Research article

Commuting Toeplitz operators on weighted harmonic Bergman spaces and hyponormality on the Bergman space of the punctured unit disk

  • Received: 16 April 2024 Revised: 08 June 2024 Accepted: 14 June 2024 Published: 20 June 2024
  • MSC : 15B05, 15B48, 47B20, 47B35, 47B47

  • We first describe commuting Toeplitz operators with harmonic symbols on weighted harmonic Bergman spaces. Then, a sufficient condition for hyponormality on weighted Bergman spaces of the punctured unit disk, when the analytic part of the symbol is a monomial, is shown.

    Citation: Houcine Sadraoui, Borhen Halouani. Commuting Toeplitz operators on weighted harmonic Bergman spaces and hyponormality on the Bergman space of the punctured unit disk[J]. AIMS Mathematics, 2024, 9(8): 20043-20057. doi: 10.3934/math.2024977

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  • We first describe commuting Toeplitz operators with harmonic symbols on weighted harmonic Bergman spaces. Then, a sufficient condition for hyponormality on weighted Bergman spaces of the punctured unit disk, when the analytic part of the symbol is a monomial, is shown.



    Finding commuting Toeplitz operators is a question of interest for researchers working on Toeplitz operators. The most important results on this subject are from the work of Axler and Cuckovic [1] in the case of the Bergman space, and Choe and Lee in the case of the harmonic Bergman space [2]. The results of Axler and Cuckovic are generalized by the authors to the case of the weighted Bergman space [3]. In the first part, we generalize most of the results of Choe and Lee to the case of weighted harmonic Bergman spaces. In the second part, we give a sufficient condition for hyponormality on newly considered Bergman spaces on the punctured unit disk. Hyponormality has seen growing interest the last few decades. The first important result is the following.

    Theorem 1.1. Let f, g be bounded and analytic in the unit disk D with fH2. If Tf+¯g is hyponormal on the Bergman space, then gH2 and |g||f| a.e in the unit circle.

    This can be found in [4]. Ahern and Cuckovic generalized this result in [5] by weakening the assumption on the derivative. Their result itself is generalized to weighted Bergman spaces in [6].

    Theorem 1.2. Let f, g be bounded and analytic in the unit disk D. Assume fH2(I), where I is an open arc of the unit circle. If Tf+¯g is hyponormal on the weighted Bergman space L2a,ω, where ω(r)=(α+1)(1r2)α, α>1, then, gH2(I) and |g||f| a.e on I.

    Denote by D the unit disk and consider the Hilbert space L2(D,dμα) of measurable functions on D such that

    (α+1)D|f(z)|2(1|z|2)α dA(z)<.

    The closed subspace of L2(D,dμα) consisting of harmonic functions is denoted by L2h. If the weighted Bergman space is denoted by L2a,ω where ω(r)=(α+1)(1r2)α, then L2h=L2a,ω¯zL2a,ω. As in the case of unweighted spaces (ω=1) [2], the reproducing kernel is given by

    K(z,w)=1(1z¯w)α+2+1(1¯zw)α+21.

    In what follows, we consider commuting Toeplitz operators on L2h with harmonic symbols φ and ψ. We denote by P the projection of L2(D,dμα) onto L2a,ω, and by Q the projection onto L2h. We have Q(φ)=P(φ)+¯P(¯φ)P(φ)(0), where P(φ)(0)=DφdA. We give a generalization of some of the results in [2], shown in the case of the unweighted harmonic Bergman space, to the case of weighted harmonic Bergman spaces with radial weights ω(r)=(α+1)(1r2)α, where α>1. Our first main result is the following theorem.

    Theorem 2.1. Let φ and ψ be in L2h and not both conjugate holomorphic. If TφTψ=TψTφ, then ψz=cφz for some constant c.

    The proof is based on the properties of the projection. Here is a summary of the proof of Theorem 2.1. Set φ=φ1+¯φ2, ψ=ψ1+¯ψ2. Then,

    TφTψ=TψTφTφ1T¯ψ2+T¯φ2Tψ1=T¯ψ2Tφ1+Tψ1T¯φ2.

    By evaluation at ¯w, and after making extensive use of some properties of the Bergman space projection we get φ1P(¯wψ1)=ψ1P(¯wφ1). By Proposition 2.2, this is reduced to ψ1=cφ1.

    We start by verifying these properties.

    Proposition 2.2. Let φL2a,ω. The projection onto L2a,ω satisfies the following properties:

    (ⅰ) z2ddzP((1|w|2)φ)+(α+2)zP((1|w|2)φ)=(α+1)zφ(z).

    (ⅱ) z2ddzP((1|w|2)¯φ)+(α+2)zP((1|w|2)¯φ)=(α+1)z¯φ(0).

    (ⅲ) P(|w|2φ)(z)=φ(z)(α+1)zα+2z0uα+1φ(u)du.

    (ⅳ) P(|w|2¯φ)=¯P(|w|2¯φ)(0)=α+1α+2¯φ(0).

    Proof. For (ⅰ) we have

    f(z)=P((1|w|2)φ)(z)=(α+1)Dφ(w)(1z¯w)α+2(1|w|2)α+1dA(w)

    and, by differentiating under the integral sign, we get

    f(z)=(α+1)(α+2)Dφ(w)¯w(1z¯w)α+3(1|w|2)α+1dA(w).

    We deduce

    (α+2)zf(z)+z2f(z)=(α+1)(α+2)Dφ(w)z(1z¯w)α+3(1|w|2)α+1dA(z)=(α+1)zφ(z),

    where the last equality holds because Kz(w)=1(1z¯w)α+3 is the reproducing kernel of the weighted Bergman space with a weight ω2(r)=(α+2)(1r2)α+1dA(z) [7]. We obtain Eq (ⅱ) in a similar manner. For Eq (ⅲ), we formally solve the differential equation

    f+(α+2)zf=(α+1)zφ,

    which gives

    P((1|w|2)φ)(z)=(α+1)zα+2z0uα+1φ(u) du,

    and

    P(|w|2φ)(z)=φ(z)(α+1)zα+2z0uα+1φ(u) du.

    Similarly, from (ⅱ), by integration we get

    P((1|w|2)¯φ)=α+1α+2¯φ(0),

    thus

    P(|w|2¯φ)=1α+2¯φ(0).

    This leads to the following properties of the projection.

    Lemma 2.3. Let φL2a,ω satisfy φ(0)=0. The following properties hold:

    1) P(¯wφ)(z)=φ(z)zα+1zα+2z0uαφ(u) du.

    2) P(w¯φ)(z)=1α+2¯φ(0).

    Proof. Since φ(0)=0, φ(z)=zψ(z). From (ⅲ) and (ⅳ) of Proposition 2.2, we get

    P(¯wφ)(z)=P(|w|2ψ)(z)=φ(z)z(α+1)zα+2z0uαφ(u) du,
    P(w¯φ)(z)=P(|w|2¯ψ)(z)=1α+2¯ψ(0)=1α+2¯φ(0).

    The following properties of the projection of L2(D,dμα) are easy to check. The first one is a consequence of the known property of Toeplitz operators TfTg=Tfg on L2a,ω if g is analytic or f is conjugate analytic.

    Lemma 2.4. For φ and ψL2a,ω, we have the following equalities:

    (ⅰ) P(¯φP(¯wψ))(z)=P(¯φ¯wψ)(z).

    (ⅱ) P(φ¯P(¯wψ)(0)=P(φw¯ψ)(0).

    Proof. We verify (ⅱ):

    P(φ¯P(¯wψ)(0)=P(φ¯P(¯wψ)),1=φ¯P(¯wψ),1=¯P(¯wψ),¯φ=φ,P(¯wψ)=P(φw¯ψ)(0).

    This leads to the following theorem.

    Theorem 2.5. Let φ and ψ be in L2h. If TφTψ=TψTφ, then, there exists a constant λ such that ψz=λφz.

    Proof. We have φ=φ1+¯φ2, and ψ=ψ1+¯ψ2, where φ1, φ2, ψ1 and ψ2L2a,ω. We may assume

    φ1(0)=φ2(0)=ψ1(0)=ψ2(0)=0.

    We evaluate both sides of the equality TφTψ=TψTφ at the function ¯w. We make use of the following equalities, where Lemmas 2.3 and 2.4 are used.

    Tψ1(¯w)=P(¯wψ1)+¯P(w¯ψ1)P(¯wψ1)(0)=P(¯wψ1)
    T¯φ2Tψ1(¯w)=P(¯φ2(P(¯wψ1)))+¯P(φ2¯P(¯wψ1))P(¯φ2P(¯wψ1))(0)
    Tφ1Tψ1(¯w)=φ1P(¯wψ1)
    Tφ1T¯ψ2(¯w)=Q(φ1¯w¯ψ2)=P(φ1¯w¯ψ2)+¯P(¯φ1wψ2)P(φ1¯w¯ψ2)(0)
    T¯φ2T¯ψ2(¯w)=¯φ2¯ψ2¯w.

    Similarly, we compute

    T¯ψ2Tφ1(¯w)=P(¯ψ2(P(¯wφ1)))+¯P(ψ2¯P(¯wφ))P(¯ψ2P(¯wφ1))(0)
    Tψ1T¯φ2(¯w)=Q(ψ1¯w¯φ2)=P(ψ1¯w¯φ2)+¯P(¯ψ1wφ2)P(ψ1¯w¯φ2)(0)
    Tψ1Tφ1(¯w)=ψ1P(¯wφ1)
    T¯ψ2T¯φ2(¯w)=¯φ2¯ψ2¯w.

    Thus, the equality

    TφTψ(¯w)=TψTφ(¯w)

    leads to

    φ1P(¯wψ1)+¯P(¯φ1wψ2)+¯P(φ2¯P(¯wψ1))=ψ1P(¯wφ1)+¯P(¯ψ1wφ2)+¯P(ψ2¯P(¯wφ1)). (2.1)

    We deduce that

    φ1P(¯wψ1)=ψ1P(¯wφ1),

    which, by Proposition 2.2, gives

    φ1(z).(ψ1(z)zα+1zα+2z0uαψ1(u)du)=ψ1(z).(φ1(z)zα+1zα+2z0uαφ1(u) du),

    which can be written as

    Φ1(z)Ψ1(z)=Φ1(z)Ψ1(z),

    where

    Φ1(z)=z0uαφ1(u)du),Ψ1(z)=z0uαψ1(u)du.

    This shows that

    Ψ1(z)=cΦ1(z),

    and, consequently,

    ψ1=cφ1

    for some constant c.

    Corollary 2.6. Let φ, ψ be in L2a,ω. The equality TφTψ=TψTφ holds if and only if ψ=cφ+d for some constants c and d.

    Our next main result is based on the following lemmas.

    Lemma 2.7. Let φ, ψ be in L2a,ω. Then we have the following identity:

    (α+1)(Dφ(w)¯ψ(w)(1|w|2)αdA(w))=D((α+2)φ(w)¯ψ(w)+¯wψ(w))(1|w|2)α+1dA(w).

    Proof. Use power series and the fact that, for φ, ψ in L2a,ω, with φ=n0φnzn, ψ=n0ψnzn, we have the inner product property [7]

    (α+1)(Dφ(w)¯ψ(w)(1|w|2)αdA(w))=n0n!Γ(α+2)Γ(n+α+2)φn¯ψn.

    Lemma 2.8. Let φ, ψ be in L2a,ω and satisfy φ(0)=ψ(0)=0. If TφT¯ψ=T¯ψTφ on L2h, then the following equality holds

    Dφ(w)¯Ψ(w)wn+α(1|w|2)α+1dA(w)=0

    for any nN, where Ψ(w)=α+1wα+1w0uαψ(u)du.

    Proof. From TφT¯ψ=T¯ψTφ, we deduce TψT¯φ=T¯φTψ, which, by evaluation at ¯w and from Eq (2.1) by taking φ=φ2, φ1=0, ψ=ψ1, and ψ2=0 gives

    P(φw¯ψ)=P(φ¯P(¯wψ)).

    This gives

    φ,P(ψ¯w)wn+α+1=P(φ¯P(¯wψ)),wn+α+1=P(φw¯ψ),wn+α+1=φw,ψwn+α+1,n0.

    Using the identity wP(ψ¯w)=ψΨ, we get

    (α+1)Dφ(w)¯(ψ(w)Ψ(w))wn+α(1|w|2)αdA(w)=(α+1)Dφ(w)¯ψ(w)wn+α|w|2(1|w|2)αdA(w),

    which can be written as

    (α+1)Dφ(w)¯ψ(w)wn+α(1|w|2)α+1dA(w)=(α+1)Dφ(w)¯Ψ(w)wn+α(1|w|2)αdA(w). (2.2)

    Using Lemma 2.7, we get

    (α+1)Dφ(w)¯Ψ(w)wn+α(1|w|2)αdA(w)=Dφ(w)((α+2)¯Ψ(w)wn+α+¯w¯(Ψ(w)wn+α))(1|w|2)α+1dA(w)=φ(w)((n+α+1)¯Ψ(w)+(α+1)¯ψ(w))¯wn+α(1|w|2)α+1dA(w).

    After a simplification using Eq (2.2), we deduce

    Dφ(w)¯(Ψ(w)wn+α)(1|w|2)α+1dA(w)=0.

    This leads to the second main result. The proof, being similar to the case of the unweighted harmonic Bergman space [2], is omitted.

    Theorem 2.9. Assume fL2h. Then, Tf is normal if and only if f(D) is contained in a line in the complex plane. If fL2a,ω, then Tf is normal if and only if f is constant.

    Let m be a nonnegative integer, β be in (m1,m], and α>1 be a real number. Let dμα,β(z)=1B(α+1,β+1)|z|2β(1|z|2)αdA(z) be the normalized Lebesgue measure on the unit disk D. The space L2(D,dμα,β) is the Hilbert space of measurable functions f on D such that

    ||f||2=D|f(z)|2dμα,β(z)<.

    We consider holomorphic functions g on D=D{0} that satisfy

    D|g(z)|2dμα,β(z)<.

    In [8], it is shown that this space, denoted by A2α,β, is a closed Hilbert space of L2(D,dμα,β). If f=nmanzn and g=nmbnzn, then the inner product is given by

    f,g=nmB(α+1,β+n+1)B(α+1,β+1)an¯bn,

    and its natural orthonormal basis is {en(z)=B(α+1,β+1)B(α+1,β+n+1)zn, nm}. We set ρn=B(α+1,β+n+1)B(α+1,β+1), nm, and recall that a Toeplitz operator Tφ, for φ bounded measurable on D, is defined on A2α,β by Tφ(f)=P(φf), where P is the orthogonal projection onto A2α,β. Hankel opertaors on A2α,β are defined by Hφ(f)=(IP)(φf). The following properties of Toeplitz operators, known in the case of the weighted Bergman space A2α (β=0), hold also on A2α,β.

    Proposition 3.1. [8] Let φ and ψ be bounded measurable on D. The following properties hold:

    (ⅰ) Tφ+ψ=Tφ+Tψ.

    (ⅱ) (Tφ)=T¯φ.

    (ⅲ) TψTφ=Tφψ if φ analytic or ψ is conjugate analytic on D.

    A bounded operator S on a Hilbert space is hyponormal if SSSS0. In what follows, we consider hyponormality of Toeplitz operator Tzq+¯ψ on A2α,β where ψ is a polynomial. We show a sufficient condition in this case. As in the case of the classical unweighted Bergman space A20,0 [4], hyponormality is expressed in various equivalent forms which are listed in the following proposition.

    Proposition 3.2. [4] The following statements are equivalent:

    (a) Tφ+¯ψ is hyponormal.

    (b) T¯ψTψTψT¯ψT¯φTφTφT¯φ.

    (c) (H¯ψ)H¯ψ(H¯φ)H¯φ.

    (d) H¯ψ=CH¯φ, where C is bounded of norm less than or equal to one.

    We consider the case α=3. Let l and q be two integers such that qm+1. Denote by N the smallest integer such that Nmax(6q+1,8,12(q21)). Our main result is the following theorem.

    Theorem 3.3. Let (λl)NlN1 be any finite sequence of complex numbers such that NlN1|λl|1, and set g=NlN1λlqlzl. Then, the operator Tzq+¯g is hyponormal on A2α,β i.e T¯gTgTgT¯gT¯zqTzqTzqT¯zq.

    In the case α=3, we have

    ρn=(β+4)(β+3))(β+2)(β+1)(β+n+4)(β+n+3)(β+n+2)(β+n+1).

    We start by computing the matrix of T¯zqTzqTzqT¯zq in the orthonomal basis {en, nm}.

    Lemma 3.4. The matrix of T¯zqTzqTzqT¯zq is diagonal and is given by

    ηi,i={ρq+iρi,ifi<qm;ρq+iρiρiρiq,ifiqm.

    Proof. We clearly have

    T¯zqTzqei,ej=0ifij,andT¯zqTzqei,ei=ρq+iρi.

    Since

    P(¯zqzi),zk=0ifkiq,

    necessarily,

    P(¯zqzi)=cziq,

    and

    P(¯zqzi),ziq=cziq,ziq=cρiq
    P(¯zqzi),ziq=ρi.

    We deduce c=ρiρiq, and

    T¯zqei,T¯zqei={ρiρiqifiqm;0ifi<qm.

    This leads to the following proposition on which our main result is based.

    Proposition 3.5. Let l and q be two integers with qm+1, lmax(6q+1,12(q21),8). Then, Tzq+δ¯zl is hyponormal if and only if |δ|ql.

    Proof. Hyponormality is equivalent to the following three inequalities:

    |δ|2ρl+iρiρq+iρi,i<qm. (3.1)
    |δ|2ρl+iρiρq+iρiρiρiq,qmi<lm. (3.2)
    |δ|2(ρl+iρiρiρil) ρq+iρiρiρiq,ilm. (3.3)

    The first inequality (3.1) takes the form

    |δ|2min{(β+l+i+4)(β+l+i+3)(β+l+i+2)(β+l+i+1)(β+q+i+4)(β+q+i+3)(β+q+i+2)(β+q+i+1), i<qm}.

    Since the right hand side of this inequality decreases with i, inequality (3.1) is equivalent to

    |δ|(β+l+qm+3)(β+l+qm+2)(β+l+qm+1)(β+l+qm)(β+2qm+3)(β+2qm+2)(β+2qm+1)(β+2qm).

    Inequality (3.2) can be written as

    |δ|2ρi+qρiqρi2ρl+iρiq,qmi<lm.

    A computation reduces the above inequality to

    |δ|2min{Ql,q,iRq,i,qmi<lm},

    where

    Ql,q,i=(β+l+i+4)(β+l+i+3)(β+l+i+2)(β+l+i+1)(β+q+i+4)(β+q+i+3)(β+q+i+2)(β+q+i+1),

    and

    Rq,i=1(1q2(β+i+4)2)(1q2(β+i+3)2)(1q2(β+i+2)2)(1q2(β+i+1)2). (3.4)

    As mentioned before, Ql,q,i is a decreasing function of i and it is not difficult to see that Rq,i is also decreasing. Thus, we get that inequality (3.2) is equivalent to

    |δ|Ql,q,lm1Rq,lm1,

    or, equivalently,

    |δ|(β+2lm+3)(β+2lm+2)(β+2lm+1)(β+2lm)Rq,lm1(β+q+lm+3)(β+q+lm+2)(β+q+lm+1)(β+q+lm).

    Since Ql,q,i is decreasing, Ql,q,lm1Ql,q,qm1. Thus, we have

    Ql,q,lm1Rq,lm1Ql,q,qm1.

    That is, if inequality (3.2) is satisfied, so is inequality (3.1). Inequality (3.3) takes the form

    |δ|2ρilρiqρi+qρiqρ2iρi+lρilρ2i,ilm.

    To simplify notation, set Ai:=(β+i+1). A computation leads to the following inequality for ilm:

    |δ|2A2i+3A2i+2A2i+1A2i(A2i+3q2)((A2i+2q2)(A2i+1q2)(A2iq2)A2i+3A2i+2A2i+1A2i(A2i+3l2)((A2i+2l2)(A2i+1l2)(A2il2)Ql,q,i.

    After simplification of

    F(i)=A2i+3A2i+2A2i+1A2i(A2i+3q2)((A2i+2q2)(A2i+1q2)(A2iq2)A2i+3A2i+2A2i+1A2i(A2i+3l2)((A2i+2l2)(A2i+1l2)(A2il2),

    we get

    F(i)=q2F1(i)q4F2(i)+q6F3(i)q8l2F1(i)l4F2(i)+l6F3(i)l8,

    where the functions F1, F2, and F3 are as follows:

    F1(i)=A2i+3A2i+2A2i+1+A2i+3A2i+2A2i+A2i+2A2i+1A2i+A2i+3A2i+1A2i;
    F2(i)=A2i+3A2i+2+A2i+3A2i+1+A2i+3A2i+A2i+2A2i+1+A2i+2A2i+A2i+1A2i;
    F3(i)=A2i+3+A2i+2+A2i+1+A2i.

    As seen before, Ql,q,i is a decreasing function of i. We will show that F(i) is also a decreasing function. A computation gives

    F(i)=E1(i)+E2(i)+E3(i)+E4(i)+E5(i)+E6(i)(l2F1(i)l4F2(i)+l6F3(i)l8)2,

    where we have

    E1(i)=l2q2(q6l6)F1(i);
    E2(i)=l4q4(l2+q2)(l2q2)F2(i);
    E3(i)=l6q6(q2l2)F3(i);
    E4(i)=l2q2(q2l2)(F1(i)F2(i)F1(i)F2(i));
    E5(i)=l2q2(l2+q2)(l2q2)(F1(i)F3(i)F1(i)F3(i));
    E6(i)=l4q4(q2l2)(F2(i)F3(i)F2(i)F3(i)).

    It suffices to verify the inequalities

    E1(i)+E2(i)0,ilm; (3.5)
    E4(i)+E5(i)0,ilm. (3.6)

    Note lmqm+12. We show inequality (3.5) first. It takes the form

    l2q2(l2+q2)F2(i)(l4+l2q2+q4)F1(i).

    From the expressions of F1(i) and F2(i), we see that, since β+i+1q for ilm,

    q2F2(i)F1(i),

    and thus inequality (3.5) holds. Inequality (3.6) can be reduced to

    (l2+q2)(F1(i)F3(i)F1(i)F3(i))F1(i)F2(i)F1(i)F2(i). (3.7)

    An elementary computation gives

    F1(i)F3(i)F1(i)F3(i)=Φ1(i)+Φ2(i)+Φ3(i)+Φ4(i)+Ψ(i),

    where

    Φ1(i)=P2,31(i)+P2,41(i)+P3,41(i);
    P2,31(i)=2(β+i+1)((β+i+2)4(β+i+3)2+(β+i+2)2(β+i+3)4);
    P2,41(i)=2(β+i+1)((β+i+2)4(β+i+4)2+(β+i+2)2(β+i+4)4);

    and

    P3,41(i)=2(β+i+1)((β+i+3)4(β+i+4)2+(β+i+3)2(β+i+4)4).

    The function Φ2(i) is defined by

    Φ2(i)=P1,32(i)+P1,42(i)+P3,42(i)

    where

    P1,32(i)=2(β+i+2)((β+i+1)4(β+i+3)2+(β+i+1)2(β+i+3)4).

    P1,42 and P3,42 are defined in a similar manner.

    The functions Φ3 and Φ4 are defined in a similar way. The function Ψ is given by

    Ψ(i)=Ψ1(i)+Ψ2(i)+Ψ3(i)+Ψ4(i)

    with

    Ψ1(i)=4(β+i+1)(β+i+2)2(β+i+3)2(β+i+4)2,

    and

    Ψ2(i)=4(β+i+2)(β+i+1)2(β+i+3)2(β+i+4)2,

    and similar definitions for Ψ3 and Ψ4. Next, we simplify the right hand side of inequality (3.7). A tedious, but elementary, computation gives

    F1(i)F2(i)F1(i)F2(i)=Δ(i)+Ω(i),

    where

    Δ(i)=Δ1(i)+Δ2(i)+Δ3(i)+Δ4(i)

    with

    Δ1(i)=2(β+i+1)[(β+i+2)4(β+i+3)4+(β+i+2)4(β+i+4)4+(β+i+3)4(β+i+4)4];
    Δ2(i)=2(β+i+2)[(β+i+1)4(β+i+3)4+(β+i+1)4(β+i+4)4+(β+i+3)4(β+i+4)4];

    and similar definitions for Δ3(i) and Δ4(i). The function Ω(i) is given by

    Ω(i)=Ω1(i)+Ω2(i)+Ω3(i)+Ω4(i)

    with Ω1(i) and Ω2(i) defined by

    Ω1(i)=Ω1,2(i)+Ω1,3(i)+Ω1,4(i),

    where

    Ω1,2(i)=2(β+i+1)((β+i+2)4(β+i+3)2(β+i+4)2);
    Ω1,3(i)=2(β+i+1)((β+i+3)4(β+i+2)2(β+i+4)2);

    and Ω1,4(i) defined in a similar way. We show

    (l2+q2)Φ1(i)Δ1(i),

    and, since l12(q21),

    (l2+q2)(β+i+2)4(β+i+3)2(β+i+2)4(β+i+3)4,

    and similarly

    (l2+q2)(β+i+2)2(β+i+3)4(β+i+2)4(β+i+3)4.

    Thus,

    (l2+q2)P2,312(β+i+1)(β+i+2)4(β+i+3)4.

    Similar inequalities lead to

    (l2+q2)Φ1(i)Δ1(i),

    and we have

    (l2+q2)Φ2(i)Δ2(i),(l2+q2)Φ3(i)Δ3(i),(l2+q2)Φ4(i)Δ4(i)

    for ilm. Next, we show (l2+q2)Ψ1(i)Ω1(i) for ilm. We clearly have

    (l2+q2)Ψ1(i)6(β+i+1)((β+i+2)4(β+i+3)2(β+i+4)2Ω1(i).

    Similarly, we have

    (l2+q2)Ψ2(i)Ω2(i),(l2+q2)Ψ3(i)Ω3(i),(l2+q2)Ψ4(i)Ω4(i).

    We deduce that inequality (3.6) holds and that F(i) is decreasing. Thus, inequality (3.3) holds if and only if |δ|limiF(i)=ql. We finally show that

    q2l2(β+2lm+3)(β+2lm+2)(β+2lm+1)(β+2lm)Rq,lm1(β+l+qm+3)(β+l+qm+2)(β+l+qm+1)(β+l+qm). (3.8)

    From inequality (3.7), we have

    Rq,lm1=1(1q2(β+lm+3)2)(1q2(β+lm+2)2)(1q2(β+lm+1)2)(1q2(β+lm)2).

    To simplify notation, we set Bi:=(β+i+3)(β+i+2)(β+i+1)(β+i). Inequality (3.8) takes the form

    1l2B2lm(B2lmBlm+qBlmq)q2Blm+qB2lm,

    and we have

    B2lmBlm+qBlmq=q2F1(lm1)q4F2(lm1)+q6F3(lm1)q8,

    where

    F1(lm1)=A2lm+2A2lm+1A2lm+A2lm+2A2lm+1A2lm1+A2lm+1A2lmA2lm1+A2lm+2A2lmA2lm1

    and, with Ai:=β+i+1 as before,

    F2(lm1)=A2lmA2lm1+A2lm+1A2lm1+A2lm+1A2lm+A2lm+2A2lm+1+A2lm+2A2lm+A2lm+2A2lm1,

    and

    F3(lm1)=A2lm1+A2lm+A2lm+1+A2lm+2.

    Since

    B2lm=A2lm+2A2lm+1A2lmA2lm1,

    we get

    F1(lm1)B2lm=1A2lm1+1A2lm+1A2lm+1+1A2lm+2
    F2(lm1)B2lm=1A2lm1A2lm+1A2lm1A2lm+1+1A2lm1A2lm+2+1A2lmA2lm+1+1A2lmA2lm+2+1A2lm+1A2lm+2
    F3(lm1)B2lm=1A2lm+1A2lmA2lm1+1A2lm+2A2lm+1A2lm+1A2lm+2A2lm+1A2lm1+1A2lm+2A2lmA2lm1.

    We see that inequality (3.8) is equivalent to

    l2(B2lmBlm+qBlmq)q2B2lm4l2A2lm+26q2l2A2lm1A2lm+4q4l2A2lm+2A2lm+1A2lmq6l2A2lm+2A2lm+1A2lmA2lm1. (3.9)

    We show that if lN, then

    4l2A2lm+26q2l2A2lm1A2lm+4q4l2A2lm+2A2lm+1A2lmq6l2A2lm+2A2lm+1A2lmA2lm11.

    We then rewrite this last inequality as

    4l2A2lm+1A2lmA2lm16q2l2A2lm+2A2lm+1+4q4l2A2lm1q6l2A2lm+2A2lm+1A2lmA2lm1.

    Since β(m1,m], by assumptions on l we have

    2l2A2lm+1A2lmA2lm1A2lm+2A2lm+1A2lmA2lm1.

    Thus, inequality (3.9) is satisfied if

    2A2lm+1A2lmA2lm16q2A2lm+2A2lm+1+4q4A2lm1q60.

    Since by assumptions on l and q,

    4q4A2lm1q6,

    it is enough to have

    A2lmA2lm13q2A2lm+20,

    and this is satisfied by assumptions on l. Hence, inequality (3.9) holds. We conclude that if qm+1 and lmax(6q+1,12(q21),8), Tzq+δ¯zl is hyponormal if and only if |δ|ql. The proof is complete.

    It is easy to see that if a and b are complex numbers satisfying |a|+|b|1, and if Tφ+¯ψ1 and Tφ+¯ψ2 are hyponormal then, Tφ+a¯ψ1+b¯ψ2 is hyponormal. This leads to our main result, where N is defined as earlier.

    Theorem 3.6. Let (λl)NlN1 be any finite sequence of complex numbers such that NlN1|λl|1, and set g=NlN1λlqlzl. Then, we have T¯gTgTgT¯gT¯zqTzqTzqT¯zq on A2α,β.

    The other results of Choe and Lee [2] can be generalized and this could be addressed in a future work. Hyponormality on the punctured disk is a very recent subject of interest and much remains to be done on this. We hope this paves the way for further works.

    All authors contributed equally and significantly to the study conception and design, material preparation, data collection and analysis. The first draft of the manuscript was written by H. Sadraoui, and all authors read and approved the final manuscript.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors would like to extend their sincere appreciation to Researchers Supporting Project number (RSPD2024R1112), King Saud University, Riyadh, Saudi Arabia.

    The authors declare that there is no conflict of interest regarding the publication of this paper.



    [1] S. Axler, Z. Cuckovic, Commuting Toeplitz operators with harmonic symbols, Integr. Equ. Oper. Theory, 14 (1991), 1–12. https://doi.org/10.1007/BF01194925 doi: 10.1007/BF01194925
    [2] B. R. Choe, Y. J. Lee, Commuting Toeplitz operators on the harmonic Bergman space, Michigan Math. J., 46 (1999), 163–174. https://doi.org/10.1307/mmj/1030132367 doi: 10.1307/mmj/1030132367
    [3] H. Sadraoui, M. Guediri, Hyponormality of Toeplitz operators on the Bergman space of an annulus, Rev. Union Mat. Argent., 61 (2020), 303–313.
    [4] H. Sadraoui, Hyponormality of Toeplitz operators and cohyponormality of composition operators, PhD Thesis, Purdue University, 1992.
    [5] P. Ahern, Z. Cuckovic, A mean value inequality with applications to Bergman space operators, Pacific J. Math., 173 (1996), 295–305.
    [6] H. Sadraoui, M. Garayev, H. Guediri, On Hyponormality of Toeplitz operators, Rocky Mountain J. Math., 51 (2021), 1821–1831. https://doi.org/10.1216/rmj.2021.51.1821 doi: 10.1216/rmj.2021.51.1821
    [7] H. Hedenmalm, B. Korenblum, K. H. Zhu, Theory of Bergman spaces, New York: Springer, 2000. https://doi.org/10.1007/978-1-4612-0497-8
    [8] N. Ghiloufi, M. Zaway, Meromorphic Bergman spaces, Ukr. Math. J., 74 (2023), 1209–1224. https://doi.org/10.1007/s11253-023-02130-9 doi: 10.1007/s11253-023-02130-9
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