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Journal of Inequalities and Applications
 
J Inequal Appl. 2017; 2017(1): 48.
Published online 2017 February 22. doi:  10.1186/s13660-017-1325-z
PMCID: PMC5321714

Weighted norm inequalities for multilinear Calderón-Zygmund operators in generalized Morrey spaces

Abstract

In this paper, the authors study the boundedness of multilinear Calderón-Zygmund singular integral operators and their commutators in generalized Morrey spaces.

Keywords: multilinear Calderón-Zygmund operators, commutators, weighted Morrey spaces

Introduction

Let T be a multilinear operator initially defined on the m-fold product of Schwartz spaces and taking values into the space of tempered distributions, i.e.,

T:𝒮(ℝn) ×  ⋯  × 𝒮(ℝn) → 𝒮(ℝn).

In [1], it is said that a function K belongs to the class m-CZK(Aε) if

  1. |K(y0,y1,,ym)|A(k,l=0m|ykyl|)mn,
  2. if |yjyj|12max0km|yjyk|,
    |K(y0,,yj,,ym)K(y0,,yj,,ym)|A|yjyj|ε(k,l=0m|ykyl|)mn+ε
    for some ε > 0 and j = 0, 1, 2…, m.

The operator T is said to be an m-linear Calderón-Zygmund operator if there exists a function K ∈ m-CZK(Aε) defined away from the diagonal y0y1y2 ⋯  = ym in (ℝn)m+1 such that

T(f1, …, fm)(x) = ∫(ℝn)mK(xy1, …, ym)f1(y1) ⋯ fm(ym) dy1 ⋯  dym
1.1

for xj=1msuppfj, and that T extends to a bounded multilinear operator from Lq1 ×  ⋯  × Lqm to Lq for some 1 ⩽ qj < ∞ with 1q=1q1++1qm.

It was shown in [1] that if 1r1++1rm=1r, then an m-linear Calderón-Zygmund operator satisfies

T:Lr1 ×  ⋯  × Lrm → Lr

when 1 < rj < ∞ for j = 1, …, m and

T:Lr1 ×  ⋯  × Lrm → Lr,∞

when 1 ≤ rj < ∞ for j = 1, …, m and at least one rj = 1. In particular,

T:L1 ×  ⋯  × L1 → L1,∞.

The theory of multiple weight associated with m-linear Calderón-Zygmund operators was developed by Lerner et al. [2]. Let 1 < pj < ∞ for j = 1, …, m, 1p=1p1++1pm and p=(p1,,pm), we say ωAp if

supB(1|B|Bvω)1/pj=1m(1|B|Bωj1pj)1/pj<,

where B is the ball in n and vω=j=1mωjp/pj. They showed that if ωAp then

T(f)Lp(vω)Cj=1mfjLpj(ωj).
1.2

If 1 ⩽ pj < ∞ for j = 1, …, m and at least one of the pj = 1, they also proved

T(f)Lp,(vω)Cj=1mfjLpj(ωj).
1.3

Let b=(b1,,bm) be a vector-valued locally integrable function. If b=(b1,,bm) in (BMO)m, we denote b(BMO)m=supj=1,,mbjBMO (see [2]). The commutator generated by an m-linear Calderón-Zygmund operator T and a (BMO)m function b is defined by

Tb(f1,,fm)=j=1mTbj(f),
1.4

where each term is the commutator of bj and T in the jth entry of T, that is,

Tbj(f)=bjT(f1,,fj,,fm)T(f1,,bjfj,,fm).

Pérez and Torres [3] proved that if b(BMO)m then

Tb:Lp1××LpmLp

for 1 < pj < ∞ and 1 < p < ∞ with 1p=1p1++1pm, where j = 1, …, m. In [2], the authors proved that if ωAp and b(BMO)m, then

Tb(f)Lp(vω)Cb(BMO)mj=1mfjLpj(ωj)
1.5

for 1 < pj < ∞ with 1p=1p1++1pm, where j = 1, …, m.

Feuto [4] introduced the generalized weighted Morrey space (Lp(ω),Lq)α. Let 1 ⩽ p ⩽ α ⩽ q ⩽ ∞ and ω be a weight. The space (Lp(ω),Lq)α was defined to be the set of all measurable functions f satisfying f(Lp(ω),Lq)α < ∞, where

f(Lp(ω),Lq)α=supr>0rf(Lp(ω),Lq)α

with

f(Lp(ω),Lq)αr:=[Rn(ω(B(y,r))1/α1/p1/qfχB(y,r)Lp(ω))qdy]1/q.

When ω ≡ 1, the space (Lp,Lq)α was introduced in [5]. If p < α and q = ∞, the space (Lp(ω),L)α is just the weighted Morrey space Lp,κ(ω) with κ = 1 − p/α defined by Komori and Shirai [6].

Similarly, the weak space (Lp,∞(ω),Lq)α is defined with

f(Lp,(ω),Lq)αr:=[Rn(ω(B(y,r))1/α1/p1/qfχB(y,r)Lp,(ω))qdy]1/p.

When p = 1, the space (L1,∞(ω),Lq)α was introduced in [4].

Feuto proved in [4] that Calderón-Zygmund singular integral operators, Marcinkiewicz operators, the maximal operators associated to Bochner-Riesz operators and their commutators are bounded on (Lp(ω),Lq)α.

In this paper, we aim to study the boundedness of multilinear singular integral operators on the product of generalized Morrey spaces. Inspired by the above mentioned works, we state our main results as follows.

Theorem 1.1

Let T be an m-linear Calderón-Zygmund operator, 1p=1p1++1pm and ωAp.

  1. If 1 < pj < ∞, j = 1, …, m and p ⩽ α < q ⩽ ∞, then
    T(f)(Lp(vω),Lq)αi=1mfi(Lpi(ωi),Lqpi/p)αpi/p.
  2. If 1 ⩽ pj < ∞, j = 1, …, m and at least one of pj = 1, p ⩽ α < q ⩽ ∞, then
    T(f)(Lp,(vω),Lq)αi=1mfi(Lpi(ωi),Lqpi/p)αpi/p.

Theorem 1.2

Let Tb be a multilinear commutator, b(BMO)m, 1p=1p1++1pm with 1 < pj < ∞ and ωAp. If p ⩽ α < q ⩽ ∞, then

Tb(f)(Lp(vω),Lq)αb(BMO)mi=1mfi(Lpi(ωi),Lqpi/p)αpi/p.

Remark 1.3

When m = 1, Theorem 1.1 is just Theorem 2.1 in [4] and Theorem 1.2 is just Theorem 2.5 in [4].

Notations and preliminaries

We first recall the definition of Ap weight. A nonnegative locally integrable function ω belongs to Ap (p > 1) if

supB(1|B|Bω(x)dx)(1|B|Bω(x)1pdx)p1<,

where p is the conjugate index of p, i.e., 1/p + 1/p = 1. We say that ω ∈ A1 if there is a constant C > 0 such that

1|B|Bω(x)dxCinfxBω(x)

for any ball B. If ω ∈ Ap, then there exists δ > 0 such that

ω(E)ω(B)(|E||B|)δ
2.1

for any measurable subset E of a ball B. Since the Ap classes are increasing with respect to p, we use the following notation A = ⋃p>1Ap. A ≲ B means A ≤ CB, where C is a positive constant independent of the main parameters. For λ > 0 and a ball B ⊂ ℝn, we write λB for the ball with same center as B and radius λ times radius of B.

Now we give the definition of Ap condition.

Definition 2.1

[2]

Let 1 ⩽ pj < ∞ for j = 1, …, m, p=(p1,,pm) and 1p=1p1++1pm. Given ω=(ω1,,ωm), set

vω=j=1mωjp/pj.

We say that ωAp if

supB(1|B|Bvω)1/pj=1m(1|B|Bωj1pj)1/pj<.

p is the conjugate index of p. When pj = 1, denote pj=, (1|B|Bωj1pj)1/pj is understood as (infBωj)−1

Obviously, if m = 1, Ap is the classical Ap class. Ap has the following characterization.

Lemma 2.2

[2]

Let ω=(ω1,,ωm). Then ωAp if and only if

ωj1pjAmpjandvωAmp,

where the condition ωj1pjAmpj is understood as ωj1/mA1 in the case pj = 1.

Lemma 2.3

[2]

Assume that ω=(ω1,,ωm) satisfies Ap condition. Then there exists a finite constant r > 1 such that ωAp/r.

In order to prove the results for commutators, we need the following properties of BMO. For b ∈ BMO, 1 < p < ∞ and ω ∈ A, we get

bBMOsupB(1|B|B|b(x)bB|pdx)1/p
2.2

and for all balls B,

(1ω(B)B|b(x)bB|pω(x)dx)1/pCbBMO.
2.3

For all nonnegative integers k, we obtain

|b2k+1B − bB| ≤ C(k + 1)∥bBMO
2.4

where ω(B) = ∫Bω(x) dx, bB=1|B|Bb(x)dx (see [4]).

Proof of the main results

Proof of Theorem 1.1

(1) Let BB(yr) be a ball of n, fifiχ2Bfiχ(2B)c and denote fiχ2B by fi0 and fiχ(2B)c by fi (i = 1, …, m), χE denotes the characteristic function of set E. For x ∈ B(yr), we have

|Tf(x)||T(f10,,fm0)(x)|+α1,,αm{0,}|T(f1α1,,fmαm)(x)|+|T(f1,,fm)(x)|=I+II+III,

where α1, …, αm are not all equal to 0 or ∞ at the same time. We first estimate III. Since 2k−1r ⩽ |x − yi| ⩽ 2k+2r, we have

III=|(Rn2B)mf1(y1)fm(ym)(i=1m|xyi|)mndy|(Rn2B)m|f1(y1)fm(ym)|(i=1m|xyi|)mndy=k=1(2k+1B2kB)m|f1(y1)fm(ym)|(i=1m|xyi|)mndyk=1i=1m2k+1B2kB|fi(yi)||xyi|ndyik=11|2k+1B|mi=1m2k+1B|fi(yi)|dyi,
3.1

the Hölder inequality gives us that

2k+1B|fi(yi)| dyi ⩽ (∫2k+1B|f(yi)|piωi(yi) dyi)1/pi(∫2k+1Bω(yi)pi/pi dyi)1/pi.
3.2

By the definition of Ap condition, we obtain

IIIk=11(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi).
3.3

For II, we just consider this case: αi = ∞ for i = 1, …, l and αj = 0 for jl + 1, …, m,

|T(f1,,fl,fl+10,,fm0)(x)|=|(Rn2B)l(2B)mlf1(y1)fm(ym)(i=1m|xyi|)mndy|(Rn2B)l(2B)ml|f1(y1)fm(ym)|(i=1m|xyi|)mndyi=l+1m2B|fi(yi)|dyik=11|2k+1B|mi=1l2k+1B2kB|fi(yi)|dyik=11|2k+1B|mi=1m2k+1B|fi(yi)|dyi.
3.4

By (3.2) and the definition of Ap condition, we have

IIk=11(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi).
3.5

Combining all the cases together, we obtain

|Tf(x)||(2B)mK(x,y1,,ym)f1(y1)fm(ym)dy|+k=11(2k+1Bvω)1/pi=1mfχ2k+1BLpi(ωi).
3.6

Taking Lp(vω) norm on the ball B(yr) on both sides of (3.6), by (1.2), we get

TfχB(y,r)Lp(vω)i=1mfiχB(y,2r)Lpi(ωi)+k=1(Bvω)1/p(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi).
3.7

Multiplying both sides of (3.7) by vω(B)1/α1/q1/p, by Lemma 2.2 and (2.1), we obtain

vω(B)1/α1/q1/pTfχB(y,r)Lp(vω)k=012nkδ(1/α1/q)vω(2k+1B)1/α1/q1/pi=1mfiχB(y,2k+1r)Lpi(ωi).
3.8

For 1p1/p+1p2/p++1pm/p=1, we have

vω(B)1/α1/q1/pTfχB(y,r)Lp(vω)Lq(Rn)k=012nkδ(1/α1/q)i=1mωi(2k+1B)p/αpi1/pip/qpifiχB(y,2k+1r)Lpi(ωi)Lqpi/p(Rn)

by the Hölder inequality. Since k=012nkδ(1/α1/q)<, we obtain the expected result

Tf(Lp(vω),Lq)αi=1mfi(Lpi(ωi),Lqpi/p)αpi/p.

(2) For λ > 0, by (3.6) and (1.3), we get

λvω(xB(y,r):|Tf(x)|>λ)1/pi=1mfiχB(y,2r)Lpi(ωi)+k=1(Bvω)1/p(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi).

That is,

TfχB(y,r)Lp,(vω)i=1mfiχB(y,2r)Lpi(ωi)+k=1(Bvω)1/p(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi).
3.9

Multiplying both sides of (3.9) by vω(B)1/α1/q1/p, we conclude as in the case (1).

Proof of Theorem 1.2

It suffices to prove Tbj. For BB(yr), x ∈ B

Tbj(f)(x)=Tbj(fχ2B)(x)+α1,,αm{0,}(bj(x)T(f1α1,,fjαj,,fmαm)T(f1α1,,bjfjαj,,fmαm)(x))+bj(x)T(f1,,fj,,fm)T(f1,,bjfj,,fm)(x)=I+II+III,

where α1, …, αm are not all equal to 0 or ∞ at the same time. We first deal with III.

|III||(bj(x)bB)T(f1,,fj,,fm)|+|T(f1,,(bjbB)fj,,fm)(x)||(bj(x)bB)|k=11(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi)+k=1|b2k+1BbB|(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi)+k=11|2k+1B|m(2k+1B)mi=1,ijm|fi(yi)fj(yj)(bj(yj)b2k+1B)|dy.
3.10

Select suitable s > 1 to raise (2k+1B)mi=1,ijm|fi(yi)fj(yj)(bj(yj)b2k+1B)|dy by the Hölder inequality and such that ωAp/s by Lemma 2.3. Then characterization Ap/s and (2.3) yield

k=11|2k+1B|m(2k+1B)mi=1,ijm|fi(yi)fj(yj)(bj(yj)b2k+1B)|dyk=11|2k+1B|m/s(i=1,ijm2k+1B|fi(yi)|sdyi)1/s×(2k+1B|(bj(yj)b2k+1B)fj(yj)|sdyj)1/sk=11|2k+1B|m/si=1,ijm(2k+1B|fi(yi)|piωi(yi)dyi)1/pi×(2k+1Bωi(yi)s/(pis)dyi)(pis)/pis×(2k+1B|bj(yj)b2k+1B|pjs/(pjs)ωi(yj)s/(pjs)dyj)1/s×(2k+1B|fj(yj)|pjωj(yj)dyj)1/pjbjBMOk=11(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi).
3.11

So we have

|III||(bj(x)bB)|k=11(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi)+k=1|b2k+1BbB|(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi)+bjBMOk=11(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi).
3.12

For II, we just consider this case: αi = ∞ for i = 1, …, l and αj = 0 for jl + 1, …, m. There are two cases:

bj(x)T(f1,,fj,,fl,fl+10,,fm0)T(f1,,bjfj,,fl,fl+10,,fm0)(x)

or

bj(x)T(f1,,fl,fl+10,,fj0,,fm)T(f1,,fl,fl+10,,bjfj0,,fm0)(x).

We just consider the following case:

|bj(x)T(f1,,fj,,fl,fl+10,,fm0)T(f1,,bjfj,,fl,fl+10,,fm0)(x)||(bj(x)bB)T(f1,,fj,,fl,fl+10,,fm0)|+|T(f1,,(bjbB)fj,,fl,fl+10,,fm0)(x)||(bj(x)bB)|k=11(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi)+k=1|b2k+1BbB|(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi)+k=1i=l+1m2B|fi(yi)|dyi|2k+1B|m(2k+1B)mi=1,ijl|fi(yi)fj(yj)(bj(yj)b2k+1B)|dy.
3.13

The estimate for

k=1i=l+1m2B|fi(yi)|dyi|2k+1B|m(2k+1B)mi=1,ijl|fi(yi)fj(yj)(bj(yj)b2k+1B)|dy

is similar to (3.11). We get

k=1i=l+1m2B|fi(yi)|dyi|2k+1B|m(2k+1B)mi=1,ijl|fi(yi)fj(yj)(bj(yj)b2k+1B)|dyk=11|2k+1B|mi=1,ijm2k+1B|fi(yi)|dyi×2k+1B|bj(yj)b2k+1B|f(yj)dyjbjBMOk=11(2k+1Bvω)1/pi=1mfχ2k+1BLpi(ωi),
3.14

so we have

|Tbj(f)(x)||Tbj(fχ2B)(x)|+|(bj(x)bB)|k=11(2k+1Bvω)1/pi=1mfχ2k+1BLpi(ωi)+k=1|b2k+1BbB|(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi)+bjBMOk=11(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi).
3.15

Take Lp(vω) norm on the ball B(yr) on both sides of (3.15). By (1.5), (2.3), (2.4), we have

Tbj(f)χB(y,r)Lp(vω)bjBMOi=1mfiχB(y,2r)Lpi(ωi)+bjBMOk=1k(Bvω)1/p(2k+1Bvω)1/pi=1mfiχ2k+1BLpi(ωi).
3.16

Multiplying both sides of (3.16) by vω(B)1/α1/q1/p, by Lemma 2.2 and (2.1), we obtain

vω(B)1/α1/q1/pTbj(f)χB(y,r)Lp(vω)k=0(k+1)bjBMO2nkδ(1/α1/q)vω(2k+1B)1/α1/q1/pi=1mfiχB(y,2k+1r)Lpi(ωi).
3.17

Next the proof is similar to Theorem 1.1 of (1), we get

Tbj(f)(Lp(vω),Lq)αb(BMO)mi=1mfi(Lpi(ωi),Lqpi/p)αpi/p.

Acknowledgements

This work is supported by NNSF-China (Grant Nos. 11171345 and 51234005).

Footnotes

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

PW put forward the ideas of the paper, and the authors completed the paper together. They also read and approved the final manuscript.

Contributor Information

Panwang Wang, moc.liamg@wgnawnap.

Zongguang Liu, nc.ude.btmuc@gzuil.

References

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