Definition and Properties of the Libera Operator on Mixed Norm Spaces

We consider the action of the operator ℒg(z) = (1 − z)−1∫z 1f(ζ)dζ on a class of “mixed norm” spaces of analytic functions on the unit disk, X = H α,ν p,q, defined by the requirement g ∈ X⇔r ↦ (1 − r)α M p(r, g (ν)) ∈ L q([0,1], dr/(1 − r)), where 1 ≤ p ≤ ∞, 0 < q ≤ ∞, α > 0, and ν is a nonnegative integer. This class contains Besov spaces, weighted Bergman spaces, Dirichlet type spaces, Hardy-Sobolev spaces, and so forth. The expression ℒg need not be defined for g analytic in the unit disk, even for g ∈ X. A sufficient, but not necessary, condition is that ∑n=0∞|g^(n)|/(n+1)<∞. We identify the indices p, q, α, and ν for which 1°ℒ is well defined on X, 2°ℒ acts from X to X, 3° the implication g∈X⇒∑n=0∞|g^(n)|/(n+1)<∞ holds. Assertion 2° extends some known results, due to Siskakis and others, and contains some new ones. As an application of 3° we have a generalization of Bernstein's theorem on absolute convergence of power series that belong to a Hölder class.


Introduction and Definitions
Let (D) denote the class of all functions holomorphic in the unit disk D 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 [4][5][6], and other papers (see [7] for further references). If | | < 1, then Λ is defined on (D), and, on classical spaces such as Hardy, Bergman, and Besov, has almost the same linear topological properties as the integration operator ( ) → ∫ 0 ( ) , and therefore is not so interesting from the functional analytic point of view (cf. [8]). So we can assume that | | = 1. In fact we can and will assume that = 1, so 2 The Scientific World Journal where ( ) = ∑ ∞ =0̂( ) ∈ (D) and ( ) = ∑ ∞ =0̂( ) ∈ (D) (see, e.g., [9]). Clearly, Λ 1 is well defined on (D), and it is easy to check that it maps (D) into (D), and The last integral is obtained from (3) by integration over the straight line joining and 1.
Definition 1. We use the symbol L to denote the operator Λ 1 : (D) → (D).

Definition 2.
We denote by L the operator L ( ) = ∫ 1 0 ( + (1 − ) ) whenever the integral converges uniformly on compact subset of D. "Uniform convergence" means that the limit is uniform with respect to . This hypothesis guarantees that L is an analytic function.
It is easy to verify the validity of the following.
Conversely, the dual of C : (D) → (D) coincides with L.
In [6], it was proved that L is defined on the Bloch space and maps it into BMOA. This assertion, which improves the earlier result that L 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 L cannot be extended to a continuous operator from (D) to (D). Moreover, L cannot be extended to a continuous operator from to (D), where is some of common spaces, for example, see [6]. ( denotes the Lebesgue measure on D). We will identify a large family of spaces that possess the same property.
Concerning these spaces, we mention the following results.
Theorem A (see [2]). The operator L acts as a bounded operator from to if and only if 1 < ≤ ∞.
Theorem B (see [4,10]). L acts as a bounded operator from B into B if and only if 0 < < 2.
Theorem C (see [3]). L acts as a bounded operator from , into , if and only if > + 2.
Theorem D (see [13]). If̂( ) ≥ 0 for all , and ∑(̂( )/( + 1)) < ∞, then L ∈ D 1 0 if and only if Here "acts" means, among other things, that L is defined on the space in the sense of Definition 2; that is, converges uniformly in | | < for all ∈ (0, 1). Condition (15) is implied, as is easily seen, by The Scientific World Journal 3 This condition is satisfied, according to Hardy's inequality, for ∈ 1 and therefore for ∈ ∪ D 1 0 , ≥ 1, because ⊂ 1 and D 1 0 ⊂ 1 . However, (15) does not imply (16) even in the case when L maps a space into itself. This can be seen from the following reformulation of Bernstein's theorem and Theorem 5 below.
Theorem 5. L acts as a bounded operator from ∞ into ∞ if and only if 0 < < 1.
(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 = 0, we see that the integral exists and is finite. However, it may happen that L maps a space into itself and that where > 0 is positive constants. For instance, as an application of Khinchin's inequality and a profound result of Kisliakov, we have the following.
On the other hand, if a space satisfies (15) for all ∈ , this does not mean that L maps into (although it is defined on ). Besides 1 , we have, for instance, the following.
Theorem F (see [13]). The space D 1 0 is contained in 1 and satisfies (16), but L does not map D 1 0 into 1 .
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 , (0 < , ≤ ∞), where > 0 when < ∞, and ≥ 0 when = ∞, the class of those ∈ (D) for which Then letting ] be a nonnegative integer, we define the space The (quasi) norm in , ,] is given by where ∑ −1 =0 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 = , ] = 0, 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.
From now on, unless specified otherwise, we suppose that It is sometimes more convenient to work with the Besov type spaces. , 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 > 0, then B , is "true" Besov spaces; if < 0, then B , = , − ; if = 0, then it is called Hardy-Bloch spaces [15]. If = ∞, then we have the space Here Δ ] 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 = 1, the space Λ , is denoted by Λ * 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.
See, for example, [16]. In particular, ∞ 1 coincides with the usual Lipschitz space consisting of those from the diskalgebra for which There are various inclusions between members of the scale , ,] . Here we mention the following. Proof. We can assume that ] = 0. Besides, we omit the proof that the inclusions are strict because we do not need this fact.
In proving (1) we use the inequality where = ] − 1/ , which is a relatively simple consequence of the Littlewood subordination principle. A generalized version of (36) immediately gives sufficient conditions for L 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 L 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 L is and is therefore nonincreasing. In this paper we consider only the functions where = ( + 1) /log ( + 2). Discussion of the general case will appear in a separate paper.

Results
Before stating our first result we give a sufficient condition for the validity of (15). This condition is not necessary (see Theorem 16).
then the integrals converge uniformly on compact subsets of D, and the operator

maps
, ,] to (D) and coincides with L on (D), and we have Proof. By the well-known theorem from complex analysis, it is enough to prove that 0 ( ) converges uniformly. Since ( , ) increases with , the condition ∈ , ,] implies that This implies, by the well-known estimate Hence, by successive integration we get ∞ ( , ) ≤ ( ), where It turns out that This implies, by Weierstrass' theorem, that 0 ( ) converges uniformly on compact subsets of D. The rest of the proof is easy.
Our first result is as follows.
. Then the following assertions are equivalent: (a) the operator L acts as a bounded operator from into ; Observe that condition (40) is independent of .  if ] ≥ 1, then for every ∈ [1, ∞], and, in particular, L maps the ordinary Lipschitz space into itself. This is, maybe, a new result.
Case 2 (Theorem C). In particular, L does not act as a bounded operator from 1 into 1 , for any > −1.
This is seen from Theorem 11 by taking = − 1; that is, = ( + 1)/ .  This is seen from Theorem 11 by taking ] = 0. This is related to a result of [19], which, when reformulated in our notation, gives the assertion (a) under the additional condition − 1 ≥ −(1/ ), or equivalently ≤ 1 and ≤ 1/(1 − ). For example, if = 1 and = 1/2, then this assertion says nothing because of the hypothesis ≤ 2, while we still have that L maps This is seen from Theorem 11 (or Remark 12): the inequality ] + 1 > ] − + 1/ holds for any > 0 and ∈ [1, ∞]. In the case ≥ 1, a direct proof can be found in [7]. In [6], assertion (a) ( ≥ 1) is proved by using the relation C * = L and the fact that C maps , (1/ + 1/ = 1) into itself. (The latter was proved in [20]; a quick proof is given in [21].) What is new here is that (a) holds for < 1.
(b) As a special case of (a) ( = = ∞) we have that L maps the Lipschitz (=Hölder) class Λ = Λ ∞ into itself. The same holds for the little space b = ℎ ,1 . In particular, when = ∞, this condition reduces to 0 < < 2; this is Theorem B. This is, maybe, new. In the case 2 < = < ∞, there exists a better result [6]: L maps B , into (the wellknown result of Littlewood and Paley states that ⫋ B , ).
Case 8. L maps the Zygmund space Λ * into itself. The same holds for * .
The implication (a)⇒(b) of Theorem 11 is valid because of the following two propositions. Then the operator L cannot be extended to a bounded operator from to (D).
As mentioned in Introduction (see (4), in page 4, or Theorem 16 below), the class of such spaces is larger.
belongs to , ,] ; the function L is well defined but L is not in , ,] .
However, if , ,] = 0, it may happen that L is well defined on This theorem can easily be deduced from (36); we will omit the proof.
There are cases when , ,] > 0 (which implies that L is well defined on ) but the assertion (c) of Theorem 16 does not hold.  By taking = = ∞ we get Bernstein's theorem.

Proof of Theorem 11
We need a variant of the Littlewood subordination principle.
Theorem G (see [22]). If : D → D is an analytic function and ∈ , then ∘ ∈ and Here ‖ ‖ denotes the norm of in , As an application we have the following lemma.
and hence which was to be proved.
and is independent of and .
In the case ≤ 1, this lemma follows from Lemma 22 and the following. where is independent of .
Proof. Fix ∈ (0, 1), and define the function V on (0, 1) by Then the desired inequality can be written as In a similar way we can prove that where 1 = const. > 0. Comparing these inequalities we get the result.
Proof of Theorem 11, ((b)⇒(a)). By Remark 24, we may assume that is finite, which implies that > 0. Let > 1. A standard application of the maximum modulus principle for analytic functions gives Then, if > 1, we have Since − − 1 = + 1/ − ] − 1 < 0, we can choose > 0 so that − − 1 + < 0. Then Combining this with the preceding inequality we get the result in the case > 1. In the case ≤ 1 the proof is similar and simpler and is omitted.

Proof of Theorem 11 ((a)⇒(b)). As noted in
which is a contradiction. This proves the desired result in the case of ℎ ,] . Since ℎ ,] ⊂ ,] , we see that the result holds for this space as well.

Remark 26.
This lemma was deduced in [25] from the case ] = 0 (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 = 0), we can make this deduction elementary.
We have where are ( , 1)-means of . It follows that where we have used Fejér's inequality ‖ ‖ 1 ≤ ‖ ‖ 1 . 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.  1, ,] if and only if On the other hand, by Lemma 27 and the property (88), which implies that is in 1, ,] , because > 1. Thus, we have proved the implication (a)⇒(b) of Theorem 11.
Case (1) is part of Proposition 13. In view of the same proposition, we can assume, in what follows, that < ∞.
The following assertion proves the desired result in Cases (2) where is independent of . This inequality remains true if ,1 is replaced by , with < 1. The function , ( < 1) belongs to (D) and the set { , : > 0, < 1} is bounded in , ,] . On the the other hand, The result follows.

12
The Scientific World Journal Theorem I. Let { } be a finite sequence, and let 0 < < ∞. Then where the "involving" constants depend only on . Let The following fact was proved in [29].
As a consequence of this lemma and Lemma 27 we have the following.
Proof. In the case = , the relation immediately follows from Theorem I. Let > . Then, by Jensen's inequality for the convex function → / , On the other hand, since ‖Δ ‖ ≥ ‖Δ ‖ , we have This proves the result in the case > . The remaining case is discussed similarly.
Proof. Consider first the case 2 < < ∞. Let Since because 2/ < 1, the sequence of partial sums of the series (1) diverges on a set ⊂ [0, 1] such that | | = 1, which follows from Theorem H. (We can assume that does not contain points where ( ) = 0 because the set of such points is denumerable). On the other hand, by Lemma 31, we have It follows that ∈ , ,] for at least one ∈ . To complete the proof we consider the polynomials (∈ (D)) It follows from Lemma B that ‖ ‖ ≤ ‖ ‖ < ∞ (in the norm of , ,] ), where is independent of . On the other hand, as noted before, the sequence is not bounded, which proves the result in the case < ∞. Remark 33. If = ∞, > 2, ] = 0, and = 0, that is, = 1 − 1/ , then we can take and apply the above approach to show that there is a function ∈ ,∞ such that̂( It is not easy to give a concrete example of such a function. A "natural" example is whose coefficients satisfy ( * * ). However, If ∈ ,∞ , then which implies (1/ − 1) ≤ 1 − 3 /2, while this implies ≤ 2, a contradiction which shows that ∉ ,∞ , for > 2.

Proof of Theorem 18
For technical reasons, we introduce the space In the proof of Proposition 34 we use the following deep result of Kisliakov [30].
where is an absolute constant.
The following proposition completes the proof of Theorem 18.