# Quasinormality and Numerical Ranges of Certain Classes of Dual Toeplitz Operators

Authors: Hocine Guediri
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

The study of dual Toeplitz operators was elaborated by Stroethoff and Zheng (2002), where various corresponding algebraic and spectral properties were established. In this paper, we characterize numerical ranges of certain classes of dual Toeplitz operators. Moreover, we
introduce the analog of Halmos' fifth classification problem for quasinormal dual Toeplitz operators. In particular, we show that there are no quasinormal dual Toeplitz operators with bounded analytic or coanalytic symbols which are not normal.


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## Body

## 1. Introduction

Let𝔻 be the unit disk of the complex plane ℂ, and let dA be the Lebesgue measure on 𝔻. The Lebesgue space of (classes of) square summable complex-valued functions is denoted by ℒ2(𝔻,dA). The Bergman space ℒa2 is the Hilbert subspace of ℒ2(𝔻,dA) consisting of analytic functions. The orthogonal complement of ℒa2 in ℒ2(𝔻,dA) is denoted by (ℒa2)⊥. The Hilbert space (ℒa2)⊥ is readily seen to be not a reproducing kernel Hilbert space. This is one of the major difficulties that occurs when dealing with this space. A second one is the fact that its elements have no standard common qualities such as analyticity harmonicity, while a lesser difficulty is the complicated form of the corresponding basis.Despite the difficulties just listed, Stroethoff and Zheng in [1, 2] have adopted new techniques to investigate various properties of a class of operators acting on (ℒa2)⊥, namely, dual Toeplitz operators. A dual Toeplitz operator is defined on (ℒa2)⊥ to be a multiplication (by the symbol) followed by a projection onto (ℒa2)⊥. Although dual Toeplitz operators are different from Toeplitz operators in many respects, they do share some properties with them. But surprisingly, dual Toeplitz operators on (ℒa2)⊥ resemble much more Hardy space Toeplitz operators than Bergman space Toeplitz operators. Lu in [3] and Cheng and Yu in [4] considered dual Toeplitz operators in higher dimensions; while Yu and Wu in [5] considered dual Toeplitz operators in the framework of Dirichlet spaces.The study of the numerical ranges of Hardy space Toeplitz operators goes back to Brown and Halmos [6]. Subsequent treatment was reconsidered in Halmos' book [7]. Later on, Klein [8] showed that the numerical range depends only on the spectrum of the given Hardy space Toeplitz operator. The Bergman space case was successfully considered only twenty years later by Thukral [9] in case of bounded harmonic symbols. More recently Choe and Lee [10],as well as Gu[11], treat higher-dimensional Bergman space analogs. The case of Bergman space Toeplitz operators with bounded radial symbols has been considered very recently by Wang and Wu [12]. The connection between spectral sets and numerical ranges was considered first by Schreiber [13]. Further investigations had been pursued by Hildebrandt [14] and Clark [15]. The subnormality, and particularly the quasinormality, of Hardy space Toeplitz operators has been discussed chronologically by Itô and Wong [16], Amemiya et al. [17], Abrahamse [18], Cowen and Long [19], Cowen [20, 21], Lee [22], and Yoshino [23]. The case of Bergman space Toeplitz operators has been discussed by Faour in [24] as well as by Jim Gleason in a recent preprint.Accordingly, in this paper, we mainly investigate qualitative properties of the numerical range of a dual Toeplitz operator. We consider various classes of such operators, such as normal and quasinormal ones. We completely describe the numerical ranges of some of them and establish the main qualitative properties of the numerical ranges of others. We also shed some light on the analog of Halmos' fifth problem on the classification of subnormal Toeplitz operators.Our paper is organized as follows: in Section2, we exhibit some preliminary concepts needed in the sequel. Section 3 mainly concerns the description of the numerical ranges of normal dual Toeplitz operators. Section 4 contains the characterization of the numerical ranges of the more general case of nonnormal dual Toeplitz operators with harmonic symbols. In Section 5, we give heuristic proofs of some results based on the concept of lines of support of the numerical range. In Section 6, we briefly discuss the connection between spectral sets and spectra of dual Toeplitz operators. Section 7 is devoted to the quasinormality of dual Toeplitz operators. For the case of dual Toeplitz operators, we introduce and adumbrate the analog of Halmos' fifth problem on the classification of subnormal Toeplitz operators.


## 2. Preliminaries

LetT be a bounded operator on a Hilbert space ℍ. Denote its spectrum by σ(T). The numerical range is a prototype of the spectrum, and it proves useful whenever information about T is needed. It is defined by 𝒲(T):={〈Tφ,φ〉,φ∈ℍ,∥φ∥=1}. The main involved features of the numerical range are as follows. 𝒲(T) is always bounded and convex, (Toeplitz-Hausdorff theorem). Its closure 𝒲(T)¯ contains the spectrum σ(T). If 𝒲(T) reduces to the singleton {λ}, then T=λI. 𝒲(T) is a linear function of T, that is, 𝒲(αT+β)=α𝒲(T)+β,∀α,β∈ℂ; hence we see that ω𝒲(𝒮f)=𝒲(𝒮ωf),forω∈ℂ, and that λ∈𝒲(𝒮f) implies that 0∈𝒲(𝒮f-λ). If 𝒲(T) is a subset of the real axis, then T must be self-adjoint. If T is normal and 𝒲(T) is closed, then the extreme points of 𝒲(T) are eigenvalues. If T is normal, the closure of 𝒲(T) is the smallest closed convex set containing the spectrum of T. For further details on the numerical ranges as well as various applications of this pioneering tool in operator theory, see [7, 25–27].Forf∈ℒ∞(𝔻), define the dual Toeplitz operator 𝒮f to be the operator on  (ℒa2)⊥ given by (2.1)𝒮f:u∈(ℒa2)⊥↦𝒮f(u)=𝒬(fu)∈(ℒa2)⊥.
Here 𝒬=I-𝒫 is the familiar orthogonal projection from ℒ2(𝔻,dA) onto (ℒa2)⊥ and 𝒫 is the orthogonal projection from ℒ2(𝔻,dA) onto the Bergman space ℒa2. Moreover, for f∈ℒ∞(𝔻) and g∈ℌ∞, (the algebra of bounded analytic functions) we have
(2.2)𝒮fg=𝒮g𝒮f,𝒮fg¯=𝒮f𝒮g¯,(2.3)𝒮f𝒮g=𝒮fg-HfHg¯*,
where Hf:u∈ℒa2↦Hfu=𝒬(fu)∈(ℒa2)⊥ is the Hankel operator. If E is a subset of ℂ, then the convex hull of E, denoted by ℋ(E), is the smallest convex set containing E. Useful properties of the convex hull are well known [26, 28]. For instance, we know that in general the convex hull of an open set is open and the convex hull of a compact set is compact. The following easy property will be also usedRemark 2.1.
IfA is a bounded subset of a finite dimensional normed space, then we have ℋ(A)¯=ℋ(A¯). Indeed, observe that A⊂A¯ implies that ℋ(A)⊂ℋ(A¯), whence ℋ(A)¯⊂ℋ(A¯)¯=ℋ(A¯). Conversely, since A¯⊂ℋ(A)¯, we infer that ℋ(A¯)⊂ℋ(ℋ(A)¯)=ℋ(A)¯, as ℋ(A)¯ is convex.The following main spectral properties of𝒮f are due to Stroethoff and Zheng [1].Theorem 2.2.
(1)  If f is in ℒ∞(𝔻), then ℛ(f)⊂σe(𝒮f)⊂σ(𝒮f)⊂ℋ(ℛ(f)).
(2) Let f be a continuous real-valued function on 𝔻, then  σ(𝒮f)=σe(𝒮f)=f(𝔻)¯.
(3)  If f is a bounded analytic or coanalytic function on 𝔻, then σ(𝒮f)=σe(𝒮f)=f(𝔻)¯.For our purpose, we now prove the following useful facts about dual Toeplitz operators.Lemma 2.3.
Letf be in ℒ∞(𝔻). Then, 𝒮f is self-adjoint if and only if f is real.Proof.
𝒮f is self-adjoint means that 𝒮f=𝒮f*. This is equivalent to the fact that f=f¯, since ∥𝒮g∥=∥g∥∞, for g∈ℒ∞(𝔻), which is equivalent to the fact that f is real-valued.Lemma 2.4.
Letf be in ℒ∞(𝔻). Then, 𝒮f≥0 if and only if f≥0.Proof.
Iff≥0, then 〈𝒮fg,g〉=〈Q(fg),g〉=〈fg,g〉=∫𝔻f(z)|g(z)|2dA(z)≥0,∀g∈(ℒa2)⊥. Conversely, suppose that 𝒮f≥0, then in particular its spectrum lies in [0,∞). By part (1) of Theorem 2.2, we obtain ℛ(f)⊂σ(𝒮f)⊂[0,∞), whence f≥0.Corollary 2.5.
(i) A bounded dual Toeplitz operator with a real spectrum must be self-adjoint.
(ii) A bounded dual Toeplitz operator with spectrum lying in the positive real half-axis must be positive.Proof.
This result follows from part (1) of Theorem2.2 and Lemmas 2.3 and 2.4.Theorem 2.6.
Letf be a nonconstant bounded harmonic real-valued function on 𝔻. Then, the operator 𝒮f has no eigenvalues.Proof.
Since in general for a constantλ we have 𝒮f-λ=𝒮f-λ and f-λ is harmonic if f is, it suffices to show that 𝒮fg=0 implies that g=0,∀g∈(ℒa2)⊥. If 𝒮fg=𝒬(fg)=0, then fg∈ℒa2. Let h∈ℌ∞(𝔻), then h¯g∈(ℒa2)⊥. Indeed, ∥h¯g∥2≤∥h∥∞∥g∥2, and 〈h¯g,u〉=〈g,hu〉=0,∀u∈ℒa2, (as hu∈ℒa2 and g∈(ℒa2)⊥). It follows that 〈fg,h¯g〉=0. Taking real parts and noticing that f is real, we see that
(2.4)ℛe(〈fg,h¯g〉)=ℛe∫𝔻fg(hg¯)dA=∫𝔻f|g|2ℛe(h)dA=0.
Since ℛe(ℌ∞) is weak *-dense in the set of bounded real harmonic functions [29], we can replace ℛe(h) with f in the latter to obtain f2|g|2=0. This implies that f2 and |g|2 have disjoint supports. However, the harmonic function f cannot vanish on a set of positive measure without being zero, whence g=0.


## 3. Characterization of the Numerical Range

An operatorT:M→M is said to be subnormal if it admits an extension S:E→E, such that M⊂E,  S is normal, and M is invariant under S. It is well known that if an operator T is subnormal, then it is convexoid, that is, 𝒲(T)¯=ℋ(σ(T)). First, the following observation is worth stressing.Proposition 3.1.
Suppose thatf∈ℌ∞(𝔻). Then 𝒮f and 𝒮f¯ are convexoid, that is, 𝒲(𝒮f)¯=ℋ(σ(𝒮f)), and 𝒲(𝒮f¯)¯=ℋ(σ(𝒮f¯)).Proof.
Letf∈ℌ∞(𝔻), then the multiplication operator ℳf¯:ℒ2(𝔻)→ℒ2(𝔻) is a normal extension of 𝒮f¯. Indeed, we have ℳf¯ℳf=ℳfℳf¯ and ℳf¯((ℒa2)⊥)={f¯g,g∈(ℒa2)⊥}⊂(ℒa2)⊥. Moreover, it is clear that 𝒮f¯ is the restriction of ℳf¯ to (ℒa2)⊥. Thus 𝒮f¯ is subnormal, whence 𝒲(𝒮f¯)¯=ℋ(σ(𝒮f¯)).
Concerning the other part of the Proposition, of course𝒮f is not necessarily subnormal, nevertheless we obtain a similar result by exploring the fact that 𝒲(𝒮f)={λ¯,λ∈𝒲(𝒮f¯)} and σ(𝒮f)={λ¯,λ∈σ(𝒮f¯)}. Hence, we obtain 𝒲(𝒮f)¯=ℋ(σ(𝒮f)).For bounded analytic or coanalytic symbols, the fact that𝒮f is convexoid comes from the subnormality of this operator as Proposition 3.1 asserts. However, the spectral inclusion theorem, (namely, part (1) of Theorem 2.2), refines this result. Indeed, making use of the spectral inclusion property, it turns out that all bounded dual Toeplitz operators are convexoid; this represents the aim of the following assertion.Proposition 3.2.
The closure of the numerical range of a bounded dual Toeplitz operator is the convex hull of its spectrum, that is,𝒲(𝒮f)¯=ℋ(σ(𝒮f)), for f∈ℒ∞(𝔻).Proof.
Consider the multiplication operatorℳf on ℒ2(𝔻,dA). It is known to be normal, whence convexoid. Thus 𝒲(ℳf)¯=ℋ(σ(ℳf)). By Problem 67 of [7], we see that σ(ℳf)=ℛ(f). By the spectral inclusion Theorem, we see that σ(ℳf)=ℛ(f)⊂σ(𝒮f). This yields the following inclusions 𝒲(ℳf)¯=ℋ(σ(ℳf))=ℋ(ℛ(f))⊂ℋ(σ(𝒮f)). Now, since ℳf is the minimal normal dilation of 𝒮f, we see that 𝒲(𝒮f)¯⊂𝒲(ℳf)¯, which is clear from the definition of the numerical range and the fact that 𝒮f is the compression of ℳf. Therefore, we obtain the first inclusion 𝒲(𝒮f)¯⊂ℋ(σ(𝒮f)).
The reverse inclusion is easier:  we haveσ(𝒮f)⊂𝒲(𝒮f)¯, which implies that ℋ(σ(𝒮f))⊂ℋ(𝒲(𝒮f)¯)=𝒲(𝒮f)¯, since 𝒲(𝒮f)¯ is convex.Remark 3.3.
In connection with Proposition3.2, we ask whether all elements of the dual Toeplitz algebra 𝒟𝒯, (the C*-algebra generated by {𝒮f,f∈ℌ∞}), are convexoid, according to the fact that it is generated by subnormal operators.Now, we are going to characterize the numerical ranges of dual Toeplitz operators with bounded harmonic symbols. First, we make the following observation.Remark 3.4.
According to part (2) of Theorem2.2, for a nonconstant bounded continuous real-valued function f on 𝔻, we infer that σ(𝒮f)=ℛ(f)=f(𝔻)¯ is an interval. As f is bounded, we deduce that σ(𝒮f)=[inff,supf]. Obviously, we have ℋ(σ(𝒮f))=[inff,supf].Lemma 3.5.
Suppose thatf is a nonconstant bounded harmonic real-valued function on 𝔻. Then one has  𝒲(𝒮f)=(inff,supf).Proof.
By Proposition3.2 and Remark 3.4, we see that 𝒲(𝒮f)¯=[inff,supf], (the continuity is redundant in the harmonicity). Since 𝒲(𝒮f) is a convex set whose closure is [inff,supf], 𝒲(𝒮f) contains all elements of [inff,supf] except possibly the extreme points inff and supf. Thus (inff,supf)⊂𝒲(𝒮f).
Now, suppose that eitherinff or supf belongs to 𝒲(𝒮f). Then it is an extreme point of 𝒲(𝒮f), which in fact must be an eigenvalue of 𝒮f. However, Theorem 2.6 tells us that such 𝒮f has no eigenvalues. Thus, we should have 𝒲(𝒮f)=(inff,supf).Parallel to Brown and Halmos [6] characterization of normal Hardy space Toeplitz operators as well as Axler and Čučković [30] one of normal Bergman space Toeplitz operators with bounded harmonic symbols, normal dual Toeplitz operators were characterized in [1] as follows: for a bounded measurable function f on 𝔻, the dual Toeplitz operator 𝒮f is normal if and only if the range of f lies on a line. Accordingly, we are able to characterize the numerical range of normal dual Toeplitz operators with bounded harmonic symbols.Theorem 3.6.
Letf be a nonconstant bounded harmonic function on 𝔻, and suppose that 𝒮f is normal. Then, there are (complex) numbers a,b such that σ(𝒮f) is the closed line segment [a,b] and 𝒲(𝒮f) is the corresponding open line segment (a,b).Proof.
Taking into account the assumption that𝒮f is normal, we are certain from the existence of (complex) constants α,β and a real-valued function g such that f=αg+β whence 𝒮f=α𝒮g+β. From the harmonicity of f and the linearity of the Laplacian, we see that g must be bounded harmonic and real-valued. Now, Lemma 3.5 asserts, therefore, that σ(𝒮g)=[m,M] and 𝒲(𝒮g)=(m,M), where m=infg and M=supg. Thus σ(𝒮f)=[a,b] and 𝒲(𝒮f)=(a,b), (line segments in ℂ), with a=αm+β, and b=αM+β.


## 4. The Numerical Range of a Nonnormal Dual Toeplitz Operator

Lemma2.4 has a nice consequence on the self-adjointness of certain dual Toeplitz operators.Theorem 4.1.
Letf∈ℒ∞(𝔻) be harmonic. Suppose that 𝒲(𝒮f) lies in the complex upper half-plane and contains some real number. Then 𝒮f must be self-adjoint.Proof.
The numerical range lies in the upper half-plane means thatℑm(〈𝒮fg,g〉)≥0, for ∥g∥=1. Taking g/∥g∥2 if necessary, we may conclude that ℑm(〈𝒮fg,g〉)≥0,∀g∈(ℒa2)⊥. Since 〈𝒮fg,g〉=∫𝔻f|g|2dA,∀g∈(ℒa2)⊥, taking imaginary parts, we obtain
(4.1)ℑm(〈𝒮fg,g〉)=∫𝔻ℑm(f)|g|2dA=〈𝒮ℑm(f)g,g〉≥0,∀g∈(ℒa2)⊥.
This happens only if ℑm(f)≥0, by Lemma 2.4. Now, suppose that a real number c∈𝒲(𝒮f), then there exists some h∈(ℒa2)⊥, with ∥h∥=1, such that 〈𝒮fh,h〉=c. Writing c in the form c=c〈h,h〉, we obtain  ∫𝔻(f-c)hh¯dA=0. Again, taking imaginary parts we obtain
(4.2)∫𝔻ℑm(f)|h|2dA=0.
As ℑm(f)≥0, we deduce that ℑm(f)|h|2=0 on 𝔻. This implies that ℑm(f) and |h|2 have disjoint supports. Since ∥h∥=1, we deduce that h≠0 on 𝔻. Thus supp(ℑm(f)) has a positive measure, that is, the harmonic function ℑm(f) must be zero on a set of nonzero measure. It follows that ℑm(f)=0 on 𝔻, whence 𝒮f is self-adjoint by Lemma 2.3.Regarding the numerical ranges of a certain class of dual Toeplitz operators, we have the following qualitative characterization.Theorem 4.2.
Letf∈ℒ∞(𝔻) be harmonic such that 𝒮f is nonnormal. Then 𝒲(𝒮f) is an open convex set.Proof.
We need only to verify that the convex set𝒲(𝒮f) is open. To see this, we proceed by contradiction and suppose that it is not open. Hence it intersects its boundary ∂𝒲(𝒮f). Let λ be one of such points, that is, λ∈𝒲(𝒮f)∩∂𝒲(𝒮f), which can be rewritten as 0∈𝒲(𝒮f-λ)∩∂𝒲(𝒮f-λ). Now, the convexity of 𝒲(𝒮f-λ) and the fact that 0∈∂𝒲(𝒮f-λ) enable us to rotate it so that it lies in the upper half-plane. This means that there exists a unimodular complex number ω=eiθ, for some θ∈[0,2π], such that ω𝒲(𝒮f-λ)=𝒲(𝒮ωf-ωλ) lies in the upper half-plane. By Theorem 4.1, 𝒮ωf-ωλ must be self-adjoint. In other words there exists a real-valued function g on 𝔻 such that ωf-ωλ=g. This implies that f=αg+β, for some constants α and β. So, we infer that 𝒮f=α𝒮g+β is normal, which contradicts the original hypothesis and completes the proof.Remark 4.3.
Note that Theorems4.1 and 4.2, (as well as their subsequent corollaries), remain still valid if one assumes merely that the symbols have harmonic imaginary parts.Corollary 4.4.
Iff is a bounded analytic or coanalytic function, then   𝒲(𝒮f)=ℋ(f(𝔻)).Proof.
Iff is constant, the fact is trivially satisfied. If f is not constant, then 𝒮f is not normal (because the range of an analytic or coanalytic function cannot lie on a line). Hence, by Theorem 4.2, 𝒲(𝒮f) is an open convex set. On the other hand, by the open mapping theorem f(𝔻) is open  whence ℋ(f(𝔻)) is an open convex set too. Now, by Proposition 3.1 and part (3) of Theorem 2.2 as well as Remark 2.1, we see that  𝒲(𝒮f)¯=ℋ(f(𝔻)¯)=ℋ(f(𝔻))¯. Since an open convex planar set is the interior of its closure, we infer that 𝒲(𝒮f) coincides with the convex hull of f(𝔻).


## 5. Aesthetic Consequences Using Lines of Support of𝒲(𝒮f)

A lineL is said to be a line of support of a planar convex set K at a boundary point P∈∂K, if P∈L and the set K is contained in the closure of one of the two half-planes into which L cuts the plane. Clearly, every point on the boundary of a planar convex set lies on a line of support. Based on the concept of lines of support of the numerical range 𝒲(𝒮f), several interesting consequences of Theorem 4.1 can be observed. Note that the underlying idea goes back to Brown and Halmos [6].Corollary 5.1.
Letf∈ℒ∞(𝔻) be harmonic. If a line of support of the numerical range 𝒲(𝒮f) of the dual Toeplitz operator 𝒮f contains a point of 𝒲(𝒮f), then it contains its whole spectrum σ(𝒮f) (and hence its entire numerical range 𝒲(𝒮f)).Corollary 5.2.
Letf∈ℒ∞(𝔻) be harmonic. If the spectrum σ(𝒮f) consists of merely a finite number of eigenvalues, then the operator 𝒮f is scalar.Proofs
Proof of Corollary5.1
With regard to Corollary5.1, clearly the line of support can be rotated along with the numerical range till the line will be horizontal and the numerical range above it. Then we translate both in such a way that the line of support will be the real axis; (these two operations correspond to a linear function of the form z→eiθz+ω, for a fixed complex number ω and a fixed θ∈[0,2π]). Then one can apply Theorem 4.1 to conclude that eiθ𝒮f+ω is self-adjoint, whence eiθσ(𝒮f)+ω⊂ℝ, and therefore σ(𝒮f) lies on the original line of support. Since 𝒲(𝒮f)⊂𝒲(𝒮f)¯=ℋ(σ(𝒮f)), by Proposition 3.2, we infer that 𝒲(𝒮f) lies on the relevant line of support.Proof of Corollary5.2
Concerning Corollary5.2, if σ(𝒮f) consists of a finite number of eigenvalues, then ℋ(σ(𝒮f)) is a polygon with vertices of some of such values. Thus, one can find at least a line of support passing through one of such vertices, which must be in fact an extreme point of 𝒲(𝒮f). Hence, making use of Corollary 5.1, we infer that all those eigenvalues of 𝒮f lie on such line. Consequently, the numerical range is a segment on this line with two extreme points. Taking the line of support perpendicular to the above line, and passing by an endpoint of the segment, and repeating the same procedure, we infer that the numerical range lies on the new line of support as well. A segment lying in two perpendicular lines must be a single point. So, σ(𝒮f) must be a singleton. By Proposition 3.2 and the properties of the numerical range, we infer that 𝒮f must be scalar.Remark 5.3.
Suppose that𝒮f is compact. We know that its spectrum consists of at most a countable number of eigenvalues, including 0. Arguing in the same manner as in the proof of Corollary 5.2, we can show that there are no nonzero compact dual Toeplitz operators with bounded harmonic symbols, which is a particular case of [1, Theorem 7.5].


## 6. Connection with Spectral Sets

A compact subset𝔼⊂ℂ containing the spectrum σ(T) of a bounded linear operator T acting on a given Hilbert space is called a k-spectral set for T if(6.1)∥ℱ(T)∥≤k∥ℱ∥𝔼
holds for every function ℱ(z) analytic in a neighborhood of 𝔼 (in [13], rational functions are used instead), where the first norm in the above inequality is the operator norm and the second one is the sup norm over 𝔼. In particular, if k=1, 1-spectral sets are simply called spectral sets. For more details on spectral sets, we refer to [13–15, 31] and the references therein. For instance, the spectrum of any subnormal operator is a spectral set. The closed unit disk is a spectral set for any contraction.Our main concern in this section is to describe the dual Toeplitz operators analog of an interesting connection, pointed out by Schreiber [13], between spectral sets and numerical ranges of Toeplitz operators.  Let us start with some trivial situations, where σ(𝒮φ) is spectral(1)
If𝒮φ is a bounded normal dual Toeplitz operator, (i.e., φ(𝔻) lies on a line in ℂ [1]), then σ(𝒮φ) is a spectral set.(2)
Ifφ is a bounded coanalytic function on the unit disk 𝔻, then σ(𝒮φ) is a spectral set, (since it is subnormal by the proof of Proposition 3.1).(3)
If the spectrum of a bounded dual Toeplitz operator𝒮φ is a disk, then σ(𝒮φ) is a spectral set too.However, we can observe that the spectra of dual Toeplitz operators with analytic symbols are also spectral sets for their corresponding operators. This follows from the following observation. If𝔼⊂ℂ is a planar subset, set 𝔼*={ω¯,ω∈𝔼}, in particular it can be easily verified that σ(T*)=σ(T)*. Also we adopt the notation f*(z)=f(z¯)¯ which is an analytic function on 𝔼* whenever f is analytic on 𝔼. Then we have the following.Lemma 6.1.
LetT be a bounded linear operator on a given Hilbert space. Then, σ(T) is a k-spectral set for T if and only if σ(T)* is a k-spectral set for its adjoint T*.Proof.
The fact thatσ(T) is a k-spectral set of T means that Inequality (6.1) holds, for any ℱ holomorphic in a neighborhood of σ(T). Notice that from the definition of the holomorphic functional calculus we have
(6.2)∥ℱ*(T*)∥=∥(ℱ(T))*∥=∥ℱ(T)∥.
On the other hand, we have
(6.3)∥ℱ*∥σ(T)*=sup{|ℱ*(w)|,w∈σ(T)*}=sup{|ℱ¯(z)|,z∈σ(T)}=∥ℱ∥σ(T).
Combining Inequality (6.1) and the last two identities, we infer that
(6.4)∥ℱ*(T*)∥≤k∥ℱ*∥σ(T*).
In order to establish the equivalence, it suffices to observe that if ℱ is holomorphic in a neighborhood of σ(T), then ℱ* is holomorphic in the conjugate of the same neighborhood which contains σ(T)* and conversely.Therefore, from the above discussion, we deduce the following.Corollary 6.2.
Ifφ is a bounded analytic function on the unit disk 𝔻, then σ(𝒮φ) is a spectral set for 𝒮φ.Similar results for coanalytic Toeplitz operators on both Hardy and Bergman spaces can also be inferred.Corollary 6.3.
(i) If φ¯∈ℌ∞(∂𝔻), then σ(Tφ) is a spectral set for Tφ defined on ℌ2(∂𝔻).
(ii) If φ is a bounded coanalytic function on the unit disk 𝔻, then σ(Tφ) is a spectral set for Tφ defined on ℒa2(𝔻).


## 7. Some Thoughts on Quasinormal Dual Toeplitz Operators

The Bergman spaceℒa2 has normalized reproducing kernel kw given by (7.1)kw(z)=1-|w|2(1-w¯z)2.
Recall that for w∈𝔻, the involutive disk automorphism φw is defined by (7.2)φw(z)=w-z1-w¯z,forz∈𝔻.
For a linear operator T on (ℒa2)⊥ and w∈𝔻, define the operator 𝕊w(T)∶=T-𝒮φwTSφ¯w.  A second application of it gives(7.3)𝕊w2(T)=T-2SφwTSφ¯w+Sφw2TSφ¯w2.
For f,g∈ℒ2(𝔻,dA), define the rank one operator (f⊗g) by (7.4)(f⊗g):h∈ℒ2(D,dA)→(f⊗g)(h)=〈h,g〉f∈ℒ2(D,dA).IfT1 and T2 are bounded linear operators on ℒ2(𝔻,dA), then for f, g∈ℒ2(𝔻,dA), we have(7.5)T1(f⊗g)T2*=(T1f)⊗(T2g).
In the sequel, we will need a formula relating the image of the product HfHg¯* under the action of the operator 𝕊w2 and the functions Hf(kw) and Hg¯(kw). For f,g∈ℒ∞(𝔻), combining (2.3), (7.3), and (7.5), (for a detailed proof see [1, 2]), we obtain(7.6)𝕊w2(HfHg¯*)=Hf(kw⊗kw)Hg¯*=(Hf(kw))⊗(Hg¯(kw)).
An operator T on a Hilbert space is called quasinormal if it commutes with T*T. It is well known that quasinormal operators are subnormal. In what follows we are going to show that there are no quasinormal dual Toeplitz operators with bounded analytic or coanalytic symbols that are not normal.Theorem 7.1.
Letf be in ℌ∞(𝔻), and suppose that 𝒮f is quasinormal. Then, the symbol f must be constant.Proof.
Iff=0, the conclusion is obvious. So, suppose that f≠0 and that 𝒮f is quasinormal, then we have
(7.7)𝒮f𝒮f*𝒮f=𝒮f*𝒮f𝒮f.
Since f is analytic, using Relations (2.2), we obtain
(7.8)𝒮f𝒮f*𝒮f=𝒮|f|2𝒮f=𝒮|f|2f-H|f|2Hf¯*,(7.9)𝒮f*𝒮f𝒮f=𝒮f¯𝒮f2=𝒮f¯f2-Hf¯Hf¯2*.
Now, (7.7)–(7.9) yield
(7.10)H|f|2Hf¯*=Hf¯Hf¯2*.
Introducing the operator 𝕊ω, by (7.6), we see that
(7.11)𝕊ω2(H|f|2Hf¯*)=H|f|2kω⊗Hf¯kω,𝕊ω2(Hf¯Hf¯2*)=Hf¯kω⊗Hf¯2kω.
Combining the latter with (7.10), we obtain   H|f|2kω⊗Hf¯kω=Hf¯kω⊗Hf¯2kω. Since k0=1, taking ω=0, we obtain   H|f|21⊗Hf¯1=Hf¯1⊗Hf¯21.  In other words, by the definition of the rank one operator, we have
(7.12)〈u,Hf¯1〉H|f|21=〈u,Hf¯21〉Hf¯1,∀u∈(ℒa2)⊥.
If Hf¯1=0, then f¯∈ℒa2. Thus f¯ must be constant, whence the result follows immediately. If Hf¯1≠0, we distinguish several cases as follows.  (i)
The caseH|f|21=0 and Hf¯21=0 cannot happen (because Hf¯1=𝒬(f¯)=f¯-f¯(0) and Hf¯21=𝒬(f¯2)=f¯2-f¯2(0) vanish simultaneously).(ii)
The similar caseH|f|21≠0 and Hf¯21=0 is impossible too, for the same reason.(iii)
The caseH|f|21=0 and Hf¯21≠0 is impossible too, otherwise 〈u,Hf¯21〉=0,for all u∈(ℒa2)⊥; thus Hf¯21∈ℒa2 too, whence Hf¯21=0 contradicting the assumption.(iv)
IfH|f|21≠0 and Hf¯21≠0, then clearly from (7.12) there exists some nonzero constant λ∈ℂ, such that(7.13)Hf¯1=λH|f|21,Hf¯1=λ¯Hf¯21.
Rephrasing (7.13), we see that 𝒬(f¯-λ|f|2)=0 and 𝒬(f¯-λ¯f¯2)=0. Thus, f¯-λ|f|2 and f¯-λ¯f¯2 are in ℒa2, (they are analytic in particular). But, since f is analytic, we see that f(1-λf)=f-λf2 is analytic too, whence it is constant. So, set f(1-λf)=μ, for some nonzero complex constant μ. On the other hand, since f¯-λ|f|2=f¯(1-λf) is analytic, multiplying by the analytic function f2, we obtain an analytic function, namely, f¯(1-λf)f2=|f|2(1-λf)f=μ|f|2. Now, the function μ|f|2, (whose range lies on a line, as |f|2 is real-valued and μ is constant), can be analytic only if it is constant; whence we infer that |f|2 is constant. Finally, it is well known that an analytic function with a constant modulus must be constant, whence f must be constant.For bounded conjugate analytic symbols the matter is much more simpler and it uses the Brown-Halmos type Theorem (namely, [1, Theorem 3.1]). Indeed, we have the following.Theorem 7.2.
Letf be in ℌ∞(𝔻)¯, and suppose that 𝒮f is quasinormal. Then, f must be constant.Proof.
Iff=0, the matter is obvious; so suppose that f≠0. Since f is coanalytic, using Relations (2.2) and (2.3), we obtain
(7.14)𝒮f𝒮f*𝒮f=𝒮f𝒮|f|2,(7.15)𝒮f*𝒮f𝒮f=𝒮|f|2𝒮f=𝒮f|f|2.
Suppose that 𝒮f is quasinormal, then from (7.7), (7.14), and (7.15), we see that 𝒮f𝒮|f|2=𝒮f|f|2. Hence 𝒮f𝒮|f|2 is a dual Toeplitz operator. By the Brown-Halmos type Theorem ([1, Theorem 3.1]), we infer that either f is analytic or |f|2 is coanalytic. If f is analytic, then it is constant since it is coanalytic by hypothesis. If the real function |f|2 is coanalytic, then it is constant; whence f is constant as well, (as a coanalytic function with constant modulus). Thus, in all cases, we infer that f is constant.Remark 7.3.
Clearly iff is constant, then 𝒮f is normal and then it is quasinormal. So, Theorems 7.1 and 7.2 can be expressed as follows.  Letfbe a bounded analytic, (or coanalytic), function. Then,𝒮fis quasinormal if and only iffis constant, that is, there are no quasinormal dual Toeplitz operators with bounded analytic or coanalytic symbols that are not normal.Corollary 7.4.
Letf be a bounded analytic or coanalytic function, and suppose that 𝒮f is quasinormal. Then, the numerical range of 𝒮f reduces to a singleton, that is, 𝒲(𝒮f)={λ} for some complex constant λ.Proof.
Just observe that from Proposition3.1 one has 𝒲(𝒮f)¯=ℋ(σ(𝒮f)). If 𝒮f is quasinormal, by Theorems 7.1 and 7.2, we infer that f=λ for some complex constant λ. The convex hull of a singleton is the set itself. Hence, we obtain 𝒲(𝒮f)¯=ℋ({λ})={λ}.It seems to be of interest to consider also the typical case of bounded harmonic symbols. It is well known that such functions can be decomposed asf=f1+f2¯, with Bloch functions f1 and f2. Here, we confine ourselves to the related case, where f=g+λg¯, for g∈ℌ∞(𝔻) and λ a complex constant.Proposition 7.5.
Suppose thatf=g+λg¯,0≠λ∈ℂ, and 0≠g∈ℌ∞(𝔻). If 𝒮f is quasinormal, then 𝒮f is normal and λ must be unimodular.Proof.
Since𝒮f is quasinormal, (7.7) holds. Taking adjoints, we obtain
(7.16)𝒮f¯(𝒮f¯𝒮f-𝒮f𝒮f¯)=0,(𝒮f¯𝒮f-𝒮f𝒮f¯)𝒮f=0.
Therefore, using the Hilbert space orthogonality relations, we see that  ker([𝒮f¯,𝒮f])=ℛan(𝒮f)¯, which must be nontrivial. Next, a couple of manipulations lead to
(7.17)[𝒮f¯,𝒮f]=(1-|λ|2)(𝒮g¯𝒮g-𝒮g𝒮g¯)=(1-|λ|2)[𝒮g¯,𝒮g].
Now, (2.3) as well as the fact that g is analytic yields [𝒮g¯,𝒮g]=-Hg¯Hg¯*, whence
(7.18)[𝒮f¯,𝒮f]=(|λ|2-1)Hg¯Hg¯*.
Since the Hankel operator Hg¯ is one-to-one (as g∈ℌ∞), we infer that Hg¯Hg¯* has a trivial kernel. This contradicts the fact that [𝒮f¯,𝒮f] has a nontrivial kernel, unless [𝒮f¯,𝒮f]=0, which happens only if |λ|2=1. Thus 𝒮f is normal and λ must be unimodular.Corollary 7.6.
Letf be as in Proposition 7.5 and suppose that 𝒮f is quasinormal. Then, 𝒲(𝒮f)=(a,b) for some complex constants a and b.Proof.
Combining Proposition7.5 and Theorem 3.6, the result follows.At this stage, we would like to conclude with a crucial point, which probably sheds some light on the fifth Halmos' problem [32]. This problem asks whether every subnormal Hardy space Toeplitz operator is either normal or analytic. In the Hardy space setting, the original general problem was weakened to whether every quasinormal Toeplitz operator is either normal or analytic, and it was completely solved positively by Amemiya et al. [17], whereas Cowen and Long [19] answered the original problem in the negative. For further results in this direction, see [16, 18, 20–23]. The Bergman space analog seems to be still pending [24]. However, for dual Toeplitz operators, a similar conjecture can be stated with slight modifications, namely: every quasinormal dual Toeplitz operator must be normal. Theorems 7.1 and 7.2, Proposition 7.5 as well as Remark 7.3 support this conjecture.



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