# Embedding of Besov Spaces and the Volterra Integral Operator

Authors: Dan Qu; Xiangling Zhu; Ruishen Qian
Journal: Journal of Mathematics
 (2021)

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## Abstract

The boundedness and compactness of the inclusion mapping from Besov spaces to tent spaces are studied in this paper. Meanwhile, the boundedness, compactness, and essential norm of the Volterra integral operatorTg from Besov spaces to a class of general function spaces are also investigated.


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## Body

## 1. Introduction

LetD be the unit disk in the complex plane ℂ and HD be the class of functions analytic in D. For 1<p<∞, the Besov space, denoted by Bp, is the space of all functions f∈HD satisfying(1)fBpp=f0p+∫Df′zp1−z2p−2dAz<∞.Let0<p<∞, −2<q<∞, and 0≤s<∞. The space Fp,q,s is the space consisting of all f∈HD such that(2)fFp,q,sp=f0p+supa∈D∫Df′zp1−z2q1−σaz2sdAz<∞,where σaz=a−z/1−a¯z. This space was first introduced by Zhao in [1]. F2,0,s is the Qs space (see [2]). F2,0,1 is the BMOA space. Fp,α,0 is called the Dirichlet-type space, denoted by Dαp. In particular, Fp,p−2,0 is the Besov space Bp. Fp,p,0 is just the classical Bergman space Ap. When s>1, from [1], we see that Fp,p−2,s is equivalent to the Bloch space, denoted by ℬ, which consists of all f∈HD such that(3)fℬ=f0+supz∈D1−z2f′z<∞.For1<q<∞ and 0<s,t<∞, let ℒℱq,q−2,s,t denote the space of all f∈HD such that(4)fLogq=supa∈D1log2/1−a2t∫Df′zq1−z2q−21−σaz2sdAz<∞.The norm forf∈ℒℱq,q−2,s,t is given by fℒℱq,q−2,s,t=f0+fLog. The space ℒℱq,q−2,s,t is a Banach space when q>1.The Volterra integral operatorTg was introduced by Pommerenke in [3]. Here,(5)Tgfz=∫0zfwg′wdw,f∈HD.In [3], Pommerenke showed that Tg is bounded on H2 if and only if g∈BMOA. Aleman and Siskakis showed that Tg is bounded on Hp(p≥1) if and only if g∈BMOA in [4]. In [5], Aleman and Siskakis proved that Tg:Ap⟶Ap is bounded (compact) if and only if g∈ℬg∈ℬ0. Recently, the operator Tg has been receiving much attention. See [4–18] and the references therein for more study of the operator Tg.For any arcI⊆∂D, the boundary of D, let I=1/2π∫Idζ denote the normalized length of I and SI be the Carleson box defined by(6)SI=z∈D:1−I≤z<1,zz∈I.Let0<s<∞ and μ be a positive Borel measure on D. We say that μ is a α-Carleson measure if(7)μCMα=supI⊂∂DμSIIα<∞.Whenα=1, it gives the classical Carleson measure. μ is said to be a vanishing α-Carleson measure if limI⟶0μSI/Iα=0. The Carleson measure is very useful in the theory of function spaces and operator theory. The famous embedding theorem says that the inclusion mapping Id:H2⟶L2dμ is bounded if and only if μ is a Carleson measure (see [19]). See [7, 20] for the study of the inclusion mapping Id:Bp⟶Lpdμ.Let0<s,q<∞,0≤t<∞, and μ be a positive Borel measure on D. Let Ts,tqμ denote the space of all μ-measurable functions f such that (see, e.g., [21])(8)supI⊂∂D1Islog2/It∫SIfzqdμz<∞.The tent spaceTs,tqμ was introduced by Liu et al. in [21]. When t=0, Ts,tqμ will be denoted by Tsqμ for the simplicity. In [21], Liu et al. studied the embedding of some Möbius invariant spaces, such as the Bloch space and the Qp space, into Ts,t2.In [12], Pau and Zhao showed that the inclusion mapping Id:Fp,p−2,s⟶Tspμ is bounded if and only if μ is a p-logarithmic s-Carleson measure. In [9], Li et al. proved that the inclusion mapping Id:Dp−1p⟶Tspμ is bounded if and only if μ is a s+1-Carleson measure. In [14], Qian and Li proved that the inclusion mapping Id:Bp⟶Tsqμ is bounded (resp. compact) if and only if μ is a q1−1/p-logarithmic s-Carleson measure (resp. vanishing q1−1/p-logarithmic s-Carleson measure) under the assumption that 1<p<q<∞ and 0<s<∞.Motivated by [14, 21], in this paper, we study the boundedness and compactness of the inclusion mapping Id:Bp⟶Ts,q1−1/pqμ. More precisely, we show that Id:Bp⟶Ts,q1−1/pqμ is bounded (resp. compact) if and only if μ is an s-Carleson measure (resp. vanishing s-Carleson measure) under the assumption that 1<p<q<∞ and 0<s<∞. As an application, we study the boundedness of the operator Tg:Bp⟶ℒℱq,q−2,s,q−q/p. Moreover, the compactness and essential norm of the operator Tg:Bp⟶ℒℱq,q−2,s,q−q/p are also investigated.In this paper, the symbolf≈g means that f≲g≲f. We say that f≲g if there exists a constant C such that f≤Cg.


## 2. Embedding the Besov SpaceBp into Ts,tqμ

We need the following equivalent description ofα-Carleson measure (see Lemma 2.2 in [12]).Lemma 1.
Let0<α,t<∞ and μ be a positive Borel measure on D. Then, μ is an α-Carleson measure if and only if(9)supa∈D∫D1−a2t1−a¯zα+tdμz<∞.
Moreover,(10)μCMα≈supa∈D∫D1−a2t1−a¯zα+tdμz.Using Lemma 3.10 in [22], we can easily obtain the following result.Lemma 2.
Let1<p<∞ and w∈D. Set(11)fwz=1log2/1−w21/plog21−w¯z,z∈D.
Then,fw∈Bp.Theorem 1.
Let1<p<q<∞, 0<s<∞, and μ be a positive Borel measure on D. Then, the inclusion mapping Id:Bp⟶Ts,q1−1/pqμ is bounded if and only if μ is an s-Carleson measure.Proof.
First, we assume thatId:Bp⟶Ts,q1−1/pqμ is bounded. For any given arc I⊆∂D, set a=1−Iη, and η is the center point of I. It is easy to see that(12)1−a¯z≈1−a2≈I,z∈SI.
Let(13)faz=1log2/1−a21/plog21−a¯z.
By Lemma2, we see that fa∈Bp. From the boundedness of Id:Bp⟶Ts,q1−1/pqμ, we have(14)faTsqμq=supI⊂∂D1Islog2/Iq1−1/p∫SIfazqdμz<∞.
By the fact thatfaz≈log2/I1−1/p when z∈SI, we get(15)supI⊂∂DμSIIs<∞.
Hence,μ is an s-Carleson measure.
Conversely, assume thatμ is an s-Carleson measure. Let f∈Bp. For any given arc I⊆∂D, set w=1−Iη, and η is the center point of I. Then,(16)1Islog2/Iq1−1/p∫SIfzqdμz≲1Islog2/Iq1−1/p∫SIfz−fwqdμz+1Islog2/Iq1−1/p∫SIfwqdμz=A+B,where(17)A=1Islog2/Iq1−1/p∫SIfz−fwqdμz,B=1Islog2/Iq1−1/p∫SIfwqdμz.
Since(18)fw≲log21−w21−1/pfBp≲log2I1−1/pfBp,we get(19)B≲μSIIsfBpq≲fBpq.
Now, we turn to estimateA. By Theorem 1 in [7], we see that μ is an s-Carleson measure if and only if Id:Dp−2+ps/qp⟶Lqμ is bounded. Note that(20)f∈Bp⊆Dp−2+ps/qp.
Then,(21)A≲1−w2s∫SIfz−fw1−w¯z2s/qqdμz≲1−|2ps/q∫Dfz−fw1−w¯z2s/q′p1−z2p−2+ps/qdAzq/p.
Since(22)fz−fw1−w¯z2s/q′=f′z1−w¯z2s/q+w¯2s/qfz−fw1−w¯z2s/q−11−w¯z4s/q,we deduce that A≲W1+W2q/p, where(23)W1=1−w2ps/q∫Df′zp1−w¯z2ps/q1−z2p−2+ps/qdAz,W2=1−w2ps/q∫Dfz−fwp1−w¯z2ps/q+p1−z2p−2+ps/qdAz.
Since(24)1−w21−z21−w¯z2=1−σwz2≤1,we get that(25)W1≲fBpp.
Making the change of variableη=σwz and combining with Proposition 4.2 in [22], we have(26)W2=1−w2ps/q∫Df°σwη−f°σw0p1−w¯σwη2ps/q+p1−σwη2p−2+ps/q×1−w221−w¯η4dAη=∫Df°σwη−f°σw0p1−η2p−2+ps/q1−w¯ηpdAη≲∫Df°σw′ηp1−η22p−2+ps/q1−w¯ηpdAη≲∫Df′σwηp1−σwη2p1−η2p−2+ps/q1−w¯ηpdAη≲∫Df′zp1−z2p1−σwz2p−2+ps/q1−w¯σwzp1−w221−w¯z4dAz≲∫Df′zp1−w2ps/q1−z22p−2+ps/q1−w¯zp+2ps/qdAz≲fBpp.
Therefore,(27)supI⊂∂D1Islog2/Iq1−1/p∫SIfzqdμz≲fBpq,which implies the desired result. The proof is completed.
We say that the inclusion mappingId:Bp⟶Ts,q1−1/pqμ is compact if(28)limn⟶∞1Islog2/Iq1−1/p∫SIfnzqdμz=0whenever I⊂∂D and fn is a bounded sequence in Bp that converges to 0 uniformly on compact subsets of D.Theorem 2.
Let1<p<q<∞ and 0<s<∞. Let μ be a nonnegative Borel measure on D such that point evaluation is a bounded functional on Ts,q1−1/pqμ. Then, the inclusion mapping Id:Bp⟶Ts,q1−1/pqμ is compact if and only if μ is a vanishing s-Carleson measure.Proof.
First, we assume thatId:Bp⟶Ts,q1−1/pqμ is compact. Let Ik be a sequence arc with limk⟶∞Ik=0. Set ak=1−Ikηk, where ηk is the midpoint of arc Ik. Take(29)fkz=1log2/1−ak21/plog21−ak¯z.
We see thatfk∈Bp, and fk converges to 0 uniformly on compact subsets of D when k⟶∞. Then, we get(30)μSIkIks≲1Ikslog2/Ikq1−1/p∫SIkfkzqdμz⟶0,as k⟶∞, which implies that μ is a vanishing s-Carleson measure.
Conversely, assume thatμ is a vanishing s-Carleson measure. From [12], we see that(31)μ−μrCMs⟶0,r⟶1.
Here,μrz=μz for z<r and μrz=0 for r≤z<1. Let fkBp≲1 and fk converge to 0 uniformly on compact subsets of D. Then,(32)1Islog2/Iq1−1/p∫SIfkzqdμz≲1Islog2/Iq1−1/p∫SIfkzqdμrz+1Islog2/Iq1−1/p∫SIfkzqdμ−μrz≲1Islog2/Iq1−1/p∫SIfkzqdμrz+μ−μrCMsfkBpq≲1Islog2/Iq1−1/p∫SIfkzqdμrz+μ−μrCMs.
Lettingk⟶∞ and then r⟶1, we have(33)limk⟶∞fkTs,q1−1/pqμ=0.
Therefore,Id:Bp⟶Ts,q1−1/pqμ is compact. The proof is completed.


## 3. Volterra Integral OperatorTg:Bp⟶ℒℱq,q−2,s,q−q/p

Lemma 3.
Let1<q<∞ and 0<s,t<∞. Then, f∈ℒℱq,q−2,s,t if and only if(34)supI⊂∂D1Islog2/It∫SIf′zq1−z2q−2+sdAz<∞.Proof.
The proof is similar to that of Proposition 1 in [15]. Thus, we omit the details of the proof.Theorem 3.
Let1<p<q<∞ and 0<s<∞. Then, Tg:Bp⟶ℒℱq,q−2,s,q−q/p is bounded if and only if g∈Fq,q−2,s.Proof.
Assume thatTg:Bp⟶ℒℱq,q−2,s,q−q/p is bounded. For any fixed arc I⊆∂D, let eiθ denote the center of I and a=1−Ieiθ. Set(35)faz=1log2/1−a21/plog21−a¯z.
By Lemma2, we have fa∈Bp for 1<p<∞. In addition, it is easy to see that(36)1−a¯z≈1−a=I,faz≈log2I1−1/p,when z∈SI. By the boundedness of Tg:Bp⟶ℒℱq,q−2,s,q−q/p, we get(37)∞>Tgfaℒℱq,q−2,s,q−q/pq≥1Islog2/Iq1−1/p∫SIfazqg′zq1−z2q−2+sdAz≈1Is∫SIg′zq1−z2q−2+sdAz,which implies that g∈Fq,q−2,s by [1].
Conversely, suppose thatg∈Fq,q−2,s. By [1], we see that dμgz=g′zq1−z2q−2+sdAz is an s-Carleson measure. Let f∈Bp. By Theorem 1, we see that Id:Bp⟶Ts,q1−1/pqμg is bounded, i.e.,(38)supI⊂∂D1Islog2/Iq1−1/p∫SITgf′zq1−z2q−2+sdAz=supI⊂∂D1Islog2/Iq1−1/p∫SIfzqg′zq1−z2q−2+sdAz=supI⊂∂D1Islog2/Iq1−1/p∫SIfzqdμgz≲fBpq<∞.
By Lemma3, we get that(39)supa∈D1log2/1−a2q1−1/p∫DTgf′zq1−z2q−21−σaz2sdAz<∞.
Therefore,Tg:Bp⟶ℒℱq,q−2,s,q−q/p is bounded.
Next, we give an estimation for the essential norm ofTg. First, we recall some definitions. The essential norm of T:X⟶Y, denoted by Te,X⟶Y, is defined by(40)Te,X⟶Y=infKT−KX⟶Y:Kis compact fromXtoY.
Here,X and Y are Banach spaces, and T:X⟶Y is a bounded linear operator. It is easy to see that T:X⟶Y is compact if and only if Te,X⟶Y=0. Let A be a closed subspace of X. Given f∈X, the distance from f to A, denoted by distXf,A, is defined by(41)distXf,A=infg∈Af−gX.Lemma 4.
(see [14]). Let 1<q<∞ and 0<s<∞. If g∈Fq,q−2,s, then(42)distFq,q−2,sg,F0q,q−2,s≈limsupr⟶1−g−grFq,q−2,s≈limsupa⟶1∫Dg′zq1−z2q−21−σaz2sdAz.
Here,grz=grz, 0<r<1,z∈D.Lemma 5.
Let1<p<q<∞ and 0<s<∞. If 0<r<1 and g∈Fq,q−2,s, then Tgr:Bp⟶ℒℱq,q−2,s,q−q/p is compact.Proof.
Givenfn⊂Bp such that fn converges to zero uniformly on any compact subset of D and supnfnBp≤1. Since g∈Fq,q−2,s⊂ℬ, we get that(43)gr′z≲gFq,q−2,s1−r2.
Hence,(44)Tgrfnℒℱq,q−2,s,q−q/pq=supa∈D1log2/1−a2q1−1/p∫Dfnzqgr′zq1−z2q−21−σaz2sdAz≲gℬq1−r2qsupa∈D∫Dfnzq1−z2q−21−σaz2sdAz≲gFq,q−2,sqfnBpq−p1−r2qsupa∈D∫Dfnzp1−z2p−21−σaz2sdAz≲gFq,q−2,sqfnBpq−p1−r2q∫Dfnzp1−z2p−2dAz≲gFq,q−2,sqfnBpq−p1−r2q∫Dfn′zp1−z2p−2dAz≲gFq,q−2,sqfnBpq1−r2q.
By the dominated convergence theorem, we get the desired result. The proof is completed.Lemma 6.
(see [23]). Let XandY be two Banach spaces of analytic functions on D. Suppose that(1)
The point evaluation functionals onY are continuous(2)
The closed unit ball ofX is a compact subset of X in the topology of uniform convergence on compact sets(3)
T:X⟶Y is continuous when X and Y are given the topology of uniform convergence on compact sets
Then,T is a compact operator if and only if for any bounded sequence fn in X such that fn converges to zero uniformly on every compact set of D, the sequence Tfn converges to zero in the norm of Y.Theorem 4.
Let1<p<q<∞ and 0<s<∞. If Tg:Bp⟶ℒℱq,q−2,s,q−q/p is bounded, then(45)Tge,Bp⟶ℒℱq,q−2,s,q−q/p≈distFq,q−2,sg,F0q,q−2,s.Proof.
LetIn⊆∂D and In⟶0 as n⟶∞. Suppose eiθn is the center of In and wn=1−Ineiθn. For each n, let(46)fwnz=1log2/1−wn21/plog21−wn¯z.
Then,fwnz≈log2/In1−1/p when z∈SIn, and fwn is bounded in Bp. Furthermore, fwn converges to zero uniformly on every compact subset of D. Given a compact operator K:Bp⟶ℒℱq,q−2,s,q−q/p, by Lemma 6, we have(47)limn⟶∞Kfwnℒℱq,q−2,s,q−q/p=0.
So,(48)Tg−K>limsupn⟶∞Tg−Kfwnℒℱq,q−2,s,q−q/p>limsupn⟶∞Tgfwnℒℱq,q−2,s,q−q/p−Kfwnℒℱq,q−2,s,q−q/p=limsupn⟶∞Tgfwnℒℱq,q−2,s,q−q/p≥limsupn⟶∞1log2/1−wn2q1−1/p∫Dfwnzqg′zq1−z2q−21−σwnz2sdAz1/q≥limsupn⟶∞1log2/1−wn2q1−1/p∫SInfwnzqg′zq1−z2q−21−σwnz2sdAz1/q>limsupn⟶∞1Ins∫SIng′zq1−z2q−2+sdAz1/q,which implies that(49)Tge,Bp⟶ℒℱq,q−2,s,q−q/p>limsupn⟶∞∫Dg′zq1−z2q−21−σwnz2sdAz1/q.
It follows from Lemma4 that(50)Tge,Bp⟶ℒℱq,q−2,s,q−q/p>distFq,q−2,sg,F0q,q−2,s.
On the contrary, by Lemma5, Tgr:Bp⟶ℒℱq,q−2,s,q−q/p is compact. Then,(51)Tge,Bp⟶ℒℱq,q−2,s,q−q/p≤Tg−Tgr=Tg−gr≈g−grFq,q−2,s.
Using Lemma4 again, we have(52)Tge,Bp⟶ℒℱq,q−2,s,q−q/p≲limsupr⟶1−g−grFq,q−2,s≈distFq,q−2,sg,F0q,q−2,s.
The proof is completed.
The following result can be deduced by Theorem4 directly.Corollary 1.
Let1<p<q<∞ and 0<s<∞. If Tg:Bp⟶ℒℱq,q−2,s,q−q/p is bounded, then Tg:Bp⟶ℒℱq,q−2,s,q−q/p is compact if and only if(53)g∈F0q,q−2,s.



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