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Maximal estimates for Bilinear Bochner-Riesz means

Jotsaroop Kaur and Saurabh Shrivastava Department of Mathematics
Indian Institute Science Education and Research
Mohali, India
[email protected] Department of Mathematics
Indian Institute Science Education and Research Bhopal
Bhopal-462066, India
[email protected]
Abstract.

We establish improved and sharp LpL^{p} estimates for the maximal bilinear Bochner-Riesz means in all dimensions n1n\geq 1. This work extends the results proved by Jeong and Lee [11]. We also recover the known results for the bilinear Bochner-Riesz means. The method of proof involves a new decomposition of the bilinear Bochner-Riesz multiplier (1|ξ|2|η|2)+α(1-|\xi|^{2}-|\eta|^{2})_{+}^{\alpha} and delicate analysis in proving LpL^{p} estimates for frequency localized square functions.

Key words and phrases:
Bilinear Bochner-Riesz means, Square function, Bochner Riesz means
2010 Mathematics Subject Classification:
Primary 42A85, 42B15, Secondary 42B25

1. Introduction

The convergence of the partial sum operators corresponding to two fold product of nn- dimensional Fourier series

|m|2+|k|2R2f^(m)g^(k)e2πi(k+m)x\displaystyle\sum_{|m|^{2}+|k|^{2}\leq R^{2}}\hat{f}(m)\hat{g}(k)e^{2\pi i(k+m)\cdot x}

to f(x)g(x)f(x)g(x) as RR\rightarrow\infty is a classical problem in Fourier analysis. Here ff and gg are 11-periodic functions with respect to each variable belonging to Lebesgue spaces Lp1([0,1]n)L^{p_{1}}([0,1]^{n}) and Lp2([0,1]n)L^{p_{2}}([0,1]^{n}) respectively, where 1p1,p21\leq p_{1},p_{2}\leq\infty.

This motivates us to consider the following more general partial sum operators

(1) |m|2+|k|2R2(1|m|2+|k|2R2)αf^(m)g^(k)e2πi(k+m)x,whereα0.\displaystyle\sum_{|m|^{2}+|k|^{2}\leq R^{2}}\left(1-\frac{|m|^{2}+|k|^{2}}{R^{2}}\right)^{\alpha}\hat{f}(m)\hat{g}(k)e^{2\pi i(k+m)\cdot x},~{}~{}\text{where}~{}~{}\alpha\geq 0.

The partial sum operators are referred to as the bilinear Bochner-Riesz means of order α\alpha.

The transference principle for bilinear multipliers establishes that the study of LpL^{p} boundedness of the bilinear Bochner-Riesz means is equivalent to studying the corresponding estimates for the bilinear Bochner-Riesz operator of index α0\alpha\geq 0 defined on n{\mathbb{R}}^{n} by

Rα(f,g)(x)=nn(1|ξ|2+|η|2R2)+αf^(ξ)g^(η)e2πix.(ξ+η)𝑑ξ𝑑η,\mathcal{B}^{\alpha}_{R}(f,g)(x)=\int_{{\mathbb{R}}^{n}}\int_{{\mathbb{R}}^{n}}\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)^{\alpha}_{+}\hat{f}(\xi)\hat{g}(\eta)e^{2\pi ix.(\xi+\eta)}d\xi d\eta,

where f,g𝒮(n)f,g\in\mathcal{S}({\mathbb{R}}^{n}) and f^\hat{f} denotes the Fourier transform of ff given by f^(ξ)=nf(x)e2πix.ξ𝑑x\hat{f}(\xi)=\int_{{\mathbb{R}}^{n}}f(x)e^{-2\pi ix.\xi}dx. When R=1R=1, we simply use the notation α\mathcal{B}^{\alpha}.

In this paper we are interested in investigating the a.e. convergence of Rα(f,g)\mathcal{B}^{\alpha}_{R}(f,g) as RR\rightarrow\infty for (f,g)(f,g) in Lp1(n)×Lp2(n)L^{p_{1}}({\mathbb{R}}^{n})\times L^{p_{2}}({\mathbb{R}}^{n}) for 1p1,p21\leq p_{1},p_{2}\leq\infty. In view of the general principle due to Stein we know that in order to study the a.e. convergence of Rα(f,g)\mathcal{B}^{\alpha}_{R}(f,g) it is enough to prove suitable LpL^{p} estimates for the corresponding maximal function

α(f,g)(x)=:supR>0|Rα(f,g)(x)|.\mathcal{B}^{\alpha}_{*}(f,g)(x)=:\sup_{R>0}|\mathcal{B}^{\alpha}_{R}(f,g)(x)|.

More precisely, for f,g𝒮(n)f,g\in\mathcal{S}({\mathbb{R}}^{n}) we are concerned with the following estimates

(2) α(f,g)Lp(n)fLp1(n)gLp2(n).\displaystyle\|\mathcal{B}_{*}^{\alpha}(f,g)\|_{L^{p}({\mathbb{R}}^{n})}\lesssim\|f\|_{L^{p_{1}}({\mathbb{R}}^{n})}\|g\|_{L^{p_{2}}({\mathbb{R}}^{n})}.

Also, the corresponding estimate for the bilinear Bochner-Riesz mean

(3) α(f,g)Lp(n)fLp1(n)gLp2(n).\displaystyle\|\mathcal{B}^{\alpha}(f,g)\|_{L^{p}({\mathbb{R}}^{n})}\lesssim\|f\|_{L^{p_{1}}({\mathbb{R}}^{n})}\|g\|_{L^{p_{2}}({\mathbb{R}}^{n})}.

Here we shall always assume that the exponents p1,p2,pp_{1},p_{2},p satisfy 1p1,p21\leq p_{1},p_{2}\leq\infty and 1p1+1p2=1p\frac{1}{p_{1}}+\frac{1}{p_{2}}=\frac{1}{p}. The notation ABA\lesssim B in the estimates above means that there is an implicit constant C>0C>0 such that ACB.A\leq CB. The constant CC may depend on α,n,p1\alpha,n,p_{1} and p2.p_{2}. We will not keep track of such constants appearing in various inequalities unless we need them in order for the proof to work. In some inequalities in the later part of paper we shall also use the notation Aa,bBA\lesssim_{a,b}B to emphasize the dependence of the implicit constant on the parameters aa and bb.

If the estimate (2) (or (3)) holds we say that the operator α\mathcal{B}^{\alpha}_{*} (or α\mathcal{B}^{\alpha}) is bounded from Lp1(n)×Lp2(n)Lp(n)L^{p_{1}}({\mathbb{R}}^{n})\times L^{p_{2}}({\mathbb{R}}^{n})\rightarrow L^{p}({\mathbb{R}}^{n}). Observe that the standard dilation arguments tell us that α:Lp1(n)×Lp2(n)Lp(n)\mathcal{B}^{\alpha}:L^{p_{1}}({\mathbb{R}}^{n})\times L^{p_{2}}({\mathbb{R}}^{n})\rightarrow L^{p}({\mathbb{R}}^{n}) if, and only if Rα:Lp1(n)×Lp2(n)Lp(n)\mathcal{B}^{\alpha}_{R}:L^{p_{1}}({\mathbb{R}}^{n})\times L^{p_{2}}({\mathbb{R}}^{n})\rightarrow L^{p}({\mathbb{R}}^{n}) where R>0R>0 and 1p1+1p2=1p\frac{1}{p_{1}}+\frac{1}{p_{2}}=\frac{1}{p}.

The question of Lp1(n)×Lp2(n)Lp(n)L^{p_{1}}({\mathbb{R}}^{n})\times L^{p_{2}}({\mathbb{R}}^{n})\rightarrow L^{p}({\mathbb{R}}^{n}) boundedness for the operators α\mathcal{B}^{\alpha} and α\mathcal{B}^{\alpha}_{*} is referred to as the bilinear Bochner-Riesz probem in general. This is one of the outstanding open problems in the theory of bilinear Fourier multipliers. In this paper we make significant progress on the bilinear Bochner-Riesz problem for α\mathcal{B}^{\alpha}_{*}. We establish new and sharp results for the maximal function α\mathcal{B}^{\alpha}_{*}.

Note that the operator Rα\mathcal{B}^{\alpha}_{R} takes the following form in the spatial coordinates.

Rα(f,g)(x):=nnKRα(y,z)f(xy)g(xz)𝑑y𝑑z,\mathcal{B}^{\alpha}_{R}(f,g)(x):=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}K^{\alpha}_{R}(y,z)f(x-y)g(x-z)dydz,

where

KRα(y,z)\displaystyle K^{\alpha}_{R}(y,z) =\displaystyle= nn(1|ξ|2+|η|2R2)+αe2πι(yξ+zη)𝑑ξ𝑑η\displaystyle\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)^{\alpha}_{+}e^{2\pi\iota(y\cdot\xi+z\cdot\eta)}d\xi d\eta
=\displaystyle= cn+αR2nJα+n(2π|(Ry,Rz)|)|(Ry,Rz)|α+n,y,zn.\displaystyle c_{n+\alpha}R^{2n}\frac{J_{\alpha+n}(2\pi|(Ry,Rz)|)}{|(Ry,Rz)|^{\alpha+n}},~{}y,z\in{\mathbb{R}}^{n}.

Here Jα+nJ_{\alpha+n} denotes the standard Bessel function of order α+n\alpha+n.

Using the asymptotic estimates of Bessel functions it is easily verified that for α>n12\alpha>n-\frac{1}{2}, the kernel KRαK^{\alpha}_{R} is an integrable function with a uniform constant with respect to RR. Consequently, the estimates (2) and (3) follow for all 1<p1,p21<p_{1},p_{2}\leq\infty and α>n12\alpha>n-\frac{1}{2}, see [1, 13] for detail. The index α=n12\alpha=n-\frac{1}{2} is referred to as the critical index for the bilinear Bochner-Riesz problem. Therefore, the bilinear Bochner-Riesz problem concerning estimates (2) and (3) needs to be investigated when 0αn120\leq\alpha\leq n-\frac{1}{2}.

When α=0\alpha=0, the operator α\mathcal{B}^{\alpha} is denoted by \mathcal{B} and it is referred to as the bilinear ball multiplier operator. In [8] Grafakos and Li proved the estimate (3) in dimension n=1n=1 when α=0\alpha=0 and 2p1,p2,p<2\leq p_{1},p_{2},p^{\prime}<\infty satisfying the Hölder relation 1p1+1p2=1p.\frac{1}{p_{1}}+\frac{1}{p_{2}}=\frac{1}{p}. Here pp^{\prime} denotes the conjugate index to pp given by 1p+1p=1\frac{1}{p}+\frac{1}{p^{\prime}}=1. As mentioned previously in this paper we shall always assume that the exponents p1,p2p_{1},p_{2} and pp satisfy the Hölder relation unless specified otherwise. We will denote such a triplet by (p1,p2,p)(p_{1},p_{2},p). In [8] authors also proved that \mathcal{B} fails to satisfy (3) for triplets (2,2,1),(2,,2)(2,2,1),(2,\infty,2) and (,2,2)(\infty,2,2) when α=0\alpha=0. We do not know of any positive result for the bilinear ball multiplier in dimension n=1n=1 for exponents lying outside the range mentioned as above. Later, Diestel and Grafakos [4] proved a negative result for boundedness of the operator \mathcal{B} when n2n\geq 2. They exploited Fefferman’s argument for the ball multiplier problem [6] to show that in dimension n2n\geq 2 the estimate (3) cannot hold true for α=0\alpha=0 if exactly one of p1,p2,pp_{1},p_{2},p^{\prime} is less than 2.2.

The study of LpL^{p} boundedness properties of the operator α\mathcal{B}^{\alpha} for α>0\alpha>0 was initiated in [1] and they proved several positive and negative results for the boundedness of the bilinear Bochner-Riesz means. In particular, they obtained almost complete result in dimension n=1n=1 proving that (3) holds for α>0\alpha>0 for all 1p1,p2,p1\leq p_{1},p_{2},p\leq\infty, see Theorem 4.1 in [1] for detail. This range of exponents is referred to as the Banach triangle in the theory of bilinear multipliers. Observe that the exponent pp has its natural range as 1/2p1/2\leq p\leq\infty when we allow p1p_{1} and p2p_{2} in the range 1p1,p2.1\leq p_{1},p_{2}\leq\infty. Later, in [18] authors extended these results, specifically in the non-Banach traingle (i.e. when p<1p<1) to an improved range of p1,p2,pp_{1},p_{2},p and α\alpha.

Recently, Jeong, Lee, and Vargas [12] further improved the range of exponents when p1,p22p_{1},p_{2}\geq 2 for the estimate (3). They introduced a new approach to address the bilinear Bochner-Riesz problem in dimension n2n\geq 2. Their approach relies on a new decompostion of the bilinear Bochner-Riesz operator into a product of square functions in the pointwise sense. This allows them to obtain new results for the estimate (3) when n2n\geq 2 and p1,p22p_{1},p_{2}\geq 2, see [Section 3, [12]] for detail. In particular, in [12] authors obtained new results by improving the lower bounds on the index α\alpha. They proved optimal result for the estimate (3) when p1=p2=2p_{1}=p_{2}=2 and α>0\alpha>0 for all n2n\geq 2, see Corollary 1.3 in  [12] (also see Proposition 4.10 in [1]). However, the results in other cases are not optimal. We would like to remark here that their method requires the condition that p1,p22p_{1},p_{2}\geq 2. We will see that with our approach we can recover these results. Moreover, our method allows us to obtain LpL^{p} estimates for α\mathcal{B}^{\alpha} when the exponents are less than 22. In particular, the estimate (3) for exponents in non-Banach triangle are new in dimension n=1n=1, see Theorem 2.2. However, for n2n\geq 2 we do not get improved results for the estimate (3) when p1,p2<2p_{1},p_{2}<2 as compared to already known results in [1, 18].

Now, coming back to the main theme of the paper let us discuss the known results for the bilinear Bochner-Riesz maximal function concerning the estimate (2). The study of LpL^{p} boundedness properties of α\mathcal{B}^{\alpha}_{*} has been initiated in [9, 10]. In [9] authors proved boundedness of maximal bilinear multiplier operator with a certain decay condition on the multiplier symbol. As an application of their result they proved estimate (2) for p1=p2=2p_{1}=p_{2}=2 and α>2n+34\alpha>\frac{2n+3}{4}. Note that this result is interesting only if 2n+34\frac{2n+3}{4} is smaller or equal to the critical index α=n12\alpha=n-\frac{1}{2}. Very recently, Jeong and Lee [11] have obtained improved results in this direction generalizing the previously known results significantly. In order to prove LpL^{p} estimates in [11] authors have exploited the decomposition of the bilinear Bochner-Riesz operators into square functions as carried out in [12]. In particular, they proved that the estimate (2) holds for α>1\alpha>1 for p1=p2=2p_{1}=p_{2}=2. Note that the condition α>1\alpha>1 is far from being sharp.

In this paper our main goal is to improve the range of p1,p2,pp_{1},p_{2},p and α\alpha in the maximal estimate (2). We establish a new decomposition of the bilinear Bochner-Riesz operators in terms of product of square functions which are closely related with the classical Bochner-Riesz square functions (see section 3 for precise detail). As a result, for each piece of the decomposition parametrised by jj, the corresponding bilinear operator can be written as superposition of product of localised linear operators Bj,βR,tfB_{j,\beta}^{R,t}f and BtδgB_{t}^{\delta}g. For more details please refer to (10). The operator Bj,βR,tB_{j,\beta}^{R,t} turns out to be a slight perturbation of the localised Bochner Riesz operator. A further careful analysis allows us to reduce the LpL^{p} boundedness of the corresponding square function to the known case of the localised square function of Bochner Riesz operator as defined in [15]. This allows us to improve the result of E. Jeong and S. Lee [11] for the boundedness of maximal Bilinear Bochner Riesz function. See (18) and equation 4.24.2 in Proposition 4.14.1 in [11] for comparision.

This approach allows us to establish new and sharp results for the estimates (2) and (3) with an improved range of exponents p1,p2,pp_{1},p_{2},p and lower bound on α\alpha. Moreover, it is valid uniformly in all dimensions n1n\geq 1. In particular, we obtain optimal results for p1=p2=2p_{1}=p_{2}=2 when n1n\geq 1, see Theorems 2.1 and 2.2. We remark that the maximal estimate  (2) in dimension n=1n=1 is not known. Our approach allows us to address this case and we obtain new and sharp results for the maximal estimate (2) in dimension n=1n=1, see Theorem 2.2.

Organization of the paper

In Section 2 we state the main results and discuss the methodology of our proofs. In this section we also make a comparision of our results with the known results. In Section 3 we present the proof of decomposition of bilinear Bochner-Riesz operator. In Section 4 we develop a discussion about Bochner-Riesz square function and some of the results stated in this section will be used in proving the main results of this paper. Section 5 is devoted to proving Theorem 2.1 concerning the boundedness of the bilinear Bochner-Riesz maximal function in dimension n2n\geq 2. Finally, in Section 6 we present the proof of Theorem 2.2 which addresses the LpL^{p} boundedness of the bilinear Bochner-Riesz maximal function in dimension n=1n=1.

2. Main results and methodology

We first set some notation that are required in order to describe our results. For 1p1\leq p\leq\infty and n1n\geq 1 denote α(p)=max{n|1p12|12,0}\alpha(p)=\max\left\{n|\frac{1}{p}-\frac{1}{2}|-\frac{1}{2},0\right\}. Note that for n=1n=1 we have α(p)=0\alpha(p)=0 for all 1p1\leq p\leq\infty. For n2,n\geq 2, define p0(n)=2+124n6kp_{0}(n)=2+\frac{12}{4n-6-k} where nkmod3,k=0,1,2n\equiv k~{}\text{mod}~{}3,k=0,1,2. Denote 𝔭n=min{p0(n),2(n+2)n}.\mathfrak{p}_{n}=\text{min}~{}\left\{p_{0}(n),\frac{2(n+2)}{n}\right\}. For n1n\geq 1 and 1p1,p21\leq p_{1},p_{2}\leq\infty we define

α(p1,p2)={α(p1)+α(p2)when𝔭np1,p2;α(p1)+(12p2112(𝔭n)1)α(𝔭n)when𝔭np1and2p2<𝔭n;(12p1112(𝔭n)1)α(𝔭n)+α(p2)when2p1<𝔭nand𝔭np2;(22p112p2112(𝔭n)1)α(𝔭n)when2p1,p2<𝔭n.\alpha_{*}(p_{1},p_{2})=\begin{cases}\alpha(p_{1})+\alpha(p_{2})&\textup{when}~{}~{}\mathfrak{p}_{n}\leq p_{1},p_{2}\leq\infty;\\ \\ \alpha(p_{1})+\left(\frac{1-2p_{2}^{-1}}{1-2(\mathfrak{p}_{n})^{-1}}\right)\alpha(\mathfrak{p}_{n})&\textup{when}~{}~{}\mathfrak{p}_{n}\leq p_{1}\leq\infty~{}\text{and}~{}2\leq p_{2}<\mathfrak{p}_{n};\\ \\ \left(\frac{1-2p_{1}^{-1}}{1-2(\mathfrak{p}_{n})^{-1}}\right)\alpha(\mathfrak{p}_{n})+\alpha(p_{2})&\textup{when}~{}~{}2\leq p_{1}<\mathfrak{p}_{n}~{}\text{and}~{}\mathfrak{p}_{n}\leq p_{2}\leq\infty;\\ \\ \left(\frac{2-2p_{1}^{-1}-2p_{2}^{-1}}{1-2(\mathfrak{p}_{n})^{-1}}\right)\alpha(\mathfrak{p}_{n})&\textup{when}~{}~{}2\leq p_{1},p_{2}<\mathfrak{p}_{n}.\end{cases}

The following theorems concerning the LpL^{p} boundedness of the bilinear Bochner-Riesz maximal function α\mathcal{B}_{*}^{\alpha} are the main results of this paper. We state the results for n=1n=1 and n2n\geq 2 separately.

Theorem 2.1.

(The case of n2n\geq 2) Let n2n\geq 2 and (p1,p2,p)(p_{1},p_{2},p) be a triplet of exponents satisfying p1,p22p_{1},p_{2}\geq 2 and 1p=1p1+1p2.\frac{1}{p}=\frac{1}{p_{1}}+\frac{1}{p_{2}}. Then for α>α(p1,p2)\alpha>\alpha_{*}(p_{1},p_{2}) the estimate  (2) holds with the implicit constant depending only on α,p1,p2\alpha,p_{1},p_{2} and nn.

Theorem 2.2.

(The case of n=1n=1.) Let α>0\alpha>0 and n=1n=1. Let 1<p1,p2<1<p_{1},p_{2}<\infty and 1p1+1p2=1p\frac{1}{p_{1}}+\frac{1}{p_{2}}=\frac{1}{p}. Then the estimate (2) holds for each of the following cases.

  1. (1)

    p1,p22p_{1},p_{2}\geq 2 and α>0\alpha>0.

  2. (2)

    1<p1<2,p221<p_{1}<2,p_{2}\geq 2 and α>1p112\alpha>\frac{1}{p_{1}}-\frac{1}{2}.

  3. (3)

    1<p2<2,p121<p_{2}<2,p_{1}\geq 2 and α>1p212\alpha>\frac{1}{p_{2}}-\frac{1}{2}.

  4. (4)

    1<p1,p2<21<p_{1},p_{2}<2 and α>1p1\alpha>\frac{1}{p}-1.

Remark 2.3.

Note that in case (4) of Theorem 2.2 we have p<1p<1. When 2/3<p<12/3<p<1 we get that 1p1<1/2\frac{1}{p}-1<1/2 which is interesting as α=1/2\alpha=1/2 is the critical index for the bilinear Bochner-Riesz problem when n=1n=1.

1/41/41/21/21/p1/p1/21/2113/23/2α\alphaα=14p\alpha=1-\frac{4}{p}α=14p\alpha=1-\frac{4}{p}
Figure 1. Range of α\alpha and pp for Lp(2)×Lp(2)Lp/2(2)L^{p}({\mathbb{R}}^{2})\times L^{p}({\mathbb{R}}^{2})\rightarrow L^{p/2}({\mathbb{R}}^{2}) boundedness of α\mathcal{B}_{*}^{\alpha}. The region with dark gray color represent the range obtained in [11].

.

2.1. Methodology of proof and comparision

We decompose the bilinear multiplier (1|ξ|2|η|2)+α(1-|\xi|^{2}-|\eta|^{2})_{+}^{\alpha} along the ξ\xi and η\eta axes. Note that due to symmetry of the multiplier symbol in ξ\xi and η\eta, we need to perform only one of the two decompositions as the other case can be dealt with in a similar fashion. This idea is motivated from the observation that the standard decomposition of the bilinear multipliers (1|ξ|2|η|2)+α(1-|\xi|^{2}-|\eta|^{2})_{+}^{\alpha} into smooth symbols supported in thin annular regions near the singularity |ξ|2+|η|2=1|\xi|^{2}+|\eta|^{2}=1 requires further decomposition of the smooth symbols into frequency localized multiplier operators using the idea of Littlewood-Paley decompositions. The standard Littlewood-Paley decomposition of these smooth bilinear multipliers poses difficulty when we try to prove estimates for frequency localised operators with supports along the coordinate axes. Due to the curvature of the sphere these Littlewood-Paley Fourier projections overlap along the coordinate axes and it becomes difficult to have a good control on the operators arising in this process. Therefore, in order to overcome this difficulty we decompose the bilinear multiplier (1|ξ|2|η|2)+α(1-|\xi|^{2}-|\eta|^{2})_{+}^{\alpha} into smooth functions supported in annular regions with respect to variables ξ\xi and η\eta separately. This approach along with an identity (6) from [21] [page 278] yields a very useful decomposition of the bilinear Bochner-Riesz multiplier (1|ξ|2|η|2)+α(1-|\xi|^{2}-|\eta|^{2})_{+}^{\alpha} into a product of frequency localized Bochner-Riesz square functions, see estimate (10) for precise detail. This approach has a big advantage in obtaining LpL^{p} estimates for the bilinear Bochner-Riesz means and its maximal function. When p1,p22p_{1},p_{2}\geq 2 this approach allows us to improve the known range of α\alpha significantly for which the estimate (2) holds. For example, Jeong and Lee [[11], Theorem 1.2] has the lower bound on α\alpha as α>α(p1)+α(p2)+1\alpha>\alpha(p_{1})+\alpha(p_{2})+1 when p1,p2pn(s)p_{1},p_{2}\geq p_{n}(s) or p1=p2=2p_{1}=p_{2}=2 whereas with our approach we reduce the lower bound to α>α(p1)+α(p2)\alpha>\alpha(p_{1})+\alpha(p_{2}). See Figure  1 for a comparision. The region shaded with dark gray color in the figure above represents the range of pp versus the index α\alpha for the Lp(2)×Lp(2)Lp/2(2)L^{p}({\mathbb{R}}^{2})\times L^{p}({\mathbb{R}}^{2})\rightarrow L^{p/2}({\mathbb{R}}^{2}) of the maximal function α\mathcal{B}_{*}^{\alpha} from [11]. We extend the estimate to the region shaded with light gray color. In the process, we obtain optimal results for boundedness of bilinear Bochner-Riesz maximal function when p1=p2=2p_{1}=p_{2}=2.

When n=1n=1 we get LpL^{p} estimates for the bilinear Bochner-Riesz maximal function for a wide range of exponents including the cases when p1p_{1} or p2p_{2} is less than 22. Indeed our results for exponents outside the Banach triangle (i.e. 1p1,p2,p1\leq p_{1},p_{2},p\leq\infty) are new even in the case of bilinear Bochner-Riesz operator. Finally, we would like to emphasise that our method is applicable uniformly to all dimensions n1n\geq 1 and provides us with simplified proofs of the existing results. We would like to remark here that this method works best when p1,p22p_{1},p_{2}\geq 2. When either of p1p_{1} and p2p_{2} is less than 22, we need a different approach in order to improve the range of α\alpha for the boundedness of Bilinear Bochner Riesz operator and its associated maximal function.

Refer to caption
Figure 2. This figure represents the standard decomposition of the Bochner- Riesz multiplier in dimension n=1n=1 in terms of smooth functions supported on annuli of width comparable to the distance from |ξ|2+|η|2=1|\xi|^{2}+|\eta|^{2}=1.
Refer to caption
Figure 3. This figure represents the decomposition of the Bochner-Riesz multiplier in dimension n=1n=1. This involves decomposing along the ξ\xi and η\eta axes.

3. Decomposition of the bilinear Bochner-Riesz multiplier

Let N20nN\geq 20n and II be a compact interval of {\mathbb{R}}. Define CN(I):={ψCc(I):supxI,0jN|djdxjψ(x)|1}C^{N}(I):=\{\psi\in C^{\infty}_{c}(I):\sup_{x\in I,0\leq j\leq N}|\frac{d^{j}}{dx^{j}}\psi(x)|\leq 1\}. Consider the partition of unity for the interval [0,1],[0,1], i.e., for t[0,1]t\in[0,1] consider

1=j2ψ(2j(1t))+ψ0(t),1=\sum_{j\geq 2}\psi(2^{j}(1-t))+\psi_{0}(t),

where ψC0[12,2]\psi\in C^{\infty}_{0}[\frac{1}{2},2] and ψ0C0[34,34].\psi_{0}\in C_{0}^{\infty}[-\frac{3}{4},\frac{3}{4}].

Let D=supx[12,2],0jN|djdxjψ(x)|D=\sup_{x\in[\frac{1}{2},2],0\leq j\leq N}|\frac{d^{j}}{dx^{j}}\psi(x)|. Then D1ψCN([12,2])D^{-1}\psi\in C^{N}([\frac{1}{2},2]).

Setting t=|ξ|2R2t=\frac{|\xi|^{2}}{R^{2}} in the above we get that

mRα(ξ,η)\displaystyle m_{R}^{\alpha}(\xi,\eta) =\displaystyle= (1|ξ|2+|η|2R2)+α\displaystyle\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}
=\displaystyle= j2ψ(2j(1|ξ|2R2))(1|ξ|2R2)+α(1|η|2R2(1|ξ|2R2)1)+α\displaystyle\sum_{j\geq 2}\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}}\right)_{+}^{\alpha}\left(1-\frac{|\eta|^{2}}{R^{2}}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)^{-1}\right)^{\alpha}_{+}
+ψ0(|ξ|2R2)(1|ξ|2+|η|2R2)+α\displaystyle+\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}
=\displaystyle= m1,Rα(ξ,η)+m0,Rα(ξ,η),\displaystyle m^{\alpha}_{1,R}(\xi,\eta)+m^{\alpha}_{0,R}(\xi,\eta),

where

m1,Rα(ξ,η)=j2ψ(2j(1|ξ|2R2))(1|ξ|2R2)+α(1|η|2R2(1|ξ|2R2)1)+αm^{\alpha}_{1,R}(\xi,\eta)=\sum_{j\geq 2}\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}}\right)_{+}^{\alpha}\left(1-\frac{|\eta|^{2}}{R^{2}}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)^{-1}\right)^{\alpha}_{+}

and

m0,Rα(ξ,η)=ψ0(|ξ|2R2)(1|ξ|2+|η|2R2)+α.m^{\alpha}_{0,R}(\xi,\eta)=\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}.

With the decomposition of the symbols as above we can write

Rα(f,g)=0,Rα(f,g)+1,Rα(f,g),\mathcal{B}^{\alpha}_{R}(f,g)=\mathcal{B}_{0,R}^{\alpha}(f,g)+\mathcal{B}_{1,R}^{\alpha}(f,g),

where

(4) i,Rα(f,g)(x)=nnmi,Rα(ξ,η)f^(ξ)g^(η)e2πix.(ξ+η)𝑑ξ𝑑η,i=0,1.\mathcal{B}_{i,R}^{\alpha}(f,g)(x)=\int_{{\mathbb{R}}^{n}}\int_{{\mathbb{R}}^{n}}m^{\alpha}_{i,R}(\xi,\eta)\hat{f}(\xi)\hat{g}(\eta)e^{2\pi ix.(\xi+\eta)}d\xi d\eta,i=0,1.

First, we deal with the bilinear operator 1,Rα(f,g)\mathcal{B}_{1,R}^{\alpha}(f,g). We further decompose the multiplier as

m1,Rα(ξ,η)=j2m~j,Rα(ξ,η),m^{\alpha}_{1,R}(\xi,\eta)=\sum\limits_{j\geq 2}\tilde{m}^{\alpha}_{j,R}(\xi,\eta),

where

m~j,Rα(ξ,η)=ψ(2j(1|ξ|2R2))(1|ξ|2R2)+α(1|η|2R2(1|ξ|2R2)1)+α.\tilde{m}^{\alpha}_{j,R}(\xi,\eta)=\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}}\right)_{+}^{\alpha}\left(1-\frac{|\eta|^{2}}{R^{2}}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)^{-1}\right)^{\alpha}_{+}.

Therefore we have

(5) 1,Rα(f,g)=j2Tj,Rα(f,g)\mathcal{B}^{\alpha}_{1,R}(f,g)=\sum_{j\geq 2}T^{\alpha}_{j,R}(f,g)

where Tj,Rα(f,g)(x)=nnm~j,Rα(ξ,η)f^(ξ)g^(η)e2πix.(ξ+η)𝑑ξ𝑑ηT^{\alpha}_{j,R}(f,g)(x)=\int_{{\mathbb{R}}^{n}}\int_{{\mathbb{R}}^{n}}\tilde{m}^{\alpha}_{j,R}(\xi,\eta)\hat{f}(\xi)\hat{g}(\eta)e^{2\pi ix.(\xi+\eta)}d\xi d\eta.

Let us denote φR(ξ)=(1|ξ|2R2)+.\varphi_{R}(\xi)=\left(1-\frac{|\xi|^{2}}{R^{2}}\right)_{+}. Now using an identity from Stein and Weiss [[21], page 278] we have the following relation.

(6) (1|η|2R2φR(ξ))+α=cαR2αφR(ξ)α0R(R2φR(ξ)t2)+β1t2δ+1(1|η|2t2)+δ𝑑t,\left(1-\frac{|\eta|^{2}}{R^{2}\varphi_{R}(\xi)}\right)^{\alpha}_{+}=c_{\alpha}R^{-2\alpha}\varphi_{R}(\xi)^{-\alpha}\int_{0}^{R}\left(R^{2}\varphi_{R}(\xi)-t^{2}\right)_{+}^{\beta-1}t^{2\delta+1}\left(1-\frac{|\eta|^{2}}{t^{2}}\right)^{\delta}_{+}dt,

where β>12\beta>\frac{1}{2}, δ>12\delta>-\frac{1}{2} and β+δ=α.\beta+\delta=\alpha.

Substituting this in the expression of m~j,Rα{\tilde{m}}^{\alpha}_{j,R} we get that

(7) m~j,Rα(ξ,η)\displaystyle{\tilde{m}}^{\alpha}_{j,R}(\xi,\eta)
=\displaystyle= cαψ(2j(1|ξ|2R2))R2α0Rj(R2φR(ξ)t2)+β1t2δ+1(1|η|2t2)+δ𝑑t,\displaystyle c_{\alpha}\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)R^{-2\alpha}\int_{0}^{R_{j}}\left(R^{2}\varphi_{R}(\xi)-t^{2}\right)_{+}^{\beta-1}t^{2\delta+1}\left(1-\frac{|\eta|^{2}}{t^{2}}\right)^{\delta}_{+}dt,

where Rj=R2j+1R_{j}=R\sqrt{2^{-j+1}}.

Notice that in the equation above due to the support of the function ψ(2j(1|ξ|2R2))\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right) the upper limit of tt in the integral is RjR_{j}. Define

(8) Bj,βR,tf(x)=nf^(ξ)ψ(2j(1|ξ|2R2))(R2φR(ξ)t2)+β1e2πix.ξ𝑑ξ\displaystyle B_{j,\beta}^{R,t}f(x)=\int_{{\mathbb{R}}^{n}}\hat{f}(\xi)\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(R^{2}\varphi_{R}(\xi)-t^{2}\right)_{+}^{\beta-1}e^{2\pi ix.\xi}d\xi

and

(9) Btδg(x)=ng^(η)(1|η|2t2)+δe2πix.η𝑑η.\displaystyle B_{t}^{\delta}g(x)=\int_{{\mathbb{R}}^{n}}\hat{g}(\eta)\left(1-\frac{|\eta|^{2}}{t^{2}}\right)^{\delta}_{+}e^{2\pi ix.\eta}d\eta.

This yields the following decompostion of the bilinear operator Tj,RαT^{\alpha}_{j,R} associated with the symbol m~j,Rα(ξ,η){\tilde{m}}^{\alpha}_{j,R}(\xi,\eta).

(10) Tj,Rα(f,g)(x)\displaystyle T^{\alpha}_{j,R}(f,g)(x) =\displaystyle= nnm~j,Rα(ξ,η)f^(ξ)g^(η)e2πix.(ξ+η)𝑑ξ𝑑η\displaystyle\int_{{\mathbb{R}}^{n}}\int_{{\mathbb{R}}^{n}}\tilde{m}^{\alpha}_{j,R}(\xi,\eta)\hat{f}(\xi)\hat{g}(\eta)e^{2\pi ix.(\xi+\eta)}d\xi d\eta
=\displaystyle= cα0RjBj,βR,tf(x)Btδg(x)t2δ+1𝑑t,\displaystyle c_{\alpha}\int_{0}^{R_{j}}B_{j,\beta}^{R,t}f(x)B_{t}^{\delta}g(x)t^{2\delta+1}dt,

where α=β+δ\alpha=\beta+\delta.

We will show that the decomposition of the operator as above yields new estimates for the bilinear Bochner-Riesz maximal function. Before going into the proofs of our main results we discuss some useful results concerning the Bochner-Riesz square function.

4. A brief discussion on the Bochner-Riesz Square function

The Bochner-Riesz square function GαG^{\alpha} is defined by

Gα(f)(x):=(0|tBtα+1f(x)|2t𝑑t)1/2=(0|𝒦tαf(x)|2dtt)1/2,\displaystyle G^{\alpha}(f)(x):=\left(\int_{0}^{\infty}|\frac{\partial}{\partial t}B_{t}^{\alpha+1}f(x)|^{2}tdt\right)^{1/2}=\left(\int_{0}^{\infty}|\mathcal{K}^{\alpha}_{t}\ast f(x)|^{2}\frac{dt}{t}\right)^{1/2},

where Btαf^(ξ)=(1|ξ|2t2)+αf^(ξ)\widehat{B_{t}^{\alpha}f}(\xi)=\left(1-\frac{|\xi|^{2}}{t^{2}}\right)^{\alpha}_{+}\hat{f}(\xi) and 𝒦tα^(ξ)=2(δ+1)|ξ|2t2(1|ξ|2t2)+α.\widehat{\mathcal{K}^{\alpha}_{t}}(\xi)=2(\delta+1)\frac{|\xi|^{2}}{t^{2}}(1-\frac{|\xi|^{2}}{t^{2}})^{\alpha}_{+}.

The square function GαG^{\alpha} was introduced by Stein [20]. The LpL^{p} boundedness

(11) Gα(f)Lp(n)fLp(n)\displaystyle\|G^{\alpha}(f)\|_{L^{p}({\mathbb{R}}^{n})}\lesssim\|f\|_{L^{p}({\mathbb{R}}^{n})}

of the square function have been studied extensively in the literature. It has various applications, in particular, it plays an improtant role in the study of maximal Bochner-Riesz functions. We refer the reader to [20, 22, 23, 2, 3, 19, 17, 16, 15] and references there in for details.

Due to the derivative with respect to tt the operator GαG^{\alpha} essentially behaves like the Bochner-Riesz operator BαB^{\alpha} of order α\alpha. The Plancherel theorem immediately yields that GαG^{\alpha} is bounded on L2(n)L^{2}({\mathbb{R}}^{n}) provided α>12\alpha>-\frac{1}{2}, see [20]. It is conjectured that for 1<p21<p\leq 2, the estimate (11) holds if, and only if α>n(1p12)12.\alpha>n(\frac{1}{p}-\frac{1}{2})-\frac{1}{2}. Whereas for the range p>2p>2 the behaviour of GαG^{\alpha} is different and it is conjectured that for p>2p>2, the estimate (11) holds if, and only if α>α(p)12.\alpha>\alpha(p)-\frac{1}{2}.

The conjecture for the range 1<p21<p\leq 2 has been resolved. We know that the estimate (11) holds if, and only if α>n(1p12)12\alpha>n(\frac{1}{p}-\frac{1}{2})-\frac{1}{2} for 1<p21<p\leq 2 and n1n\geq 1, see [22, 23, 17] for detail. Further, in dimensions n=1,2n=1,2, the conjecture has been proved to hold for the range p2p\geq 2 , see [23] and [2] for the case of n=1n=1 and n=2n=2 respectively. When n3n\geq 3 and p>2p>2 the sufficient part of the conjecture is not known completely. There are many interesting developments on the conjecture and we refer to [3, 19, 17, 16, 15] for more detail. The most recent develpoment in this direction is due to  Lee [15], where he proved the following result.

Theorem 4.1.

[15] Let n2n\geq 2. The estimate (11) holds for pmin{𝔭n,2(n+2)n}p\geq\min\{\mathfrak{p}_{n},\frac{2(n+2)}{n}\} and α>n(121p)1\alpha>n(\frac{1}{2}-\frac{1}{p})-1.

The LpL^{p} estimates for the square function mentioned as above will be used in order to prove our results. Indeed, we shall need LpL^{p} boundedness of certain local variants of the square function. We consider the following setting.

For ψCN+1([12,2])\psi\in C^{N+1}([\frac{1}{2},2]) and 0<ν<1160<\nu<\frac{1}{16} define

Bν,tψf(x)=nψ(ν1(1|ξ|2t2))f^(ξ)e2πix.ξ𝑑ξ.B^{\psi}_{\nu,t}f(x)=\int_{{\mathbb{R}}^{n}}\psi\left(\nu^{-1}\left(1-\frac{|\xi|^{2}}{t^{2}}\right)\right)\hat{f}(\xi)e^{2\pi ix.\xi}d\xi.

Also, for ϕCN([1,1])\phi\in C^{N}([-1,1]) and 0<ν<1160<\nu<\frac{1}{16} consider the operator

ϕ(t2|D|2ν)f(x)=nϕ(t2|ξ|2ν)f^(ξ)e2πix.ξ𝑑ξ.\phi\left(\frac{t^{2}-|D|^{2}}{\nu}\right)f(x)=\int_{{\mathbb{R}}^{n}}\phi\left(\frac{t^{2}-|\xi|^{2}}{\nu}\right)\hat{f}(\xi)e^{2\pi ix.\xi}d\xi.

Jeong, Lee and Vargas [Lemma 2.62.6 [12]] proved that for any ϵ>0\epsilon>0, there exists N1N\geq 1 such that for ϕCN([1,1])\phi\in C^{N}([-1,1]) the following holds

(12) (1/22|ϕ(t2|D|2ν)f()|2𝑑t)1/2Lp(n)ϵ,Nν(1/2α(p))νϵfLp(n)\left\|\left(\int_{1/2}^{2}\left|\phi\left(\frac{t^{2}-|D|^{2}}{\nu}\right)f(\cdot)\right|^{2}dt\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim_{\epsilon,N}\nu^{(1/2-\alpha(p))}\nu^{-\epsilon}\|f\|_{L^{p}({\mathbb{R}}^{n})}

when p>𝔭np>\mathfrak{p}_{n}.

We need the following version of the result stated above.

Lemma 4.2.

Let n2,0<ν<116n\geq 2,0<\nu<\frac{1}{16} and ϵ>0\epsilon>0. Then for p𝔭np\geq\mathfrak{p}_{n} and p=2,p=2, there exists N1N\geq 1 such that for all ψCN+1([1/2,2])\psi\in C^{N+1}([1/2,2]) the following holds

(13) (0|Bν,tψf()|2dtt)1/2Lp(n)ϵ,Nν(1/2α(p))νϵfLp(n)\left\|\left(\int_{0}^{\infty}\left|B^{\psi}_{\nu,t}f(\cdot)\right|^{2}\frac{dt}{t}\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim_{\epsilon,N}\nu^{(1/2-\alpha(p))}\nu^{-\epsilon}\|f\|_{L^{p}({\mathbb{R}}^{n})}

where the implicit constant depends on ϵ\epsilon and NN.

In Lemma 4.2 we can take the same value of NN as in Lemma 2.62.6 in [12]. The proof of the lemma above may be completed using Lemma 2.62.6 in [12] with minor modifications. We present the proof here for an easy reference and completeness.

Proof.

Note that when p=2p=2 the inequality (13) is a consequence of Plancherel theorem.

Next, let ψr(x):=r(N+1)ψ(rx)\psi_{r}(x):=r^{-(N+1)}\psi(rx) where r>2r>2 is fixed such that ψrCN+1([1,1])\psi_{r}\in C^{N+1}([-1,1]). First, consider the operator Bν,tψrB^{\psi_{r}}_{\nu,t}.

When p>2p>2 we know that by a standard Littlewood Paley decomposition, it is enough to prove the inequality (13) for 12<t<2\frac{1}{2}<t<2. Consider

Bν,tψrf(x)\displaystyle B^{\psi_{r}}_{\nu,t}f(x) =\displaystyle= Bν,tψrf(x)ψr(t2|D|241ν)f(x)+ψr(t2|D|241ν)f(x)\displaystyle B^{\psi_{r}}_{\nu,t}f(x)-\psi_{r}\left(\frac{t^{2}-|D|^{2}}{4^{-1}\nu}\right)f(x)+\psi_{r}\left(\frac{t^{2}-|D|^{2}}{4^{-1}\nu}\right)f(x)
=\displaystyle= ψr(t2|D|241ν)f(x)+1/4t2ψr~(t2|D|2νs)f(x)dss,\displaystyle\psi_{r}\left(\frac{t^{2}-|D|^{2}}{4^{-1}\nu}\right)f(x)+\int_{1/4}^{t^{2}}\widetilde{\psi_{r}}\left(\frac{t^{2}-|D|^{2}}{\nu s}\right)f(x)\frac{ds}{s},

where ψr~(x)=r(N+1)rxψ(rx)\widetilde{\psi_{r}}(x)=r^{-(N+1)}rx\psi^{\prime}(rx).

It is easy to verify that ψr~\widetilde{\psi_{r}} belongs to CN([1,1])C^{N}([-1,1]). Using Minkowski’s integral inequality and triangle inequality in the identity above, we get that

(1/22|Bν,tψrf()|2dtt)1/2Lp(n)\displaystyle\left\|\left(\int_{1/2}^{2}\left|B^{\psi_{r}}_{\nu,t}f(\cdot)\right|^{2}\frac{dt}{t}\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})} \displaystyle\lesssim (1/22|ψr(t2|D|241ν)f()|2𝑑t)1/2Lp(n)\displaystyle\left\|\left(\int_{1/2}^{2}\left|\psi_{r}\left(\frac{t^{2}-|D|^{2}}{4^{-1}\nu}\right)f(\cdot)\right|^{2}dt\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}
+1/44(1/22|ψr(t2|D|2νs)f()|2𝑑t)1/2Lp(n)dss.\displaystyle+\int_{1/4}^{4}\left\|\left(\int_{1/2}^{2}\left|\psi_{r}\left(\frac{t^{2}-|D|^{2}}{\nu s}\right)f(\cdot)\right|^{2}dt\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\frac{ds}{s}.

From (12) we know that for all s[1/4,4]s\in[1/4,4] the following holds.

(1/22|ψr(t2|D|2νs)f()|2𝑑t)1/2Lp(n)ϵ,N(νs)(1/2α(p))νϵfLp(n).\left\|\left(\int_{1/2}^{2}\left|\psi_{r}\left(\frac{t^{2}-|D|^{2}}{\nu s}\right)f(\cdot)\right|^{2}dt\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim_{\epsilon,N}(\nu s)^{(1/2-\alpha(p))}\nu^{-\epsilon}\|f\|_{L^{p}({\mathbb{R}}^{n})}.

Therefore, we get that

(1/22|Bν,tψrf()|2dtt)1/2Lp(n)ϵ,Nν(1/2α(p))νϵfLp(n)\left\|\left(\int_{1/2}^{2}\left|B^{\psi_{r}}_{\nu,t}f(\cdot)\right|^{2}\frac{dt}{t}\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim_{\epsilon,N}\nu^{(1/2-\alpha(p))}\nu^{-\epsilon}\|f\|_{L^{p}({\mathbb{R}}^{n})}

for p>𝔭np>\mathfrak{p}_{n}.

The standard interpolation argument between p=2p=2 and p>𝔭np>\mathfrak{p}_{n} yields that the inequality as above holds for p=𝔭np=\mathfrak{p}_{n}.

Further, it is easy to verify that rN+1Bν,tψrf(x)=Br1ν,tψf(x)r^{N+1}B^{\psi_{r}}_{\nu,t}f(x)=B^{\psi}_{r^{-1}\nu,t}f(x). Note that for a fixed r>2r>2 we can choose ν\nu small enough so that r1ν<1/16r^{-1}\nu<1/16. This completes the proof of the inequality (13). ∎

Next, we consider the maximal function supt>0|Bν,tψf(x)|\sup_{t>0}|B^{\psi}_{\nu,t}f(x)| associated with the frequency localised Bochner-Riesz operators. The square function estimate proved in Lemma 4.2 gives us the following estimate

(14) supt>0|Bν,tψf|Lp(n)ϵ,Nνϵνα(p)fLp(n)\|\sup_{t>0}|B^{\psi}_{\nu,t}f|\|_{L^{p}({\mathbb{R}}^{n})}\lesssim_{\epsilon,N}\nu^{-\epsilon}\nu^{\alpha(p)}\|f\|_{L^{p}({\mathbb{R}}^{n})}

for p=2p=2 and p𝔭n.p\geq\mathfrak{p}_{n}.

Note that if mC0([1/2,2])m\in C^{\infty}_{0}([1/2,2]) is such that A1mCN([1/2,2])A^{-1}m\in C^{N}([1/2,2]) for some A>0A>0, then from (14) we get that

(15) supt>0|Bν,tmf|Lp(n)ϵ,NAνϵνα(p)fLp(n)\|\sup_{t>0}|B^{m}_{\nu,t}f|\|_{L^{p}({\mathbb{R}}^{n})}\lesssim_{\epsilon,N}A\nu^{-\epsilon}\nu^{\alpha(p)}\|f\|_{L^{p}({\mathbb{R}}^{n})}

holds for the same range of pp as in (14).

Let hh be a smooth function supported on a compact interval II such that C1hCN(I)C^{-1}h\in C^{N}(I) for some C>0C>0. In what follows we shall use the statement that hk(x)=xkh(x),k1h_{k}(x)=x^{k}h(x),k\geq 1 and h(x)h(x) behave similarly. This would mean that the corresponding square function and the maximal function for the operator Bν,thkB^{h_{k}}_{\nu,t} defined above are bounded for the same range of pp as that of Bν,thB^{h}_{\nu,t} with the same constants except for an extra factor of k!k!.

Next, we make use of a technique from [[21], page 277-278] to derive the LpL^{p} boundedness of the operator fsupR>0(R10R|Btδf()|2𝑑t)1/2f\rightarrow\sup_{R>0}\left(R^{-1}\int_{0}^{R}|B_{t}^{\delta}f(\cdot)|^{2}dt\right)^{1/2}. This idea gives us the following estimate.

Lemma 4.3.

The operator

fsupR>0(R10R|Btδf()|2𝑑t)1/2f\rightarrow\sup_{R>0}\left(R^{-1}\int_{0}^{R}|B_{t}^{\delta}f(\cdot)|^{2}dt\right)^{1/2}

is bounded on Lp(n)L^{p}({\mathbb{R}}^{n}) for p=2p=2 and p𝔭np\geq\mathfrak{p}_{n} where δ>α(p)1/2\delta>\alpha(p)-1/2.

Proof.

The case of p=2p=2 is easy as usual. When p>2p>2 we write

Btδf=k=1d(Btδ+k1fBtδ+kf)+Btδ+df,B_{t}^{\delta}f=\sum_{k=1}^{d}\left(B_{t}^{\delta+k-1}f-B_{t}^{\delta+k}f\right)+B_{t}^{\delta+d}f,

where dd is chosen so that δ+d>n12\delta+d>\frac{n-1}{2}.

This implies that

(0R|Btδf(x)|2𝑑t)1/2k=1d(0R|Btδ+kf(x)Btδ+k1f(x)|2𝑑t)1/2+(0R|Btδ+df(x)|2𝑑t)1/2.\left(\int_{0}^{R}|B_{t}^{\delta}f(x)|^{2}dt\right)^{1/2}\leq\sum_{k=1}^{d}\left(\int_{0}^{R}|B_{t}^{\delta+k}f(x)-B_{t}^{\delta+k-1}f(x)|^{2}dt\right)^{1/2}+\left(\int_{0}^{R}|B_{t}^{\delta+d}f(x)|^{2}dt\right)^{1/2}.

Since δ+d>n12\delta+d>\frac{n-1}{2} the convolution kernel of BtδB_{t}^{\delta} is an integrable function and consequently supt>0|Btδf|c(δ+d,n)f(x)\sup_{t>0}|B_{t}^{\delta}f|\leq c(\delta+d,n)\mathcal{M}f(x), where ()\mathcal{M}(\cdot) is the Hardy-Littlewood Maximal function. Therefore supR>0(R10R|Btδ+df(x)|2𝑑t)1/2\sup_{R>0}\left(R^{-1}\int_{0}^{R}|B_{t}^{\delta+d}f(x)|^{2}dt\right)^{1/2} is bounded on Lp(n),1<pL^{p}({\mathbb{R}}^{n}),1<p\leq\infty.

Next, consider the term

supR>0(R10R|Btδ+kf(x)Btδ+k1f(x)|2𝑑t)1/2\sup_{R>0}\left(R^{-1}\int_{0}^{R}|B_{t}^{\delta+k}f(x)-B_{t}^{\delta+k-1}f(x)|^{2}dt\right)^{1/2}

for a fixed 1kd1\leq k\leq d. Clearly the above can be dominated by

(0|Btδ+kf(x)Btδ+k1f(x)|2t1𝑑t)1/2\left(\int_{0}^{\infty}|B_{t}^{\delta+k}f(x)-B_{t}^{\delta+k-1}f(x)|^{2}t^{-1}dt\right)^{1/2}

which is nothing but the square function for Bochner-Riesz means of order δ+k1\delta+k-1. Invoking Theorem 4.1 from [15] we get the desired Lp(n)L^{p}({\mathbb{R}}^{n}) estimates for p𝔭np\geq\mathfrak{p}_{n} when δ>α(p)1/2\delta>\alpha(p)-1/2 for all k1k\geq 1. This completes the proof. ∎

5. Proof of Theorem 2.1: Boundedness of bilinear Bochner-Riesz maximal function

This section is devoted to proving Theorem 2.1. We shall prove the estimates for pj=2p_{j}=2 or pi𝔭n,i,j=1,2p_{i}\geq\mathfrak{p}_{n},~{}i,j=1,2 and the remaining cases in Theorem 2.1 will be obtained by interpolation arguments.

Recall that from section 3 we have the following decomposition of the bilinear Bochner-Riesz operator Rα\mathcal{B}^{\alpha}_{R}.

Rα(f,g)=0,Rα(f,g)+1,Rα(f,g).\mathcal{B}^{\alpha}_{R}(f,g)=\mathcal{B}_{0,R}^{\alpha}(f,g)+\mathcal{B}_{1,R}^{\alpha}(f,g).

In order to prove Theorem 2.1 it is enough to prove the desired estimates for the bilinear maximal functions

i,α(f,g)(x)=supR>0|i,Rα(f,g)(x)|,i=0,1.\mathcal{B}^{\alpha}_{i,*}(f,g)(x)=\sup_{R>0}|\mathcal{B}^{\alpha}_{i,R}(f,g)(x)|,~{}i=0,1.

We will deal with both the maximal functions separately. Let us first consider the case of i=1i=1.

5.1. Boundedness of the bilinear maximal function 1,α(f,g)\mathcal{B}^{\alpha}_{1,*}(f,g)

Recall that from section 3 we have the following decomposition

1,Rα(f,g)(x)=j2Tj,Rα(f,g)(x),\mathcal{B}^{\alpha}_{1,R}(f,g)(x)=\sum_{j\geq 2}T^{\alpha}_{j,R}(f,g)(x),

where Tj,Rα(f,g)(x)=nnm~j,Rα(ξ,η)f^(ξ)g^(η)e2πix.(ξ+η)𝑑ξ𝑑ηT^{\alpha}_{j,R}(f,g)(x)=\int_{{\mathbb{R}}^{n}}\int_{{\mathbb{R}}^{n}}\tilde{m}^{\alpha}_{j,R}(\xi,\eta)\hat{f}(\xi)\hat{g}(\eta)e^{2\pi ix.(\xi+\eta)}d\xi d\eta. Therefore, it is enough to consider the maximal function

Tj,α(f,g)(x)=supR>0|Tj,Rα(f,g)(x)|for  each j2.T^{\alpha}_{j,*}(f,g)(x)=\sup_{R>0}|T^{\alpha}_{j,R}(f,g)(x)|~{}\text{for~{} each~{}}j\geq 2.

Applying Cauchy Schwartz inequality we get that

|Tj,Rα(f,g)(x)|cα(0Rj|Bj,βR,tf(x)t2δ+1|2𝑑t)1/2(0Rj|Btδg(x)|2𝑑t)1/2.|T_{j,R}^{\alpha}(f,g)(x)|\leq c_{\alpha}\left(\int_{0}^{R_{j}}|B_{j,\beta}^{R,t}f(x)t^{2\delta+1}|^{2}dt\right)^{1/2}\left(\int_{0}^{R_{j}}|B_{t}^{\delta}g(x)|^{2}dt\right)^{1/2}.

Making a change of variable tRtt\rightarrow Rt in the integral (0Rj|Bj,βR,tf(x)t2δ+1|2𝑑t)1/2\left(\int_{0}^{R_{j}}|B_{j,\beta}^{R,t}f(x)t^{2\delta+1}|^{2}dt\right)^{1/2} we get that

0Rj|Bj,βR,tf(x)t2δ+1|2𝑑t=R4α102j+1|Sj,βR,tf(x)t2δ+1|2𝑑t,\int_{0}^{R_{j}}|B_{j,\beta}^{R,t}f(x)t^{2\delta+1}|^{2}dt=R^{4\alpha-1}\int_{0}^{\sqrt{2^{-j+1}}}|{S}_{j,\beta}^{R,t}f(x)t^{2\delta+1}|^{2}dt,

where

Sj,βR,tf(x)=nψ(2j(1|ξ|2R2))(1|ξ|2R2t2)+β1f^(ξ)e2πix.ξ𝑑ξ.{S}_{j,\beta}^{R,t}f(x)=\int_{{\mathbb{R}}^{n}}\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)_{+}^{\beta-1}\hat{f}(\xi)e^{2\pi ix.\xi}d\xi.

Finally, we get that

Tj,α(f,g)(x)\displaystyle T^{\alpha}_{j,*}(f,g)(x) \displaystyle\leq 2j/4supR>0(02j+1|Sj,βR,tf(x)t2δ+1|2𝑑t)1/2\displaystyle 2^{-j/4}\sup_{R>0}\left(\int_{0}^{\sqrt{2^{-j+1}}}|{S}_{j,\beta}^{R,t}f(x)t^{2\delta+1}|^{2}dt\right)^{1/2}
(supR>0Rj10Rj|Btδg(x)|2𝑑t)1/2.\displaystyle\left(\sup_{R>0}R_{j}^{-1}\int_{0}^{R_{j}}|B_{t}^{\delta}g(x)|^{2}dt\right)^{1/2}.

We have the following result for the maximal function involving the operator Sj,βR,t{S}_{j,\beta}^{R,t} in the inequality above.

Theorem 5.1.

Let n2.n\geq 2. For p𝔭np\geq\mathfrak{p}_{n} or p=2p=2 and β>α(p)+1/2,\beta>\alpha(p)+1/2, the following estimate holds

(17) supR>0(02j+1|Sj,βR,tf()t2δ+1|2𝑑t)1/2Lp(n)2j(α(p)α+14+ϵ)fLp(n).\displaystyle\left\|\sup_{R>0}\left(\int_{0}^{\sqrt{2^{-j+1}}}|{S}_{j,\beta}^{R,t}f(\cdot)t^{2\delta+1}|^{2}dt\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim 2^{j(\alpha(p)-\alpha+\frac{1}{4}+\epsilon)}\|f\|_{L^{p}({\mathbb{R}}^{n})}.

We postpone the proof of Theorem 5.1 to the next subsection. Assuming Theorem 5.1 we complete the estimate for the maximal function 1,α\mathcal{B}^{\alpha}_{1,*}.

Note that the Hölder inequality in the estimate (5.1) yields that

Tj,α(f,g)Lp(n)\displaystyle\|T^{\alpha}_{j,*}(f,g)\|_{L^{p}({\mathbb{R}}^{n})}
\displaystyle\lesssim 2j/4supR>0(02j+1|Sj,βR,tf()t2δ+1|2𝑑t)1/2Lp1(n)(supR>0Rj10Rj|Btδg()|2𝑑t)1/2Lp2(n),\displaystyle 2^{-j/4}\left\|\sup_{R>0}\left(\int_{0}^{\sqrt{2^{-j+1}}}|{S}_{j,\beta}^{R,t}f(\cdot)t^{2\delta+1}|^{2}dt\right)^{1/2}\right\|_{L^{p_{1}}({\mathbb{R}}^{n})}\left\|\left(\sup_{R>0}R_{j}^{-1}\int_{0}^{R_{j}}|B_{t}^{\delta}g(\cdot)|^{2}dt\right)^{1/2}\right\|_{L^{p_{2}}({\mathbb{R}}^{n})},

where δ+β=α.\delta+\beta=\alpha.

Invoking the estimates from Theorem 5.1 and Lemma 4.3 we get that

(18) Tj,α(f,g)Lp(n)\displaystyle\|T^{\alpha}_{j,*}(f,g)\|_{L^{p}({\mathbb{R}}^{n})} n,α,p1,p2\displaystyle\lesssim_{n,\alpha,p_{1},p_{2}} 2j(αα(p1))fLp1(n)gLp2(n)\displaystyle 2^{-j(\alpha-\alpha(p_{1}))}\|f\|_{L^{p_{1}}({\mathbb{R}}^{n})}\|g\|_{L^{p_{2}}({\mathbb{R}}^{n})}

where β>α(p1)+1/2\beta>\alpha(p_{1})+1/2 for p1=2p_{1}=2 or p1𝔭np_{1}\geq\mathfrak{p}_{n} and δ>α(p2)1/2\delta>\alpha(p_{2})-1/2 for p2=2p_{2}=2 or p2𝔭np_{2}\geq\mathfrak{p}_{n}, which is same as saying that α>α(p1)+α(p2)=α(p1,p2)\alpha>\alpha(p_{1})+\alpha(p_{2})=\alpha_{*}(p_{1},p_{2}) when pi=2p_{i}=2 or pi𝔭np_{i}\geq\mathfrak{p}_{n} for i=1,2i=1,2.

The decomposition (5) along with the estimate above implies that for α>α(p1,p2)\alpha>\alpha_{*}(p_{1},p_{2}) we have

(19) 1,α(f,g)Lp(n)\displaystyle\|\mathcal{B}^{\alpha}_{1,*}(f,g)\|_{L^{p}({\mathbb{R}}^{n})} n,α,p1,p2\displaystyle\lesssim_{n,\alpha,p_{1},p_{2}} fLp1(n)gLp2(n).\displaystyle\|f\|_{L^{p_{1}}({\mathbb{R}}^{n})}\|g\|_{L^{p_{2}}({\mathbb{R}}^{n})}.

This proves the desired estimate for the maximal function 1,α(f,g)\mathcal{B}^{\alpha}_{1,*}(f,g) under the assumption that Theorem 5.1 holds.

5.2. Proof of Theorem 5.1

We set β=γ+1\beta=\gamma+1 in Sj,βR,tf{S}_{j,\beta}^{R,t}f for convenience.

Note that when t2[0,2j1)t^{2}\in[0,2^{-j-1}) we can rewrite the function

ψ(2j(1|ξ|2R2))(1|ξ|2R2t2)+γ=ψ(2j(1|ξ|2R2))(1|ξ|2R2t2)γ.\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)_{+}^{\gamma}=\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)^{\gamma}.

Moreover, in this range of t,t, the multiplier has no singularity. Indeed it behaves like 2jγ2^{-j\gamma} times a smooth function supported in the annulus of width of the order R2jR\sqrt{2^{-j}}.

With this observation in mind, we split the interval [0,2j+1][0,2^{-j+1}], the range of t2t^{2} for the multiplier (1|ξ|2R2t2)+β1\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)_{+}^{\beta-1}, into two subintervals [0,2j1ϵ0][0,2^{-j-1-\epsilon_{0}}] and [2j1ϵ0,2j+1][2^{-j-1-\epsilon_{0}},2^{-j+1}] and deal with the corresponding operators separately. Here we have taken 0<ϵ0<<10<\epsilon_{0}<<1 to be a fixed number.

Case I: When t[0,2j1ϵ0]t\in[0,\sqrt{2^{-j-1-\epsilon_{0}}}]

For γ>1\gamma>-1 and p=2p=2 or p𝔭np\geq\mathfrak{p}_{n}, we have

(20) supR>0(02j1ϵ0|Sj,γ+1R,tf(x)t2δ+1|2𝑑t)1/2Lp(n)\displaystyle\left\|\sup_{R>0}\left(\int_{0}^{\sqrt{2^{-j-1-\epsilon_{0}}}}|{S}_{j,\gamma+1}^{R,t}f(x)t^{2\delta+1}|^{2}dt\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})} \displaystyle\lesssim 2j(δ+γ+3/4α(p))fLp(n).\displaystyle 2^{-j(\delta+\gamma+3/4-\alpha(p))}\|f\|_{L^{p}({\mathbb{R}}^{n})}.

We first consider the range 1<γ<0.-1<\gamma<0. Set γ=ρ\gamma=-\rho, where ρ>0\rho>0. By Taylor’s expansion we can write

(21) (1|ξ|2R2)ρ(1t21|ξ|2R2)ρ=2jρ(2j(1|ξ|2R2))ρk0Γ(ρ+k)Γ(ρ)k!(2jt22j(1|ξ|2R2))k.\left(1-\frac{|\xi|^{2}}{R^{2}}\right)^{-\rho}\left(1-\frac{t^{2}}{1-\frac{|\xi|^{2}}{R^{2}}}\right)^{-\rho}=2^{j\rho}\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)^{-\rho}\sum_{k\geq 0}\frac{\Gamma(\rho+k)}{\Gamma(\rho)k!}\left(\frac{2^{j}t^{2}}{2^{j}(1-\frac{|\xi|^{2}}{R^{2}})}\right)^{k}.

Observe that the series above converges as t21|ξ|2R22j1ϵ02j+1<1.\frac{t^{2}}{1-\frac{|\xi|^{2}}{R^{2}}}\leq 2^{-j-1-\epsilon_{0}}2^{j+1}<1. Therefore, using (21) we get that

(22) Sj,γ+1R,tf\displaystyle{S}_{j,\gamma+1}^{R,t}f =\displaystyle= 2jρk0Γ(ρ+k)Γ(ρ)k!(2jt2)kB2j,Rψkf.\displaystyle 2^{j\rho}\sum_{k\geq 0}\frac{\Gamma(\rho+k)}{\Gamma(\rho)k!}\left(2^{j}t^{2}\right)^{k}B^{\psi^{k}}_{2^{-j},R}f.

Denote ψk(x):=xkρψ(x)\psi^{k}(x):=x^{-k-\rho}\psi(x). Observe that ψkC0([1/2,2])\psi^{k}\in C^{\infty}_{0}([1/2,2]) and it satisfies the estimate

supx[1/2,2],0lN|dlψkdxl|C(ρ)2N+kkN+1\sup_{x\in[1/2,2],0\leq l\leq N}\left|\frac{d^{l}\psi^{k}}{dx^{l}}\right|\leq C(\rho)2^{N+k}k^{N+1}

for N20nN\geq 20n. Therefore, the corresponding maximal function is bounded on Lp(n)L^{p}({\mathbb{R}}^{n}) with an additional factor of 2N+kkN2jα(p)2^{N+k}k^{N}2^{j\alpha(p)}, see the estimate (15) for detail. More precisely, we get that

supR>0|B2j,Rψkf|Lp(n)N,p12N+kkN2jα(p)fLp(n)\|\sup_{R>0}|B^{\psi^{k}}_{2^{-j},R}f|\|_{L^{p}({\mathbb{R}}^{n})}\lesssim_{N,p_{1}}2^{N+k}k^{N}2^{j\alpha(p)}\|f\|_{L^{p}({\mathbb{R}}^{n})}

where p=2p=2 or p𝔭np\geq\mathfrak{p}_{n}.

Using the Minkowski’s integral inequality and the fact that 2jt221ϵ02^{j}t^{2}\leq 2^{-1-\epsilon_{0}} we have

supR>0(02j1ϵ0|Sj,γ+1R,tf()t2δ+1|2𝑑t)1/2Lp(n)\displaystyle\left\|\sup_{R>0}\left(\int_{0}^{\sqrt{2^{-j-1-\epsilon_{0}}}}|{S}_{j,\gamma+1}^{R,t}f(\cdot)t^{2\delta+1}|^{2}dt\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})} \displaystyle\lesssim (02j1ϵ0supR>0|Sj,γ+1R,tf|Lp(n)2t4δ+2𝑑t)1/2\displaystyle\left(\int_{0}^{\sqrt{2^{-j-1-\epsilon_{0}}}}\|\sup_{R>0}|{S}_{j,\gamma+1}^{R,t}f|\|_{L^{p}({\mathbb{R}}^{n})}^{2}t^{4\delta+2}dt\right)^{1/2}
\displaystyle\lesssim 2j(δ+1/2)2jγ2j/42jα(p)\displaystyle 2^{-j(\delta+1/2)}2^{-j\gamma}2^{-j/4}2^{j\alpha(p)}
×k0Γ(ρ+k)Γ(ρ)k!2N+kkN2kϵ0kfLp(n).\displaystyle\times\sum_{k\geq 0}\frac{\Gamma(\rho+k)}{\Gamma(\rho)k!}2^{N+k}k^{N}2^{-k-\epsilon_{0}k}\|f\|_{L^{p}({\mathbb{R}}^{n})}.

By the asymptotic formula of Gamma function we know that Γ(ρ+k)Γ(ρ)k!kρ\frac{\Gamma(\rho+k)}{\Gamma(\rho)k!}\approx k^{\rho} and hence the series in the expression above converges. This gives us the desired estimate.

Note that when γ0,\gamma\geq 0, we prove the desired estimates for γ\gamma to be non-negative integers. For γ=0,\gamma=0, we know that

fsupR>0|Sj,γ+1R,tf|f\rightarrow\sup_{R>0}|{S}_{j,\gamma+1}^{R,t}f|

is bounded on Lp(n)L^{p}({\mathbb{R}}^{n}) with the operator norm bounded by a constant multiple of 2jα(p)2^{j\alpha(p)} for p=2p=2 or p𝔭np\geq\mathfrak{p}_{n} (see [14] for detail). Therefore, we get that

supR>0(02j1ϵ0|Sj,1R,tf(x)t2δ+1|2𝑑t)1/2Lp(n)2j(δ+1/2)2j/42jα(p)fLp(n).\left\|\sup_{R>0}\left(\int_{0}^{\sqrt{2^{-j-1-\epsilon_{0}}}}|{S}_{j,1}^{R,t}f(x)t^{2\delta+1}|^{2}dt\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim 2^{-j(\delta+1/2)}2^{-j/4}2^{j\alpha(p)}\|f\|_{L^{p}({\mathbb{R}}^{n})}.

Note that the term 2j/42^{-j/4} occurs because of integration in the variable tt.

Next, when γ=1\gamma=1 we write

(23) ψ(2j(1|ξ|2R2))(1|ξ|2R2t2)=2j(2j(1|ξ|2R2)ψ(2j(1|ξ|2R2)))t2ψ(2j(1|ξ|2R2)).\displaystyle\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)=2^{-j}\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\right){\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}-}t^{2}\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right).

Since the function ϕ(1|ξ|2R2):=2j(1|ξ|2R2)ψ(2j(1|ξ|2R2))\phi\left(1-\frac{|\xi|^{2}}{R^{2}}\right):=2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right) behaves like ψ(2j(1|ξ|2R2))\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right), the identity as above gives us that

|Sj,2R,tf(x)|2j|B2j,Rϕf(x)|+t2|B2j,Rψf(x)|.|{S}_{j,2}^{R,t}f(x)|\leq 2^{-j}|B^{\phi}_{2^{-j},R}f(x)|+t^{2}|B^{\psi}_{2^{-j},R}f(x)|.

Using the identity above along with the estimate (15) we get that

supR>0(02j1ϵ0|Sj,2R,tf(x)t2δ+1|2𝑑t)1/2Lp(n)2j(δ+1/2)2j2j/42jα(p1)fLp(n).\left\|\sup_{R>0}\left(\int_{0}^{\sqrt{2^{-j-1-\epsilon_{0}}}}|{S}_{j,2}^{R,t}f(x)t^{2\delta+1}|^{2}dt\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim 2^{-j(\delta+1/2)}2^{-j}2^{-j/4}2^{j\alpha(p_{1})}\|f\|_{L^{p}({\mathbb{R}}^{n})}.

When 0<γ<10<\gamma<1, we write γ=ζ+1\gamma=\zeta+1 for ζ(1,0)\zeta\in(-1,0).

We can write

ψ(2j(1|ξ|2R2))(1|ξ|2R2t2)γ=2jΨ1(|ξ|2R2)(1|ξ|2R2t2)ζt2Ψ2(|ξ|2R2)(1|ξ|2R2t2)ζ,\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)^{\gamma}=2^{-j}\Psi_{1}\left(\frac{|\xi|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)^{\zeta}-t^{2}\Psi_{2}\left(\frac{|\xi|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)^{\zeta},

where Ψ1(|ξ|2R2)=2j(1|ξ|2R2)ψ(2j(1|ξ|2R2))\Psi_{1}\left(\frac{|\xi|^{2}}{R^{2}}\right)=2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right) and Ψ2(|ξ|2R2)=ψ(2j(1|ξ|2R2))\Psi_{2}\left(\frac{|\xi|^{2}}{R^{2}}\right)=\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)

The boundedness of the square function corresponding to term Ψ2(|ξ|2R2)(1|ξ|2R2t2)ζ\Psi_{2}\left(\frac{|\xi|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)^{\zeta} is dealt with in a similar way as in the case of 1<γ<0-1<\gamma<0 as above. Further, note that Ψ1(|ξ|2R2)\Psi_{1}\left(\frac{|\xi|^{2}}{R^{2}}\right) behaves in a similar manner as ψ(2j(1|ξ|2R2))\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right). Therefore the corresponding inequality (20) can be proved as in the case of γ=0\gamma=0. When γ=m\gamma=m or m+1m+1 for m2m\geq 2, (20) follows in the exact same manner as in the case γ=1\gamma=1. We write split the multiplier using binomial expansion of (1|ξ|2R2t2)m\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)^{m} as in (23) and get the desired estimate (20) as done for γ=1\gamma=1. When γ(m,m+1)\gamma\in(m,m+1), we write γ=ζ+m+1,ζ(1,0)\gamma=\zeta+m+1,\zeta\in(-1,0) and follow the same procedure as in the case when γ(0,1)\gamma\in(0,1) above.

Case II: When t[2j1ϵ0,2j+1]t\in[\sqrt{2^{-j-1-\epsilon_{0}}},\sqrt{2^{-j+1}}]

Note that in this case t2j/2.t\approx 2^{-j/2}. We rewrite

ψ(2j(1|ξ|2R2))(1|ξ|2R2t2)+γ=(1t2)γψ(2j(1|ξ|2R2))(1|ξ|2R2(1t2))+γ\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}}-t^{2}\right)_{+}^{\gamma}=(1-t^{2})^{\gamma}\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}(1-t^{2})}\right)_{+}^{\gamma}

Making a change of variable 1t2=s21-t^{2}=s^{2} in 2j(δ+1/2)2j/4(2j1ϵ02j+1|Sj,γ+1R,tf(x)|2t𝑑t)1/2,2^{-j(\delta+1/2)}2^{j/4}\left(\int_{\sqrt{2^{-j-1-\epsilon_{0}}}}^{\sqrt{2^{-j+1}}}|{S}_{j,\gamma+1}^{R,t}f(x)|^{2}tdt\right)^{1/2}, we get the following operator

Sj,γ+1R,sf(x)=nψ(2j(1|ξ|2R2))(1|ξ|2s2R2)+γf^(ξ)e2πix.ξ𝑑ξ,{S^{\prime}}_{j,\gamma+1}^{R,s}f(x)=\int_{{\mathbb{R}}^{n}}\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{s^{2}R^{2}}\right)_{+}^{\gamma}\hat{f}(\xi)e^{2\pi ix.\xi}d\xi,

where t2[2j1ϵ0,2j+1]t^{2}\in[2^{-j-1-\epsilon_{0}},2^{-j+1}] and 1t2=s21-t^{2}=s^{2}.

Therefore, we need to establish LpL^{p} boundedness of the following square function

supR>0(s1s2|Sj,γ+1R,sf(x)|2s𝑑s)1/2,\sup_{R>0}\left(\int_{s_{1}}^{s_{2}}|{S^{\prime}}_{j,\gamma+1}^{R,s}f(x)|^{2}sds\right)^{1/2},

where s1=12j+1s_{1}=\sqrt{1-2^{-j+1}} and s2=12j1ϵ0.s_{2}=\sqrt{1-2^{-j-1-\epsilon_{0}}}.

Note that we can ignore the term ss inside the integral as s1s\approx 1. Let M>100M^{\prime}>100 be a large number. We shall deal with the cases jMj\geq M^{\prime} and 2j<M2\leq j<M^{\prime} separately.

Proposition 5.2.

For jMj\geq M^{\prime} and 0<ϵ<10<\epsilon<1 we have the following estimate

supR>0(s1s2|Sj,γ+1R,sf()|2𝑑s)1/2Lp(n)ϵ2jγ2j(α(p)1/2)2ϵjfLp(n),\left\|\sup_{R>0}\left(\int^{s_{2}}_{s_{1}}|{S^{\prime}}_{j,\gamma+1}^{R,s}f(\cdot)|^{2}ds\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim_{\epsilon}2^{-j\gamma}2^{j(\alpha(p)-1/2)}2^{\epsilon j}\|f\|_{L^{p}({\mathbb{R}}^{n})},

when p𝔭np\geq\mathfrak{p}_{n} or p=2p=2 and γ>α(p)1/2\gamma>\alpha(p)-1/2.

Proposition 5.3.

When γ>α(p)1/2\gamma>\alpha(p)-1/2 and p=2p=2 or p𝔭np\geq\mathfrak{p}_{n} and 0<ϵ<10<\epsilon<1

supR>0(s1s2|Sj,γ+1R,sf(x)|2𝑑s)1/2Lp(n)C(M,γ,ϵ)fLp(n)\left\|\sup_{R>0}\left(\int^{s_{2}}_{s_{1}}|{S^{\prime}}_{j,\gamma+1}^{R,s}f(x)|^{2}ds\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\leq C(M^{\prime},\gamma,\epsilon)\|f\|_{L^{p}({\mathbb{R}}^{n})}

for all 2jM2\leq j\leq M^{\prime}.

Notice that Theorem 5.1 follows from Propositions 5.2 and 5.3 with γ=β1\gamma=\beta-1. Therefore, it remains to prove the Propositions 5.2 and 5.3.

Proof of Proposition 5.2

We perform suitable decomposition of the multiplier

ψ(2j(1|ξ|2R2))(1|ξ|2s2R2)+γ\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{s^{2}R^{2}}\right)_{+}^{\gamma}

and reduce our problem to already known forms of the square functions.

For convenience, we set R=1R=1 in the multiplier above and write

(24) (1|ξ|2s2)+γ\displaystyle\left(1-\frac{|\xi|^{2}}{s^{2}}\right)_{+}^{\gamma} =\displaystyle= k22kγψ~(2k(1|ξ|2s2))+ψ~0(|ξ|2s2)\displaystyle\sum_{k\geq 2}2^{-k\gamma}\tilde{\psi}\left(2^{k}\left(1-\frac{|\xi|^{2}}{s^{2}}\right)\right)+\tilde{\psi}_{0}\left(\frac{|\xi|^{2}}{s^{2}}\right)

where ψ~C0([1/2,2])\tilde{\psi}\in C^{\infty}_{0}([1/2,2]) and ψ~0C0([0,3/4])\tilde{\psi}_{0}\in C^{\infty}_{0}([0,3/4]). Using the above decomposition (24) we get

(25) ψ(2j(1|ξ|2R2))(1|ξ|2s2)+γ\displaystyle\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{s^{2}}\right)_{+}^{\gamma} =k22kγψ~(2k(1|ξ|2s2))ψ(2j(1|ξ|2R2))\displaystyle=\sum_{k\geq 2}2^{-k\gamma}\tilde{\psi}\left(2^{k}\left(1-\frac{|\xi|^{2}}{s^{2}}\right)\right)\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)
+ψ~0(|ξ|2s2)ψ(2j(1|ξ|2R2))\displaystyle+\tilde{\psi}_{0}\left(\frac{|\xi|^{2}}{s^{2}}\right)\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)

Let M>0M>0 be such that M1ψ~CN([12,2])M^{-1}\tilde{\psi}\in C^{N}([\frac{1}{2},2]). Note that ψ(2j(1|ξ|2))ψ~(2k(1|ξ|2s2))=0\psi\left(2^{j}\left(1-|\xi|^{2}\right)\right)\tilde{\psi}\left(2^{k}\left(1-\frac{|\xi|^{2}}{s^{2}}\right)\right)=0 whenever s2(12k1)<12j+1,s^{2}(1-2^{-k-1})<1-2^{-j+1}, i.e. when 2k<2j1s222j(1s2).2^{k}<2^{j-1}\frac{s^{2}}{2-2^{j}(1-s^{2})}. For s[s1,s2]s\in[s_{1},s_{2}] we have s2221ϵ0s222j(1s2)\frac{s^{2}}{2-2^{-1-\epsilon_{0}}}\leq\frac{s^{2}}{2-2^{j}(1-s^{2})}. Therefore, if 2k<2j1s2221ϵ02^{k}<2^{j-1}\frac{s^{2}}{2-2^{-1-\epsilon_{0}}}, the product of terms corresponding to jj and kk in (25)vanishes. This tells us that in the decomposition (25) we need to consider only those terms for which kj2k\geq j-2.

Next, we decompose the annular region {ξ:s2(12k+1)|ξ|2s2(12k1)}\{\xi:s^{2}(1-2^{-k+1})\leq|\xi|^{2}\leq s^{2}(1-2^{-k-1})\} which is nothing but the support of ψ~(2k(1|ξ|2s2))\tilde{\psi}\left(2^{k}\left(1-\frac{|\xi|^{2}}{s^{2}}\right)\right), into annuli of smaller width such that the function ψ(2j(1|ξ|2))\psi\left(2^{j}\left(1-|\xi|^{2}\right)\right), when restricted to each of these annuli, behaves like a constant.

Let φCc([1,1])\varphi\in C_{c}^{\infty}([-1,1]) be such that mφ(xm)=1\sum_{m\in{\mathbb{Z}}}\varphi(x-m)=1 on .{\mathbb{R}}. Let L=sup0lN,x[1,1]|dlφdxl|.L=\sup_{0\leq l\leq N,x\in[-1,1]}\left|\frac{d^{l}\varphi}{dx^{l}}\right|. For 0<ν<10<\nu<1 we can rewrite the sum above as mφ(ν1(xνm))1\sum_{m\in{\mathbb{Z}}}\varphi(\nu^{-1}(x-\nu m))\equiv 1 on .{\mathbb{R}}.

We restrict this identity to the interval [0,1][0,1], i.e., we consider 0k[ν1]+1φ(ν1(xνm))=1,x[0,1].\sum_{0\leq k\leq[\nu^{-1}]+1}\varphi(\nu^{-1}(x-\nu m))=1,~{}x\in[0,1]. Using suitable translation and dilation of the function we can get the same identity on the interval [s2(12k+1),s2(12k1)][s^{2}(1-2^{-k+1}),s^{2}(1-2^{-k-1})], i.e., we have

0m[ν1]+1φ(ν1(ba)1(xa(ba)νm))=1,\sum_{0\leq m\leq[\nu^{-1}]+1}\varphi(\nu^{-1}(b-a)^{-1}(x-a-(b-a)\nu m))=1,

where a=s2(12k+1)a=s^{2}(1-2^{-k+1}) and b=s2(12k1).b=s^{2}(1-2^{-k-1}).

Let ν=232kϵ\nu=\frac{2}{3}2^{-k\epsilon} for 0<ϵ<10<\epsilon<1 and note that (ba)ν=2(1+ϵ)ks2.(b-a)\nu=2^{-(1+\epsilon)k}s^{2}.

Putting x=|ξ|2x=|\xi|^{2} in the sum above and multiplying by ψ~(2k(1|ξ|2s2))\tilde{\psi}\left(2^{k}\left(1-\frac{|\xi|^{2}}{s^{2}}\right)\right) on both sides, we have the following decomposition

ψ~(2k(1|ξ|2s2))=0m[ν1]+1φ(2(1+ϵ)k(|ξ|2s21+2k+12(1+ϵ)km))ψ~(2k(1|ξ|2s2)).\tilde{\psi}\left(2^{k}\left(1-\frac{|\xi|^{2}}{s^{2}}\right)\right)=\sum_{0\leq m\leq[\nu^{-1}]+1}\varphi\left(2^{(1+\epsilon)k}\left(\frac{|\xi|^{2}}{s^{2}}-1+2^{-k+1}-2^{-(1+\epsilon)k}m\right)\right)\tilde{\psi}\left(2^{k}\left(1-\frac{|\xi|^{2}}{s^{2}}\right)\right).

Taking the one dimensional Fourier transform of ψ~\tilde{\psi} we have

ψ~(2k(1|ξ|2R2s2))=ψ~^(μ)e2πi(2k(1|ξ|2s2))μ𝑑μ.\tilde{\psi}\left(2^{k}\left(1-\frac{|\xi|^{2}}{R^{2}s^{2}}\right)\right)=\int_{{\mathbb{R}}}\widehat{\tilde{\psi}}(\mu)e^{2\pi i\left(2^{k}\left(1-\frac{|\xi|^{2}}{s^{2}}\right)\right)\mu}d\mu.

We rewrite the exponential term in the integral above in the following way

e2πi(2k(1|ξ|2s2)μ)=e2πi2k(2k+12k(1+ϵ)m)μe2πi2ϵk2(1+ϵ)k(|ξ|2s21+2k+12(1+ϵ)km)μ.e^{2\pi i\left(2^{k}\left(1-\frac{|\xi|^{2}}{s^{2}}\right)\mu\right)}=e^{2\pi i2^{k}(2^{-k+1}-2^{-k(1+\epsilon)}m)\mu}e^{-2\pi i2^{-\epsilon k}2^{(1+\epsilon)k}\left(\frac{|\xi|^{2}}{s^{2}}-1+2^{-k+1}-2^{-(1+\epsilon)k}m\right)\mu}.

Note that 2k(2k+1+2k(1+ϵ)m)2^{k}(2^{-k+1}+2^{-k(1+\epsilon)}m) is uniformly bounded for all m[ν1]+1m\leq[\nu^{-1}]+1 and k2k\geq 2. Denote

ξk,ms=(|ξ|2s21+2k+12(1+ϵ)km).\xi_{k,m}^{s}=\left(\frac{|\xi|^{2}}{s^{2}}-1+2^{-k+1}-2^{-(1+\epsilon)k}m\right).

Using Taylor’s series expansion we get

e2πi2kξk,msμ=q=0N1(1)qq!(2πi2ϵk)q2q(1+ϵ)k(ξk,ms)qμq+rN(2kξk,msμ),e^{-2\pi i2^{k}\xi_{k,m}^{s}\mu}=\sum_{q=0}^{N-1}\frac{(-1)^{q}}{q!}(2\pi i2^{-\epsilon k})^{q}2^{q(1+\epsilon)k}\left(\xi_{k,m}^{s}\right)^{q}\mu^{q}+r_{N}\left(2^{k}\xi_{k,m}^{s}\mu\right),

where rNr_{N} is the remainder term.

For 0qN0\leq q\leq N the remainder term rNr_{N} satisfies

(26) |dqrNdsq(s)|sNq.\left|\frac{d^{q}r_{N}}{ds^{q}}(s)\right|\leq s^{N-q}.

Let us write 2k(2k+1+2k(1+ϵ)m)=dk,mϵ2^{k}(2^{-k+1}+2^{-k(1+\epsilon)}m)=d^{\epsilon}_{k,m}. Using the expansion and notation as above we get that

ψ~(2k(1|ξ|2s2))\displaystyle\tilde{\psi}\left(2^{k}\left(1-\frac{|\xi|^{2}}{s^{2}}\right)\right) =\displaystyle= m[ν1]+1q=0N1(1)qq!(2πi2ϵk)qdqψ~dxq(dk,mϵ)φq(2(1+ϵ)kξk,ms)\displaystyle\sum_{m\leq[\nu^{-1}]+1}\sum_{q=0}^{N-1}\frac{(-1)^{q}}{q!}(2\pi i2^{-\epsilon k})^{q}\frac{d^{q}\tilde{\psi}}{dx^{q}}(d^{\epsilon}_{k,m})\varphi_{q}(2^{(1+\epsilon)k}\xi^{s}_{k,m})
+m[ν1]+1φ(2(1+ϵ)kξk,ms)ψ~^(μ)e2πidk,mϵμrN(2kξk,msη)𝑑μ,\displaystyle+\sum_{m\leq[\nu^{-1}]+1}\varphi\left(2^{(1+\epsilon)k}\xi_{k,m}^{s}\right)\int_{{\mathbb{R}}}\widehat{\tilde{\psi}}(\mu)e^{2\pi id^{\epsilon}_{k,m}\mu}~{}r_{N}\left(2^{k}\xi_{k,m}^{s}\eta\right)d\mu,

where φq(x)=xqφ(x)\varphi_{q}(x)=x^{q}\varphi(x).

Observe that the condition on ψ~\tilde{\psi} and the fact that dk,mϵd^{\epsilon}_{k,m} is uniformly bounded in kk and mm imply that

sup0qN1|dqψ~dxq(dk,mϵ)|M,\sup_{0\leq q\leq N-1}\left|\frac{d^{q}\tilde{\psi}}{dx^{q}}(d^{\epsilon}_{k,m})\right|\leq M,

where MM is independent of kk and m[ν1]+1m\leq[\nu^{-1}]+1.

Define

(27) URs,Rφq,j(k,m)f(x)=nφq(2(1+ϵ)kξk,mRs)ψ(2j(1|ξ|2R2))f^(ξ)e2πix.ξ𝑑ξ,U^{\varphi_{q},j}_{Rs,R}(k,m)f(x)=\int_{{\mathbb{R}}^{n}}\varphi_{q}\left(2^{(1+\epsilon)k}\xi_{k,m}^{Rs}\right)\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\hat{f}(\xi)e^{2\pi ix.\xi}d\xi,
(28) PRs,Rj,k(μ,m)f(x)=nφ(2(1+ϵ)kξk,mRs)ψ(2j(1|ξ|2R2))rN(2kξk,mRsμ)f^(ξ)e2πix.ξ𝑑ξ.P_{Rs,R}^{j,k}(\mu,m)f(x)=\int_{{\mathbb{R}}^{n}}\varphi\left(2^{(1+\epsilon)k}\xi_{k,m}^{Rs}\right)\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)r_{N}\left(2^{k}\xi_{k,m}^{Rs}\mu\right)\hat{f}(\xi)e^{2\pi ix.\xi}d\xi.

The decomposition of the symbol corresponding to the operator Sj,γ+1R,sS^{\prime R,s}_{j,\gamma+1} (replace ξ\xi by ξ/R\xi/R) as carried out in the discussion above gives us that

Sj,γ+1R,sf(x)\displaystyle{S^{\prime}}_{j,\gamma+1}^{R,s}f(x) =\displaystyle= kj22kγm[ν1]+1q=0N1(1)qq!(2πi2ϵk)qdqψ~dxq(dk,mϵ)URs,Rφq,j(k,m)f(x)\displaystyle\sum_{k\geq j-2}2^{-k\gamma}\sum_{m\leq[\nu^{-1}]+1}\sum_{q=0}^{N-1}\frac{(-1)^{q}}{q!}(2\pi i2^{-\epsilon k})^{q}\frac{d^{q}\tilde{\psi}}{dx^{q}}(d^{\epsilon}_{k,m})U^{\varphi_{q},j}_{Rs,R}(k,m)f(x)
+\displaystyle+ kj22kγm[ν1]+1ψ~^(μ)e2πidk,mϵμPRs,Rj,k(μ,m)f(x)𝑑μ.\displaystyle\sum_{k\geq j-2}2^{-k\gamma}\sum_{m\leq[\nu^{-1}]+1}\int_{{\mathbb{R}}}\widehat{\tilde{\psi}}(\mu)e^{2\pi id^{\epsilon}_{k,m}\mu}P_{Rs,R}^{j,k}(\mu,m)f(x)d\mu.

Therefore, in order to prove Proposition 5.2 we need to deal with the operators URs,Rφq,j(k,m)U^{\varphi_{q},j}_{Rs,R}(k,m) and PRs,Rj,k(μ,m)P_{Rs,R}^{j,k}(\mu,m). We first consider the operator URs,Rφq,j(k,m)U^{\varphi_{q},j}_{Rs,R}(k,m) with q=0q=0 and note that the terms with 1qN11\leq q\leq N-1 can be dealt with similarly as the function φq\varphi_{q} behaves in the same way for all 0qN10\leq q\leq N-1. Also, φq\varphi_{q} is a smooth function supported in [1,1][-1,1] and it satisfies sup0lN,x[1,1]|dlφqdxl|Lq!\sup_{0\leq l\leq N,x\in[-1,1]}|\frac{d^{l}\varphi_{q}}{dx^{l}}|\leq Lq!.

In order to prove the required estimates we need to perform another decomposition for the operator URs,Rφ0,j(k,m)U^{\varphi_{0},j}_{Rs,R}(k,m). We decompose it into operators whose multipliers are supported in smaller annular regions of width approximately 2(1+ϵ)k2^{-(1+\epsilon)k}. This process leaves a remainder term which can be dealt with easily.

As previously, we work with R=1R=1 and then replace ξ\xi by ξ/R\xi/R. We perform a similar decomposition to the function ψ(2j(1|ξ|2))\psi(2^{j}(1-|\xi|^{2})) into smooth functions having their supports in the annulus {ξ:|ξ|2[a+s22(1+ϵ)k(m1),a+s22(1+ϵ)k(m+1)]}\{\xi:|\xi|^{2}\in[a+s^{2}2^{-(1+\epsilon)k}(m-1),a+s^{2}2^{-(1+\epsilon)k}(m+1)]\}, where a=s2(12k+1).a=s^{2}(1-2^{-k+1}).

Consider ψ(2j(1|ξ|2))=ψ^(μ)e2πi(2j(1|ξ|2))μ𝑑μ\psi(2^{j}(1-|\xi|^{2}))=\int_{{\mathbb{R}}}\hat{\psi}(\mu)e^{2\pi i(2^{j}(1-|\xi|^{2}))\mu}d\mu and write

e2πi(2j(1|ξ|2))μ=e2πi2j(1as22(1+ϵ)km)μe2πi2js2(|ξ|2s21+2k+12(1+ϵ)km)μ.e^{2\pi i(2^{j}(1-|\xi|^{2}))\mu}=e^{2\pi i2^{j}(1-a-s^{2}2^{-(1+\epsilon)k}m)\mu}e^{-2\pi i2^{j}s^{2}(\frac{|\xi|^{2}}{s^{2}}-1+2^{-k+1}-2^{-(1+\epsilon)k}m)\mu}.

Next we write 2js2ξk,ms=2j2(1+ϵ)ks22(1+ϵ)kξk,ms2^{j}s^{2}\xi^{s}_{k,m}=2^{j}2^{-(1+\epsilon)k}s^{2}2^{(1+\epsilon)k}\xi^{s}_{k,m} and use Taylor’s series expansion of e2πi2js2ξk,msμe^{-2\pi i2^{j}s^{2}\xi^{s}_{k,m}\mu} to get that

φ(2(1+ϵ)kξk,ms)ψ(2j(1|ξ|2))\displaystyle\varphi\left(2^{(1+\epsilon)k}\xi_{k,m}^{s}\right)\psi(2^{j}(1-|\xi|^{2})) =\displaystyle= l0N11l!s22d(1+ϵ)l2jϵlφl(2(1+ϵ)kξk,ms)dlψdxl(2j(1as22(1+ϵ)km))\displaystyle\sum_{l\geq 0}^{N-1}\frac{1}{l!}s^{2}2^{-d(1+\epsilon)l}2^{-j\epsilon l}\varphi_{l}\left(2^{(1+\epsilon)k}\xi_{k,m}^{s}\right)\frac{d^{l}\psi}{dx^{l}}\left(2^{j}(1-a-s^{2}2^{-(1+\epsilon)k}m)\right)
+φ(2(1+ϵ)kξk,ms)ψ^(μ)e2πi(2j(1as22(1+ϵ)km)μ)rN(2js2ξk,msμ)𝑑μ.\displaystyle+\varphi\left(2^{(1+\epsilon)k}\xi_{k,m}^{s}\right)\int_{{\mathbb{R}}}\hat{\psi}(\mu)e^{2\pi i\left(2^{j}(1-a-s^{2}2^{-(1+\epsilon)k}m)\mu\right)}r_{N}\left(2^{j}s^{2}\xi_{k,m}^{s}\mu\right)d\mu.

Recall that here k=j+dk=j+d and φl(x)=xlφ(x)\varphi_{l}(x)=x^{l}\varphi(x).

Consider the following operators

VRsφl(k,m)f(x)=nφl(2(1+ϵ)kξk,mRs)f^(ξ)e2πix.ξ𝑑ξ,V^{\varphi_{l}}_{Rs}(k,m)f(x)=\int_{{\mathbb{R}}^{n}}\varphi_{l}\left(2^{(1+\epsilon)k}\xi_{k,m}^{Rs}\right)\hat{f}(\xi)e^{2\pi ix.\xi}d\xi,
QN,R,sk(μ)f(x)=nφ(2(1+ϵ)kξk,mRs)rN(2js2ξk,mRsμ)f^(ξ)e2πix.ξ𝑑ξ.Q_{N,R,s}^{k}(\mu)f(x)=\int_{{\mathbb{R}}^{n}}\varphi\left(2^{(1+\epsilon)k}\xi_{k,m}^{Rs}\right)r_{N}\left(2^{j}s^{2}\xi_{k,m}^{Rs}\mu\right)\hat{f}(\xi)e^{2\pi ix.\xi}d\xi.

Finally, replace ξ\xi by ξ/R\xi/R in the decomposition of URs,Rφ0,j(k,m)U^{\varphi_{0},j}_{Rs,R}(k,m) as above to get that

(30) URs,Rφ0,j(k,m)f(x)=l0N11l!s22d(1+ϵ)l2jϵlψl(2j(1as22(1+ϵ)km))VRsφl(k,m)f(x)+XN,R,skf(x),U^{\varphi_{0},j}_{Rs,R}(k,m)f(x)=\sum_{l\geq 0}^{N-1}\frac{1}{l!}s^{2}2^{-d(1+\epsilon)l}2^{-j\epsilon l}\psi^{l}(2^{j}(1-a-s^{2}2^{-(1+\epsilon)k}m))V^{\varphi_{l}}_{Rs}(k,m)f(x)+X_{N,R,s}^{k}f(x),

where

XN,R,skf(x)=ψ^(μ)e2πi2j(1as22(1+ϵ)km)μQN,R,sk(μ)f(x)𝑑μ.X_{N,R,s}^{k}f(x)=\int_{{\mathbb{R}}}\hat{\psi}(\mu)e^{2\pi i2^{j}(1-a-s^{2}2^{-(1+\epsilon)k}m)\mu}Q_{N,R,s}^{k}(\mu)f(x)d\mu.

It is easy to see that

sup0lN1|dlψdxl(2j(1as22(1+ϵ)km))|D,\sup_{0\leq l\leq N-1}\left|\frac{d^{l}\psi}{dx^{l}}(2^{j}(1-a-s^{2}2^{-(1+\epsilon)k}m))\right|\leq D,

where DD is independent of j,kj,k, s[s1,s2]s\in[s_{1},s_{2}] and m[ν1]+1m\leq[\nu^{-1}]+1.

Now we have the following estimate for the operator VRsφl(k,m)V^{\varphi_{l}}_{Rs}(k,m).

Lemma 5.4.

Let pp and α(p)\alpha(p) be as in Proposition 5.2. Then the following estimate holds.

supR>0(s1s2|VRsφl(k,m)f()|2𝑑s)1/2Lp(n)C(p)l!2(1+ϵ)(α(p)1/2)kfLp(n).\left\|\sup_{R>0}\left(\int_{s_{1}}^{s_{2}}|V^{\varphi_{l}}_{Rs}(k,m)f(\cdot)|^{2}ds\right)^{1/2}\right\|_{L^{p}({\mathbb{R}}^{n})}\leq C(p)l!2^{(1+\epsilon)(\alpha(p)-1/2)k}\|f\|_{L^{p}({\mathbb{R}}^{n})}.
Proof.

Recall that

φl(2(1+ϵ)kξk,mRs)=φl(2(1+ϵ)k(|ξ|2R2s21+2k+12(1+ϵ)km)).\varphi_{l}\left(2^{(1+\epsilon)k}\xi_{k,m}^{Rs}\right)=\varphi_{l}\left(2^{(1+\epsilon)k}\left(\frac{|\xi|^{2}}{R^{2}s^{2}}-1+2^{-k+1}-2^{-(1+\epsilon)k}m\right)\right).

Denote (12k+12(1+ϵ)km)=ck,mϵ(1-2^{-k+1}-2^{-(1+\epsilon)k}m)=c_{k,m}^{\epsilon}. Note that for fixed ϵ>0\epsilon>0 and k10k\geq 10, ck,mϵc_{k,m}^{\epsilon} has a uniform lower and upper bound in kk and for all m[ν1]+1m\leq[\nu^{-1}]+1. Since 2(1+ϵ)kξk,ms=2(1+ϵ)kck,mϵ(1|ξ|2s2ck,mϵ),2^{(1+\epsilon)k}\xi^{s}_{k,m}=-2^{(1+\epsilon)k}c_{k,m}^{\epsilon}\left(1-\frac{|\xi|^{2}}{s^{2}c_{k,m}^{\epsilon}}\right), We can rewrite

φl(2(1+ϵ)kξk,mRs)=φl(ϱ1(1|ξ|2R2s2ck,mϵ)),\varphi_{l}\left(2^{(1+\epsilon)k}\xi_{k,m}^{Rs}\right)=\varphi_{l}\left(-\varrho^{-1}\left(1-\frac{|\xi|^{2}}{R^{2}s^{2}c_{k,m}^{\epsilon}}\right)\right),

where ϱ1=2(1+ϵ)kck,mϵ\varrho^{-1}=2^{(1+\epsilon)k}c_{k,m}^{\epsilon}.

After making a change of variable sR1(ck,mϵ)1ss\rightarrow R^{-1}(c_{k,m}^{\epsilon})^{-1}s in (s1s2|VRsφl(k,m)f()|2𝑑s)1/2\left(\int_{s_{1}}^{s_{2}}|V^{\varphi_{l}}_{Rs}(k,m)f(\cdot)|^{2}ds\right)^{1/2} observe that it can be dominated by

supR>0(R1Rck,mϵs1Rck,mϵs2|Bϱ,Rck,mϵsφlf(x)|2𝑑s)1/2.\sup_{R>0}\left(R^{-1}\int_{Rc^{\epsilon}_{k,m}s_{1}}^{Rc^{\epsilon}_{k,m}s_{2}}|B_{\varrho,Rc^{\epsilon}_{k,m}s}^{\varphi_{l}}f(x)|^{2}ds\right)^{1/2}.

Further, the quantity above can be dominated by the following square function

(0|Bϱ,sφl,f(x)|2s1𝑑s)1/2.\left(\int_{0}^{\infty}|B_{\varrho,s}^{\varphi_{l},}f(x)|^{2}s^{-1}ds\right)^{1/2}.

The boundedness of the square function as above can be deduced from [15] with the desired bound, see Lemma 4.2 for detail. ∎

Next, we have the following estimate for the operator XN,RskX^{k}_{N,Rs}.

Lemma 5.5.

For 1<p1<p\leq\infty, the following holds.

supR>0,s[s1,s2]|XN,R,skf|Lp(n)2(1+ϵ)(α(p)1/2)kfLp(n).\left\|\sup_{R>0,~{}s\in[s_{1},s_{2}]}|X^{k}_{N,R,s}f|\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim 2^{(1+\epsilon)(\alpha(p)-1/2)k}\|f\|_{L^{p}({\mathbb{R}}^{n})}.
Proof.

Recall the definition of XN,R,skX^{k}_{N,R,s} and observe that it is enough to prove that

(31) supR>0,s[s1,s2]|QN,R,sk(μ)f|Lp(n)(1+|μ|)N2jϵN2d(1+ϵ)N2(1+ϵ)knfLp(n)\left\|\sup_{R>0,~{}s\in[s_{1},s_{2}]}|Q_{N,R,s}^{k}(\mu)f|\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim(1+|\mu|)^{N}2^{-j\epsilon N}2^{-d(1+\epsilon)N}2^{(1+\epsilon)kn}\|f\|_{L^{p}({\mathbb{R}}^{n})}

for all 1<p1<p\leq\infty.

Let k=j+dk=j+d and define

(32) Ms,μ(ξ)=2jϵN2d(1+ϵ)Nφ(2(1+ϵ)kξk,ms)rN(2js2ξk,msμ).M_{s,\mu}(\xi)=2^{j\epsilon N}2^{d(1+\epsilon)N}\varphi\left(2^{(1+\epsilon)k}\xi_{k,m}^{s}\right)r_{N}\left(2^{j}s^{2}\xi_{k,m}^{s}\mu\right).

We also define MR,s,μ(ξ)=Ms,μ(ξ/R)M_{R,s,\mu}(\xi)=M_{s,\mu}(\xi/R). Note that QN,R,sk(μ)f=2jϵN2d(1+ϵ)Nf(1MR,s,μ)Q_{N,R,s}^{k}(\mu)f=2^{-j\epsilon N}2^{-d(1+\epsilon)N}f*(\mathcal{F}^{-1}M_{R,s,\mu}). Here 1f\mathcal{F}^{-1}f denotes the inverse Fourier transform of the function ff.

From the estimates on rNr_{N} in (26), it is easy to see that |βξβMs,μ(ξ)|cL(1+|μ|)N2(1+ϵ)k|β||\frac{\partial^{\beta}}{\partial\xi^{\beta}}M_{s,\mu}(\xi)|\leq cL(1+|\mu|)^{N}2^{(1+\epsilon)k|\beta|} for all |β|N.|\beta|\leq N. In particular, when β=0\beta=0, we have

|Ms,μ(ξ)|2jϵN2d(1+ϵ)N|φ(2(1+ϵ)kξk,ms)||2js2ξk,msμ|N.|M_{s,\mu}(\xi)|\leq 2^{j\epsilon N}2^{d(1+\epsilon)N}|\varphi\left(2^{(1+\epsilon)k}\xi_{k,m}^{s}\right)|~{}|2^{j}s^{2}\xi_{k,m}^{s}\mu|^{N}.

We know that φ(2(1+ϵ)kξk,ms)\varphi\left(2^{(1+\epsilon)k}\xi_{k,m}^{s}\right) is non-zero for |ξk,ms|2(1+ϵ)k|\xi_{k,m}^{s}|\leq 2^{-(1+\epsilon)k} and φL\|\varphi\|_{\infty}\leq L. This gives us the desired estimate for βξβMs,μ(ξ)\frac{\partial^{\beta}}{\partial\xi^{\beta}}M_{s,\mu}(\xi) when β=0\beta=0.

Note that the function Ms,μ(ξ)M_{s,\mu}(\xi) is supported in a cube of sidelength at most 22 for all s[s1,s2]s\in[s_{1},s_{2}], m[ν1]+1m\leq[\nu^{-1}]+1 and μ.\mu.

An integration by parts argument implies that

(33) |1Ms,μ(x)|cL(1+|μ|)N(1+2(1+ϵ)k|x|)n1|\mathcal{F}^{-1}M_{s,\mu}(x)|\leq cL(1+|\mu|)^{N}(1+2^{-(1+\epsilon)k}|x|)^{-n-1}

holds uniformly in ss. The estimate above yields the pointwise estimate

supR>0,s[s1,s2]|QN,R,sk(μ)f(x)|(1+|μ|)N2jϵN2d(1+ϵ)N2k(1+ϵ)nf(x).\sup_{R>0,~{}s\in[s_{1},s_{2}]}|Q_{N,R,s}^{k}(\mu)f(x)|\lesssim(1+|\mu|)^{N}2^{-j\epsilon N}2^{-d(1+\epsilon)N}2^{k(1+\epsilon)n}\mathcal{M}f(x).

Choosing NN large enough we get the desired estimate. ∎

Combining the results from Lemmata 5.4 and 5.5 and putting them in (30) we get that

(34) supR>0(s1s2|URs,Rφq,j(k,m)f()|2𝑑s)1/2p2(1+ϵ)(α(p)1/2)kfp\left\|\sup_{R>0}\left(\int_{s_{1}}^{s_{2}}|U^{\varphi_{q},j}_{Rs,R}(k,m)f(\cdot)|^{2}ds\right)^{1/2}\right\|_{p}\lesssim 2^{(1+\epsilon)(\alpha(p)-1/2)k}\|f\|_{p}

for the same range of pp as in Lemma 5.4.

Now we deal with the remaining part in (5) which is given by

m[ν1]+1ψ~^(μ)e2πidk,mϵμPRs,Rj,k(μ,m)f(x)𝑑μ.\sum_{m\leq[\nu^{-1}]+1}\int_{{\mathbb{R}}}\widehat{\tilde{\psi}}(\mu)e^{2\pi id^{\epsilon}_{k,m}\mu}P_{Rs,R}^{j,k}(\mu,m)f(x)d\mu.

Recall from (28) that

PRs,Rj,k(μ,m)f(x)=nφ(2(1+ϵ)kξk,mRs)ψ(2j(1|ξ|2R2))rN(2kξk,mRsμ)f^(ξ)e2πix.ξ𝑑ξ.P_{Rs,R}^{j,k}(\mu,m)f(x)=\int_{{\mathbb{R}}^{n}}\varphi\left(2^{(1+\epsilon)k}\xi_{k,m}^{Rs}\right)\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)r_{N}\left(2^{k}\xi_{k,m}^{Rs}\mu\right)\hat{f}(\xi)e^{2\pi ix.\xi}d\xi.

Fix m[ν1]+1m\leq[\nu^{-1}]+1 and decompose ψ(2j(1|ξ|2R2))\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right) into a sum of smooth functions supported in an annulus of width of the order 2(1+ϵ)k2^{-(1+\epsilon)k} and a remainder term. This is similar to the previous case. We get that

(35) PRs,Rj,k(μ,m)f(x)=l=0N11l!s22d(1+ϵ)l2jϵlψj(k,m,l)Ds,Rl,k(μ,m)f(x)\displaystyle P_{Rs,R}^{j,k}(\mu,m)f(x)=\sum_{l=0}^{N-1}\frac{1}{l!}s^{2}2^{-d(1+\epsilon)l}2^{-j\epsilon l}\psi_{j}(k,m,l)D_{s,R}^{l,k}(\mu,m)f(x)
+nψ^(μ)sNR(μ,μ)φ(2(1+ϵ)kξk,mRs)f^(ξ)e2πix.ξe2πi2j(1as22(1+ϵ)km)μ𝑑μ𝑑ξ,\displaystyle+\int_{{\mathbb{R}}}\int_{{\mathbb{R}}^{n}}\hat{\psi}(\mu^{\prime})s_{N}^{R}(\mu,\mu^{\prime})\varphi(2^{(1+\epsilon)k}\xi_{k,m}^{Rs})\hat{f}(\xi)e^{2\pi ix.\xi}e^{2\pi i2^{j}(1-a-s^{2}2^{-(1+\epsilon)k}m)\mu^{\prime}}d\mu^{\prime}d\xi,

where ψj(k,m,l)=dlψdxl(2j(1as22(1+ϵ)km)),sNR(μ,μ)=rN(2kξk,mRsμ)rN(2js2ξk,mRsμ)\psi_{j}(k,m,l)=\frac{d^{l}\psi}{dx^{l}}(2^{j}(1-a-s^{2}2^{-(1+\epsilon)k}m)),~{}s_{N}^{R}(\mu,\mu^{\prime})=r_{N}(2^{k}\xi_{k,m}^{Rs}\mu)r_{N}(2^{j}s^{2}\xi_{k,m}^{Rs}\mu^{\prime}) and

(36) Ds,Rl,k(μ,m)f(x)=nφl(2(1+ϵ)kξk,mRs)rN(2kξk,mRsμ)e2πix.ξf^(ξ)𝑑ξD_{s,R}^{l,k}(\mu,m)f(x)=\int_{{\mathbb{R}}^{n}}\varphi_{l}(2^{(1+\epsilon)k}\xi_{k,m}^{Rs})r_{N}(2^{k}\xi_{k,m}^{Rs}\mu)e^{2\pi ix.\xi}\hat{f}(\xi)d\xi

for 0lN10\leq l\leq N-1

It is not difficult to see that the maximal function (with respect to RR ) associated with ψ^(μ)e2πidk,mϵμDs,Rl,k(μ,m)f(x)𝑑μ\int_{{\mathbb{R}}}\hat{\psi}(\mu)e^{2\pi id^{\epsilon}_{k,m}\mu}D_{s,R}^{l,k}(\mu,m)f(x)d\mu can be dealt with in a similar manner as we did for the maximal function supR>0s[s1,s2]|XN,R,skf|\sup_{R>0~{}s\in[s_{1},s_{2}]}|X_{N,R,s}^{k}f| in Lemma 5.5. Moreover, the bound is independent of k,mk,m and ss and grows at most like l!l! in ll.

Define E(μ,μ)=e2πidk,mϵμe2πi2j(1as22(1+ϵ)km)μE(\mu,\mu^{\prime})=e^{2\pi id^{\epsilon}_{k,m}\mu}e^{2\pi i2^{j}(1-a-s^{2}2^{-(1+\epsilon)k}m)\mu^{\prime}} and consider

(37) IN,Rskf(x)=2ψ~^(μ)ψ^(μ)E(μ,μ)TNRs(μ,μ)f(x)𝑑μ𝑑μ,I_{N,Rs}^{k}f(x)=\int_{{\mathbb{R}}^{2}}\widehat{\tilde{\psi}}(\mu)\hat{\psi}(\mu^{\prime})E(\mu,\mu^{\prime})T_{N}^{Rs}(\mu,\mu^{\prime})f(x)d\mu d\mu^{\prime},

where

TNRs(μ,μ)f(x)=nsNR(μ,μ)φ(2(1+ϵ)kξk,mRs)e2πix.ξf^(ξ)𝑑ξ.T_{N}^{Rs}(\mu,\mu^{\prime})f(x)=\int_{{\mathbb{R}}^{n}}s_{N}^{R}(\mu,\mu^{\prime})\varphi(2^{(1+\epsilon)k}\xi_{k,m}^{Rs})e^{2\pi ix.\xi}\hat{f}(\xi)d\xi.

For 1<p1<p\leq\infty, we have the following estimate for IN,RskfI_{N,Rs}^{k}f

(38) supR>0,s[s1,s2]|IN,Rskf|Lp(n)2jϵN2d(1+ϵ)N2k(1+ϵ)N2(1+ϵ)knfLp(n)\displaystyle\left\|\sup_{R>0,~{}s\in[s_{1},s_{2}]}|I_{N,Rs}^{k}f|\right\|_{L^{p}({\mathbb{R}}^{n})}\lesssim 2^{-j\epsilon N}2^{-d(1+\epsilon)N}2^{-k(1+\epsilon)N}2^{(1+\epsilon)kn}\|f\|_{L^{p}({\mathbb{R}}^{n})}

Now from (37) we know that it is enough to show that

supR>0,s[s1,s2]|TNRs(μ,μ)f|Lp(n)(1+|μ1|+|μ2|)N2jϵN2d(1+ϵ)N2k(1+ϵ)N2(1+ϵ)knfLp(n).\left\|\sup_{R>0,~{}s\in[s_{1},s_{2}]}|T_{N}^{Rs}(\mu,\mu^{\prime})f|\right\|_{L^{p}({\mathbb{R}}^{n})}\leq(1+|\mu_{1}|+|\mu_{2}|)^{N}2^{-j\epsilon N}2^{-d(1+\epsilon)N}2^{-k(1+\epsilon)N}2^{(1+\epsilon)kn}\|f\|_{L^{p}({\mathbb{R}}^{n})}.

Define

Mμ,μR,s(ξ)=2jϵN2d(1+ϵ)N2k(1+ϵ)NsNR(μ,μ)φ(2(1+ϵ)kξk,mRs).M_{\mu,\mu^{\prime}}^{R,s}(\xi)=2^{j\epsilon N}2^{d(1+\epsilon)N}2^{k(1+\epsilon)N}s_{N}^{R}(\mu,\mu^{\prime})\varphi(2^{(1+\epsilon)k}\xi_{k,m}^{Rs}).

It can be verified that Mμ,μR,s(ξ)=Mμ,μ1,s(ξ/R)M_{\mu,\mu^{\prime}}^{R,s}(\xi)=M_{\mu,\mu^{\prime}}^{1,s}(\xi/R). Further, the inverse Fourier transform of Mμ,μ1,s(ξ)M_{\mu,\mu^{\prime}}^{1,s}(\xi) satisfies estimate similar to the one in (33). Therefore, the boundedness of the corresponding maximal function follows in a similar fashion as in Lemma 5.5.

By the estimate (38) and the explanation provided as earlier we get the desired estimate for the operator

ψ~^(μ)e2πidk,mϵμPRs,Rj,k(μ,m)f(x)𝑑μ.\int_{{\mathbb{R}}}\widehat{\tilde{\psi}}(\mu)e^{2\pi id^{\epsilon}_{k,m}\mu}P_{Rs,R}^{j,k}(\mu,m)f(x)d\mu.

Combining all the results together in (5) we complete the proof of Proposition 5.2. ∎

Proof of Proposition 5.3

Note that

Sj,γ+1R,sf(x)\displaystyle{S^{\prime}}_{j,\gamma+1}^{R,s}f(x) =\displaystyle= nψ(2j(1|ξ|2R2))(1|ξ|2R2s2)+γf^(ξ)e2πix.ξ𝑑ξ\displaystyle\int_{{\mathbb{R}}^{n}}\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}}{R^{2}s^{2}}\right)_{+}^{\gamma}\hat{f}(\xi)e^{2\pi ix.\xi}d\xi
=\displaystyle= B2j,RψBRsγf(x),\displaystyle B^{\psi}_{2^{-j},R}B_{Rs}^{\gamma}f(x),

where

B2j,Rψ(f)(x)=nψ(2j(1|ξ|2R2))f^(ξ)e2πix.ξ𝑑ξ.B^{\psi}_{2^{-j},R}(f)(x)=\int_{{\mathbb{R}}^{n}}\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right)\hat{f}(\xi)e^{2\pi ix.\xi}d\xi.

As done in Lemma 4.3 we have that

(supR>0s1s2|B2j,RψBRsγf(x)|2𝑑s)1/2k=1n0(supR>0s1s2|B2j,Rψ(BRsγ+kBRsγ+k1)f(x)|2𝑑s)1/2\displaystyle\left(\sup_{R>0}\int_{s_{1}}^{s_{2}}|B^{\psi}_{2^{-j},R}B_{Rs}^{\gamma}f(x)|^{2}ds\right)^{1/2}\leq\sum_{k=1}^{n_{0}}\left(\sup_{R>0}\int_{s_{1}}^{s_{2}}|B^{\psi}_{2^{-j},R}\left(B_{Rs}^{\gamma+k}-B_{Rs}^{\gamma+k-1}\right)f(x)|^{2}ds\right)^{1/2}
+supR>0(s1s2|B2j,RψBRsγ+n0f(x)|2𝑑s)1/2,\displaystyle+\sup_{R>0}\left(\int_{s_{1}}^{s_{2}}|B^{\psi}_{2^{-j},R}B_{Rs}^{\gamma+n_{0}}f(x)|^{2}ds\right)^{1/2},

where n0n_{0} is such that γ+n0>n12.\gamma+n_{0}>\frac{n-1}{2}.

For a fixed 1kn01\leq k\leq n_{0} let us first consider

(supR>0s1s2|B2j,Rψ(BRsγ+kBRsγ+k1)f(x)|2𝑑s)1/2\left(\sup_{R>0}\int_{s_{1}}^{s_{2}}|B^{\psi}_{2^{-j},R}\left(B_{Rs}^{\gamma+k}-B_{Rs}^{\gamma+k-1}\right)f(x)|^{2}ds\right)^{1/2}

After applying the change of variable sRss\rightarrow Rs the quantity above can be dominated by the following square function

(supR>00|B2j,Rψ(Bsγ+kBsγ+k1)f(x)|2s1𝑑s)1/2.\left(\sup_{R>0}\int_{0}^{\infty}|B^{\psi}_{2^{-j},R}\left(B_{s}^{\gamma+k}-B_{s}^{\gamma+k-1}\right)f(x)|^{2}s^{-1}ds\right)^{1/2}.

Since ψ(2j(1|ξ|2R2))\psi\left(2^{j}\left(1-\frac{|\xi|^{2}}{R^{2}}\right)\right) is a compactly supported smooth function. An integration by parts argument tells us that the L1L^{1} norm of it’s inverse Fourier transform is bounded by 2jn2^{jn}. Therefore, we get the pointwise estimate. supR>0|B2j,Rψf(x)|2jnf(x)\sup_{R>0}|B^{\psi}_{2^{-j},R}f(x)|\leq 2^{jn}\mathcal{M}f(x) and consequently we have that

(supR>00|B2j,Rψ(Bsγ+kBsγ+k1)f(x)|2s1𝑑s)1/22jn(0|(Bsγ+kBsγ+k1)f(x)|2s1𝑑s)1/2.\left(\sup_{R>0}\int_{0}^{\infty}|B^{\psi}_{2^{-j},R}\left(B_{s}^{\gamma+k}-B_{s}^{\gamma+k-1}\right)f(x)|^{2}s^{-1}ds\right)^{1/2}\leq 2^{jn}\left(\int_{0}^{\infty}|\mathcal{M}\left(B_{s}^{\gamma+k}-B_{s}^{\gamma+k-1}\right)f(x)|^{2}s^{-1}ds\right)^{1/2}.

Invoking the vector valued boundedness of the Hardy-Littlewood Maximal function, see [5] for detail, the right hand side quantity in the inequality above is bounded on Lp(n,0fs()s1𝑑s)L^{p}({\mathbb{R}}^{n},\int_{0}^{\infty}f_{s}(\cdot)s^{-1}ds) for all 1<p.1<p\leq\infty. This yields

(supR>00|B2j,Rψ(Bsγ+kBsγ+k1)f(x)|2s1𝑑s)1/2p\displaystyle\left\|\left(\sup_{R>0}\int_{0}^{\infty}|B^{\psi}_{2^{-j},R}\left(B_{s}^{\gamma+k}-B_{s}^{\gamma+k-1}\right)f(x)|^{2}s^{-1}ds\right)^{1/2}\right\|_{p}
\displaystyle\leq 2jn(0|(Bsγ+kBsγ+k1)f(x)|2s1𝑑s)1/2p.\displaystyle 2^{jn}\left\|\left(\int_{0}^{\infty}|\left(B_{s}^{\gamma+k}-B_{s}^{\gamma+k-1}\right)f(x)|^{2}s^{-1}ds\right)^{1/2}\right\|_{p}.

We can also deduce the above vector valued inequality for the Hardy Lttlewood Maximal function from the result of Feffereman and Stein [7] as here we are dealing with a separable Hilbert space valued functions and any separable Hilbert space is isomorphic to l2()l^{2}({\mathbb{N}}). The quantity on the right side in the inequality above is nothing but the square function for the Bochner-Riesz means, which it is bounded on Lp(n)L^{p}({\mathbb{R}}^{n}) for γ>α(p)1/2\gamma>\alpha(p)-1/2 and p𝔭np\geq\mathfrak{p}_{n} or p=2.p=2.

Next, we consider the maximal function

supR>0(s1s2|B2j,RψBRsγ+n0f(x)|2𝑑s)1/2.\sup_{R>0}\left(\int_{s_{1}}^{s_{2}}|B^{\psi}_{2^{-j},R}B_{Rs}^{\gamma+n_{0}}f(x)|^{2}ds\right)^{1/2}.

Again a change of variable ssRs\rightarrow sR in the above gives

supR>0(R1Rs1Rs2|B2j,RψBsγ+n0f(x)|2𝑑s)1/2.\sup_{R>0}\left(R^{-1}\int_{Rs_{1}}^{Rs_{2}}|B^{\psi}_{2^{-j},R}B_{s}^{\gamma+n_{0}}f(x)|^{2}ds\right)^{1/2}.

Since B2j,Rψf(x)2jnf(x)B^{\psi}_{2^{j},R}f(x)\leq 2^{jn}\mathcal{M}f(x) a.e. we get that

|Bsγ+n0B2j,Rψf(x)|2jnn|Ks(xy)|f(y)𝑑y,|B_{s}^{\gamma+n_{0}}B^{\psi}_{2^{-j},R}f(x)|\leq 2^{jn}\int_{{\mathbb{R}}^{n}}|K_{s}(x-y)|\mathcal{M}f(y)dy,

where KsK_{s} is the kernel associated with the operator Bsγ+n0B_{s}^{\gamma+n_{0}} and hence the LpL^{p} boundedness of sups>0,R>0|Bsγ+n0B2j,Rψf|\sup_{s>0,R>0}|B_{s}^{\gamma+n_{0}}B^{\psi}_{2^{-j},R}f| follows for 1<p1<p\leq\infty as γ+n0\gamma+n_{0} is large enough to guarantee that Bsγ+n0B_{s}^{\gamma+n_{0}} is dominated by the Hardy-Littlewood maximal function in a pointwise a.e. sense. ∎

5.3. Boundedness of the maximal function 0,α(f,g)\mathcal{B}^{\alpha}_{0,*}(f,g)

In this subsection we prove the boundedness of the maximal function 0,α(f,g).\mathcal{B}^{\alpha}_{0,*}(f,g). In particular, we show that for n2n\geq 2 and α>α(p1,p2)\alpha>\alpha_{*}(p_{1},p_{2}), the following holds

(39) 0,α(f,g)|Lp(n)fLp1(n)gLp2(n),\displaystyle\|\mathcal{B}^{\alpha}_{0,*}(f,g)|\|_{L^{p}({\mathbb{R}}^{n})}\lesssim\|f\|_{L^{p_{1}}({\mathbb{R}}^{n})}\|g\|_{L^{p_{2}}({\mathbb{R}}^{n})},

where pi=2p_{i}=2 or pi𝔭np_{i}\geq\mathfrak{p}_{n} for i=1,2i=1,2 and 1p=1p1+1p2\frac{1}{p}=\frac{1}{p_{1}}+\frac{1}{p_{2}}.

We decompose the multiplier m0,R(ξ,η)m_{0,R}(\xi,\eta) using the partition of identity in the η\eta variable. This is same as we did earlier in the ξ\xi variable. We get that

m0,R(ξ,η)=j2ψ0(|ξ|2R2)ψ(2j(1|η|2R2))(1|η|2R2)+α(1|ξ|2R2(1|η|2R2)1)+α\displaystyle m_{0,R}(\xi,\eta)=\sum_{j\geq 2}\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi\left(2^{j}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}\left(1-\frac{|\xi|^{2}}{R^{2}}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)^{-1}\right)^{\alpha}_{+}
+ψ0(|ξ|2R2)ψ0(|η|2R2)(1|ξ|2+|η|2R2)+α.\displaystyle+\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi_{0}\left(\frac{|\eta|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}.

We further split ψ0(x)=ψ01(x)+ψ02(x)\psi_{0}(x)=\psi_{0}^{1}(x)+\psi_{0}^{2}(x) such that support of ψ01\psi_{0}^{1} is contained in the interval [0,3/16][0,3/16] and support of ψ02\psi_{0}^{2} is contained in the interval [332,34][\frac{3}{32},\frac{3}{4}]. This gives us the follwoing decomposition of m0,Rm_{0,R}

m0,R(ξ,η)=j2ψ0(|ξ|2R2)ψ(2j(1|η|2R2))(1|η|2R2)+α(1|ξ|2R2(1|η|2R2)1)+α\displaystyle m_{0,R}(\xi,\eta)=\sum_{j\geq 2}\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi\left(2^{j}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}\left(1-\frac{|\xi|^{2}}{R^{2}}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)^{-1}\right)^{\alpha}_{+}
(40) +ψ0(|ξ|2R2)ψ01(|η|2R2)(1|ξ|2+|η|2R2)+α+ψ0(|ξ|2R2)ψ02(|η|2R2)(1|ξ|2+|η|2R2)+α.\displaystyle+\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi_{0}^{1}\left(\frac{|\eta|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}+\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi_{0}^{2}\left(\frac{|\eta|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}.

Note that ψ0(|ξ|2R2)ψ01(|η|2R2)(1|ξ|2+|η|2R2)+α\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi_{0}^{1}\left(\frac{|\eta|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha} is a smooth function supported in a ball of radius 1516R\sqrt{\frac{15}{16}}R. Therefore, it can be easily proved that the maximal function

(41) supR>0|n×nψ0(|ξ|2R2)ψ01(|η|2R2)(1|ξ|2+|η|2R2)+αf^(ξ)g^(η)e2πix.(ξ+η)𝑑ξ𝑑η|\sup_{R>0}\left|\int_{{\mathbb{R}}^{n}\times{\mathbb{R}}^{n}}\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi_{0}^{1}\left(\frac{|\eta|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}\hat{f}(\xi)\hat{g}(\eta)e^{2\pi ix.(\xi+\eta)}d\xi d\eta\right|

is bounded from Lp1(n)×Lp2(n)Lp(n)L^{p_{1}}({\mathbb{R}}^{n})\times L^{p_{2}}({\mathbb{R}}^{n})\rightarrow L^{p}({\mathbb{R}}^{n}) for all 1<pi,i=1,21<p_{i}\leq\infty,i=1,2.

Next, we deal with the maximal functions corresponding to the remaining terms in the decomposition (5.3).

Let KK be a large number such that for all jKj\geq K, the support of (1|ξ|2R2(1|η|2R2)1)+α\left(1-\frac{|\xi|^{2}}{R^{2}}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)^{-1}\right)^{\alpha}_{+} is contained in |ξ|R8|\xi|\leq\frac{R}{8} where η\eta belongs to the support of the function ψ(2j(1|η|2R2)).\psi\left(2^{j}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)\right). We know that ψ0(|ξ|2R2)1\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\equiv 1 when |ξ|R8.|\xi|\leq\frac{R}{8}. Consider the operator

T~R,Kα(f,g)(x)=n×nm~R,Kα(ξ,η)f^(ξ)g^(η)e2πix.(ξ+η)𝑑ξ𝑑η,\tilde{T}^{\alpha}_{R,K}(f,g)(x)=\int_{{\mathbb{R}}^{n}\times{\mathbb{R}}^{n}}\tilde{m}_{R,K}^{\alpha}(\xi,\eta)\hat{f}(\xi)\hat{g}(\eta)e^{2\pi ix.(\xi+\eta)}d\xi d\eta,

where

m~R,Kα(ξ,η)=jKψ0(|ξ|2R2)ψ(2j(1|η|2R2))(1|η|2R2)+α(1|ξ|2R2(1|η|2R2)1)+α.\tilde{m}_{R,K}^{\alpha}(\xi,\eta)=\sum_{j\geq K}\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi\left(2^{j}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}\left(1-\frac{|\xi|^{2}}{R^{2}}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)^{-1}\right)^{\alpha}_{+}.

The boundedness of the maximal function supR>0|T~R,Kα(f,g)|\sup_{R>0}|\tilde{T}^{\alpha}_{R,K}(f,g)| can be obtained in a similar fashion as in the previous section for the maximal function supR>0|jMTj,Rα(f,g)|.\sup_{R>0}|\sum_{j\geq M^{\prime}}T^{\alpha}_{j,R}(f,g)|. Therefore, we get that supR>0|T~R,Kα(f,g)|\sup_{R>0}|\tilde{T}^{\alpha}_{R,K}(f,g)| is bounded from Lp1(n)×Lp2(n)Lp(n)L^{p_{1}}({\mathbb{R}}^{n})\times L^{p_{2}}({\mathbb{R}}^{n})\rightarrow L^{p}({\mathbb{R}}^{n}) when α>α(p1,p2)\alpha>\alpha_{*}(p_{1},p_{2}) and pi=2p_{i}=2 or pi𝔭np_{i}\geq\mathfrak{p}_{n} for i=1,2i=1,2.

Now it remains to prove the desired estimate for the terms corresponding to 2j<K2\leq j<K. We consider

mR(ξ,η)=j2Kψ0(|ξ|2R2)ψ(2j(1|η|2R2))(1|ξ|2+|η|2R2)+α\displaystyle m^{\prime}_{R}(\xi,\eta)=\sum_{j\geq 2}^{K}\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi\left(2^{j}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}
+ψ0(|ξ|2R2)ψ02(|η|2R2)(1|ξ|2+|η|2R2)+α.\displaystyle+\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi_{0}^{2}\left(\frac{|\eta|^{2}}{R^{2}}\right)\left(1-\frac{|\xi|^{2}+|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}.

Let T~Rα(f,g)\tilde{T}^{\alpha}_{R}(f,g) be the bilinear operator corresponding to the multiplier mRm^{\prime}_{R}. i.e.,

T~Rα(f,g)(x)=nnmR(ξ,η)f^(ξ)g^(η)e2πix.(ξ+η)𝑑ξ𝑑η.\tilde{T}^{\alpha}_{R}(f,g)(x)=\int_{{\mathbb{R}}^{n}}\int_{{\mathbb{R}}^{n}}m^{\prime}_{R}(\xi,\eta)\hat{f}(\xi)\hat{g}(\eta)e^{2\pi ix.(\xi+\eta)}d\xi d\eta.

Note that for each fixed 2j<K2\leq j<K it is enough to deal with the operator associated with the multipiler

m(ξR,ηR)=ψ0(|ξ|2R2)ψ(2j(1|η|2R2))(1|η|2R2)+α(1|ξ|2R2(1|η|2R2)1)+α.m\left(\frac{\xi}{R},\frac{\eta}{R}\right)=\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi\left(2^{j}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)\right)\left(1-\frac{|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}\left(1-\frac{|\xi|^{2}}{R^{2}}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)^{-1}\right)^{\alpha}_{+}.

This is precisely the situation in section 3 when we decompose the bilinear Bochner-Riesz multiplier. We perform the same decomposition to the multiplier (1|ξ|2R2(1|η|2R2)1)+α\left(1-\frac{|\xi|^{2}}{R^{2}}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)^{-1}\right)^{\alpha}_{+} as we did in (6) with the roles of ξ\xi and η\eta interchanged. This gives us an expression similar to (10) with the roles of ff and gg interchanged and an extra multiplier term ψ0(|ξ|2R2)\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right) with f^\hat{f}. Define

(42) B1,Rψ0f(x)=nψ0(|ξ|2R2)f^(ξ)e2πix.ξ𝑑ξ.B^{\psi_{0}}_{1,R}f(x)=\int_{{\mathbb{R}}^{n}}\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\hat{f}(\xi)e^{2\pi ix.\xi}d\xi.

Observe that it is enough to prove that the maximal function (supR>0Rj10Rj|B1,Rψ0Btδf(x)|2𝑑t)1/2\left(\sup_{R>0}R_{j}^{-1}\int_{0}^{R_{j}}|B^{\psi_{0}}_{1,R}B_{t}^{\delta}f(x)|^{2}dt\right)^{1/2} is bounded on Lp1(n)L^{p_{1}}({\mathbb{R}}^{n}) for δ>α(p1)1/2\delta>\alpha(p_{1})-1/2 and p1=2p_{1}=2 or p1𝔭n.p_{1}\geq\mathfrak{p}_{n}. The term corresponding to the function gg can be dealt with by using the similar method as we did to prove estimate (20) and Proposition 5.3. Again following the same technique as in Lemma 4.3, we get the following estimate

(supR>0Rj10Rj|B1,Rψ0Btδf(x)|2𝑑t)1/2\displaystyle\left(\sup_{R>0}R_{j}^{-1}\int_{0}^{R_{j}}|B^{\psi_{0}}_{1,R}B_{t}^{\delta}f(x)|^{2}dt\right)^{1/2} \displaystyle\leq k=1n0(supR>0Rj10Rj|B1,Rψ0(Btδ+kBtδ+k1)f(x)|2𝑑t)1/2\displaystyle\sum_{k=1}^{n_{0}}\left(\sup_{R>0}R_{j}^{-1}\int_{0}^{R_{j}}|B^{\psi_{0}}_{1,R}\left(B_{t}^{\delta+k}-B_{t}^{\delta+k-1}\right)f(x)|^{2}dt\right)^{1/2}
+supR>0(Rj10Rj|B1,Rψ0Btδ+n0f(x)|2𝑑t)1/2,\displaystyle+\sup_{R>0}\left(R_{j}^{-1}\int_{0}^{R_{j}}|B^{\psi_{0}}_{1,R}B_{t}^{\delta+n_{0}}f(x)|^{2}dt\right)^{1/2},

where n0n_{0} is such that δ+n0>n12\delta+n_{0}>\frac{n-1}{2}.

Moreover, we proceed as in Proposition 5.3 to get that the maximal function

(supR>0Rj10Rj|B1,Rψ0Btδf(x)|2𝑑t)1/2\left(\sup_{R>0}R_{j}^{-1}\int_{0}^{R_{j}}|B^{\psi_{0}}_{1,R}B_{t}^{\delta}f(x)|^{2}dt\right)^{1/2}

is bounded on Lp1(n)L^{p_{1}}({\mathbb{R}}^{n}) for δ>α(p1)1/2\delta>\alpha(p_{1})-1/2 and p1=2p_{1}=2 or p1𝔭np_{1}\geq\mathfrak{p}_{n}. We skip the details here to avoid a repetition.

Let us now consider the term corresponding to the multiplier

M~(ξR,ηR)=ψ0(|ξ|2R2)ψ02(|η|2R2)(1|η|2R2)+α(1|ξ|2R2(1|η|2R2)1)+α\widetilde{M}\left(\frac{\xi}{R},\frac{\eta}{R}\right)=\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi_{0}^{2}\left(\frac{|\eta|^{2}}{R^{2}}\right)\left(1-\frac{|\eta|^{2}}{R^{2}}\right)_{+}^{\alpha}\left(1-\frac{|\xi|^{2}}{R^{2}}\left(1-\frac{|\eta|^{2}}{R^{2}}\right)^{-1}\right)^{\alpha}_{+}

Indeed, using (6) with the roles of ξ\xi and η\eta interchanged, we get that

M~(ξR,ηR)=ψ0(|ξ|2R2)ψ02(|η|2R2)R2α0uR(R2φR(η)t2)+β1t2δ+1(1|ξ|2t2)+δ𝑑t,\widetilde{M}\left(\frac{\xi}{R},\frac{\eta}{R}\right)=\psi_{0}\left(\frac{|\xi|^{2}}{R^{2}}\right)\psi_{0}^{2}\left(\frac{|\eta|^{2}}{R^{2}}\right)R^{-2\alpha}\int_{0}^{uR}\left(R^{2}\varphi_{R}(\eta)-t^{2}\right)_{+}^{\beta-1}t^{2\delta+1}\left(1-\frac{|\xi|^{2}}{t^{2}}\right)^{\delta}_{+}dt,

where φR(η)=(1|η|2R2)+\varphi_{R}(\eta)=\left(1-\frac{|\eta|^{2}}{R^{2}}\right)_{+} and u=2932u=\sqrt{\frac{29}{32}}.

This part is handled using the same strategy as in section 5. After applying the Cauchy Schwartz inequality and making a change of variable tRtt\rightarrow Rt in the integral involving the function gg our job is reduced to proving the Lp2(n)L^{p_{2}}({\mathbb{R}}^{n}) boundedness of the operator

supR>00u|HR,tβg()|2𝑑t,\sup_{R>0}\int_{0}^{u}|H_{R,t}^{\beta}g(\cdot)|^{2}dt,

where HR,tβg(x)=nψ02(|η|2R2)(1t2|η|2R2)+β1g^(η)e2πix.η𝑑ηH_{R,t}^{\beta}g(x)=\int_{{\mathbb{R}}^{n}}\psi_{0}^{2}\left(\frac{|\eta|^{2}}{R^{2}}\right)\left(1-t^{2}-\frac{|\eta|^{2}}{R^{2}}\right)_{+}^{\beta-1}\hat{g}(\eta)e^{2\pi ix.\eta}d\eta.

The term with the function ff is dealt with in a similar way as in the previous case. Note that HR,tβgH_{R,t}^{\beta}g is similar to Sj,Rβ,tfS^{\beta,t}_{j,R}f with the term ψ02(|η|2R2)\psi_{0}^{2}\left(\frac{|\eta|^{2}}{R^{2}}\right) on the Fourier transform side in place of ψ(2j(1||2R2))\psi\left(2^{j}\left(1-\frac{|\cdot|^{2}}{R^{2}}\right)\right) in Sj,Rβ,tfS^{\beta,t}_{j,R}f. Since the function ψ02(|η|2R2)\psi_{0}^{2}\left(\frac{|\eta|^{2}}{R^{2}}\right) is compactly supported and smooth with its support lying away from the origin, we can use arguments similar to the ones used in proving the estimate (20) and Proposition 5.3 to deduce that the maximal function

supR>0|0u|HR,tβg()|2𝑑t|\sup_{R>0}\left|\int_{0}^{u}|H_{R,t}^{\beta}g(\cdot)|^{2}dt\right|

is bounded on Lp2(n)L^{p_{2}}({\mathbb{R}}^{n}) for β>α(p2)+1/2\beta>\alpha(p_{2})+1/2 and p2=2p_{2}=2 or p2𝔭np_{2}\geq\mathfrak{p}_{n}.

Combining the estimates together we get the desired result for the maximal function 0,α(f,g).\mathcal{B}^{\alpha}_{0,*}(f,g).

6. Proof of Theorem 2.2: The case of dimension n=1n=1

In this section we address the LpL^{p} boundedness of the maximal function α(f,g)\mathcal{B}^{\alpha}_{*}(f,g) in dimension n=1n=1.

Recall that in view of the decomposition of the Bochner-Riesz operator given in section 3, it is enough to establish the desired LpL^{p} estimates for the square functions in the following inequality.

T,Rα(f,g)Lp()\displaystyle\|T_{*,R}^{\alpha}(f,g)\|_{L^{p}({\mathbb{R}})} \displaystyle\lesssim 2j4supR>0(02j+1|Sj,βR,tf()t2δ+1|2𝑑t)1/2Lp1()\displaystyle 2^{-\frac{j}{4}}\left\|\sup_{R>0}\left(\int_{0}^{\sqrt{2^{-j+1}}}|{S}_{j,\beta}^{R,t}f(\cdot)t^{2\delta+1}|^{2}dt\right)^{1/2}\right\|_{L^{p_{1}}({\mathbb{R}})}
(supR>0Rj10Rj|Btδg(x)|2𝑑t)1/2Lp2(),\displaystyle\left\|\left(\sup_{R>0}R_{j}^{-1}\int_{0}^{R_{j}}|B_{t}^{\delta}g(x)|^{2}dt\right)^{1/2}\right\|_{L^{p_{2}}({\mathbb{R}})},

where α=β+δ\alpha=\beta+\delta.

Invoking the LpL^{p} boundedness of the square function for Bochner Riesz operator in dimension n=1n=1 (see [23] Theorem B, page 361) and using the techniques of Lemma 4.3, it is easy to verify that the estimate

(43) (supR>0Rj10Rj|Btδg(x)|2𝑑t)1/2Lp2()gLp2()\left\|\left(\sup_{R>0}R_{j}^{-1}\int_{0}^{R_{j}}|B_{t}^{\delta}g(x)|^{2}dt\right)^{1/2}\right\|_{L^{p_{2}}({\mathbb{R}})}\lesssim\|g\|_{L^{p_{2}}({\mathbb{R}})}

holds for δ>1/2\delta>-1/2 when 2p2<2\leq p_{2}<\infty and for δ>1p21\delta>\frac{1}{p_{2}}-1 when 1<p2<21<p_{2}<2.

Next, we have the following result for the other square function.

Theorem 6.1.

The inequality

(44) supR>0(02j+1|Sj,βR,tf()t2δ+1|2𝑑t)1/2Lp1()2jα~+j/4fLp1()\left\|\sup_{R>0}\left(\int_{0}^{\sqrt{2^{-j+1}}}|{S}_{j,\beta}^{R,t}f(\cdot)t^{2\delta+1}|^{2}dt\right)^{1/2}\right\|_{L^{p_{1}}({\mathbb{R}})}\lesssim 2^{-j\tilde{\alpha}+j/4}\|f\|_{L^{p_{1}}({\mathbb{R}})}

holds where α=β+δ,α~=min{α,δ+1/2}\alpha=\beta+\delta,\tilde{\alpha}=\text{min}\{\alpha,\delta+1/2\}, β>1/2\beta>1/2 for 2p2<2\leq p_{2}<\infty and β>1p1\beta>\frac{1}{p_{1}} for 1<p1<21<p_{1}<2.

Note that the choice of δ\delta in (43) implies that α~>0\tilde{\alpha}>0 for all 1<p2<1<p_{2}<\infty.

The following lemma provides us with an important kernel estimate in dimension n=1n=1. This estimate is used in the proof of Theorem 6.1.

Lemma 6.2.

Let ψC0([1/2,2])\psi\in C^{\infty}_{0}([1/2,2]) and Kj(x):=ψ(2j(1ξ2))e2πixξ𝑑ξK_{j}(x):=\int_{{\mathbb{R}}}\psi(2^{j}(1-\xi^{2}))e^{2\pi ix\xi}d\xi. Then for all j2j\geq 2 the following estimate holds

|Kj(x)|C2j(1+|2jx|)2,|K_{j}(x)|\lesssim C2^{-j}(1+|2^{-j}x|)^{-2},

where C=supx,0k2|dkψdxk|C=\sup_{x\in{\mathbb{R}},~{}0\leq k\leq 2}\left|\frac{d^{k}\psi}{dx^{k}}\right|.

Remark 6.3.

We believe that the estimate in the lemma above should be known in the existing literature. However, we could not find it. Therefore, we have decided to present its proof here for the sake of completeness.

Proof.

Let jj be large enough so that 2j<<1/82^{-j}<<1/8. Let χ\chi be an even smooth function supported in [1/4,1/4][-1/4,1/4] and identically one in [1/8,1/8][-1/8,1/8]. We can write

Kj(x)=χ(ξ+1)ψ(2j(1ξ2))e2πixξ𝑑ξ+χ(1ξ)ψ(2j(1ξ2))e2πixξ𝑑ξ.K_{j}(x)=\int_{{\mathbb{R}}}\chi(\xi+1)\psi(2^{j}(1-\xi^{2}))e^{2\pi ix\xi}d\xi+\int_{{\mathbb{R}}}\chi(1-\xi)\psi(2^{j}(1-\xi^{2}))e^{2\pi ix\xi}d\xi.

Observe that it is enough to show that K~j(x)=χ(1ξ)ψ(2j(1ξ2))e2πixξ𝑑ξ\tilde{K}_{j}(x)=\int_{{\mathbb{R}}}\chi(1-\xi)\psi(2^{j}(1-\xi^{2}))e^{2\pi ix\xi}d\xi satisfies the desired estimate.

We perform change of variable 1ξξ1-\xi\rightarrow\xi to the expression above and then one more time change variable 2jξξ2^{j}\xi\rightarrow\xi to get that

K~j(x)=e2πix2jχ(2jξ)ψ(ξ(22jξ))e2πi2jxξ𝑑ξ.\tilde{K}_{j}(x)=e^{2\pi ix}2^{-j}\int_{{\mathbb{R}}}\chi(2^{-j}\xi)\psi(\xi(2-2^{-j}\xi))e^{-2\pi i2^{-j}x\xi}d\xi.

Observe that ψ(ξ(22jξ))χ(2jξ)\psi(\xi(2-2^{-j}\xi))\chi(2^{-j}\xi) is a smooth function supported in [2j2,2j2][-2^{j-2},2^{j-2}]. Now we see that the function ψ(ξ(22jξ))χ(2jξ)\psi(\xi(2-2^{-j}\xi))\chi(2^{-j}\xi) is of compact support with its support lying inside a set independent of jj. We use an integration by parts argument to prove our claim.

We already know that support of ψ\psi is [1/2,2][1/2,2]. Let us look at the function u(ξ)=ξ(22jξ)u(\xi)=\xi(2-2^{-j}\xi). It defines a parabola with its zeros at 0 and 2j+12^{j+1}. One can show that 2j2^{j} is the point where it attains its maximum and between 0 to 2j2^{j} it is a strictly increasing function.

Further note that u(2)=4(12j)>2u(2)=4(1-2^{-j})>2 and u(14)=12(12j3)<1/2u(\frac{1}{4})=\frac{1}{2}(1-2^{-j-3})<1/2 for all j8j\geq 8. Therefore {ξ:u(ξ)[1/2,2]}[1/4,2]\{\xi:u(\xi)\subset[1/2,2]\}\subset[1/4,2]. Hence the support of ψ(ξ(22jξ))χ(2jξ)\psi(\xi(2-2^{-j}\xi))\chi(2^{-j}\xi) is a subset of [1/4,2][1/4,2] for all j8j\geq 8.

For y0y\neq 0, consider

χ(2jξ)ψ(ξ(22jξ))e2πiyξ𝑑ξ\displaystyle\int_{{\mathbb{R}}}\chi(2^{-j}\xi)\psi(\xi(2-2^{-j}\xi))e^{-2\pi iy\xi}d\xi
=\displaystyle= 12πiyχ(2jξ)ψ(ξ(22jξ))(ddξe2πiyξ)𝑑ξ\displaystyle-\frac{1}{2\pi iy}\int_{{\mathbb{R}}}\chi(2^{-j}\xi)\psi(\xi(2-2^{-j}\xi))\left(\frac{d}{d\xi}e^{-2\pi iy\xi}\right)d\xi
=\displaystyle= 12πiy(2jχ(2jξ)ψ(ξ(22jξ))e2πiyξ𝑑ξ+χ(2jξ)(22j+1ξ)ψ(ξ(22jξ))e2πiyξ𝑑ξ).\displaystyle\frac{1}{2\pi iy}\left(\int_{{\mathbb{R}}}2^{-j}\chi^{\prime}(2^{-j}\xi)\psi(\xi(2-2^{-j}\xi))e^{-2\pi iy\xi}d\xi+\int_{{\mathbb{R}}}\chi(2^{-j}\xi)(2-2^{-j+1}\xi)\psi^{\prime}(\xi(2-2^{-j}\xi))e^{-2\pi iy\xi}d\xi\right).

Notice that the term

|2jχ(2jξ)ψ(ξ(22jξ))e2πiyξ𝑑ξ+χ(2jξ)(22j+1ξ)ψ(ξ(22jξ))e2πiyξ𝑑ξ|\left|\int_{{\mathbb{R}}}2^{-j}\chi^{\prime}(2^{-j}\xi)\psi(\xi(2-2^{-j}\xi))e^{-2\pi iy\xi}d\xi+\int_{{\mathbb{R}}}\chi(2^{-j}\xi)(2-2^{-j+1}\xi)\psi^{\prime}(\xi(2-2^{-j}\xi))e^{-2\pi iy\xi}d\xi\right|

is bounded by a uniform constant CC with respect to jj.

Similarly, the integration by parts argument applied to each of the integrals above gives us that

|χ(2jξ)ψ(ξ(22jξ))e2πiyξ𝑑ξ|C(1+y2)1.\left|\int_{{\mathbb{R}}}\chi(2^{-j}\xi)\psi(\xi(2-2^{-j}\xi))e^{-2\pi iy\xi}d\xi\right|\lesssim C(1+y^{2})^{-1}.

This completes the proof. ∎

Proof of Theorem 6.1: We shall skip detail in the proof and point out only the key steps as the proof uses the same strategy as in the previous sections.

Recall that

Sj,βR,tf(x)=ψ(2j(1ξ2R2))(1t2ξ2R2)+β1f^(ξ)e2πixξ𝑑ξ.{S}_{j,\beta}^{R,t}f(x)=\int_{{\mathbb{R}}}\psi\left(2^{j}\left(1-\frac{\xi^{2}}{R^{2}}\right)\right)\left(1-t^{2}-\frac{\xi^{2}}{R^{2}}\right)_{+}^{\beta-1}\hat{f}(\xi)e^{2\pi ix\xi}d\xi.

It is easy to verify that the kernel estimate from Lemma 6.2 implies that

(45) supR>0|B2j,Rψf|Lp1()CfLp1()\|\sup_{R>0}|B^{\psi}_{2^{-j},R}f|\|_{L^{p_{1}}({\mathbb{R}})}\leq C\|f\|_{L^{p_{1}}({\mathbb{R}})}

holds for all j2j\geq 2 and 1<p11<p_{1}\leq\infty.

Fix 0<ϵ0<<10<\epsilon_{0}<<1 and consider the case when 0t2<2j1ϵ00\leq t^{2}<2^{-j-1-\epsilon_{0}}. In this case we proceed the same way as in Case 1 of Theorem 5.1. Using the estimate (45) we get that

supR>0(02j1ϵ0|Sj,βR,tf()t2δ+1|2𝑑t)1/2Lp1()2jα+j/4fLp1()\left\|\sup_{R>0}\left(\int_{0}^{\sqrt{2^{-j-1-\epsilon_{0}}}}|{S}_{j,\beta}^{R,t}f(\cdot)t^{2\delta+1}|^{2}dt\right)^{1/2}\right\|_{L^{p_{1}}({\mathbb{R}})}\lesssim 2^{-j\alpha+j/4}\|f\|_{L^{p_{1}}({\mathbb{R}})}

holds for all β>1/2\beta>1/2.

Next, for 2j1ϵ0<t22j+12^{-j-1-\epsilon_{0}}<t^{2}\leq 2^{-j+1}, we use the estimate (45) again and follow the idea of Proposition 5.3 to deduce the following estimate

supR>0(2j1ϵ02j+1|Sj,βR,tf()t2δ+1|2𝑑t)1/2Lp1()2j(δ+1/2)+j/4fLp1()\left\|\sup_{R>0}\left(\int^{\sqrt{2^{-j+1}}}_{\sqrt{2^{-j-1-\epsilon_{0}}}}\left|{S}_{j,\beta}^{R,t}f(\cdot)t^{2\delta+1}\right|^{2}dt\right)^{1/2}\right\|_{L^{p_{1}}({\mathbb{R}})}\lesssim 2^{-j(\delta+1/2)+j/4}\|f\|_{L^{p_{1}}({\mathbb{R}})}

for all β>1/2\beta>1/2 if 2p12\leq p_{1} and β>1/p1\beta>1/p_{1} if 1<p1<21<p_{1}<2.

Acknowledgement

The second author acknowledges the financial support by Science and Engineering Research Board (SERB), Government of India, under the grant MATRICS: MTR/2017/000039/Math.

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