On the Boundedness of the Fractional Bergman Operators

and Applied Analysis 3 Clearly, ω < 0. As α + 1 > 0, we can find two numbers s and r such that 0 < r < α + 1 q , 0 < s < α + 1 p󸀠 with r > s. (21) Let t fl − (α + 2) /p 󸀠 + s − r ω (22) so that 1 − t = r − s − (α + 2) /q ω . (23) We observe that the operatorP+α,γ can be represented as P + α,γf (z) = ∫ H f (w)K (z, w) dVα (w) , (24) whereK(z, w) = 1/|z − w|2+α−γ. Let us define φ1 (w) = (Iw)−s , φ2 (w) = (Iw)−r . (25) Applying Okikiolu’s test toP+α,γ we first obtain ∫ H K (z, w)tp󸀠 φp󸀠 1 (w) (Iw)α dV (w) = ∫ H V +α |z − w|t(2+α−γ)p dV (w) . (26) From our choice of s we have −sp󸀠 + α > −1. Using the definitions of ω and t, we obtain t (2 + α − γ) p󸀠 + sp󸀠 − α − 2 = (2 + α) (1 − γ 2 + α) tp 󸀠 + sp󸀠 − α − 2 = −ωtp󸀠 + sp󸀠 − α − 2 = (α + 2 p󸀠 + r − s)p 󸀠 + sp󸀠 − α − 2 = rp󸀠 > 0. (27) Hence we obtain from the above observations and Lemma 3 that ∫ H K (z, w)tp󸀠 φp󸀠 1 (w) (Iw)α dV (w) = Cy−rp󸀠 = Cφ2 (z)p󸀠 . (28) In the same way, we first have ∫ H K (z, w)(1−t)q φq 2 (z) (Iz)α dV (z) = ∫ H (Iz)−rq+α |z − w|(1−t)(2+α−γ)q dV (z) . (29) From our choice of s, we have −rq + α > −1. From the definition of ω and 1 − t, we obtain (1 − t) (2 + α − γ) q + rq − α − 2 = −ω (1 − t) q + rq − α − 2 = (2 + α q + s − r) q + rq − α − 2 = sq > 0. (30) Hence we obtain from the above observations and Lemma 3 that ∫ H K (z, w)(1−t)q φq 2 (z) (Iz)α dV (w) = C (Iw)−sq = Cφ1 (w)q . (31) The proof is complete. Proof of Theorem 1. It is obvious that (b) ⇒ (a). That (a) ⇒ (c) is Lemma 5. That (c) ⇒ (b) is Lemma 7. The proof is complete. Proof of Proposition 2. Clearly, (b) ⇒ (a). That (a) ⇒ (c) is Lemma 6. That (c) ⇒ (b) follows from Lemma 7 and the fact that the boundedness of P+α,γ from Lp(H, dVα) to Lq(H, dVα) implies the boundedness of P+ α from Lp(H, dVα) to Lq(H, dVqγ+α). The proof is complete. Conflicts of Interest The author declares that there are no conflicts of interest regarding the publication of this paper.


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 H be the upper-half plane, that is, the set { =  +  ∈ C :  > 0}.We denote by    (H) the Lebesgue space   (H,   ), that is, the space of all functions  such that For  > −1 and 1 <  < ∞, the weighted Bergman space    (H) is the subspace of    (H) consisting of analytic functions.It is well known that the Bergman space  2  (H) (−1 <  < ∞) is a reproducing kernel Hilbert space with kernel    () =   (, ) = 1/( − ) 2+ .That is, for any  ∈  2  (H), 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    (H) if and only if 1 <  < ∞ (see, e.g., [3]).
The fractional Bergman operator P , is defined by where 0 ≤  < 2 + .The corresponding positive operator will be denoted by P + , 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  = 0, P , is just the Bergman projection.
We have the following necessary and sufficient condition for the boundedness of P , and P + , .
Then the following conditions are equivalent: (a) The operator P , is bounded from   (H,   ) to   (H,   ).
(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].Proposition 2. Let  > −1, 0 ≤  < 2+, and 1 <  ≤  < ∞.Then the following conditions are equivalent: (b) The operator  +  is bounded from   (H,   ) to   (H,  + ).
(c) The following relation holds: For the proof of the sufficient part, we will use the offdiagonal Schur test due to Okikiolu [7].

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  > 0, belongs to   (H,  ] ), if and only if ] > −1 and  > (] + 2)/.In this case, The proof of the sufficient parts in our results is based on the following off-diagonal Schur-type test.
The following is obtained as above.
Then the operator P + , is bounded from   (H,   ) to   (H,   ).
Let us put Clearly,  < 0. As  + 1 > 0, we can find two numbers  and  such that with  > .
We observe that the operator P + , can be represented as where (, ) = 1/| − | 2+− .Let us define Applying Okikiolu's test to P + , we first obtain | − | (2+−)   () . (26) From our choice of  we have −  +  > −1.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 − +  > −1.From the definition of  and 1 − , we obtain ( The proof is complete.