R E S E A R C H
Open Access
On commutators of certain fractional type
integrals with Lipschitz functions
Wenting Hu
1, Yongming Wen
1and Huoxiong Wu
1**Correspondence: [email protected]
1School of Mathematical Sciences,
Xiamen University, Xiamen, China
Abstract
In this paper, we study the commutators generated by Lipschitz functions and fractional type integral operators with kernels of the form
Kα(x,y) =
κ
1(x–A1y)κ
2(x–A2y)· · ·κ
m(x–Amy),where 0≤
α
=α
1+· · ·+α
m<n, eachκ
isatisfies the (n–α
i)-order fractional sizecondition and a generalized fractional Hörmander condition,Aiis invertible, and
Ai–Ajis invertible fori=j, 1≤i,j≤m. We establish the corresponding sharp maximal
function estimates and obtain the weighted Coifman type inequalities, weighted Lp(wp)→Lq(wq) estimates, and the weighted endpoint estimates for such
commutators.
MSC: Primary 42B25; secondary 42B20
Keywords: Commutators; Lipschitz functions; Generalized fractional Hörmander condition; Fractional size condition; Weighted estimates; Coifman type inequality
1 Introduction and main results
Letn,m∈N, 0≤α<n. For any locally integrable bounded functionf, define
Tα,mf(x) :=
RnKα(x,y)f(y)dy, (1)
where
Kα(x,y) =k1(x–A1y)k2(x–A2y)· · ·km(x–Amy),
α=α1+· · ·+αm, and for each 1≤i≤m,kisatisfies (n–αi)-order fractional size condition,
Aiis a matrix such that
(H) Aiis invertible andAi–Ajis invertible fori=j,1≤i,j≤m.
Clearly,Tα,1 =Iα, the Riesz potential, for m= 1,A1 is then-order identity matrix, and k1(x–A1y) = 1/|x–y|α. For generalmand certaink
i,T0,mbehaves like a singular integral
operator andTα,mhas been studied in [1–10]. In particular, Riveros and Urciuolo [5,6,11]
considered eachkias a rough fractional kernel, and eachkisatisfies anLαi,γi-Hörmander
regular condition, or more generalki∈Hα,γi, that is, for allx∈Rnand|x|<R,
∞
j=1
2jRn–αKα(·–x) –Kα(·)χB(x,2j+1R)\B(x,2jR)γ
i,B(x,2mR)<∞.
They showed that these operators are bounded fromLpintoLq, for 1 <p≤q<∞, 1/q=
1/p–α/n. In [12], Ibañez-Firnkorn and Riveros analyzed operators of the form (1) with conditions of regularity more general than theLα,γ-Hörmander condition and a fractional size condition. Before giving the definitions of the fractional size conditionSn–αi,Ψiand the
generalized fractional Hörmander conditionHn–αi,Ψi,k, we first recall the definitions and
properties for Young function.
A functionΨ : [0,∞)→[0,∞) is said to be a Young function ifΨ is continuous, con-vex, nondecreasing and satisfiesΨ(0) = 0 andlimt→∞Ψ(t) =∞. Forf∈L1loc(Rn) and each
Young functionΨ, we can induce an average of the Luxemburg norm of a functionf in the ballBdefined by
fΨ,B:=inf
λ> 0 : 1
|B|
B
Ψ
|f (x)|
λ dx≤1
,
and a fractional maximal operatorMα,Ψ (0≤α<n) defined by
Mα,Ψf(x) :=sup
Bx|B|
α/nf
Ψ,B,
and we denoteM0,Ψ byMΨ, the Orlicz maximal operator. In particular, forΨ(t) =t,fΨ,B:=|B|–1
B|f(x)|dxandMα,Ψ =Mα, the fractional max-imal operator; for Ψ(t) =tr with 1 <r<∞, f
Ψ,B=fr,B:= (|B|–1
B|f(x)|rdx)1/r and
Mα,Ψ =Mα,r, andM0,rf :=supBxfr,B:=M(fr)1/r.
Next, we recall the definitions of the fractional size condition and the generalized frac-tional Hörmander condition. Normally, we use|x| ∼sto represents<|x| ≤2s. For the Young functionΨ, we write
fΨ,|x|∼s=fχ|x|∼sΨ,B(0,2s).
For 0≤α<n, the functionKαis said to satisfy the fractional size condition if there exists a constantC> 0 such that
KαΨ,|x|∼s≤Csα–n.
And we denoteKα∈Sα,Ψ in this case. WhenΨ(t) =t, we writeSα,Ψ =Sα. Observe that if
Kα∈Sα, then there exists a constantC> 0 such that
|x|∼s
Kα(x)dx≤Csα.
We say that the functionKαsatisfies theLα,Ψ,k-Hörmander condition denoted byKα∈
Hα,Ψ,kif there exist constantscΨ > 1 andCΨ> 0 such that, for allxandR>cΨ|x|,
∞
j=1
WhenΨ(t) =tr, 1≤r<∞, we simply writeH
α,r,k instead ofHα,Ψ,k. See [13] or [14] for
more details.
In this paper, we consider thek-order commutatorsTαk,m,bgenerated by Lipschitz func-tions and the operatorTα,m, whereki∈Sn–αi,Ψi∩Hn–αi,Ψi,k, and fork∈N∪ {0},
Tαk,m,b(f)(x) =
Rn
b(x) –b(y)kKα(x,y)f(y)dy.
Clearly,T0
α,m,b=Tα,m.
Also, we consider the following condition for the weights: there existsC> 0 such that
ω(Aix)≤Cω(x), a.e.x∈Rn (2)
for all 1≤i≤m. Letω=log|1x|, |x| ≤e–1,
1, |x|>e–1 . It is easy to check thatω∈A1and satisfies (2) (see
[15]).
In [14], Gallo, Ibañez-Firnkorn, and Riveros obtained the weighted estimates for this kind of operator and certain weights satisfying (2). Precisely as for the classical fractional integral operatorIαwith 0 <α<n, or the singular integral operator withα= 0, they proved the Lp(Rn,ωp)→Lq(Rn,ωq) boundedness ofT
α,m for weightsω∈A(p,q), 1 <p<n/α,
1/q= 1/p–α/n, and 0≤α<n. In [15], forb∈BMO, Ibañez-Firnkorn and Riveros obtained the weighted Coifman type estimates, weightedLp(ωp)→Lq(ωq) estimates, and weighted
BMO estimates as well as two-weighted inequalities. Inspired by these results, we consider the weighted boundedness ofTk
α,m,bforb∈ ˙Λβand a weightedΛ˙βestimate for weights in the classA(n/(α+kβ)r,∞). Our results can be formulated as follows.
Theorem 1.1 For0 <β< 1, 0≤α<n,k∈N∪{0},m∈N,and1≤i≤m,let b∈ ˙Λβ,Ψi
be Young functions and0≤αi<n such thatα1+· · ·+αn=n–α.Let Tα,mbe the integral
operator defined by(1)and Tαk,m,b be the k-order commutator of Tα,m.Suppose that the
matrices Aisatisfy hypothesis(H)and ki∈Sn–αi,Ψi∩Hn–αi,Ψi,k.Moreover,forα= 0,suppose that T0,m is strong type(q∗,q∗)for some1 <q∗<∞.Letϕk(t) =tlog(e+t)k,φ be a Young
function satisfyingΨ–1
1 (t)· · ·Ψm–1(t)φ–1(t)ϕk–1(t)t for t≥t0,some t0> 0.Then there
ex-ists0 <C=C(n,α,A1, . . . ,Am)such that,for0 <δ≤1and f ∈L∞c (Rn),
Mδ
Tαk,m,bf
(x)≤C
k–1
l=0 bk–l
˙
ΛβM(k–l)β
Tαl,m,bf
(x) +CbkΛ˙
β m
i=0
Mα+kβ,φf
A–1i x.
Theorem 1.2 Under the assumptions of Theorem1.1,for1≤r<p<pl≤q<∞,k,l∈N,
1/q= 1/pl– (k–l)β/n, 1/q= 1/p– (α+kβ)/n,there exists0 <C=C(n,α,A1, . . . ,Am)such
that,for f∈L∞c (Rn)andωr∈A(p/r,q/r),
Tαk,m,bfLq(ωq)≤Cb
k
˙
Λβ m
i=1
k
l=0
Rn
Mα+lβ,φf(x)
plωpl(A
ix)dx
1/pl
. (3)
Furthermore,ifωr∈A(p/r,q/r)and satisfying(2),then
Tαk,m,bfLq(ωq)≤CbkΛ˙
β k
l=0
Theorem 1.3 Let0≤α<n, 1 <p<n/(α+kβ), 1/q= 1/p– (α+kβ)/n andφbe a Young function such thatη–1(t)t(α+nkβ) φ–1(t)for every t> 0,whereφ1+
sn
n–(α+kβ) ∈B sn
n–(α+kβ)for every s>r(n– (α+kβ))/(n– (α+kβ)r).Then,under the hypotheses of Theorem1.2,forωr∈
A(p/r,q/r),
Tαk,m,bfLq(ωq)≤CbkΛ˙
βfL
p(ωp).
Theorem 1.4 Under the hypotheses of Theorem1.3,ifωr∈A(n/(α+kβ)r,∞)and satisfies
(2),then there exists C> 0such that,for f ∈L∞c (Rn),
Tαk,m,bfω≤Cb
k
˙
ΛβfωLn/α+kβ,
where
Tαk,m,bfω=sup
B
ωχB∞
1
|B|
B
Tαk,m,bf(y) – 1
|B|
B
Tαk,m,bf(z)dzdy .
The rest of this paper is organized as follows. In Sect.2we recall some relevant def-initions and previous results that are needed to state the other results, which appear in Sect.1. The proofs of sharp maximal functions estimates and Coifman type inequalities are given in Sect.3. Finally, the weightedLp(wp)→Lq(wq) estimates and the weighted
endpoint estimates are presented in Sect.4.
2 Preliminaries
In this section we present some relevant concepts and previous results, which will be used in our proofs.
2.1 The generalized Hölder inequality and the fractionalBpcondition
Now, we present some extra properties for Young functions. For more details of these topics, see [16] or [17].
The functionΨ¯ is called the complementary of the functionΨ if the generalized Hölder inequality holds:
fgL1,B≤2fΨ,BgΨ¯,B.
IfΨ1, . . . ,Ψm,φare Young functions satisfyingΨ1–1(t)· · ·Ψm–1(t)φ–1(t)tfort≥t0, some
t0> 0, then
f1· · ·fmgL1,B≤cf1Ψ1,B· · · fmΨm,Bgφ,B,
where the functionφis called the complementary of the functionsΨ1, . . . ,Ψm.
In 2013, Cruz-Uribe and Moen [18] introduced the fractionalBpcondition: for 1 <p<
n/αand 1/q= 1/p–α/n, a Young functionφ∈Bα
p if
∞
1
φ(t)q/p
tq
And they proved that ifφ∈Bα
p, thenMα,φ:Lp(dx)→Lq(dx) and
Mα,φLp→Lq≤
∞
1
φ(t)q/p
tq
dt t
1/q
.
2.2 The Lipschitz function spaces
For a locally integrable functionf defined inRn, we sayf belongs to the spaceΛ˙β(Rn),
0 <β< 1, if there exists a constantC> 0 such that
sup
Bx
1
|B|1+β/n
B
f(x) –fBdx<∞.
The smallest boundCsatisfying upper inequality is taken to be the norm off denoted by
fΛ˙β. HereBis a ball inRn, and
fB=
1
|B|
B
f(x)dx.
Lemma 2.1 If f ∈ ˙Λβ,then (1) for everyx,y∈Rn,
f(x) –f(y)≤ fΛ˙β|x–y|
β ;
(2) for any ballB,
sup
x∈B
f(x) –fB≤CfΛ˙
β|B|
β/n;
(3) forB⊂B∗,
|fB–fB∗| ≤ fΛ˙βB
∗β/n
.
In particular,ifAiare matrices satisfying(H),B˜is a measurable set,and
˜
Bi=A–1i B,˜ 1≤i≤m,then
|fB˜–f(mi=1B˜i)∪ ˜B| ≤CfΛ˙β| ˜B|
β/n,
and forB=B(cB,R),the ball centered atcBwith radiusR,andBj:=B(cB, 2jR),j∈N,
|fB–fBj| ≤CjfΛ˙βB jβ/n
.
2.3 Weights and maximal operators
A weight functionωis in the Muckenhoupt classAp for 1 <p<∞if there existsC> 1
such that, for any ballB,
1
|B|
B
ω(x)dx 1 |B|
B
ω(x)1–pdx
where 1/p+ 1/p= 1 and the infimum ofCsatisfying the above inequality is denoted by [ω]Ap. We defineA∞=
1≤p<∞Ap. Whenp= 1,ω∈A1if there existsC> 1 such that, for
almost everyx,
Mω(x)≤Cω(x),
and the infimum ofCsatisfying the above inequality is denoted by [ω]A1.
A weight functionωbelongs toA(p,q) for 1 <p<q<∞if there existsC> 1 such that
1
|B|
B
ω(x)qdx 1/q
1
|B|
B
ω(x)–pdx 1/p
≤C,
where 1/p+ 1/p= 1 and the infimum ofCsatisfying the above inequality is denoted by [ω]Ap,q. Whenp= 1,ωis inA(1,q) with 1 <q<∞if there existsC> 1 such that
1
|B|
B
ω(x)qdx esssup
x∈B
1
ω(x) ≤C,
and the infimum ofCsatisfying the above inequality is denoted by [ω]A1,q.
Remark2.1 For 1≤r<p<p0≤p,ω∈A(p,q), by Hölder’s inequality, we can knowω∈ A(p0,q) andω∈A(p,p0). Forωr∈A(p/r,q/r), we also can knowω∈A(p,q).
The sharp maximal function is defined by
Mf(x) =sup
Bx
1
|B|
B
f(y) – 1
|B|
B
f(z)dzdy.
A locally integrable functionf has bounded mean oscillation (f ∈BMO) ifMf(x)∈L∞ and the normfBMO=Mf
∞. Observe that the BMO norm is equivalent to
fBMO=Mf∞∼sup
B
inf
a∈C
1
|B|
B
f(y) –ady.
There is also a weighted version of BMO, which is denoted by BMO(ω), which is described by the semi-norm
fω=sup
B
ωχB∞
1
|B|
B
f(y) – 1
|B|
B
f(z)dzdy .
It is easy to check that
fωωMf∞.
Proposition 2.1([19]) LetΨ be a Young function.Then,for all x∈Rnand r> 1,there
exists a constant Crsuch that
Proposition 2.2 ([15]) Let Ψ be a Young function and A be an invertible matrix. Set
ωA(x) =ω(Ax).Then
Mα,Ψ(ωA)
A–1x≤cA,nMα,Ψ(ω)(x)
for almost every x∈Rn.
2.4 Previous results
In this subsection, we illustrate some known results for the operatorTα,m, which will be
used below, see [12] for more details.
Theorem 2.1([12]) Let0≤α<n,m∈N,and Tα,mbe the integral operator defined by(1).
For1≤i≤m,letΨibe Young functions, 0≤αi<n such thatα1+· · ·+αm=n–α.Also
suppose ki∈Sn–αi,Ψi∩Hn–αi,Ψiand that matrices Aisatisfy hypothesis(H). Ifα= 0,suppose T0,mis of strong type(q∗,q∗)for some1 <q∗<∞.
Ifφ is the complementary of the functionsΨ1, . . . ,Ψm,then there exists C> 0such that,
for0 <δ≤1and f ∈L∞c (Rn),
Mδ(Tα,mf)(x) :=M
|Tα,mf|δ
(x)1/δ≤C
m
i=1 Mα,φf
A–1i x.
Theorem 2.2([12]) Let0≤α<n,m∈N,and Tα,mbe the integral operator defined by(1).
For1≤i≤m,letΨibe Young functions, 0≤αi<n such thatα1+· · ·+αm=n–α.Also
suppose ki∈Sn–αi,Ψi∩Hn–αi,Ψiand that matrices Aisatisfy hypothesis(H). Whenα= 0,suppose that T0,mis of strong type(q∗,q∗)for some1 <q∗<∞.
Let0 <p<∞.Ifφ is the complementary of the functionsΨ1, . . . ,Ψm,then there exists
C> 0such that,forω∈A∞and f ∈L∞c (Rn),
Rn
Tα,mf(x) p
ω(x)dx≤C
m
i=1
Rn
Mα,φf(x)
p
ω(Aix)dx,
whenever the left-hand side is finite.
3 Sharp maximal function estimates and Coifman type inequality
This section is devoted to the proofs of Theorems1.1and1.2.
Proof of Theorem1.1 We just consider the casem= 2 andk= 1, i.e.,T1
α,2,b= [b,Tα,2], and
we will just write [b,Tα] for simplicity. The general case is proved in an analogous way. Letf be a bounded function with compact support, 0 <δ≤1. Forx∈Rn, letB=B(c
B,R)
be a ball that contains x, centered atcB with radiusR. We writeB˜ =B(cB, 2R), and for
1≤i≤2, setB˜i=A–1i B˜. Letf1=fχ2i=1B˜iandf2=f–f1. Suppose thata:=Tα((bB˜∪ ˜B1∪ ˜B2– b)f)(cB) <∞. For 0 <δ≤1, we write
[b,Tα](f)(x) =b(x) –bB˜∪ ˜B1∪ ˜B2
Tαf(x) +Tα
(bB˜∪ ˜B1∪ ˜B2–b)f
(x).
And from the inequality|tδ–sδ|1/δ≤ |t–s|and Jensen’s inequality, we get
1
|B|
B
[b,Tα](f)δ(y) –aδdy
≤
1
|B|
B
[b,Tα](f)(y) –aδdy 1/δ
≤|B|1
B
b(y) –bB˜∪ ˜B1∪ ˜B2
Tαf(y)dy
+ 1
|B|
B
Tα
(bB˜∪ ˜B1∪ ˜B2–b)f1
(y)dy
+ 1
|B|
B
Tα
(bB˜∪ ˜B1∪ ˜B2–b)f2
(y) –Tα
(bB˜∪ ˜B1∪ ˜B2–b)f2
(cB)dy
=:I+II+III. (4)
ForI, by Lemma2.1, we have
I≤CbΛ˙β|B|
β/n
1
|B|
B
Tαf(y)dy
≤CbΛ˙βMβ(Tαf)(x). (5)
ForII, we know
II= 1
|B|
B
˜
B1∪ ˜B2
Kα(y,z)b(z) –bB˜∪ ˜B1∪ ˜B2f1(z)dz dy
≤ 2
i=1
1
|B|
˜
Bi
b(z) –b˜
B∪ ˜B1∪ ˜B2f1(z)
B
K(y,z)dy dz.
We estimate only the first summand, that is,z∈ ˜B1, since the case z∈ ˜B2is analogous.
Observe that
B
Kα(y,z)dy≤
{y∈B:|y–A1z|≤|y–A2z|}
Kα(y,z)dy+
{y∈B:|y–A2z|≤|y–A1z|}
Kα(y,z)dy.
Forj∈N, let us consider the set
Cj1:=y∈B:|y–A1z| ≤ |y–A2z|,|y–A1z| ∼2–j–1R.
Notice that ify∈Bandz∈ ˜B1, then|y–A1z| ≤3R< 4R. Thus,
{y∈B:|y–A1z|≤|y–A2z|}
Kα(y,z)dy
≤
∞
j=–2
C1 j
Kα(y,z)dy
≤
∞
j=–2
|B(A1z, 2–jR)|
|B(A1z, 2–jR)|
B(A1z,2–jR)
Kα(y,z)χ{y:|y–A
1z|∼2–j–1R}dy
≤C
∞
j=–2
BA1z, 2–jRk1(·–A1z)χ{y:|y–A
1z|∼2–j–1R}Ψ1,B(A1z,2–jR)
×k2(·–A2z)χ{y:|y–A1z|∼2–j–1R}
≤C
∞
j=–2
BA1z, 2–jRk1(·–A1z)Ψ1,|y–A
1z|∼2–j–1R
×k2(·–A2z)Ψ
2,|y–A1z|∼2–j–1R.
And fory∈C1
j, we have|y–A2z| ≥ |y–A1z| ≥2–j–1R. Byk2∈Sn–α2,Ψ2 andk1∈Sn–α1,Ψ1,
we get
k2(·–A2z)Ψ2,|y–A
1z|∼2–j–1R≤
k≥0
k2(·–A2z)Ψ2,|y–A
2z|∼2–j+k–1R
≤
k≥0
k2(·)Ψ2,|y|∼2–j+k–1R
≤C
k≥0
2–j+k–1R–α2≤C2–jR–α2,
and
k1(·–A1z)Ψ1,|y–A1z|∼2–j–1R≤C
2–j–1R–α1≤
C2–jR–α1
.
Consequently,
{y∈B:|y–A1z|≤|y–A2z|}
Kα(y,z)dy≤C
∞
j=–2
2–jRn–α1–α2≤CRα.
Similarly,
{y∈B:|y–A2z|≤|y–A1z|}
Kα(y,z)dy≤CRα.
Then
II≤CRα 2
i=1
1
|B|
˜
Bi
b(z) –b˜
B∪ ˜B1∪ ˜B2f1(z)dz
≤CbΛ˙β
2
i=1 Rα+β 1
| ˜Bi|
˜
Bi
f(z)dz
≤CbΛ˙β
2
i=1
Rα+βf φ,B˜i
≤CbΛ˙β
2
i=1
Mα+β,φf
A–1i x. (6)
ForIII, we have
III= 1
|B|
B
(B˜1∪ ˜B2)c
Kα(y,z) –Kα(cB,z)b(z) –b˜
B∪ ˜B1∪ ˜B2f(z)dz dy
≤ 2
l=1
1
|B|
B
Zl
where
Zl= (B1˜ ∪ ˜B2)c∩z:|cB–Alz| ≤ |cB–Arz|,r=l, 1≤r≤2
.
Let us estimate|Kα(y,z) –Kα(cB,z)|fory∈Bandz∈Zl:
Kα(y,z) –Kα(cB,z)≤k1(y–A1z) –k1(cB–A1z)k2(y–A2z)
+k2(y–A2z) –k2(cB–A2z)k1(cB–A1z).
For simplicity we estimate the first summand of|Kα(y,z) –Kα(cB,z)|, the other one follows
in an analogous way. Forj∈N, let
Dlj:=z∈Zl:|cB–Alz| ∼2j+1R
.
Observe that Dlj⊂ {z:|cB–Alz| ∼2j+1R} ⊂A–1l B(cB, 2j+2R) =:B˜l,j. Using the generalized
Hölder inequality, we have
Zl
k1(y–A1z) –k1(cB–A1z)k2(y–A2z)b(z) –bB˜∪ ˜B1∪ ˜B2f(z)dz
≤
∞
j=1
Dlj
k1(y–A1z) –k1(cB–A1z)k2(y–A2z)b(z) –bB˜∪ ˜B1∪ ˜B2f(z)dz
≤
∞
j=1 | ˜Bl,j|
| ˜Bl,j|
˜
Bl,j
k1(y–A1z) –k1(cB–A1z)k2(y–A2z)χDl
jχ{z:|cB–Alz|∼2j +1R}
×b(z) –bB˜l,j+|bB˜l,j–bB˜l|+|bB˜l–bB˜∪ ˜B1∪ ˜B2|f(z)dz
≤
∞
j=1 | ˜Bl,j|
| ˜Bl,j|
˜
Bl,j
k1(y–A1z) –k1(cB–A1z)k2(y–A2z)χDl
jχ{z:|cB–Alz|∼2j+1R} ×CbΛ˙β| ˜Bl,j|
β/n+Cjb
˙
Λβ| ˜Bl,j|
β/n+Cb
˙
Λβ|B|
β/nf(z)dz
≤CbΛ˙β
∞
j=1
| ˜Bl,j|1+β/nj
1
| ˜Bl,j|
˜
Bl,j
k1(y–A1z) –k1(cB–A1z)
×k2(y–A2z)χDl
jχ{z:|cB–Alz|∼2j+1R}
f(z)dz
≤CbΛ˙β
∞
j=1
| ˜Bl,j|1+β/njk1(y–A1·) –k1(cB–A1·)
χDl j
Ψ1,|cB–Alz|∼2j+1R
×k2(y–A2·)χDl
jΨ2,|cB–Alz|∼2j+1Rfφ,B˜l,j.
Note that|cB–Alz|/2≤ |y–Alz| ≤2|cB–Alz|, and if|cB–Alz| ∼2j+1R, then 2jR≤ |y–
Alz| ≤2j+2R. Thus,
kl(y–Al·)χDl
jΨl,|cB–Alz|∼2j+1R ≤kl(y–Al·)Ψ
+kl(y–Al·)Ψ
l,|y–Alz|∼2j+1R ≤kl(·)Ψl,|x|∼2jR+kl(·)Ψl,|x|∼2j+1R
≤C2jR–αl
,
where the last inequality holds sincekl∈Sn–αl,Ψl. Also, by the hypothesis,
kl(cB–Al·)χDljΨl,|cB–Alz|∼2j+1R≤C
2j+1R–αl
.
Forr=l, observe that ifz∈Dl
j, then|cB–Arz| ≥ |cB–Alz| ≥2j+1R. We decomposeDlj=
k≥j(Dlj)k,r, where
Dljk,r=z∈Dlj:|cB–Arz| ∼2k+1R
.
Since (Dl
j)k,r⊂ {z:|cB–Arz| ∼2k+1R}andkr∈Sn–αr,Ψr, we have
kr(y–Ar·)χDljΨr,|cB–Alz|∼2j+1R≤
k≥j
kr(y–Ar·)χ(Dlj)k,rΨr,|cB–Alz|∼2j+1R
≤
k≥j
kr(y–Ar·)χ(Dl
j)k,rΨr,|cB–Arz|∼2k+1R
≤
k≥j
kr(y–Ar·)Ψr,|c
B–Arz|∼2k+1R
≤
k≥j
kr(·)Ψr,|x|∼2kR+kr(·)Ψr,|x|∼2k+1R
≤C
k≥j
2kR–αr ≤C2jR–αr.
By the same arguments, we can get
kr(cB–Ar·)χDljΨr,|cB–Alz|∼2j+1R≤
k≥j
kr(cB–Ar·)χ(Dlj)k,rΨr,|cB–Arz|∼2k+1R
≤C
k≥j
2kR–αr≤C2jR–αr.
As a result, no matterl= 1 orl= 2, we have
k2(y–A2·)χDljΨ2,|cB–Alz|∼2j+1R≤C
2jR–α2
, k1(cB–A1·)χDljΨ1,|cB–Alz|∼2j+1R≤C
2jR–α1
.
Hence,
∞
j=1
| ˜Bl,j|1+β/njk1(y–A1·) –k1(cB–A1·)
χDl
jΨ1,|cB–Alz|∼2j+1R
×k2(y–A2·)χDl
≤C
∞
j=1
2jRn+β–α2jk1(y–A1·) –k1(cB–A1·)
χDl
jΨ1,|cB–Alz|∼2j+1Rfφ,B˜l,j
=C
∞
j=1
2jRα+βfφ,B˜l,j
2jRn–α–α2jk1(y–A1·) –k1(cB–A1·)
χDl
jΨ1,|cB–Alz|∼2j+1R
=C
∞
j=1
2jRα+βfφ,B˜l,j
2jRα1jk1(y–A1·) –k1(cB–A1·)
χDl j
Ψ1,|cB–Alz|∼2j+1R.
So, whenl= 1, fromk1∈Hn–α1,Ψ1,1, we can get
∞
j=1
2jRα+βfφ,B˜l,j
2jRα1jk1(y–A1·) –k1(cB–A1·)
χDl j
Ψ1,|cB–Alz|∼2j+1R
×k2(y–A2·)χDl j
Ψ2,|cB–Alz|∼2j+1R
≤CMα+β,φf
A–11 x
∞
j=1
2jRα1jk1(y–A1·) –k1(cB–A1·)
χDl j
Ψ1,|cB–Alz|∼2j+1R
≤CMα+β,φf
A–11 x.
Forl= 2, note that
k1(y–A1·) –k1(cB–A1·)
χD2
jΨ1,|cB–A2z|∼2j+1R
≤
k≥j
k1(y–A1·) –k1(cB–A1·)
χ(D2
j)k,1Ψ1,|cB–A1z|∼2k+1R,
we have
∞
j=1
2jRα1
jk1(y–A1·) –k1(cB–A1·)
χD2 j
Ψ1,|cB–A2z|∼2j+1R
≤
∞
j=1
2jRα1
j
k≥j
k1(y–A1·) –k1(cB–A1·)
χ(D2
j)k,1Ψ1,|cB–A1z|∼2k+1R
≤
∞
j=1
2jRα1
k≥j
(2kR)α1
(2kR)α1kk1(y–A1·) –k1(cB–A1·)
χ(D2
j)k,1Ψ1,|cB–A1z|∼2k+1R
≤
∞
j=1
k≥j
(2jR)α1
(2kR)α1
2kRα1kk1(y–A1·) –k1(cB–A1·)
χ(D2
j)k,1Ψ1,|cB–A1z|∼2k+1R
≤ ∞ k=1 k j=1 2–α1k–j
2kRα1
kk1(y–A1·) –k1(cB–A1·)
χ(D2
j)k,1Ψ1,|cB–A1z|∼2k+1R
≤
∞
k=1
2kRα1
kk1(y–A1·) –k1(cB–A1·)
χ(D2
where the last inequality follows from thatk1∈Hn–α1,Ψ1,1. Hence,
∞
j=1
2jRα+βfφ,B˜l,j
2jRα1jk1(y–A1·) –k1(cB–A1·)
χD2
jΨ1,|cB–Alz|∼2j+1R
≤CMα+β,φf
A–12 x
∞
j=1
2jRα1jk1(y–A1·) –k1(cB–A1·)
χD2
jΨ1,|cB–Alz|∼2j+1R
≤CMα+β,φf
A–12 x.
Then
III≤CbΛ˙β
2
i=1
Mα+β,φf
A–1i x. (7)
Summing up (4)–(7), we know that
MδTα1,m,bf
(x)≤CbΛ˙βMβ(Tαf)(x) +CbΛ˙β
2
i=0
Mα+β,φf
A–1i x.
For the case α= 0, we repeat the same argument to inequality (4) and get the desired conclusion.
For the general case, from the definition ofTαk,m,b, we know that, for anyλ,
Tαk,m,b(f)(y)
=
Rn
b(y) –b(z)kKα(y,z)f(z)dz
=
Rn
b(y) –λ+λ–b(z)kKα(y,z)f(z)dz
=
k
i=0
Rncki
b(y) –λiλ–b(z)k–iKα(y,z)f(z)dz
=
k
i=0
b(y) –λi
Rncki
λ–b(z)k–iKα(y,z)f(z)dz
=
Rn
λ–b(z)kKα(y,z)f(z)dz
+
k
i=1 cki
b(y) –λi
Rn
λ–b(y) +b(y) –b(z)k–iKα(y,z)f(z)dz
=
k
i=1 cki
b(y) –λi
k–i
j=0 ckj
λ–b(y)j
Rn
b(y) –b(z)k–i–jKα(y,z)f(z)dz
+Tα,m
λ–b(·)kf(y)
=
k
i=1
k–i
j=0 ckij
b(y) –λi+j
Rn
+Tα,m
λ–b(·)kf(y)
=Tα,m
λ–b(·)kf(y) +
k–1
l=0 ckl
b(y) –λk–lTαl,m,bf(y).
Letλ:=bB˜∪ ˜B1∪ ˜B2∪···∪ ˜Bm,a:=Tα((b–bB˜∪ ˜B1∪ ˜B2∪···∪ ˜Bm)f2)(cB). We write
1
|B|
B
Tαk,m,b(f)(y) –a
δ
dy)1/δ
≤
1
|B|
B k–1 i=0
b(y) –λk–iTαi,m,bf(y)
δ dy 1/δ + 1
|B|
B
Tα
λ–b(·)kf1)(y)δdy 1/δ + 1 |B| B Tα
λ–b(·)kf2)(y) –Tα
λ–b(·)kf2)(cB)
δ
dy 1/δ
=:IV+V+VI.
To estimateIV, by Hölder’s inequality and Lemma2.1, we obtain
IV≤ k–1 l=0 1 |B| B
b(y) –λk–lTαl,m,bf(y)
δ dy 1/δ ≤ k–1 l=0
CbkΛ˙–l
β|B|
(k–l)β/n
1
|B|
B
Tαl,m,bf(y)δdy 1/δ
≤
k–1
l=0
Cbk–l
˙
Λβ|B|
(k–l)β/n 1
|B|
B
Tαl,m,bf(y)dy
≤C
k–1
l=0 bk–l
˙
ΛβM(k–l)β
Tαl,m,bf
(x).
The termsVandVIare analogous to the ones in the casem= 2 andk= 1, we can get
V≤CbkΛ˙
β m
i=1
Mα+kβ,φf
A–1i x,
VI≤CbkΛ˙
β m
i=1
Mα+kβ,φf
A–1i x.
Then we conclude
MδTαk,m,bf(x)≤C k–1
l=0 bk–l
˙
ΛβM(k–l)β
Tαl,m,bf
(x) +CbkΛ˙
β m
i=1
Mα+kβ,φf
A–1i x.
Theorem1.1is proved.
Proof of Theorem1.2 By the extrapolation result Theorem 1.1 in [20], we need only to show that (3) is true for some 0 <q∗<∞and allωr∈A(p/r,q∗/r) with (n–α)/n<q∗<∞.
Without loss of generality, we may assumebΛ˙β= 1. We will prove the desired conclusion
by induction.
Whenk= 0,Tα0,m,b=Tα,m. Aski∈Hn–αi,Ψi,0=Hn–αi,Ψi, Theorem 3.3 in [15] tells us that
Rn
Tα,mf(x) q∗
ωq∗(x)dx≤C
m
i=1
Rn
Mα,φf(x)
q∗
ωq∗(Aix)dx.
Now, for anyk∈N, we assume that the results hold for all 0≤j≤k–1, and let us see how to derive the casek. Forωr∈A(p/r,q∗/r), by Remark2.1, we know thatω∈A(p,q∗). Then
ωq∗∈A
q∗. By Lemma 5.1 in [12], we haveTα,mfLq∗(ωq∗)<∞. Thereforeωr∈A(p/r,q∗/r)
andb∈L∞, we have
Tαk,m,bfLq∗(ωq∗)=
k
j=0
ck,jbk–jTα,m
bjf
Lq∗(ωq∗)
<∞.
Besides, forp<pl≤q∗,ω∈A(p,q∗) impliesω∈A(pl,q∗), and 1/q∗= 1/pl– (k–l)β/n
impliesq∗=pkwhenl=k. Then there existsC> 0 such that
M(k–l)β
Tαl,m,bfLq∗(ωq∗)≤CT
l
α,m,bfLpl(ωpl).
By the induction hypothesis, for 0≤l≤k– 1 and 1/pl= 1/qj– (l–j)β/n, we get
Tαl,m,bfLpl(ωpl)≤Cb
l
˙
Λβ m
i=1
l
j=0
Rn
Mα+jβ,φf(x)
qj
ωqj(A
ix)dx
1/qj
.
Since 1/q∗= 1/pl– (k–l)β/nand 1/pl= 1/qj– (l–j)β/n, which impliespl=qjwhenl=j,
we have
Tαk,m,bfLq∗(ωq∗)
≤
Rn
MTαk,m,bfδ(x)q∗/δωq∗(x)dx 1/q∗
≤
Rn
Mδ
Tαk,m,bf
(x)q∗ωq∗(x)dx 1/q∗
≤C
k–1
l=0 bkΛ˙–l
βM(k–l)β
Tαl,m,bf(x)Lq∗(ωq∗)
+CbkΛ˙
β m
i=1
Rn
Mα+kβ,φf
A–1i xq∗ωq∗(x)dx 1/q∗
≤CbkΛ˙
β m
i=1
k–1
l=0
l
j=0
Rn
Mα+jβ,φf(x)
qj
ωqj(A
ix)dx
1/qj
+CbkΛ˙
β m
i=1
Rn
Mα+kβ,φf
≤CbkΛ˙
β m
i=1
k–1
j=0
Rn
Mα+jβ,φf(x)
qj
ωqj(A
ix)dx
1/qj
+CbkΛ˙
β m
i=1
Rn
Mα+kβ,φf(x)
q∗ωq∗(A ix)dx
1/q∗
≤Cbk
˙
Λβ m
i=1
k–1
l=0
Rn
Mα+lβ,φf(x)plωpl(Aix)dx
1/pl
+CbkΛ˙
β m
i=1
Rn
Mα+kβ,φf(x)
pkωpk(A
ix)dx
1/pk
≤Cbk
˙
Λβ m
i=1
k
l=0
Rn
Mα+lβ,φf(x) plωpl(A
ix)dx
1/pl
.
Namely,
Tαk,m,bfLq∗(ωq∗)≤Cb
k
˙
Λβ m
i=1
k
l=0
Rn
Mα+lβ,φf(x) plωpl(A
ix)dx
1/pl
. (8)
For the general case, if b ∈ ˙Λβ, for any N ∈ N, we define bN = bχ{x:–N<b(x)<N} +
Nχ{x:b(x)≥N}–Nχ{x:b(x)≤–N}, thenbNΛ˙β≤cbΛ˙β. Using convergence theorems, for
de-tails see [21], we conclude that (8) holds for anyb∈ ˙Λβ andωr∈A(p/r,q∗/r). Thus, as mentioned, using the extrapolation results obtained in [20], (3) holds for all 0 <q<∞,
b∈ ˙Λβ, andωr∈A(p/r,q/r). Ifωsatisfies (2), we have
Tαk,m,bfLq(ωq)≤Cb
k
˙
Λβ m
i=1
k
l=0
Rn
Mα+lβ,φf(x) plωpl(A
ix)dx
1/pl
≤Cbk
˙
Λβ k
l=0
Mα+lβ,φfLpl(ωpl),
which completes the proof of Theorem1.2.
4 The weighted inequalities of commutators
This section is concerned with the proofs of Theorems1.3and1.4. For the proof of Theo-rem1.3, we need the Coifman inequality (3) and the boundedness of the maximal operator, given in [22] (see Theorem 2.6). Let us begin with the following previous result.
Theorem 4.1([22]) Let0≤α<n,ωbe a weight, 1≤r<p<n/(α+kβ),and1/q= 1/p– (α+kβ)/n.Letφbe a Young function such thatφ1+
ρ(α+kβ)
n–(α+kβ) ∈B ρn
n–(α+kβ) for everyρ>r(n–
(α+kβ))/(n– (α+kβ)r),and letηbe a Young function such thatη–1(t)t(α+kβ)/nφ–1(t)for every t> 0.Ifωr∈A(p/r,q/r),then M
α+kβ,φis bounded from Lp(ωp)to Lq(ωq). Now we are in a position to prove Theorem1.3.
Proof of Theorem1.3 For 1/q= 1/pl– (k–l)β/nand 1/q= 1/p– (α+kβ)/n, we can know
ω∈A(p,q); moreover,ω∈A(pl,q). Therefore, from Theorem4.1, we know thatMα+lβ,φis bounded fromLp(ωp) intoLpl(ωpl). Then, by Theorem1.2andωsatisfies (2), we have
Tαk,m,bfLq(ωq)≤CbkΛ˙
β k
l=0
Mα+lβ,φfLpl(ωpl)≤CbkΛ˙βfL
p(ωp).
This proves the conclusion of Theorem1.3.
Finally, we give the proof of Theorem1.4.
Proof of Theorem1.4 Following the ideas in [21], forωr∈A((α+nkβ)r,∞), we know
ωMα+kβ,rf∞≤ fωn/(α+kβ). (9)
Note thatωr∈A((α+nkβ)r,∞) impliesωr∈A((k–nl)βr,∞), we have
ωM(k–l)β,rTαl,m,bf∞≤Tαl,m,bfωn/(k–l)β. (10)
Now, by (9), (10), (2), Proposition2.1, Theorems1.1and1.3, we get
Tk
α,m,bfωωM
Tαk,m,bf∞
≤C
k–1
l=0 bk–l
˙
ΛβωM(k–l)β
Tαl,m,bf∞
+CbkΛ˙
β m
i=1
ωMα+kβ,φf
A–1
i ·∞
≤Cκr1
k–1
l=0 bkΛ˙–l
βωM(k–l)β,r
Tαl,m,bf∞
+Cκr2b
k
˙
Λβ m
i=1
ωMα+kβ,rf
A–1i ·∞
≤Cκr1
k–1
l=0 bkΛ˙–l
βωT l
α,m,bfn/(k–l)β
+Cκr2b
k
˙
Λβ m
i=1
ωfA–1i ·n/(α+kβ)
≤Cκr1
k–1
l=0 bk–l
˙
Λβb l
˙
Λβωfn/(α+kβ)
+Cκr2b
k
˙
Λβ m
i=1
ωfA–1i ·n/(α+kβ)
≤Cκr1b
k
˙
Λβωfn/(α+kβ)
+Cκr2b
k
˙
Λβ m
i=1
≤Cbk
˙
ΛβfωLn/α+kβ.
This completes the proof of Theorem1.4.
Acknowledgements
The authors thank the referees cordially for their valuable suggestions and comments.
Funding
This project is supported by the National Natural Science Foundation of China (Grant Nos. 11771358, 11871101).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript. Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 12 February 2019 Accepted: 25 July 2019
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