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
[1] | Qian Ding . Commuting Toeplitz operators and H-Toeplitz operators on Bergman space. AIMS Mathematics, 2024, 9(1): 2530-2548. doi: 10.3934/math.2024125 |
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[10] | Salma Aljawi, Kais Feki, Hranislav Stanković . Jointly $ A $-hyponormal $ m $-tuple of commuting operators and related results. AIMS Mathematics, 2024, 9(11): 30348-30363. doi: 10.3934/math.20241464 |
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 f′∈H2. If Tf+¯g is hyponormal on the Bergman space, then g′∈H2 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 f′∈H2(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)(1−r2)α, α>−1, then, g′∈H2(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)(1−r2)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(1−z¯w)α+2+1(1−¯zw)α+2−1. |
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)(1−r2)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 ∂ψ∂z=c∂φ∂z 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=TψTφ⇔Tφ1T¯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 φ∈L2a,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+2∫z0uα+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)(1−z¯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(1−z¯w)α+3(1−|w|2)α+1dA(w). |
We deduce
(α+2)zf(z)+z2f′(z)=(α+1)(α+2)∫Dφ(w)z(1−z¯w)α+3(1−|w|2)α+1dA(z)=(α+1)zφ(z), |
where the last equality holds because Kz(w)=1(1−z¯w)α+3 is the reproducing kernel of the weighted Bergman space with a weight ω2(r)=(α+2)(1−r2)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+2∫z0uα+1φ(u) du, |
and
P(|w|2φ)(z)=φ(z)−(α+1)zα+2∫z0uα+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 φ∈L2a,ω satisfy φ(0)=0. The following properties hold:
1) P(¯wφ)(z)=φ(z)z−α+1zα+2∫z0uαφ(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+2∫z0uαφ(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 ψ∈L2a,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 ∂ψ∂z=λ∂φ∂z.
Proof. We have φ=φ1+¯φ2, and ψ=ψ1+¯ψ2, where φ1, φ2, ψ1 and ψ2∈L2a,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)z−α+1zα+2∫z0uαψ1(u)du)=ψ1(z).(φ1(z)z−α+1zα+2∫z0uαφ1(u) du), |
which can be written as
Φ′1(z)Ψ1(z)=Φ1(z)Ψ′1(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=∑n≥0φnzn, ψ=∑n≥0ψnzn, we have the inner product property [7]
(α+1)(∫Dφ(w)¯ψ(w)(1−|w|2)αdA(w))=∑n≥0n!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 n∈N, where Ψ(w)=α+1wα+1∫w0uαψ(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⟩,n≥0. |
Using the identity wP(ψ¯w)=ψ−Ψ, 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 f∈L2h. Then, Tf is normal if and only if f(D) is contained in a line in the complex plane. If f∈L2a,ω, then Tf is normal if and only if f is constant.
Let m be a nonnegative integer, β be in (m−1,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=∑n≥−manzn and g=∑n≥−mbnzn, then the inner product is given by
⟨f,g⟩=∑n≥−mB(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, n≥−m}. We set ρn=B(α+1,β+n+1)B(α+1,β+1), n≥−m, 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)=(I−P)(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 S∗S−SS∗≥0. 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¯ψTψ−TψT¯ψ≤T¯φTφ−TφT¯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 q≥m+1. Denote by N the smallest integer such that N≥max(6q+1,8,12(q2−1)). Our main result is the following theorem.
Theorem 3.3. Let (λl)N≤l≤N1 be any finite sequence of complex numbers such that ∑N≤l≤N1|λl|≤1, and set g=∑N≤l≤N1λlqlzl. Then, the operator Tzq+¯g is hyponormal on A2α,β i.e T¯gTg−TgT¯g≤T¯zqTzq−TzqT¯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¯zqTzq−TzqT¯zq in the orthonomal basis {en, n≥−m}.
Lemma 3.4. The matrix of T¯zqTzq−TzqT¯zq is diagonal and is given by
ηi,i={ρq+iρi,ifi<q−m;ρq+iρi−ρiρi−q,ifi≥q−m. |
Proof. We clearly have
⟨T¯zqTzqei,ej⟩=0ifi≠j,and⟨T¯zqTzqei,ei⟩=ρq+iρi. |
Since
⟨P(¯zqzi),zk⟩=0ifk≠i−q, |
necessarily,
P(¯zqzi)=czi−q, |
and
⟨P(¯zqzi),zi−q⟩=⟨czi−q,zi−q⟩=cρi−q |
⟨P(¯zqzi),zi−q⟩=ρi. |
We deduce c=ρiρi−q, and
⟨T¯zqei,T¯zqei⟩={ρiρi−qifi≥q−m;0ifi<q−m. |
This leads to the following proposition on which our main result is based.
Proposition 3.5. Let l and q be two integers with q≥m+1, l≥max(6q+1,12(q2−1),8). Then, Tzq+δ¯zl is hyponormal if and only if |δ|≤ql.
Proof. Hyponormality is equivalent to the following three inequalities:
|δ|2ρl+iρi≤ρq+iρi,i<q−m. | (3.1) |
|δ|2ρl+iρi≤ρq+iρi−ρiρi−q,q−m≤i<l−m. | (3.2) |
|δ|2(ρl+iρi−ρiρi−l)≤ ρq+iρi−ρiρi−q,i≥l−m. | (3.3) |
The first inequality (3.1) takes the form
|δ|2≤min{(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<q−m}. |
Since the right hand side of this inequality decreases with i, inequality (3.1) is equivalent to
|δ|≤√(β+l+q−m+3)(β+l+q−m+2)(β+l+q−m+1)(β+l+q−m)(β+2q−m+3)(β+2q−m+2)(β+2q−m+1)(β+2q−m). |
Inequality (3.2) can be written as
|δ|2≤ρi+qρi−q−ρi2ρl+iρi−q,q−m≤i<l−m. |
A computation reduces the above inequality to
|δ|2≤min{Ql,q,iRq,i,q−m≤i<l−m}, |
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−(1−q2(β+i+4)2)(1−q2(β+i+3)2)(1−q2(β+i+2)2)(1−q2(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,l−m−1Rq,l−m−1, |
or, equivalently,
|δ|≤√(β+2l−m+3)(β+2l−m+2)(β+2l−m+1)(β+2l−m)Rq,l−m−1(β+q+l−m+3)(β+q+l−m+2)(β+q+l−m+1)(β+q+l−m). |
Since Ql,q,i is decreasing, Ql,q,l−m−1≤Ql,q,q−m−1. Thus, we have
√Ql,q,l−m−1Rq,l−m−1≤√Ql,q,q−m−1. |
That is, if inequality (3.2) is satisfied, so is inequality (3.1). Inequality (3.3) takes the form
|δ|2≤ρi−lρi−qρi+qρi−q−ρ2iρi+lρi−l−ρ2i,i≥l−m. |
To simplify notation, set Ai:=(β+i+1). A computation leads to the following inequality for i≥l−m:
|δ|2≤A2i+3A2i+2A2i+1A2i−(A2i+3−q2)((A2i+2−q2)(A2i+1−q2)(A2i−q2)A2i+3A2i+2A2i+1A2i−(A2i+3−l2)((A2i+2−l2)(A2i+1−l2)(A2i−l2)Ql,q,i. |
After simplification of
F(i)=A2i+3A2i+2A2i+1A2i−(A2i+3−q2)((A2i+2−q2)(A2i+1−q2)(A2i−q2)A2i+3A2i+2A2i+1A2i−(A2i+3−l2)((A2i+2−l2)(A2i+1−l2)(A2i−l2), |
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(q6−l6)F′1(i); |
E2(i)=l4q4(l2+q2)(l2−q2)F′2(i); |
E3(i)=l6q6(q2−l2)F′3(i); |
E4(i)=l2q2(q2−l2)(F′1(i)F2(i)−F1(i)F′2(i)); |
E5(i)=l2q2(l2+q2)(l2−q2)(F′1(i)F3(i)−F1(i)F′3(i)); |
E6(i)=l4q4(q2−l2)(F′2(i)F3(i)−F2(i)F′3(i)). |
It suffices to verify the inequalities
E1(i)+E2(i)≤0,i≥l−m; | (3.5) |
E4(i)+E5(i)≤0,i≥l−m. | (3.6) |
Note l−m≥q−m+1≥2. We show inequality (3.5) first. It takes the form
l2q2(l2+q2)F′2(i)≤(l4+l2q2+q4)F′1(i). |
From the expressions of F1(i) and F2(i), we see that, since β+i+1≥q for i≥l−m,
q2F′2(i)≤F′1(i), |
and thus inequality (3.5) holds. Inequality (3.6) can be reduced to
(l2+q2)(F′1(i)F3(i)−F1(i)F′3(i))≤F′1(i)F2(i)−F1(i)F′2(i). | (3.7) |
An elementary computation gives
F′1(i)F3(i)−F1(i)F′3(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
F′1(i)F2(i)−F1(i)F′2(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 l≥12(q2−1),
(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)P2,31≤2(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 i≥l−m. Next, we show (l2+q2)Ψ1(i)≤Ω1(i) for i≥l−m. We clearly have
(l2+q2)Ψ1(i)≤6(β+i+1)((β+i+2)4(β+i+3)2(β+i+4)2≤Ω1(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|≤√limi→∞F(i)=ql. We finally show that
q2l2≤(β+2l−m+3)(β+2l−m+2)(β+2l−m+1)(β+2l−m)Rq,l−m−1(β+l+q−m+3)(β+l+q−m+2)(β+l+q−m+1)(β+l+q−m). | (3.8) |
From inequality (3.7), we have
Rq,l−m−1=1−(1−q2(β+l−m+3)2)(1−q2(β+l−m+2)2)(1−q2(β+l−m+1)2)(1−q2(β+l−m)2). |
To simplify notation, we set Bi:=(β+i+3)(β+i+2)(β+i+1)(β+i). Inequality (3.8) takes the form
1≤l2B2l−m(B2l−m−Bl−m+qBl−m−q)q2Bl−m+qB2l−m, |
and we have
B2l−m−Bl−m+qBl−m−q=q2F1(l−m−1)−q4F2(l−m−1)+q6F3(l−m−1)−q8, |
where
F1(l−m−1)=A2l−m+2A2l−m+1A2l−m+A2l−m+2A2l−m+1A2l−m−1+A2l−m+1A2l−mA2l−m−1+A2l−m+2A2l−mA2l−m−1 |
and, with Ai:=β+i+1 as before,
F2(l−m−1)=A2l−mA2l−m−1+A2l−m+1A2l−m−1+A2l−m+1A2l−m+A2l−m+2A2l−m+1+A2l−m+2A2l−m+A2l−m+2A2l−m−1, |
and
F3(l−m−1)=A2l−m−1+A2l−m+A2l−m+1+A2l−m+2. |
Since
B2l−m=A2l−m+2A2l−m+1A2l−mA2l−m−1, |
we get
F1(l−m−1)B2l−m=1A2l−m−1+1A2l−m+1A2l−m+1+1A2l−m+2 |
F2(l−m−1)B2l−m=1A2l−m−1A2l−m+1A2l−m−1A2l−m+1+1A2l−m−1A2l−m+2+1A2l−mA2l−m+1+1A2l−mA2l−m+2+1A2l−m+1A2l−m+2 |
F3(l−m−1)B2l−m=1A2l−m+1A2l−mA2l−m−1+1A2l−m+2A2l−m+1A2l−m+1A2l−m+2A2l−m+1A2l−m−1+1A2l−m+2A2l−mA2l−m−1. |
We see that inequality (3.8) is equivalent to
l2(B2l−m−Bl−m+qBl−m−q)q2B2l−m≥4l2A2l−m+2−6q2l2A2l−m−1A2l−m+4q4l2A2l−m+2A2l−m+1A2l−m−q6l2A2l−m+2A2l−m+1A2l−mA2l−m−1. | (3.9) |
We show that if l≥N, then
4l2A2l−m+2−6q2l2A2l−m−1A2l−m+4q4l2A2l−m+2A2l−m+1A2l−m−q6l2A2l−m+2A2l−m+1A2l−mA2l−m−1≥1. |
We then rewrite this last inequality as
4l2A2l−m+1A2l−mA2l−m−1−6q2l2A2l−m+2A2l−m+1+4q4l2A2l−m−1−q6l2≥A2l−m+2A2l−m+1A2l−mA2l−m−1. |
Since β∈(m−1,m], by assumptions on l we have
2l2A2l−m+1A2l−mA2l−m−1≥A2l−m+2A2l−m+1A2l−mA2l−m−1. |
Thus, inequality (3.9) is satisfied if
2A2l−m+1A2l−mA2l−m−1−6q2A2l−m+2A2l−m+1+4q4A2l−m−1−q6≥0. |
Since by assumptions on l and q,
4q4A2l−m−1≥q6, |
it is enough to have
A2l−mA2l−m−1−3q2A2l−m+2≥0, |
and this is satisfied by assumptions on l. Hence, inequality (3.9) holds. We conclude that if q≥m+1 and l≥max(6q+1,12(q2−1),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)N≤l≤N1 be any finite sequence of complex numbers such that ∑N≤l≤N1|λl|≤1, and set g=∑N≤l≤N1λlqlzl. Then, we have T¯gTg−TgT¯g≤T¯zqTzq−TzqT¯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.
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