Abstract
We first explore conditions under which every weighted composition-differentiation operator on the Hardy space is completely continuous. We then discuss necessary and sufficient conditions for these operators to be Hilbert–Schmidt on the derivative Hardy space .
1. Introduction
Let be a Banach space consisting of analytic functions on a domain in the complex plane. For simplicity, we shall assume that the underlying domain is the unit disk.
The boundary of the unit disk will be denoted by
Let be an analytic self-mapping on the unit disk; this means that is an analytic function that maps the unit disk into the unit disk. The composition operator with symbol , denoted by , is defined as follows:
This operator was first introduced by Nordgren [1] in the framework of the classical Hardy space . Assuming that the symbol function is an inner function (an analytic self-map of the unit disk whose radial limit equals 1 almost everywhere on ), Nordgren studied the boundedness of , computed its norm, described the spectrum of the composition operator, and finally related properties of this operator to the existence of fixed points of the Poisson integral of . It turned out that this operator plays a prominent role in the operator theory of function spaces. There are a bulk of papers on this topic, and to mention just a few, we should cite the papers [2–10] and the references therein; see also the books written on this topic [11, 12]. Among other properties, several authors have obtained conditions that ensure compactness, Fredholmness, and Hilbert–Schmidtness of the composition operator. It is well known that the composition operator is bounded on the Hardy space and on the Bergman space where is a positive number (see [12]).
For an analytic function , the weighted composition operator is defined by
Similarly, we can define the composition-differentiation operator by
Continuing in this way, we define the weighted composition-differentiation operator as follows:
We should mention that in most cases, the functional Banach space equals either the Hardy space or the Bergman space for . According to [[13], Corollary 3.2], for a univalent self-map of the unit disk, the operator on the Hardy space is bounded if and only if
Moreover, is compact on if and only if
The weighted composition-differentiation operator was recently studied in [2, 3]. In [2], the authors found necessary and sufficient conditions for to be Hilbert–Schmidt both on the Hardy space and on the Bergman space. We also consider the operator byand investigate its complete continuity as well as its Hilbert–Schmidtness on the derivative Hardy space .
In this paper, we first focus on the nonreflexive Hardy space and try to find conditions under which the weighted composition-differentiation operator is completely continuous. We shall provide characterizations for the complete continuity of this operator in terms of the boundary behaviors of and . More precisely, we prove that is completely continuous if and only if almost everywhere in . Similar problem will be solved for the operator (Theorems 1 and 2).
The second topic to investigate is the Hilbert–Schmidt operators on a closed subspace of ; that is, the derivative Hardy space consisting of all analytic functions in the unit disk for which or
We will prove that is Hilbert–Schmidt on if and only if
Similar problem for the operator will also be discussed (Theorems 3 and 4).
2. Preliminaries
An analytic function on the unit disk is said to belong to the Hardy space if
For , the Hardy space is a Banach space of analytic functions, and for , it is a Hilbert space with the following inner product:whereis the boundary function of (for the existence of boundary function , see [14], Chap. 17). It is easy to see that for with Taylor series , the norm of is given by
The next space we study is the space of all analytic functions in the unit disk for which or
The space is called the derivative Hardy space. The norm in is defined by the following relation:
It is clear that is a closed subspace of , and if , is a Hilbert space of analytic functions. A computation shows that for , we have
This space equipped with the above norm was studied by Korenblum in 1972 [15]. We should mention that many authors use the expressionfor the norm of (see for instance [16,17]). We shall use this norm in §3.
Let be a Banach space. An operator is compact if for every bounded sequence in , has a convergent subsequence. On the other hand, the operator is called completely continuous if the weak convergence of in implies
It is well known that every compact operator is completely continuous. We remark that for , the Hardy space is reflexive, meaning that it is isometrically isomorphic with its dual. It is known that on reflexive Banach spaces, an operator is compact if and only if it is completely continuous. In this paper, we concentrate on the nonreflexive Banach space and the composition-differentiation operator on . We will find conditions on the functions and to ensure that the operator is completely continuous on . We also consider and investigate its complete continuity (see Theorems 1 and 2).
The second problem to be discussed is the conditions under which the abovementioned differentiation operators are Hilbert–Schmidt. We recall that if is a separable Hilbert space and if is an operator on , then is said to be Hilbert–Schmidt provided thatwhere is an orthonormal basis in . In the next section, we take and try to find the condition for the operators and to be Hilbert–Schmidt.
3. Main Results
In the following theorem, we shall characterize the complete continuity of the composition-differentiation operator in terms of and .
Theorem 1. Let and be a self-map on . Assume that is bounded on . Then, is completely continuous on if and only if almost everywhere in .
Proof. Let be completely continuous, and let denote the unit circle. Assume that and let be its -th Fourier coefficient. By the Riemann–Lebesgue lemma, we havewhere is the normalized arc-length measure on . This means that converges to zero weakly in and hence weakly in . Since is completely continuous, it follows thatOn the other hand, for each ,Therefore, the integral on the left-hand side must be zero, from which it follows that almost everywhere in .
Conversely, let be a weak null sequence in . It follows that uniformly on compact subsets of . Using this fact together with the assumption that almost everywhere in , we conclude thatIt now follows that converges to zero in measure in (see [18], page 74). Moreover, the boundedness of on implies that in the weak topology of and hence in the weak topology of . Finally, we invoke the fact that weak convergence of a given sequence together with its convergence in measure implies its norm convergence (see [19], page 295), that is, as .
Theorem 2. Let and be a self-map on . Assume that is bounded on . Then, is completely continuous on if and only if almost everywhere in
Proof. Let be completely continuous, and let denote the unit circle. As in the proof of Theorem 1, we use the Riemann–Lebesgue lemma to conclude that converges to zero weakly in and hence weakly in . Since is completely continuous, it follows thatOn the other hand, on , we have , so that for each ,Therefore, the integral on the left-hand side must be zero, from which it follows that almost everywhere in .
Conversely, let be a sequence in that converges to zero weakly. It follows that uniformly on compact subsets of . Using this fact together with the assumption that almost everywhere in , we conclude thatThis implies that converges to zero in measure in (see [18], page 74). Moreover, the boundedness of on implies that in the weak topology of and hence in the weak topology of . Again, we recall that weak convergence together with convergence in measure implies norm convergence (see [7], page 295); therefore, as .
In the following theorems, we discuss the conditions under which the composition-differentiation operator on is Hilbert–Schmidt. We consider the following norm:It is easy to verify that with respect to this norm, the vectors form an orthonormal basis for .
Theorem 3. Let be a self-map on . Then, is Hilbert–Schmidt on if and only if
Proof. Assume that condition (1) holds. We first note that for ,Since , the series converges, so that the seriesconverges if and only ifIt is easy to see that for , we haveThis implies thatwhich means that is Hilbert–Schmidt on .
On the other hand, assume that is Hilbert–Schmidt on . By the above computations and equality (2), we havewhich is the desired result.
Corollary 4. Let be a self-map on and be an analytic function on . Then, is Hilbert–Schmidt on if and only if
Proof. The proof is similar to that of the preceding theorem.
Theorem 5. Let be a self-map on . Then, is Hilbert–Schmidt on if
Proof. First, note thatSince , the seriesconverges, so that the seriesconverges if and only ifOn the other hand,where for the last term, we used the identitySince , it follows from (44) that is Hilbert–Schmidt if (39) holds.
Theorem 6. Let be a self-map on such that is Hilbert–Schmidt on . Then,
Proof. Assume that is Hilbert–Schmidt. We note thatfrom which the result follows.
Data Availability
Data sharing is not applicable to this article as no data sets were generated or analyzed during the current study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Authors’ Contributions
All the authors have contributed equally to this work.