Abstract
This article examines the norms of composition operators from the weighted harmonic Bloch space to the weighted harmonic Zygmund space . The critical norm is on the open unit disk. We first give necessary and sufficient conditions where the composition operator between and is bounded. Secondly, we will study the compactness case of the composition operator between and . Finally, we will estimate the essential norms of the composition operator between and .
1. Introduction
Operator theory for spaces of analytic functions has been described in various settings, and there is a rich volume of studies in the academic literature that focus on the operator theory of spaces related to analytic functions on the unit disk. These studies delve into diverse environments, and the references will be highlighted below.
In [1], the second author discusses the essential norms of Stević-Sharma operators from general Banach spaces into Zygmund-type spaces, and in [2], the authors characterize the bounded and compact Stević-Sharma operator from a general class of Banach function spaces into Zygmund-type space. In [3], the authors show a new essential norm estimate of composition operators from weighted Bloch space into -Bloch spaces. Cowen and MacCluer in [4] investigated composition operators on spaces of analytic functions. In [5], the necessary and sufficient conditions for the compactness and boundedness of product operator from to Zygmund spaces were characterized.
Yet, there is a noticeable lack of investigations that offer a comprehensive look into the harmonic setting. We would like to highlight several of these references below. Characterization composition operators on some Banach spaces of harmonic mappings were discussed in [6]. Colonna in [7] discussed the Bloch constant of bounded harmonic mappings. Lusky studied the weighted spaces of harmonic and holomorphic functions in [8] and then in [9] determined the isomorphism classes of weighted spaces of harmonic and analytic functions. Characterization of the harmonic Bloch space and the harmonic Besov spaces by an oscillation in [10]. Jordá and Zarco studied the weighted Banach spaces of harmonic functions and the isomorphisms on weighted Banach spaces of harmonic and holomorphic functions in [11, 12].
This paper is part of a series of works that address several different properties of composition operators between weighted Banach spaces of harmonic mappings. We discussed the boundedness, compactness, and the essential norm of composition operators from the space of bounded harmonic mappings into the harmonic Zygmund space in [13]. Bakhit et al. in [14] discussed the boundedness, compactness, and the essential norm of composition operators from harmonic Lipschitz space into .
A harmonic mapping is a complex-valued function with simply connected domain such that
Here, let be the space containing all analytic functions on the unit disk and be the space of harmonic mappings, while be a compact subset of the unit disk . Further, let be the space of all bounded mappings equipped with the norm
The harmonic mapping always can invariably be represented in the form , where both . Up to an additive constant, this representation attains uniqueness. For the scope of our study, we will focus on harmonic mappings with the domain . Therefore,
See [15], as an excellent reference on the harmonic function theory.
The composition operator induced by analytic self-maps (or conjugate analytic self-maps) can be expressed as
Surely, this operator preserves harmonicity (see [6]).
In this work, we begin with some preliminaries that we use to derive the main results. We continue our research in [13, 14] by focusing on the boundedness and the compactness of from harmonic -Bloch space into the weighted harmonic Zygmund space . We conclude by estimating the essential norm from into .
Let and be two normed linear spaces. Then, the linear operator is bounded if there exists a positive constant such that
Further, the operator is compact if every bounded set in whose closure is compact, while the essential norm of is its distance from the compact operators in the operator norm. Then, the essential norm of is given by , where is a compact operator.
1.1. The Harmonic -Bloch Space
For , the harmonic -Bloch space contains all which is defined such that
If is represented as , with , the harmonic -Bloch seminorm can be characterized as
The quantity gives a Banach space structure on (see [16]).
The little harmonic -Bloch space is considered as the subspace of consisting of such that
1.2. The Weighted Harmonic Zygmund Space
For consists of all mappings such that
Define
Obviously, is a norm on , and is a Banach space. For , is with the harmonic Zygmund space (see [13]).
Remark 1. Let ; then, is simplified to and . Thus, for all , the collection of all in is the classical -Zygmund space , and both norms are identical.
For , let be represented as , with . For given , let us define
The following lemma will help to characterize the boundedness of . Its proof is the proof of Theorem 19 in [16].
Lemma 2. Given and , let be represented as , with . Then, (1) if and only if (2) if and only if
Let be a fixed point, and let . For any , we consider the test functions defined as
Moreover, it is evident that uniformly on . Recall that the power series representation of is
For all and , by direct calculations, we know that
Then, we have
As before, for all ,
Then, we have
Thus, it can be demonstrated that and , for every .
Throughout this article, the notation means that , where is a constant. Therefore, the notation means that and are equivalent, when .
2. Boundedness
In this section, we work on the boundedness of the operator .
Theorem 3. Let and let . Then, is bounded if and only if
Proof. Let the sequence , for and . The sequence is bounded in with supremum norms (the authors in Theorem 2.9 of [17] have demonstrated that ). If is bounded, then for each and , we have
Therefore,
Conversely, suppose that (20) holds and set
Since , we see that
For any and represented as , with , note that . Therefore, because , we see that
For any and , we note that
Now, multiplying the above expressions (26) and (27) by , we have
By Lemma 2, we know that
Since can be expressed as , with , we obtain
where and . To prove that is a bounded operator, it suffices to show that both quantities and are finite. For since , we have
Then,
Moreover, we know that ,
For since , so we have
Thus,
By the linearity of the test functions in (14), for and , we have
From (16), for and , we obtain
Next, for , we let
By equation (38), for , we obtain
Moreover,
Thus, from (39), we obtain
Moreover, from (40), we obtain
Now we let ; then, if in (41), we have
Conversely, if we let in (35), we have
From (43) and (44), it follows that the quantity is finite.
Similarly, for , we let . Then, if in (42), we have
If we let in (32), we have
Therefore, the quantity is finite, and the proof is complete.
3. Compactness
In this section, we focus on discussing the compactness of the operator . The proof of the following lemma is a slight modification of the proof of Proposition 3.11 in [4] (the case of Banach spaces of analytic functions).
Lemma 4. Let be bounded operator; then, is compact if and only if as , for any bounded sequence in converges to zero uniformly on .
The following theorem shows that the compactness of can be characterized in terms of the sequence , where , for and when .
Theorem 5. Let be bounded operator, where . Then, is compact if and only if
Proof. First, we consider the sequence , for and . Since the sequence is bounded in and converges to zero uniformly on , if is compact, then is a bounded operator, and (47) holds by Lemma 4.
On the other hand, assume that is a bounded operator and
Now, we define a sequence in with , and uniformly on , as .
By Lemma 4, to prove that is compact, it is sufficient to show that
Next we suppose ( is an upper bound for ). For , then there is such that
For and , let us use (14) the power series representation of the test function ; then, we have
Moreover,
Next, for any , let be sufficiently close to such that ; thus,
For , since is arbitrary, so it follows that
From (41), we know
Further, from (42), we know
Using (55), we obtain
Moreover, by (56), we obtain
Thus, sufficiently close to if , for any . Then,
For any and , by using Lemma 2, if , then we have . From (32) and (35), we know
From (32) and (35) in the proof of Theorem 2, we know
We know that from Cauchy’s estimates, all the sequences , , , and are convergent to zero on compact subsets of the unit disk . Thus, using (61), for any if , we obtain
which implies that
Therefore, and . Thus, we obtain
From Lemma 4, we verify that is compact.
4. Essential Norm
In this section, our emphasis shifts to a comprehensive discussion regarding the essential norms of the operator . First, we define
Theorem 6. For , let be bounded operator. Then,
Proof. First, for and , by the test function (13), we will prove that Fix , since for all and converges uniformly to on . Then, for a compact operator , we have Thus, Hence, we obtain Next, to prove that , we define the sequence such that , for . We also define For all , it can be seen that and uniformly on . Moreover, by simple calculation, we have Since is a compact operator, by Lemma 4, we have Similarly, we have Thus, Hence, we obtain Secondly, we prove that Now, we consider the operator , for any such that Without a doubt, uniformly on , as . Moreover, is compact on and For any sequence such that as , we obtain But the definition of the essential norm says Thus, we only need to demonstrate that Let such that ; then, It is clear that On the other hand, we consider Now, we let be large enough and , for all . Then, Since is bounded, from Theorem 2, we see that Moreover, all the limits, are uniformly on . Then, we have Hence, by the above equations, we have Next, assume , and we have To find estimates of the quantities , and , we define Because and , for all and and by Lemma 2, we have Consequently, Similarly, we see that By direct calculation, , for all . Because , Thus, we obtain Similarly, we see that By the inequalities (94)–(98), we obtain Hence, by applying (90) and (99), we determine that Finally, we prove that Now, we only need to prove that From (93), we see that Similarly, Further, for (96), we see that Similarly, Therefore, by the inequalities (103)–(106), we get The proof now is complete.
Theorem 7. For , let be bounded. Then,
Proof. First, we prove that Recall that the sequence , for and when . Then, , and converges uniformly to 0 on . Therefore, by Lemma 4, we see that Hence, Therefore, Next, we prove that Since is bounded, then by Theorem 2 Now assume the test function with in (14), for . By linearity of the composition operator, for any fixed positive integer , we obtain Then, for all positive integer and , we get Hence, Since is bounded, then we have By (112) and (118), we fulfilled the desired result.
Data Availability
The research conducted in this paper does not make use of separate data.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
The authors extend their appreciation to the Deanship of Scientific Research at Northern Border University, Arar, KSA, for funding this research work through the project number NBU-FFR-2024-2182-01.