A Note on Discontinuous Functions with Continuous Second Iterate ()
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).