Received 6 May 2015; accepted 26 June 2015; published 30 June 2015

ABSTRACT

This paper investigates four classes of functions with a single discontinuous point. We give the sufficient and necessary conditions under which the second order iterates are continuous functions. Furthermore, the sufficient conditions for the continuity of the even order iterates with finitely many discontinuous points are obtained.

Keywords:

Iteration, Discontinuous Point, Continuous Function

1. Introduction

For a nonempty set X and, the n-th iterate of a self-mapping is defined by and for all inductively. As a nonlinear operator, iteration usually amplifies the complexity

of functions [1] - [7] , computing the n-th iterate of functions is complicated, even for simple functions (see [8] - [12] ). On the other hand, iteration can turn complex functions into simple ones. Recently, the following problem was first formulated by X. Liu, L. Liu and W. Zhang: what are discontinuous functions whose iterates of a certain order are continuous? This question, together with three classes of discontinuous functions defined on compact interval, was answered in the affirmative in [13] . That is, suppose that with a single discontinuous point (removable discontinuous point, jumping discontinuous or oscillating discontinuous), the authors respectively gave the sufficient and necessary conditions under which the second order iterates are continuous functions.

The purpose of this paper is to study the discontinuous functions defined on open interval. For four classes of discontinuous functions with unique discontinuous point, we obtain the sufficient and necessary conditions for functions being continuous ones under second iterate, which are easily verified respectively. As corollaries, the sufficient conditions for the continuity of the even order iterates with finitely many discontinuous points are obtained. Our results are illustrated by examples in Section 3 .

2. Main Results

In this section the main results for the continuity of are stated. Throughout the paper we let

Theorem 1. Suppose that has unique removable discontinuous point. Let

(1)

Then is continuous on I if and only if the following conditions are fulfilled:

(A₁)

(A₂)

Proof. (Þ) Assume that is continuous on I, the removable discontinuous point of f is continuous point of under iteration. Whether defined by (1) is continuous point of f or not, we have

(2)

On the other hand, using the definition of and the continuity of,

(3)

Thus (2) and (3) lead to (A₁). For an indirect proof of (A₂), assume that for Then

which contradicts the continuity of on I and gives a proof to (A₂).

(Ü) It follows from (A₁)

implying that is continuous at. The condition (A₂), i.e., , shows that all points are continuous points of. Therefore is continuous on I. This completes the proof. W

Corollary 1. Suppose that has finitely many removable discontinuous points. If the following conditions

()

()

are fulfilled for all where Then is continuous on I for arbitrary integer.

Proof. By using the sufficiency of Theorem 1, the assumption () implies that is continuous on those points and () guarantees that all points are continuous points of.

Thus is continuous on I. Since the composition of continuous functions is continuous, we get the continuity of for all integers inductively. This completes the proof. W

Theorem 2. Suppose that has unique jumping discontinuous point. Let and

Then is continuous on I if and only if the following conditions are fulfilled:

(B₁)

(B₂)

Proof. (Þ) In view of the definitions of and the continuity of, we get

(4)

and

(5)

Clearly, (4) and (5) yield (B₁). Suppose the contrary to (ii), there is for some. The limit

is nonexistence since is a jumping discontinuous point of f, which contradicts the fact that is conti- nuous at the point. This contradiction proves (B₂).

(Ü) The condition (B₁) implies

(6)

and

(7)

Thus, (6) and (7) lead to

which implies that the jumping discontinuous point of f change into the continuous point of. Using the similar argument as the sufficiency for (B₂) in Theorem 1, we can prove that all points are continuous points of. Thus is continuous on I. That is, we prove the sufficiency. This completes the proof. W

Corollary 2. Suppose that has finitely many jumping discontinuous points If the following conditions

()

()

are fulfilled for all where Then is continuous on I for arbitrary integer

Proof. The discussion is similar as that of Corollary 1. By using the sufficiency of Theorem 2, the assumption () implies that is continuous on those points and () implies that are all continuous points of. Thus is continuous on I. Consequently, we obtain the continuity of for all integers inductively. This completes the proof. W

Theorem 3. Suppose that has unique oscillating discontinuous point. Then is conti- nuous on I if and only if the following conditions are fulfilled:

(C₁) on a neighborhood,

(C₂).

Proof. (Þ) We first show that the condition (C₁) holds. Suppose the contrary, for any there exists a

corresponding point satisfying Put then for there is such that implying is discontinuous at, a

contradiction. This gives a proof to (C₁). To prove (C₂), by reduction to absurdity, we assume that there is such that Note that is oscillating discontinuous point of f, the limit

is nothingness, which contradicts the continuity of. Therefore, the claim (C₂) is proved.

(Ü) From the assumption (C₁) we see that

implying the oscillating discontinuous point of f is a continuous point of. On the other hand, one can use the similar argument as the sufficiency for the condition (C₂) in Theorem 1 and prove that all points are continuous points of. This completes the proof. W

Corollary 3. Suppose that has finitely many oscillating discontinuous points If the following conditions

()on a neighborhood,

()

are fulfilled for all. Then is continuous on I for arbitrary integer

Proof. The discussion is similar as that of Corollary 1. By using the sufficiency of Theorem 3, the second iterate is continuous on by () and is continuous on all points from (), thus is continuous on I. Consequently, we have the continuity of for all integers inductively. This completes the proof. W

Theorem 4. Suppose that has unique infinite discontinuous point. Then is continuous on if and only if the following conditions are fulfilled:

(D₁)

(D₂)

Proof. (Þ) Note that is continuous on, then the infinite discontinuous point of f is a continuous point of, i.e.,

which shows the limit exists and is equivalent to. This implies the result (D₁). To prove (D₂), suppose the contrary, there exists a point such that Since is infinite disconti- nuous point of f, the limit

is infinite, which contradicts the continuity of. Thus, the necessary proof of (D₂) is completed.

(Ü) From the assumption (D₁) and the fact one can see that

implying the infinite discontinuous point of f is a continuous point of. If (D₂) holds, then all real numbers are continuous points of. This completes the proof. W

Corollary 4. Suppose that has unique infinite discon- tinuous point, where Then is continuous on if and only if the follow- ing conditions are fulfilled:

()

()

Proceeding similarly as Theorem 4 one can show this corollary.

Corollary 5. Suppose that has finitely many infinite discontinuous points If the following conditions

()

()

are fulfilled for all Then is continuous on for arbitrary integer

Proof. We obtain the result by using the similar argument as Corollary 1. In view of the sufficiency of Theorem 4, the second iterate is continuous on those points from () and is continuous on

all points from (), thus is continuous on I. Then we have the continuity of for all integers inductively. This completes the proof. W

3. Examples

In this section we demonstrate our theorems with examples.

Example 1. Consider the mapping defined by

Clearly, is the unique removable discontinuous point of. By simple calculation, we have

Moreover, the set is not include the point -2. By using the sufficiency of Theorem 1, we obtain the continuity of on.

Example 2. Consider the mapping defined by

Clearly, is the unique jumping discontinuous point of. By calculating we have

and is not include the points -1. Then is continuous on using the sufficiency of Theorem 2.

Example 3. Consider the mapping defined by

Clearly, is an oscillating discontinuous point of. By calculating, is not include 5. Moreover, for. Thus the function is continuous on by the sufficiency of Theorem 3.

Example 4. Consider the mapping defined by

Clearly, is an infinite discontinuous point of. By calculating we have

and is not include 2, then is continuous on by the sufficiency of Theorem 4.

Acknowledgements

We thank the Editor and the referee for their comments. Project supported by Shandong Provincial Natural Science Foundation of China (ZR2014AL003), Scientific Research Fund of Sichuan Provincial Education Departments (12ZA086), Scientific Research Fund of Shandong Provincial Education Department (J12L59) and Doctoral Fund of Binzhou University (2013Y04).

References