# Hyponormality on a Weighted Bergman Space

Authors: Houcine Sadraoui; Borhen Halouani; Mubariz T. Garayev; Adel AlShehri
Journal: Journal of Function Spaces
 (2020)

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## Abstract

A bounded Hilbert space operatorT is hyponormal if T∗T−TT∗ is a positive operator. We consider the hyponormality of Toeplitz operators on a weighted Bergman space. We find a necessary condition for hyponormality in the case of a symbol of the form f+g¯ where f and g are bounded analytic functions on the unit disk. We then find sufficient conditions when f is a monomial.


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## Body

## 1. Introduction

LetDR denote the disk of radius R in the complex plane, dμαz=1/πR21−z/R2αdAz where dAz is the Lebesgue measure on DR and α>−1. L2DR,dμz denotes the Hilbert space of complex-valued functions on DR that are square integrable with respect to μ. We write f2=∫DRfz2dμz. When f is analytic on DR, we have
(1)fz=∑n=0∞anzn,f2=∑n=0∞R2nn!Γα+1Γn+α+2an2.Denote byAα2 the space of analytic functions on DR such that f2<∞. It is known that Aα2 is a Hilbert space [1], and an orthonormal basis is given by enz=Γn+α+2/Rnn!Γα+1zn. The Toeplitz operator with symbol f on Aα2 is defined by Tfk=Pfk where f is bounded and measurable on DR, k is in Aα2, and P is the orthogonal projection of L2DR,dμαz onto Aα2. Hankel operators are defined by Hfk=I−Pfk, f and k as before. Recall that a bounded operator T on a Hilbert space is hyponormal if T∗T−TT∗ is a positive operator. Hyponormality on the Hardy space is studied by Cowen in [2, 3]. Unweighted Bergman spaces on the unit disk α=0,R=1 and Toeplitz operators on these spaces were considered in [4, 5] and [6]. Hyponormality was first considered by Sadraoui in [7]. An improvement of the necessary condition is due to Ahern and Cuckovic [8]. A new necessary condition in a special case is found by Cuckovic and Curto in [9]. Many results on hyponormality on weighted Bergman spaces treat special cases. We cite for example [1, 10]. In this work for simplicity’s sake, we consider the case α=1 and R=1. Under a smoothness assumption, we give a fairly general necessary condition for the hyponormality of Toeplitz operators with a symbol of the form f+g¯ where f and g are bounded and analytic. We also give sufficient conditions when f is a monomial and g a polynomial without assuming R=1.We begin by recalling some general properties of Toeplitz operators.


## 2. Some General Properties

We assumef, g are in L∞DR. Then, we have:
(1)
Tf+g=Tf+Tg(2)
Tf∗=Tf¯(3)
Tf¯Tg=Tf¯g if f or g analytic on DRThe use of these properties leads to describing hyponormality in more than one form. These are easy to prove by using Douglas lemma [11].Proposition 1.
Letf,g be bounded and analytic on DR. Then, the following are equivalent: [label=)]
(i)
Tf+g¯ is hyponormal(ii)
Hg¯∗Hg¯≤Hf¯∗Hf¯(iii)
I−Pg¯k≤I−Pf¯k for any k in Aα2(iv)
g¯k2−Pg¯k2≤f¯k2−Pf¯k2 for any k in Aα2(v)
Hg¯=CHf¯ where C is of norm less than or equal to oneWe also need the following lemmas.Lemma 2.
Fors and t integers, we have
(2)Pz¯tzs=R2ts!Γs−t+α+2Γs+α+2s−t!zs−t,ifs≥t0,ifs<t.Proof.
We have(3)Pz¯tzs,zm=zs,zm+t=1πR2∫02π∫0Rrs+m+t+1eis−t−mθ1−rR2αdrdθ.We see thatPz¯tzs,zm=0 if s<t. For s≥t, Pz¯tzs,zm≠0 if and only if m=s−t. In this case, we get
(4)Pz¯tzs,zs−t=2R2∫0Rr2s+11−rR2αdr=2R2s∫01u2s+11−u2αdu=R2s∫01xs1−xαdx=R2sβs+1,α+1=R2sΓs+1Γα+1Γs+α+2.On the other hand, ifPz¯tzs=λzs−t, then
(5)Pz¯tzs,zs−t=λzs−t2,so λ=Pz¯tzs,zs−t/zs−t2. A computation shows
(6)zs−t2=R2s−ts−t!Γα+1Γs−t+α+2,so λ=R2ts!Γs−t+α+2/s−t!Γs+α+2, and the result follows.Lemma 3.
Iff=∑n=0∞anzn is bounded analytic on the disk DR, then the i,j th entry of the matrix of Hf¯∗Hf¯ with respect to the orthonormal basis Γn+α+2/Rnn!Γα+1znn≥0 is given by:
(7)Γj+α+2Rjj!RiΓi+α+2i!∑m≥j−i,m≥0am+i−jam¯R2mm+i!Γm+i+α+2−j!i!Rj−iΓj+α+2Γi+α+2∑0≤m,i−j≤m≤iR2mamam+j−i¯Γi−m+α+2i−m!.Proof.
We have:(8)f¯fΓj+α+2Rjj!Γα+1zj=Γj+α+2Rjj!Γα+1∑n,m∞anam¯zn+jz¯m=Γj+α+2Rjj!Γα+1∑p≥j,m≥0ap−jam¯zm¯zp.Thus, by Lemma2,
(9)Tf¯TfΓj+α+2Rjj!Γα+1zj=Γj+α+2Rjj!Γα+1∑p≥j,p≥m,m≥0ap−jam¯R2mp!Γp−m+α+2Γp+α+2p−m!zp−m=Γj+α+2Rjj!Γα+1∑p≥j,m≥0,p≥map−jam¯R2mp!Γp−m+α+2Γp+α+2p−m!zp−m.Thus, noting thatzp−m,zi≠0 if p=m+i, we get
(10)Tf¯TfΓj+α+2Rjj!Γα+1zj,Γi+α+2Rii!Γα+1zi=Γj+α+2Rjj!Γα+1Γi+α+2Rii!Γα+1∑m≥j−i,m≥0am+i−jam¯R2mm+i!Γi+α+2Γm+i+α+2i!R2ii!Γα+1Γi+α+2=Γj+α+2Rjj!RiΓi+α+2i!∑m≥j−i,m≥0am+i−jam¯R2mm+i!Γm+i+α+2.We also have by Lemma2(11)Pf¯Γj+α+2Rjj!Γα+1zj=Γj+α+2Rjj!Γα+1P∑0∞an¯z¯nzj=Γj+α+2Rjj!Γα+1∑0jan¯R2nj!Γj−n+α+2Γj+α+2.j−n!zj−n.Then,(12)Tf¯Γj+α+2Rjj!Γα+1zj,Tf¯Γi+α+2Rii!Γα+1zi=Γj+α+2Γi+α+2RiRjj!Γα+2i!Γα+1∑m=1∞aman¯Pz¯nzj,Pz¯mzi.The above inner product is nonzero ifi−m=j−n≥0. We get by Lemma 2(13)TfTf¯Γj+α+2Rjj!Γα+1zj,Γi+α+2Rii!Γα+1zi=j!i!Rj−iΓj+α+2Γi+α+2∑0≤m,i−j≤m≤iR2mamam+j−i¯Γi−m+α+2i−m!.WhenR=1, we set D1=D. We obtain in this case the following corollary.Corollary 4.
Iff=∑n=0+∞anzn is bounded and analytic on the unit disk D, and if α=1. Then, the matrix of Hf¯∗Hf¯ with respect to the orthonormal basisn+2n+1znn≥0 is given by
(14)Δi,j=j+2j+1i+2i+1∑m≥j−i,m≥0am+i−jam¯1m+i+2m+i+1−1j+2j+1i+2i+1∑0≤m,i−j≤m≤iamam+j−i¯i−m+2i−m+1.Recall that the Hardy spaceH2 is defined by
(15)H2=f=∑n=0+∞anznanalyticontheunitdisksuchthat∑n=0+∞an2<+∞.In the following theorem, the Toeplitz operatorTf′2 on H2 is not necessarily bounded. However, its matrix Γi,ji,j in the usual orthonormal basis of H2 is defined.The following theorem is an extension of Lemma2.4.2 [7] to the weighted Bergman space.Theorem 5.
Assumef bounded analytic on the unit disk D and f′∈H2. Let Γi,j be the matrix of the Toeplitz operator on H2,Tf′2. Then, n2Δi+n,j+n⟶2Γi,j as n⟶∞.Proof.
The symmetric matrix ofHf¯∗Hf¯ with respect to the orthonormal basis n+2n+1znn≥0 is given by (set j=i+p in Corollary 4 with p≥0.)
(16)Δi,i+p=i+p+2i+p+1i+2i+1∑m≥pam−pam¯1m+i+2m+i+1−1i+p+2i+p+1i+2i+1∑0≤m≤iamam+p¯i−m+2i−m+1.To simplify, setCn,i,p=i+n+1i+n+2i+n+p+1i+n+p+2. Then,
(17)Δi+n,i+n+p=Cn,i,p∑m≥pam−pam¯1m+i+n+2m+i+n+1−1Cn,i,p∑0≤m≤i+namam+p¯i+n−m+2i+n−m+1for any integer n. Set m=p+l in the first sum, and m=l in the second. With some obvious notations, we get
(18)Δi+n,i+n+p=Cn,i,p∑l≥0alap+l¯1p+l+i+n+2p+l+i+n+1−1Cn,i,p∑0≤l≤i+nalap+l¯i+n−l+2i+n−l+1=∑l≤i+nalap+l¯Cn,i,p−i+n−l+2i+n−l+1p+l+i+n+2p+l+i+n+1Cn,i,pp+l+i+n+2p+l+i+n+1+Cn,i,p∑l≥i+n+1alap+l¯1p+l+i+n+2p+l+i+n+1=∑l≤i+nAn,l,p,ialap+l¯+∑l≥i+n+1Bn,l,p,ialap+l¯.A computation shows that(19)An,l,p,i=ll+pi+n+12+i+n+22+lpl+p2n+2i+3+l2l+p2Cn,i,pp+l+i+n+2p+l+i+n+1.It is not difficult to see that(20)n2An,l,p,ialap+l¯→n→∞2ll+palap+l¯,and that the following inequality holds:
(21)n2An,l,p,ialap+l¯≤5ll+palap+l≤5/2l2al2+l+p2al+p2for l≤i+n. We have that ∑l2al2+l+p2al+p2<∞ since f′∈H2. Writing the sum ∑l≤i+nn2An,l,p,ialap+l¯ as an integral, where the measure is the counting measure, and using the dominated convergence theorem, we get that
(22)∑l≤i+nn2An,l,p,ialap+l¯→n→∞2∑ll+palal+p¯.Clearly,Cn,i,p≤i+n+p+1i+n+p+2. So we have for l≥i+n+1,
(23)n2Bn,l,p,ialal+p¯=n2Cn,i,pp+l+i+n+2p+l+i+n+1alal+p¯≤ll+palal+pfor l≥i+n+1. Thus,
(24)∑l≥i+n+1n2Bn,l,p,ialal+p≤∑l≥i+n+1ll+palal+p≤12∑l≥i+n+1l2al2+12∑l≥i+n+1l+p2al+p2.This shows that(25)∑l≥i+n+1n2Bn,l,p,ialap+l¯→n→∞0.We deduce thatn2Δi+n,i+n+p→n→∞2∑ll+palal+p¯=2Γi,i+p.


## 3. The Results

Using the previous theorem leads us to our main result which extends Theorem5.4.3 [7].Theorem 6.
Letf and g be bounded and analytic on the unit disk D, and such that f′∈H2. If Tf+g¯ is hyponormal on A12 then g′∈H2, and g′≤f′ a.e. on the unit circle.Proof.
Setg=∑nbnzn. All diagonal terms of the matrix of Hf¯∗Hf¯−Hg¯∗Hg¯ are positive; thus, using the same notation as in the previous proof, we have
(26)∑l≤i+nn2An,l,0,ial2+∑l≥i+n+1n2Bn,l,0,ial2≥∑l≤i+nn2An,l,0,ibl2+∑l≥i+n+1n2Bn,l,0,ibl2.We have seen that the left hand side tends to2∑l2al2, a finite sum since f′∈H2. Writing ∑l≤i+nn2An,l,0,ibl2 as an integral with respect to the counting measure, noticing that χ0,..i+nln2An,l,0,ibl2→n→∞2l2bl2, and using Fatou’s lemma, we deduce that 2∑l2bl2≤2∑l2al2 and g′∈H2. If Ωi,j denotes the matrix of Hf¯∗Hf¯−Hg¯∗Hg¯, then n2Ωi+n,j+n⟶Λi,j, where Λi,j is the matrix of the Toeplitz operator on H2, Tf′2−g′2. Since the matrix of Hf¯∗Hf¯−Hg¯∗Hg¯ is positive by hypothesis, it follows that the matrix of Tf′2−g′2 is also positive. A Toeplitz matrix is positive if its symbol is positive [12]. It follows that f′≥g′a.e on the unit circle.To obtain sufficient conditions whenf is a monomial, we need the following lemma.Lemma 7.
Letq be a positive integer, the matrix of Hz¯q∗Hz¯q is diagonal and given by:
(27)di=R2qq+i!Γi+α+2i!Γq+i+α+2,ifi<qR2qq+i!Γi+α+2i!Γq+i+α+2−i!Γi−q+α+2i−q!Γi+α+2,ifi≥q.In what follows, we will takeα=1. It follows from Proposition 1 that, if p>q, the hyponormality of Tzq+λzp¯ is equivalent to the following inequalities:
(28)λ2i+2i+1p+i+2p+i+1≤R2q−pi+2i+1q+i+2q+i+1i<q;(29)λ2i+2i+1p+i+2p+i+1≤R2q−pi+2i+1q+i+2q+i+1−i−q+2i−q+1i+2i+1q≤i<p;(30)λ2i+2i+1p+i+2p+i+1−i−p+2i−p+1i+2i+1≤R2q−pi+2i+1q+i+2q+i+1−i−q+2i−q+1i+2i+1p≤i.Lemma 8.
The inequality (28) holds if and only if λ≤Rq−pp+q+1p+q/2q2q+1.Proof.
Setgi=p+i+2p+i+1/q+i+2q+i+1. Then, (28) is equivalent to λ≤Rq−pmingii<q. Using logarithmic differentiation, it is easy to see that g decreases with i, the minimum is thus assumed at i=q−1; hence, the result follows.Lemma 9.
The inequality (29) holds if and only if
(31)λ≤2qp+1Rq−p2p+12p2+2p+1−q2pp+qp+q+1.Proof.
A computation shows that inequality (29) is equivalent to
(32)λ2≤q2R2q−pp+i+2p+i+12i2+6i+5−q2q+i+1q+i+2i+12i+22q≤i<p.Using logarithmic differentiation, we can see that the right hand side of the last inequality is a decreasing function ofi. Thus, the minimum is assumed at i=p−1, and this proves the lemma.A similar argument shows the following.Lemma 10.
The inequality (30) holds if and only if λ≤q/pRq−p.Proposition 11.
Ifq<p, then Tzq+λzp¯ is hyponormal if and only if λ≤q/pRq−p.Proof.
Using elementary computations and the previous lemma, it is easy to see that ifλ≤q/pRq−p, then inequalities (28) and (29) are satisfied. This proves the result.Remark 12.
The result remains true ifp=q.Denote byU1 the unit ball of A12⊥.Definition 13.
Forf∈A12, we denote by Gf the set
(33)Gf=g∈A12:supl∈U1g¯,k¯l≤supl∈U1f¯,k¯lforanyk∈H∞.We see, from the density ofH∞ in A12 and Proposition 1, that when g and f are in H∞, g∈Gf is equivalent to Tf+g¯ is hyponormal. The following proposition lists some properties of Gf [7]. We provide the proof for the sake of completeness.Proposition 14.
Forf∈A12 the following holds: [label=()]
(a)
Gf is convex and balanced(b)
Ifg∈Gf, then g+c is in Gf for any complex number c(c)
f∈Gf.(d)
Gf is closed in the weak topology of L2DR,dμ1zProof.
Only (d) needs to be verified. Assumegμ is a net in Gf and gμ⟶g. We have for l0 in U1, g¯,k¯l0=limgμ¯,k¯l0 and gμ¯,k¯l0≤supgμ¯,k¯l,l∈U1≤supf¯,k¯l,l∈U1. The result follows by taking the limit and taking the supremum on U1.
Using the two previous propositions, we obtain the following.Theorem 15.
Letλn be a sequence of complex numbers satisfying ∑q∞λn≤1 where q is a positive integer. Then, Tzq+∑q∞λnq/nRq−nzn¯ is hyponormal.We now consider the hyponormality ofTzq+λzp¯ when p<q. In this case, hyponormality is equivalent to the following three inequalities:
(34)λ2≤R2q−pp+i+2p+i+1q+i+2q+i+1i<p,λ2≤1p2R2q−pp+i+2p+i+1i+22i+12q+i+2q+i+12i2+6i+5−p2p≤i<q,λ2≤q2p2R2q−pp+i+2p+i+12i2+6i+5−q2q+i+2q+i+12i2+6i+5−p2q≤i.A computation, using logarithmic differentiation, allows us to see that the right hand side of each of these inequalities increases withi. This leads to the following inequalities:
(35)∣λ∣≤Rq−pp+2p+1q+2q+1;∣λ∣≤Rq−pp+1p+2p22p+1p+q+1p+q+2p+5;∣λ∣≤Rq−pqpq+5p+q+1p+q+222q2+6q+5−p22q+1.Set(36)Ap,q=p+2p+1q+2q+1,Bp,q=p+1p+2p22p+1p+q+1p+q+2p+5,Cp,q=qpq+5p+q+1p+q+222q2+6q+5−p22q+1.Write(37)Bp,q=2p+12p+2p+1p2+4p+4p+q+1p+q+2p2p+5.Sincegi=p+i+2p+i+1/q+i+2q+i+1 increases with i, we get
(38)p+1p+2q+1q+2≤2p+12p+2p+q+1p+q+2.Also,p+1p2+4p+4≥p2p+5. Thus, Ap,q<Bp,q for p<q. We have
(39)Cp,q=q2p+q+2p+q+1q+1q+52p2q+12q+12q2+6q+5−p2,and we can readily verify that for p<q, we have
(40)q2q+1q+5≥p22q2+6q+5−p2.As before,gi=p+i+2p+i+1/q+i+2q+i+1 increases with i if p<q. We get g0≤gq which leads to the inequality
(41)p+2p+1q+1q+2≤p+q+2p+q+12q+12q+1.Combining the last inequality with inequality (40), we deduce the following:Proposition 16.
Ifp<q then Tzq+λzp¯ is hyponormal if and only if λ≤Rq−pp+2p+1/q+2q+1.Remark 17.
The result holds also ifp=q.From the properties ofGzq, noted in Proposition 14, we deduce the following corollaries.Corollary 18.
Letλ1,λ2⋯.,λq be complex numbers satisfying ∑1qλi≤1. Then, Tzq+∑1qλnRq−nn+2n+1/q+2q+1zn¯ is hyponormal.Corollary 19.
Letλn be a sequence of complex numbers satisfying ∑1∞λi≤1. Then, Tzq+∑1qλnRq−nn+2n+1/q+2q+1zn¯+∑q+1∞λnq/nRq−nzn¯ is hyponormal.



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