HYPONORMALITY OF TOEPLITZ OPERATORS ON THE BERGMAN SPACE OF AN ANNULUS

A bounded operator S on a Hilbert space is hyponormal if S∗S− SS∗ is positive. In this work we find necessary conditions for the hyponormality of the Toeplitz operator Tf+g on the Bergman space of the annulus {1/2 < |z| < 1}, where f and g are analytic and f satisfies a smoothness condition.


Introduction
A bounded operator S on a Hilbert space is hyponormal if S * S − SS * is positive. Hyponormality of Toeplitz operators has been studied by many authors. Hyponormality of these operators on the Hardy space was considered in [3,4]. Hyponormality of these operators with a symbol of the form g 1 + g 2 on the Bergman space of the unit disk was first considered in [8]. Therein a necessary condition was proved, which was later improved in [1]. Some special cases are treated in [7]. A sufficient condition when g 1 is a monomial and g 2 is a polynomial is proved in [9]. An improvement of the necessary condition in the case when g 1 and g 2 are binomials is given in [5]. Basic material on Toeplitz operators on the Bergman space of the unit disk can be found in [2]. In this work we consider hyponormality of Toeplitz operators on the Bergman space of an annulus.

HOUCINE SADRAOUI AND MOHAMMED GUEDIRI
A 2 1/2 , and h is in A 2 1/2 . The Hankel operators on the space A 2 1/2 are defined by H f (h) = (I − P )(hf ). The space A 2 1/2 has an orthonormal basis given by the union of the sets e n = 3(n + 1) , and We consider hyponormality of Toeplitz operators with a symbol of the form f = g 1 + g 2 , where g 1 and g 2 are bounded analytic functions on C 1/2 . We begin by recalling some known properties of Toeplitz operators.

Some basic properties
Lemma 2.1. Let f and g be bounded and measurable on C 1/2 . The following properties hold: The next proposition is easy to prove and its proof is omitted. Proposition 2.2. Let g 1 and g 2 be polynomials. The following are equivalent: where K is an operator of norm less than one.
The following lemma provides computations that will be needed. Lemma 2.3. The projection P on A 2 1/2 satisfies the following relations:

First main result
We begin with a matrix computation.

Proof.
We have .

HOUCINE SADRAOUI AND MOHAMMED GUEDIRI
Similarly, we get Proof. An elementary computation shows that The first sum in the above expression of β i+p,i can be written as h i (k) dµ(k), where dµ is the counting measure. It is easy to see that for i sufficiently large, is integrable with respect to the counting measure.
By the dominated convergence theorem we obtain: Also, for i large, there exists a constant C such that Thus lim i→∞ i 2 a i+1 a i+p+1 R i,p = 0. Finally, it is not difficult to see that i 2 |S i,k,p | ≤ k(k + p). Using the dominated convergence theorem we obtain We deduce that lim i→∞ i 2 β i+p,i = k(k + p)a k a k+p and recognize this last limit as being equal to γ i+p,i , where (γ i,j ) is the matrix of the Hardy space Toeplitz operator T |f | 2 .
We are led to the following necessary condition for hyponormality.
If T f +g is hyponormal then g ∈ H 2 and |g | ≤ |f | a.e. on the unit circle.
We deduce that 1≤k≤i, i 2 |b k | 2 Q i,k,0 ≤ i 2 β i,i . Since lim i→∞ i 2 Q i,k,0 = k 2 , writing the left hand side of this last inequality as an integral with respect to the counting measure and using Fatou's lemma we get k 2 |b k | 2 ≤ k 2 |a k | and g ∈ H 2 . From the previous lemma, lim i→∞ i 2 θ i+p,i = λ i+p,i , where (λ i,j ) denotes the matrix of the Hardy space Toeplitz operator T |f | 2 −|g | 2 . Hyponormality and a property of Toeplitz matrices [6] lead to |g | ≤ |f | a.e. on the unit circle. Proof. Only the necessary condition needs to be shown. Normality implies that |g | = |f | on the unit circle. Thus f and g have the same finite number of zeros (if any) with the same multiplicity. We thus have |f | |g | = |g | |f | = 1 on the unit circle. By the maximum principle, g = cf with |c| = 1. We get g = cf .

HOUCINE SADRAOUI AND MOHAMMED GUEDIRI
Proof. We have Similarly, , e −i and denote by (ζ i,j ) the matrix of the Toeplitz operator T |f 1/2 | 2 on the Hardy space of the unit disk, where f 1/2 (z) = a k z k 2 k . We can show the following lemma.
A computation shows that lim i→∞ i 2 Q i,p,k = 1 2 2k+p . As in the proof of the previous theorem we can show that We see that this last limit is equal to ζ i,i+p . We also show that We deduce that We can also see that |g 1/2 | ≤ |f 1/2 | a.e. on the unit circle is equivalent to |g | ≤ |f | a.e. on {z : |z| = 1/2}.

Second main result
We now put f = Proof. We have a l a k 3(i + 1) (j + 1) a k a l 3(i + 1) We get, using the same notations as before, A computation shows that lim i→∞ i 2 β i+p,i = 1≤k,k−p a k a k−p k(k − p).
We recognize the general element ξ m+p,m of the matrix of the Toeplitz operator T | f | on the Hardy space of the unit disk with f defined by f (z) = ∞ 1 a k z k . Obviously the condition | g (e iθ )| ≤ | f (e iθ )| a.e. on the unit circle is the same as |g | ≤ |f | a.e. on the unit circle. The condition f ∈ H 2 is equivalent to k 2 |a k | 2 < ∞ and this is satisfied if f = ∞ 1 a k 1 z k is bounded on C 1/2 . Using similar methods we obtain the following theorem. If we set f 2 (z) = 2 k a k z k , then f 2 ∈ H 2 is equivalent to k 2 2 2k |a k | 2 < ∞. In this case, |g 2 | ≤ |f 2 | a.e. on the unit circle is equivalent to |g | ≤ |f | a.e. on {z : |z| = 1/2}. Let (ρ i,j ) denote the matrix of the Hardy space Toeplitz operator T |f 2 | 2 . Using the same notations we can show the following lemma, the proof of which is omitted. We obtain our second main result.