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Nguyen Manh Hung 1 and Erdal Karapinar 2 and Nguyen Van Luong 1
Recommended by Stefan Siegmund
1, Department of Natural Sciences, Hong Duc University, Thanh Hoa 41000, Vietnam
2, Department of Mathematics, Atilim University, 06586 Incek, Ankara, Turkey
Received 28 February 2012; Revised 19 April 2012; Accepted 19 April 2012
1. Introduction and Preliminaries
The notion of coupled fixed point was introduced by Guo and Lakshmikantham [1] in 1987. Later, Bhaskar and Lakshmikantham [2] defined the notions of mixed monotone mapping and proved some coupled fixed point theorems for the mixed monotone mappings. In this pioneer paper [2], they also discussed the existence and uniqueness of solution for a periodic boundary value problem. We start with recalling these basic concepts.
Definition 1.1 (see [2]).
Let (X,...) be a partially ordered set and F:X×X[arrow right]X . The mapping F is said to have the mixed monotone property if F(x,y) is monotone nondecreasing in x and is monotone nonincreasing in y , that is, for any x,y∈X , [figure omitted; refer to PDF]
Definition 1.2 (see [2]).
An element (x,y)∈X×X is called a coupled fixed point of the mapping F:X×X[arrow right]X if [figure omitted; refer to PDF]
The main results of Bhaskar and Lakshmikantham in [2] are the following theorems.
Theorem 1.3 (see [2]).
Let (X,...) be a partially ordered set and suppose there exists a metric d on X such that (X,d) is a complete metric space. Let F:X×X[arrow right]X be a continuous mapping having the mixed monotone property on X . Assume that there exists a k∈[0,1) with [figure omitted; refer to PDF] for all x[succeeds, =]u and y...v . If there exist two elements x0 ,y0 ∈X with [figure omitted; refer to PDF] then there exist x,y∈X such that [figure omitted; refer to PDF]
Theorem 1.4 (see [2]).
Let (X,...) be a partially ordered set and suppose there exists a metric d on X such that (X,d) is a complete metric space. Assume that X has the following property:
(i) if a nondecreasing sequence {xn }[arrow right]x , then xn ...x for all n ,
(ii) if a nonincreasing sequence {yn }[arrow right]y , then y...yn for all n .
Let F:X×X[arrow right]X be a mapping having the mixed monotone property on X . Assume that there exists a k∈[0,1) with [figure omitted; refer to PDF] for all x[succeeds, =]u and y...v . If there exist two elements x0 ,y0 ∈X with [figure omitted; refer to PDF] then there exist x,y∈X such that [figure omitted; refer to PDF]
Afterwards, a number of coupled coincidence/fixed point theorems and their application to integral equations, matrix equations, and periodic boundary value problem have been established (e.g., see [3-28] and references therein). In particular, Lakshmikantham and Ciric [7] established coupled coincidence and coupled fixed point theorems for two mappings F:X×X[arrow right]X and g:X[arrow right]X , where F has the mixed g -monotone property and the functions F and g commute, as an extension of the fixed point results in [2]. Choudhury and Kundu in [15] introduced the concept of compatibility and proved the result established in [7] under a different set of conditions. Precisely, they established their result by assuming that F and g are compatible mappings. For the sake of completeness, we remind these characterizations.
Definition 1.5 (see [7]).
Let (X,...) be a partially ordered set and let F:X×X[arrow right]X and g:X[arrow right]X are two mappings. We say F has the mixed g -monotone property if F(x,y) is g -nondecreasing in its first argument and is g -nonincreasing in its second argument, that is, for any x,y∈X , [figure omitted; refer to PDF]
Definition 1.6 (see [7]).
An element (x,y)∈X×X is called a coupled coincident point of the mappings F:X×X[arrow right]X and g:X[arrow right]X if [figure omitted; refer to PDF]
Definition 1.7 (see [15]).
The mappings F and g where F:X×X[arrow right]X , g:X[arrow right]X are said to be compatible if [figure omitted; refer to PDF] where {xn } and {yn } are sequences in X such that lim n[arrow right]∞ F(xn ,yn )=lim n[arrow right]∞ gxn =x and lim n[arrow right]∞ F(yn ,xn )=lim n[arrow right]∞ gyn =y for all x,y∈X are satisfied.
Luong and Thuan [11] slightly extended the concept of compatible mappings into the context of partially ordered metric spaces, namely, O -compatible mappings and proved some coupled coincidence point theorems for such mappings in partially ordered generalized metric spaces.
The concept of O -compatible mappings is stated as follows.
Definition 1.8 (cf. [11]).
Let (X,...,d) be a partially ordered metric space. The mappings F:X×X[arrow right]X and g:X[arrow right]X are said to be O -compatible if [figure omitted; refer to PDF] where {xn } and {yn } are sequences in X such that {gxn } , {gyn } are monotone and [figure omitted; refer to PDF] for all x,y∈X are satisfied.
Let (X,...,d) be a partially metric space. If F:X×X[arrow right]X and g:X[arrow right]X are compatible then they are O -compatible. However, the converse is not true. The following example shows that there exist mappings that are O -compatible but not compatible.
Example 1.9 (see [11]).
Let X={0}∪[1/2,2] with the usual metric d(x,y)=|x-y| , for all x,y∈X . We consider the following order relation on X : [figure omitted; refer to PDF] Let F:X×X[arrow right]X be given by [figure omitted; refer to PDF] and g:X[arrow right]X be defined by [figure omitted; refer to PDF] Then F and g are O -compatible but not compatible.
Indeed, let {xn },{yn } in X such that {gxn },{gyn } are monotone and [figure omitted; refer to PDF] for some x,y∈X . Since F(xn ,yn )=F(yn ,xn )∈{0,1} for all n , x=y∈{0,1} . The case x=y=1 is impossible. In fact, if x=y=1 . Then since {gxn },{gyn } are monotone, gxn =gyn =1 for all n...5;n1 , for some n1 . That is xn ,yn ∈[1/2,1] for all n...5;n1 . This implies F(xn ,yn )=F(yn ,xn )=0 , for all n...5;n1 , which is a contradiction. Thus x=y=0 . That implies gxn =gyn =0 for all n...5;n2 , for some n2 . That is xn =yn =0 for all n...5;n2 . Thus, for all n...5;n2 , [figure omitted; refer to PDF] Hence [figure omitted; refer to PDF] hold. Therefore F and g are O -compatible.
Now let {xn },{yn } in X be defined by [figure omitted; refer to PDF] We have [figure omitted; refer to PDF] but [figure omitted; refer to PDF] Thus, F and g are not compatible.
Implicit relation on metric spaces has been used in many articles (see, e.g., [29-31] and references therein). In this paper, we use the following implicit relation to prove a coupled coincidence point theorem for mappings F:X×X[arrow right]X and g:X[arrow right]X , where F has the mixed g -monotone property and F,g are O -compatible.
Let Φ denote all functions [straight phi]:...+ [arrow right]...+ which satisfy
(i) [straight phi] is continuous,
(ii) [straight phi](t)<t for each t>0 .
Obviously, if [straight phi]∈Φ then [straight phi](0)=0 .
Let ... denote all continuous functions H:(...+)5 [arrow right]... which satisfy
(H1) H(t1 ,t2 ,t3 ,t4 ,t5 ) is nonincreasing in t2 and t5 ,
(H2) there exists a function [straight phi]∈Φ such that [figure omitted; refer to PDF]
It is easy to check that the following functions are in ... :
(i) H1 (t1 ,t2 ,t3 ,t4 ,t5 )=t1 -αt2 -βt3 -γt4 -θt5 , where α,β,γ,θ are nonnegative real numbers satisfying 2α+β+γ+2θ<1 ;
(ii) H2 (t1 ,t2 ,t3 ,t4 ,t5 )=t1 -αmax {t2 /2,t3 ,t4 ,t5 /2} , where α∈(0,1) ;
(iii): H3 (t1 ,t2 ,t3 ,t4 ,t5 )=t1 -[straight phi](max {t3 ,t4 }) , where [straight phi]∈Φ .
In this paper, we prove a coupled coincidence point theorem for mappings satisfying such implicit relations.
2. Coupled Coincidence Point Theorem
Now we are going to prove our main result.
Theorem 2.1.
Let (X,d,...) be a partially ordered complete metric space. Suppose F:X×X[arrow right]X and g:X[arrow right]X are mappings such that F has the mixed g -monotone property. Assume that there exists H∈... such that [figure omitted; refer to PDF] for all x,y,u,v∈X with gx[succeeds, =]gu and gy...gv . Suppose F(X×X)⊆g(X) , g is continuous and g is O -compatible with F . Suppose either
(a) F is continuous or;
(b) X has the following property:
(i) if a nondecreasing sequence {xn }[arrow right]x , then gxn ...gx for all n ,
(ii) if a nonincreasing sequence {yn }[arrow right]y , then gy...gyn for all n .
If there exist two elements x0 ,y0 ∈X with [figure omitted; refer to PDF] then F and g have a coupled coincidence point in X .
Proof.
Let x0 ,y0 ∈X be such that gx0 ...F(x0 ,y0 ) and gy0 [succeeds, =]F(y0 ,x0 ) . Since F(X×X)⊆g(X) , we construct the sequences {xn } and {yn } in X as follows: [figure omitted; refer to PDF] By using the mathematical induction and the mixed g -monotone property of F , we can show that [figure omitted; refer to PDF]
If there is some n0 ∈...* such that gxn0 =gxn0 +1 and gyn0 =gyn0 +1 then [figure omitted; refer to PDF] that means (xn0 ,yn0 ) is a coupled coincidence point of F and g . Thus we may assume that max {d(gxn+1 ,gxn ),d(gyn+1 ,gyn )}>0 for all n .
Since gxn+1 [succeeds, =]gxn and gyn+1 ...gyn , from (2.1), we have [figure omitted; refer to PDF] or [figure omitted; refer to PDF] By the properties of H , we have [figure omitted; refer to PDF] which implies that [figure omitted; refer to PDF] Similarly, one can show that [figure omitted; refer to PDF] From (2.9) and (2.10), we have [figure omitted; refer to PDF] which implies [figure omitted; refer to PDF] This means that {dn :=max {d(gxn+1 ,gxn ),d(gyn+1 ,gyn )}} is a decreasing sequence of positive real numbers. So there is a d...5;0 such that [figure omitted; refer to PDF] We will show that d=0 . Assume, to the contrary, that d>0 . Taking n[arrow right]∞ in (2.11), we have [figure omitted; refer to PDF] which is a contradiction. Thus d=0 .
In what follows, we will show that {gxn } and {gyn } are Cauchy sequences. Suppose, to the contrary that at least one of {gxn } or {gyn } is not a Cauchy sequence. This means that there exists an [straight epsilon]>0 for wich we can find subsequences {gxn(k) },{gxm(k) } of {gxn } and {gyn(k) },{gym(k) } of {gyn } with n(k)>m(k)...5;k such that [figure omitted; refer to PDF] Further, corresponding to m(k) , we can choose n(k) in such a way that it is the smallest integer with n(k)>m(k)...5;k and satisfies (2.15). Then [figure omitted; refer to PDF] Using the triangle inequality and (2.16), we have [figure omitted; refer to PDF] From (2.15) and (2.17), we have [figure omitted; refer to PDF] Letting k[arrow right]∞ in the inequalities above and using (2.13) we get [figure omitted; refer to PDF] By the triangle inequality [figure omitted; refer to PDF] From the last two inequalities and (2.15), we have [figure omitted; refer to PDF] Again, by the triangle inequality, [figure omitted; refer to PDF] Therefore, [figure omitted; refer to PDF] From (2.21) and (2.23), we have [figure omitted; refer to PDF] Taking k[arrow right]∞ in the inequalities above and using (2.13), we get [figure omitted; refer to PDF] From (2.19) and (2.25), the sequences {d(gxn(k) ,gxm(k) )} , {d(gyn(k) ,gym(k) )} , {d(gxn(k)-1 ,gxm(k)-1 )} , and {d(gyn(k)-1 ,gym(k)-1 )} have subsequences converging to [straight epsilon]1 , [straight epsilon]2 , [straight epsilon]3 and [straight epsilon]4 , respectively, and max {[straight epsilon]1 ,[straight epsilon]2 }=max {[straight epsilon]3 ,[straight epsilon]4 }=[straight epsilon]>0 . We may assume that [figure omitted; refer to PDF]
We first assume that [straight epsilon]1 =max {[straight epsilon]1 ,[straight epsilon]2 }=[straight epsilon] . Since n(k)>m(k) , gxn(k)-1 [succeeds, =]gxm(k)-1 and gyn(k)-1 ...gym(k)-1 . From (2.1), we have [figure omitted; refer to PDF] or [figure omitted; refer to PDF] or [figure omitted; refer to PDF] Letting k[arrow right]∞ , we have [figure omitted; refer to PDF] Thus, [figure omitted; refer to PDF] which implies [straight epsilon]=[straight epsilon]1 ...4;[straight phi](max {[straight epsilon]3 ,[straight epsilon]4 })=[straight phi]([straight epsilon])<[straight epsilon] . That is a contradiction.
Using the same argument as above for the case [straight epsilon]2 =max {[straight epsilon]1 ,[straight epsilon]2 }=[straight epsilon] , we also get a contradiction. Thus {gxn } and {gyn } are Cauchy sequences. Since X is complete, there exist x,y∈X such that [figure omitted; refer to PDF] Thus [figure omitted; refer to PDF] Since F and g are O -compatible, from (2.33), we have [figure omitted; refer to PDF] [figure omitted; refer to PDF] Now, suppose that assumption (a) holds. We have [figure omitted; refer to PDF] Taking the limit as n[arrow right]∞ in (2.36) and by (2.32), (2.34) and the continuity of F and g we get d(gx,F(x,y))=0 .
Similarly, we can show that d(gy,F(y,x))=0 . Therefore, gx=F(x,y) and gy=F(y,x) .
Finally, suppose that assumption (b) holds. Since {gxn } is nondecreasing sequence and gxn [arrow right]x and {gyn } is nonincreasing sequence and gyn [arrow right]y , by the assumption, we have ggxn ...gx and ggyn [succeeds, =]gy for all n .
Since g is continuous, from (2.32), (2.34), and (2.35) we have [figure omitted; refer to PDF] We have [figure omitted; refer to PDF] Letting n[arrow right]∞ and using (2.37), we have [figure omitted; refer to PDF] which implies that d(gx,F(x,y))...4;[straight phi](max {0,0})=0 . Hence gx=F(x,y) . Similarly, one can show that gy=F(y,x) .
Thus proved that F and g have a coupled coincidence point in X .
Example 2.2 (see, e.g., [11]).
Let (X,d,...) , F and g be defined as in Example 1.9. Then
(i) X is complete and X has the property
(a) if a nondecreasing sequence {xn }[arrow right]x , then gxn ...gx for all n ,
(b) if a nonincreasing sequence {yn }[arrow right]y , then gy...gyn for all n ;
(ii) F(X×X)={0,1}⊂{0}∪[1/2,1]=g(X) ;
(iii): g is continuous and g and F are O -compatible;
(iv) there exist x0 =0,y0 =1 such that gx0 ...F(x0 ,y0 ) and gy0 [succeeds, =]F(y0 ,x0 ) ;
(v) F has the mixed g -monotone property. Indeed, for every y∈X , let x1 ,x2 ∈X such that gx1 ...gx2
(a) if gx1 =gx2 then x1 ,x2 =0 or x1 ,x2 ∈[1/2,1] or x1 ,x2 ∈(1,3/2] or x1 ,x2 ∈(3/2,2] . Thus, [figure omitted; refer to PDF] otherwise F(x1 ,y)=1=F(x2 ,y) ,
(b) if gx1 [precedes]gx2 , then gx1 =0 and gx2 =1 , that is, x1 =0 and x2 ∈[1/2,1] . Thus [figure omitted; refer to PDF] therefore, F is the g -nondecreasing in its first argument. Similarly, F is the g -nonincreasing in its second argument;
(vi) for x,y,u,v∈X , if gx[succeeds, =]gu and gy...gv then d(F(x,y),F(u,v))=0 . Indeed,
(a) if gx[succeeds]gu and gy[precedes]gv then y=u=0 and x,v∈[1/2,1] . Thus d(F(x,y),F(u,v))=d(0,0)=0 ,
(b) if gx=gu and gy[precedes]gv then y=0 and v∈[1/2,1] . Thus if x=u=0 or x,u∈[1/2,1] then d(F(x,y),F(u,v))=d(0,0)=0 , otherwise d(F(x,y),F(u,v))=d(1,1)=0 . Similarly, if gx[succeeds]gu and gy=gv then d(F(x,y),F(u,v))=0 ,
(c) if gx=gu and gy=gv then both x,u are in one of the sets {0} , [1/2,1] , (1,3/2] or (3/2,2] and both y,v are also in one of the sets {0} , [1/2,1] , (1,3/2] or (3/2,2] . Thus d(F(x,y),F(u,v))=d(0,0)=0 if x=u=0 or x,u∈[1/2,1] and y=v=0 or y,v∈[1/2,1] , otherwise, d(F(x,y),F(u,v))=d(1,1)=0 .
Therefore, all the conditions of Theorem 2.1 are satisfied with H(t1 ,t2 ,t3 ,t4 ,t5 )=t1 -max {t3 ,t4 }/2 . Applying Theorem 2.1, we conclude that F and g have a coupled coincidence point.
Note that, we cannot apply the result of Choudhury and Kundu [15], the result of Choudhury et al. [32] as well as the result of Lakshmikantham and Ciric [7] to this example.
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Abstract
We prove a coupled coincidence point theorem for mappings F : X×X[arrow right]X and g : X[arrow right]X , where F has the mixed g-monotone property, in partially ordered metric spaces via implicit relations. Our result extends and improves several results in the literature. Examples are also given to illustrate our work.
You have requested "on-the-fly" machine translation of selected content from our databases. This functionality is provided solely for your convenience and is in no way intended to replace human translation. Show full disclaimer
Neither ProQuest nor its licensors make any representations or warranties with respect to the translations. The translations are automatically generated "AS IS" and "AS AVAILABLE" and are not retained in our systems. PROQUEST AND ITS LICENSORS SPECIFICALLY DISCLAIM ANY AND ALL EXPRESS OR IMPLIED WARRANTIES, INCLUDING WITHOUT LIMITATION, ANY WARRANTIES FOR AVAILABILITY, ACCURACY, TIMELINESS, COMPLETENESS, NON-INFRINGMENT, MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Your use of the translations is subject to all use restrictions contained in your Electronic Products License Agreement and by using the translation functionality you agree to forgo any and all claims against ProQuest or its licensors for your use of the translation functionality and any output derived there from. Hide full disclaimer