Duality of Large Fock Spaces in Several Complex Variables and Compact Localization Operators

Liu, Youqi; Wang, Xiaofeng

doi:https://doi.org/10.1155/2021/6619459

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Research Article | Open Access

Volume 2021 | Article ID 6619459 | https://doi.org/10.1155/2021/6619459

Duality of Large Fock Spaces in Several Complex Variables and Compact Localization Operators

Youqi Liu¹and Xiaofeng Wang¹

Academic Editor: Kunyu Guo

Received30 Nov 2020

Revised09 Jan 2021

Accepted13 Jan 2021

Published08 Feb 2021

Abstract

In this paper, dual spaces of large Fock spaces with are characterized. Also, algebraic properties and equivalent conditions for compactness of weakly localized operators are obtained on .

1. Introduction

Let be the -dimensional complex Euclidean space. Let denote the Lebesgue volume measure on . For any two points and in , we write and .

For each and r >0,denotes the Euclidean ball centered at with radius .

Let ∆ denote the Laplacian operator. Suppose is a plurisubharmonic function (see [1]). We say that belongs to the weight class if satisfies the following statements:(A)There exists such that for (B)For any and , satisfies the reverse-Hölder inequalityfor some ;(C)The eigenvalues of are comparable, i.e., for every , there exists such that where

Suppose , . The space consists of all Lebesgue measurable functions on for which

is the set of all Lebesgue measurable functions on with

Let be the family of all entire functions on . The large Fock space is defined as

is a Banach space under if , and is a quasi-Banach space with distance if . Assume that , then is the classical Fock space which has been studied in [2–4] for example. Also, the weight function on with the restriction that belongs to , where and . See [5, 6] for more details.

Particularly, is a reproducing kernel Hilbert space. That is, for any , there exists a unique function so that , where

We say that the function is the reproducing kernel of . It is well known that the orthogonal projection is given by

As we know if and is the conjugate exponents of , then the dual space of can be identified with by the integral pairing defined by (9). In general, for , no less important than the Hahn-Banach theorem is the Bergman projection to explore the dual spaces of . However, there are some differences for these quasi-Banach spaces . To do this, we will mainly apply Hörmander’s solution of the equation and the Lebesgue dominated convergence theorem to consider the duality of .

The “weakly localized” operators were introduced for the first time in [7], and the authors studied the compactness of these operators on the Bergman space A^p and weighted the Bargmann-Fock space with . In fact, this kind of operators is interesting since these weakly localized operators contain Toeplitz operators which are induced by bounded symbols. Indeed, Toeplitz operators are a kind of significant operators, and these Toeplitz operators induced by diverse functions enjoy abundant properties, see more in [8, 9]. As a further research, Hu, Lv, and Wick characterized the compactness of these weakly localized operators on generalized Fock spaces with , see [5]. Besides, in generalized Bergman space setting [10], there are two questions: whether Toeplitz operators induced by bounded symbols are weakly localized operators? Would these weakly localized operators form an algebra?

This paper is devoted to consider the compactness of these weakly localized operators on large Fock spaces with . To ensure the validity of these fascinating operators, we show these localization operators contain Toeplitz operators induced by bounded symbols on , see Theorem 16. Meanwhile, we also give affirmative answer about the second question on our Fock spaces, see Theorem 15.

Notice that although in the one-dimensional case, the diverse weight function gives another Bergman metric, and the resulting Bergman disk will be changed. Furthermore, there is no inclusion relation between and if . The above properties are much different from [5], so we have to apply more techniques to discuss the compactness of weakly localized operators in case . For case , the ideas to study compact weakly localized operators in [7] are not entirely applicable to the situation we are discussing. Hence, we finally combine the skills in [5, 7] to consider the compactness of these operators on . Eventually, when , we bring new consequences even if is the generalized Fock space in [5].

This paper is organized as follows. In Section 2, we give some lemmas which will play key roles in our proofs. In Section 3, we show some properties of projection and dual spaces of large Fock spaces when . In Section 4, we conclude the algebraic properties and boundedness of localization operators. Finally, in Section 5, we consider the compactness of weakly localized operators on our Fock spaces.

Throughout this paper, we write for two quantities and if there is a constant such that . Furthermore, means that both and are satisfied.

2. Preliminaries and Basic Estimates

In this section, we will give some useful estimates for our proofs. For , set

In the following, we write instead of for short.

By [11] (see also [12]), we have the following consequences.

Lemma 1. Let be as defined in (2). Then, the function satisfies the following properties:(A)There exists such that(B)The function is Lipschitz, that is(C)For ), there holds(D)There exist such that

Let , we write and . In fact, it is easily obtained from estimate (14) that there is some constant such that , where for any . That is for every , we have whenever . Besides, (14) and the triangle inequality give and so that

Given , there is a sequence in such that covers , and the balls are pairwise disjoint. We say the sequence is an -lattice. For the -lattice and , there exists some integer such that any in belongs to at most balls of . That is, for every ,

Now, we are going to state the properties of the reproducing kernel . Let , and it follows from [11–13] that(A)For , there are constants such that(B)For , there exists such that(C)For , there holds

With the help of Lemmas 1 and 2 in [12], we get the following lemma.

Lemma 2. Given and , there exists such that for

For and , we write . Let . It is directly from ([14], Lemma 2.7) that we have the next estimate.

Lemma 3. For any , , , , and , there is a constant such thatand whenever .

We will write for the normalized reproducing kernel at , where and .

Lemma 4. Let. Then, for every , we haveas.

Proof. By joining (18) and (20), we have

Here, the last step is from the estimate (12). Thus, the assertion follows from Lemma 3 with for any fixed .

The next lemma is immediately from ([12], Lemma 4) (see also ([11], Lemma 2) for any .

Lemma 5. For 0<p<∞, there is a constant C>0 such that for each and , we have

For r>0 and some domain , write . Let be the Euclidean distance, and we have the following lemma.

Lemma 6. For , , and , there is some constant (depending on , , and ) such that for any domain and ,

Proof. Consider the -lattice in . For and , we get whenever . By letting , it follows from (16) thatwhere . Also notice that for positive and . Let be sufficiently small so that . Thus, the above inequality, (17) and (25) showwhich completes the proof.

3. Bergman Projection and Duality

The paper [12] points out that the Bergman projection is bounded on for . And there is no answer to whether on . In what follows, we use the classical theorem to prove that the projection is an identity operator on .

Theorem 7 ([15], Theorem 4.2.6). Let be a pseudo-convex open set in , a plurisubharmonic function in, and . Ifis in locally in and , then the equation has a solution such thatFor , we let be the conjugate exponent of such that .

Theorem 8. If and , then .

Proof. Suppose that satisfying if , if , if and

Set where . It follows that for any ,

Because of (15), there are so that whenever . Indeed, by choosing sufficiently small, we obtain when is large enough.

If , then by Lemma 6, we have

This together with (18), Lemma 2 and Fubini’s theorem give

We now let . Notice that estimate (18) and Lemma 2 indicate

So, s inequality and Fubini’s theorem show

And then, for , we get

This combined with (25) means that

as .

On the other hand, applying Theorem 7 with to the solution of in , we have

Hence, for and let be sufficiently large, it follows immediately from esitmates (25), (30), and Lemma 2 that

By combining the above estimate and (37), we finally obtain

as . This ends the proof.

We now proceed to identify the dual space of when . Arguing as in [16], we let

Theorem 9. Suppose . Then, .

Proof. For any , consider where . Then, (25) says

The above inqueality shows that is a bounded linear functional on and .

For , define where is a bounded linear functional on . Pick an such that . For some and every , using Cauchy’s estimates, we have

We note that for any fixed , the function is in . Fix and , and we get

Thus Lebesgue dominated convergence theorem indicates

Hence, for any , we obtain since

and . The result is

To complete the proof, it only remains to show that

Let be an -lattice. For , we consider

Since is compact, there exists so that . Moreover, we see that because it is actually a finite sum of analytic functions. And there is such that . It follows from (43) that

goes to 0 by letting , where denotes the characteristic function for the ball . This means that

It is clear that, for each fixed , is in . Hence, by the following estimate,and the Lebesgue dominated convergence theorem we deduce

Furthermore, we claim that

as . Here, the last assertion follows from fact that whenever . To see this, by letting and , we then get . Indeed, (18) gives us a dominating function, and it is from Lemma 2 that the function is in since for any fixed . Then, the desired assertion holds by Lebesgue dominated convergence theorem again. So, we have

Therefore, we have by Theorem 8, (51) and (55) that

This finishes the proof.

Theorem 10. Suppose . Then, under the pairing

Proof. If , define

For any , s inequality gives

This means that is a bounded linear functional on and .

On the other hand, let be a bounded linear functional. The Hahn-Banach extension theorem implies that can be extended to a bounded linear functional on . It follows from the duality theory of that there is a function such that and

Set , then since is bounded. Also, note that Theorem 8 shows for . So, (60) indicates

This completes the proof.

Corollary 11. Suppose . Then, the linear span of all reproducing kernel functions is dense in .

Proof. Let . It is immediately from Theorem 8 and the proof of Theorem 9, for any , that

Next, we assume that . By Theorem 10 and the Hahn-Banach theorem, it suffices to show that for any , we have if satisfies . This follows from the fact that for every .

4. Localization Operators

In this section, we will explore some properties of weakly localized operators on our Fock spaces. In particular, we will show the algebraic properties of these localization operators.

Before stating weakly localized operators, we consider firstly the following proposition.

Proposition 12. Suppose . Then,

Proof. It is from (11) that there exists some such that for each . Fix , and we have

For every , let be sufficiently large and let , and it follows that estimate (18) together with (64) gives

is a dimensionless. For fixed , we get , where . Hence, Theorem 9 together with (19) shows thatwhich is the desired estimate.

Now, with the above preparations, we are ready for the definition of weakly localized operators.

Definition 13. Let . A linear operator on is called weakly localized for ifwhere

Recall that, for , is the conjugate exponent of so that .

Definition 14. Suppose . A linear operator on is called weakly localized for if

Next, we are going to answer the questions raised at the beginning of the paper in our Fock spaces. In fact, each set of these weakly localized operators on is an algebra.

Theorem 15. Suppose . Then, weakly localized operators on form an algebra.

Proof. Suppose operators and are weakly localized. So, it remains to show that is a weakly localized operator because the linear combination of two weakly localized operators is also a weakly localized operator.

We let . It follows from (68) that there is some such that

where and any (when , estimate (71) gives an analogous representation). For , if then and by the triangle inequality, we have . That is, whenever .

By joining Lemma 6 and Fubini’s theorem, we get

Since and are weakly localized, hence

Therefore, by combining and , we get

where the constant does not depend on . This means

when . Meanwhile, we also get

whenever .

On the other hand, let now , by Fubini’s theorem, and we have

We split again the above integral on into the corresponding integrals on and , then

all go to 0 as . This ends the proof since others are obvious.

Let denote the algebra generated by weakly localized operators for . Let be a Toeplitz operator (see [8]) on , where is called a symbol function. Then, each contains some special Toeplitz operators.

Theorem 16. Suppose and . Then, Toeplitz operator .

Proof. We first suppose . Clearly, it suffices to prove thatconverges to 0 as .

Since for any fixed , hence Lemma 4 gives that

goes to 0 whenever .

Now, assume that . It is easily obtained from (18), (20), and Lemma 2 that

Thus, we only need to show

as . In fact, if . This together with (82) indicates

Therefore, the desired conclusion follows when . This ends the proof.

Remark. Moreover, Theorem 16 indicates that the identity operator is also in . Namely, each algebra possesses an unit.

We next consider the boundedness of operator for .

Theorem 17. If and , then is bounded on .

Proof. First, we see that

Let and let

Estimate (20) combined with Lemma 6 yields

So, we conclude by Fubini’s theorem that

We now assume that . Set

By Fubini’s theorem and s inequality, we obtainwhich completes the proof.

Now, it follows from Theorem 17 that each is analogous to a Banach algebra.

Theorem 18. Suppose and . Then, .

Proof. Suppose . For every , by the proof of Theorem 15, we see that

Since , then Theorem 17 says and are bounded on . Thus, the above estimate implies . This completes the proof since the supremum of is no more than times .

Theorem 19. If, then is closed under the operator norm on .

Proof. See Lemma 2.6 of [5]. We omit the details.

5. Equivalent Conditions for Compactness

For this section, we use the ideas in [5, 7] to characterize compactness of weakly localized operators on large Fock spaces. Indeed, it is more complex than [5] because Bergman metric works in a different way than in Euclidean metric.

We begin with the following preparations. Recall that, for fixed , there is an -lattice such that covers . Let . It follows that is also a covering of and . We write , and it is from estimate (16) that we consider

In what follows, we always define and as above. Also, there is some constant such that

Lemma 20. If and , then for every , there exists sufficiently large such that for the covering (associated to r), we obtain

Proof. Since , then there is some sufficiently large such that

Define . Then, Lemma 6 indicates that

Notice that

This together with, for some , estimate (20), Lemma 6, and Proposition 12 shows

We claim first that if and . Furthermore, applying (18), (93), Lemma 6, and Fubini’s theorem, we obtain

Since on , thus is well defined.

Thus it only remains to prove that .

In fact, (16) gives where . We then choose a satisfying .

Note that for any , there exists so that by estimate (12). For fixed , it follows that whenever is sufficiently large. So, for any , there is some such that

whenever . Thus, the triangle inequality implies . Further, the above inequality concludes that .

Next, we assume that . For any fixed , there is a satisfying . Since is a Lipschitz function, then where . Notice that , thus it allows us to let . It follows that . For and , we can pick an appropriate so that . Hence,

shows .

Therefore, the desired assertion holds, and the proof is finished.

Lemma 21. If and , then for every , there exists sufficiently large such that for the covering (associated to r), we obtain

Proof. By (71), for any and , there is some such thatand (70) shows

We also consider . For any fixed , we assume that . Note first that if , then where . For , we have and since is a Lipschitz function. Suppose , so , that is . It follows from Lemma 20 that

This together with s inequality and Fubini’s theorem implieswhich proves the desired result.

Lemma 22. Suppose . For any bounded linear operator on , there is some constant such thatwhere .

Proof. First, let . Suppose and , define

Since on , then by (85) we get

By combining the above estimate and Lemma 6, we obtain

Notice that there is a such that (we can let be small enough), then for , we get by (14). Now, by (18), (25), Lemma 2, and Fubini’s theorem, we have

Therefore, we deduce

Now, . It comes from the above proof that

For fixed and , let . Note that , we then get and by (16), where . Furthermore, there exists some such that . It follows from estimate (14) we have if and if (here we also assume is small enough so that ). Hence, by estimates (18), (20), (25), (93), Minkowski’s inequality, and Lemma 6, we obtain

This completes the proof.

Theorem 23. Suppose and . Then, the following conditions are equivalent (if):(A);(B) for any ;(C).

Proof. Suppose condition (A) holds. If , then (25) gives

Similarly, when , we obtain

Then (A) implies (C).

Because is clear, so it remains to prove that the implication . If and , then, by (68), there is some such that

Note that by the definition of function , we get . Then, (B) shows

whenever is large enough. By hypothesis, holds when .

Suppose . It follows from (20), (25), and Theorem 17 that

Therefore, joining condition and (71), we deduce

for sufficiently large . This completes the proof.

It is similar to ([17], Lemma 3.2) that we have the following assertion about relatively compact. That is, for every , there is some such that

if and only if a bounded subset is relatively compact. In what follows, we call the set of compact operators on .

Theorem 24. If and , then

Proof. We omit the details for , see ([5], Lemma 2.11).

If , then s inequality, Fubini’s theorem, (18), and Lemma 2 implyconverges to 0 as . This finishes the proof.

Due to that can be viewed as a Toeplitz operator induced by , then is compact (the reason is similar to ([18], Lemma 3.1)).

Theorem 25. Suppose . Then, there exists such thatwhere means the essential norm of a bounded operator on .

Proof. Since for , then , and we always assume that . Thus, Lemma 21 shows there is some satisfying

Because of is a compact operator where , thenwhere . For the rest of the task, we show

Let and . Note that , hence

where . It follows that

Furthermore, we have

Let be a sequence with such that

where . Fix , for , and there is a such that by (16). By joining the proof of Lemma 22 and s inequality, we deduce

where is independent of . This finishes the proof.

Theorem 26. Suppose and . Then, if and only if .

Proof. () For any , by Lemma 20, we get

Consider , then is a compact operator on , where is any positive integer. Hence,

With our assumption, there is an such that for . Since whenever is large enough, then Lemma 22 indicates for . Also, when , follows immediately from Theorems 23 and 25.

() The case is similar to the following discussion of .

Consider and . With the help of Theorem 23, we will finish the proof if as . Recall first that, for , by (25), we have

Since , estimate (18), , and Proposition 12, then

goes to 0 as , where .

On the other hand, Theorems 17 and 24 conclude that

as . Altogether gives thatwhich ends the proof since as .

Data Availability

No data are used.

Conflicts of Interest

The authors declare that they have no competing interests.

Acknowledgments

This research was supported by the National Natural Science Foundation of China (Grant No. 11971125).

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Copyright © 2021 Youqi Liu and Xiaofeng Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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