big ] \\&\quad \times \big (\phi ^{(1)}(x_1-u_1)-\phi ^{(1)}(x_1)\big ) \phi ^{(2)}(u_2,u_3) \widetilde{\phi }\Big (\frac{u_1}{10}\Big )\text {d}u_3\text {d}u_2\text {d}u_1 \\&\quad +\iiint \big [{\mathcal {K}}(u_1,x_2-u_2,x_3-u_3)-{\mathcal {K}}(u_1,x_2,x_3)\big ] \phi ^{(1)}(x_1)\phi ^{(2)}(u_2,u_3) \widetilde{\phi }\\&\qquad \Big (\frac{u_1}{10}\Big )\text {d}u_3\text {d}u_2\text {d}u_1. \end{aligned}$$

Note that ${\mathcal {K}}(u_1,x_2-u_2,x_3-u_3)-{\mathcal {K}}(u_1,x_2,x_3)=\Delta _{x_2,-u_2} \mathcal {K}(u_1,x_2,x_3-u_3) + \Delta _{x_3,-u_3} \mathcal {K}(u_1,x_2,x_3).$ Thus, using the condition (RR) on ${\mathcal {K}}$, the smoothness of $\phi ^{(1)}$ for the first integral, and the cancellation conditions (C3.b) with $\beta =1, \gamma =0$ and $\beta =0, \gamma =1,$ respectively, for the second integral, and applying the dominated convergence theorem, we obtain

$$\begin{aligned} |S|&\lesssim \int _{|u_1|\le 10}\int _{|u_2|\le 1}\int _{|u_3|\le 1} \bigg (\frac{|u_2|^{\theta _1}}{|u_1| |x_2|^{1+{\theta _1}} |x_3|}+\frac{|u_3|^{\theta _1}}{|u_1| |x_2| |x_3|^{1+{\theta _1}}}\Big )\\&\quad \times \Big (\Big |\frac{u_1x_2}{x_3}\Big | +\Big |\frac{x_3}{u_1x_2}\Big |\Big )^{-\lambda }|u_1|\text {d}u_3\text {d}u_2\text {d}u_1\\&\quad + \int _{|u_2| \le 1}\int _{|u_3| \le 1} \bigg (\frac{|u_2|^{\theta _1}}{|x_2|^{1+{\theta _1}} |x_3|}+\frac{|u_3|^{\theta _1}}{|x_2| |x_3|^{1+{\theta _1}}}\Big )\Big (\Big |\frac{4x_2}{x_3}\Big |+\Big |\frac{x_3}{4x_2}\Big | \Big )^{-\lambda }\text {d}u_3\text {d}u_2\\&\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }} \end{aligned}$$

Case 6$|x_1|< 3, |x_2| \ge 3, |x_3|< 3.$ Note that

$$\begin{aligned} S&= \iiint {\mathcal {K}}(u_1,x_2-u_2,x_3-u_3) \big (\phi ^{(1)}(x_1-u_1) -\phi ^{(1)}(x_1)\big ) \phi ^{(2)}(u_2,u_3) \widetilde{\phi }\\&\qquad \Big (\frac{u_1}{10}\Big ) \text {d}u_3\text {d}u_2\text {d}u_1\\&\quad +\iiint {\mathcal {K}}(u_1,x_2-u_2,x_3-u_3) \phi ^{(1)}(x_1)\phi ^{(2)}(u_2,u_3) \widetilde{\phi }\Big (\frac{u_1}{10}\Big )\text {d}u_3\text {d}u_2\text {d}u_1. \end{aligned}$$

By the condition (RR) with $\alpha =\beta =\gamma =0$ and the smoothness condition of $\phi ^{(1)}$ on the first integral and the condition (C3.b) with $\beta =\gamma =0$ for the second integral, we have

$$\begin{aligned} |S|&\lesssim \int _{|u_1|\le 10}\int _{|u_2|\le 1}\int _{|u_3|\le 1} \frac{1}{|u_1| |x_2| |x_3-u_3|}\Big (\Big |\frac{u_1x_2}{x_3-u_3}\Big |+\Big |\frac{x_3-u_3}{u_1x_2}\Big | \Big )^{-\lambda }\\&\qquad |u_1|\text {d}u_3\text {d}u_2\text {d}u_1\\&\quad +\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{1}{|x_2| |x_3-u_3|}\Big (\Big |\frac{x_2}{x_3-u_3}\Big |+\Big |\frac{x_3-u_3}{x_2}\Big | \Big )^{-\lambda }\text {d}u_3\text {d}u_2\\&\lesssim \frac{1}{|x_2|^{1+\lambda }}\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 7$|x_1|< 3, |x_2| < 3, |x_3| \ge 3$. The required estimate follows directly from the condition (RR):

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 1}\int _{|u_2| \le 1}\int _{|u_3| \le 1} \frac{1}{|x_1-u_1| |x_2-u_2| |x_3|}\Big |\frac{x_3}{(x_1-u_1) (x_2-u_2)}\Big |^{-\lambda }\text {d}u_3\text {d}u_2\text {d}u_1\\&\lesssim \frac{1}{|x_3|^{1+\lambda }}\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

Case 8$|x_1|< 3, |x_2|< 3, |x_3| < 3$. Inserting

$$\begin{aligned}&\phi ^{(1)}(x_1-u_1)\phi ^{(2)}(x_2-u_2,x_3-u_3)\\&\quad = [\phi ^{(1)}(x_1-u_1)-\phi ^{(1)}(x_1)][\phi ^{(2)}(x_2-u_2,x_3-u_3) -\phi ^{(2)}(x_2,x_3)]\\&\quad \qquad + \phi ^{(1)}(x_1)[\phi ^{(2)}(x_2-u_2,x_3-u_3)-\phi ^{(2)}(x_2,x_3)]\\&\qquad + [\phi ^{(1)}(x_1-u_1)-\phi ^{(1)}(x_1)]\phi ^{(2)}(x_2,x_3) +\phi ^{(1)}(x_1)\phi ^{(2)}(x_2,x_3), \end{aligned}$$

we write

$$\begin{aligned} S&=\iiint {\mathcal {K}}(u_1,u_2,u_3) \times \Big \{[\phi ^{(1)}(x_1-u_1)-\phi ^{(1)}(x_1)][\phi ^{(2)}(x_2-u_2,x_3-u_3)\\&\qquad -\phi ^{(2)}(x_2,x_3)]\\&\quad + \phi ^{(1)}(x_1)[\phi ^{(2)}(x_2-u_2,x_3-u_3)-\phi ^{(2)}(x_2,x_3)] + [\phi ^{(1)}(x_1-u_1)\\&\qquad -\phi ^{(1)}(x_1)]\phi ^{(2)}(x_2,x_3)\\&\quad +\phi ^{(1)}(x_1)\phi ^{(2)}(x_2,x_3)\Big \}\widetilde{\phi } \Big (\frac{u_1}{10}\Big ) \widetilde{\phi }\Big (\frac{u_2}{10}\Big )\widetilde{\phi } \Big (\frac{u_3}{10}\Big )\text {d}u_3\text {d}u_2\text {d}u_1 \end{aligned}$$

as four integrals. Using the condition (RR) with $\alpha =\beta =\gamma =0$ and the smoothness condition of $\phi ^{(1)}$ for the first integral, the cancellation conditions (C3.b), (C3.c), and (C3.a) for the last three integrals, we obtain

$$\begin{aligned} |S|&\lesssim \int _{|u_1| \le 4}\int _{|u_2| \le 4}\int _{|u_3| \le 4} \frac{1}{|u_1| |u_2| |u_3|}\Big (\Big |\frac{u_1u_2}{u_3}\Big |+\Big |\frac{u_3}{u_1u_2}\Big | \Big )^{-{\theta _2}}\\&\quad |u_1|(|u_2|+|u_3|)\text {d}u_3\text {d}u_2\text {d}u_1\\&\quad +\int _{|u_2| \le 4}\int _{|u_3| \le 4} \frac{1}{|u_2| |u_3|}\Big (\Big |\frac{4u_2}{u_3}\Big |+\Big |\frac{u_3}{4u_2}\Big | \Big )^{-{\theta _2}}(|u_2|+|u_3|)\text {d}u_3\text {d}u_2\\&\quad +\int _{|u_1| \le 4} \frac{1}{|u_1|} |u_1|\text {d}u_1+1 \\&\lesssim 1\lesssim \frac{1}{(1+|x_1|)^{1+\lambda } (1+|x_2|)^{1+\lambda } (1+|x_3|)^{1+\lambda }}. \end{aligned}$$

This completes the proof of Lemma 3.3. $\square$

The proof of part (b) of Theorem 1.4 follows from part (a). Indeed, the conditions (RR) and (C2.a)–(C2.c) imply the conditions (C3.a)–(C3.c). To see this, inserting

$$\begin{aligned} \widetilde{\phi }( x_1,x_2,x_3)&=\big [(\widetilde{\phi }(x_1,x_2,x_3)-\widetilde{\phi }(0,x_2,x_3)) -(\widetilde{\phi }(x_1,0,0)-\widetilde{\phi }(0,0,0))\big ]\\&\quad +(\widetilde{\phi }(x_1,0,0)-\widetilde{\phi }(0,0,0)) +(\widetilde{\phi }(0,x_2,x_3)-\widetilde{\phi }(0,x_2,0))\\&\quad +(\widetilde{\phi }(0,x_2,0)-\widetilde{\phi }(0,0,0))+\widetilde{\phi }(0,0,0) \end{aligned}$$

into $\iiint \mathcal {K}(x_1,x_2,x_3)\widetilde{\phi }( x_1, x_2, x_3)\text {d}x_1\text {d}x_2\text {d}x_3,$ we obtain that

$$\begin{aligned}&\iiint \mathcal {K}(x_1,x_2,x_3)\widetilde{\phi }(R_1 x_1, R_2 x_2, R_1R_2 x_3)\text {d}x_1\text {d}x_2\text {d}x_3\\&\qquad =\lim _{\epsilon _1,\epsilon _2,\epsilon _3\rightarrow 0}\iiint _{E(\epsilon ,R_1,R_2)}\mathcal {K}(x_1,x_2,x_3)\widetilde{\phi }(R_1 x_1, R_2 x_2, R_1R_2 x_3)\text {d}x_1\text {d}x_2\text {d}x_3. \end{aligned}$$

Let $E(\epsilon ,R_1,R_2)=\{x\in {\mathbb {R}}^3: \epsilon _1 \le |x_1|\le \frac{1}{R_1},\epsilon _2 \le |x_2|\le \frac{1}{R_2},\epsilon _3 \le |x_3|\le \frac{1}{R_1R_2}\}$. Then we have

$$\begin{aligned}&\Bigg |\iiint _{E(\epsilon ,R_1,R_2)}\mathcal {K}(x_1,x_2,x_3)\widetilde{\phi }(R_1 x_1, R_2 x_2, R_1R_2 x_3)\text {d}x_1\text {d}x_2\text {d}x_3\Bigg |\\&\qquad \le \Bigg |\iiint _{E(\epsilon ,R_1,R_2)}\mathcal {K}(x_1,x_2,x_3)\Big \{[\widetilde{\phi }(R_1 x_1, R_2 x_2, R_1R_2 x_3)-\widetilde{\phi }(0,R_2 x_2,R_1R_2 x_3)] \\&\qquad \quad -[\widetilde{\phi }(R_1 x_1,0,0)-\widetilde{\phi }(0,0,0)]\Big \} \text {d}x_1\text {d}x_2\text {d}x_3\Bigg |\\&\qquad \quad +\Bigg |\iiint _{E(\epsilon ,R_1,R_2)}\mathcal {K}(x_1,x_2,x_3)(\widetilde{\phi }(R_1 x_1,0,0)-\widetilde{\phi }(0,0,0))\text {d}x_1\text {d}x_2\text {d}x_3\Bigg |\\&\qquad \quad +\Bigg |\iiint _{E(\epsilon ,R_1,R_2)}\mathcal {K}(x_1, x_2, x_3)(\widetilde{\phi }(0,R_2 x_2,R_1R_2 x_3)-\widetilde{\phi }(0,R_2 x_2,0)) \text {d}x_1\text {d}x_2\text {d}x_3\Bigg |\\&\qquad \quad +\Bigg |\iiint _{E(\epsilon ,R_1,R_2)}\mathcal {K}(x_1, x_2, x_3)(\widetilde{\phi }(0,R_2 x_2,0)-\widetilde{\phi }(0,0,0))\text {d}x_1\text {d}x_2\text {d}x_3\Bigg |\\&\qquad \quad +\Bigg |\iiint _{E(\epsilon ,R_1,R_2)}\mathcal {K}(x_1, x_2, x_3)\widetilde{\phi }(0,0,0)\text {d}x_1\text {d}x_2\text {d}x_3\Bigg |\\&\qquad \lesssim \int _{|x_3|\le \frac{1}{R_1R_2}}\int _{|x_2|\le \frac{1}{R_2}}\int _{|x_1|\le \frac{1}{R_1}}\frac{1}{|x_1| |x_2| |x_3|}\Big (\Big |\frac{x_1x_2}{x_3}\Big |+\Big |\frac{x_3}{x_1x_2}\Big |\Big )^{-{\theta _2}}\\&\qquad \quad \times |R_1x_1|(|R_2x_2|+|R_1R_2x_3|)\text {d}x_1\text {d}x_2\text {d}x_3\\&\qquad \quad +\int _{|x_1| \le \frac{1}{R_1}}\frac{1}{|x_1|}|R_1x_1|\text {d}x_1\\&\qquad \quad +\int _{|x_3| \le \frac{1}{R_1R_2}}\int _{{\mathbb {R}}}\frac{1}{|x_2| |x_3|}\Big (\Big |\frac{x_2}{R_1x_3}\Big |+\Big |\frac{R_1x_3}{x_2} \Big |\Big )^{-{\theta _2}}|R_1R_2x_3|\text {d}x_2\text {d}x_3\\&\qquad \quad +\Bigg |\int _{|x_2| \le \frac{1}{R_2}}\frac{1}{|x_2|}|R_2x_2|\text {d}x_2 +1 \lesssim 1, \end{aligned}$$

where we apply (RR) and (C2.c) for the first and second term, respectively, (C2.b) for the third and fourth term, and (C2.a) for the last term above, which imply that ${\mathcal {K}}$ satisfies (C3.a).

Similarly, for any $0 \le \beta +\gamma \le 1$, n.b.f. $\widetilde{\phi }$ on ${\mathbb {R}}$ and $R>0,$ we can write

$$\begin{aligned}&\bigg |\Delta _{x_2,h_2}^\beta \Delta _{x_3,h_3}^\gamma \mathcal {K}(x_1,x_2,x_3) \widetilde{\phi }(R x_1)\text {d}x_1\bigg |\\&\quad =\lim \limits _{\epsilon \rightarrow 0}\bigg |\int _{\epsilon \le |x_1| \le \frac{1}{R}} \Delta _{x_2,h_2}^\beta \Delta _{x_3,h_3}^\gamma \mathcal {K}(x_1, x_2, x_3)\widetilde{\phi }(R x_1)\text {d}x_1\bigg |. \end{aligned}$$

and

$$\begin{aligned}&\bigg |\int _{\epsilon \le |x_1| \le \frac{1}{R}} \Delta _{x_2,h_2}^\beta \Delta _{x_3,h_3}^\gamma \mathcal {K}(x_1,x_2,x_3)\widetilde{\phi }(R x_1)\text {d}x_1\bigg |\\&\quad \le \bigg |\int _{\epsilon \le |x_1| \le \frac{1}{R}} \Delta _{x_2,h_2}^\beta \Delta _{x_3,h_3}^\gamma \mathcal {K}(x_1,x_2,x_3)(\widetilde{\phi }(R x_1) -\widetilde{\phi }(0))\text {d}x_1\bigg |\\&\qquad +\bigg |\int _{\epsilon \le |x_1| \le \frac{1}{R}} \Delta _{x_2,h_2}^\beta \Delta _{x_3,h_3}^\gamma \mathcal {K}(x_1,x_2,x_3)\widetilde{\phi }(0)\text {d}x_1\bigg |\\&\quad \lesssim \int _{ |x_1| \le \frac{1}{R}} \frac{|h_2|^{\beta \theta _1}|h_3|^{\gamma \theta _1}}{|x_1| |x_2|^{{\beta \theta _1}+1} |x_3|^{{\gamma \theta _1}+1}}\Big (\Big |\frac{x_1x_2}{x_3}\Big |+\Big |\frac{x_3}{x_1x_2} \Big |\Big )^{-{\theta _2}}|Rx_1|\text {d}x_1\\&\qquad \quad +\frac{|h_2|^{\beta \theta _1}|h_3|^{\gamma \theta _1}}{|x_2|^{{\beta \theta _1}+1} |x_3|^{{\gamma \theta _1}+1}}\Big (\Big |\frac{x_2}{Rx_3}\Big |+\Big |\frac{Rx_3}{x_2}\Big |\Big )^{-{\theta _2}}\\&\qquad \quad + \frac{|h_2|^{\beta \theta _1}|h_3|^{\gamma \theta _1}}{|x_2|^{{\beta \theta _1}+1} |x_3|^{{\gamma \theta _1}+1}}\Big (\Big |\frac{\epsilon x_2}{ x_3}\Big |+\Big |\frac{ x_3}{\epsilon x_2}\Big |\Big )^{-{\theta _2}}\\&\qquad \lesssim \frac{|h_2|^{\beta \theta _1}|h_3|^{\gamma \theta _1}}{|x_2|^{{\beta \theta _1}+1} |x_3|^{{\gamma \theta _1}+1}}\Big (\Big |\frac{x_2}{Rx_3}\Big |+\Big |\frac{Rx_3}{x_2}\Big |\Big )^{-{\theta _2}}\\&\quad \qquad + \frac{|h_2|^{\beta \theta _1}|h_3|^{\gamma \theta _1}}{|x_2|^{{\beta \theta _1}+1} |x_3|^{{\gamma \theta _1}+1}}\Big (\Big |\frac{x_2}{\epsilon x_3}\Big |+\Big |\frac{\epsilon x_3}{x_2}\Big |\Big )^{-{\theta _2}}. \end{aligned}$$

Taking $\epsilon \rightarrow 0$, then (C3.b) is obtained.

Finally we verify (C3.c). For any $0\le \alpha \le 1$, n.b.f. $\widetilde{\phi }$ on ${\mathbb {R}}^2$ and $R_1,R_2>0,$ we write

$$\begin{aligned}&\bigg |\iint \Delta _{x_1,h_1}^\alpha \mathcal {K}(x_1,x_2,x_3)\widetilde{\phi }(R_1 x_2, R_2x_3)\text {d}x_2\text {d}x_3\bigg |\\&\qquad =\lim _{\epsilon _1,\epsilon _2 \rightarrow 0} \bigg |\int _{\epsilon _2\le |x_3|\le \frac{1}{R_2}}\int _{\epsilon _1\le |x_2|\le \frac{1}{R_1}}\Delta _{x_1,h_1}^\alpha \mathcal {K}(x_1,x_2,x_3) \widetilde{\phi }(R_1 x_2, R_2x_3)\text {d}x_2\text {d}x_3\bigg |. \end{aligned}$$

and

$$\begin{aligned}&\bigg |\int _{\epsilon _2\le |x_3|\le \frac{1}{R_2}}\int _{\epsilon _1\le |x_2|\le \frac{1}{R_1}}\Delta _{x_1,h_1}^\alpha \mathcal {K}(x_1,x_2,x_3)\widetilde{\phi }(R_1 x_2, R_2x_3)\text {d}x_2\text {d}x_3\bigg |\\&\qquad \le \bigg |\int _{\epsilon _2\le |x_3|\le \frac{1}{R_2}}\int _{\epsilon _1\le |x_2|\le \frac{1}{R_1}}\Delta _{x_1,h_1}^\alpha \mathcal {K}(x_1,x_2,x_3)\big (\widetilde{\phi }(R_1 x_2, R_2x_3) -\widetilde{\phi }(0, 0)\big )\text {d}x_2\text {d}x_3\bigg |\\&\qquad \quad +\bigg |\int _{\epsilon _2\le |x_3|\le \frac{1}{R_2}}\int _{\epsilon _1\le |x_2|\le \frac{1}{R_1}}\Delta _{x_1,h_1}^\alpha \mathcal {K}(x_1,x_2,x_3)\widetilde{\phi }(0, 0)\text {d}x_2\text {d}x_3\bigg |\\&\qquad \lesssim \int _{ |x_3|\le \frac{1}{R_2}}\int _{ |x_2|\le \frac{1}{R_1}} \frac{|h_1|^{\alpha \theta _1}}{|x_1|^{{\alpha \theta _1}+1} |x_2| |x_3|}\Big (\Big |\frac{x_1x_2}{x_3}\Big |+\Big |\frac{x_3}{x_1x_2} \Big |\Big )^{-{\theta _2}}(|R_1x_2|+|R_2x_3|)\text {d}x_2\text {d}x_3\\&\qquad \quad +\frac{|h_1|^{\alpha \theta _1}}{|x_1|^{{\alpha \theta _1}+1} }\\&\qquad \lesssim \frac{|h_1|^{\alpha \theta _1}}{|x_1|^{{\alpha \theta _1}+1} }. \end{aligned}$$

Thus (C3.c) is obtained. This completes the proof of part (b) and hence Theorem 1.4 is concluded.

4 Proof of Theorem 1.6

In this section, we provide the proof of Theorem 1.6. We first study the singular integral $T_{{\mathfrak {z}}}$ with the convolution kernel as in (1.6), see Theorem 4.1 below. Then we consider the singular integral T with the convolution kernel as in (1.7), see Remark 4.5.

Theorem 4.1

Suppose that $\phi ^{(1)}$ and $\phi ^{(2)}$ are defined as in (3.1) and

$$\begin{aligned} \mathcal {K}(x_1, x_2, x_3)=\sum _{j,k}2^{2j+2k}\phi ^{(1)}(2^jx_1) \phi ^{(2)}(2^kx_2,2^{j+k}x_3). \end{aligned}$$

Then

$$\begin{aligned} |\partial _{x_1}^\alpha \partial _{x_2}^\beta \partial _{x_3}^\gamma \mathcal {K}(x_1, x_2, x_3)| \le \frac{C_{\alpha ,\beta ,\gamma ,{\theta _2}}}{|x_1|^{\alpha +1} |x_2|^{\beta +1} |x_3|^{\gamma +1}} \Big (\Big |\frac{x_1x_2}{x_3}\Big |+\Big |\frac{x_3}{x_1x_2}\Big |\Big )^{-{\theta _2}} \end{aligned}$$

(4.1)

for all $\alpha ,\beta ,\gamma \ge 0$ and $0<{\theta _2}<1;$

$$\begin{aligned} \bigg |\int _{\delta _1 \le |x_1|\le r_1}\int _{\delta _2 \le |x_2|\le r_2}\int _{\delta _3 \le |x_3|\le r_3} \mathcal {K}(x_1, x_2, x_3)\text {d}x_1\text {d}x_2\text {d}x_3\bigg | \le C \end{aligned}$$

(4.2)

uniformly for all $\delta _1,\delta _2,\delta _3,r_1,r_2,r_3>0;$

$$\begin{aligned}&\bigg |\int _{\delta \le |x_1|\le r}\partial _{x_2}^\beta \partial _{x_3}^\gamma \mathcal {K}(x_1, x_2, x_3)\text {d}x_1\bigg |\\&\qquad \le \frac{C_{\beta ,\gamma ,{\theta _2}}}{|x_1|^{\beta +1} |x_3|^{1+\gamma }}\Bigg (\frac{1}{\big (|\frac{rx_2}{x_3}|+|\frac{x_3}{rx_2}| \big )^{{\theta _2}}}+\frac{1}{\big (|\frac{\delta x_2}{x_3}|+|\frac{x_3}{\delta x_2}| \big )^{{\theta _2}}}\Bigg ),\nonumber \end{aligned}$$

(4.3)

for all $\delta , r>0, \beta , \gamma \ge 0$ and $0<{\theta _2}<1;$

$$\begin{aligned} \bigg |\int _{\delta _1 \le |x_2|\le r_1}\int _{\delta _2 \le |x_3|\le r_2}\partial _{x_1}^\alpha \mathcal {K}(x_1, x_2, x_3)\text {d}x_2 \text {d}x_3 \bigg | \le \frac{C_{\alpha }}{|x_1|^{\alpha +1}} \end{aligned}$$

(4.4)

uniformly for all $\delta _1,\delta _2,r_1,r_2>0$ and $\alpha \ge 0.$

To show Theorem 4.1, we need the following simple lemmas.

Lemma 4.2

Suppose $a,b,c>0$ with $b> a$ and $r_1,r_2,r_3>0$. Then, for all $0<\varepsilon <1,$

$$\begin{aligned} \sum _j 2^{ja}\frac{1}{(1+2^jr_1)^b}\frac{1}{(r_2+2^jr_3)^c}\le C_\varepsilon r_1^{-a}r_2^{-c}\Big (1+\frac{r_3}{r_1r_2}\Big )^{-(a\wedge c) +1-\varepsilon }. \end{aligned}$$

Proof

We first write

$$\begin{aligned}&\sum _j 2^{ja}\frac{1}{(1+2^jr_1)^b}\frac{1}{(r_2+2^jr_3)^c} = \sum _j 2^{ja}\frac{1}{(1+2^jr_1)^b}\frac{1}{\big (1+2^j\frac{r_3}{r_2}\big )^c}\frac{1}{r_2^c}\\&\qquad \lesssim \sum _{j:2^j>r_1^{-1},2^j>r_2r_3^{-1}} 2^{ja}\frac{1}{(2^jr_1)^b}\frac{1}{\big (2^j\frac{r_3}{r_2}\big )^c}\frac{1}{r_2^c} +\sum _{j:2^j>r_1^{-1},2^j\le r_2r_3^{-1}}2^{ja}\frac{1}{(2^jr_1)^b}\frac{1}{r_2^c}\\&\qquad \qquad +\sum _{j:2^j\le r_1^{-1},2^j>r_2r_3^{-1}} 2^{ja}\frac{1}{\big (2^j\frac{r_3}{r_2}\big )^c}\frac{1}{r_2^c} +\sum _{j:2^j\le r_1^{-1},2^j\le r_2r_3^{-1}} 2^{ja}\frac{1}{r_2^c}\\&\qquad := I+ II+ III + IV. \end{aligned}$$

For term I,  we observe that

$$\begin{aligned} I\lesssim \Big (1+\frac{r_3}{r_1r_2}\Big )^{a-b-c}r_1^{-a}r_2^{-c} \Big (\frac{r_3}{r_1r_2}\Big )^{b-a} \lesssim r_1^{-a}r_2^{-c}\Big (1+\frac{r_3}{r_1r_2}\Big )^{-c}. \end{aligned}$$

For II, since $r_3<r_1r_2$,

$$\begin{aligned} II \lesssim r_1^{-a}r_2^{-c} \lesssim r_1^{-a}r_2^{-c}\Big (1+\frac{r_3}{r_1r_2}\Big )^{-c}. \end{aligned}$$

For III,  note that $r_3>r_1r_2$. We consider three cases. In the first case where $a>c$, we obtain

$$\begin{aligned} III \lesssim r_1^{-a}r_2^{-c}\Big (\frac{r_3}{r_1r_2}\Big )^{-c} \lesssim r_1^{-a}r_2^{-c}\Big (1+\frac{r_3}{r_1r_2}\Big )^{-c}. \end{aligned}$$

If $a<c$, then

$$\begin{aligned} III \lesssim r_1^{-a}r_2^{-c}\Big (\frac{r_3}{r_1r_2}\Big )^{-a} \lesssim r_1^{-a}r_2^{-c}\Big (1+\frac{r_3}{r_1r_2}\Big )^{-a}. \end{aligned}$$

When $a=c$, we have

$$\begin{aligned} III \lesssim r_3^{-c}\log \Big (\frac{r_3}{r_1r_2}\Big )\lesssim r_1^{-a}r_2^{-c}\Big (1+\frac{r_3}{r_1r_2}\Big )^{-a+1-{\theta _2}}. \end{aligned}$$

Finally, for term IV,  we have

$$\begin{aligned} IV \lesssim \Big (\frac{r_1r_2}{r_3+r_1r_2}\Big )^ar_1^{-a}r_2^{-c}= r_1^{-a}r_2^{-c}\Big (1+\frac{r_3}{r_1r_2}\Big )^{-a}. \end{aligned}$$

These estimates yield the required bound and Lemma 4.2 is proved. $\square$

Lemma 4.3

For any $N>0$, $r>0$ and $k \in \mathbb {Z},$ we have

$$\begin{aligned} \int _{\{x_1\in {\mathbb {R}}:|x_1|\le r\}} \frac{1}{(1+2^k|x_1|)^N} \text {d}x_1 \lesssim \frac{r}{1+2^kr} \end{aligned}$$

and

$$\begin{aligned} \int _{\{x_1\in {\mathbb {R}}: |x_1|> r\}} \frac{1}{(1+2^k|x_1|)^N} \text {d}x_1 \lesssim \frac{2^{-k}}{(1+2^kr)^{N-1}}. \end{aligned}$$

Proof

We consider two cases. For the case $r \le 2^{-k}$, we clearly have $\int _{|x_1|\le r} \frac{1}{(1+2^k|x_1|)^N} \text {d}x_1 \lesssim r$. The second inequality follows from

$$\begin{aligned} \int _{{\mathbb {R}}} \frac{1}{(1+2^k|x_1|)^N} \text {d}x_1 = 2^{-k}\int _{{\mathbb {R}}} \frac{1}{(1+|x_1|)^N} \text {d}x_1\lesssim 2^{-k}. \end{aligned}$$

(4.5)

If $r > 2^{-k}$, then the first inequality follows again from (4.5) while the second follows from

$$\begin{aligned} \int _{|x_1> r} \frac{1}{(1+2^k|x_1|)^N} \text {d}x_1 \le \int _{|x_1|> r} \frac{1}{(2^k|x_1|)^N} \text {d}x_1 \lesssim \frac{2^{-k}}{(1+2^kr)^{N-1}}. \end{aligned}$$

The proof of Lemma 4.3 is finished. $\square$

We now return to show Theorem 4.1.

Proof of Theorem 4.1

We prove the regularity estimate (4.1) first. By the definition of ${\mathcal {K}}$ and the conditions on $\phi ^{(1)}$ and $\phi ^{(2)},$ we have

$$\begin{aligned} |\partial _{x_1}^\alpha \partial _{x_2}^\beta \partial _{x_3}^\gamma {\mathcal {K}}(x_1, x_2, x_3)| \lesssim \sum _{j,k}\frac{2^{2j+2k+j(\alpha +\gamma )+k(\beta +\gamma )}}{(1+2^j|x_1|)^{3+\alpha +\gamma }(1+2^k|x_2|+2^{j+k}|x_3|)^3}. \end{aligned}$$

Note that

$$\begin{aligned} \sum _{k} \frac{2^{2k+k(\beta +\gamma )}}{(1+2^k|x_2|+2^{j+k}|x_3|)^3}&= \sum _{k:2^k \le (|x_2|+2^j|x_3|)^{-1}}\frac{2^{2k+k(\beta +\gamma )}}{(1+2^k|x_2|+2^{j+k}|x_3|)^3}\\ \nonumber&\qquad + \sum _{k:2^k > (|x_2|+2^j|x_3|)^{-1}}\frac{2^{2k+k(\beta +\gamma )}}{(1+2^k|x_2|+2^{j+k}|x_3|)^3}\\&\lesssim \frac{1}{(|x_2|+2^j|x_3|)^{2+\beta +\gamma }}. \end{aligned}$$

Inserting this estimate into the above inequality, we obtain

$$\begin{aligned} |\partial _{x_1}^\alpha \partial _{x_2}^\beta \partial _{x_3}^\gamma {\mathcal {K}}(x_1, x_2, x_3)|&\lesssim \sum _{j} \frac{2^{2j+j(\alpha +\gamma )}}{(1+2^j|x_1|)^{3+\alpha +\gamma } (|x_2|+2^j|x_3|)^{2+\beta +\gamma }}\\&\lesssim \frac{1}{|x_1|^{\alpha +\gamma +2} |x_2|^{\beta +\gamma +2} \big (1+|\frac{x_3}{x_1x_2}| \big )^{(\alpha \wedge \beta )+\gamma +1+{\theta _2}}}, \end{aligned}$$

where we apply Lemma 4.2 with $a=2+\alpha +\gamma , b=3+\alpha +\gamma , c=2+\beta +\gamma , r_1=|x_1|, r_2=|x_2|$ and $r_3=|x_3|$ in the last inequality. This implies the required estimate.

We now show the cancellation conditions (4.2)–(4.4). To verify (4.3), we observe that

$$\begin{aligned}&\bigg |\int _{\delta \le |x_1|\le r}\partial _{x_2}^\beta \partial _{x_3}^\gamma {\mathcal {K}}(x_1, x_2, x_3)\text {d}x_1\bigg |\\&\qquad \lesssim \sum _{j,k}2^{2j+2k+k\beta +(j+k)\gamma }\bigg |\int _{\delta \le |x_1|\le r}\phi ^{(1)}(2^jx_1) (\partial _{x_2}^\beta \partial _{x_3}^\gamma \phi ^{(2)})(2^k x_2,2^{j+k}x_3)\text {d}x_1\bigg |. \end{aligned}$$

Note that for all $N\ge 2,$

$$\begin{aligned} \bigg |\int _{|x_1|\le r}\phi ^{(1)}(2^jx_1)\text {d}x_1\bigg | \lesssim r, \end{aligned}$$

and by the vanishing condition of $\phi ^{(1)},$

$$\begin{aligned} \bigg |\int _{|x_1|\le r}\phi ^{(1)}(2^jx_1)\text {d}x_1\bigg |&=\bigg |\int _{|x_1|> r}\phi ^{(1)}(2^jx_1)\text {d}x_1\bigg |\\&\le C_N\bigg |\int _{|x_1|> r}\frac{1}{(2^j|x_1|)^N}\text {d}x_1\bigg | \le C_N 2^{-jN}r^{1-N}. \end{aligned}$$

Therefore,

$$\begin{aligned} \bigg |\int _{|x_1|\le r}\phi ^{(1)}(2^jx_1)\text {d}x_1\bigg | \le C_N\frac{r}{(1+2^jr)^N}, \end{aligned}$$

which implies

$$\begin{aligned} \bigg |\int _{\delta \le |x_1|\le r}\partial _{x_2}^\beta \partial _{x_3}^\gamma {\mathcal {K}}(x_1, x_2, x_3)\text {d}x_1\bigg |&\lesssim \sum _{j,k} \frac{2^{2j+2k+j\gamma +k(\beta +\gamma )} r}{(1+2^jr)^{3+\gamma } (1+2^k|x_2|+2^{j+k}|x_3|)^3} \\&\quad +\sum _{j,k}\frac{2^{2j+2k+j\gamma +k(\beta +\gamma )} \delta }{(1+2^j\delta )^{3+\gamma } (1+2^k|x_2|+2^{j+k}|x_3|)^3}. \end{aligned}$$

Summing over k first yields that the two summations above are dominated by

$$\begin{aligned} \sum _{j}\big (\frac{r}{(1+2^jr)^{3+\gamma }}+\frac{\delta }{(1+2^j\delta )^{3+\gamma }}\big ) \frac{2^{2j+j\gamma }}{(|x_2|+2^j|x_3|)^{2+\beta +\gamma }}. \end{aligned}$$

Applying Lemma 4.2 with $a=2+\gamma , b=3+\gamma , c=2+\gamma +\beta , r_1=r$ or $\delta$, $r_2=|x_2|, r_3=|x_3|$, we obtain

$$\begin{aligned}&\bigg |\int _{\delta \le |x_1|\le r}\partial _{x_2}^\beta \partial _{x_3}^\gamma {\mathcal {K}}(x_1, x_2, x_3)\text {d}x_1\bigg |\\&\qquad \lesssim \frac{1}{r^{\gamma +1} |x_2|^{\beta +\gamma +2} \big (1+|\frac{x_3}{rx_2}| \big )^{\gamma +1+{\theta _2}}}+\frac{1}{\delta ^{\gamma +1} |x_2|^{\beta +\gamma +2} \big (1+|\frac{x_3}{\delta x_2}| \big )^{\gamma +1+{\theta _2}}} \end{aligned}$$

which implies the desired cancellation condition (4.3).

To show the cancellation condition (4.4), we start with

$$\begin{aligned}&\bigg |\int _{\delta _1 \le |x_2|\le r_1}\int _{\delta _2 \le |x_3|\le r_2}\partial _{x_1}^\alpha {\mathcal {K}}(x_1, x_2, x_3)\text {d}x_2 \text {d}x_3 \bigg | \\&\qquad \lesssim \sum _{j,k}2^{2j+2k+j\alpha }\bigg |\int _{|x_2|\le r_1}\int _{|x_3|\le r_2}(\partial _{x_1}^\alpha \phi ^{(1)})(2^jx_1)\phi ^{(2)}(2^k x_2,2^{j+k}x_3)\text {d}x_2 \text {d}x_3\bigg |\\&\qquad \quad + \sum _{j,k}2^{2j+2k+j\alpha }\bigg |\int _{|x_2|\le \delta _1}\int _{|x_3|\le \delta _2}(\partial _{x_1}^\alpha \phi ^{(1)})(2^jx_1)\phi ^{(2)}(2^k x_2,2^{j+k}x_3)\text {d}x_2 \text {d}x_3 \bigg |. \end{aligned}$$

By the vanishing condition of $\phi ^{(2)}$,

$$\begin{aligned}&\int _{|x_2|\le r_1}\int _{|x_3|\le r_2}\phi ^{(2)}(2^k x_2,2^{j+k}x_3)\text {d}x_2\text {d}x_3\\&\quad =\int _{|x_2|> r_1}\int _{|x_3|\le r_2}\phi ^{(2)}(2^k x_2,2^{j+k}x_3)\text {d}x_2\text {d}x_3\\&\qquad +\int _{|x_2|\le r_1}\int _{|x_3|> r_2}\phi ^{(2)}(2^kx_2,2^{j+k}x_3)\text {d}x_2\text {d}x_3\\&\qquad +\int _{|x_2|> r_1}\int _{|x_3|> r_2}\phi ^{(2)}(2^kx_2,2^{j+k}x_3)\text {d}x_2\text {d}x_3. \end{aligned}$$

Applying the size condition of $\phi ^{(2)}$ and Lemma 4.3, we obtain

$$\begin{aligned}&\bigg |\int _{|x_2|\le r_1}\int _{|x_3|\le r_2}\phi ^{(2)}(2^k x_2,2^{j+k}x_3)\text {d}x_2\text {d}x_3\bigg |\\&\quad \lesssim \frac{2^{-k}}{(1+2^kr_1)^{3}}\frac{r_2}{1+2^{j+k}r_2}+ \frac{r_1}{1+2^{k}r_1}\frac{2^{-j-k}}{(1+2^{j+k}r_2)^{3}} +\frac{2^{-k}}{(1+2^kr_1)^{3}}\frac{2^{-j-k}}{(1+2^{j+k}r_2)^{3}}. \end{aligned}$$

On other hand, by the size condition on $\phi ^{(2)}$,

$$\begin{aligned} \bigg |\int _{|x_2|\le r_1}\int _{|x_3|\le r_2}\phi ^{(2)}(2^k x_2,2^{j+k}x_3)\text {d}x_2d x_3\bigg | \le \frac{r_1}{1+2^kr_1}\frac{r_2}{1+2^{j+k}r_2}. \end{aligned}$$

Therefore,

$$\begin{aligned}&\sum _{j,k}2^{2j+2k+j\alpha }\Big | (\partial _{x_1} \phi ^{(1)})(2^jx_1)\int _{|x_2|\le r_1}\int _{|x_3|\le r_2} \phi ^{(2)}(2^kx_2,2^{j+k}x_3)(x_2,x_3)\text {d}x_2\text {d}x_3\Big |\\&\quad \lesssim \sum _{j,k}\frac{2^{2j+2k+j\alpha }}{(1+2^j|x_1|)^{3+\alpha }} \min \bigg \{\bigg (\frac{2^{-k}}{(1+2^kr_1)^3}\frac{r_2}{1+2^{j+k}r_2} +\frac{r_1}{1+2^{k}r_1}\frac{2^{-j-k}}{(1+2^{j+k}r_2)^{3}}\\&\qquad + \frac{2^{-k}}{(1+2^kr_1)^{3}}\frac{2^{-j-k}}{(1+2^{j+k}r_2)^{3}}\bigg ), \frac{r_1}{1+2^kr_1}\frac{r_2}{1+2^{j+k}r_2}\bigg \}. \end{aligned}$$

Summing over k first and considering the four cases: (i) $2^{k}\le r_1^{-1}$ and $2^k \le 2^{-j}r_2^{-1}$; (ii) $2^{k}\le r_1^{-1}$ and $2^k > 2^{-j}r_2^{-1}$; (iii) $2^{k}> r_1^{-1}$ and $2^k \le 2^{-j}r_2^{-1}$; (iv) $2^{k}> r_1^{-1}$ and $2^k > 2^{-j}r_2^{-1}$, we obtain that the last summation above is dominated by

$$\begin{aligned} \sum _{j}\frac{2^{2j+j\alpha }}{(1+2^j|x_1|)^{3+\alpha }}2^{-j}, \end{aligned}$$

which yields the cancellation condition (4.4).

Finally, the cancellation (4.2) follows directly from the following estimates.

$$\begin{aligned}&\bigg |\int _{\delta _1 \le |x_1|\le r_1}\int _{\delta _2 \le |x_2|\le r_2}\int _{\delta _3 \le |x_3|\le r_3} {\mathcal {K}}(x_1, x_2, x_3)\text {d}x_1\text {d}x_2\text {d}x_3\bigg |\\&\quad \lesssim \bigg |\sum _{j,k}2^{2j+2k}\int _{\delta _1\le |x_1|\le r_1} \phi ^{(1)}(2^jx_1)\text {d}x_1 \\&\qquad \times \int _{\delta _2\le |x_2|\le r_2}\int _{\delta _3\le |x_3|\le r_3} \phi ^{(2)}(2^kx_2,2^{j+k}x_3)(x_2,x_3)\text {d}x_2 \text {d}x_3\bigg |\\&\quad \lesssim \sum _{j}2^{2j}\bigg (\frac{r_1}{(1+2^jr_1)^3}2^{-j} +\frac{\delta _1}{(1+2^j\delta _1)^3}2^{-j}\bigg )\lesssim 1. \end{aligned}$$

The proof of Theorem 4.1 is complete. $\square$

As mentioned in Sect. 1, a special class of singular integral operators $T_{\mathfrak {z}}$ considered by Ricci and Stein [28] is of the form $T_{\mathfrak {z}}f=f* \mathcal {K}$, where

$$\begin{aligned} \mathcal {K}(x_1, x_2, x_3) =\sum _{k,j\in {{\mathbb {Z}}}}2^{2(k+j)} \phi \Big (2^jx_1,2^k x_2,2^{j+k}x_3\Big ) \end{aligned}$$

and the function $\phi$ is supported in an unit cube in ${\mathbb {R}}^3$ and satisfies a certain amount of uniform smoothness with cancellation conditions

$$\begin{aligned} \int _{{{\mathbb {R}}}^2}\phi (x_1, x_2, x_3)\text {d}x_1\text {d}x_2=\int _{{{\mathbb {R}}}^2} \phi (x_1, x_2, x_3)\text {d}x_2\text {d}x_3=\int _{{{\mathbb {R}}}^2}\phi (x_1, x_2, x_3)\text {d}x_3\text {d}x_1=0. \end{aligned}$$

Fefferman and Pipher [13] show that the above cancellation conditions are necessary for the $L^2$ boundedness for singular integral $T_{\mathfrak {z}}$. Moreover, if $\phi$ satisfies the above cancellation conditions, then $\phi$ can be decomposed by $\phi =\phi _1+\phi _2,$ where $\phi _1$ and $\phi _2$ have the following cancellation conditions

$$\begin{aligned} \int _{{\mathbb {R}}}\phi _1(x_1, x_2, x_3)\text {d}x_1=\int _{{{\mathbb {R}}}^2} \phi _1(x_1, x_2, x_3)\text {d}x_2\text {d}x_3=0 \end{aligned}$$

and

$$\begin{aligned} \int _{{\mathbb {R}}}\phi _2(x_1, x_2, x_3)\text {d}x_2=\int _{{{\mathbb {R}}}^2} \phi _2(x_1, x_2, x_3)\text {d}x_1\text {d}x_3=0. \end{aligned}$$

This means that the operator $T_{\mathfrak {z}}$ studied by Ricci and Stein can be decomposed as $T_{\mathfrak {z}}=T_{\mathfrak {z},1}+T_{\mathfrak {z},2},$ where the kernels of $T_{\mathfrak {z},1}$ and $T_{\mathfrak {z},2}$ are given, respectively, by

$$\begin{aligned} {\mathcal {K}}_1(x_1, x_2, x_3)=\sum _{j,k\in {{\mathbb {Z}}}}2^{2(k+j)} \phi _1\Big (2^jx_1,2^kx_2,2^{j+k}x_3 \Big ) \end{aligned}$$

and

$$\begin{aligned} {\mathcal {K}}_2(x_1, x_2, x_3)=\sum _{j,k\in {{\mathbb {Z}}}}2^{2(k+j)}\phi _2\Big (2^jx_1,2^kx_2,2^{j+k}x_3 \Big ). \end{aligned}$$

Theorem 4.1 shows that the kernel ${\mathcal {K}}_1$ satisfies the regularity (RR) and cancellation conditions (C2.a)–(C2.c) while the kernel ${\mathcal {K}}_2$ satisfies the regularity (RR) and cancellation conditions (C$2^\prime .$a)–(C$2^\prime .$c). Therefore, these operators ${\mathcal {K}}_1$ and ${\mathcal {K}}_2$ belong to our class.

Remark 4.4

Actually, based on the proof of Theorem 4.1, we note that the kernel

$$\begin{aligned} \mathcal {K}(x_1, x_2, x_3)=\sum _{j,k\in {\mathbb {Z}}} 2^{2j+2k}\phi ^{(1)}(2^jx_1) \phi ^{(2)}(2^kx_2,2^{j+k}x_3) \end{aligned}$$

as in Theorem 4.1 satisfies the following stronger conditions:

$$\begin{aligned}&|\partial _{x_1}^\alpha \partial _{x_2}^\beta \partial _{x_3}^\gamma \mathcal {K}(x_1, x_2, x_3)| \le \frac{C_{\alpha ,\beta ,\gamma ,{\theta _2}}}{|x_1|^{\alpha +\gamma +2} |x_2|^{\beta +\gamma +2} \big (1+|\frac{x_3}{x_1x_2}| \big )^{(\alpha \wedge \beta )+\gamma +1+{\theta _2}}};\\&\bigg |\int _{\delta _1 \le |x_1|\le r_1}\int _{\delta _2 \le |x_2|\le r_2}\int _{\delta _3 \le |x_3|\le r_3} \mathcal {K}(x_1, x_2, x_3)\text {d}x_1\text {d}x_2\text {d}x_3\bigg | \le C \end{aligned}$$

uniformly for all $\delta _1,\delta _2,\delta _3,r_1,r_2,r_3>0;$

$$\begin{aligned}&\bigg |\int _{\delta \le |x_1|\le r}\partial _{x_2}^\beta \partial _{x_3}^\gamma \mathcal {K}(x_1, x_2, x_3)\text {d}x_1\bigg |\\&\quad \le C_{\beta ,\gamma ,{\theta _2}}\bigg (\frac{1}{r^{\gamma +1} |x_2|^{\beta +\gamma +2} \big (1+|\frac{x_3}{rx_2}| \big )^{\gamma +1+{\theta _2}}}+\frac{1}{\delta ^{\gamma +1} |x_2|^{\beta +\gamma +2} \big (1+|\frac{x_3}{\delta x_2}| \big )^{\gamma +1+{\theta _2}}}\bigg ) \end{aligned}$$

for all $\delta , r>0, \beta , \gamma \ge 0$ and $0<{\theta _2}<1$ and

$$\begin{aligned} \bigg |\int _{\delta _1 \le |x_2|\le r_1}\int _{\delta _2 \le |x_3|\le r_2}\partial _{x_1}^\alpha \mathcal {K}(x_1, x_2, x_3)\text {d}x_2 \text {d}x_3 \bigg | \le \frac{C_{\alpha }}{|x_1|^{\alpha +1}} \end{aligned}$$

uniformly for all $\delta _1,\delta _2,r_1,r_2>0$ and $\alpha \ge 0.$

Remark 4.5

In [23], Nagel and Wainger considered the $L^2$ boundedness of certain singular integral operators on ${\mathbb {R}}^n$ whose kernels have appropriate homogeneities with respect to a multi-parameter group of dilations, generated by a finite number of diagonal matrices. In particular, they considered the following two-parameter dilation group

$$\begin{aligned} \delta (s, t)(x_1, x_2, x_3)=(sx_1, tx_2, s^\alpha t^\beta x_3) \end{aligned}$$

(4.6)

acting on ${\mathbb {R}}^3$ for $s, t, \alpha ,\beta >0.$ They defined a singular kernel ${\mathcal {K}}$ by

$$\begin{aligned} {\mathcal {K}}(x_1, x_2, x_3) = \text {sgn}(x_1 x_2) \bigg \{\frac{|x_1|^{\alpha -1} |x_2|^{\beta -1}}{|x_1|^{2\alpha } |x_2|^{2\beta }+x_3^2}\bigg \} \end{aligned}$$

and proved that convolution with ${\mathcal {K}}$ is bounded on $L^2({\mathbb {R}}^3)$.

It is easy to see that when $\alpha =\beta =1, {\mathcal {K}}(x_1, x_2, x_3)$ satisfies all conditions in Corollary 1.2 and Theorem 1.3. Therefore, by Theorem 1.3, the convolution singular integral operator ${\mathcal {K}}*f$ with $\alpha =\beta =1$ is also bounded on $L^p({\mathbb {R}}^3)$ for $1<p<\infty ,$ where ${\mathcal {K}}*f$ is defined by the limit of ${\mathcal {K}}_\epsilon ^N*f$ in the $L^p, 1<p<\infty ,$ norm. It is worthwhile to point out that the theory we are developing here can easily be generalized to the ‘anisotropic’ case (adapted to $\delta (s, t)$ in (4.6)). The details are left to the interested reader.