1、Vol.43(2023)No.5?J.of Math.(PRC)TWO-DIMENSIONAL MAXIMAL OPERATOR OFVILENKIN-LIKE SYSTEM ON HARDY SPACESZHANG Xue-ying,WANG Chao-yue,ZHANG Chuan-zhou,XIAO Jun(College of Science,Wuhan University of Science and Technology,Wuhan 430065,China)Abstract:In this paper,we research the boundedness of two-dim
2、ensional maximal operatorof Vilenkin-like system on Hardy spaces.By means of atomic decomposition,the two-dimensionalmaximal operator Tf:=sup26nm62|f Pn,m|is bounded from Hpto Lp,where 0 p 12and 0.As an application,we prove the boundedness of two-dimensional operator f=sup26nm62|n,mf|(n+1)(m+1)1/p2.
3、By a counterexample,we also prove that two dimensional maximaloperator f=supn,mN|n,mf|(n+1)(m+1)1/2p1is not bounded from Hpto Lp,where 0 p 12.Theresults as above generalize the known conclusions in Walsh system or in Vilenkin system.Keywords:Vilenkin-like system;maximal operator;Dirichlet kernels;Fe
4、j er kernels2010 MR Subject Classification:42C10;60G46Document code:AArticle ID:0255-7797(2023)05-0398-111 IntroductionThe weak type inequality for maximal operator of Fej er means for trigonometric systemcan be found in Zygmund 1,in Schipp 2 for Walsh system and in P al,Simon 3 forbounded Vilenkin
5、system.Later,Schipp 2 showed that maximal operator f:=supn|nf|is of weak type(1,1),from which the a.e.convergencefollows by standard argument.Schippsresult implies by interpolation also the boundedness of:Lp Lp(1 p 6).Thisfails to hold for p=1,but Fujii 4 proved that is bounded from the dyadic Hardy
6、 spaceH1to L1(see also Simon 5).Fujiis results were extended by Wesiz 6,7 to Hpspacesfor 1/2 p 6 1,in the two-dimensional case,too.Simon 8 gave a counterexample,whichshows that boundedness of does not hold for 0 p 1/2.The counterexample for when p=1/2 is due to Goginava 9.Goginava 10 proved that the
7、 maximal operator defined by f=supnN|nf|log2(n+1)is bounded from the Hardy space H1/2to the space L1/2for Walsh system.He also proved,Received date:2022-06-26Accepted date:2022-07-15Foundation item:Supported by National Natural Science Foundation of China(11871195).Biography:Zhang Xueying(1980),fema
8、le,born at Hebi,Henan,associate professor,major indyadic Harmonic analysis,E-mail:.Corresponding author:Zhang ChuanzhouNo.5Two-dimensional maximal operator of vilenkin-like system on Hardy spaces399that for any nondecreasing function :N 1,),satisfying the conditionlimnlog2(n+1)(n)=+,the maximal oper
9、atorsupnN|nf|(n)is not bounded from the Hardy space H1/2to the space L1/2.Tephnadze 11 generalizedthis result and proved the boundedness ofsupnN|nf|(n+1)1/p2is bounded from the martingale Hardy space Hpto the space Lp,where nf is n-th Fej ermean with respect to bounded Vilenkin system for 0 p 1/2.In
10、 this paper the two-dimensional case will be investigated with respect to Vilenkin-like system.We show that the boundedness of some maximal operators.Throughout thispaper,we denote the set of integers and the set of non-negative integers by Z and N,respectively.We use c,cp,Cpto denote constants and
11、may denote different constants atdifferent occurrences.2 Definitions and NotationsLet m:=(m0,m1,mk,)be sequence of natural numbers such that mk 2(k N).For all k N we denote by Zmkthe mk-th discrete cyclic group.Let Zmkbe representedby 0,1,mk1.Suppose that each(coordinate)set has the discrete topolog
12、y and themeasure kwhich maps every singleton of Zmkto 1/mk(uk(Zmk)=1)for k N.Let Gmdenote the complete direct product of Zmks equipped with product topology and productmeasure,then Gmforms a compact Abelian group with Haar measure 1.The elements ofGmare sequences of the form(x0,x1,xk,),where xk Zmkf
13、or every k N and thetopology of the group Gmis completely determined by the setsIn(0):=(x0,x1,xk,)Gm:xk=0(k=0,n 1)(I0(0):=Gm).The Vilenkin space Gmis said to be bounded if the generating system m isbounded.We assume q=supimi .Let M0:=1 and Mk+1:=mkMkfor k N,it is so-called the generalized powers.The
14、n every n N can be uniquely expressed as n=Pk=0nkMk,0 nk mk,nk N.The sequence(n0,n1,)is called the expansion of n with respect to m.We often usethe following notations:|n|:=maxk N:nk6=0(that is,M|n|n M|n|+1)andn(k)=Pj=knjMj.400Journal of MathematicsVol.43For k N and x Gmdenote rkthe k-th generalized
15、 Rademacher function:rk(x):=exp(2xkmk)(x Gm,:=1,k N).It is known that for x Gm,n Nmn1Xi=0rin(x)=(0if xn6=0,mnif xn=0.(2.1)Now we define the nbyn:=Yk=0rnkk(n N).Then n:n N is a complete orthonormal system with respect to.We introduce the so-called Vilenkin-like(or)system(see 12).Let functions n,kj:Gm
16、 C(n,j,k N)satisfy for all x,y Gm:(1)kjis measurable with respect to jand kj(x+y)=kj(x)kj(y);(2)|kj|=kj(0)=k0=0j=1(j,k N);(3)n:=Yj=0n(j)j(n N).Let n:=nn(n N).The system :=n:n N is called a Vilenkin-like(or)system.Define Dirichlet kernels and Fej er kernels with respect to Vilenkin-like system andVil
17、enkin system as follows.Dn(y,x)=n1Xk=0k(y)k(x),Dn(x)=n1Xk=0k(x),Kn(y,x)=1nn1Xk=0Dk(y,x),Kn(x)=1nn1Xk=0Dk(x).Its well known thatDMn(y,x)=DMn(y x)=(Mnif y x In,0if y x GmIn.(2.2)Moreover for y,x Gm,Dn(y,x)=n(y)n(x)Dn(y x)=n(y)n(x)(Xj=0DMj(y x)mj1Xk=mjnjrkj(y x).(2.3)Since kj(x+y)=kj(x)kj(y)and rj(x+y)
18、=rj(x)rj(y),we haven(y)n(x)=n(y x+x)n(x)=n(y x)n(x)n(x)=n(y x)|n(x)|2=n(y x)n(0).(2.4)No.5Two-dimensional maximal operator of vilenkin-like system on Hardy spaces401Thus we obtainDn(y,x)=Dn(y x,0)andKn(y,x)=Kn(y x,0).(2.5)Now we define n,m(x,y):=n(x)m(y),(x,y Gm).If f L1then the numberf(n,m):=E(fn,m
19、)is said to be the(n,m)-th coefficient of f with respect to system.Denote by Sn,mf the(n,m)-th partial sum of the Fourier series of a martingale f withrespect to character system,namely,Sn,mf:=n1Xk=0m1Xl=0f(k,l)k,l.It is easy to see thatSMn,Mmf=fn,m.Let Fn,m(n,m N)be the-algebra generated by the rec
20、tangles In,m(x,y):=In(x)Im(y),(x,y Gm).A sequence of integrable functions f=(fn,m;n,m N)is said to bea martingale if fn,mis Fn,mmeasurable for all n,m N and SMn,Mmfk,l=fn,mfor alln,m,k,l N such that n 6 k and m 6 l.We say that a martingale f=(fn,m;n,m N)is Lp-bounded if kfkp:=supn,mkfn,mkp.The set o
21、f the Lp-bounded martingales will be denoted by Lp(G2m).The diagonal maximal function of a martingale f=(fn,m;n,m N)is defined byf:=supnN|fn,n|.It is easy to see that in case when f is an integrable real valued function given on G2m,the above maximal functions can be computed for all x,y Gmbyf(x,y)=
22、supnN1|In,n(x,y)|ZIn,n(x,y)f|.Define the spaces Hp(G2m)of Hardy type as the set of martingales f such thatkfkHp(G2m):=kfkp.The martingale Hardy spaces Hp(G2m)(0 p 6 1)have atomic characterizations.Abounded measurable function a defined on G2mis a p-atom if a 1 or there exists a dyadicsquare I such t
23、hatsupp a I,kak6|I|1/p,Z Za 0.We shall say also that a is supported on I.Then a martingale f=(fn,m;n,m N)is inHp(G2m)if there exists a sequence(ak,k N)of p-atoms and a sequence(k,k N)of realnumbers such thatPk=0|k|p andXk=0kSMn,Mnak=fn,n(n N).(2.6)402Journal of MathematicsVol.43Moreover,cpinf(Pk=0|k
24、|p)1/p6 kfkHp6 Cpinf(Pk=0|k|p)1/p,where the infimum is takenover all decompositions of f of the form(2.6).Next we will consider the boudedness of operator f and f in the two-dimensionalVilenkin-like system,where f=sup26nm62|n,mf|(n+1)(m+1)1/p2,f=supn,mN|n,mf|(n+1)(m+1)1/2p1.3 Some LemmasLemma 3.1(13
25、)Suppose that the operator T is sublinear and for 0 0 such thatZGmI|Ta|p6 Cp,(3.1)for every p-atom a Hpsupported on the dyadic interval I.If T is bounded from LsintoLsfor some 1 6 s 6,thenkTfkp CpkfkHp(f Hp L1).If(3.1)is true,T is called p-quasi-local.Lemma 3.2(13)Let 0 p 1,1 s and assume that the s
26、ublinear operatorT is p-quasi-local and(Ls,Ls)-bounded.Then T:Hu,v Lu,vis bounded for all p u sand 0 v 6.Especially,T is of weak type(1,1).Further we assume that for all n N the kernel Pn Lis given such that supnkPnk1.If we consider the maximal operatorTf:=supn|f Pn|(f L1),then T:L Lis evidently bou
27、nded.Therefore,if T is p-quasi-local for some 0 p MNZIN|Pn(x t,0)|dt)pdx 6 Cp1MN(n N)(3.2)implies the p-quasi-locality of T.ProofIndeed,to prove(3.1)let a be a p-atom supported on the interval I.Withoutloss of generality we can assume that I=INfor some N N.Then aPn=0 holds for allNo.5Two-dimensional
28、 maximal operator of vilenkin-like system on Hardy spaces403n=0,.,MN1,since the functions k(k=0,.,MN1)are constant on I.Therefore,Ta=supnMN|a Pn|and thusZGmIN(Ta(x)pdx=ZGmIN(supnMN|ZINa(t)Pn(x t,0)dt|)pdx6kakpZGmIN(supnMNZIN|Pn(x t,0)|dt)pdx6MNZGmIN(supnMNZIN|Pn(x t,0)|dt)pdx.(3.3)Hence,(3.1)follows
29、 from(3.2)and(3.3).Lemma 3.4(14)Let z Ik,lN,k=0,N 2,l=k+1,N 1 and n MN.ThenZIN|Kn(z t,0)|d(t)6cMlMknMN.(3.4)Let z Ik,NN,k=0,N 1 and n MN.ThenZIN|Kn(z t,0)|d(t)6cMkMN,(3.5)where c is an absolute constant andIk,lN=(IN(0,0,xk6=0,0,0,xl6=0,xl+1,xN1,)if k l N,IN(0,0,xk6=0,xk+1=0,xN1=0,xN)if l=N.Lemma 3.5
30、(14)Let 2 A N+,k s 0 define TbyTf:=sup26nm62|f Pn,m|.4 Formulations of Main Results404Journal of MathematicsVol.43Theorem 4.1Assume(3.2)for a given 0 p 6 1.Then Tis p-quasi-local.ProofIt is enough to prove(3.6)with a suitable r N.To this end let a L(G2m)be a p-atom.We can assume that a is supported
31、on the dyadic square IN INfor someN N.Furthermore,it follows that a Pn,m=0 when n,m MNor m MN.In the first casem MNr,while in the second case n MNrfollows.In other words,we get the estimateTa 6supn,mMNr|a Pn,m|,where r N is determined by r 1 6 MNr|ZINZINa(u,v)Pn(u x,0)Pm(v y,0)dudv|6M2pNsupn,mMNrZIN
32、|Pn(u x,0)|duZIN|Pm(v y,0)|dv.(4.1)Therefore,to verity(3.6)it is enough to show thatZG2m(INrINr)(supn,mMNrZIN|Pn(u x,0)|duZIN|Pm(v y,0)|dv)pdxdy 6CpM2N.(4.2)To this end let us decompose the double integral in question as follows:ZG2m(INrINr)(supn,mMNrZIN|Pn(u x,0)|duZIN|Pm(v y,0)|dv)pdxdy=ZGmINrZINr
33、(supn,mMNrZIN|Pn(u x,0)|duZIN|Pm(v y,0)|dv)pdxdy+ZINrZGmINr(supn,mMNrZIN|Pn(u x,0)|duZIN|Pm(v y,0)|dv)pdxdy+ZGmINrZGmINr(supn,mMNrZIN|Pn(u x,0)|duZIN|Pm(v y,0)|dv)pdxdy=:A1+A2+A3.(4.3)Here A1can be estimated in the following way:A16ZGmINr(supnMNrZIN|Pn(u x,0)|du)pdxZIN(supmZG|Pm(v y,0)|dv)pdy6ZGmINr
34、(supnMNrZIN|Pn(u x,0)|du)pdx|IN|(supmkPmk1)p6Cp1MNZGmINr(supnMNrZINr|Pn(u x,0)|du)pdx.(4.4)Thus we getA16 Cp1MN1MNr6CpM2N.No.5Two-dimensional maximal operator of vilenkin-like system on Hardy spaces405The estimate A26CpM2Ncan be derived similarly.Finally,applying(3.2)twice theestimationA36(ZGmINr(su
35、pkMNrZIN|Pn(u x,0)|du)pdx)26 Cp1M2N(4.5)follows,which proves Theorem 4.1.Theorem 4.2Let f=sup26nm62|n,mf|(n+1)(m+1)1/p2.Then for all 0 p 1/2 we havek fkp6 Cpkfkp(f Lp(G2m).ProofLet Pn(x,0)=nPk=01(n+1)1/p2Kn(x,0).By Theorem 4.1,it is enough to prove(3.2)for Pn(x,0).Let z Ik,lN,0 6 k 0 we getsupnMN1(n
36、+1)1/p2ZI|Kn(z t,0)|dt c1M1/p2NMlMknMN cMlMkM1/pN.(4.6)Thus we obtainZGmIN(supnMNZIN|Pn(x t,0)|dt)pdx=ZGmIN(supnMN1(n+1)1/p2ZIN|Kn(x t,0)|dt)pdx=N2Xk=0N1Xl=k+1mj1Xxj=0,jl+1,N1ZIk,lN(supnMN1(n+1)1/p2ZIN|Kn(x t,0)|dt)pd(z)+N1Xk=0ZIk,NN(supnMN1(n+1)1/p2ZIN|Kn(x t,0)|dt)pd(z)cN2Xk=0N1Xl=k+1mlmN 1MN(MlMk
37、M1/pN)p+N1Xk=01MN(MNMkM1/pn)pcN2Xk=0N1Xl=k+1(MlMk)pMlMN+N1Xk=01M2N(MNMk)p=c1MN(N2Xk=0N1Xl=k+11M12pl(MlMk)pM2pl+N1Xk=01M12pN(MNMk)pM2pN)=c1MN(N2Xk=0N1Xl=k+112(12p)l+N1Xk=012N(12p)=c1MN(N2Xk=012(12p)k+N2N(12p)cMN,(4.7)which complete the proof of Theorem 4.2.406Journal of MathematicsVol.43By Lemma 3.2
38、and Theorem 4.2,we easily get Theorem 4.3,we omit the proof.Theorem 4.3Let 0 p 1/2.Then :Hu,v(G2m)Lu,v(G2m)is bounded for allp u and 0 v 6.Especially,is of weak type(1,1).Theorem 4.4Let 0 p 1/2.Then the two dimensional maximal operator defined by f=supn,mN|n,mf|(n+1)(m+1)1/2p1is not bounded from Hp(
39、G2m)to Lp(G2m).ProofLet A N andfA(x,y):=(DM2A+1(x,0)DM2A(x,0)(DM2A+1(y,0)DM2A(y,0).It is simple to calculatefA(i,j)=(1,if i,j=M2A,M2A+1,M2A+1 10,otherwise(4.8)andSk,l(fA;x,y)=(Dk(x,0)DM2A(x,0)(Dl(y,0)DM2A(y,0),if k,l=M2A,M2A+1,M2A+1 1fA(x,y),if k,l M2A+10,otherwise.(4.9)We havefA=supk|SMk,Mk(fA;x,y)
40、|=|fA(x,y)|,kfAkHp=kfAkp=kfAkp=?ZGm?DM2A+1(x,0)DM2A(x,0)?pdxZGm?DM2A+1(y,0)DM2A(y,0)?pdy?1/p=?ZGm?DM2A+1(x)DM2A(x)?pdx?2/p=?ZI2A+1?DMA+1(x)DM2A(x)?pdx+ZI2AI2A+1?DM2A+1(x)DM2A(x)?pdx?2/pm2A1M2A+1Mp2A+(m2A 1)pMp2AM2A+12/pcM2(11/p)2A.SinceDi+MA(x,0)DMA(x,0)=MA(x)Dk(x,0)(4.10)No.5Two-dimensional maximal
41、 operator of vilenkin-like system on Hardy spaces407we have f=supn,mN|n,mf|(n+1)(m+1)1/2p1|fAKnA,nA(nA+1)(1/p2)|=1(nA)2(nA+1)(1/p2)|nA1Xi=0nA1Xj=0Si,jfA|=1(nA)2(nA+1)(1/p2)|nA1Xi=M2A+1nA1Xj=M2A+1Si,jfA|=1(nA)2(nA+1)(1/p2)|M2A+11Xi=M2A+1M2A+11Xj=M2A+1(Di(x,0)DMA(x,0)(Dj(y,0)DMA(y,0)|=1(nA)2(nA+1)(1/p
42、2)|nA11Xi=1nA11Xj=1(Di+MA(x,0)DMA(x,0)(Dj+MA(y,0)DMA(y,0)|=(nA1)2(nA)2(nA+1)(1/p2)|KnA1(x,0)KnA1(y,0)|.(4.11)Let q=supimi.For every l=1,14logq(A1/2p)1(A is supposed to belarge enough)let klbe the smallest natural numbers,for which M2AA1/2p1q2l/p M22klM2AA1/2p1q(2l2)/pholds.Suppose x,y Ikl,kl+12A:=I2
43、A(0,0,z2kl6=0,z2kl+16=0,z2s+1,z2A1),then byLemma 3.5 we have f(nA1)2(nA)2(nA+1)(1/p2)|KnA1(x,0)KnA1(y,0)|(M2klM2kl+1)2(nA)2(nA+1)(1/p2)c(M2klM2kl+1)2(M2A)2(M2A)(1/p2)1(M2A)(1/p2)A1/2pq4l/p.Thusk fkpp(1(M2A)(12p)Aq4l)(14logq(A1/2p)1Xl=1m2kl+31Xx2kl+3=0m2A11Xx2A1=0|Ikl,kl+12A|)2(1(M2A)(12p)Aq4l)(14log
44、q(A1/2p)1Xl=1m2kl+3m2A1M2A)21(M2A)(12p)Aq4l(14logq(A1/2p)1Xl=11M2kl)21(M2A)(12p)(logqAM2A)2=1M22p2A(logqA)2.(4.12)Thenk fkppkfAkpHp1M22p2A(logqA)2cM2p22A=(logqA)2.(4.13)408Journal of MathematicsVol.43Thus the proof of Theorem 4.4 is complete.References1 Zygmund A.Trigonometric seriesM.Cambridge Univ
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50、kin-like system on HardyspacesJ.Acta Mathematica Scientia(Chinese Series),2022,42(5):12941305?Hardy?,!,$#&%,()+*-,-.0/01+20143,506)+*430065)78:94:0;=?HardyACBD?-EFHG0I4J=KML/N0OPKRQHS0T.U4V4W-O4X0Y4Z0,0$_0?-L0/0N-OTf:=sup26nm62|f Pn,m|0a-bHardyAHBHpcLpQ4SCK-dfehg0 p-?+L-/-N+O f=sup26nm62|n,mf|(n+1)(