Boundedness and Compactness of Hankel Operators on Large Fock Space

Abstract

We introduce the BMO spaces and use them to characterize complex-valued functions such that the big Hankel operators and are both bounded or compact from a weighted large Fock space into a weighted Lebesgue space when .

1. Introduction

Let denote the set of complex numbers. For a fixed positive integer , let be the -dimensional Euclidean space. Given and , let denote the Euclidean ball defined as

Let be the Laplace operator and denote the Lebesgue volume measure on . Suppose is a -plurisubharmonic function. We say that belongs to the weight class , if satisfies the following statements: (A)There exists such that for (B) satisfies the reverse-Hlder’s inequality for some (C)The eigenvalues of are comparable; i.e., there exists such that where

Let . For , the weighted Lebesgue space is the set of those measurable functions on such that

For , the space consists of all measurable functions on such that

The weighted Fock space consists of all analytic functions on such that .

The weight class is first introduced by Dall’Ara in [1] where the pointwise estimation of the Bergman kernel is given. Recently, the research on the weighted Fock space has been gradually improved. Lv [2] obtained characterizations on the boundedness of the Bergman projection. Arroussi and Tong [3] gave the -norm estimation of the Bergman kernel and studied the boundedness, compactness, and Schatten class membership of the weighted composition operators. Liu and Wang [4] consider the duality of large Fock spaces and obtained equivalent conditions for compactness of weakly localized operators. The weight class contains abundant functions. For example, when is replaced by , is the classical Fock space (see [5]). The weight function satisfying in [6, 7] is the member of . It is interesting that the weight class also contains some nonradial weights. An example of nonradial weight is given in [3].

For , is a Banach space. In particular, is a Hilbert space. Let be the Bergman kernel of and denote for simplicity. We know that . The Bergman projection can be represented as

Moreover, is bounded from to and for any when . See [2, 4] for more details. From [4], we know that is dense in . To let make sense on , we naturally consider those measurable functions in the symbol class defined as

It is obvious that the space of essentially bounded measurable functions on is contained in . Given some symbol function , one defines the big Hankel operator as

Using the property of reproducing kernel, it has the form of

Hankel operators on Fock spaces and Bergman spaces have been studied by many mathematicians during the past few decades; see [5, 8] for instance. The main problem we consider in this paper is to characterize the simultaneous boundedness and compactness of Hankel operators and from to with in terms of suitable defined spaces and spaces. There are exclusive references for this issue both in the setting of Bergman spaces and Fock spaces. See [9–12] for Hankel operators that are simultaneously bounded or compact between the same exponents and [13–16] between different exponents.

This paper is organized as follows. In Section 2, we will give some basic useful results including the -norm as well as pointwise estimation of the reproducing kernel. The space of will be established in Section 3. The decomposition theorem of is given. We will use the space to characterize simultaneously bounded Hankel operators and in Section 4. The space is introduced in Section 5, and it has similar results of the space . Compact Hankel operators are characterized in Section 6.

Throughout this paper, we will use to denote a positive constant whose value may change from line to line, but does not depend on functions being considered. We denote if there is a positive constant such that . stands for and .

2. Preliminaries

In this section, we are going to give some basic definitions and results, which will be frequently used. For , set

Given , write . We will call a radius function.

Lemma 1. Let . Then, the function satisfies the following properties: (A) is bounded on , that is (B)For , and , there holds

Proof. See [3].☐

Let be the normalized reproducing kernel in , that is

We are going to give some estimations of the reproducing kernel. One can find the proof in [3].

Lemma 2. Let , then one has (A)For (B)There is some such that(C)The set is bounded in and uniformly on every compact subsets of when .

In the following argument, we will always assume to be a constant as in (B) of Lemma 2. Notice that , so inequality (15) always holds when . Before giving the pointwise estimation of the Bergman kernel, we need to introduce a distance function associated with the radius function . In fact, we are interested with the Riemannian distance. To describe it explicitly, given , we define where the infimum is taken as which varies over the collection of piecewise -curves with and . We define the ball of -metric as

The following lemma gives the pointwise estimation of .

Lemma 3. There is a constant such that

Proof. See Theorem 20 of [1].
In what follows, we will always assume to be a constant as in Lemma 3.☐

3. The Space BMO

For , let be the set of all -th locally Lebesgue integrable functions on . Suppose and , the averaging function is defined as where is the volume of . For , the -th mean oscillation of at is given by

Suppose and , the space consists of those functions such that

The following proposition reveals that the function in (23) can be replaced by any complex-valued function in .

Proposition 4. Let , , and . Then, if and only if there exists a complex-valued function on such that where is a constant.

Proof. If , then (25) holds with and .
Conversely, if (25) is satisfied, by the triangle inequality for the -norm, we obtain For , by the Hölder’s inequality, one can get For , it is obvious that So we have which implies This completes the proof.☐

Let be the set of all analytic functions on .

Lemma 5. Let , , and . If for any , there exist functions and constants such that then .

Proof. The approach is similar to [13]. It is well known that, if is pluriharmonic, there exists some pluriharmonic function such that . Theorem 1 from [17] tells us there is some constant such that Hence, for any , by change of variables, we know that, if is pluriharmonic, there exists a pluriharmonic function such that and For , set . By the triangle inequality, we have Since is real-valued, we get Notice that , by (33), one has By the triangle inequality, there holds Similarly, there holds Set . By the triangle inequality again, we have From this and Proposition 4, the desired result follows immediately.☐

To arrive at the decomposition theorem, we need to introduce two subspaces of : one is , the other is .

For a continuous function on and radius , let

The function is called the oscillation of at . For , let denote the space of all continuous functions on such that

Before we give a characterization of functions , we need the following lemma.

Lemma 6. There holds for and .

Proof. If , take the curve . For , . By (15), there holds It follows that which makes .
If , take any piecewise -curve connecting and , and let be the minimum time such that . By (15), there holds This implies that With the fact that , one has . It follows that . The proof is completed.☐

Proposition 7. Suppose . Then, the space is independent of . Moreover, a continuous function on belongs to if and only if for any , there exists a constant such that where .

Proof. Suppose . Set . According to Lemma 6, we get and whenever .
If , then for any , there holds Take , then the inequality (47) holds in the case of .
If , we replace points on the geodesic in the -metric from to in such a way: for and . Note that it is possible to avoid using the completeness of the -metric. More specifically, we do not have to use geodesics. Instead, we can simply use a curve from and whose length in the -metric differs slightly from . Since , one can get It follows that By the triangle inequality, there holds Take , then the inequality (47) comes from combining (50) and (51).
Conversely, suppose (47) holds. Set . According to Lemma 6, we have whenever . By (47), there holds It follows that This completes the proof.☐

From the above proposition, we have proved that is independent of the radius for any . So one can simply write for short.

For and , let denote the space of all locally -th integrable functions on such that

Given a Borel measure , a radius , and , we define the averaging function with respect to as follows:

Suppose is a Borel measure. Let and ; we define the -Berezin transform with respect to as follows:

In particular, for a function on , set , then

For , we call a positive Borel measure on a Carleson measure if there is a positive constant such that for every function . This definition means that the identity mapping is bounded. The following lemma characterizes the Carleson measure, and one can find it from Theorem 3.2 in [3] and its proof.

Lemma 8. Let be a finite positive Borel measure on . Assume . Then, the following conditions are all equivalent: (A)The measure is a Carleson measure for (B)The function for (C)The function for . Moreover

Set . Then, we have and . The following lemma comes immediately.

Lemma 9. Let and . Then, the following conditions are all equivalent: (A)The measure is a Carleson measure for (B)The function for (C)The function for , i.e., . Moreover

From the previous lemma, one can see that the space is independent of the choice of . So we will write for short.

Lemma 10. Let , then there holds

Proof. See [2].☐

Theorem 11. Let , , and . Then, the following conditions are all equivalent: (A)(B) admits a decomposition where and (C)The function is bounded. Moreover

Proof. : suppose . For , we have . In fact, let . Since , using inequality (15), we know that Applying the triangle inequality, we get Set and . For and , by the triangle inequality and Hölder’s inequality, one can get For and , there also holds Thus . By Proposition 7, we have . Moreover, Next, we prove the part with regard to . By the triangle inequality, one can get For , using the Hölder’s inequality, there holds For , the above inequality is obvious. So we conclude that for all . Thus, . By Lemma 9, we have . Moreover, (B) (C): set where and . By the triangle inequality, there holds By (17) and Lemma 3, we have Since as for any , there exists a constant such that By Proposition 7, Lemma 10, (74), and (75), one can get For , by Hölder’s inequality and (76), we obtain The same is true for (77) when . By (73), (76), and (77), there holds Next, we prove the part with regard to . For , applying Hölder’s inequality, we obtain For , inequality (79) is obvious. By the triangle inequality, (79), and Lemma 9, there holds