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The Scientific World Journal
Volume 2014 (2014), Article ID 590656, 15 pages
Research Article

Definition and Properties of the Libera Operator on Mixed Norm Spaces

Faculty of Mathematics, University of Belgrade, Studentski Trg 16, P.O. Box 550, 11001 Beograd, Serbia

Received 7 August 2013; Accepted 23 October 2013; Published 20 February 2014

Academic Editors: H. Jafari and Y. Wang

Copyright © 2014 Miroslav Pavlovic. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


We consider the action of the operator on a class of “mixed norm” spaces of analytic functions on the unit disk, , defined by the requirement , where , , , and is a nonnegative integer. This class contains Besov spaces, weighted Bergman spaces, Dirichlet type spaces, Hardy-Sobolev spaces, and so forth. The expression need not be defined for analytic in the unit disk, even for . A sufficient, but not necessary, condition is that . We identify the indices , , , and for which is well defined on , acts from to , the implication holds. Assertion extends some known results, due to Siskakis and others, and contains some new ones. As an application of we have a generalization of Bernstein’s theorem on absolute convergence of power series that belong to a Hölder class.

1. Introduction and Definitions

Let denote the class of all functions holomorphic in the unit disk of the complex plane. In [1], Libera introduced the operator and showed its importance in the theory of univalent functions. In particular, it was shown in [1] that this operator transforms the class of star-like functions into itself. Since then many papers were published devoted to this aspect of the Libera operator. The “generalized” Libera operator was introduced and studied from the functional analytic point of view by Siskakis in [2, 3], then in [46], and other papers (see [7] for further references). If , then is defined on , and, on classical spaces such as Hardy, Bergman, and Besov, has almost the same linear topological properties as the integration operator , and therefore is not so interesting from the functional analytic point of view (cf. [8]). So we can assume that . In fact we can and will assume that , so whenever the integral is somehow defined. This definition of requires further explanation because the integral need not be defined for (e.g., ).

Endowed with the topology of uniform convergence on compact subsets of , the class becomes a complete locally convex space. The dual of is equal to , where means that is holomorphic in a neighborhood of (depending on ). The duality pairing is given by where and (see, e.g., [9]). Clearly, is well defined on , and it is easy to check that it maps into , and The last integral is obtained from (3) by integration over the straight line joining and 1.

Definition 1. We use the symbol to denote the operator .

Definition 2. We denote by the operator whenever the integral converges uniformly on compact subset of . “Uniform convergence” means that the limit is uniform with respect to . This hypothesis guarantees that is an analytic function.

It is easy to verify the validity of the following.

Proposition 3. The dual operator coincides with the Cesàro operator, , defined on by Conversely, the dual of coincides with .

In [6], it was proved that is defined on the Bloch space and maps it into BMOA. This assertion, which improves the earlier result that maps Bloch space into itself (e.g., [4, 10]), was deduced from a result of Nowak [11] and Proposition 3.

However, as it can easily be seen, the operator cannot be extended to a continuous operator from to . Moreover, cannot be extended to a continuous operator from to , where is some of common spaces, for example, see [6]. ( denotes the Lebesgue measure on ). We will identify a large family of spaces that possess the same property.

In this paper we consider, in particular, the spaces listed in the following definition.

Definition 4. Bloch type spaces: Weighted Hardy spaces: where Weighted Bergman spaces: Dirichlet type spaces:

The space is closely related to in that , where denotes the set of all which can be represented as , , , with (see [12]).

Concerning these spaces, we mention the following results.

Theorem A (see [2]). The operator acts as a bounded operator from to if and only if .

Theorem B (see [4, 10]). acts as a bounded operator from into if and only if .

Theorem C (see [3]). acts as a bounded operator from into if and only if .

Theorem D (see [13]). If for all , and , then if and only if

Here “acts” means, among other things, that is defined on the space in the sense of Definition 2; that is, for all .

Condition (15) is implied, as is easily seen, by This condition is satisfied, according to Hardy’s inequality, for and therefore for , , because and . However, (15) does not imply (16) even in the case when maps a space into itself. This can be seen from the following reformulation of Bernstein’s theorem and Theorem 5 below.

Theorem E (Zygmund, [14], Ch. VI ). Let . Then the implication (16) holds if and only if .

Theorem 5. acts as a bounded operator from into if and only if .

(The proof is very easy, although the theorem is a special case of Theorem 11.)

What we can deduce from condition (15) is that the series Namely, taking , we see that the integral exists and is finite. However, it may happen that maps a space into itself and that where is positive constants. For instance, as an application of Khinchin’s inequality and a profound result of Kisliakov, we have the following.

Theorem 6. Let . Then there is a function such that (although maps into ).

Proof. See the proof of Proposition 34, Case (1), .

On the other hand, if a space satisfies (15) for all , this does not mean that maps into (although it is defined on ). Besides , we have, for instance, the following.

Theorem F (see [13]). The space is contained in and satisfies (16), but does not map into .

One of the aims of the present paper is to extend Theorems A, B, C, 5, and 6 to a large scale of “mixed norm” spaces.

Definition 7. We denote by (, ), where when , and when , the class of those for which Then letting be a nonnegative integer, we define the space The (quasi) norm in is given by where should be interpreted as equal to zero.

It is well known and easy to prove that these spaces are complete.

The norm (22) is not the most natural one but is convenient for technical reasons. For instance, in the case , , a more (but not most) natural norm is given by This norm is equivalent to that given by (22) because of the maximum modulus principle for analytic functions.

The space is specific in that the set, , of all analytic polynomials is not dense in it. The closure of in coincides with the “little oh” space

From now on, unless specified otherwise, we suppose that

It is sometimes more convenient to work with the Besov type spaces.

Definition 8. Let and choose any integer such that . The space is defined as If , we can assume that . It is well known that this definition is independent of ; this follows immediately from Lemma A below. If , then is “true” Besov spaces; if , then ; if , then it is called Hardy-Bloch spaces [15]. If , then we have the space and its subspace Lipschitz Spaces. It is known that the space () coincides with the (Lipschitz) space consisting of those in the -Hardy space for which Here denotes the symmetric th difference with step , and denotes the norm in . This result is essentially due to Hardy and Littlewood and Zygmund (see [16] for a simple proof of a generalized variant). We also have In the case and , the space is denoted by and is called the Zygmund space. The corresponding “little” space is denoted by . These spaces were introduced by Zygmund via symmetric differences. In [17], connections of and with Besov spaces were established.
Hardy-Sobolev Spaces. Let be an integer. The Hardy-Sobolev space consists of those such that ; that is, . It is known that See, for example, [16]. In particular, coincides with the usual Lipschitz space consisting of those from the disk-algebra for which
There are various inclusions between members of the scale . Here we mention the following.

Proposition 9. If , then (a), for ,(b), where , where , and ,(c), where , and where .

Proof. We can assume that . Besides, we omit the proof that the inclusions are strict because we do not need this fact.(a)If , then () because increases with .(b)It is well known that for (see, e.g., [18, Corollary 5.1.2]).(c)This follows from (b) and the density of in .

Let We will determine the indices , , , and for which: (1) acts from into ;(2) acts from to but not to [ and and ; see Theorem 16];(3) acts from into but the implication (16) does not hold ( and (either and , or and ); see Theorem 18);(4)the operator cannot be extended to a bounded operator from to (, and ( or ); see Theorem 16).

Observe that, since in (3), the inequality is equivalent to which has sense because due to the condition .

In proving (1) we use the inequality where , which is a relatively simple consequence of the Littlewood subordination principle. A generalized version of (36) immediately gives sufficient conditions for to map to . In order to prove that these conditions are necessary we analyze membership in of functions with nonnegative, nonincreasing coefficients and apply this to the Libera transform of functions with positive coefficients.

Let be a function of the form and define by Definition 2. This definition is correct if and only if because a series with positive coefficients is Abel summable if and only if it is summable in the ordinary sense. If (38) is satisfied, then the sequence of the Taylor coefficients of is and is therefore nonincreasing. In this paper we consider only the functions where . Discussion of the general case will appear in a separate paper.

In considering (2) and (4) we use, besides functions with positive coefficients, a deep theorem of Kolmogorov and Khinchin, while in the case of (3) we need another deep result, due to Kisliakov. By use of these theorems we can say much more about (3). Namely, if the implication does not hold, then there exists a function such that , and in some cases or even for some (see Theorem 5 and Theorem 41).

2. Results

Before stating our first result we give a sufficient condition for the validity of (15). This condition is not necessary (see Theorem 16).

Proposition 10. If , where ,, and then the integrals converge uniformly on compact subsets of , and the operator maps to and coincides with on , and we have

Proof. By the well-known theorem from complex analysis, it is enough to prove that converges uniformly. Since increases with , the condition implies that This implies, by the well-known estimate , that Hence, by successive integration we get , where It turns out that It is not difficult to check that if , the , where . This implies, by Weierstrass’ theorem, that converges uniformly on compact subsets of . The rest of the proof is easy.

Our first result is as follows.

Theorem 11. Let or . Then the following assertions are equivalent: (a)the operator acts as a bounded operator from into ;(b)Condition (40) is satisfied.

Observe that condition (40) is independent of .

Remark 12. In other words, the theorem says the following (excluding Hardy and Hardy-Sobolev spaces): let or ,. Then acts from to if and only if , or what is the same, if and only if and .
Theorem 11 contains some known and some new results as special cases.

Case 1. Theorem 11 covers the case when ,. In particular, (a) maps into if and only if . This is Theorem A;if , then(b) maps Hardy-Sobolev space into , for every , and, in particular, maps the ordinary Lipschitz space into itself.

This is, maybe, a new result.

Case 2 (Theorem C). In particular, does not act as a bounded operator from into , for any .

This is seen from Theorem 11 by taking ; that is, .

Case 3. (a) maps the Dirichlet space into itself if and only if . In particular, maps if and only if .

Another case: (b) maps into itself if and only if .

These facts are, maybe, new.

Case 4. (a) maps into itself if and only if and .

This is seen from Theorem 11 by taking . This is related to a result of [19], which, when reformulated in our notation, gives the assertion (a) under the additional condition , or equivalently and . For example, if and , then this assertion says nothing because of the hypothesis , while we still have that maps into itself if and only if .

If , and , , then (a) says that maps into if and only if . (Compare Case 1.) In particular maps into if and only (this is Theorem 5). Also maps if and only if .

Case 5. (a) maps into itself for every , , and . The same holds for the little space .

This is seen from Theorem 11 (or Remark 12): the inequality holds for any and . In the case , a direct proof can be found in [7]. In [6], assertion (a) is proved by using the relation and the fact that maps () into itself. (The latter was proved in [20]; a quick proof is given in [21].) What is new here is that (a) holds for .

(b) As a special case of (a) we have that maps the Lipschitz (=Hölder) class into itself.

Case 6. maps -Bloch type spaces into itself if and only if .

The same holds for the little space .

In particular, when , this condition reduces to ; this is Theorem B.

Case 7. (a) maps the Hardy-Bloch space if and only if .

This is, maybe, new. In the case , there exists a better result [6]: maps into (the well-known result of Littlewood and Paley states that ).

Case 8. maps the Zygmund space into itself. The same holds for .

The implication (a)(b) of Theorem 11 is valid because of the following two propositions.

Proposition 13. Let (1) , , and , or (2) and ().
Then the operator cannot be extended to a bounded operator from to .

As mentioned in Introduction (see (4), in page 4, or Theorem 16 below), the class of such spaces is larger.

Remark 14. In (1), the condition is necessary since is either a Hardy space or a Hardy-Sobolev space , which is contained in .

Proposition 15. If and , then the function belongs to ; the function is well defined but is not in .

However, if , it may happen that is well defined on (but, by Proposition 15, does not map the space into itself). The following theorem together with Proposition 10 characterizes those such that can be extended to a bounded operator from to .

Theorem 16. Let () or , and . Then the following four conditions are equivalent: (a)the operator can be extended to a bounded operator from to ;(b) acts as a bounded operator from to ;(c)If , then ;(d), , and .

A natural question arises from this theorem: under condition (d), find a quasi-Banach such that . It turns out that we can take, for instance,

Theorem 17. Let , , (i.e., ), and . Then maps to , where is defined by (49).

This theorem can easily be deduced from (36); we will omit the proof.

There are cases when (which implies that is well defined on ) but the assertion (c) of Theorem 16 does not hold.

Theorem 18. Let () and . Then assertion (c) of Theorem 16 does not hold if and only if one of the following two conditions is satisfied: (1),, and ,(2),, and .

Remark 19. In the case of Besov type spaces, this theorem says the following: let . The implication does not hold if and only if either (1) , , and or (2) , , and .

Theorem 18 can be used to get a generalization of Bernstein’s theorem to the case of the Besov spaces.

Theorem 20. Let . The implication does not hold if and only if either (1) , , and , or (2) , , and .

Proof. This follows from Theorem 18, the definition of Besov type spaces (Definition 8), and the equivalence .

By taking we get Bernstein’s theorem.

3. Proof of Theorem 11

We need a variant of the Littlewood subordination principle.

Theorem G (see [22]). If is an analytic function and , then and Here denotes the norm of in , As an application we have the following lemma.

Lemma 21. If and are positive real numbers such that and if ,, then

Proof. The case is easy. Let . Let ,, , and . Then which was to be proved.

Lemma 22. If is well defined (see Definition 2), then where .

Remark 23. The integral in (55) diverges if and only if and .

Proof. Let . We have and hence Applying the Minkowski inequality and Lemma 21 with , we obtain Substituting and taking , which was to be proved.

Remark 24. Before going further note that an immediate consequence of this lemma is the validity of implication (b)(a) of Theorem 11 in the case . Then we use this fact and the following two ones to show that maps into . The set (all polynomials) is dense in and maps into .

Lemma 25. If and is well defined, then where and is independent of and .

In the case , this lemma follows from Lemma 22 and the following.

Sublemma 25.1. Let , , and let be a nondecreasing function. Thenwhere is independent of .

Proof. Fix , and define the function on by Then the desired inequality can be written as Let , where is a nonnegative integer. Then In a similar way we can prove that where const. . Comparing these inequalities we get the result.

In the case , Lemma 25 is a consequence of the following fact.

Sublemma 25.2. Let ,, and , a measurable function defined on . Thenwhere is independent of .

Proof. Let . Then, by Hölder’s inequality , Since because , we have that the last integral is less than The result follows.

Proof of Theorem 11, ((b)(a)). By Remark 24, we may assume that is finite, which implies that . Let . A standard application of the maximum modulus principle for analytic functions gives Then, if , we have Since , we can choose so that . Then Combining this with the preceding inequality we get the result in the case . In the case the proof is similar and simpler and is omitted.

Proof of Theorem 11 ((a)(b)). As noted in Section 2, it is enough to prove Propositions 13 and 15.

Proof of Proposition 13. We have
Case (1). By Proposition 9 we have , where . Since , it is enough to consider the case of . Let ,. It is clear that . A simple calculation shows that the set is bounded in . Hence, if has an extension to a bounded operator from to , then the set is bounded because the functional is bounded on . However, which is a contradiction. This proves the desired result in the case of . Since , we see that the result holds for this space as well.
Case (2). Let and choose so that . This implies that . Then it is easy to check that , which together with the Case (1) gives the result.

Proof of Proposition 15. The proof of Proposition 15 is more delicate. First note that, by Proposition 9, where and that see (34). Therefore, it is enough to prove that the function defined by (48) belongs to while does not belong to .
It is easier to prove that . Namely, since the coefficients of are nonnegative, we see that is in if and only if which, by a theorem of -integrability of power series with positive coefficients (see [23, Theorem  1]), is equivalent to Here Since (), the latter is equivalent to where are coefficients of . Since , we get the equivalent condition This condition is not satisfied because and and hence where the function is well defined because , which implies
In proving that we use a sequence constructed in the following way (see, e.g., [24]).
Let be a function on such that (1) for ,(2) for ,(3) is decreasing and positive on the interval .
Let , and let , and, for ,
These polynomials have the following properties: (Here denotes the Hadamard product).
In [25, Lemma 2.1], the following characterization of was proved.

Lemma A. Let , and , , and let be a nonnegative integer. A function is in if and only ifand we have . In the case of (resp., ) this is interpreted as (resp. ).

Remark 26. This lemma was deduced in [25] from the case (which is relatively easy to discuss) by using some nontrivial results of Hardy and Littlewood [26] and of Flett (see [27]). By a successive application of Lemma 27 below (case ), we can make this deduction elementary.

Lemma 27. If is a positive integer, where is a complex sequence, and then there is a constant depending only on and such that

Proof. The proof can be reduced to two cases: (1) , and (2) , . We will consider only Case (1); Case (2) is discussed similarly.
Let be a function on such that and for . Then we have as where for . Fix and let where We have where are -means of . It follows that where we have used Fejér’s inequality . Now we use Lagrange’s inequality in the form and the easily proved inequality to get which proves the desired result in one direction. To prove the reverse inequality we write as Applying the above case, we get which completes the proof.

Proof of Proposition 15. As noted above it is enough to prove that , where . In this case, by Lemma A, the function belongs to if and only if On the other hand, by Lemma 27 and the property (88), which implies that is in , because .
Thus, we have proved the implication (a)(b) of Theorem 11.

4. Proof of Theorem 16

It is clear that (b) implies (a). We have already noted that (c) implies (b). The following assertion shows that (d) implies (c).

Proposition 28. If ,,, and , then

Proof. Since , it is enough to consider the case . Let Since , which follows from (86) and the fact for , we see that Also the relation implies that , and hence whenever . In particular for (see (79)). We use Hardy’s inequality and Lemma (107) to conclude that if , then Hence, by Hölder’s inequality,