#### Abstract

We give a necessary and sufficient condition for the boundedness of the Bergman fractional operators.

#### 1. Introduction and Statement of the Result

We are interested in this note in the boundedness of the Bergman fractional operator. The Bergman fractional operator has been shown recently to be quite useful in understanding off-diagonal questions for the Bergman operator (see [1, 2]). Our aim here is to provide a necessary and sufficient condition for the boundedness of this operator. In the next lines, we provide some notions and definitions needed in the sequel.

Let be the upper-half plane, that is, the set . We denote by the Lebesgue space , that is, the space of all functions such that

For and , the weighted Bergman space is the subspace of consisting of analytic functions. It is well known that the Bergman space () is a reproducing kernel Hilbert space with kernel . That is, for any , the following representation holds:where, for simplicity, we write . The positive Bergman operator is defined by

Note that the boundedness of implies the boundedness of . It is an elementary exercise to prove that the Bergman projection is bounded on if and only if (see, e.g., [3]).

The fractional Bergman operator is defined bywhere . The corresponding positive operator will be denoted by and can be seen as the upper-half space analogue of the Riesz potential also known as fractional operator (see [4]). Note also that, for , is just the Bergman projection.

We have the following necessary and sufficient condition for the boundedness of and .

Theorem 1. *Let , , and . Then the following conditions are equivalent: *(a)*The operator is bounded from to .*(b)*The operator is bounded from to .*(c)*The following relation holds:*

Unlike the case of the unit ball (see [5]), the above result can not be deduced from the boundedness of the families of Bergman-type operators considered in [2, 6].

We remark that the boundedness of the operator from to implies the boundedness of from to , where . It follows that we have the following.

Proposition 2. *Let , , and . Then the following conditions are equivalent: *(a)*The operator is bounded from to .*(b)*The operator is bounded from to .*(c)*The following relation holds:*

For the proof of the sufficient part, we will use the off-diagonal Schur test due to Okikiolu [7].

#### 2. Proof of Theorem 1 and Proposition 2

We start by recalling the following easy fact (see [3]).

Lemma 3. *Let be real. Then the function , with , belongs to , if and only if and . In this case,*

The proof of the sufficient parts in our results is based on the following off-diagonal Schur-type test.

Lemma 4 (Okikiolu [7]). *Let be positive numbers such that and Let be a complex-valued function measurable on and suppose that there exist , measurable functions and , and nonnegative constants and such thatIf is given by where , then is bounded and for each , *

We prove the following.

Lemma 5. *Let , , and . If the operator is bounded from to , then *

*Proof. *We assume that the operator is bounded from to . Let and associate to any function , the function defined by . Then it is easy to see that It follows also from an easy change of variables that Hence It follows from the above considerations and the boundedness of that there exists a constant such that, for any , That is, As the latter holds for any and any , we must have That is, .

The following is obtained as above.

Lemma 6. *Let , , and . If the operator is bounded from to , then *

We next prove that condition (5) is sufficient for the boundedness of the fractional operator in the case .

Lemma 7. *Let , , and . Assume that Then the operator is bounded from to .*

*Proof. *We are assuming that .

Let us put Clearly, . As , we can find two numbers and such that Letso thatWe observe that the operator can be represented as where . Let us define Applying Okikiolu’s test to we first obtain From our choice of we have . Using the definitions of and , we obtain Hence we obtain from the above observations and Lemma 3 that In the same way, we first have From our choice of , we have . From the definition of and , we obtainHence we obtain from the above observations and Lemma 3 that The proof is complete.

*Proof of Theorem 1. *It is obvious that . That is Lemma 5. That is Lemma 7. The proof is complete.

*Proof of Proposition 2. *Clearly, . That is Lemma 6. That follows from Lemma 7 and the fact that the boundedness of from to implies the boundedness of from to . The proof is complete.

#### Conflicts of Interest

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