Explicit ℓ² Decoupling in ${\mathbb{R}}^n$, Part I: Bounds for Constants in an Alternate Formulation

Abstract

Z.K. Li established the first explicit decoupling estimate for the parabola in the range $4<p<6$. Such explicit bounds are essential in number theory applications. Following the approach of J. Bourgain and C. Demeter, we derive an explicit bound for the implicit constant in an alternate form of decoupling in J. Bourgain and C. Demeter’s work. This is the first part of our three-part project to extend Z.K. Li’s result on the explicit bound for the decoupling constant to the n-dimensional case.

1. Introduction

The ${\mathsf{\ell }}^2$ decoupling theorem for the paraboloid, established by J. Bourgain and C. Demeter [1], has found numerous applications in harmonic analysis, number theory, and PDE, cf. [2,3,4,5,6,7,8,9,10]. In particular, the explicit $ϵ$-dependence of the decoupling constant is crucial for extracting numerically sharp bounds in Vinogradov-type problems. Z.K. Li obtained the first such explicit bound in the two-dimensional case, covering the range $4<p<6$ for the parabola, cf. [11]. L. Guth, D. Maldague, and H. Wang obtained a better result in the 2-dimensional case when $p=6$ in [12]. Extending Z.K. Li’s result to the n-dimensional paraboloid is a natural next step and is essential for advancing the corresponding moment curve decoupling theory.

We aim to follow the approach of J. Bourgain and C. Demeter in [13] to get an explicit expression of the decoupling constant in the n-dimensional case, clarifying its dependence on $ϵ$. This paper is the first part of our three-part project for this purpose, and presents the explicit bound of an implicit constant in an alternate decoupling form. In the forthcoming part II (about 30 pp.), estimates about the implicit constant in the multilinear-to-linear decoupling recursion theorem will be presented. The final goal will be addressed in the planned Part III. In the present paper (Part I), we establish only an intermediate explicit estimate for the constant in an alternate decoupling formulation, not the full explicit n-dimensional decoupling theorem.

1.1. Preliminaries

First we introduce key notations, definitions and fundamental results of the ${\ell }^2$ decoupling theorem that will be used in the sequel.

Let ${\mathbb{P}}^{n-1}$ be the truncated elliptic paraboloid in ${\mathbb{R}}^n$, i.e., let

$${\mathbb{P}}^{n-1}:=\left\{({\xi }_1,\dots ,{\xi }_{n-1},{\xi }_1^2+\dots +{\xi }_{n-1}^2)|0\le {\xi }_i\le 1\right\}.$$

Let $\mathbb{N}$ be the set of natural numbers. For each $r\in \mathbb{R}\setminus 0$, set $r^{-\mathbb{N}}=\left\{r^{-m}|m\in \mathbb{N}\right\}$. Let $Q\subset {[0,1]}^{n-1}$ be a cube with side length $l\left(Q\right)\in 2^{-\mathbb{N}}$. Given $\alpha \in 2^{-\mathbb{N}}\cap (0,l\left(Q\right))$, we denote by ${\mathrm{Part}}_{\alpha }\left(Q\right)$ the unique partition of Q by using cubes $Q_{\alpha }$ with side length $l\left(Q_{\alpha }\right)=\alpha$. For $\delta <1$ and $Q\subset {\left[0,1\right]}^{n-1}$, we define the $\delta$-neighborhood $N_{\delta }\left(Q\right)$ of ${\mathbb{P}}^{n-1}$ above Q as

$$N_{\delta }\left(Q\right)=\left\{\left({\xi }_1,\dots ,{\xi }_{n-1},{\xi }_1^2+\dots +{\xi }_{n-1}^2+t\right)|{\xi }_i\in Q,0\le t\le \delta \right\}.$$

For each $f\in L^1\left({\mathbb{R}}^n\right)$ and each measurable set $S\subset {\mathbb{R}}^n$, we write

$$f_S\left(x\right)={\int }_S\widehat{f}\left(\xi \right)e(x·\xi )\mathrm{d}\xi$$

for the Fourier restriction of f on S, where $e\left(z\right)=e^{2\pi iz}$. A positive weight $\omega \left(x\right)$ is a function in ${\mathbb{R}}^n$ satisfying

$$\omega \left(x\right)\ge 0\mathrm{and}{\int }_{{\mathbb{R}}^n}\omega \left(x\right)\mathrm{d}x<+\infty .$$

For a positive weight $v:{\mathbb{R}}^n\to [0,\infty )$ and $f:{\mathbb{R}}^n\to \mathbb{C}$ satisfying $f{v}^{1p}\in L^p$, set the weighted integral

$${\parallel f\parallel }_{L^p\left(v\right)}={\left({\int }_{{\mathbb{R}}^n}{\left|f\left(x\right)\right|}^{p}v\left(x\right)\mathrm{d}x\right)}^{{\displaystyle \frac{1}{p}}}.$$

For a cube $B\in {\mathbb{R}}^n$, we shall use the shorthand notation

$$f_{L^p\left(B\right)}=f_{L^p\left({\chi }_B\right)}={\left({\int }_B{\left|f\right|}^p\right)}^{{\displaystyle \frac{1}{p}}}.$$

We write $B_R$ for a cube whose side length is R. A cube $B_R\subset {\mathbb{R}}^n$ centered at c is denoted by $B\left(c,R\right)$. Let

$${\omega }_{B\left(c,R\right)}\left(x\right)={\left(1+{\displaystyle \frac{|x-c|}{R}}\right)}^{-100n},$$

and for $E\ge 100n$, set

$${\omega }_{B\left(c,R\right),E}\left(x\right)={\left(1+{\displaystyle \frac{|x-c|}{R}}\right)}^{-E}.$$

For any cube $Q\subset {[0,1]}^{n-1}$ and $g\in L^1\left(Q\right)$, set

$$E_{Q}g\left(x\right)={\int }_{Q}g({\xi }_1,\dots ,{\xi }_{n-1})e\left({\xi }_1{x}_1+\dots +{\xi }_{n-1}{x}_{n-1}+\left({\xi }_1^2+\dots +{\xi }_{n-1}^2\right)x_n\right)\mathrm{d}\xi ,$$

where $\xi =\left({\xi }_1,\dots ,{\xi }_{n-1}\right)$ and $x=\left(x_1,\dots ,x_n\right)$. We shall use the shorthand notation

$$Eg=E_{{[0,1]}^{n-1}}g.$$

For $2\le p\le \infty$ and $\delta \in 4^{-\mathbb{N}}$, let ${Dec}_n(\delta ,p,E)$ be the optimal constant such that the inequality

$${\parallel Eg\parallel }_{L^p\left({\omega }_{B_{{\delta }^{-1}},E}\right)}\le {Dec}_n(\delta ,p,E){\left(\sum _{Q\in {Part}_{{\delta }^{1/2}}\left({[0,1]}^{n-1}\right)}{∥E_{Q}g∥}_{L^p\left({\omega }_{B_{{\delta }^{-1}},E}\right)}^2\right)}^{{\displaystyle \frac{1}{2}}},\forall g:{[0,1]}^{n-1}\to \mathbb{C}$$

holds for all cubes $B_{{\delta }^{-1}}$, and ${Dec}_n(\delta ,p)$ be the optimal constant such that the inequality

$${\parallel Eg\parallel }_{L^p\left({\omega }_{B_{{\delta }^{-1}}}\right)}\le {Dec}_n(\delta ,p){\left(\sum _{Q\in {Part}_{{\delta }^{1/2}}\left({[0,1]}^{n-1}\right)}{∥E_{Q}g∥}_{L^p\left({\omega }_{B_{{\delta }^{-1}}}\right)}^2\right)}^{{\displaystyle \frac{1}{2}}},\forall g:{[0,1]}^{n-1}\to \mathbb{C}$$

holds for all cubes $B_{{\delta }^{-1}}$.

For a cube $B\in {\mathbb{R}}^n$ and a function $f:{\mathbb{R}}^n\to \mathbb{C}$ satisfying $f\in L^p\left(B\right)$, we set

$$f_{L_{♯}^p\left(B\right)}={\left({\displaystyle \frac{1}{\left|B\right|}}{\int }_B{\left|f\right|}^p\right)}^{{\displaystyle \frac{1}{p}}},$$

where $\left|B\right|$ is the Lebesgue measure of B. We denote by $I_R$ the interval $[-R/2,R/2]$. As in [14], we denote by $\sigma \left(S^{n-1}\right)$ the surface measure of the unit sphere in ${\mathbb{R}}^n$. A tiling is a collection of subsets (the tiles) of ${\mathbb{R}}^n$ such that the union of all tiles is exactly ${\mathbb{R}}^n$, and that the interiors of any two distinct tiles are disjoint. For a set A, we denote by $card\left(A\right)$ the cardinality of A. Let $\left[n\right]=\mathbb{Z}\cap [1,n]$, where $\mathbb{Z}$ is the set of integers. We denote by $\mathrm{supp}f$ the support of a function f. Regarding basic properties of the Fourier transform, we refer to [15,16].

In 2016, J. Bourgain proved the following ${\ell }^2$ decoupling theorem in [1].

Theorem 1. 

When $2\le p\le {\displaystyle \frac{2(n+1)}{n-1}}$, we have

$${Dec}_n(\delta ,p)\le C{\delta }^{-ϵ},$$

where the implicit constant C depends on ϵ, p, and n, but is independent of δ.

Theorem 5.1 in [13] states that there exists a sufficiently large constant ${\Gamma }_n\left(E\right)$ such that for all $F\ge {\Gamma }_n\left(E\right)$, $p\ge 2$, $f:{\mathbb{R}}^n\to \mathbb{C}$ with Fourier transform supported in $N_{1/R}\left({[0,1]}^{n-1}\right)$, and cube $B_R$, there exists a constant C depending on n and E such that

$${\parallel f\parallel }_{L^p\left({\omega }_{B_R,E}\right)}\le C·{Dec}_n\left(R^{-1},p,F\right){\left(\sum _{Q\in {Part}_{R^{-1/2}}\left({[0,1]}^{n-1}\right)}{∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B_R,E}\right)}^2\right)}^{1/2}.$$

1.2. Main Results and a Roadmap

Denote by $\mathcal{S}\left({\mathbb{R}}^n\right)$ the Schwartz space on ${\mathbb{R}}^n$. Let $\eta \in \mathcal{S}\left(\mathbb{R}\right)$ be a Schwartz function satisfying

$$\eta \left(x\right)=1,\forall x\in [-n,n] \mathrm{and} \mathrm{supp}\eta \subset [-(n+1),n+1].$$

(1)

Let $M_j\in \mathcal{S}\left(\mathbb{R}\right)$ be a Schwartz function with a compact support satisfying

$$M_j\left(t\right)=t^j,\forall t\in [0,{\displaystyle \frac{1}{2}}] \mathrm{and} {∥{\displaystyle \frac{{\mathrm{d}}^k}{{\mathrm{d}}^{tk}}}{M}_j∥}_{L^{\infty }\left(\mathbb{R}\right)}\le {\tilde{C}}_k,$$

(2)

where ${\tilde{C}}_k$ is a constant depending on k but independent of j. For any $s\in \mathbb{N}$, set

$$\rho \left(s\right)=sup\left\{|x^l{\eta }^{\left(m\right)}\left(x\right)\left|\right|l,m\in [0,s]\cap \mathbb{N}\right\},$$

(3)

where $\eta \left(x\right)$ is defined as in (1).

Our main result derives an explicit lower bound for ${\Gamma }_n\left(E\right)$ and an explicit upper bound for C as follows.

Theorem 2. 

Let $p\ge 2$, $R^{-1}\in 4^{-\mathbb{N}}$, and $f:{\mathbb{R}}^n\to \mathbb{C}$ be a function satisfying $\mathrm{supp}\widehat{f}\subset N_{1/R}\left({[0,1]}^{n-1}\right)$. Consider $\eta \left(x\right)\in \mathcal{S}\left(\mathbb{R}\right)$ as (1), $M_j\left(t\right)\in \mathcal{S}\left(\mathbb{R}\right)$ as (2), and $\rho \left(s_j\right)$ as satisfying (3). Then for any $E\ge 100n$, $F\ge 2(n-1)⌈E⌉+(n-1)(p-1)+2$, and each cube $B_R\subset {\mathbb{R}}^n$, we have

$${\parallel f\parallel }_{L^p\left({\omega }_{B_R,E}\right)}\le C_0{Dec}_n\left(R^{-1},p,F\right)·{\left(\sum _{Q\in {Part}_{R^{-1/2}}\left({[0,1]}^{n-1}\right)}{∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B_R,E}\right)}^2\right)}^{1/2},$$

where $C_0=max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{{\displaystyle \frac{20n⌈E⌉}{p}}}{\left(9n\right)}^{{\displaystyle \frac{F}{p}}}{\left(30n\right)}^E$.

To clarify the meaning and motivation of the constants in the above alternate formulation, we provide a theorem on recursion that we have obtained in our ongoing Part II as follows.

Theorem 3. 

Assume that ${Dec}_2\left(\delta ,p,{\Gamma }_2\left(10E\right)\right)\le D_{ϵ}·{\delta }^{-ϵ}$. Let $K\ge 11$ satisfy $\sqrt{K}\in {\mathbb{N}}^{+}$. For any $m\ge 1$ and $R\ge K^{2^m}$ with ${\displaystyle \frac{\sqrt{R}}{K}}\in {\mathbb{N}}^{+}$, if $ϵ\le 3$, then

$$\begin{array}{cc}\hfill {\parallel Eg\parallel }_{L^p\left({\omega }_{B_R,E}\right)}& \le {\left(6\sqrt{3}\right)}^E·3\left(362·A+C_0·D_{ϵ}·{10}^{10E}\right)·{\left(6\sqrt{3}\right)}^{{\displaystyle \frac{10E}{p}}}·{\left(\sum _{\alpha \in \in {\mathrm{Part}}_{K^{-1}}{[0,1]}^2}{E_{\alpha }g}_{L^p\left({\omega }_{B_R,E}\right)}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\left(6\sqrt{3}\right)}^E·3·C_0·D_{ϵ}·{10}^{10E}·{\left(6\sqrt{3}\right)}^{{\displaystyle \frac{10E}{p}}}·K^{ϵ}·{\left(\sum _{\beta \in \in {\mathrm{Part}}_{{\displaystyle \frac{1}{\sqrt{k}}}}{[0,1]}^2}{E_{\beta }g}_{L^p\left({\omega }_{B_R,E}\right)}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\left(6\sqrt{3}\right)}^E·3·\left(A+C_0·{Dec}_3(R^{-1},p,K^{-2},m,E)·{10}^{10E}\right)·K^{16}\hfill \\ & ·{\left(\sum _{\Delta \in {\mathrm{Part}}_{{\displaystyle \frac{1}{\sqrt{R}}}{[0,1]}^2}}{E_{\Delta }g}_{L^p({\omega }_{B_R},E)}^2\right)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Note that the constant $C_0$ in the above theorem is exactly the same as the constant in Theorem 2, and it is the crucial step in the multilinear-to-linear decoupling recursion.

Our paper is organized as follows: Section 2 collects several auxiliary lemmas concerning weight functions, convolution estimates, and derivative bounds that will be needed in the proof of the main result. Section 3 presents the proof of Theorem 2 through a five-step argument, ultimately leading to the explicit constant $C_0$. A brief discussion of the result is given in Section 4.

2. Auxiliary Lemmas

Lemma 1. 

Let $\mathcal{W}$ be the set of all positive weights in ${\mathbb{R}}^n$. Fix $R>0$ and E. Let $O_1,O_2:\mathcal{W}\to [0,\infty ]$ be maps satisfying the following properties:

(W1) 

there exists $C\in {\mathbb{R}}^{+}$, such that $O_1\left({\chi }_B\right)\le C·O_2\left({\omega }_{B,E}\right)$, $\forall B=B(c,R)\subset {\mathbb{R}}^n$;

(W2) 

$O_1(\alpha u+\beta v)\le \alpha O_1\left(u\right)+\beta O_1\left(v\right)$, $\forall u,v\in \mathcal{W}$, $\forall \alpha ,\beta >0$;

(W3) 

$O_2(\alpha u+\beta v)\ge \alpha O_2\left(u\right)+\beta O_2\left(v\right)$, $\forall u,v\in \mathcal{W}$, $\forall \alpha ,\beta >0$;

(W4) 

if $u\le v$, then $O_i\left(u\right)\le O_i\left(v\right)$, $\forall i\in 1,2$.

Then

$$O_1\left({\omega }_{B,E}\right)\le C{\left(6\sqrt{n}\right)}^E{O}_2\left({\omega }_{B,E}\right),\forall B=B_R\subset {\mathbb{R}}^n.$$

Proof. 

Let $\mathcal{B}$ be a tiling of ${\mathbb{R}}^n$ satisfying ${[0,R]}^n\in \mathcal{B}$, whose tiles are cubes with the form $B^{\prime }=B(c_{B^{\prime }},R)$. First we claim that

$${\omega }_{B,E}\left(x\right)\le {\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^E\sum _{B^{\prime }\in \mathcal{B}}{\chi }_{B^{\prime }}\left(x\right){\omega }_{B,E}\left(c_{B^{\prime }}\right)$$

(4)

and

$$\sum _{B^{\prime }\in \mathcal{B}}{\omega }_{B^{\prime },E}\left(x\right){\omega }_{B,E}\left(c_{B^{\prime }}\right)\le 3·{\displaystyle \frac{2^{2E}}{E-n-1}}{\omega }_{B,E}\left(x\right)$$

(5)

hold for each $x\in {\mathbb{R}}^n$.

To prove (4), note that $|c_{B^{\prime }}-x|\le (\sqrt{n}/2)R,\forall x\in B^{\prime }$. Thus, we only need to prove

$${\omega }_{B,E}\left(x\right)\le {\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^E{\omega }_{B,E}\left(c_{B^{\prime }}\right),$$

which follows directly from

$${\displaystyle \frac{1+\frac{|c_{B^{\prime }}-c_B|}{R}}{1+\frac{|x-c_B|}{R}}}\le {\displaystyle \frac{1+\frac{|c_{B^{\prime }}-x|+|x-c_B|}{R}}{1+\frac{|x-c_B|}{R}}}\le 1+{\displaystyle \frac{\sqrt{n}}{2}}.$$

For (5), it suffices to prove that

$$\sum _{B^{\prime }\in \mathcal{B}}{\left[{\displaystyle \frac{1+\frac{|c_B-x|}{R}}{\left(1+\frac{|c_{B^{\prime }}-x|}{R}\right)\left(1+\frac{|c_{B^{\prime }}-c_B|}{R}\right)}}\right]}^E\le 3·{\displaystyle \frac{2^{2E}}{E-n-1}},\forall x\in {\mathbb{R}}^n.$$

With the substitution $u=c_B/R$ and $v=x/R$, the inequality above is equivalent to

$$I:=\sum _{B^{\prime }\in \mathcal{B}}{\left[{\displaystyle \frac{1+\left|u-v\right|}{\left(1+|\frac{c_{B^{\prime }}}{R}-u|\right)\left(1+|\frac{c_{B^{\prime }}}{R}-v|\right)}}\right]}^E\le 3·{\displaystyle \frac{2^{2E}}{E-n-1}},\forall x\in {\mathbb{R}}^n.$$

Let $\left|u-v\right|=A$. Define

$${\mathcal{B}}_1=\left\{B^{\prime }\in \mathcal{B}\mid \left|{\displaystyle \frac{c_{B^{\prime }}}{R}}-u\right|<{\displaystyle \frac{A}{2}}\right\} \mathrm{and} {\mathcal{B}}_2=\left\{B^{\prime }\in \mathcal{B}\mid \left|{\displaystyle \frac{c_{B^{\prime }}}{R}}-u\right|\ge {\displaystyle \frac{A}{2}}\right\}.$$

Note that each $B^{\prime }$ in ${\mathcal{B}}_1$ satisfies $|c_{B^{\prime }}/R-v|\ge |u-v|-|c_{B^{\prime }}/R-u|\ge A/2$. Thus,

$$\begin{array}{cc}\hfill I=& \left(\sum _{B^{\prime }\in {\mathcal{B}}_1}+\sum _{B^{\prime }\in {\mathcal{B}}_2}\right){\left[{\displaystyle \frac{1+|u-v|}{\left(1+|\frac{c_{B^{\prime }}}{R}-u|\right)\left(1+|\frac{c_{B^{\prime }}}{R}-v|\right)}}\right]}^E\hfill \\ \hfill \le & \sum _{B^{\prime }\in {\mathcal{B}}_1}{\left({\displaystyle \frac{1+A}{1+\frac{A}{2}}}\right)}^E{\left({\displaystyle \frac{1}{1+|\frac{c_{B^{\prime }}}{R}-u|}}\right)}^E+\sum _{B^{\prime }\in {\mathcal{B}}_2}{\left({\displaystyle \frac{1+A}{1+\frac{A}{2}}}\right)}^E{\left({\displaystyle \frac{1}{1+|\frac{c_{B^{\prime }}}{R}-v|}}\right)}^E\hfill \\ \hfill \le & 2^E\sum _{B^{\prime }\in {\mathcal{B}}_2}{\left({\displaystyle \frac{1}{1+|\frac{c_{B^{\prime }}}{R}-u|}}\right)}^E+2^E\sum _{B^{\prime }\in {\mathcal{B}}_2}{\left({\displaystyle \frac{1}{1+|\frac{c_{B^{\prime }}}{R}-v|}}\right)}^E\hfill \\ \hfill \le & 2^{E+1}(1+{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{{(2i+1)}^n}{{(1+\frac{2i-1}{2})}^E}})\hfill \\ \hfill \le & 2^{E+1}(1+2^n{\int }_{{\displaystyle \frac{1}{2}}}^{+\infty }{\displaystyle \frac{1}{x^{E-n}}}x)\le 3·{\displaystyle \frac{2^{2E}}{E-n-1}}.\hfill \end{array}$$

(6)

We provide an interpretation of (6). Let $\mathcal{B}/R=\left\{B^{\prime }/R|B^{\prime }\in \mathcal{B}\right\}$. Then $\mathcal{B}/R$ is a tiling of ${\mathbb{R}}^n$ with integer lattice points as vertices. By applying a suitable translation if necessary, we may assume that $v\in {[0,1]}^n$. Let

$${\mathcal{B}}^{\left(i\right)}=\left\{{\displaystyle \frac{B^{\prime }}{R}}\in {\displaystyle \frac{\mathcal{B}}{R}}|{\displaystyle \frac{B^{\prime }}{R}}\in {[-i,i+1]}^n\setminus {(1-i,i)}^n\right\}.$$

We have $\mathcal{B}/R={[0,1]}^n\bigsqcup {\mathcal{B}}^{\left(1\right)}\bigsqcup {\mathcal{B}}^{\left(2\right)}\dots$. Moreover,

$$card\left({\mathcal{B}}^{\left(i\right)}\right)={(2i+1)}^n-{(2i-1)}^n\le {(2i+1)}^n.$$

For any $B^{\prime }/R\in \mathcal{B}/R$, we have $|c_{B^{\prime }}/R-v|\ge 2i-1/2$. Thus, the inequality (6) holds.

Now we return to the proof of the lemma. By (4), (W4), (W2), and (W1), we have

$$\begin{array}{cc}\hfill O_1\left({\omega }_{B,E}\right)\le & O_1\left({\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^E\sum _{B^{\prime }\in \mathcal{B}}{\chi }_{B^{\prime }}\left(x\right){\omega }_{B,E}\left(c_{B^{\prime }}\right)\right)\hfill \\ \hfill \le & {\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^E\sum _{B^{\prime }\in \mathcal{B}}{\omega }_{B,E}\left(c_{B^{\prime }}\right)O_1\left({\chi }_{B^{\prime }}\left(x\right)\right)\hfill \\ \hfill \le & C{\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^E\sum _{B^{\prime }\in \mathcal{B}}{\omega }_{B,E}\left(c_{B^{\prime }}\right)O_2\left({\omega }_{B,E}\right)\hfill \end{array}$$

By (5), (W3), and (W4), we have

$$\begin{array}{cc}\hfill C{\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^E\sum _{B^{\prime }\in \mathcal{B}}{\omega }_{B,E}\left(c_{B^{\prime }}\right)O_2\left({\omega }_{B,E}\right)\le & C{\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^E{O}_2\left(\sum _{B^{\prime }\in \mathcal{B}}{\omega }_{B,E}\left(c_{B^{\prime }}\right){\omega }_{B,E}\right)\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill \le & C{\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^E·3·{\displaystyle \frac{2^{2E}}{E-n-1}}{O}_2\left({\omega }_{B,E}\right)\hfill \\ \hfill \le & C{\left(6\sqrt{n}\right)}^E{O}_2\left({\omega }_{B,E}\right). \square \hfill \end{array}$$

(8)

Remark 1. 

As Bourgain noted in Remark 4.2 in [13], for any $f:{\mathbb{R}}^n\to \mathbb{C}$ and positive weight ν, the map $O_1\left(v\right):={\left|\right|f\left|\right|}_{L^p\left(v\right)}^p$ satisfies (W2) and (W4). Also, for any $p\ge 2$, $f_i:{\mathbb{R}}^n\to \mathbb{C}$, and positive weight ν satisfying $f_i{v}^{1/p}\in L^p$, the map $O_2\left(v\right):={\left({\sum }_i{∥f_i∥}_{L^p\left(v\right)}^2\right)}^{p/2}$ satisfies (W3) and (W4) by Minkowski’s inequality.

Lemma 2. 

Let $f:\mathbb{R}\to \mathbb{C}$ be a smooth function. We have

$${\left[f\left(x^2\right)\right]}^{\left(n\right)}={\displaystyle \sum _{k=0}^{⌊{\displaystyle \frac{n}{2}}⌋}}{\displaystyle \frac{n!}{(n-2k)!k!}}{\left(2x\right)}^{n-2k}{f}^{(n-k)}\left(x^2\right).$$

(9)

Proof. 

By Faà di Bruno’s Formula (cf. [17]), for two functions g and f with a sufficient number of derivatives, we have

$${\left[f\left(g\left(x\right)\right)\right]}^{\left(n\right)}=\sum {\displaystyle \frac{n!}{b_1!\dots b_n!}}{f}^{\left(l\right)}\left(g\left(x\right)\right)\prod _{\begin{array}{c}1\le i\le n\\ b_i>0\end{array}}{\left({\displaystyle \frac{g^{\left(i\right)}\left(x\right)}{i!}}\right)}^{b_i},$$

(10)

where the sum is over all different solutions in nonnegative integers $b_1,\dots ,b_n$ of

$$b_1+2b_2+\dots +n{b}_n=n,$$

and $l:=b_1+\dots +b_n$. Let $g\left(x\right)=x^2$. We have $g^{\left(i\right)}\left(x\right)=0$, $\forall i\ge 3$. Hence,

$$\prod _{\begin{array}{c}1\le i\le n\\ b_i>0\end{array}}{\left({\displaystyle \frac{g^{\left(i\right)}\left(x\right)}{i!}}\right)}^{b_i}\ne 0$$

if and only if $b_i=0$, $\forall i\ge 3$. Consequently, $b_1+2b_2+\dots +n{b}_n=n$ reduces to $b_1+2b_2=n$, and correspondingly we have $l=b_1+b_2$. Let $k=b_2$. Then all integers $0\le k\le ⌊n/2⌋$ are in one-to-one correspondence with the nonnegative integer solutions $(b_1,b_2)$ of $b_1+2b_2=n$. In this case, we obtain $b_1=n-2k$ and $l=n-2k+k=n-k$. Substituting into (10) yields (9). □

Remark 2. 

Let $g\left(t\right)=f(-at+b)$. Then $g\left(x^2\right)=f(-a{x}^2+b)$. By denoting $I(n,k)=2^{n-2k}n!/\left(\right(n-2k)!k!)$, it follows from (9) that

$$\begin{array}{cc}\hfill {\left[f(-a{x}^2+b)\right]}^{\left(n\right)}={\left[g\left(x^2\right)\right]}^{\left(n\right)}=& {\displaystyle \sum _{k=0}^{⌊{\displaystyle \frac{n}{2}}⌋}}I(n,k)x^{n-2k}{g}^{(n-k)}\left(x^2\right)\hfill \\ \hfill =& {\displaystyle \sum _{k=0}^{⌊{\displaystyle \frac{n}{2}}⌋}}I(n,k)x^{n-2k}{(-a)}^{n-k}{f}^{(n-k)}(-a{x}^2+b)\hfill \end{array}$$

(11)

and

$${\displaystyle \sum _{k=0}^{⌊{\displaystyle \frac{n}{2}}⌋}}I(n,k)={\displaystyle \sum _{k=0}^{⌊{\displaystyle \frac{n}{2}}⌋}}{C}_n^{n-2k}{C}_{2k}^{k}k!2^{n-2k}\le n!2^n{(1+2)}^n=6^{n}n!.$$

(12)

In the rest of this section, the cube $B_R$ is centered at the origin.

Lemma 3. 

We have

$${\omega }_{B_R,E}\ast \left({\displaystyle \frac{1}{{\left(R^{\prime }\right)}^n}}{\omega }_{B_{R^{\prime }},E}\right)\le 6^E\sigma \left(S^{n-1}\right)B(E-n,n){\omega }_{B_R,E},\forall R^{\prime }\in \left(0,R\right],$$

(13)

where $B(E-n,n)$ is the Beta function.

Proof. 

We have the following equivalent forms of (13):

$$\begin{array}{cc}\hfill & \int {\left({\displaystyle \frac{1}{1+\frac{\left|x-t\right|}{R}}}\right)}^E·{\displaystyle \frac{1}{{\left(R^{\prime }\right)}^n}}·{\left({\displaystyle \frac{1}{1+\frac{\left|t\right|}{R^{\prime }}}}\right)}^E\mathrm{d}t\le 6^E\sigma \left(S^{n-1}\right)B(E-n,n){\left({\displaystyle \frac{1}{1+\frac{\left|x\right|}{R}}}\right)}^E,\forall x\in {\mathbb{R}}^n\hfill \\ \hfill \iff & \int {\left({\displaystyle \frac{1+\frac{\left|x\right|}{R}}{1+\frac{\left|x-t\right|}{R}}}\right)}^E·{\displaystyle \frac{1}{{\left(R^{\prime }\right)}^n}}·{\left({\displaystyle \frac{1}{1+\frac{\left|t\right|}{R^{\prime }}}}\right)}^E\mathrm{d}s\le 6^E\sigma \left(S^{n-1}\right)B(E-n,n),\forall x\in {\mathbb{R}}^n\hfill \\ \hfill \stackrel{y={\displaystyle \frac{x}{R}}}{\iff }& \int {\left({\displaystyle \frac{1+\left|y\right|}{1+|y-\frac{t}{R}|}}\right)}^E·{\displaystyle \frac{1}{{\left(R^{\prime }\right)}^n}}·{\left({\displaystyle \frac{1}{1+\frac{\left|t\right|}{R^{\prime }}}}\right)}^E\mathrm{d}s\le 6^E\sigma \left(S^{n-1}\right)B(E-n,n),\forall y\in {\mathbb{R}}^n\hfill \\ \hfill \stackrel{\theta ={\displaystyle \frac{R^{\prime }}{R}}}{\underset{s={\displaystyle \frac{t}{R}}}{\iff }}& {\displaystyle \frac{1}{{\theta }^n}}\int {\left({\displaystyle \frac{1+\left|y\right|}{1+\left|y-s\right|}}\right)}^E·{\left({\displaystyle \frac{1}{1+\frac{\left|s\right|}{\theta }}}\right)}^E\mathrm{d}s\le 6^E\sigma \left(S^{n-1}\right)B(E-n,n),\forall y\in {\mathbb{R}}^n.\hfill \end{array}$$

We distinguish two cases.

Case 1. $\left|y\right|\le 1/2$. By Corollary 2.51 in [14], we have

$$\begin{array}{cc}\hfill \int {\left({\displaystyle \frac{1+\left|y\right|}{1+\left|y-s\right|}}\right)}^E·{\left({\displaystyle \frac{1}{1+\frac{\left|s\right|}{\theta }}}\right)}^E\mathrm{d}s\le & \int {\left({\displaystyle \frac{1+\frac{1}{2}}{1+\left|s\right|-\frac{1}{2}}}\right)}^E·{\left({\displaystyle \frac{1}{1+\frac{\left|s\right|}{\theta }}}\right)}^E\mathrm{d}s\hfill \\ \hfill =& {\left({\displaystyle \frac{3}{2}}\right)}^E\sigma \left(S^{n-1}\right){\int }_0^{+\infty }{\left({\displaystyle \frac{1}{r+\frac{1}{2}}}\right)}^E·{\left({\displaystyle \frac{1}{1+\frac{r}{\theta }}}\right)}^E{r}^{n-1}\mathrm{d}r.\hfill \end{array}$$

By the substitution $r=\theta u$, we get

$$\begin{array}{cc}\hfill & {\left({\displaystyle \frac{3}{2}}\right)}^E\sigma \left(S^{n-1}\right){\int }_0^{+\infty }{\left({\displaystyle \frac{1}{r+\frac{1}{2}}}\right)}^E·{\left({\displaystyle \frac{1}{1+\frac{r}{\theta }}}\right)}^E{r}^{n-1}\mathrm{d}r\hfill \\ \hfill =& {\left({\displaystyle \frac{3}{2}}\right)}^E\sigma \left(S^{n-1}\right){\theta }^n{\int }_0^{+\infty }{\displaystyle \frac{1}{{\left(\theta u+\frac{1}{2}\right)}^E}}{\displaystyle \frac{u^{n-1}}{{\left(1+u\right)}^E}}\mathrm{d}u\le 3^E\sigma \left(S^{n-1}\right){\theta }^n{\int }_0^{+\infty }{\displaystyle \frac{u^{n-1}}{{\left(1+u\right)}^E}}\mathrm{d}u.\hfill \end{array}$$

Hence,

$$\int {\left({\displaystyle \frac{1+\left|y\right|}{1+\left|y-s\right|}}\right)}^E·{\left({\displaystyle \frac{1}{1+\frac{\left|s\right|}{\theta }}}\right)}^E\mathrm{d}s\le 3^E\sigma \left(S^{n-1}\right){\theta }^n{\int }_0^{+\infty }{\displaystyle \frac{u^{n-1}}{{\left(1+u\right)}^E}}\mathrm{d}u.$$

(14)

Thus, (13) holds.

Case 2. $\left|y\right|>1/2$. Consider

$$\begin{array}{cc}\hfill \int {\left({\displaystyle \frac{1+\left|y\right|}{1+\left|y-s\right|}}\right)}^E·{\left({\displaystyle \frac{1}{1+\frac{\left|s\right|}{\theta }}}\right)}^E\mathrm{d}s=& \left({\int }_{\left|s\right|<{\displaystyle \frac{3}{4}}\left|y\right|}+{\int }_{\left|s\right|\ge {\displaystyle \frac{3}{4}}\left|y\right|}\right){\left({\displaystyle \frac{1+\left|y\right|}{1+\left|y-s\right|}}\right)}^E·{\left({\displaystyle \frac{1}{1+\frac{\left|s\right|}{\theta }}}\right)}^E\mathrm{d}s.\hfill \end{array}$$

(15)

Let A and B be the first and the second summand in the right-hand side of the equality above, respectively. On the one hand,

$$\begin{array}{c}\hfill A\le {\int }_{\left|s\right|<{\displaystyle \frac{3}{4}}\left|y\right|}{\left({\displaystyle \frac{1+\left|y\right|}{1+\frac{1}{4}\left|y\right|}}\right)}^E·{\left({\displaystyle \frac{1}{1+\frac{\left|s\right|}{\theta }}}\right)}^E\mathrm{d}s\le 4^E\int {\left({\displaystyle \frac{1}{1+\frac{\left|s\right|}{\theta }}}\right)}^E\mathrm{d}s.\end{array}$$

Let $s=\theta u$. We have

$$\begin{array}{cc}\hfill 4^E\int {\left({\displaystyle \frac{1}{1+\frac{\left|s\right|}{\theta }}}\right)}^E\mathrm{d}s=& 4^E{\theta }^n\int {\left({\displaystyle \frac{1}{1+\left|u\right|}}\right)}^E\mathrm{d}u=4^E{\theta }^n\sigma \left(S^{n-1}\right){\int }_0^{+\infty }{\displaystyle \frac{r^{n-1}}{{\left(1+r\right)}^E}}\mathrm{d}r.\hfill \end{array}$$

Hence, we have

$$A\le 4^E{\theta }^n\sigma \left(S^{n-1}\right){\int }_0^{+\infty }{\displaystyle \frac{r^{n-1}}{{\left(1+r\right)}^E}}\mathrm{d}r.$$

(16)

On the other hand, by $\left|s\right|\ge 3/4\left|y\right|$ and $\theta \le 1$, we have

$$1+{\displaystyle \frac{\left|s\right|}{\theta }}\ge {\displaystyle \frac{\left|s\right|}{\theta }}\ge {\displaystyle \frac{3\left|y\right|}{4\theta }},$$

and therefore

$${\displaystyle \frac{1}{1+\left|s\right|/\theta }}\le {\displaystyle \frac{4\theta }{3\left|y\right|}}.$$

Thus, we have

$$\begin{array}{cc}\hfill B& \le {\int }_{\left|s\right|\ge {\displaystyle \frac{3}{4}}\left|y\right|}{\left({\displaystyle \frac{1+\left|y\right|}{1+\frac{3}{4\theta }\left|y\right|}}\right)}^E·{\left({\displaystyle \frac{1}{1+\left|y-s\right|}}\right)}^E\mathrm{d}s\hfill \\ \hfill & \le {\int }_{\left|s\right|\ge {\displaystyle \frac{3}{4}}\left|y\right|}{\left(4\theta {\displaystyle \frac{1+\left|y\right|}{3\left|y\right|}}\right)}^E·{\left({\displaystyle \frac{1}{1+\left|y-s\right|}}\right)}^E\mathrm{d}s\le \int {\left(4\theta \right)}^E·{\left({\displaystyle \frac{1}{1+\left|y-s\right|}}\right)}^E\mathrm{d}s.\hfill \end{array}$$

By the substitution $u=s-y$, we get

$$\begin{array}{cc}\hfill \int {\left(4\theta \right)}^E·{\left({\displaystyle \frac{1}{1+\left|y-s\right|}}\right)}^E\mathrm{d}s& =4^E{\theta }^E\int {\left({\displaystyle \frac{1}{1+\left|u\right|}}\right)}^E\mathrm{d}u\le 4^E{\theta }^n\sigma \left(S^{n-1}\right)\int {\displaystyle \frac{r^{n-1}}{{\left(1+r\right)}^E}}\mathrm{d}r,\hfill \end{array}$$

which implies that

$$B\le 4^E{\theta }^n\sigma \left(S^{n-1}\right)\int {\displaystyle \frac{r^{n-1}}{{\left(1+r\right)}^E}}\mathrm{d}r.$$

(17)

By (15)–(17), and $E\ge 100n$, we have (13) again. □

Corollary 1. 

We have

$${\omega }_{I_R,E}\ast \left({\displaystyle \frac{1}{R^{\prime }}}{\omega }_{I_{R^{\prime }},E}\right)\le 6^E{\omega }_{I_R,E},\forall R^{\prime }\in \left(0,R\right],$$

(18)

Proof. 

Note that $\sigma \left(S^0\right)=2$, $B(E-1,1)={\displaystyle \frac{1}{E-1}}$, and ${\displaystyle \frac{2}{E-1}}\le 1$. □

Corollary 2. 

We have

$${\omega }_{B_R,E}\ast \left({\displaystyle \frac{1}{{\left(R^{\prime }\right)}^n}}{\chi }_{B_{R^{\prime }}}\right)\le {(1+{\displaystyle \frac{\sqrt{n}}{2}})}^E{6}^E\sigma \left(S^{n-1}\right)B(E-n,n){\omega }_{B_R,E},\forall R^{\prime }\in \left(0,R\right].$$

(19)

Proof. 

When $x\in B_{R^{\prime }}$, we have $\left|x\right|\le \sqrt{{(R^{\prime }/2)}^{2}n}=R^{\prime }\sqrt{n}/2$. Thus,

$${\omega }_{B_{R^{\prime }},E}\left(x\right)={\left(1+{\displaystyle \frac{\left|x\right|}{R^{\prime }}}\right)}^{-E}\ge {\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^{-E}.$$

From (13), it follows that

$$\begin{array}{cc}\hfill {\omega }_{B_R,E}\ast \left({\displaystyle \frac{1}{{\left(R^{\prime }\right)}^n}}{\chi }_{B_{R^{\prime }}}\right)\le & {(1+{\displaystyle \frac{\sqrt{n}}{2}})}^E{\omega }_{B_R,E}\ast \left({\displaystyle \frac{1}{{\left(R^{\prime }\right)}^n}}{\omega }_{B_{R^{\prime }},E}\right)\hfill \\ \hfill \le & {(1+{\displaystyle \frac{\sqrt{n}}{2}})}^E{6}^E\sigma \left(S^{n-1}\right)B(E-n,n){\omega }_{B_R,E}.\hfill \end{array}$$

□

Corollary 3. 

We have

$${\chi }_{B_R}\ast \left({\displaystyle \frac{1}{{\left(R^{\prime }\right)}^n}}{\omega }_{B_{R^{\prime }},E}\right)\le {(1+{\displaystyle \frac{\sqrt{n}}{2}})}^E{6}^E\sigma \left(S^{n-1}\right)B(E-n,n){\omega }_{B_R,E},\forall R^{\prime }\in \left(0,R\right].$$

(20)

Proof. 

Similarly to the proof of (19), just note that when $x\in B_R$, we have

$$\left|x\right|\le \sqrt{{(R/2)}^{2}n}=R\sqrt{n}/2$$

and

$${\omega }_{B_R,E}\left(x\right)={\left(1+{\displaystyle \frac{\left|x\right|}{R}}\right)}^{-E}\ge {\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^{-E}.$$

□

From (18)–(20), we have the following corollary by setting $n=1$.

Corollary 4. 

For any $R^{\prime }\in \left(0,R\right]$, we have

$${\omega }_{I_R,E}\ast \left({\displaystyle \frac{1}{R^{\prime }}}{\chi }_{I_{R^{\prime }}}\right)\le 9^E{\omega }_{I_R,E} and {\chi }_{I_R}\ast \left({\displaystyle \frac{1}{R^{\prime }}}{\omega }_{I_{R^{\prime }},E}\right)\le 9^E{\omega }_{I_R,E}.$$

(21)

Lemma 4. 

Let $x=(x_1,\dots ,x_n)$. If ${\sum }_{i=1}^n{E}_i\le E$, where $E_i>0$, $\forall i\in \left[n\right]$, then

$${\omega }_{B_R,E}\left(x\right)\le {\displaystyle \prod _{i=1}^n}{\omega }_{I_R,E_i}\left(x_i\right).$$

(22)

Proof. 

Clearly,

$$\begin{array}{cc}\hfill {\omega }_{B_R,E}\left(x\right)={\left(1+{\displaystyle \frac{\sqrt{{x_1}^2+\dots +{x_n}^2}}{R}}\right)}^{-E}& \le {\displaystyle \prod _{i=1}^n}{\left(1+{\displaystyle \frac{\sqrt{{x_1}^2+\dots +{x_n}^2}}{R}}\right)}^{-E_i}\hfill \\ \hfill & \le {\displaystyle \prod _{i=1}^n}{\left(1+{\displaystyle \frac{\sqrt{{x_i}^2}}{R}}\right)}^{-E_i}={\displaystyle \prod _{i=1}^n}{\omega }_{I_R,E_i}\left(x_i\right).\hfill \end{array}$$

□

Lemma 5. 

Let $x=(x_1,\dots ,x_n)$. Then

$${\displaystyle \prod _{k=1}^n}{\omega }_{I_R,E}\left(x_k\right)\le {\omega }_{B_R,E}\left(x\right).$$

(23)

Proof. 

Note that

$${\displaystyle \prod _{k=1}^n}(1+{\displaystyle \frac{|x_k|}{R}})\ge 1+{\displaystyle \frac{{\sum }_{i=1}^n\left|x_i\right|}{R}}\ge 1+{\displaystyle \frac{\sqrt{{\sum }_{i=1}^n{x}_i^2}}{R}}=1+{\displaystyle \frac{\left|x\right|}{R}},$$

which implies that

$${\displaystyle \prod _{k=1}^n}{\omega }_{I_R,E}\left(x_k\right)\le {\omega }_{B_R,E}\left(x\right).$$

□

Lemma 6. 

Let $u_i\in \left[0,1\right],\forall i\in \left[n-1\right]$, and $u=(u_1,\dots ,u_{n-1})$. Then

$${\displaystyle \frac{1}{\sqrt{6n}}}\left|y\right|\le \sqrt{y_n^2+{\sum }_{j=1}^{n-1}{\left(y_j\pm 2y_n{u}_j\right)}^2}\le \sqrt{6n}\left|y\right|,\forall y=\left(y_1,\dots ,y_n\right)\in {\mathbb{R}}^n.$$

(24)

Proof. 

For each $i\in \left[n-1\right]$, set

$${\delta }_i={\displaystyle \frac{4u_i^2}{4u_i^2+\frac{1}{n}}}\in \left[0,{\displaystyle \frac{4}{4+\frac{1}{n}}}\right].$$

Consider the matrices

$$\left(\begin{array}{ccc}{\delta }_i& & 2u_i\\ 2u_i& & {\displaystyle \frac{1}{n}}+4u_i^2\end{array}\right) and \left(\begin{array}{ccc}{\delta }_i& & -2u_i\\ -2u_i& & {\displaystyle \frac{1}{n}}+4u_i^2\end{array}\right).$$

We have ${\delta }_i\ge 0$, $1/n+4u_i^2\ge 0$, and

$${\delta }_i({\displaystyle \frac{1}{n}}+4u_i^2)-4u_i^2=0.$$

Thus, both matrices are positive semi-definite. Then

$$I_i:={\delta }_i{y}_i^2+\left({\displaystyle \frac{1}{n}}+4u_i^2\right)y_n^2\pm 4u_i{y}_i{y}_n\ge 0.$$

From

$$\left(1-{\delta }_i\right)\ge 1-{\displaystyle \frac{4}{\frac{1}{n}+4}}={\displaystyle \frac{1}{1+4n}}\ge {\displaystyle \frac{1}{6n}},$$

it follows that

$$\begin{array}{cc}\hfill {\displaystyle \sum _{i=1}^{n-1}}{\left(y_i\pm 2y_n{u}_i\right)}^2+y_n^2=& {\displaystyle \sum _{i=1}^{n-1}}{y}_i^2+\left(1+4{\displaystyle \sum _{i=1}^{n-1}}{u}_i^2\right)y_n^2\pm 4y_n{\displaystyle \sum _{i=1}^{n-1}}{u}_i{y}_i\hfill \\ \hfill =& {\displaystyle \sum _{i=1}^{n-1}}\left(1-{\delta }_i\right)y_i^2+{\displaystyle \frac{1}{n}}{y}_n^2+{\displaystyle \sum _{i=1}^{n-1}}{I}_i\ge {\displaystyle \frac{1}{6n}}{\left|y\right|}^2.\hfill \end{array}$$

Moreover, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{i=1}^{n-1}}{\left(y_i\pm 2y_n{u}_i\right)}^2+y_n^2=& {\left|y\right|}^2+4y_n^2{\displaystyle \sum _{i=1}^{n-1}}{u}_i^2\pm 4y_n{\displaystyle \sum _{i=1}^{n-1}}{u}_i{y}_i\hfill \\ \hfill \le & {\left|y\right|}^2+4y_n^2(n-1)+2{\displaystyle \sum _{i=1}^{n-1}}\left(y_i^2+y_n^2\right)\le 6n{\left|y\right|}^2.\hfill \end{array}$$

□

For any $y=(y_1,\dots ,y_n)\in {\mathbb{R}}^n$, set $y^{\prime }=(y_1,\dots ,y_{n-1})$.

Corollary 5. 

Let u be as in Lemma 6. Then

$${\omega }_{B_R,E}(y^{\prime }+2y_{n}u,y_n)\le {\left(\sqrt{6n}\right)}^E{\omega }_{B_R,E}\left(y\right),\forall y=(y_1,\dots ,y_n)\in {\mathbb{R}}^n.$$

(25)

Proof. 

By (24), we have

$$\begin{array}{cc}\hfill {\omega }_{B_R,E}(y^{\prime }+2y_{n}u,y_n)& ={(1+{\displaystyle \frac{\sqrt{{\sum }_{j=1}^{n-1}{\left(y_j\pm 2y_n{u}_j\right)}^2+y_n^2}}{R}})}^{-E}\hfill \\ \hfill & \le {(1+{\displaystyle \frac{\left|y\right|}{{\left(\sqrt{6n}\right)}^{E}R}})}^{-E}\le {\left(\sqrt{6n}\right)}^E{\omega }_{B_R,E}\left(y\right).\hfill \end{array}$$

□

In the rest of this work, we assume that $R^{-1}\in 4^{-\mathbb{N}}$ and $f:{\mathbb{R}}^n\to \mathbb{C}$ is a function satisfying $\mathrm{supp}\widehat{f}\subset N_{1/R}\left({[0,1]}^{n-1}\right)$.

Lemma 7. 

For $f:{\mathbb{R}}^n\to \mathbb{C}$, the following inequalities hold:

$${\left({\displaystyle \frac{1}{2\sqrt{n}}}\right)}^F{∥f∥}_{L^p\left({\omega }_{B_R,F}\right)}^p\le \int {∥f∥}_{L_{♯}^p\left(B(y,R)\right)}^p{\omega }_{B_R,F}\left(y\right)\mathrm{d}y\le {\left(2\sqrt{n}\right)}^F{∥f∥}_{L^p\left({\omega }_{B_R,F}\right)}^p.$$

(26)

Proof. 

Clearly,

$$\begin{array}{cc}\hfill \int {∥f∥}_{L_{♯}^p\left(B(y,R)\right)}^p{\omega }_{B_R,F}\left(y\right)\mathrm{d}y=& \int {\displaystyle \frac{1}{\left|B(y,R)\right|}}[{\int }_{B(y,R)}{\left|f\left(x\right)\right|}^{p}x]{\omega }_{B_R,F}\left(y\right)\mathrm{d}y\hfill \\ \hfill =& {\displaystyle \frac{1}{R^n}}\int \int {\left|f\left(x\right)\right|}^p{\omega }_{B_R,F}\left(y\right){\chi }_{B(y,R)}\left(x\right)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill =& {\displaystyle \frac{1}{R^n}}\int {\left|f\left(x\right)\right|}^p[\int {\omega }_{B_R,F}\left(y\right){\chi }_{B(y,R)}\left(x\right)\mathrm{d}y]\mathrm{d}x\hfill \\ \hfill =& {\displaystyle \frac{1}{R^n}}\int {\left|f\left(x\right)\right|}^p\left[{\int }_{B(x,R)}{\omega }_{B_R,F}\left(y\right)\mathrm{d}y\right]\mathrm{d}x.\hfill \end{array}$$

Let $y=x+s$, we have

$${\int }_{B(x,R)}{\omega }_{B_R,F}\left(y\right)\mathrm{d}y={\int }_{B(0,R)}{\omega }_{B_R,F}(x+s)\mathrm{d}s={\int }_{B(0,R)}{\displaystyle \frac{1}{{\left(1+\frac{\left|x+s\right|}{R}\right)}^F}}\mathrm{d}s.$$

By the substitution $s=Rt$, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{{\omega }_{B_R,F}\left(x\right)}}·{\displaystyle \frac{1}{R^n}}{\int }_{B(0,R)}{\displaystyle \frac{1}{{\left(1+\frac{\left|x+s\right|}{R}\right)}^F}}\mathrm{d}s=& {\displaystyle \frac{1}{R^n}}{\int }_{B(0,R)}{\left({\displaystyle \frac{1+\frac{\left|x\right|}{R}}{1+\frac{\left|x+s\right|}{R}}}\right)}^F\mathrm{d}s={\int }_{B(0,1)}{\left({\displaystyle \frac{1+\frac{\left|x\right|}{R}}{1+|\frac{x}{R}+t|}}\right)}^F\mathrm{d}t.\hfill \end{array}$$

Hence,

$$\int {∥f∥}_{L_{♯}^p\left(B(y,R)\right)}^p{\omega }_{B_R,F}\left(y\right)\mathrm{d}y=\int {\left|f\left(x\right)\right|}^p{\omega }_{B_R,F}\left[{\int }_{B(0,1)}{\left({\displaystyle \frac{1+\frac{\left|x\right|}{R}}{1+|\frac{x}{R}+t|}}\right)}^F\mathrm{d}t\right]\mathrm{d}x.$$

Next, let $z=x/R$. Since $t\in B(0,1)$, we have

$${\displaystyle \frac{1+\left|z\right|}{1+\left|z+t\right|}}\ge {\displaystyle \frac{1+\left|z\right|}{1+\left|z\right|+\left|t\right|}}\ge {\displaystyle \frac{1+\left|z\right|}{1+\left|z\right|+\frac{\sqrt{n}}{2}}}\ge {\displaystyle \frac{1+\left|z\right|}{1+\frac{\sqrt{n}}{2}}}\ge {\displaystyle \frac{1}{2\sqrt{n}}}.$$

Furthermore,

$${\displaystyle \frac{1+\left|z\right|}{1+\left|z+t\right|}}\le \left\{\begin{array}{cc}1+\sqrt{n},\hfill & 0\le \left|z\right|\le \sqrt{n},\hfill \\ {\displaystyle \frac{1+\left|z\right|}{1+\left|z\right|-\left|t\right|}}\le {\displaystyle \frac{1+\left|z\right|}{1+\left|z\right|-\frac{\sqrt{n}}{2}}}\le {\displaystyle \frac{1+\sqrt{n}}{1+\frac{\sqrt{n}}{2}}}\le 2\sqrt{n},\hfill & \left|z\right|\ge \sqrt{n}.\hfill \end{array}\right.$$

The proof is complete. □

For any $s\in {\mathbb{R}}^{n-1}$ and $j\in \mathbb{N}$, let

$$g_j\left(s\right)={\int }_0^{{\displaystyle \frac{1}{R}}}\widehat{f}\left({s,\left|s\right|}^2+t\right){\left({\displaystyle \frac{Rt}{2}}\right)}^j\mathrm{d}t.$$

For any cube $Q\subset {[0,1]}^{n-1}$, by the definition of $E_Q{g}_j\left(x\right)$, we have

$$\begin{array}{cc}\hfill E_Q{g}_j\left(x\right)=& {\int }_{Q}e\left(s·x^{\prime }+{\left|s\right|}^2{x}_n\right)g_j\left(s\right)\mathrm{d}s={\int }_Q\mathrm{d}s{\int }_0^{{\displaystyle \frac{1}{R}}}e\left(s·x^{\prime }+{\left|s\right|}^2{x}_n\right)\widehat{f}\left({s,\left|s\right|}^2+t\right){\left({\displaystyle \frac{Rt}{2}}\right)}^j\mathrm{d}t.\hfill \end{array}$$

(27)

Take the cube $Q={\prod }_{i=1}^{n-1}\left[u_i,u_i+R^{-1/2}\right]\subset {[0,1]}^{n-1}$. Set $u=\left(u_1,\dots ,u_{n-1}\right)$. Define

$$A_Q=\left(\begin{array}{cc}{E}_{n-1}& 2u^{\mathrm{T}}\\ 0& 1\end{array}\right) \mathrm{and} b_Q=\left({u,\left|u\right|}^2\right),$$

(28)

where $E_{n-1}$ is the identity matrix of order $n-1$. Clearly, for any $\xi \in {\mathbb{R}}^n$, $\xi A_Q+b_Q\in N_{1/R}\left(Q\right)$ if and only if $\xi \in N_{1/R}{[0,1/\sqrt{R}]}^{n-1}$.

Lemma 8. 

Let $\eta ,x\in {\mathbb{R}}^n$. We have

$$\begin{array}{cc}\hfill E_Q{g}_j\left(x\right)=& e\left(x^{\prime }·u+{\left|u\right|}^2{x}_n\right)\hfill \\ \hfill & ·{\int }_{N_{\frac{1}{R}}{\left[0,{\displaystyle \frac{1}{\sqrt{R}}}\right]}^{n-1}}e\left[x^{\prime }·{\eta }^{\prime }+x_n\left(|{\eta }^{\prime }{|}^2+2u·{\eta }^{\prime }\right)\right]\widehat{f}\left(\eta A_Q+b_Q\right){\left({\displaystyle \frac{R\left({\eta }_n-{\left|{\eta }^{\prime }\right|}^2\right)}{2}}\right)}^j\mathrm{d}\eta .\hfill \end{array}$$

(29)

Proof. 

By (27) and applying the transformation

$$\left({s,\left|s\right|}^2+t\right)=\eta A_Q+b_Q=\left({\eta }^{\prime }+u,2{\eta }^{\prime }·u^{\prime }+{\eta }_n+{\left|u\right|}^2\right),$$

we have

$$\begin{array}{cc}\hfill E_Q{g}_j\left(x\right)=& {\int }_{N_{\frac{1}{R}}{\left[0,{\displaystyle \frac{1}{\sqrt{R}}}\right]}^{n-1}}e\left[x^{\prime }·\left(u+{\eta }^{\prime }\right)+{\left|u+{\eta }^{\prime }\right|}^2{x}_n\right]\widehat{f}\left(\eta A_Q+b_Q\right){\left({\displaystyle \frac{R\left({\eta }_n-{\left|{\eta }^{\prime }\right|}^2\right)}{2}}\right)}^j{\mathrm{d}}_{\eta 1}\dots {\mathrm{d}}_{\eta n}\hfill \\ \hfill =& e\left(x^{\prime }·u+{\left|u\right|}^2{x}_n\right)\hfill \\ \hfill & ·{\int }_{N_{\frac{1}{R}}{\left[0,{\displaystyle \frac{1}{\sqrt{R}}}\right]}^{n-1}}e\left[x^{\prime }·{\eta }^{\prime }+x_n\left(|{\eta }^{\prime }{|}^2+2u·{\eta }^{\prime }\right)\right]\widehat{f}\left(\eta A_Q+b_Q\right){\left({\displaystyle \frac{R\left({\eta }_n-{\left|{\eta }^{\prime }\right|}^2\right)}{2}}\right)}^j\mathrm{d}\eta ,\hfill \end{array}$$

which completes the proof. □

Lemma 9. 

For any $y\in {\mathbb{R}}^n$ and $j\in \mathbb{N}$, set

$$h_j\left(y\right)={\int }_{N_{\frac{1}{R}}{\left[0,{\displaystyle \frac{1}{\sqrt{R}}}\right]}^{n-1}}e\left[{\eta }^{\prime }·y^{\prime }+y_n{\left|{\eta }^{\prime }\right|}^2\right]\widehat{f}\left(\eta A_Q+b_Q\right){\left({\displaystyle \frac{R\left({\eta }_n-{\left|{\eta }^{\prime }\right|}^2\right)}{2}}\right)}^j\mathrm{d}\eta .$$

(30)

We have

$${∥E_Q{g}_j∥}_{L^p\left({\omega }_{B_R,F}\right)}\le {\left(\sqrt{6n}\right)}^{{\displaystyle \frac{F}{p}}}{∥h_j∥}_{L^p\left({\omega }_{B_R,F}\right)}.$$

(31)

Proof. 

From (29) and (24), it follows that

$$\begin{array}{cc}\hfill {∥E_Q{g}_j∥}_{L^p\left({\omega }_{B_R,F}\right)}^p=& {\int \left|h_j\left(x^{\prime }+2x_{n}u,x_n\right)\right|}^p{\omega }_{B_R,F}\left(x\right)\mathrm{d}x\hfill \\ \hfill =& {\int \left|h_j\left(y\right)\right|}^p{\displaystyle \frac{1}{{\left(1+\frac{\sqrt{|y^{\prime }-2y_{n}u{|}^2+y_n^2}}{R}\right)}^F}}\mathrm{d}y\hfill \\ \hfill \le & {\int \left|h_j\left(y\right)\right|}^p{\displaystyle \frac{1}{{\left(1+\frac{\sqrt{\frac{1}{6n}}\left|y\right|}{R}\right)}^F}}\mathrm{d}y\hfill \\ \hfill \le & {\int \left|h_j\left(y\right)\right|}^p{\left({\displaystyle \frac{\sqrt{6n}}{1+\frac{\left|y\right|}{R}}}\right)}^F\mathrm{d}y={\left(\sqrt{6n}\right)}^F\int {\left|h_j\left(y\right)\right|}^p{\left({\displaystyle \frac{1}{1+\frac{\left|y\right|}{R}}}\right)}^F\mathrm{d}y.\hfill \end{array}$$

□

Lemma 10. 

For a cube $Q\subset {[0,1]}^{n-1}$, set $F=f_{N_{1/R}\left(Q\right)}$. We have

$$\mathrm{supp}\widehat{F}\subset N_{\frac{1}{R}}\left(Q\right) and \widehat{F}\left(x\right)=\widehat{f}\left(x\right),\forall x\in N_{\frac{1}{R}}\left(Q\right).$$

(32)

Proof. 

Take a test function $h\in \mathcal{S}\left({\mathbb{R}}^n\right)$. We have

$$\begin{array}{cc}\hfill \int \widehat{F}h=\int F\widehat{h}=& {\int }_{{\mathbb{R}}^n}\left[{\int }_{N_{\frac{1}{R}}\left(Q\right)}\widehat{f}\left(\xi \right)e(x·\xi )\mathrm{d}\xi \right]\widehat{h}\left(x\right)\mathrm{d}x\hfill \\ \hfill =& {\int }_{N_{\frac{1}{R}}\left(Q\right)}\widehat{f}\left(\xi \right)\left[\int \widehat{h}\left(x\right)e(x·\xi )\mathrm{d}x\right]\mathrm{d}\xi ={\int }_{N_{\frac{1}{R}}\left(Q\right)}\widehat{f}\left(\xi \right)h\left(\xi \right)\mathrm{d}\xi ,\hfill \end{array}$$

which implies that $\mathrm{supp}\widehat{F}\subset N_{\frac{1}{R}}\left(Q\right)$. Furthermore, by

$$\begin{array}{cc}\hfill \int \left(\widehat{F}-\widehat{f}\right)h=& {\int }_{N_{\frac{1}{R}}\left(Q\right)}\widehat{f}\left(\xi \right)h\left(\xi \right)\mathrm{d}\xi -{\int }_{N_{\frac{1}{R}}{\left[0,1\right]}^{n-1}}\widehat{f}\left(\xi \right)h\left(\xi \right)\mathrm{d}\xi \hfill \\ \hfill =& {\int }_{N_{\frac{1}{R}}{\left[0,1\right]}^{n-1}\setminus N_{\frac{1}{R}}\left(Q\right)}\widehat{f}\left(\xi \right)h\left(\xi \right)\mathrm{d}\xi ,\hfill \end{array}$$

we have $\mathrm{supp}\left(\widehat{F}-\widehat{f}\right)\in N_{\frac{1}{R}}{\left[0,1\right]}^{n-1}\setminus N_{\frac{1}{R}}\left(Q\right)$, which means that $\widehat{F}=\widehat{f}$ in $N_{\frac{1}{R}}\left(Q\right)$. □

For any $y,z\in {\mathbb{R}}^n$ and any $j\in \mathbb{N}$, define

$${\psi }_j(y,z)={\int }_{N_{\frac{1}{R}}{\left[0,{\displaystyle \frac{1}{\sqrt{R}}}\right]}^{n-1}}\widehat{f}\left(\eta A_Q+b_Q\right){\left({\displaystyle \frac{R\left({\eta }_n-{\left|{\eta }^{\prime }\right|}^2\right)}{2}}\right)}^{j}e\left(\eta ·y\right)e\left(z_n\left(|{\eta }^{\prime }{|}^2-{\eta }_n\right)\right)\mathrm{d}\eta ,$$

(33)

where $A_Q$ and $b_Q$ are defined as in (28). Let $\eta \in \mathcal{S}\left(\mathbb{R}\right)$ be a Schwartz function satisfying (1), and $M_j\in \mathcal{S}\left(\mathbb{R}\right)$ be a Schwartz function with a compact support and satisfying (2).

Lemma 11. 

Let $Q\subset {[0,1]}^{n-1}$ be a cube and $F=f_{N_{1/R}\left(Q\right)}$. We have

$${\psi }_j(y,z)=\int \widehat{F}(\xi A_Q+b_Q)m_j\left(\xi \right)e\left({\xi }^{\prime }·y^{\prime }+{\xi }_n\left(y_n-z_n\right)\right)\mathrm{d}\xi ,\forall y,z\in {\mathbb{R}}^n,$$

(34)

where

$$m_j\left(\xi \right)=e\left(\right|{\xi }^{\prime }{|}^2{z}_n)M_j\left({\displaystyle \frac{R({\xi }_n-|{\xi }^{\prime }{|}^2)}{2}}\right)\eta \left(R^{{\displaystyle \frac{1}{2}}}{\xi }_1\right)\dots \eta \left(R^{{\displaystyle \frac{1}{2}}}{\xi }_{n-1}\right)\eta \left(R{\xi }_n\right).$$

(35)

Proof. 

Note that $m_j$ is actually a function of z. For notational simplicity, this dependence will be omitted in what follows.

Note that when $\widehat{F}(\xi A_Q+b_Q)\ne 0$, we have $\xi A_Q+b_Q\in N_{1/R}\left(Q\right)$ by (32). It means $\xi \in N_{1/R}{\left[0,1\sqrt{/R}\right]}^{n-1}$. Thus,

$${\displaystyle \frac{R\left({\xi }_n-{\left|{\xi }^{\prime }\right|}^2\right)}{2}}\in \left[0,{\displaystyle \frac{1}{2}}\right] \mathrm{and} {\xi }_i\in \left[0,{\displaystyle \frac{1}{\sqrt{R}}}\right],\forall i\in \left[n-1\right],$$

which implies that

$$0\le R{\xi }_n\le 1+R{\left|{\xi }^{\prime }\right|}^2\le 1+(n-1)=n.$$

It follows that

$$\begin{array}{cc}\hfill & \int \widehat{F}\left(\xi A_Q+b_Q\right)m_j\left(\xi \right)e\left({\xi }^{\prime }·y^{\prime }+{\xi }_n\left(y_n-z_n\right)\right)\mathrm{d}\xi \hfill \\ \hfill =& {\int }_{N_{\frac{1}{R}}{\left[0,{\displaystyle \frac{1}{\sqrt{R}}}\right]}^{n-1}}\widehat{F}\left(\xi A_Q+b_Q\right){\left[{\displaystyle \frac{R\left({\xi }_n-{\left|{\xi }^{\prime }\right|}^2\right)}{2}}\right]}^{j}e(\xi ·y)e\left(z_n\left(|{\xi }^{\prime }{|}^2-{\xi }_n\right)\right)\mathrm{d}\xi ={\psi }_j(y,z).\hfill \end{array}$$

□

Lemma 12. 

For any $s\in \mathbb{N}$, let $\rho \left(s\right)$ be defined as in (3). For any $s_1,\dots ,s_n\in \mathbb{N}$, we have

$$\left|{\partial }_{{\xi }_1}^{s_1}\dots {\partial }_{{\xi }_n}^{s_n}{m}_j\left(\xi \right)\right|\le C_0(s_1,\dots ,s_n){(\sqrt{R}+{\displaystyle \frac{|z_n|}{\sqrt{R}}})}^{s_1+\dots +s_{n-1}}{R}^{s_n},$$

(36)

where $m_j\left(\xi \right)$ is defined as in (35),

$$C_0(s_1,\dots ,s_n)=max{\tilde{C}}_0,\dots ,{\tilde{C}}_{{\sum }_{i=1}^n{s}_i}\left({\displaystyle \prod _{j=1}^n}\rho \left(s_j\right)\right){({\displaystyle \prod _{k=1}^{n-1}}{s}_k!)}^2{\left(14n\right)}^{{\displaystyle \sum _{i=1}^n}{s}_i},$$

and ${\tilde{C}}_0,\dots ,{\tilde{C}}_{{\sum }_{i=1}^n{s}_i}$ are given by (2).

Proof. 

By $\mathrm{supp}\eta \subset [-(n+1),n+1]$, for each $\xi \in {\mathbb{R}}^n$ satisfying that there exists $i\in [n-1]$ such that $|{\xi }_i|>(n+1)/\sqrt{R}$, we have $m_j\left(\xi \right)=0$. Thus, without loss of generality, we may assume that $|{\xi }_i|\le (n+1)\sqrt{R}$, $\forall i\in [n-1]$. Observing that

$$\begin{array}{cc}\hfill {\partial }_{{\xi }_1}^{s_1}\dots {\partial }_{{\xi }_n}^{s_n}{m}_j\left(\xi \right)=& \sum _{\alpha +\beta =s_n}\left\{{\displaystyle \frac{s_n!}{\alpha !\beta !}}{R}^{\alpha }{\left({\displaystyle \frac{R}{2}}\right)}^{\beta }{\eta }^{\left(\alpha \right)}\left(R{\xi }_n\right)\right.\hfill \\ \hfill & ·\sum _{u_{n-1}+v_{n-1}+w_{n-1}=s_{n-1}}\left[{\displaystyle \frac{s_{n-1}!}{u_{n-1}!v_{n-1}!w_{n-1}!}}{R}^{{\displaystyle \frac{v_{n-1}}{2}}}{\partial }_{{\xi }_{n-1}}^{u_{n-1}}e\left(z_n{\xi }_{n-1}^2\right)·{\eta }^{\left(v_{n-1}\right)}\left(R^{{\displaystyle \frac{1}{2}}}{\xi }_{n-1}\right)\right.\hfill \\ \hfill & \dots \sum _{u_1+v_1+w_1=s_1}\left({\displaystyle \frac{s_1!}{u_1!v_1!w_1!}}{R}^{{\displaystyle \frac{v_1}{2}}}{\partial }_{{\xi }_1}^{u_1}e\left(z_n{\xi }_1^2\right)·{\eta }^{\left(v_1\right)}\left(R^{{\displaystyle \frac{1}{2}}}{\xi }_1\right)\right.\hfill \\ \hfill & \left.\left.\left.·{\partial }_{{\xi }_1}^{w_1}\dots {\partial }_{{\xi }_{n-1}}^{w_{n-1}}{M}_j^{\left(\beta \right)}\left({\displaystyle \frac{R({\xi }_n-|{\xi }^{\prime }{|}^2)}{2}}\right)\right)\dots \right]\right\},\hfill \end{array}$$

(37)

by (11) and the definition of $M_j\left(t\right)$, we have

$$\begin{array}{cc}\hfill & |{\partial }_{{\xi }_1}^{w_1}\dots {\partial }_{{\xi }_{n-1}}^{w_{n-1}}{M}_j^{\left(\beta \right)}\left({\displaystyle \frac{R({\xi }_n-|{\xi }^{\prime }{|}^2)}{2}}\right)|\hfill \\ \hfill =& \left|{\displaystyle \sum _{k_{n-1}=0}^{⌊{\displaystyle \frac{w_{n-1}}{2}}⌋}}\dots {\displaystyle \sum _{k_1=0}^{⌊{\displaystyle \frac{w_1}{2}}⌋}}I(w_{n-1},k_{n-1})\dots I(w_1,k_1){\xi }_{n-1}^{w_{n-1}-2k_{n-1}}\dots {\xi }_1^{w_1-2k_1}\right.\hfill \\ \hfill & \left.·{(-{\displaystyle \frac{R}{2}})}^{w_{n-1}-k_{n-1}}\dots {(-{\displaystyle \frac{R}{2}})}^{w_1-k_1}{M}_j^{(\beta +{\displaystyle \sum _{i=1}^{n-1}}(w_i-k_i))}\left({\displaystyle \frac{R({\xi }_n-|{\xi }^{\prime }{|}^2)}{2}}\right)\right|\hfill \\ \hfill \le & max\left\{{\tilde{C}}_1,\dots ,{\tilde{C}}_{{\sum }_{i=1}^n{s}_i}\right\}{\displaystyle \prod _{r=1}^{n-1}}{\displaystyle \sum _{k_r=0}^{⌊{\displaystyle \frac{w_r}{2}}⌋}}(I(w_r,k_r)|{\xi }_r{|}^{w_r-2k_r}{\left({\displaystyle \frac{R}{2}}\right)}^{w_r-k_r})\hfill \\ \hfill \le & max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{{\sum }_{i=1}^n{s}_i}\right\}{\displaystyle \prod _{r=1}^{n-1}}{\displaystyle \sum _{k_r=0}^{⌊{\displaystyle \frac{w_r}{2}}⌋}}\left(I(w_r,k_r){\left({\displaystyle \frac{n+1}{\sqrt{R}}}\right)}^{w_r-2k_r}{\left({\displaystyle \frac{R}{2}}\right)}^{w_r-k_r}\right)\hfill \\ \hfill \le & max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{{\sum }_{i=1}^n{s}_i}\right\}{\displaystyle \prod _{r=1}^{n-1}}\left({(n+1)}^{w_r}{R}^{{\displaystyle \frac{w_r}{2}}}{\displaystyle \sum _{k_r=0}^{⌊{\displaystyle \frac{w_r}{2}}⌋}}I(w_r,k_r)\right).\hfill \end{array}$$

By (12), we get

$$\begin{array}{cc}\hfill {\displaystyle \prod _{r=1}^{n-1}}\left({(n+1)}^{w_r}{R}^{{\displaystyle \frac{w_r}{2}}}{\displaystyle \sum _{k_r=0}^{⌊{\displaystyle \frac{w_r}{2}}⌋}}I(w_r,k_r)\right)\le & {\left(6(n+1)\sqrt{R}\right)}^{{\displaystyle \sum _{j=1}^{n-1}}{w}_j}{\displaystyle \prod _{j=1}^{n-1}}{s}_j!.\hfill \end{array}$$

Thus,

$$\left|{\partial }_{{\xi }_1}^{w_1}\dots {\partial }_{{\xi }_{n-1}}^{w_{n-1}}{M}_j^{\left(\beta \right)}\left({\displaystyle \frac{R({\xi }_n-|{\xi }^{\prime }{|}^2)}{2}}\right)\right|\le max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{{\sum }_{i=1}^n{s}_i}\right\}{\left(6(n+1)\sqrt{R}\right)}^{{\displaystyle \sum _{j=1}^{n-1}}{w}_j}{\displaystyle \prod _{j=1}^{n-1}}{s}_j!.$$

(38)

For any $i\in [n-1]$, we have

$$\begin{array}{cc}\hfill \left|{\partial }_{{\xi }_i}^{u_i}e\left(z_n{\xi }_i^2\right)·{\eta }^{\left(v_i\right)}\left(R^{{\displaystyle \frac{1}{2}}}{\xi }_i\right)\right|=& |{\eta }^{\left(v_i\right)}\left(R^{{\displaystyle \frac{1}{2}}}{\xi }_i\right){\displaystyle \sum _{k=1}^{⌊{\displaystyle \frac{u_i}{2}}⌋}}{\displaystyle \frac{u_i!}{(u_i-2k)!k!}}{\left(2{\xi }_i\right)}^{u_i-2k}{\left(2\pi i{z}_n\right)}^{u_i-k}e\left(z_n{\xi }_i^2\right)|\hfill \\ \hfill \le & {\displaystyle \sum _{k=1}^{⌊{\displaystyle \frac{u_i}{2}}⌋}}{\displaystyle \frac{u_i!}{(u_i-2k)!k!}}|{\eta }^{\left(v_i\right)}(R^{{\displaystyle \frac{1}{2}}}{\xi }_i)\left|\right|2R^{{\displaystyle \frac{1}{2}}}{\xi }_i{|}^{u_i-2k}{R}^{-{\displaystyle \frac{1}{2}}(u_i-2k)}{\left|2\pi z_n\right|}^{u_i-k}\hfill \\ \hfill \le & \rho \left(s_i\right){\displaystyle \sum _{k=0}^{⌊{\displaystyle \frac{u_i}{2}}⌋}}{\displaystyle \frac{u_i!}{(u_i-2k)!k!}}{\left({\displaystyle \frac{2}{\sqrt{R}}}\right)}^{u_i-2k}{\left|2\pi z_n\right|}^{u_i-k}\hfill \\ \hfill =& \rho \left(s_i\right){\displaystyle \sum _{k=0}^{⌊{\displaystyle \frac{u_i}{2}}⌋}}{\displaystyle \frac{u_i!}{(u_i-k)!k!}}{\displaystyle \frac{(u_i-k)!}{(u_i-2k)!}}{\left({\displaystyle \frac{\sqrt{R}}{2}}\right)}^k{\left({\displaystyle \frac{4\pi |z_n|}{\sqrt{R}}}\right)}^{u_i-k}\hfill \\ \hfill \le & \rho \left(s_i\right)s_i!{({\displaystyle \frac{\sqrt{R}}{2}}+{\displaystyle \frac{4\pi |z_n|}{\sqrt{R}}})}^{u_i}.\hfill \end{array}$$

(39)

By (37)–(39), we have

$$\begin{array}{cc}\hfill & {\partial }_{{\xi }_1}^{s_1}\dots {\partial }_{{\xi }_n}^{s_n}{m}_j\left(\xi \right)\hfill \\ \hfill \le & max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{{\sum }_{i=1}^n{s}_i}\right\}\left({\displaystyle \prod _{i=1}^n}\rho \left(s_i\right)\right){({\displaystyle \prod _{j=1}^{n-1}}{s}_j!)}^2\hfill \\ \hfill & ·\sum _{\alpha +\beta =s_n}\left\{{\displaystyle \frac{s_n!}{\alpha !\beta !}}{R}^{\alpha }{\left({\displaystyle \frac{R}{2}}\right)}^{\beta }\right.\hfill \\ \hfill & ·\sum _{u_{n-1}+v_{n-1}+w_{n-1}=s_{n-1}}\left[{\displaystyle \frac{s_{n-1}!}{u_{n-1}!v_{n-1}!w_{n-1}!}}{R}^{{\displaystyle \frac{v_{n-1}}{2}}}{({\displaystyle \frac{\sqrt{R}}{2}}+{\displaystyle \frac{4\pi |z_n|}{\sqrt{R}}})}^{u_{n-1}}{\left(6(n+1)\sqrt{R}\right)}^{w_{n-1}}\right.\hfill \\ \hfill & \left.\left.\dots \sum _{u_1+v_1+w_1=s_1}\left({\displaystyle \frac{s_1!}{u_1!v_1!w_1!}}{R}^{{\displaystyle \frac{v_1}{2}}}{({\displaystyle \frac{\sqrt{R}}{2}}+{\displaystyle \frac{4\pi |z_n|}{\sqrt{R}}})}^{u_1}{\left(6(n+1)\sqrt{R}\right)}^{w_1}\right)\dots \right]\right\}.\hfill \end{array}$$

Finally, by

$$\begin{array}{cc}\hfill & \sum _{\alpha +\beta =s_n}\left\{{\displaystyle \frac{s_n!}{\alpha !\beta !}}{R}^{\alpha }{\left({\displaystyle \frac{R}{2}}\right)}^{\beta }\right.\hfill \\ \hfill & ·\sum _{u_{n-1}+v_{n-1}+w_{n-1}=s_{n-1}}\left[{\displaystyle \frac{s_{n-1}!}{u_{n-1}!v_{n-1}!w_{n-1}!}}{R}^{{\displaystyle \frac{v_{n-1}}{2}}}{({\displaystyle \frac{\sqrt{R}}{2}}+{\displaystyle \frac{4\pi |z_n|}{\sqrt{R}}})}^{u_{n-1}}{\left(6(n+1)\sqrt{R}\right)}^{w_{n-1}}\right.\hfill \\ \hfill & \left.\left.\dots \sum _{u_1+v_1+w_1=s_1}\left({\displaystyle \frac{s_1!}{u_1!v_1!w_1!}}{R}^{{\displaystyle \frac{v_1}{2}}}{({\displaystyle \frac{\sqrt{R}}{2}}+{\displaystyle \frac{4\pi |z_n|}{\sqrt{R}}})}^{u_1}{\left(6(n+1)\sqrt{R}\right)}^{w_1}\right)\dots \right]\right\}\hfill \\ \hfill =& {\left({\displaystyle \frac{3R}{2}}\right)}^{s_n}{((6(n+1)+{\displaystyle \frac{3}{2}})\sqrt{R}+{\displaystyle \frac{4\pi |z_n|}{\sqrt{R}}})}^{{\displaystyle \sum _{j=1}^{n-1}}{s}_j}\le {\left(14n\right)}^{{\displaystyle \sum _{i=1}^n}{s}_i}{R}^{s_n}{(\sqrt{R}+{\displaystyle \frac{|z_n|}{\sqrt{R}}})}^{{\displaystyle \sum _{j=1}^{n-1}}{s}_j},\hfill \end{array}$$

we complete the proof. □

Lemma 13. 

Let $z_n$ be consistent with the choice in (35). For any $x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n$ and $s_1,\dots ,s_n\in \mathbb{N}$, set

$${\varphi }_k\left(x_k\right)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{\sqrt{R}}}{\left(1+{\displaystyle \frac{|x_k|}{\sqrt{R}+\frac{|z_n|}{\sqrt{R}}}}\right)}^{-s_k},\hfill & k\in [n-1],\hfill \\ {\displaystyle \frac{1}{R}}{\left(1+{\displaystyle \frac{|x_n|}{R}}\right)}^{-s_n},\hfill & k=n.\hfill \end{array}\right.$$

We have

$$|\widehat{m_j}\left(x\right)|\le C_0(s_1,\dots ,s_n){\left(4n\right)}^n{2}^{{\displaystyle \sum _{i=1}^n}{s}_i}{\displaystyle \prod _{j=1}^n}{\varphi }_j\left(x_j\right),$$

(40)

where $m_j$ is defined as in (35).

Proof. 

Actually, when $k\ne n$, ${\varphi }_k$ are also functions of z. These dependences will be omitted in what follows.

Let $a_k=|x_k| /(\sqrt{R}+|z_n|/R)$ for each $k\in [n-1]$, and $a_n=|x_n|/R$. Let $I=\left\{k\in \left[n\right]|a_k\ge 1\right\}$. Clearly, we have

$$\mathrm{supp}{m}_j\subset {[-(n+1)/\sqrt{R},(n+1)/\sqrt{R}]}^{n-1}\times [-(n+1)/R,(n+1)/R]=:\mathsf{\Omega }.$$

By (36),

$$\left|\left(\prod _{k\in I}{\partial }_{{\xi }_k}^{s_k}\right)m_j\left(\xi \right)\right|\le C_0(s_1,\dots ,s_n){(\sqrt{R}+{\displaystyle \frac{|z_n|}{R}})}^{\sum _{k\in I\cap [n-1]}{s}_k}{R}^{\sum _{k\in I\cap \left\{n\right\}}{s}_k}$$

and

$$\widehat{\prod _{k\in I}{\partial }_{{\xi }_k}^{s_k}{m}_j}\left(x\right)=\left(\prod _{k\in I}{\left(2\pi i{x}_k\right)}^{s_k}\right)\widehat{m_j}\left(x\right),$$

we obtain

$$\begin{array}{cc}\hfill |\widehat{m_j}\left(x\right)|\le & \left(\prod _{k\in I}{|x_k|}^{-s_k}\right)|\int \left(\prod _{k\in I}{\partial }_{{\xi }_k}^{s_k}\right)m_j\left(\xi \right)e(-x\xi )\mathrm{d}\xi |\hfill \\ \hfill \le & (\prod _{k\in I}|x_k{|}^{-s_k}){\int }_{\mathsf{\Omega }}\left|\left(\prod _{k\in I}{\partial }_{{\xi }_k}^{s_k}\right)m_j\left(\xi \right)\right|\mathrm{d}\xi \hfill \\ \hfill \le & (\prod _{k\in I}|x_k{|}^{-s_k})C_0(s_1,\dots ,s_n){(\sqrt{R}+{\displaystyle \frac{|z_n|}{R}})}^{\sum _{k\in I\cap [n-1]}{s}_k}{R}^{\sum _{k\in I\cap \left\{n\right\}}{s}_k}{\left({\displaystyle \frac{2(n+1)}{\sqrt{R}}}\right)}^{n-1}{\displaystyle \frac{2(n+1)}{R}}\hfill \\ \hfill =& C_0(s_1,\dots ,s_n){\left[2(n+1)\right]}^n{\left({\displaystyle \frac{1}{\sqrt{R}}}\right)}^{n-1}{\displaystyle \frac{1}{R}}\left(\prod _{k\in I\cap [n-1]}{\left({\displaystyle \frac{1}{a_k}}\right)}^{s_k}\right)\left(\prod _{k\in I\cap \left\{n\right\}}{\left({\displaystyle \frac{1}{a_k}}\right)}^{s_k}\right)\hfill \\ \hfill \le & C_0(s_1,\dots ,s_n){\left[2(n+1)\right]}^n{\left({\displaystyle \frac{1}{\sqrt{R}}}\right)}^{n-1}{\displaystyle \frac{1}{R}}\prod _{k\in I}{2}^{s_k}{\left({\displaystyle \frac{1}{1+a_k}}\right)}^{s_k}\hfill \\ \hfill \le & C_0(s_1,\dots ,s_n){\left[2(n+1)\right]}^n{\left({\displaystyle \frac{1}{\sqrt{R}}}\right)}^{n-1}{\displaystyle \frac{1}{R}}\left(\prod _{k\in I}{2}^{s_k}{\left({\displaystyle \frac{1}{1+a_k}}\right)}^{s_k}\right)\left(\prod _{k\in \left[n\right]\setminus I}{2}^{s_k}{\left({\displaystyle \frac{1}{1+a_k}}\right)}^{s_k}\right)\hfill \\ \hfill \le & C_0(s_1,\dots ,s_n){\left(4n\right)}^n{2}^{{\displaystyle \sum _{i=1}^n}{s}_i}{\displaystyle \prod _{j=1}^n}{\varphi }_j\left(x_j\right),\hfill \end{array}$$

which implies (40). □

Lemma 14. 

We have

$$\int |\widehat{m_j}\left(x\right)|x\le C_m{(1+{\displaystyle \frac{|z_n|}{R}})}^{n-1},$$

(41)

where $C_m=max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{2n}\right\}{\left(\rho \left(2\right)\right)}^n{\left(30n\right)}^{3n}$ and $m_j$ is defined as in (35).

Proof. 

It follows from (40) that

$$|\widehat{m_j}\left(x\right)|\le C_0(2,\dots ,2){\left(4n\right)}^n{2}^{2n}{\displaystyle \frac{1}{R}}{\left(1+{\displaystyle \frac{|x_n|}{R}}\right)}^{-2}{\displaystyle \prod _{k=1}^{n-1}}{\displaystyle \frac{1}{\sqrt{R}}}{\left(1+{\displaystyle \frac{|x_k|}{\sqrt{R}+\frac{|z_n|}{\sqrt{R}}}}\right)}^{-2}.$$

(42)

We have

$$\begin{array}{cc}\hfill C_0(2,\dots ,2){\left(4n\right)}^n{2}^{2n}& =max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{2n}\right\}{\left(\rho \left(2\right)\right)}^n{2}^{2(n-1)}{\left(14n\right)}^{2n}{\left(4n\right)}^n{2}^{2n}\hfill \\ \hfill & \le max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{2n}\right\}{\left(\rho \left(2\right)\right)}^n{\left({\left(28n\right)}^{2}16n\right)}^n,\hfill \end{array}$$

(43)

$$\int {\displaystyle \frac{1}{R}}{\left(1+{\displaystyle \frac{|x_n|}{R}}\right)}^{-2}\mathrm{d}{x}_n=2{\int }_0^{\infty }{\left({\displaystyle \frac{1}{1+t}}\right)}^2\mathrm{d}t=2,$$

(44)

and

$$\begin{array}{cc}\hfill \int {\displaystyle \frac{1}{\sqrt{R}}}{\left(1+{\displaystyle \frac{|x_k|}{\sqrt{R}+\frac{|z_n|}{\sqrt{R}}}}\right)}^{-2}\mathrm{d}{x}_k=& 2{\int }_0^{\infty }{(1+{\displaystyle \frac{t}{1+\frac{|z_n|}{\sqrt{R}}}})}^{-2}\mathrm{d}t\hfill \\ \hfill =& -2(1+{\displaystyle \frac{|z_n|}{R}}){{\displaystyle \frac{1}{1+\frac{t}{1+\frac{|z_n|}{R}}}}}_0^{+\infty }=2(1+{\displaystyle \frac{|z_n|}{R}}).\hfill \end{array}$$

(45)

From (42)–(45), it follows that

$$\begin{array}{cc}\hfill \int |\widehat{m_j}\left(\xi \right)|\xi \le & max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{2n}\right\}{\left(\rho \left(2\right)\right)}^n{\left({\left(28n\right)}^{2}16n\right)}^n{2}^n{(1+{\displaystyle \frac{|z_n|}{R}})}^{n-1}\hfill \\ \hfill \le & max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{2n}\right\}{\left(\rho \left(2\right)\right)}^n{\left(30n\right)}^{3n}{(1+{\displaystyle \frac{|z_n|}{R}})}^{n-1},\hfill \end{array}$$

which implies (41). □

3. The Main Theorem

We now recall our main theorem.

Theorem 4. 

Let $p\ge 2$, $R^{-1}\in 4^{-\mathbb{N}}$, and $f:{\mathbb{R}}^n\to \mathbb{C}$ be a function satisfying $\mathrm{supp}\widehat{f}\subset N_{1/R}\left({[0,1]}^{n-1}\right)$. Consider $\eta \left(x\right)\in \mathcal{S}\left(\mathbb{R}\right)$ as (1),$M_j\left(t\right)\in \mathcal{S}\left(\mathbb{R}\right)$ as (2), and $\rho \left(s_j\right)$ as satisfying (3). Then for all $E\ge 100n$, $F\ge 2(n-1)⌈E⌉+(n-1)(p-1)+2$, and each cube $B_R\subset {\mathbb{R}}^n$, we have

$${\parallel f\parallel }_{L^p\left({\omega }_{B_R,E}\right)}\le C_0{Dec}_n\left(R^{-1},p,F\right)·{\left(\sum _{Q\in {Part}_{R^{-1/2}}\left({[0,1]}^{n-1}\right)}{∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B_R,E}\right)}^2\right)}^{1/2},$$

where $C_0=max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{{\displaystyle \frac{20n⌈E⌉}{p}}}{\left(9n\right)}^{{\displaystyle \frac{F}{p}}}{\left(30n\right)}^E$.

Proof. 

By Lemma 1 and Remark 1, we only need to prove that there exists a constant $C_1$, such that

$${\parallel f\parallel }_{L^p\left(B_R\right)}\le C_1{Dec}_n\left(R^{-1},p,F\right){\left(\sum _{Q\in {Part}_{R^{-1/2}}\left({[0,1]}^{n-1}\right)}{∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B_R,E}\right)}^2\right)}^{1/2},$$

(46)

and

$${\left(6\sqrt{n}\right)}^E{C}_1\le C_0.$$

(47)

In the rest of the proof, for $Q={\prod }_{i=1}^{n-1}[u_i,u_i+R^{-1/2}]\in {Part}_{R^{-1/2}}\left({[0,1]}^{n-1}\right)$, we always assume that the corresponding $A_Q$ and $b_Q$ are defined as in (28).

• Step 1: First, we show that, without loss of generality, the center of the cube may be assumed to be at the origin.

Noting that

$$\widehat{f(·+c)}\left(\xi \right)=e\left(c\xi \right)\widehat{f}\left(\xi \right),$$

we have

$$\begin{array}{cc}\hfill f_{N_{\frac{1}{R}}\left(Q\right)}(y+c)=& {\int }_{N_{1/R}\left(Q\right)}\widehat{f}\left(\xi \right)e\left((y+c)\xi \right)\mathrm{d}\xi \hfill \\ \hfill =& {\int }_{N_{\frac{1}{R}}\left(Q\right)}\widehat{f(·+c)}\left(\xi \right)e\left(y\xi \right)\mathrm{d}\xi =f{(·+c)}_{N_{\frac{1}{R}}\left(Q\right)}\left(y\right),\hfill \end{array}$$

which implies that

$${∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B(c,R),E}\right)}^p=\int {\left|f_{N_{\frac{1}{R}}\left(Q\right)}\left(x\right)\right|}^p{\omega }_{B(c,R),E}\left(x\right)\mathrm{d}x\stackrel{x=y+c}{=}\int {\left|f_{N_{\frac{1}{R}}\left(Q\right)}(y+c)\right|}^p{\omega }_{B(0,R),E}\left(y\right)\mathrm{d}y.$$

Simultaneously,

$${\left|\right|f\left|\right|}_{L^p\left(B(c,R)\right)}^p={\int }_{B(c,R)}{\left|f\left(x\right)\right|}^p\mathrm{d}x={\int }_{B(0,R)}{\left|f(y+c)\right|}^p\mathrm{d}y={\left|\right|f(·+c)\left|\right|}_{L^p\left(B(0,R)\right)}^p.$$

Thus, without loss of generality, we may assume that the center of $B(c,R)$ is at the origin, and denote the cube by $B_R$.

• Step 2: We show that the validity of (46) is guaranteed by

$${∥E_Q{g}_j∥}_{L^p\left({\omega }_{B_R,F}\right)}\le C_2{∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B_R,E}\right)},\forall Q\in {Part}_{R^{-{\displaystyle \frac{1}{2}}}}\left({[0,1]}^{n-1}\right),$$

(48)

where $C_1={\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^{{\displaystyle \frac{F}{p}}}{e}^{2\pi }{C}_2$.

Set $x=\left(x_1,\dots ,x_n\right)$, $\xi =\left({\xi }_1,\dots ,{\xi }_n\right)$, $x^{\prime }=\left(x_1,\dots ,x_{n-1}\right)$, and $s=\left(s_1,\dots ,s_{n-1}\right)$. By applying the transformation

$${\xi }_i=s_i,{\xi }_n={\left|s\right|}^2+t,\forall i\in [n-1],$$

we have

$$\begin{array}{cc}\hfill f\left(x\right)=& {\int }_{N_{1/R}\left({[0,1]}^{n-1}\right)}\widehat{f}\left(\xi \right)e(\xi ·x)\mathrm{d}\xi \hfill \\ \hfill =& {\int }_{{[0,1]}^{n-1}}s{\int }_0^{{\displaystyle \frac{1}{R}}}\widehat{f}\left({s,\left|s\right|}^2+t\right)e\left(x^{\prime }·s+{\left|s\right|}^2{x}_n\right)e\left(t{x}_n\right)\mathrm{d}t\hfill \\ \hfill =& \sum _{j\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^j}{j!}}{\left({\displaystyle \frac{2i{x}_n}{R}}\right)}^j{\int }_{{[0,1]}^{n-1}}\mathrm{d}s{\int }_0^{{\displaystyle \frac{1}{R}}}\widehat{f}\left({s,\left|s\right|}^2+t\right)e\left(x^{\prime }·s+{\left|s\right|}^2{x}_n\right){\left({\displaystyle \frac{Rt}{2}}\right)}^j\mathrm{d}t.\hfill \end{array}$$

For the last equality, since $\widehat{f}$ is bounded, we have

$$sup|\widehat{f}|\sum _{j\ge 0}{\displaystyle \frac{\left(2\pi \right|x_n{\left|\right)}^j}{j!(j+1)R^{j+1}}}\le {\displaystyle \frac{sup|\widehat{f}|}{R}}{e}^{{\displaystyle \frac{2\pi |x_n|}{R}}}<\infty .$$

Thus, Fubini’s theorem justifies the interchange of the sum and the integral. Combining with (27), we obtain

$$\left|f\left(x\right)\right|\le \sum _{j\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^j}{j!}}\left|E{g}_j\left(x\right)\right|,\forall x\in B_R,$$

which, together with Minkowski’s inequality and (48), implies that

$$\begin{array}{cc}\hfill {∥f∥}_{L^p\left(B_R\right)}\le & \sum _{j\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^j}{j!}}{∥E{g}_j∥}_{L^p\left(B_R\right)}\hfill \\ \hfill \le & {\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^{{\displaystyle \frac{F}{p}}}\sum _{j\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^j}{j!}}{∥E{g}_j∥}_{L^p\left({\omega }_{B_R,F}\right)}\hfill \\ \hfill \le & {\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^{{\displaystyle \frac{F}{p}}}{Dec}_n\left(R^{-1},p,F\right)\sum _{j\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^j}{j!}}{\left(\sum _{Q\in {Part}_{R^{-1/2}}\left({[0,1]}^{n-1}\right)}{∥E_Q{g}_j∥}_{L^p\left({\omega }_{B_R,F}\right)}^2\right)}^{1/2}\hfill \\ \hfill \le & {\left(1+{\displaystyle \frac{\sqrt{n}}{2}}\right)}^{{\displaystyle \frac{F}{p}}}{Dec}_n\left(R^{-1},p,F\right)e^{2\pi }{C}_2{\left(\sum _{Q\in {Part}_{R^{-1/2}}\left([0,1]\right)}{∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B_R,E}\right)}^2\right)}^{1/2}\hfill \\ \hfill =& C_1{\left(\sum _{Q\in {Part}_{R^{-1/2}}\left([0,1]\right)}{∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B_R,E}\right)}^2\right)}^{1/2}.\hfill \end{array}$$

Thus, (48) implies (46) as claimed.

• Step 3: We show that for any $Q\in {Part}_{R^{-{\displaystyle \frac{1}{2}}}}\left({[0,1]}^{n-1}\right)$ and ${\psi }_j(y,z)$ defined as in (33), the inequality

$$\int {∥{\psi }_j(y,z)∥}_{L_{♯}^p\left(B(z,R)\right)}^p{\omega }_{B_R,F}\left(z\right)z\le C_3^p{∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B_R,E}\right)}^p$$

(49)

implies (48), where $C_3$ is the constant satisfying $C_2={\left(2\sqrt{6}n\right)}^{{\displaystyle \frac{F}{p}}}{e}^{2\pi }{C}_3$.

In fact, by

$$\begin{array}{cc}\hfill e\left({\eta }^{\prime }·y^{\prime }+{\left|{\eta }^{\prime }\right|}^2{y}_n\right)=& e\left(\eta ·y\right)e\left(y_n\left(|{\eta }^{\prime }{|}^2-{\eta }_n\right)\right)\hfill \\ \hfill =& e\left(\eta ·y\right)e\left(z_n\left(|{\eta }^{\prime }{|}^2-{\eta }_n\right)\right)e\left(\left(y_n-z_n\right)\left(|{\eta }^{\prime }{|}^2-{\eta }_n\right)\right)\hfill \end{array}$$

and

$$e\left(\left(y_n-z_n\right)\left(|{\eta }^{\prime }{|}^2-{\eta }_n\right)\right)=\sum _{k\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^k}{k!}}{\left[{\displaystyle \frac{2i\left(z_n-y_n\right)}{R}}\right]}^k{\left[{\displaystyle \frac{R\left({\eta }_n-{\left|{\eta }^{\prime }\right|}^2\right)}{2}}\right]}^k,$$

when $y\in B(z,R)$, we have

$$\begin{array}{cc}\hfill |h_j\left(y\right)|\le & \sum _{k\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^k}{k!}}\left|{\int }_{N_{\frac{1}{R}}{\left[0,{\displaystyle \frac{1}{\sqrt{R}}}\right]}^{n-1}}\widehat{f}\left(\eta A_Q+b_Q\right){\left({\displaystyle \frac{R\left({\eta }_n-{\left|{\eta }^{\prime }\right|}^2\right)}{2}}\right)}^{k+j}e\left(\eta ·y\right)e\left(z_n\left(|{\eta }^{\prime }{|}^2-{\eta }_n\right)\right)\mathrm{d}\eta \right|\hfill \\ \hfill =& \sum _{k\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^k}{k!}}\left|{\psi }_{k+j}(y,z)\right|,\hfill \end{array}$$

(50)

where $h_j\left(y\right)$ is defined as in (30). We briefly justify the interchange of the integral and the sum in the inequality above. Denote by V the volume of $N_{1/R}{\left[0,1/\sqrt{R}\right]}^{n-1}$. On the domain of integration we have

$$0\le {\displaystyle \frac{R({\eta }_n-|{\eta }^{\prime }{|}^2)}{2}}\le {\displaystyle \frac{1}{2}},$$

and when $y\in B(z,R)$, $|y_n-z_n|\le R$. Then

$$\begin{array}{cc}\hfill & \sum _{k\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^k}{k!}}{\int }_{N_{\frac{1}{R}}{\left[0,{\displaystyle \frac{1}{\sqrt{R}}}\right]}^{n-1}}\left|\widehat{f}\left(\eta A_Q+b_Q\right){\left({\displaystyle \frac{R\left({\eta }_n-{\left|{\eta }^{\prime }\right|}^2\right)}{2}}\right)}^{k+j}e\left(\eta ·y\right)e\left(z_n\left(|{\eta }^{\prime }{|}^2-{\eta }_n\right)\right)\right|\mathrm{d}\eta \hfill \\ \hfill \le & \sum _{k\ge 0}V·{\displaystyle \frac{{\left(2\pi R\right)}^k}{k!}}{\left({\displaystyle \frac{1}{2}}\right)}^{j+k}{\displaystyle \frac{sup|\widehat{f}|}{R^{n+1}}}<\infty .\hfill \end{array}$$

Thus, Fubini’s theorem justifies it.

Now, by (31), (26), (50), Minkwoski’s inequality, and (49), we have

$$\begin{array}{cc}\hfill {∥E_Q{g}_j∥}_{L^p\left({\omega }_{B_R,F}\right)}\le & {\left(\sqrt{6n}\right)}^{{\displaystyle \frac{F}{p}}}{∥h_j∥}_{L^p\left({\omega }_{B_R,F}\right)}\hfill \\ \hfill \le & {\left(\sqrt{6n}\right)}^{{\displaystyle \frac{F}{p}}}{\left(2\sqrt{n}\right)}^{{\displaystyle \frac{F}{p}}}{\left(\int {∥h_j∥}_{L_{♯}^p\left(B(z,R)\right)}^p{\omega }_{B_R,F}\left(z\right)\mathrm{d}z\right)}^{{\displaystyle \frac{1}{p}}}\hfill \\ \hfill \le & {\left(2\sqrt{6}n\right)}^{{\displaystyle \frac{F}{p}}}{\left[\int {\left(\sum _{k\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^k}{k!}}{∥{\psi }_{k+j}(·,z)∥}_{L_{♯}^p\left(B(z,R)\right)}\right)}^p{\omega }_{B_R,F}\left(z\right)\mathrm{d}z\right]}^{{\displaystyle \frac{1}{p}}}\hfill \\ \hfill \le & {\left(2\sqrt{6}n\right)}^{{\displaystyle \frac{F}{p}}}\sum _{k\ge 0}{\displaystyle \frac{{\left(2\pi \right)}^k}{k!}}{\left(\int {∥{\psi }_{k+j}(·,z)∥}_{L_{♯}^p\left(B(z,R)\right)}^p{\omega }_{B_R,F}\left(z\right)z\mathrm{d}\right)}^{{\displaystyle \frac{1}{p}}}\hfill \\ \hfill \le & {\left(2\sqrt{6}n\right)}^{{\displaystyle \frac{F}{p}}}{e}^{2\pi }{C}_3{∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B_R,E}\right)}=C_2{∥f_{N_{\frac{1}{R}}\left(Q\right)}∥}_{L^p\left({\omega }_{B_R,E}\right)}.\hfill \end{array}$$

Thus, (49) implies (48) as claimed.

Step 4: We show that to prove (49), it suffices to verify, for each $Q\in {Part}_{R^{-1/2}}\left({[0,1]}^{n-1}\right)$, the inequality below:

$$\int \left(\right|\widehat{m_j}|\ast R^{-n}{\chi }_{B_R})(\eta A_Q^{\mathrm{T}}+z_nϵ-z){(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)z\le C_4^p{\omega }_{B_R,E}\left(\eta \right),\forall \eta \in {\mathbb{R}}^n,$$

(51)

where $ϵ=(0,\dots ,0,1)$, $C_4^p=C_m^{1-p}{C}_3^p$, i.e., $C_3=C_4{C}_m^{{\displaystyle \frac{1}{q}}}$, and $C_m$ is defined as in Lemma 14.

For $F=f_{N_{1/R}}\left(Q\right)$, note that

$$\begin{array}{cc}\hfill \widehat{\widehat{F}(·A_Q+b_Q)}\left(\xi \right)=& e(\xi ·b_Q{A}_Q^{-1})\widehat{\widehat{F}(·A_Q)}\left(\xi \right)\hfill \\ \hfill =& e(\xi ·b_Q{A}_Q^{-1})\widehat{\widehat{F}}\left(\xi {\left(A_Q^{-1}\right)}^{\mathrm{T}}\right)=e(\xi ·b_Q{A}_Q^{-1})F(-\xi {\left(A_Q^{-1}\right)}^{\mathrm{T}}).\hfill \end{array}$$

From (34) it follows that

$$\begin{array}{cc}\hfill \left|{\psi }_j(y,z)\right|=& |\int \widehat{F}(\xi A_Q+b_Q)m_j\left(\xi \right)e({\xi }^{\prime }·y^{\prime }+{\xi }_n(y_n-z_n))\mathrm{d}\xi |\hfill \\ \hfill =& |\int \widehat{F}(\xi A_Q+b_Q)m_j\left(\xi \right)e(\xi ·(y-z_nϵ))\mathrm{d}\xi |\hfill \\ \hfill =& |\widehat{F}\widehat{(·A_Q+b_Q)m_j}(·)(-y+z_nϵ)|\hfill \\ \hfill =& |(\widehat{F}\widehat{(·A_Q+b_Q)}\ast \widehat{m_j})(-y+z_nϵ)|\hfill \\ \hfill =& \left|\int e(\xi ·b_Q{A}_Q^{-1})F(-\xi {\left(A_Q^{-1}\right)}^{\mathrm{T}})\widehat{m_j}(-y+z_nϵ-\xi )\mathrm{d}\xi \right|.\hfill \end{array}$$

By performing the transformation $\eta =-\xi {\left(A_Q^{-1}\right)}^{\mathrm{T}}$, we have

$$\begin{array}{cc}\hfill & |\int e(\xi ·b_Q{A}_Q^{-1})F(-\xi {\left(A_Q^{-1}\right)}^{\mathrm{T}})\widehat{m_j}(-y+z_nϵ-\xi )\mathrm{d}\xi |\hfill \\ \hfill =& |\int e(-\left(\eta A_Q^{\mathrm{T}}\right)·b_Q{A}_Q^{-1})F\left(\eta \right)\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\mathrm{d}\eta |\hfill \\ \hfill \le & \int \left|F\left(\eta \right)\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\right|\mathrm{d}\eta \hfill \\ \hfill =& \int \left|F\right(\eta \left)\right|{\left|\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\right|}^{\frac{1}{p}}{\left|\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\right|}^{1-{\displaystyle \frac{1}{p}}}\mathrm{d}\eta .\hfill \end{array}$$

Let $1/q=1-1/p$. By Hölder’s inequality, we obtain

$$\begin{array}{cc}\hfill & \int \left|F\left(\eta \right)\right|{\left|\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\right|}^{{\displaystyle \frac{1}{p}}}{\left|\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\right|}^{1-{\displaystyle \frac{1}{p}}}\mathrm{d}\eta \hfill \\ \hfill \le & {\left({\int \left|F\left(\eta \right)\right|}^p\right|\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\left|\mathrm{d}\eta \right)}^{{\displaystyle \frac{1}{p}}}{(\int |\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\left|\mathrm{d}\eta \right)}^{{\displaystyle \frac{1}{q}}}\hfill \\ \hfill =& {(\int |\widehat{m_j}\left(\eta \right)\left|\mathrm{d}\eta \right)}^{{\displaystyle \frac{1}{q}}}{\left({\int \left|F\left(\eta \right)\right|}^p\right|\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\left|\mathrm{d}\eta \right)}^{{\displaystyle \frac{1}{p}}}.\hfill \end{array}$$

By (41), we have

$$\begin{array}{cc}\hfill & {\left|\right|{\psi }_j(·,z)\left|\right|}_{L_{♯}^p\left(B(z,R)\right)}^p\hfill \\ \hfill \le & {(\int |\widehat{m_j}\left(\eta \right)\left|\mathrm{d}\eta \right)}^{{\displaystyle \frac{p}{q}}}{\displaystyle \frac{1}{R^n}}{\int }_{B(z,R)}\left[{\int \left|F\left(\eta \right)\right|}^p\right|\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\left|\mathrm{d}\eta \right]\mathrm{d}y\hfill \\ \hfill \le & C_m^{{\displaystyle \frac{p}{q}}}{\displaystyle \frac{1}{R^n}}{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\int }_{B(z,R)}\left[{\int \left|F\left(\eta \right)\right|}^p\right|\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\left|\mathrm{d}\eta \right]\mathrm{d}y.\hfill \end{array}$$

(52)

Thus,

$$\begin{array}{cc}\hfill & \int {\left|\right|{\psi }_j(·,z)\left|\right|}_{L_{♯}^p\left(B(z,R)\right)}^p{\omega }_{B_R,F}\left(z\right)\mathrm{d}z\hfill \\ \hfill \le & {\displaystyle \frac{C_m^{\frac{p}{q}}}{R^n}}\int {(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\left\{{\int }_{B(z,R)}\left[{\int \left|F\left(\eta \right)\right|}^p\right|\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\left|\mathrm{d}\eta \right]\mathrm{d}y\right\}\mathrm{d}z\hfill \\ \hfill =& {\displaystyle \frac{C_m^{\frac{p}{q}}}{R^n}}\int {\left|F\left(\eta \right)\right|}^p\{\int [\int \left|\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})\right|{\chi }_{B(z,R)}\left(y\right)\mathrm{d}y\left]{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}z\right\}\mathrm{d}\eta .\hfill \end{array}$$

Let $Y=-y+z_nϵ+\eta A^{\mathrm{T}}$. Then

$$\begin{array}{cc}\hfill \int |\widehat{m_j}(-y+z_nϵ+\eta A_Q^{\mathrm{T}})|{\chi }_{B(z,R)}\left(y\right)\mathrm{d}y=& \int |\widehat{m_j}\left(Y\right)|{\chi }_{B(0,R)}(z_nϵ+\eta A_Q^{\mathrm{T}}-Y-z)\mathrm{d}Y\hfill \\ \hfill =& \left(\right|\widehat{m_j}|\ast {\chi }_{B(0,R)})(z_nϵ+\eta A_Q^{\mathrm{T}}-z).\hfill \end{array}$$

Thus, by (51), we have

$$\begin{array}{cc}\hfill & \int {\left|\right|{\psi }_j(·,z)\left|\right|}_{L_{♯}^p\left(B(z,R)\right)}^p{\omega }_{B_R,F}\left(z\right)\mathrm{d}z\hfill \\ \hfill \le & C_m^{p-1}\int {\left|F\left(\eta \right)\right|}^p\left\{\int \left[\right(\left|\widehat{m_j}\right|\ast {\displaystyle \frac{1}{R^n}}{\chi }_{B(0,R)}\left)(z_nϵ+\eta A_Q^{\mathrm{T}}-z)\right]{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}z\right\}\mathrm{d}\eta \hfill \\ \hfill \le & C_m^{p-1}\int {\left|F\left(\eta \right)\right|}^p{C}_4^p{\omega }_{B_R,E}\left(\eta \right)\mathrm{d}\eta \hfill \\ \hfill =& C_m^{p-1}{C}_4^p{\left|\right|f_{N_{\frac{1}{R}}\left(Q\right)}\left|\right|}_{L^p\left({\omega }_{B_R,E}\right)}^p=C_3^p{\left|\right|f_{N_{\frac{1}{R}}\left(Q\right)}\left|\right|}_{L^p\left({\omega }_{B_R,E}\right)}^p.\hfill \end{array}$$

Hence, (49) holds.

Our main theorem shows that the inequality holds when

$$F\ge 2(n-1)⌈E⌉+(n-1)(p-1)+2,$$

where the term $(n-1)(p-1)$ actually comes from Hölder’s inequality and (52). Note that in [13], for $n=2$, Theorem 5.1 shows that the result holds under the condition $F\ge 2E+2$. According to our result, this would instead require $F\ge 2⌈E⌉+2+(p-1)$. We have not found a way to eliminate p by the proof in [13], and therefore we can only provide a slightly looser bound.

• Step 5: We start to prove (51).

Take $s_1=\dots =s_n=⌈E⌉$. By (40), we have

$$(\left|\widehat{m_j}\right|\ast {\displaystyle \frac{1}{R^n}}{\chi }_{B_R})\left(\xi \right)\le C_0(⌈E⌉,\dots ,⌈E⌉){\left(4n\right)}^n{2}^{n⌈E⌉}{\displaystyle \prod _{k=1}^n}({\varphi }_k\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})\left({\xi }_k\right).$$

(53)

When $k=n$, from (21), it follows that

$$({\varphi }_n\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})\left({\xi }_n\right)={\displaystyle \frac{1}{R}}({\omega }_{I_R,⌈E⌉}\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})\left({\xi }_n\right)\le {\displaystyle \frac{1}{R}}{9}^{⌈E⌉}{\omega }_{I_R,⌈E⌉}\left({\xi }_n\right).$$

(54)

When $k\in [n-1]$, consider the cover of $\left[0,+\infty \right)$

$$\left[0,+\infty \right)=[0,R]\cup \left({\displaystyle \bigcup _{i=0}^{⌊{log}_2{R}^{{\displaystyle \frac{1}{2}}}⌋}}\left(2^{i}R,2^{i+1}R\right]\right)\cup \left(\bigcup _{i\in \mathbb{N}}\left(2^i{R}^{{\displaystyle \frac{3}{2}}},2^{i+1}{R}^{{\displaystyle \frac{3}{2}}}\right]\right).$$

Case 1. $|z_n|\le R$. We have

$${\varphi }_k\left(x_k\right)\le {\displaystyle \frac{1}{\sqrt{R}}}{(1+{\displaystyle \frac{|x_k|}{2\sqrt{R}}})}^{-⌈E⌉}\le {\displaystyle \frac{1}{\sqrt{R}}}{2}^{⌈E⌉}{\omega }_{I_{\sqrt{R}},⌈E⌉}\left(x_k\right).$$

From (21), it follows that

$$({\varphi }_k\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})\left({\xi }_k\right)\le 2^{⌈E⌉}{\displaystyle \frac{1}{R}}{9}^{⌈E⌉}{\omega }_{I_R,⌈E⌉}\left({\xi }_k\right)={\displaystyle \frac{1}{R}}{18}^{⌈E⌉}{\omega }_{I_R,⌈E⌉}\left({\xi }_k\right).$$

(55)

Case 2. $2^{i}R<\left|z_n\right|\le 2^{i+1}R,i\in \left[⌊{log}_2{R}^{{\displaystyle \frac{1}{2}}}⌋\right]\cup 0$. We have

$$\begin{array}{cc}\hfill {\varphi }_k\left(x_k\right)& \le {\displaystyle \frac{1}{\sqrt{R}}}{(1+{\displaystyle \frac{|x_k|}{(1+2^{i+1})\sqrt{R}}})}^{-⌈E⌉}\hfill \\ \hfill & \le {\displaystyle \frac{1}{\sqrt{R}}}{(1+{\displaystyle \frac{|x_k|}{2^{i+2}\sqrt{R}}})}^{-⌈E⌉}\le {\displaystyle \frac{1}{\sqrt{R}}}{4}^{⌈E⌉}{(1+{\displaystyle \frac{|x_k|}{2^i\sqrt{R}}})}^{-⌈E⌉}=2^i{4}^{⌈E⌉}{\displaystyle \frac{1}{2^i\sqrt{R}}}{\omega }_{I_{2^i\sqrt{R}},⌈E⌉}\left(x_k\right).\hfill \end{array}$$

Note that $2^i\sqrt{R}\le 2^{{log}_2{R}^{{\displaystyle \frac{1}{2}}}}\sqrt{R}=R$. By (21), we obtain

$$\begin{array}{cc}\hfill ({\varphi }_k\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})\left({\xi }_k\right)& \le 2^i{4}^{⌈E⌉}{\displaystyle \frac{1}{R}}[{\chi }_{I_R}\ast \left({\displaystyle \frac{1}{2^i\sqrt{R}}}{\omega }_{I_{2^i\sqrt{R}},⌈E⌉}\right)]\left({\xi }_k\right)\hfill \\ \hfill & \le 2^i{4}^{⌈E⌉}{\displaystyle \frac{1}{R}}{9}^{⌈E⌉}{\omega }_{I_R,⌈E⌉}\left({\xi }_k\right)=2^i{\displaystyle \frac{{36}^{⌈E⌉}}{R}}{\omega }_{I_R,⌈E⌉}\left({\xi }_k\right).\hfill \end{array}$$

(56)

Case 3. $2^i{R}^{{\displaystyle \frac{3}{2}}}<\left|z_n\right|\le 2^{i+1}{R}^{{\displaystyle \frac{3}{2}}},i\in \mathbb{N}$. We have

$${\varphi }_k\left(x_k\right)\le {\displaystyle \frac{1}{\sqrt{R}}}{(1+{\displaystyle \frac{|x_k|}{\sqrt{R}+2^{i+1}R}})}^{-⌈E⌉}\le {\displaystyle \frac{1}{\sqrt{R}}}{(1+{\displaystyle \frac{|x_k|}{2^{i+2}R}})}^{-⌈E⌉}={\displaystyle \frac{1}{\sqrt{R}}}{\omega }_{I_{2^{i+2}R},⌈E⌉}\left(x_k\right).$$

Note that $2^{i+2}R>R$. Thus, similarly to the case above, applying (21), we obtain

$$({\varphi }_k\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})\left({\xi }_k\right)\le {\displaystyle \frac{1}{\sqrt{R}}}{9}^{⌈E⌉}{\omega }_{I_{2^{i+2}R},⌈E⌉}\left({\xi }_k\right).$$

(57)

By (53) and (54), we have

$$\begin{array}{cc}\hfill J:=& (\left|\widehat{m_j}\right|\ast {\displaystyle \frac{1}{R^n}}{\chi }_{B(0,R)})(\eta A_Q^{\mathrm{T}}+z_nϵ-z)\hfill \\ \hfill \le & C_0(⌈E⌉,\dots ,⌈E⌉){\left(4n\right)}^n{2}^{n⌈E⌉}({\varphi }_n\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})\left({\eta }_n\right){\displaystyle \prod _{k=1}^{n-1}}({\varphi }_k\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})({\eta }_k+2{\eta }_n{u}_k-z_k)\hfill \\ \hfill \le & {\displaystyle \frac{1}{R}}{9}^{⌈E⌉}{C}_0(⌈E⌉,\dots ,⌈E⌉){\left(4n\right)}^n{2}^{n⌈E⌉}{\omega }_{I_R,⌈E⌉}\left({\eta }_n\right){\displaystyle \prod _{k=1}^{n-1}}({\varphi }_k\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})({\eta }_k+2{\eta }_n{u}_k-z_k)\hfill \\ \hfill =:& A\left({\eta }_n\right){\displaystyle \prod _{k=1}^{n-1}}({\varphi }_k\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})({\eta }_k+2{\eta }_n{u}_k-z_k).\hfill \end{array}$$

(58)

Let $z=(z^{\prime },z_n)$. We have

$$\begin{array}{cc}\hfill & \int J·{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}z\hfill \\ \hfill =& \int \int J·{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}{z}_n\mathrm{d}{z}^{\prime }\hfill \\ \hfill \le & \int {\int }_{|z_n|\le R}J·{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}{z}_n\mathrm{d}{z}^{\prime }\hfill \\ \hfill & +{\displaystyle \sum _{i=0}^{⌊{log}_2{R}^{{\displaystyle \frac{1}{2}}}⌋}}\int {\int }_{2^{i}R<\left|z_n\right|\le 2^{i+1}R}J·{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}{z}_n\mathrm{d}{z}^{\prime }\hfill \\ \hfill & +{\displaystyle \sum _{j=0}^{+\infty }}\int {\int }_{2^j{R}^{{\displaystyle \frac{3}{2}}}<\left|z_n\right|\le 2^{j+1}{R}^{{\displaystyle \frac{3}{2}}}}J·{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}{z}_n\mathrm{d}{z}^{\prime }.\hfill \end{array}$$

Let $U_1$, $U_2$, and $U_3$ be the first, the second, and the third summand in the inequality above, respectively. For $U_1$, when $(n-1)⌈E⌉+(n-1)(p-1)\le F$, by (22), we obtain

$$\begin{array}{cc}\hfill & {(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\hfill \\ \hfill \le & {(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{I_R,(n-1)(p-1)}\left(z_n\right){\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}\left(z_k\right)={\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}\left(z_k\right).\hfill \end{array}$$

(59)

It follows from (55) and (58) that

$$J\le A\left({\eta }_n\right){\displaystyle \prod _{k=1}^{n-1}}({\varphi }_k\ast {\displaystyle \frac{1}{R}}{\chi }_{I_R})({\eta }_k+2{\eta }_n{u}_k-z_k)\le {\displaystyle \prod _{k=1}^{n-1}}{\displaystyle \frac{1}{R}}{18}^{⌈E⌉}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k-z_k).$$

Thus, by (59), we have

$$\begin{array}{cc}\hfill & {\int }_{|z_n|\le R}J·{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}{z}_n\hfill \\ \hfill \le & A\left({\eta }_n\right){\int }_{|z_n|\le R}\left({\displaystyle \prod _{k=1}^{n-1}}{\displaystyle \frac{1}{R}}{18}^{⌈E⌉}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k-z_k)\right)\left({\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}\left(z_k\right)\right)\mathrm{d}{z}_n\hfill \\ \hfill =& A\left({\eta }_n\right){18}^{⌈E⌉(n-1)}2R{\displaystyle \prod _{k=1}^{n-1}}({\displaystyle \frac{1}{R}}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k-z_k)·{\omega }_{I_R,⌈E⌉}\left(z_k\right)).\hfill \end{array}$$

Meanwhile, by (18), we have

$$\begin{array}{cc}\hfill U_1\le & \int A\left({\eta }_n\right){18}^{⌈E⌉(n-1)}2R{\displaystyle \prod _{k=1}^{n-1}}({\displaystyle \frac{1}{R}}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k-z_k)·{\omega }_{I_R,⌈E⌉}\left(z_k\right))\mathrm{d}{z}^{\prime }\hfill \\ \hfill \le & A\left({\eta }_n\right){18}^{⌈E⌉(n-1)}2R·6^{⌈E⌉(n-1)}{\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k).\hfill \end{array}$$

(60)

For $U_2$, let ${\Phi }_i=[2^{i}R,2^{i+1}R]$. When

$$(n-1)⌈E⌉+((n-1)(p-1)+(n+1))\le F,$$

applying a similar approach as in (59), we have

$${(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\le {\omega }_{I_R,n+1}\left(z_n\right){\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}\left(z_k\right).$$

(61)

Now, it follows from (56), (58) and (61) that

$$\begin{array}{cc}\hfill & {\int }_{|z_n|\in {\Phi }_i}J·{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}{z}_n\hfill \\ \hfill \le & A\left({\eta }_n\right){36}^{⌈E⌉(n-1)}{2}^{i(n-1)}\left({\displaystyle \prod _{k=1}^{n-1}}{\displaystyle \frac{{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k-z_k)}{R}}{\omega }_{I_R,⌈E⌉}\left(z_k\right)\right){\int }_{|z_n|\in {\Phi }_i}{\omega }_{I_R,n+1}\left(z_n\right)\mathrm{d}{z}_n,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill {\int }_{|z_n|\in {\Phi }_i}{\omega }_{I_R,n+1}\left(z_n\right)\mathrm{d}{z}_n={\int }_{|z_n|\in {\Phi }_i}{(1+{\displaystyle \frac{|z_n|}{R}})}^{-(n+1)}\mathrm{d}{z}_n=& 2R{\int }_{2^i<t\le 2^{i+1}}{(1+t)}^{-(n+1)}\mathrm{d}t\hfill \\ \hfill \le & 2R·2^i{(1+2^i)}^{-(n+1)}\le 2R{\left(2^i\right)}^{-n}.\hfill \end{array}$$

By (18), we have

$$\begin{array}{cc}\hfill U_2\le & {\displaystyle \sum _{i=0}^{⌊{log}_2{R}^{{\displaystyle \frac{1}{2}}}⌋}}\int A\left({\eta }_n\right){36}^{⌈E⌉(n-1)}{2}^{i(n-1)}\left({\displaystyle \prod _{k=1}^{n-1}}{\displaystyle \frac{{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k-z_k)}{R}}{\omega }_{I_R,⌈E⌉}\left(z_k\right)\right)2R{\left(2^i\right)}^{-n}\mathrm{d}{z}^{\prime }\hfill \\ \hfill \le & A\left({\eta }_n\right){36}^{⌈E⌉(n-1)}4R·6^{⌈E⌉(n-1)}{\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k).\hfill \end{array}$$

(62)

For $U_3$, when

$$(n-1)⌈E⌉+((n-1)⌈E⌉+2+(n-1)(p-1))\le F,$$

applying a similar approach as in (59), we have

$${(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\le {\omega }_{I_R,(n-1)⌈E⌉+2}\left(z_n\right){\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}\left(z_k\right).$$

(63)

Let ${\mathrm{\Psi }}_j=[2^j{R}^{{\displaystyle \frac{3}{2}}},2^{j+1}{R}^{{\displaystyle \frac{3}{2}}}]$. From (57), (58) and (63), it follows that

$$\begin{array}{cc}\hfill & {\int }_{\left|z_n\right|\in {\mathsf{\Psi }}_j}J·{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}{z}_n\hfill \\ \hfill \le & A\left({\eta }_n\right){\int }_{\left|z_n\right|\in {\mathsf{\Psi }}_j}\left({\displaystyle \prod _{k=1}^{n-1}}{\displaystyle \frac{1}{\sqrt{R}}}{9}^{⌈E⌉}{\omega }_{I_{2^{j+2}R},⌈E⌉}({\eta }_k+2{\eta }_n{u}_k-z_k)\right){\omega }_{I_R,(n-1)⌈E⌉+2}\left(z_n\right)\hfill \\ \hfill & ·\left({\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}\left(z_k\right)\right)\mathrm{d}{z}_n\hfill \\ \hfill \le & A\left({\eta }_n\right){\left({\displaystyle \frac{9^{⌈E⌉}}{\sqrt{R}}}\right)}^{n-1}{R}^{n-1}{\displaystyle \prod _{k=1}^{n-1}}\left({\omega }_{I_{2^{j+2}R},⌈E⌉}({\eta }_k+2{\eta }_n{u}_k-z_k){\displaystyle \frac{1}{R}}{\omega }_{I_R,⌈E⌉}\left(z_k\right)\right)\hfill \\ \hfill & ·{\int }_{\left|z_n\right|\in {\mathsf{\Psi }}_j}{\omega }_{I_R,(n-1)⌈E⌉+2}\left(z_n\right){\mathrm{d}}_{zn},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill {\int }_{\left|z_n\right|\in {\mathsf{\Psi }}_j}{\omega }_{I_R,(n-1)⌈E⌉+2}\left(z_n\right)\mathrm{d}{z}_n=& {\int }_{\left|z_n\right|\in {\mathsf{\Psi }}_j}{(1+{\displaystyle \frac{|z_n|}{R}})}^{-(n-1)⌈E⌉-2}\mathrm{d}{z}_n\hfill \\ \hfill =& 2R{\int }_{2^j\sqrt{R}<t\le 2^{j+1}\sqrt{R}}{(1+t)}^{-(n-1)⌈E⌉-2}\mathrm{d}t\hfill \\ \hfill \le & 2R·2^j\sqrt{R}{\left(2^j\sqrt{R}\right)}^{-(n-1)⌈E⌉-2}=2R{\left(2^j\sqrt{R}\right)}^{-(n-1)⌈E⌉-1}.\hfill \end{array}$$

Thus, we obtain

$$\begin{array}{cc}\hfill U_3\le & {\displaystyle \sum _{j=0}^{+\infty }}\int A\left({\eta }_n\right){\left({\displaystyle \frac{9^{⌈E⌉}}{\sqrt{R}}}\right)}^{n-1}{R}^{n-1}·2R{\left(2^j\sqrt{R}\right)}^{-(n-1)⌈E⌉-1}\hfill \\ \hfill & ·\left({\displaystyle \prod _{k=1}^{n-1}}\left({\omega }_{I_{2^{j+2}R},⌈E⌉}({\eta }_k+2{\eta }_n{u}_k-z_k){\displaystyle \frac{1}{R}}{\omega }_{I_R,⌈E⌉}\left(z_k\right)\right)\right)\mathrm{d}{z}^{\prime }\hfill \\ \hfill =& {\displaystyle \sum _{j=0}^{+\infty }}A\left({\eta }_n\right)9^{⌈E⌉(n-1)}{R}^{{\displaystyle \frac{n-1}{2}}}2R{\left(2^j\sqrt{R}\right)}^{-(n-1)⌈E⌉-1}\hfill \\ \hfill & ·{\displaystyle \prod _{k=1}^{n-1}}({\omega }_{I_{2^{j+2}R},⌈E⌉}\ast \left({\displaystyle \frac{1}{R}}{\omega }_{I_R,⌈E⌉}\right))({\eta }_k+2{\eta }_n{u}_k).\hfill \end{array}$$

By (18) and the condition $2^{j+2}R>R$, we have

$${\omega }_{I_{2^{j+2}R},⌈E⌉}\ast \left({\displaystyle \frac{1}{R}}{\omega }_{I_R,⌈E⌉}\right)\le 6^{⌈E⌉}{\omega }_{I_{2^{j+2}R},⌈E⌉}.$$

Thus,

$$U_3\le {\displaystyle \sum _{j=0}^{+\infty }}A\left({\eta }_n\right)9^{⌈E⌉(n-1)}{R}^{{\displaystyle \frac{n-1}{2}}}2R{\left(2^j\sqrt{R}\right)}^{-(n-1)⌈E⌉-1}{\displaystyle \prod _{k=1}^{n-1}}{6}^{⌈E⌉}{\omega }_{I_{2^{j+2}R},⌈E⌉}({\eta }_k+2{\eta }_n{u}_k).$$

Moreover, from

$${\omega }_{I_{2^{j+2}R},⌈E⌉}\left(x\right)={(1+{\displaystyle \frac{\left|x\right|}{2^{j+2}R}})}^{-⌈E⌉}\le 2^{(j+2)⌈E⌉}{\omega }_{I_R,⌈E⌉}\left(x\right),$$

it follows that

$$\begin{array}{cc}\hfill U_3\le & {\displaystyle \sum _{j=0}^{+\infty }}A\left({\eta }_n\right)9^{⌈E⌉(n-1)}{R}^{{\displaystyle \frac{n-1}{2}}}2R{\left(2^j\sqrt{R}\right)}^{-(n-1)⌈E⌉-1}{2}^{(j+2)⌈E⌉(n-1)}\hfill \\ \hfill & ·{\displaystyle \prod _{k=1}^{n-1}}{6}^{⌈E⌉}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k)\hfill \\ \hfill =& {\displaystyle \sum _{j=0}^{+\infty }}{2}^{-j}A\left({\eta }_n\right)9^{⌈E⌉(n-1)}{R}^{{\displaystyle \frac{n-1}{2}}}2R·R^{{\displaystyle \frac{-(n-1)⌈E⌉-1}{2}}}{6}^{⌈E⌉(n-1)}{2}^{2⌈E⌉(n-1)}{\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k)\hfill \\ \hfill =& 4R·A\left({\eta }_n\right){(9\times 6\times 4)}^{⌈E⌉(n-1)}{R}^{{\displaystyle \frac{-(n-1)(⌈E⌉-1)-1}{2}}}{\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k)\hfill \\ \hfill \le & 4R·A\left({\eta }_n\right){(36\times 6)}^{⌈E⌉(n-1)}{\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k).\hfill \end{array}$$

(64)

Now, since $E\ge 100n$, we have

$$\begin{array}{cc}\hfill (n-1)⌈E⌉+(n-1)(p-1)\le & (n-1)⌈E⌉+((n-1)(p-1)+(n+1))\hfill \\ \hfill \le & 2(n-1)⌈E⌉+2+(n-1)(p-1).\hfill \end{array}$$

Thus, when

$$2(n-1)⌈E⌉+2+(n-1)(p-1)\le F,$$

the inequalities (60), (62) and (64) hold simultaneously. Under this condition, recalling the definition of $A\left({\eta }_n\right)$ as in (58), we obtain

$$\begin{array}{cc}\hfill \int J·{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}z\le & 12R·A\left({\eta }_n\right){(36\times 6)}^{⌈E⌉(n-1)}{\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k)\hfill \\ \hfill =& 12·9^{⌈E⌉}{C}_0(⌈E⌉,\dots ,⌈E⌉){\left(4n\right)}^n{2}^{n⌈E⌉}{(36\times 6)}^{⌈E⌉(n-1)}\hfill \\ \hfill & ·{\omega }_{I_R,⌈E⌉}\left({\eta }_n\right){\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k).\hfill \end{array}$$

From (23), it follows that

$${\omega }_{I_R,⌈E⌉}\left({\eta }_n\right){\displaystyle \prod _{k=1}^{n-1}}{\omega }_{I_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k)\le {\omega }_{B_R,⌈E⌉}(\eta ·A_Q^{\mathrm{T}}).$$

The inequality (25) implies that

$${\omega }_{B_R,⌈E⌉}({\eta }_k+2{\eta }_n{u}_k,{\eta }_n)\le {\sqrt{6n}}^{⌈E⌉}{\omega }_{B_R,⌈E⌉}\left(\eta \right).$$

Hence, combining the above three inequalities, we get

$$\begin{array}{cc}\hfill & \int J·{(1+{\displaystyle \frac{|z_n|}{R}})}^{(n-1)(p-1)}{\omega }_{B_R,F}\left(z\right)\mathrm{d}z\hfill \\ \hfill \le & 12·9^{⌈E⌉}{C}_0(⌈E⌉,\dots ,⌈E⌉){\left(4n\right)}^n{2}^{n⌈E⌉}{(36\times 6)}^{⌈E⌉(n-1)}{\sqrt{6n}}^{⌈E⌉}{\omega }_{B_R,⌈E⌉}\left(\eta \right).\hfill \end{array}$$

By Stirling’s inequality

$$⌈E⌉!\le \sqrt{2\pi ⌈E⌉}{\left({\displaystyle \frac{⌈E⌉}{e}}\right)}^{⌈E⌉}{e}^{{\displaystyle \frac{1}{12⌈E⌉}}},$$

we have

$$\begin{array}{cc}\hfill & 12·9^{⌈E⌉}{C}_0(⌈E⌉,\dots ,⌈E⌉){\left(4n\right)}^n{2}^{n⌈E⌉}{(36\times 6)}^{⌈E⌉(n-1)}{\sqrt{6n}}^{⌈E⌉}\hfill \\ \hfill =& 12·9^{⌈E⌉}{\left(4n\right)}^n{2}^{n⌈E⌉}{(36\times 6)}^{⌈E⌉(n-1)}max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{(⌈E⌉!)}^{2(n-1)}{\left(14n\right)}^{n⌈E⌉}{\sqrt{6n}}^{⌈E⌉}\hfill \\ \hfill \le & max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{e}^{n⌈E⌉(ln(36\times 6\times 2\times 14n)+{\displaystyle \frac{ln\left(\sqrt{6}n\right)}{n}})+ln12+⌈E⌉ln9+nln4n+2(n-1)ln(⌈E⌉!)}\hfill \\ \hfill \le & max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{e}^{n⌈E⌉(13+lnn)+3+3⌈E⌉+nln4n+2(n-1)ln\left(\sqrt{2\pi ⌈E⌉}{\left({\displaystyle \frac{⌈E⌉}{e}}\right)}^{⌈E⌉}{e}^{{\displaystyle \frac{1}{12⌈E⌉}}}\right)}.\hfill \end{array}$$

Moreover,

$$\begin{array}{cc}\hfill & n⌈E⌉(13+lnn)+3+3⌈E⌉+nln4n+2(n-1)ln\left(\sqrt{2\pi ⌈E⌉}{\left({\displaystyle \frac{⌈E⌉}{e}}\right)}^{⌈E⌉}{e}^{{\displaystyle \frac{1}{12⌈E⌉}}}\right)\hfill \\ \hfill \le & n⌈E⌉(13+lnn)+3+3⌈E⌉+nln4n+2n[ln\sqrt{2\pi ⌈E⌉}+⌈E⌉(ln⌈E⌉-1)+{\displaystyle \frac{1}{12⌈E⌉}}]\hfill \\ \hfill \le & 2n[⌈E⌉{\displaystyle \frac{13+lnn}{2}}+1+⌈E⌉+{\displaystyle \frac{ln4n}{2}}+ln\sqrt{2\pi ⌈E⌉}+⌈E⌉(ln⌈E⌉-1)+{\displaystyle \frac{1}{12⌈E⌉}}]\hfill \\ \hfill \le & 2n\left[⌈E⌉({\displaystyle \frac{13+lnn}{2}}+ln⌈E⌉+3)\right]\hfill \\ \hfill \le & 2n⌈E⌉·10ln⌈E⌉=20n⌈E⌉ln⌈E⌉.\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill & 12·9^{⌈E⌉}{C}_0(⌈E⌉,\dots ,⌈E⌉){\left(4n\right)}^n{2}^{n⌈E⌉}{(36\times 6)}^{⌈E⌉(n-1)}{\sqrt{6n}}^{⌈E⌉}\hfill \\ \hfill \le & max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{20n⌈E⌉}.\hfill \end{array}$$

Let $C_4^p=max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{20n⌈E⌉}$. At this stage, we restate the definitions mentioned before as follows:

1.

$C_4^p=max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{20n⌈E⌉}$;

2.

$C_m=max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{2n}\right\}{\left(\rho \left(2\right)\right)}^n{\left(30n\right)}^{3n}$;

3.

$C_3=C_4{C}_m^{{\displaystyle \frac{1}{q}}}$;

4.

$C_2={\left(2\sqrt{6}n\right)}^{{\displaystyle \frac{F}{p}}}{C}_3{e}^{2\pi }$;

5.

$C_1={(1+{\displaystyle \frac{\sqrt{n}}{2}})}^{{\displaystyle \frac{F}{p}}}{e}^{2\pi }{C}_2$.

Then we show that such a choice of $C_1$ guarantees the validity of (47). We have

$$\begin{array}{cc}\hfill C_3=& {(max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{20n⌈E⌉})}^{{\displaystyle \frac{1}{p}}}{(max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{2n}\right\}{\left(\rho \left(2\right)\right)}^n{\left(30n\right)}^{3n})}^{{\displaystyle \frac{1}{q}}}\hfill \\ \hfill \le & max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{{\displaystyle \frac{20n⌈E⌉}{p}}}{\left(30n\right)}^{{\displaystyle \frac{3n}{q}}},\hfill \end{array}$$

which shows that

$$C_2\le {\left(2\sqrt{6}n\right)}^{{\displaystyle \frac{F}{p}}}{e}^{2\pi }max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{{\displaystyle \frac{20n⌈E⌉}{p}}}{\left(30n\right)}^{{\displaystyle \frac{3n}{q}}},$$

$$C_1\le {\left[\sqrt{6}n(2+\sqrt{n})\right]}^{{\displaystyle \frac{F}{p}}}{e}^{4\pi }max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{{\displaystyle \frac{20n⌈E⌉}{p}}}{\left(30n\right)}^{{\displaystyle \frac{3n}{q}}},$$

and

$$\begin{array}{cc}\hfill {\left(6\sqrt{n}\right)}^E{C}_1\le & {\left(6\sqrt{n}\right)}^E{\left(9n\right)}^{{\displaystyle \frac{F}{p}}}{e}^{4\pi }max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{{\displaystyle \frac{20n⌈E⌉}{p}}}{\left(30n\right)}^{{\displaystyle \frac{3n}{q}}}\hfill \\ \hfill \le & max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{{\displaystyle \frac{20n⌈E⌉}{p}}}{\left(9n\right)}^{{\displaystyle \frac{F}{p}}}{e}^{4\pi }{\left(6\sqrt{n}\right)}^E{\left(30n\right)}^{3n}.\hfill \end{array}$$

Since $E\ge 100n$, we have

$$\begin{array}{c}\hfill e^{4\pi }{\left(6\sqrt{n}\right)}^E{\left(30n\right)}^{3n}=e^{4\pi +Eln\left(6\sqrt{n}\right)+3nln\left(30n\right)}\le e^{Eln\left(30n\right)}={\left(30n\right)}^E.\end{array}$$

Finally, we obtain

$${\left(6\sqrt{n}\right)}^E{C}_1\le max\left\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n⌈E⌉}\right\}{\left(\rho \left(⌈E⌉\right)\right)}^n{⌈E⌉}^{{\displaystyle \frac{20n⌈E⌉}{p}}}{\left(9n\right)}^{{\displaystyle \frac{F}{p}}}{\left(30n\right)}^E.$$

Hence, $C_1$ is a constant satisfying (47). □

4. Conclusions

In this paper, we have established an explicit upper bound for the implicit constant in an alternate form of the ${\ell }^2$ decoupling inequality on the n-dimensional truncated paraboloid. More precisely, for any $p\ge 2$, $E\ge 100n$, and

$$F\ge 2(n-1)\lceil E\rceil +(n-1)(p-1)+2,$$

we proved that the inequality

$${\parallel f\parallel }_{L^p\left({\omega }_{B_R,E}\right)}\le C_0{Dec}_n(R^{-1},p,F){\left(\sum _Q{\parallel f_{N_{1/R}\left(Q\right)}\parallel }_{L^p\left({\omega }_{B_R,E}\right)}^2\right)}^{1/2}$$

holds for every function f with Fourier support in $N_{1/R}\left({[0,1]}^{n-1}\right)$ and every cube $B_R\subset {\mathbb{R}}^n$, where the constant

$$C_0=max\{{\tilde{C}}_0,\dots ,{\tilde{C}}_{n\lceil E\rceil }\}{\left(\rho \left(\lceil E\rceil \right)\right)}^n{\lceil E\rceil }^{{\displaystyle \frac{20n\lceil E\rceil }{p}}}{\left(9n\right)}^{{\displaystyle \frac{F}{p}}}{\left(30n\right)}^E$$

is completely explicit in terms of n, p, E, F, and the auxiliary functions $\eta$ and $M_j$.

Our result constitutes the first part of a larger project aimed at obtaining fully explicit decoupling estimates in ${\mathbb{R}}^n$, extending the two-dimensional work of Z.K. Li [11] to higher dimensions. The explicit control of constants obtained here is essential for number-theoretic applications such as Vinogradov-type mean value theorems, where the dependence of the decoupling constant on $ϵ$ must be tracked precisely.