Abstract

For and let be holomorphic mean Lipschitz spaces on the unit ball in . It is shown that, if the space is a multiplicative algebra. If , then the space is not a multiplicative algebra. We give some sufficient conditions for a holomorphic function to be a pointwise multiplier of .

1. Introduction

Let and be two function spaces. We call a pointwise multiplier from to if for every . The collection of all pointwise multipliers from to is denoted by . When , we let .

Multipliers arise in the theory of differential equations. Coefficients of differential operators can be naturally considered as multipliers. The same is true for symbols of more general pseudodifferential operators.

To give some motivations for our study, we recall studies on multipliers of Sobolev spaces. Strichartz [1] was the first who studied on multipliers of Sobolev spaces. Let be a bounded domain in with Lipschitz boundary. Let . For , let be the Sobolev spaces over . Given and in , one cannot in general expect that their product will belong to . However, if , then there exists a constant depending on , and , such that [13] This implies that . Since contains constant functions, . Thus, we have

In the complex case, multipliers on Hardy-Sobolev spaces on the unit ball in were studied by [4, 5] for and [6] for . Let denote the open unit ball in . Let and be a positive integer. Let be the Hardy-Sobolev space of order . In papers [46], it was proved that Complete characterization of multipliers on other Hardy-Sobolev spaces of nonregular cases remains open, but Beatrous and Burbea [7] gave some sufficient conditions for functions to be pointwise multipliers in these nonregular cases. Ortega and Fàbrega [8] introduced a family of nonisotropic tent-Sobolev spaces to characterize multipliers in some Hardy-Sobolev spaces of nonregular cases. Usually, characterization of multipliers of nonregular cases is difficult.

Many authors have studied properties of multipliers for several function spaces (see [912] for Dirichlet-type spaces, [13, 14] for Bloch-type spaces, [4] for Bergman-Sobolev spaces, [15] for mixed norm spaces, [16] for spaces, [17] for spaces, and [18] for the BMO space).

For points and in , we write Let denote the unit sphere in . The normalized Lebesgue measure on will be denoted by . Let denote the space of all holomorphic functions in . Given , , and , we define When , we write For , the Hardy space consists of all functions such that See [19] for basic information about the Hardy spaces.

We denote by the radial derivative of in defined by We consider the space of holomorphic functions on such that for . We define the norm of as follows: It can be shown that the norm is independent of the choice of ; see [20]. When , this is exactly the classical holomorphic Lipschitz space ; see [19].

We adapt the first order mean variation defined as follows: where denotes the group of all unitary operators on , denotes the identity of , and . Then, we have for (see [20]). This justifies our usage of the term holomorphic mean Lipschitz space for with . Now, for , we consider the second-order mean variation defined as follows: It was shown in [21] that, if , and ,   then

Theorem 1.1. Let and . (i) If (regular), then is a multiplicative algebra. (ii) If (nonregular), then is not a multiplicative algebra.

By (ii) of Theorem 1.1, the space is not a multiplicative algebra. We give some sufficient conditions for a holomorphic function to be a pointwise multiplier of as follows. We do not know if our sufficient condition is also necessary.

Theorem 1.2. Let . Then, .

Throughout the paper, we write or for nonnegative quantities and whenever there is a constant (independent of the parameters in and ) such that . Similarly, we write if and .

2. Auxiliary Embedding Results

The ball algebra is the class of all functions that are continuous on the closed ball and that are holomorphic in its interior .

Proposition 2.1. Let and . (i)There is a function in the ball algebra that is not in . (ii) If , then there is a function in that is not in . (iii) If , then .

Remark 2.2. By (ii) and (iii), we can see that if and only if .

Proof. (i) Let be a sequence of Ryll and Wojtaszczyk [22] homogeneous polynomials in the unit sphere of such that has degree and Let This function was constructed in [7]. In fact, it was shown in [7] that this function is not contained in any Hardy-Sobolev space. Thus, the result of (i) of Proposition 2.1 follows, since every mean Lipschitz space is contained in a Hardy-Sobolev space.
Since the series converges uniformly on , its sum is therefore in the ball algebra. It is enough to prove that for . If , then However, since the polynomials are orthogonal, for any , Take Then, where By (2.3) and (2.6), it is a contradiction. Thus, for .
(ii) It is clear, if we consider the function with .
(iii) Let . Let be the greatest integer less than and . Let . By the Cauchy's integral formula, we have Making a change of variables and replacing by , we get
By (2.10) and Hölder's inequality, we have Take . Then, we obtain Therefore, if .

Proposition 2.3. Let and . (i). (ii).

Proof. (i) Let and , where is the greatest integer less than . By the fundamental theorem of calculus and Minkowski's inequality, we have Applying this repeatedly and using Fubini's theorem, we obtain
Let . By (2.11), we have For , we take . Then, For , we have
By (2.14) and (2.17), we have since . Thus, we get the result.
(ii) If in (2.18), then, for any , we have

Remark 2.4. The obvious question: Is contained in ? In the case , this was proved by Bourdon et al. in [23]. However, their method does not work in higher dimensions.
We have some observations by a Carleson measure. There is a characterization for functions by a Carleson measure such that if and only if is a Carleson measure, where is the volume measure on (see [19]). Even though we cannot prove that is a Carleson measure for , we have some weak results such that is a Carleson measure for with . For the proof, let and . By (2.17), we have . This means that Hence, we have Thus, is a Carleson measure (see [19]). However, the embedding problem is still open.

3. Regular Cases

We need an elementary variant of Hölder's inequality.

Lemma 3.1. Let with . Then, for and , the product is in and

Theorem 3.2. If , then is a multiplicative algebra.

Proof. Let be the greatest integer less than and . Let . We will prove that We note that Since , by (iii) of Proposition 2.1, we have .
Let . Then, applied to the sphere of radius (see [24], Theorem 1 on page 69, for such inequalities). Thus, for , we have Therefore, by a variant of Hölder's inequality, That is,

4. Nonregular Cases

A function is a multiplier for if the multiplication operator is continuous from to itself. The space of those multipliers will be denoted by .

Let . Since , we have . Thus, Hence,

The following is a special case of Lemma 5.1 in [7].

Lemma 4.1 (see [7]). Let . Then, is bounded pointwise by its multiplier norm.

Corollary 4.2. Let . Then, we have

Proof. Let . Since , we have . Thus, Hence, Thus, by Lemma 4.1, we get . By (i) of Proposition 2.1, there is a function . Since , it follows that . Thus, we get the result.

Theorem 4.3. Let . Then, is not a multiplicative algebra.

Proof. If is multiplicative, then By (ii) of Proposition 2.1, this is a contradiction.

5. Some Sufficient Conditions for a Nonregular Case

Lemma 5.1. Let . Let be the integer part of and . Let . Then, for , we have

Proof. We proved this inequality at (2.17) in the proof of (i) of Proposition 2.3.

Lemma 5.2. Let and . Then, we have

Proof. We proved this inequality at (2.14) in the proof of (i) of Proposition 2.3.

Theorem 5.3. Let . Then, .

Proof. Let and . Let be the integer part of and . It is enough to prove that
We note that Thus,
Let . Then, By Proposition 2.1, we have Choose with . Then, Also, Thus,
Now, let .
If , then, since and ,
If , we choose such that Then, By (ii) of Proposition 2.3, . By a variant of Hölder's inequality, we have
Now, we consider the case . Since is decreasing in the parameter , it is enough to consider for sufficiently near so that . We choose such that Then, Since , we can choose such that
By a variant of Hölder's inequality, we have By Lemmas 5.2 and 5.1, Here, Moreover, by Lemmas 5.2 and 5.1 again, we have Thus, it follows that where Thus, This completes the proof.

Acknowledgment

The author wishes to thank the referee for some very helpful suggestions and comments that improved the presentation of the paper. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (NRF-2009-0073976).