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 $H^2$. In particular, we discuss the kernels of the operators $T_{\bar f/ f}$ and $T_{\bar I\bar f/ f}$, where $f$ is an outer function in $H^2$ and $I$ is inner such that $I(0)=0$. We also obtain results on 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 + (ϕf ), 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 (T), 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 Recall that a function f ∈ H 1 is called rigid if and only if no other functions in H 1 , except for positive scalar multiples of f have the same argument as f a.e. on ∂D.
It is known 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 ⊖ IH 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 1−Ib 0 , where (b 0 , a) is a special pair and a 1−b 0 2 is a rigid function in H 1 . Then we also have ker TĪf f = f K I . In the recent paper [5] the authors considered the Toeplitz operator Tḡ g where g ∈ H ∞ is outer. Among other results, they described all outer functions g such that ker Tḡ g = K I . In Section 2 we describe all such functions g for which ker Tḡ and get a generalization of results obtained in [13] and [4] for the case when pairs are rational.
2. 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. [16]).
The finite dimensional kernels of Toeplitz operators were described by Nakazi [15]. 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 [18, 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 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 [16] 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. Under the above notations Hayashi's theorem reads as follows: In the above mentioned paper [5] the authors described all outer functions g ∈ H ∞ for which ker Tḡ 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 [5], we use the following result due to H. Helson [11].
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 [8] (see also [6]). For an outer function f in H 2 the weighted Hardy space is defined as follows and, for an inner function I, K 2 I (|f | 2 ) = K I (|f | 2 ) is given by [8]).
Proof of Theorem 1. Assume that ker Tḡ g = f K I . Then Since f ∈ ker TĪf f , the last equalities imply that 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) = cf (z), where c is a unimodular constant. Replacing cI 0 by I 0 , we get It then follows f K I = ker Tḡ 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 h = f I 0 ψ 0 , where ψ 0 ∈ H 2 0 (|f | 2 ), if and only if h = f Iψ, where ψ ∈ H 2 0 . 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 is an isometry of H 2 (µ) onto H(b) ( [19], [3]). Furthermore, if (b, a) is a pair and f = a 1−b , then the operator T 1−b Tf is an isometry of H 2 into H(b). Its range is all of H(b) if and only if the pair (b, a) is special.
Proof of Theorem 2. Since the pair (b, a) is special, the operator T 1−b Tf is an isometry of H 2 onto H(b). Moreover, since I divides b, T f acts as an isometry on K I and T 1−b Tf (f K I ) = K I [18]. Hence where TIf  pair (b 0 , a) is special but f 2 0 is not rigid, and describe the space ker TĪf Our example is a slight modification of the one given in [17], see also [3, vol. 2, p. 494]. The corresponding functions f and f 0 are defined by taking a(z) = 1 2 (1 + z), b 0 (z) = 1 2 z(1−z), and I(z) = zB(z), where B(z) is a Blaschke produkt with zero sequence {r n } ∞ n=1 lying in (−1, 0) and converging to −1. It has been proved in [17] (see also [3, 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 [13, Thm. 4.1.] (see also [4]) that M(a) is a closed subspace of H(b 0 ) and Thus we see that Our aim is to prove that (7) ker TĪf where the function g ∈ H 2 is given by g = f k −1 (I + 1), with k −1 (z) = (1 + z) −1 , z ∈ D.
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 Since I(r n ) = 0, (i) and (ii) in the Lemma yield Next, since T I : H(b 0 ) → H(Ib 0 ) = H(b) is an isometry and T 1−b Tf is an isometry of H 2 onto H(b), we see that {f k rn (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 rn (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 Remark. One can check directly that the function g = f k −1 (I + 1) is in ker TĪf

Indeed, we have
TĪf f (f k −1 (I + 1)) = P + fĪ I + 1 1 + z = P + fz (Ī + 1) To see that the functions (f k rn I − b 0 (r n )f k rn ) are orthogonal to f K I note that a function h ∈ H 2 is in K I if and only if h = h − IP + (Īh). So, we have to check that for any h ∈ H 2 , f k rn I − b 0 (r n )f k rn , f (h − IP + (Īh)) = 0.
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. Let H 0 (b) denote the orthogonal complement of M(a) in 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). Necessary and sufficient conditions for H 0 (b) to have finite dimension are given in Chapter X of [19]. The space H 0 (b) depends on the spectrum of the restriction of the operator Y * to H 0 (b) which actually equals Y * 0 . The spectrum of Y * 0 is contained in the unit circle. More exactly, if |z 0 | = 1 and k is a positive integer, then ker(Y −z 0 ) k ⊂ H 0 (b) and the dimension of H 0 (b) is N if and only if the operator Y * 0 has eigenvalues z 1 , z 2 , . . . , z s with their algebraic multiplicities n 1 , . . . , n s and N = n 1 + n 2 + · · · + n s .
In this section we consider the case when the eigenspaces correspondings to eigenvalues z 1 , z 2 , . . . , z s are one dimensional.
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 .
The following theorem is proved in Chapter X of [19].
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 .
The space H 0 (b) is described by means of an operator A λ 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 (10) g(z) = F λ (z) 1 − z 0 z = F λ (z)k z 0 (z).
We also note that since b has the angular derivative in the sense of Carathéodory at z 0 , the function where b(z 0 ) is the nontangential limit of b in z 0 , is in H(b). Using Sarason's theorem cited above we show the following