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Research | Open | Published:

Some new theorems of expanding mappings without continuity in cone metric spaces

The Erratum to this article has been published in Fixed Point Theory and Applications 2014 2014:178

Abstract

In this paper, fixed point theorems for one mapping and common fixed point theorems for two mappings satisfying generalized expansive conditions are obtained. The mappings are not necessarily continuous and the cone is not normal. These results improve and generalize several well-known comparable results in (Aage and Salunke, in Acta Mathematica Sinica, English Series 27(6):1101-1106, 2011). Moreover, examples are given to support our new results.

MSC:54H25, 47H10.

1 Introduction and preliminaries

Recently, Huang and Zhang [1] introduced the concept of a cone metric space as a generalization of a metric space. They proved the properties of sequences in cone metric spaces and obtained various fixed point theorems for contractive mappings. Afterwords, Abbas and Jungck [2] established the common fixed points for two mappings without exploiting the notion of continuity. Since then, common fixed point theorems in cone metric spaces have been proved for mappings satisfying different contractive conditions by many authors (see [38]). But there are few results about expanding mappings. Chintaman and Jagannath [9] introduced several meaningful fixed point theorems for one expanding mapping. However, the mapping depended strongly on continuity. In this paper, we delete the continuity of the mappings and obtain some fixed point theorems for one expanding mapping and introduce common fixed point theorems for two expanding mappings, which satisfy generalized expansive conditions in nonnormal cone metric spaces. These results improve and generalize some important known results in [9, 10].

We recall some definitions of cone metric spaces and some of their properties [1]. Let E be a real Banach space and P be a subset of E. θ denotes the zero element of E and intP denotes the interior of P. The subset P is called a cone if and only if:

  1. (i)

    P is closed, nonempty and P{θ},

  2. (ii)

    a,bR, a,b0, x,yPax+byP,

  3. (iii)

    xP and xPx=θ.

Given a cone PE, we define a partial ordering ≤ with respect to P by xy if and only if yxP. We will write x<y if xy and xy, while xy will stand for yxintP. A cone P is called normal if there is a number K>0 such that for all x,yP,

θxyimpliesxKy.

The least positive number satisfying the above inequality is called the normal constant of P.

Definition 1.1 ([1])

Let X be a nonempty set. Suppose that the mapping d:X×XE satisfies the following:

  1. (d1)

    θd(x,y) for all x,yX and d(x,y)=θ if and only if x=y;

  2. (d2)

    d(x,y)=d(y,x) for all x,yX;

  3. (d3)

    d(x,y)d(x,z)+d(z,y) for all x,y,zX.

Then d is called a cone metric on X and (X,d) is called a cone metric space. It is clear that the cone metric space is more general than a metric space.

Definition 1.2 ([1])

Let (X,d) be a cone metric space. Then we say that { x n } is

  1. (i)

    a Cauchy sequence if for every cE with cθ, there is N such that for all n,m>N, d( x n , x m )c;

  2. (ii)

    a convergent sequence if for every cE with cθ, there is N such that for all m>N, d( x m ,x)c for some fixed x in X.

A cone metric space X is said to be complete if every Cauchy sequence in X is convergent in X.

Lemma 1.3 ([11])

The limit of a convergent sequence in a cone metric space is unique.

2 Main results

In this section, we prove some fixed point theorems for expanding mappings without continuity in the following theorems.

Theorem 2.1 Let (X,d) be a complete cone metric space. Suppose the mapping f:XX is onto and satisfies

d(fx,fy) a 1 d(x,y)+ a 2 d(x,fx)+ a 3 d(y,fy)+ a 4 d(x,fy)+ a 5 d(y,fx)
(2.1)

for all x,yX, where a i (i=1,2,3,4,5) satisfies a 1 + a 2 + a 3 >1 and a 3 1+ a 4 . Then f has a fixed point.

Proof Since f is an onto mapping, for each x 0 X, there exists f x 1 = x 0 . Continuing this process, we can define { x n } by x n =f x n + 1 , n=0,1,2, . Without loss of generality, we suppose that x n 1 x n for all n1. According to (2.1), we have

d ( x n , x n 1 ) = d ( f x n + 1 , f x n ) a 1 d ( x n + 1 , x n ) + a 2 d ( x n + 1 , f x n + 1 ) + a 3 d ( x n , f x n ) + a 4 d ( x n + 1 , f x n ) + a 5 d ( x n , f x n + 1 ) = a 1 d ( x n + 1 , x n ) + a 2 d ( x n + 1 , x n ) + a 3 d ( x n , x n 1 ) + a 4 d ( x n + 1 , x n 1 ) + a 5 d ( x n , x n ) .

By d( x n + 1 , x n 1 )d( x n + 1 , x n )d( x n 1 , x n ), the above inequality implies that

d( x n + 1 , x n ) 1 a 3 + a 4 a 1 + a 2 + a 4 d( x n , x n 1 ).

Let h= 1 a 3 + a 4 a 1 + a 2 + a 4 . By a 3 1+ a 4 and a 1 + a 2 + a 3 >1, we know a 1 + a 2 + a 4 >1 a 3 + a 4 0 and h[0,1). Hence, we get

d( x n + 1 , x n )hd( x n , x n 1 ).

So, by the triangle inequality, for any n>m, we see

d ( x n , x m ) d ( x n , x n 1 ) + d ( x n 1 , x n 2 ) + + d ( x m + 1 , x m ) ( h n 1 + h n 2 + + h m ) d ( x 1 , x 0 ) h m 1 h d ( x 1 , x 0 ) .

Thus, as h[0,1), we can choose a natural number N 0 such that h m 1 h d( x 1 , x 0 )c for each cθ and m> N 0 . Hence, we see

d( x n , x m )cfor all n>m> N 0 .

Therefore, { x n } is a Cauchy sequence in (X,d).

Since X is complete, there exists qX such that f x n + 1 = x n q as n. Consequently, we can find a pX such that fp=q. Now, we show that p=q. Substituting x=p, y= x n + 1 in (2.1), we get

d ( q , x n ) = d ( f p , f x n + 1 ) a 1 d ( p , x n + 1 ) + a 2 d ( p , f p ) + a 3 d ( x n + 1 , f x n + 1 ) + a 4 d ( p , f x n + 1 ) + a 5 d ( x n + 1 , f p ) .

For the second and fourth term on the right-hand side, we have d(p,q)d(p, x n + 1 )d(q, x n + 1 ) and d(p, x n )d(p, x n + 1 )d( x n , x n + 1 ). For the left-hand side, d(q, x n )d(q, x n + 1 )+d( x n + 1 , x n ). It follows that

( a 1 + a 2 + a 4 )d(p, x n + 1 )(1+ a 2 a 5 )d(q, x n + 1 )+(1 a 3 + a 4 )d( x n + 1 , x n ).

Now, we have

d ( p , x n + 1 ) 1 + a 2 a 5 a 1 + a 2 + a 4 d ( q , x n + 1 ) + 1 a 3 + a 4 a 1 + a 2 + a 4 d ( x n + 1 , x n ) 1 + a 2 a 5 a 1 + a 2 + a 4 d ( q , x n + 1 ) + d ( x n + 1 , x n ) .

If 1+ a 2 a 5 >0 for each cθ, we can choose a natural number N 1 such that d( x n + 1 , x n ) c 2 and d(q, x n + 1 ) ( a 1 + a 2 + a 4 ) c 2 ( 1 + a 2 a 5 ) for n N 1 . Thus, we obtain

d(p, x n + 1 ) c 2 + c 2 =c.

If 1+ a 2 a 5 0 for n N 1 ,

d(p, x n + 1 )d( x n + 1 , x n ) c 2 c.

Therefore, x n + 1 p. From Lemma 1.3, we see p=q. The conclusion is true. □

Taking some particular value of a i (i=1,2,3,4,5) in Theorem 2.1, we obtain several new results in the following.

Corollary 2.2 Let (X,d) be a complete cone metric space. Suppose the mapping f:XX is onto and satisfies

d(fx,fy)kd(x,y)+ld(x,fx)+pd(y,fy)

for all x,yX, where p1 and k+l+p>1. Then f has a fixed point.

Corollary 2.3 Let (X,d) be a complete cone metric space. Suppose the mapping f:XX is onto and satisfies

d(fx,fy)kd(x,y)+ld(fx,y)

for all x,yX, where k, l are constants and k>1. Then f has a fixed point.

Remark 2.4 Obviously, in our theorem and its corollaries above, we delete the continuity of the mappings which is essential in the results of [9]. Moreover, in Corollary 2.2 we delete k1, l>1, which is essential in Theorem 2.6 in [9]. In Corollary 2.3 we delete l0, which is essential in Theorem 2.5 in [9]. Theorem 2.3 in [9] is a special case of Theorem 2.1 with a 1 = a 4 = a 5 =0, a 2 = a 3 =K, and f is continuous.

Now, we introduce some common fixed point theorems for two expanding mappings which satisfy generalized expansive conditions without continuity of the mappings.

Theorem 2.5 Let (X,d) be a complete cone metric space. Suppose mappings f,g:XX are onto and satisfy

d(fx,gy) a 1 d(x,y)+ a 2 d(x,fx)+ a 3 d(y,gy)+ a 4 d(x,gy)+ a 5 d(y,fx)
(2.2)

for all x,yX, where a i (i=1,2,3,4,5) satisfies a 1 + a 2 + a 3 >1 and a 2 1+ a 5 , a 3 1+ a 4 . Then f and g have a common fixed point.

Proof Suppose x 0 is an arbitrary point in X. Since f, g are onto, there exist x 1 , x 2 X such that x 0 =g x 1 , x 1 =f x 2 . Continuing this process, we can define { x n } by x 2 n =g x 2 n + 1 , x 2 n + 1 =f x 2 n + 2 , n=0,1,2, . By (2.2), we have

d ( x 2 n + 1 , x 2 n ) = d ( f x 2 n + 2 , g x 2 n + 1 ) a 1 d ( x 2 n + 2 , x 2 n + 1 ) + a 2 d ( x 2 n + 2 , f x 2 n + 2 ) + a 3 d ( x 2 n + 1 , g x 2 n + 1 ) + a 4 d ( x 2 n + 2 , g x 2 n + 1 ) + a 5 d ( x 2 n + 1 , f x 2 n + 2 ) = a 1 d ( x 2 n + 2 , x 2 n + 1 ) + a 2 d ( x 2 n + 2 , x 2 n + 1 ) + a 3 d ( x 2 n + 1 , x 2 n ) + a 4 d ( x 2 n + 2 , x 2 n ) + a 5 d ( x 2 n + 1 , x 2 n + 1 ) .

Since d( x 2 n + 2 , x 2 n )d( x 2 n + 2 , x 2 n + 1 )d( x 2 n , x 2 n + 1 ), the above inequality implies that

(1 a 3 + a 4 )d( x 2 n , x 2 n + 1 )( a 1 + a 2 + a 4 )d( x 2 n + 1 , x 2 n + 2 ).

Similarly, it can be shown that

d ( x 2 n 1 , x 2 n ) = d ( f x 2 n , g x 2 n + 1 ) a 1 d ( x 2 n , x 2 n + 1 ) + a 2 d ( x 2 n , f x 2 n ) + a 3 d ( x 2 n + 1 , g x 2 n + 1 ) + a 4 d ( x 2 n , g x 2 n + 1 ) + a 5 d ( x 2 n + 1 , f x 2 n ) = a 1 d ( x 2 n , x 2 n + 1 ) + a 2 d ( x 2 n , x 2 n 1 ) + a 3 d ( x 2 n + 1 , x 2 n ) + a 4 d ( x 2 n , x 2 n ) + a 5 d ( x 2 n + 1 , x 2 n 1 ) ,

which also implies that

(1 a 2 + a 5 )d( x 2 n 1 , x 2 n )( a 1 + a 3 + a 5 )d( x 2 n , x 2 n + 1 ).

Let M= 1 a 3 + a 4 a 1 + a 2 + a 4 , N= 1 a 2 + a 5 a 1 + a 3 + a 5 . From a 1 + a 2 + a 3 >1 and a 2 1+ a 5 , a 3 1+ a 4 , we see a 1 + a 2 + a 4 >1 a 3 + a 4 0 and a 1 + a 3 + a 5 >1 a 2 + a 5 0. Thus, h=MN[0,1). Now, by induction we have

d ( x 2 n + 2 , x 2 n + 1 ) M d ( x 2 n + 1 , x 2 n ) M N d ( x 2 n , x 2 n 1 ) M 2 N d ( x 2 n 1 , x 2 n 2 ) M h n d ( x 1 , x 0 )

and

d( x 2 n + 1 , x 2 n )Nd( x 2 n , x 2 n 1 ) h n d( x 1 , x 0 ).

Hence, for any n>m, we deduce

d ( x 2 n + 1 , x 2 m + 1 ) d ( x 2 n + 1 , x 2 n ) + d ( x 2 n , x 2 n 1 ) + + d ( x 2 m + 2 , x 2 m + 1 ) ( i = m + 1 n h i + M i = m n 1 h i ) d ( x 1 , x 0 ) ( h m + 1 1 h + M h m 1 h ) d ( x 1 , x 0 ) = ( N + 1 ) M h m 1 h d ( x 1 , x 0 ) .

In an analogous way, we gain

and

d( x 2 n + 1 , x 2 m )(M+1) h m 1 h d( x 1 , x 0 ).

Thus, for n>m>0,

d( x n , x m )max { ( N + 1 ) M h m 1 h , ( M + 1 ) h m 1 h } d( x 1 , x 0 )= λ m d( x 1 , x 0 ),

where λ m 0 as m.

For each cθ, choose δ>0 such that cxintP, where x<δ, i.e., xc. For this δ, we can choose a natural number N 2 such that λ m d( x 1 , x 0 )<δ for m> N 2 . Thus, we get

d( x n , x m ) λ m d( x 1 , x 0 )cfor all n>m> N 2 .

Therefore, { x n } is a Cauchy sequence in (X,d).

As X is complete, there exists qX such that x n q as n. It is equivalent to x 2 n =g x 2 n + 1 q, x 2 n + 1 =f x 2 n + 2 q as n. Since f, g are onto, there exist u,pX such that fu=gp=q. Now, we show that u=p=q. By (2.2), we have

d ( f x 2 n + 2 , g p ) a 1 d ( x 2 n + 2 , p ) + a 2 d ( x 2 n + 2 , f x 2 n + 2 ) + a 3 d ( p , g p ) + a 4 d ( x 2 n + 2 , g p ) + a 5 d ( p , f x 2 n + 2 ) ,

that is,

d( x 2 n + 1 ,q) a 1 d( x 2 n + 2 ,p)+ a 2 d( x 2 n + 2 , x 2 n + 1 )+ a 3 d(p,q)+ a 4 d( x 2 n + 2 ,q)+ a 5 d(p, x 2 n + 1 ).

From the fact that d(p,q)d(p, x 2 n + 2 )d(q, x 2 n + 2 ), d(p, x 2 n + 1 )d(p, x 2 n + 2 )d( x 2 n + 1 , x 2 n + 2 ) and d( x 2 n + 1 ,q)d( x 2 n + 1 , x 2 n + 2 )+d( x 2 n + 2 ,q), we get

( a 1 + a 3 + a 5 )d(p, x 2 n + 2 )(1+ a 3 a 4 )d( x 2 n + 2 ,q)+(1 a 2 + a 5 )d( x 2 n + 1 , x 2 n + 2 ).

Now, we have

d ( p , x 2 n + 2 ) 1 + a 3 a 4 a 1 + a 3 + a 5 d ( x 2 n + 2 , q ) + 1 a 2 + a 5 a 1 + a 3 + a 5 d ( x 2 n + 1 , x 2 n + 2 ) 2 a 1 + a 3 + a 5 d ( x 2 n + 2 , q ) + d ( x 2 n + 1 , x 2 n + 2 ) .

For each cθ, we can choose a natural number N 3 such that d( x 2 n + 1 , x 2 n + 2 ) c 2 and d( x 2 n + 2 ,q) ( a 1 + a 3 + a 5 ) c 4 for n N 3 . Hence, we obtain d(p, x 2 n + 2 ) c 2 + c 2 =c, i.e., x 2 n + 2 p. By Lemma 1.3, we know p=q, gq=q. Similarly, we also have

d ( f u , g x 2 n + 1 ) a 1 d ( u , x 2 n + 1 ) + a 2 d ( u , f u ) + a 3 d ( x 2 n + 1 , g x 2 n + 1 ) + a 4 d ( u , g x 2 n + 1 ) + a 5 d ( x 2 n + 1 , f u ) .

As in the previous proof, it is not difficult to get q=u, i.e., fq=q. Therefore, fq=gq=q. □

Corollary 2.6 Let (X,d) be a complete cone metric space. Suppose mappings f,g:XX are onto and satisfy

d(fx,gy)αd(x,y)+β [ d ( x , f x ) + d ( y , g y ) ] +γ [ d ( x , g y ) + d ( y , f x ) ]

for all x,yX, where β1+γ and α+2β>1. Then f and g have a common fixed point.

Corollary 2.7 Let (X,d) be a complete cone metric space. Suppose mappings f,g:XX are onto and satisfy

d(fx,gy)kd(x,y)

for all x,yX, where k>1 is a constant. Then f and g have a unique common fixed point.

Corollary 2.8 Let (X,d) be a complete cone metric space. Suppose the mapping f:XX is onto and satisfies

d ( f p x , f q y ) kd(x,y)

for all x,yX, where p, q are positive integers and k>1 is a constant. Then f has a unique fixed point.

Proof Let f= f p , g= f q . Since f is an onto mapping, f= f p , g= f q are onto mappings, the conditions of Corollary 2.7 are satisfied. □

Remark 2.9 In Corollary 2.8, we obtain Corollary 2.2 in [9] when we take p=q.

Now, we present the following examples. In Example 1, we gain a fixed point for one expanding mapping of the situation when Corollary 2.2 can be applied, while the results in [9] cannot. In Example 2, we obtain the common fixed point for two expanding mappings in a cone metric space.

Example 1 Let X=[1,+), E= C R 2 ([0,1]) with x= x + x and P={xE:x(t)0,t[0,1]} (this cone is not normal). Define d:X×XE by d(x,y)=|xy|φ, where φ:[0,1]R such that φ(t)= e t . Consider the mapping

f(x)={ 2 x 2 x , 1 x < 2 ; 5 2 x , x 2 ,

which implies that f is onto in X. Taking k=2, l= 1 4 , p= 1 4 , for 1x<2, all the conditions of Corollary 2.2 are fulfilled. Indeed, since 0< 1 2 x+ 1 2 y+ 3 2 <2x+2y1, we have

d ( f x , f y ) = | 2 x 2 x ( 2 y 2 y ) | e t = | ( x y ) ( 2 x + 2 y 1 ) | e t | x y | ( 1 2 x + 1 2 y + 3 2 ) e t = 2 | x y | e t + 1 2 | ( x y ) ( x + y 1 ) | e t 2 | x y | e t + 1 4 | 2 x 2 x 2 | e t 1 4 | 2 y 2 y 2 | e t .

For x2, since f(x) is increasing in x, we have

d(fx,fy)= | 5 2 x 5 2 y | e t 2|xy| e t + 1 4 | x 5 2 x | e t 1 4 | y 5 2 y | e t .

Therefore, we can apply Corollary 2.2 and conclude that f has a (unique) fixed point 0 in X. Since f is not continuous in X and l<1, Theorem 2.6 in [9] is not applicable. Hence, our theorems have improved and generalized the main results in [9].

Example 2 Let X={1,2,3} and d:X×X R 2 be defined by d(x,y)=(0,0) for x=y and

d(2,3)=d(3,2)=(0,0),d(2,1)=d(1,2)=(1,1),d(1,3)=d(3,1)=(1,1).

Then (X,d) is a complete cone metric space. Further, define mappings f,g:XX as follows:

f(x)={ 1 , x = 1 , 3 , x = 2 , 2 , x = 3 ; g(x)={ 1 , x = 1 , 2 , x = 2 , 3 , x = 3 ,

which implies that f, g are onto in X. Note that

d(fx,gy) a 1 d(x,y)+ a 2 d(x,fx)+ a 3 d(y,gy)+ a 4 d(x,gy)+ a 5 d(y,fx)

for all x,yX by taking a 1 = 1 7 , a 2 = 2 7 , a 3 = 11 7 , a 4 = 5 7 , a 5 = 3 7 . Thus, all the conditions of Theorem 2.5 are fulfilled. Then f and g have a unique common fixed point 1 in X.

Remark 2.10 Obviously, in the above two examples, we obtain the (common) fixed point which essentially needs the structure of a cone metric and not an ordinary metric on X. Then the results in a metric space in [10] cannot be applied to these examples.

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Acknowledgements

The authors thank the editor and the referees for their valuable comments and suggestions which improve greatly the quality for this paper. The research was supported by the Foundation of Education Ministry, Hubei Province, China (No: D20102502).

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Correspondence to Shaoyuan Xu.

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The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

An erratum to this article can be found online at 10.1186/1687-1812-2014-178.

An erratum to this article is available at http://dx.doi.org/10.1186/1687-1812-2014-178.

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Keywords

  • cone metric space
  • expanding mapping
  • common fixed point