# On the Boundedness of the Fractional Bergman Operators

Authors: Benoît F. Sehba
Journal: Abstract and Applied Analysis
 (2017)

---

## Abstract

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


---

## Body

## 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.LetH be the upper-half plane, that is, the set z=x+iy∈C:y>0. We denote by LαpH the Lebesgue space LpH,yαdxdy, that is, the space of all functions f such that(1)fp,αp≔∫Hfx+iypyαdxdy<∞.Forα>-1 and 1<p<∞, the weighted Bergman space Aαp(H) is the subspace of Lαp(H) consisting of analytic functions. It is well known that the Bergman space Aα2(H) (-1<α<∞) is a reproducing kernel Hilbert space with kernel Kzα(w)=Kα(z,w)=1/z-w¯2+α. That is, for any f∈Aα2(H), the following representation holds:(2)fw=Pαfw=f,Kwαα=∫HfzKαw,zdVαz,where, for simplicity, we write dVα(x+iy)=yαdxdy. The positive Bergman operator Pα+ is defined by(3)Pα+fw=∫HfzKαw,zdVαz.Note that the boundedness ofPα+ implies the boundedness of Pα. It is an elementary exercise to prove that the Bergman projection Pα is bounded on Lαp(H) if and only if 1<p<∞ (see, e.g., [3]).The fractional Bergman operatorPα,γ is defined by(4)Pα,γfw=∫Hfzw-z¯2+α-γdVαz,where 0≤γ<2+α. The corresponding positive operator will be denoted by Pα,γ+ 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 γ=0, Pα,γ is just the Bergman projection.We have the following necessary and sufficient condition for the boundedness ofPα,γ and Pα,γ+.Theorem 1.
Letα>-1, 0≤γ<2+α, and 1<p≤q<∞. Then the following conditions are equivalent: (a)
The operatorPα,γ is bounded from LpH,dVα to Lq(H,dVα).(b)
The operatorPα,γ+ is bounded from Lp(H,dVα) to Lq(H,dVα).(c)
The following relation holds:(5)1p-1q=γ2+α.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 operatorPα,γ+ from Lp(H,dVα) to Lq(H,dVα) implies the boundedness of Pα+ from LpH,dVα to LqH,dVβ, where β=qγ+α. It follows that we have the following.Proposition 2.
Letα>-1, 0≤γ<2+α, and 1<p≤q<∞. Then the following conditions are equivalent: (a)
The operatorPα is bounded from Lp(H,dVα) to Lq(H,dVqγ+α).(b)
The operatorPα+ is bounded from Lp(H,dVα) to Lq(H,dVqγ+α).(c)
The following relation holds:(6)1p-1q=γ2+α.For the proof of the sufficient part, we will use the off-diagonal Schur test due to Okikiolu [7].


## 2. Proof of Theorem1 and Proposition 2

We start by recalling the following easy fact (see [3]).Lemma 3.
Letα be real. Then the function f(z)=z+it/i-α, with t>0, belongs to Lp(H,dVν), if and only if ν>-1 and α>ν+2/p. In this case,(7)fp,νp=Cα,p,qt-pα+ν+2.The proof of the sufficient parts in our results is based on the following off-diagonal Schur-type test.Lemma 4 (Okikiolu [7]).
Letp,r,q be positive numbers such that 1<p≤r and 1/p+1/q=1. Let K(x,y) be a complex-valued function measurable on X×Y and suppose that there exist 0<t≤1, measurable functions φ1:X→(0,∞) and φ2:Y→(0,∞), and nonnegative constants M1 and M2 such that(8)∫XKx,ytqφ1qydμy≤M1qφ2qxa.eonY,∫YKx,y1-trφ2rxdνx≤M2rφ1rya.eonX.If T is given by (9)Tfx=∫XfyKx,ydμy,where f∈Lp(X,dμ), then T:Lp(X,dμ)→Lr(Y,dν) is bounded and for each f∈Lp(X,dμ), (10)TfLrY,dν≤M1M2fLpX,dμ.We prove the following.Lemma 5.
Letα>-1, 0≤γ<2+α, and 1≤p≤q<∞. If the operator Pα,γ is bounded from Lp(H,dVα) to Lq(H,dVα), then (11)1p-1q=γ2+α.Proof.
We assume that the operatorPα,γ is bounded from Lp(H,dVα) to Lq(H,dVα). Let R>0 and associate to any function f, the function fR defined by fR(z)=f(Rz). Then it is easy to see that (12)fRp,α=R-2+α/pfp,α.It follows also from an easy change of variables that (13)Pα,γfRw=R-γPαfRw.Hence (14)Pα,γfRq,α=R-γ-2+α/qPα,γfq,α.It follows from the above considerations and the boundedness of Pα,γ that there exists a constant C>0 such that, for any f∈Lp(H,dVα), (15)R-γ-2+α/qPα,γfq,α=Pα,γfRq,α≤CfRp,α=R-2+α/pfp,α.That is, (16)R-γ-2+α/q+2+α/pPα,γfq,α≤Cfp,α.As the latter holds for any f∈LpH,dVα and any R>0, we must have (17)-γ-2+αq+2+αp=0.That is, 1/p-1/q=γ/(2+α).The following is obtained as above.Lemma 6.
Letα>-1, 0≤γ<2+α, and 1≤p≤q<∞. If the operator Pα is bounded from Lp(H,dVα) to Lq(H,dVqγ+α), then (18)1p-1q=γ2+α.We next prove that condition (5) is sufficient for the boundedness of the fractional operator in the case p>1.Lemma 7.
Letα>-1, 0≤γ<2+α, and 1<p≤q<∞. Assume that (19)1p-1q=γ2+α.Then the operator Pα,γ+ is bounded from Lp(H,dVα) to Lq(H,dVα).Proof.
We are assuming that1/p-1/q=γ/(2+α).
Let us put(20)ω=-2+α1p′+1q=-2+α1-γ2+α.Clearly, ω<0. As α+1>0, we can find two numbers s and r such that (21)0<r<α+1q,0<s<α+1p′withr>s.
Let(22)t≔-α+2/p′+s-rωso that(23)1-t=r-s-α+2/qω.We observe that the operator Pα,γ+ can be represented as (24)Pα,γ+fz=∫HfwKz,wdVαw,where K(z,w)=1/z-w¯2+α-γ. Let us define (25)φ1w=Iw-s,φ2w=Iw-r.Applying Okikiolu’s test to Pα,γ+ we first obtain (26)∫HKz,wtp′φ1p′wIwαdVw=∫Hv-sp′+αz-w¯t2+α-γp′dVw.From our choice of s we have -sp′+α>-1. Using the definitions of ω and t, we obtain (27)t2+α-γp′+sp′-α-2=2+α1-γ2+αtp′+sp′-α-2=-ωtp′+sp′-α-2=α+2p′+r-sp′+sp′-α-2=rp′>0.Hence we obtain from the above observations and Lemma 3 that (28)∫HKz,wtp′φ1p′wIwαdVw=Cy-rp′=Cφ2zp′.In the same way, we first have (29)∫HKz,w1-tqφ2qzIzαdVz=∫HIz-rq+αz-w¯1-t2+α-γqdVz.From our choice of s, we have -rq+α>-1. From the definition of ω and 1-t, we obtain(30)1-t2+α-γq+rq-α-2=-ω1-tq+rq-α-2=2+αq+s-rq+rq-α-2=sq>0.Hence we obtain from the above observations and Lemma 3 that (31)∫HKz,w1-tqφ2qzIzαdVw=CIw-sq=Cφ1wq.
The proof is complete.Proof of Theorem1.
It is obvious that(b)⇒(a). That (a)⇒(c) is Lemma 5. That (c)⇒(b) is Lemma 7. The proof is complete.Proof of Proposition2.
Clearly,(b)⇒(a). That (a)⇒(c) is Lemma 6. That (c)⇒(b) follows from Lemma 7 and the fact that the boundedness of Pα,γ+ from Lp(H,dVα) to Lq(H,dVα) implies the boundedness of Pα+ from Lp(H,dVα) to Lq(H,dVqγ+α). The proof is complete.



---

*Source: 8363478-2017-05-10.xml*