Research Article | Open Access
Houcine Sadraoui, Borhen Halouani, Mubariz T. Garayev, Adel AlShehri, "Hyponormality on a Weighted Bergman Space", Journal of Function Spaces, vol. 2020, Article ID 8398012, 7 pages, 2020. https://doi.org/10.1155/2020/8398012
Hyponormality on a Weighted Bergman Space
A bounded Hilbert space operator is hyponormal if is a positive operator. We consider the hyponormality of Toeplitz operators on a weighted Bergman space. We find a necessary condition for hyponormality in the case of a symbol of the form where and are bounded analytic functions on the unit disk. We then find sufficient conditions when is a monomial.
Let denote the disk of radius in the complex plane, where is the Lebesgue measure on and . denotes the Hilbert space of complex-valued functions on that are square integrable with respect to . We write . When is analytic on , we have
Denote by the space of analytic functions on such that . It is known that is a Hilbert space , and an orthonormal basis is given by . The Toeplitz operator with symbol on is defined by where is bounded and measurable on , is in , and is the orthogonal projection of onto . Hankel operators are defined by , and as before. Recall that a bounded operator on a Hilbert space is hyponormal if is a positive operator. Hyponormality on the Hardy space is studied by Cowen in [2, 3]. Unweighted Bergman spaces on the unit disk and Toeplitz operators on these spaces were considered in [4, 5] and . Hyponormality was first considered by Sadraoui in . An improvement of the necessary condition is due to Ahern and Cuckovic . A new necessary condition in a special case is found by Cuckovic and Curto in . Many results on hyponormality on weighted Bergman spaces treat special cases. We cite for example [1, 10]. In this work for simplicity’s sake, we consider the case and . Under a smoothness assumption, we give a fairly general necessary condition for the hyponormality of Toeplitz operators with a symbol of the form where and are bounded and analytic. We also give sufficient conditions when is a monomial and a polynomial without assuming .
We begin by recalling some general properties of Toeplitz operators.
2. Some General Properties
We assume , are in . Then, we have: (1)(2)(3) if or analytic on
The use of these properties leads to describing hyponormality in more than one form. These are easy to prove by using Douglas lemma .
Proposition 1. Let be bounded and analytic on . Then, the following are equivalent: [label=)] (i) is hyponormal(ii)(iii) for any in (iv) for any in (v) where is of norm less than or equal to one
We also need the following lemmas.
Lemma 2. For and integers, we have
Proof. We have
We see that if . For , if and only if . In this case, we get
On the other hand, if , then so . A computation shows so , and the result follows.
Lemma 3. If is bounded analytic on the disk , then the th entry of the matrix of with respect to the orthonormal basis is given by:
Proof. We have:
Thus, by Lemma 2,
Thus, noting that if , we get
We also have by Lemma 2
The above inner product is nonzero if . We get by Lemma 2
When , we set . We obtain in this case the following corollary.
Corollary 4. If is bounded and analytic on the unit disk , and if . Then, the matrix of with respect to the orthonormal basis
is given by
Recall that the Hardy space is defined by
In the following theorem, the Toeplitz operator on is not necessarily bounded. However, its matrix in the usual orthonormal basis of is defined.
Theorem 5. Assume bounded analytic on the unit disk and . Let be the matrix of the Toeplitz operator on . Then, as .
Proof. The symmetric matrix of with respect to the orthonormal basis is given by (set in Corollary 4 with .)
To simplify, set . Then, for any integer . Set in the first sum, and in the second. With some obvious notations, we get
A computation shows that
It is not difficult to see that and that the following inequality holds: for . We have that since . Writing the sum as an integral, where the measure is the counting measure, and using the dominated convergence theorem, we get that
Clearly, . So we have for , for . Thus,
This shows that
We deduce that .
3. The Results
Theorem 6. Let and be bounded and analytic on the unit disk , and such that . If is hyponormal on then , and a.e. on the unit circle.
Proof. Set . All diagonal terms of the matrix of are positive; thus, using the same notation as in the previous proof, we have
We have seen that the left hand side tends to , a finite sum since . Writing as an integral with respect to the counting measure, noticing that , and using Fatou’s lemma, we deduce that and . If denotes the matrix of , then , where is the matrix of the Toeplitz operator on , . Since the matrix of is positive by hypothesis, it follows that the matrix of is also positive. A Toeplitz matrix is positive if its symbol is positive . It follows that on the unit circle.
To obtain sufficient conditions when is a monomial, we need the following lemma.
Lemma 7. Let be a positive integer, the matrix of is diagonal and given by:
In what follows, we will take . It follows from Proposition 1 that, if , the hyponormality of is equivalent to the following inequalities:
Lemma 8. The inequality (28) holds if and only if .
Proof. Set . Then, (28) is equivalent to . Using logarithmic differentiation, it is easy to see that decreases with , the minimum is thus assumed at ; hence, the result follows.
Lemma 9. The inequality (29) holds if and only if
Proof. A computation shows that inequality (29) is equivalent to
Using logarithmic differentiation, we can see that the right hand side of the last inequality is a decreasing function of . Thus, the minimum is assumed at , and this proves the lemma.
A similar argument shows the following.
Lemma 10. The inequality (30) holds if and only if .
Proposition 11. If , then is hyponormal if and only if .
Remark 12. The result remains true if .
Denote by the unit ball of .
Definition 13. For , we denote by the set
We see, from the density of in and Proposition 1, that when and are in , is equivalent to is hyponormal. The following proposition lists some properties of . We provide the proof for the sake of completeness.
Proposition 14. For the following holds: [label=()] (a) is convex and balanced(b)If , then is in for any complex number (c).(d) is closed in the weak topology of
Proof. Only (d) needs to be verified. Assume is a net in and . We have for in , and . The result follows by taking the limit and taking the supremum on .
Using the two previous propositions, we obtain the following.
Theorem 15. Let be a sequence of complex numbers satisfying where is a positive integer. Then, is hyponormal.
We now consider the hyponormality of when . In this case, hyponormality is equivalent to the following three inequalities:
A computation, using logarithmic differentiation, allows us to see that the right hand side of each of these inequalities increases with . This leads to the following inequalities:
Since increases with , we get
Also, . Thus, for . We have and we can readily verify that for , we have
As before, increases with if . We get which leads to the inequality
Combining the last inequality with inequality (40), we deduce the following:
Proposition 16. If then is hyponormal if and only if .
Remark 17. The result holds also if .
From the properties of , noted in Proposition 14, we deduce the following corollaries.
Corollary 18. Let be complex numbers satisfying . Then, is hyponormal.
Corollary 19. Let be a sequence of complex numbers satisfying . Then, is hyponormal.
No data were used to support this study.
Conflicts of Interest
The authors declare that are no conicts of interest regarding the publication of this paper.
The authors would like to extend their sincere appreciation to the Deanship of scientic research at King Saud University for its funding of this work through Research Group Project no RG-1439-68.
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