On kernels of Toeplitz operators

We apply the theory of de Branges–Rovnyak spaces to describe kernels of some Toeplitz operators on the classical Hardy space H2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$H^2$$\end{document}. In particular, we discuss the kernels of the operators Tf¯/f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$T_{{\bar{f}}/ f}$$\end{document} and TI¯f¯/f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$T_{{\bar{I}}{\bar{f}}/ f}$$\end{document}, where f is an outer function in H2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$H^2$$\end{document} and I is inner such that I(0)=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$I(0)=0$$\end{document}. We also obtain a result on the structure of de Branges–Rovnyak spaces generated by nonextreme functions.


Introduction
Let H 2 denote the standard Hardy space on the unit disk D. For ϕ ∈ L ∞ (∂D) the Toeplitz operator on H 2 is given by T ϕ f = P + (ϕ f ), where P + is the orthogonal projection of L 2 (∂D) onto H 2 . We will denote by M(ϕ) the range of T ϕ equipped with the range norm, that is, the norm that makes the operator T ϕ a coisometry of H 2 onto M(ϕ). For a nonconstant function b in the unit ball of H ∞ the de Branges-Rovnyak space H(b) is the image of H 2 under the operator (1 − T b Tb) 1/2 with the corresponding range norm. The norm and the inner product in H(b) will be denoted by · b and ·, · b . The space H(b) is a Hilbert space with the reproducing kernel In the case when b is an inner function the space H(b) is the well-known model space If the function b fails to be an extreme point of the unit ball in H ∞ , that is, when log(1 − |b|) ∈ L 1 (∂D), we will say simply that b is nonextreme. In this case one can define an outer function a whose modulus on ∂D equals 1 − |b| 2 1/2 . Then we say that the functions b and a form a pair (b, a). By the Herglotz representation theorem there exists a positive measure μ on ∂D such that Moreover the function a if and only if f 2 is rigid ( [20]). Spaces H(b) for nonextreme b have been studied in [2,3,5,15,16,22], and [23].
The kernels of Toeplitz operators have been studied since the late 80's. We mention that two recent survey articles [1,8] and the book [4] contain a number of results on this topic. The Hayashi theorem [12] (see also [21]) states that the kernel of a Toeplitz operator T ϕ is a subspace of H 2 of the form ker T ϕ = f K I , where K I = H 2 I H 2 is the model space corresponding to the inner function I such that I (0) = 0 and f is an outer function of unit H 2 norm that acts as an isometric multiplier from K I onto f K I . Moreover, f can be expressed as f = a Among other results, they described all outer functions g such that ker Tḡ g = K I . In Sect. 2 we describe all such functions g for which ker Tḡ a) is a special pair, f = a 1−b and b = I b 0 , where I as above, then f K I ⊂ ker TĪf f . In the next two sections we study the space ker TĪf f f K I and show that it is isometrically isomorphic to the orthogonal complement of M(a) in the de Branges-Rovnyak space H(b 0 ). We also give an example of a function f for which the space ker TĪf f f K I is one dimensional. In the last section we discuss the orthogonal complement of M(a) in H(b) and get a generalization of results obtained in [15] and [5] for the case when pairs are rational.

The kernel of Tḡ g
It is known that if g is an outer function in H 2 , then the kernel of Tḡ g is trivial if and only if g 2 is rigid (see e.g. [18]).
The finite dimensional kernels of Toeplitz operators were described by Nakazi [17]. Nakazi's theorem says that dim ker T ϕ = n if and only if there exists an outer function f ∈ H 2 such that f 2 is rigid and ker T ϕ = { f p : p ∈ P n−1 }, where P n−1 denotes the set of all polynomials of degree at most n − 1.
Consider the following example.
Then the kernel of Tḡ g is trivial for α ∈ (− 1 2 , 1 2 ] and dimension of the kernel of Tg g is n for α ∈ (n− 1 2 , n+ 1 2 ], n = 1, 2, . . ., and In the general case the kernels of Toeplitz operators are characterized by Hayashi's theorem. To state this theorem we need some notation. We note that an outer function f having unit norm in H 2 ( f 2 = 1) can be written as where a is an outer function, b is a function from the unit ball of H ∞ such that |a| 2 + |b| 2 = 1 a.e. on ∂D. Following Sarason [20, p. 156] we call (b, a) the pair associated with f . Note also that b is a nonextreme point of the closed unit ball of H ∞ and is given by Let S denote the unilateral shift operator on H 2 , i.e., S = T z . A closed subspace M of H 2 is said to be nearly S * -invariant if for every f ∈ M vanishing at 0, we also have S * f ∈ M. It is known that the kernels of Toeplitz operators are nearly S * -invariant.

Hitt's Theorem The closed subspace M of H 2 is nearly S * -invariant if and only if there exists a function f of unit norm and a model space K I = H 2 I H 2 such that M = T f K I , where I is an inner function vanishing at the origin, and T f acts isometrically on K I .
It has been proved by D. Sarason [18] that T f acts isometrically on K I if and only if I divides b (the first function in the pair associated with f ). Consequently, the function f in Hitt's theorem can be written as The function 1+b 0 (z) 1−b 0 (z) has a positive real part and is the Herglotz integral of a positive measure on ∂D up to an additive imaginary constant, Clearly b 0 is also a nonextreme point of the closed unit ball of H ∞ and |a| 2 +|b 0 | 2 = 1 a.e. on ∂D. We remark that in view of (2) the pair (b, a) associated with an outer function f ∈ H 2 is special, while the pair (b 0 , a) need not to be special. We also put Under the above notations Hayashi's theorem reads as follows: In the above mentioned paper [6] the authors described all outer functions g ∈ H ∞ for which ker Tḡ

Hayashi's Theorem
where I is an inner function not necessarily satisfying I (0) = 0.
We prove the following where I 1 and I 2 are inner and I 1 − I 2 is outer.
Recall that the Smirnov class N + consists of those holomorphic functions in D that are quotients of functions in H ∞ in which the denominators are outer functions.
In the proof of Theorem 1, similarly to [6], we use the following result due to H. Helson [13].
Helson's Theorem The functions f ∈ N + that are real almost everywhere on ∂D can be written as where I 1 and I 2 are inner and I 1 − I 2 is outer.
We also apply a description of kernels in terms of S * -invariant subspaces K p I (| f | p ) of weighted Hardy spaces (in the case when p = 2) considered by A. Hartmann and K. Seip in their paper [10] (see also [7]). For an outer function f in H 2 the weighted Hardy space is defined as follows and, for an inner function I , [10]). Since f ∈ ker TĪf f , the last equalities imply that

Proof of Theorem 1. Assume that ker
where I 0 is an inner function such that I 0 (0) = 0, and h ∈ H 2 is outer. This means that | f (z)| = |h(z)| a.e. on |z| = 1 and consequently h(z) = c f (z), where c is a unimodular constant. Replacing cI 0 by I 0 , we get It then follows which implies I = I 0 up to a unimodular constant. Indeed, these equalities imply that an analytic function h can be written in the form Since |ψ 0 | = |ψ| a.e. on |z| = 1 and ψ 0 ∈ N + , we see that also ψ 0 ∈ H 2 0 . Hence K I = K I 0 . Consequently, equality (4) can be written as which means that the function g f (1+I ) is real a.e. on ∂D. Since this function is in the Smirnov class N + , our claim follows from Helson's theorem. To prove the other implication it is enough to observe that if In the proof of this theorem we use some ideas from Sarason's proof of Hayashi's theorem. If a positive measure μ on the unit circle ∂D is as in (1) and H 2 (μ) is the closure of the polynomials in L 2 (μ), then an operator V b given by

The complement of fK I in ker
where  H(b). On the other hand, Since We remark that the orthogonal complement of M(a) in H(b) is discussed in Sect. 5.

The example
Let, as in the previous sections, f be an outer function in H 2 and let (b, a) pair (b 0 , a) is special and f 2 0 is rigid. Moreover, if the pair (b 0 , a) is special and f 2 0 is rigid, then  (b, a) is special and f 2 is rigid but the converse implication fails ( [18, p. 158] pair (b 0 , a) is special but f 2 0 is not rigid, and describe the space ker TĪf f f K I .
Our example is a slight modification of the one given in [19, p. 491], see also [4, vol. 2, p. 494]. The corresponding functions f and f 0 are defined by taking a(z) is a Blaschke product with zero sequence {r n } ∞ n=1 lying in (−1, 0) and converging to −1. It has been proved in [19, pp. 491-492] (see also [4, vol. 2, pp. 494-496] that f 2 is rigid while f 2 0 is not. Notice that the pair (b 0 , a) is rational and the point −1 is the only zero of the function a. It then follows from [15, Theorem 4.1] (see also [5]) that M(a) is a closed subspace of H(b 0 ) and Thus we see that

Moreover (6) implies that
Our aim is to prove that ker TĪf where the function g ∈ H 2 is given by For λ in D let k λ denote the kernel function in H 2 for the functional of evaluation at λ, k λ (z) = (1 −λz) −1 . In the proof of (7) we will apply the following Lemma [9, Lemma 2] Since I (r n ) = 0, (i) and (ii) in the Lemma yield Hence It follows from [4, Theorem 21.1] that is an isometry and T 1−b Tf is an isometry of H 2 onto H(b), we see that { f k r n (I − b 0 (r n ))} n∈N is a bounded sequence in H 2 . So it contains a subsequence that converges weakly, say, to a function g ∈ H 2 . Without loss of generality, we may assume that the sequence { f k r n (I − b 0 (r n ))} itself converges weakly to g. Then for any point z ∈ D, Now observe that since in H 2 strongly. Finally, passing to the limit in (8) gives which proves (7).

Remark
One can check directly that the function g = f k −1 (I +1) is in ker TĪf f f K I .

Indeed, we have
TĪf f ( f k −1 (I + 1)) = P + fĪ I + 1 1 + z To see that the functions ( f k r n I − b 0 (r n ) f k r n ) are orthogonal to f K I note that a function h ∈ H 2 is in K I if and only if h = h − I P + (Ī h). So, we have to check that for any h ∈ H 2 , Since the functions {k λ , λ ∈ D} are dense in H 2 , it is enough to show that for any λ ∈ D, Finally, the last equality follows from

A remark on orthogonal complement of M(a) in H(b)
In this section we continue to assume that b is nonextreme. Recall that if the pair (b, a) is special and f 2 = a  H(b). Let Y be the restriction of the shift operator S to H(b) and let Y 0 be the compression of Y to the subspace H 0 (b). Then the spectrum of Y 0 is contained in the unit circle. Moreover, if z 0 ∈ ∂D and k is a positive integer, then ker(Y * −z 0 ) k , which actually equals ker(Y * 0 −z 0 ) k , lies in H 0 (b). The necessary and sufficient conditions for H 0 (b) to have finite dimension are given in Chapter X of [21] (see also [4,Theorem 29.11]). In particular, the dimension of H 0 (b) is N if and only if the operator Y 0 has distinct eigenvalues z 1 , z 2 , …, z s with their algebraic multiplicities n 1 , …, n s , N = n 1 + n 2 + · · · + n s . Thenz 1 ,z 2 , …,z s are the eigenvalues of Y * 0 with the same multiplicities, i.e., dim ker(Y 0 − z j ) n j = dim ker(Y * 0 −z j ) n j and H 0 (b) is the direct sum of the subspaces ker(Y * 0 −z j ) n j , j = 1, 2, . . . , s. On the other hand, if z 0 is a point of ∂D and b has an angular derivative in the sense of Carathéodory at z 0 , then the function given by where b(z 0 ) is the nontangential limit of b at z 0 , is in H(b) (see [21,5], [4,Theorem 21.1]). In this section we actually show that k b z 0 is in H 0 (b). Here we consider the case when the eigenspaces corresponding to eigenvalues z 1 , z 2 , …, z s are one dimensional and show that then the space H 0 (b) is spanned by the functions k b z 1 , k b z 2 , …, k b z s . For |λ| = 1 let μ λ denote the measure for which equality in (1) holds when b is replaced byλb. If we put F λ (z) = a 1−λb , then the Radon-Nikodym derivative of the absolutely continuous component of μ λ is |F λ | 2 . Note also that H(b) = H(λb).
In the proof of our main result in this section we use the following theorem proved in [21, X-13].
Sarason's Theorem Let z 0 be a point of ∂D and λ a point of ∂D such that the measure μ λ is absolutely continuous. The following conditions are equivalent.
(i)z 0 is an eigenvalue of Y * .
(iii) The function b has an angular derivative in the sense of Carathéodory at z 0 .
In view of remark in Sect. 3 under the assumption of Sarason's Theorem the operator T 1−λb T F λ is an isometry of H 2 onto H(b). Let A λ be an operator on H 2 that intertwines T 1−λb T F λ with the operator Y * , i.e., The operator A λ is given by It follows from the proof of Sarason's theorem that if one of conditions (i)-(iii) holds true, then the space ker(A λ −z 0 ) is one dimensional and is spanned by the function g(z) = F λ (z) 1 −z 0 z = F λ (z)k z 0 (z).
We also note that condition (iii) in Sarason's Theorem is equivalent to the fact that the function k b z 0 given by (9) is in H(b).