Bounded Subsets of Classes of Holomorphic Functions

Abstract

Some characterizations of boundedness in will be described, where are -algebras which consist of holomorphic functions defined by maximal functions.

1. Introduction

Let be a positive integer. The space of -complex variables is denoted by . The unit polydisk is denoted by and the distinguished boundary is . The unit ball is denoted by and is its boundary. In this paper, denotes the unit polydisk or the unit ball for and denotes for or for . The normalized (in the sense that ) Lebesgue measure on is denoted by .

The Hardy space on is denoted by . The Nevanlinna class on is defined as the set of all holomorphic functions on such thatholds. It is known that has a finite nontangential limit, denoted by , almost everywhere on .

The Smirnov class is defined as the set of all which satisfy the equalityDefine a metricfor . With the metric is an -algebra. Recall that an -algebra is a topological algebra in which the topology arises from a complete metric.

The Privalov class , , is defined as the set of all holomorphic functions on such thatholds. It is well-known that is a subalgebra of ; hence every has a finite nontangential limit almost everywhere on . Under the metric defined byfor , becomes an -algebra (cf. [1]).

Now we define the class . For , the class is defined as the set of all holomorphic functions on such thatwhere is the maximal function. The class with in the case was introduced by Kim in [2]. As for and , the class was considered in [3, 4]. For , define a metricwhere . With this metric is also an -algebra (see [5]).

It is well-known that the following inclusion relations hold:Moreover, it is known that [6].

A subset of a linear topological space is said to be bounded if for any neighborhood of zero in there exists a real number , such that . Yanagihara characterized bounded subsets of in the case [7]. As for with in the case , Kim described some characterizations of boundedness (see [2]). For and , these characterizations were considered by Meštrović [8]. As for with in the case , Subbotin investigated the properties of boundedness [1].

In this paper, we consider some characterizations of boundedness in with in the case .

2. The Results

Theorem 1. Let . is bounded if and only if
(i) there exists such thatfor all ;
(ii) for each there exists such thatfor any measurable set with the Lebesgue measure .

Proof. ⁡
Necessity. Let be a bounded subset of . We put .
(i) For any , there is a number such thatfor all and . It follows thatfor all and . Sinceusing the elementary inequalitywe haveThus (i) is satisfied.
(ii) For given , we take as and as above. Next take such thatThen, for each set with and for every , we obtainTherefore, the condition (ii) is satisfied.
Sufficiency. Letbe a neighborhood of in . Take such thatThen, there is such that (ii) is satisfied. For , we can find so thatby Chebyshev’s inequality. We haveChoose such that . Then, using inequality (14) andwe obtain, for every ,Therefore we get , which shows is a bounded subset of .
The proof of the theorem is complete.

Remark 2. We note that the characterization of boundedness in has the same conditions as the characterization of boundedness in the Smirnov class in the case ([7], Theorem ), the class with in the case ([2], Theorem ), and the Privalov class ([1], Theorem ). On the other hand, we see that (ii) implies (i) in Theorem 1. Suppose that (ii) holds. Then there is a positive integer such thatfor any measurable set with . There are measurable sets such that , for , and for every . Thenholds (cf. [8] (Theorem and Remark )).

Next we show a standard example of a bounded set of . The following theorem is easily proved in the same way of [1] (p.236) and [2] (Theorem ); therefore, we do not prove it here.

Theorem 3. Let . If , then form a bounded set in .

Let and we set . Subbotin proved an equivalent condition that a subset is bounded. The following is a theorem by Subbotin.

Theorem 4 (see [1]). Let . A subset is bounded if and only if the following two conditions are satisfied:
(i) There exists such that for all .
(ii) For each there exists such thatfor any measurable set with the Lebesgue measure .

As shown in [1, 3], for any the class coincides with the class and the metrics and are equivalent. Therefore the topologies induced by these metrics are identical on the set .

The following theorem is clear; therefore the proof may be omitted.

Theorem 5. Let . A subset is bounded if and only if the following two conditions are satisfied:
(i) There exists such thatfor all .
(ii) For each there exists such thatfor any measurable set with the Lebesgue measure .

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The author is partly supported by the Grant for Assist KAKEN from Kanazawa Medical University (K2017-6).