# Necessary and Sufficient Conditions for Boundedness of Commutators of the General Fractional Integral Operators on Weighted Morrey Spaces

Authors: Zengyan Si; Fayou Zhao
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We prove thatb is in Lipβ(ω) if and only if the commutator [b,L-α/2] of the multiplication operator by b and the general fractional integral operator L-α/2 is bounded from the weighted Morrey space Lp,k(ω) to Lq,kq/p(ω1-(1-α/n)q,ω), where 0<β<1, 0<α+β<n,1<p<n/(α+β), 1/q=1/p-(α+β)/n, 0≤k<p/q, ωq/p∈A1, and rω>(1-k)/(p/(q-k)), and here rω denotes the critical index of ω for the reverse Hölder condition.


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## Body

## 1.  Introduction and Main Results

Suppose thatL is a linear operator on L2(ℝn) which generates an analytic semigroup e-tL with a kernel pt(x,y) satisfying a Gaussian upper bound, that is,
(1.1)|pt(x,y)|≤Ctn/2e-c(|x-y|2/t)
for x,y∈ℝn and all t>0. Since we assume only upper bound on heat kernel pt(x,y) and no regularity on its space variables, this property (1.1) is satisfied by a class of differential operator, see [1] for details.For0<α<n, the general fractional integral L-α/2 of the operator L is defined by
(1.2)L-α/2f(x)=1Γ(α/2)∫0∞‍e-tLfdtt-α/2+1(x).Note that, ifL=-Δ is the Laplacian on ℝn, then, L-α/2 is the classical fractional integral Iα which plays important roles in many fields. Let b be a locally integrable function on ℝn, the commutator of b and L-α/2 is defined by
(1.3)[b,L-α/2]f(x)=b(x)L-α/2f(x)-L-α/2(bf)(x).For the special case ofL=-Δ, many results have been produced. Paluszyński [2] obtained that b∈Lipβ(ℝn) if the commutator [b,Iα] is bounded from Lp(ℝn) to Lr(ℝn), where 1<p<r<∞,0<β<1 and 1/p-1/r=(α+β)/n with p<n/(α+β). Shirai [3] proved that b∈Lipβ(ℝn) if and only if the commutator [b,Iα] is bounded from the classical Morrey spaces Lp,λ(ℝn) to Lq,λ(ℝn) for 1<p<q<∞,0<α,0<β<1, and 0<α+β=(1/p-1/q)(n-λ)<n or Lp,λ(ℝn) to Lq,μ(ℝn) for 1<p<q<∞,0<α,0<β<1,0<α+β=(1/p-1/q)<n,0<λ<n-(α+β)p, and μ/q=λ/p. Wang [4] established some weighted boundedness of properties of commutator [b,Iα] on the weighted Morrey spaces Lp,k under appropriated conditions on the weight ω, where the symbol b belongs to (weighted) Lipschitz spaces. The weighted Morrey space was first introduced by Komori and Shirai [5]. For the general case, Wang [6] proved that if b∈Lipβ(ℝn), then the commutator [b,L-α/2] is bounded from Lp(ωp) to Lq(ωq), where 0<β<1,0<α+β<n,1<p<n/(α+β),1/p-1/q=(α+β)/n, and ωq∈A1.The purpose of this paper is to give necessary and sufficient conditions for boundedness of commutators of the general fractional integrals withb∈Lipβ(ω) (the weighted Lipschitz space). Our theorems are the following.Theorem 1.1.
Let0<β<1, 0<α+β<n,1<p<n/(α+β), 1/q=1/p-(α+β)/n,0≤k<min{p/q,pβ/n}, and ωq∈A1. Then one has the following. (a)
Ifb∈Lipβ(ℝn), then [b,L-α/2] is bounded from Lp,k(ωp,ωq) to Lq,kq/p(ωq);(b)
If[b,L-α/2] is bounded from Lp,k(ωp,ωq) to Lq,kq/p(ωq), then b∈Lipβ(ℝn).Theorem 1.2.
Let0<β<1, 0<α+β<n,1<p<n/(α+β), 1/q=1/p-(α+β)/n,0≤k<p/q,ωq/p∈A1, and rω>(1-k)/(p/(q-k)), where rω denotes the critical index of ω for the reverse Hölder condition. Then one has the following. (a)
Ifb∈Lipβ(ω), then [b,L-α/2] is bounded from Lp,k(ω) to Lq,kq/p(ω1-(1-α/n)q,ω);(b)
If[b,L-α/2] is bounded from Lp,k(ω) to Lq,kq/p(ω1-(1-α/n)q,ω), then b∈Lipβ(ω).Our results not only extend the results of [4] from (-Δ) to a general operator L, but also characterize the (weighted) Lipschitz spaces by means of the boundedness of [b,L-α/2] on the weighted Morrey spaces, which extend the results of [4, 6]. The basic tool is based on a modification of sharp maximal function ML# introduced by [7].Throughout this paper all notation is standard or will be defined as needed. Denote the Lebesgue measure ofB by |B| and the weighted measure of B by ω(B), where ω(B)=∫B‍ω(x)dx. For a measurable set E, denote by χE the characteristic function of E. For a real number p, 1<p<∞, let p' be the dual of p such that 1/p+1/p'=1. The letter C will be used for various constants, and may change from one occurrence to another.


## 2. Some Preliminaries

A nonnegative functionω defined on ℝn is called weight if it is locally integral. A weight ω is said to belong to the Muckenhoupt class Ap(ℝn) for 1<p<∞, if there exists a positive constant C such that
(2.1)(1|B|∫B‍ω(x)dx)(1|B|∫B‍ω(x)-1/(p-1)dx)p-1≤C,
for every ball B⊂ℝn. The class A1(ℝn) is defined replacing the above inequality by
(2.2)(1|B|∫B‍ω(x)dx)≤Cessinfx∈Bω(x).
When p=∞,ω∈A∞, if there exist positive constants δ and C such that given a ball B and E is a measurable subset of B, then
(2.3)ω(E)ω(B)≤C(|E||B|)δ.A weight functionω belongs to Ap,q for 1<p<q<∞ if for every ball B in ℝn, there exists a positive constant C which is independent of B such that
(2.4)(1|B|∫B‍ω(x)qdx)1/q(1|B|∫B‍ω(x)-p′dx)1/p′≤C.From the definition ofAp,q, we can get that
(2.5)ω∈Ap,q,iffωq∈A1+q/p′.
Since ωq/p∈A1, then by (2.5), we have ω1/p∈Ap,q.A weight functionω belongs to the reverse Hölder class RHr if there exist two constants r>1 and C>0 such that the following reverse Hölder inequality,
(2.6)(1|B|∫B‍ω(x)rdx)1/r≤C1|B|∫B‍ω(x)dx,
holds for every ball B in ℝn.It is well known that ifω∈Ap with 1≤p<∞, then there exists r>1 such that ω∈RHr. It follows from Hölder inequality that ω∈RHr implies ω∈RHs for all 1<s<r. Moreover, if ω∈RHr,r>1, then we have ω∈RHr+ε for some ϵ>0. We thus write rw=sup{r>1:ω∈RHr} to denote the critical index of ω for the reverse Hölder condition. For more details on Muckenhoupt class Ap,q, we refer the reader to [8–10].Definition 2.1 (see [5]).
Let1≤p<∞ and 0≤k<1. Then for two weights μ and ν, the weighted Morrey space is defined by
(2.7)Lp,k(μ,ν)={f∈Llocp(μ):∥f∥Lp,k(μ,ν)<∞},
where
(2.8)∥f∥Lp,k(μ,ν)=supB(1ν(B)k∫B‍|f(x)|pμ(x)dx)1/p,
and the supremum is taken over all balls B in ℝn.Ifμ=ν, then we have the classical Morrey space Lp,k(μ) with measure μ. When k=0, then Lp,0(μ)=Lp(μ) is the Lebesgue space with measure μ.Definition 2.2 (see [11]).
Let1≤p<∞, 0<β<1, and ω∈A∞. A locally integral function b is said to be in Lipβp(ω) if
(2.9)∥b∥Lipβp(ω)=supB1ω(B)β/n(1ω(B)∫B‍|b(x)-bB|pω(x)1-pdx)1/p≤C<∞,
where bB=|B|-1∫B‍b(y)dy and the supremum is taken over all ball B⊂Rn. When p=1, we denote Lipβp(ω) by Lipβ(ω).Obviously, for the caseω=1, then the Lipβp(ω) space is the classical Lipβp space.Remark 2.3.
Letω∈A1, García-Cuerva [11] proved that the spaces ∥f∥Lipβp(ω) coincide, and the norms of ||·||Lipβp(ω) are equivalent with respect to different values of provided that 1≤p<∞.Given a locally integrable functionf and β, 0≤β<n, define the fractional maximal function by
(2.10)Mβ,rf(x)=supx∈B(1|B|1-βr/n∫B‍|f(y)|rdy)1/r,r≥1,
when 0<β<n. If β=0 and r=1, then M0,1f=Mf denotes the usual Hardy-Littlewood maximal function.Letω be a weight. The weighted maximal operator Mω is defined by
(2.11)Mωf(x)=supx∈B1ω(B)∫B‍|f(y)|dy.
The fractional weighted maximal operator Mβ,r,ω is defined by
(2.12)Mβ,r,ωf(x)=supx∈B(1ω(B)1-βr/n∫B‍|f(y)|rω(y)dy)1/r,
where 0≤β<n and r≥1. For any f∈Lp(ℝn),p≥1, the sharp maximal function ML#f associated the generalized approximations to the identity {e-tL,t>0} is given by Martell [7] as follows:
(2.13)ML#f(x)=supx∈B1|B|∫B‍|f(y)-e-tBLf(y)|dy,
where tB=rB2 and rB is the radius of the ball B. For 0<δ<1, we introduce the δ-sharp maximal operator ML,δ# as
(2.14)ML,δ#f=ML#(|f|δ)1/δ,
which is a modification of the sharp maximal operator M# of Stein and Murphy [9]. Set Mδf=M(|f|δ)1/δ. Using the same methods as those of [9, 12], we can get the following.Lemma 2.4.
Assume that the semigroupe-tL has a kernel pt(x,y) which satisfies the upper bound (1.1). Let λ>0 and f∈Lp(ℝn) for some 1<p<∞. Suppose that ω∈A∞, then for every 0<η<1, there exists a real number γ>0 independent of γ,f such that one has the following weighted version of the local good λ inequality, for η>0, A>1,
(2.15)ω{x∈ℝn:Mδf>Aλ,ML,δ#f(x)≤γλ}≤ηω{x∈ℝn:Mδf(x)>λ},
where A>1 is a fixed constant which depends only on n.Ifμ,ν∈A∞,1<p<∞,0≤k<1, then
(2.16)∥f∥Lp,k(μ,ν)≤∥Mδf∥Lp,k(μ,ν)≤C∥ML,δ#f∥Lp,k(μ,ν).
In particular, when μ=ν=ω and ω∈A∞, we have
(2.17)∥f∥Lp,k(ω)≤∥Mδf∥Lp,k(ω)≤C∥ML,δ#f∥Lp,k(ω).


## 3. Proof of Theorem1.1

To prove Theorem1.1, we need the following lemmas.Lemma 3.1 (see [1]).
Assume that the semigroupe-tL has a kernel pt(x,y) which satisfies the upper bound (1.1). Then for 0<α<1, the difference operator L-α/2-e-tLL-α/2 has an associated kernel Kα,t(x,y) which satisfies
(3.1)Kα,t(x,y)≤C|x-y|n-αt|x-y|2.Lemma 3.2 (see [4]).
Let0<α+β<n, 1<p<n/(α+β),1/q=1/p-(α+β)/n, and ω∈A1. Then for every 0<k<p/q and 1<r<p, one has
(3.2)∥Mα+β,rf∥Lq,kq/p(ωq)≤C∥f∥Lp,q(ωp,ωq).Lemma 3.3 (see [5]).
Let0<β<n, 1<p<n/β,1/s=1/p-β/n, and ω∈Ap,s. Then for every 0<k<p/s, one has
(3.3)∥Mβ,1f∥Ls,ks/p(ωs)≤C∥f∥Lp,k(ωp,ωs).Lemma 3.4 (see [4]).
Let0<α+β<n, 1<p<n/(α+β),1/q=1/p-α/n, 1/s=1/q-β/n, and ωq∈A1. Then for every 0<k<p/s, one has
(3.4)∥Mβ,1f∥Ls,ks/p(ωs)≤C∥f∥Lq,kq/p(ωq,ωs).Lemma 3.5.
Let0<α+β<n, 1<p<n/(α+β),1/q=1/p-α/n, 1/s=1/q-β/n, and ωq∈A1. Then for every 0<k<pβ/n, one has
(3.5)∥L-α/2f∥Lq,kq/p(ωq,ωs)≤C∥f∥Lp,k(ωp,ωs).Proof.
Since the semigroupe-tL has a kernel pt(x,y) which satisfies the upper bound (1.1), it is easy to check that L-α/2f(x)≤CIα(|f|)(x) for all x∈ℝn. Together with the result (cf. [4]), that is,
(3.6)∥Iαf∥Lq,kq/p(ωq,ωs)≤C∥f∥Lp,k(ωp,ωs),
we can get the desired result.Remark 3.6.
Using the boundedness property ofIα, we also know L-α/2 is bounded from L1 to weak Ln/(n-α). It is easy to check that Lemmas 3.2–3.5 also hold when k=0.The following lemma plays an important role in the proof of Theorem1.1.Lemma 3.7.
Let0<δ<1,0<α<n,0<β<1, and b∈Lipβ(ℝn). Then for all r>1 and for all x∈ℝn, one has
(3.7)ML,δ#([b,L-α/2]f)(x)≤C∥b∥Lipβ(ℝn)(Mβ,1(L-α/2f)(x)+Mα+β,rf(x)+Mα+β,1f(x)).The same method of proof as that of Lemma4.7 (see below), one omits the details.Proof of Theorem1.1.
We first prove(a). We only prove Theorem 1.1 in the case 0<α<1. For the general case 0<α<n, the method is the same as that of [1]. We omit the details.
For0<α+β<n and 1<p<n/(α+β), we can find a number r such that 1<r<p. By (2.17) and Lemma 3.7, we obtain the following:
(3.8)∥[b,L-α/2]f∥Lq,kq/p(ωq)≤C∥ML,δ#([b,L-α/2]f)∥Lq,kq/p(ωq)≤C∥b∥Lipβ(ℝn)(∥Mβ,1(L-α/2f)∥Lq,kq/p(ωq)+∥Mα+β,rf∥Lq,kq/p(ωq)+∥Mα+β,1f∥Lq,kq/p(ωq)∥Mβ,1(L-α/2f)∥Lq,kq/p(ωq)).
Let 1/q1=1/p-α/n and 1/q=1/q1-β/n. Since ωq∈A1, then by (2.5), we have ω∈Ap,q. Since 0<k<min{p/q,pβ/n}, by Lemmas 3.2–3.5, we yield that
(3.9)∥[b,L-α/2]f∥Lq,kq/p(ωq)≤C∥b∥Lipβ(ℝn)(∥L-α/2f∥Lq1,kq1/p(ωq1,ωq)+∥f∥Lp,k(ωp,ωq))≤C∥b∥Lipβ(ℝn)∥f∥Lp,k(ωp,ωq).
Now we prove(b). Let L=-Δ be the Laplacian on ℝn, then L-α/2 is the classical fractional integral Iα. Let k=0 and weight ω≡1, then Lp,k(ωp,ωq)=Lp and Lq,kq/p(ωq,ω)=Lq. From [2], the (Lp,Lq) boundedness of [b,Iα] implies that b∈Lipβ(ℝn).
Thus Theorem1.1 is proved.


## 4. Proof of Theorem1.2

We also need some Lemmas to prove Theorem1.2.Lemma 4.1 (see [4]).
Let0<α+β<n,1<p<n/(α+β),1/q=1/p-α/n,1/s=1/q-β/n, and ωs/p∈A1. Then if 0<k<p/s and rω>1/(p/(q-k)), one has
(4.1)∥Mβ,1f∥Ls,ks/p(ωs/p,ω)≤C∥f∥Lq,kq/p(ωq/p,ω).Lemma 4.2 (see [4]).
Let0<α<n,1<p<n/α,1/q=1/p-α/n, and ωq/p∈A1. Then if 0<k<p/q and rω>(1-k)/(p/(q-k)), one has
(4.2)∥Mα,1f∥Lq,kq/p(ωq/p,ω)≤C∥f∥Lp,k(ω).Lemma 4.3 (see [4]).
Let0<α<n,1<p<n/α,1/q=1/p-α/n, 0<k<p/q, and ω∈A∞. For any 1<r<p, one has
(4.3)∥Mα,r,ωf∥Lq,kq/p(ωq/p,ω)≤C∥f∥Lp,k(ω).Lemma 4.4.
Let0<α<n,1<p<n/α,1/q=1/p-α/n, and ωq/p∈A1. Then if 0<k<p/q and rω>(1-k)/(p/(q-k)), one has
(4.4)∥L-α/2f∥Lq,kq/p(ωq/p,ω)≤C∥f∥Lp,k(ω).Proof.
As before, we know thatL-α/2f(x)≤CIα(|f|)(x) for all x∈ℝn. Using the boundedness property of Iα on weighted Morrey space (cf. [4]), we have
(4.5)∥L-α/2f∥Lq,kq/p(ωq/p,ω)≤∥Iαf∥Lq,kq/p(ωq/p,ω)≤C∥f∥Lp,k(ω),
where 1<p<n/α and 1/q=1/p-α/n.Remark 4.5.
It is easy to check that the above lemmas also hold fork=0.Lemma 4.6.
Assume that the semigroupe-tL has a kernel pt(x,y) which satisfies the upper bound (1.1), and let b∈Lipβ(ω),ω∈A1. Then, for every function f∈Lp(ℝn),p>1,x∈ℝn, and 1<r<∞, one has
(4.6)supx∈B1|B|∫B‍|e-tBL(b(y)-b2B)f(y)|dy≤C∥b∥Lipβ(ω)ω(x)Mβ,r,ωf(x).Proof.
Fixf∈Lp(ℝn),1<p<∞ and x∈B. Then,
(4.7)1|B|∫B‍|e-tBL((b-b2B)f)(y)|dy≤1|B|∫B‍∫ℝn‍|ptB(y,z)||(b(z)-b2B)f(z)|dzdy≤1|B|∫B‍∫2B|ptB(y,z)|‍|(b(z)-b2B)f(z)|dzdy≤≤+1|B|∫B‍∑k=1∞‍∫2k+1B∖2kB‍|ptB(y,z)||(b(z)-b2B)f(z)|dzdy≐ℳ+𝒩.
It follows from y∈B and z∈2B that
(4.8)|ptB(y,z)|≤CtB-n/2≤C1|2B|.
Thus, Hölder's inequality and Definition 2.2 lead to the following:
(4.9)ℳ≤C1|2B|∫2B‍|(b(z)-b2B)f(z)|dz≤C1|2B|(∫2B‍|b(z)-b2B|r'ω(z)1-r'dz)1/r'(∫2B‍|f(z)|rω(z)dz)1/r≤C∥b∥Lipβ(ω)1|2B|ω(2B)β/n+1/r′ω(2B)1/r(1ω(2B)∫2B‍|f(z)|rω(z)dz)1/r≤C∥b∥Lipβ(ω)1|2B|ω(2B)β/n+1(1ω(2B)∫2B‍|f(z)|rω(z)dz)1/r≤C∥b∥Lipβ(ω)ω(x)(1ω(2B)1-βr/n∫2B‍|f(z)|rω(z)dz)1/r≤C∥b∥Lipβ(ω)ω(x)Mβ,r,ωf(x).
Moreover, for any y∈B and z∈2k+1B∖2kB, we have |y-z|≥2k-1rB and |ptB(y,z)|≤C(e-c22(k-1)2(k+1)n/|2k+1B|)(4.10)𝒩=1|B|∫B‍∑k=1∞‍∫2k+1B∖2kB‍|ptB(y,z)||(b(z)-b2B)f(z)|dzdy≤C∑k=1∞‍e-c22(k-1)2(k+1)n|2k+1B|∫2k+1B‍|(b(z)-b2B)f(z)|dz≤C∑k=1∞‍e-c22(k-1)2(k+1)n|2k+1B|∫2k+1B‍|(b(z)-b2k+1B)f(z)|dz≤+C∑k=1∞‍e-c22(k-1)2(k+1)n|2k+1B|∫2k+1B‍|(b2k+1B-b2B)f(z)|dz≐𝒩1+𝒩2.
We will estimate the values of terms 𝒩1 and 𝒩2, respectively.
Using Hölder's inequality and Remark2.3, we have the following:
(4.11)𝒩1≤C∑k=1∞‍e-c22(k-1)2(k+1)n|2k+1B|≤×(∫2k+1B‍|b(z)-b2k+1B|r'ω(z)1-r'dz)1/r'(∫2k+1B‍|f(z)|rω(z)dz)1/r≤C∑k=1∞‍2(k+1)ne-c22(k-1)≤×∥b∥Lipβ(ω)ω(2k+1B)|2k+1B|(1ω(2k+1B)1-βr/n∫2k+1B‍|f(z)|rω(z)dz)1/r≤C∥b∥Lipβ(ω)ω(x)Mβ,r,ωf(x).
By a simple calculation, we have(4.12)|b2k+1B-b2B|≤Ckω(x)∥b∥Lipβ(ω)ω(2k+1B)β/n.
Since ω∈A1, by the Hölder inequality, we get
(4.13)𝒩2≤C∑k=1∞‍2(k+1)ne-c22(k-1)kω(2k+1B)β/n|2k+1B|ω(x)∥b∥Lipβ(ω)∫2k+1B‍|f(z)|dz≤C∑k=1∞‍k2(k+1)ne-c22(k-1)ω(x)∥b∥Lipβ(ω)ω(2k+1B)β/n|2k+1B|1/r(∫2k+1B‍|f(z)|rdz)1/r=C∑k=1∞‍k2(k+1)ne-c22(k-1)=×ω(x)∥b∥Lipβ(ω)(ω(2k+1B)|2k+1B|1ω(2k+1B)1-βr/n∫2k+1B‍|f(z)|rdz)1/r≤C∑k=1∞‍k2(k+1)ne-c22(k-1)ω(x)∥b∥Lipβ(ω)(1ω(2k+1B)1-βr/n∫2k+1B‍|f(z)|rω(z)dz)1/r≤C∥b∥Lipβ(ω)ω(x)Mβ,r,ωf(x).
Thus, Lemma4.6 is proved.Lemma 4.7.
Let0<δ<1, 0<α<1,ω∈A1, and b∈Lipβ(ω). Then for all r>1 and for all x∈ℝn, one has
(4.14)ML,δ#([b,L-α/2]f)(x)≤C∥b∥Lipβ(ω)×(ω(x)1+β/nMβ,1(L-α/2f)(x)+ω(x)1-α/nMα+β,r,ωf(x)+ω(x)1+β/nMα+β,1f(x)).Proof.
For any givenx∈ℝn, fix a ball B=B(x0,rB) which contains x. We decompose f=f1+f2, where f1=fχ2B. Observe that
(4.15)[b,L-α/2]f=(b-b2B)L-α/2f-L-α/2((b-b2B)f1)-L-α/2((b-b2B)f2)e-tBL([b,L-α/2]f)=e-tBL[(b-b2B)L-α/2f-L-α/2((b-b2B)f1)-L-α/2((b-b2B)f2)].
Then
(4.16)(1|B|∫B‍|[b,L-α/2]f(y)-e-tBL([b,L-α/2]f)(y)|δdy)1/δ≤C(1|B|∫B‍|(b(y)-b2B)L-α/2f(y)|δdy)1/δ≤+C(1|B|∫B‍|L-α/2((b-b2B)f1)(y)|δdy)1/δ≤+C(1|B|∫B‍|e-tBL((b-b2B)L-α/2f)(y)|δdy)1/δ≤+C(1|B|∫B‍|e-tBL(L-α/2((b-b2B)f1))(y)|δdy)1/δ≤+C(1|B|∫B‍|(L-α/2-e-tBLL-α/2)((b-b2B)f2)(y)|δdy)1/δ≐I+II+III+IV+V.We are going to estimate each term, respectively. Fix0<δ<1 and choose a real number τ such that 1<τ<2 and τ'δ<1. Since ω∈A1, then it follows from Hölder's inequality that
(4.17)I≤C(1|B|∫B‍|(b(y)-b2B)|τδdy)1/τδ(∫B‍|L-α/2f(y)|τ'δdy)1/τ'δ≤C(1|B|∫2B‍|(b(y)-b2B)|dy)(∫B‍|L-α/2f(y)|dy)≤C∥b∥Lipβ(ω)1|2B|ω(2B)1+β/n(∫B‍|L-α/2f(y)|dy)≤C∥b∥Lipβ(ω)ω(x)1+β/nMβ,1(L-α/2f)(x).ForII, using Hölder's inequality, Kolmogorov's inequality (see page 485 [8],) and Remark 3.6, then we deduce that
(4.18)II≤C1|B|∫B‍|L-α/2((b-b2B)f1)(y)|dy≤C1|B||B|α/n∥L-α/2(b-b2B)f1∥Ln/(n-α),∞≤C1|B|1-α/n∫B‍|(b(y)-b2B)f1(y)|dy≤C1|B|1-α/n(∫2B‍|f(y)|rw(y)dy)1/r(∫2B‍|b(y)-b2B|r'w(y)-r'/rdy)1/r'≤C∥b∥Lipβ(ω)Mα+β,r,ωf(x)(ω(2B)|2B|)1-α/n≤C∥b∥Lipβ(ω)ω(x)1-α/nMα+β,r,ωf(x),
where we have used the condition that ω∈A1.Using Hölder's inequality and Lemma4.6, we obtain that
(4.19)III≤C∥b∥Lipβ(ω)ω(x)Mβ,r,ω(L-α/2f)(x).ForIV, using the estimate in II, we get
(4.20)IV≤C|B|∫B‍∫2B‍|ptB(y,z)||L-α/2((b-b2B)f)(z)|dzdy≤C|2B|∫2B‍|L-α/2((b-b2B)f)(z)|dz≤C∥b∥Lipβ(ω)ω(x)1-α/nMα+β,r,ωf(x).By virtue of Lemma3.1, we have the following:
(4.21)V≤C|B|∫B‍∫(2B)c‍|Kα,tB(y,z)||(b(z)-b2B)f(z)|dzdy≤C∑k=1∞‍∫2krB≤|x0-z|<2k+1rB‍1|x0-z|n-αrB2|x0-z|2|(b(z)-b2B)f(z)|dz≤C∑k=1∞‍2-2k1|2k+1B|1-α/n∫2k+1B‍|(b(z)-b2B)f(z)|dz≤C∑k=1∞‍2-2k1|2k+1B|1-α/n∫2k+1B‍|(b(z)-b2k+1B)f(z)|dz≤+C∑k=1∞‍2-2k|b2k+1B-b2B|1|2k+1B|1-α/n∫2k+1B‍|f(z)|dz≐VI+VII.
Making use of the same argument as that of II, we have
(4.22)VI≤C∥b∥Lipβ(ω)ω(x)1-α/nMα+β,r,ωf(x).Note thatω∈A1,
(4.23)|b2k+1B-b2B|≤Ckω(x)∥b∥Lipβ(ω)ω(2k+1B)β/n.
So, the value of VII can be controlled by
(4.24)C∥b∥Lipβ(ω)ω(x)1+β/nMα+β,1f(x).Combining the above estimates forI–V, we finish the proof of Lemma 4.7.Proof of Theorem1.2.
We first prove(a). As before, we only prove Theorem 1.2 in the case 0<α<1. For 0<α+β<n and 1<p<n/(α+β), we can find a number r such that 1<r<p. By Lemma 4.7, we obtain the following:
(4.25)∥[b,L-α/2]f∥Lq,kq/p(ω1-(1-α/n)q,ω)≤C∥ML,δ#([b,L-α/2]f)∥Lq,kq/p(ω1-(1-α/n)q,ω)≤C∥b∥Lipβ(ω)(∥ω(·)1+β/nMβ,1(L-α/2f)∥Lq,kq/p(ω1-(1-α/n)q,ω)≤+∥ω(·)1-α/nMα+β,r,ωf∥Lq,kq/p(ω1-(1-α/n)q,ω)≤+∥ω(·)1+β/nMα+β,1f∥Lq,kq/p(ω1-(1-α/n)q,ω))≤C∥b∥Lipβ(ω)(∥Mβ,1(L-α/2f)∥Lq,kq/p(ωq/p,ω)≤+∥Mα+β,r,ωf∥Lq,kq/p(ω)+∥Mα+β,1f∥Lq,kq/p(ωq/p,ω)).
Let 1/q1=1/p-α/n and 1/q=1/q1-β/n. Lemmas 4.1–4.4 yield that
(4.26)∥[b,L-α/2]f∥Lq,kq/p(ω1-(1-α/n)q,ω)≤C∥b∥Lipβ(ω)(∥L-α/2f∥Lq1,kq1/p(ωq1/p,ω)+∥f∥Lp,k(ω))≤C∥b∥Lipβ(ω)∥f∥Lp,k(ω).
Now we prove (b). LetL=-Δ be the Laplacian on ℝn, then L-α/2 is the classical fractional integral Iα. We use the same argument as Janson [13]. Choose Z0∈ℝn so that |Z0|=3. For x∈B(Z0,2),|x|-α+n can be written as the absolutely convergent Fourier series, |x|-α+n=∑m∈ℤn‍amei〈νm,x〉 with ∑m‍|am|<∞ since |x|-α+n∈C∞(B(Z0,2)). For any x0∈ℝn and ρ>0, let B=B(x0,ρ) and BZ0=B(x0+Z0ρ,ρ),
(4.27)∫B‍|b(x)-bBZ0|dx=1|BZ0|∫B‍|∫BZ0‍(b(x)-b(y))dy|dx=1ρn∫B‍s(x)(∫BZ0‍(b(x)-b(y))|x-y|-α+n|x-y|n-αdy)dx,
where s(x)=sgn(∫BZ0‍(b(x)-b(y))dy). Fix x∈B and y∈BZ0, then (y-x)/ρ∈BZ0,2, hence,
(4.28)ρ-α+nρn∫B‍s(x)(∫BZ0‍(b(x)-b(y))|x-y|-α+n(|x-y|ρ)n-αdy)dx=ρ-α∑m∈Zn‍am∫B‍s(x)(∫BZ0‍(b(x)-b(y))|x-y|n-αei〈νm,y/ρ〉dy)e-i〈νm,x/ρ〉dx≤ρ-α|∑m∈Zn‍|am|∫B‍s(x)[b,L-α/2](χBZ0ei〈νm,·/ρ〉)(x)χB(x)e-i〈νm,x/ρ〉dx|≤ρ-α∑m∈Zn‍|am|∥[b,L-α/2](χBZ0ei〈νm,·/ρ〉)∥Lq,0(ω1-(1-α/n)q,ω)(∫B‍ω(x)q′(1/q′-α/n)dx)1/q'≤Cρ-α∑m∈Zn‍|am|∥χBZ0∥Lp,0(ω)(∫B‍ω(x)q′(1/q′-α/n)dx)1/q'≤Cω(B)1/p+1/q′-α/n=Cω(B)1+β/n.
This implies that b∈Lipβ(ω). Thus, (b) is proved.



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*Source: 929381-2012-09-04.xml*