Boundedness of iterated spherical average

Abstract

The iterated spherical average \mathrm{\Delta }(A_1{)}^N is an important operator in harmonic analysis, and has very important applications in approximation theory and probability theory, where \mathrm{\Delta } is the Laplacian, A_1 is the unit spherical average and (A_1{)}^N is its iteration. In this paper, we mainly study the sufficient and necessary conditions for the boundedness of this operator in Besov-Lipschitz space, and prove the boundedness of the operator in Triebel-Lizorkin space. Moreover, we use above conclusions to improve the existing results of the boundedness of this operator in L^p space.

Keywords

Iterated spherical average / Besov-Lipschitz space / Triebel-Lizorkin space / $ L^{p} $ space

1 Introduction

The average operator of functions f on the unit sphere is defined as

A_1(f)(x)={\int }_{{\mathbb{S}}^{n-1}}f(x-y^{\mathrm{\prime }})\mathrm{d}\sigma (y^{\mathrm{\prime }}),

where {\mathbb{S}}^{n-1} is the unit sphere in {\mathbb{R}}^n, \mathrm{d}\sigma is the normalized surface Lebesgue measure on {\mathbb{R}}^n. The generation of this operator can be traced back to the 1970s. It has a profound background and wide applications in harmonic analysis (see [10, 9]). Moreover, it is very important in the study of random walks in high dimensional spaces, which was originated by Pearson [8] about 120 years ago. The definition of random walks is the N-steps uniform walk in {\mathbb{R}}^n starts at the origin and consists of N independent steps of length 1, each of which is taken into a uniformly random direction. By some calculations, the probability density function p_N(\frac{n-2}{2},x) of uniform walk in {\mathbb{R}}^n is the Fourier inverse of (A_1{)}^N[2]. Besides, this operator also plays a significant role in the approximation theory [1, 4]. In order to obtain some equivalent forms of the K-functional in L^p({\mathbb{R}}^n) spaces, Belinsky, Dai and Ditzian studied the operator (A_1{)}^N in [1] and obtained the following result.

Theorem A [1]　 Let 1⩽p⩽\mathrm{\infty }, n⩾2 and N>\frac{2(n+2)}{n-1}. The inequality

\begin{array}{l}{\Vert \mathrm{\Delta }(A_1{)}^N(f)\Vert }_{L^p({\mathbb{R}}^n)}⪯\Vert f{\Vert }_{L^p({\mathbb{R}}^n)}\end{array}

holds for all f\in L^p({\mathbb{R}}^n).

In the proof of Theorem A, they proposed a meaningful question: what was the smallest positive integer N to guarantee the inequality

(1)\begin{array}{l}{\Vert \mathrm{\Delta }(A_1{)}^N(f)\Vert }_{L^1({\mathbb{R}}^n)}⪯\Vert f{\Vert }_{L^1({\mathbb{R}}^n)}.\end{array}

This question was completely solved by Fan, Lou and Wang. In [3], they obtained the following theorem.

Theorem B [3]　 Let n\ne 3,5, and N be positive integers. The inequality

\begin{array}{l}{\Vert \mathrm{\Delta }(A_1{)}^N(f)\Vert }_{L^1({\mathbb{R}}^n)}⪯\Vert f{\Vert }_{L^1({\mathbb{R}}^n)}\end{array}

holds if and only if N>\frac{n+3}{n-1}.Let n=3,5, and N be positive integers. The inequality

\begin{array}{l}{\Vert \mathrm{\Delta }(A_1{)}^N(f)\Vert }_{L^1({\mathbb{R}}^n)}⪯\Vert f{\Vert }_{L^1({\mathbb{R}}^n)}\end{array}

holds if and only if N⩾\frac{n+3}{n-1}.

In the proof of Theorem B, they used the fact that iterate steps N is an positive integer. But, if we extend the iterate steps N to real number by the Fourier transforms, the conclusion of Theorem B is not sharp. On the other hand, there is no regularity in L^1 space. To study the boundedness of (A_1{)}^N for N\in \mathbb{R} and the influence of the regularity of function space on the boundedness of this operator, Huang [6] studied the boundedness of (A_1{)}^N on modulation space and obtained the sufficient and necessary conditions of the boundedness of \mathrm{\Delta }(A_1{)}^N on the modulation space.

Theorem C [6]　 Let \sigma =2-\frac{n-1}{2}N,1⩽p_i,q_i⩽\mathrm{\infty },s_i\in \mathbb{R}(i=1,2). When q_1⩽q_2, the iterated spherical average \mathrm{\Delta }(A_1{)}^N is bounded from M_{p_1,q_1}^{s_1}({\mathbb{R}}^n) to M_{p_2,q_2}^{s_2}({\mathbb{R}}^n) if and only if

p_1⩽p_2,s_1⩾s_2+\sigma .

When q_1>q_2, the iterated spherical average \mathrm{\Delta }(A_1{)}^N is bounded from M_{p_1,q_1}^{s_1}({\mathbb{R}}^n) to M_{p_2,q_2}^{s_2}(R^n) if and only if

p_1⩽p_2,s_1+\frac{n}{q_1}⩾s_2+\sigma +\frac{n}{q_2}.

On the other hand, Besov-Lipschitz space B_{p,q}^s and Triebel-Lizorkin space F_{p,q}^s are very important spaces in the theory of function space. They can not only characterize many function spaces, such as L^p({\mathbb{R}}^n) space, Sobolev space, Hardy space, but also have important applications in the fields of partial differential equations and time-frequency analysis [5, 7, 11, 12]. Thus, it is significant to study the boundedness of \mathrm{\Delta }(A_1{)}^N in Besov space and Triebel space. Besides, by the Littlewood-Paley theorem, when , the Triebel space is equivalent to L^P({\mathbb{R}}^n) space, that is,

(2)\begin{array}{l}{F}_{p,2}^0({\mathbb{R}}^n)\approx L^p({\mathbb{R}}^n).\end{array}

Therefore, the boundedness of \mathrm{\Delta }(A_1{)}^N in Triebel space can be used to study the boundedness of this operator in L^P({\mathbb{R}}^n) space. In this paper, we study the boundedness of \mathrm{\Delta }(A_1{)}^N in Besov space and Triebel space, and improve the boundedness of \mathrm{\Delta }(A_1{)}^N in L^P({\mathbb{R}}^n) space by the conclusion in Triebel space. The following theorems are our main results.

Theorem 1　 Let {\sigma }_1=2+\frac{n}{2}-\frac{n-1}{2}N, 1⩽p_i,q⩽\mathrm{\infty },s_i\in \mathbb{R}(i=1,2). If

p_1⩽p_2,s_1-\frac{n}{p_1}⩾s_2+{\sigma }_1-\frac{n}{p_2},

then \mathrm{\Delta }(A_1{)}^N is bounded from B_{p_1,q}^{s_1}({\mathbb{R}}^n) to B_{p_2,q}^{s_2}({\mathbb{R}}^n).

Theorem 2　 Let {\sigma }_2=2-\frac{n}{2}-\frac{n-1}{2}N, 1⩽p_i,q⩽\mathrm{\infty },s_i\in \mathbb{R}(i=1,2). If the iterated spherical average \mathrm{\Delta }(A_1{)}^N is bounded from B_{p_1,q}^{s_1}({\mathbb{R}}^n) to B_{p_2,q}^{s_2}({\mathbb{R}}^n), then the following conditions must be hold:

p_1⩽p_2,s_1⩾s_2+{\sigma }_2.

Theorem 3　 Let {\sigma }_1=2+\frac{n}{2}-\frac{n-1}{2}N, 1⩽p_i,q⩽\mathrm{\infty },s_i\in \mathbb{R}(i=1,2). If

p_1⩽p_2,s_1-\frac{n}{p_1}>s_2+{\sigma }_1+\frac{n}{p_2},

then \mathrm{\Delta }(A_1{)}^N is bounded from F_{p_1,q}^{s_1}({\mathbb{R}}^n) to F_{p_2,q}^{s_2}({\mathbb{R}}^n).

Corollary 1　 Let . When N>\frac{n+4}{n-1}, \mathrm{\Delta }(A_1{)}^N is bounded in L^p(R^n) space, that is

\parallel \mathrm{\Delta }(A_1{)}^N(f){\parallel }_{L^p(R^n)}⪯\parallel f{\parallel }_{L^p(R^n)}.

Remark 1　We can find that the conclusion in Corollary 1 is better than Theorem A, since the smallest iterate steps N in Corollary 1 is smaller than Theorem A.

Remark 2　We can obtain the necessary conditions of boundedness of \mathrm{\Delta }(A_1{)}^N in Triebel space by a similar method, and the conclusion is also similar as Theorem 2. The proof follows the same pattern so that we leave the proof to the reader.

Throughout this paper, we use the inequality A⪯B to mean that there is a positive number C independent of all main variables such that A⩽CB, and use the notation A\simeq B to mean A⪯B and B⪯A.

2 Preliminaries and lemmas

In this section, we will give the definitions of Besov space and Triebel space, and discuss some basic properties of Besov space and Triebel space. Also, we will prove some estimates and lemmas which will be used in our proof.

Definition 1 [11] (Besov space)　Let {\phi }_1(\xi ) be a radial Schwartz function on {\mathbb{R}}^n which satisfy {\phi }_1(\xi )\in C_0^{\mathrm{\infty }}({\mathbb{R}}^n) and \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}{\phi }_1\subset \{\xi :\frac{1}{2}⩽|\xi |⩽2\}. When \frac{2}{3}⩽|\xi |⩽\frac{3}{2}, {\phi }_1(\xi )\equiv 1. Assume {\phi }_0(\xi )\in C_0^{\mathrm{\infty }}({\mathbb{R}}^n), \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}{\phi }_0\subset \{\xi :|\xi |⩽2\} and satisfy

\sum _{k=0}^{+\mathrm{\infty }}{\phi }_k\equiv 1,

where {\phi }_k(\xi )=\phi (2^{-k}\xi ), k\in {\mathbb{N}}^{+}. Define

{\mathrm{\Delta }}_{k}f=({\phi }_k\hat{f}{)}^{\vee }.

Let s\in \mathbb{R}, . The Besov space is defined as follows:

Definition 2 [11] (Triebel space)　Let s\in R, . The Triebel space is defined as follows:

Proposition 1 [7]　 Let , s_i\in \mathbb{R}(i=1,2). When

p_1⩽p_2,s_1-\frac{n}{p_1}⩾s_2-\frac{n}{p_2},

we have

(3)\begin{array}{l}{B}_{p_1,q}^{s_1}\subset B_{p_2,q}^{s_2},F_{p_1,q}^{s_1}\subset F_{p_2,q}^{s_2}.\end{array}

Proposition 2 [7]　 If 1⩽p,q_i⩽\mathrm{\infty }(i=1,2) and s\in \mathbb{R}, then \mathrm{\forall }\epsilon >0, we have

(4)\begin{array}{l}{F}_{p,q_1}^{s+\epsilon }\subset F_{p,q_2}^s.\end{array}

The Fourier multiplier m(D) is a linear operator acting on the test function f, which is defined as:

\begin{array}{l}\widehat{m(D)f}(\xi )=m(\xi )\hat{f}(\xi ).\end{array}

The function m(\xi ) is called the symbol or multiplier of m(D). By Fourier transform, it is known that m(D) is a convolution operator

m(D)=(m(\xi ){)}^{\vee }\ast f.

By the Young inequality, we have

\begin{array}{l}\Vert m(D)f{\Vert }_{L^p}⪯\Vert (m(\xi ){)}^{\vee }(x){\Vert }_{L^1}\Vert f{\Vert }_{L^p}\end{array}

for any 1⩽p⩽\mathrm{\infty }. Thus, to obtain the boundedness of m(D) in L^p space, we only need to estimate \parallel (m(\xi ){)}^{\vee }(x){\parallel }_{L^1}. For this purpose, we need the following Bernstein multiplier theorem to estimate \Vert (m(\xi ){)}^{\vee }(x){\Vert }_{L^1}.

Lemma 1 [12] (Bernstein multiplier theorem)　 Assume and {\mathrm{\partial }}^{\gamma }m(\xi )\in L^2 for all multi-indices \gamma with |\gamma |⩽\left[n\left(\frac{1}{p}-\frac{1}{2}\right)\right]+1. We have

(5)\begin{array}{l}{\Vert (m(\xi ){)}^{\vee }(x)\Vert }_{L^p}⪯{\displaystyle \sum _{|\gamma |⩽\left[n\left(\frac{1}{p}-\frac{1}{2}\right)\right]+1}\Vert {\mathrm{\partial }}^{\gamma }m(\xi ){\Vert }_{L^2}.}\end{array}

By Fourier transform, ignoring constant factors, m(D) is a Fourier multiplier as follows:

\begin{array}{l}\widehat{\mathrm{\Delta }(A_1{)}^{N}f}=|\xi {|}^2{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^N\hat{f}(\xi ),\end{array}

where

V_{\delta }(r)=\frac{J_{\delta }(r)}{r^{\delta }}

and J_{\delta }(r) is the Bessel function of order \delta which has following properties.

Lemma 2 [9]　 When \delta >-\frac{1}{2}, we have

(6)\begin{array}{l}{V}_{\delta }(r)=O(1),when|r|⪯1,\end{array}

(7)\frac{\mathrm{d}{V}_{\delta }(r)}{\mathrm{d}r}=-r{V}_{\delta +1}(r).

Lemma 3 [9]　 Let r>1 and \delta >-\frac{1}{2}. For any L\in {\mathbb{Z}}^{+} and r\in [1,\mathrm{\infty }), we have

(8)J_{\delta }(r)=\sqrt{\frac{2}{\pi r}}\cos \left(r-\frac{\delta \pi }{2}-\frac{\pi }{4}\right)+\sum _{j=1}^L{a}_j{\mathrm{e}}^{\mathrm{i}r}{r}^{-\frac{1}{2}-j}+\sum _{j=1}^L{b}_j{\mathrm{e}}^{-\mathrm{i}r}{r}^{-\frac{1}{2}-j}+E(r),

where a_j and b_j are constants for all j, and E(r) is a C^{\mathrm{\infty }} function satisfying

\begin{array}{l}\left|E^{(k)}(r)\right|⪯r^{-\frac{1}{2}-L-1}\end{array}

for any k=0,1,2,\dots .

3 Proof of main results

We firstly prove Theorem 1. By the definition of Besov space, we need to estimate \Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^{N}f{\Vert }_{L^p}. For this purpose, we obtain the following lemma.

Lemma 4　 Let 1⩽p⩽\mathrm{\infty }, {\sigma }_1=2+\frac{2}{n}-\frac{n-1}{2}N. Then we have the following estimate

\begin{array}{l}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^{N}f{\Vert }_{L^p(R^n)}⪯2^{k{\sigma }_1}\Vert {\mathrm{\Delta }}_{k}f{\Vert }_{L^p(R^n)}.\end{array}

Proof　 \mathrm{\forall }k\in \mathbb{N}, {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^{N}f is a convolution operator m_k(D)(f)={\mathrm{\Omega }}_k(x)\ast f, where

(9){\mathrm{\Omega }}_k(x)={\int }_{{\mathbb{R}}^n}{\phi }_k(\xi )\mid \xi {\mid }^2{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^N{\mathrm{e}}^{\mathrm{i}\xi \cdot x}\mathrm{d}\xi .

By the orthogonality of dyadic decomposition, when |l-k|⩾2, we have{\phi }_l(\xi ){\phi }_k(\xi )=0. Since

\sum _{k\in {\mathbb{Z}}^n}{\mathrm{\Delta }}_k=I,

by Young's inequality and Minkowski's inequality, we have

\begin{array}{rl}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^{N}f{\Vert }_{L^p}& ={\Vert \sum _{l\in \mathbb{N}}{\mathrm{\Delta }}_l{\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^{N}f\Vert }_{L^p}\\ & ⪯\sum _{l\in \mathbb{N},\mid l-k\mid ⩽1}\Vert {\mathrm{\Delta }}_l\mathrm{\Delta }(A_1{)}^N{\mathrm{\Delta }}_{k}f{\Vert }_{L^p}\\ & ⪯\sum _{l\in \mathbb{N},\mid l-k\mid ⩽1}{\Vert {\left({\phi }_l(\xi )\cdot \mid \xi {\mid }^2{\left(V_{\frac{n-2}{2}}(\mid \xi \mid )\right)}^N\right)}^{\vee }\ast {\mathrm{\Delta }}_{k}f\Vert }_{L^p}\\ & ⪯\sum _{l\in \mathbb{N},\mid l-k\mid ⩽1}{\Vert {\left({\phi }_l(\xi )\cdot \mid \xi {\mid }^2{\left(V_{\frac{n-2}{2}}(\mid \xi \mid )\right)}^N\right)}^{\vee }\Vert }_{L^1}\Vert {\mathrm{\Delta }}_{k}f{\Vert }_{L^p}\\ & =\sum _{l\in \mathbb{N},\mid l-k\mid ⩽1}\Vert {\mathrm{\Omega }}_l{\Vert }_{L^1}\Vert {\mathrm{\Delta }}_{k}f{\Vert }_{L^p}.\end{array}

Thus, we only need to estimate

\sum _{l\in \mathbb{N},\mid l-k\mid ⩽1}\Vert {\mathrm{\Omega }}_l(x){\Vert }_{L^1}.

For every k\in \mathbb{N}, the number of l which satisfy l\in \mathbb{N}:|l-k|⩽1 is at most 3. So, we only need to estimate \Vert {\mathrm{\Omega }}_l(x){\Vert }_{L^1} when l\simeq k. When , by condition (7), we have |{\mathrm{\Omega }}_l(x)|⪯1, when . If |x|⩾100, without losing generality, assume |x_1|⩾\frac{x}{n}. By the derivative of V_{\delta }(t) (see (8)) and taking integration by part on {\xi }_1, we have

|{\mathrm{\Omega }}_l(x)|⪯\frac{1}{\mid x_1{\mid }^{n+1}}⪯\frac{1}{\mid x{\mid }^{n+1}}.

So, when , \Vert {\mathrm{\Omega }}_l(x){\Vert }_{L^1}⪯1, by the fact that l\simeq k. Then, we choose L=1 in Lemma 3. We have the following asymptotic form of V_{\delta }(r):

V_{\delta }(r)=r^{-\delta -\frac{1}{2}}\sqrt{\frac{2}{\pi }}\cos \left(r-\frac{\delta \pi }{2}-\frac{\pi }{4}\right)+O(r^{-\delta -\frac{3}{2}})

for . So, when |k|>100, l\simeq k and \xi \in \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}{\phi }_l(\xi ), we have

(10)|V_{\delta }(|\xi |){|}^N⪯|\xi {|}^{(-\delta -\frac{1}{2})N}\simeq (2^l{)}^{(-\delta -\frac{1}{2})N}\simeq 2^{-(\delta +\frac{1}{2})Nk}.

Then, by the chain rule and the derivative formula of V_{\delta }(t), we obtain

\begin{array}{rl}\frac{\mathrm{\partial }}{\mathrm{\partial }{\xi }_i}{\left(V_{\delta }(|\xi |)\right)}^N& =-{\left(V_{\delta }(|\xi |)\right)}^{N-1}|\xi |\cdot V_{\delta +1}(|\xi |)\cdot \frac{{\xi }_i}{|\xi |}\\ & =-{\left(V_{\delta }(|\xi |)\right)}^{N-1}{V}_{\delta +1}(|\xi |)\cdot {\xi }_i.\end{array}

By the asymptotic form of V_{\delta }(r), we obtain that

(11)\left|\frac{\mathrm{\partial }}{\mathrm{\partial }{\xi }_i}(V_{\delta }(|\xi |){)}^N\right|⪯|\xi {|}^{(-\delta -\frac{1}{2})(N-1)}\cdot |\xi {|}^{(-\delta -\frac{3}{2})}\cdot |\xi |⪯|\xi {|}^{(-\delta -\frac{1}{2})N}\simeq 2^{-(\delta +\frac{1}{2})Nk}

for \xi \in \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}{\phi }_l(\xi ). By (10) and (11), when \delta >-\frac{1}{2} and \xi \in \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}{\phi }_l(\xi ), V_{\delta }(|\xi |{)}^N and \frac{\mathrm{\partial }}{\mathrm{\partial }{\xi }_i}{\left(V_{\delta }(|\xi |)\right)}^N share the same upper bound which is 2^{-(\delta +\frac{1}{2})Nl}. Since

{\mathrm{\partial }}^{\gamma }(|\xi {|}^2)\{\begin{array}{ll}⪯|\xi {|}^{2-|\gamma |},& |\gamma |\le 2;\\ =0,& |\gamma |>2,\end{array}

we can obtain that

(12)\begin{array}{rl}\Vert {\mathrm{\Omega }}_l(x){\Vert }_{L^1}& =\sum _{|\gamma |⩽[\frac{n}{2}]+1}{\Vert {\mathrm{\partial }}^{\gamma }({\phi }_l(\xi )\cdot |\xi {|}^2{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^N)\Vert }_{L^2}\\ & ⪯\sum _{|\gamma |⩽[\frac{n}{2}]+1}\sum _{{\gamma }_1+{\gamma }_2+{\gamma }_3=\gamma }{\Vert {\mathrm{\partial }}^{{\gamma }_1}{\phi }_l(\xi )\cdot {\mathrm{\partial }}^{{\gamma }_2}|\xi {|}^2\cdot {\mathrm{\partial }}^{{\gamma }_3}{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^N\Vert }_{L^2(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}{\phi }_l(\xi ))}\\ & ⪯2^{\frac{kn}{2}}\cdot 2^{2k}\cdot 2^{-\frac{n-1}{2}Nk}\\ & =2^{(2+\frac{n}{2}-\frac{n-1}{2}N)k}.\end{array}

By the definition of Besov space and above estimate, we have

\begin{array}{rl}\Vert \mathrm{\Delta }(A_1{)}^{N}f{\Vert }_{B_{p_2,q}^{s_2}}& ={\left(\sum _{k=0}^{+\mathrm{\infty }}{2}^{k{s}_{2}q}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1^N)f{\Vert }_{p_2}^q\right)}^{\frac{1}{q}}\\ & ={\left(\sum _{k=0}^{100}{2}^{k{s}_{2}q}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1^N)f{\Vert }_{p_2}^q+\sum _{k>100}{2}^{k{s}_{2}q}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1^N)f{\Vert }_{p_2}^q\right)}^{\frac{1}{q}}\\ & ⪯{\left(\sum _{k=0}^{100}{2}^{k{s}_{2}q}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1^N)f{\Vert }_{p_2}^q\right)}^{\frac{1}{q}}+{\left(\sum _{k>100}{2}^{k{s}_{2}q}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1^N)f{\Vert }_{p_2}^q\right)}^{\frac{1}{q}}\\ & ⪯{\left(\sum _{k=0}^{100}{2}^{(s_2+2+\frac{n}{2}-\frac{n-1}{2}N)qk}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1^N)f{\Vert }_{p_2}^q\right)}^{\frac{1}{q}}\\ & +{\left(\sum _{k>100}{2}^{(s_2+2+\frac{n}{2}-\frac{n-1}{2}N)qk}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1^N)f{\Vert }_{p_2}^q\right)}^{\frac{1}{q}}\\ & ⪯\Vert f{\Vert }_{B_{p_2,q}^{s_2+2+\frac{n}{2}-\frac{n-1}{2}N}}.\end{array}

By the embedding property (3) of Besov space, when p_1⩽p_2,s_1-\frac{n}{p_1}⩾s_2+{\sigma }_1-\frac{n}{p_2}, we can easily obtain that

\begin{array}{l}\Vert \mathrm{\Delta }(A_1{)}^{N}f{\Vert }_{B_{p_2,q}^{s_2}}⪯\Vert f{\Vert }_{B_{p_2,q}^{s_2+{\sigma }_1}}⪯\Vert f{\Vert }_{B_{p_2,q}^{s_1}}.\end{array}

Theorem 1 is proved.

Next, we will prove Theorem 2. From Lemma 3, we can find that the main term in asymptotic expansion of Bessel function is a trigonometric function which has zero points in every semiperiod. Thus, we can't estimate the dual operator of the iterated spherical average at these zero points. To solve this problem, we need the following lemmas.

Lemma 5　 Assume i\in {\mathbb{Z}}^{+} which is large enough. Then there exists a constant {\epsilon }_0>0, such that

(13)V_{\frac{n-1}{2}}(|\xi |)⩾{\epsilon }_0|\xi {|}^{-\frac{n-1}{2}}

for |\xi |\in [i\pi +\frac{n\pi }{4}-\frac{3}{4}\pi +\frac{\pi }{20},(i+1)\pi +\frac{n\pi }{4}-\frac{3}{4}\pi -\frac{\pi }{20}].

Proof　Choosing L=1 in Lemma 3, we have

\begin{array}{rl}{V}_{\frac{n-2}{2}}(r)& =r^{-\frac{n-1}{2}}\sqrt{\frac{2}{\pi }}\cos \left(r-\frac{n\pi }{4}+\frac{\pi }{4}\right)+O(r^{-\frac{n+1}{2}})\\ & =r^{-\frac{n-1}{2}}\sqrt{\frac{2}{\pi }}\sin \left(r-\frac{n\pi }{4}+\frac{3\pi }{4}\right)+O(r^{-\frac{n+1}{2}}).\end{array}

Define

u(r)=\sin \left(r-\frac{n\pi }{4}+\frac{3\pi }{4}\right)

and {\epsilon }_0=\frac{1}{2}\sin \frac{\pi }{20}. When r-\frac{n\pi }{4}+\frac{3\pi }{4}\in [i\pi +\frac{\pi }{20},(i+1)\pi -\frac{\pi }{20}], we have

|u(r)|⩾2{\epsilon }_0.

Moreover, when r is large enough, it is easy to obtain

O(r^{-\frac{n+1}{2}})⩽{\epsilon }_0{r}^{-\frac{n-1}{2}}.

Thus, when |\xi |\in [i\pi +\frac{n\pi }{4}-\frac{3\pi }{4}+\frac{\pi }{20},(i+1)\pi +\frac{n\pi }{4}-\frac{3\pi }{4}-\frac{\pi }{20}] and i is large enough, we have

|V_{\frac{n-2}{2}}(|\xi |)|⩾{\epsilon }_0|\xi {|}^{-\frac{n-1}{2}}.

Lemma 6　 Let 1⩽p⩽\mathrm{\infty }, {\sigma }_2=2-\frac{n}{2}-\frac{n-1}{2}N. For k⩾2, if the smooth function \{f_k\} satisfy

\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\widehat{f_k}(\xi )\subset \left\{\xi :|\xi |\in \left[i_k\pi +\frac{n\pi }{4}-\frac{3}{4}\pi +\frac{\pi }{20},(i_k+1)\pi +\frac{n\pi }{4}-\frac{3}{4}\pi -\frac{\pi }{20}\right]\right\},

where i_k\in {\mathbb{Z}}^{+}, and [i_k\pi +\frac{n\pi }{4}-\frac{3}{4}\pi ,(i_k+1)\pi +\frac{n\pi }{4}-\frac{3}{4}\pi ]\subseteq (\frac{2}{3}\cdot 2^k,\frac{3}{2}\cdot 2^k), then we can obtain

\begin{array}{l}{2}^{k{\sigma }_2}\Vert f_k{\Vert }_{L^p}⪯\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^N{f}_k{\Vert }_{L^p}.\end{array}

Proof　By the assumption of \{f_k\}, it is obvious that |\xi |\simeq 2^k when \xi \in \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\hat{f}(\xi ). Thus, by (12) and Lemma 5, when \xi \in \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\hat{f}(\xi ), we have

(14){\epsilon }_0|\xi {|}^{-\frac{n-1}{2}}⩽\left|\left(V_{\frac{n-2}{2}}(|\xi |)\right)\right|⪯|\xi {|}^{-\frac{n-1}{2}},

that is, |V_{\frac{n-2}{2}}(|\xi |)|\simeq 2^{-\frac{n-1}{2}k}. Then, by the chain rule and the derivative formula of V_{\delta }(t), we can obtain

(15)\begin{array}{rl}\left|\frac{\mathrm{\partial }}{\mathrm{\partial }{\xi }_i}{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^{-N}\right|& =\left|-{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^{-(N+1)}{V}_{\frac{n-2}{2}}(|\xi |)\cdot {\xi }_i\right|\\ & ⪯|\xi {|}^{(-\frac{n-1}{2})(-N-1)}|\xi {|}^{-\frac{n+1}{2}}|\xi |\\ & \simeq 2^{\frac{n-1}{2}Nk}.\end{array}

Therefore, when \xi \in \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\hat{f}(\xi ), |(V_{\frac{n-2}{2}}(|\xi |)){|}^{-N} and |\frac{\mathrm{\partial }}{\mathrm{\partial }{\xi }_i}(V_{\frac{n-2}{2}}(|\xi |){)}^{-N}| share the same upper bound which is 2^{\frac{n-1}{2}Nk}.

By Definition 1, {\phi }_k(\xi )\equiv 1 when \frac{2}{3}\cdot 2^k⩽|\xi |⩽\frac{3}{2}\cdot 2^k. So, when k is large enough, we can choose some suitable i_k\in Z^{+} such that [i_k\pi +\frac{n\pi }{4}-\frac{3}{4}\pi ,(i_k+1)\pi +\frac{n\pi }{4}-\frac{3}{4}\pi ]\subseteq (\frac{2}{3}\cdot 2^k,\frac{3}{2}\cdot 2^k). So, we have

{\phi }_k(\xi )\widehat{f_k}(\xi )=\widehat{f_k}(\xi ),

when \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\widehat{f_k}(\xi )\subset \{\xi :|\xi |\in [i_k\pi +\frac{n\pi }{4}-\frac{3}{4}\pi +\frac{\pi }{20},(i_k+1)\pi +\frac{n\pi }{4}-\frac{3}{4}\pi -\frac{\pi }{20}]\}. Thus, by (13), (14), Young’s inequality and Bernsteins Multiplier Theorem, we can obtain

\begin{array}{rl}\Vert f_k{\Vert }_{L^p}& =\Vert (\widehat{f_k}{)}^{\vee }{\Vert }_{L^p}\\ & ={\Vert ({\phi }_k(\xi ){\phi }_k(\xi )\cdot |\xi {|}^{-2}{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^{-N}\cdot |\xi {|}^2{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^N\widehat{f_k}{)}^{\vee }\Vert }_{L^p}\\ & ={\Vert ({\phi }_k(\xi )|\xi {|}^{-2}{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^{-N}\cdot {\phi }_k(\xi )|\xi {|}^2{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^N\widehat{f_k}{)}^{\vee }\Vert }_{L^p}\\ & ⪯{\Vert ({\phi }_k(\xi )|\xi {|}^{-2}{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^{-N}{)}^{\vee }\Vert }_{L^1}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^N{f}_k{\Vert }_{L^p}\\ & ⪯\sum _{|\gamma |⩽[\frac{n}{2}]+1}{\Vert {\mathrm{\partial }}^{\gamma }({\phi }_k(\xi )|\xi {|}^{-2}{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^{-N})\Vert }_{L^2}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^N{f}_k{\Vert }_{L^p}\\ & ⪯\sum _{\begin{array}{c}|\gamma |⩽[\frac{n}{2}]+1\\ {\gamma }_1+{\gamma }_2+{\gamma }_3=\gamma \end{array}}{\Vert {\mathrm{\partial }}^{{\gamma }_1}{\phi }_k(\xi )\cdot {\mathrm{\partial }}^{{\gamma }_2}|\xi {|}^{-2}\cdot {\mathrm{\partial }}^{{\gamma }_3}{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^{-N}\Vert }_{L^2}\\ & \cdot \Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^N{f}_k{\Vert }_{L^p}\\ & ⪯\sum _{\begin{array}{c}|\gamma |⩽[\frac{n}{2}]+1\\ {\gamma }_1+{\gamma }_2=\gamma \end{array}}{\Vert {\mathrm{\partial }}^{{\gamma }_1}{\phi }_k(\xi )\cdot |\xi {|}^{-2}\cdot {\mathrm{\partial }}^{{\gamma }_2}{\left(V_{\frac{n-2}{2}}(|\xi |)\right)}^{-N}\Vert }_{L^2}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^N{f}_k{\Vert }_{L^p}\\ & \simeq 2^{(-2+\frac{n}{2}+\frac{n-1}{2}N)k}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^N{f}_k{\Vert }_{L^p},\end{array}

which means

2^{(2-\frac{n}{2}-\frac{n-1}{2}N)k}\Vert f_k{\Vert }_{L^p}⪯\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^N{f}_k{\Vert }_{L^p}.

Lemma 6 is proved.

Now, we continue to prove Theorem 2. Let f\in S({\mathbb{R}}^n) satisfy \hat{f}(\xi )\in C_0^{\mathrm{\infty }}(B^{-}), where . Then choose {\xi }_k=(i_k\pi +\frac{n\pi }{4}-\frac{3}{4}\pi +\frac{\pi }{2},0,\dots ,0), where i_k is defined in the proof of Lemma 6. For , we define smooth function f_{k,\lambda } by

{\hat{f}}_{k,\lambda }(\xi )=\hat{f}\left(\frac{\xi -{\xi }_k}{\lambda }\right).

It is obvious that \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\widehat{f_{k,\lambda }}(\xi )\subset \{\xi :|\xi |\in [i_k\pi +\frac{n\pi }{4}-\frac{3\pi }{4}+\frac{\pi }{20},(i_k+1)\pi +\frac{n\pi }{4}-\frac{3\pi }{4}-\frac{\pi }{20}]\}. By the definition of f_{k_i,\lambda }, we can obtain

\begin{array}{l}{\mathrm{\Delta }}_k{f}_{k,\lambda }=f_{k,\lambda };\end{array}

\begin{array}{l}{\mathrm{\Delta }}_j{f}_{k,\lambda }=0,j\ne k.\end{array}

By Lemma 6, we have

\begin{array}{rl}\Vert \mathrm{\Delta }(A_1{)}^N{f}_{k_i,\lambda }{\Vert }_{B_{p_2,q}^{s_2}}& =2^{k{s}_2}\Vert {\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^N{f}_{k_i,\lambda }{\Vert }_{L^{p_2}}\\ & ⪰2^{k(s_2+2-\frac{n}{2}-\frac{n-1}{2}N)}\Vert f_{k_i,\lambda }{\Vert }_{L^{p_2}}\\ & ⪰2^{k(s_2+{\sigma }_2)}{\lambda }^{n(1-\frac{1}{p_2})}.\end{array}

Moreover, by some simple computation, we have

\begin{array}{l}\Vert f_{k_i,\lambda }{\Vert }_{B_{p_1,q}^{s_1}}=2^{k{s}_1}\Vert f_{k_i,\lambda }{\Vert }_{L^{p_1}}\simeq 2^{k{s}_1}{\lambda }^{n(1-\frac{1}{p_1})}.\end{array}

By the assumption that \mathrm{\Delta }(A_1{)}^N is bounded on Besov space, we can obtain \Vert \mathrm{\Delta }(A_1{)}^N{f}_{k_i,\lambda }{\Vert }_{B_{p_2,q}^{s_2}}⪯\Vert f_{k,\lambda }{\Vert }_{B_{p_1,q}^{s_1}}. Therefore, when , we have

\begin{array}{l}{2}^{k(s_2+{\sigma }_1)}{\lambda }^{n(1-\frac{1}{p_2})}⩽2^{k{s}_1}{\lambda }^{n(1-\frac{1}{p_1})}.\end{array}

Fixed k\in {\mathbb{Z}}^{+}, we have

{\lambda }^{n(1-\frac{1}{p_2})}⪯{\lambda }^{n(1-\frac{1}{p_1})}.

Then let \lambda \to 0, we can obtain

p_1⩽p_2.

On the other hand, when fixed \lambda \in (0,\frac{1}{2}) and letting k\to \mathrm{\infty }, we have

2^{k(s_2+{\sigma }_1)}⩽2^{k{s}_1},

which means

s_1⩾s_2+{\sigma }_2.

Theorem 2 is proved.

Now, we prove the boundedness of \mathrm{\Delta }(A_1{)}^N on Triebel space. We firstly assume s=0, 1⩽p=q⩽\mathrm{\infty }, by the definition of Triebel space and orthogonality of \{{\phi }_k{\}}_{k=1}^{\mathrm{\infty }}, we can obtain

\begin{array}{rl}\Vert \mathrm{\Delta }(A_1{)}^{N}f{\Vert }_{F_{p,p}^0}^p& ={\Vert {\left(\sum _{k=0}^{+\mathrm{\infty }}|{\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^{N}f{|}^p\right)}^{\frac{1}{p}}\Vert }_{L^p({\mathbb{R}}^n)}^p\\ & ={\Vert {\left(\sum _{k=0}^{+\mathrm{\infty }}|\sum _{j\in {\mathbb{Z}}^{+}\cup \{0\}}{\mathrm{\Delta }}_j{\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^{N}f{|}^p\right)}^{\frac{1}{p}}\Vert }_{L^p({\mathbb{R}}^n)}^p\\ & ={\Vert {\left(\sum _{k=0}^{+\mathrm{\infty }}|\sum _{j=(k-1)\vee 0}^{j=k+1}{\mathrm{\Delta }}_j{\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^{N}f{|}^p\right)}^{\frac{1}{p}}\Vert }_{L^p({\mathbb{R}}^n)}^p\end{array}

\begin{array}{rl}& ⩽{\int }_{R^n}\sum _{k=0}^{+\mathrm{\infty }}\sum _{j=(k-1)\vee 0}^{j=k+1}|{\mathrm{\Delta }}_j{\mathrm{\Delta }}_k\mathrm{\Delta }(A_1{)}^{N}f{|}^p\mathrm{d}\xi \\ & ⪯\sum _{k=0}^{+\mathrm{\infty }}\sum _{j=(k-1)\vee 0}^{j=k+1}{\int }_{{\mathbb{R}}^n}|{\mathrm{\Omega }}_j(x)\ast {\mathrm{\Delta }}_{k}f{|}^p\mathrm{d}\xi \\ & =\sum _{k=0}^{+\mathrm{\infty }}\sum _{j=(k-1)\vee 0}^{j=k+1}\Vert {\mathrm{\Omega }}_j(x)\ast {\mathrm{\Delta }}_{k}f{\Vert }_{L^p({\mathbb{R}}^n)}^p,\end{array}

where {\mathrm{\Omega }}_j(x) is defined in (9). By (11), when |j-k|⩽1, we have

\Vert {\mathrm{\Omega }}_j{\Vert }_{L^1}⪯2^{(2+\frac{n}{2}-\frac{n-1}{2}N)j}\simeq 2^{(2+\frac{n}{2}-\frac{n-1}{2}N)k}.

Combining the above estimates, we have

\begin{array}{rl}\Vert \mathrm{\Delta }(A_1{)}^N{\Vert }_{F_{p,p}^0}^p& ⪯\sum _{k=0}^{+\mathrm{\infty }}\sum _{j=k-1}^{j=k+1}\Vert {\mathrm{\Omega }}_j(x)\ast {\mathrm{\Delta }}_{k}f{\Vert }_{L^p({\mathbb{R}}^n)}^p\\ & ⪯\sum _{k=0}^{+\mathrm{\infty }}\sum _{j=k-1}^{j=k+1}\Vert {\mathrm{\Omega }}_j{\Vert }_{L^1}^p\Vert {\mathrm{\Delta }}_{k}f{\Vert }_{L^p({\mathbb{R}}^n)}^p\\ & ⪯\sum _{k=0}^{+\mathrm{\infty }}{2}^{(2+\frac{n}{2}-\frac{n-1}{2}N)kp}\Vert {\mathrm{\Delta }}_{k}f{\Vert }_{L^p({\mathbb{R}}^n)}^p\\ & =\sum _{k=0}^{+\mathrm{\infty }}{\int }_{R^n}{2}^{(2+\frac{n}{2}-\frac{n-1}{2}N)kp}|{\mathrm{\Delta }}_{k}f{|}^p\mathrm{d}\xi \\ & \simeq {\int }_{R^n}\left(\sum _{k=0}^{+\mathrm{\infty }}{2}^{(2+\frac{n}{2}-\frac{n-1}{2}N)kp}|{\mathrm{\Delta }}_{k}f{|}^p\right)\mathrm{d}\xi \\ & =\Vert f{\Vert }_{F_{p,p}^{2+\frac{n}{2}-\frac{n-1}{2}N}}^p.\end{array}

Therefore, when s_1-\frac{n}{p_1}>s_2-\frac{n}{p_2}+{\sigma }_1, there exists \epsilon >0 such that s_1-\frac{n}{p_1}>s_2-\frac{n}{p_2}+{\sigma }_1+2\epsilon. By Proposition 1 and Proposition 2, we can obtain

\begin{array}{rl}\Vert \mathrm{\Delta }(A_1{)}^{N}f{\Vert }_{F_{p_2,q_2}^{s_2}}& ⪯\Vert \mathrm{\Delta }(A_1{)}^{N}f{\Vert }_{F_{p_2,p_2}^{s_2+\epsilon }}\\ & ⪯\Vert f{\Vert }_{F_{p_2,p_2}^{s_2+{\sigma }_1+\epsilon }}\\ & ⪯\Vert f{\Vert }_{F_{p_2,q_1}^{s_2+{\sigma }_1+2\epsilon }}\\ & ⪯\Vert f{\Vert }_{F_{p_1,q_1}^{s_1}}.\end{array}

Theorem 3 is proved.

Finally, we prove Corollary 1. The condition N>\frac{n+4}{n-1} yields,

By choosing 1⩽p_1=p_2=p⩽\mathrm{\infty }, q_1=q_2=2, s_1=s_2=0 in Theorem 3, we have

\Vert \mathrm{\Delta }(A_1{)}^N(f){\Vert }_{F_{p,2}^0({\mathbb{R}}^n)}⪯\Vert f{\Vert }_{F_{p,2}^0({\mathbb{R}}^n)}.

By the equivalence between Triebel and L^p space (see (2)), we can obtain

\Vert \mathrm{\Delta }(A_1{)}^N(f){\Vert }_{L^p({\mathbb{R}}^n)}⪯\Vert f{\Vert }_{L^p({\mathbb{R}}^n)}.

Corollary 1 is proved.

References

[1] Belinsky E, Dai F, Ditzian Z. Multivariate approximating average. J Approx Theory 2003; 125: 85–105

[2] Borwein J, Straub A, Vignat C. Densities of short uniform random walks in higher dimensions. J Math Anal Appl 2016; 437(1): 668–707

[3] Fan D, Lou Z, Wang Z. A note on iterated sperical average on Lebesgue spaces. Nonlinear Anal 2019; 180: 170–183

[4] Fan D, Zhao F. Approximation properties of combination of multivariate averages on Hardy spaces. J Approx Theory 2017; 223: 77–95

[5] GrafakosL. Modern Fourier Analysis. 2nd ed. Graduate Texts in Mathematics, Vol 250. New York: Springer, 2009

[6] Huang Q. Boundedness of iterated spherical average on modulation spaces. Nonlinear Analysis 2020; 199: 111968

[7] MiaoC. Harmonic Analysis and Its Applications on Partial Differential Equations. Beijing: Science Press, 1999 (in Chinese)

[8] Pearson K. The problem of random walk. Nature 1905; 72: 342

[9] Stein E M. Maximal functions: spherical means. Proc Nat Acad Sci USA 1976; 73(7): 2174–2175

[10] SteinE M. Harmonic Analysis: Real-variable Methods, Orthogonality, and Oscillatory Integrals. Princeton N J: Princeton Univ Press, 1993

[11] TriebelH. Theory of Function Spaces, Monographs in Mathematcis, Vol 78. Basel: Birkhäuser, 1983

[12] WangBHuoZHaoCGuoZ. Harmonic Analysis Method for Nonlinear Evolution Equations I. Hackensack N J: World Scientific, 2011

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