Abstract

The boundedness and compactness of the weighted differentiation composition operator from the mixed-norm space to the nth weighted-type space on the unit disk are characterized.

1. Introduction

Throughout this paper will denote the open unit disk in the complex plane , the class of all holomorphic functions on , and the space of all bounded holomorphic functions on with the norm

The mixed norm space ,โ€‰โ€‰,โ€‰โ€‰, consists of all such that where

A positive continuous function on is called weight. Let be a weight and . The th weighted-type space on , denoted by , consists of all such that The space was recently introduced by this author in [1] as an extension of several weighted-type spaces which attracted a lot of attention in last few decades. For instance, when , the space becomes the weighted-type space (see, e.g., [2โ€“4]), when the Bloch-type space (see, e.g., [5โ€“7]), and for , the Zygmund-type space . Some information on Zygmund-type spaces on and some operators on them can be found, for example, in [8โ€“10] and on the unit ball, for example, in [11, 12].

The quantity is a seminorm on the th weighted-type space and a norm on , where is the set of all polynomials whose degrees are less than or equal to . A natural norm on the th weighted-type space is introduced as follows: With this norm the th weighted-type space becomes a Banach space.

The little th weighted-type space, denoted by , is a closed subspace of consisting of those for which

An analytic self-map induces the composition operator on , defined by for (see, e.g., [8, 13โ€“16]).

Let be an analytic self-map of , , and . Then the weighted differentiation composition operator, denoted by , is defined on by

Recently there has been some interest in studying some particular cases of operator (see, e.g., [17โ€“25]). For some other products of linear operators on spaces of holomorphic functions see also recent papers [11, 26โ€“32].

Here we study the boundedness and compactness of the operator from to th weighted-type spaces, where .

Throughout this paper, constants are denoted by ; they are positive and may differ from one occurrence to the other. The notation means that there is a positive constant such that .

2. Auxiliary Results

Here we quote some auxiliary results which will be used in the proofs of the main results. The first lemma can be proved in a standard way (see, e.g., in [13, Propositin 3.11] or in [15, Lemm 3]).

Lemma 2.1. Assume that , , , , is an analytic self-map of and . Then the operator is compact if and only if is bounded and for any bounded sequence in which converges to zero uniformly on compact subsets of , in as

The next lemma is known, but we give a proof of it for the benefit of the reader.

Lemma 2.2. Assume that , , and Then there is a positive constant independent of such that

Proof. By the monotonicity of the integral means, using the well-known asymptotic formula and Therem 7.2.5 in [33], we have that from which the result follows.

The following lemma can be found in [34].

Lemma 2.3. For and one has

A proof of the next lemma can be found in [35, Lemma ].

Lemma 2.4. Assume and Then

The following formula where the second sum is over all nonnegative integers satisfying and , is attributed to Faร  di Bruno [36]. By using Bell polynomials it can be written as follows: For the last sum can go from since however we will keep the summation since for the only existing term is equal to 1 and we will use it.

The Leibnitz formula along with (2.6) yields Hence we have the next result.

Lemma 2.5. Assume that and is an analytic self-map of . Then

3. The Boundedness and Compactness of

This section characterizes the boundedness and compactness of the operator .

Theorem 3.1. Suppose that , , , is an analytic self-map of the unit disk, , and is a weight. Then the operator is bounded if and only if for each Moreover if is bounded, then the following asymptotic relation holds

Proof. First assume that is bounded; then there exists a constant such that for all .
For a fixed , , and constants , set where By [33, Theore 1.4.10], we get Applying Lemma 2.3, we have that Therefore and moreover .
Now we show that for each there are constants , such that By differentiating function , for each (3.8) becomes
Applying Lemma 2.4 with and where , we see that the determinant of system (3.9) is different from zero, as claimed.
By , denote the corresponding family of functions which satisfy (3.8) with . Then, for each fixed inequality (3.3) along with (2.9) and (3.8) implies that for each
From (3.10) it follows that for each
Let Then clearly
By formula (2.9) applied to the function we get which along with the boundedness of the operator and (3.13) implies that
Now assume that we have proved that for and a
Applying (2.9) to the function โ€‰ and noticing that for , we get From (3.17), the boundedness of the operator , the fact that , the triangle inequality, noticing that is the coefficient at , and finally using hypothesis (3.16) we get Hence by induction, (3.18) holds for each
From (3.18), for each fixed Inequalities (3.11) and (3.19) imply Now assume that (3.1) holds. Then for any , by (2.9) and Lemma 2.2 we have We also have that for each Using (3.23), (3.24), and (3.1) it follows that the operator is bounded.
From (3.23) and (3.20) the asymptotic relation (3.2) follows.

Theorem 3.2. Suppose that , , , is an analytic self-map of the unit disk, and is a weight. Then the operator is bounded if and only if is bounded and for each

Proof. The boundedness of clearly implies that is bounded. Applying (2.9) to the function and using the assumption it follows that as which is (3.25) for
Assume that we have proved the following inequalities: for and a .
Applying formula (2.9) to the function we get (3.17). From (3.17), by using the boundedness of function , the triangle inequality, noticing that the coefficient at is independent of , and finally using hypothesis (3.27), we easily obtain Hence by induction we get that (3.25) holds for each
Now assume that is bounded and (3.25) holds for each . For each polynomial we have as
From (3.29) we have that, for each polynomial , . The set of all polynomials is dense in , so we have that for each there is a sequence of polynomials such that as Thus the boundedness of implies Hence , from which the boundedness of follows, completing the proof of the theorem.

Theorem 3.3. Suppose that , , , is an analytic self-map of the unit disk, and is a weight. Then the operator is compact if and only if is bounded and for each

Proof. First assume that is bounded and (3.31) holds. By Theorem 3.1 we have that for each , (3.1) holds.
Let be a sequence in such that and converges to uniformly on compact subsets of as By the assumption, for any , there is a , such that for each and We have
Now we estimate , , and : where in (3.34) we have used the fact that from uniformly on compact subsets of as it follows that for each , uniformly on compact subsets of as
The fact that as is proved similarly; so we omit it.
By Lemma 2.2 and (3.32) we have that
From (3.34), (3.35), and (3.36) we obtain From this and applying Lemma 2.1 the implication follows.
Now assume that is compact; then clearly is bounded. Let be a sequence in such that as . If such a sequence does not exist, then the conditions in (3.31) automatically hold.
Let , be as in Theorem 3.1. Then the sequences are bounded and uniformly on compact subsets of as . Since is compact, we have that for each On the other hand, from (3.10) we obtain which along with as and (3.38) implies that for each , from which (3.31) holds in this case.

4. The Compactness of the Operator

The compactness of is characterized here. The proof of the next lemma is similar to the proof of the corresponding result in [14].

Lemma 4.1. Suppose that and is a radial weight such that A closed set in is compact if and only if it is bounded and satisfies

Theorem 4.2. Suppose that , , , is an analytic self-map of the unit disk, and is a radial weight such that Then the operator is compact if and only if for each

Proof. First assume that is compact. Then it is bounded and since the test functions in (3.12) belong to , we have that (3.25) holds. Beside this the operator is compact too, so that (3.31) holds. Hence, if from (3.25) for each we get as , hence we obtain (4.2) in this case.
Now assume Let be a sequence such that as Then from (3.31) we have that for every , there is an such that for each when , and from (3.25) there exists a such that for
Therefore, when and , we have that On the other hand, if and from (4.5) we obtain Combining the last two inequalities we obtain (4.2), as desired.
Now assume that (4.2) holds. Taking the supremum in (3.22) over in the unit ball of , then letting is such obtained inequality and using (4.2) we get Hence by Lemma 4.1 the compactness of the operator follows.