Boundedness of differential transforms for one-sided fractional Poisson-type operator sequence
aa r X i v : . [ m a t h . C A ] A ug BOUNDEDNESS OF DIFFERENTIAL TRANSFORMS FOR ONE-SIDEDFRACTIONAL POISSON-TYPE OPERATOR SEQUENCE
ZHANG CHAO, TAO MA AND JOS´E L. TORREA
Abstract.
Let P ατ f be given by P ατ f ( t ) = 14 α Γ( α ) Z + ∞ τ α e − τ / (4 s ) s α f ( t − s ) ds, τ > , t ∈ R , < α < . It is known that the function U α ( t, τ ) = P ατ f ( t ) is a classical solution to the extension problem − D left U α + 1 − ατ U ατ + U αττ = 0 , in R × (0 , ∞ )and lim τ → + P ατ f ( t ) = f ( t ) , a.e. and in L p ( R , w )-norm , w ∈ A − p . In this paper, we analyze the convergence speed of a series related with P ατ f by discussing thebehavior of the family of operators T αN f ( t ) = N X j = N v j ( P αa j +1 f ( t ) − P αa j f ( t )) , N = ( N , N ) ∈ Z with N < N , where { v j } j ∈ Z is a bounded number sequence, and { a j } j ∈ Z is a ρ -lacunary sequence of positivenumbers, that is, 1 < ρ ≤ a j +1 /a j , for all j ∈ Z . We shall show the boundedness of the maximaloperator T ∗ f ( t ) = sup N | T αN f ( t ) | , t ∈ R , in the one-sided weighted Lebesgue spaces L p ( R , ω )( ω ∈ A − p ), 1 < p < ∞ . As a consequence weinfer the existence of the limit, in norm and almost everywhere, of the family T αN f for functions in L p ( R , ω ). Results for L ( R , ω )( ω ∈ A − ), L ∞ ( R ) and BMO ( R ) are also obtained.It is also shown that the local size of T ∗ f , for functions f having local support, is the same withthe order of a singular integral. Moreover, if { v j } j ∈ Z ∈ ℓ p ( Z ), we get an intermediate size betweenthe local size of singular integrals and Hardy-Littlewood maximal operator. Introduction
Let P ατ f be given by P ατ f ( t ) = 14 α Γ( α ) Z + ∞ τ α e − τ / (4 s ) s α f ( t − s ) ds, τ > , t ∈ R , < α < . (1.1)This is a fractional Poisson-type operator on the line, which can be found in [3]. It is known thatthe Poisson-type operator appeared when solving the extension problem, see [5, 12, 13]. In [3], theauthors showed that P ατ is a classical solution to a version of extension problem for the given initialdata f in a weighted space L p ( w ), where w satisfies the one-sided A p condition. Moreover, in thisextension problem, they proved that the fractional derivatives on the line are Dirichlet-to-Neumann Key words: differential transforms, heat semigroup, fractional Poisson operator.The first author was supported by the Natural Science Foundation of Zhejiang Province(Grant No. LY18A010006),the first Class Discipline of Zhejiang - A (Zhejiang Gongshang University- Statistics) and the State Scholarship Fund(No.201808330097). The second author was supported by National Natural Science Foundation of China(Grant Nos.11671308, 11431011) and the independent research project of Wuhan University (Grant No. 2042017kf0209). Thethird author was supported by grant MTM2015-66157-C2-1-P (MINECO/FEDER) from Government of Spain. operators. Precisely, it is shown that for functions f ∈ L p ( R , w ) , w ∈ A − p , < p < ∞ , the function U α ( t, τ ) = P ατ f ( t ) is a classical solution to the extension problem − D left U α + 1 − ατ U ατ + U αττ = 0 , in R × (0 , ∞ ) , lim τ → + P ατ f ( t ) = f ( t ) , a.e. and in L p ( R , w )-norm.Moreover, for c α := 4 α − / Γ( α )Γ(1 − α ) > − c α lim τ → + τ − α U ατ ( t, τ ) = ( D left ) α f ( t ) , in the distributional sense . In the above formulas, D left f ( t ) = lim s → − f ( t ) − f ( t − s ) s and ( D left ) α f ( t ) = 1Γ( − α ) Z ∞ f ( t − s ) − f ( t ) s α +1 ds. By A − p we denote the class of lateral weights introduced by E.Sawyer [11], see (2.2) and (2.3).The purpose of this note is to give some extra information about the convergence of the family {P ατ f } τ> . In order to do this, we shall discuss the behavior of the series X j ∈ Z v j ( P αa j +1 f ( t ) − P αa j f ( t )) , where { v j } j ∈ Z is a sequence of bounded numbers and { a j } j ∈ Z is a ρ -lacunary sequence of positivenumbers, that is, 1 < ρ ≤ a j +1 /a j , for all j ∈ Z . This way to analyze convergence of sequences wasconsidered by Jones and Rosemblatt for ergodic averages(see [7]), and latter by Bernardis et al. fordifferential transforms(see [2]).For each N ∈ Z , N = ( N , N ) with N < N , we define the sum(1.2) T αN f ( t ) = N X j = N v j ( P αa j +1 f ( t ) − P αa j f ( t )) . We shall consider the maximal operator T ∗ α f ( t ) = sup N | T αN f ( t ) | , t ∈ R . Along the paper, we shall denote T ∗ to be T ∗ α for simply . (1.3)The supremum are taken over all N = ( N , N ) ∈ Z with N < N .In order to prove the results, we shall use the vector-valued Calder´on-Zygmund theory in anessential way. In the proof of the maximal operator T ∗ , we shall use a kind of Cotlar’s lemma thatin some sense is parallel to the classical Cotlar’s inequality used to control the maximal operator ofthe truncations in the Calder´on-Zygmund theory. Looking to the first set of our results, the readercould have the impression that the operator T ∗ is of the same size of the maximal operator M − . In this line of thought we present a series of results contained in Theorem 1.3 and Theorem 1.4 inwhich it is shown that the size of T ∗ acting over functions of compact support is in fact of the orderof a singular integral. At this point we want to observe the analogy of our operators with martingaletransforms. On the other hand if we consider the sequence of Rademacher functions { r j } j ∈ Z , byKintchine’s inequality we have (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:16) X j ∈ Z |P αa j +1 f ( · ) − P αa j f ( · ) | (cid:17) / (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p ( R ) ≤ K p (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X j ∈ Z r j ( · )( P αa j +1 f ( · ) − P αa j f ( · )) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p (Ω) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p ( R ) . In other words, as a by product of our results we get the boundedness of the operator (cid:16) X j ∈ Z |P αa j +1 f ( · ) − P αa j f ( · ) | (cid:17) / in the same spaces that we get for operator T ∗ . Finally, in Theorem 1.4 it is also shown that if weassume the sequence { v j } j ∈ Z ∈ ℓ p ( Z ), then the local behavior of T ∗ is approaching to the maximaloperator as p → + . Now we present our main results.
Theorem 1.1.
Let < α < , { v j } j ∈ Z a sequence of bounded numbers and { a j } j ∈ Z a ρ -lacunarysequence of positive numbers. Let T ∗ be defined in (1.3) .(a) For any < p < ∞ and ω ∈ A − p , there exists a constant C depending on p, ρ, α, ω and k v k l ∞ ( Z ) such that k T ∗ f k L p ( R ,ω ) ≤ C k f k L p ( R ,ω ) , for all functions f ∈ L p ( R , ω ) . (b) For any ω ∈ A − , there exists a constant C depending on ρ, α, ω and k v k l ∞ ( Z ) such that ω ( {−∞ < t < + ∞ : | T ∗ f ( t ) | > λ } ) ≤ C λ k f k L ( R ,ω ) , λ > , for all functions f ∈ L ( R , ω ) . (c) Given f ∈ L ∞ ( R ) , then either T ∗ f ( t ) = ∞ for all t ∈ R , or T ∗ f ( t ) < ∞ for a.e. t ∈ R . And inthis later case, there exists a constant C depending on ρ , α and k v k l ∞ ( Z ) such that k T ∗ f k BMO ( R ) ≤ C k f k L ∞ ( R ) . (d) Given f ∈ BM O ( R ) , then either T ∗ f ( t ) = ∞ for all t ∈ R , or T ∗ f ( t ) < ∞ for a.e. t ∈ R . Andin this later case, there exists a constant C depending on ρ , α and k v k l ∞ ( Z ) such that (1.4) k T ∗ f k BMO ( R ) ≤ C k f k BMO ( R ) . We have denoted by L p ( R , ω ) , ≤ p < ∞ , the Lebesgue space of measurable functions satisfying Z R | f ( t ) | p ω ( t ) dt < ∞ , and L ∞ ( R ) the space of measurable functions such that ess sup t ∈ R | f ( t ) | < ∞ . Both of them are withthe obvious norms. Also, we define BM O ( R ) as the space of measurable functions such that for anyinterval B , 1 | B | Z B (cid:12)(cid:12)(cid:12) f ( t ) − f B (cid:12)(cid:12)(cid:12) dt ≤ C < ∞ , and k f k BMO ( R ) = sup B | B | Z B (cid:12)(cid:12)(cid:12) f ( t ) − f B (cid:12)(cid:12)(cid:12) dt , where f B = 1 | B | Z B f ( t ) dt . For more details, see [6].The proof of the last theorem contains three steps:(A) We prove the following uniform boundedness of the family of operators T αN : from L p ( R , ω )into L p ( R , ω ), 1 < p < ∞ , from L ( R , ω ) into weak- L ( R , ω ), from L ∞ ( R ) into BM O ( R ),and from BM O ( R ) into BM O ( R ), see Theorem 2.6.(B) The following pointwise Cotlar’s type inequalitysup − M ≤ N The dichotomy results announced in Theorem 1.1, parts ( c ) and ( d ), about L ∞ ( R ) and BM O ( R )are motivated, in part, by the existence of a bounded function f such that T ∗ f ( t ) = ∞ as the followingtheorem shows. Theorem 1.3. There exist bounded sequence { v j } j ∈ Z , ρ -lacunary sequence { a j } j ∈ Z and f ∈ L ∞ ( R ) such that T ∗ f ( t ) = ∞ for all t ∈ R . This last theorem also says that the operator T ∗ is essentially bigger than the operator P ∗ f ( t ) =sup τ P / τ f ( t ) which is bounded in L p ( R , ω ) , < p < ∞ , and in L ∞ ( R ), see [3].On the other hand, if f = χ (0 , and H is the Hilbert transform, it is easy to see that 1 r Z − r H ( f )( x ) dx ∼ log er as r → + . In general, this is the growth of a singular integral applied to a bounded function atthe origin. The following theorem shows that the growth of the function T ∗ f for bounded function f at the origin is of the same order of a singular integral operator. Theorem 1.4. (a) Let { v j } j ∈ Z ∈ l p ( Z ) for some ≤ p ≤ ∞ . For every f ∈ L ∞ ( R ) with support inthe unit ball B = B (0 , , for any ball B r ⊂ B with r < , there exists a constant C > suchthat | B r | Z B r | T ∗ f ( t ) | dt ≤ C (cid:18) log 2 r (cid:19) /p ′ k v k l p ( Z ) k f k L ∞ ( R ) . (b) When < p < ∞ , for any ε > , there exist a ρ -lacunary sequence { a j } j ∈ Z , a sequence { v j } j ∈ Z ∈ ℓ p ( Z ) and a function f ∈ L ∞ ( R ) with support in the unit ball B = B (0 , , satisfying the followingstatement: for any ball B r ⊂ B with r < , there exists a constant C > such that | B r | Z B r | T ∗ f ( t ) | dt ≥ C (cid:18) log 2 r (cid:19) / ( p − ε ) ′ k v k l p ( Z ) k f k L ∞ ( R ) . (c) When p = ∞ , there exist a ρ -lacunary sequence { a j } j ∈ Z , a sequence { v j } j ∈ Z ∈ l ∞ ( Z ) and f ∈ L ∞ ( R ) with support in the unit ball B = B (0 , , satisfying the following statements: for any ball B r ⊂ B with r < , there exists a constant C > such that | B r | Z B r | T ∗ f ( t ) | dt ≥ C (cid:18) log 2 r (cid:19) k v k l ∞ ( Z ) k f k L ∞ ( R ) . In the statements above, p ′ = pp − , and if p = 1 , p ′ = ∞ . Some related results about the local behavior of variation operators can be found in [4]. Onedimensional results about the variation of some convolutions operators can be found in [8].The organization of the paper is as follows. In Section 2, we will get the kernel estimates tosee that the kernel K αN is a vector-valued Caldr´on-Zygmund kernel, and then we can get the uniformboundedness of T αN , i.e. Theorem 2.6. And with a Cotlar’s inequality, we can get the proof of Theorem1.1 in Section 3. In Section 4, we will give the proof of Theorem 1.3 and Theorem 1.4.Throughout this paper, the symbol C in an inequality always denotes a constant which may dependon some indices, but never on the functions f in consideration.2. Uniform L p boundedness of the operators T αN We shall need the following lemma. Lemma 2.1. Let < α < . Then for any complex number z with Rez > and | arg z | ≤ π/ , wehave Z ∞ e − z u e − z u duu α = z − α Z ∞ e − r e − z /r r − α dr. IFFERENTIAL TRANSFORMS FOR FRACTIONAL POISSON TYPE OPERATOR SEQUENCE 5 Proof. Let ϕ = arg z . Assume that 0 ≤ ϕ ≤ π/ 4. The case − π/ ≤ ϕ ≤ ϕ := { z = re iϕ : 0 < r < ∞} . And then let C denote the sector in the real part of the complex plane, with 0 ≤ arg z ≤ ϕ buttruncated at c ε : | z | = ε and C R : | z | = ε . In fact, the boundary of C consists four parts: C ε , C R ,ray ϕ and positive half part of the real line.Let us consider the complex function F ( u ) = e − z /u e − uz u α , which is holomorphic function when u = 0. Thus, by the Cauchy theorem, we have Z C F ( u ) du = 0.We first calculate (cid:12)(cid:12)(cid:12)(cid:12)Z C ε F ( u ) du (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z ϕ e − z / ( e iθ ε ) e − z εe iθ ε α e iαθ iεe iθ dθ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z ϕ e −| z | e i ( ϕ − θ ) /ε e −| z | εe i ( ϕ θ ) ε α e iαθ iεe iθ dθ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Since ϕ < π/ ϕ − θ < π/ ϕ + θ < π/ 2. Hence (cid:12)(cid:12)(cid:12)(cid:12)Z C ε F ( u ) du (cid:12)(cid:12)(cid:12)(cid:12) ≤ ε − α Z ϕ e −| z | cos( ϕ − θ ) /ε e −| z | ε cos( ϕ + θ ) dθ → , as ε → 0. Similarly, along the curve C R , we have (cid:12)(cid:12)(cid:12)(cid:12)Z C ε F ( u ) du (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z ϕ e −| z | cos( ϕ − θ ) /R e −| z | R cos( ϕ + θ ) R − α dθ. If ϕ < π/ (cid:12)(cid:12)(cid:12)(cid:12)Z C R F ( u ) du (cid:12)(cid:12)(cid:12)(cid:12) ≤ e − C z R R − α Z ϕ e − cos( ϕ − θ ) dθ → , as R → ∞ . But for the case ϕ = π/ ϕ + θ can be π/ 2, then we can not take the limit as above.However, we have (cid:12)(cid:12)(cid:12)(cid:12)Z C R F ( u ) du (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z π e −| z | cos( π − θ ) /R e −| z | R cos( π + θ ) R − α dθ ≤ Z π e −| z | R cos( π + θ ) R − α dθ ≤ Z π e −| z | R sin( π − θ ) R − α dθ ≤ Z π e −| z | R sin ω R − α dω ≤ Z π e −| z | π Rω R − α dω, where we have changed variable ω = π/ − θ and used the inequality 2 ω/π ≤ sin ω . Thus we have (cid:12)(cid:12)(cid:12)(cid:12)Z C R F ( u ) du (cid:12)(cid:12)(cid:12)(cid:12) ≤ π | z | R − α Z ∞ e − u du ≤ CR − α → , R → ∞ . Therefore, we conclude that (cid:12)(cid:12)(cid:12)(cid:12)Z C R F ( u ) du (cid:12)(cid:12)(cid:12)(cid:12) = 0 for | arg z | ≤ π/ Z ∞ F ( u ) du = Z ray ϕ F ( u ) du. Taking u = sz , we have Z ∞ F ( u ) du = Z Ray ϕ e − /s e − sz s α z α z ds = z − α Z ∞ e − r e − z /r r − α dr. Then this lemma is completely proved. (cid:3) ZHANG CHAO, TAO MA AND JOS´E L. TORREA Remark 2.2. Notice that the integral y s s Γ( s ) Z ∞ e − y / (4 τ ) e − τ ( iρ + λ ) dττ s , ρ ∈ R , λ ≥ , < s < . is absolutely convergent. Uniform L -boundedness. It is known that, see [3], the Fourier transform of P ατ f is d P ατ f ( ρ ) = 1Γ( α ) Z ∞ e − r e − iρτ / r ˆ f ( ρ ) drr − α . By b f ( ρ ) we denote the Fourier transform of the function f , that is, b f ( ρ ) = 1(2 π ) / Z R f ( x ) e − ixρ dx, ρ ∈ R . Theorem 2.3. There is a constant C , depending on α and k v k l ∞ ( Z ) , such that sup N k T αN f k L ( R ) ≤ C k f k L ( R ) . Proof. Let f ∈ L ( R ). Using the Plancherel theorem, we have k T αN f k L ( R ) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X j = N v j ( P αa j +1 f − P αa j f ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ( R ) ≤ C k v k l ∞ ( Z ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X j = −∞ Z a j +1 a j (cid:12)(cid:12)(cid:12) ∂ τ d P ατ f (cid:12)(cid:12)(cid:12) dτ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ( R ) . Observe that, ∂ τ d P ατ f ( ρ ) = C∂ τ Z ∞ e − r e − τ r ( iρ ) b f ( ρ ) drr − α = C Z ∞ e − r τ ( iρ ) e − τ r ( iρ ) b f ( ρ ) drr − α . Note that the Fourier transform above is well defined, see Remark 2.2. Then we deduce that k T αN f k L ( R ) ≤ C (cid:13)(cid:13)(cid:13)(cid:13) b f ( ρ ) Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ e − r τ ( iρ ) e − τ r ( iρ ) drr − α (cid:12)(cid:12)(cid:12)(cid:12) dτ (cid:13)(cid:13)(cid:13)(cid:13) L ( R ) . Changing variable z = τ √ iρ , by using Lemma 2.1 , we have (cid:12)(cid:12)(cid:12) Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ e − r τ ( iρ ) e − τ r ( iρ ) drr − α (cid:12)(cid:12)(cid:12)(cid:12) dτ = Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ e − r z e − z r drr − α (cid:12)(cid:12)(cid:12)(cid:12) dz = 2 − α Z ∞ (cid:12)(cid:12)(cid:12)(cid:12) z α Z ∞ e − z u e − z u duu α (cid:12)(cid:12)(cid:12)(cid:12) dz . Since | arg z | = π/ 4, we have | e − z / (2 u ) | ≤ e − c | z | /u and | e − z u/ | ≤ e − c | z | u , where c = √ / 4. Then (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ z α Z ∞ e − z /u e − z u duu α dz (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z ∞ | z | α Z ∞ e − c | z | /u e − c | z | u duu α dz ≤ Z ∞ | z | α − Z ∞ e − c | z | /v e − cv dvv α dz = Z ∞ | p iρ | α τ α − Z ∞ e − c ( |√ iρ | τ ) /v e − cv dvv α dτ = Z ∞ Z ∞ ( | p iρ | τ ) α − e − c ( mτ ) /v d ( | p iρ | τ ) e − cv dvv α = Z ∞ Z ∞ τ α − e − cτ /v dτ e − cv dvv α ≤ C Z ∞ e − cv dv ≤ C. Then the proof of the theorem is complete. (cid:3) IFFERENTIAL TRANSFORMS FOR FRACTIONAL POISSON TYPE OPERATOR SEQUENCE 7 Uniform L p -boundedness. Let us come back to the definition of the operators T αN , see (1.2). By using the formula (1.1), wehave T αN f ( t ) = N X j = N v j ( P αa j +1 f ( t ) − P αa j f ( t ))= 14 α Γ( α ) N X j = N v j Z + ∞ a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α f ( t − s ) ds = Z + ∞ K αN ( s ) f ( t − s ) ds = Z t −∞ K αN ( t − s ) f ( s ) ds, where(2.1) K αN ( s ) = 14 α Γ( α ) N X j = N v j a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α . The kernel K αN ( s ) is supported in (0 , + ∞ ) . Our study of T αN will be related to the one-sidedCalder´on-Zygmund operators. In particular, we shall look for Lebesgue estimates with absolutecontinuous measures w ( x ) dx , where w is a weight in any of the classes A ± p defined by E. Sawyer, see[11]. This classes were introduced in relation with the boundedness of the one-sided Hardy-Littlewoodmaximal operator M − defined by M − f ( t ) = sup ε> ε Z − ε | f ( t + s ) | ds. We recall the results that we shall use related with weights for M − :(1) The operator M − is of weak type (1 , 1) with respect to the measure ω ( t ) dt if and only if ω ∈ A − ,i.e., there exists C such that(2.2) M + ω ≤ Cω a.e., where M + is the right-sided Hardy Littlewood maximal operator defined as M + f ( t ) = sup ε> ε Z ε | f ( t + s ) | ds. (2) The operator M − is bounded in L p ( ω ), 1 < p < ∞ , if and only if ω ∈ A − p , i.e., if there exists C such that for any three points a < b < c (2.3) Z ba ω − p ′ ! p ′ (cid:18)Z cb ω (cid:19) p ≤ C ( c − a ) , where 1 p + 1 p ′ = 1 . For more details about the one-sided weights, see [1, 2, 11]. Theorem 2.4. Let K αN be the kernel defined in (2.1). For any s = 0 , there exists constant C depending on α and k v k l ∞ ( Z ) (but not on N ) such thati) | K αN ( s ) | ≤ Cs ,ii) | ∂ s K αN ( s ) | ≤ Cs . The proof of Theorem 2.4 involves an estimate we will repeat several times, so we formulate it inthe following remark. ZHANG CHAO, TAO MA AND JOS´E L. TORREA Remark 2.5. Along the paper, we shall use frequently the estimate x A e − x/B ≤ Ce − x/B ′ with x, A, B, B ′ , C > . Proof of Theorem 2.4. For i ), we have | K αN ( s ) | ≤ C ∞ X j = −∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = C ∞ X j = −∞ (cid:12)(cid:12)(cid:12) a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) (cid:12)(cid:12)(cid:12) s α . Observe that, by Remark 2.5, ∞ X j = −∞ (cid:12)(cid:12)(cid:12) a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) (cid:12)(cid:12)(cid:12) = ∞ X j = −∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z a j +1 a j ∂ u (cid:16) u α e − u / (4 s ) (cid:17) du (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z ∞ (cid:12)(cid:12)(cid:12)(cid:12) (2 αu α − − u α +1 s ) e − u / (4 s ) (cid:12)(cid:12)(cid:12)(cid:12) du ≤ C Z ∞ (cid:12)(cid:12)(cid:12)(cid:12) ( u α − + u α +1 s ) e − u / (4 s ) (cid:12)(cid:12)(cid:12)(cid:12) du ≤ C √ s (cid:16) Z ∞ ( √ s ) α − (cid:18) u √ s (cid:19) α − e − ( u/ √ s ) d u √ s + s α − / Z ∞ (cid:18) u √ s (cid:19) α +1 e − ( u/ √ s ) d u √ s (cid:17) ≤ Cs α . Then | K αN ( s ) | ≤ Cs . This proves i) .For ii) , we can write K αN ( s ) = C N X j = N v j (cid:16) a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) (cid:17) = C N X j = N s α v j Z a j +1 a j (cid:18) αu α − − u α +1 s (cid:19) e − u s du. The partial derivative ∂ s K αN ( s ) consists two parts. The first part is I = C N X j = N s α v j Z a j +1 a j (cid:18) u α +1 s + (cid:18) αu α − − u α +1 s (cid:19) u s (cid:19) e − u s du = C N X j = N s α v j Z a j +1 a j (cid:18) ( α + 1) u α +1 s − u α +3 s (cid:19) e − u s du. And the second part is II = C N X j = N ∂ s (cid:18) s α (cid:19) v j Z a j +1 a j (cid:18) αu α − − u α +1 s (cid:19) e − u s du = C N X j = N (cid:18) − αs α (cid:19) v j Z a j +1 a j (cid:18) αu α − − u α +1 s (cid:19) e − u s du. Then by using Remark 2.5 again, we have | I | ≤ C s α Z ∞ (cid:12)(cid:12)(cid:12)(cid:12) ( α + 1) u α +1 s − u α +3 s (cid:12)(cid:12)(cid:12)(cid:12) e − u s du ≤ C s α s α − ≤ Cs , and | II | ≤ C (cid:12)(cid:12)(cid:12) αs α (cid:12)(cid:12)(cid:12) Z ∞ (cid:12)(cid:12)(cid:12) (2 αu α − − u α +1 s ) e − u s du (cid:12)(cid:12)(cid:12) ≤ C s α s α ≤ Cs . IFFERENTIAL TRANSFORMS FOR FRACTIONAL POISSON TYPE OPERATOR SEQUENCE 9 Combining the estimates I and II , we have | ∂ s K αN ( s ) | ≤ Cs . All the estimates above are true uniform for N . The proof of the Theorem 2.4 is complete. (cid:3) From Theorems 2.3, 2.4, and standard Calder´on-Zygmund theory, we can get the uniform estimatein L p ( R , w ) (1 < p < ∞ , w ∈ A p ) of the operators T αN . Here, A p denotes the classical Muckenhoupt A p weights, see [10]. However, to the one-side nature of the kernel, we can apply Theorem 2.1 in [1]to get the uniform boundedness in L p ( R , w ) of the operators T αN with w ∈ A − p in the following. Theorem 2.6. Let T αN be the family of operators defined in (1.2) , we have the following statements.(a) For any < p < ∞ and ω ∈ A − p , there exists a constant C depending on p, α , k v k l ∞ ( Z ) and ω (noton N ) such that k T αN f k L p ( R ,ω ) ≤ C k f k L p ( R ,ω ) , for all functions f ∈ L p ( R , ω ) . (b) For any ω ∈ A − , there exists a constant C depending on α , k v k l ∞ ( Z ) and ω (not on N ) such that ω ( { t ∈ R : | T αN f ( t ) | > λ } ) ≤ C λ k f k L ( R ,ω ) , λ > , for all functions f ∈ L ( R , ω ) . (c) There exists a constant C depending on α and k v k l ∞ ( Z ) (not on N ) such that k T αN f k BMO ( R ) ≤ C k f k L ∞ ( R ) , for all functions f ∈ L ∞ ( R ) . (d) There exists a constant C depending on α and k v k l ∞ ( Z ) (not on N ) such that k T αN f k BMO ( R ) ≤ C k f k BMO ( R ) , for all functions f ∈ BM O ( R )) . The constants C appeared above all are independent with N. As we have said before the proof of ( a ) and ( b ) in the theorem above is obtained by using Theorem2.1 in [1]. On the other hand the proof of ( c ) and ( d ) are standard in the Calder´on-Zygmund theoryand it can be found in [9].3. Boundedness of the maximal operator T ∗ In this section, we will give the proof of Theorem 1.1 related to the boundedness of the maximaloperator T ∗ . The next proposition, parallel to Proposition 3.2 in [2], shows that, without lost ofgenerality, we may assume that(3.1) 1 < ρ ≤ a j +1 a j ≤ ρ , j ∈ Z . Proposition 3.1. Given a ρ -lacunary sequence { a j } j ∈ Z and a multiplying sequence { v j } j ∈ Z ∈ l ∞ ( Z ) ,we can define a ρ -lacunary sequence { η j } j ∈ Z and { ω j } j ∈ Z ∈ l ∞ ( Z ) verifying the following properties:(i) < ρ ≤ η j +1 /η j ≤ ρ , k ω j k l ∞ ( Z ) = k v j k l ∞ ( Z ) .(ii) For all N = ( N , N ) there exists N ′ = ( N ′ , N ′ ) with T αN = ˜ T αN ′ , where ˜ T αN ′ is the operatordefined in (1.2) for the new sequences { η j } j ∈ Z and { ω j } j ∈ Z . Proof. We follow closely the ideas in the proof of Proposition 3.2 in [2]. We include it at here forcompleteness.Let η = a , and let us construct η j for positive j as follows (the argument for negative j isanalogous). If ρ ≥ a /a ≥ ρ, define η = a . In the opposite case where a /a > ρ , let η = ρa . It verifies ρ ≥ η /η = ρ ≥ ρ. Further, a /η ≥ ρ a /ρa = ρ. Again, if a /η ≤ ρ , then η = a . If this is not the case, define η = ρ a ≤ a . By the same calculations as before, η , η , η are partof a lacunary sequence satisfying (3.1). To continue the sequence, either η = a (if a /η ≤ ρ ) or η = ρ η (if a /η > ρ ). Since ρ > , this process ends at some j such that η j = a . The rest ofthe elements η j are built in the same way, as the original a k plus the necessary terms put in betweentwo consecutive a j to get (3.1).Let J ( j ) = { k : a j − < η k ≤ a j } , and ω k = v j if k ∈ J ( j ). Then v j ( P αa j +1 f ( t ) − P αa j f ( t )) = X k ∈ J ( j ) ω k ( P αa k +1 f ( t ) − P αa k f ( t )) . If M = ( M , M ) is the number such that η M +1 = a N +1 and η M = a N , then we get T αN f ( t ) = N X j = N v j ( P αa j +1 f ( t ) − P αa j f ( t )) = M X k = M ω k ( P αη k +1 f ( t ) − P αη k f ( t )) = ˜ T αM f ( t ) , where ˜ T αM is the operator defined in (1.2) related with sequences { η k } k ∈ Z , { ω k } k ∈ Z , α and M =( M , M ). (cid:3) It follows from this proposition that it is enough to prove all the results of this article in the caseof a ρ -lacunary sequence satisfying (3.1). For this reason, in the rest of the article we assume that { a j } j ∈ Z satisfies (3.1) without saying it explicitly.In order to prove Theorem 1.1, we need a Cotlar’s type inequality to control the operator T ∗ bysome one-sided Hardy-Littlewood maximal operators.For any M ∈ Z + , let T ∗ M f ( t ) = sup − M ≤ N For each q ∈ (1 , + ∞ ) , there exists a constant C depending on q, k v k l ∞ ( Z ) , α and ρ such that for every M ∈ Z + , T ∗ M f ( t ) ≤ C (cid:8) M − ( T α − M,M f )( t ) + M − q f ( t ) (cid:9) , −∞ < t < + ∞ , where M − q f ( t ) = sup ε> (cid:18) ε Z − ε | f ( t + s ) | q ds (cid:19) q . Proof. Since the operators T αN are given by convolutions, they are invariant under translations, andtherefore it is enough to prove the theorem for t = 0 . Observe that, for N = ( N , N ) ,T αN f ( t ) = T αN ,M f ( t ) − T αN +1 ,M f ( t ) , with − M ≤ N < N ≤ M. Then, it suffices to estimate (cid:12)(cid:12) T αm,M f (0) (cid:12)(cid:12) for | m | ≤ M with constantsindependent of m and M. Let us split f as f ( t ) = f ( t ) χ ( − a m +1 , ( t ) + f ( t ) χ ( −∞ , − a m +1 ] ( t ) + f ( t ) χ (0 , + ∞ ) =: f ( t ) + f ( t ) + f ( t ) , for −∞ < t < + ∞ . First, notice that T αm,M f (0) = 0 . Then, we have (cid:12)(cid:12) T αm,M f (0) (cid:12)(cid:12) ≤ (cid:12)(cid:12) T αm,M f (0) (cid:12)(cid:12) + (cid:12)(cid:12) T αm,M f (0) (cid:12)(cid:12) =: I + II. IFFERENTIAL TRANSFORMS FOR FRACTIONAL POISSON TYPE OPERATOR SEQUENCE 11 For I , by the mean value theorem, we have I = (cid:12)(cid:12) T αm,M f (0) (cid:12)(cid:12) = C α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z + ∞ M X j = m v j a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α f ( − s ) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C α k v k l ∞ ( Z ) M X j = m a αj +1 e − a j +1 / (4 s ) + a αj e − a j / (4 s ) s α | f ( − s ) | ds ≤ C α,v Z + ∞ M X j = m a j +1 + 1 a j ! | f ( − s ) | ds ≤ C α,v ( ρ + 1) Z + ∞ M X j = m a j +1 | f ( − s ) | ds (since ρ ≤ a j +1 a j ≤ ρ ) ≤ C α,v,ρ a m +1 Z + ∞ M X j = m a m +1 a j | f ( − s ) | ds ≤ C α,v,ρ a m +1 Z + ∞ (cid:16) ρ + M X j = m ρ j − m ) (cid:17) | f ( − s ) | ds ≤ C α,v,ρ a m +1 Z + ∞ ( ρ − ρ α (cid:16) ρ + + ∞ X j =0 ρ j (cid:17) | f ( − s ) | ds ≤ C α,v,ρ a m +1 Z − a m +1 | f ( s ) | ds ≤ C α,ρ,v M − q f (0) . For part II , II = (cid:12)(cid:12) T αm,M f (0) (cid:12)(cid:12) = 1 a m Z − a m (cid:12)(cid:12) T αm,M f (0) (cid:12)(cid:12) du ≤ a m Z − a m (cid:12)(cid:12) T α − M,M f ( u ) (cid:12)(cid:12) du + 1 a m Z − a m (cid:12)(cid:12) T α − M,M f ( u ) (cid:12)(cid:12) du + 1 a m Z − a m (cid:12)(cid:12) T αm,M f ( u ) − T αm,M f (0) (cid:12)(cid:12) du + 1 a m Z − a m (cid:12)(cid:12) T α − M,m − f ( u ) (cid:12)(cid:12) du =: A + A + A + A . (If m = − M , we understand that A = 0.) It is clear that A ≤ M − ( T α − M,M f )(0) . For A , by the uniform boundedness of T αN in Theorem 2.6, we get A ≤ a m Z − a m (cid:12)(cid:12) T α − M,M f ( u ) (cid:12)(cid:12) q du ! /q ≤ C (cid:18) a m Z R | f ( u ) | q du (cid:19) /q = C a m Z − a m | f ( u ) | q du ! /q ≤ C M − q f (0) . For the third term A , with − a m ≤ u ≤ 0, by the mean value theorem and Theorem 2.4, we have (cid:12)(cid:12) T αm,M f ( u ) − T αm,M f (0) (cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z u −∞ K αm,M ( u − s ) f ( s ) ds − Z −∞ K αm,M ( − s ) f ( s ) ds (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z u −∞ (cid:12)(cid:12) K αm,M ( u − s ) − K αm,M ( − s ) (cid:12)(cid:12) | f ( s ) | ds + (cid:12)(cid:12)(cid:12)(cid:12)Z u K αm,M ( − s ) f ( s ) ds (cid:12)(cid:12)(cid:12)(cid:12) = Z − a m +1 −∞ (cid:12)(cid:12) K αm,M ( u − s ) − K αm,M ( − s ) (cid:12)(cid:12) | f ( s ) | ds = + ∞ X j = m +1 Z − a j − a j +1 (cid:12)(cid:12) K αm,M ( u − s ) − K αm,M ( − s ) (cid:12)(cid:12) | f ( s ) | ds = + ∞ X j = m +1 Z − a j − a j +1 (cid:12)(cid:12)(cid:12) ∂ t K αm,M ( t ) (cid:12)(cid:12) t = ξ j (cid:12)(cid:12)(cid:12) | u | | f ( s ) | ds ( a j − a m ≤ ξ j ≤ a j +1 ) ≤ C + ∞ X j = m +1 Z − a j − a j +1 | u || ξ j | | f ( s ) | ds ≤ C + ∞ X j = m +1 a m (cid:0) a j − a m (cid:1) Z − a j +1 | f ( s ) | ds ≤ C + ∞ X j = m +1 a m a j · ρ ( ρ − a j +1 Z − a j +1 | f ( s ) | ds ≤ C + ∞ X j = m +1 ρ j − m ) M − f (0) ≤ C M − q f (0) . Then, A = 1 a m Z − a m (cid:12)(cid:12) T αm,M f ( u ) − T αm,M f (0) (cid:12)(cid:12) du ≤ C M − q f (0) . For the latest one, A , we have A = 1 a m Z − a m (cid:12)(cid:12) T α − M,m − f ( u ) (cid:12)(cid:12) du ≤ a m Z − a m Z − a m +1 −∞ (cid:12)(cid:12) K α − M,m − ( u − s ) f ( s ) (cid:12)(cid:12) dsdu. Then, we consider the inner integral appeared in the above inequalities first. Since − a m ≤ u ≤ −∞ < s ≤ − a m +1 and the sequence { a j } j ∈ Z is ρ -lacunary sequence, we have | u − s | ∼ | s | . From thisand by the mean value theorem, we get Z − a m +1 −∞ (cid:12)(cid:12) K α − M,m − ( u − s ) f ( s ) (cid:12)(cid:12) ds = + ∞ X k = m +1 Z − a k − a k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) m − X j = − M v j a αj +1 e − a j +1 / (4( u − s )) − a αj e − a j / (4( u − s )) ( u − s ) α f ( s ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ds ≤ + ∞ X k = m +1 Z − a k − a k +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) m − X j = − M v j ( a j +1 − a j ) ξ α − j e − ξ j / (4( u − s )) ( u − s ) α f ( s ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ds ( a j ≤ ξ j ≤ a j +1 ) ≤ C k v k l ∞ ( Z ) + ∞ X k = m +1 Z − a k − a k +1 m − X j = − M (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ρ α ( ρ − a αj e − a j / (4 s ) s α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | f ( s ) | ds ≤ C ρ,v,α + ∞ X k = m +1 a k Z − a k − a k +1 m − X j = − M a αj a αk | f ( s ) | ds IFFERENTIAL TRANSFORMS FOR FRACTIONAL POISSON TYPE OPERATOR SEQUENCE 13 ≤ C ρ,v,α + ∞ X k = m +1 a k +1 Z − a k − a k +1 m − X j = − M ρ − α ( k − j ) | f ( s ) | ds ≤ C ρ,v,α + ∞ X k = m +1 ρ − α ( k − m +1) a k +1 Z − a k − a k +1 | f ( s ) | ds ≤ C ρ,v,α + ∞ X k = m +1 ρ α ( k − m +1) a k +1 Z − a k +1 | f ( s ) | ds ≤ C ρ,v,α + ∞ X k = m +1 ρ α ( k − m +1) M − f (0) ≤ C ρ,v,α M − q f (0) . Hence, A ≤ C M − q f (0) . Combining the estimates above for A , A , A and A , we get II ≤ M − ( T α − M,M f )(0) + C M − q f (0) . And, then we have (cid:12)(cid:12) T αm,M f (0) (cid:12)(cid:12) ≤ C (cid:0) M − ( T α − M,M f )(0) + M − q f (0) (cid:1) . As the constants C appeared above all depend on k v k l ∞ ( Z ) , ρ and α , not on m, M , we complete theproof. (cid:3) Now we can start the proof of Theorem 1.1. Proof of Theorem 1.1. For each ω ∈ A − p , choose 1 < q < p < ∞ such that ω ∈ A − p/q . Then, it is wellknown that the maximal operators M − and M − q are bounded in L p ( R , ω ). On the other hand, byTheorem 2.6, the operators T αN are uniformly bounded in L p ( R , ω ) with ω ∈ A − p . Hence k T ∗ M f k L p ( ω ) ≤ C (cid:16)(cid:13)(cid:13) M − ( T α − M,M f ) (cid:13)(cid:13) L p ( ω ) + (cid:13)(cid:13) M − q f (cid:13)(cid:13) L p ( ω ) (cid:17) ≤ C (cid:16)(cid:13)(cid:13) T α − M,M f (cid:13)(cid:13) L p ( ω ) + k f k L p ( ω ) (cid:17) ≤ C k f k L p ( ω ) . Note that the constants C appeared above do not depend on M . Consequently, letting M increaseto infinity, we get the proof of the L p boundedness of T ∗ . This completes the proof of part ( a ) of thetheorem.In order to prove ( b ), we consider the ℓ ∞ ( Z )-valued operator T f ( t ) = { T αN f ( t ) } N ∈ Z . Since kT f ( t ) k ℓ ∞ ( Z )) = T ∗ f ( t ), by using ( a ) we know that the operator T is bounded from L p ( R , ω )into L pℓ ∞ ( Z ) ( R , ω ), for every 1 < p < ∞ and ω ∈ A − p . The kernel of the operator T is given by K α ( t ) = { K αN ( t ) } N ∈ Z . By Theorem 2.4 and the vector valued version of Theorem 2.1 in [1], weget that the operator T is bounded from L ( R , ω ) into weak- L ℓ ∞ ( Z ) ( R , ω ) for ω ∈ A − . Hence, as kT f ( t ) k ℓ ∞ ( Z ) = T ∗ f ( t ), we get the proof of ( b ).For ( c ) , we shall prove that if f ∈ L ∞ ( R ) and there exists t ∈ R such that T ∗ f ( t ) < ∞ , then T ∗ f ( t ) < ∞ for a.e. t ∈ R . Given t = t . Set f = f χ ( t − | t − t | , t +4 | t − t | ) and f = f − f . Note that T ∗ is L p -bounded for any 1 < p < ∞ . Then T ∗ f ( t ) < ∞ , because f ∈ L p ( R ), for any 1 < p < ∞ . On the other hand, as the kernel K N is supported in R + , we have (cid:12)(cid:12)(cid:12) T αN f ( t ) − T αN f ( t ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z t −∞ K αN ( t − s ) f ( s ) ds − Z t −∞ K αN ( t − s ) f ( s ) ds (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z t − | t − t |−∞ ( K αN ( t − s ) − K αN ( t − s )) f ( s ) ds (cid:12)(cid:12)(cid:12) ≤ Z t − | t − t |−∞ | ∂ s K αN ( ξ ( s )) | | t − t | | f ( s ) | ds ( t − s ≤ ξ ( s ) ≤ t − s ) ≤ C Z t − | t − t |−∞ | t − t | ( t − s ) | f ( s ) | ds ≤ C k f k L ∞ ( R ) < ∞ . Hence k T αN f ( t ) − T αN f ( t ) k l ∞ ( Z ) ≤ C k f k L ∞ ( R ) and therefore T ∗ f ( t ) = k T αN f ( t ) k l ∞ ( Z ) ≤ C < ∞ . For the L ∞ − BM O boundedness, we will prove itlater.( d ) Let t be one point in R such that T ∗ f ( t ) < ∞ . Set I = [ t − | t − t | , t + 4 | t − t | ] with t = t . And we decompose f to be f = ( f − f I ) χ I + ( f − f I ) χ I c + f I =: f + f + f . Note that T ∗ is L p -bounded for any 1 < p < ∞ . Then T ∗ f ( t ) < ∞ , because f ∈ L p ( R ), for any1 < p < ∞ . And T αN f = 0, since P αa j f = f for any j ∈ Z . On the other hand, as the kernel K N issupported in R + , we have (cid:12)(cid:12)(cid:12) T αN f ( t ) − T αN f ( t ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z t −∞ K αN ( t − s ) f ( s ) ds − Z t −∞ K αN ( t − s ) f ( s ) ds (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z t − | t − t |−∞ ( K αN ( t − s ) − K αN ( t − s )) f ( s ) ds (cid:12)(cid:12)(cid:12) ≤ Z t − | t − t |−∞ | ∂ s K αN ( ξ ( s )) | | t − t | | f ( s ) | ds ( t − s ≤ ξ ( s ) ≤ t − s ) ≤ C Z t − | t − t |−∞ | t − t | ( t − s ) | f ( s ) | ds ≤ C + ∞ X k =2 | t − t | Z t − k | t − t | t − k +1 | t − t | | f ( s ) − f I || t − s | ds ≤ C + ∞ X k =2 | t − t | (2 k +1 | t − t | ) Z t − k | t − t | t − k +1 | t − t | | f ( s ) − f I | ds ≤ C + ∞ X k =2 | t − t | (2 k +1 | t − t | ) Z t +2 k +1 | t − t | t − k +1 | t − t | | f ( s ) − f I | ds = C + ∞ X k =2 − ( k +1) k +1 | t − t | Z I k +1 | f ( s ) − f I | ds ≤ C + ∞ X k =2 − ( k +1) k +1 | t − t | Z I k +1 (cid:12)(cid:12) f ( s ) − f I k +1 (cid:12)(cid:12) + k X l =2 (cid:12)(cid:12) f I l +1 − f I l (cid:12)(cid:12)! ds IFFERENTIAL TRANSFORMS FOR FRACTIONAL POISSON TYPE OPERATOR SEQUENCE 15 ≤ C + ∞ X k =2 − ( k +1) k +1 | t − t | Z I k +1 (cid:16)(cid:12)(cid:12) f ( s ) − f I k +1 (cid:12)(cid:12) + 2 k k f k BMO ( R ) (cid:17) ds ≤ C + ∞ X k =2 − ( k +1) (1 + 2 k ) k f k BMO ( R ) ≤ C k f k BMO ( R ) , where I k +1 = [ t − k +1 | t − t | , t + 2 k +1 | t − t | ] for any k ∈ N . Hence k T αN f ( t ) − T αN f ( t ) k l ∞ ( Z ) ≤ C k f k BMO ( R ) and therefore T ∗ f ( t ) = k T αN f ( t ) k l ∞ ( Z ) ≤ C < ∞ . Now, we shall prove the estimate (1.4) for functions such that T ∗ f ( t ) < ∞ a.e. For any h > t such that T ∗ f ( t ) < ∞ , consider the integral I = ( t , t + h ) and f I = 1 h Z I f ( t ) dt. Wehave T ∗ f I ( t ) = 0 . Let f ( t ) = f ( t ) + f ( t ) + f I , where f ( t ) = ( f ( t ) − f I ) χ ( t − h,t +4 h ) ( t ) and f ( t ) = ( f ( t ) − f I ) χ ( −∞ ,t − h ) ( t ) + ( f ( t ) − f I ) χ ( t +4 h, + ∞ ) ( t ). Then,1 h Z t + ht | T ∗ f ( t ) − ( T ∗ f ) I | dt = 1 h Z t + ht (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) h Z t + ht ( T ∗ f ( t ) − T ∗ f ( s )) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt ≤ h Z t + ht Z t + ht | T ∗ f ( t ) − T ∗ f ( s ) | dsdt = 1 h Z t + ht Z t + ht (cid:12)(cid:12)(cid:12) k T αN f ( t ) k l ∞ ( Z ) − k T αN f ( s ) k l ∞ ( Z ) (cid:12)(cid:12)(cid:12) dsdt ≤ h Z t + ht Z t + ht k T αN f ( t ) − T αN f ( s ) k l ∞ ( Z ) dsdt ≤ h Z t + ht Z t + ht k T αN f ( t ) − T αN f ( s ) k l ∞ ( Z ) dsdt + 1 h Z t + ht Z t + ht k T αN f ( t ) − T αN f ( s ) k l ∞ ( Z ) dsdt =: A + B. The H¨older inequality and L -boundedness of T ∗ imply that A ≤ h Z t + ht k T αN f ( t ) k l ∞ ( Z ) dt + 1 h Z t + ht k T αN f ( s ) k l ∞ ( Z ) ds ≤ h Z t + ht k T αN f ( t ) k l ∞ ( Z ) dt ! / + h Z t + ht k T αN f ( s ) k l ∞ ( Z ) ds ! / ≤ C h / k f k L ( R ) ≤ C k f k BMO ( R ) . For B , since t ≤ t, s ≤ t + h and the support of f is ( −∞ , t − h ) S ( t + 4 h, + ∞ ), we have (cid:12)(cid:12)(cid:12) T αN f ( t ) − T αN f ( s ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z t −∞ K αN ( t − u ) f ( u ) du − Z t −∞ K αN ( s − u ) f ( u ) du (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z t − h −∞ ( K αN ( t − u ) − K αN ( s − u )) f ( u ) du (cid:12)(cid:12)(cid:12) ≤ Z t − h −∞ | ∂ u K αN ( ξ ( u )) | | t − s | | f ( u ) | du ( t − u ≤ ξ ( u ) ≤ t − u ) ≤ C Z t − h −∞ | t − s | ( t − u ) | f ( u ) | du ≤ C + ∞ X k =2 Z t − k ht − k +1 h h | f ( u ) − f I || t − u | du ≤ C + ∞ X k =2 h (2 k +1 h ) Z t − k ht − k +1 h | f ( u ) − f I | du ≤ C + ∞ X k =2 h (2 k +1 h ) Z t +2 k +1 ht − k +1 h | f ( u ) − f I | du = C + ∞ X k =2 − ( k +1) k +1 h Z I k +1 | f ( u ) − f I | du ≤ C + ∞ X k =2 − ( k +1) k +1 h Z I k +1 (cid:12)(cid:12) f ( u ) − f I k +1 (cid:12)(cid:12) + k X l =2 (cid:12)(cid:12) f I l +1 − f I l (cid:12)(cid:12)! du ≤ C + ∞ X k =2 − ( k +1) k +1 h Z I k +1 (cid:16)(cid:12)(cid:12) f ( u ) − f I k +1 (cid:12)(cid:12) + 2 k k f k BMO ( R ) (cid:17) du ≤ C + ∞ X k =2 − ( k +1) (1 + 2 k ) k f k BMO ( R ) ≤ C k f k BMO ( R ) , where I k +1 denotes the interval [ t − k +1 h, t + 2 k +1 h ]. Hence, we have B ≤ C k f k BMO ( R ) . Then bythe arbitrary of t and h > 0, we proved k T ∗ f k BMO ( R ) ≤ C k f k BMO ( R ) . For the second part of ( c ), we can deduce it from the BM O -boundedness of T ∗ and the inclusion of L ∞ ( R ) ⊂ BM O ( R ) . This completes the proof of Theorem 1.1. (cid:3) Now we shall prove Theorem 1.2. Proof of Theorem 1.2. First, we shall see that if ϕ is a test function, then T αN ϕ ( t ) converges for all t ∈ R . In order to prove this, it is enough to see that for any ( L, M ) with 0 < L < M , the series A = M X j = L v j ( P αa j +1 ϕ ( t ) − P αa j ϕ ( t )) and B = − L X j = − M v j ( P αa j +1 ϕ ( t ) − P αa j ϕ ( t ))converge to zero, when L, M → + ∞ . By the mean value theorem, following the arguments in theproof of Theorem 3.2, we have | A | ≤ C α k v k l ∞ ( Z ) Z ∞ M X j = L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ξ α − j e − ξ j / (4 s ) ( a j +1 − a j ) s α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | ϕ ( t − s ) | ds, ( ∃ a j ≤ ξ j ≤ a j +1 ) ≤ C α,v Z + ∞ ρ α ( ρ − M X j = L a αj e − a j / (4 s ) s α | ϕ ( t − s ) | ds, (since ρ ≤ a j +1 a j ≤ ρ ) ≤ C α,v,ρ Z + ∞ M X j = L Ca j | ϕ ( t − s ) | ds ≤ C α,v,ρ a L M X j = L a L a j Z + ∞ | ϕ ( t − s ) | ds IFFERENTIAL TRANSFORMS FOR FRACTIONAL POISSON TYPE OPERATOR SEQUENCE 17 ≤ C α,v,ρ ρ ρ − k ϕ k L ( R ) a L −→ , as L, M → + ∞ . On the other hand, as the integral of the kernels are zero, we can write B = C α Z + ∞ − L X j = − M v j a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α ( ϕ ( t − s ) − ϕ ( t )) ds = C α (cid:26)Z + Z ∞ (cid:27) − L X j = − M v j a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α ( ϕ ( t − s ) − ϕ ( t )) ds =: B + B . Proceeding as in the case A , and by using the fact that ϕ is a test function, we have | B | = C α (cid:12)(cid:12)(cid:12) Z − L X j = − M v j a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α ( ϕ ( t − s ) − ϕ ( t )) ds (cid:12)(cid:12)(cid:12) ≤ C α k ϕ ′ k L ∞ ( R ) Z − L X j = − M v j a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α ds ≤ C α,ϕ k v k l ∞ ( Z ) Z ρ α ( ρ − − L X j = − M a αj e − a j / (4 s ) s α ds ≤ C α,ϕ,v,ρ a α − L − L X j = − M a αj a α − L Z s α ds ≤ C α,ϕ,v,ρ ρ α ρ α − a α − L −→ , as L, M → + ∞ . On the other hand, | B | ≤ C α,ρ k v k l ∞ ( Z ) k ϕ k L ∞ ( R ) Z ∞ − L X j = − M a αj s α ds ≤ C α,v,ϕ,ρ − L X j = − M a αj Z ∞ s α ds ≤ C α,v,ϕ,ρ a α − L − L X j = − M a αj a α − L ≤ C α,ϕ,v,ρ ρ α ρ α − a α − L −→ , as L, M → + ∞ . As the set of test functions is dense in L p ( R ), by Theorem 1.1 we get the a.e. convergence forany function in L p ( R ). Analogously, since L p ( R ) ∩ L p ( R , ω ) is dense in L p ( R , ω ), we get the a.e. convergence for functions in L p ( R , ω ) with 1 ≤ p < ∞ . By using the dominated convergence theorem,we can prove the convergence in L p ( R , ω )-norm for 1 < p < ∞ , and also in measure. (cid:3) Proofs of Theorems 1.3 and 1.4 In this section, we will give the proof of Theorems 1.3 and 1.4. Proof of Theorem 1.3. Let f be the function defined by f ( s ) = X k ∈ Z ( − k χ ( − a k +1 , − a k ] ( s ) , where a > f ( a j s ) = ( − j f ( s ) . Let a j = a j . Then P αa j f ( t ) = 14 α Γ( α ) Z + ∞ a αj e − a j / (4 s ) s α f ( t − s ) ds = 14 α Γ( α ) Z + ∞ e − / (4 u ) u α f ( t − a j u ) duu . So P αa j f (0) = 14 α Γ( α ) Z + ∞ e − / (4 u ) u α f ( − a j u ) duu = ( − j α Γ( α ) Z + ∞ e − / (4 u ) u α f ( − u ) duu . We observe that Z + ∞ e − / (4 u ) u α (cid:12)(cid:12) f ( − u ) (cid:12)(cid:12) duu ≤ Z + ∞ e − / (4 u ) u α duu = 4 α Γ( α ) < ∞ . Hence lim R → + ∞ Z + ∞ R e − / (4 u ) u α f ( − u ) duu = 0 and lim ε → + Z ε e − / (4 u ) u α f ( − u ) duu = 0 . On the other hand, lim a → + ∞ Z a e − / (4 u ) u α f ( − u ) duu = lim a → + ∞ Z a e − / (4 u ) u α duu = C > . Hence we canchoose a > Z a e − / (4 u ) u α f ( − u ) duu = Z a e − / (4 u ) u α duu > (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z /a e − / (4 u ) u α duu (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z + ∞ a e − / (4 u ) u α duu (cid:12)(cid:12)(cid:12)(cid:12) > (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z /a e − / (4 u ) u α f ( − u ) duu (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z + ∞ a e − / (4 u ) u α f ( − u ) duu (cid:12)(cid:12)(cid:12)(cid:12) . In other words, with the a > C > Z + ∞ e − / (4 u ) u α f ( − u ) duu = C . Hence (cid:12)(cid:12)(cid:12) P αa j f (0) − P αa j +1 f (0) (cid:12)(cid:12)(cid:12) = 2 C α Γ( α ) > . Therefore we have X j ∈ Z (cid:12)(cid:12)(cid:12) P αa j +1 f (0) − P αa j f (0) (cid:12)(cid:12)(cid:12) = ∞ . By using (4.1) and changing variable we get P αa j f ( t ) = 14 α Γ( α ) Z + ∞ e − / (4 u ) u α f ( t − a j u ) duu = ( − j α Γ( α ) Z + ∞ e − / (4 u ) u α f (cid:18) ta j − u (cid:19) duu . Then P αa j +1 f ( t ) − P αa j f ( t )= ( − j +1 α Γ( α ) n Z + ∞ e − / (4 u ) u α f (cid:18) ta j +1) − u (cid:19) duu + Z + ∞ e − / (4 u ) u α f (cid:18) ta j − u (cid:19) duu o . (4.3)By the dominated convergence theorem, we know thatlim h → Z + ∞ e − / (4 u ) u α f ( h − u ) duu = Z + ∞ e − / (4 u ) u α f ( − u ) duu = C > , where C is the constant appeared in (4.2). So, there exists 0 < η < , such that, for | h | < η , Z + ∞ e − / (4 u ) u α f ( h − u ) duu ≥ Z + ∞ e − / (4 u ) u α f ( − u ) duu = C . Then, for each t ∈ R , we can choose j ∈ Z such that | t | a j < η (there are infinite j satisfying thiscondition), and we have Z + ∞ e − / (4 u ) u α f (cid:18) ta j +1) − u (cid:19) duu + Z + ∞ e − / (4 u ) u α f (cid:18) ta j − u (cid:19) duu ≥ C > . IFFERENTIAL TRANSFORMS FOR FRACTIONAL POISSON TYPE OPERATOR SEQUENCE 19 Choosing v j = ( − j +1 , j ∈ Z , by (4.3) we have, for any t ∈ R ,T ∗ f ( t ) ≥ X | taj | <η ( − j +1 (cid:0) P αa j +1 f ( t ) − P αa j f ( t ) (cid:1) = 14 α Γ( α ) X | taj | <η (cid:18)Z + ∞ e − / (4 u ) u α f (cid:18) ta j +1) − u (cid:19) duu + Z + ∞ e − / (4 u ) u α f (cid:18) ta j − u (cid:19) duu (cid:19) = ∞ . We complete the proof of Theorem 1.3. (cid:3) Also, we will give the proof of Theorem 1.4 which gives a local growth characterization of theoperator T ∗ with f ∈ L ∞ ( R n ). Proof of Theorem 1.4. First, we prove the theorem in the case 1 < p < ∞ . Since 2 r < , we knowthat B \ B r = ∅ . Let f ( t ) = f ( t ) + f ( t ), where f ( t ) = f ( t ) χ B r ( t ) and f ( t ) = f ( t ) χ B \ B r ( t ). Then | T ∗ f ( t ) | ≤ | T ∗ f ( t ) | + | T ∗ f ( t ) | . By Theorem 1.1,1 | B r | Z B r | T ∗ f ( t ) | dt ≤ (cid:18) | B r | Z B r | T ∗ f ( t ) | dt (cid:19) / ≤ C (cid:18) | B r | Z R | f ( t ) | dt (cid:19) / ≤ C k f k L ∞ ( R ) . We also know that, for any j ∈ Z , (4.4) Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ds ≤ Z ∞ a αj +1 e − a j +1 / (4 s ) + a αj e − a j / (4 s ) s α ds = 2 · α Γ( α ) . Then, by H¨older’s inequality, (4.4) and Fubini’s Theorem, for 1 < p < ∞ and any N = ( N , N ), wehave (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X j = N v j (cid:16) P αa j +1 f ( t ) − P αa j f ( t ) (cid:17)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C N X j = N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) v j Z ∞ a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s )) s α f ( t − s ) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C k v k l p ( Z ) N X j = N Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | f ( t − s ) | ds ! p ′ /p ′ ≤ C k v k l p ( Z ) (cid:16) N X j = N n Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | f ( t − s ) | p ′ ds o × n Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ds o p ′ /p (cid:17) /p ′ ≤ C k v k l p ( Z ) N X j = N Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | f ( t − s ) | p ′ ds /p ′ ≤ C k v k l p ( Z ) Z ∞ ∞ X j = −∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a αj +1 e − a j +1 / (4 s ) − a αj e − a j / (4 s ) s α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | f ( t − s ) | p ′ ds /p ′ ≤ C k v k l p ( Z ) (cid:18)Z ∞ | s | | f ( t − s ) | p ′ ds (cid:19) /p ′ ≤ C k v k l p ( Z ) Z R | t − s | | f ( s ) | p ′ ds. For s ∈ B \ B r and t ∈ B r , we have r ≤ | t − s | ≤ 2. Then, we get1 | B r | Z B r | T ∗ f ( t ) | dt ≤ C | B r | Z B r (cid:18)Z R | t − s | | f ( s ) | p ′ ds (cid:19) /p ′ dt ≤ C k f k L ∞ ( R ) | B r | Z B r Z r ≤| t − s |≤ | t − s | ds ! /p ′ dt ∼ (cid:16) log 2 r (cid:17) /p ′ k f k L ∞ ( R ) . Hence, 1 | B r | Z B r | T ∗ f ( t ) | dt ≤ C (cid:18) (cid:16) log 2 r (cid:17) /p ′ (cid:19) k f k L ∞ ( R ) ≤ C (cid:16) log 2 r (cid:17) /p ′ k f k L ∞ ( R ) . For the case p = 1 and p = ∞ , the proof is similar and easier. Then we get the proof of ( a ) . For ( b ) , when 1 < p < ∞ , for any 0 < ε < p − 1, let f ( t ) = X k = −∞ ( − k χ ( − a k , − a k − ] ( t ) and a j = a j , with a > f is contained in [ − , , and { a j } j ∈ Z is a ρ -lacunarysequence with ρ = a > . We observe that (cid:12)(cid:12)(cid:12)(cid:12)Z + ∞ e − / (4 u ) u α f ( − u ) duu (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z + ∞ e − / (4 u ) u α duu = 4 α Γ( α ) < ∞ . Hence lim R → + ∞ Z + ∞ R e − / (4 u ) u α f ( − u ) duu = 0 and lim ε → Z ε e − / (4 u ) u α f ( − u ) duu = 0 . Also there exists a constant C > a → + ∞ Z a − e − / (4 u ) u α f ( − u ) duu = lim a → + ∞ Z a − e − / (4 u ) u α duu = C. So we can choose a > Z a − e − / (4 u ) u α f ( − u ) duu = Z a − e − / (4 u ) u α duu ≥ Z /a e − / (4 u ) u α duu + Z + ∞ a − e − / (4 u ) u α duu ! > (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z /a e − / (4 u ) u α f ( − u ) duu (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z + ∞ a − e − / (4 u ) u α f ( − u ) duu (cid:12)(cid:12)(cid:12)(cid:12)! . Therefore, there exists a constant C > Z + ∞ e − / (4 u ) u α f ( − u ) duu = C > < Z /a e − / (4 u ) u α duu + Z + ∞ a − e − / (4 u ) u α duu ≤ C . On the other hand, by the dominated convergence theorem, we havelim h → Z + ∞ e − / (4 u ) u α f ( h − u ) duu = Z + ∞ e − / (4 u ) u α f ( − u ) duu = C > , IFFERENTIAL TRANSFORMS FOR FRACTIONAL POISSON TYPE OPERATOR SEQUENCE 21 where C is the constant appeared in (4.5). So, there exists 0 < η < , such that, for | h | < η , (4.7) Z + ∞ e − / (4 u ) u α f ( h − u ) duu ≥ Z + ∞ e − / (4 u ) u α f ( − u ) duu = C . It can be checked that f ( a j t ) = ( − j f ( t ) + ( − j − j X k =1 ( − k χ ( − a k , − a k − ] ( t )when j ≤ . We will always assume j ≤ P αa j f ( t ) = 14 α Γ( α ) Z + ∞ e − / (4 u ) u α f ( t − a j u ) duu = ( − j α Γ( α ) Z + ∞ e − / (4 u ) u α ( f (cid:18) ta j − u (cid:19) + − j X k =1 ( − k χ ( − a k , − a k − ] (cid:18) ta j − u (cid:19)) duu . Then P αa j +1 f ( t ) − P αa j f ( t )= ( − j +1 α Γ( α ) n Z + ∞ e − / (4 u ) u α f (cid:18) ta j +1) − u (cid:19) duu + Z + ∞ e − / (4 u ) u α f (cid:18) ta j − u (cid:19) duu (4.8) + Z + ∞ e − / (4 u ) u α − j − X k =1 ( − k χ ( − a k , − a k − ] (cid:18) ta j +2 − u (cid:19) duu + Z + ∞ e − / (4 u ) u α − j X k =1 ( − k χ ( − a k , − a k − ] (cid:18) ta j − u (cid:19) duu o . For given η as above, let 2 r < r < η and r ∼ a J η for a certain negative integer J .If J ≤ j ≤ 0, we have ra j < η . And, for any − r ≤ t ≤ r we have − · χ [ a − , + ∞ ) ( u ) ≤ − j − X k =1 ( − k χ ( − a k , − a k − ] (cid:18) ta j +2 − u (cid:19) ≤ χ [ a − , + ∞ ) ( u )and − · χ [ a − , + ∞ ) ( u ) ≤ − j X k =1 ( − k χ ( − a k , − a k − ] (cid:18) ta j − u (cid:19) ≤ χ [ a − , + ∞ ) ( u ) . Hence, for the third and fourth integrals in (4.8), by (4.6) we have Z + ∞ e − / (4 u ) u α − j − X k =1 ( − k χ ( − a k , − a k − ] (cid:18) ta j +2 − u (cid:19) duu + Z + ∞ e − / (4 u ) u α − j X k =1 ( − k χ ( − a k , − a k − ] (cid:18) ta j − u (cid:19) duu (4.9) ≥ ( − Z + ∞ a − e − / (4 u ) u α duu ≥ − C . So, for any t ∈ [ − r, r ] and J ≤ j ≤ 0, combining (4.8), (4.7) and (4.9), we have (cid:12)(cid:12)(cid:12) P αa j +1 f ( t ) − P αa j f ( t ) (cid:12)(cid:12)(cid:12) ≥ C α · (cid:18) C − C (cid:19) = C · C > . We choose the sequence { v j } j ∈ Z ∈ ℓ p ( Z ) given by v j = ( − j +1 ( − j ) − p − ε , then for N = ( J , , wehave12 r Z [ − r,r ] | T ∗ f ( t ) | dt ≥ r Z [ − r,r ] | T αN f ( t ) | dt ≥ α Γ( α ) 12 r Z [ − r,r ] 0 X j = J (cid:16) C · C · ( − j ) − p − ε (cid:17) dt ≥ C p,ε,α · C · ( − J ) p − ε ) ′ ∼ (cid:18) log 2 r (cid:19) p − ε ) ′ . For ( c ), let v j = ( − j +1 , a j = a j with a > < η < b ). Considerthe same function f as in ( b ) . Then, k v k l ∞ ( Z ) = 1 and k f k L ∞ ( R ) = 1 . By the same argument as in ( b ),with N = ( J , 0) and 0 < α < 1, we have12 r Z [ − r,r ] | T ∗ f ( t ) | dt ≥ r Z [ − r,r ] | T αN f ( t ) | dt ≥ α Γ( α ) 12 r Z [ − r,r ] 0 X j = J C dt ≥ C α Γ( α ) · ( − J ) ∼ log 2 r . (cid:3) Acknowledgments. Zhang Chao is grateful to the Department of Mathematics at UniversidadAut´onoma de Madrid for its hospitality during the period of this research. References [1] H. Aimar, L. Forzani and F.J. Mart´ın-Reyes, On weighted inequalities for singular integrals. Proc. Amer. Math.Soc. 125 (1997), 2057–2064. 7, 9, 13[2] A.L. Bernardis, M. Lorente, F.J. Mart´ın-Reyes, M.T. Mart´ınez, A. de la Torre and J.L. Torrea, Differential trans-forms in weighted spaces, J. Fourier Anal. Appl. 12 (2006), 83-103. 2, 7, 9[3] A. Bernardis, F.J. Mart´ın-Reyes, P.R. Stinga and J.L. Torrea, Maximum principles, extension problem and inversionfor nonlocal one-sided equations. J. Differential Equations . 260 (2016) 6333-6362. 1, 4, 6[4] J.J. Betancor, R. Crescimbeni and J.L. Torrea, The ρ -variation of the heat semigroup in the Hermitian setting:behaviour in L ∞ . Proceedings of the Edinburgh Mathematical Society. 54 (2011), 569-585. 4[5] L. Caffarelli and L. Silvestre, An extension problem related to the fractional Laplacian, Comm. Partial DifferentialEquations. 32 (2007), 1245–1260. 1[6] J. Duoandikoetxea, Fourier analysis , Translated and revised from the 1995 Spanish original by David Cruz-Uribe.Graduate Studies in Mathematics, Volume 29, American Mathematical Society, Providence, RI, 2001. 3[7] R.L. Jones and J. Rosenblatt, Differential and ergodic transforms, Math. Ann. 323 (2002), 525-546. 2[8] T. Ma, J.L. Torrea and Q. Xu, Weighted variation inequalities for differential operators and singular integrals. J.Funct. Anal. 268 (2015), 376–416. 4[9] T. Mart´ınez, J.L. Torrea and Q. Xu, Vector-valued Littlewood–Paley–Stein theory for semigroups, Adv. Math. Adv. inMath. 62 (1986), 7-48. 9[11] E. Sawyer, Weighted inequalities for the one-sided Hardy-Littlewood maximal functions. Trans. Amer. Math. Soc. 297 (1986), 53–61. 2, 7[12] P. R. Stinga and J.L. Torrea, Extension problem and Harnack’s inequality for some fractional operators, Comm.Partial Differential Equations. 35 (2010), 2092–2122. 1[13] P.R. Stinga and J.L. Torrea, Regularity theory and extension problem for fractional nonlocal parabolic equationsand the master equation. SIAM J. Math. Anal. 49 (2017), 3893–3924. 1 School of Statistics and Mathematics, Zhejiang Gongshang University, Hangzhou 310018, P.R. China E-mail address : [email protected] School of Mathematics and Statistics, Wuhan University, Wuhan 430072, P.R. China E-mail address : [email protected] Departamento de Matem´aticas, Facultad de Ciencias, Universidad Aut´onoma de Madrid, 28049 Madrid,Spain E-mail address ::