# Compact Operators on the Bergman Spaces with Variable Exponents on the Unit Disc ofC

Authors: Dieudonne Agbor
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

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## Abstract

We study the compactness of some classes of bounded operators on the Bergman space with variable exponent. We show that via extrapolation, some results on boundedness of the Toeplitz operators with generalL1 symbols and compactness of bounded operators on the Bergman spaces with constant exponents can readily be extended to the variable exponent setting. In particular, if S is a finite sum of finite products of Toeplitz operators with symbols from class BT, then S is compact if and only if the Berezin transform of S vanishes on the boundary of the unit disc.


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## Body

## 1. Introduction and Statement of Results

Variable Lebesgue spaces are a generalization of the Lebesgue spaces that allow the exponents to be a measurable function and thus the exponent may vary. These spaces have many properties similar to the normal Lebesgue spaces, but they also differ in surprising and subtle ways. For this reason, the variable Lebesgue spaces have an intrinsic interest, but they are also very important in applications to partial differential equations and variational integrals with nonstandard growth conditions. See [1] for more details on the variable Lebesgue spaces.LetΔ denote the unit disc in C and dA the normalized Lebesgue measure on Δ. For 1≤p<∞, the Bergman space Ap=Ap(Δ,dA) is the space of all analytic functions, f, on Δ such that (1)fpp=∫ΔfzpdAz<∞.Let P be the Bergman projection from L2 onto A2. Then P is an integral operator given by(2)Pfz=∫Δfw1-zw¯2dAw,for each z∈Δ and f∈L2. Here, the function K(z,w)=Kw(z)=1/1-zw¯2 is the reproducing kernel for A2. For f∈L∞, the Toeplitz operator with symbol f is defined on Ap by (3)Tfg=Pfg,g∈Ap.Toeplitz operators are amongst the most widely studied classes of concrete operators and have attracted a lot of interest in recent years. The behaviour of these operators on the Hardy spaces, Bergman spaces, and Fock spaces has been studied widely and a lot of results are available in the literature. The characterization of compactness has been studied in [2–8] just to cite a few.GivenΩ⊂Rn, a measurable function p:Ω→[1,∞) will be called a variable exponent. If p is a variable exponent then we denote (4)p+=pΩ+≔esssupx∈Ωpx,p-=pΩ-≔essinfx∈Ωpx.Let P(Ω) denote the set of all variable exponents for which p+<∞.For a complex-valued measurable functionϕ:Ω→C, we define the modular ρp(·) by (5)ρp·ϕ≔∫Ωϕxpxdxand the norm(6)ϕp·≔infλ>0:ρp·ϕλ≤1.Let p(·)∈P(Ω). Then the Lebesgue variable exponent space Lp(·) is the set of all complex-valued measurable functions ϕ:Ω→C for which ρp(·)(ϕ)<∞. If we equip Lp(·) with the norm given in (6), then Lp(·) becomes a Banach space. We note here that the condition ρp(·)(ϕ)<∞ is not enough in general to define the variable exponent Lebesgue space (e.g., see chapter 2 of [1]).It is known (e.g., see chapter 2 of [1]) that the dual of Lp(·) is Lp′(·), where 1/p(·)+1/p′(·)=1. A straightforward computation shows that(7)p′·+=p-′,p′·-=p+′.For simplicity, we will omit one set of parenthesis and write the left-hand side of each equality as p′(·)+ and p′(·)-. Throughout this work, we shall use p′(·) as the conjugate exponent of p(·) and if p is a constant in (1,∞) we shall use p′ as the conjugate exponent of p. In other words, to study these spaces, some regularity conditions are imposed on the exponents. A function p:Ω→C is said to be log-Hölder continuous on Ω if there exists a positive constant Clog such that (8)px-py≤Cloglog⁡1/x-y,for all x,y∈Ω with |x-y|<1/2. It follows that (9)px-py≤2lCloglog⁡2l/x-y,for all x,y∈Ω with |x-y|<l. We denote by Plog(Ω) the exponents in P(Ω) that are log-Hölder continuous on Ω. For p(·)∈Plog(Ω) and a given measurable function, f, define (10)fp·∗=sup∫Ωfxgxdx:gp·≤1.Theorem 2.34 of [1] shows that there exist constants C1 and C2, depending on p(·), such that(11)C1fp·≤fp·∗≤C2fp·.The next result which establishes a relationship between the Lebesgue spaces with exponents p-, p+, and p(·) will be very useful in the rest of the work. It is Corollary 2.50 of [1].Lemma 1.
Supposep(·)∈Plog(Ω) and |Ω|<∞. Then there exist constants c1 and c2 such that (12)c1fp-≤fp·≤c2fp+.The study of variable exponent Bergman space,Ap(·), which is the space of analytic functions in Lp(·), has been introduced in [9]. There it was shown, amongst other things, that the Bergman projector P is bounded from Lp(·) onto Ap(·). Also in [10], the authors studied Carleson measures in such spaces.In this paper, we will extend the results in [3, 7] on boundedness and compactness of operators for the Bergman spaces with constant exponents to the Bergman spaces with variable exponents.Forz∈Δ, let φz be the analytic map of Δ onto Δ given by φz(w)=(z-w)/(1-z¯w). We define the operator Uz on A2 by (13)Uzf=f∘φzφz′,z∈Δ.Then Uz is a unitary operator on A2. We shall show later that Uz is bounded on Ap(·). For S, a bounded operator on Ap(·), we define Sz by Sz=UzSUz.IfS is a bounded operator on Ap(·), then the Berezin transform of S is the function S~ on Δ defined by (14)S~z=Skz,kz,where kz(w)=(1-|z|2)Kz is the normalized Bergman kernel which also belongs to Ap(·) and 〈,〉 is the inner product of A2. We let (15)f~z≔Tf~z=fkz,kz=∫Δfwkzw2dAwand set (16)BT≔f∈L1:fBT=supz∈Δf~z<∞.Our first result gives some conditions for the boundedness of Toeplitz operators with L1 symbols on the variable Bergman spaces.Theorem 2.
Supposep(·)∈Plog(Δ), 1<p0≤p-≤p+<∞, and p1=min(p0,p0′). Suppose f∈L1 is such that(17)C1=supz∈ΔPf∘φzq<∞,C2=supz∈ΔPf¯∘φzq<∞for all q>(p1+1)/(p1-1). Then Tf is bounded on Ap(·).We note here that this result was proved in [3] in the Bergman spaces Ap, where p is a constant. We also have the following result on compactness.Theorem 3.
Supposep(·)∈Plog(Δ), 1<p0≤p-≤p+<∞, and p1=min(p0,p0′). If S is a bounded operator on Ap(·) such that (18)C1=supz∈ΔSz1q<∞,C2=supz∈ΔSz∗1q<∞for some q>(p1+1)/(p1-1), then the following are equivalent: (1)
S is compact on Ap(·),(2)
S~(z)→0 as z→∂Δ,(3)
for everys∈[1,q), Sz1s→0 as z→∂Δ,(4)
Sz11→0 as z→∂Δ.This theorem is well known in the Bergman spaces with constant exponents; for example, see [3, 7]. However, the techniques here are different from those used in either of the papers for both the proof of boundedness and compactness. This is because their proofs depend on the use of Schur’s test which does not hold in the variable Lebesgue space. However, using the Muckenhoupt weights we were able to develop some Schur-like tests from where we obtain the theory that builds upon the Rubio de Francia theory of extrapolation from the theory of weighted norm inequalities. The advantage of this approach is that it quickly yields to sufficient conditions for these operators to be bounded on variable Lebesgue spaces. Through such techniques, we are also able to obtain some norm estimates for bounded operators on the space Ap(·).Similar to the work of Miao and Zheng [7], we consider the case of the algebra of Toeplitz operators generated by symbols in the class BT. To be precise, we have the following.Theorem 4.
Supposep(·)∈Plog(Δ) and S is a finite sum of finite products of Toeplitz operators with symbols in the class BT. Then S is compact on Ap(·) if and only if S~(z)→0 as z→∂Δ.This paper is organized as follows: in Section2, we will study some basic concepts on the Muckenhoupt weights. Section 3 deals with the variable Bergman spaces and the proof of Theorem 2. In Section 4, we study some norm estimates on these spaces and in Section 5 we give the proof of the compactness results.


## 2. Muckenhoupt WeightsA1

Definition 5.
LetΩ be a set. Then the function d:Ω×Ω→R+ is said to be a pseudodistance on Ω if it satisfies the following: (1)
d(x,y)=0 if and only if x=y;(2)
d(x,y)=d(y,x);(3)
there exists a positive constantK≥1, such that, for all x,y,z∈Ω, (19)dx,y≤Kdx,z+dz,y.Forx∈Ω and r>0, the set B(x,r)={y∈Ω:d(x,y)<r} is called a pseudoball with centre x and radius r. If μ is a measure on Ω, then the triple (Ω,d,μ) is called a homogeneous space if Ω is endowed with the topology generated by the collection {B(x,r):x∈Ω,r>0} (that is, the topology generated by the pseudoballs) and μ satisfies the doubling property; there exists a constant δ such that, for all x∈Ω and r>0, we have (20)0<μBx,2r≤δμBx,r<∞.We now turn our attention to the case when Ω=Δ. By lemma 2.2 of [11], it is shown that the distance function d given on Δ by (21)dz,w=z-w+1-zw¯zwifz,w∈Δ∗,z+wotherwiseis a pseudodistance on Δ, where Δ∗=Δ∖{0}. It is known (see [12]) that, at the boundary of Δ, d becomes the Koranyi distance. Also by Lemma 2 of [12], we have that for any pseudoball B(w,r), w≠0, and r∈(0,2) we have that (22)Bw,r≈r2, where B denotes the Lebesgue area measure of set B. Also observe that the pseudoball B(0,1)=Δ. It is known that (see [12]) (Δ,d,dA) is a homogeneous space.Letf be a locally integrable function in Δ. Then the Hardy-Littlewood maximal function relative to the pseudodistance d is given by (23)Mfz=supB1B∫BfwdAw,where the supremum is taken over all pseudoballs containing z.Suppose0<ω(z)<∞ almost everywhere on Δ. Then we say that ω is in the Muckenhoupt weight A1 if (24)ωA1=esssupz∈ΔMωzωz<∞.There are two equivalent definitions which are useful in practice. First, ω∈A1, if for almost every z∈Δ,(25)Mωz≤ωA1ωz.It follows that if ω∈A1 then (26)ωA1ωz≥Mωz≥ωΔ,and thus(27)1≤ωA1ωzωΔ.Alternatively, w∈A1 if for every pseudoball B we have that(28)1B∫BωzdAz≤ωA1essinfu∈Bωu.For more details on the Muckenhoupt weights, see chapter 9 of [13] or chapter 4 of [1].We will need some results on extrapolation. The following proposition is Theorem5.24 of [1].Proposition 6.
LetΩ∈Rn and suppose there is some p0≥1 and the family F such that for all ω∈A1,(29)∫ΩFxp0ωxdx≤C0∫ΩGxp0ωxdx,F,G∈F.Given p(·)∈Plog(Ω), if 1≤p0≤p-≤p+<∞ and the maximal operator is bounded on L(p(·)/p0)′(Ω), then(30)Fp·≤Cp·Gp·,F,G∈F,where Cp(·)=CC0 and C is some positive constant depending on the dimension of Ω.The following is Theorem3.16 of [1].Proposition 7.
Letp∈Plog(Ω). Then the Hardy-Littlewood maximal operator function is bounded in Lp(·)(Ω) and we have (31)Mfp·≤Cfp·.


## 3. Variable Exponent Bergman Spaces

Givenp(·)∈Plog(Δ), we define the variable exponent Bergman space Ap(·) as the space of all analytic functions on Δ that belong to the variable exponent Lebesgue space Lp(·) with respect to the area measure dA on the unit disc. With this definition Ap(·) is a closed subspace of Lp(·). By Theorem 4.4 of [9], the Bergman projection, P, given by (2) is bounded from Lp(·) onto Ap(·) for any p(·)∈Plog(Δ). It is, thus, necessary to study the behaviour of Toeplitz operators on such spaces. Similar to the definition of Toeplitz operators on the Bergman spaces with constant exponent, we define the Toeplitz operator with symbol f∈L∞ on Ap(·) by (32)Tfg=Pfg,g∈Ap·.Lemma 8.
The operatorUz is bounded on Ap(·) for p(·)>1.Proof.
Let1<p0≤p-≤p+<∞ and ω∈A1. Then (33)∫ΔUzfξp0ωξdAξ≤∫ΔUzfξp0MωξdAξsince ω(ξ)≤Mω(ξ) for almost every ξ∈Δ. Now, for 0<ϵ<ω(Δ) there is a pseudoball B containing ξ such that(34)Mωξ≤ωBB+ϵ≤ωΔB+ωΔ.It follows that(35)∫ΔUzfξp0MωξdAξ≤ωΔB+ωΔ∫ΔUzfξp0dAξ.Also (36)∫ΔUzfξp0dAξ=∫Δf∘φzξφz′ξp0dAξ=∫Δfξφz′φzξp0φz′ξ2dAξ=∫Δfξp0φz′φzξp0-2dAξ≤2p01-z2p0-2∫Δfξp0dAξ≤ωA12p0ωΔ1-z2p0-2∫Δfξp0ωξdAξ,where the last inequality comes from (27). This shows that (37)∫ΔUzfξp0ωξdAξ≤B+1ωA12p0B1-z2p0-2∫Δfξp0ωξdAξ.It follows that the family {(Uzf,f):f∈Lp0} satisfies inequality (29). Also, by Proposition 7 the maximal function Mf belongs to L(p(·)/p0)′. Thus by Proposition 6  Uz is bounded on Ap(·).Remark 9.
We just want to give an alternative argument to obtain the estimate (35), and this argument has different effects and may be useful in applications.
We recall that ifω is locally integrable in Δ, then(38)limr→01Bz,r∫Bz,rωz-ωξdAξ=0.The proof of this statement can easily be adapted from that of Theorem 1.3 of [14]. We use this statement as follows:
Letω∈A1 and p0>1. Then for any ϵ>0, we can find R>0 such that (39)1Bz,r∫Bz,rωz-ωξdAξ<ϵωΔ,for all r∈(0,R). Now, if we fix such R then for 0<r<R we have (40)∫ΔUzfξp0ωξdAξ=∫ΔUzfξp0ωξ-1Bξ,r∫Bξ,rωζdAζ+1Bξ,r∫Bξ,rωζdAζdAξ≤∫ΔUzfξp01Bξ,r∫Bξ,rωξ-ωζdAζdAξ+∫ΔUzfξp01Bξ,r∫Bξ,rωζdAζdAξ≤ωΔϵ+Cr∫ΔUzfξp0dAξ,where the constant Cr does not depend on ω∈A1, and thus (35) holds. We may also use this same argument in some parts of the proofs of Theorem 2 and Proposition 17 by replacing similar statements that give rise to the estimates (34) in the proof of Lemma 8.Let(41)Tfz=∫ΔfwKz,wdAw,z∈Δ,where K:Δ×Δ→C is a kernel function. We give a Schur-type lemma that will be useful in our work.Lemma 10.
Letω∈A1, p>1, and 1/p+1/p′=1. If there exist positive constants C1 and C2 that depends on [ω]A1 and not ω, and a nonnegative measurable function, h, such that(42)∫ΔKξ,zhzp′dAz≤C1hξp′for almost every ξ∈Δ and(43)ωΔ-1∫ΔKξ,zhξpωξdAz≤C2hzpfor almost every z∈Δ, then(44)∫ΔTfzpωzdAz≤C1C2∫ΔfzpωzdAz.Proof.
Using Hölder’s inequality, we have(45)Tfz≤∫ΔKξ,zhξhξ-1fξdAξ≤C1′hz∫ΔKξ,zhξ-pfξpdAξ1/p≤C1′hz∫ΔKξ,zhξ-pfξpωA1ωξωΔdAξ1/p,where the second inequality comes from (42) and the third inequality is from (27). Also, Fubini’s theorem gives (46)∫ΔTfzpωzdAz≤C1∫ΔωA1hzpωΔ∫ΔKξ,zhξ-pfξpωξdAξωzdAz=C1ωA1∫Δfξphξ-p∫ΔKξ,zhzpωzωΔdAzωξdAw≤C1C2ωA1∫ΔfzpωzdAz,where we have used (43) to get the last inequality.Proposition 11.
Letp(·)∈Plog(Δ), p∈(1,∞) be such that 1<p≤p-≤p+<∞. Suppose the function K:Δ×Δ→C satisfies the hypothesis of Lemma 10. Then there is a constant C such that (47)Tfp·≤CC1C2fp·,for all f∈Lp(·).Proof.
Letω∈A1. Then by Lemma 10, we obtain (29) for the family {(|Tf|,|f|):f∈Lp0(Δ)}, where 1<p0≤p-≤p+<∞. Also by Proposition 7, the maximal function Mf belongs to L(p(·)/p0)′(Δ) and thus by Proposition 6 we get the required estimates.In the application of Lemma10, we may assume that essinfz∈Δω(z)≥1 for ω∈A1, as the following lemma shows.Lemma 12.
Letω∈A1 be such that essinfz∈Δω(z)<1 and μ=δω for any δ>1 such that essinfz∈Δμ(z)≥1. If the hypothesis of Lemma 10 holds for the weight μ, then the conclusion of Lemma 10 holds for the weight ω.Proof.
By (27), we have that 1≤([μ]A1/μΔ)μ(z) for almost every z∈Δ. Now by Proposition 9.1.5 of [13], we have that [μ]A1=[δω]A1=[ω]A1. It follows that 1≤([ω]A1/μΔ)μ(z) for almost every z∈Δ. Thus we have that the constants C1=[ω]A1 and C2 are independent of μ and, hence, independent of ω. Now since the hypothesis of Lemma 10 holds for the weight μ, we have that (48)∫ΔTfzpμzdAz≤C1C2fμ1/ppp.Thus, (49)δ∫ΔTfzpωzdAz≤δC1C2fω1/ppp.It follows that (50)∫ΔTfzpωzdAz≤C1C2fω1/ppp,which gives the result.The next lemma will be used frequently and is well known; see, for example, Lemma3.10 of [15] for the proof.Lemma 13.
Supposea<1 and a+b<2. Then (51)supz∈Δ∫ΔdAw1-w2a1-z¯wb<∞.Proof of Theorem2.
Letg∈Ap(·). Then(52)Tfgz=Tfg,Kz=g,Tf¯Kz=∫ΔgξTf¯Kz¯ξdAξ.Now, let h(z)=(1-z2)-ϵ for(53)max⁡2qp0,2qp0′<ϵ<min⁡1q′p0′,1q′p0.Then using the identity(54)Tf¯Kzξ=Pf¯∘φzφzξKzw,z,ξ∈Δwe have(55)∫ΔTf¯Kzξhξp0′dAξ=∫ΔPf¯∘φzφzξKzξ1-z2-ϵp0′dAξ=11-z2ϵp0′∫ΔPf¯∘φzξdAξ1-ξ2ϵp0′1-z¯ξ2-2ϵp0′≤Pf¯∘φzq1-z2ϵp0′∫ΔdAξ1-ξ2ϵp0′q′1-z¯ξ2-2ϵp0′q′1/q′.Also, (56)supz∈Δ∫ΔdAξ1-ξ2ϵp0′q′1-z¯ξ2-2ϵp0′q′<∞provided that ϵ<1/p0′q′ and ϵp0′q′+2q′-2ϵp0′q′<2. That is, 2/p0′q<ϵ<1/p0′q′ which holds from the choice of ϵ.
Now, observe that(57)Tf¯Kzξ=Tf¯Kz,Kξ=Kz,TfKξ=TfKξz¯.Thus, for each ω∈A1, we have(58)∫ΔTf¯Kzξhzp0ωzωΔdAz=∫ΔTfKξzhzp0ωzωΔdAz≤1ωΔ∫ΔTfKξzhzp0MωzdAzsince ω(z)≤Mω(z) for almost every z∈Δ. For any 0<ϵ1<ω(Δ), there is a pseudoball B containing z such that (59)Mωz≤ωBB+ϵ1≤ωΔB+ϵ1≤ωΔB+ωΔ.Substitute this in (58) to obtain (60)∫ΔTfKξzhzp0ωzωΔdAz≤B-1+1∫ΔTfKξzhzp0dAz.Using the identity (54), we have (61)∫ΔTf¯Kzξhξp0ωξωΔdAξ≤C′∫ΔTfKξz11-z2ϵp0dAz=C′∫ΔPf¯∘φξφξzKξz1-z2-ϵp0dAξ=C′1-ξ2-ϵp0∫ΔPf¯∘φξzKξz1-z2-ϵp0dAξ=C′1-ξ2-ϵp0∫ΔPf¯∘φξzdAz1-z2ϵp01-z¯ξ2-2ϵp0,where the last equality is from the change of variable z=φξ. By Hölder’s inequality, we have that (62)∫ΔPf¯∘φξzdAz1-z2ϵp01-z¯ξ2-2ϵp0≤Pf¯∘φzq∫ΔdAξ1-w2ϵp0q′1-z¯ξ2-2ϵp0q′1/q′,supξ∈Δ∫ΔdAξ1-ξ2ϵp0q′1-z¯ξ2-2ϵp0q′<∞if ϵ<1/p0q′ and ϵp0q′+2q′-2ϵp0q′<2; that is, 2/qp0<ϵ<1/p0q′. Now if ϵ>0 is chosen to satisfy (53) we see that the hypothesis of Lemma 10 is satisfied and thus for every ω∈A1 we have(63)∫ΔTfgzpωzdAz≤Csupz∈ΔPf∘φzqsupz∈ΔPf¯∘φzq∫ΔgzpωzdAz=CC1C2∫ΔgzpωzdAz,where the constant C does not depend on ω but on [ω]A1. Finally we apply Proposition 11 to {(|Tfg|,|g|):g∈Lp0} to obtain the result.Forf∈BT it is easy to show that (64)supz∈ΔPf∘φzq<∞,supz∈ΔPf¯∘φzq<∞for any q>0. Thus, from Theorem 2, we immediately have the following.Corollary 14.
Supposef∈BT. Then (1)
Tf is bounded on Ap(·),(2)
Tf∘φz is bounded on Ap(·), and (65)Tf∘φzp·≤CfBT.Proof.
(
1
) is an immediate consequence of Theorem 2.
(
2
) follows from the fact that f∘φzBT=fBT and Tf∘φzp(·)≤Cf∘φzBT, which is given by assertion (1).We also have the following estimate for operators in the Toeplitz algebra. To be precise, we have the following.Lemma 15.
LetS=Tf1⋯Tfn, fi∈BT, i=1,…,n. Then(66)supz∈ΔSz1p·<∞,supz∈ΔSz∗1p·<∞for any P∈Plog(Δ).Proof.
By assertion(2) of Corollary 14, we have that (67)Sz1p·=Tf1∘φz⋯Tfn∘φzp·≤Cf1BT⋯fnBT.Also, since each fj¯∈BT and fj¯BT=fjBT, we have that (68)Sz1p·=Tf1¯∘φz⋯Tfn¯∘φzp·≤Cf1BT⋯fnBT.This completes the proof of the lemma.


## 4. Norm Estimates

Lemma 16.
Let1<p0<∞ and p1=min⁡(p0,p0′) and suppose that S is a bounded operator on Ap(·) and q>(p1+1)/(p1-1). If ϵ>0 satisfies (53), then for all ω∈A1 we have the following:(69)∫ΔSKzξ1-ξ2ϵp0′dAξ≤C1Sz1q1-z2ϵp0′for all z∈Δ and(70)ωΔ-1∫ΔSKzξ1-z2ϵp0ωzdAz≤C2Sξ∗1q1-ξ2ϵp0for all ξ∈Δ, where the constant C2 does not depend on ω.Proof.
Fixz∈Δ. Then (71)SKz=SUz1z2-1=UzSz1z2-1=Sz1∘φzφz′z2-1,where the second equality comes from the definition of Sz and the third equality from the definition of Uz. Thus, (72)∫ΔSKzξ1-ξ2ϵp0′dAξ=11-z2∫ΔSz1∘φzξφz′ξ1-ξ2ϵp0′dAξ=11-z2ϵp0′∫ΔSz1ξ1-ξ2ϵp0′1-z¯ξ2-2ϵp0′dAξ≤CSz1qfzq′1-z2ϵp0′,where fz(ξ)=1/(1-ξ2)ϵp0′|1-z¯ξ|2-2ϵp0′. By the choice of ϵ, we have that supz∈Δfzq′<∞ and (69) holds.
To prove (70), replace S by S∗ in (69), interchange ξ and ξ in (69), and then use the equation(73)S∗Kξz=S∗Kξ,Kz=Kξ,SKz=SKzξ¯.Finally, we use the same argument as in the proof of Theorem 2 to obtain that there is a pseudoball B containing z such that ω(z)≤Mω(z)≤ω(Δ)(|B|-1+1) and thus (74)ωΔ-1∫ΔSKzξ1-z2ϵp0ωzdAz≤B-1+1∫ΔS∗Kξz1-z2ϵp0dAz.A similar argument as the one used to obtain the estimate (69) will give us (70).Proposition 17.
Letp∈Plog(Δ), 1<p0≤p-≤p+<∞, and p1=min(p0,p0′) and suppose that S is a bounded operator on Ap(·). If (75)C1=supz∈ΔSz1q<∞,C2=supz∈ΔSz∗1q<∞for some q>(p1+1)/(p1-1), then there is a constant C such that (76)Sp·≤CC1C2.Proof.
Forf∈Ap(·) and w∈Δ, we have(77)Sfξ=Sf,Kξ=f,S∗Kξ=∫ΔfzS∗Kξz¯dAz=∫ΔfzSKzξdAz,where the last equation follows from (69). Given that ϵ>0 that satisfies (53), we have by (69) that (78)∫ΔSKzξ1-ξ2ϵp0′dAw≤CSz1q1-z2ϵp0′.In a similar manner, we use (73) and (70) to get that (79)ωΔ-1∫ΔSKzξ1-z2ϵp0ωzdAz=ωΔ-1∫ΔS∗Kξz1-z2ϵp0ωzdAz≤CSξ∗1q1-ξ2ϵp0.∫ΔSfzpωzdAz≤C∫ΔfzpωzdAzfor all ω∈A1 where the constant C depends on [ω]A1 and not on ω. We now apply Proposition 11 to get the required result.Lemma 18.
(a)Kzp(·) is equivalent to (1-z2)-2+2/p(z) for all z∈Δ.
(b)Kz/Kzp(·)→0 weakly in Ap(·) as z→∂Δ.Proof.
The assertion (a) is just Theorem3.5 of [10].
(b) Iff∈Ap′(·), then assertion (a) implies (80)f,KzKzp·≅fz1-z2-2+2/pz.Thus if f is a bounded function in Ap′(·), then 〈f,Kz/Kzp·〉→0 as z→∂Δ. The assertion follows from the fact that polynomials are dense in Ap′(·).


## 5. Compact Operators onAp(·)

Theorem 19.
Letp∈Plog(Δ) be such that 1<p-≤p+<∞ and suppose that(81)∫ΔK·,zp·p′·+dAz<∞.Then the operator T given by (41) is compact on Lp(·)(Δ).Proof.
Firstly we observe that if (81) holds and g∈Lp′(·), then the function w↦F(w), where (82)Fw=∫ΔKz,wgzdAzbelongs to Lp′(·)(Δ). Indeed, (83)∫ΔFwp′·+dAw=∫Δ∫ΔgzKz,wdAzp′·+dAw≤Cp·∫Δgp′·p′·+K·,wp·p′·+dAw<∞.It follows from Lemma 1 that Fp′(·)≤Fp′(·)+<∞. Thus for g∈Lp′(·), we see that F∈Lp′(·). Now, suppose (φn) is a bounded sequence in Lp(·) such that φn→φ weakly in Lp(·). For ϵ>0 and any g∈Lp′(·), we can find that N∈N such that for n≥N we have(84)∫Δφnw-φwg¯wdAw<ϵ.We will show that Tφn-Tφp(·)→0 as n→∞. Now, given ϵ, we fix N such that (84) holds. It follows for any n≥N and (11) that (85)Tφn-Tφp·≤Cp·Tφn-Tφp·∗=sup⁡∫ΔTφn-Tφwg¯wdAw:gp′·≤1=sup⁡∫Δ∫Δφnw-φwKw,zdAwg¯zdAz:gp′·≤1=sup⁡∫Δφnw-φw∫ΔKz,wgzdAz¯dAw:gp′·≤1=sup⁡∫Δφnw-φwFw¯dAw:gp′·≤1<ϵ.Thus Tφn-Tφp(·)→0 as n→∞. Finally, it is shown in Corollary 2.81 of [1] that the variable Lebesgue space Lp(·) is reflexive if and only if 1<p-≤p+<∞. We thus conclude that T is compact, since Lp(·) is reflexive.We will need the power series formula for the Berezin transform of the bounded operatorS on A2. From the definition of the reproducing kernel, we get that(86)kzw=1-z2∑n=0∞n+1wnz¯nfor z,w∈Δ. To compute S~(z)=〈Skz,kz〉, we first compute Skz by applying S to both sides of (86) and then take the inner product with kz, again using (86), to obtain(87)S~z=1-z22∑m,j=0∞j+1m+1Swj,wmz¯jzm.Lemma 20.
SupposeS is a bounded operator on Ap(·) such that (88)supz∈ΔSz1q<∞for some q>1. Then S~(z)→0 as z→∂Δ if and only if for every t∈[1,q), Sz1t→0 as z∈∂Δ.Proof.
Suppose for everyt∈[1,q), Sz1t→0 as z→∂Δ. In particular, Sz11→0 as z∈∂Δ. Thus (89)S~z=Sz1,1≤Sz11⟶0as z∈∂Δ.
SupposeS~(z)→0 as z→∂Δ. Fix t∈[1,q). We will show that Sz1t→0 as z∈∂Δ.
Forz∈Δ, j,m=0,1,…, an easy computation shows that(90)∫ΔSzwjwm¯dAw=∫ΔSUzwjUzwm¯dAw.Since (91)Uzwj=φzwjφz′w=-φzwj1-z2Kzw,we have that(92)∫ΔSUzwjUzwm¯dAw=1-z22∫ΔSφzjKzwφzwmKzw¯dAw.It follows from (90) and (92) and Hölder’s inequality that(93)Szwj,wm=∫ΔSzwjwm¯dAw=1-z22∫ΔSφzjKzwφzwmKzw¯dAw≤Cp·1-z22SφzjKzp·φzmKzp′·≤Cp·1-z22Sp·φzjKzp·φzmKzp′·.Now, let (94)S1=λ>0:∫ΔKzwλpwdAw≤1,S2=λ>0:∫ΔφzwKzwλpwdAw≤1.Then S1⊂S2 and thus infS1≥infS2. This shows that (95)φzjKzp·≤Kzp·for any p(·)∈Plog(Δ). This and Lemma 18(a) show that (96)Szwj,wm≤Cp·1-z22Sp·Kzp·Kzp′·≤CSp·;that is, 〈Szwj,wm〉 is uniformly bounded in z∈Δ and j,m=0,1,….
Now, we will show that for every nonnegative integern(97)Sz1,wn⟶0asz⟶∂Δ.If this is not true, then there is a sequence zk∈Δ such that (98)Szk1,wn⟶a0naszk⟶∂Δfor some nonzero constant a0n and n≥1. Since 〈Szwj,wm〉 is uniformly bounded, we may assume without loss of generality that for each j and m(99)Szwj,wm⟶ajmasz⟶∂Δ,for some constant ajm. For z,ξ∈Δ, we have(100)S~φzξ=Sz~ξ=1-ξ22∑j,m=0∞j+1m+1Szwj,wmξ¯jξm,where the second equality comes from (87). Also, note that the power series in (100) converges uniformly for each ξ∈Δ.
For eachξ∈Δ, we know that φzk(ξ)→∂Δ as zk→∂Δ. Thus S~φz(ξ)→0 as zk→∂Δ for each ξ∈Δ. Replacing z by zk in (100) and taking the limit as zk→∂Δ, we get (101)1-ξ22∑j,m=0∞j+1m+1ajmξ¯jξm=0for each ξ∈Δ. If (102)fξ=∑j,m=0∞j+1m+1ajmξ¯jξm,then f(ξ)=0 for all ξ∈Δ. This gives (103)∂m∂ξm∂j∂ξ¯jf0=0for each j and m. On the other hand, we have (104)∂m∂ξm∂j∂ξ¯jf0=j+1!m+1!ajmfor each j and m. In particular, a0n=0 which is a contradiction. Hence, we obtain (105)limz→∂ΔSz1,wn=0.For ξ∈Δ, we have (106)Sz1ξ=∑n=0∞n+1Sz1,wnξn. It is clear that for each fixed ξ∈Δ, the power series above converges uniformly for z∈Δ. This gives (107)limz→∂ΔSz1ξ=0for each ξ∈Δ. It follows that (108)limz→∂ΔSz1ξt=0for each ξ∈Δ. If s=q/t then s>1 and (109)∫ΔSz1ξtsdAξ=Sz1qq≤supz∈ΔSz1qq<∞.This shows that {|(Sz1)|t}z∈Δ is uniformly integrable. By Vitali’s Theorem or Exercise 11 on pages 133-134 of [16], we have that (110)limz→∂ΔSz1t·=0.This completes the proof of the lemma.Proof of Theorem3.
IfS is compact on Ap(·), then by Lemma 18(b), (111)SKzKzp·,KzKzp′·⟶0as z→∂Δ. Now by Lemma 18(a), we see that S~(z) is equivalent to 〈SKz/Kzp·,Kz/Kzp′·〉 for z∈Δ. Thus S~(z)→0 as z→∂Δ.
SupposeS~(z)→0 as z→∂Δ. By Lemma 20, we have that Sz1t→0 as z→∂Δ for every t∈[1,q). We will show that S is compact on Ap(·). Fix t∈(1,q) in the rest of the proof.
Forf∈Ap(·), we have that (112)Sfw=∫ΔfzSKzwdAz.
For0<r<1, we define an operator S[r] on Ap(·) by(113)Srfw=∫rΔfzSKzwdAz.Then S[r] is an integral operator with kernel (SKz)(w)χrΔ(z). We will first show that the operator S[r] is compact on Ap(·). By Theorem 19, we only need to show that (114)∫ΔSKzχrΔzp·p′·+dAz=∫rΔSKzp·p′·+dAz<∞.But (115)∫rΔSKzp·p′·+dAz≤∫rΔSp·p′·+Kzp·p′·+dAz≤CSp·p′·+∫rΔdAz1-z22-2/pzp′·+≤Cr2Sp·p′·+1-r22p′·+.This shows that (116)∫rΔSKzp·p′·+dAz<∞,and thus S[r] is compact on Ap(·). Hence, to prove that S is compact, we only need to show that S-Srp(·)→0 as r→1-.
Ifr∈(0,1) then S-S[r] is the integral operator with kernel (117)SKzwχΔ∖rΔzas can be seen from (77) and (113). The proof of Proposition 17 indicates that (118)S-Srp·≤CC1C2,where (119)C1=sup⁡Sz1t:r≤z<1,C2=sup⁡Sz∗1t:z∈Δ.We have shown that C1→0 as r→1- and the hypothesis of the theorem shows that C2<∞. Thus, S-Srp(·)→0 as r→1-, which completes the proof.Proof of Theorem4.
SupposeS is a finite sum of operators of the form Tf1⋯Tfn, where each fj∈BT. By Corollary 14 and Lemma 15, we have that S is bounded on Ap(·) and (120)supz∈ΔSz1q<∞,supz∈ΔSz∗1q<∞for all q>0. The conclusion then follows from Theorem 3.



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