R E S E A R C H
Open Access
Some sharp inequalities for multilinear
integral operators
Daqing Lu
**Correspondence:
College of Mathematics, Changsha University of Science and Technology, Changsha, 410077, P.R. China
Abstract
In this paper, some sharp inequalities for certain multilinear operators related to the Littlewood-Paley operator and the Marcinkiewicz operator are obtained. As an application, we obtain the (Lp,Lq)-norm inequalities and Morrey spaces boundedness for the multilinear operators.
MSC: 42B20; 42B25
Keywords: multilinear operator; Littlewood-Paley operator; Marcinkiewicz operator; Morrey space; BMO
1 Introduction and results
In this paper, we study some multilinear operators related to some integral operators, whose definitions are as follows.
Fixn>δ≥. We denote(x) ={(y,t)∈Rn+
+ :|x–y|<t}and the characteristic function of(x) byχ(x). Suppose thatmjare the positive integers (j= , . . . ,l),m+· · ·+ml=m andAjare the functions onRn(j= , . . . ,l). Let
Rmj+(Aj;x,y) =Aj(x) –
|α|≤mj α!D
α
Aj(y)(x–y)α.
Definition Letε> andψbe a fixed function which satisfies the following properties: () Rnψ(x)dx= ,
() |ψ(x)| ≤C( +|x|)–(n+–δ),
() |ψ(x+y) –ψ(x)| ≤C|y|ε( +|x|)–(n++ε–δ)when|y|<|x|.
The multilinear Littlewood-Paley operator is defined by
SAψ(f)(x) =
(x)
FtA(f)(x,y)dy dt
tn+
/ ,
where
FtA(f)(x,y) =
Rn
l
j=Rmj+(Aj;x,z)
|x–z|m ψt(y–z)f(z)dz
andψt(x) =t–n+δψ(x/t) fort> . SetF
t(f)(y) =f∗ψt(y). We also define that
Sψ(f)(x) =
(x)
Ft(f)(y) dy dt
tn+
/ ,
which is the Littlewood-Paley operator (see []). LetH be the Hilbert spaceH={h:h= ( Rn+
+ |h(y,t)|
dy dt/tn+)/<∞}. Then for
each fixedx∈Rn,FtA(f)(x,y) may be viewed as a mapping from (, +∞) toH, and it is clear that
SAψ(f)(x) =χ(x)FtA(f)(x,y), Sψ(f)(x) =χ(x)Ft(f)(y).
Definition Let < γ ≤ and be homogeneous of degree zero on Rn with
Sn– (x)dσ(x) = . Assume that ∈Lipγ(Sn–), that is, there exists a constantM>
such that for anyx,y∈Sn–,| (x) – (y)| ≤M|x–y|γ. The multilinear Marcinkiewicz
operator is defined by
μAS(f)(x) =
(x)
FA t(f)(x,y)
dy dt
tn+
/ ,
where
FtA(f)(x,y) =
|y–z|≤t
l
j=Rmj+(Aj;x,z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz.
Set
Ft(f)(y) =
|y–z|≤t
(y–z)
|y–z|n––δf(z)dz.
We also define that
μS(f)(x) =
(x)
Ft(f)(y)dy dt
tn+
/ ,
which is the Marcinkiewicz operator (see []). LetHbe the Hilbert spaceH={h:h= ( Rn+
+ |h(y,t)|
dy dt/tn+)/<∞}, then for
each fixedx∈Rn, FA
t (f)(x,y) may be viewed as a mapping from (, +∞) toH, and it is clear that
μAS(f)(x) =χ(x)FtA(f)(x,y), μS(f)(x) =χ(x)Ft(f)(y). Note that whenm= ,SA
sharp inequalities for the multilinear integral operatorsSA
ψandμAS whenDαAj∈BMO(Rn) for allα with |α|=mj. As an application, we obtain the (Lp,Lq)-norm inequalities and Morrey spaces boundedness for the multilinear operators.
First, let us introduce some notations. Throughout this paper,Qwill denote a cube of
Rnwith sides parallel to the axes. For any locally integrable functionf, the sharp function off is defined by
f#(x) =sup
Q x
|Q|
Q
f(y) –fQdy,
where, and in what follows,fQ=|Q|–
Qf(x)dx. It is well-known that (see [, ])
f#(x) =sup
Q x
inf
c∈C
|Q|
Q
f(y) –cdy.
We say thatf belongs toBMO(Rn) iff#belongs toL∞(Rn) andfBMO=f#L∞. For ≤
p<∞and ≤δ<n, let
Mδ,p(f)(x) =sup Q x
|Q|–pδ/n
Q
f(y)pdy /p
;
we write thatMμ(f) =Mnμ,(f), which is the fractional maximal operator.
Fixedλ> . For ≤p<∞, let
fLp,λ= sup x∈Rn,d>
dλ
B(x,d)
f(y)p
dy /p
,
whereB(x,d) ={y∈Rn:|x–y|<d}. The Morrey spaces are defined by (see [–])
Lp,λRn=f ∈Lloc Rn:fLp,λ<∞
.
As the Morrey space may be considered as an extension of the Lebesgue space, it is nat-ural and important to study the boundedness of the multilinear integral operator on the Morrey space.
We shall prove the following theorems.
Theorem Let DαA
j∈BMO(Rn)for allαwith|α|=mjand j= , . . . ,l.
() Then there exists a constantC> such that for anyf ∈C∞(Rn), <r<n/δand
x∈Rn,
SAψ(f)
# (x)≤C
l
j=
|αj|=mj
DαjA jBMO
Mδ,r(f)(x);
() If <p<n/δand/p– /q=δ/n,thenSA
ψ is bounded fromLp(Rn)toLq(Rn),that is,
SA
ψ(f)Lq≤C l
j=
|αj|=mj
DαjA jBMO
() If <p<n/δ, <λ<n–pδ,/q= /p–δ/(n–λ),thenSA
ψis bounded fromLp,λ(Rn)
toLq,λ(Rn),that is,
SAψ(f)Lq,λ≤C l
j=
|αj|=mj
DαjA jBMO
fLp,λ.
Theorem Let DαA
j∈BMO(Rn)for allαwith|α|=mjand j= , . . . ,l.
() Then there exists a constantC> such that for anyf ∈C∞(Rn), <r<n/δand
x∈Rn,
μAS(f)#(x)≤C
l
j=
|αj|=mj
DαjA jBMO
Mδ,r(f)(x);
() If <p<n/δand/p– /q=δ/n,thenμAS is bounded fromLp(Rn)toLq(Rn),that is,
μAS(f)Lq≤C l
j=
|αj|=mj
DαjA jBMO
fLp;
() If <p<n/δ, <λ<n–pδ,/q= /p–δ/(n–λ),thenμA
S is bounded fromLp,λ(Rn)
toLq,λ(Rn),that is,
μAS(f)Lq,λ≤C l
j=
|αj|=mj
DαjA jBMO
fLp,λ.
Remark The conclusions of Theorems and are completely the same. Thus, they ex-plain that the Littlewood-Paley and Marcinkiewicz operators have the many similar bond-edness properties.
2 Proofs of theorems
To prove the theorems, we need the following lemmas.
Lemma [] Let A be a function on Rnand DαA∈Lq(Rn)for allαwith|α|=m and some
q>n.Then
Rm(A;x,y)≤C|x–y|m
|α|=m
| ˜Q(x,y)|
˜
Q(x,y)
DαA(z)qdz /q
,
whereQ is the cube centered at x and having side length˜ √n|x–y|.
Lemma [] Suppose that≤r<p<n/δand/q= /p–δ/n.Then
Mδ,r(f)Lq≤CfLp.
Lemma [, ] Let <p<∞and <λ<n.Then the following estimates hold: (a) M(f)Lp,λ≤Cf#Lp,λ;
Lemma Let <p<n/δand/q= /p–δ/n.Then SψandμSare all bounded from Lp(Rn)
to Lq(Rn).
Proof ForSψ, by Minkowski inequality and the condition ofψ, we have
Sψ(f)(x)≤
Rn
f(z)
(x)
ψt(y–z) dy dt
t+n
/
dz
≤C
Rn
f(z)
∞
|x–y|≤t
t–n+δ
( +|y–z|/t)n+–δ
dy dt t+n
/
dz
≤C
Rn
f(z)
∞
|x–y|≤t
n+t–n
(t+|y–z|)n+–δ dy dt
/
dz,
noting that t+|y–z| ≥t+|x–z|–|x–y| ≥t+|x–z|when|x–y| ≤tand
∞
t dt
(t+|x–z|)n+–δ =C|x–z|
–n+δ
,
we obtain
Sψ(f)(x)≤C
Rn
f(z)
∞
t dt
(t+|x–z|)n+–δ
/
dz
=C
Rn
|f(z)|
|x–z|n–δ dz.
ForμS, note that|x–z| ≤t,|y–z| ≥ |x–z|–t≥ |x–z|– twhen|x–y| ≤t,|y–z| ≤t, we have
μS(f)(x)≤
Rn |x–y|≤t
| (y–z)f(z)|
|y–z|n––δ
χ(z)(y,t)
dy dt tn+
/
dz
≤C
Rn
f(z)
|x–y|≤t
χ(z)(y,t)t–n–
(|x–z|– t)n––δdy dt
/
dz
≤C
Rn
|f(z)|
|x–z|/
∞
|x–z|/
dt
(|x–z|– t)n–
/
dz
≤C
Rn
|f(z)|
|x–z|n–δ dz.
Thus, the lemma follows from [].
Proof of Theorem () It suffices to prove forf ∈C∞(Rn) and some constantC
, the fol-lowing inequality holds:
|Q|
Q
SAψ(f)(x) –Cdx≤C l
j=
|αj|=mj
DαjA jBMO
Mδ,r(f)(x).
Without loss of generality, we may assumel= . Fix a cubeQ=Q(x,d) andx˜∈Q. Let
˜
Q= √nQandA˜j(x) =Aj(x) –|α|=mj
α!(D
αAj)
˜
DαA˜
j=DαAj– (DαAj)Q˜ for|α|=mj. We write, forf=fχQ˜ andf=fχRn\ ˜Q,
FtA(f)(x,y) =
Rn
j=Rmj+(A˜j;x,z)
|x–z|m ψt(y–z)f(z)dz
=
Rn
j=Rmj+(A˜j;x,z)
|x–z|m ψt(y–z)f(z)dz
+
Rn
j=Rmj(A˜j;x,z)
|x–z|m ψt(y–z)f(z)dz
–
|α|=m α!
Rn
Rm(A˜;x,z)(x–z)α
|x–z|m D
αA˜
(z)ψt(y–z)f(z)dz
–
|α|=m α!
Rn
Rm(A˜;x,z)(x–z)α
|x–z|m D
αA˜(z)ψ
t(y–z)f(z)dz
+
|α|=m,|α|=m α!α!
Rn
(x–z)α+αDαA˜(z)DαA˜(z)
|x–z|m ψt(y–z)f(z)dz,
then
SψA(f)(x) –S
˜
A
ψ(f)(x)
=χ(x)FtA(f)(x,y)–χ(x)FtA˜(f)(x,y)
≤χ(x)FtA(f)(x,y) –χ(x)F
˜
A
t (f)(x,y)
≤χ(x)
Rn
j=Rmj(A˜j;x,z)
|x–z|m ψt(y–z)f(z)dz
+χ(x)
|α|=m α!
Rn
Rm(A˜;x,z)(x–z)α
|x–z|m D
αA˜(z)ψt(y–z)f(z)dz
+χ(x)
|α|=m α!
Rn
Rm(A˜;x,z)(x–z)α
|x–z|m D
αA˜
(z)ψt(y–z)f(z)dz
+χ(x)
|α|=m,|α|=m α!α!
Rn
(x–z)α+αDαA˜(z)DαA˜(z)
|x–z|m ψt(y–z)f(z)dz
+χ(x)FtA˜(f)(x,y) –χ(x)FtA˜(f)(x,y) :=I(x) +I(x) +I(x) +I(x) +I(x),
thus,
|Q|
Q
SAψ(f)(x) –S
˜
A
ψ(f)(x)dx
≤
|Q|
Q
I(x)dx+ C
|Q|
Q
I(x)dx+ C
|Q|
Q
I(x)dx
+ C
|Q|
Q
I(x)dx+
|Q|
Q
I(x)dx
Now, let us estimateI,I,I,IandI, respectively. First, forx∈Qandz∈ ˜Q, by Lemma , we get
Rmj(A˜j;x,z)≤C|x–y| mj
|αj|=mj
DαjA jBMO.
Thus, by the (Lr,Lq)-boundedness ofSψ, for <r<n/δand /q= /r–δ/n, we obtain
I≤C
j=
|αj|=mj
DαjA jBMO
|Q|
Q
Sψ(f)(x)dx
≤C
j=
|αj|=mj
DαjA jBMO
|Q|
Q
|Sψ(f)(x)|qdx
/q
≤C
j=
|αj|=mj
DαjA jBMO
|Q|–/q
Q
f(x)rdx /r
≤C
j=
|αj|=mj
DαjA jBMO
Mδ,r(f)(x˜).
ForI, denotingr=pqfor <p<n/δ,q> , /q+ /q= and /s= /p–δ/n, we have, by Hölder’s inequality,
I≤C
|α|=m
DαA BMO
|α|=m
|Q|
Q
Sψ
DαA˜
f
(x)dx
≤C
|α|=m
DαA BMO
|α|=m
|Q|
Rn
Sψ
DαA˜
f
(x)sdx /s
≤C
|α|=m
DαA BMO
|α|=m
|Q|–/s
Rn
DαA˜(x)f (x)
p
dx /p
≤C
|α|=m
DαA BMO
×
|α|=m
|Q|
˜
Q
DαA˜(x)pqdx
/pq
|Q|–rδ/n
˜
Q
f(x)pqdx /pq
≤C
j=
|α|=mj
DαAjBMO
Mδ,r(f)(x˜).
ForI, similar to the proof ofI, we get
I≤C
j=
|α|=mj
DαAjBMO
Similarly, forI, denotingr=pq for <p<n/δ,q,q,q> , /q+ /q+ /q= and /s= /p–δ/n, we obtain
I≤C
|α|=m,|α|=m
|Q|
Q
Sψ DαA˜
DαA˜f
(x)dx
≤C
|α|=m,|α|=m
|Q|
Rn
Sψ
DαA˜
DαA˜f
(x)sdx /s
≤C
|α|=m,|α|=m
|Q|–/s
Rn
DαA˜(x)DαA˜(x)f(x)pdx
/p
≤C
|α|=m,|α|=m
|Q|
˜
Q
DαA˜(x)pqdx
/pq
|Q|
˜
Q
DαA˜(x)pqdx
/pq
× |
Q|–rδ/n
˜
Q
f(x)pqdx /pq
≤C
j=
|α|=mj
DαA
jBMO
Mδ,r(f)(x˜).
ForI, we write
χ(x)FtA˜(f)(x,y) –χ(x)FtA˜(f)(x,y)
=
Rn(χ(x)– χ(x))
j=Rmj(A˜j;x,z)
|x–z|m ψt(y–z)f(z)dz
+χ(x)
Rn
|x–z|m –
|x–z|m
j=
Rmj(A˜j;x,z)ψt(y–z)f(z)dz
+χ(x)
Rn
Rm(A˜;x,z) –Rm(A˜;x,z)
Rm(A˜;x,z)
|x–z|m
ψt(y–z)f(z)dz
+χ(x)
Rn
Rm(A˜;x,z) –Rm(A˜;x,z)
Rm(A˜;x,z)
|x–z|m
ψt(y–z)f(z)dz
–
|α|=m α!
Rn
Rm(A˜;x,z)(x–z)αχ(x)
|x–z|m –
Rm(A˜;x,z)(x–z)αχ(x)
|x–z|m
×DαA˜(z)ψt(y–z)f(z)dz
–
|α|=m α!
Rn
Rm(A˜;x,z)(x–z)αχ(x)
|x–z|m –
Rm(A˜;x,z)(x–z)αχ(x)
|x–z|m
×DαA˜(z)ψt(y–z)f(z)dz
+
|α|=m,|α|=m α!α!
Rn
(x–z)α+αχ
(x)
|x–z|m –
(x–z)α+αχ(x)
|x–z|m
By Lemma and the following inequality (see [])
|bQ–bQ| ≤Clog
|Q|/|Q|
bBMO forQ⊂Q,
we know that, forx∈Qandz∈k+Q˜ \kQ˜,
Rm(A˜;x,z)≤C|x–z|m
|α|=m
DαA
BMO+D
αA
˜
Q(x,z)–
DαA
˜
Q
≤Ck|x–z|m
|α|=m
DαA
BMO.
Note that|x–z| ∼ |x–z|forx∈Qandz∈Rn\ ˜Q, we obtain, similar to the proof of Lemma ,
I()≤
Rn
Rn+ +
j=|Rmj(A˜j;x,z)||ψt(y–z)||f(z)|
|x–z|m
×χ(x)(y,t) –χ(x)(y,t)
dy dt
tn+
/ dz
≤C
Rn
j=|Rmj(A˜j;x,z)||f(z)|
|x–z|m
×
(x)
t–ndy dt (t+|y–z|)n+–δ –
(x)
t–ndy dt (t+|y–z|)n+–δ
/dz
≤C
Rn
j=|Rmj(A˜j;x,z)||f(z)|
|x–z|m
×
|y|≤t
(t+|x+y–z|)n+–δ –
(t+|x+y–z|)n+–δ
dy dttn–
/
dz
≤C
Rn
j=|Rmj(A˜j;x,z)||f(z)|
|x–z|m
|y|≤t
|x–x|t–ndy dt (t+|x+y–z|)n+–δ
/
dz
≤C
Rn
j=Rmj(A˜j;x,z)|f(z)||x–x|/
|x–z|m+n+/–δ
dz
≤C
j=
|α|=mj
DαAjBMO
∞
k=
k+Q˜\kQ˜k
|x–x|/
|x–z|n+/–δ
f(z)dz
≤C
j=
|α|=mj
Dα
AjBMO
∞
k=
k–k/
|kQ˜|–δ/n
kQ˜
f(z)dz
≤C
j=
|α|=mj
DαAjBMO
Mδ,r(f)(x˜);
I()≤C
Rn
|x–x|
|x–z|m+n+–δ
j=
Rmj(A˜j;x,z)f(z)dz
≤C
j=
|α|=mj
DαAjBMO
∞
k=
k+Q˜\kQ˜k
|x–x|
|x–z|n+–δ
≤C
j=
|α|=mj
DαAjBMO
∞
k=
k–k
|kQ˜|–δ/n
kQ˜
f(z)dz
≤C
j=
|α|=mj
DαAjBMO
Mδ,r(f)(x˜).
ForI()andI(), by the formula (see [])
Rm(A˜;x,z) –Rm(A˜;x,z) =
|β|<m β!Rm–|β|
DβA˜;x,x
(x–z)β
and Lemma , we have
Rm(A˜;x,z) –Rm(A˜;x,z)≤C
|β|<m
|α|=m
|x–x|m–|β||x–z||β|DαABMO.
Thus, similar to the proof of Lemma ,
I()≤C
j=
|α|=mj
DαA
jBMO
∞
k=
k+Q˜\kQ˜
k |x–x| |x–z|n+–δ
f(y)dy
≤C
j=
|α|=mj
DαAjBMO
Mδ,r(f)(x˜);
I()≤C
j=
|α|=mj
DαA
jBMO
Mδ,r(f)(x˜).
Similarly, we get
I()≤C
|α|=m
Rn
Rm(A˜;x,z)(x–z)αχ(x)
|x–z|m –
Rm(A˜;x,z)(x–z)αχ(x)
|x–z|m
×ψt(y–z)
DαA˜
(z)f(z)dz
≤C
|α|=m
DαABMO
|α|=m
∞
k=
k–k/+ –k
×
|kQ˜|
kQ˜
DαA˜ (y)r
dy
/r
|kQ˜|–rδ/n
kQ˜
f(y)rdy /r
≤C
j=
|α|=mj
DαAjBMO
Mδ,r(f)(x˜);
I()
≤C
j=
|α|=mj
Dα
AjBMO
ForI(), takingq,q> such that /r+ /q+ /q= , then
I()≤C
|α|=m,|α|=m
Rn
(x–z)α+αχ
(x)
|x–z|m –
(x–z)α+αχ(x)
|x–z|m
ψt(y–z)
×DαA˜(z)DαA˜(z)f(z)dz
≤C
|α|=m,|α|=m
∞
k=
k–k/+ –k
|kQ˜|–pδ/n
kQ˜
f(y)rdy /r
×
|kQ˜|
kQ˜
DαA˜ (y)
q
dy /q
|kQ˜|
kQ˜
DαA˜ (y)
q
dy /q
≤C
j=
|α|=mj
Dα
AjBMO
Mδ,r(f)(x˜).
Thus
I ≤C
j=
|α|=mj
DαA
jBMO
Mδ,r(f)(x˜).
We choose <r<pin (), then () follows from Lemma . For (), taking <r<min(p, (n– λ)/pδ) in () and by Lemma , we obtain
SψA(f)Lq,λ≤CM
SψA(f)Lq,λ≤CSAψ(f)
# Lq,λ
≤C
j=
|α|=mj
Dα
AjBMO
Mδ,r(f)Lq,λ
≤C
j=
|α|=mj
DαAjBMO
Mrδ/n
|f|r/rLq,λ
≤C
j=
|α|=mj
DαA
jBMO
Mrδ/n
|f|r/rLq/r,λ
≤C
j=
|α|=mj
DαAjBMO
|f|r/rLp/r,λ
≤C
j=
|α|=mj
DαAjBMO
fLp,λ.
This completes the proof of Theorem .
Proof of Theorem It is only to prove (). LetQ,Q˜,A˜j(x),fandfbe the same as the proof of Theorem . We write
FtA(f)(x,y)
=
Rn
j=Rmj+(A˜j;x,z)
|x–z|m
(y–z)
+
Rn
j=Rmj(A˜j;x,z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz
–
|α|=m α!
Rn
Rm(A˜;x,z)(x–z)αDαA˜(z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz
–
|α|=m α!
Rn
Rm(A˜;x,z)(x–z)αDαA˜(z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz
+
|α|=m,|α|=m α!α!
Rn
(x–z)α+αDαA˜(z)DαA˜(z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz,
then
|Q|
Q
μAS(f)(x) –μSA˜(f)(x)dx
≤
|Q|
Q
χ(x)FtA(f)(x,y) –χ(x)FtA˜(f)(x,y)dx
≤| Q|
Q
χ(x)
Rn
j=Rmj(A˜j;x,z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz
dx
+
|Q|
Q
χ(x)
|α|=m α!
×
Rn
Rm(A˜;x,z)(x–z)αDαA˜(z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz
dx
+
|Q|
Q
χ(x)
|α|=m α!
×
Rn
Rm(A˜;x,z)(x–z)αDαA˜(z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz
dx
+
|Q|
Q
χ(x)
|α|=m,|α|=m α!α!
×
Rn
(x–z)α+αDαA˜
(z)DαA˜(z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz
dx
+
|Q|
Q
χ(x)F
˜
A
t (f)(x,y) –χ(x)F
˜
A
t (f)(x,y)dx
:=J+J+J+J+J.
Similar to the proof of Theorem , we get
J≤C
j=
|αj|=mj
DαjA jBMO
|Q|
Q
μS(f)(x)dx
≤C
j=
|αj|=mj
DαjA jBMO
|Q|
Q
μS(f)(x)qdx /q
≤C
j=
|αj|=mj
DαjA jBMO
J≤C
|α|=m
DαA BMO
|α|=m
|Q|
Q
μS
DαA˜
f
(x)dx
≤C
|α|=m
DαA BMO
|α|=m
|Q|
Rn
μS
DαA˜
f
(x)sdx /s
≤C
j=
|α|=mj
DαAjBMO
Mδ,r(f)(x˜);
J≤C
j=
|α|=mj
DαAjBMO
Mδ,r(f)(x˜);
J≤C
|α|=m,|α|=m
|Q|
Q
μS
DαA˜
DαA˜f
(x)dx
≤C
|α|=m,|α|=m
|Q|
Rn
μS
DαA˜
DαA˜f
(x)sdx /s
≤C
j=
|α|=mj
DαAjBMO
Mδ,p(f)(x˜).
ForJ, we write
χ(x)FtA˜(f)(x,y) –χ(x)FtA˜(f)(x,y)
=
Rn(χ(x)– χ(x))
j=Rmj(A˜j;x,z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz
+χ(x)
Rn
|x–z|m –
|x–z|m
j=
Rmj(A˜j;x,z)
(y–z)
|y–z|n––δf(z)dz
+χ(x)
Rn
Rm(A˜;x,z) –Rm(A˜;x,z)
Rm(A˜;x,z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz
+χ(x)
Rn
Rm(A˜;x,z) –Rm(A˜;x,z)
Rm(A˜;x,z)
|x–z|m
(y–z)
|y–z|n––δf(z)dz
–
|α|=m α!
Rn
Rm(A˜;x,z)(x–z)αχ(x)
|x–z|m –
Rm(A˜;x,z)(x–z)αχ(x)
|x–z|m
×| (y–z) y–z|n––δD
αA˜
(z)f(z)dz
–
|α|=m α!
Rn
Rm(A˜;x,z)(x–z)αχ(x)
|x–z|m –
Rm(A˜;x,z)(x–z)αχ(x)
|x–z|m
×| (y–z) y–z|n––δD
αA˜(z)f(z)dz
+
|α|=m,|α|=m α!α!
Rn
(x–z)α+αχ
(x)
|x–z|m –
(x–z)α+αχ(x)
|x–z|m
× (y–z) |y–z|n––δD
Then, similar to the proof of Lemma and Theorem , we get
J ≤C
j=
|α|=mj
DαAjBMO
Mδ,r(f)(x˜).
The same argument as the proof of Theorem will give the proof of () and (), we omit
the details and finish the proof.
Competing interests
The author declares that he has no competing interests.
Received: 21 September 2012 Accepted: 10 May 2013 Published: 4 October 2013 References
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doi:10.1186/1029-242X-2013-445