Almost Everywhere Convergence of Riesz Means Related to Schrödinger Operator with Constant Magnetic Fields

Abstract

We study almost everywhere convergence for Riesz means related to Schrödinger operator with constant magnetic fields. Through researching the weighted norm estimates for the maximal operator with power-weight functions, we obtain the desired result, which is similar to the work given by Anthony Carbery, Jose L. Rubio de Francia, and Luis Vega.

1. Introduction

The magnetic Schrödinger operator (MSO) with constant magnetic fields 𝐻_𝑏 in ℝ^𝑛 is of the form 𝐻_𝑏=−∇+𝑖𝐵𝑧2²,𝑧∈𝑅^𝑛, (1) where 𝐵 is a real antisymmetric matrix. If 𝐵 is degenerated (this requires 𝑛=2𝑑+𝑙 to be odd, and 2𝑑 is the rank of the matrix 𝐵), then its eigenvalues have the form ±𝑖𝑏_𝑗, 𝑗=1,2,…,𝑑. In properly chosen coordinates 𝑧=(𝑧₁,𝑧₂,…,𝑧_𝑛)=(𝑥₁,𝑦₁,𝑥₂,𝑦₂,…,𝑥_𝑑,𝑦_𝑑,𝑧_2𝑑+1,…,𝑧_2𝑑+𝑙), the operator becomes 𝐻_𝑏=−^𝑑_𝑗=1𝜕_𝑥𝑗𝑏+𝑖^𝑗2𝑦_𝑗²+𝜕_𝑦𝑗𝑏+𝑖^𝑗2𝑥_𝑗−Δ_𝑙, (2) where Δ_𝑙=∑^𝑙_𝑗=1𝜕²/𝜕𝑧²_2𝑑+𝑗 is the Laplacian in 𝑅^𝑙. The spectrum of 𝐻_𝑏 is in the semiaxis starting from the point 𝑐=∑𝑏_𝑗, and its spectral expansion is continuous (see [1]).

Let 𝑏_𝑗’s be positive. Denote by 𝐸_𝑡 the spectral function of 𝐻_𝑏. It is an integral operator with a kernel 𝑒_𝑡(𝑧,𝑧^′), which is skew-translation invariant; that is, 𝑒_𝑡(𝑧+𝜔,𝑧^′+𝜔)=𝑒_𝑡(𝑧,𝑧^′)exp𝑖⟨𝐵(𝑧−𝑧^′),𝜔⟩.

For 𝛽>0, set the Riesz means of order 𝛽 as 𝑆^𝛽_𝜆𝑓(𝑥)=^∞₀𝑡1−𝜆^𝛽₊𝑑𝐸_𝑡𝑓(𝑥)=^∞_𝑐𝑡1−𝜆^𝛽₊𝑑𝐸_𝑡𝑓(𝑥), (3) and the kernel of 𝑆^𝛽_𝜆 as 𝑠^𝛽_𝜆𝑧,𝑧^′=𝑠^𝛽𝜆,𝑧,𝑧^′=^∞₀𝑡1−𝜆^𝛽₊𝑑𝑒_𝑡𝑧,𝑧^′, (4) with the same skew-translation invariance. We will denote by 𝑆^𝛽_∗ the corresponding maximal operator; that is, 𝑆^𝛽_∗𝑓(𝑥)=sup_0<𝜆<∞𝑆^𝛽_𝜆𝑓(𝑥). (5)

As an indispensable part in harmonic analysis, many people investigate the convergence of Bochner-Riesz means for Fourier transform in norm and almost everywhere, which is defined as 𝑇^𝛽_𝑅𝑓^∧(𝜉)=1−|𝜉|²𝑅^𝛽₊𝑓(𝜉). (6) Since the convergence of 𝑇^𝛽_𝑅 in 𝐿^𝑝-norm is equivalent to the boundedness of 𝑇^𝛽₁ in 𝐿^𝑃(ℝ^𝑛), persons look for the 𝐿^𝑝 boundedness of it. For 2𝑛/(𝑛+1+2𝛽)<𝑝<2𝑛/(𝑛−1−2𝛽) and 0≤𝛽≤(𝑛−1/2), Carleson, Cordoba, and Fefferman turn out the boundedness in ℝ² (see [2–4]). When 𝑛>2, it is only proven for 𝛽≥(𝑛−1)/2(𝑛+1) (see [4, 5]). Returning the problem about almost everywhere convergence of 𝑇^𝛽_𝑅, Carbery has finished it for 𝛽>0 and 2𝑛/(𝑛+1+2𝛽)<𝑝<2𝑛/(𝑛−1−2𝛽) in two dimensions in 1983 (see [6]). For higher dimensions, it is completed by Christ only for 𝛽≥(𝑛−1)/2(𝑛+1) (see [7]). In the special case, 𝛽=0, Fefferman studies the 𝐿²-boundeness of 𝑆⁰_𝜆 for 𝑛≥2 (see [8]). Not until 1988 was it solved by Carbery et al. for 𝛽>0, 𝑛≥2, and 2≤𝑝<2𝑛/(𝑛−1−2𝛽) (see [9]).

In [1], Rozenblum and Tashchiyan investigated the 𝐿^𝑝-norm convergence for Riesz means for Schrödinger operator with constant magnetic fields. They showed that under the restriction theorem similar to one for Fourier transform in [5], it is of the same results as Bochner-Riesz means in ℝ^𝑛. However, very few results are considered for almost everywhere convergence of Riesz means for Schrödinger operator with constant magnetic fields. In the paper, we are interested in it. Through researching the boundedness of the maximal operator in 𝐿(ℝ^𝑛,|𝑥|^𝛼𝑑𝑥), we get the desired result.

2. Main Results

Theorem 1. Set 𝛽>0 and 𝑛≥2. Write 𝑝_𝛽=2𝑛/(𝑛−1−2𝛽). If 𝑓∈𝐿^𝑝(ℝ^𝑛) with 2≤𝑝<𝑝_𝛽, then lim_𝜆→∞𝑆^𝛽_𝜆𝑓(𝑥)=𝑓(𝑥) almost everywhere.

Usually, we replace the almost everywhere convergence of Riesz means with 𝐿^𝑝 estimates for the maximal operator. However, we only need to think about weighted 𝐿² estimates for the maximal operator 𝑆^𝛽_∗, as follows. In fact, based on the idea in [9], for 2≤𝑝<𝑝_𝛽, there exists a number 𝛼 with 0≤𝛼<1+2𝛽 such that 𝐿^𝑝⊆𝐿²+𝐿²(|𝑥|^𝛼). We have gotten 𝐿² boundedness of the maximal operator in another paper.

Theorem 2. Suppose that 𝛽>0 and 0≤𝛼<1+2𝛽≤𝑛. Then, 𝑆^𝛽_∗𝑓(𝑥)²|𝑥|^−𝛼𝑑𝑥≤𝐶_𝛼,𝛽𝑓(𝑥)²|𝑥|^−𝛼𝑑𝑥. (7)

In order to prove the theorem, we introduce the following lemmas, which are the essential tools. In the following lemmas, we reduce the maximal operator to the one generated by a multiplier with compact support. Since it is easy to see that the boundedness of the maximal operator generated by the multipliers is independent of the index 𝛽, the dimension 𝑛=2𝑑+𝑙 will not play an important role in the following estimates.

Lemma 3 (see [9]). If  0<𝜀<1/2, then _{||𝑥|−1|≤𝜀}𝑓(𝑥)²𝑑𝑥≤𝐶_𝛼𝐵_𝛼(𝜀)𝑓(𝑥)²|𝑥|^𝛼𝑑𝑥, (8) where 𝐵_𝛼(𝜀)=𝜀^𝛼𝜀,𝜀log𝜀,0≤𝛼<1,log𝜀,𝛼=1,𝜀,1<𝛼<𝑛. (9)

For a small 𝛿>0, let 𝑚^𝛿(𝑡) be a 𝐶^∞ function supported in [1−𝛿,1] and satisfy 0≤𝑚^𝛿(𝑡)≤1,𝐷^𝑘𝑚^𝛿(𝑡)≤𝐶𝛿^−𝑘∀𝑘∈ℕ. (10) Define 𝐾^𝛿_𝜆𝑓(𝑥)=𝑚^𝛿(𝜆𝑡)𝑑𝐸_𝑡𝑓(𝑥),𝐾^𝛿_∗𝑓(𝑥)=sup_𝜆>0𝐾^𝛿_𝜆𝑓(𝑥). (11) Let 𝜙(𝑡)∈𝐶^∞(ℝ) supported in [−1,2] and 𝜙(𝑡)=1 for 0≤𝑡≤1. Set ℎ_𝑙(𝑡)=𝜙(𝑡)𝜙2,𝑙=0,_−𝑙𝑡−𝜙2^1−𝑙𝑡,𝑙≥1. (12)

Denoting by 𝜒_𝐸 the character function of the set 𝐸, write 𝐻_𝑗=𝜒𝑥,𝑦^[0,1)𝜒𝑥−𝑦,𝑗=0,_{[2𝑗−1,2𝑗)}𝑥−𝑦,𝑗≥1. (13) Let 𝑘^𝛿(𝑡) be a function with Fourier transform as 𝑘^𝛿_𝜆(𝑡)=𝑚^𝛿(𝜆𝑡). (14) Set 𝑘_𝜆^{𝛿,𝑙,𝑗}(𝑡)=𝑘^𝛿_𝜆(𝑡)ℎ_𝑙𝛿²2^−𝑗𝑡𝜆,𝐸^𝑗_𝑡𝑓(𝑥)=_ℝ𝑛𝑒_𝑡𝑥,𝑦𝐻_𝑗𝑥,𝑦𝑓𝑦𝑑𝑦. (15) Accordingly, define the operator 𝐾_𝜆^{𝛿,𝑙,𝑗} as 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓(𝑥)=𝑘_𝜆^{𝛿,𝑙,𝑗}(𝑡)𝑑𝐸^𝑗_𝑡𝑓(𝑥), (16) where 𝑘_𝜆^{𝛿,𝑙,𝑗} is the Fourier transform of 𝑘_𝜆^{𝛿,𝑙,𝑗}. Apparently, since 𝐾^𝛿_𝜆𝑓(𝑥)=𝑘^𝛿_𝜆(𝑡)𝑑𝐸_𝑡𝑓(𝑥), (17) we decompose 𝐾^𝛿_𝜆𝑓(𝑥)=𝑘^𝛿_𝜆(𝑡)𝑑𝐸_𝑡𝑓(𝑥)=^∞∞_𝑗=0𝑙=0𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓(𝑥)=^∞∞_𝑗=0𝑙=0𝑘_𝜆^{𝛿,𝑙,𝑗}(𝑡)𝑑𝐸^𝑗_𝑡𝑓(𝑥). (18)

Lemma 4. For 𝜆>0, one has 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓(𝑥)²₂≤𝐶2^{−2𝑀(𝑗+𝑙)}𝛿^2𝑀𝑓²₂, (19) where the constant 𝐶 is independent of 𝜆 and 𝛿.

Proof. With the method similar to the proof of Lemma  4 in [9], we write ℎ(𝑡)=𝜙(𝑡)−𝜙(2𝑡) and expand 𝑚^𝛿 into a Taylor series around 𝜆𝑡. Then, 𝑘_𝜆^{𝛿,𝑙,𝑗}(𝑡)=𝑚^𝛿𝜆2𝑡−^{−(𝑗+𝑙)}𝛿²𝑟𝜆ℎ(𝑟)=𝑑𝑟𝑚^𝛿𝜆𝑡−2^{−(𝑗+𝑙)}𝛿²𝑟ℎ(𝑟)=𝑅𝑑𝑟_𝑀(𝑡,𝑟)ℎ(𝑟)𝑑𝑟, (20) where the remainder 𝑅_𝑀 satisfies 𝑅_𝑀(≤𝐷𝑡,𝑟)_𝑀𝑚^𝛿2^{−(𝑗+𝑙)}𝛿²𝑟^𝑀≤2^{−𝑀(𝑙+𝑗)}𝛿^𝑀|𝑟|^𝑀. (21)
But ℎ is a Schwartz function and can be integrated against |𝑟|^𝑀. Hence, 𝑘_𝜆^{𝛿,𝑙,𝑗}(𝑡)≤𝐶_𝑀2^{−𝑀(𝑗+𝑙)}𝛿^𝑀. (22)
Since 𝐸_ℝ is a resolution of the identity, we see 𝐸_ℝ𝑓(𝑥)=^∞₀𝑑𝐸_𝑡𝑓(𝑥)=𝑓(𝑥). (23)
Denote by 𝑒_ℝ(𝑥,𝑦) the kernel of 𝐸_ℝ. For almost all 𝑥₀∈ℝ^𝑛, we let 𝑆={𝑦∈ℝ^𝑛∶𝑒_ℝ(𝑥₀,𝑦)𝑓(𝑦)>0}. Decompose 𝑓(𝑥)=𝑓(𝑥)𝜒_𝑆(𝑥)+𝑓(𝑥)𝜒_ℝ𝑛/𝑆(𝑥)=𝑓₁(𝑥)+𝑓₂(𝑥). (24)
We have 𝐸^𝑗_ℝ𝑓𝑥₀=_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝐻_𝑗𝑥₀𝑓𝑦=,𝑦𝑑𝑦_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝐻_𝑗𝑥₀,𝑦𝑓₁𝑦+𝑑𝑦_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝐻_𝑗𝑥₀,𝑦𝑓₂𝑦≤𝑑𝑦_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝐻_𝑗𝑥₀,𝑦𝑓₁𝑦+𝑑𝑦_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝐻_𝑗𝑥₀,𝑦𝑓₂𝑦=𝑑𝑦_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝐻_𝑗𝑥₀,𝑦𝑓₁𝑦+𝑑𝑦_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝐻_𝑗𝑥₀,𝑦𝑓₂𝑦≤𝑑𝑦_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝑓₁𝑦+𝑑𝑦_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝑓₂𝑦=𝑑𝑦_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝑓₁𝑦+𝑑𝑦_ℝ𝑛𝑒_ℝ𝑥₀,𝑦𝑓₂𝑦≤𝑓𝑑𝑦₁𝑥₀+𝑓₂𝑥₀≤𝐶𝑓𝑥₀. (25)
It is easy to show 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓(𝑥)²₂=𝑘_𝜆^{𝛿,𝑙,𝑗}(𝑡)𝑑𝐸^𝑗_𝑡𝑓(𝑥)²₂=𝑑𝐸^𝑗_𝑡𝑘_𝜆^{𝛿,𝑙,𝑗}𝐻_𝑏𝑓(𝑥)²₂=𝐸^𝑗_ℝ𝑘_𝜆^{𝛿,𝑙,𝑗}𝐻_𝑏𝑓,𝐸^𝑗_ℝ𝑘_𝜆^{𝛿,𝑙,𝑗}𝐻_𝑏𝑓≤𝑘_𝜆^{𝛿,𝑙,𝑗}𝐻_𝑏𝑓²₂≤^∞₀𝑘_𝜆^{𝛿,𝑙,𝑗}(𝑡)²𝑑𝐸_𝑡𝑓,𝑓≤𝐶_𝑀2^{−2𝑀(𝑗+𝑙)}𝛿^2𝑀∞₀𝑑𝐸_𝑡𝑓,𝑓≤𝐶_𝑀2^{−2𝑀(𝑗+𝑙)}𝛿^2𝑀𝑓²₂. (26)

Lemma 5. If  0<𝛼<𝑛 and 𝜆>0, then 𝑘^𝛿_𝜆(𝑡)𝑑𝐸_𝑡𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼𝛿𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼, (27) where 𝐶_𝛼 is independent of 𝛿 and 𝜆.

Proof. Suppose that 𝑓 is supported in {|𝑥|≤𝐶2^𝑗}. Write 𝑓=𝜒_{{0≤|𝑥|≤1/4}}(𝑥)𝑓(𝑥)+𝜒_{{𝐶<|𝑥|≤𝐶2𝑗}}(𝑥)𝑓(𝑥)+𝜒_{{1/4<|𝑥|≤𝐶}}(𝑥)𝑓(𝑥)=𝑓₁+𝑓₂+𝑓₃. If 𝐶≤1/4, then 𝑓₃=0. Since 𝐾^𝛿_𝜆𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼1/2=𝐾^𝛿_𝜆𝑓₁+𝑓₂+𝑓₃(𝑥)²𝑑𝑥|𝑥|^𝛼1/2≤𝐾^𝛿_𝜆𝑓₁²𝑑𝑥|𝑥|^𝛼1/2+𝐾^𝛿_𝜆𝑓₂²𝑑𝑥|𝑥|^𝛼1/2+𝐾^𝛿_𝜆𝑓₃²𝑑𝑥|𝑥|^𝛼1/2, (28) we only need to prove that 𝑘^𝛿_𝜆(𝑡)𝑑𝐸_𝑡𝑓_𝑖(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼𝛿𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼,(𝑖=1,2,3). (29) For the case of 𝑖=2, with Lemma 4, it follows that 𝐾_𝜆^{𝛿,𝑗,𝑙}𝑓₂(𝑥)²𝑑𝑥≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀𝑓₂²₂≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀_{𝐶≤|𝑥|≤𝐶2𝑗}𝑓(𝑥)²𝑑𝑥≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀×^𝑗−1_{𝑘=0𝐶2𝑘≤|𝑥|≤𝐶2𝑘+1}𝑓(𝑥)²𝑑𝑥. (30) It is easy to see that ^𝑗−1_{𝑘=0𝐶2𝑘<|𝑥|≤𝐶2𝑘+1}𝑓(𝑥)²𝑑𝑥≤𝐶^𝑗−1_𝑘=02^−𝑘𝛼_{𝐶2𝑘<|𝑥|≤𝐶2𝑘+1}𝑓(𝑥)²|𝑥|^𝛼𝑓(𝑥𝑑𝑥≤𝐶)²|𝑥|^𝛼𝑑𝑥. (31) Thus, we have 𝐾_𝜆^{𝛿,𝑗,𝑙}𝑓₂(𝑥)²𝑑𝑥≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀𝑓(𝑥)²|𝑥|^𝛼𝑑𝑥. (32) Choosing 𝑀>1, we get 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓₂(𝑥)²𝑑𝑥≤𝐶_𝛼2^−𝑙2^−𝑗𝑀𝛿𝑓(𝑥)²|𝑥|^𝛼𝑑𝑥. (33)
On the other hand, 𝐸_𝑡 is self-adjoint. So, ⟨𝐸_𝑡=𝑒𝑓,𝑔⟩_𝑡𝑓𝑦(𝑥)=𝑓𝑦𝑒𝑥,𝑦𝑑𝑦𝑔𝑑𝑥_𝑡𝑥,𝑦𝑔(𝑥)𝑑𝑥𝑑𝑦=⟨𝑓,𝐸_𝑡=𝑦𝑔⟩𝑓𝑒_𝑡𝑦,𝑥𝑔(𝑥)𝑑𝑥𝑑𝑦. (34) Hence, 𝑒_𝑡𝑥,𝑦=𝑒_𝑡𝑦,𝑥. (35) With 𝐻_𝑗𝑥,𝑦=𝐻_𝑗𝑦,𝑥, (36) we get 𝑒^𝑗_𝑡𝑥,𝑦=𝑒_𝑡𝑥,𝑦𝐻_𝑗𝑥,𝑦=𝑒_𝑡𝑦,𝑥𝐻_𝑗𝑦,𝑥=𝑒^𝑗_𝑡𝑦,𝑥, (37) and it implies that 𝐾_𝜆^{𝛿,𝑙,𝑗} is also self-adjoint; that is, ⟨𝐸^𝑗_𝑡𝑓,𝑔⟩=⟨𝑓,𝐸^𝑗_𝑡𝑔⟩. (38) Therefore, by duality, 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓₂(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼2^−𝑙2^−𝑗𝑀𝛿𝑓(𝑥)²𝑑𝑥≤𝐶_𝛼2^−𝑙2^{𝑗(𝛼−𝑀)}𝛿𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼. (39) Taking 𝑀>𝛼+1, we can establish the inequality 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓₂(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼2^−𝑙2^−𝑗𝛿𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼. (40)
Nextly, we consider 𝑖=1. By the definition of 𝐾_𝜆^{𝛿,𝑙,𝑗} and 𝑓₁, we see that the kernel of 𝐾_𝜆^{𝛿,𝑙,𝑗} is supported in 𝑥,𝑦∶2^𝑗−1≤𝑥−𝑦≤2^𝑗 (41) and 𝑓(𝑦) is supported in 𝑦≤10≤4. (42) So, the support of 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓(𝑥) is contained in 𝑥∈ℝ^𝑛∶2^𝑗+14≥|𝑥|≥2^𝑗−1−14≥2⁰⁻¹−14=14. (43) With Lemma 4, we have 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓₁(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓₁(𝑥)²4^𝛼𝐾𝑑𝑥≤𝐶_𝜆^{𝛿,𝑙,𝑗}𝑓₁(𝑥)²𝑑𝑥≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀𝑓₁²₂≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀𝑓²₂≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀2^𝑗𝛼𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶2^−𝑙2^−𝑗𝛿𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼, (44) where 𝑀>𝛼+1.
At last, we turn to the case of 𝑖=3. Similar to the aforementioned, we have 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓₃(𝑥)²𝑑𝑥≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀𝑓₃²₂≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀_{1/4≤|𝑥|≤𝐶}𝑓(𝑥)²𝑑𝑥≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀14^−𝛼_{1/4≤|𝑥|≤𝐶}𝑓(𝑥)²14^𝛼𝑑𝑥≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀4^𝛼_{1/4≤|𝑥|≤𝐶}𝑓(𝑥)²|𝑥|^𝛼𝑑𝑥≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀𝑓(𝑥)²|𝑥|^𝛼𝑑𝑥. (45) By duality again, we see 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓₃(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀𝑓(𝑥)²𝑑𝑥≤𝐶2^−𝑙2^−𝑗𝑀𝛿^𝑀2^𝑗𝛼𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶2^−𝑙2^{−𝑗(𝑀−𝛼)}𝛿^𝑀𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼. (46) Through we choose 𝑀>𝛼+1, it is not difficult to get 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓₃(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶2^−𝑙2^−𝑗𝛿𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼. (47) Combining (28), (40),(44) with (47), when 𝑓 is supported in {|𝑥|≤𝐶2^𝑗}, we come to the conclusion 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶2^−𝑙2^−𝑗𝛿𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼. (48)
Now, we hope to establish (48) for all 𝑓∈𝐿²(ℝ^𝑛). Decompose 𝑓=∑_𝑖∈ℤ^𝑛𝜒_𝑄𝑖𝑓=∑_𝑖∈ℤ^𝑛𝑓_𝑖, where {𝑄_𝑖} are disjoint cubes of common side 10⋅2^𝑗 with 𝑄₀ centered at 0. Since {𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓_𝑖(𝑥)} have essentially disjoint supports, it suffices to prove (48) for every 𝑓_𝑖. When 𝑖=0, we have proved it. If 𝑖>1, then 10<2+𝑙10⋅2^𝑗<|𝑥|^−𝛼<12+𝑙+110⋅2^𝑗(𝑙≥0). (49) Therefore, we only need to confirm 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓(𝑥)²𝑑𝑥≤𝐶_𝛼2^−𝑙2^−𝑗𝛿𝑓(𝑥)²𝑑𝑥. (50) In fact, it follows from Lemma 4 that 𝐾_𝜆^{𝛿,𝑙,𝑗}𝑓(𝑥)²𝑑𝑥≤𝐶2^{−2𝑀(𝑗+𝑙)}𝛿^2𝑀𝑓(𝑥)²𝑑𝑥≤𝐶2^−𝑙2^−𝑗𝛿𝑓(𝑥)²𝑑𝑥. (51)
At present, we complete the proof of Lemma 5.

Lemma 6. For 𝛿>0, 𝑘∈ℤ, and 0≤𝛼<𝑛, one gets _ℝ𝑛^2𝑘_2𝑘−1𝐾^𝛿_𝜆𝑓(𝑥)²𝑑𝜆𝜆𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼𝛿𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼. (52)

Proof. Applying Minkowski and Cauchy-Schwartz’s inequalities into the left hand side of (52), we get _ℝ𝑛^2𝑘_2𝑘−1𝐾^𝛿_𝜆𝑓(𝑥)𝑑𝜆𝜆²|𝑥|^−𝛼𝑑𝑥≤𝐶^2𝑘_{2𝑘−1ℝ𝑛}𝐾^𝛿_𝜆𝑓(𝑥)²|𝑥|^−𝛼𝑑𝑥^1/2𝑑𝜆𝜆²≤𝐶^2𝑘_{2𝑘−1ℝ𝑛}𝐾^𝛿_𝜆𝑓(𝑥)²|𝑥|^−𝛼𝑑𝑥𝑑𝜆𝜆^1/2×^2𝑘_2𝑘−1𝑑𝜆𝜆^1/22≤𝐶^2𝑘_{2𝑘−1ℝ𝑛}𝐾^𝛿_𝜆𝑓(𝑥)²|𝑥|^−𝛼𝑑𝑥𝑑𝜆𝜆. (53) Now, it suffices to prove that _ℝ𝑛𝐾^𝛿_𝜆𝑓(𝑥)²|𝑥|^−𝛼𝑑𝑥≤𝐶_𝛼𝛿_ℝ𝑛𝑓(𝑥)²|𝑥|^−𝛼𝑑𝑥, (54) where 𝐶_𝛼𝛿 is uniform in 2^𝑘−1≤𝜆≤2^𝑘. For 0≤𝛼<𝑛, it is equivalent to _ℝ𝑛𝐾^𝛿_𝜆𝑔(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼𝛿_ℝ𝑛𝑔(𝑥)²𝑑𝑥|𝑥|^𝛼. (55) It is just as the result of Lemma 5.

Now, we come back to the proof of Theorem 2.

Proof. As in [3], we can decompose 𝑡1−𝜆^𝛽₊=^∞_𝑘=02^−𝑘𝛽𝑚^2−𝑘𝑡𝜆. (56) Thus, 𝑆^𝛽_∗𝑓(𝑥)≤^∞_𝑘=02^−𝑘𝛽𝐾^2−𝑘_∗𝑓(𝑥). (57) Consequently, we consider _ℝ𝑛𝐾^𝛿_∗𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼𝛿_ℝ𝑛𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼. (58) Let 𝐺^𝛿𝑓(𝑥)=^∞₀𝐾^𝛿_𝜆𝑓(𝑥)²𝑑𝜆𝜆^1/2 (59) and 𝐺^𝛿 be defined in the same way but using instead of 𝑚^𝛿 the function 𝑚^𝛿(𝜆)=𝛿𝜆𝑑𝑚^𝛿(𝜆)𝜆 (60) which satisfies the same estimates as 𝑚^𝛿. Then, by the fundamental theorem in calculus and Hölder’s inequality, we have 𝐾^𝛿_∗𝑓(𝑥)²≤^∞₀2𝐾^𝛿_𝜆𝑓(𝑥)𝑑𝐾^𝛿_𝜆𝑓(𝑥)𝑑𝜆𝑑𝜆≤2𝛿^−1∞₀𝐾^𝛿_𝜆𝑓(𝑥)𝜆^1/2𝜆𝛿𝑑𝐾^𝛿_𝜆𝑓(𝑥)/𝑑𝜆𝜆^1/2𝑑𝜆≤2𝛿^−1∞₀𝐾^𝛿_𝜆𝑓(𝑥)²𝑑𝜆𝜆^1/2×^∞₀𝜆𝛿𝑑𝐾^𝛿_𝜆𝑓(𝑥)𝑑𝜆²𝑑𝜆𝜆^1/2=2𝛿⁻¹𝐺^𝛿𝑓(𝑥)𝐺𝑓(𝑥). (61)
Take a Schwartz function 𝜓 such that 𝜓(0)=0 and 𝜓(𝑡)=1 if 1/2≤𝑡≤2 and 𝜓_𝑘(𝑡)=𝜓(2^𝑘𝑡). Then, when 2^𝑘−1≤𝜆≤2^𝑘, we have 𝐾^𝛿_𝜆𝑓(𝑥)=𝑚^𝛿(𝜆𝑡)𝜓_𝑘(𝑡)𝑑𝐸_𝑡𝑓(𝑥). (62) Using Lemma 6, we get _ℝ𝑛^2𝑘_2𝑘−1𝐾^𝛿_𝜆𝑓(𝑥)²𝑑𝜆𝜆𝑑𝑥|𝑥|^𝛼=_ℝ𝑛^2𝑘_2𝑘−1𝑚^𝛿(𝜆𝑡)𝜓_𝑘(𝑡)𝑑𝐸_𝑡𝑓(𝑥)²𝑑𝜆𝜆𝑑𝑥|𝑥|^𝛼=_ℝ𝑛^2𝑘_2𝑘−1𝑚^𝛿(𝜆𝑡)𝑑𝐸_𝑡𝜓_𝑘𝐻_𝑏𝑓(𝑥)²𝑑𝜆𝜆𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼𝛿𝜓_𝑘𝐻_𝑏𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼=𝐶_𝛼𝛿𝜓_𝑘(𝑡)𝑑𝐸_𝑡𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼. (63)
From Theorem 3.1 in page 411 of [10] and the density of 𝐿^∞_𝑐 in 𝐿^𝑝(𝑤), we induce that ^∞_𝑘=−∞𝜓_𝑘(𝑡)𝑑𝐸_𝑡𝑓(𝑥)^21/2 (64) is bounded in 𝐿²(𝑑𝑥/|𝑥|^𝛼). As a result, 𝐺^𝛿𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼=^∞₀𝐾^𝛿_𝜆𝑓(𝑥)²𝑑𝜆𝜆𝑑𝑥|𝑥|^𝛼=^∞_𝑘=−∞^2𝑘_2𝑘−1𝐾^𝛿_𝜆𝑓(𝑥)²𝑑𝜆𝜆𝑑𝑥|𝑥|^𝛼=^∞_𝑘=−∞^2𝑘_2𝑘−1𝐾^𝛿_𝜆𝑓(𝑥)²𝑑𝜆𝜆𝑑𝑥|𝑥|^𝛼≤^∞_𝑘=−∞𝐶_𝛼𝛿𝜓_𝑘(𝑡)𝑑𝐸_𝑡𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼𝛿^∞_𝑘=−∞𝜓_𝑘(𝑡)𝑑𝐸_𝑡𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼𝛿𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼. (65) At last, with (61) and Hölder’s inequality, we come to the result that 𝐾^𝛿_∗𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼≤𝐶_𝛼𝑓(𝑥)²𝑑𝑥|𝑥|^𝛼. (66)

Acknowledgments

The authors deeply thank the referee for reviewing their paper. The first author was supported by the NSF of China (71201051, 71031004, and 71073047), NSF of Hunan Province (2012RS4028) and Postdoctoral Science Foundation of China (2012M521513). The second author was supported by the NSF of China (10771054) and NSF of Hunan Province (09JJ5002).