A Note on Discontinuous Functions with Continuous Second Iterate

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.


Introduction
For a nonempty set X and n ∈  , the n-th iterate of a self-mapping : f X X → is defined by ( ) ( ) ( ) ( ) 0 f x x = for all x X ∈ 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 [ ] [ ] : 0,1 0,1 f → 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 .

Main Results
In this section the main results for the continuity of 2 f are stated.
Then 2 f is continuous on I if and only if the following conditions are fulfilled: f is continuous on I, the removable discontinuous point r x of f is continuous point of 2 f under iteration.Whether r y defined by ( 1) is continuous point of f or not, we have ( ) ( ) On the other hand, using the definition of r c and the continuity of 2 f , ( ) ( ) ( ) Thus ( 2) and (3) lead to (A 1 ).For an indirect proof of (A 2 ), assume that ( ) which contradicts the continuity of 2 f on I and gives a proof to (A 2 ).(⇐) It follows from (A 1 ) implying that 2 f is continuous at r x .The condition (A 2 ), i.e., ( ) , shows that all points \ r x I x ∈ are continuous points of 2 f .Therefore 2 f is continuous on I.This completes the proof. Corollary 1. Suppose that : f I I → has finitely many removable discontinuous points 1 2 , , , m x x x  . If the following conditions Thus 2 f is continuous on I. Since the composition of continuous functions is continuous, we get the continuity of 2n f for all integers 1 n ≥ inductively.This completes the proof. Theorem 2. Suppose that : f I I → has unique jumping discontinuous point j x .Let ( ) : lim and : lim .
Then 2 f is continuous on I if and only if the following conditions are fulfilled: and the continuity of 2 f , we get and Clearly, ( 4) and ( 5) yield (B 1 ).Suppose the contrary to (ii), there is ( ) x is a jumping discontinuous point of f, which contradicts the fact that 2 f is conti- nuous at the point x .This contradiction proves (B 2 ).
(⇐) The condition (B 1 ) implies and Thus, ( 6) and ( 7) lead to which implies that the jumping discontinuous point j x of f change into the continuous point of 2 f .Using the similar argument as the sufficiency for (B 2 ) in Theorem 1, we can prove that all points \ j x I x ∈ are continuous points of 2  f .Thus 2 f is continuous on I.That is, we prove the sufficiency.This completes the proof. Corollary 2. Suppose that : f I I → has finitely many jumping discontinuous points 1 2 , , , .
 where ( ) We first show that the condition (C 1 ) holds.Suppose the contrary, for any 0 δ > there exists a corresponding point ( ) .
contradiction.This gives a proof to (C 1 ).To prove (C 2 ), by reduction to absurdity, we assume that ( ) is nothingness, which contradicts the continuity of 2 f .Therefore, the claim (C 2 ) is proved.(⇐) From the assumption (C 1 ) we see that implying the oscillating discontinuous point o x of f is a continuous point of 2 f .On the other hand, one can use the similar argument as the sufficiency for the condition (C 2 ) in Theorem 1 and prove that all points \ o x I x ∈ are continuous points of 2 f .This completes the proof. Corollary 3. Suppose that : f I I → has finitely many oscillating discontinuous points 1 2 , , , .
lim lim , which shows the limit ( ) →∞ exists and is equivalent to ( ) f x .This implies the result (D 1 ).To prove (D 2 ), suppose the contrary, there exists a point is infinite, which contradicts the continuity of 2 f .Thus, the necessary proof of (D 2 ) is completed.(⇐) From the assumption (D 1 ) and the fact ( ) ( ) lim constant or lim constant , f is continuous on I. Then we have the continuity of 2n f for all integers 1 n ≥ inductively.This completes the proof.

Examples
In this section we demonstrate our theorems with examples.
Example 1.Consider the mapping ( ) ( ) x F .By simple calculation, we have ( ) ( )  F is continuous on ( ) We obtain the result by using the similar argument as Corollary 1.In view of the sufficiency of Theorem 4, the second iterate2 By using the sufficiency of Theorem 1, we obtain the continuity of2 1F on ( )