# Equivalent Characterization on Besov Space

Authors: Cong He; Jingchun Chen
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

In this paper, we give an equivalent characterization of the Besov space. This reveals the equivalent relation between the mixed derivative norm and single-variable norm. Fourier multiplier, real interpolation, and Littlewood-Paley decomposition are applied.


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## Body

## 1. Introduction

In Sobolev spaces, it is known that∥f∥H2ℝ2~∥f∥L2ℝ2+∑i=12∥∂2f/∂xi2∥L2ℝ2, where ∥f∥H2ℝ2≕∥f∥L2ℝ2+∥∂x1∂x2f∥L2ℝ2+∥∂x12f∥L2ℝ2+∥∂x22f∥L2ℝ2. Note that on the right hand side of the definition ∥f∥H2ℝ2, it contains the mixed derivative norm ∥∂x1∂x2f∥L2ℝ2. This mixed derivative norm would make the calculation more complicated or even infeasible to estimate partial differential equations with some anisotropy property, like Vlasov-Poisson equation [1, 2], in fractional Sobolev space [3]. So, separating variables becomes necessary and meaningful.In this paper, we aim to prove∥f∥Bp,rsℝn~∑j=1n∥f∥Bp,r,xjsℝn which realizes the separation, i.e., the right hand side does not contain the “mixed derivative” term, it only contains fractional derivative with respect to a single variable for each term. Thus, when it comes to estimate ∥f∥Bp,rsℝn in solving partial differential equations, it is equivalent to estimate ∥f∥Bp,r,xjsℝn individually. For the other equivalent characterizations for Besov spaces, refer to [4–7] and the references therein.


## 2. Preliminaries

We first recall definitions on Besov spaces, see [8]. Given f∈S which is the Schwartz function, its Fourier transform Ff=f^ is defined by
(1)f^ξ=12πn/2∫ℝne−ix·ξfxdx,and its inverse Fourier transform is defined by F−1fx=f^−x.We considerφ∈S satisfying suppφ⊂ξ∈ℝn:1/2≤∣ξ∣≤2. Setting φjξ=φ2−jξ with j=1,2,⋯,, we can adjust the normalization constant in front of φ and choose φ0∈S satisfying suppφ0⊂ξ∈ℝn:∣ξ∣≤2, such that
(2)∑j=0∞φjξ=1,∀ξ∈ℝn.We observe(3)suppφj∩suppφj′=∅if∣j−j′∣≥2.Givenf∈S′, we denote Δjf=F−1φjFf. For s,p,r∈ℝ×1,∞×1,∞, then we define the inhomogeneous Besov space by
(4)Bp,rs=f∈S′:fBp,rs=∑j=0∞2jsΔjfpljr<∞,with the usual interpretation for p=∞ or r=∞. Throughout this paper, all the function spaces are defined on Euclidean space ℝn; we will omit it whenever there is no confusion.Next, we would like to present some known results which will be used later. The first one is the unit decomposition.Lemma 1 (see [8], page 145).
Assume thatn≥2, and take φ as in the definition of Besov space. Then, there exist functions χj∈Sℝnj=1,⋯,n, such that
(5)∑j=1nχ^j=1onsuppφ=ξ:1/2≤∣ξ∣≤2,suppχ^j⊂ξ∈ℝn:∣ξj∣≥3n−1j=1,⋯,n.
Next, we recall the real interpolation characterization for Besov spaces.Lemma 2 (see [8], page 142).
Suppose1≤p,q≤∞,0<θ<1,s=1−θs0+θs1, where s0≠s1. We have
(6)Hps0,Hps1θ,q=Bp,qs.Remark 3.
We also have(7)Hp,xjs0,Hp,xjs1θ,q=Bp,q,xjs.
Its proof can be repeated the process of Lemma2 completely.


## 3. Equivalent Characterization

Now, we are in the position to state and prove our theorems. Firstly, we apply the Fourier multiplier [9] to prove that Hpsℝn=⋂j=1nHp,xjsℝn directly; Hps space has an advantage that the factor 1+ξ2s/2 is positive everywhere, which is fundamentally important when applying the Fourier multiplier theorem.For the sake of brevity, we denote(8)ξ=1+ξ21/2.We have the following equivalent norm theorem in Sobolev spaces.Theorem 4.
Suppose1<p<∞,s>0. We have
(9)Hps=⋂j=1nHp,xjs,where
(10)∥f∥Hps=∥F−1ξsf^∥Lp,∥f∥Hp,xjs=∥F−1ξjsf^∥Lp.Proof.
On the one hand, iff∈Hps, i.e., ∥F−1ξsFf∥p<∞ where ξ=1+ξ21/2. Note that, for any j=1,⋯,n, we have
(11)∥F−1ξjsFf∥p=∥F−1ξjsξsξsFf∥p.
Next, we just need to show thatm1ξ=ξjs/ξs is an Lp multiplier. To prove the assertion, we introduce an auxiliary function on ℝn+1 defined by
(12)m~1ξ,t=t2+ξj2t2+ξ2s/2.
It is easy to verify thatm~1 is homogeneous of degree 0 and smooth on ℝn+1\0. The derivatives ∂βm~1 are homogeneous of degree −∣β∣ and satisfy
(13)∣∂βm~1ξ,t∣≤Cβξ,t−∣β∣,withCβ=sup∣θ∣=1∂βm~1θ,whenever ξ,t≠0 and β is a multiindex of n+1 variables. In particular, taking β=α,0, we obtain
(14)∣∂αm~1ξ,t∣≤Cαt2+ξ2−∣α∣/2,and setting t=1, we deduce that ∣∂αm1ξ∣≤Cα1+ξ2−∣α∣/2≤Cαξ−∣α∣, which implies that m1ξ is an Lp Fourier multiplier by the Mihlin-Ho¨mander theorem [9] (page 446).
On the other hand, assumef∈⋂j=1nHp,xjs, that is, ∥F−1ξjsFf∥p<∞,foranyj=1,⋯,n. Note that,
(15)∥F−1ξsFf∥p=∥F−1ξs∑j=1nξjs∑j=1nξjsFf∥p.
Similarly, we can verify thatm2ξ=ξs/∑j=1nξjs is an Lp Fourier multiplier which finishes the proof of Theorem 4.We return to prove the equivalent characterization on Besov spaces. However, we cannot do the same trick as inHps space since φjξ is not positive everywhere as ξ. Fortunately, we have Hps0,Hps1θ,q=Bp,qs, see Lemma 2. This observation is favourable to prove the equivalent relation in one direction; however, for the other direction, we need a more delicate technique, in fact, we establish an identity by applying the Littlewood-Paley decomposition [10], which is very important in our proof. In what follows, A≲B means there exists a constant c independent of the main parameters such that A≤cB.A~B means A≲B and B≲A.Theorem 5.
Suppose1<p<∞,1≤q≤∞,s>0. We have
(16)Bp,qs=⋂j=1nBp,q,xjs,where ∥f∥Bp,q,xjs=∑k=0∞2ksq∥F−1φkjFf∥Lpq1/q and φkj is the dyadic block of the unit decomposition for the jth variable as in the definition of Besov spaces.Proof.
We split the proof into the following two steps:
Step I. To prove(17)Bp,qs⊂⋂j=1nBp,q,xjs
Assumef∈Bp,qs, by the real interpolation Lemma 2, we have
(18)∥f∥Bp,qs~∥f∥Hps0,Hps1θ,q~∫0∞t−θinff=f0+f1f0∈Hps0,f1∈Hps1∥f0∥Hps0+t∥f1∥Hps1qdtt1/q,where 0<θ<1,s=1−θs0+θs1, and we applied the equivalent norm for the interpolation space Hps0,Hps1θ,q, see [8] ((3) page 39 and (5) page 40).
By Remark3, we obtain, for any j=1,⋯,n,(19)∥f∥Bp,q,xjs~∥f∥Hp,xjs0,Hp,xjs1θ,q~∫0∞t−θinff=f0+f1,f0∈Hp,xjs0,f1∈Hp,xjs1∥f0∥Hp,xjs0+t∥f1∥Hp,xjs1qdtt1/q∫0∞t−θinff=f0+f1,f0∈Hps0,f1∈Hps1∥f0∥Hps0+t∥f1∥Hps1qdtt1/q,combining (18) and (19), it follows that
(20)∥f∥Bp,q,xjs≲∥f∥Bp,qs,the arbitrariness of j implies that (17) holds.
Step II. To prove(21)⋂j=1nBp,q,xjs⊂Bp,qs
Forn=1, it is trival.
Forn≥2, we need the following key claim.Claim.
There exists a positive integerm depending on n only such that
(22)∑∣l−k∣≤mφkχ^j,kφlj=φkχ^j,k,where
(23)φkξ=φ2−kξ,χ^j,kξ=χ^j2−kξ,φljξ=φ2−lξj,which is the dyadic block for jth variable, φk is the usual dyadic block as in the definition of Besov spaces, and χj is the same as in Lemma 1.Proof of Claim.
By Lemma1, we have φk=∑j=1nφkχ^j,k.
Note(24)∑l=0∞φkχ^j,kφlj=φkχ^j,k.
In order to getφkχ^j,kφlj≠0, for any chosen j and k, we must have
(25)2k−1≤∣ξ∣≤2k+1,2l−1≤ξj≤2l+1,∣ξj∣≥2k3n−1,which implies that ∣l−k∣≤m with m=log23n+1, ending the proof of the claim. With this claim in mind, we get
(26)2ks∥F−1φkF−1f∥Lp=2ks∥∑j=1nF−1φkχ^j,kFf∥Lp≤2ks∑j=1n∥F−1φkχ^j,kFf∥Lp≤2ks∑j=1n∥∑∣l−k∣≤mF−1φkχ^j,kφljFf∥Lp≤2ks∑j=1n∑∣l−k∣≤m∥F−1φkχ^j,kφljFf∥Lp≤2ms∑j=1n2ls∑∣l−k∣≤m∥F−1φljFf∥Lp,where we used the fact that φkχ^j,k is the Fourier multiplier.
With Young’s inequality [11], taking the lq norm on both sides of (26) yields that
(27)∥f∥Bp,qs≤Cn,s,q∑j=1n∥f∥Bp,q,xjs,which implies (21) holds; thus, we complete the proof our main theorem.Remark 6.
The methods could be adapted to the weighted Sobolev spaces and weighted Besov space, or even in the anisotropic function space.



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*Source: 6688250-2021-04-26.xml*