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Weighted Lp boundedness of maximal operators with rough kernels

  • In this paper, we study the weighted spaces Lp(ω,Rd) boundedness of certain class of maximal operators when their kernels belong to the space Lq(Sd1), q>1. Our results in this paper are improvements and extensions of some previously known results.

    Citation: Hussain Al-Qassem, Mohammed Ali. Weighted Lp boundedness of maximal operators with rough kernels[J]. AIMS Mathematics, 2024, 9(9): 25966-25978. doi: 10.3934/math.20241269

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  • In this paper, we study the weighted spaces Lp(ω,Rd) boundedness of certain class of maximal operators when their kernels belong to the space Lq(Sd1), q>1. Our results in this paper are improvements and extensions of some previously known results.



    Let Sd1 be the unit sphere in the d-dimensional Euclidean space Rd(d2) which is equipped with the normalized Lebesgue surface measure dσ=dσ().

    Let h:R+C be a radial function satisfying

    hL2(R+,dss)=(0|h(s)|2dss)1/21,

    and let be a homogeneous function of degree zero on Rd with L1(Sd1) and

    Sd1(x)dσ(x)=0, (1.1)

    where x=x/|x| for xRd{0}.

    For a Schwartz function f(Rd), we consider the maximal operator M,P given by

    M,P(f)(x)=suphL2(R+,dss)| RdeiP(u)f(xu)(u)h(|u|)|u|ddu|, (1.2)

    where P:RdR is a real-valued polynomial.

    We notice that if P(y)0, then the operator M,P is reduced to be the classical maximal operator denoted by M, which was introduced by Chen and Lin in [1]. The authors of [1] proved the boundedness of M on Lp(Rd) for 2d/(2d1)<p< if C(Sd1), and they showed that the range of p is the best possible. This result was extended in [2] in which the author confirmed the Lp boundedness of M for all p[2,) whenever L(logL)1/2(Sd1), and that the condition L(logL)1/2(Sd1) is optimal in the sense that the operator M may fail to be bounded on L2(Rd) when L(logL)r(Sd1) for any r(0,1/2). On the other hand, the author of [3] proved that M is bounded on Lp(Rd) for p2 if lies in the block spaces B(0,1/2)q(Sd1) with q>1, and they also proved that if the kernel belongs to B(0,r)q(Sd1) for some r(1,1/2), then M may not be bounded in L2(Rd). In [4], the author generalized the above results. In fact, he proved that M,P is bounded on Lp(Rd) for all p2 provided that B(0,1/2)q(Sd1)L(logL)1/2(Sd1). Subsequently, the investigation of the boundedness of M,P on Lp(Rd) under various conditions has attracted the attention of many authors: For background information [5,6,7,8], importance and the development [9,10,11], and recent advances and studies [12,13].

    On the other hand, in [14] Y. Ding and H. Qingzheng proved the weighted Lp boundedness of M as described in the following theorem.

    Theorem A. Let d2. Assume L2(Sd1) satisfies (1.1). Then,

    M(f)Lp(ω, Rd)CpfLp(ω, Rd), (1.3)

    if p and ω satisfy one of the following conditions:

    (a) 2p< and ωAp/2(Rd);

    (b) 2d/(2d1)<p<2, ω(x)=|x|α, and 12(1d)(2p)<α< 12(2dp2dp), where Ap is the Muckenhoupt's weight class, and the weighted Lp( ω,Rd) with ω0 is defined by

    Lp( ω,Rd)={f:fLp(ω,Rd)=(Rd|f(y)|pω(y)dy)1/p<}.

    Subsequently, Al-Qassem in [15] generalized the above result as in the following theorem:

    Theorem B. Suppose that Lq(Sd1) for some q>1 and it satisfies (1.1). Then,

    M(f)Lp(ω,Rd)CpfLp(ω,Rd),

    if p and ω satisfy one of the following conditions:

    (a) δp< and ωAp/δ;

    (b) 2dδ/(2d+dδ2)<p<2, ω(x)=|x|α, 12(1d)(2p)<α< 12(2dp2dp), where δ=max{2,q} and q is the dual exponent of q.

    In view of the results in [4] concerning the Lp boundedness of M,P and of the results in [15] concerning the weighted Lp boundedness of M, it is natural to ask wether the weighted Lp boundedness of M,P holds under the same conditions as assumed in Theorem B. We shall obtain an answer to this question in the affirmative as described in the following theorem.

    Theorem 1.1. Let Lq(Sd1) with q>1. Suppose that P:RdR is a polynomial of degree k, then the estimate

    M,P(f)Lp(ω,Rd)CpfLp(ω,Rd) (1.4)

    holds for δp< and ωAp/δ, where δ=max{2,q}.

    Now let us give some results which follow as a consequence of Theorem 1.1. For γ(1,), we let Lγ(R+,dss) be the set of all measurable functions h:R+R such that

    hLγ(R+,dss)=(0|h(s)|γdss)1/γ1.

    Consider the maximal operator M(γ),P given by

    M(γ),P(f)(x)=suphLγ(R+,dss)| RdeiP(u)f(xu)(u)h(|u|)|u|ddu|, (1.5)

    where P:RdR is a real-valued polynomial, fS(Rd) and 1γ2.

    The study of the boundedness of the operator M(γ),P started in [1] in which the authors proved that if C(Sd1) and hLγ(R+,dss) for some 1γ2, then the Lp(Rd) boundedness of the operator M(γ),0 is satisfied for (γd)<p<. For more information about the investigation of M(γ),P, under various conditions and some past studies, readers are referred to see [16,17,18] and the references therein. In this work, an extension and improvement over the result in [1] shall be obtained by proving the weighted Lp of M(γ),P when the condition C(Sd1) is replaced by the weaker condition Lq(Sd1) with q>1. Precisely, we have the following:

    Theorem 1.2. Let Lq(Sd1) with q>1. Let ωAp/δ and hLγ(R+,dss) with 1γ2. Then, we have

    M(γ),P(f)Lp(ω,Rd)CpfLp(ω,Rd) (1.6)

    for (δγ)/2p<.

    Concerning the boundedness of a certain class of oscillatory singular integrals, we have the following:

    Theorem 1.3. Assume that Lq(Sd1) with q>1. Let ωAp/δ and hLγ(R+,dss) for some 1<γ2. Then, the oscillatory singular integral operator T(γ),P given by

    T(γ),P(f)(x)=p.v.RdeiP(u)f(xu)(u)h(|u|)|u|ddu,

    is bounded on Lp(ω,Rd) for (δγ)/2p<, and it is bounded on Lp(ω,Rd) for 1<p(δγ2) and ω1pAp/δ.

    For background information and related work about the operator, see [19,20,21,22,23,24].

    We point out that the generalized Marcinkiewicz operator concerning the operator M(γ),P is given by

    M(γ),P(f)(x)=(R+|1s |u|seiP(u)f(xu)(u)|u|d+1du|γdss)1/γ. (1.7)

    As an immediate consequence of the fact

    M(γ),P(f)(x)CM(γ),P(f)(x)

    for 1γ2, we obtain the following result:

    Theorem 1.4. Let , ω, P, and γ be given as in Theorem 1.2. Then, the generalized Marcinkiewicz integral M(γ),P is bounded on Lp(ω,Rd) for (δγ)/2p< with 1<γ2.

    It is clear that for the special case P=0 and γ=2, the operator M(2),0 reduces to the classical Marcinkiewicz integral operator, which was introduced in [25], in which the author proved that the operator is bounded on Lp(Rd) only for 1<p2 whenever Lipη(Sd1) for some 0<η2. Thereafter, the study of the operator M(γ),P under several conditions has been discussed by many mathematicians (see, for instance [4,26,27,28,29,30]).

    Throughout the rest of the paper, the letter C stands for a positive constant which is independent of the essential variables and its value is not necessary the same at each occurrence.

    In this section, we give some preliminary lemmas to prove our main results. Let us start with the following lemma, which is found in [4].

    Lemma 2.1. Let Lq(Sd1), q>1 be a homogeneous function of degree zero. Suppose that

    P(x)=|η|kληxη,

    is a polynomial of degree k>1 such that |u|k is not one of its terms. For jZ, define Ij,:RdR by

    Ij,(ξ)=2(j1)2(j+1)|Sd1(u)ei[P(su)+suξ]dσ(u)|2dss. (2.1)

    Then, there exist constants C>0 and 0<ϵ<1 such that

    supξRdIj,(ξ)C2(j+1)/4q(|η|=m|λη|)ϵ/q.

    We need the following lemma from [15].

    Lemma 2.2. Let Lq(Sd1) for some q>1 and ωAp/q(R+) with 1<p<. Assume that the maximal function M is given by

    Mf(x)=supjZ2j|u|2(j+1)|f(xu)||(u)||u|ddu.

    Then there exists a positive constant Cp such that

    M(f)Lp(ω,Rd)CpfLp(ω,Rd)

    for any fLp(ω,Rd) with qp<.

    The next lemma can be proved by employing the same argument as in the proof of Theorem 1.1 in [15].

    Lemma 2.3. Let ωAp/δ and Lq(Sd1) with q>1. Then, there is a constant Cp>0 such that

    M(f)Lp(ω,Rd)CpfLp(ω,Rd) (2.2)

    for all δp<.

    Proof. Let {ψj}jZ be a smooth partition of unity in (0,) with the following properties:

    ψjC,supp ψj[2(j+1),2(j1)],0ψj1, jZψj(s)=1, and |dkψj(s)dsk|Cksk. (2.3)

    For jZ, define the operator Υj in Rd by

    ^(Υj(f))(ξ)=ψj(|ξ|))ˆf(ξ)forξRd.

    Then, for fS(Rd), we have that

    M(f)(x)kZG,k(f)(x), (2.4)

    where

    G,k(f)(x)=(jZ2(j1)2(j+1)|Sd1(Υk+jf)(xsu)(u)dσ(u))|2dss)1/2.

    By following the same argument utilized in the proof of Theorem 1.1 in [15], along with invoking Lemma 2.1, we obtain that

    G,k(f)Lp(ω,Rd)Cp2τ|k|fLp(ω,Rd), (2.5)

    for some constant τ(0,1) and for all δp<. Consequently, by (2.4) and (2.5), we get (2.2) for all δp<.

    Proof of Theorem 1.1. We shall use some of the ideas from [4]. Precisely, we use the induction on the degree of the polynomial P. It is clear that if the degree of P is 0, then by Lemma 2.3 we get

    M,P(f)Lp(ω,Rd)Cp fLp(ω,Rd) (3.1)

    for all δp<. Now, if the degree of P is 1, then we deduce that there is cRd so that P(u)=cu. Hence, if we set g(u)=eiP(u)f(u), then by (3.1) we get that

    M,P(f)Lp(ω,Rd)M,P(g)Lp(ω,Rd)Cp fLp(ω,Rd).

    Next, suppose that (1.4) holds for any polynomial P whose degree is less than or equal to k1. We need to prove that the inequality (1.4) is also satisfied for any polynomial of degree k+1. Let

    P(u)=|η|k+1ληuη

    be a polynomial of degree k+1. Without loss of generality, we may assume that P does not contain |u|k+1 as one of its terms, and |η|=k+1|λη|=1.

    For jZ, let {ψj} and Υj be chosen as those in (2.3). Set

    Γ(s)=0j=ψj(s) andΓ0(s)=j=1ψj(s).

    Then, Γ(s)+Γ0(s)=1, supp(Γ(s))[21,), and supp(Γ0(s))(0,1]. Hence, we get by Minkowski's inequality that

    M,P(f)(x)M,P,(f)(x)+M,P,0(f)(x), (3.2)

    where

    M,P,(f)(x)=( 21| Γ(s)Sd1eiP(su)f(xsu)(u)dσ(u)|2dss)1/2,

    and

    M,P,0(f)(x)=( 10| Γ0(s)Sd1eiP(su)f(xsu)(u)dσ(u)|2dss)1/2.

    Let us estimate M,P,(f)Lp(ω,Rd). Define

    M,P,,j(f)(x)=(2(j1)2(j+1) | Sd1eiP(su)f(xsu)(u)dσ(u)|2dss)1/2.

    Then, by the generalized Minkowski's inequality, we have

    M,P,(f)(x)0j=M,P,,j(f)(x). (3.3)

    Case 1. When q2. In this case, we have 2p< and ωAp/2. Let us consider first the case p>2. By duality, there is gL(p/2)(ω1(p/2),Rd) such that gL(p/2)(ω1(p/2),Rd)1 and

    M,P,,j(f)2Lp(ω,Rd)=Rd 41 | Sd1eiP(2(j+1)su)(u)f(x2(j+1)su)dσ(u)|2dss|g(x)|dx2Lq(Sd1)Rd 41(Sd1|f(x2(j+1)su)|qdσ(u))2/qdss|g(x)|dx2Lq(Sd1)Rd 41(Sd1|f(x2(j+1)su)|2dσ(u))dss|g(x)|dx.

    Hence, by Hölder's inequality, we get

    M,P,a,j(f)2Lp(ω,Rd)CRd |f(y)|2 41 Sd1 |g(y+2(j+1)su)|dσ(u)dssdyCp|f|2L(p/2)(ω,Rd)M(˜g)L(p/2)(ω1(p/2),Rd)Cpf2Lp(ω,Rd)˜gL(p/2)(ω1(p/2),Rd),

    where ˜g(y)=g(y) and M(f) is the Hardy-Littlewood maximal function. Thus,

    M,P,,j(f)Lp(ω,Rd)CpfLp(ω,Rd), (3.4)

    for 2<p< and ωAp/2.

    Now, for the case p=2 and ωA1, we have

    M,P,,j(f)2L2(ω,Rd)=Rd 41 | Sd1eiP(2(j+1)su)(u)f(x2(j+1)su)dσ(u)|2dssω(x)dx2qRn|f(x)|2(41Sd1ω(x+2(j+1)su)dσ(u)dss)ω(x)dxCRn|f(x)|2M(˜ω)(x)dx, with ˜ω(x)=ω(x)CRn|f(x)|2ω(x)dx=Cf2L2(ω,Rd), (3.5)

    where the last inequality is obtained by the fact that Mω(x)Cω(x) for a.e.xRd.

    Since for any ωAp/2 there exists α>0 such that ω1+αAp/2, by (3.4) and (3.5), we get that

    M,P,,j(f)Lp(ω1+α,Rd)CpfLp(ω1+α,Rd), (3.6)

    for 2<p< and ωAp/2.

    Nowwe will obtain a sharp unweighted L2 estimate of M,P,,j(f). By Fubini's theorem, Plancherel's theorem and Lemma 2.1 we get

    M,P,,j(f)L2(Rd)=(Rd|ˆf(ξ)|2Ij,(ξ)dξ)1/2C2(j+1)8qfL2(Rd). (3.7)

    Thus, using the Stein-Weiss interpolation theorem with change of measure [31], we may interpolate between (3.6) and (3.7) to obtain

    M,P,,j(f)Lp(ω,Rd)Cp2ε(j+1)8qfLp(ω,Rd) (3.8)

    for 2p<, ωAp/2, and for some ε(0,1). Consequently, by (3.3) and (3.8), we conclude that

    M,P,(f)Lp(ω,Rd)CpfLp(ω,Rd) (3.9)

    for 2p< and ωAp/2.

    Case 2. When 1<q<2. In this case, we have qp< and ωAp/q. Since p>2, by duality, there exists FL(p/2)(ω1(p/2),Rd) such that FL(p/2)(ω1(p/2),Rd)1 and

    M,P,,j(f)2Lp(ω,Rd)=Rd 41 | Sd1eiP(2(j+1)su)(u)f(x2(j+1)su)dσ(u)|2dss|F(x)|dxqLq(Sd1)Rd 41(Sd1|(u)|2q|f(x2(j+1)su)|2dσ(u))dss|F(x)|dx.

    Hence, by Hölder's inequality, we get

    M,P,,j(f)2Lp(ω,Rd)CRd |f(y)|2 41 Sd1 |(u)|2q|F(y+2(j+1)su)|dσ(u)dssdyC|f|2L(p/2)(ω,Rd)M(2q)(˜F)L(p/2)(ω1(p/2),Rd)Cpf2Lp(ω,Rd)˜FL(p/2)(ω1(p/2),Rd),

    where ˜F(y)=F(y). The last inequality holds since (p/2)>q/(2q) and by invoking Lemma 2.2. Therefore, we have

    M,P,,j(f)Lp(ω,Rd)CpfLp(ω,Rd) (3.10)

    for qp< and ωAp/q. By the last inequality and (3.3), we have that

    M,P,(f)Lp(ω,Rd)CpfLp(ω,Rd) (3.11)

    for δp< and ωAp/δ.

    Now, let us estimate the M,P,0(f)Lp(ω,Rd). Take Q(x)=|η|kληxη, and let M,Q,0(f) and M,P,Q,0(f) be given by

    M,Q,0(f)(x)=(10| Sd1eiQ(sw)f(xsu)(u)dσ(u)|2dss)1/2,

    and

    M,P,Q,0(f)(x)=(10| Sd1(eiP(su)eiQ(su))f(xsu)(u)dσ(u)|2dss)1/2.

    By Minkowski's inequality, we deduce that

    M,P,0(f)(x)M,Q,0(f)(x)+M,P,Q,0(f)(x). (3.12)

    Since the degree of the polynomial Q is less than or equal to k, we have that

    M,Q,0(f)Lp(ω,Rd)Cp fLp(ω,Rd) (3.13)

    for all δp< and ωAp/δ. By noticing that

    |eiP(su)eiQ(su)|s(d+1)||η|=d+1ληuη|s(d+1)

    and using the Cauchy-Schwartz inequality, we obtain

    M,P,Q,0(f)(x)C(10 Sd1s2(k+1)|(u)f(xsu)|2dσ(u)dss)1/2(=12j2(k+1))2+12Sd1|(u)f(xsu)|2dσ(u)dss)1/2.

    Therefore, by following the same arguments as above, we obtain that

    M,P,Q,0(f)Lp(ω,Rd)Cp fLp(ω,Rd) (3.14)

    for all δp< and ωAp/δ. Hence, by (3.13) and (3.14), we deduce that

    M,P,0(f)Lp(ω,Rd)Cp fLp(ω,Rd). (3.15)

    Consequently, by (3.2), (3.9), (3.11) and (3.15), the proof of Theorem 1.1 is complete.

    Proof of Theorem 1.2. By duality, it is easy to get that

    M(γ),P(f)(x)=( 0| Sd1eiP(sv)f(xsv)(v)dσ(v)|γdss)1/γ

    for all 1<γ2. Hence,

    M(γ),P(f)Lp(ω,Rd)=S(f)Lp(Lγ(R+,dss),ω,Rd),

    where S:Lp(ω,Rd)Lp(Lγ(R+,dss),ω,Rd) is a linear operator given by

    S(f)(x,s)=Sd1eiP(sv)f(xsv)(v)dσ(v).

    Now, if γ=1, fL(Rd) and hL1(R+,dss), then we have that

    | 0Sd1eiP(su)f(xsu)(u)h(s)dσ(u)dss|fL(Rd)L1(Sd1)hL1(R+,dss),

    and, hence,

    M(1),P(f)L(Rd)CfL(Rd)

    which, in turn, implies

    M(1),P(f)L(Rd)=S(f)L(L(R+,dss),Rd)CfL(Rd).

    Since L(Rd,ω)=L(Rd), we have

    M(1),P(f)L(ω,Rd)=S(f)L(L(R+,dss),ω,Rd)CfL(ω,Rd). (3.16)

    On the other hand, by Theorem 1.1 we get

    M(2),P(f)Lp(ω,Rd)=M,P(f)Lp(ω,Rd)=S(f)Lp(L2(R+,dss),ω,Rd)CpfLp(ω,Rd) (3.17)

    for δp<. Therefore, by applying the interpolation theorem for the Lebesgue mixed normed spaces to (3.16) and (3.17), we deduce that

    M(γ),P(f)Lp(ω,Rd)CpfLp(ω,Rd)

    for all (δγ)/2p< with 1<γ2.

    Proof of Theorem 1.3. To begin, we notice that (T(γ),Pf)(x)=limε0T(γ),P,εf(x), where T(γ),P,ε is the truncated singular integral operator given by

    T(γ),P,εf(x)=|u|>εeiP(u)f(xu)(u)h(|u|)|u|ddu. (3.18)

    By Hölder's inequality, we deduce

    |T(γ),P,εf(x)|ε|h(s)||Sd1eiP(sv)f(xsv)(v)dσ(v)|dsshLγ(R+,dr/r)(0|Sd1eiP(sv)f(xsv)(v)dσ(v)|γdss)1/γ.

    Hence,

    |T(γ),P,ε(f)(x)|hLγ(R+,dss)M(γ),P(f)(x). (3.19)

    Therefore, by Theorem 1.2, we get that T(γ),P is bounded on Lp(ω,Rd) for (δγ)/2p< and ωAp/δ. On the other hand, by a standard duality argument, we get that T(γ),P is bounded on Lp(ω,Rd) for 1<p(δγ2) and ω1pAp/δ. The proof is complete.

    In this work, we studied the mapping properties of the maximal integral operators M(γ),P. In fact, we proved the weighted space Lp(ω,Rd) boundedness of M(γ),P for all (δγ)/2p< whenever ωAp/δ, Lq(Sd1), and 1γ2. Then, as consequence of the this result, we confirmed the weighted Lp(ω,Rd) boundedness of the operators T(γ),P and M(γ),P. The results of this paper are substantial extensions and improvements of the main results in [4] and [15].

    Mohammed Ali: Writing-original draft, Formal Analysis, Commenting; Hussain Al-Qassem: Writing-original draft, Commenting. All authors have read and approved the final version of the manuscript for publication.

    The authors would like to express their gratitude to the editor for handling the full submission of the manuscript.

    The authors declare that they have no conflicts of interest in this paper.



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