# Toeplitz Operators whose Symbols Are Borel Measures

Authors: Jaehui Park
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we are concerned with Toeplitz operators whose symbols are complex Borel measures. When a complex Borel measureμ on the unit circle is given, we give a formal definition of a Toeplitz operator Tμ with symbol μ, as an unbounded linear operator on the Hardy space. We then study various properties of Tμ. Among them, there is a theorem that the domain of Tμ is represented by a trichotomy. Also, it was shown that if the domain of Tμ contains at least one polynomial, then Tμ is densely defined. In addition, we give evidence for the conjecture that Tμ with a singular measure μ reduces to a trivial linear operator.


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## Body

## 1. Introduction

A classical Toeplitz operator is the compression of a multiplication operator on the Lebesgue spaceL2T of the unit circle T to the Hardy space H2T. The study of Toeplitz operators seems to have originated from the paper of Toeplitz [1]. In the paper [2], he used Toeplitz matrices to characterize nonnegative continuous functions on the unit circle in terms of their Fourier coefficients. The remarkable paper of Brown and Halmos [3] started the systematic study of spectral properties of Toeplitz operators. Since then, the theory of Toeplitz operators has been studied in various ways. Recently, the theory of Toeplitz operators has been studied in a variety of settings and connections with other fields. One direction is to deal with Toeplitz operators on reproducing kernel spaces like Bergman spaces, Dirichlet spaces, or Fock spaces (cf. [4–8]). Another direction is to study Toeplitz operators with operator-valued symbols (cf. [9–11]). Also, truncated Toeplitz operators have attracted attention. A systematic approach on truncated Toeplitz operators can be found in the paper of Sarason in 2007 [12]. In that paper, he has used “compatible” measures to describe bounded truncated Toeplitz operators. The boundedness of infinite Hankel matrices is also related to the compatibility of measures: the infinite Hankel matrix of the moment of a nonnegative Carleson measure is bounded and vice versa [13]. (For related recent studies, see [14].) These works inspired us to consider Toeplitz operators whose symbols are measures. The Toeplitz operators whose symbols are measures have been studied in the setting of Bergman spaces and other spaces (cf. [15], chapter 7).In this paper, we consider Toeplitz operators on the Hardy space, whose symbols are measures. In this study, unbounded Toeplitz operators arise naturally. When studying unbounded Toeplitz operators, it was usually considered that the symbols come fromL2T. In 2008, Sarason [16] treated not only the case of L2T-symbols but the case of analytic functions on the open unit disk D. It is natural to attempt to extend the symbols of Toeplitz operators to measures, because the initial reasearch for them was related to the moment problem. As mentioned before, Toeplitz and Hankel operators associated with measures can be seen in the papers [13] and [12]. In this paper, we provide an explicit definition of Toeplitz operators whose symbols are complex Borel measures and then consider their unbounded operator theory. As the study on Toeplitz operators whose symbols are functions shows the interplay between function theory and operator theory, the study on Toeplitz operators whose symbols are measures is also expected to show the interplay between measure theory and operator theory.Our consideration for the symbol of a Toeplitz operator, denoted byTμ, is a complex Borel measure μ on the unit circle. When we study an unbounded linear operator, we usually assume that its domain is dense, i.e., the operator is densely defined. Hence, one may ask if Tμ is densely defined, i.e., the domain is dense in H2. Toeplitz operators with L2-symbols are always densely defined. Unlike when the symbol is a function, it does not seem easy to answer the question. Nonetheless, we will show that the domain of Tμ is represented by a trichotomy (Theorem 8). In particular, we can show that if the domain of Tμ contains at least one polynomial, then Tμ is densely defined (Proposition 10). We also give evidence for the conjecture that the cases of singular measures induce trivial linear operators (Theorem 15).The organization of this paper is as follows. In Section2, we give notations, definitions, and preliminary facts, which will be used in the sequel. In Section 3, we give a formal definition of Toeplitz operators whose symbols are complex Borel measures on T and then investigate their properties in the viewpoint of unbounded linear operator theory.


## 2. Preliminaries

LetT be the unit circle in the complex plane. Let m be the normalized Lebesgue measure on T, so that mT=1. For 1≤p≤∞, we write LpT=LpT,m for the Lebesgue space on T and HpT for the Hardy space on T. Note that HpT is a closed subspace of LpT.LetD be the open unit disk and let D¯ be the closed unit disk in the complex plane. Let CAD denote the disk algebra, i.e., the set of all continuous functions on D¯ which is analytic on D.For1≤p≤∞, we write HpD for the Hardy space on D. Two spaces HpD and HpT are identified via nontangential limits and Poisson integral. Thus, we often write Hp to denote the both of them. The norm in LpT (or HpD) will be denoted by ∥·∥p and the inner product in L2T (or H2D) will be denoted by ·,·. We refer the reader to the texts [17–19] and [20] for details of Hardy spaces.The shift operator and its adjoint are one of the most interesting operators on the Hardy space. For convenience, we define them onHD, the class of all analytic functions on D. For f∈HD, define
(1)Sfz=zfzz∈D,S∗fz=fz−f0zz∈D.The operatorsS and S∗ are often called the unilateral shift and the backward shift, respectively. We refer the reader to the text [21] which treats the shift operator in great detail.One of the most remarkable theorems in analysis is Beurling’s theorem (cf. [18, 20, 22]), which characterizes all S-invariant subspaces of H2. (We use the term “subspace” for a closed linear subspace.) For a nonzero subspace M of H2, M is S-invariant if and only if
(2)M=θH2=θf:f∈H2,for some inner function θ∈H∞. A bounded analytic function θ on D is called an inner function if its radial limit θ∗eit=limr→1−θreit has a unit modulus for almost all eit∈T. If an inner function has no zero in D, we call it a singular inner function.LetMT be the set of all complex (finite) Borel measures on T. Note that MT is a Banach space with the total variation norm ∥μ∥=∣μ∣T, where ∣μ∣ is the total variation measure of μ. We may regard the normalized Lebesgue measure m as a finite positive Borel measure. Hence, m∈MT. We write BT for the σ-algebra of all Borel sets in T. We say μ is singular if μ⊥m.Suppose thatμ∈MT. For any function f∈L1T,∣μ∣, let f·μ denote the complex Borel measure on T defined by
(3)f·μE=∫EfdμE∈BT.Then,∣f·μ∣=∣f∣·∣μ∣. Hence, ∥f·μ∥=∥f∥L1T,∣μ∣. In particular, for every f∈CT, the measure f·μ is defined and ∥f·μ∥≤∥f∥∞∥μ∥.Forμ∈MT, the nth Fourier–Stieltjes coefficient of μ is given by
(4)μ^n=∫Tζ¯ndμζn∈ℤ.For anyμ∈MT, the bilateral sequence μ^=μ^nn∈ℤ is bounded and the mapping μ↦μ^ is a bounded linear transformation from MT into ℓ∞ℤ. Note that the mapping μ↦μ^ is one-to-one, and hence, a measure μ∈MT is completely determined by its Fourier–Stieltjes coefficients. By the theorem of F. and M. Riesz, if μ∈MT is analytic, i.e., μ^n=0 for all n≤0, then μ≪m and dμ/dm∈H1T; in other words, μ=f·m for some f∈H1T.For the definition of Toeplitz operators whose symbols are measures, we use the Cauchy transform as the “projection” of measures. For this reason, we use the notationPμ instead of Kμ for the Cauchy transform of μ. We refer the reader to the text [23] for thorough treatments of the Cauchy transform. For μ∈MT, the analytic function Pμ on D, given by
(5)Pμz=∫T11−ζ¯zdμζ=∑n=0∞μ^nznz∈D,is called the Cauchy transform of μ. Clearly, the mapping P is a linear transformation from MT into HD. We may regard f∈L1T as the absolutely continuous measure f·m∈MT. Hence, we denote Pf·m by Pf, i.e.,
(6)Pfz=∫Tfζ1−ζ¯zdmζ=∑n=0∞f^nznz∈D.(Clearly,f·m^n=f^n.) As we have identified H2D with H2T, the mapping P may be regarded as the orthogonal projection of L2T onto H2T (the so-called Riesz projection).Letφ∈L2T. The Toeplitz operatorTφwith symbolφ is the linear operator on H2 with domain
(7)DTφ=f∈H2D:Pφf∈H2D,given by
(8)Tφf=Pφff∈DTφ.(Recall that every function inH2D may be identified with its nontangential limit function which belongs to H2T.) Clearly, CAD⊆DTφ. Hence, Tφ is densely defined. Also, Tφ is closed. Observe that
(9)φzj,zi=φ,zi−j=φ^i−j,for every i,j∈ℕ∪0. Hence, the matrix representation of Tφ with respect to the orthonormal basis 1,z,z2,⋯ is
(10)φ^0φ^−1φ^−2⋯φ^1φ^0φ^−1⋯φ^2φ^1φ^0⋯⋮⋮⋮⋱.A matrix of this form is called aToeplitz matrix; in other words, an infinite matrix αi,ji,j≥0 is called a Toeplitz matrix if
(11)αi,j=αi+1,j+1,for every i,j∈ℕ∪0.For a bilateral sequences=snn∈ℤ of complex numbers, we denote by Ts the infinite Toeplitz matrix corresponding to s, i.e., Ts is the infinite matrix whose i,j-entry is si−j. Note that if φ∈L2T, then the matrix representations of Tφ is Tφ^. For n∈ℕ∪0, we denote by Tns the n+1×n+1 Toeplitz matrix corresponding to s, i.e.,
(12)Tns=s0s−1⋯s−ns1s0⋯s−n+1⋮⋮⋱⋮snsn−1⋯s0.


## 3. The Main Results

Letμ be a complex Borel measure on T. For any function f∈CAD, f·μ is a complex Borel measure on T, and hence, the Cauchy transform Pf·μ is an analytic function on D. Define
(13)DTμ=f∈CAD:Pf·μ∈H2D.It is easy to show thatDTμ is a linear manifold of H2D. Now define
(14)Tμf=Pf·μf∈DTμ.Then,Tμ is a linear operator on H2D with domain DTμ.Definition 1.
The operatorTμ is called the Toeplitz operator with symbol μ.We begin with the following:Proposition 2.
Suppose thatμ≪m and the Radon–Nikodym derivative φ=dμ/dm belongs to L2T. Then, DTμ=CAD and
(15)Tμf=Tφf,for every f∈CAD.Proof.
Suppose thatμ=φ·m, where φ∈L2T. Let f be an arbitrary function in CAD. Then,
(16)Pf·μz=∫Tfζ1−ζ¯zdμζ=∫Tfζφζ1−ζ¯zdmζ=Pφfz,for every z∈D, and so, Pf·μ=Pφf. Since φf∈L2T, it follows that Pf·μ∈H2D. Hence, f∈DTμ and
(17)Tμf=Pf·μ=Pφf=Tφf.
This completes the proof.Proposition2 shows that the notion of Tμ is a kind of generalization of the Toeplitz operators whose symbols are L2-functions.Remark 3.
(a)
Toeplitz operators withL1-symbols: every function φ∈L1T would be regarded as the absolutely continuous measure φ·m∈MT. Hence, we may use Definition 1 to define Toeplitz operators with L1-symbols: if φ∈L1T and μ=φ·m, then(18)DTμ=f∈CAD:Pφf∈H2D,(19)Tμf=Pφf,for f∈DTμ.
(b)
Toeplitz operators withH1-symbols: let φ∈H1T and put μ=φ·m∈MT. For every f∈CAD, φf∈H1T. Hence, Pφf=φf (if we view φ in the right-hand side as a function in H1D). It follows that(20)DTμ=f∈CAD:φf∈H2D,(21)Tμf=φf,for f∈DTμ. This shows that a Toeplitz operator with H1-symbol behaves as a multiplication. Notice that the action of Tμ is the same as that of Tφ defined in ([16], Section 5). (In that paper, the domain of Tφ is given by DTφ=f∈H2D:φf∈H2D.) Moreover, since φ is of Smirnov class, φ=b/a for some a,b∈H∞D such that a is an outer function, a0>0, and a2+b2=1 on T. In this case, DTφ=aH2D (cf. [16]). It follows that
(22)DTμ=DTφ∩CAD=aH2D∩CAD.
Sincea is an outer function, it follows that aH2D is dense in H2D.
Question: is aH2∩CAD dense in H2?
We give some concrete examples.Example 4.
(a)
Letφ be the analytic function on D such that φz2=1−z−1 and φ0=1. Then, φ∈H1D but φ∉H2D. Put μ=φ·m. By Remark 3, (b), we have(23)DTμ=f∈CAD:φf∈H2D.
How large is the domainDTμ? Suppose that g∈CAD and g1≠0. Then, there exists a constant c>0 such that ∣g∣≥c on a neighborhood of ζ=1. It follows that φg∉H2D. Hence, g∉DTμ. This shows that
(24)DTμ⊆f∈CAD:f1=0.
On the other hand, ifr>0 and if ψr is the function in CAD which satisfies ψrz1/r=1−z and ψr0=1, then, for every g∈CAD,
(25)∥φψrg∥22=∫Tφζ2ψrζ2gζ2dmζ=∫T1−ζ2r∣1−ζ∣gζ2dmζ≤∥g∥∞2·∫T1−ζ2r−1dmζ=∥g∥∞22π∫−ππ1−eit2r−1dt≤∥g∥∞22π∫−ππt2r−1dt=∥g∥∞2ππ2r2r,and hence, φψrg∈H2D, i.e., ψrg∈DTμ. It follows that
(26)⋃r>0ψrCAD⊆DTμ.
Sinceψ1=1−z, we have
(27)1−z·CAD⊆DTμ.
In particular,DTμ contains all polynomials vanishing at ζ=1.
(b)
Letμ=δ1 be the unit point mass concentrated at ζ=1. Note that the measure μ is discrete. Observe that, for f∈CAD,(28)Pf·μz=∫Tfζ1−ζ¯zdμζ=f11−zz∈D.
Since1/1−z=∑n=0∞zn, the function 1/1−z does not belong to H2D. It follows that Pf·μ∈H2D if and only if f1=0. Therefore,
(29)DTμ=f∈CAD:f1=0.
Also, we have(30)Tμf=0,for all f∈DTμ. Hence, Tμ is trivial, i.e., Tμf=0 for all f∈DTμ. Consequently, Tμ is bounded (on DTμ). Notice that DTμ does not contain the constant function 1. We show later (see Remark 11) that DTμ is dense in H2D.
(c)
The Cantor middle-third measure: let C denote the Cantor ternary set and let φ be the Cantor function, i.e., for x=∑j=1∞aj/3j∈C,(31)φx=∑j=1∞aj/22j,and φx=supφy:y<x,y∈C for x∉C. Then, φ is continuous and monotonically increasing. Hence, there exists a positive Borel measure μ on T such that
(32)μe2πiθ:0≤θ<t=φt0≤t≤1.
The measureμ (the so-called Cantor middle-third measure) is a typical example of a singular continuous measure. We refer the reader to the papers [24] and [25] which treat measures of the Cantor type. It is known that
(33)μ^n=−1n∏j=1∞cos2πn3jn∈ℤ.
Hence,(34)μ∧n2=∏j=1∞1−sin22πn3jn∈ℤ.
Since0≤sin22πn/3j<1 for each j and ∑j=1∞sin22πn/3j<∞, it follows that μ^n≠0. Note also that μ^−n=μ^n and μ^3n=μ^n for every n∈ℤ. We may here ask the following questions:
(a)
What isDTμ? Is DTμ dense in H2D?(b)
What isTμ? Is Tμ trivial?
We next ask:when is the domainDTμdense inH2D? It does not seem easy to answer this question in general. The following lemma is used to derive some properties of DTμ which are helpful to determine the density of DTμ in H2D. Recall that S is the shift operator on HD, i.e., if f∈HD, then Sfz=zfz for z∈D.We then have the following:Lemma 5.
For everyμ∈MT and f∈CAD,
(35)PSf·μ=SPf·μ+PSf·μ0.Proof.
For eachz∈D,
(36)PSf·μz−PSf·μ0=∫Tζfζ1−ζ¯zdμζ−∫Tζfζdμζ=∫Tζ¯z1−ζ¯zζfζdμζ=z∫Tfζ1−ζ¯zdμζ=zPf·μz=SPf·μz.The following proposition gives an important information for the domain ofTμ.Proposition 6.
Letμ∈MT and let α be a complex number such that ∣α∣≠1. Then, the following statements hold:
(a)
Forf∈CAD, f∈DTμ if and only if S−αf∈DTμ(b)
Forf∈H2D, f∈clH2DTμ if and only if S−αf∈clH2DTμProof.
(a) Suppose thatf∈CAD. Then, by Lemma 5,
(37)PS−αf·μ=PSf·μ−Pαf·μ=SPf·μ+PSf·μ0−αPf·μ=S−αPf·μ+PSf·μ0.
Hence,PS−αf·μ∈H2D if and only if S−αPf·μ∈H2D. Since Pf·μ∈HD and ∣α∣≠1, it follows that Pf·μ∈H2D if and only if S−αPf·μ∈H2D. Therefore, f∈DTμ if and only if S−αf∈DTμ. This proves (a).
(b) Suppose thatf∈H2D and f∈clH2DTμ. Then, there exists a sequence fj in DTμ such that ∥f−fj∥2⟶0. Since S−α is a bounded operator on H2D, we have
(38)∥S−αf−S−αfj∥2=∥S−αf−fj∥2⟶0.
By (a), eachS−αfj belongs to DTμ. It follows that S−αf∈clH2DTμ.
Conversely, suppose thatf∈H2D and S−αf∈clH2DTμ. Then, there exists a sequence gj in DTμ such that
(39)∥S−αf−gj∥2⟶0.
We want to show thatf∈clH2DTμ. To see this we consider two cases.Case 1.
(∣α∣<1). Assume first that gjα=0 for all j. Then,
(40)gj=S−αfj,where fj∈CAD. Since gj∈DTμ, it follows from (a) that fj∈DTμ. Note that the approximate point spectrum of the operator S on H2D is σapS=T (cf. [26]). Since α does not belong to T, the operator S−α is bounded below on H2D. It follows that there exists a constant c>0 such that
(41)∥S−αf−gj∥2=∥S−αf−fj∥2≥c·∥f−fj∥2for all j. This implies that ∥f−fj∥2→0. Therefore, f∈clH2DTμ.In the case thatgjα≠0 for some j, we may assume that g1α≠0. Note that gj⟶S−αf weakly. Hence, gjz⟶S−αfz for each z∈D. In particular, we have
(42)gjα⟶0.Now put(43)hj=gj−gjαg1αg1j=1,2,3,⋯.Then,hj∈DTμ and hjα=0 for all j. Observe that
(44)∥S−αf−hj∥2≤∥S−αf−gj∥2+∣gjαg1α∣g12.It follows that(45)∥S−αf−hj∥2→0.Hence, by the preceding paragraph, we conclude thatf∈clH2DTμ.Case 2.
(∣α∣>1). The operator S−α on H2D is invertible. Hence,
(46)∥f−S−α−1gj∥2⟶0.SinceS−α−1=−∑n=0∞Sn/αn+1 and DTμ is S-invariant by (a), each S−α−1gj belongs to clH2DTμ. It follows that f∈clH2DTμ, and the proof is complete.Remark 7.
If we takeα=0 in Proposition 6, then the linear subspaces DTμ and its closure clH2DTμ are S-invariant. Also, the equality in Lemma 5 can be rewritten as S∗PSf·μ=Pf·μ. Consequently, we have S∗TμSf=Tμf for every f∈DTμ.As a consequence of Proposition6, we derive the following theorem which describes the domain DTμ. Recall that an inner function is said to be singular if it has no zero in the unit disk.Theorem 8.
Letμ∈MT. Then, one of the following holds:
(i)
DTμ=0(ii)
DTμ is dense in H2D(iii)
clH2DTμ=θH2D, where θ is a singular inner functionProof.
By Proposition6, clH2DTμ is an S-invariant subspace of H2D. It follows from Beurling’s theorem that
(47)clH2DTμ=θH2D,where θ is an inner function or θ=0. If θ=0, then the case (i) occures. If θ is a nonzero constant function, case (ii) occurs. Now, suppose that θ is nonconstant. We show that θ has no zero in D. To see this, choose any nonzero function f in DTμ. Fix an arbitrary point α of D and let n be the multiplicity of the zero of f at α. Then,
(48)fz=z−αngzz∈D,where g∈CAD and gα≠0. Hence, by a repeated application of Proposition 6(a), we have
(49)g∈DTμ⊆θH2D.
It follows thatg=θh for some h∈H2D. Thus, θα cannot be 0. Since α was arbitrary, we conclude that θ has no zero in D. Therefore θ is a singular inner function.Remark 9.
Unfortunately, we cannot find a concrete example for the third case. It would be possible that the third case never occurs.
The following proposition is another consequence of Proposition6 which gives a sufficient condition for the domain DTμ to be dense in H2D.Proposition 10.
IfclH2DTμ contains a polynomial, then DTμ is dense in H2D.Proof.
Suppose thatclH2DTμ contains a polynomial. Then, by Proposition 6, (b), there exists a polynomial p∈clH2DTμ, all of whose zeros are in T, such that p0=1. Let ζ1,⋯,ζN∈T be the zeros of p, listed according to their multiplicities. Then,
(50)pz=1−ζ1¯z⋯1−ζN¯z.
Choose a sequencekn in ℕ such that kn+1>Nkn (e.g., kn=N+1n). For each n∈ℕ, define
(51)pnz=1n∑j=1n1−ζ1¯zkj⋯1−ζN¯zkj.
All of them are polynomials, divisible byp. Since clH2DTμ is S-invariant, the polynomials pn belong to DTμ. It follows by a direct computation that
(52)∥1−pn∥22≤nn2N1N1+⋯+NNNN,for every n∈ℕ. This implies that pn⟶1 in H2D. Therefore, the constant function 1 belongs to clH2DTμ. Since clH2DTμ is S-invariant, we conclude that clH2DTμ=H2D; in other words, DTμ is dense in H2D.Remark 11.
Proposition10 shows that the domains DTμ, presented in (a) and (b) of Example 4, are dense in H2D, because they contain the polynomial pz=1−z. The proof of Proposition 10 shows that every polynomial, all of whose zeros are in T, is an outer function.In order to consider the matrix representation of a linear operator onH2D, it is necessary that its domain contains all polynomials. Let us interpret the condition that DTμ contains all polynomials. Note that this is equivalent to the condition that DTμ contains any polynomial which does not vanish on T, by Proposition 6, (a).Lemma 12.
Letμ∈MT. Then, the following are equivalent:
(i)
DTμ contains all polynomials, or equivalently, DTμ contains the constant function 1(ii)
μ≪m and dμ/dm∈H2T+H01T¯Proof.
i⇒ii: suppose that the constant function 1 belongs to DTμ. Then, Pμ=P1·μ∈H2D. Let ψ denote the nontangential limit function of Pμ. Since Pμ=∑n=0∞μ^nzn, it follows that ψ^n=μ^n for all n∈ℕ∪0. Put ν=μ−ψ·m. Then, ν∈MT and
(53)ν^n=μ^n−ψ^n=0,for all n∈ℕ∪0. It follows from the F. and M. Riesz theorem that ν≪m and ν=χ·m for some χ∈H01T¯. Thus, we have μ=ν+ψ·m=χ+ψ·m. This proves (ii).
ii⇒i: suppose that (ii) holds so that μ=ψ+χ·m for some ψ∈H2T and χ∈H01T¯. Then, μ^n=ψ^n for all n∈ℕ∪0. Hence, we have
(54)∑n=0∞μ∧n2<∞.
SincePμ=∑n=0∞μ^nzn, it follows that P1·μ=Pμ∈H2D. Clearly, the constant function 1 belongs to CAD. Therefore, 1∈DTμ. Now, Proposition 6, (a), implies that DTμ contains all polynomials.Corollary 13.
Letμ∈MT be a real measure. Then, DTμ=CAD if and only if μ≪m and dμ/dm∈L2T.Proof.
Suppose thatDTμ=CAD. Then, μ≪m and μ=ψ+χ·m for some ψ∈H2T and χ∈H01T¯ by Lemma 12. Since μ is a real measure, we have
(55)μ^−n=∫Tz¯−ndμ=∫Tz¯ndμ¯=μ^n¯,for every n∈ℤ. Thus, χ^−n=ψ^n¯ for every n∈ℕ. Since ψ∈H2T, we have
(56)∑n=−∞−1χ∧n2=∑n=1∞ψ∧n2<∞.
It follows thatχ∈H02T¯. Therefore, dμ/dm=ψ+χ∈L2T.
The converse is a part of Proposition2.On the other hand, we would like to conjecture the following:Conjecture 14.
Every Toeplitz operator with a singular symbol is trivial.We give evidence for Conjecture14 by using the known fact about the Cauchy transform. Let E be a closed subset of T and let
(57)FE=g∈H2D:g=Pμforsomeμ∈ME.Then, it is known thatFE=0 if and only if mE=0 (cf. [23], Theorem 5.5.2).We then have the following:Theorem 15.
Ifμ∈MT is singular and msuppμ=0, then Tμ is trivial.Proof.
LetE=suppμ. By assumption, mE=0. Thus, FE=0. Suppose that f∈DTμ, i.e., f∈CAD and Pf·μ∈H2D. Note that suppf·μ⊆suppμ=E. Hence, f·μ∈ME. So the function Pf·μ∈H2D belongs to FE=0. It follows that Pf·μ=0. We have shown that Pf·μ∈H2D implies Pf·μ=0. In other words,
(58)f∈DTμTμf=0.
ThereforeTμ is trivial (on its domain).Remark 16.
Conjecture14 seems to be known when μ is a positive singular measure. Indeed, if μ is a positive singular measure, then its Poisson integral is the real part of 1+θ/1−θ for some inner function θ (cf. [23], Remark 9.1.4). Now, if f∈CAD and Pf·μ∈H2D, then the function g=1−θPf·μ belongs to H2D!θH2D (cf. [27], Chapter III), and hence, θg¯∈zH2D. Since 1−θ is the outer H2-function, it follows that
(59)Pf·μ¯=g1−θ¯=g¯1−θ¯=−θg¯1−θ,which implies that Pf·μ¯∈zH2D. Therefore, Pf·μ=0.The Cantor-middle-third measureμ in Example 4, (c), is a singular continuous measure, and its support is the Cantor set (in T) whose Lebesgue measure is 0. Hence, Theorem 15 implies that Tμ is trivial.We have seen that the Toeplitz operatorTμ in Example 4, (b), is a densely defined trivial linear operator. This result can be extended to the case that μ has a finite support. In this case, the fact that Tμ is trivial may follow from Theorem 15. However, we give a direct proof and also show that Tμ is densely defined.Proposition 17.
Letμ∈MT be a discrete measure whose support is a finite set. Then, the Toeplitz operator Tμ is a densely defined trivial linear operator with domain
(60)DTμ=f∈CAD:fζ=0foreveryζ∈suppμ.Proof.
Suppose thatsuppμ consists of N distinct points ζ1,⋯,ζN of T. Then,
(61)μ=c1δζ1+⋯+cNδζN,where c1,⋯,cN are nonzero complex numbers and δζ is the unit point mass concentrated at ζ.
We first show that(62)DTμ=f∈CAD:fζ1=⋯=fζN=0.
For anyf∈CAD,
(63)Pf·μz=∑j=1NcjPf·δζjz=∑j=1Ncjfζj1−ζj¯zz∈D.
It follows that(64)f∈CAD:fζ1=⋯=fζN=0⊆DTμ.
Conversely, letf∈DTμ. Then, Pf·μ∈H2D. For each j, put
(65)Fjζ=cjfζj1−ζj¯ζζ∈T.
Then,F=∑j=1NFj is the nontangential limit function of Pf·μ. Thus, F∈H2T. Choose disjoint open arcs Ij⊆T with ζj∈Ij. Fix an index j0 and let χ denote the characteristic function of Ij0. Then, χ·F∈L2T. Also, χ·Fj∈L∞T for each j≠j0. Hence,
(66)χ·Fj0=χ·F−∑j≠j0χ·Fj∈L2T.
Since1−χ·Fj0∈L∞T, it follows that
(67)Fj0=χ·Fj0+1−χ·Fj0∈L2T.
This implies thatfζj0=0, because otherwise, Fj0∉L2T. Since j0 was arbitrary, we have fζj=0 for each j. It follows that
(68)DTμ⊆f∈CAD:fζ1=⋯=fζN=0.
This proves (62). In particular, DTμ contains the polynomial pz=z−ζ1⋯z−ζN. Hence, by Proposition 10, DTμ is dense in H2D.
Equations (62) and (63) imply that Tμf=0 for all f∈DTμ, i.e., Tμ is trivial. This completes the proof.Example 18.
Letμ∈MT be a discrete measure whose support has only finitely many limit points, for example,
(69)μ=∑n=1∞12nδζn,where ζn=eπi/2n. By an argument similar to the proof of Proposition 17, we may show that
(70)DTμ=f∈CAD:fζ=0foreveryζ∈suppμ,and Tμf=0 for all f∈DTμ. Hence, Tμ is trivial. Note that every polynomial has only finitely many zeros. It follows that DTμ cannot contain any polynomial. Nevertheless, DTμ contains a nonzero function by Fatou’s theorem for CAD, which says that, for any given closed set K⊆T with mK=0, there exists a function in CAD which vanishes precisely on K (cf. [19]). Hence by Theorem 8, DTμ is dense in H2D or clH2DTμ=θH2D for some singular inner function θ. But it does not seem easy to determine whether DTμ is dense in H2D or not.To each Toeplitz operatorTμ, there corresponds an infinite Toeplitz matrix Tμ^. In general, however, it is a bit awkward to call Tμ^ as the matrix representation of Tμ, because the domain DTμ may not contain the monomials zn. Nevertheless, often, information about Tμ gives information about Tμ^. The following is one of such example.Corollary 19.
Letμ∈MT be a discrete measure whose support consists of N points of T. Then,
(71)detTnμ^=0,for all n≥N.Proof.
Suppose thatμ is the discrete measure given by (61). Then, the domain DTμ is given by (62). Choose any polynomial p in DTμ whose degree is N (e.g., pz=z−ζ1⋯z−ζN). Write p=∑k=0Nakzk. Since Tμzk=∑n=0∞μ^n−kzn, it follows that
(72)0=Tμp=∑k=0NakTμzk=∑k=0Nak∑n=0∞μ^n−kzn=∑n=0∞∑k=0Nakμ^n−kzn.
Hence, we have(73)∑k=0Nakμ^n−k=0,for all n≥0. Now, let n≥N and put
(74)x=a0⋯aN0⋯0T∈ℂn+1.
Then, by (73), Tnμ^x=0, i.e., x∈kerTnμ^. Since x≠0, the square matrix Tnμ^ is not invertible, or equivalently, detTnμ^=0.Lastly, we may ask: what is the adjoint ofTμ? To answer this question, we need the following:Lemma 20.
Letμ∈MT. Then,
(75)Tμf,g=∫Tfg¯dμ,for every f∈DTμ and g∈CAD.Proof.
Suppose thatf∈DTμ and g∈CAD. Then, Tμf∈H2D. Write Tμf=∑n=0∞anzn and g=∑n=0∞bnzn. Then,
(76)Tμf,g=∑n=0∞anbn¯.
Observe that, for eachz∈D,
(77)Tμfz=∫Tfζ1−ζ¯zdμζ=∫Tfζ∑n=0∞ζ¯nzndμζ=∑n=0∞∫Tfζζ¯ndμζzn.
Hence, we have(78)an=∫Tfζζ¯ndμζ.
Observe that, for each0<r<1,
(79)gr=∑n=0∞bnrnzn∈CAD.
It follows that(80)Tμf,gr=∑n=0∞anbn¯rn=∑n=0∞∫Tfζζ¯nbn¯rndμζ=∫Tfζ∑n=0∞bnrnζn¯dμζ=∫Tfgr¯dμ.
If we letr⟶1, then ∥g−gr∥∞⟶0, and hence, Tμf,gr⟶Tμf,g and ∫Tfgr¯dμ→∫Tfg¯dμ. This proves (75).Assume thatμ∈MT and DTμ is dense in H2D. Then, the adjoint Tμ∗ of Tμ can be defined; the domain of Tμ∗ is
(81)DTμ∗=g∈H2D:∃h∈H2Ds.t.Tμf,g=f,h∀f∈DTμ,and, for each g∈DTμ∗, Tμ∗g is the (unique) element of H2D such that
(82)Tμf,g=f,Tμ∗g,for every f∈DTμ.Ifφ∈L∞T, then Tφ∗=Tφ¯. Hence, it is reasonable to expect that the adjoint of Tμ is the Toeplitz operator induced by the “complex conjugation” of μ. For μ∈MT, define
(83)μ¯E=μE¯E∈BT.Then,μ¯∈MT. Of course, μ∈MT is a real measure if and only if μ¯=μ. Note that
(84)μ¯^n=μ^−n¯,for every n∈ℤ.We now have the following:Proposition 21.
Letμ∈MT. Assume that DTμ is dense in H2D. Then,
(85)Tμ¯⊆Tμ∗,that is DTμ¯⊆DTμ∗ and Tμ¯=Tμ∗ on DTμ¯.Proof.
Letg∈DTμ¯. By Lemma 20, it follows that
(86)Tμf,g=∫Tfg¯dμ=∫Tgf¯dμ¯¯=f,Tμ¯g,for every f∈DTμ. It follows that g∈DTμ∗ and Tμ∗g=Tμ¯g. Therefore, we conclude that
(87)DTμ¯⊆DTμ∗,and Tμ∗g=Tμ¯g for every g∈DTμ¯. This completes the proof.Ifμ∈MT, and T is the restriction of the Toeplitz operator Tμ to clH2DTμ, then T is a densely defined linear operator. In this case, T∗ is a linear operator from H2D onto clH2DTμ. By the same argument as the proof of Proposition 21, we have DTμ¯⊆DT∗ and T∗g=Tμ¯g for g∈DTμ¯.We also have the following:Proposition 22.
Letμ∈MT be positive. Then, the following hold:
(a)
Tμ is positive, i.e., μf,f≥0 for all f∈DTμ(b)
kerTμ=f∈CAD:fζ=0foreveryζ∈suppμProof.
(a) Suppose thatμ≥0. Then, by Lemma 20, we have
(88)Tμf,f=∫Tf2dμ≥0,for every f∈DTμ.
(b) Suppose thatμ∈MT is positive. If f∈kerTμ, then ∫Tf2dμ=Tμf,f=0. Hence, f=0μ-a.e. on T. We show that f=0 on suppμ. Assume to the contrary that fζ0≠0 for some ζ0∈suppμ. Since f∈CAD, there exist a constant ε>0 and an open arc I⊆T with center ζ0 such that ∣fζ∣≥ε for all ζ∈I. Since ζ0∈suppμ, we have μI>0. It follows that
(89)∫Tf2dμ≥∫If2dμ≥ε·μI>0,which is a contradiction. Hence, fζ=0 for all ζ∈suppμ. Therefore,
(90)kerTμ⊆f∈CAD:f=0onsuppμ.
The reverse inclusion is trivial.The operatorTμ may be positive even though μ is complex. For example, for any complex number α, the measure α·δ1 is trivial, and hence, it is positive.We conclude with a remark on the boundedness ofTμ. It is well known (cf. [3]) that for φ∈L2T, Tφ is bounded if and only if φ∈L∞T, in which case, ∥Tφ∥=∥φ∥∞. If μ≥0 and Tμ is bounded, then
(91)∫Tf2dμ≤c·∥f∥22f∈DTμ.Let us call a positive measureμ∈MT a compatible measure if μ satisfies (91) for all f∈CAD. The word “compatible” comes from the paper [12]. One can show that the following statements are equivalent:
(i)
μ is a compatible measure(ii)
μ≪m and dμ/dm∈L∞T(iii)
DTμ contains all polynomials and Tμ is boundedIf these conditions are satisfied and ifφ=dμ/dm, then DTμ=CAD and
(92)Tμf=Tφf,for every f∈CAD. In (iii), we cannot reduce the condition that DTμ contains all polynomials to the condition that DTμ is dense in H2D: there is a measure μ∈MT which is not compatible such that Tμ is densely defined and bounded (see Example 4, (b)).



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