International Journal of Analysis
Volume 2017 (2017), Article ID 9134768, 4 pages
https://doi.org/10.1155/2017/9134768

Research Article

Bounded Subsets of Smirnov and Privalov Classes on the Upper Half Plane

Yasuo Iida

Department of Mathematics, Kanazawa Medical University, Uchinada, Ishikawa 920-0293, Japan

Correspondence should be addressed to Yasuo Iida; yiida@kanazawa-med.ac.jp

Received 30 August 2017; Accepted 13 November 2017; Published 19 December 2017

Academic Editor: Ahmed Zayed

Copyright © 2017 Yasuo Iida. 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.

Abstract.

Some characterizations of boundedness in and    will be described, where denote the Smirnov class and the Privalov class on the upper half plane , respectively.

1. Introduction

Let and denote the unit disk and the unit circle in , respectively. The Privalov class , , is defined as the set of all holomorphic functions on , satisfying where denotes normalized Lebesgue measure on . The notion of was introduced by Privalov [1] and has been explored by several authors (see [2–4]). Letting , we have the Nevanlinna class . It is well-known that each function in has the nontangential limit   a.e. and that (and, hence, for ) is subharmonic if is holomorphic. Define a metric for . With the metric becomes an -algebra [2]. Recall that an -algebra is a topological algebra in which the topology arises from a complete metric.

We denote the Smirnov class by , which consists of all holomorphic functions on such that    for some ,  , where the right side denotes the Poisson integral of on . It is known that if , belongs to if and only ifUnder the metric for , the class is also an -algebra (see [5]).

For , the class is defined as the set of all holomorphic functions on such thatwhere is the maximal function. The class was introduced by Kim in [6]. As for , the class was considered in [7, 8]. For , define a metric where . With this metric is also an -algebra (see [9]).

It is well-known that   , where denotes the Hardy space on . Moreover, it is known that    [6].

Mochizuki [10] introduced the Nevanlinna class and the Smirnov class on the upper half plane : the class is the set of all holomorphic functions on satisfying and the set of all holomorphic functions on satisfying    for some , , where the right side denotes the Poisson integral of on . It is well-known that each function in has the nontangential limit (a.e. ). Let . Then if and only if (see [10]). Moreover, under the metricthe class becomes an -algebra [10].

The class    is defined as the set of all holomorphic functions on such thatwhere . The class with was introduced by Ganzhula in [11]. As for , Efimov and Subbotin investigated this class [12]. For , define a metric where . With this metric is also an -algebra (see [11, 12]).

In [13], the class was introduced, analogous to ; that is, we denote by the set of all holomorphic functions on such that Each has the nontangential limit for a.e. , and under the metric,the class becomes an -algebra [13].

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 [14]. As for with , Kim described some characterizations of boundedness (see [6]). For , these characterizations were considered by Meštrović [15]. As for with , Ganzhula investigated the properties of boundedness [11] and Efimov characterized bounded subsets of in the case [16]. In recent paper [17], the author described bounded subsets of in the case .

In this paper, we consider some characterizations of boundedness in and .

2. The Results

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

Proof. We follow [16, Theorem ].
Necessity. Let be a bounded subset of .
(i) For any number there exists , , such thatfor all . Utilizing the inequality   , it follows that, from (16),for all . Therefore, condition (i) holds.
(ii) For any number , we take as . Choose a number , , such that equality (16) holds for all . Then for any measurable set , using Minkowski’s inequality, we have the estimateIf we take as , thenfor all and any measurable set , . Thus condition (ii) holds.
Sufficiency. Let conditions (i) and (ii) hold for a subset of , . Consider a neighborhoodTake as . According to (ii), there exists a number such thatfor all and any measurable set , . Next there exists a finite constant such that condition (i) holds for all . Applying Chebyshev’s inequality to the Lebesgue measure of the set for , the following estimate is valid:Then we may assume and in inequality (21); that is, for all . Therefore, for any number and all , we have the following:where , , and . By using the elementary inequality    to the second integral in (23), using (21) and takingwe have the following estimateTherefore, and the set is bounded in by definition.
The proof of the theorem is complete.

Next we consider some characterizations of boundedness in . Proof of the following theorem can be obtained by taking in the whole proof of Theorem 1; therefore, this proof may be omitted.

Theorem 2. is bounded if and only if
(i) there exists such that for 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).

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