# Wavelets, Sobolev Multipliers, and Application to Schrödinger Type Operators with Nonsmooth Potentials

Authors: Pengtao Li; Qixiang Yang; Yueping Zhu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We employ Meyer wavelets to characterize multiplier spaceXr,pt(ℝn) without using capacity. Further, we introduce logarithmic Morrey spaces Mr,pt,τ(ℝn) to establish the inclusion relation between Morrey spaces and multiplier spaces. By fractal skills, we construct a counterexample to show that the scope of the index τ of Mr,pt,τ(ℝn) is sharp. As an application, we consider a Schrödinger type operator with potentials in Mr,pt,τ(ℝn).


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## Body

## 1. Introduction

As useful tools, multipliers on the spaces of differential functions are applied to the study of various problems in harmonic analysis and differential equations. A functiong is called a multiplier from a Sobolev space Ht+r,p(ℝn) to another Sobolev space Ht,p(ℝn) if for every function f∈Ht+r,p(ℝn), the product fg∈Ht,p(ℝn). We denote by Xr,pt(ℝn) the class of all such functions g. In the famous book [1], Maz'ya and Shaposhnikova gave many characterizations of different kinds of multiplier spaces. For different indices r, t, and p, the multiplier spaces can adapt to the needs of different problems. Further, for different indices r, t, and p, they need different skills to deal with different multiplier spaces. In this paper, we consider multiplier spaces Xr,pt(ℝn) defined as follows.Definition 1 (see [1]).
Givent≥0, r>0 and 1<p<n/(r+t), the multiplier space Xr,pt(ℝn) is defined as the set of all the functions f(x) such that
(1)∥f∥Xr,pt(ℝn):=sup∥g∥Ht+r,p(ℝn)≤1∥fg∥Ht,p(ℝn)<∞.For a compact sete⊂ℝn, the capacity cap(e,Ht,p) is defined by
(2)cap(e,Ht,p)=inf{∥u∥Ht,p(ℝn)p:u∈𝒮,u≥1one},
where 𝒮 denotes the Schwartz class of rapidly decreasing smooth functions on ℝn.Lemma 2 (see [1]).
Givenr>0 and t≥0. (i)
For1<p<n/(r+t), f∈Xr,pt(ℝn) if and only if
(3)supe⊂ℝn(∥(-Δ)t/2f∥Lp(e)(cap(e,Ht+r,p))1/p+∥f∥Lp(e)(cap(e,Hr,p))1/p)<∞.(ii)
For1<p<n/r and any cube Q with a length less than 1, the capacity cap(Q,Hr,p) is less than C|Q|1-pr/n.Our motivation is based on the following consideration. For complicated compact sets, it is very difficult to compute the capacity. The main aim of this paper is to establish a relation betweenXr,pt(ℝn) and the Morrey spaces Mr,pt(ℝn), t≥0, and r>0 by wavelets. See also Liang et al. [2, 3], Triebel [4], and Yuan et al. [5] for further information on wavelet characterization of Morrey spaces. The special case t=0 has been studied by Yang [6] and Yang and Zhou [7]. We point out that the result for t>0 is not a simple generalization. For t=0, the Sobolev space H0,p(ℝn) becomes Lebesgue space Lp(ℝn). It is well known that
(4)∥f∥Lp~∥M(f)∥Lp,1<p<∞,
where M is the Hardy-Littlewood maximal operator. Yang and Zhou [7] used this equivalence to characterize Xr,p0(ℝn). See also Yang [6]. However, (4) does not hold for Ht,p(ℝn),t>0.For the caset>0, it is necessary to make some progress in technique. The difficulty is to deal with the impact of the maximal operator M on the frequencies when we split a product of two functions. To overcome this difficulty, we introduce an almost local operator Tt. See Definition 12. Let r>0, t≥0 with t+r<1<p<n/(r+t). In Theorem 21, we characterize Xr,pt(ℝn) by Meyer wavelets without using capacity. Also our method can be applied to study the relation between multiplier spaces and Morrey spaces.Lemma2 implies that Xr,pt(ℝn)⊂Mr,pt(ℝn). For the converse imbedding, Fefferman [8] established the following relation:
(5)Mr,qt(ℝn)⊂Xr,pt(ℝn),q>p>1.Letf∈Mr,pt(ℝn) and g∈Ht+r,p(ℝn). From the counterexample in Theorem 33, we can see that the product fg may produce a logarithmic type blowup on the fractal sets with Hausdorff dimension n-p(r+t). To eliminate this defect, we introduce a logarithmic type Morrey space Mr,pt,τ(ℝn) and prove that, for τ>1/p′,
(6)Mr,pt,τ(ℝn)⊂Xr,pt(ℝn),
where r>0, t≥0, and 1<p<n/(r+t). See Section 4.1. In (6), the scope of τ is (1/p′,∞), where 1/p+1/p′=1. In Section 4.2, our counterexample implies that, for 0<τ≤1/p′, there exists some function f∈Mr,pt,τ(ℝn), but f∉Xr,pt(ℝn). See Section 4.2 for the details. Theorems 24 and 33 illustrate the difference between Morrey spaces and multiplier spaces.Remark 3.
For the caset=0, some similar counterexamples have been obtained. Lemarié-Rieusset [9] gave a counterexample to show that Xr,20(ℝn)≠Mr,20(ℝn), where n-2r is an integer. Recently, Lemarié-Rieusset [10] and Yang and Zhou [7] constructed some counterexamples for t=0 and 1<p<n/r. The counterexamples there are independent of the wavelet characterization of Xr,pt(ℝn). Our counterexample depends on the wavelet characterization of multiplier space, Theorem 24, and fractal skills. See Theorem 33.As an application, we apply our results obtained in Section4 to the Schrödinger operator L=I-Δ+V, where V is the potential function. Maz'ya and Verbitsky [11] considered the multipliers from H1,2(ℝn) to H-1,2(ℝn). For a Schrödinger operator L=I-Δ+V, they got many sufficient and necessary conditions such that V is a multiplier from H1,2(ℝn) to H-1,2(ℝn). For more information, we refer the reader to Jiang et al. [12], Lemarié-Rieusset [9], Maz'ya and Shaposhnikova [1], Maz'ya and Verbitsky [11, 13], Yang and Yang [14], Yang et al. [15–17], Yang and Zhou [7], and the references therein.Givenr>0, t≥0, 1<p<n/(r+t), and τ>1/p′, we consider the following equation:
(7)(I+(-Δ)r/2+V)f=g,
where g∈Ht,p(ℝn) and V∈Mr,pt,τ(ℝn). If V is a function of Hölder class, one usual method to deal with (7) is the boundedness of Calderón-Zygmund operators. As a function in Mr,pt,τ(ℝn), V may not be a L∞ function. In Section 5, by Theorem 32, we prove that if V∈Mr,pt,τ(ℝn), (7) has an unique solution in the Sobolev space Ht+r,p(ℝn).The rest of this paper is organized as follows. In Section2, we state some notations and known results which will be used throughout this paper. In Section 3, we give a wavelet characterization of Xr,pt(ℝn). In Section 4, we introduce a class of logarithmic Morrey spaces Mr,pt,τ(ℝn) such that Mr,pt,τ(ℝn)⊂Xr,pt(ℝn). Further, we construct a counterexample to prove the sharpness of the scope of the index τ. In the last section, we consider an application to PDE problem.Notations. U≈V represents that there is a constant c>0 such that c-1V≤U≤cV whose right inequality is also written as U≲V. Similarly, if V≥cU, we denote V≳U.


## 2. Some Preliminaries

In this section, we state some notations, knowledge, and preliminary lemmas which will be used in the sequel. Firstly, we recall some background knowledge of wavelets and multiresolution analysis.For anyj∈ℕ and k=(k1,k2,…,kn)∈ℤn, let Qj,k=∏s=1n[2-jks,2-j(ks+1)] and denote by Ω the set of all dyadic cubes Qj,k. For arbitrary set Q, we denote by Q~ the 2M+2-multiple of Q. Finally, let χ, χ~ be the characteristic functions of the unit cube Q0 and Q~0, respectively.We will adopt real-valued tensor product wavelets to study the multiplier spaces in this paper. Let{Vj1,j∈ℤ} be an orthogonal multiresolution in L2(ℝ) with the scaling function Φ0(x). Denote by Wj1 the orthogonal complement space of Vj1 in Vj+11; that is, Wj1=Vj+11⊖Vj1. Let {Φ1(x-k),k∈ℤ} be an orthogonal basis in W01. For ɛ=(ɛ1,…,ɛn)∈{0,1}n, denote Φɛ(x)=∏i=1nΦɛi(xi).In the proof, we use only Meyer wavelets and regular Daubechies wavelets. We say a Daubechies wavelet is regular if it has sufficient vanishing moment until orderm and Φɛ(x)∈C0m([-2M,2M]n), where the regularity exponent m is large enough and M is determined by m; see [18, 19] for more details. For any ɛ∈{0,1}n, j∈ℕ, and k∈ℤn, we denote Φj,kɛ(x)=2jn/2Φɛ(2jx-k). For j∈ℤ and k∈ℤn, let χ(2jx-k) and χ~(2jx-k) be the characteristic functions on Qj,k and Q~j,k, respectively. For simplicity, we denote by χj,k and χ~j,k for short.In addition we define(8)Λn={(ɛ,j,k):ɛ∈{0,1}n,j∈ℕ,k∈ℤn,ɛ≠0,ifj>0}.
For fixed tempered distribution f, if we use wavelets which are sufficiently regular, then we can define fj,kɛ=〈f,Φj,kɛ〉. And the wavelet representation f=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ holds in the sense of distribution.Let{Vj,j∈ℤn} be the orthogonal multiresolution in L2(ℝn) with the scaling function Φ0(x). Denote by Wj the orthogonal complement space of Vj in Vj+1; that is, Wj=Vj+1⊖Vj. Denote by Pj and Qj the projection operators from L2(ℝn) to Vj and Wj, respectively. Dobynski got a decomposition of the product of two functions f and g, which is similar to Bony's paraproduct (see [20]). Denote
(9)Λ~n={(ε,ε′,j,k,k′),ε,ε′∈{0,1}n∖{0},k2j≥0,k,k′∈ℤn,(ε,k)≠(ε′,k′)}.
By the projection operators Pj and Qj, we divide the product fg into the following terms:
(10)f(x)g(x)=P0(f)P0(g)+∑j≥0‍Pj(f)Qj(g)+∑j≥0‍Qj(f)Pj(g)+∑Λ~n‍fj,kɛgj,k′ɛ′Φj,kɛ(x)Φj,k′ɛ′(x)+∑Λn,ɛ≠0‍fj,kɛgj,kɛ(Φj,kɛ(x))2.
To facilitate our use, we make a modification to (10) and use special wavelets for different cases. Let N be a positive integer. We decompose the product fg as
(11)fg=∑j=1∞‍[Pj+1(f)Pj+1(g)-Pj(f)Pj(g)]+P0(f)P0(g)=∑j=N∞‍[Qj(f)Qj(g)+Pj(f)Qj(g)+Qj(f)Pj(g)]+PN(f)PN(g),
and the term ∑j=N∞Qj(f)Pj(g) can be decomposed as
(12)∑j=N∞‍Qj(f)Pj(g)=∑j=N∞‍Qj(f)∑t=1N‍Qj-t(g)+∑j=N∞‍Qj(f)Pj-N(g)=∑j=0∞‍∑t=1N‍Qj+t(f)Qj(g)+∑j=0∞‍Qj+N(f)Pj(g).In 1970s, Triebel introduced Triebel-Lizorkin spacesFpr,q(ℝn) ([21]). Many function spaces can be seen as the special cases for Fpr,q(ℝn). For example, F1r,2(ℝn) is the fractional Hardy space. For 1<p<∞, Fpr,2(ℝn) are the Sobolev spaces Hr,p(ℝn). For p=∞, F∞-r,2(ℝn) is the space BMOr(ℝn) defined as
(13)BMOr(ℝn):=(I-Δ)-r/2BMOloc(ℝn),
where I and Δ denote the unit operator and the Laplace operator, respectively. Here BMOloc(ℝn) denotes the set of all measurable functions f with
(14)sup1/2≤l(Q)≤1|fQ|+sup0<l(Q)<1/2|Q|-1∫Q‍|f(x)-fQ|dx<∞,
where fQ denotes the mean value of f on Q. See also Section 3.1 of [1].For1≤p<∞ and r∈ℝ, it is well known that (Fpr,2(ℝn))′=Fp′-r,2(ℝn). The following lemma gives a characterization of Fpr,2(ℝn) via Meyer wavelets and regular Daubechies wavelets. For the proof, we refer the reader to Chapters 5 and 6 of Meyer [18]. See also Yuan et al. [22] and Yang [23].Lemma 4.
Given|r|<m, let {Φj,kɛ} be Meyer wavelets or m-regular Daubechies wavelets. Then,(i)
for1≤p<∞,
(15)g(x)=∑(ɛ,j,k)∈Λn‍gj,kɛΦj,kɛ(x)∈Fpr,2(ℝn)⟺∥(∑(ɛ,j,k)∈Λn‍22j(r+n/2)|gj,kɛ|2χj,k(·))1/2∥Lp<∞⟺∥(∑(ɛ,j,k)∈Λn‍22j(r+n/2)|gj,kɛ|2χ~j,k(·))1/2∥Lp<∞,(ii)
g(x)=∑(ɛ,j,k)∈Λngj,kɛΦj,kɛ(x)∈F∞r,2(ℝn) if and only if there exists p∈(1,∞) such that for Q∈Ω,
(16)∥(∑ɛ∈En,Qj,k⊂Q‍22j(r+n/2)|gj,kɛ|2χj,k(·))1/2∥Lp≤C|Q|1/p.The wavelet characterizations of function spaces have been studied by many authors. In [18, Chapters 5 and 6], Meyer established wavelet characterizations for many function spaces, for example, Hardy space, BMO spaces, Besov spaces, and Bloch space. Yang et al. [24] used wavelets to characterize Lorentz type Triebel-Lizorkin spaces and Lorentz type Besov spaces. For the wavelet characterization of Besov type Morrey spaces and Triebel-Lizorkin type Morrey spaces, We refer to Yang and Yuan [25, 26], Yang [23], and Yuan et al. [22].Morrey spaces were introduced by Morrey in 1938 and played an important role in the research of partial differential equations. Xiao [27] established a relation between the homogeneous Morrey space ℒ2,n-2α(ℝn) and Q-type space Qα(ℝn) by heat semigroup and the fractional integrals. In the recent 20 years, Q-type spaces are studied extensively. See Essèn et al. [28], Dafni and Xiao [29, 30], Peng and Yang [31], Wu and Xie [32], Yang [23], and Yuan et al. [22].Defineft,Q as
(17)ft,Q=1|Q|∫Q‍(I-Δ)t/2f(x)dx.
The Morrey spaces Mr,pt(ℝn) are defined as follows.Definition 5.
Given1≤p<∞ and r,t≥0, the Morrey space Mr,pt(ℝn) is defined as the set of all measurable functions f such that sup|Q|=1|ft,Q|≤C and (18)∫Q‍|(I-∆)t/2f(x)-ft,Q|pdx≤C|Q|1-p(r+t)/n,
where Q is any cube in ℝn with |Q|≤1.Morrey spacesMr,pt(ℝn) can be also characterized by wavelets. We state it as the following theorem and refer to Yuan et al. [22] for the proof.Theorem 6.
Givent∈ℝ, 1<p<∞ and 0≤p(r+t)<n,
(19)f(x)=∑(ɛ,j,k)∈Λn‍fj,kɛΦj,kɛ(x)∈Mr,pt(ℝn)
if and only if for any Q∈Ω with |Q|≤1(20)∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≤C|Q|1-p(r+t)/n.The following lemma can be obtained by Lemmas2 and 4 immediately.Lemma 7.
Givenr>0, t≥0 and 1<p<n/(r+t), then Xr,pt(ℝn)⊂Mr,pt(ℝn).Now we give two lemmas aboutBMOr(ℝn).Lemma 8.
Givenr>0, t≥0 and 1<p<n/(r+t), then Mr,pt(ℝn)⊂BMOr(ℝn).Proof.
For any dyadic cubeQ, we have
(21)∫(∑ɛ∈En,Qj,k⊂Q‍2jn-2jr|fj,kɛ|2χj,k(x))p/2dx≲|Q|p(r+t)/n∫‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲|Q|.For anyɛ∈{0,1}n, j∈ℕ and k∈ℤn, we denote fj,kɛ=〈f,Φj,kɛ〉. We can get the following result.Lemma 9.
Suppose thatr>0 and f∈BMOr(ℝn). The wavelet coefficients of f satisfy
(22)|fj,kɛ|≲2(r-n/2)j,∀ɛ∈1{0,1}n,j∈ℕ,k∈ℤn.Proof.
Takej∈ℕ and k∈ℤn. We consider two cases ɛ∈En and ɛ=0 separately.(i)
Forɛ∈En, by Lemma 4, we get
(23)∫‍(2jn-2jr|fj,kɛ|2χj,k(x))p/2dx≲2-jn.It is easy to see that|fj,kɛ|≲2j(r-n/2).(ii)
Forɛ=0,
(24)fj,k0=〈∑(ɛ′,j′,k′)∈Λn‍fj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉=〈∑(ɛ′,j′,k′)∈Λn:j′<j‍fj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉.Because|∑(ɛ′,j′,k′)∈Λn:j′<jfj′,k′ɛ′Φj′,k′ɛ′(x)|≲2rj, we have
(25)|fj,k0|≲〈2rj,|Φj,k0(x)|〉≲2j(r-n/2).LetΨ1 and Ψ2 be two functions such that
(26)Ψi∈C0μ([-2M+1,2M+1]n),i=1,2,∫xαΨi(x)dx=0,where|α|≤μ,i=1,2.
Write aj,k,j′,k′=〈Ψj,k1,Ψj′,k′2〉. The following lemma can be found in Chapter 8 of Meyer [18] or Chapter 6 of Yang [23].Lemma 10 (see [18, Chapter 8, Lemma 1]).
Let|μ|≤m. For |s|<μ, the coefficients aj,k,j′,k′ satisfy the following condition:
(27)|aj,k,j′,k′|≲2-|j-j′|(n/2+s)(2-j+2-j′2-j+2-j′+|k2-j-k′2-j′|)n+s.By Lemma4, the boundedness of Calderón-Zygmund operators on Hr,p(ℝn) is equivalent to the following lemma. We refer the reader to [18, 23, 33] for the proof.Lemma 11.
Suppose thats>|r| and g(x)=∑(ɛ,j,k)∈Λngj,kɛΦj,kɛ(x)∈Hr,p(ℝn). Let g~j,kɛ=∑(ɛ,j,k)∈Λnaj,k,j′,k′ɛ,ɛ′gj′,k′ɛ. If the coefficients aj,k,j′,k′ɛ,ɛ′ satisfy (27), then
(28)∫(∑(ɛ,j,k)∈Λn‍2j(n+2r)|g~j,kɛ|2χj,k(x))p/2dx≲∫(∑(ɛ,j,k)∈Λn‍2j(n+2r)|gj,kɛ|2χj,k(x))p/2dx.We say thatT is a local operator if there exists some constant C>1 such that, for all x∈ℝn and r>0, T maps a distribution with the support B(x,r) to another distribution supported on the ball B(x,Cr). If t/2 is not a nonnegative integer, the operator (I-Δ)t/2 is not a local operator. Now we use wavelets to construct some special fractional differential operators Tt, which are almost local operators and will be used in the proof of our main result.Definition 12.
Fort≥0 and h(x)=∑(ɛ,j,k)∈Λnhj,kɛΦj,kɛ(x), we call that Tt is an operator associated to the kernel Kt(x,y)=∑(ɛ,j,k)∈Λn2-jtΦj,kɛ(x)Φj,kɛ(y) if
(29)Tth(x)=∑(ɛ,j,k)∈Λn‍2-jthj,kɛΦj,kɛ(x).It is easy to prove thatT0 is the identity operator and ∥Tth∥Lp=∥h∥H-t,p for 1<p<∞. Furthermore, we have the following.Lemma 13.
Supposet≥0. For any Qj,k∈Ω and x∈Qj,k,
(30)2j(n/2-t)|hj,k0|≤CMTth(x),
where M is the Hardy-Littlewood maximal operator.Proof.
Ift=0, the proof was given by Meyer [18]. Now we consider the case t>0. It is easy to verify that
(31)|T-tΦ0(x)|=|∫‍K-t(x,y)Φ0(y)dy|≤C(1+|x|)-n-t.
By the fact that t>0, we have
(32)2j(n/2-t)hj,k0=2j(n/2-t)〈Tth,T-tΦj,k0〉=2jn/2〈Tth,(T-tΦ0)j,k〉.
Hence, we can get
(33)|2j(n/2-t)hj,k0|=2jn/2|〈Tth,(T-tΦ0)j,k〉|≲2jn/2∫‍|Tth(x)|2jn/2dx(1+|2jx-k|)n+t≲2jn(∑l=1∞‍∫|2jx-k|≤1‍|Tth(x)|dx+∑l=1∞‍∫2l-1≤|2jx-k|≤2l‍|Tth(x)|dx(1+|2jx-k|)n+t)≲2jn(2-jnM(Tth)(x)+∑l=1∞‍2-ltM(Tth)(x)2-jn)≲M(Tt)h(x).
This completes the proof of Lemma 13.In the rest of this section, we give a decomposition of Sobolev spaces associated with combination atoms. For|r|<m and g(x)=∑(ɛ,j,k)∈Λngj,kɛΦj,kɛ(x), denote
(34)Srg(x)=(∑(ɛ,j,k)∈Λn‍2j(2r+n)|gj,kɛ|2χj,k(x))1/2,
and, for r=0, denote also Sg(x)=S0g(x).Definition 14.
Letr∈ℝ and λ>0. For arbitrary measurable set E with finite measure, we say that g(x) is a (r,λ,E)-combination atom if supp(Srg)⊂E and Srg(x)≤λ. If E is a dyadic cube, then we say g(x) is a (r,λ,E)-atom.In the next theorem, we give a combination atom decomposition of Sobolev spaces. This result is a generalization of that of Yang [34].Theorem 15.
If1<p<∞, |r|<m and ∥g∥Hr,p≤1, there exists a series of (r,2v,Ev)-combination atoms gv(x) such that ∑v∈ℕ2pv|Ev|≤C.Proof.
Denote(35)S~rg(x)=(∑(ɛ,j,k)∈Λn‍2j(2r+n)|gj,kɛ|2χ~j,k(x))1/2.
For v≥1, let Ev={x:S~rg(x)>2v}. By wavelet characterization of Sobolev spaces, we have ∑v∈ℕ2pv|Ev|≲1. Let Ev=⋃lQv,l, where Qv,l are disjoint maximal dyadic cubes with |Qv,l|≤1. Let 𝔉v,l be the set of dyadic cubes contained in Qv,l but not in Ev+1, 𝔉v=⋃l𝔉v,l and 𝔉0=Ω∖⋃v≥1𝔉v. Let E0={x∈Q,Q∈𝔉0}, and we can write also E0=⋃lQ0,l, where Q0,l are disjoint maximal dyadic cubes in Ω. The related set 𝔉0,l is defined as 𝔉0,l={Q⊂Q0,landQ∈𝔉0}.
For anyv≥0, we write
(36)gv,l(x)=∑Qj,k∈𝔉v,l‍gj,kɛΦj,kɛ(x),gv(x)=∑Qj,k∈𝔉v‍gj,kɛΦj,kɛ(x).
Then gv(x) is a desired combination atom. This completes the proof.


## 3. Wavelet Characterization of the Multiplier Spaces

In this section, we use Meyer wavelets to characterizeXr,pt(ℝn). Let Φ be a function in C0∞(B(0,1)) satisfying Φ(x)≥0 and ∫‍Φ(x)dx=1. For any g∈Ht,p(ℝn), define (37)gj,kΦ=〈g,2jnΦ(2j·-k)〉.
The function space Sr,pΦ,t(ℝn) is defined as follows.Definition 16.
Givenr>0, t≥0 and r+t<1<p<n/(r+t), we say f∈Sr,pΦ,t(ℝn) if f(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x) and
(38)∫‍(∑(ɛ,j,k)∈Λn‍2j(n+2t)|gj,kΦ|2|fj,kɛ|2χj,k(x))p/2dx≲∥g∥Hr+t,p,
where g∈Hr+t,p(ℝn) and ∥g∥Hr+t,p≤1.Now we give a wavelet characterization ofXr,pt(ℝn). Let Φ0 and Φɛ,ɛ∈En, be the scaling function and wavelet functions, respectively. For (ɛ,j,k), (ɛ′,j′,k′), (ɛ′′,j′,k′)∈Λn and l∈ℤn, let
(39)aj,k,l,j′,k′ɛ,ɛ′=〈Φj,k+l0Φj,kɛ,Φj′,k′ɛ′〉,aj,k,0,j′,k′ɛ,ɛ,ɛ′′,0=〈(Φj,kɛ)2-2jnΦ(2j·-k),Φj′,k′ɛ′′〉.
Furthermore, for 0≤s≤N, ɛ′∈En, l∈ℤn, and s+|ɛ-ɛ′|+|l|≠0, let
(40)aj,k,l,j′,k′ɛ,ɛ′,ɛ′′,s=〈Φj,kɛ′Φj+s,2sk+lɛ,Φj′,k′ɛ′′〉.
By the same method of [18, Chapter 8, Lemma 1], we could prove the following lemma.Lemma 17.
There exist sufficient big integersN, N1, and N2 such that min{N,N1,N2}>8n+8m and the following estimates hold. (i)
If(ɛ,j,k), (ɛ′,j′,k′)∈Λn, l∈ℤn, and j≥j′, then
(41)|aj,k,l,j′,k′ɛ,ɛ′|≲(1+|l|)-N12nj′/2+j′-j(1+|k′-2j′-jk|)-N2.(ii)
If(ɛ,j,k), (ɛ′′,j′,k′)∈Λn, 0≤s≤N, ɛ′∈En, l∈ℤn, and j≥j′, then
(42)|aj,k,l,j′,k′ɛ,ɛ′,ɛ′′,s|≲(1+|l|)-N12nj′/2+j′-j(1+|k′-2j′-jk|)-N2.(iii)
If(ɛ,j,k), (ɛ′,j′,k′)∈Λn, l∈ℤn, and j<j′, then
(43)|aj,k,l,j′,k′ɛ,ɛ′|≲(1+|l|)-N12-nj′/2+nj+N(j-j′)(1+|k-2j-j′k′|)-N2.(iv)
If(ɛ,j,k), (ɛ′′,j′,k′)∈Λn, 0≤s≤N, ɛ′∈En, l∈ℤn, and j<j′, then
(44)|aj,k,l,j′,k′ɛ,ɛ′,ɛ′′,s|≲(1+|l|)-N12-nj′/2+nj+N(j-j′)(1+|k-2j-j′k′|)-N2.


### 3.1. Decomposition of Products via Multilinear Analysis

At first, we give a wavelet decomposition of the product offh. Let Φ0 and {Φε,ε∈En} be the scaling function and wavelet functions of Meyer wavelets, respectively. There exists an integer N≥3 such that
(45)∫xαΦ0(x)Φε(2Nx-k)dx=0∫1foranyk∈ℤn,α∈ℕn,ε∈En.
Denote
(46)Λɛ,n={(s,ε′,l),0≤s≤N,ε′∈En,l∈ℤn,|l|≤2(M+2+s)n,andifs=0,(0,ɛ,0)≠(0,ɛ′,0)},ℤNn={l=(l1,…,ln)∈ℤn,0≤li≤2N-1,i=1,…,n}.
For ɛ∈En, l∈ℤn, |l|≤2(M+2)n, and (ɛ′,l)∈Λɛ,n, we denote
(47)T1,ɛ(x)=∑j∈ℕ,k∈ℤn‍∑l∈ℤn‍fj,k+l0hj,kεΦj,k+l0(x)Φj,kε(x),T2,0,ɛ,ɛ(x)=∑j∈ℤ,k∈ℤn‍∑l∈ℤn,l≠0‍fj,k+lɛhj,kɛΦj,k+lε(x)Φj,kε(x)+∑j∈ℕ,k∈ℤn‍fj,kεhj,kε[(Φj,kε(x))2-2jnΦ(2jx-k)],T3,ɛ(x)=∑j∈ℕ,k∈ℤn‍∑l∈ℤNn‍∑l′∈ℤn‍fj+N,2Nk+lεhj,k+l′0×Φj,k+l′0(x)Φj+N,2Nk+lε(x),T4,ɛ(x)=∑j∈ℕ,k∈ℤn‍fj,kεhj,kε2jnΦ(2jx-k),T5(x)=∑k∈ℤn‍∑l∈ℤn‍f0,k0h0,k+l0Φj,k0(x)Φ0,k+l0(x).
For ɛ,ɛ′∈En, 0<s≤N or s=0, and ε≠ε′, denote
(48)T2,s,ε,ε′(x)=∑j∈ℕ,k∈ℤn‍∑l∈ℤn‍fj+s,2sk+lε′hj,kεΦj+s,2sk+lε′(x)Φj,kε(x).
By (11) and (12), we can decompose fh as follows:
(49)f(x)h(x)=∑ɛ∈En‍T1,ε(x)+∑0≤s≤N,ɛ,ɛ′∈En‍T2,s,ε,ε′(x)+∑ɛ∈En‍T3,ε(x)+∑ɛ∈En‍T4,ε(x)+T5(x):=∑i=15‍Ti(x).
If g∈Ht+r,p(ℝn), write g(x)=∑(ε,j,k)∈Λngj,kεΦj,kε(x). For ε,ε′∈En, and 0≤s≤N, we define
(50)T1,ε=∫‍T1,ε(x)g(x)dx,T2,s,ε,ε′=∫‍T2,s,ε,ε′(x)g(x)dx.
Let
(51)T1,1,ε,l=∑j∈ℕ,k∈ℤn‍∑j≥j′≥0,ε′,k′∈ℤn‍|fj,k+l0||hj,kε||aj,k,l,j′,k′ε,ε′||gj′,k′ε′|,T1,2,ε,l=∑j∈ℕ,k∈ℤn‍∑j<j′,ε′,k′∈ℤn‍|fj,k+l0||hj,kε||aj,k,l,j′,k′ε,ε′||gj′,k′ε′|,T2,1,s,ε,ε′,l=∑j∈ℕ,k∈ℤn‍∑j≥j′>0,ɛ′′,k′∈ℤn‍|fj+s,2sk+lε′||hj,kε||aj,k,l,j′,k′ε,ε′,ε′′,s||gj′,k′ε′′|,T2,2,s,ε,ε′,l=∑j∈ℕ,k∈ℤn‍∑j<j′,ɛ′′,k‍|fj+s,2sk+lε′||hj,kε||aj,k,l,j′,k′∈ℤnε,ε′,ε′′,s||gj′,k′ε′′|.
It is easy to see that
(52)|T1,ε|≲∑l∈ℤn‍(T1,1,ε,l+T1,2,ε,l),|T2,s,ε,ε′|≲∑l∈ℤn‍(T2,1,s,ε,ε′,l+T2,2,s,ε,ε′,l).


### 3.2. Several Technical Lemmas

Now we estimate the quantitiesT1,1,ɛ,l, T1,2,ɛ,l, T2,1,s,ɛ,ɛ′,l, and T2,2,s,ɛ,ɛ′,l separately. Let
(53)Stgj′(x)=∑ɛ′,k′‍2j′(n/2+t)|gj′,k′ɛ′|χ(2j′x-k′).
For fixed x, there is only one k′ such that χ(2j′x-k′)≠0 and the number of ɛ′ is finite. Then
(54)Stgj′(x)=(∑ɛ′,k′‍22j′(n/2+t)|gj′,k′ɛ′|2χ(2j′x-k′))1/2.LetM be the Hardy-Littlewood maximal function. Then, if j<j′ and x∈Qj,k, we have
(55)∑k′‍(1+|k-2j-j′k′|)-N22j′(-n/2+t)gj′,k′ɛ′≲2-jnMStgj′(x).Lemma 18.
Givenr>0, t≥0, and t+r<1<p<n/(t+r). If f∈BMOr(ℝn), then (i)
forg∈Ht+r,p and h∈H-t,p′,
(56)|T1,ɛ|+|T2,s,ɛ,ɛ′|≲∥g∥Ht+r,p∥h∥H-t,p′,(ii)
forg∈Ht+r,p and h∈H-t,p′,
(57)|〈T5,g〉|≲∥g∥Ht+r,p∥h∥H-t,p′.Proof.
(1) Forj′≤j, we have
(58)T1,1,ɛ,l=∑j,k‍∑j≥j′,ɛ′,k′‍|fj,k+l0||hj,kɛ||aj,k,l,j′,k′ɛ,ɛ′||gj′,k′ɛ′|.
By Lemma17, we know
(59)|aj,k,l,j′,k′ɛ,ɛ′|≲(1+|l|)-N1(1+|k′-2j′-jk|)-N22nj′/2+j′-j.
By Lemma9, f∈BMOr(ℝn) implies |fj,kɛ|≤C2(r-n/2)j. Now we can get
(60)∑l∈ℤn‍T1,1,ɛ,l≲∑l∈ℤn‍∑k,j≥j′‍∫‍23nj′/2+(j′-j)+(r-n/2)j∑k′‍(∑ɛ′‍|gj′,k′ɛ′|2)1/2×|hj,kɛ|χ(2j′x-k′)(1+|l|)N1(1+|k′-2j′-jk|)N2dx≲∑j≥j′‍∫‍23nj′/2+(j′-j)+(r-n/2)j×∑k′‍(∑ɛ′‍22j′(n/2+t+r)|gj′,k′ɛ′|2)1/22-j′(t+r+n/2)×∑k‍|hj,kɛ|χ(2j′x-k′)(1+|k′-2j′-jk|)N22j(-n/2-t)2j(n/2+t)dx≲∫∑j≥j′‍23nj′/2-j′(t+r+n/2)+(j′-j)+(r-n/2)j+j(n/2+t)×2-j′nSt+rgj′(x)M(S-thj)(x)dx.
Because0<t+r<1,
(61)∑l∈ℤn‍T1,1,ɛ,l≲∑j≥j′‍2(j′-j)+(r+t)j-j′(r+t)×∫St+rgj′(x)M(S-thj)(x)dx≲∑j≥j′‍2(j′-j)(1-r-t)∥g∥Ht+r,p∥h∥H-t,p′≲∥g∥Ht+r,p∥h∥H-t,p′.
Now we estimate the term(62)T2,1,s,ɛ,ɛ′,l(x)=∑j,k‍∑j≥j′>0‍|fj,kɛ′||hj+s,2sk+lɛ||aj,k,l,j′,k′ɛ,ɛ′,ɛ′′,s||gj′,k′ɛ′′|.
Becausef∈BMOr(ℝn), by Lemma 9, we have |fj,k+lɛ′|≲2(r-n/2)j. By Lemma 17,
(63)|aj,k,l,j′,k′ɛ,ɛ′,ɛ′′,s|≲(1+|l|)-N1(1+|k′-2j′-jk|)-N22nj′/2+(j′-j).
We can get, similarly,(64)∑l∈ℤNn‍T2,1,s,ɛ,ɛ′,l(x)≲∑l∈ℤNn‍∑j,k‍∑j≥j′>0‍2(r-n/2)j2nj′/2+j′-j∑j,k‍∑j≥j′>0‍×|hj+s,2sk+lε||gj′,k′ε′′|(1+|l|)N1(1+|k′-2j′-jk|)N2≲∑l∈ℤNn∑j≥j′>0∫23nj′/2+(j′-j)+(r-n/2)j∑j,k‍∑j≥j′>0‍×∑ε′′,k′‍|gj′,k′ε′||hj+s,2sk+lε|χ(2j′x-k′)(1+|l|)N1(1+|k′-2j′-jk|)N2dx≲∑j≥j′>0‍2(j′-j)(1-t-r)∫‍St+r(gj′)(x)MS-thj+s(x)dx≲∥St+rgj′∥Lp∥MS-thj+s∥Lp′≲∥g∥Ht+r,p∥h∥H-t,p′.
(2) Ifj′>j, the estimates of T1,2,ɛ,l and T2,2,s,ɛ,ɛ′,l are easier than those of T1,1,ɛ,l and T2,1,s,ɛ,ɛ′,l. For example, we estimate the term
(65)T1,2,ɛ,l=∑j,k‍∑0≤j<j′,ɛ′,k′‍|fj,k+l0||hj,kɛ||aj,k,l,j′,k′ɛ,ɛ′||gj′,k′ɛ′|.
Becausef∈BMOr(ℝn), by Lemma 9, |fj,k+l0|≲2(r-n/2)j. By Lemma 17,
(66)|aj,k,l,j′,k′ɛ,ɛ′|≲(1+|l|)-N1(1+|k-2j-j′k′|)-N22-nj′/2+nj+N(j-j′).
Because2nj∫‍χj,k(x)dx=1, we can obtain
(67)∑l∈ℤn‍T1,2,ɛ,l≲∑l∈ℤn‍∑j,k‍∑0≤j<j′,ɛ′,k′‍∫‍2(r-n/2)j2-nj′/2+nj+N(j-j′)2nj×|hj,kε||gj′,k′ε′|χj,k(x)(1+|l|)N1(1+|k-2j-j′k′|)N2dx≲∑ε,j,k‍∫2(r-n/2)2-nj′/2+nj+N(j-j′)2nj2-nj2-j′(-n/2+t+r)×MSt+r(gj′)(x)|hj,kε|χj,k(x)dx≲∑0≤j<j′∫2(r-n/2)j2-(nj′/2)+nj+N(j-j′)2-j(n/2-t)×2-j′(-n/2+t+r)MSt+r(gj′)(x)S-t(hj)(x)dx≲∥g∥Ht+r,p∥h∥H-t,p′.
The estimate forT2,2,s,ɛ,ɛ′,l(x) can be obtained similarly. By the same methods used in (1) and (2), we can get the estimate of the term T5. We omit the details.Now we consider the termT3,ɛ. We have the following claim.Lemma 19.
Givenr>0, t≥0, and t+r<1<p<n/(t+r), if f∈BMOr(ℝn), then
(68)|〈T3,ɛ,g〉|≲∥g∥Ht+r,p∥h∥H-t,p′.Proof.
In fact, forl∈ℤNn, (ɛ,j,k)∈Λn and l′∈ℤn, let
(69)gj+N,2Nk+lɛ,l′=2-nj/2〈Φj,k+l′0Φj+N,2Nk+lɛ,g〉.
We have
(70)|〈T3,ɛ,g〉|=|∑(ɛ,j,k)∈Λn‍∑l∈ℤNn,l′∈ℤn‍2jn/2hj,k+l′0gj+N,2Nk+lɛ,l′fj+N,2Nk+lɛ|≲∑l∈ℤNn‍∑(ɛ,j,k)∈Λn‍(∑l′∈ℤn‍2jn/2|hj,k+l′0||gj+N,2Nk+lɛ,l′|)×|fj+N,2Nk+lɛ|.
Because (F1r,2(ℝn))′=BMOr(ℝn), by Lemmas 9 and 13, we have
(71)|〈T3,ɛ,g〉|≲∑l∈ℤNn‍∥f∥BMOr∫(∑(ɛ,j,k)∈Λn‍2j(n+2r+t)≲∑l∈ℤNn‍∥f∥BMOr×(∑l′∈ℤn‍2j(n/2-t)|hj,k+l′0||gj+N,2Nk+lɛ,l′|)2≲∑l∈ℤNn‍∥f∥BMOr×χj,k(x)∑(ɛ,j,k)∈Λn‍)1/2dx≲∥f∥BMOr∫(∑(ɛ,j,k)∈Λn‍2j(n+2r+2t)≲∥f∥BMOrgi×(∑l′∈ℤn‍(1+|l′|)N+n|gj+N,2Nk+lɛ,l′|)2≲∥f∥BMOrgi×χj,k(x)∑(ɛ,j,k)∈Λn‍)1/2MTth(x)dx.
Because
(72)|gj+N,2Nk+lɛ,l′|=2-nj/2|〈Φj,k+l′0Φj+N,2Nk+l′ɛ∑(ɛ′,j′,k′)∈Λn‍gj′,k′ɛ′Φj′,k′ɛ′〉|=2-nj/2|〈∑(ɛ′,j′,k′)‍Φj,k+l′0Φj+N,2Nk+l′ɛ=552-nji×∑(ɛ′,j′,k′)∈Λn,|j′-j-N|≤2‍gj′,k′ɛ′Φj′,k′ɛ′〉|,
we can get
(73)(∑(ɛ,j,k)∈Λn‍2(n+2t+2t)jhi×(∑l′∈ℤn‍(1+|l′|)N+n|gj+N,2Nk+lɛ,l′|)2χj,k(x))1/2≲∑|j′-j-N|≤2‍MSt+rgj′(x).
Hence, we have
(74)∫((∑l′∈ℤn‍2j(n/2-t)|hj,k+l′0∥gj+N,2Nk+lɛ,l′|)2∑(ɛ,j,k)∈Λn‍2j(n+2r+t)×(∑l′∈ℤn‍2j(n/2-t)|hj,k+l′0∥gj+N,2Nk+lɛ,l′|)2χj,k(x))p/2dx≲∫(∑j≥0‍(∑|j′-j-N|≤2‍MSt+rgj′(x))2)p/2dx≲∥g∥Ht+r,p.
Finally, we obtain, by Hölder's inequality,
(75)|〈T3,ɛ,g〉|≲∥f∥BMOr∥g∥Ht+r,p∥MTth∥p′≲∥f∥BMOr∥g∥Ht+r,p∥h∥H-t,p′.
This completes the proof of Lemma 19.In order to deal with the term(76)T4,ɛ(x)=∑(ɛ,j,k)∈Λn‍fj,kɛhj,kɛ2jnΦ(2jx-k),
we need the following estimate.Lemma 20.
Givenr>0, t≥0 with t+r<1<p<n/(r+t), f∈Sr,pΦ,t(ℝn) if and only if
(77)|〈T4,ɛ,g〉|≲∥g∥Hr+t,p∥h∥H-t,p′.Proof.
In fact, forg∈Hr+t,p(ℝn),
(78)〈T4,ɛ,g〉=∑(ɛ,j,k)∈Λn‍fj,kɛgj,kΦhj,kɛ.
By Lemma 4, using Hölder's inequality, we have
(79)|〈T4,ɛ,g〉|≲∑(ɛ,j,k)∈Λn,j≥0‍2jn|gj,kΦ|2|fj,kɛ|2χj,k(x)dx≲∫(∑(ɛ,j,k)∈Λn,j≥0‍2j(n+2t)|gj,kΦ|2|fj,kɛ|2χj,k(x))1/2×(∑(ɛ,j,k)∈Λn,j≥0‍2j(n-2t)|hj,kɛ|2χj,k(x))1/2dx≲(∫(∑(ɛ,j,k)∈Λn,j≥0‍2j(n+2t)|gj,kΦ|2|fj,kɛ|2×χj,k(x)∑∈Λn,j≥0‍)p/2dx)1/p×(∫‍(∑(ɛ,j,k)∈Λn,j≥0‍2j(n-2t)|hj,kɛ|2χj,k(x))p′/2dx)1/p′≲∥f∥Sr,pΦ,t∥h∥H-t,p′.
Because f∈Sr,pΦ,t(ℝn), we can see that |〈T4,ɛ,g〉|≲∥g∥Ht+r,p∥h∥H-t,p′.
Conversely, let(80)τj,kɛ=2jtfj,kɛgj,kΦ,τ(x)=∑(ɛ,j,k)∈Λn,j≥0‍τj,kɛΦj,kɛ(x).
Denote h=|τ|p-2τ¯. For h(x)=∑(ɛ,j,k)∈Λnhj,kɛΦj,kɛ(x), we write ht(x) as
(81)ht(x)=∑(ɛ,j,k)∈Λn,j≥0‍2jthj,kɛΦj,kɛ(x):=∑(ɛ,j,k)∈Λn‍(ht)j,kɛΦj,kɛ(x).
It is easy to see that h∈Lp′(ℝn) is equivalent to ht∈H-t,p′(ℝn).
By the wavelet characterization ofHt,p(ℝn), we get
(82)∫(∑(ɛ,j,k)∈Λn,j≥0‍2j(n+2t)|gj,kΦ|2|fj,kɛ|2χj,k(x))p/2dx=∫‍(∑(ɛ,j,k)∈Λn,j≥0‍2jn|τj,kɛ|2χj,k(x))p/2dx=∫‍|τ|pdx=∫‍τhdx=∑(ɛ,j,k)∈Λn‍2j(t+n/2)fj,kɛgj,kΦhj,kɛ=∑(ɛ,j,k)∈Λn‍2jn/2fj,kɛgj,kΦ(ht)j,kɛ.
Further, we can deduce that
(83)∫(∑(ɛ,j,k)∈Λn,j≥0‍2j(n+2t)|gj,kΦ|2|fj,kɛ|2χj,k(x))p/2dx=|〈T4,ɛ,g〉|≲∥g∥Ht+r,p∥ht∥H-t,p′≲∥g∥Ht+r,p∥τ∥Lpp/p′.
Hence ∥τ∥Lp<∞ and f∈Sr,pΦ,t(ℝn). This completes the proof of this lemma.


### 3.3.  Wavelet Characterization of Sobolev Multipliers

Theorem 21.
Lett≥0, r>0 with t+r<1<p<n/(t+r). Then,
(84)Xr,pt(ℝn)=Mr,pt(ℝn)∩Sr,pΦ,t(ℝn).Proof.
We first assumef∈Mr,pt(ℝn)∩Sr,pΦ,t(ℝn). Let g∈Ht+r,p(ℝn) and h∈H-t,p′(ℝn). By the decomposition given in Section 3.1, we have
(85)f(x)h(x)=∑ɛ∈En‍T1,ɛ(x)+∑0≤s≤N,ɛ,ɛ′∈En‍T2,s,ɛ,ɛ′(x)+∑ɛ∈En‍T3,ɛ(x)+∑ɛ∈En‍T4,ɛ(x)+T5(x).
Notice that f∈Mr,pt(ℝn) implies that f∈BMOr(ℝn). By Lemmas 18–20, we obtain that, for f∈Mr,pt(ℝn)∩Sr,pΦ,t(ℝn), fh∈H-t-r,p′(ℝn) and
(86)∥fh∥H-t-r,p′≲∥f∥Mr,pt∩Sr,pΦ,t∥h∥H-t,p′,
which gives
(87)∥f∥Xr,pt=sup∥g∥Hr+t,p≤1sup∥h∥H-t,p′|〈fg,h〉|=sup∥g∥Hr+t,p≤1sup∥h∥H-t,p′|〈fh,g〉|≤sup∥g∥Hr+t,p≤1sup∥h∥H-t,p′∥fh∥H-t-r,p′∥g∥Ht+r,p≤∥f∥Mr,pt∩Sr,pΦ,t.
Hence, we can get Mr,pt(ℝn)∩Sr,pΦ,t(ℝn)⊂Xr,pt(ℝn).
Conversely, assume thatf∈Xr,pt(ℝn). By Lemma 7, we have f∈Mr,pt(ℝn)⊂BMOr(ℝn). We apply Lemmas 18 and 19 again to deduce
(88)|〈T4,ɛ,g〉|≤|T1,ɛ|+|T2,s,ɛ,ɛ′|+|〈T3,ɛ,g〉|+|〈T5,g〉|≤∥g∥Hr+t,p∥h∥H-t,p′.
Then, Lemma 20 implies that f∈Sr,pΦ,t(ℝn). This completes the proof of this theorem.Now we characterizeXr,pt(ℝn) by the fractional integration. Let Ψ(x)≥0, Ψ(x)∈C0∞(B(0,2)) with Ψ(x)=1 on B(0,1). It is well known that g~∈Ht+r,p(ℝn) if and only if there exists g∈Lp(ℝn) such that
(89)g~(x)=Jt+rg(x)=∫‍g(y)Ψ(|x-y|)|x-y|t+r-ndy.
For g∈Lp(ℝn), let
(90)gj,k=∫‍(2-j+|y-2-jk|)t+r-ng(y)dy.Definition 22.
Givenr>0, t≥0 and 1<p<n/(r+t), then f(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x)∈Qr,pt(ℝn) if and only if
(91)∫‍(∑(ɛ,j,k)∈Λn‍2j(n+2t)|gj,k|2|fj,kɛ|2χj,k(x))p/2dx≲1,
where g∈Lp(ℝn) and g(x)≥0.Letf(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x) and h(x)=∑(ɛ,j,k)∈Λnhj,kɛΦj,kɛ(x). Define
(92)TΦ(f,h)(x)=∑j∈ℕ‍∑k∈ℤn‍2jnfj,kɛhj,kɛΦ(2jx-k).
Similar to Lemma 20, we can get the following.Lemma 23.
Givent≥0, r>0 and t+r<1<p<n/(t+r), then f∈Sr,pΦ,t(ℝn) if and only if
(93)|〈TΦ(f,h),g〉|≲∥g∥Hr+t,p∥h∥H-t,p′,
where g∈Ht+r,p(ℝn) and h∈H-t,p′(ℝn).Now we give another characterization ofXr,pt(ℝn).Theorem 24.
Givent≥0, r>0 and t+r<1<p<n/(t+r), then
(94)Xr,pt(ℝn)=Mr,pt(ℝn)∩Qr,pt(ℝn).Proof.
By modifying the coefficients such thatfj,kɛhj,kɛ=|fj,kɛ||hj,kɛ|, we could suppose that g(y)≥0. By Theorem 21 (ii) and Lemma 23, we know that f∈Sr,pΦ,t(ℝn) is equivalent to
(95)∬∑(ɛ,j,k)∈Λn‍|fj,kɛ∥hj,kɛ|2jnΦ∑(ɛ,j,k)∈Λn‍×(2jx-k)Ψ(|x-y|)g(y)|x-y|n-(t+r)dydx≲∥g∥Lp∥h∥H-t,p′,∀g≥0.
By a simple calculation, we get(96)∫2jnΦ(2jx-k)Ψ(|x-y|)|x-y|t+r-ndx≃(2-j+|y-2-jk|)(t+r-n).
So (95) is equivalent to the following inequality:
(97)∫∑(ɛ,j,k)∈Λn‍|fj,kɛ∥hj,kɛ|(2-j+|y-2-jk|)t+r-ng(y)dy≲∥g∥Lp∥h∥H-t,p′,∀g≥0.
That is to say(98)∬∑(ɛ,j,k)∈Λn‍2jn|fj,kɛ∥hj,kɛ|χj,k×(x)g(y)(2-j+|y-2-jk|)n-(t+r)dxdy≲∥g∥Lp∥h∥H-t,p′,∀g≥0.
Let(99)S-th(x)=(∑(ɛ,j,k)∈Λn‍2j(n-2t)|hj,kɛ|2χj,k(x))1/2.
We have∥S-th∥p′≃eq∥h∥H-t,p′. Then, the inequality (98) is equivalent to
(100)∫(∑(ɛ,j,k)∈Λn‍2j(n+2t)|fj,kɛ|2|gj,k|2χj,k(x))p/2dx≲∥g∥Lpp,∀g≥0.
The above inequality is equivalent tof∈Qr,pt(ℝn).


## 3.1. Decomposition of Products via Multilinear Analysis

At first, we give a wavelet decomposition of the product offh. Let Φ0 and {Φε,ε∈En} be the scaling function and wavelet functions of Meyer wavelets, respectively. There exists an integer N≥3 such that
(45)∫xαΦ0(x)Φε(2Nx-k)dx=0∫1foranyk∈ℤn,α∈ℕn,ε∈En.
Denote
(46)Λɛ,n={(s,ε′,l),0≤s≤N,ε′∈En,l∈ℤn,|l|≤2(M+2+s)n,andifs=0,(0,ɛ,0)≠(0,ɛ′,0)},ℤNn={l=(l1,…,ln)∈ℤn,0≤li≤2N-1,i=1,…,n}.
For ɛ∈En, l∈ℤn, |l|≤2(M+2)n, and (ɛ′,l)∈Λɛ,n, we denote
(47)T1,ɛ(x)=∑j∈ℕ,k∈ℤn‍∑l∈ℤn‍fj,k+l0hj,kεΦj,k+l0(x)Φj,kε(x),T2,0,ɛ,ɛ(x)=∑j∈ℤ,k∈ℤn‍∑l∈ℤn,l≠0‍fj,k+lɛhj,kɛΦj,k+lε(x)Φj,kε(x)+∑j∈ℕ,k∈ℤn‍fj,kεhj,kε[(Φj,kε(x))2-2jnΦ(2jx-k)],T3,ɛ(x)=∑j∈ℕ,k∈ℤn‍∑l∈ℤNn‍∑l′∈ℤn‍fj+N,2Nk+lεhj,k+l′0×Φj,k+l′0(x)Φj+N,2Nk+lε(x),T4,ɛ(x)=∑j∈ℕ,k∈ℤn‍fj,kεhj,kε2jnΦ(2jx-k),T5(x)=∑k∈ℤn‍∑l∈ℤn‍f0,k0h0,k+l0Φj,k0(x)Φ0,k+l0(x).
For ɛ,ɛ′∈En, 0<s≤N or s=0, and ε≠ε′, denote
(48)T2,s,ε,ε′(x)=∑j∈ℕ,k∈ℤn‍∑l∈ℤn‍fj+s,2sk+lε′hj,kεΦj+s,2sk+lε′(x)Φj,kε(x).
By (11) and (12), we can decompose fh as follows:
(49)f(x)h(x)=∑ɛ∈En‍T1,ε(x)+∑0≤s≤N,ɛ,ɛ′∈En‍T2,s,ε,ε′(x)+∑ɛ∈En‍T3,ε(x)+∑ɛ∈En‍T4,ε(x)+T5(x):=∑i=15‍Ti(x).
If g∈Ht+r,p(ℝn), write g(x)=∑(ε,j,k)∈Λngj,kεΦj,kε(x). For ε,ε′∈En, and 0≤s≤N, we define
(50)T1,ε=∫‍T1,ε(x)g(x)dx,T2,s,ε,ε′=∫‍T2,s,ε,ε′(x)g(x)dx.
Let
(51)T1,1,ε,l=∑j∈ℕ,k∈ℤn‍∑j≥j′≥0,ε′,k′∈ℤn‍|fj,k+l0||hj,kε||aj,k,l,j′,k′ε,ε′||gj′,k′ε′|,T1,2,ε,l=∑j∈ℕ,k∈ℤn‍∑j<j′,ε′,k′∈ℤn‍|fj,k+l0||hj,kε||aj,k,l,j′,k′ε,ε′||gj′,k′ε′|,T2,1,s,ε,ε′,l=∑j∈ℕ,k∈ℤn‍∑j≥j′>0,ɛ′′,k′∈ℤn‍|fj+s,2sk+lε′||hj,kε||aj,k,l,j′,k′ε,ε′,ε′′,s||gj′,k′ε′′|,T2,2,s,ε,ε′,l=∑j∈ℕ,k∈ℤn‍∑j<j′,ɛ′′,k‍|fj+s,2sk+lε′||hj,kε||aj,k,l,j′,k′∈ℤnε,ε′,ε′′,s||gj′,k′ε′′|.
It is easy to see that
(52)|T1,ε|≲∑l∈ℤn‍(T1,1,ε,l+T1,2,ε,l),|T2,s,ε,ε′|≲∑l∈ℤn‍(T2,1,s,ε,ε′,l+T2,2,s,ε,ε′,l).


## 3.2. Several Technical Lemmas

Now we estimate the quantitiesT1,1,ɛ,l, T1,2,ɛ,l, T2,1,s,ɛ,ɛ′,l, and T2,2,s,ɛ,ɛ′,l separately. Let
(53)Stgj′(x)=∑ɛ′,k′‍2j′(n/2+t)|gj′,k′ɛ′|χ(2j′x-k′).
For fixed x, there is only one k′ such that χ(2j′x-k′)≠0 and the number of ɛ′ is finite. Then
(54)Stgj′(x)=(∑ɛ′,k′‍22j′(n/2+t)|gj′,k′ɛ′|2χ(2j′x-k′))1/2.LetM be the Hardy-Littlewood maximal function. Then, if j<j′ and x∈Qj,k, we have
(55)∑k′‍(1+|k-2j-j′k′|)-N22j′(-n/2+t)gj′,k′ɛ′≲2-jnMStgj′(x).Lemma 18.
Givenr>0, t≥0, and t+r<1<p<n/(t+r). If f∈BMOr(ℝn), then (i)
forg∈Ht+r,p and h∈H-t,p′,
(56)|T1,ɛ|+|T2,s,ɛ,ɛ′|≲∥g∥Ht+r,p∥h∥H-t,p′,(ii)
forg∈Ht+r,p and h∈H-t,p′,
(57)|〈T5,g〉|≲∥g∥Ht+r,p∥h∥H-t,p′.Proof.
(1) Forj′≤j, we have
(58)T1,1,ɛ,l=∑j,k‍∑j≥j′,ɛ′,k′‍|fj,k+l0||hj,kɛ||aj,k,l,j′,k′ɛ,ɛ′||gj′,k′ɛ′|.
By Lemma17, we know
(59)|aj,k,l,j′,k′ɛ,ɛ′|≲(1+|l|)-N1(1+|k′-2j′-jk|)-N22nj′/2+j′-j.
By Lemma9, f∈BMOr(ℝn) implies |fj,kɛ|≤C2(r-n/2)j. Now we can get
(60)∑l∈ℤn‍T1,1,ɛ,l≲∑l∈ℤn‍∑k,j≥j′‍∫‍23nj′/2+(j′-j)+(r-n/2)j∑k′‍(∑ɛ′‍|gj′,k′ɛ′|2)1/2×|hj,kɛ|χ(2j′x-k′)(1+|l|)N1(1+|k′-2j′-jk|)N2dx≲∑j≥j′‍∫‍23nj′/2+(j′-j)+(r-n/2)j×∑k′‍(∑ɛ′‍22j′(n/2+t+r)|gj′,k′ɛ′|2)1/22-j′(t+r+n/2)×∑k‍|hj,kɛ|χ(2j′x-k′)(1+|k′-2j′-jk|)N22j(-n/2-t)2j(n/2+t)dx≲∫∑j≥j′‍23nj′/2-j′(t+r+n/2)+(j′-j)+(r-n/2)j+j(n/2+t)×2-j′nSt+rgj′(x)M(S-thj)(x)dx.
Because0<t+r<1,
(61)∑l∈ℤn‍T1,1,ɛ,l≲∑j≥j′‍2(j′-j)+(r+t)j-j′(r+t)×∫St+rgj′(x)M(S-thj)(x)dx≲∑j≥j′‍2(j′-j)(1-r-t)∥g∥Ht+r,p∥h∥H-t,p′≲∥g∥Ht+r,p∥h∥H-t,p′.
Now we estimate the term(62)T2,1,s,ɛ,ɛ′,l(x)=∑j,k‍∑j≥j′>0‍|fj,kɛ′||hj+s,2sk+lɛ||aj,k,l,j′,k′ɛ,ɛ′,ɛ′′,s||gj′,k′ɛ′′|.
Becausef∈BMOr(ℝn), by Lemma 9, we have |fj,k+lɛ′|≲2(r-n/2)j. By Lemma 17,
(63)|aj,k,l,j′,k′ɛ,ɛ′,ɛ′′,s|≲(1+|l|)-N1(1+|k′-2j′-jk|)-N22nj′/2+(j′-j).
We can get, similarly,(64)∑l∈ℤNn‍T2,1,s,ɛ,ɛ′,l(x)≲∑l∈ℤNn‍∑j,k‍∑j≥j′>0‍2(r-n/2)j2nj′/2+j′-j∑j,k‍∑j≥j′>0‍×|hj+s,2sk+lε||gj′,k′ε′′|(1+|l|)N1(1+|k′-2j′-jk|)N2≲∑l∈ℤNn∑j≥j′>0∫23nj′/2+(j′-j)+(r-n/2)j∑j,k‍∑j≥j′>0‍×∑ε′′,k′‍|gj′,k′ε′||hj+s,2sk+lε|χ(2j′x-k′)(1+|l|)N1(1+|k′-2j′-jk|)N2dx≲∑j≥j′>0‍2(j′-j)(1-t-r)∫‍St+r(gj′)(x)MS-thj+s(x)dx≲∥St+rgj′∥Lp∥MS-thj+s∥Lp′≲∥g∥Ht+r,p∥h∥H-t,p′.
(2) Ifj′>j, the estimates of T1,2,ɛ,l and T2,2,s,ɛ,ɛ′,l are easier than those of T1,1,ɛ,l and T2,1,s,ɛ,ɛ′,l. For example, we estimate the term
(65)T1,2,ɛ,l=∑j,k‍∑0≤j<j′,ɛ′,k′‍|fj,k+l0||hj,kɛ||aj,k,l,j′,k′ɛ,ɛ′||gj′,k′ɛ′|.
Becausef∈BMOr(ℝn), by Lemma 9, |fj,k+l0|≲2(r-n/2)j. By Lemma 17,
(66)|aj,k,l,j′,k′ɛ,ɛ′|≲(1+|l|)-N1(1+|k-2j-j′k′|)-N22-nj′/2+nj+N(j-j′).
Because2nj∫‍χj,k(x)dx=1, we can obtain
(67)∑l∈ℤn‍T1,2,ɛ,l≲∑l∈ℤn‍∑j,k‍∑0≤j<j′,ɛ′,k′‍∫‍2(r-n/2)j2-nj′/2+nj+N(j-j′)2nj×|hj,kε||gj′,k′ε′|χj,k(x)(1+|l|)N1(1+|k-2j-j′k′|)N2dx≲∑ε,j,k‍∫2(r-n/2)2-nj′/2+nj+N(j-j′)2nj2-nj2-j′(-n/2+t+r)×MSt+r(gj′)(x)|hj,kε|χj,k(x)dx≲∑0≤j<j′∫2(r-n/2)j2-(nj′/2)+nj+N(j-j′)2-j(n/2-t)×2-j′(-n/2+t+r)MSt+r(gj′)(x)S-t(hj)(x)dx≲∥g∥Ht+r,p∥h∥H-t,p′.
The estimate forT2,2,s,ɛ,ɛ′,l(x) can be obtained similarly. By the same methods used in (1) and (2), we can get the estimate of the term T5. We omit the details.Now we consider the termT3,ɛ. We have the following claim.Lemma 19.
Givenr>0, t≥0, and t+r<1<p<n/(t+r), if f∈BMOr(ℝn), then
(68)|〈T3,ɛ,g〉|≲∥g∥Ht+r,p∥h∥H-t,p′.Proof.
In fact, forl∈ℤNn, (ɛ,j,k)∈Λn and l′∈ℤn, let
(69)gj+N,2Nk+lɛ,l′=2-nj/2〈Φj,k+l′0Φj+N,2Nk+lɛ,g〉.
We have
(70)|〈T3,ɛ,g〉|=|∑(ɛ,j,k)∈Λn‍∑l∈ℤNn,l′∈ℤn‍2jn/2hj,k+l′0gj+N,2Nk+lɛ,l′fj+N,2Nk+lɛ|≲∑l∈ℤNn‍∑(ɛ,j,k)∈Λn‍(∑l′∈ℤn‍2jn/2|hj,k+l′0||gj+N,2Nk+lɛ,l′|)×|fj+N,2Nk+lɛ|.
Because (F1r,2(ℝn))′=BMOr(ℝn), by Lemmas 9 and 13, we have
(71)|〈T3,ɛ,g〉|≲∑l∈ℤNn‍∥f∥BMOr∫(∑(ɛ,j,k)∈Λn‍2j(n+2r+t)≲∑l∈ℤNn‍∥f∥BMOr×(∑l′∈ℤn‍2j(n/2-t)|hj,k+l′0||gj+N,2Nk+lɛ,l′|)2≲∑l∈ℤNn‍∥f∥BMOr×χj,k(x)∑(ɛ,j,k)∈Λn‍)1/2dx≲∥f∥BMOr∫(∑(ɛ,j,k)∈Λn‍2j(n+2r+2t)≲∥f∥BMOrgi×(∑l′∈ℤn‍(1+|l′|)N+n|gj+N,2Nk+lɛ,l′|)2≲∥f∥BMOrgi×χj,k(x)∑(ɛ,j,k)∈Λn‍)1/2MTth(x)dx.
Because
(72)|gj+N,2Nk+lɛ,l′|=2-nj/2|〈Φj,k+l′0Φj+N,2Nk+l′ɛ∑(ɛ′,j′,k′)∈Λn‍gj′,k′ɛ′Φj′,k′ɛ′〉|=2-nj/2|〈∑(ɛ′,j′,k′)‍Φj,k+l′0Φj+N,2Nk+l′ɛ=552-nji×∑(ɛ′,j′,k′)∈Λn,|j′-j-N|≤2‍gj′,k′ɛ′Φj′,k′ɛ′〉|,
we can get
(73)(∑(ɛ,j,k)∈Λn‍2(n+2t+2t)jhi×(∑l′∈ℤn‍(1+|l′|)N+n|gj+N,2Nk+lɛ,l′|)2χj,k(x))1/2≲∑|j′-j-N|≤2‍MSt+rgj′(x).
Hence, we have
(74)∫((∑l′∈ℤn‍2j(n/2-t)|hj,k+l′0∥gj+N,2Nk+lɛ,l′|)2∑(ɛ,j,k)∈Λn‍2j(n+2r+t)×(∑l′∈ℤn‍2j(n/2-t)|hj,k+l′0∥gj+N,2Nk+lɛ,l′|)2χj,k(x))p/2dx≲∫(∑j≥0‍(∑|j′-j-N|≤2‍MSt+rgj′(x))2)p/2dx≲∥g∥Ht+r,p.
Finally, we obtain, by Hölder's inequality,
(75)|〈T3,ɛ,g〉|≲∥f∥BMOr∥g∥Ht+r,p∥MTth∥p′≲∥f∥BMOr∥g∥Ht+r,p∥h∥H-t,p′.
This completes the proof of Lemma 19.In order to deal with the term(76)T4,ɛ(x)=∑(ɛ,j,k)∈Λn‍fj,kɛhj,kɛ2jnΦ(2jx-k),
we need the following estimate.Lemma 20.
Givenr>0, t≥0 with t+r<1<p<n/(r+t), f∈Sr,pΦ,t(ℝn) if and only if
(77)|〈T4,ɛ,g〉|≲∥g∥Hr+t,p∥h∥H-t,p′.Proof.
In fact, forg∈Hr+t,p(ℝn),
(78)〈T4,ɛ,g〉=∑(ɛ,j,k)∈Λn‍fj,kɛgj,kΦhj,kɛ.
By Lemma 4, using Hölder's inequality, we have
(79)|〈T4,ɛ,g〉|≲∑(ɛ,j,k)∈Λn,j≥0‍2jn|gj,kΦ|2|fj,kɛ|2χj,k(x)dx≲∫(∑(ɛ,j,k)∈Λn,j≥0‍2j(n+2t)|gj,kΦ|2|fj,kɛ|2χj,k(x))1/2×(∑(ɛ,j,k)∈Λn,j≥0‍2j(n-2t)|hj,kɛ|2χj,k(x))1/2dx≲(∫(∑(ɛ,j,k)∈Λn,j≥0‍2j(n+2t)|gj,kΦ|2|fj,kɛ|2×χj,k(x)∑∈Λn,j≥0‍)p/2dx)1/p×(∫‍(∑(ɛ,j,k)∈Λn,j≥0‍2j(n-2t)|hj,kɛ|2χj,k(x))p′/2dx)1/p′≲∥f∥Sr,pΦ,t∥h∥H-t,p′.
Because f∈Sr,pΦ,t(ℝn), we can see that |〈T4,ɛ,g〉|≲∥g∥Ht+r,p∥h∥H-t,p′.
Conversely, let(80)τj,kɛ=2jtfj,kɛgj,kΦ,τ(x)=∑(ɛ,j,k)∈Λn,j≥0‍τj,kɛΦj,kɛ(x).
Denote h=|τ|p-2τ¯. For h(x)=∑(ɛ,j,k)∈Λnhj,kɛΦj,kɛ(x), we write ht(x) as
(81)ht(x)=∑(ɛ,j,k)∈Λn,j≥0‍2jthj,kɛΦj,kɛ(x):=∑(ɛ,j,k)∈Λn‍(ht)j,kɛΦj,kɛ(x).
It is easy to see that h∈Lp′(ℝn) is equivalent to ht∈H-t,p′(ℝn).
By the wavelet characterization ofHt,p(ℝn), we get
(82)∫(∑(ɛ,j,k)∈Λn,j≥0‍2j(n+2t)|gj,kΦ|2|fj,kɛ|2χj,k(x))p/2dx=∫‍(∑(ɛ,j,k)∈Λn,j≥0‍2jn|τj,kɛ|2χj,k(x))p/2dx=∫‍|τ|pdx=∫‍τhdx=∑(ɛ,j,k)∈Λn‍2j(t+n/2)fj,kɛgj,kΦhj,kɛ=∑(ɛ,j,k)∈Λn‍2jn/2fj,kɛgj,kΦ(ht)j,kɛ.
Further, we can deduce that
(83)∫(∑(ɛ,j,k)∈Λn,j≥0‍2j(n+2t)|gj,kΦ|2|fj,kɛ|2χj,k(x))p/2dx=|〈T4,ɛ,g〉|≲∥g∥Ht+r,p∥ht∥H-t,p′≲∥g∥Ht+r,p∥τ∥Lpp/p′.
Hence ∥τ∥Lp<∞ and f∈Sr,pΦ,t(ℝn). This completes the proof of this lemma.


## 3.3.  Wavelet Characterization of Sobolev Multipliers

Theorem 21.
Lett≥0, r>0 with t+r<1<p<n/(t+r). Then,
(84)Xr,pt(ℝn)=Mr,pt(ℝn)∩Sr,pΦ,t(ℝn).Proof.
We first assumef∈Mr,pt(ℝn)∩Sr,pΦ,t(ℝn). Let g∈Ht+r,p(ℝn) and h∈H-t,p′(ℝn). By the decomposition given in Section 3.1, we have
(85)f(x)h(x)=∑ɛ∈En‍T1,ɛ(x)+∑0≤s≤N,ɛ,ɛ′∈En‍T2,s,ɛ,ɛ′(x)+∑ɛ∈En‍T3,ɛ(x)+∑ɛ∈En‍T4,ɛ(x)+T5(x).
Notice that f∈Mr,pt(ℝn) implies that f∈BMOr(ℝn). By Lemmas 18–20, we obtain that, for f∈Mr,pt(ℝn)∩Sr,pΦ,t(ℝn), fh∈H-t-r,p′(ℝn) and
(86)∥fh∥H-t-r,p′≲∥f∥Mr,pt∩Sr,pΦ,t∥h∥H-t,p′,
which gives
(87)∥f∥Xr,pt=sup∥g∥Hr+t,p≤1sup∥h∥H-t,p′|〈fg,h〉|=sup∥g∥Hr+t,p≤1sup∥h∥H-t,p′|〈fh,g〉|≤sup∥g∥Hr+t,p≤1sup∥h∥H-t,p′∥fh∥H-t-r,p′∥g∥Ht+r,p≤∥f∥Mr,pt∩Sr,pΦ,t.
Hence, we can get Mr,pt(ℝn)∩Sr,pΦ,t(ℝn)⊂Xr,pt(ℝn).
Conversely, assume thatf∈Xr,pt(ℝn). By Lemma 7, we have f∈Mr,pt(ℝn)⊂BMOr(ℝn). We apply Lemmas 18 and 19 again to deduce
(88)|〈T4,ɛ,g〉|≤|T1,ɛ|+|T2,s,ɛ,ɛ′|+|〈T3,ɛ,g〉|+|〈T5,g〉|≤∥g∥Hr+t,p∥h∥H-t,p′.
Then, Lemma 20 implies that f∈Sr,pΦ,t(ℝn). This completes the proof of this theorem.Now we characterizeXr,pt(ℝn) by the fractional integration. Let Ψ(x)≥0, Ψ(x)∈C0∞(B(0,2)) with Ψ(x)=1 on B(0,1). It is well known that g~∈Ht+r,p(ℝn) if and only if there exists g∈Lp(ℝn) such that
(89)g~(x)=Jt+rg(x)=∫‍g(y)Ψ(|x-y|)|x-y|t+r-ndy.
For g∈Lp(ℝn), let
(90)gj,k=∫‍(2-j+|y-2-jk|)t+r-ng(y)dy.Definition 22.
Givenr>0, t≥0 and 1<p<n/(r+t), then f(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x)∈Qr,pt(ℝn) if and only if
(91)∫‍(∑(ɛ,j,k)∈Λn‍2j(n+2t)|gj,k|2|fj,kɛ|2χj,k(x))p/2dx≲1,
where g∈Lp(ℝn) and g(x)≥0.Letf(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x) and h(x)=∑(ɛ,j,k)∈Λnhj,kɛΦj,kɛ(x). Define
(92)TΦ(f,h)(x)=∑j∈ℕ‍∑k∈ℤn‍2jnfj,kɛhj,kɛΦ(2jx-k).
Similar to Lemma 20, we can get the following.Lemma 23.
Givent≥0, r>0 and t+r<1<p<n/(t+r), then f∈Sr,pΦ,t(ℝn) if and only if
(93)|〈TΦ(f,h),g〉|≲∥g∥Hr+t,p∥h∥H-t,p′,
where g∈Ht+r,p(ℝn) and h∈H-t,p′(ℝn).Now we give another characterization ofXr,pt(ℝn).Theorem 24.
Givent≥0, r>0 and t+r<1<p<n/(t+r), then
(94)Xr,pt(ℝn)=Mr,pt(ℝn)∩Qr,pt(ℝn).Proof.
By modifying the coefficients such thatfj,kɛhj,kɛ=|fj,kɛ||hj,kɛ|, we could suppose that g(y)≥0. By Theorem 21 (ii) and Lemma 23, we know that f∈Sr,pΦ,t(ℝn) is equivalent to
(95)∬∑(ɛ,j,k)∈Λn‍|fj,kɛ∥hj,kɛ|2jnΦ∑(ɛ,j,k)∈Λn‍×(2jx-k)Ψ(|x-y|)g(y)|x-y|n-(t+r)dydx≲∥g∥Lp∥h∥H-t,p′,∀g≥0.
By a simple calculation, we get(96)∫2jnΦ(2jx-k)Ψ(|x-y|)|x-y|t+r-ndx≃(2-j+|y-2-jk|)(t+r-n).
So (95) is equivalent to the following inequality:
(97)∫∑(ɛ,j,k)∈Λn‍|fj,kɛ∥hj,kɛ|(2-j+|y-2-jk|)t+r-ng(y)dy≲∥g∥Lp∥h∥H-t,p′,∀g≥0.
That is to say(98)∬∑(ɛ,j,k)∈Λn‍2jn|fj,kɛ∥hj,kɛ|χj,k×(x)g(y)(2-j+|y-2-jk|)n-(t+r)dxdy≲∥g∥Lp∥h∥H-t,p′,∀g≥0.
Let(99)S-th(x)=(∑(ɛ,j,k)∈Λn‍2j(n-2t)|hj,kɛ|2χj,k(x))1/2.
We have∥S-th∥p′≃eq∥h∥H-t,p′. Then, the inequality (98) is equivalent to
(100)∫(∑(ɛ,j,k)∈Λn‍2j(n+2t)|fj,kɛ|2|gj,k|2χj,k(x))p/2dx≲∥g∥Lpp,∀g≥0.
The above inequality is equivalent tof∈Qr,pt(ℝn).


## 4. Logarithmic Morrey Spaces and Multipliers

### 4.1. A Logarithmic Condition

Lemma7 implies that Xr,pt(ℝn)⊂Mr,pt(ℝn). In this section, we consider the reverse inclusion relation. At first we introduce a logarithmic Morrey spaces.Definition 25.
Fix1<p<n/(r+t) and τ≥0. We say f(x)∈Mr,pt,τ(ℝn) if sup|Q|=1|ft,Q|≤C and
(101)∫Q‍|(I-Δ)t/2f(x)-ft,Q|pdx≲|1-log2|Q||-pτ|Q|1-p(r+t)/n,
for any cube Q with |Q|≤1.Similar to Theorem6, we have the following wavelet characterization of Mr,pt,τ(ℝn).Theorem 26.
Givent,τ≥0, r>0, 1<p<n/(r+t), f(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x) belongs to Mr,pt,τ(ℝn) if and only if
(102)∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p(r+t)/n,
where Q∈Ω with |Q|≤1.Proof.
Similar to that of Theorem6, the proof of this theorem can be obtained by the characterization of Triebel-Lizorkin spaces. See Lemma 4. We omit the detail.In [8], Fefferman established the following relation:
(103)Mr,qt(ℝn)⊂Xr,pt(ℝn)⊂Mr,pt(ℝn),
where q>p>1. In this section, we use wavelet characterization to give a logarithmic type inclusion. Let r>0, t≥0, 1<p<n/(r+t), and τ>1/p′. In Theorem 32, we prove that Mr,pt,τ(ℝn) is a subspace of Xr,pt(ℝn). Hence,
(104)Mr,pt,τ(ℝn)⊂Xr,pt(ℝn)⊂Mr,pt(ℝn)=Mr,pt,0(ℝn).Lemma 27.
Givenτ>0, r>0, t≥0, and 1<p<n/(r+t), if f∈Mr,pt,τ(ℝn), then
(105)|fj,kɛ|≲2j(r-n/2)(1+j)-τ.Proof.
Becausef∈Mr,pt,τ(ℝn), then for Q∈Ω,
(106)∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p(r+t)/n.
We have(107)∫Q(∑ɛ∈En,Qj,k⊂Q‍2j(n-2r)|fj,kɛ|2χj,k(x))p/2dx‍≲|Q|p(r+t)/n∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲|Q|p(r+t)/n|Q|1-p(r+t)/n(1-log2|Q|)-τp≲|Q|(1-log2|Q|)-τp.
Forɛ∈En, j∈ℕ, and k∈ℤn, take Q=Qj,k. By Lemma 4, we get
(108)∫(2jn-2jr|fj,kɛ|2χj,k(x))p/2dx≲2-jn(1-log22-jn)-τp≲2-jn(1+j)-τp,
which gives
(109)|fj,kɛ|≲2j(r-n/2)(1+j)-τ.
Whenɛ=0,
(110)fj,k0=〈∑(ɛ′,j′,k′)∈Λn‍fj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉=〈∑(ɛ′,j′,k′)∈Λn:j′<j‍fj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉.
Since(111)|∑(ɛ′,j′,k′)∈Λn:j′<j‍fj′,k′ɛ′Φj′,k′ɛ′(x)|≲2(r-n/2)j(1+j)-τ,
we have(112)|fj,k0|≲〈2rj(1+j)-τ,|Φj,k0|〉≲2(r-n/2)j(1+j)-τ.Letβ=(β1,β2,…,βn), βi∈ℕ+, where ℕ+ denotes the set of all positive integers. Denote by fβ(x) the derivative (∂/∂x)βf. We have the following two lemmas.Lemma 28.
Givenτ>0, r>0, t≥0, and 1<p<n/(r+t), if f∈Mr,pt,τ(ℝn), the derivative fβ∈Mr+|β|,pt-|β|,τ(ℝn), where |β|=∑i=1nβi.Proof.
Iff∈Mr,pt,τ(ℝn) and f(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x), by Theorem 26, we have
(113)∫(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p(r+t)/n.
Denote by fj,kɛ,β the wavelet coefficients of fβ(x). We can get fj,kβ,ɛ=2j|β|fj,kɛ and
(114)∫(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t-2|β|)|fj,kβ,ɛ|2χj,k(x))p/2dx≲∫‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p[(r+|β|)+(t-|β|)]/n.
This implies that fβ(x)∈Mr+|β|,pt-|β|,τ(ℝn).To get the sufficient condition for multiplier spaces, we need to consider carefully the relationship of different dyadic cubes relative to combination atoms. Because of this reason, we use always the same Daubechies wavelets in the rest of this section. For a cubeQ and a measurable set E, if |Q⋂‍E|=|Q|, we say that Q is a cube in the set E.For(m,2u,Eu)-combination atoms gu(x) defined in Definition 14, the measure of Eu is finite. Hence, the number of biggest dyadic cubes in Eu is finite. We denote the number of biggest dyadic cubes in Eu by i1. Denote by Fu,1 the set {i∈ℕ,i=1,…,i1}. If i∈Fu,1, we denote such cube by Qu,1,i. The volume of Qu,1,i is denoted by 2-nju,1; that is, |Qu,1,i|=2-nju,1. Denote Eu,1=Eu∖(⋃i∈Fu,1Qu,1,i).The measure ofEu,1 is finite. We denote the number of biggest dyadic cubes in Eu,1 by i2. Denote by Fu,2 the set {i∈ℕ,i=1,…,i2}. If i∈Fu,2, we denote such cube by Qu,2,i. The volume of Qu,2,i is denoted by 2-nju,2; that is, |Qu,2,i|=2-nju,2. Denote Eu,2=Eu,1∖(⋃i∈Fu,2Qu,2,i).We continue this process until there exists somes such that Eu,s+1 is empty. For s′≥s+1, we denote is′=0, and Fu,s′ and Eu,s′ are empty sets. Otherwise we continue until infinity. Then Eu=⋃s≥1,i∈Fu,sQu,s,i and gu(x)=∑s≥1,i∈Fu,sgu,s,i(x), where gu,s,i(x)=∑Qj,k⊂Qu,s,igj,kɛΦj,kɛ(x).To compute the norm off(x)gu(x), we need to find out a special set of dyadic cubes denoted by {Qu,s,i,k}(s,i,k)∈Hu such that suppgu⊂⋃(s,i,k)∈HuQu,s,i,k. gu(x) is nearly L∞ function on Qu,s,i,k and satisfies the estimate of Lemma 29. We divide such process into the following three steps.Step
1. For all s≥1, if  i∈Fu,s and k∈ℤn with |k|<n2(M+3)n, we denote (i,k)∈Gu,s. Denote Fu={(s,i,k),s≥1,(i,k)∈Gu,s}. For (s,i,k)∈Fu, we denote Qu,s,i,k=2-ju,sk+Qu,s,i. For s≥1, we denote E~u,s0=⋃(i,k)∈Gu,sQu,s,i,k. In the next step, we choose a special subcover to the support of gu(x).Step
2. We define now Hu,s(s≥1), E~u,s1(s≥1) and Gu,s1(s≥2).Fors=1, denote Hu,1=Gu,1 and E~u,s1=E~u,s0. For s=2, denote (i,k)∈Gu,21, if there exists 0≤j≤ju,1, l∈ℤn such that Qj,l⊂⋃|k-k′|≤n2(M+2)nQu,2,i,k′ and 〈gu,Φj,l0(x)〉≠0. We know that ⋃(i,k)∈Gu,21Qu,2,i,k⊂E~u,11. Denote Hu,2=Gu,2∖Gu,21 and E~u,21=⋃(i,k)∈Hu,2Qu,2,i,k.Fors=3, denote (i,k)∈Gu,31, if there exists 0≤j≤ju,2, l∈ℤn such that Qj,l⊂⋃|k-k′|≤n2(M+2)nQu,2,i,k′ and 〈gu(x),Φju,s,l0(x)〉≠0. We know that ⋃(i,k)∈Gu,31Qu,3,i,k⊂⋃1≤s≤2E~u,s1. Denote Hu,3=Gu,3∖Gu,31 and E~u,31=⋃(i,k)∈Hu,3Qu,3,i,k.We continue this process until infinity. Fors≥2, maybe, a part of Gu,s1, Hu,s, and E~u,s1 is empty set.Step
3. Let Hu={(s,i,k),s≥1,(i,k)∈Hu,s}. It is easy to see that the support of gu(x) is contained in ⋃s≥1E~u,s1. For a (t+r,2u,Eu)-combination atom gu(x) and gj,k0=〈gu,Φj,k0〉, we have the following estimate.Lemma 29.
Givenr>0, t≥0, 1<p<n/(t+r) and s≥1, for (i,m′)∈Hu,s and Qj,k⊂Qu,s,i,m′, we have |gj,k0|≲2u-nj/22-(t+r)ju,s.Proof.
By the definition ofHu,s, we have
(115)gj,k0=〈∑(ɛ′,j′,k′)∈Λn‍gj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉=〈∑(ɛ′,j′,k′)∈Λn:j′≥ju,s‍gj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉.
Becausegu is a (m,2u,Eu)-combination atom, St+r(gu)(x)≤C2u. Hence, for every (ɛ′,j′,k′), |gj′,k′ɛ′|≲2u2-j′(r+t+n/2). We can obtain
(116)|gj,k0|=|〈∑(ɛ′,j′,k′)∈Λn:j′≥ju,s‍gj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉|≲2u∫∑j′≥ju,s‍2-j′(r+t)2-jn′/22j′n/22jn/2×|Φɛ′(2j′x-k′)||Φ0(2jx-k)|dx≲2u2-jn/22-ju,s(r+t).Theorem 30.
Suppose thatτ>1/p′, t≥0, r>0, and 1<p<n/(r+t). If f∈Mr,pt,τ(ℝn) and gu is a (t+r,2u,Eu)-combination atom, then, for s≥1, (i,m′)∈Hu,s, and Q=Qu,s,i,m′, we have
(117)∥fgu∥Ht,p(Q)≲(1+ju,s)-τ2u|Q|1/p.Proof.
First, fort≥0, we prove ∥fgu∥Lp(Q)≤Cju,s-τ2u|Q|1/p. Let f(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x) and gu(x)=∑(ɛ,j,k)∈Λngj,kɛΦj,kɛ(x). Denote by Λn′ the set
(118){(ɛ,ɛ′,j,k,l),ɛ,ɛ′∈En,(ɛ,k)≠(ɛ′,k+l),j∈ℤ,k,l∈ℤn,|l|≤n2(M+2)n}.
For any dyadic cube Q⊂Eu, by the formulas (11) and (12), we decompose fgu as
(119)T1(x)=∑(ɛ,j,k)∈Λn,j=ju,s‍∑|l|≤n2(M+2)n‍fj,k+l0gj,k0Φj,k+l0(x)Φj,k0(x),T2(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+l0gj,kɛΦj,k+l0(x)Φj,kɛ(x),T3(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+lɛgj,k0Φj,k+lɛ(x)Φj,k0(x),T4(x)=∑(ɛ,ɛ′,j,k,l)∈Λ′n,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+lɛ′×gj,kɛΦj,k+lɛ′(x)Φj,kɛ(x),T5(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kɛgj,kɛ((Φj,kɛ(x))2-2nj/2Φj,k0(x)),T6(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kɛgj,kɛ2nj/2Φj,k0(x).
The rest of the proof is divided into three steps.Step
1. For t≥0, we estimate ∥Ti∥Lp(Q),i=1,2,…,6. By Lemma 4, we obtain
(120)f=∑(ɛ,j,k)∈Λn‍fj,kɛΦj,kɛ(x)∈Lp⟺∥(∑(ɛ,j,k)∈Λn‍2jn|fj,kɛ|2χ(2j·-k))1/2∥Lp.
Let
(121)S0(f)(x)=(∑(ɛ,j,k)∈Λn‍2jn|fj,kɛ|2χj,k(x))1/2.
We have ∥f∥Lp≃∥S0(f)∥Lp.(1) Becausef∈Mr,pt,τ, by Lemma 27,
(122)|fju,s,k0|≲(1+ju,s)-τ2(r-n/2)ju,s.
By Lemma 29, we have |gju,s,k0|≲2u-(r+t+n/2)ju,s. Hence we can get
(123)S0(T1)(x)=(∑|Qj,k|=|Qju,s,k|‍22ju,sn|fju,s,k0|2|gju,s,k0|2χ(2ju,sx-k))1/2≲(∑|Qj,k|=|Qju,s,k|‍22ju,sn(1+ju,s)-2τ22(r-n/2)ju,s×22u-2(r+t+n/2)ju,s22ju,s(n/2+t)χ(2ju,sx-k)∑|Qju,s,k|‍)1/2.
Because ju,s≥0, we have
(124)S0(T1)(x)≲(1+ju,s)-τ2u(∑|Qj,k|=|Qju,s,k|‍χj,k(x))1/2,∥T1∥Lp(Q)=∥S0(T1)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.(2) Now we estimate(125)T2(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+l0gj,kɛΦj,k+l0(x)Φj,kɛ(x).
Because j≥ju,s≥0, we have
(126)|fj,k0|≲(1+ju,s)-τ2(r-n/2)j≤(1+ju,s)-τ2(r+t-n/2)j.
Let
(127)Λn,2Q={(ɛ,j,k)∈Λn∣j≥ju,s,∀|l|≤n2(M+2)n,|supp(Φj,k+l0(x)Φj,kɛ(x))∩Q|≠0}.
Then, we can get
(128)S0(T2)(x)=(∑(ɛ,j,k)∈Λn,2Q‍|fj,k+l0|2|gj,kɛ|2χj,k(x))1/2≲(∑(ɛ,j,k)∈Λn,2Q‍22jn(1+j)-2τ×22j(r+t-n/2)|gj,kɛ|2χj,k(x)∑∈Λn,2Q‍)1/2≲(1+ju,s)-τSt+r(gu)(x).
Because gu(x) is a (t+r,2u,Eu)-combination atom,
(129)∥S0(T2)∥Lp(Q)≲(1+ju,s)-τ∥St+r(gu)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.(3) Sincegu(x) is a (t+r,2u,Eu)-combination atom, for s≥1, (i,m′)∈Hu,s, j≥0, and Q=Qu,s,i,m′,
(130)|gj,k0|≲2u-nj/22-(r+t)j=2u-nj/2|Q|(r+t)/n.
Let
(131)Λn,3Q={(ɛ,j,k)∈Λn∣j≥ju,s,∀|l|≤n2(M+2)n,i52|supp(Φj,k0Φj,k+lɛ)∩Q|≠0}.
We have, by ju,s≥0,
(132)S0(T3)(x)=(∑(ɛ,j,k)∈Λn,3Q‍2jn|gj,k0|2|fj,kɛ|2χj,k(X))1/2≲2u|Q|(r+t)/n(∑(ɛ,j,k)∈Λn,3Q‍2j(n+2t)|fj,kɛ|2χj,k(x))1/2.
By the fact that j≥ju,s and f∈Mr,pt,τ(ℝn), we get
(133)∥S0(T3)∥Lp(Q)≲2u|Q|(r+t)/n(1-log2|Q|)-τ|Q|1/p-(r+t)/n≲2u(1+ju,s)-τ|Q|1/p.(4) Now we estimate the termT4(x). Let
(134)Λn,4Q={(ɛ,j,k)∈Λn∣j≥ju,s,∀(ɛ,ɛ′,j,k,l)∈Λ′n,|supp(Φj,k+lɛ′(x)Φj,kɛ(x))∩Q|≠0}.
Because f∈Mr,pt,τ(ℝn), we have
(135)|fj,kɛ|≲(1+j)-τ2j(r-n/2)≲(1+j)-τ2j(r+t-n/2),(136)S0(T4)(x)=(∑(ɛ,j,k)∈Λn,4Q‍2jn(1+j)-2τ×2j(2r+2t-n)2jn|gj,kɛ|2χj,k(x)∑∈Λn,4Q‍)1/2≲(1+ju,s)-τSt+r(gu)(x).
Then we can get, by the fact that gu is a (t+r,2u,Eu)-combination atom,
(137)∥S0(T4)∥Lp(Q)≲(1+ju,s)-τ∥St+r(gu)∥Lp(Q)≲2u(1+ju,s)-τ|Q|1/p.(5) Now we estimate the term(138)T5(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kɛgj,kɛ((Φj,kɛ(x))2-2nj/2Φj,k0(x)).
Because the function (Φj,kɛ)2-2nj/2Φj,k0 plays the role as that of Φj,kɛ(x), we have
(139)∥S0(T5)∥Lp(Q)≲2u(1+ju,s)-τ|Q|1/p.(6) To estimate the termT6(x), we take h∈Lp′(Q). Let
(140)Λn,6Q={|supp(Φj,k+l0(x)Φj,kɛ(x))∩Q|(ɛ,j,k)∈Λn∣j≥ju,s,i|supp(Φj,k+l0(x)Φj,kɛ(x))∩Q|≠0}.
By the orthogonality of the wavelet function, we have
(141)〈T6,h〉=〈∑(ɛ,j,k)∈Λn,6Q‍fj,kɛgj,kɛ2jn/2Φj,k0,h〉=∑(ɛ,j,k)∈Λn,6Q‍fj,kɛgj,kɛ2nj/2hj,k0.
Then,
(142)|〈T6,h〉|≲∫∑(ɛ,j,k)∈Λn,6Q‍2nj|fj,kɛ∥gj,kɛ|2nj/2|hj,k0|χj,k(x)dx≲∫∑(ɛ,j,k)∈Λn,6Q‍2nj|fj,kɛ∥gj,kɛ|χj,k(x)M(h)(x)dx.
By Hölder's inequality and j≥ju,s≥0, it can be deduced that
(143)∑(ɛ,j,k)∈Λn,6Q‍2nj|fj,kɛ||gj,kɛ|χj,k(x)≲(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j|fj,kɛ|2χj,k(x))1/2×(∑(ɛ,j,k)∈Λn,6Q‍2(n+2t+2r)j|gj,kɛ|2χj,k(x))1/2≲2u|Q|(r+t)/n(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j|fj,kɛ|2χj,k(x))1/2.
Finally, we can get
(144)|〈T6,h〉|≲2u|Q|(r+t)/n∥(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j≲2u|Q|(r+t)/n2000gi×|fj,kɛ|2χj,k(x)∑∈Λn,6Q‍)1/2∥Lp(Q)∥h∥Lp′(Q)≲2u|Q|(r+t)/n|1-log2|Q∥-τ|Q|1/p-(r+t)/n≲2u|Q|1/p(1+ju,s)-τ.
Because h takes over all functions in Lp′(Q), it is obvious that
(145)∥S0(T6)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.Step
2. Assume that 0≤t<1. Let g(x)=∑(ɛ,j,k)∈Λngj,kɛΦj,kɛ(x) and
(146)St(g)(x)=(∑(ɛ,j,k)∈Λn‍2j(2t+n)|gj,kɛ|2χj,k(x))1/2.We need to prove that(147)∥St(fgu)∥Lp(Q)≲2u(1+ju,s)-τ|Q|1/p.The index setsΛn,iQ,i=1,2,…,6, are the same as Step 1.(1) For the termT1, we have
(148)St(T1)(x)=(∑|Qj,k|=|Qju,s,k|‍2ju,s(n+2t)2ju,sn==00i×|fju,s,k0|2|gju,s,k0|2χ(2ju,sx-k)∑|Qju,s,k|‍)1/2≲(∑|Qj,k|=|Qju,s,k|‍2ju,s(n+2t)2ju,sn(1+ju,s)-2τ22(r-n/2)ju,s×22u-2(r+t+n/2)ju,s22ju,s(n/2+t)χ(2ju,sx-k)∑=|Qju,s,k|‍)1/2≲(1+ju,s)-τ2u(∑|Qj,k|=|Qju,s,k|‍χj,k(x))1/2.
Then, we have
(149)∥T1∥Ht,p(Q)=∥St(T1)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.(2) For the term(150)T2(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+l0gj,kɛΦj,k+l0(x)Φj,kɛ(x),
we have
(151)St(T2)(x)=(∑(ɛ,j,k)∈Λn,2Q‍2j(n+2t)2jn|fj,k+l0|2|gj,kɛ|2χj,k(x))1/2≲(∑(ɛ,j,k)∈Λn,2Q‍2j(n+2t)2jn(1+j)-2τ×22j(r-n/2)|gj,kɛ|2χj,k(x)∑∈Λn,2Q‍)1/2≲(1+ju,s)-τSt+r(gu)(x).
Because gu(x) is a (t+r,2u,Eu)-combination atom, we have
(152)∥St(T2)∥Lp(Q)≲(1+ju,s)-τ∥St+r(gu)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.(3) Becausegu(x) is a (t+r,2u,Eu)-combination atom, for s≥1, (i,m′)∈Hu,s, and Q=Qu,s,i,m′, we can get
(153)|gj,k0|≲2u-nj/22-(r+t)j=2u-nj/2|Q|(r+t)/n,St(T3)(x)≲2u(∑(ɛ,j,k)∈Λn,3Q‍2j(n+2t)2-nj2-(r+t)j|fj,kɛ|2χj,k(x))1/2≲2u|Q|(r+t)/n(∑(ɛ,j,k)∈Λn,3Q‍2j(n+2t)|fj,kɛ|2χj,k(x))1/2.
From the fact that j≥ju,s and f∈Mr,pt,τ(ℝn), we deduce that
(154)∥St(T3)∥Lp(Q)≲2u|Q|(r+t)/n(-log2|Q|)-τ|Q|1/p-(r+t)/n≲2uju,s-τ|Q|1/p.(4) For the termT4(x), because f∈Mr,pt,τ(ℝn) and gu is a (t+r,2u,Eu)-combination atom, we have |fj,kɛ|≲(1+j)-τ2j(r-n/2) and
(155)St(T4)(x)=(∑(ɛ,j,k)∈Λn,4Q‍2j(n+2t)(1+j)-2τSt(T4)(x)ddddhhhhhh×2j(2r-n)2jn|gj,kɛ|2χj,k(x)∑(ɛ,j,k)∈Λn,4Q‍)1/2≲(1+ju,s)-τSt+r(gu)(x).
Then, we can get
(156)∥St(T4)∥Lp(Q)≲(1+ju,s)-τ∥St+r(gu)∥Lp(Q)≲2u(1+ju,s)-τ|Q|1/p.(5) Now we estimate the term(157)T5(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kɛgj,kɛ((Φj,kɛ(x))2-2nj/2Φj,k0(x)).
Because the function (Φj,kɛ(x))2-2nj/2Φj,k0(x) plays the role as that of Φj,kɛ(x), we have
(158)∥St(T5)∥Lp(Q)≲2u(1+ju,s)-τ|Q|1/p.(6) For the termT6(x), we take h∈H-t,p′(Q). By the orthogonality of the wavelet functions, we have
(159)〈T6,h〉=〈∑(ɛ,j,k)∈Λn,6Q‍fj,kɛgj,kɛ2jn/2Φj,k0,h〉=∑(ɛ,j,k)∈Λn,6Q‍fj,kɛgj,kɛ2nj/2hj,k0.
We can get
(160)|〈T6,h〉|≲|∫‍∑(ɛ,j,k)∈ΛnQ‍2nj/2gj,kɛfj,kɛ2jnhj,k0χj,k(x)dx|≲∫∑(ɛ,j,k)∈ΛnQ‍2nj|gj,kɛ||fj,kɛ|M(h)(x)χj,k(x)dx.
By Hölder's inequality, we have
(161)∑(ɛ,j,k)∈Λn,6Q‍2nj|fj,kɛ||gj,kɛ|χj,k(x)≲(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j|fj,kɛ|2χj,k(x))1/2×(∑(ɛ,j,k)∈Λn,6Q‍2(n+2t+2r)j|gj,kɛ|2χj,k(x))1/2≲St+r(gu)(x)(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j|fj,kɛ|2χj,k(x))1/2.
Because j≥ju,s implies that 2-j(r+t)≤2-ju,s(r+t), we can get
(162)|〈T6,h〉|≲2u∥(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j|fj,kɛ|2χj,k(x))1/2∥Lp(Q)×∥h∥Lp′(Q)≲2u|Q|(r+t)/n|1-log2|Q||-τ|Q|1/p-(r+t)/n≲2u|Q|1/p(1+ju,s)-τ.Becauseh takes over all functions in H-t,p′(Q), we can get
(163)∥St(T6)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.This completes the proof for the case0≤t<1.Step
3. Now we consider the case t≥1. In this case, there exists an integer [t] such that [t]≤t<[t]+1. For any α=(α1,α2,…,αn)∈ℕn with |α|=∑i=1nαi=[t], the derivative ∂α/∂xα of the product fgu can be represented as
(164)∂α∂xα(fgu)=∑γ,β‍Cα,β,γ(∂β∂xβ)f(x)∂γ∂xγgu(x),
where |α|=|β|+|γ|. Denote (∂β/∂xβ)f(x) and (∂γ/∂xγ)gu(x) by fβ(x) and gu,γ(x), respectively. Applying the conclusion in Step 1, we only need to prove
(165)∥fβgu,γ∥Ht-[t],p≲(1+ju,s)-τ2u|Q|1/p,
that is, ∥St-[t](fβgu,γ)∥Lp≲2u(1+ju,s)-τ|Q|1/p.Ifgu is a (t+r,2u,Eu)-combination atom, then
(166)gu,γ(x)=∑Qj,k∈ℱv,l‍gj,kɛ(∂∂x)γ[2nj/2Φj,kɛ(x)]=∑Qj,k∈ℱv,l‍gj,kɛ2nj/22j|γ|(∂γ∂xγΦɛ)(2jx-k).
Hence,
(167)St-|γ|(gu,γ)(x)=(∑(ɛ,j,k)∈Λn‍2j(n+2t-2|γ|)|gj,kɛ|22j|γ|χj,k(x))1/2=(∑(ɛ,j,k)∈Λn‍2j(n+2t)|gj,kɛ|2χj,k(x))1/2=St(gu)(x).
On the other hand, if f∈Mr,pt,τ(ℝn), then
(168)∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p(r+t)/n
and (fβ)j,kɛ=2j|β|fj,kɛ. We have
(169)∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t-2|β|)|fj,kβ,ɛ|2χj,k(x))p/2dx=∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t-2|β|)22j|β||fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p(t-|β|+r+|β|)/n;
that is, fβ(x)∈Mr+|β|,pt-|β|,τ(ℝn).For any cubeQ, the function fβ(x)=∑(ɛ,j,k)∈Λnfj,kβ,ɛΦj,kɛ(x), and any dyadic cube Q⊂Su, we divide the product fβ·gu,γ into the following parts:
(170)T1β,γ(x)=∑(ɛ,j,k)∈Λn,j=ju,s‍∑|l|≤n2(M+2)n‍fj,k+lβ,0gj,kγ,0Φj,k+l0(x)Φj,k0(x),T2β,γ(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+lβ,0gj,kγ,ɛΦj,k+l0(x)Φj,kɛ(x),T3β,γ(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+lβ,ɛgj,kγ,0Φj,k+lɛ(x)Φj,k0(x),T4β,γ(x)=∑(ɛ,ɛ′,j,k,l)∈Λ′n,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+lβ,ɛ′gj,kγ,ɛΦj,k+lɛ′(x)Φj,kɛ(x),T5β,γ(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kβ,ɛgj,kγ,ɛ((Φj,kɛ(x))2-2nj/2Φj,k0(x)),T6β,γ(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kβ,ɛgj,kγ,ɛ2nj/2Φj,k0(x).
Similar to the method used in the case 0≤t<1, we can complete the proof of the case t≥1. This completes the proof of this theorem.By Theorem30, we can get the following lemma.Lemma 31.
Givenr>0, t≥0, 1<p<n/(r+t), and τ>1/p′, if f∈Mr,pt,τ(ℝn) and gu is a (t+r,2u,Eu)-combination atom, then
(171)∥fgu∥Ht,p≲(1+u)-τ2u|Eu|1/p.Theorem 32.
Givenr>0, t≥0, 1<p<n/(r+t), and τ>1/p′, then Mr,pt,τ(ℝn)⊂Xr,pt(ℝn).Proof.
By Lemma31, we have
(172)∥fg∥Ht,p≲∑u≥0‍(1+u)-τ2u|Eu|1/p≲(∑u≥0‍(1+u)-p′τ)1/p′(∑u≥0‍2pu|Eu|)1/p≲∥g∥Ht+r,p.


### 4.2. The Sharpness forMr,pt,τ(ℝn)

In this section, applying our wavelet characterization of multiplier spaces and fractal theory, we prove that the scope of the index ofMr,pt,τ(ℝn) obtained in Theorem 32 is sharp for r+t<1. Precisely, by Meyer wavelets, we construct a counterexample to show that Theorem 32 is not true for the case 0≤τ≤1/p′.Our key idea is to construct a group of setsCs composed by special dyadic cubes and fractal set H with Hausdorff dimension n-(t+r)p. Denote by Ss the union ⋃Q∈CsQ and H=⋂s≥1Ss. By the above dyadic cubes Ss,s≥0, we construct a special Lp function g(x), which is bounded on Ss∖Ss+1 for all s≥1. The fractional integration Ir+tg(x) bumps on the fractal set H. Then we construct a multiplier f(x) such that its wavelet coefficients are based on these special dyadic cubes Cs for all s≥1. Applying our wavelet characterization of multiplier spaces, we prove that the product of the above multiplier f(x) and the function Ir+tg(x) will go out to the desired space Ht,p(ℝn).Theorem 33.
If0≤τ≤1/p′, there exists f∈Mr,pt,τ(ℝn) such that f∉Xr,pt(ℝn).Proof.
We use Meyer wavelets and suppose thatε=(1,1,…,1) and Φε(0)>0. First of all, we construct a group of self-similar cubes {Qu} such that the limitation is a set with Hausdorff measure n-(t+r)p.
We construct two series of integers{τs}s≥1 and {vs}s≥1 such that
(173)V1=max{[4nn-(t+r)p],2(M+2)n},1≤τs<[4nn-(t+r)p]≤vs≤max{[5nn-(t+r)p],2(M+2)n}.
Denote σs=∑1≤i≤sτi and us=∑1≤i≤svi. We take τs such that there exists C>0 satisfying that
(174)2nσs-(n-(t+r)p)us≥2-nu1,lims→+∞2nσs-(n-(t+r)p)us=C,n-(t+r)p=lims→+∞nσsus.
For τs and vs, denote ls=(ls,1,ls,2,…,ls,n)∈ℤs,0n if ls∈ℤn, and for i=1,2,…,n, we have
(175)0≤ls,i<2τs-1or0≤ls,i+2τs-1-2vs<2τs-1.
Denote ℤ1n=ℤ1,0n, and, for s≥2, denote ks∈ℤsn, if there exists ks-1∈ℤs-1n and ls∈ℤs,0n such that ks=2vsks-1+ls.
We divide the unit dyadic cubeQ0=[0,1]n into 2nv1 dyadic cubes. Then, we reserve the 2nτ1 dyadic cubes which are near the 2n vertex points; that is, we reserve C1={Qv1,l1:l1∈ℤ1,0n} and denote x∈S1, if there exists l1∈ℤ1,0n such that x∈Qv1,l1.
For the dyadic cubeQv1,l1∈C1, we divide it into 2nv2 dyadic cubes, and we reserve the 2nτ2 dyadic cubes which are near the 2n vertex points; that is, we reserve
(176)C2,l1={Qu2,2v2l1+l2,l2∈ℤ1,0n}.
For s=2, denote
(177)C2={Qu2,k2,k2=2v2l1+l2,l1,l2∈ℤ1,0n}
and denote x∈S2 if there exists k2∈ℤ2n such that x∈Qu2,k2.
We continue this process until infinity, and we get a series of dyadic cubesCs and sets Ss. We know that |Ss|=2n(σs-us) and the limitation of {Ss} is a fractal set H with Hausdorff dimension n-(t+r)p.
Fors≥1, let
(178)fs(x)=∑Qus,ks⊂Cs‍fus,ksɛΦus,ksε(x),
where fus,ksɛ=s-1/p′2-tus2(t+r)us. Let f(x)=∑s≥1fs(x). Applying the wavelet characterization of Morrey spaces, f∈Mr,pt,1/p′(ℝn). In fact, for s,l≥1 and any cube Q with 2-nus+1≤|Q|<2-nus, we have
(179)|Q∩(Ss+l∖Ss+l+1)|≲2-n(σs+l-σs)-nus.
Hence, we get
(180)∫Q‍(∑Qus,ks⊂Q‍2us(n+2t)|fus,ksɛ|2χ(2usx-ks))p/2dx≲∫Q∖Ss+1‍(∑Qus,ks⊂Q‍2us(n+2t)s-2/p′×2-2tus22(r+t)usχ(2usx-ks)∑Qus,ks⊂Q‍)p/2dx+∑l≥1‍∫Q∩(Ss+l∖SS+l+1)‍(∑Qus,ks⊂Q‍2us(n+2t)s-2/p′2-2tus22(r+t)us×χ(2usx-ks)∑Qus,ks⊂Q‍)p/2dx≲s-p/p′2(t+r)pus2-nus+∑l≥1‍(s+l)-p/p′2(t+r)pus+l2-n(σs+l-σs)-nus≲(1-log2|Q|)-p/p′|Q|1-(t+r)p/n.
Let δ be a sufficient small positive real number. For x∈[0,1]n∖S1, then g(x)=1. For s≥1 and x∈Ss∖Ss+1, we take
(181)g(x)=s-1/p[log2(1+s)]-(1+δ)/p2(t+r)us.
Then, we have g∈Lp([0,1]n). In fact,
(182)∫[0,1]n‍|g(x)|pdx=∫[0,1]n∖S1‍gp(x)dx+∑s≥1‍∫Ss∖Ss+1‍gp(x)dx≲1+∑s≥1‍∫Ss∖Ss+1‍s-1[log2(1+s)]-(1+δ)2(t+r)pusdx≲∑s≥1‍s-1[log2(1+s)]-(1+δ)2(t+r)pus2n(σs-us)≲1.
Now we estimate the coefficients|gus,ks|. We divide the estimate into two cases.
(1) Fordist(Qus,ks,Ss)≤2-us,
(183)gj,k=∫Ss‍g(y)(2-us+|y-2-usks|)(t+r-n)dy+∑1≤l≤s-1‍∫Sl∖Sl+1‍l-1/p[log2(1+l)]-(1+δ)/p2n(ul-σl)/p×(2-ul+|y-2-ulkl|)(t+r-n)dy≳s-1/p[log2(1+s)]-(1+δ)/p+∑1≤l≤s-1‍s-1/p[log2(1+s)]-(1+δ)/p≳s1/p′[log2(1+s)]-(1+δ)/p.
(2) Fordist(Qus,ks,Ss)>2-us, there exists l<s such that Qus,ks⊂Sl-1∖Sl. It is easy to see that gj,k is equivalent to l-1/p′[log2(1+l)]-(1+δ)/p.
Finally, we have(184)∫(∑(ɛ,j,k)∈Λn‍2j(n+2t)|gj,k|2|fj,kɛ|2χj,k(x))p/2dx≳∑s≥2‍(∑dist(Qus,ks,Ss)≤2-us‍2us(n+2t)|gus,ks|2≳00∑s≥2‍×|fus,ksɛ|2χ(2usx-ks)∑dist(Qus,ks,Ss)≤2-us‍)p/2dx≳∑s≥2‍(∑dist(Qus,ks,Ss)≤2-us‍2us(n+2t)s2/p′∑s≥200‍ki×[log2(1+s)]-2(1+δ)/ps-2/p′∑s≥2‍h00i×2-2tus+2(t+r)usχ(2usx-ks)∑dist(Qus,ks,Ss)≤2-us‍)p/2dx≳∑s≥2‍[log2(1+s)]-(1+δ)2(t+r)pus2n(σs-us)≳∑s≥2‍[log2(1+s)]-(1+δ)=∞.
This completes the proof.


## 4.1. A Logarithmic Condition

Lemma7 implies that Xr,pt(ℝn)⊂Mr,pt(ℝn). In this section, we consider the reverse inclusion relation. At first we introduce a logarithmic Morrey spaces.Definition 25.
Fix1<p<n/(r+t) and τ≥0. We say f(x)∈Mr,pt,τ(ℝn) if sup|Q|=1|ft,Q|≤C and
(101)∫Q‍|(I-Δ)t/2f(x)-ft,Q|pdx≲|1-log2|Q||-pτ|Q|1-p(r+t)/n,
for any cube Q with |Q|≤1.Similar to Theorem6, we have the following wavelet characterization of Mr,pt,τ(ℝn).Theorem 26.
Givent,τ≥0, r>0, 1<p<n/(r+t), f(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x) belongs to Mr,pt,τ(ℝn) if and only if
(102)∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p(r+t)/n,
where Q∈Ω with |Q|≤1.Proof.
Similar to that of Theorem6, the proof of this theorem can be obtained by the characterization of Triebel-Lizorkin spaces. See Lemma 4. We omit the detail.In [8], Fefferman established the following relation:
(103)Mr,qt(ℝn)⊂Xr,pt(ℝn)⊂Mr,pt(ℝn),
where q>p>1. In this section, we use wavelet characterization to give a logarithmic type inclusion. Let r>0, t≥0, 1<p<n/(r+t), and τ>1/p′. In Theorem 32, we prove that Mr,pt,τ(ℝn) is a subspace of Xr,pt(ℝn). Hence,
(104)Mr,pt,τ(ℝn)⊂Xr,pt(ℝn)⊂Mr,pt(ℝn)=Mr,pt,0(ℝn).Lemma 27.
Givenτ>0, r>0, t≥0, and 1<p<n/(r+t), if f∈Mr,pt,τ(ℝn), then
(105)|fj,kɛ|≲2j(r-n/2)(1+j)-τ.Proof.
Becausef∈Mr,pt,τ(ℝn), then for Q∈Ω,
(106)∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p(r+t)/n.
We have(107)∫Q(∑ɛ∈En,Qj,k⊂Q‍2j(n-2r)|fj,kɛ|2χj,k(x))p/2dx‍≲|Q|p(r+t)/n∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲|Q|p(r+t)/n|Q|1-p(r+t)/n(1-log2|Q|)-τp≲|Q|(1-log2|Q|)-τp.
Forɛ∈En, j∈ℕ, and k∈ℤn, take Q=Qj,k. By Lemma 4, we get
(108)∫(2jn-2jr|fj,kɛ|2χj,k(x))p/2dx≲2-jn(1-log22-jn)-τp≲2-jn(1+j)-τp,
which gives
(109)|fj,kɛ|≲2j(r-n/2)(1+j)-τ.
Whenɛ=0,
(110)fj,k0=〈∑(ɛ′,j′,k′)∈Λn‍fj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉=〈∑(ɛ′,j′,k′)∈Λn:j′<j‍fj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉.
Since(111)|∑(ɛ′,j′,k′)∈Λn:j′<j‍fj′,k′ɛ′Φj′,k′ɛ′(x)|≲2(r-n/2)j(1+j)-τ,
we have(112)|fj,k0|≲〈2rj(1+j)-τ,|Φj,k0|〉≲2(r-n/2)j(1+j)-τ.Letβ=(β1,β2,…,βn), βi∈ℕ+, where ℕ+ denotes the set of all positive integers. Denote by fβ(x) the derivative (∂/∂x)βf. We have the following two lemmas.Lemma 28.
Givenτ>0, r>0, t≥0, and 1<p<n/(r+t), if f∈Mr,pt,τ(ℝn), the derivative fβ∈Mr+|β|,pt-|β|,τ(ℝn), where |β|=∑i=1nβi.Proof.
Iff∈Mr,pt,τ(ℝn) and f(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x), by Theorem 26, we have
(113)∫(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p(r+t)/n.
Denote by fj,kɛ,β the wavelet coefficients of fβ(x). We can get fj,kβ,ɛ=2j|β|fj,kɛ and
(114)∫(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t-2|β|)|fj,kβ,ɛ|2χj,k(x))p/2dx≲∫‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p[(r+|β|)+(t-|β|)]/n.
This implies that fβ(x)∈Mr+|β|,pt-|β|,τ(ℝn).To get the sufficient condition for multiplier spaces, we need to consider carefully the relationship of different dyadic cubes relative to combination atoms. Because of this reason, we use always the same Daubechies wavelets in the rest of this section. For a cubeQ and a measurable set E, if |Q⋂‍E|=|Q|, we say that Q is a cube in the set E.For(m,2u,Eu)-combination atoms gu(x) defined in Definition 14, the measure of Eu is finite. Hence, the number of biggest dyadic cubes in Eu is finite. We denote the number of biggest dyadic cubes in Eu by i1. Denote by Fu,1 the set {i∈ℕ,i=1,…,i1}. If i∈Fu,1, we denote such cube by Qu,1,i. The volume of Qu,1,i is denoted by 2-nju,1; that is, |Qu,1,i|=2-nju,1. Denote Eu,1=Eu∖(⋃i∈Fu,1Qu,1,i).The measure ofEu,1 is finite. We denote the number of biggest dyadic cubes in Eu,1 by i2. Denote by Fu,2 the set {i∈ℕ,i=1,…,i2}. If i∈Fu,2, we denote such cube by Qu,2,i. The volume of Qu,2,i is denoted by 2-nju,2; that is, |Qu,2,i|=2-nju,2. Denote Eu,2=Eu,1∖(⋃i∈Fu,2Qu,2,i).We continue this process until there exists somes such that Eu,s+1 is empty. For s′≥s+1, we denote is′=0, and Fu,s′ and Eu,s′ are empty sets. Otherwise we continue until infinity. Then Eu=⋃s≥1,i∈Fu,sQu,s,i and gu(x)=∑s≥1,i∈Fu,sgu,s,i(x), where gu,s,i(x)=∑Qj,k⊂Qu,s,igj,kɛΦj,kɛ(x).To compute the norm off(x)gu(x), we need to find out a special set of dyadic cubes denoted by {Qu,s,i,k}(s,i,k)∈Hu such that suppgu⊂⋃(s,i,k)∈HuQu,s,i,k. gu(x) is nearly L∞ function on Qu,s,i,k and satisfies the estimate of Lemma 29. We divide such process into the following three steps.Step
1. For all s≥1, if  i∈Fu,s and k∈ℤn with |k|<n2(M+3)n, we denote (i,k)∈Gu,s. Denote Fu={(s,i,k),s≥1,(i,k)∈Gu,s}. For (s,i,k)∈Fu, we denote Qu,s,i,k=2-ju,sk+Qu,s,i. For s≥1, we denote E~u,s0=⋃(i,k)∈Gu,sQu,s,i,k. In the next step, we choose a special subcover to the support of gu(x).Step
2. We define now Hu,s(s≥1), E~u,s1(s≥1) and Gu,s1(s≥2).Fors=1, denote Hu,1=Gu,1 and E~u,s1=E~u,s0. For s=2, denote (i,k)∈Gu,21, if there exists 0≤j≤ju,1, l∈ℤn such that Qj,l⊂⋃|k-k′|≤n2(M+2)nQu,2,i,k′ and 〈gu,Φj,l0(x)〉≠0. We know that ⋃(i,k)∈Gu,21Qu,2,i,k⊂E~u,11. Denote Hu,2=Gu,2∖Gu,21 and E~u,21=⋃(i,k)∈Hu,2Qu,2,i,k.Fors=3, denote (i,k)∈Gu,31, if there exists 0≤j≤ju,2, l∈ℤn such that Qj,l⊂⋃|k-k′|≤n2(M+2)nQu,2,i,k′ and 〈gu(x),Φju,s,l0(x)〉≠0. We know that ⋃(i,k)∈Gu,31Qu,3,i,k⊂⋃1≤s≤2E~u,s1. Denote Hu,3=Gu,3∖Gu,31 and E~u,31=⋃(i,k)∈Hu,3Qu,3,i,k.We continue this process until infinity. Fors≥2, maybe, a part of Gu,s1, Hu,s, and E~u,s1 is empty set.Step
3. Let Hu={(s,i,k),s≥1,(i,k)∈Hu,s}. It is easy to see that the support of gu(x) is contained in ⋃s≥1E~u,s1. For a (t+r,2u,Eu)-combination atom gu(x) and gj,k0=〈gu,Φj,k0〉, we have the following estimate.Lemma 29.
Givenr>0, t≥0, 1<p<n/(t+r) and s≥1, for (i,m′)∈Hu,s and Qj,k⊂Qu,s,i,m′, we have |gj,k0|≲2u-nj/22-(t+r)ju,s.Proof.
By the definition ofHu,s, we have
(115)gj,k0=〈∑(ɛ′,j′,k′)∈Λn‍gj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉=〈∑(ɛ′,j′,k′)∈Λn:j′≥ju,s‍gj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉.
Becausegu is a (m,2u,Eu)-combination atom, St+r(gu)(x)≤C2u. Hence, for every (ɛ′,j′,k′), |gj′,k′ɛ′|≲2u2-j′(r+t+n/2). We can obtain
(116)|gj,k0|=|〈∑(ɛ′,j′,k′)∈Λn:j′≥ju,s‍gj′,k′ɛ′Φj′,k′ɛ′,Φj,k0〉|≲2u∫∑j′≥ju,s‍2-j′(r+t)2-jn′/22j′n/22jn/2×|Φɛ′(2j′x-k′)||Φ0(2jx-k)|dx≲2u2-jn/22-ju,s(r+t).Theorem 30.
Suppose thatτ>1/p′, t≥0, r>0, and 1<p<n/(r+t). If f∈Mr,pt,τ(ℝn) and gu is a (t+r,2u,Eu)-combination atom, then, for s≥1, (i,m′)∈Hu,s, and Q=Qu,s,i,m′, we have
(117)∥fgu∥Ht,p(Q)≲(1+ju,s)-τ2u|Q|1/p.Proof.
First, fort≥0, we prove ∥fgu∥Lp(Q)≤Cju,s-τ2u|Q|1/p. Let f(x)=∑(ɛ,j,k)∈Λnfj,kɛΦj,kɛ(x) and gu(x)=∑(ɛ,j,k)∈Λngj,kɛΦj,kɛ(x). Denote by Λn′ the set
(118){(ɛ,ɛ′,j,k,l),ɛ,ɛ′∈En,(ɛ,k)≠(ɛ′,k+l),j∈ℤ,k,l∈ℤn,|l|≤n2(M+2)n}.
For any dyadic cube Q⊂Eu, by the formulas (11) and (12), we decompose fgu as
(119)T1(x)=∑(ɛ,j,k)∈Λn,j=ju,s‍∑|l|≤n2(M+2)n‍fj,k+l0gj,k0Φj,k+l0(x)Φj,k0(x),T2(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+l0gj,kɛΦj,k+l0(x)Φj,kɛ(x),T3(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+lɛgj,k0Φj,k+lɛ(x)Φj,k0(x),T4(x)=∑(ɛ,ɛ′,j,k,l)∈Λ′n,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+lɛ′×gj,kɛΦj,k+lɛ′(x)Φj,kɛ(x),T5(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kɛgj,kɛ((Φj,kɛ(x))2-2nj/2Φj,k0(x)),T6(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kɛgj,kɛ2nj/2Φj,k0(x).
The rest of the proof is divided into three steps.Step
1. For t≥0, we estimate ∥Ti∥Lp(Q),i=1,2,…,6. By Lemma 4, we obtain
(120)f=∑(ɛ,j,k)∈Λn‍fj,kɛΦj,kɛ(x)∈Lp⟺∥(∑(ɛ,j,k)∈Λn‍2jn|fj,kɛ|2χ(2j·-k))1/2∥Lp.
Let
(121)S0(f)(x)=(∑(ɛ,j,k)∈Λn‍2jn|fj,kɛ|2χj,k(x))1/2.
We have ∥f∥Lp≃∥S0(f)∥Lp.(1) Becausef∈Mr,pt,τ, by Lemma 27,
(122)|fju,s,k0|≲(1+ju,s)-τ2(r-n/2)ju,s.
By Lemma 29, we have |gju,s,k0|≲2u-(r+t+n/2)ju,s. Hence we can get
(123)S0(T1)(x)=(∑|Qj,k|=|Qju,s,k|‍22ju,sn|fju,s,k0|2|gju,s,k0|2χ(2ju,sx-k))1/2≲(∑|Qj,k|=|Qju,s,k|‍22ju,sn(1+ju,s)-2τ22(r-n/2)ju,s×22u-2(r+t+n/2)ju,s22ju,s(n/2+t)χ(2ju,sx-k)∑|Qju,s,k|‍)1/2.
Because ju,s≥0, we have
(124)S0(T1)(x)≲(1+ju,s)-τ2u(∑|Qj,k|=|Qju,s,k|‍χj,k(x))1/2,∥T1∥Lp(Q)=∥S0(T1)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.(2) Now we estimate(125)T2(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+l0gj,kɛΦj,k+l0(x)Φj,kɛ(x).
Because j≥ju,s≥0, we have
(126)|fj,k0|≲(1+ju,s)-τ2(r-n/2)j≤(1+ju,s)-τ2(r+t-n/2)j.
Let
(127)Λn,2Q={(ɛ,j,k)∈Λn∣j≥ju,s,∀|l|≤n2(M+2)n,|supp(Φj,k+l0(x)Φj,kɛ(x))∩Q|≠0}.
Then, we can get
(128)S0(T2)(x)=(∑(ɛ,j,k)∈Λn,2Q‍|fj,k+l0|2|gj,kɛ|2χj,k(x))1/2≲(∑(ɛ,j,k)∈Λn,2Q‍22jn(1+j)-2τ×22j(r+t-n/2)|gj,kɛ|2χj,k(x)∑∈Λn,2Q‍)1/2≲(1+ju,s)-τSt+r(gu)(x).
Because gu(x) is a (t+r,2u,Eu)-combination atom,
(129)∥S0(T2)∥Lp(Q)≲(1+ju,s)-τ∥St+r(gu)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.(3) Sincegu(x) is a (t+r,2u,Eu)-combination atom, for s≥1, (i,m′)∈Hu,s, j≥0, and Q=Qu,s,i,m′,
(130)|gj,k0|≲2u-nj/22-(r+t)j=2u-nj/2|Q|(r+t)/n.
Let
(131)Λn,3Q={(ɛ,j,k)∈Λn∣j≥ju,s,∀|l|≤n2(M+2)n,i52|supp(Φj,k0Φj,k+lɛ)∩Q|≠0}.
We have, by ju,s≥0,
(132)S0(T3)(x)=(∑(ɛ,j,k)∈Λn,3Q‍2jn|gj,k0|2|fj,kɛ|2χj,k(X))1/2≲2u|Q|(r+t)/n(∑(ɛ,j,k)∈Λn,3Q‍2j(n+2t)|fj,kɛ|2χj,k(x))1/2.
By the fact that j≥ju,s and f∈Mr,pt,τ(ℝn), we get
(133)∥S0(T3)∥Lp(Q)≲2u|Q|(r+t)/n(1-log2|Q|)-τ|Q|1/p-(r+t)/n≲2u(1+ju,s)-τ|Q|1/p.(4) Now we estimate the termT4(x). Let
(134)Λn,4Q={(ɛ,j,k)∈Λn∣j≥ju,s,∀(ɛ,ɛ′,j,k,l)∈Λ′n,|supp(Φj,k+lɛ′(x)Φj,kɛ(x))∩Q|≠0}.
Because f∈Mr,pt,τ(ℝn), we have
(135)|fj,kɛ|≲(1+j)-τ2j(r-n/2)≲(1+j)-τ2j(r+t-n/2),(136)S0(T4)(x)=(∑(ɛ,j,k)∈Λn,4Q‍2jn(1+j)-2τ×2j(2r+2t-n)2jn|gj,kɛ|2χj,k(x)∑∈Λn,4Q‍)1/2≲(1+ju,s)-τSt+r(gu)(x).
Then we can get, by the fact that gu is a (t+r,2u,Eu)-combination atom,
(137)∥S0(T4)∥Lp(Q)≲(1+ju,s)-τ∥St+r(gu)∥Lp(Q)≲2u(1+ju,s)-τ|Q|1/p.(5) Now we estimate the term(138)T5(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kɛgj,kɛ((Φj,kɛ(x))2-2nj/2Φj,k0(x)).
Because the function (Φj,kɛ)2-2nj/2Φj,k0 plays the role as that of Φj,kɛ(x), we have
(139)∥S0(T5)∥Lp(Q)≲2u(1+ju,s)-τ|Q|1/p.(6) To estimate the termT6(x), we take h∈Lp′(Q). Let
(140)Λn,6Q={|supp(Φj,k+l0(x)Φj,kɛ(x))∩Q|(ɛ,j,k)∈Λn∣j≥ju,s,i|supp(Φj,k+l0(x)Φj,kɛ(x))∩Q|≠0}.
By the orthogonality of the wavelet function, we have
(141)〈T6,h〉=〈∑(ɛ,j,k)∈Λn,6Q‍fj,kɛgj,kɛ2jn/2Φj,k0,h〉=∑(ɛ,j,k)∈Λn,6Q‍fj,kɛgj,kɛ2nj/2hj,k0.
Then,
(142)|〈T6,h〉|≲∫∑(ɛ,j,k)∈Λn,6Q‍2nj|fj,kɛ∥gj,kɛ|2nj/2|hj,k0|χj,k(x)dx≲∫∑(ɛ,j,k)∈Λn,6Q‍2nj|fj,kɛ∥gj,kɛ|χj,k(x)M(h)(x)dx.
By Hölder's inequality and j≥ju,s≥0, it can be deduced that
(143)∑(ɛ,j,k)∈Λn,6Q‍2nj|fj,kɛ||gj,kɛ|χj,k(x)≲(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j|fj,kɛ|2χj,k(x))1/2×(∑(ɛ,j,k)∈Λn,6Q‍2(n+2t+2r)j|gj,kɛ|2χj,k(x))1/2≲2u|Q|(r+t)/n(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j|fj,kɛ|2χj,k(x))1/2.
Finally, we can get
(144)|〈T6,h〉|≲2u|Q|(r+t)/n∥(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j≲2u|Q|(r+t)/n2000gi×|fj,kɛ|2χj,k(x)∑∈Λn,6Q‍)1/2∥Lp(Q)∥h∥Lp′(Q)≲2u|Q|(r+t)/n|1-log2|Q∥-τ|Q|1/p-(r+t)/n≲2u|Q|1/p(1+ju,s)-τ.
Because h takes over all functions in Lp′(Q), it is obvious that
(145)∥S0(T6)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.Step
2. Assume that 0≤t<1. Let g(x)=∑(ɛ,j,k)∈Λngj,kɛΦj,kɛ(x) and
(146)St(g)(x)=(∑(ɛ,j,k)∈Λn‍2j(2t+n)|gj,kɛ|2χj,k(x))1/2.We need to prove that(147)∥St(fgu)∥Lp(Q)≲2u(1+ju,s)-τ|Q|1/p.The index setsΛn,iQ,i=1,2,…,6, are the same as Step 1.(1) For the termT1, we have
(148)St(T1)(x)=(∑|Qj,k|=|Qju,s,k|‍2ju,s(n+2t)2ju,sn==00i×|fju,s,k0|2|gju,s,k0|2χ(2ju,sx-k)∑|Qju,s,k|‍)1/2≲(∑|Qj,k|=|Qju,s,k|‍2ju,s(n+2t)2ju,sn(1+ju,s)-2τ22(r-n/2)ju,s×22u-2(r+t+n/2)ju,s22ju,s(n/2+t)χ(2ju,sx-k)∑=|Qju,s,k|‍)1/2≲(1+ju,s)-τ2u(∑|Qj,k|=|Qju,s,k|‍χj,k(x))1/2.
Then, we have
(149)∥T1∥Ht,p(Q)=∥St(T1)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.(2) For the term(150)T2(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+l0gj,kɛΦj,k+l0(x)Φj,kɛ(x),
we have
(151)St(T2)(x)=(∑(ɛ,j,k)∈Λn,2Q‍2j(n+2t)2jn|fj,k+l0|2|gj,kɛ|2χj,k(x))1/2≲(∑(ɛ,j,k)∈Λn,2Q‍2j(n+2t)2jn(1+j)-2τ×22j(r-n/2)|gj,kɛ|2χj,k(x)∑∈Λn,2Q‍)1/2≲(1+ju,s)-τSt+r(gu)(x).
Because gu(x) is a (t+r,2u,Eu)-combination atom, we have
(152)∥St(T2)∥Lp(Q)≲(1+ju,s)-τ∥St+r(gu)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.(3) Becausegu(x) is a (t+r,2u,Eu)-combination atom, for s≥1, (i,m′)∈Hu,s, and Q=Qu,s,i,m′, we can get
(153)|gj,k0|≲2u-nj/22-(r+t)j=2u-nj/2|Q|(r+t)/n,St(T3)(x)≲2u(∑(ɛ,j,k)∈Λn,3Q‍2j(n+2t)2-nj2-(r+t)j|fj,kɛ|2χj,k(x))1/2≲2u|Q|(r+t)/n(∑(ɛ,j,k)∈Λn,3Q‍2j(n+2t)|fj,kɛ|2χj,k(x))1/2.
From the fact that j≥ju,s and f∈Mr,pt,τ(ℝn), we deduce that
(154)∥St(T3)∥Lp(Q)≲2u|Q|(r+t)/n(-log2|Q|)-τ|Q|1/p-(r+t)/n≲2uju,s-τ|Q|1/p.(4) For the termT4(x), because f∈Mr,pt,τ(ℝn) and gu is a (t+r,2u,Eu)-combination atom, we have |fj,kɛ|≲(1+j)-τ2j(r-n/2) and
(155)St(T4)(x)=(∑(ɛ,j,k)∈Λn,4Q‍2j(n+2t)(1+j)-2τSt(T4)(x)ddddhhhhhh×2j(2r-n)2jn|gj,kɛ|2χj,k(x)∑(ɛ,j,k)∈Λn,4Q‍)1/2≲(1+ju,s)-τSt+r(gu)(x).
Then, we can get
(156)∥St(T4)∥Lp(Q)≲(1+ju,s)-τ∥St+r(gu)∥Lp(Q)≲2u(1+ju,s)-τ|Q|1/p.(5) Now we estimate the term(157)T5(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kɛgj,kɛ((Φj,kɛ(x))2-2nj/2Φj,k0(x)).
Because the function (Φj,kɛ(x))2-2nj/2Φj,k0(x) plays the role as that of Φj,kɛ(x), we have
(158)∥St(T5)∥Lp(Q)≲2u(1+ju,s)-τ|Q|1/p.(6) For the termT6(x), we take h∈H-t,p′(Q). By the orthogonality of the wavelet functions, we have
(159)〈T6,h〉=〈∑(ɛ,j,k)∈Λn,6Q‍fj,kɛgj,kɛ2jn/2Φj,k0,h〉=∑(ɛ,j,k)∈Λn,6Q‍fj,kɛgj,kɛ2nj/2hj,k0.
We can get
(160)|〈T6,h〉|≲|∫‍∑(ɛ,j,k)∈ΛnQ‍2nj/2gj,kɛfj,kɛ2jnhj,k0χj,k(x)dx|≲∫∑(ɛ,j,k)∈ΛnQ‍2nj|gj,kɛ||fj,kɛ|M(h)(x)χj,k(x)dx.
By Hölder's inequality, we have
(161)∑(ɛ,j,k)∈Λn,6Q‍2nj|fj,kɛ||gj,kɛ|χj,k(x)≲(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j|fj,kɛ|2χj,k(x))1/2×(∑(ɛ,j,k)∈Λn,6Q‍2(n+2t+2r)j|gj,kɛ|2χj,k(x))1/2≲St+r(gu)(x)(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j|fj,kɛ|2χj,k(x))1/2.
Because j≥ju,s implies that 2-j(r+t)≤2-ju,s(r+t), we can get
(162)|〈T6,h〉|≲2u∥(∑(ɛ,j,k)∈Λn,6Q‍2(n-2t-2r)j|fj,kɛ|2χj,k(x))1/2∥Lp(Q)×∥h∥Lp′(Q)≲2u|Q|(r+t)/n|1-log2|Q||-τ|Q|1/p-(r+t)/n≲2u|Q|1/p(1+ju,s)-τ.Becauseh takes over all functions in H-t,p′(Q), we can get
(163)∥St(T6)∥Lp(Q)≲(1+ju,s)-τ2u|Q|1/p.This completes the proof for the case0≤t<1.Step
3. Now we consider the case t≥1. In this case, there exists an integer [t] such that [t]≤t<[t]+1. For any α=(α1,α2,…,αn)∈ℕn with |α|=∑i=1nαi=[t], the derivative ∂α/∂xα of the product fgu can be represented as
(164)∂α∂xα(fgu)=∑γ,β‍Cα,β,γ(∂β∂xβ)f(x)∂γ∂xγgu(x),
where |α|=|β|+|γ|. Denote (∂β/∂xβ)f(x) and (∂γ/∂xγ)gu(x) by fβ(x) and gu,γ(x), respectively. Applying the conclusion in Step 1, we only need to prove
(165)∥fβgu,γ∥Ht-[t],p≲(1+ju,s)-τ2u|Q|1/p,
that is, ∥St-[t](fβgu,γ)∥Lp≲2u(1+ju,s)-τ|Q|1/p.Ifgu is a (t+r,2u,Eu)-combination atom, then
(166)gu,γ(x)=∑Qj,k∈ℱv,l‍gj,kɛ(∂∂x)γ[2nj/2Φj,kɛ(x)]=∑Qj,k∈ℱv,l‍gj,kɛ2nj/22j|γ|(∂γ∂xγΦɛ)(2jx-k).
Hence,
(167)St-|γ|(gu,γ)(x)=(∑(ɛ,j,k)∈Λn‍2j(n+2t-2|γ|)|gj,kɛ|22j|γ|χj,k(x))1/2=(∑(ɛ,j,k)∈Λn‍2j(n+2t)|gj,kɛ|2χj,k(x))1/2=St(gu)(x).
On the other hand, if f∈Mr,pt,τ(ℝn), then
(168)∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t)|fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p(r+t)/n
and (fβ)j,kɛ=2j|β|fj,kɛ. We have
(169)∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t-2|β|)|fj,kβ,ɛ|2χj,k(x))p/2dx=∫Q‍(∑ɛ∈En,Qj,k⊂Q‍2j(n+2t-2|β|)22j|β||fj,kɛ|2χj,k(x))p/2dx≲(1-log2|Q|)-pτ|Q|1-p(t-|β|+r+|β|)/n;
that is, fβ(x)∈Mr+|β|,pt-|β|,τ(ℝn).For any cubeQ, the function fβ(x)=∑(ɛ,j,k)∈Λnfj,kβ,ɛΦj,kɛ(x), and any dyadic cube Q⊂Su, we divide the product fβ·gu,γ into the following parts:
(170)T1β,γ(x)=∑(ɛ,j,k)∈Λn,j=ju,s‍∑|l|≤n2(M+2)n‍fj,k+lβ,0gj,kγ,0Φj,k+l0(x)Φj,k0(x),T2β,γ(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+lβ,0gj,kγ,ɛΦj,k+l0(x)Φj,kɛ(x),T3β,γ(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+lβ,ɛgj,kγ,0Φj,k+lɛ(x)Φj,k0(x),T4β,γ(x)=∑(ɛ,ɛ′,j,k,l)∈Λ′n,j≥ju,s‍∑|l|≤n2(M+2)n‍fj,k+lβ,ɛ′gj,kγ,ɛΦj,k+lɛ′(x)Φj,kɛ(x),T5β,γ(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kβ,ɛgj,kγ,ɛ((Φj,kɛ(x))2-2nj/2Φj,k0(x)),T6β,γ(x)=∑(ɛ,j,k)∈Λn,j≥ju,s‍fj,kβ,ɛgj,kγ,ɛ2nj/2Φj,k0(x).
Similar to the method used in the case 0≤t<1, we can complete the proof of the case t≥1. This completes the proof of this theorem.By Theorem30, we can get the following lemma.Lemma 31.
Givenr>0, t≥0, 1<p<n/(r+t), and τ>1/p′, if f∈Mr,pt,τ(ℝn) and gu is a (t+r,2u,Eu)-combination atom, then
(171)∥fgu∥Ht,p≲(1+u)-τ2u|Eu|1/p.Theorem 32.
Givenr>0, t≥0, 1<p<n/(r+t), and τ>1/p′, then Mr,pt,τ(ℝn)⊂Xr,pt(ℝn).Proof.
By Lemma31, we have
(172)∥fg∥Ht,p≲∑u≥0‍(1+u)-τ2u|Eu|1/p≲(∑u≥0‍(1+u)-p′τ)1/p′(∑u≥0‍2pu|Eu|)1/p≲∥g∥Ht+r,p.


## 4.2. The Sharpness forMr,pt,τ(ℝn)

In this section, applying our wavelet characterization of multiplier spaces and fractal theory, we prove that the scope of the index ofMr,pt,τ(ℝn) obtained in Theorem 32 is sharp for r+t<1. Precisely, by Meyer wavelets, we construct a counterexample to show that Theorem 32 is not true for the case 0≤τ≤1/p′.Our key idea is to construct a group of setsCs composed by special dyadic cubes and fractal set H with Hausdorff dimension n-(t+r)p. Denote by Ss the union ⋃Q∈CsQ and H=⋂s≥1Ss. By the above dyadic cubes Ss,s≥0, we construct a special Lp function g(x), which is bounded on Ss∖Ss+1 for all s≥1. The fractional integration Ir+tg(x) bumps on the fractal set H. Then we construct a multiplier f(x) such that its wavelet coefficients are based on these special dyadic cubes Cs for all s≥1. Applying our wavelet characterization of multiplier spaces, we prove that the product of the above multiplier f(x) and the function Ir+tg(x) will go out to the desired space Ht,p(ℝn).Theorem 33.
If0≤τ≤1/p′, there exists f∈Mr,pt,τ(ℝn) such that f∉Xr,pt(ℝn).Proof.
We use Meyer wavelets and suppose thatε=(1,1,…,1) and Φε(0)>0. First of all, we construct a group of self-similar cubes {Qu} such that the limitation is a set with Hausdorff measure n-(t+r)p.
We construct two series of integers{τs}s≥1 and {vs}s≥1 such that
(173)V1=max{[4nn-(t+r)p],2(M+2)n},1≤τs<[4nn-(t+r)p]≤vs≤max{[5nn-(t+r)p],2(M+2)n}.
Denote σs=∑1≤i≤sτi and us=∑1≤i≤svi. We take τs such that there exists C>0 satisfying that
(174)2nσs-(n-(t+r)p)us≥2-nu1,lims→+∞2nσs-(n-(t+r)p)us=C,n-(t+r)p=lims→+∞nσsus.
For τs and vs, denote ls=(ls,1,ls,2,…,ls,n)∈ℤs,0n if ls∈ℤn, and for i=1,2,…,n, we have
(175)0≤ls,i<2τs-1or0≤ls,i+2τs-1-2vs<2τs-1.
Denote ℤ1n=ℤ1,0n, and, for s≥2, denote ks∈ℤsn, if there exists ks-1∈ℤs-1n and ls∈ℤs,0n such that ks=2vsks-1+ls.
We divide the unit dyadic cubeQ0=[0,1]n into 2nv1 dyadic cubes. Then, we reserve the 2nτ1 dyadic cubes which are near the 2n vertex points; that is, we reserve C1={Qv1,l1:l1∈ℤ1,0n} and denote x∈S1, if there exists l1∈ℤ1,0n such that x∈Qv1,l1.
For the dyadic cubeQv1,l1∈C1, we divide it into 2nv2 dyadic cubes, and we reserve the 2nτ2 dyadic cubes which are near the 2n vertex points; that is, we reserve
(176)C2,l1={Qu2,2v2l1+l2,l2∈ℤ1,0n}.
For s=2, denote
(177)C2={Qu2,k2,k2=2v2l1+l2,l1,l2∈ℤ1,0n}
and denote x∈S2 if there exists k2∈ℤ2n such that x∈Qu2,k2.
We continue this process until infinity, and we get a series of dyadic cubesCs and sets Ss. We know that |Ss|=2n(σs-us) and the limitation of {Ss} is a fractal set H with Hausdorff dimension n-(t+r)p.
Fors≥1, let
(178)fs(x)=∑Qus,ks⊂Cs‍fus,ksɛΦus,ksε(x),
where fus,ksɛ=s-1/p′2-tus2(t+r)us. Let f(x)=∑s≥1fs(x). Applying the wavelet characterization of Morrey spaces, f∈Mr,pt,1/p′(ℝn). In fact, for s,l≥1 and any cube Q with 2-nus+1≤|Q|<2-nus, we have
(179)|Q∩(Ss+l∖Ss+l+1)|≲2-n(σs+l-σs)-nus.
Hence, we get
(180)∫Q‍(∑Qus,ks⊂Q‍2us(n+2t)|fus,ksɛ|2χ(2usx-ks))p/2dx≲∫Q∖Ss+1‍(∑Qus,ks⊂Q‍2us(n+2t)s-2/p′×2-2tus22(r+t)usχ(2usx-ks)∑Qus,ks⊂Q‍)p/2dx+∑l≥1‍∫Q∩(Ss+l∖SS+l+1)‍(∑Qus,ks⊂Q‍2us(n+2t)s-2/p′2-2tus22(r+t)us×χ(2usx-ks)∑Qus,ks⊂Q‍)p/2dx≲s-p/p′2(t+r)pus2-nus+∑l≥1‍(s+l)-p/p′2(t+r)pus+l2-n(σs+l-σs)-nus≲(1-log2|Q|)-p/p′|Q|1-(t+r)p/n.
Let δ be a sufficient small positive real number. For x∈[0,1]n∖S1, then g(x)=1. For s≥1 and x∈Ss∖Ss+1, we take
(181)g(x)=s-1/p[log2(1+s)]-(1+δ)/p2(t+r)us.
Then, we have g∈Lp([0,1]n). In fact,
(182)∫[0,1]n‍|g(x)|pdx=∫[0,1]n∖S1‍gp(x)dx+∑s≥1‍∫Ss∖Ss+1‍gp(x)dx≲1+∑s≥1‍∫Ss∖Ss+1‍s-1[log2(1+s)]-(1+δ)2(t+r)pusdx≲∑s≥1‍s-1[log2(1+s)]-(1+δ)2(t+r)pus2n(σs-us)≲1.
Now we estimate the coefficients|gus,ks|. We divide the estimate into two cases.
(1) Fordist(Qus,ks,Ss)≤2-us,
(183)gj,k=∫Ss‍g(y)(2-us+|y-2-usks|)(t+r-n)dy+∑1≤l≤s-1‍∫Sl∖Sl+1‍l-1/p[log2(1+l)]-(1+δ)/p2n(ul-σl)/p×(2-ul+|y-2-ulkl|)(t+r-n)dy≳s-1/p[log2(1+s)]-(1+δ)/p+∑1≤l≤s-1‍s-1/p[log2(1+s)]-(1+δ)/p≳s1/p′[log2(1+s)]-(1+δ)/p.
(2) Fordist(Qus,ks,Ss)>2-us, there exists l<s such that Qus,ks⊂Sl-1∖Sl. It is easy to see that gj,k is equivalent to l-1/p′[log2(1+l)]-(1+δ)/p.
Finally, we have(184)∫(∑(ɛ,j,k)∈Λn‍2j(n+2t)|gj,k|2|fj,kɛ|2χj,k(x))p/2dx≳∑s≥2‍(∑dist(Qus,ks,Ss)≤2-us‍2us(n+2t)|gus,ks|2≳00∑s≥2‍×|fus,ksɛ|2χ(2usx-ks)∑dist(Qus,ks,Ss)≤2-us‍)p/2dx≳∑s≥2‍(∑dist(Qus,ks,Ss)≤2-us‍2us(n+2t)s2/p′∑s≥200‍ki×[log2(1+s)]-2(1+δ)/ps-2/p′∑s≥2‍h00i×2-2tus+2(t+r)usχ(2usx-ks)∑dist(Qus,ks,Ss)≤2-us‍)p/2dx≳∑s≥2‍[log2(1+s)]-(1+δ)2(t+r)pus2n(σs-us)≳∑s≥2‍[log2(1+s)]-(1+δ)=∞.
This completes the proof.


## 5. An Application to Schrödinger Type Operators

LetL=I-Δ+V be a Schrödinger operator. Maz'ya and Verbitsky [11] established many sufficient and necessary conditions such that V is a multiplier from H1,2(ℝn) to H-1,2(ℝn). Their results can be used to study the Schrödinger operators L. In Section 4, Theorem 32 gives a relation between Morrey spaces and Sobolev multipliers. In this section, we give an application of the wavelet characterization of Xr,pt(ℝn) to the Schrödinger type operator I+(-Δ)r/2+V.ForV∈Mr,pt,τ(ℝn), τ>1/p′, and g∈Ht,p(ℝn), we want to find a solution f∈Ht+r,p(ℝn) to the equation
(185)(I+(-Δ)r/2+V)f(x)=g(x).Remark 34.
Fixedr>0, t≥0, and 1<p<n/(r+t). (i)
If there exists aδ>0 such that ∥V∥Cr+t+δ is sufficiently small, according to the continuity of Calderón-Zygmund operator (I+(-Δ)r/2)(I+(-Δ)r/2+V)-1, (185) can be solved easily. But if we consider a nonsmooth potential V∈Mr,pt,τ(ℝn), applying the same proof in Lemmas 8 and 9, it is possible that V is not a L∞ function.(ii)
The conditionτ>1/p′ cannot be weaken to τ≥1/p′. In fact, according to our counterexample in Section 5, if r+t<1, there exists some V∈Mr,pt,1/p′(ℝn) such that the operator I+(-Δ)r/2+V is not continuous from Ht+r,p(ℝn) to Ht,p(ℝn).Now, we use our sufficient condition of multiplier spacesXr,pt(ℝn) to get the solution of (185). We need the following two operators. For t,r>0, let
(186)Tt,r=[I+(-Δ)t/2,I+(-Δ)r/2],St,r=[I+(-Δ)t/2]VTt,r-1.Lemma 35.
Givenr>0, t≥0, 1<p<n/(r+t), and τ>1/p′, if V∈Mr,pt,τ(ℝn), the operator St,r is bounded from Lp(ℝn) to Lp(ℝn) with the operator norm less than Ct,r,τ,p∥V∥Mr,pt,τ, where Ct,r,τ,p denotes a constant associated with t, r, τ, and p.Proof.
By Theorem32, for any f∈Lp(ℝn), we have
(187)∥St,rf∥Lp≲∥[I+(-Δ)t/2]VTt,r-1f∥Lp≲∥VTt,r-1f∥Ht,p≲Cr,t,τ,p∥V∥Mr,pt,τ∥Tt,r-1f∥Ht+r,p≲Cr,t,τ,p∥V∥Mr,pt,τ∥f∥Lp.
Then, St,r is a bounded operator on Lp(ℝn) with the norm less than Cr,t,τ,p∥V∥Mr,pt,τ.In the following lemma, we prove that(I+St,r) is invertible in Lp(ℝn) and the inverse (I+St,r)-1 can be written formally as ∑n=0∞(-1)nSt,rn.Lemma 36.
Givenr>0, t≥0, 1<p<n/(r+t), and τ>1/p′, if ∥V∥Mr,pt,τ<1/Cr,t,τ,p, the operator I+St,r is invertible in Lp(ℝn).Proof.
By Lemma35, St,r is bounded on Lp(ℝn). Hence, for any f∈Lp(ℝn),
(188)∥∑n=0∞‍(-1)nSt,rnf∥Lp≲∑n=0∞‍∥St,rnf∥Lp≲∑n=0∞‍(Cr,t,τ,p∥V∥Mr,pt,τ)n∥f∥Lp.
If∥V∥Mr,pt,τ<1/Cr,t,τ,p, the above series is convergent in Lp(ℝn). Further,
(189)(I+St,r)[∑n=0∞‍(-1)nSt,rn]=∑n=0∞‍(-1)nSt,rn-∑n=1∞‍(-1)nSt,rn=I.
Similarly, we can also get
(190)[∑n=0∞‍(-1)nSt,rn](I+St,r)=I;
that is, the operator I+St,r is invertible in Lp(ℝn).Theorem 37.
Givenr>0, t≥0, 1<p<n/(r+t), and τ>1/p′, if ∥V∥Mr,pt,τ<1/Cr,t,τ,p, then, for g∈Ht,p(ℝn), there exists a unique solution f∈Hr+t,p(ℝn) for (185).Proof.
Becauseg∈Ht,p(ℝn), we have g~=(I+(-Δ)t/2)g∈Lp(ℝn). By Lemma 36, the operator (I+St,r) is invertible in Lp(ℝn). Hence, we can get that there exists a unique solution to the following equation in Lp(ℝn):
(191)(I+St,r)f~=g~,
where g~∈Lp(ℝn). Hence, for the above g~∈Lp(ℝn),
(192)f~=(I+St,r)-1g~=(I+St,r)-1(I+(-Δ)t/2)g(x)∈Lp(ℝn)
is a solution to (191). Write f=(I+(-Δ)r/2)-1(I+(-Δ)t/2)-1f~. Then, f~∈Lp(ℝn) is equivalent to f∈Hr+t,p(ℝn). It is easy to verify that f is a solution to (185). This completes the proof.



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*Source: 193420-2013-11-24.xml*