Compact Operators on the Bergman Spaces with Variable Exponents on the Unit Disc of

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 general 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 is a finite sum of finite products of Toeplitz operators with symbols from class , then is compact if and only if the Berezin transform of vanishes on the boundary of the unit disc.

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.

Let denote the unit disc in and the normalized Lebesgue measure on . For , the Bergman space is the space of all analytic functions, , on such that Let be the Bergman projection from onto . Then is an integral operator given byfor each and . Here, the function is the reproducing kernel for . For , the Toeplitz operator with symbol is defined on by 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.

Given , a measurable function will be called a variable exponent. If is a variable exponent then we denote Let denote the set of all variable exponents for which .

For a complex-valued measurable function , we define the modular by and the normLet . Then the Lebesgue variable exponent space is the set of all complex-valued measurable functions for which . If we equip with the norm given in (6), then becomes a Banach space. We note here that the condition 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 is , where . A straightforward computation shows thatFor simplicity, we will omit one set of parenthesis and write the left-hand side of each equality as and . Throughout this work, we shall use as the conjugate exponent of and if is a constant in we shall use as the conjugate exponent of . In other words, to study these spaces, some regularity conditions are imposed on the exponents. A function is said to be log-Hölder continuous on if there exists a positive constant such that for all with . It follows that for all with . We denote by the exponents in that are log-Hölder continuous on . For and a given measurable function, , define Theorem of [1] shows that there exist constants and , depending on , such thatThe next result which establishes a relationship between the Lebesgue spaces with exponents , , and will be very useful in the rest of the work. It is Corollary of [1].

Lemma 1. Suppose and . Then there exist constants and such that

The study of variable exponent Bergman space, , which is the space of analytic functions in , has been introduced in [9]. There it was shown, amongst other things, that the Bergman projector is bounded from onto . 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.

For , let be the analytic map of onto given by . We define the operator on by Then is a unitary operator on . We shall show later that is bounded on . For , a bounded operator on , we define by .

If is a bounded operator on , then the Berezin transform of is the function on defined by where is the normalized Bergman kernel which also belongs to and is the inner product of . We let and set Our first result gives some conditions for the boundedness of Toeplitz operators with symbols on the variable Bergman spaces.

Theorem 2. Suppose , , and . Suppose is such thatfor all . Then is bounded on .

We note here that this result was proved in [3] in the Bergman spaces , where is a constant. We also have the following result on compactness.

Theorem 3. Suppose , , and . If is a bounded operator on such that for some , then the following are equivalent: (1) is compact on ,(2) as ,(3)for every , as ,(4) as .

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 .

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 . To be precise, we have the following.

Theorem 4. Suppose and is a finite sum of finite products of Toeplitz operators with symbols in the class . Then is compact on if and only if as .

This paper is organized as follows: in Section 2, 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 Weights

Definition 5. Let be a set. Then the function is said to be a pseudodistance on if it satisfies the following: (1) if and only if ;(2);(3)there exists a positive constant , such that, for all ,

For and , the set is called a pseudoball with centre and radius . If is a measure on , then the triple is called a homogeneous space if is endowed with the topology generated by the collection (that is, the topology generated by the pseudoballs) and satisfies the doubling property; there exists a constant such that, for all and , we have We now turn our attention to the case when . By lemma 2.2 of [11], it is shown that the distance function given on by is a pseudodistance on , where . It is known (see [12]) that, at the boundary of , becomes the Koranyi distance. Also by Lemma of [12], we have that for any pseudoball , , and we have that where denotes the Lebesgue area measure of set . Also observe that the pseudoball . It is known that (see [12]) is a homogeneous space.

Let be a locally integrable function in . Then the Hardy-Littlewood maximal function relative to the pseudodistance is given by where the supremum is taken over all pseudoballs containing .

Suppose almost everywhere on . Then we say that is in the Muckenhoupt weight if There are two equivalent definitions which are useful in practice. First, , if for almost every ,It follows that if then and thusAlternatively, if for every pseudoball we have thatFor more details on the Muckenhoupt weights, see chapter of [13] or chapter of [1].

We will need some results on extrapolation. The following proposition is Theorem of [1].

Proposition 6. Let and suppose there is some and the family such that for all ,Given , if and the maximal operator is bounded on , thenwhere and is some positive constant depending on the dimension of .

The following is Theorem of [1].

Proposition 7. Let . Then the Hardy-Littlewood maximal operator function is bounded in and we have

3. Variable Exponent Bergman Spaces

Given , we define the variable exponent Bergman space as the space of all analytic functions on that belong to the variable exponent Lebesgue space with respect to the area measure on the unit disc. With this definition is a closed subspace of . By Theorem of [9], the Bergman projection, , given by (2) is bounded from onto for any . 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 on by

Lemma 8. The operator is bounded on for .

Proof. Let and . Then since for almost every . Now, for there is a pseudoball containing such thatIt follows thatAlso where the last inequality comes from (27). This shows that It follows that the family satisfies inequality (29). Also, by Proposition 7 the maximal function belongs to . Thus by Proposition 6   is bounded on .

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 if is locally integrable in , thenThe proof of this statement can easily be adapted from that of Theorem of [14]. We use this statement as follows:
Let and . Then for any , we can find such that for all Now, if we fix such then for we have where the constant does not depend on , 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.

Letwhere is a kernel function. We give a Schur-type lemma that will be useful in our work.

Lemma 10. Let , , and . If there exist positive constants and that depends on and not , and a nonnegative measurable function, , such thatfor almost every andfor almost every , then

Proof. Using Hölder’s inequality, we have where the second inequality comes from (42) and the third inequality is from (27). Also, Fubini’s theorem gives where we have used (43) to get the last inequality.

Proposition 11. Let , be such that . Suppose the function satisfies the hypothesis of Lemma 10. Then there is a constant such that for all .

Proof. Let . Then by Lemma 10, we obtain (29) for the family , where . Also by Proposition 7, the maximal function belongs to and thus by Proposition 6 we get the required estimates.

In the application of Lemma 10, we may assume that for , as the following lemma shows.

Lemma 12. Let be such that and for any such that . If the hypothesis of Lemma 10 holds for the weight , then the conclusion of Lemma 10 holds for the weight .

Proof. By (27), we have that for almost every . Now by Proposition of [13], we have that It follows that for almost every . Thus we have that the constants and are independent of and, hence, independent of . Now since the hypothesis of Lemma 10 holds for the weight , we have that Thus, It follows that which gives the result.

The next lemma will be used frequently and is well known; see, for example, Lemma of [15] for the proof.

Lemma 13. Suppose and . Then

Proof of Theorem 2. Let . ThenNow, let forThen using the identitywe haveAlso, provided that and . That is, which holds from the choice of .
Now, observe that Thus, for each , we havesince for almost every . For any , there is a pseudoball containing such that Substitute this in (58) to obtain Using the identity (54), we have where the last equality is from the change of variable . By Hölder’s inequality, we have that if and ; that is, Now if is chosen to satisfy (53) we see that the hypothesis of Lemma 10 is satisfied and thus for every we havewhere the constant does not depend on but on . Finally we apply Proposition 11 to to obtain the result.

For it is easy to show that for any . Thus, from Theorem 2, we immediately have the following.

Corollary 14. Suppose . Then (1) is bounded on ,(2) is bounded on , and

Proof. is an immediate consequence of Theorem 2.
follows from the fact that and , which is given by assertion .

We also have the following estimate for operators in the Toeplitz algebra. To be precise, we have the following.

Lemma 15. Let , , . Thenfor any .

Proof. By assertion of Corollary 14, we have that Also, since each and , we have that This completes the proof of the lemma.

4. Norm Estimates

Lemma 16. Let and and suppose that is a bounded operator on and . If satisfies (53), then for all we have the following:for all andfor all , where the constant does not depend on .

Proof. Fix . Then where the second equality comes from the definition of and the third equality from the definition of . Thus, where . By the choice of , we have that and (69) holds.
To prove (70), replace by in (69), interchange and in (69), and then use the equationFinally, we use the same argument as in the proof of Theorem 2 to obtain that there is a pseudoball containing such that and thus A similar argument as the one used to obtain the estimate (69) will give us (70).

Proposition 17. Let , , and and suppose that is a bounded operator on If for some , then there is a constant such that

Proof. For and , we havewhere the last equation follows from (69). Given that that satisfies (53), we have by (69) that In a similar manner, we use (73) and (70) to get that for all where the constant depends on and not on . We now apply Proposition 11 to get the required result.

Lemma 18. (a) is equivalent to for all .
(b) weakly in as .

Proof. The assertion (a) is just Theorem of [10].
(b) If , then assertion (a) implies Thus if is a bounded function in , then as . The assertion follows from the fact that polynomials are dense in .

5. Compact Operators on

Theorem 19. Let be such that and suppose thatThen the operator given by (41) is compact on .

Proof. Firstly we observe that if (81) holds and , then the function , where belongs to . Indeed, It follows from Lemma 1 that Thus for , we see that . Now, suppose is a bounded sequence in such that weakly in . For and any , we can find that such that for we haveWe will show that as . Now, given , we fix such that (84) holds. It follows for any and (11) that Thus as . Finally, it is shown in Corollary of [1] that the variable Lebesgue space is reflexive if and only if . We thus conclude that is compact, since is reflexive.

We will need the power series formula for the Berezin transform of the bounded operator on . From the definition of the reproducing kernel, we get thatfor . To compute , we first compute by applying to both sides of (86) and then take the inner product with , again using (86), to obtain

Lemma 20. Suppose is a bounded operator on such that for some . Then as if and only if for every , as .

Proof. Suppose for every , as . In particular, as . Thus as .
Suppose as . Fix . We will show that as .
For , , an easy computation shows thatSince we have thatIt follows from (90) and (92) and Hölder’s inequality thatNow, let Then and thus . This shows that for any . This and Lemma 18(a) show that that is, is uniformly bounded in and
Now, we will show that for every nonnegative integer If this is not true, then there is a sequence such that for some nonzero constant and . Since is uniformly bounded, we may assume without loss of generality that for each and for some constant . For , we havewhere the second equality comes from (87). Also, note that the power series in (100) converges uniformly for each .
For each , we know that as . Thus as for each . Replacing by in (100) and taking the limit as , we get for each . If then for all . This gives for each and . On the other hand, we have for each and . In particular, which is a contradiction. Hence, we obtain For , we have It is clear that for each fixed , the power series above converges uniformly for . This gives for each . It follows that for each . If then and This shows that is uniformly integrable. By Vitali’s Theorem or Exercise 11 on pages 133-134 of [16], we have that This completes the proof of the lemma.

Proof of Theorem 3. If is compact on , then by Lemma 18(b), as . Now by Lemma 18(a), we see that is equivalent to for . Thus as .
Suppose as . By Lemma 20, we have that as for every . We will show that is compact on . Fix in the rest of the proof.
For , we have that For , we define an operator on byThen is an integral operator with kernel . We will first show that the operator is compact on . By Theorem 19, we only need to show that But This shows that and thus is compact on . Hence, to prove that is compact, we only need to show that as .
If then is the integral operator with kernel as can be seen from (77) and (113). The proof of Proposition 17 indicates that where We have shown that as and the hypothesis of the theorem shows that . Thus, as , which completes the proof.

Proof of Theorem 4. Suppose is a finite sum of operators of the form , where each . By Corollary 14 and Lemma 15, we have that is bounded on and for all . The conclusion then follows from Theorem 3.

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.