Open Access

Some new results for single-valued and multi-valued mixed monotone operators of Rhoades type

Fixed Point Theory and Applications20132013:73

https://doi.org/10.1186/1687-1812-2013-73

Received: 11 May 2012

Accepted: 1 March 2013

Published: 28 March 2013

Abstract

In (2008), Zhang proved the existence of fixed points of mixed monotone operators along with certain convexity and concavity conditions. In this paper, mixed monotone single-valued and multi-valued operators of Rhoades type are defined and two fixed point theorems are proved.

MSC:47H10, 47H07.

Keywords

mixed monotone operatorRhoades typemulti-valuedincreasing inward mappings L -function

1 Introduction and preliminaries

In (1987), mixed monotone operators were introduced by Guo and Lakshmikantham [1]. Then many authors studied them in Banach spaces and obtained lots of interesting results (see [2, 3] and [48]).

On the other hand, in (2001), Rhoades [9] introduced a new fixed point theorem as a generalization of Banach fixed point theorem.

Theorem 1.1 (Rhoades [9])

Let ( X , d ) be a complete metric space. Suppose that T : X X is a single-valued mapping that satisfies
d ( T x , T y ) d ( x , y ) ψ ( d ( x , y ) )
(1)

for each x , y X , where ψ : [ 0 , + ) [ 0 , + ) is continuous, nondecreasing and ψ 1 ( 0 ) = { 0 } (i.e., weakly contractive mappings). Then T has a fixed point.

In this paper, a weak mixed monotone single-valued and multi-valued operator of Rhoades type is defined. Then two fixed point theorems for this kind of operators are proved.

Let E be a real Banach space. The zero element of E is denoted by θ. A subset P of E is called a cone if and only if:

  • P is closed, nonempty and P { θ } ,

  • a , b R , a , b 0 and x , y P imply that a x + b y P ,

  • x P and x P imply that x = θ .

Given a cone P E , a partial ordering ≤ with respect to P is defined by x y if and only if y x P . We write x < y to indicate that x y but x y , while x y stands for y x int P , where intP denotes the interior of P. The cone P is called normal if there exists a number K > 0 such that θ x y implies x K y for every x , y E . The least positive number satisfying this is called the normal constant of P.

Assume that u 0 , v 0 E and u 0 v 0 . The set { x E : u 0 x v 0 } is denoted by [ u 0 , v 0 ] .

Now, we recall the following definitions from [2, 3].

Definition 1.1 Let P be a cone of a real Banach space E. Suppose that D P and α ( , + ) . An operator A : D D is said to be α-convex (α-concave) if it satisfies A ( t x ) t α A x ( A ( t x ) t α A x ) for ( t , x ) ( 0 , 1 ) × D .

Definition 1.2 Let E be an ordered Banach space and D E . An operator is called mixed monotone on D × D if A : D × D E and A ( x 1 , y 1 ) A ( x 2 , y 2 ) for any x 1 , x 2 , y 1 , y 2 D , where x 1 x 2 and y 2 y 1 . Also, x D is called a fixed point of A if A ( x , x ) = x .

Let C ( E ) be a collection of all closed subsets of E.

Definition 1.3 For two subsets X, Y of E, we write

  • X Y if for all x X , there exists y Y such that x y ,

  • x X if there exists z X such that x z ,

  • X x if for all z X , z x .

Definition 1.4 Let D be a nonempty subset of E. T : D C ( E ) is called increasing (decreasing) upward if u , v D , u v and x T ( u ) imply there exists y T ( v ) such that x y ( x y ). Similarly, T : D C ( E ) is called increasing (decreasing) downward if u , v D , u v and y T ( v ) imply there exists x T ( u ) such that x y ( x y ). T is called increasing (decreasing) if T is an increasing (decreasing) upward and downward.

Definition 1.5 Let D be a nonempty subset of E. A multi-valued operator T : D × D C ( E ) is said to be mixed monotone upward if T ( x , y ) is increasing upward in x and decreasing upward in y, i.e.,

(A1) for each y D and any x 1 , x 2 D with x 1 x 2 , if u 1 T ( x 1 , y ) , then there exists a u 2 T ( x 2 , y ) such that u 1 u 2 ;

(A2) for each x D and any y 1 , y 2 D with y 1 y 2 , if v 1 T ( x , y 1 ) , then there exists a v 2 T ( x , y 2 ) such that v 1 v 2 .

Definition 1.6 x D is called a fixed point of T if x T ( x , x ) .

Definition 1.7 [10]

A function Ψ : [ 0 , 1 ) × P × P × E E is called an L -function if Ψ ( t , x , y , 0 ) = 0 , Ψ ( t , x , y , s ) 0 for s 0 , and Ψ ( t , x , y , z ) < z for all ( t , x , y , z ) [ 0 , 1 ) × P × P × E .

In 2011, Khojasteh and Razani [10] extended the results given by Zhang [6]. Also, in 2011 Khojasteh and Razani [11] introduced the concept of integral with respect to a cone. We recall the following definitions and lemmas of cone integration and refer to [11, 12] for their proofs.

Definition 1.8 [11]

Suppose that P is a cone in E. Let a , b E and a < b . Define
[ a , b ] : = { x E : x = t b + ( 1 t ) a  for some  t [ 0 , 1 ] }
(2)
and
[ a , b ) : = { x E : x = t b + ( 1 t ) a  for some  t [ 0 , 1 ) } .
(3)

Definition 1.9 [11]

The set { a = x 0 , x 1 , , x n = b } is called a partition for [ a , b ] if and only if the intervals { [ x i 1 , x i ) } i = 1 n are pairwise disjoint and [ a , b ] = { i = 1 n [ x i 1 , x i ) } { b } . Denote P [ a , b ] as the collection of all partitions of [ a , b ] .

Definition 1.10 [12]

For each partition Q of [ a , b ] and each increasing function ϕ : [ a , b ] E , we define cone lower summation and cone upper summation as
L n Con ( ϕ , Q ) = i = 0 n 1 ϕ ( x i ) x i x i + 1
(4)
and
U n Con ( ϕ , Q ) = i = 0 n 1 ϕ ( x i + 1 ) x i x i + 1 ,
(5)

respectively. Also, we denote Δ ( Q ) = sup { x i x i 1 , x i Q } .

Definition 1.11 [12]

Suppose that P is a cone in E. ϕ : [ a , b ] E is called an integrable function on [ a , b ] with respect to a cone P or, to put it simply, a cone integrable function if and only if for all partition Q of [ a , b ] ,
lim Δ ( Q ) 0 L n Con ( ϕ , Q ) = S Con = lim Δ ( Q ) 0 U n Con ( ϕ , Q ) ,

where S Con must be unique.

We show the common value S Con by
a b ϕ ( x ) d P ( x ) or to simplicity a b ϕ d p .

We denote the set of all cone integrable functions ϕ : [ a , b ] E by L 1 ( [ a , b ] , E ) .

Lemma 1.1 [11]

Let M be a subset of P. The following conditions hold:
  1. (1)

    If [ a , b ] [ a , c ] M , then a b f d p a c f d p for f L 1 ( M , P ) .

     
  2. (2)

    a b ( α f + β g ) d p = α a b f d p + β a b g d p for f , g L 1 ( M , P ) and α , β R .

     

Remark 1.1 [[13], Remark 1.2]

Let P be a cone of E, and let u P . If for each ϵ int ( P ) , 0 u ϵ , then u = 0 .

2 Main results

In this section, we introduce some new fixed point theorems in the class of mixed monotone operators. Due to this, the following definition is presented.

Definition 2.1 A mixed monotone operator A : D × D E is said to be a Weak Mixed Monotone single-valued operator of Rhoades type (WM2R property for short) if
A ( t x , y ) A ( x , t y ) Ψ ( t , x , y , A ( x , t y ) )
(6)

for all ( x , y ) D × D , where Ψ : [ 0 , 1 ) × P × P × E E is an L -function.

Theorem 2.1 Let P be a cone of E, let S be a completely ordered closed subset of E with S 0 = S { θ } int P and let λ S S for all λ [ 0 , 1 ] . Let u 0 , v 0 S 0 , A : P × P E be a weak mixed monotone operator of Rhoades type with A ( ( [ θ , v 0 ] S ) × ( [ θ , v 0 ] S ) ) S satisfying the following conditions:
  1. (I)

    there exists r 0 > 0 such that u 0 r 0 v 0 ,

     
  2. (II)

    A ( u 0 , v 0 ) u 0 v 0 A ( v 0 , u 0 ) ,

     
  3. (III)

    for u , v [ u 0 , v 0 ] S with A ( u , v ) u v , there exists u S such that u A ( u , v ) u v ; similarly, for u , v [ u 0 , v 0 ] S with u v A ( v , u ) , there exists v S such that u v A ( v , u ) v .

     

Then A has at least one fixed point x [ u 0 , v 0 ] S .

Proof By the above condition (III), there exists u 1 S such that u 0 A ( u 1 , v 0 ) u 1 v 0 . Then there exists v 1 S such that u 1 v 1 A ( v 1 , u 1 ) v 0 . Likewise, there exists u 2 S such that u 1 A ( u 2 , v 1 ) u 2 v 1 . Then there exists v 2 S such that u 2 v 2 A ( v 2 , u 2 ) v 1 . In general, there exists u n S such that u n 1 A ( u n , v n 1 ) u n v n 1 . Then there exists v n S such that u n v n A ( v n , u n ) v n 1 ( n = 1 , 2 , ).

Take r n = sup { r ( 0 , 1 ) : u n r v n } , thus 0 < r 0 < r 1 < < r n < r n + 1 < < 1 and lim n r n = sup { r n : n = 0 , 1 , 2 , } = r ( 0 , 1 ] . Since r n + 1 > r n = sup { r ( 0 , 1 ) : u n r v n } , thus u n r n + 1 v n . In addition, S is completely ordered and λ S S for all λ [ 0 , 1 ] , then u n < r n + 1 v n . Now, one can prove r = 1 . Otherwise, r ( 0 , 1 ) .

Since u n < r n + 1 v n and r n + 1 < r , hence u n < r v n , and we have
A ( u n + 1 , v n + 1 ) A ( 1 r u n + 1 , r v n + 1 ) A ( u n + 1 , v n + 1 ) Ψ ( r , 1 r u n + 1 , v n + 1 , A ( u n + 1 , v n + 1 ) ) < A ( u n + 1 , v n + 1 ) ,
(7)
which is a contradiction. Thus, r = 1 . Let ϵ 0 be given. Choose δ > 0 such that ϵ + N δ ( 0 ) P , where N δ ( 0 ) = { y E : y < δ } . Since r n 1 , one can choose a natural number N 1 such that ( 1 r n ) v 1 N δ ( 0 ) for all n N 1 . Therefore ( 1 r n ) v 1 ϵ . Also, v n v 1 and
0 < v n u n ( 1 r n ) v n ( 1 r n ) v 1 ϵ .
(8)

By Remark 1.1, lim n u n = lim n v n .

For all n , p 1 , applying the same argument, we have
0 < v n v n + p v n u n ϵ .
(9)
Also,
0 < u n + p u n v n u n ϵ .
(10)

Hence, { u n } and { v n } are Cauchy sequences in E, then there exist u , v E such that u n u , v n v ( n ) and u = v . Write x = u = v .

It is easy to see u 0 u n u v n v 0 for all n = 1 , 2 ,  . In addition, S is closed, then u [ u n , v n ] S [ u 0 , v 0 ] S ( n = 0 , 1 , 2 , ).

Finally, by the mixed monotone property of A,
u n 1 A ( u n , v n ) A ( x , x ) A ( u n , v n ) u n 1 .
(11)

On taking limit on both sides of (11), when n , we have A ( x , x ) = x . This means x is a fixed point of A in [ u 0 , v 0 ] S . □

Corollary 2.1 Let P be a cone of E, let S be a completely ordered closed subset of E with S 0 = S { θ } int P and let λ S S for all λ [ 0 , 1 ] . Let u 0 , v 0 S 0 , A : P × P E satisfy
y t x ϕ d P t y x ϕ d P Ψ ( t , x , y , t y x ϕ d P )
(12)
for all ( x , y ) D × D , where Ψ : [ 0 , 1 ) × P × P × E E is an L -function, and let ϕ : P P be a non-vanishing, cone integrable mapping on each [ a , b ] P such that for each ϵ 0 , 0 ϵ ϕ d p 0 and the mapping θ ( x ) = 0 x ϕ d P for ( x 0 ) has a continuous inverse at zero. Also, A ( ( [ θ , v 0 ] S ) × ( [ θ , v 0 ] S ) ) S satisfies the following conditions:
  1. (I)

    there exists r 0 > 0 such that u 0 r 0 v 0 ,

     
  2. (II)

    A ( u 0 , v 0 ) u 0 v 0 A ( v 0 , u 0 ) ,

     
  3. (III)

    for u , v [ u 0 , v 0 ] S with A ( u , v ) u v , there exists u S such that u A ( u , v ) u v ; similarly, for u , v [ u 0 , v 0 ] S with u v A ( v , u ) , there exists v S such that u v A ( v , u ) v .

     

Then A has at least one fixed point x [ u 0 , v 0 ] S .

Proof

Define
A ( x , y ) = y x ϕ d P .

A is a mixed monotone operator, and one can easily see that all conditions of Theorem 2.1 hold. Thus we obtain the desired result. □

3 M3R property

In this section, we introduce a new fixed point theorem in the class of multi-valued mixed monotone operators. Due to this, the following definition is given.

Definition 3.1 A mixed monotone operator T : D × D C ( E ) is said to be a Mixed Monotone Multi-valued operator of Rhoades type (M3R property for short) if
T ( t x , y ) T ( x , t y ) Ψ ( t , x , y , T ( t x , y ) )
(13)

for each ( x , y ) D × D , where Ψ : [ 0 , 1 ) × P × P × E E is an L -function.

Theorem 3.1 Let P be a cone of E, let S be a completely ordered closed subset of E with S 0 = S { θ } int P and let λ S S for all λ [ 0 , 1 ] . Let u 0 , v 0 S 0 , T : P × P C ( E ) be a mixed monotone multi-valued operator of Rhoades type with T ( ( [ θ , v 0 ] S ) × ( [ θ , v 0 ] S ) ) S satisfying the following conditions:
  1. (I)

    there exists r 0 > 0 such that u 0 r 0 v 0 ,

     
  2. (II)

    T ( u 0 , v 0 ) u 0 v 0 T ( v 0 , u 0 ) ,

     
  3. (III)

    for u , v [ u 0 , v 0 ] S with T ( u , v ) u v , there exists u S such that u T ( u , v ) u v ; similarly, for u , v [ u 0 , v 0 ] S with u v T ( v , u ) , there exists v S such that u v T ( v , u ) v .

     

Then T has at least one fixed point x [ u 0 , v 0 ] S .

Proof By the above condition (III), there exists u 1 S such that u 0 T ( u 1 , v 0 ) u 1 v 0 . Then there exists v 1 S such that u 1 v 1 T ( v 1 , u 1 ) v 0 . Likewise, there exists u 2 S such that u 1 T ( u 2 , v 1 ) u 2 v 1 . Then there exists v 2 S such that u 2 v 2 T ( v 2 , u 2 ) v 1 . In general, there exists u n S such that u n 1 T ( u n , v n 1 ) u n v n 1 . Then there exists v n S such that u n v n T ( v n , u n ) v n 1 ( n = 1 , 2 , ).

Take r n = sup { r ( 0 , 1 ) : u n r v n } , thus 0 < r 0 < r 1 < < r n < r n + 1 < < 1 , and lim n r n = sup { r n : n = 0 , 1 , 2 , } = r ( 0 , 1 ] . Since r n + 1 > r n = sup { r ( 0 , 1 ) : u n r v n } , thus u n r n + 1 v n . In addition, S is completely ordered and λ S S for all λ [ 0 , 1 ] , then u n < r n + 1 v n . Now, one can prove r = 1 . Otherwise, r ( 0 , 1 ) . We claim
T ( u n + 1 , v n + 1 ) T ( ( 1 / r ) u n + 1 , r v n + 1 ) .
(14)

Suppose that x T ( u n + 1 , v n + 1 ) is arbitrary. We have u n + 1 ( 1 / r ) u n + 1 . If x 1 = u n + 1 , x 2 = ( 1 / r ) u n + 1 and y = v n + 1 , then by (A1) of Definition 1.5, there exists z T ( ( 1 / r ) u n + 1 , v n + 1 ) such that x z . Thus, T ( u n + 1 , v n + 1 ) T ( ( 1 / r ) u n + 1 , v n + 1 ) .

Also, if y 1 = r v n + 1 , y 2 = v n + 1 and x = ( 1 / r ) u n + 1 , then for w T ( ( 1 / r ) u n + 1 , r v n + 1 ) , there exists h T ( ( 1 / r ) u n + 1 , v n + 1 ) such that w h . It means that
T ( ( 1 / r ) u n + 1 , v n + 1 ) T ( ( 1 / r ) u n + 1 , r v n + 1 ) .
(15)
Thus,
T ( u n + 1 , v n + 1 ) T ( ( 1 / r ) u n + 1 , r v n + 1 ) T ( u n + 1 , v n + 1 ) Ψ ( 1 r , u n + 1 , r v n + 1 , T ( u n + 1 , v n + 1 ) ) T ( u n + 1 , v n + 1 ) ,
(16)
and this is a contradiction. Therefore, r = 1 . Let ϵ 0 be given. Choose δ > 0 such that ϵ + N δ ( 0 ) P , where N δ ( 0 ) = { y E : y < δ } . Since r n 1 , one can choose a natural number N 1 such that ( 1 r n ) v 1 N δ ( 0 ) for all n N 1 . Therefore ( 1 r n ) v 1 ϵ . Also, v n v 1 and
0 < v n u n ( 1 r n ) v n ( 1 r n ) v 1 ϵ .
(17)

By Remark 1.1, lim n u n = lim n v n .

For all n , p 1 , applying the same argument, we have
0 < v n v n + p v n u n ϵ .
(18)
Also,
0 < u n + p u n v n u n ϵ .
(19)

Hence, { u n } and { v n } are Cauchy sequences in E, then there exist u , v E such that u n u , v n v ( n ) and u = v . Write x = u = v .

It is easy to see that u n T ( u n + 1 , v n + 1 ) T ( x , x ) T ( v n + 1 , u n + 1 ) v n for all n = 1 , 2 ,  . Thus, there exists z n T ( x , x ) such that u n z n v n . By taking limit on both sides of (17),
0 < z n u n ( 1 r n ) v n ( 1 r n ) v 1 ϵ .
(20)
So, z n x . Since T has closed values, then x T ( x , x ) and
x [ u n , v n ] S [ u 0 , v 0 ] S .

 □

Remark 3.1 One can see easily that Theorem 2.1 should be included as a corollary of Theorem 3.1.

Example 3.1 Let E = R , P = [ 0 , + ) and S = P . Then S 0 = int ( P ) = ( 0 , + ) .

Define A : [ 0 , + ) × [ 0 , + ) R as
A ( x , y ) = { x y , ( x , y ) ( 0 , 0 ) , 0 , ( x , y ) = ( 0 , 0 ) .
A is a mixed monotone operator. Now suppose that Ψ : [ 0 , 1 ) × P × P × E E is as Ψ ( t , x , y , s ) = ( 1 t 2 ) s . Then Ψ is an L -function. Moreover,
A ( t x , y ) A ( x , t y ) Ψ ( t , x , y , A ( x , t y ) )
for each x , y S 0 . Also, by taking u 0 = 1 2 , v 0 = 3 2 and r 0 = 1 4 , we have
  1. (I)

    u 0 r 0 v 0 ,

     
  2. (II)

    A ( u 0 , v 0 ) = 1 3 u 0 v 0 A ( v 0 , u 0 ) = 3 ,

     
  3. (III)

    for u , v [ u 0 , v 0 ] S with A ( u , v ) u v , there exists u S such that u A ( u , v ) u v ; similarly, for u , v [ u 0 , v 0 ] S with u v A ( v , u ) , there exists v S such that u v A ( v , u ) v .

     
For further explanation on (III), since A ( u 0 , v 0 ) = 1 3 u 0 v 0 , by (III) there exists u 1 S such that u 0 A ( u 1 , v 0 ) u 1 v 0 . It means that 1 2 u 1 3 2 u 1 3 2 . Thus u 1 must be greater than 3 4 . Therefore we can set u 1 = 3 4 + 1 2 . Similarly, since 7 8 = u 1 v 0 = 3 2 A ( v 0 , u 1 ) = 12 7 , thus by (III) there exists v 1 S such that u 1 v 1 A ( v 1 , u 1 ) v 0 . It means that v 1 must be less than 21 16 . We can set v 1 = 21 16 + 1 2 . By the continuity of such ways, we can consider the following reflexive sequences:
u 0 = 1 2 , v 0 = 3 2 , u n = u n 1 v n 1 + 1 2 and v n = v n 1 u n + 1 2 ,
which satisfy (I), (II) and (III) (see Figure 1). Moreover, u n 1 and v n 1 and A ( 1 , 1 ) = 1 .
Figure 1

A ( u 0 , v 0 ) u 0 u 1 u n 1 v n v 1 v 0 A ( v 0 , u 0 ) .

4 Application

The following result is given by Zhang [6] and is obtained by our main result.

Corollary 4.1 Let P be a normal cone of E, let S be a completely ordered closed subset of E with S 0 = S { θ } int P and let λ S S for all λ [ 0 , 1 ] . Let u 0 , v 0 S 0 , A : P × P E be a mixed monotone operator with A ( ( [ θ , v 0 ] S ) × ( [ θ , v 0 ] S ) ) S and A ( u 0 , v 0 ) u 0 v 0 A ( v 0 , u 0 ) . Assume that there exists a function ϕ : ( 0 , 1 ) × ( [ u 0 , v 0 ] S ) × ( [ u 0 , v 0 ] S ) ( 0 , + ) such that A ( t x , y ) ϕ ( t , x , y ) A ( x , t y ) , where 0 < ϕ ( t , x , x ) < t for all ( t , x , y ) ( 0 , 1 ) × ( [ u 0 , v 0 ] S ) × ( [ u 0 , v 0 ] S ) . Suppose that
  1. (I)

    for u , v [ u 0 , v 0 ] S with A ( u , v ) u v , there exists u S such that u A ( u , v ) u v ; similarly, for u , v [ u 0 , v 0 ] S with u v A ( v , u ) , there exists v S such that u v A ( v , u ) v .

     
  2. (II)

    there exists an element w 0 [ u 0 , v 0 ] S such that ϕ ( t , x , x ) ϕ ( t , w 0 , w 0 ) for all ( t , x ) ( 0 , 1 ) × ( [ u 0 , v 0 ] S ) , and lim s t ϕ ( s , w 0 , w 0 ) < t for all t ( 0 , 1 ) .

     

Then A has at least one fixed point x [ u 0 , v 0 ] S .

Proof Set Ψ ( t , x , y , z ) = ( 1 ϕ ( t , x , y ) ) z . Then Ψ is an L -function, and we have
A ( t x , y ) ϕ ( t , x , y ) A ( x , t y ) = A ( x , t y ) Ψ ( t , x , y , A ( x , t y ) ) .

Thus, by Theorem 2.1 the desired result is obtained. □

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Arak Branch, Islamic Azad University

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© Khojasteh; licensee Springer. 2013

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