Weighted Composition Operators from Logarithmic Bloch-Type Spaces to Bloch-Type Spaces

Abstract

The boundedness and compactness of the weighted composition operators from logarithmic Bloch-type spaces to Bloch-type spaces are studied here.

1. Introduction

Let be the unit disc in the complex plane , the normalized Lebesgue area measure on , the class of all holomorphic functions on , and the space of bounded holomorphic functions on with the norm

The logarithmic Bloch-type space , was recently introduced in [1]. The space consists of all such that

(1.1)

The norm on is introduced as follows:

(1.2)

When , becomes the -Bloch space . For -Bloch and other Bloch-type spaces, see, for example, [1–9], as well as the related references therein. For , is the logarithmic Bloch space, which appeared in characterizing the multipliers of the Bloch space (see [3, 9]).

The little logarithmic Bloch-type space , consists of all such that

(1.3)

The following theorem summarizes the basic properties of the logarithmic Bloch-type spaces. Here, as usual, for fixed

Theorem 1 A (see [1]).

The following statements are true.

(a)The logarithmic Bloch-type space is Banach with the norm given in (1.2).

(b) is a closed subset of

(c)Assume Then if and only if

(d)The set of all polynomials is dense in .

(e)Assume then for each , . Moreover

(1.4)

A positive continuous function on is called weight.

The Bloch-type space consists of all such that

(1.5)

where is a weight. With the norm

(1.6)

the Bloch-type space becomes a Banach space.

The little Bloch-type space is a subspace of consisting of all such that

(1.7)

Let be a holomorphic self-map of and . For the corresponding weighted composition operator is defined by

(1.8)

It is of interest to provide function-theoretic characterizations for when and induce bounded or compact weighted composition operators on spaces of holomorphic functions. For some classical results mostly on composition operators, see, for example, [10]. For some recent related results, mostly in or related to Bloch-type or weighted-type spaces, see, for example, [4, 10–46] and the references therein.

Here we study the boundedness and compactness of the weighted composition operator from the logarithmic Bloch-type space and the little logarithmic Bloch-type space to the Bloch-type or the little Bloch-type space.

In 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 . We say that , if both and hold.

2. Auxiliary Results

In this section we quote several auxiliary results which will be used in the proofs of the main results.

Lemma 2.1.

Assume , then the following statements are true.

(a)Assume then the function

(2.1)

is increasing on the interval

(b)The function

(2.2)

is increasing on the interval

Proof.

• (a) We have

  

  (2.3)

Since when and , and the function is decreasing on the interval , we have

(2.4)

from which this statement follows.

The proof of (b) is similar, hence it is omitted.

The next lemma regarding the point evaluation functional on follows from [1, Lemma  3] and some elementary asymptotic relationship, such as

(2.5)

Lemma 2.2.

Let Then

(2.6)

for some independent of

The proof of the following lemma is similar to [25, Lemma  2.1], so we omit it.

Lemma 2.3.

Assume is a weight. A closed set in is compact if and only if it is bounded and

(2.7)

Remark 2.4.

If in Lemma 2.3 we assume that is not closed, then the word compact can be replaced by relatively compact.

The next characterization of compactness is proved in a standard way (see, e.g., the proofs of the corresponding lemmas in [10, 30, 47–49]). Hence we omit it.

Lemma 2.5.

Assume that , is a holomorphic self-map of and is a weight. Let be one of the following spaces , and one of the spaces , . Then the operator is compact if and only if is bounded and for every bounded sequence converging to uniformly on compacts of one has

(2.8)

Some concrete examples of the functions belonging to logarithmic Bloch-type spaces can be found in the next lemma.

Lemma 2.6.

The following statements are true.

(a)Assume that and then

(2.9)

where and is a nonconstant function belonging to

(b)Assume that and then

(2.10)

where and is a nonconstant function belonging to

(c)Assume that then

(2.11)

where and is a nonconstant function belonging to

Moreover, for each , it holds that belong to the corresponding space, and for fixed and

(2.12)

Proof. (a) Let be fixed. Then we have

(2.13)

(2.14)

where in (2.13) we have used that and in (2.14) we have used the fact that the function in (2.1) is increasing on the interval .

From (2.13), since , and by Lemma 2.1(a), we have that

(2.15)

as , from which it follows that as desired.

• (b) For fixed , we have

  

  (2.16)

(2.17)

where in (2.16) we have used the assumption , while in (2.17), as in (a), we have used the fact that the function in (2.1) is increasing on the interval .

From (2.16), and by Lemma 2.1(a), we obtain

(2.18)

as . Hence finishing the proof of this statement.

• (c) We have

  

  (2.19)

(2.20)

where we have used the assumption and the fact that function (2.1) is increasing on .

From (2.19), Lemma 2.1(a), and since we obtain

(2.21)

as that is,

Estimations (2.12) follow from (2.14), (2.17), (2.20) and by using the following facts

(2.22)

we finish the proof of the lemma.

Remark 2.7.

Note that from Lemmas 2.2 and 2.6 the functions defined in (2.9)–(2.11) have maximal growths in the corresponding logarithmic Bloch-type spaces.

3. Boundedness and Compactness of the Operator

This section studies the boundedness and compactness of the weighted composition operator .

Case 1.

, .

Theorem 3.1.

Assume , , is an analytic self-map of the unit disk, , and is a weight. Then the operator is bounded if and only if

(3.1)

(3.2)

Proof.

First assume that (3.1) and (3.2) hold. Then, by Lemma 2.2 and the definition of , we have

(3.3)

(3.4)

Applying (3.1) and (3.2) in (3.4), the boundedness of follows.

Now assume the operator is bounded. By taking the test functions and (which obviously belong to ), we obtain

(3.5)

(3.6)

From (3.5) and (3.6), and since the function is bounded, it follows that

(3.7)

For set

(3.8)

We have that ,

(3.9)

and as an easy consequence of Lemma 2.6(a), and for each

Using these facts and the boundedness of , for the test functions , where and , we get

(3.10)

From (3.10) it follows that

(3.11)

On the other hand, by using (3.7) and Lemma 2.1(b), we have

(3.12)

Hence, (3.11) and (3.12) imply (3.2).

Let

(3.13)

Then

(3.14)

and by Lemma 2.6(a) we get , and for every . Using the boundedness of , the test functions , and equalities (3.14) we get

(3.15)

for each , .

From (3.2), (3.5), (3.15), and using the fact that

(3.16)

condition (3.1) follows.

Theorem 3.2.

Assume , , is an analytic self-map of the unit disk, , and is a weight. Then the operator is compact if and only if is bounded

(3.17)

(3.18)

Proof.

Suppose that is compact. Then it is clear that is bounded. If , then (3.17) and (3.18) are vacuously satisfied. Hence assume that . Let be a sequence in such that as , and where is defined in (3.8). Then , uniformly on compacts of as , and

(3.19)

Hence from (3.10) and Lemma 2.5 we have that

(3.20)

from which (3.18) follows.

Let , where is defined in (3.13). Then and uniformly on compact subsets of as . Since is compact, we see that

(3.21)

From (3.15) we have

(3.22)

which along with (3.16), (3.18), and (3.21) implies

(3.23)

On the other hand, we have

(3.24)

for some positive . From (3.23) and (3.24), equality (3.17) follows.

Conversely, assume that is bounded and (3.17) and (3.18) hold. From the proof of Theorem 3.1 we know that

(3.25)

On the other hand, from (3.17) and (3.18) we have that, for every , there is a , such that

(3.26)

whenever .

Assume is a sequence in (or ) such that and converges to uniformly on compact subsets of as Let . Then from (3.25), (3.26), and by Lemma 2.2, it follows that

(3.27)

Therefore

(3.28)

Since converges to zero on compact subsets of as , by the Weierstrass theorem it follows that the sequence also converges to zero on compact subsets of as , in particular and . Using these facts and letting in the last inequality, we obtain that

(3.29)

Since is an arbitrary positive number it follows that the last limit is equal to zero. Applying Lemma 2.5, the implication follows.

Theorem 3.3.

Assume , , is an analytic self-map of the unit disk, , and is a weight. Then is bounded if and only if is bounded

(3.30)

(3.31)

Proof.

First assume that is bounded. Then, it is clear that is bounded, and as usual by taking the test functions and and using the fact , we obtain (3.30) and (3.31).

Conversely, assume that the operator is bounded, and condition (3.31) holds.

Then, for each polynomial , we have

(3.32)

from which along with conditions (3.30) and (3.31) it follows that . Since according to Theorem A the set of all polynomials is dense in we see that for every there is a sequence of polynomials such that

(3.33)

From this and by the boundedness of the operator we have that

(3.34)

as Hence and consequently is bounded.

Remark 3.4.

Note that Theorem 3.3 holds for all and

Theorem 3.5.

Assume , , is an analytic self-map of the unit disk, , and is a weight. Then the operator is compact if and only if

(3.35)

(3.36)

Proof.

If is compact, then it is bounded so that conditions (3.30) and (3.31) hold. On the other hand, is compact, which implies that (3.17) and (3.18) hold.

By (3.18) we have that, for every , there exists an such that

(3.37)

when . From (3.31), there exists a such that

(3.38)

when , and where is the function in Lemma 2.1(b).

Therefore, when and , we have that

(3.39)

On the other hand, if and , from (3.38) and Lemma 2.1(b) we have

(3.40)

Combining (3.39) and (3.40), we obtain (3.36). Similarly, from (3.17) and (3.30) is obtained (3.35), as claimed.

Conversely, assume that (3.35) and (3.36) hold. First note that (3.35) implies (3.30). Indeed if (3.30) did not hold then there would be a sequence and a such that

(3.41)

and From this and the continuity of the function

(3.42)

we would have that

(3.43)

which is a contradiction with (3.35).

For any , we have

(3.44)

Using conditions (3.30), (3.35), and (3.36) in (3.44), it follows that for each , moreover the set

(3.45)

is bounded in .

Taking the supremum in (3.44) over the unit ball of the space then letting and using conditions (3.30), (3.35), and (3.36), we obtain

(3.46)

from which along with Lemma 2.3 the compactness of the operator follows.

Case 2.

, .

Theorem 3.6.

Assume that is an analytic self-map of the unit disk, , and is a weight. Then the operator is bounded if and only if

(3.47)

Proof.

The proof of the theorem is similar to the proof of Theorem 3.1. The sufficiency follows by using the triangle inequality in (3.3) and then the third inequality in Lemma 2.2 and the definition of the space .

For the necessity it is enough to follow the lines of the corresponding part of the proof of Theorem 3.1 and use the test functions ,

(3.48)

(3.49)

which belong to (for the functions in (3.48) and (3.49) it easily follows by Lemma 2.6(b), where is the function in (2.10). We omit the details.

The proofs of the following two theorems are similar to the proofs of Theorems 3.2 and 3.5, where the test functions in (3.48) and (3.49) are used as well as the lemmas in Section 2. Hence their proofs are omitted.

Theorem 3.7.

Assume 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

(3.50)

Theorem 3.8.

Assume that is an analytic self-map of the unit disk, , and is a weight. Then the operator is compact if and only if

(3.51)

Case 3.

.

The following results were proved in [15]. Hence we quote them for the benefit of the reader, and without any proof.

Theorem 3.9.

Assume that is an analytic self-map of the unit disk, , and is a weight. Then the operator is bounded if and only if

(3.52)

Theorem 3.10.

Assume 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

(3.53)

Theorem 3.11.

Assume that is an analytic self-map of the unit disk, , and is a weight. Then the operator is compact if and only if

(3.54)

Case 4.

, or and

Here we consider the cases , or and .

Theorem 3.12.

Assume that , or and , , is a weight, and is a holomorphic self-map of Then is bounded if and only if and condition (3.2) holds.

Proof.

The sufficiency follows by using the first inequality in Lemma 2.2 and the definition of the space in (3.3).

For the necessity, by using the test functions we first get conditions (3.5) and (3.7). To get (3.2) for the case and we use the test functions

(3.55)

Note that ,

(3.56)

and similar to Lemma 2.6(b), and for each

Hence for the family we get

(3.57)

from which along with (3.5) and the assumption , easily follows (3.2) in this case.

When , condition (3.2) follows as in Theorem 3.1, by using the test functions in (3.8).

Theorem 3.13.

Assume that , or and , , is a weight, and is a holomorphic self-map of , and is bounded. Then is compact if

(3.58)

and condition (3.18) holds.

Proof.

The proof is similar to the corresponding parts of the proofs of Theorems 3.2 and 3.7, so is omitted.

Remark 3.14.

Note that if , or and and is compact, then condition (3.18) is proved as in Theorems 3.2 and 3.7, by using the test functions in (3.8) and (3.48). If then condition (3.58) is vacuously satisfied. At the moment, we are not sure if the compactness implies condition (3.58) in the case . Hence for the interested readers we leave this as an open problem.

The following theorem is proved as the corresponding part of Theorem 3.5.

Theorem 3.15.

Assume that , or and , , is a weight, and is a holomorphic self-map of . Then the operator is compact if and condition (3.36) holds.

Remark 3.16.

Note that if is compact, then clearly .