#### Abstract

In this paper, we attempt to investigate a new harmonic mapping class denoted by the Banach space in the unit disk. Several characteristics of the class are examined. We additionally describe some findings for the little harmonic space, which really is a closed subspace of . We also talk about the boundedness and compactness of the composition operator in the space.

#### 1. Introduction

A complex-valued function of is said to be a harmonic mapping of an open subset of the complex plane provided that its related Laplace equation is satisfied on . That is,where stands for the mixed complex second partial derivative of h.

A harmonic mapping in the unit disk admits the canonical decomposition where and are analytic functions, and . Analytic functions are harmonic mappings because their real and imaginary components are harmonic according to the Cauchy-Riemann equations. We shall assume that all the functions under consideration are defined on . In the complex plane , be its boundary. Let denote the normalized Lebesgue area measure on , and indicate the normalized Lebesgue area measure on . signifies the family of all holomorphic functions on . signifies the family of all harmonic functions on . Let be the Möbius transformations be defined by for each .

It is worth noting that and therefore . It has the following useful property:

We look at the linear structure of the harmonic equivalents of the -Bloch spaces and the space in particular. We also apply to the harmonic setting and certain well-known features that apply to the analytic case.

A bounded linear operator is said to be bounded below if there exists , such that for all .

Remark 1. A closed range operator is a linear operator that is bounded below (see Proposition 6.4 in ). If the operator is injective, the opposite is true and then the converse also holds (see Theorem 5.17.2 in ).

The operator is defined on the space of complex-valued functions with domain , given an analytic self-map of , and the composition operator induced by is defined as the operator:

It is obvious that such an operator preserves harmonic mappings immediately. Assuming that analytic functions are clearly harmonic, the idea of how to extend the harmonic mappings is interesting. The norm on the larger space fits with the Banach space structures of known spaces of analytic functions . When confined to the elements of , the norm is .

The organization of this work will be as follows: we dedicate Section 2 to study some basic facts about the harmonic spaces and the concept of -spaces. In Section 3, we performed characterization of boundedness and compactness of a composition operator of for each of these spaces.

We will use the notation to indicate that the quantities are comparable; that is, there exist two positive constants that satisfy . Similarly, if only the first or second inequality holds, we say or .

#### 2. Harmonic Spaces

In this section, we present some useful properties and auxiliary results to discuss .

For analytic in , the Bloch constant of is

We remember that an analytic function belongs to , if and only if (see ) with norm

Colonna (see ) shows that the Bloch constant of can be expressed in terms of the moduli of the derivatives of and as follows:

Colonna (see ) presented the following theorem.

Theorem 1. With h as above, , if and only if and are in . Furthermore,

For , the harmonic -Bloch space is the collection of all such that , where

The harmonic little -Bloch space is defined as the subspace of consisting of the mappings such that

This space is a continuation of Zhu’s harmonic mappings of the (analytic) -Bloch space . Thus, representing as with and , we can see that and . Hence, , and

Consequently, a harmonic mapping is associated with , if and only if the unique functions and are analytic on and with and are in the space (see [2, 5] for additional details on the spaces ). The space is the classical Bloch space for , and the appropriate harmonic extension will be represented by . Automorphisms of the unit disk are of the form

In particular, if , then we obtain the involution automorphism that exchanges 0 and .

The Green function of with pole at is denoted by . The pseudohyperbolic disk with a (pseudohyperbolic) center and a (pseudohyperbolic) radius is represented by .

Now, we introduce the concept of space which generalizes the concept of space as follows.

Definition 1. For , a function is said to be in the class ifThe norm of is identified asIt is obvious that the correspondence is ongoing is a norm on , which we refer to as the harmonic .

Because the Green function is conformally invariant, it can be used in a variety of applications for all and , and each class is almost self-evident. is conformally invariant in the following sense: if , then for all .

Remark 2. reduces to when is an analytic function, and . Thus, the analytic space is the set of analytic functions in the harmonic space, and the respective norms coincide. However, due to the presence of the mixed product resulting from squaring the quantity , this choice of the norm does not allow a modification that may lead to a (complex) inner product in this situation. As a result, we will propose an alternative norm on that solves the problem while also extending the analytic norm. We will see that the above-mentioned standard is, in fact, valid. Both the old and new norms are Möbius invariant.

Notice that, for the proof of the main results, we need the following lemma.

Lemma 1. Let , and . Then, , if and only if

proof 1. Proof. By the inequalities ; as well asit suffices to showBecause is a nondecreasing function of , one haswhich impliesThis, together with equation (17), applies to , and henceWe use the following lemma due to Songxiao .

Lemma 2. Let . Then,(1)On compact sets, every bounded sequence in is uniformly bounded(2)For any sequence in such that uniformly on compact sets

We state and prove the following corollary.

Corollary 1. If is the real part and imaginary part of an analytic function , then

proof 1. Let , then can be written as . Thus,Likewise, if , then can be written as . So

Theorem 2. : Let and . Then, if and only if . Moreover, if , thenThe above estimate holds even without the assumption , as shown in the proof.

proof 1. Assume and let . Then, . So, using the inequalityThis derives from the fact that for ,we haveTherefore, andUsing the 2-th root, we haveThen, we derive the upper estimate using the inequality , which holds even without the assumption .
Conversely, assume . Then, observing thatwe obtainwhich is finite. Thus, andHence, noting by (17) thatWhen we combine these two inequalities, we getSince . We deduce thatand then we get the desired result.
The relationship between the seminorm of a harmonic mapping and the seminorms of the corresponding real and imaginary parts is then examined.

Proposition 1. Let be harmonic on and let and . Then, , if and only if . Moreover,

proof 1. Assume that are true. The upper estimate arises immediately from the triangle inequality of the norm due to linearity. Assume and notice thatgives uswhere , and .
As a result, we haveIndeed, when equation (39) is squared, the left-hand side yieldsAs a result, we get by canceling opposite terms, ignoring the last word, and merging like terms. , which is the square of the right-hand side of the equation (39).
By first expressing the partials with respect to and in terms of the partials with respect to and , then computing the modulus, and lastly applying the inequality equation (39), we can now compute in terms of and .where we applied the inequality in the last stepMultiplying by and taking the supremum overall , as well as by a real-valued harmonic function , we getUsing inequality equation (42) once more, we haveWhen equations (43) and (44) are combined, the result isshowing that and are in , and that the lower estimate is correct.

Theorem 3. is a Banach space under the harmonic norm.

proof 1. We simply need to prove completeness now that we have established that is a normed linear space. Assume that in is a Cauchy sequence. The analytic functions are defined by the lower norm estimate in Theorem 2 for each , are in such that and , and the sequences and are Cauchy. Since is complete, these sequences converge in the norm to some analytic functions and , respectively.
Define . Then, by the upper estimate in Theorem 2, andas . Thus, in , proving the completeness of .

Theorem 4. , for . In addition, there is a constant that is purely dependent on , such that for ,

proof 1. Let be a harmonic on , and and be analytic on , so that and . Then, according to Theorem 2, . As a result of Lemma 1.1  andfor some positive constant only dependent on . Using Theorem 2, we see that . Finally, using again Theorem 2, we see thatThe Möbius invariance of the analytic spaces is then extended to their harmonic counterparts.

Theorem 5. is Möbius invariant Banach space for .

proof 1. Rotations clearly have no effect on the seminorm. As a result, we just need to show that for each
and ,Given that is the inverse of , the identities areandhold at each , and making the change of variables , we haveas desired.

Theorem 6. For , there is a constant such that for all and all ,

proof 1. Let and be fixed. Assume because the result is plainly valid for . The triangle inequality and by Proposition 1 in  then show that there exists a positive constant such thatwhere Theorem 2 was used to get the last inequality. Taking as an example, the following is the outcome.
In a similar way as Lemma 2.9 in , we state and prove the following theorem in harmonic space.

Theorem 7. Let . Then,(a)Every bounded sequence in is uniformly bounded on compact subsets of (b)For any sequence in such that , the sequence converges to 0 uniformly on compact subsets of

proof 1. To prove , set . Let is compact in and let . Then, by Theorem 5,Therefore, in is uniformly bounded on .
Next, to prove , let be compact in and let be as above. Since , by the definition of norm in , it follows that as .
By Theorem 6, and the triangle inequality, we getTherefore, uniformly on .

#### 3. Boundedness and Compactness Operator Norm

The following characterization of boundedness and compactness of a composition operator of sending one subclass to another is obtained by connecting the norm of a harmonic mapping (such that ) with the norms of the associated analytic functions and .

Corollary 2. Let and satisfyThen, if and only if .

We need the following lemmas in the sequel.

Lemma 3 (see ). Let and suppose that belongs to the Hadamard gap class. Then, , if and only if

Lemma 4. There exist two functions such that

proof 1. For a large number , choose a gap seriesThen, apply Lemma 3 to infer that holds for all , where is a constant. Furthermore, let us verifyObserve that for any ,And then, fix a with , and put .
Thus,If is large enough, then for one hasand hence . Since it is easy to establishit remains to deal with the third term . Noting thatnamely, in , the quotient of two successive terms is not greater than the ratio of the first two terms, and one finds that the series of is controlled by the geometric series having the same first two terms. Accordingly, equation (65) is applied to produceThe preceding estimates for , and implySimilarly, one can show thatIn a completely similar manner, one can prove that ifthen for all (owing to Lemma 3) andwe gotSimilarly, one can show thatBy addition equations (69), (70), (73), and (74), we haveOur lemma is therefore proved.

Theorem 8. Let and be holomorphic. Then,(i)The operator is bounded if and only if(ii)The operator is compact if and only if and(iii)The operator , is bounded if and only if(iv)The operator is compact if and only if and

proof 1. (i)consists of a simple computation and a reformulation of the definition of boundedness from Lemmas 1 and 4.(ii)Let and let and be analytic on such that and .By Lemma 1, we are required to show that if a sequence converges to 0 uniformly on compact subsets of , then converges to 0. For each , set . So tends to 0 uniformly on . And hence Lemma 4, for a given , there is an integer such that as ,Meanwhile, one may deduce from equation (70) and the evolution of the derivatives of functions that for any , there exists a , such that .