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BY-NC-ND 3.0 license Open Access Published by De Gruyter Open Access September 26, 2016

Parabolic Marcinkiewicz integrals on product spaces and extrapolation

  • Mohammed Ali EMAIL logo and Mohammed Al-Dolat
From the journal Open Mathematics

Abstract

In this paper, we study the the parabolic Marcinkiewicz integral MΩ,hρ1,ρ2 on product domains Rn × Rm (n, m ≥ 2). Lp estimates of such operators are obtained under weak conditions on the kernels. These estimates allow us to use an extrapolation argument to obtain some new and improved results on parabolic Marcinkiewicz integral operators.

MSC 2010: 40B20; 40B15; 40B25

1 Introduction and the main result

Let RN (N = n or m), N ≥ 2 be the N-dimensional Euclidean space, and let SN−1 be the unit sphere in RN equipped with the normalized Lebesgue surface measure = Also, let pʹ denote the exponent conjugate to p; that is 1/p+1/pʹ = 1.

Let αi(i = 1,2,…, N) be fixed real numbers such that αi ≥ 1. Define the function F: RN × R+R by F(z,ρ)=i=1Nzi2ρ2αi. It is clear that for each fixed zRN, the function F(z, ρ) is a decreasing functions in ρ > 0.

The unique solution of the equation F(z, ρ) = 1 is denoted by ρ(z). In [1], Fabes and Riviére showed that ρ(z) is a metric space in RN, and (RN ρ) is called the mixed homogeneity space related to {αi}i=0N.

For λ > 0, let Aλ be the diagonal N × N matrix

Aλ=diag{λα1,λα2,,λαN}=λαl00λαN.

The change of variables related to the space (RN, ρ) is given by the transformation x1=ρα1cosϑ1cosϑN2cosϑN1

x2=ρα2cosϑ1cosϑN2sinϑN1

xN1=ραN1cosϑ1sinϑ2,

xN=ραNsinϑ1

where x =(x1x2 xN) ∈ RN Thus, dx=ρα1JN(x)dρdσ(x), where ρα−1JN(xʹ) is the Jacobian of the above transforms,

α=k=1NαkandJN=k=1Nαk(xk)2.

It was shown in [1] that JN(xʹ) is a C(SN−1 function in the variable xʹ ∈ SN−1, and that a real constant LN ≥ 1 exists so that 1 ≤ JN (xʹ) ≤ LN.

Let KΩ,ρ(u) = Ω(u)ρ(u)1−α where #x03A9; is a real valued and measurable function on RN with Ω ∈ L1 (SN−1) that satisfies the conditions

Ω(Aλz)=Ω(z),λ>0andsN1Ω(z)JN(z)dσ(z)=0.

The parabolic Marcinkiewicz integral μΩ, which was introduced by Ding, Xue and Yabuta in [2], is defined by

μΩf(z)=0|FΩ,t(z)|2dtt31/2,

where

FΩ,t(z)=ρ(u)<_tKΩ,ρ(u)f(zu)du.

In particular, the authors of [2] proved that the parabolic Littlewood-Paley operator μΩ is bounded for p ∈ (1, ∞) provided that Ω ∈ Lq (SN−1) for q > 1. Subsequently, the study of the Lp boundedness of μΩ under various conditions on the function Ω has been studied by many authors (see for example [37]).

We point out that the class of the operators μΩ is related to the class of the parabolic singular integral operators

TΩf(z)=p.v.RNΩ(u)ρ(u)αf(zu)du.

The class of the operators TΩ belongs to the class of singular Radon transforms, which was studied by by many mathematicians (we refer the readers, in particular, to [1, 8]).

Although some open problems related to the boundedness of parabolic Marcinklewicz integral in the one- parameter setting remain open, the investigation of Lp estimates of the Marcinkiewicz integral on product spaces has been started (see for example [9, 10].)

Let αi, βj be fixed real numbers with αi, βj ≥ 1 (i = 1,2,…, n and j = 1, 2, …, m). For τ1 = a1 + ib1, τ2 = a2 + ib2 (a1, b1, a2, b2Rwitha1, a2 > 0), let KΩ,hρ1,ρ2(x,y)=Ω(x,y)h(ρ1(x),ρ2(y))ρ1(x)τ1αρ2(y)τ2β, where α=i=1nαi,β=j=1mβj,h is a measurable function on R+ × R+, and Ω is a real valued and measurable function on Rn × Rm with Ω ∈ L1(SN−1 × Sm−1) satisfying the conditions

Ω(Aλ1x,Aλ2y)=Ω(x,y),λ1,λ2>0,(1)
sn1Ω(x,.)Jn(x)dσ(x)=sm1Ω(.,y)Jm(y)dσ(y)=0.(2)

We define the parabolic Marcinkiewicz integral operator MΩ,hρ1,ρ2 for fS(Rn × Rm) by

MΩ,hρ1,ρ2f(x,y)=00|Fts(x,y)|2dtdsts1/2,(3)

where

Ft,s(x,y)=1tτ1sτ2ρ1(u)<_tρ2(v)sKΩhρ1,ρ2,(u,v)f(xu,yv)dudv.

By specializing to the case h = 1 and τ1 = τ2 = 1, plus considering the cases α1 = … = αn = 1 and β1 = … = βm = 1, we obtain that ρ1(u) = |u|, ρ2(v) = |v|, α = n, β = m, and (Rn × Rm, ρ1, ρ2 = (Rn × Rm, |·|, |·|). In this case MΩ,hρ1,ρ2 (denoted by MΩ is just the classical Marcinkiewicz integral on product domains, which was studied by many mathematicians. For instance, the author of [11] gave the L2 boundedness of MΩ if ΩL(logL)2(Sn1×Sm1). Later, it was verified in [12] that MΩ is bounded for all 1 < p < ∞ provided that ΩL(logL)2(Sn1×Sm1). This result was improved (for p = 2) in [13] in which the author established that MΩ is bounded on L2(Rn × Rm) for all ΩL(logL)(Sn1×Sm1). Recently, Al-Qaseem et al found in [14] that the boundedness of MΩ is obtained under the condition ΩL(logL)(Sn1×Sm1) for 1 < p < ∞. Furthermore, they proved that the exponent 1 is the best possible.

Al-Qassem in [15], found that MΩ,h is bounded on Rn × Rm(1 < p < ∞) provided that h is a bounded radial function and Ω is a function in certain block space Bq(0,0)(Sn1×Sm1) for q > 1. He also established the optimality of the condition in the sense that the space Bq(0,0)(Sn1×Sm1) cannot be replaced by Bq(0,ε)(Sn1×Sm1) for any −1 < ε < 0.

On the other hand, Al-Salman in [9] extended the result in [14]. In fact, he proved that MΩ,1ρ1,ρ2 is bounded in Lp(Rn × Rm)(1 < p < ∞) provided that ΩL(logL)(Sn1×Sm1).

We point out that the parabolic singular integral operator on product domains of the form

TΩ,hf(x,y)=p.v.Rn×RmΩ(u,v)h(ρ1(u),ρ2(v))ρ1(u)αρ2(v)βf(xu,yv)dudv

is being under investigation by one of our graduate students. In fact, he shall prove the Lp boundedness of TΩ, h when ΩLq(Sn1×Sm1) for some q > 1 and h ∈ Δγ(R+ × R+) for some γ > 1, where Δγ(R+ × R+) (for γ > 1) denotes the collection of all measurable functions h: R+ × R+C satisfying

supR1,R2>01R1R20R10R2|h(t,s)|γdtds1/γ<.

In view of the result in [4]; that is the parabolic Marcinkiewicz integral in the one-parameter setting, defined as in (3), is bounded on Lp(Rn) and the results concerning the classical Marcinkiewicz in the two-parameter setting, a question arises naturally Does the Lp boundedness of the operators MΩ,hρ1,ρ2 hold under the conditions when Ω belongs to the space L(log L)(Sn−1} × Sm−1) or whether it belongs to the block space Bq(0,0)(Sn1×Sm1) and h ∈ Δγ(R+ × R+) for some γ, q > 1?

We shall obtain an affirmative answer to this question, as described in the following theorems.

Theorem 1.1

Let 003E ∈ Lq(Sn−1 × Sm−1) for some 1 < q ≤ 2 andh ∈ Δγ(R+ × R+) for some γ > 1. Then there exists a constantCp (independent of Ω, h, γ, andq) such that

MΩ,hρ1,ρ2(f)Lp(Rn×Rm)CpA(γ)q1hΔγ(R+×R+)ΩLq(Sn1×Sm1)fLp(Rn×Rm)

for |1/p−1/2| < min{1/2, 1γʹ}, whereA(γ)=γifγ>2,(γ1)1if1<γ2.

The conclusion from Theorem 1.1 and the application of an extrapolation method as in [16, 17] lead to the following theorem.

Theorem 1.2

Suppose that Ω satisfies (1)–(2) andh ∈ Δγ(R+ × R+) for some γ > 1.

  1. IfΩBq(0,0)(Sn1×Sm1)for someq > 1, then

    MΩ,hρ1,ρ2(f)Lp(Rn×Rm)CpA(γ)hΔγ(R+×R+)fLp(Rn×Rm)1+ΩBq(0.0)(sn1×sm1)

    for |1/p−1/2| < min{1/2, 1/γʹ};

  2. ifΩL(logL)(Sn1×Sm1), then

    MΩ,hρ1,ρ2(f)Lp(Rn×Rm)CpA(γ)hΔγ(R+×R+)fLp(Rn×Rm)(1+ΩL(logL)(Sn1×Sm1))

    for |1/p−1/2| < min{1/2, 1/γʹ}.

Here and henceforth, the letter C denotes a bounded positive constant that may vary at each occurrence but is independent of the essential variables.

2 Preparation

In this section, we give some auxiliary lemmas used in the sequel. We shall recall the following lemma due to Ricci and Stem.

Lemma 2.1

([18]). Suppose that λisandαisare fixed real numbers, andΓ(t)=(λ1tα1,,λNtαN)is a function fromR+ to RNFor suitablef, letMτbe the maximal operator defined onRNby

MΓf(x)=suph>01h|0hf(xΓ(t)dt|

forxRNThen for 1 < p ≤ ∞, there exists a constantCp > 0 such that

MΓfLp(RN)CpfLp(RN).

The constantCpis independent ofλisandf.

Lemma 2.2

Suppose that ais,bis,αis, andβisare fixed real numbers. LetΓ(t)=(a1tα1,,antαn)andΛ(t)=(b1tβ1,,bmtβm), and letMΓ,Λbe the maximal operator defined onRn × Rm by

MΓ,Λf(x,y)=suph1,h2>01h1h2|0h10h2f(xΓ(t),yΛ(r))dtdr|

for (x, y) ∈ Rn × RmThen for 1 < p ≤ ∞, there exists a constantCp > 0 (independent ofais,bis and f) such that

MΓ,ΛfLp(Rn×Rm)CpfLp(Rn×Rm)

for allfLpRn × Rm).

It is easy to prove this lemma by using Lemma 2.1 and the inequality MΓ,Λf(x,y)MΛMΓf(x,y), where MΓf(x,y)=MΓf(.,y)(x),MΛf(x,y)=MΛf(x,)(y), and ∘ denotes the composition of operators.

Lemma 2.3

([5]). Suppose that 0 ≤ v ≤ 1. Let m denote the distinct numbers ofi}. Then foru, ξ ∈ RN, we have

12eiAλu.ξdλλC|uξ|vm,

whereCis independent ofuand ξ.

Let θ ≥ 2. For a measurable function h: R+ × R+C and Ω: Sn−1 × Sm−1R, we define the family of measures σΩ, h, t, s: t, sR+} and its corresponding maximal operators σΩ,h,t and MΩ, h θ on Rn × Rm by

Rn×RmfdσΩ,h,t,s=1tτ1sτ21/2tρ1(u)t1/2sρ2(v)sf(u,v)h(ρ1(u),ρ2(v))Ω(u,v)ρ1(u)ατ1ρ2(v)βτ2dudv,
σΩ,hf(x,y)=supt,sR+σΩ,h,t,sf(x,y),

and

MΩ,h,θf(x,y)=supi,jZθiθi+1θθj+1σΩ,h,t,sf(x,y)dtdsts,

where |σΩ, h, t, s| is defined in the same way as σΩ, h, t, s, but with replacing h by |h| and Ω by |Ω|. We write ‖σt, s‖ for the total variation of σt, s and a±r = min{ar, a−r}.

In order to obtain Theorem 1.1, we need to prove the following lemmas.

Lemma 2.4

Suppose that Ω ∈ Lq(Sn−1 × Sm−1) for someq > 1 and satisfies the cancellation conditions (1)-(2). Fort, s > 0, let

Gt,s(ρ1,ρ2)=sn1×sm1ei{Atρ1xξ+Asρ2yη}Ω(x,y)Jn(x)Jm(y)dσ(x)dσ(y).

Then there are constantsCandwwith0<w<min{12,m12q,m22q,m1α,m2β}such that

1/211/21|Gt,s(ρ1,ρ2)|2dρ1dρ2ρ1ρ2C|Atξ|±wm1q|Asη|±wm2qΩLq(Sn1×Sm1)2,(4)

wherem1, m2denote the distinct numbers ofi}, {βj}, respectively

Since Lq(Sn−1 × Sm−1)⊆ L2(Sn−1} × Sm−1) for q ≥ 2, it is enough to prove this lemma only for the case 1 < q ≤ 2. By Schwarz inequality, we get that

1/211/21|Gt,s(ρ1,ρ2)|2dρ1dρ2ρ1ρ2csm1sn1×sn1H(ξ,x,u)×Ω(x,y)Ω(u,y)¯Jn(x)Jn(u)dσ(x)dσ(u))Jm(y)dσ(y),

where H(ξ,x,u)=12eiAt2ρ1ξ(xu)dρ1ρ1. Let ζ=At2ξ|At2ξ| Thanks to Lemma 2.3, we conclude that

H(ξ,x,y)C|At2ξ(xu)|w/m1C2αw/m1(|ζ(xu)||Atξ|)w/m1C|Atξ|w/m1(|ζ(xu)|)w/m1

for any w with 0<w<min{12,m1α}. Hence, by Hölder’s inequality we get

1/211/21|Gt,s(ρ1,ρ2)|2dρ1dρ2ρ1ρ2C|Atξ|wm1qΩLq(Sn1×Sm1)2×sn1×sn1|ζ(xu)|wqm1dσ(x)dσ(y)1/q.

By choosing 0<2wqm1<1, we get that the last integral is finite, and so

1/211/21|Gt,s(ρ1,ρ2)|2dρ1dρ2ρ1ρ2C|Atξ|wqm1ΩLq(Sn1×Sm1)2.(5)

Similarly, we derive

1/211/21|Gt,s(ρ1,ρ2)|2dρ1dρ2ρ1ρ2C|Asη|wqm2ΩLq(Sn1×Sm1)2(6)

The other estimates in (4) can be reached by using the cancelation property of Ω; by a simple change of variable, we have that

1/211/21|Gt,s(ρ1,ρ2)|2dρ1dρ2ρ1ρ2C1/211/21sn1×sm1|eiAtρξx1||Ω(x,y)Jn(x)Jm(y)|dσ(x)dσ(y)2dρ1dρ2ρ1ρ2C|Atξ|2ΩL1(Sn1×Sm1)2,

which when combined with the trivial estimate 1/211/21Gt,s(ρ1,ρ2)2dρ1dρ2ρ1ρ2CΩL1(Sn1×Sm1)2 gives

1/211/21|Gt,s(ρ1,ρ2)|2dρ1dρ2ρ1ρ2C|Atξ|wqm1ΩLq(Sn1×Sm1)2(7)

In the same manner, we obtain

1/211/21Gt,s(ρ1,ρ2)2dρ1dρ2ρ1ρ2CAsηwqm2ΩLq(Sn1×Sm1)2(8)

Therefore, by (5)–(8), the proof of this lemma is complete. □

Lemma 2.5

Let Ω∈ Lq(Sn-1× Sm-1) for someq > 1, h ∈ Δγ(R+× R+) for some γ > 1 and θ = 2q′ γ′Then there are constantsCandw (as in Lemma 2.4) such that

σΩ,h,t,sChΔγR+×R+ΩLq(sn1×sm1);(9)
θiθi+1θθj+1σ^Ω,h,t,s(ξ,η)2dtdstsCln2(θ)hΔγ(R+×R+)2ΩLq(Sn1×Sm1)2×|Aθiξ|±2wm1qκ|Aθjη|±2wm2qκ(10)

hold for alli, jZ, whereκ=max{2,γ}The constantCis independen of i, j ξ, η, q, and θ.

Proof

By using the definition of σΩ, h, t, s, it is easy to show that (9) holds. By Hölder's inequality and a simple change of variables, we have that

σ^Ω,h,t,s(ξ,η)C12tt12ssh(ρ1,ρ2)sn1×Sm1ei{Aρ1xξ+Aρ2yη}Jn(x)Jm(y)×Ω(x,y)dσ(x)dσ(y)|dρ1dρ1ρ1ρ2CΔγR+×R+1/211/21Gts(ρ1,ρ2)γdρ1dρ2ρ1ρ21/γ.

If 1 < γ ≤ 2, then we conclude

σ^Ω,h,t,s(ξ,η)hΔγ(R+×R+)ΩL1(Sn1×Sm1)(12/γ)1/211/21Gt,s(ρ1,ρ2)2dρ1dρ2ρ1ρ21/γ,

and if γ > 2, then by Hölder's inequality, we deduce

σ^Ω,h,t,s(ξ,η)hΔγ(R+×R+)1/211/21Gt,s(ρ1,ρ2)2dρ1dρ2ρ1ρ21/2.

Thus, in either case we have

σ^Ω,h,t,s(ξ,η)ChΔγ(R+×R+)ΩL1(Sn1×Sm1)(κ2)/γ1/211/21Gt,s(ρ1,ρ2)2dρ1dρ2ρ1ρ21/κ,

where κ=max{2,γ} hence, by Lemma 2.4 we obtain

σ^Ω,h,t,s(ξ,η)2ChΔγ(R+×R+)2ΩLq(Sn1×Sm1)2Atξ±2wm1κq|Asη|±2wm2κq.

Since θj≤t≤θj+1 and θi≤s≤θi+1, we immediately get that

σ^Ω,h,t,s(ξ,η)2ChΔγ(R+×R+)2ΩLq(Sn1×Sm1)2Atξ±2wm1κq|Asη|±2wm2κq(11)

and

σ^Ω,h,t,s(ξ,η)2ChΔγ(R+×R+)2ΩLq(Sn1×Sm1)2Aθj+1ξ+2wm1κqAθi+1ξ+2wm2κqCθmax{α..αnβ1..βm}2wκq(1m1+1m2)1hΔγ(R+×R+)2ΩLq(Sn1×Sm1)2Aθjξ+2wm1κq×Aθiξ+2wm2κqC22wmax{α1..αnβ1.jm}(1m1+1m2)hΔγ(R+×R+)2ΩLq(Sn1×Sm1)2×Aθjξ+2wm1κqAθiξ+2wm2κqChΔγ(R+×R+)2ΩLq(Sn1×Sm1)2Aθjξ+2wm1κqAθiξ+2wm2κq.(12)

Therefore, when we combine (11) with (12), we deduce

θiθi+1θθj+1|σ^Ω,h,t,s(ξ,η)|2dtdstsCln2(θ)hΔγ(R+×R+)2ΩLq(Sn1×Sm1)2×|Aθiξ|±2wm1qκ|Aθjη|±2wm2qκ.

Lemma 2.6

Assume that Ω∈L1(Sn−1×Sm−1) andh∈Δγ(R+×R+) for some γ′ > 1. Then for anyf∈Lp(Rn×Rm) with γ′<p≤∞ there exists a cons tant C such that

σΩ,h(f)Lp(Rn×Rm)CphΔγ(R+×R+)ΩL1(Sn1×Sm1)fLp(Rn×Rm).
Proof

By Hölder's inequality, we have

||σΩ,h,t,s|f(x,y)|ChΔγ(R+×R+)ΩL1(Sn1×Sm1)1/γ(1tst2ts2ssn1×sm1Ω(ιr,v)|×|f(xAρ1u,yAρ2υ)|γdσ(u)dσ(υ)dρ1dρ2)1/γ.

Thus, by using Minkowski’s inequality for integrals and Lemma 2.2 we get

σΩ,hfLp(Rn×Rm)ChΔγ(R+×R+)ΩL1(Sn1×Sm1)1/γ×sn1×sm1Ω(u,υ)MΓ,Λ(|f|γ)L(p/γ)(Rn×Rm)dσ(u)dσ(υ)1/γChΔγ(R+×R+)ΩL1(Sn1×Sm1)MΓ,Λ(|f|)Lp(Rn×Rm)CphΔγ(R+×R+)ΩL1(Sn1×Sm1)fLp(Rn×Rm).

The following lemma can be obtained by applying the arguments (with only minor modifications) used in [4, 19].

Lemma 2.7

Let Ω∈Lq(Sn−1 × Sm−1) for some 1 > q ≤ 2 andθ = 2γʹ. Assume that h ∈ Δγ (R+ × R+) for some γ > 1. Then for any functions {gi, j(.,.), i, jZ} onRn × Rm, there exists a constantCrsuch that

(ijZθjθj+1θiθi+1σΩ,h,t,sgi,j2dsdtst)12Lr(Rn×Rm)crA(γ)q1hΔγ(R+×R+)ΩLq(Sn1×Sm1)ijZ|gij|21/2Lr(Rn×Rm)

for anyrsatisfiying |1/r−1/2| < min{1/2, 1/γʹ}.

Proof

On one hand, let us consider the case 1 < γ ≤ 2. So, |1/r−1/2| < 1/γʹ If 2r<2γ2γ; then by duality, there exists a non-negative function ϕ ∈ L(r/2)′}(Rn × Rm) with ϕL(r/2)(Rn×Rm)1 so that

i,jZθjθj+1θiθi+1σΩ,h,t,sgi,j2dtdsts1/2Lr(Rn×Rm)2=Rn×Rmi,jZθjθj+1θiθi+1σΩ,h,t,sgij(yj,y)2dtdstsϕ(x,y)dxdy.

By Schwarz's inequality, we have

σΩ,h,t,sgi,j2ChΔγ(R+×R+)γΩLq(sn1×sn1)×12tt12ssSn1×Sm1gi,j(xAρ1u,yAρ2υ)2Ω(u,v)h(ρ1,ρ2)2νdσ(u)dσ(υ)dρ1dρ2ρ1ρ2.

Hence, by a simple change of variables we derive that

i,jZθjθj+1θiθi+1σΩ,h,t,sgi,j2dtdsts1/2Lr(Rn×Rm)2ChΔγ(R+×R+)γ×ΩLq(Sn1×Sm1)Rn×RmjZgi,j(x,y)2MΩ,h2γ,θϕ~(x,y)dxdyChΔγ(R+×R+)γΩLq(Sn1×Sm1)ijZgi,j2L(r/2)(Rn×Rm)MΩ|h|2γθϕ~L(r/2)(Rn×Rm),

where ϕ~(x,y)=ϕ(x,y). As h ∈ Δγ(R+ × R+), we have |h|2γΔγ2γ(R+×R+), and since (r2)>(γ2γ), then by Hölder's inequality and Lemma 2.6, we obtain that

i,jZθjθj+1θiθi+1|σΩ,h,t,sgi,j|2dtdsts1/2Lr(Rn×Rm)2Cln2(θ)hΔγ(R+×R+)γΩLq(Sn1×Sm1)i,jZ|gi,j|21/2Lr(Rn×Rm)2×σΩ,|h|2γϕ~L(r/2)(Rn×Rm)Cp1[(γ1)(q1)]2hΔγ(R+×R+)2ΩLq(Sn1×Sm1)2i,jZ|gi,j|21/2Lr(Rn×Rm)2.

However, if 2γ3γ2<r<2, then by the duality, there are functions φ = φi, j(x, y, t,s) defined on Rn × Rm × R+ × R+ with φi,jL2([θi,θi+1]×[θj,θj+1],dtdstsl2Lr(Rn×Rm)1 such that

i,jZθjθj+1θiθi+1|σΩ,h,t,Sgi,j|2dtdsts1/2Lr(Rn×Rm)=Rn×Rmi,jZθjθj+1θiθi+1(σΩ,h,t,Sgi,j(x,y))φi,j(x,y,t,s)dtdstsddYCpln2(θ)(Υ(φ))1/2Lr(Rn×Rm)i,jZ|gi,j|21/2Lr(Rn×Rm),(13)

where

Υ(φ)(x,y)=i,jZθjθj+1θiθi+1|σΩ,h,t,sφi,j(x,y,t,s)|2dtdsts.

As r2>1, then again by the duality, there is a function ψL(r/2)(Rn×Rm) with ψL(r/2)(Rn×Rm)1 such that

(Υ(φ))1/2Lr(Rn×Rm)2=i,jεzRn×Rmθjθj+1θiθi+1|σΩ,h,t,sφi,j(x,y,t,s)|2dtdstsψ(x,y)dxdyChΔγ(R+×R+)γΩLq(Sn1×Sm1)σΩ,|h|2γ(ψ)L(r/2)(Rn×Rm)×i,jZθjθj+1θiθi+1φi,j(,,t,s)2dtdstsL(r/2)(Rn×Rm)ChΔγ(R+×R+)2ΩLqSn1×Sm12.

Thus, by the last inequality and (13); and since ln(θ)c(γ1)(q1), our estimate holds for 2γ3γ2r<2. Using the same technique as above gives the conclusion of this Lemma for the case γ ≥ 2. Consequently, the proof is complete. □

3 Proof of the main result

We prove Theorem 1.1 by applying the same approaches as in [5, 14], which have their roots in [20]. Let us assume that h ∈ Δγ(R+ × R+) for some γ > 1. Then by Minkowskl’s inequality, we get that

MΩ,hρ1,ρ2f(x,y)=R+×R+i,j=01tρ1sρ22i1t<ρ1(u)2it2j1s<ρ2(v)2jS×f(xu,yv)KΩ,hρ1,ρ2(u,v)dudv2dtdsts1/2i,j=0R+×R+1tρ1s022i1t<ρ1(u)2it2j1s<ρ2(v)2jS×f(xu,yv)KΩ,hρ1,ρ2(u,v)dudv2dtdsts1/22a1+a2(2a11)(2a21)R+×R+σΩ,h,tsf(x,y)|2dtdsts1/2.(14)

Take θ = 2qʹγʹ For iZ, let {Γi} be a smooth partition of unity in (0, ∞) adapted to the interval Ii=[(θi1),(θi+1)]. Specifically, we require the following:

ΓiC,0Γi1,iZΓi2(t)=1,suppΓiIi,anddκΓi(t)dtkCktk,

where Ck is independent of the lacunary sequence θi; iZ}. Define the multiplier operators Mi, j on Rn × Rm by (Mi,jf^)(ξ,η)=Γi(ρ1(ξ))Γj(ρ2(η))f^(ξ,η). Then for any fS(Rn × Rm) and i, jZ, we have f(x,y)=k,lZ(Mi+k,j+lf)(x,y). Therefore, by Minkowski’s inequality we obtain

R+×R+|σΩh,t,sf(x,y)|2dtdsts1/2Ck,lZSk,lf(x,y),(15)

where

Sk,lf(x,y)=00|Yk,l(x,y,t,s)|2dtdsts1/2,Yk,l(x,y,t,s)=i,jZσΩh,t,sMi+k,j+lf(x,y)χ[θi,θi+1)×[θj,θj+1)(t,s).

Thus, it suffices to show for v > 0, that

Sk,l(f)Lp(Rn×Rm)CpA(γ)q12v2(|k|+|l|)hΔγ(R+×R+)ΩLq(Sn1×Sm1)fLp(Rn×Rm)

holds for any p with |1/p−1/2| < min{1/γʹ, 1/2}.

Let us first consider the L2 boundedness of Sk, l(f). By using Plancherels theorem, Fubinis theorem, Lemma 2.5, plus the approaches used in [10], we obtain that

Sk,l(f)L2(Rn×Rm)2i,jzΔi+k,j+lθiθi+1θjθj+1|σ^Ω,h,t,s(ξ,η)|2dtdstsf^(ξ,η)2dξdηCpln2(θ)hΔγ(R+×R+)2ΩLq(Sn1×Sm1)2×ijzΔi+kj+l|Aθjξ|±2wκm1q|Aθiη|±2wκm2q|f^(ξ,η)|dξdηCpln2(θ)2δ(|k|+|l|)hΔγ(R+×R+)2ΩLq(Sn1×Sm1)2i,jzΔi+k,j+lf^(ξ,η)2dξdηCpA(γ)q122δ(|k|+|l|)hΔγ(R+×R+)2ΩLq(Sn1×Sm1)2fL2(Rn×Rm)2,(16)

where Δi,j={(ξ,η)Rn×Rm:(ρ1(ξ),ρ2(η))Ii×Ij} and 0 < δ < 1.

Now, let us compute the Lp-norm of Sk, l(f) for p = r. By using Lemma 2.7, and applying the Littlewood-Paley theory plus [[10], (3.20) pp. 1242] see also [[9], Proposition 2.1]), we obtain

Sk,l(f)Lr(Rn×Rm)CijZθiθi+1θjθj+1(|σΩh,t,sMi+k,j+lf|)2dtdsts1/2Lr(Rn×Rm)CrA(γ)q1hΔγ(R+×R+)ΩLq(Sn1×Sm1)i,jZ|Mi+k,j+lf|21/2Lr(Rn×Rm)CrA(γ)q1hΔγ(R+×R+)ΩLq(Sn1×Sm1)fLr(Rn×Rm).

By Interpolation between the last inequality and (16), we reach (16). Consequently, the proof of Theorem 1.1 is complete.

Acknowledgement

The authors would like to thank the referees for their careful reading and valuable comments. Also, the authors would like to acknowledge Dr Hussain Al-Qassem for his suggestions on this note.

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Received: 2016-3-27
Accepted: 2016-8-25
Published Online: 2016-9-26
Published in Print: 2016-1-1

© Ali and Al-Dolat, published by De Gruyter Open

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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