A Note on Discontinuous Functions with Continuous Second Iterate

http://dx.doi.org/10.4236/jamp.2015.36082

A Note on Discontinuous Functions with

Continuous Second Iterate

Pingping Zhang

1

_{, Xiaobing Gong}

2

1_{Department of Mathematics, Binzhou University, Shandong, China} 2_{Department of Mathematics, Neijiang Normal University, Sichuan, China}

Email: [email protected], [email protected]

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

Copyright © 2015 by authors and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

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 func-tions. Furthermore, the sufficient conditions for the continuity of the even order iterates with fi-nitely many discontinuous points are obtained.

Keywords

Iteration, Discontinuous Point, Continuous Function

1. Introduction

For a nonempty set X and n∈, the n-th iterate of a self-mapping f :X →X is defined by fn

( )

x = f

(

fn−1

( )

x

)

and f0

( )

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 f : 0,1

[ ] [ ]

→ 0,1 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.

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 f2 are stated. Throughout the paper we let I:=

( )

0,1 .

Theorem 1. Suppose that f :I→I has unique removable discontinuous point x_r. Let

( )

( )

: and : lim .

r

r r r

x x

c f x y f x

→

= = (1)

Then f2 is continuous on I if and only if the following conditions are fulfilled:

(A1) f c

( )

r =limx→y_r f x

( )

,

(A2) xr∉f I x

(

\ r

)

.

Proof. (⇒) Assume that 2

f is continuous on I, the removable discontinuous point x_r of f is continuous point of 2

f under iteration. Whether y_r defined by (1) is continuous point of f or not, we have

( )

( )

2

lim lim .

r r

x→x f x =x→y f x (2) On the other hand, using the definition of c_r and the continuity of f2,

( )

2

( )

2

( )

lim .

r

r r

x x

f c f x f x

→

= = (3)

Thus (2) and (3) lead to (A1). For an indirect proof of (A2), assume that xr= f x

( )

 for x∈I x\ r. Then

( )

( )

( )

( )

2 2

lim lim ,

r r

x→xf x =x→x f x ≠ f x = f x

which contradicts the continuity of f2 on I and gives a proof to (A2).

(⇐) It follows from (A1)

( )

( )

( )

( )

2 2

lim lim ,

r r

r r

x→x f x =x→y f x = f c = f x

implying that f2 is continuous at x_r. The condition (A2), i.e., xr∉f I x

(

\ r

)

, shows that all points x∈I x\ r

are continuous points of f2. Therefore f2 is continuous on I. This completes the proof. 

Corollary 1. Suppose that f :I→I has finitely many removable discontinuous points x x₁, ₂,,xm. If the following conditions

(A₁)

( )

lim

( )

,

i

i x y

f c = _→ f x

(A₂) xi∉f I

(

\

{

x x₁, ₂,,xm

}

)

,

are fulfilled for all i=1, 2,, ,m where :

( )

, : lim

( )

.

i

i i i x x

c = f x y = → f x Then f2n is continuous on I for

arbitrary integer n≥1.

Proof. By using the sufficiency of Theorem 1, the assumption (A₁) implies that 2

f is continuous on those

points x x₁, ₂,,xm and (A2) guarantees that all points x∈I\

{

x x1, 2,,xm

}

are continuous points of

2 f . 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 n≥1 inductively. This completes the proof. 

Theorem 2. Suppose that f :I→I has unique jumping discontinuous point x_j. Let cj:= f x

( )

j and

( )

( )

0 0

0: lim and 0: lim .

j j

j j

x x x x

y f x y f x

− +

− = _→ + = _→

Then 2

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

Proof. (⇒) In view of the definitions of cj,yj−₀,yj+₀ and the continuity of

2

f , we get

( )

( )

( )

( )

0 0

2 2

lim lim ,

j j

j j

x x x y

f c f x f x f x

− −

→ →

= = = (4)

and

( )

( )

( )

( )

0 0

2 _lim 2 _{lim .}

j j

j j

x x x y

f c f x f x f x

+ +

→ →

= = = (5)

Clearly, (4) and (5) yield (B1). Suppose the contrary to (ii), there is xj= f x

( )

for some x∈I x\ j. The limit

( )

( )

2

lim lim

j

x→x f x =x→x f x

is nonexistence since x_j is a jumping discontinuous point of f, which contradicts the fact that f2 is conti- nuous at the point x. This contradiction proves (B2).

(⇐) The condition (B1) implies

( )

( )

( )

( )

0 0

2 2

lim lim ,

j j j j

x→x− f x x→y− f x f c f x

= = = (6)

and

( )

( )

( )

( )

0 0

2 2

lim lim .

j j j j

x→x+ f x =x→y+ f x = f c = f x

(7)

Thus, (6) and (7) lead to

( )

( )

2 2

lim ,

j j

x→x f x = f x

which implies that the jumping discontinuous point x_j of f change into the continuous point of f2. Using the similar argument as the sufficiency for (B2) in Theorem 1, we can prove that all points x∈I x\ j are continuous points of f2. Thus f2 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 x x₁, ₂,,xm. If the following conditions

(B₁)

( )

( )

( )

0 0

lim _i lim _i ,

i x y x y

f c = → ₋ f x = → ₊ f x

(B₂) x_i∉f I

(

\

{

x x₁, ₂,,x_m

}

)

,

are fulfilled for all i=1, 2,, ,m where ci:= f x

( )

i , yi−0: lim= x→ −xi 0 f x

( )

, yi+0: lim= x→ +xi 0 f x

( )

. Then 2n

f is continuous on I for arbitrary integer n≥1.

Proof. The discussion is similar as that of Corollary 1. By using the sufficiency of Theorem 2, the assumption (B₁) implies that f2 is continuous on those points

x

₁

,

x

₂

,...,

x

_m and (B₂) implies that x∈I\

{

x x₁, ₂,,xm

}

are all continuous points of f2. Thus f2 is continuous on I. Consequently, we obtain the continuity of f2n

for all integers n≥1 inductively. This completes the proof. 

Theorem 3. Suppose that f :I →I has unique oscillating discontinuous point x_o. Then f2 is conti- nuous on I if and only if the following conditions are fulfilled:

(C1)

( )

( )

2 2

o

f x ≡ f x on a neighborhood U x

( )

_o ,

(C2) xo∉f I x

(

\ o

)

.

Proof. (⇒) We first show that the condition (C1) holds. Suppose the contrary, for any δ >0 there exists a

corresponding point y_δ ∈U x

(

_o,δ

)

satisfying f2

( )

y_δ ≠ f2

( )

x_o . Put ε =0 f2

( )

yδ − f2

( )

xo , then for

0

δ > there is yδ∈U x

(

o,δ

)

such that

( )

( )

2 2

0,

o

f y_δ − f x =ε implying 2

f is discontinuous at x_o, a

contradiction. This gives a proof to (C1). To prove (C2), by reduction to absurdity, we assume that

(

\

)

,

o o

x ∈f I x there is xˆ∈I x\ o such that xo= f x

( )

ˆ . Note that xo is oscillating discontinuous point of f, the limit

( )

( )

2

lim lim

o

is nothingness, which contradicts the continuity of f2. Therefore, the claim (C2) is proved.

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

( )

( )

( )

2 2 2

lim lim ,

o o o o

x→x f x =x→x f x = f x

implying the oscillating discontinuous point x_o of f is a continuous point of f2. 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

\ o

x∈I x are continuous points of f2. This completes the proof. 

Corollary 3. Suppose that f :I→I has finitely many oscillating discontinuous points x x₁, ₂,,xm. If the following conditions

(C₁) f2

( )

x ≡ f2

( )

x_i on a neighborhood U x

( )

_i ,

(C₂) xi∉f I

(

\

{

x x₁, ₂,,xm

}

)

,

are fulfilled for all i=1, 2,,m. Then f2n is continuous on I for arbitrary integer n≥1.

Proof. The discussion is similar as that of Corollary 1. By using the sufficiency of Theorem 3, the second iterate f2 is continuous on x x₁, ₂,,xm by (C1) and is continuous on all points x∈I\

{

x x1, 2,,xm

}

from (C₂), thus f2 is continuous on I. Consequently, we have the continuity of f2n for all integers n≥1 inductively. This completes the proof. 

Theorem 4. Suppose that f :→ has unique infinite discontinuous point x_i. Then f2 is continuous on  if and only if the following conditions are fulfilled:

(D1)

( )

( )

2

lim constant,

i x

f x = →∞ f x =

(D2) xi∉f

(

\xi

)

.

Proof. (⇒) Note that f2 is continuous on , then the infinite discontinuous point x_i of f is a continuous point of f2, i.e.,

( )

( )

( )

2 2

lim lim ,

i

i

x x x

f x f x f x

→ →∞

= =

which shows the limit limx→∞ f x

( )

exists and is equivalent to

( )

2

i

f x . This implies the result (D1). To prove

(D2), suppose the contrary, there exists a point x∈ \xi

 _{ such that}

_{( )}

_.

i

x = f x Since x_i is infinite disconti- nuous point of f, the limit

( )

( )

2

lim lim

i

x→xf x =x→x f x is infinite, which contradicts the continuity of 2

f . Thus, the necessary proof of (D2) is completed.

(⇐) From the assumption (D1) and the fact

( )

( )

2

limx→x_i f x =limx→∞ f x , one can see that

( )

( )

2 2

lim ,

i

i x x

f x f x A

→

= =

implying the infinite discontinuous point x_i of f is a continuous point of f2. If (D2) holds, then all real

numbers x≠x_i are continuous points of f2. This completes the proof. 

Corollary 4. Suppose that f :

(

a,+∞ →

) (

a,+∞

)

(

or :f

(

−∞,b

) (

→ −∞,b

)

)

has unique infinite discon-

tinuous point x_i, where a b, ∈. Then f2 is continuous on

(

a,+∞

) (

(

or −∞,b

)

)

if and only if the follow- ing conditions are fulfilled:

(D₁) 2

( )

_lim

( )

_{constant or}

(

2

( )

_lim

( )

_{constant ,}

)

i x i x

f x = _→+∞ f x = f x = _→−∞ f x =

(D₂) xi∉f

(

(

a,+∞

)

\xi

)

(

or xi∉f

(

(

−∞,b

)

\xi

)

)

.

Proceeding similarly as Theorem 4 one can show this corollary.

Corollary 5. Suppose that f :→ has finitely many infinite discontinuous points x x₁, ₂,,xm. If the following conditions

(D₁) f2

( )

xi =limx→∞ f x

( )

=constant,

are fulfilled for all i=1, 2,, .m Then f2n is continuous on  for arbitrary integer n≥1.

Proof. We obtain the result by using the similar argument as Corollary 1. In view of the sufficiency of Theorem 4, the second iterate f2 is continuous on those points x x₁, ₂,,xm from (D1) and is continuous on

all points x∈I\

{

x x₁, ₂,,xm

}

from (D2), thus 2

f is continuous on I. Then we have the continuity of 2n

f

for all integers n≥1 inductively. This completes the proof. 

3. Examples

In this section we demonstrate our theorems with examples.

Example 1. Consider the mapping F₁:

(

−3,3

) (

→ −3,3

)

defined by

( )

(

)

(

]

(

)

1 1

2, 3, 2 , 2

1, 2, 3, 2, 1 , 2, 1, 3 .

x x x F x x x x  _{+ ∈ − −}  − = − =  + ∈ − −   _{∈ −} 

Clearly, x₁= −2 is the unique removable discontinuous point of F₁. By simple calculation, we have

( )

( )

1 1 1 1

2

: 2 1, : lim 1,

x

c F y F x

→−

= − = − = =

( )

( )

1

1 1 lim 1 2.

x y

F c F x

→

= =

Moreover, the set ₁

(

(

3,3 \

) { }

2

)

1, 2

{ }

1 2

F − − ⊂_ _ −

  is not include the point −2. By using the sufficiency of

Theorem 1, we obtain the continuity of F₁2 on

(

−3,3

)

.

Example 2. Consider the mapping F₂:

(

−3, 4

) (

→ −3, 4

)

defined by

( )

(

]

( )

2

1 7

, 3, 1 ,

2 2

3 21 3

, 1, ,

5 10 2

3 6 12, , 2 ,

2 3 6, 2, 4 .

x x x x F x x x x x  _{+ ∈ − −}      _{+ ∈ −}   _ _ =   _{ } − + ∈  _{ }     − ∈ 

Clearly, x₂ = −1 is the unique jumping discontinuous point of F₂. By calculating we have

( )

( )

( )

1 2 1 0 2 1 0 2

1 0 1 0

3

: 1 3, : lim 3, : lim ,

2

x x

c F y₋ F x y₊ F x

→− − →− +

= − = = = = =

( )

( )

( )

1 0 1 0

2 1 lim 2 lim 2 3,

x y x y

F c F x F x

− +

→ →

= = =

and F₂

(

(

−3, 4 \

) { }

−1

)

=

[

0, 6

)

is not include the points −1. Then F₂2 is continuous on

(

−3, 4

)

using the sufficiency of Theorem 2.

Example 3. Consider the mapping F₃: 4,8

( ) ( )

→ 4,8 defined by

( )

[

)

3

9 7 sin 2, 4, ,

2

1 9

7 sin , , 6 ,

5 2

7 sin1, 6,8 .

x

F x x

Clearly, x₃=5 is an oscillating discontinuous point of F₃. By calculating, F₃

(

( ) { }

4,8 \ 5

)

⊂

[ ]

6,8 is not include 5. Moreover, F₃2

( )

x =F₃2

( )

5 = +7 sin1 for 9, 6

2 x∈  _

 . Thus the function F3 is continuous on

( )

4,8 by the sufficiency of Theorem 3.

Example 4. Consider the mapping F₄:

(

−∞ +∞ → −∞ +∞,

) (

,

)

defined by

( )

4

1 . 2 F x

x

= − −

Clearly, x₄ =2 is an infinite discontinuous point of F₄. By calculating we have

( )

( )

2

4 2 lim 4 0

x

F F x

→∞

= =

and F₄

(

(

−∞ +∞,

) { }

\ 2

)

⊆ −∞

(

, 0

)

is not include 2, then F₄ 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

[1] Lesmoir-Gordon, N. and Edney, W. (2006) Introducing Fractal Geometry. Icon Books, Cambridge.

[2] Lesmoir-Gordon, N. (2010) The Colours of Infinity: The Beauty and Power of Fractals. Springer, London.

http://dx.doi.org/10.1007/978-1-84996-486-9

[3] Mandelbrot, B. (2004) Fractals and Chaos: The Mandelbrot Set and Beyond. Springer, New York.

http://dx.doi.org/10.1007/978-1-4757-4017-2

[4] Zhang, P.P. (2012) Formal and Analytic Solutions for a Quadric Iterative Functional Equation. Electronic Journal of Differential Equations, 46, 1-9.

[5] Zhang, P. and Gong, X. (2014) Existence of Solutions for Iterative Differential Equaitons. Electronic Journal of Diffe-rential Equations, 7, 1-10.

[6] Zhang, P. and Mi, L. (2009) Analytic Solutions of a Second Order Iterative Functional Differential Equation. Applied Mathematics and Computation, 210, 277-283. http://dx.doi.org/10.1016/j.amc.2008.12.007

[7] Zhang, P. and Zhang, Q. (2010) Local Invertible Analytic Solutions of a First Order Iterative Functional Differential Equation. Acta Mathematica Sinica Chinese Series, 53, 409-416.

[8] Sun, D. (2004) Iteration of Quasi-Polynomial of Degree Two (in Chinese). J. Math., 24, 237-240.

[9] Targonski, G. (1981) Topics in Iterative Theory. Vandenhoeck and Ruprecht, Götingen.

[10] Wu, Z. and Sun, D. (2006) The Iteration of Quasi-Polynomials mappings (in Chinese). Acta Math. Sci. A., 26, 493-497.

[11] Xu, L. and Xu, S. (2006) On Iteratio of Linear Fractional Function and Applications (in Chinese). Math. Pract. Theory., 36, 225-228.

[12] Yu, Z., Yang, L. and Zhang, W. (2012) Discussion on Polynomials Having Polynomial Iterative Roots. Journal of Symbolic Computation, 47, 1154-1162. http://dx.doi.org/10.1016/j.jsc.2011.12.038