Projections in Toeplitz algebra
aa r X i v : . [ m a t h . F A ] J u l PROJECTIONS IN TOEPLITZ ALGEBRA
HUI DAN , XUANHAO DING , KUNYU GUO ,and YUANQI SANG ∗ Abstract.
Motivated by Barr´ıa-Halmos’s [6, Question 19] and Halmos’s [22,Problem 237], we explore projections in Toeplitz algebra on the Hardy space.We show that the product of two Toeplitz (Hankel) operators is a projectionif and only if it is the projection onto one of the invariant subspaces of theshift (backward shift) operator. As a consequence one obtains new proofsof criterion for Toeplitz operators and Hankel operators to be partial isome-tries. Furthermore, we completely characterize when the self-commutator of aToeplitz operator is a projection. This provides a class of nontrivial projectionsin Toeplitz algebra. Introduction
Let D be the open disk in the complex plane and T its boundary. The Hardyspace H is the subspace of L = L ( T ) consisting of functions whose Fouriercoefficients corresponding to negative integers vanish. A function ϑ ∈ H iscalled an inner function if | ϑ ( e iθ ) | = 1 a.e.For ϕ in L ∞ = L ∞ ( T ) , the Toeplitz operator T ϕ with symbol ϕ and the Hankeloperator H ϕ with symbol ϕ are defined on H as the following: T ϕ f = P ( ϕf ) ,H ϕ f = ( I − P )( ϕf ) , f ∈ H , where P is the orthogonal projection of L onto H . The Toeplitz algebra T L ∞ isthe C ∗ − algebra generated by { T φ , φ ∈ L ∞ } . We say that a bounded operator Q on a Hilbert space is a projection if Q satisfies Q = Q ∗ = Q . The study of projections, and applications of such study to illuminate structureof C ∗ − algebras, have been an enduring theme in operator algebra. In particular,progresses on projections in Toeplitz algebra will shed new light on the structureof T L ∞ , for instance, compact perturbation or essential commutant problem [4,13, 19, 11, 30], when a Hankel operator is in T L ∞ [5, 10], is Ces`aro operator in T L ∞ [23]?.etc.In [6, Question 19], J. Barr´ıa and P. R. Halmos raised a problem:“Which projections belong to T L ∞ ?” Date : . ∗ Corresponding author.2010
Mathematics Subject Classification.
Key words and phrases. projection, Toeplitz algebra, invariant subspace.
They remarked that although the question is vague, it “might give a hint to asuitably general context in which Toeplitz theory can be embedded”, and in whichproblems in Toeplitz theory become “more manageable”. To better understandthis problem, we first observe that if T is a finite rank diagonal operator withdiagonal entries equal to 0 or 1, then T belongs to T L ∞ , by the formula I − T z n +1 T ¯ z n +1 = z n ⊗ z n ( n ≥ . Are there any other projections in T L ∞ ? For aunital C ∗ -algebra, the projections 0 and I are trivial. The purpose of the currentpaper is find more nontrivial projections in T L ∞ , and classify them is some sense.It is easy to see that all finite sums of finite products of Toeplitz operatorsform a dense set in T L ∞ . For J. Barr´ıa and P. R. Halmos’ problem, we shouldfind a condition for the operator P mi =1 Q nj =1 T ϕ ij to be a projection. According tothe solving process of zero product problem of Toeplitz operators [20, 17, 2], wethink that it maybe difficult when n and m are large. S. Axler made an importantobservation in [6, (14)]: the projection onto a invariant subspace of T z belongsto T L ∞ , and by Beurling’s theorem, it equals T θ T ¯ θ for some inner function θ. Inspired by this, we will initially consider that for which functions f and g , T f T g is a projection? In section 3, we find that if T f T g is a projection, it must bethe projection onto a invariant subspace of T z . This result covers the result of A.Brown and R. Douglas in [8].The central role in this work is played by the following theorem (see[14, 7.11]or[15,Theorem 2]):
Symbol mapping.
Every operator in T L ∞ is of the form T = T f + S, f ∈ L ∞ , S ∈ S where S is the semicommutator ideal generated by all semicommutators T fg − T f T g , f, g ∈ L ∞ . Since a Toeplitz operator is a projection if and only if it is 0 or I [9, Corollary5]. In the view of the symbol mapping theorem and the following importantformula T fg − T f T g = H ∗ ¯ f H g , f, g ∈ L ∞ , (1.1)in what follows we shall consider that for which functions f and g , H ∗ ¯ f H g is aprojection?Let ϑ be a nonconstant inner function, the corresponding model space K ϑ isdefined to be K ϑ = H ⊖ ϑH . Moreover, K ϑ is a nontrivial invariant subspace of T ∗ z . In section 4, we showthat if H ∗ ¯ f H g is a projection, then it must be a projection onto a model space.This result covers the decription of the partially isometric Hankel operators [25,Theorem 2.6].For an operator T on a separable Hilbert space H , the self-commutator of T isdefine by T ∗ T − T T ∗ . The study of self-commutator has attracted much interest.For example, every self-adjoint operator on an infinite dimensional Hilbert spaceis the sum of two self-commutators [21] and Berger-Shaw’s theorem [7], etc. P. R.Halmos [22, Problem 237] asked that can T ∗ T − T T ∗ be a projection, and, if so,how? He also proved that if T is an abnormal operator (i.e., operators that have HORT TITLE 3 no normal direct summands) and k T k = 1, such that self-commutator of T is aprojection, then T is an isometry. It is still an interesting question for Toeplitzoperator. Note that the self-commutator of T f is in T L ∞ . In section 5, we give the necessary and sufficient condition for the self-commutatorof T f to be a projection when T f remains unrestricted. There are several difficul-ties in proving this result. One is that the symbol mapping theorem is fail to getthe information of symbol f , since the corresponding symbol of T ∗ f T f − T f T ∗ f iszero. Another is to obtain the range of T ∗ f T f − T f T ∗ f . We overcome these obstaclesby linking hyponormal Toeplitz operators and truncated Toeplitz operators.In section 6, we describe the C ∗ − algebra generated by T u T ¯ u for all inner func-tions u. We can now state our main results.
Theorem 3.4 If f, g ∈ L ∞ ( T ) , then the following statements are equivalent.(1) T f T g is a nontrivial projection;(2) T f T g is a projection, and its range is a nontrivial invariant subspace ofthe shift operator T z ;(3) There exist a nonconstant inner function θ and a nonzero constant a suchthat f = aθ and g = ¯ θa . Theorem 4.1 If f, g ∈ L ∞ ( T ) , then the following statements are equivalent.(1) H ∗ ¯ f H g is a nontrivial projection operator;(2) The range of H ∗ ¯ f H g is a model space K θ , where θ is an inner function;(3) ¯ f + ¯ µ ¯ θ, g + ¯ θµ ∈ H , where µ ∈ C \ { } . Theorem 5.8 If ϕ ∈ L ∞ ( T ) , then T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is a nontrivial projection operatorif and only if one of following conditions holds(1) The range of T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is a model space, and ϕ = aθ + b ¯ θ + c, where θ is an inner function, a, b and c are constant with | a | − | b | = 1;(2) The range of T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is not a model space, and ϕ = uv + ¯ v + c, where u is inner, c is constant, v ∈ H with | v | = Re ( uh + 1)( h ∈ H ).2. Self-adjointness of T f T g + T φ T ψ As a preparation, we obtain a necessary and sufficient condition for self-adjointnessof T f T g + T φ T ψ . The main tool is finite rank operators.Given vectors f and g in a separable Hilbert space H , we define the rank-oneoperator f ⊗ g mapping H into itself by( f ⊗ g ) h = h h, g i f. (2.1)The following properties of rank-one operators are well known. Lemma 2.1.
Given vectors f and g in a separable Hilbert space H . (1) If f ⊗ g = 0 if and only if either f = 0 or g = 0; (2) ( f ⊗ g ) ∗ = g ⊗ f ; (3) For bounded operators A and B, A ( f ⊗ g ) B = ( Af ) ⊗ ( B ∗ g ) . Lemma 2.2.
Given vectors f and g in a separable Hilbert space. If nonzerooperator f ⊗ g is self-adjoint if and only if there is a nonzero real constant λ suchthat f = λg. HUI DAN, XUANHAO DING, KUNYU GUO, and YUANQI SANG
Proof.
Assume that f ⊗ g is self-adjoint, we have f ⊗ g = g ⊗ f, and therefore( f ⊗ g ) g = ( g ⊗ f ) g, h g, g i f = h g, f i g,f = h g, f ih g, g i g. If h g, f i = 0 , then f is the zero vector. By Lemma 2.1(1), this contradict that f ⊗ g is a nonzero operator. Let λ = h g,f ih g,g i = 0 . Substituting f = λg into f ⊗ g = g ⊗ f,λg ⊗ g = ¯ λg ⊗ g. (2.2)Hence, λ is a nonzero real number. The converse follows easily from (2.2). (cid:3) Lemma 2.3.
Given vectors f, g, φ and ψ in a separable Hilbert space. If operator f ⊗ g + φ ⊗ ψ is zero if and only if one of following statement hold(1) either f or g is the zero vector and either φ or ψ is the zero vector;(2) f, g, φ and ψ are all nonzero vectors, f = λφ and ψ = − ¯ λg, λ is a nonzeroconstant.Proof. If one of four vectors f, g, φ and ψ is zero, it is easy to see condition (1)hold, by Lemma 2.1(1).Suppose that f, g, φ and ψ are all nonzero vectors and f ⊗ g = − φ ⊗ ψ, we have( f ⊗ g ) g = − ( φ ⊗ ψ ) g h g, g i f = − h g, ψ i φf = − h g, ψ ih g, g i φ. Let λ = − h g,ψ ih g,g i , since f is a nonzero vector, λ = 0 . Write f = λφ, we have, f ⊗ g + φ ⊗ ψ = λφ ⊗ g + φ ⊗ ψ = φ ⊗ (¯ λg + ψ ) = 0 . Since φ is a nonzero vector and Lemma 2.1, ¯ λg + ψ = 0 . It is easy to check thatthe converse is true. (cid:3)
Lemma 2.4.
Given vectors f, g, φ and ψ in a separable Hilbert space H . If f ⊗ g + φ ⊗ ψ is self-adjoint, then { f, g } is linearly dependent if and only if { φ, ψ } islinearly dependent.Proof. If one of { f, g, φ, ψ } is a nonzero vector, by Lemma 2.2, { f, g } and { φ, ψ } are both linearly dependent.Assume that f, g, φ and ψ are four nonzero vectors and { f, g } is linearly de-pendent, then there exist a nonzero constant λ, such that f = λg. (2.3)Since f ⊗ g + φ ⊗ ψ is self-adjoint, f ⊗ g + φ ⊗ ψ = g ⊗ f + ψ ⊗ φ, (2.4) HORT TITLE 5
Substituting (2.3) into (2.4), we have( λ − ¯ λ ) g ⊗ g = ψ ⊗ φ − φ ⊗ ψ. (2.5)If λ is real, Lemma 2.1(2) implies ψ ⊗ φ is self-adjoint, by Lemma 2.2, we have { φ, ψ } is linearly dependent.When λ = ¯ λ, assume that { φ, ψ } is linearly independent, by Gram-Schmidtprocedure, there exist two nonzero vectors x and y such that h x, φ i = 1 , h x, ψ i = 0 , h y, ψ i = 1 , h y, φ i = 0 . Applying operator equation (2.5) to x and y give( λ − ¯ λ ) h x, g i g = ψ, ( λ − ¯ λ ) h y, g i g = − φ. Since φ and ψ are nonzero vectors,( λ − ¯ λ ) h x, g i 6 = 0 , ( λ − ¯ λ ) h y, g i 6 = 0 . This contradicts our assumption ( { φ, ψ } is linearly independent). The rest ofproof is the same as the above reasoning. (cid:3) Lemma 2.5.
Given nonzero vectors f, g, φ and ψ in a separable Hilbert space. f ⊗ g + φ ⊗ ψ is a nonzero self-adjoint operator if and only if one of followingstatement holds(1) f = λg and φ = µψ, where λ, µ ∈ R \ { } ; (2) f = λg, φ = µψ, and ψ = − ag, where λ, µ, a ∈ C \ { } , Im ( λ ) =0 , Im ( µ ) = 0 , | a | Im ( µ ) Im ( λ ) = − . (3) Both { f, g } and { φ, ψ } are linearly independent, φ = a f + a gψ = a f + a g, where a , a , a , a ∈ C . a ¯ a , ¯ a a ∈ R , ¯ a a − a ¯ a = 1 . Proof.
By Lemma 2.4, there are two cases to consider.
Case I
Assume that { f, g } and { φ, ψ } are both linearly dependent, there are twononzero constants λ and µ such that f = λg, φ = µψ. (2.6)Since f ⊗ g + φ ⊗ ψ is self-adjoint, f ⊗ g + φ ⊗ ψ = g ⊗ f + ψ ⊗ φ (2.7)Substituting (2.6) into (2.7), we have( λ − ¯ λ ) g ⊗ g = (¯ µ − µ ) ψ ⊗ ψ. (2.8) HUI DAN, XUANHAO DING, KUNYU GUO, and YUANQI SANG
This means that λ = ¯ λ if and only if ¯ µ = µ. If Im ( λ ) = Im ( µ ) = 0 , then f ⊗ g + φ ⊗ ψ = λg ⊗ g + µφ ⊗ φ, and f ⊗ g + φ ⊗ ψ is a self-adjoint operator. If Im ( λ ) and Im ( µ ) both are nonzero, (2.8) becomes g ⊗ g + Im ( µ ) Im ( λ ) ψ ⊗ ψ = 0 . By Lemma 2.3, we have g = ¯ a Im ( µ ) Im ( λ ) ψ, ψ = − ag, a ∈ C \ { } , and | a | Im ( µ ) Im ( λ ) = − . Case II
If both { f, g } and { φ, ψ } are linearly independent, by Gram-Schmidt procedure,there exist two nonzero vectors x and y such that h y, ψ i = 1 , h y, φ i = 0 , h x, φ i = 1 , h x, ψ i = 0 . Applying operator equation (2.7) to x and y give φ = −h y, g i f + h y, f i g,ψ = h x, g i f − h x, f i g. Let a = −h y, g i , a = h y, f i , a = h x, g i and a = −h x, f i . Write (cid:18) φψ (cid:19) = (cid:18) a a a a (cid:19) (cid:18) fg (cid:19) . (2.9)Substituting (2.9) into (2.7), we have f ⊗ g + ( a f + a g ) ⊗ ( a f + a g )= g ⊗ f + ( a f + a g ) ⊗ ( a f + a g ) . After simplifying we get (cid:0) ( a ¯ a − ¯ a a ) f + ( a ¯ a − ¯ a a − g (cid:1) ⊗ f = (cid:0) (¯ a a − a ¯ a ) g + (¯ a a − a ¯ a − f (cid:1) ⊗ g. Since { f, g } is linearly independent and Lemma 2.3, a ¯ a − ¯ a a =0 , ¯ a a − a ¯ a =0 , ¯ a a − a ¯ a =1 . The converse follows immediately from the above reasoning. (cid:3)
Define an operator V on L by V f ( w ) = wf ( w ) HORT TITLE 7 for f ∈ L . It is easy to check that V is anti-unitary. The operator V satisfiesthe following properties [29, Lemma 2.1]: V = I,V P V = ( I − P ) ,V H f V = H ∗ f . Lemma 2.6. If f and g are in L ∞ , then T ¯ z T f T g T z − T f T g = ( V H ¯ f ⊗ ( V H g . Proof.
By the following identity: I − T z T ¯ z = 1 ⊗ , we have T ¯ z T f T g T z = T ¯ z T f (1 ⊗ T z T ¯ z ) T g T z = T ¯ z T f (1 ⊗ T g T z + T ¯ z T f T z T ¯ z T g T z = T ¯ z T f (1 ⊗ T g T z + T f T g =( T ¯ zf ⊗ ( T ¯ z ¯ g
1) + T f T g . On the other hand, one easily verifies that T ¯ zf P ¯ zf P V ¯ f = V P − ¯ f = V H ¯ f , Thus, T ¯ z T f T g T z − T f T g = ( V H ¯ f ⊗ ( V H g . (cid:3) Next, we present a proof of the result of K. Stroethoff [28, Theorem 4.4].
Lemma 2.7. If f, g, φ and ψ are in L ∞ ( T ) , then T f T g + T φ T ψ is a Toeplitzoperator if and only if ( V H ¯ f ⊗ ( V H g
1) + (
V H ¯ φ ⊗ ( V H ψ
1) = 0 if and only if one of the following cases holds:(1) either ¯ f or g is analytic and either ¯ φ or ψ is analytic;(2) f − λφ ∈ H , ψ + λg ∈ H , where λ ∈ C \ { } .In this case, T f T g + T φ T ψ = T fg + φψ . Proof.
By [9, Theorem 6] and Lemma 2.6 we get that T f T g + T φ T ψ is a Toeplitzoperator if and only if T ¯ z ( T f T g + T φ T ψ ) T z = T f T g + T φ T ψ . if and only if ( V H ¯ f ⊗ ( V H g
1) + (
V H ¯ φ ⊗ ( V H ψ
1) = 0 . (2.10)If (2.10) holds, Lemma 2.3 yields(1) either ¯ f or g is analytic and either ¯ φ or ψ is analytic;or(2) f − λφ ∈ H , ψ + λg ∈ H , where λ is a constant. HUI DAN, XUANHAO DING, KUNYU GUO, and YUANQI SANG
Conversely, if either ¯ f or g is analytic and either ¯ φ or ψ is analytic, by [9,Theorem 8], we have T f T g + T φ T ψ = T fg + φψ . An easy computation gives T f T g + T φ T ψ = T f T g − T fg + T fg + T φψ − T φψ + T φ T ψ = − H ∗ ¯ f H g − H ∗ ¯ φ H ψ + T fg + φψ (2.11)If f − λφ ∈ H , ψ + λg ∈ H , where λ is a constant, then − H ∗ ¯ f H g − H ∗ ¯ φ H ψ = − H ∗ ¯ λ ¯ φ H g − H ∗ φ H − λg = − λH ∗ ¯ φ H g + λH ∗ ¯ φ H g = 0 . (cid:3) Lemma 2.8. If f, g, φ and ψ are in L ∞ , then T f T g + T φ T ψ is not a Toeplitzoperator and is self-adjoint if and only if one of the following cases holds:(1) either ¯ f or g ∈ H , ¯ φ / ∈ H and ψ / ∈ H , ¯ φ − aψ ∈ H , a ∈ R \ { } , f g + φψ is real-valued.(2) either ¯ φ or ψ ∈ H , ¯ f / ∈ H and g / ∈ H , ¯ f − bg ∈ H , b ∈ R \ { } , f g + φψ is real-valued.(3) ¯ f , g, ¯ φ and ψ are not in H , f g + φψ is real-valued.(a) ¯ f − λg ∈ H and ¯ φ − µψ ∈ H . where λ, µ ∈ C \ { } ; (i) Im ( λ ) = Im ( µ ) = 0; (ii) Im ( λ ) = 0 and Im ( µ ) = 0 , ψ + cg ∈ H , c ∈ C \{ } , | c | Im ( µ ) Im ( λ ) = − . (b) ¯ φ − ¯ a ¯ f − ¯ a g ∈ H , and ψ − ¯ a ¯ f − ¯ a g ∈ H , where a , a , a and a are constant, a ¯ a and ¯ a a are real numbers, ¯ a a − a ¯ a = 1 . Proof.
Assume that T f T g + T φ T ψ is not a Toeplitz operator and is self-adjoint, wehave T f T g + T φ T ψ = T ¯ g T ¯ f + T ¯ ψ T ¯ φ ,T ¯ z T f T g T z + T ¯ z T φ T ψ T z = T ¯ z T ¯ g T ¯ f T z + T ¯ z T ¯ ψ T ¯ φ T z . By symbol map [14, 7.11] and [15, Theorem 2], we have f g + φψ is real-valued.By Lemma 2.6 and [9, Theorem 6] we get( V H ¯ f ⊗ ( V H g
1) + (
V H ¯ φ ⊗ ( V H ψ V H g ⊗ ( V H ¯ f
1) + (
V H ψ ⊗ ( V H ¯ φ , and T f T g + T φ T ψ = T ¯ z T f T g T z + T ¯ z T φ T ψ T z . Hence, (
V H ¯ f ⊗ ( V H g
1) + (
V H ¯ φ ⊗ ( V H ψ V H ¯ f V H g V H ¯ φ V H ψ f ∈ H or g ∈ H , and ¯ φ / ∈ H and ψ / ∈ H . Thus
HORT TITLE 9 ( V H ¯ φ ⊗ ( V H ψ
1) is a nonzero self-adjoint operator. By Lemma 2.2, we have¯ φ − aψ ∈ H , a ∈ R \ { } . Similarly, if either
V H ¯ φ V H ψ V H ¯ f V H g φ or ψ ∈ H , ¯ f / ∈ H and g / ∈ H , ¯ f − bg ∈ H , b ∈ R \ { } . If V H ¯ f , V H g , V H ¯ φ , and V H ψ f − λg ∈ H and ¯ φ − µψ ∈ H . where λ, µ ∈ C \ { } ;(i) λ and µ are real;(ii) Im ( λ ) = 0 and Im ( µ ) = 0 , ψ + cg ∈ H , c ∈ C \ { } , | c | Im ( µ ) Im ( λ ) = − . (II) ¯ φ − ¯ a ¯ f − ¯ a g ∈ H , ψ − ¯ a ¯ f − ¯ a g ∈ H , where a , a , a and a are constant, a ¯ a and ¯ a a are real numbers, ¯ a a − a ¯ a = 1 . To verify condition (1), an easy computation gives T f T g + T φ T ψ = T fg + φψ − T φψ + T φ T ψ = T fg + φψ − H ∗ ¯ φ H ψ = T fg + φψ − aH ∗ ψ H ψ ,T fg + φψ − aH ∗ ψ H ψ is self-adjoint, and condition (1) is verified.Condition (2) is verified in the same way as condition (1).To verify condition (3)(a)(i), using (5.3) we obtain T f T g + T φ T ψ = T fg + φψ − H ∗ ¯ f H g − H ∗ ¯ φ H ψ = T fg + φψ − λH ∗ g H g − µH ∗ ψ H ψ . therefore, T fg + φψ − λH ∗ g H g − µH ∗ ψ H ψ is self-adjoint, and Condition (3)(a)(i) isverified.To verify condition (3)(a)(ii): ¯ f , g, ¯ φ and ψ are not in H , f g + φψ is real-valued.¯ f − λg ∈ H and ¯ φ − µψ ∈ H . where λ, µ ∈ C \ { } ; Im ( λ ) = 0 and Im ( µ ) = 0 ,ψ + cg ∈ H , c ∈ C \ { } , | c | Im ( µ ) Im ( λ ) = − . Again using (5.3) we obtain T f T g + T φ T ψ = T fg + φψ − H ∗ ¯ f H g − H ∗ ¯ φ H ψ = T fg + φψ − ¯ λH ∗ g H g − ¯ µH ∗ ψ H ψ = T fg + φψ − ¯ λH ∗ g H g − ¯ µ | c | H ∗ g H g = T fg + φψ − (¯ λ + ¯ µ | c | ) H ∗ g H g . Since | c | Im ( µ ) Im ( λ ) = − , ¯ λ + ¯ µ | c | is a real constant, T fg + φψ − (¯ λ + ¯ µ | c | ) H ∗ g H g isself-adjoint, Condition (3)(a)(ii) is verified.To verify condition (3)(b): ¯ φ − ¯ a ¯ f − ¯ a g and ψ − ¯ a ¯ f − ¯ a g are in H , where a , a , a and a are constant, a ¯ a and ¯ a a are real numbers, ¯ a a − a ¯ a = 1 . Again using (5.3) we obtain T f T g + T φ T ψ = T fg + φψ − H ∗ ¯ f H g − H ∗ ¯ φ H ψ = T fg + φψ − H ∗ ¯ f H g − ( a H ∗ ¯ f + a H ∗ g )(¯ a H ¯ f + ¯ a H g )= T fg + φψ − a ¯ a H ∗ ¯ f H ¯ f − a ¯ a H ∗ g H g − (1 + a ¯ a ) H ∗ ¯ f H g − a ¯ a H ∗ g H ¯ f = T fg + φψ − a ¯ a H ∗ ¯ f H ¯ f − a ¯ a H ∗ g H g − ¯ a a H ∗ ¯ f H g − a ¯ a H ∗ g H ¯ f .T fg + φψ − a ¯ a H ∗ ¯ f H ¯ f − a ¯ a H ∗ g H g − ¯ a a H ∗ ¯ f H g − a ¯ a H ∗ g H ¯ f is self-adjoint,Condition (3)(b) is verified. (cid:3) The product of two Toeplitz operators is a projection
The following Lemma is well known. (see[25, Corollary 1.9,Theorem 2.3,The-orem 2.4])
Lemma 3.1.
Let R be a Hankel operator on H . (1) ker R is an invariant subspace of T z ;(2) R has nontrivial kernel if and only if the symbol of R has the form ¯ θφ where θ is some inner function and φ ∈ H ∞ . Furthermore:(a) ker H ¯ θφ = θH , ker H ∗ ¯ θφ = zθH ; (b) closure n Range ( H ∗ ¯ θφ ) o = (ker H ¯ θφ ) ⊥ = H ⊖ θH = K θ ; (c) closure (cid:8) Range ( H ¯ θφ ) (cid:9) = zK θ . Lemma 3.2.
Let f, g ∈ L ∞ ( T ) . If T f T g is a nontrivial idempotent, then f g = 1 a.e. on T . Proof.
Suppose T f T g is a nontrivial idempotent, namely, ( T f T g ) = T f T g . By sym-bol map [14, Theorem 7.11], we have ( f g ) = f g. Then there exists a measurablesubset E of T such that ( f g )( e iθ ) = (cid:26) , e iθ / ∈ E, , e iθ ∈ E. If m ( E ) > , then there exists a subset E of E with positive measure, suchthat either f | E = 0 or g | E = 0 . If f | E = 0 , by Guo’ Lemma [20, Lemma 1], then ker T f = ker T ¯ f = { } . Since T f T g is a nontrivial idempotent, ker T f T g = { } . For any nonzero vector x ∈ ker T f T g , we have T g x = 0 , hence ker T g = { } . By Coburn’ Lemma [14, 7.24],ker T ∗ g = ker T ¯ g = { } . Since T f T g is a nontrivial idempotent, ( T f T g ) ∗ = T ¯ g T ¯ f isalso a nontrivial idempotent. Hence ker T ¯ g T ¯ f = { } , it is a contradiction.If g | E = 0 , same considerations apply to T ¯ g T ¯ f , we can also get a contradiction.Hence m ( E ) = 0 . (cid:3) Lemma 3.3.
If a Toeplitz operator is a projection, it must be 0 and . Proof.
By [9, Corollary 5], the only idempotent Toeplitz operators are 0 and1 . (cid:3) Hence, a Toeplitz operator cannot be a nontrivial projection.
HORT TITLE 11
Theorem 3.4. If f, g ∈ L ∞ ( T ) , then the following statements are equivalent.(1) T f T g is a nontrivial projection;(2) T f T g is a projection, and its range is a nontrivial invariant subspace ofthe shift operator T z ; (3) There exist a nonconstant inner function θ and a nonzero constant a suchthat f = aθ and g = ¯ θa .Proof. (1) ⇒ (2) : Suppose T f T g is a nontrivial projection, by Lemma 2.8 (2), wehave ¯ f = λg + h, λ ∈ R \ { } , h ∈ H . T f T g is a nontrivial projection if and onlyif I − T f T g is a nontrivial projection. By Lemma 3.2, we have I = T fg . Hence I − T f T g = T fg − T f T g = H ∗ ¯ f H g = (cid:0) λH ∗ g + H h (cid:1) H g = λH ∗ g H g , thus λH ∗ g H g is a nontrivial projection.By Lemma 3.1 (1), ker H g is an invariant subspace of shift operator T z . More-over, ker H g = ker H ∗ g H g = ker λH ∗ g H g = ker( I − T f T g ) = Range ( T f T g ) . Therefore, the range of T f T g is a nontrivial invariant subspace of the shift operator T z . (2) ⇒ (3) : By Beurling’s theorem [14, 6.11], Range ( T f T g ) = θH for somenonconstant inner function θ. T θ T ¯ θ is the orthogonal projection of L onto θH . Hence T f T g = T θ T ¯ θ . By Lemma 2.7, we have f − aθ ∈ H , ¯ θ − ag ∈ H , a ∈ C \ { } . Note that T a f T ag = T f T g , let F , a f = θ + ¯ ϕ,G , ag = ψ + ¯ θ, (3.1)where ϕ and ψ are in H ∞ . Since Lemma 3.2,
F G = 1 , ¯ θF θG = 1 and θ ¯ F θG isan inner function.If θ ¯ F θG = 1 , then Re (1 − θ ¯ F θG ) > , by [24, Part A. 4.2.2], we have 1 − θ ¯ F θG is outer. Using (3.1), then1 − θ ¯ F θG = 1 − θ ( θ + ϕ ) θ ( θ + ψ )= 1 − (1 + θϕ )(1 + θψ )= − θ ( ϕ + ψ + ϕψ ) , it is a contradiction. Hence θ ¯ F θG = 1 . Note that θ ¯ F and θG are in H ∞ , by[14, 6.20], θ ¯ F and θG are outer functions. Since θ ¯ F θG = 1 = ¯ θF θG, θ ¯ F = θG ,θ ¯ F = ( θG ) , and θ ¯ F and θG are real-valued functions in H ∞ , there exists a nonzeroreal constant c such that F = cθ and G = ¯ θc . Combining this with (3.1), we arrive at( c − θ = ¯ ϕ and ( 1 c − θ = ψ. Since θ is not a constant, c = 1 , it follows that F = θ and G = ¯ θ. From (3.1), we have f = aθ and g = 1 a ¯ θ. (3) ⇒ (1) : Suppose f = aθ and g = ¯ θa .Then T f T g = T θ T ¯ θ . Hence T f T g is a nontrivial projection operator. (cid:3) Remark 3.5.
Widom [14, 7.46] proved that the spectrum of a Toeplitz operatoris a connected subset of complex plane. It is natural to ask whether the spectrumof the product of two Toeplitz operator is connected? Since the spectrum ofa projection operator is { , } , by Theorem 3.3, the answer to the question isnegative. Lemma 3.6. [16, Theorem 7.22]
Let H and H be Hilbert spaces, and A anoperator in L ( H , H ) Then the following are equivalent:(1) A is a partial isometry;(2) A ∗ is a partial isometry;(3) AA ∗ is an orthogonal projection, AA ∗ = P (ker A ∗ ) ⊥ ;(4) A ∗ A is an orthogonal projection, A ∗ A = P (ker A ) ⊥ . Using Theorem 3.4, we present a new proof of the result of A. Brown and R.Douglas [8].
Corollary 3.7. If f ∈ L ∞ then the following statements are equivalent.(1) T f is a partial isometry;(2) T ∗ f is a partial isometry;(3) either f or ¯ f is inner.Proof. Using Lemma 3.6 and Theorem 3.4. (cid:3)
HORT TITLE 13 The product of two Hankel operators is a projection
Theorem 4.1. If f, g ∈ L ∞ then the following statements are equivalent.(1) H ∗ ¯ f H g is a nontrivial projection operator;(2) The range of H ∗ ¯ f H g is a model space K θ , where θ is an inner function;(3) ¯ f + ¯ µ ¯ θ, g + ¯ θµ ∈ H , where µ ∈ C \ { } .Proof. (1) ⇒ (2) : We can suppose T fg − T f T g is a nontrivial projection because H ∗ ¯ f H g = T fg − T f T g . By Lemma 2.8(2), we have¯ f = λg + h, λ ∈ R \ { } , h ∈ H . (4.1)Moreover, T fg − T f T g = H ∗ ¯ f H g = (cid:0) λH ∗ g + H h (cid:1) H g = λH ∗ g H g , Hence, ker( H ∗ ¯ f H g ) = ker( λH ∗ g H g ) = ker H g . By Lemma 3.1 (1), ker H g is an invariant subspace of shift operator T z , by Beurl-ing’s theorem [14, 6.11], ker( H ∗ ¯ f H g ) = θH for some nonconstant inner function θ. T θ T ¯ θ is the orthogonal projection of L onto θH . Hence λH ∗ g H g = I − T θ T ¯ θ λ ( T ¯ gg − T ¯ g T g ) = I − T θ T ¯ θ T λ | g | − = T λ ¯ g T g − T θ T ¯ θ Since projection operator is positive, λ > . By Lemma 2.7, we have λ ¯ g + µθ ∈ H , ¯ θ + µg ∈ H , µ ∈ C \ { } . (4.2)Hence, ( λ − | µ | ) g ∈ H , If λ = | µ | , then g ∈ H and H g = 0 . By assumption T fg − T f T g = λH ∗ g H g is anontrivial projection, so λ = | µ | . Using (4.2), we have g + ¯ θµ ∈ H . (4.3)Combining (4.3) with (4.1) gives f + µθ ∈ H . (4) ⇒ (1) : Suppose f + µθ ∈ H , g + ¯ θµ ∈ H , µ ∈ C \ { } . Then H ∗ ¯ f H g = H ∗ ¯ µ ¯ θ H ¯ θµ = H ∗ ¯ θ H ¯ θ = I − T θ T ¯ θ . I − T θ T ¯ θ is the projection onto K θ . (cid:3) Next, we derive an alternative proof of [25, Theorem 2.6].
Corollary 4.2. If f ∈ L ∞ then H f is a partial isometry if and only if ¯ f is inner.Proof. Using Lemma 3.6 and Theorem 4.1. (cid:3) Projection as self-commutators of Toeplitz operators
The problem 237 in Paul R.Halmos’s famous text: A Hilbert space problembook [22] states: can T ∗ T − T T ∗ be a projection and, if so, how? He discuss thefollowing two cases.(a) If T is an abnormal operator of norm 1, such that T ∗ T − T T ∗ is a projection,then T is an isometry.(b) Does the statement remain true if the norm condition is not assumed?In particular, if T is a Toeplitz operator. Let f ∈ L ∞ ( T ) , we consider thatwhen is T ∗ f T f − T f T ∗ f a projection?Define Q = T ∗ f T f − T f T ∗ f . Example 5.1.
Corresponding case (a), we next show that if there is a constant λ such that k T f + λ k ≤ Q is a nontrivial projection, then T f + λ is an isometry.Note that T ∗ f + λ T f + λ − T f + λ T ∗ f + λ = T ∗ f T f − T f T ∗ f , λ ∈ C . Using the idea of [22, Solution 237] and k T f k = k f k ∞ we have k h k ≥ k T f + λ h k = h T ∗ f + λ T f + λ h, h i = h T f + λ T ∗ f + λ h, h i + h Qh, h i = k T ∗ f + λ h k + k Qh k . Replace h by Qx ( x ∈ H ) in the above formula, we have T ∗ f + λ Q = 0 and T f + λ isquasinormal. A Theorem in [3] tells us that a quasinormal Toeplitz operator iseither normal or analytic and f + λ = cθ, where c is a constant and θ is an innerfunction. Q is a nontrivial projection, we have f + λ = cθ. Hence, Q = T ∗ f T f − T f T ∗ f = T | f | − T f T ¯ f = H ∗ ¯ f H ¯ f = | c | H ∗ ¯ θ H ¯ θ . Since Q is an idempotent, | c | = 1 . By [9, Corollary 3], T f + λ is an isometry if andonly if f + λ is an inner function. In this case, Q = H ∗ ¯ θ H ¯ θ is the projection ontomodel space K θ . Example 5.2.
Let us recall Abrahamse’s theorem [1]. If(1) f or ¯ f is of bounded type;(2) T f is hyponormal;(3) ker Q is invariant for T f . HORT TITLE 15 then T f is normal or analytic.Using the above theorem, if(1) f or ¯ f is of bounded type;(2) Q is a nontrivial projection;(3) ker Q is invariant for T f , then f is analytic. Hence, Q = T ∗ f T f − T f T ∗ f = T | f | − T f T ¯ f = H ∗ ¯ f H ¯ f . By Theorem 4.1, there is a constant c such that f = θ + c, where θ is an innerfunction. In this case, Q = H ∗ ¯ θ H ¯ θ is the projection onto model space K θ . From the above two examples, we need to consider two things: if Q is a non-trivial projection,1. when is the range of Q a model space?2. is the range of Q necessarily a model space? Lemma 5.3. If ϕ ∈ L ∞ then T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is the projection on to a model space K θ if and only if ϕ = aθ + b ¯ θ + c, where a, b and c are constant with | a | − | b | = 1 . Proof. If ϕ = aθ + b ¯ θ + c, where a, b and c are constant with | a | − | b | = 1 , then T ∗ ϕ T ϕ − T ϕ T ∗ ϕ = T ∗ ϕ T ϕ − T ϕ ¯ ϕ + T ϕ ¯ ϕ − T ϕ T ∗ ϕ = H ∗ ¯ ϕ H ¯ ϕ − H ∗ ϕ H ϕ =( | a | − | b | ) H ∗ ¯ θ H ¯ θ = H ∗ ¯ θ H ¯ θ . Conversely, suppose T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is the projection on to a model space K θ , then T ∗ ϕ T ϕ − T ϕ T ∗ ϕ = I − T θ T ¯ θ . Write ϕ = f + ¯ g, f and g in H , using Lemma 2.6, we have( V H ¯ g ⊗ ( V H ¯ g − ( V H ¯ f ⊗ ( V H ¯ f
1) = − ( V H ¯ θ ⊗ ( V H ¯ θ . (5.1) Case 1.
Assume that (cid:8) H ¯ g , H ¯ f (cid:9) is linearly dependent, there are two constants k and k such that k H ¯ g k H ¯ f . If k is not zero, let λ = − k k , then ¯ g − λ ¯ f ∈ H and T ∗ ϕ T ϕ − T ϕ T ∗ ϕ = H ∗ ¯ ϕ H ¯ ϕ − H ∗ ϕ H ϕ = H ∗ ¯ f H ¯ f − H ∗ ¯ g H ¯ g =(1 − | λ | ) H ∗ ¯ f H ¯ f . Then (1 − | λ | ) H ∗ ¯ f H ¯ f = H ∗ ¯ θ H ¯ θ is a projection, and 1 − | λ | > . By Theorem 4.1,we have f + µ √ −| λ | θ ∈ H , µ is unimodular constant. Therefore, ϕ = − µ √ −| λ | θ − λµ √ −| λ | ¯ θ + c, where c is a constant. Let a = − µ √ −| λ | and b = − λµ √ −| λ | , we have ϕ = aθ + b ¯ θ + c, (5.2)where | a | − | b | = 1 . If k is not zero, repeating the previous reasoning, we can prove the sameequality (5.2) hold. Case 2.
Assume that (cid:8) H ¯ g , H ¯ f (cid:9) is linearly independent. Since V is anti-unitary, (cid:8) V H ¯ g , V H ¯ f (cid:9) is linearly independent, by Gram-Schmidt procedure, there exista nonzero function x in span (cid:8) V H ¯ g , V H ¯ f (cid:9) such that h V H ¯ g , x i = 1 , h V H ¯ f , x i = 0 . Applying operator equation (5.1) to x gives V H ¯ g − h x , V H ¯ θ i V H ¯ θ ,H ¯ g − h V H ¯ θ , x i H ¯ θ . Let b = −h x , V H ¯ θ i , thus g − bθ ∈ H , and g − bθ is a constant.Similarly, there exists a constant a such f − aθ is a constant. Therefore, T ∗ ϕ T ϕ − T ϕ T ∗ ϕ = H ∗ ¯ f H ¯ f − H ∗ ¯ g H ¯ g =( | a | − | b | ) H ∗ ¯ θ H ¯ θ . and | a | − | b | = 1 . (cid:3) Recall the definition of truncated Toeplitz operator. For ϕ in L ( T ) , the trun-cated Toeplitz operator A ϑϕ is densely defined on K ϑ by A ϑϕ f = ( P − T ¯ ϑ T ϑ )( ϕf ) . The algebraic properties of truncated Toeplitz operator will paly key role in thefollowing Lemma.
Lemma 5.4. If ϕ ∈ L ∞ then T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is a nontrivial projection operatorand its range is not a Model space if and only if ϕ = uv + ¯ v + a, where u is inner, v ∈ H with | v | − ∈ uH + uH and a is constant.Proof. Assume that T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is a projection. Since projection is positive, T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is positive and T ϕ is hyponormal. We recall the characterizationof Hyponormality of Toeplitz operators form Carl C. Conwen [12]. The theoremcan be stated as follows:If ϕ is in L ∞ ( T ) , where ϕ = f + ¯ g for f and g in H , then T ϕ is hyponormal ifand only if g = c + T ¯ u f (5.3)for some constant c and some function u in H ∞ with k u k ∞ ≤ . HORT TITLE 17
According to Conwen’s Theorem, if Q is a nontrivial projection, using (5.3),we have Q = H ∗ ¯ ϕ H ¯ ϕ − H ∗ ϕ H ϕ = H ∗ ¯ f H ¯ f − H ∗ ¯ g H ¯ g = H ∗ ¯ f H ¯ f − H ∗ T ¯ u f H T ¯ u f = H ∗ ¯ f H ¯ f − H ∗ P ¯ uf + P − ¯ uf H P ¯ uf + P − ¯ uf = H ∗ ¯ f H ¯ f − H ∗ u ¯ f H u ¯ f = H ∗ ¯ f H ¯ f − H ∗ ¯ f S ¯ u S u H ¯ f = H ∗ ¯ f ( I − S ¯ u S u ) H ¯ f . where S u x = P − ( ux ) , x ∈ ( H ) ⊥ . Since k u k ∞ ≤ k S u k = k u k ∞ , S u is a contraction, we have I − S ¯ u S u ispositive, and ker( I − S ¯ u S u ) = ker( I − S ¯ u S u ) / . We claim that if I − S ¯ u S u is not injective, then u is an inner function. To seethis, let x be a nonzero vector such that ( I − S ¯ u S u ) x = 0 . Hence, h ( I − S ¯ u S u ) x, x i = h x, x i − h S u x, S u x i = k x k − k S u x k = 0and Z T | x | dm = k x k = k S u x k = k P − ux k ≤ k ux k = Z T | ux | dm. Since k u k ∞ ≤ , | ux | − | x | = ( | u | − | x | ≤ . But R T ( | u | − | x | dm ≥ , thus ( | u | − | x | = 0 .a.e on T . Hence, | u | = 1 .a.e on T , and u is an inner function.Write Q = H ∗ ¯ f ( I − S ¯ u S u ) / ( I − S ¯ u S u ) / H ¯ f =(( I − S ¯ u S u ) / H ¯ f ) ∗ ( I − S ¯ u S u ) / H ¯ f , note that ker Q = ker(( I − S ¯ u S u ) / H ¯ f ) . According to the above claim, we havethat if u is not an inner function, then ker( I − S ¯ u S u ) / = ker( I − S ¯ u S u ) = { } and ker Q = ker H ¯ f . By Lemma 3.1(1), ker H ¯ f is an invariant subspace of T z . Hence, the range of Q is a model space, it is a contradiction.It remains to consider the case that u be an inner function. Write Q = H ∗ ¯ f ( I − S ¯ u S u ) H ¯ f = H ∗ ¯ f ( S ¯ uu − S ¯ u S u ) H ¯ f = H ∗ ¯ f H ¯ u H ∗ ¯ u H ¯ f . By Gu’s theorm [18, Theorem 1.1], for two Hankel operatrs H ¯ u and H ¯ f , eitherker H ∗ ¯ u H ¯ f = ker H ¯ f or ker H ∗ ¯ f H ¯ u = ker H ¯ u . If ker H ∗ ¯ u H ¯ f = ker H ¯ f , then ker Q = ker H ∗ ¯ u H ¯ f = ker H ¯ f . By Lemma 3.1(1),ker H ¯ f is an invariant subspace of T z . Hence, the range of Q is a model space, itis a contradiction.By Lemma 3.6, H ∗ ¯ f H ¯ u H ∗ ¯ u H ¯ f is an orthogonal projection, then H ∗ ¯ u H ¯ f H ∗ ¯ f H ¯ u isan orthogonal projection.If ker H ∗ ¯ f H ¯ u = ker H ¯ u = uH (Lemma 3.1(2)(a)), then H ∗ ¯ u H ¯ f H ∗ ¯ f H ¯ u f = H ∗ ¯ u H ¯ u . (5.4)Using the property V, we have V H ∗ ¯ u H ¯ f H ∗ ¯ f H ¯ u V = H ¯ u H ∗ ¯ f H ¯ f H ∗ ¯ u ,V H ∗ ¯ u H ¯ u V = H ¯ u H ∗ ¯ u . Hence H ¯ u H ∗ ¯ f H ¯ f H ∗ ¯ u = H ¯ u H ∗ ¯ u . (5.5)Note that ker H ∗ ¯ u = zuH (Lemma 3.1(2)(a)) and zH ⊖ zuH = zK u = ¯ uK u . For every h ∈ K u , we have H ∗ ¯ u ¯ uh = P ( u ¯ uh ) = h, and h H ¯ u H ∗ ¯ f H ¯ f H ∗ ¯ u ¯ uh, ¯ uh i = h H ¯ u H ∗ ¯ u ¯ uh, ¯ uh i , h H ∗ ¯ f H ¯ f H ∗ ¯ u ¯ uh, H ∗ ¯ u ¯ uh i = h H ∗ ¯ u ¯ uh, H ∗ ¯ u ¯ uh i , h H ∗ ¯ f H ¯ f h, h i = h h, h i . Hence P K u ( H ∗ ¯ f H ¯ f ) | K u = I K u . where P K u is the orthogonal projection onto K u and I K u is the identity operatoron K u . An easy computation gives P K u H ∗ ¯ f H ¯ f h = P K u P f ( I − P ) ¯ f h = P K u f ( I − P ) ¯ f h = P K u f ¯ f h − P K u f P ¯ f h = P K u f ¯ f h − P K u f ( P − uP ¯ u + uP ¯ u ) ¯ f h = P K u f ¯ f h − P K u f ( P K u + uP ¯ u ) ¯ f h = P K u f ¯ f h − P K u f P K u ¯ f h = A u | f | h − A uf A u ¯ f h. Hence A u | f | − A uf A u ¯ f = I K u ,A uf A u ¯ f = A u | f | − . (5.6)Since f is analytic, using N. A. Sedlock’ theorem [27, Theorem 5.2] leads to A uf = cI K u , where c is a constant, and A uf − c is the zero operator, then f − c ∈ uH [26, Theorem 3.1]. There is a function v ∈ H , such that f = c + uv. Since(5.3), ϕ = uv + ¯ v + a, where a is a constant. HORT TITLE 19
Substituting f = c + uv into (5.4), we have H ∗ ¯ u H ¯ u ¯ v H ∗ ¯ u ¯ v H ¯ u = H ∗ ¯ u H ¯ u . (5.7)Repeating the above reasoning form (5.5) to (5.6), we obtain A uuv A u ¯ v ¯ u = A u | v | − . Note that A uuv = A uf − c = 0 , hence A u | v | − is zero operator, using [26, Theorem3.1] again, we have | v | − ∈ uH + uH . Conversely, if ϕ = uv + ¯ v + c, where u is inner, v ∈ H with | v | − ∈ uH + uH and c is constant, by Lemma 5.3, the range of Q is not a model space. An easycomputation gives T ∗ ϕ T ϕ − T ϕ T ∗ ϕ = H ∗ ¯ ϕ H ¯ ϕ − H ∗ ϕ H ϕ = H ∗ ¯ u ¯ v H ¯ u ¯ v − H ∗ ¯ v H ¯ v = T uv ¯ u ¯ v − T uv T ¯ u ¯ v − ( T v ¯ v − T v T ¯ v )= T | v | − T uv T ¯ u ¯ v − ( T | v | − T v T ¯ v )= T v T ¯ v − T uv T ¯ u ¯ v = T v T ¯ v − T v T u T ¯ u T ¯ v = T v ( I − T u T ¯ u ) T ¯ v = T v H ∗ ¯ u H ¯ u T ¯ v . (5.8)Note that T v H ∗ ¯ u H ¯ u T ¯ v = ( H ¯ u T ¯ v ) ∗ H ¯ u T ¯ v is positive, must be self-adjoint.It remains to show that T v H ∗ ¯ u H ¯ u T ¯ v is an idempotent. Since v is analytic, T v H ∗ ¯ u H ¯ u T ¯ v T v H ∗ ¯ u H ¯ u T ¯ v = T v H ∗ ¯ u H ¯ u T | v | H ∗ ¯ u H ¯ u T ¯ v , let | v | = uh + ¯ u ¯ h + 1 , h, h ∈ H , for every k in K u , we have H ∗ ¯ u H ¯ u T | v | k = H ∗ ¯ u H ¯ u P ( uh + ¯ u ¯ h + 1) k = H ∗ ¯ u H ¯ u P ( uhk + ¯ u ¯ h k + k )= H ∗ ¯ u H ¯ u ( uhk + k )= k. Since
Range ( H ∗ ¯ u H ¯ u ) = K u , T v H ∗ ¯ u H ¯ u T ¯ v T v H ∗ ¯ u H ¯ u T ¯ v = T v H ∗ ¯ u H ¯ u T ¯ v . (cid:3) Remark 5.5.
In fact, k h k ∞ = 1 . Since ker Q is nontrivial, there is a nonzerovector x such that, H ∗ ¯ f H ¯ f x = H ∗ f H f x = 0 , k H ¯ f x k = k H f x k and k H f x k = k S h H ¯ f x k ≤ k S h kk H ¯ f x k . Hence k S h k = k h k ∞ ≥ . Lemma 5.6. If v ∈ H and u is inner, | v | − ∈ uH + uH if and only if thereis a function h ∈ H such that | v | = Re ( uh + 1) . Proof.
Since Re ( uh + 1) = 12 ( uh + 1 + ¯ u ¯ h + 1)= u ( 12 h ) + ¯ u ( 12 ¯ h ) + 1 , | v | = Re ( uh + 1) implies | v | − ∈ uH + uH . Suppose | v | − ∈ uH + uH , then there exist F, G ∈ H such that | v | − uF + ¯ uG, and uF + ¯ uG is real-vauled, uF + ¯ uG = ¯ uF + uG. Hence, u ( F − G ) = ¯ u ( F − G ) . The left-hand side of the above equation is analytic, the right-hand side is conju-gate analytic, u ( F − G ) is equals to a constant λ. If λ is not zero, then u λ ( F − G ) =1 , and u is outer [14, 6.20], that is a contradiction. Thus λ = 0 , F = G, and | v | = Re ( u (2 F ) + 1) . (cid:3) Remark 5.7.
The set Θ = { v : v ∈ H , | v | − ∈ uH + uH } is not empty. Itis easy to see that if v is inner, v ∈ Θ . Using (5.8), we have Q = T v T ¯ v − T uv T ¯ u ¯ v , and the range of Q is vH ⊖ vuH = vK u . Moreover, u ± ∈ Θ . The following theorem summarizes Lemma 5.3, Lemma 5.4 and Lemma 5.6.
Theorem 5.8. If ϕ ∈ L ∞ then T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is a nontrivial projection operatorif and only if one of following conditions holds(1) The range of T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is a model space, and ϕ = aθ + b ¯ θ + c, where θ is an inner function, a, b and c are constant with | a | − | b | = 1; (2) The range of T ∗ ϕ T ϕ − T ϕ T ∗ ϕ is not a model space, and ϕ = uv + ¯ v + c, where u is inner, c is constant, v ∈ H with | v | = Re ( uh + 1)( h ∈ H ) . Further discussion
Now we study the C ∗ − algebra T u generated by { T u T ¯ u : u is an inner function } . Since the symbol mapping of every element in T u is constant, T u is a proper subal-gebra of T L ∞ . The following theorem will give some information of the structureof T u . Theorem 6.1. T u is irreducible and contains all compact operators.Proof. Suppose that T u is reducible. Then there exists a nontrivial projection E which commutes with each T u T ¯ u for all inner function u. If u is a M¨ o biustransform u = ϕ z ( w ) = z − w − ¯ zw , and k z denote the normalized reproducing kernel at z : k z ( w ) = √ −| z | − ¯ zw . We havethe following identity: I − k z ⊗ k z = T ϕ z T ϕ z , (6.1) HORT TITLE 21 the identity can be found in [31, p.480]. Hence, E ( k z ⊗ k z ) = ( k z ⊗ k z ) E ( Ek z ) ⊗ k z = k z ⊗ ( Ek z ) h Ek z , k z i Ek z = h Ek z , Ek z i k z k Ek z k Ek z = k Ek z k k z . If Ek z is not a zero vector, we have Ek z = k z . Thus unit disc D is the disjointunion of two sets, say D = Σ ∪ Σ , where Σ = { z ∈ D : Ek z = 0 } andΣ = { z ∈ D : Ek z = k z } . So, at least one of Σ and Σ is an uncountable set. atleast of { k z : z ∈ Σ } and { k z : z ∈ Σ } is dense in H . Hence, E is zero operatoror identical operator, which is a contradiction. Using (6.1), we have T u containsat least one nonzero compact operator. By [14, 5,39], T u contains all compactoperators. (cid:3) Acknowledgement . References [1] M. Abrahamse, Subnormal Toeplitz operators and functions of bounded type, Duke Math-ematical Journal 43 (1976) 597–604.[2] A. Aleman, D. Vukoti´c, Zero products of Toeplitz operators, Duke Mathematical Journal148 (2009) 373–403.[3] I. Amemiya, T. Ito, T. K. Wong, On quasinormal Toeplitz operators, Proceedings of theAmerican Mathematical Society (1975) 254–258.[4] S. Axler, S. Y. A. Chang, D. Sarason, Products of Toeplitz operators, Integral Equationsand Operator Theory 1 (1978) 285–309.[5] J. Barr´ıa, On Hankel operators not in the Toeplitz algebra, Proceedings of the AmericanMathematical Society (1996) 1507–1511.[6] J. Barr´ıa, P. Halmos, Asymptotic Toeplitz operators, Transactions of the American Math-ematical Society 273 (1982) 621–630.[7] C. Berger, B. Shaw, Selfcommutators of multicyclic hyponormal operators are always traceclass, Bulletin of the American Mathematical Society 79 (1973) 1193–1199.[8] A. Brown, R. Douglas, Partially isometric Toeplitz operators, Proceedings of the AmericanMathematical Society 16 (1965) 681–682.[9] A. Brown, P. Halmos, Algebraic Properties of Toeplitz operators., Journal f¨ur die reineund angewandte Mathematik 213 (1964) 89–102.[10] X. Chen, F. Chen, Hankel operators in the set of essential Toeplitz operators, Acta Math-ematica Sinica 6 (1990) 354–363.[11] X. Chen, K. Guo, K. Izuchi, D. Zheng, Compact perturbations of Hankel operators, Journalf¨ur die reine und angewandte Mathematik 2005 (2005) 1–48.[12] C. C. Cowen, Hyponormality of Toeplitz operators, Proceedings of the American Mathe-matical Society 103 (1988) 809–812.[13] K. R. Davidson, On operators commuting with Toeplitz operators modulo the compactoperators, Journal of Functional Analysis 24 (1977) 291–302.[14] R. G. Douglas,
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