Research Article | Open Access

Volume 2013 |Article ID 878253 | https://doi.org/10.1155/2013/878253

Jonald P. Fenecios, Emmanuel A. Cabral, "Left Baire-1 Compositors and Continuous Functions", International Journal of Mathematics and Mathematical Sciences, vol. 2013, Article ID 878253, 3 pages, 2013. https://doi.org/10.1155/2013/878253

# Left Baire-1 Compositors and Continuous Functions

Revised23 Sep 2013
Accepted26 Sep 2013
Published06 Nov 2013

#### Abstract

We showed that the class of left Baire-1 compositors is precisely the class of continuous functions. This answers a characterization problem posed in the work by Zhao (2007, page 550) and settles in the affirmative the conjecture in the work by Fenecios and Cabral (2012, page 43). Moreover, based on the above result we provide a new proof that the class of functions where is the class of all positive continuous functions and is the class of all positive constants as defined in the work by Bakowska and Pawlak (2010) is exactly the class of all continuous functions.

#### 1. Introduction

Let be a complete separable metric space. In the classical sense a function where is the real line is Baire-1 if, for every open set , is an set in . Equivalently, a function is Baire-1 if, for every closed set in , the restriction has a point of continuity in . In fact, the set of points of continuity of is a residual subset of .

Recently, Lee et al.  discovered a new equivalent definition of Baire-1 functions. A function is Baire-1 if and only if for each there is a positive function such that for any Various investigations have been done on the class of Baire-1 functions as well as on its subclasses using the - characterization. See, for instance, .

Let us recall in  that the notion of the left Baire-1 compositors was introduced as a natural counterpart of the notion of right Baire-1 compositors. A function is a left Baire-1 compositor if for every Baire-1 function the composition of functions is Baire-1. Unlike the right Baire-1 compositors there is no known characterization for the left Baire-1 functions. It was shown in  that there exists a function with finite set of discontinuity which is not a left Baire-1 compositor. This observation leads to the conjecture that left Baire-1 functions must have very nice properties. We proved that this is indeed the case: the class of the left Baire-1 compositors is precisely the class of all continuous functions defined on . Moreover, we use this fact to reestablish that where is the class of all positive continuous functions, is the class of all positive real constants, and is the class of all continuous functions on as defined in .

#### 2. Main Result

Let us denote the by . For easy reference, we present the following useful results.

In [4, Proposition 1], it was shown that is Baire-1 if and only if for every positive continuous function there is a corresponding positive function such that for any On the other hand, following  a function belongs to if for every there exists a function such that for any where and are families of positive functions defined on .

The following theorem was proved in .

Theorem 1. The following holds: where , , and are the families of all positive lower semicontinuous, functions all positive constant functions, and all continuous functions, respectively, defined on .

Another important result is due to Alinayat in . We will state it as a theorem.

Theorem 2. If is continuous, then for each there exists a positive continuous function such that for any

We are now ready to prove our main result. To make the result as general as possible we consider real-valued Baire-1 functions defined in the Euclidean space for with the standard Euclidean norm.

Theorem 3. The following statements are equivalent.(1) is continuous.(2)For every Baire-1 function , is Baire-1.That is, is a left Baire-1 compositor.

Proof. It is easy to see that if is continuous, then is a left Baire-1 compositor. Suppose is a left Baire-1 compositor. If, on the contrary, we suppose has a discontinuity point , we may assume without loss of generality that is the only point of discontinuity of . Then may take one of the following forms:(I) with or ,(II) (III)Note that the functions and are continuous functions on their own domain. We will tackle Case I. Let be an enumeration of the cross product . Let such that First, we need to show that is Baire-1. We will adapt a well-known technique in showing that is continuous on the complement of . If , then is discontinuous at . Let and be an arbitrary element of . We can find a natural number such that . For such , choose such that for each . Suppose . Then either or for some . If , then clearly . If for some , then . Hence, is continuous on the complement of . If is any nonempty perfect set in , then is uncountable. Hence, has at least a point of continuity in . Consequently, is Baire-1. Next, we show that is not Baire-1 so that is not a left Baire-1 compositor, a contradiction to our hypothesis. Now, It is straightforward to show that is discontinuous everywhere in some set of the second category in . Case II may be dealt with similarly with given by Following the same line of reasoning in Case I, one can show that is discontinuous everywhere in some set of the second category in . The remaining case can be proved similarly. All these arguments show that is not Baire-1. Therefore, must be a continuous function.

A part of the statement of the next result is contained in . This inclusion is in Theorem 1. With our main result, Theorem 1 (Theorem 11 in ) admits a very straightforward proof.

Corollary 4. Let be the family of all positive continuous functions, the family of all positive constant functions, the family of all positive lower semicontinuous functions, and the family of all continuous functions defined on . Then

Proof. The inclusion is immediate from Theorem 2. Let . We will show that is a left Baire-1 compositor so that by Theorem 3   is continuous. Let be any Baire-1 function and let . Then there is a positive continuous function such that for any Since is Baire-1, there is a positive function such that for any It follows that for any Thus, . Since , then it follows that The proof is complete.

#### Acknowledgments

The authors are thankful to the referee for the helpful suggestions that led to the improvement of the current paper. Moreover, the first author wishes to thank the Department of Science and Technology through the Science Education Institute for the financial support while this research was ongoing.

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