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Common fixed point and coincidence point of generalized contractions in ordered metric spaces

Abstract

Let (X,) be a partially ordered set and d be a complete metric on X. The notion of f-contractive for a set-valued mapping due to Latif and Beg is extended through an implicit relation. Coincidence and fixed point results are obtained for mappings satisfying generalized contractions in a partially ordered metric space X. Our results improve and extend several known results in the existing literature.

MSC:47H10, 47H04, 47H07.

1 Introduction and preliminaries

Let (X,d) be a metric space, P(X) be the class of all non-empty subsets of X, Cl(X) be the class of all non-empty closed subsets of X and CB(X) be the class of all non-empty closed and bounded subsets of X. For A,BCB(X), let

D(A,B):=max { sup b B d ( b , A ) , sup a A d ( a , B ) } ,

where

d(a,B):= inf b B d(a,b).

D is called the Hausdorff metric induced by d.

Let f:XX be any single-valued mapping. A mapping F:XCB(X) is f-contractive if there exists a real number κ with 0<κ<1 such that

D(Fx,Fy)κd(fx,fy)for all x,yX.

By introducing the notion of f-contractiveness for set-valued mappings, Kaneko [1] obtained a result which gives the existence of a coincidence point in a metric space. An improved version of Kaneko’s result was obtained by Latif and Beg [2]. They proved the following.

Theorem 1.1 [[2], Theorem 2.6]

Let f:XX be a continuous map with f(X) complete. Suppose that F:XCB(X) is a f-contractive map such that F(X)f(X). Then there exists x 0 X such that f x 0 F x 0 .

In [3], Kaneko and Sessa proved the following coincidence point result.

Theorem 1.2 [3]

Let (X,d) be a complete metric space f:XX and F:XCB(X) be compatible continuous mappings such that F(X)f(X) and

D(Fx,Fy)κmax { d ( f x , f y ) , d ( f x , F x ) , d ( f y , F y ) , 1 / 2 [ d ( f x , F y ) + d ( f y , F x ) ] }

for all x,yX, where 0κ<1.

Then there exists x 0 X such that f x 0 F x 0 .

A mapping f:XX is called K-mapping if there exists a real number κ with 0κ< 1 2 such that

d(fx,fy)κ [ d ( x , f x ) + d ( y , f y ) ] for all x,yX.

Definition 1.3 Let F:XP(X) be a mapping. A point xX is said to be a fixed point of F if xFx.

Definition 1.4 Let F:XP(X) and f:XX be mappings. A point xX is said to be a coincidence point of F and f if fxFx.

Definition 1.5 A partial order is a binary relation over a set X which satisfies the following conditions:

  1. 1.

    xx (reflexivity);

  2. 2.

    If xy and yx, then x=y (antisymmetry);

  3. 3.

    If xy and yz, then xz (transitivity)

for all x, y and z in X.

A set with a partial order is called a partially ordered set.

Let (X,) be a partially ordered set and x,yX. Elements x and y are said to be comparable elements of X if either xy or yx.

Recently, there have been so many exciting developments in the field of existence of a fixed point in partially ordered sets (see [420] and the references cited therein). This trend was started by Ran and Reurings in [11] where they extended the Banach contraction principle in partially ordered sets with some application to a matrix equation. Ran and Reurings [11] proved the following seminal result.

Theorem 1.6 [11]

Let (X,) be a partially ordered set such that every pair x,yX has an upper and lower bound. Let d be a complete metric on X and f:XX be a continuous monotone (either order-preserving or order-reversing) mapping. Suppose that the following conditions hold:

  1. 1.

    There exists κ(0,1) with

    d ( f ( x ) , f ( y ) ) κd(x,y) for all xy.
  2. 2.

    There exists x 0 X with x 0 f x 0 or f x 0 x 0 .

Then f is a Picard operator (PO), that is, f has a unique fixed point x X and for each xX,

lim n f n x= x .

Theorem 1.6 was further extended and refined in [79, 12, 15, 19]. These results are hybrid of the two fundamental classical theorems: Banach’s fixed point theorem (see [21]) and Tarski’s [22] fixed point theorem. Our aim in this paper is to introduce a generalization of f-contractiveness through an implicit relation. This implicit relation is then used to obtain the existence of coincidence and common fixed points for a pair of single-valued mapping and set-valued mapping on a partially ordered metric space. In Section 2, we prove a coincidence point theorem where we use an implicit relation only for comparable elements of a partially ordered set X. Our result generalizes/extends [13, 23, 24] work to partial ordered sets. Section 3 deals with the existence of a common fixed point by using the notion of k set-valued mapping, which improves the results of Latif and Beg [2] to partially ordered sets.

We will make use of the following lemma in the proof of our result in the next section.

Lemma 1.7 [25]

Let A,BCB(X) and aA. Then for ϵ>0, there exists an element bB such that d(a,b)<D(A,B)+ϵ.

Throughout the next two sections, we take (X,) as a partially ordered set with a complete metric d.

2 Implicit relation and coincidence points

Implicit relations in metric spaces have been considered by several authors in connection with solving nonlinear functional equations (see, for instance, [1618, 26] and the references cited therein). First, we give some implicit relations for subsequent use.

Let R + be the set of nonnegative real numbers and T be the set of continuous real-valued functions T: R + 6 R satisfying the following conditions:

T 1 : T( t 1 , t 2 ,, t 6 ) is non-increasing in t 2 , t 3 ,, t 6 .

T 2 : there exists a real number κ with 0<κ<1 and ε>0 such that the inequalities

uw+ϵ

and

T(w,v,v,u,u+v,0)0

imply

wκv.

T 3 : T(w,0,0,v,v,0)0 implies wκv.

Next, we give some examples for T satisfying T 1 - T 3 .

Example 2.1 T( t 1 ,, t 6 )= t 1 α(max{ t 2 , t 3 , t 4 ,1/2( t 5 + t 6 )}), where 0<α<1.

T 1 : is obvious.

T 2 : Let u>0, then choose ϵ>0 such that αu+ϵ<u (this is possible since 0<α<1). As T(w,v,v,u,u+v,0)0, therefore wα(max{u,v})0. Now let uw+ϵ. If uv, then uαu+ϵ<u. Hence a contradiction. Thus u<v and wαv. If u=0, then wαv. Thus T 2 is satisfied.

T 3 : Since T(w,0,0,v,v,0)0, therefore wαv0. It further implies that wαv.

Example 2.2 T( t 1 ,, t 6 )= t 1 α(max{ t 2 , t 3 + t 4 2 , t 5 + t 6 2 }), where 0<α<1.

T 1 : is obvious.

T 2 : Let u>0, then choose ϵ>0 so that αu+ϵ<u. Since T(w,v,v,u,u+v,0)0, therefore wα(max{u,v})0. Now let uw+ϵ. If uv, then uαu+ϵ<u. Hence a contradiction. Thus u<v and wαv. If u=0, then wαv. Thus T 2 is satisfied.

T 3 : Since T(w,0,0,v,v,0)0, therefore w α 2 v0. It further implies that w α 2 vκv.

Example 2.3 T( t 1 ,, t 6 )= t 1 αmax{ t 2 , t 3 , t 4 }(1α)(a t 5 +b t 6 ), where 0<α<1 , 0a,b<1/2.

T 1 : is obvious.

T 2 : Let u>0, then choose ϵ>0 so that (α+2a(1α))u+ϵ<u (this is possible since 0<α+2a(1α)<1). Since T(w,v,v,u,u+v,0)0, therefore wαmax{u,v}(1α)a(u+v)0. Now let uw+ϵ. If uv, then u[α+2a(1α)]u+ϵ<u, hence a contradiction. Thus u<v and w(α+2a(1α))v. Thus T 2 is satisfied with κ=α+2a(1α).

T 3 : Since T(w,0,0,v,v,0)0, therefore wαv(1α)av0. It further implies that w[α+a(1α)]vκv.

Theorem 2.4 Let f:XX and F:XCB(X) satisfying

T ( D ( F x , F y ) , d ( f x , f y ) , d ( f x , F x ) , d ( f y , F y ) , d ( f x , F y ) , d ( f y , F x ) ) 0
(A)

for all comparable elements x, y of X and for some TT. If the following conditions are satisfied:

  1. 1.

    F(X)f(X) and f(X) is closed;

  2. 2.

    If fqFx, then xq;

  3. 3.

    If y n F x n is such that y n y=fp, then x n p for all n,

then there exists p with fpFp.

Proof Let x 0 X, then by using assumptions 1 and 2, we can choose x 1 X with x 0 x 1 such that y 0 =f x 1 F x 0 . Since y 0 F x 0 , then for any ε>0, from Lemma 1.7, there exists y 1 F x 1 such that

d( y 0 , y 1 )D(F x 0 ,F x 1 )+ε.
(2.1)

Using assumptions 1 and 2, since y 1 F x 1 f(X), there exists x 2 X such that y 1 =f x 2 F x 1 and so x 1 x 2 .

Now using (A), we have

T ( D ( F x 0 , F x 1 ) , d ( f x 0 , f x 1 ) , d ( f x 0 , F x 0 ) , d ( f x 1 , F x 1 ) , d ( f x 0 , F x 1 ) , d ( f x 1 , F x 0 ) ) 0.

Using the facts that d(f x 0 ,F x 0 )d(f x 0 , y 0 ), d(f x 1 ,F x 1 )d( y 0 , y 1 ),

d(f x 0 ,F x 1 )d(f x 0 , y 1 )d(f x 0 , y 0 )+d( y 0 , y 1 ),d ( f x 1 , F ( x 0 ) ) d( y 0 , y 0 ),

and by T 1 , we have

T ( D ( F x 0 , F x 1 ) , d ( f x 0 , y 0 ) , d ( f x 0 , y 0 ) , d ( y 0 , y 1 ) , d ( f x 0 , y 0 ) + d ( y 0 , y 1 ) , 0 ) 0,

that is,

T(w,v,v,u,u+v,0)0,

where w=D(F x 0 ,F x 1 ), v=d(f x 0 , y 0 ), u=d( y 0 , y 1 ). By using T 2 , we have (wκv),

D(F x 0 ,F x 1 )κd(f x 0 , y 0 ).
(2.2)

Using (2.2) in (2.1), we have

d( y 0 , y 1 )κd(f x 0 , y 0 )+ϵ.

Since y 1 F x 1 , then for κ>0, from Lemma 1.7, there exists y 2 F x 2 such that

d( y 1 , y 2 )D(F x 1 ,F x 2 )+κ.
(2.3)

Using assumptions 1 and 2, since y 2 F x 2 f(X), there exists x 3 X such that y 2 =f x 3 F x 2 and so x 2 x 3 .

Now, since x 1 x 2 , by using (A), we have

T ( D ( F x 1 , F x 2 ) , d ( f x 1 , f x 2 ) , d ( f x 1 , F x 1 ) , d ( f x 2 , F x 2 ) , d ( f x 1 , F x 2 ) , d ( f x 2 , F x 1 ) ) 0,

by T 1 we have

T ( D ( F x 1 , F x 2 ) , d ( y 0 , y 1 ) , d ( y 0 , y 1 ) , d ( y 1 , y 2 ) , d ( y 0 , y 1 ) + d ( y 1 , y 2 ) , 0 ) 0,

that is,

T(w,v,v,u,u+v,0)0,

where w=D(F x 1 ,F x 2 ), v=d( y 0 , y 1 ), u=d( y 1 , y 2 ). Therefore, by using T 2 , we have (wκv)

D(F x 1 ,F x 2 )κd( y 0 , y 1 ).
(2.4)

And so from (2.3) and (2.4), we have

d( y 1 , y 2 )D(F x 1 ,F x 2 )+κκd( y 0 , y 1 )+κ.
(2.5)

Again, since y 2 F x 2 , then for κ>0, from Lemma 1.7, there exists y 3 F x 3 such that

d( y 2 , y 3 )D(F x 2 ,F x 3 )+ κ 2 .
(2.6)

Using assumptions 1 and 2, since y 3 F x 3 f(X), there exists x 4 X such that y 3 =f x 4 F x 3 and so x 3 x 4 .

Now, since x 2 x 3 , by using (A) we have

T ( D ( F x 2 , F x 3 ) , d ( f x 2 , f x 3 ) , d ( f x 2 , F x 2 ) , d ( f x 3 , F x 3 ) , d ( f x 2 , F x 3 ) , d ( f x 3 , F x 2 ) ) 0,

by T 1 we have

T ( D ( F x 2 , F x 3 ) , d ( y 1 , y 2 ) , d ( y 1 , y 2 ) , d ( y 2 , y 3 ) , d ( y 1 , y 2 ) + d ( y 2 , y 3 ) , 0 ) 0,

that is,

T(w,v,v,u,u+v,0)0,

where w=D(F x 2 ,F x 3 ), v=d( y 1 , y 2 ), u=d( y 2 , y 3 ). Therefore, by using T 2 , we have (wκv)

D(F x 2 ,F x 3 )κd( y 1 , y 2 ).
(2.7)

Now, using (2.5), (2.6) and (2.7), we have

d( y 2 , y 3 )κd( y 1 , y 2 )+ κ 2 κ 2 d( y 0 , y 1 )+2 κ 2 .
(2.8)

Continuing in this way, we obtain a sequence { x n } with x n x n + 1 such that y n =f x n + 1 F x n for n0 and

(2.9)
(2.10)

Hence, { y n } is a Cauchy sequence. So, there exists a point y in the complete metric space X such that

lim n y n = lim n f( x n + 1 )=y lim n F( x n ).

Now, since f(X) is closed, there exists pX such that y=fpf(X) and by assumption 3, x n p for all n.

Now, using (A), we have

T ( D ( F x n , F p ) , d ( f x n , f p ) , d ( f x n , F x n ) , d ( f p , F p ) , d ( f x n , F p ) , d ( f p , F x n ) ) 0.

Now, taking limit as n and using T 2 , also the facts that d(f x n ,fp)=d( y n 1 ,y)0, d(f x n ,F x n )d( y n 1 , y n )0, d(fp,F x n )d(y, y n )0, we have

T ( lim n D ( F x n , F p ) , 0 , 0 , d ( y , F p ) , d ( y , F p ) , 0 ) 0,

that is,

T(w,0,0,v,v,0).

By using T 3 , we get

lim n D(F x n ,Fp)κ ( d ( y , F p ) ) .

Next, since y n F x n ,

d( y n ,Fv)D(F x n ,Fp),

taking limit as n, we obtain

d(y,Fp) lim n D(F x n ,Fp)κd(y,Fp).

From above, we have κ>1, a contradiction. So, d(y,Fp)=0. Therefore, d(y,Fp)=0 and fp=y F p ¯ =Fp. □

Remark 2.5 In assumptions 2 and 3 of Theorem 2.4, we need only comparability of the elements. Theorem 2.4 with Example 2.2 partially improve [[27], Theorem 3.10].

Corollary 2.6 Let f:XX and F:XCB(X) satisfy

D(Fx,Fy)κd(fx,fy)

for some κ with 0<κ<1 and for all comparable elements x, y of X.

Also, assume that the following conditions are satisfied:

  1. 1.

    F(X)f(X) and f(X) is closed.

  2. 2.

    If fqFx, then xq.

  3. 3.

    If y n F x n is such that y n y=fp, then x n p for all n.

Then there exists p such that fpFp.

Proof Let T( t 1 ,, t 6 ):= t 1 κ t 2 , then it is obvious that TT. Therefore, the proof is complete from Theorem 2.4. □

Corollary 2.7 Let f:XX and F:XCB(X) satisfy

D(Fx,Fy)κmax { d ( f x , f y ) , d ( f x , F x ) , d ( f y , F y ) , 1 / 2 [ d ( f x , F y ) + d ( f y , F x ) ] }

for some κ with 0<κ<1 and for all comparable elements x, y of X. Also, assume that the following conditions are satisfied:

  1. 1.

    F(X)f(X) and f(X) is closed.

  2. 2.

    If fqFx, then xq.

  3. 3.

    If y n F x n is such that y n y=fp, then x n p for all n.

Then fvFv.

Remark 2.8 Corollary 2.6 extends Latif and Beg [[2], Theorem 2.6], the result of Kaneko and Sessa [1] in a partially ordered set. Corollary 2.7 also extends the results of [13, 24] to a partially ordered set.

3 Common fixed points

In this section, we define K set-valued mappings on partially ordered metric spaces, and by using the definitions, we obtain the existence of common fixed points. Let A and B be two non-empty subsets of (X,), the relations between A and B are denoted and defined as follows:

A 1 B: if for every aA there exists bB such that ab.

Definition 3.1 Let M be a non-empty subset of X and F:MP(X). A mapping F is said to be K set-valued if there exists 0κ< 1 2 , and for any xM, u x Fx there exists a u y Fy with u x u y such that

d( u x , u y )κ [ d ( x , u x ) + d ( y , u y ) ]

for all yM with xy.

For K set-valued mappings, we just required comparability of the elements, but order does not matter.

Theorem 3.2 Let M be a non-empty closed subset of X and F:MCl(M) be a K set-valued mapping satisfying:

  1. 1.

    There exists x 0 in M such that { x 0 } 1 F x 0 ;

  2. 2.

    If x n x is a sequence in M whose consecutive terms are comparable, then x n x for all n.

Then there exists xM with xFx.

Proof Let x 0 M. Then by assumption 1, there exists x 1 F x 0 such that x 0 x 1 . Now, since F is a K set-valued mapping, there is x 2 F x 1 with x 1 x 2 such that

d( x 1 , x 2 )κ [ d ( x 0 , x 1 ) + d ( x 1 , x 2 ) ] ,

which gives

(1κ)d( x 1 , x 2 )κd( x 0 , x 1 ),

and consequently

d( x 1 , x 2 ) κ 1 κ d( x 0 , x 1 ).

Continuing in this way, we obtain a sequence { x n } whose consecutive terms are comparable and

d( x n , x n + 1 ) [ κ 1 κ ] n d( x 0 , x 1 ).

Take 0h= κ 1 κ <1, then we have

d( x n , x n + 1 ) h n d( x 0 , x 1 ).
(3.1)

Next, we will show that { x n } is a Cauchy sequence in X. Let m>n. Then

d ( x n , x m ) d ( x n , x n + 1 ) + d ( x n + 1 , x n + 2 ) + d ( x n + 2 , x n + 3 ) + + d ( x m 1 , x m ) [ h n + h n + 1 + h n + 2 + + h m 1 ] d ( x 0 , x 1 ) = h n [ 1 + h + h 2 + + h m n 1 ] d ( x 0 , x 1 ) = h n 1 h m n 1 h d ( x 0 , x 1 ) < h n 1 h d ( x 0 , x 1 )

because h(0,1), 1 h m n <1.

Therefore, d( x n , x m )0 as n, which further implies that { x n } is a Cauchy sequence. So, there exists some point (say) x in the complete metric space X such that x n x. By using assumption 2, x n x for all n.

Further, since M is closed, xM. Now we want to show that xFx.

Since x n F x n 1 with x n 1 x n also x n x for all n and F is a k set-valued mapping, there exists u n Fx with x n u n such that

d( x n , u n )κ [ d ( x n 1 , x n ) + d ( x , u n ) ] .
(3.2)

Now

d(x, u n )d(x, x n )+d( x n , u n ),

and using (3.2), we have

d(x, u n ) 1 1 κ [ d ( x , x n ) + κ d ( x n 1 , x n ) ] .

Letting n, we obtain u n x.

As u n Fx and Fx is closed, so xFx. □

Example 3.3 Let M={(0,0),(0, 1 4 ),(0, 1 4 ),( 1 4 , 1 4 )} be a subset of X= R 2 with usual order defined as follows: for (u,v),(x,y)X, (u,v)(x,y) if and only if ux,yv. Let d be a metric on X defined as follows:

d(x,y)=d ( ( x 1 , x 2 ) , ( y 1 , y 2 ) ) :=max { x 1 y 1 , x 2 y 2 }

for all x,yX, so that (X,d) is a complete metric space. Define F:MCl(M) as follows:

F(x,y)={ { ( 0 , 0 ) } if  x y , { ( 0 , 0 ) , ( 0 , 1 4 ) } if  x < y ,

(0, 1 4 )(0,0)(0, 1 4 ) and ( 1 4 , 1 4 )(0, 1 4 ).

Consider x=(0, 1 4 )(0,0)=y for u x =(0,0)Fx, there exists u y =(0,0)Fy such that u x u y .

Next 0=d((0,0),(0,0))κ[d((0, 1 4 ),(0,0))+d((0,0),(0,0))]= κ 4 .

Similarly, for other comparable elements of M, one can see that F is a K set-valued mapping.

Further (0,0)M is such that {0,0} 1 F(0,0). Also, assumption 2 of Theorem 3.2 is satisfied and (0,0) is the fixed point of F.

Theorem 3.4 Let M be a non-empty closed subset of X and F n :MCl(M) be a sequence of mappings satisfying the following:

(B): For any two mappings F i , F j and for any xM, u x F i x, there exists a u y F j y with u x u y such that

d( u x , u y )κ [ d ( x , u x ) + d ( y , u y ) ]

for all yM with xy and for some 0κ< 1 2 .

Assume that the following conditions also hold:

  1. 1.

    For each x 0 M, { x 0 } 1 F 1 x 0 .

  2. 2.

    If x n x is a sequence in X whose consecutive terms are comparable, then x n x for all n.

Then there exists xM with x F n x.

Proof Let x 0 M. Then by assumption 1, there exists x 1 F 1 x 0 such that x 0 x 1 . Now, using (B), there is x 2 F 2 x 1 with x 1 x 2 such that

d( x 1 , x 2 )κ [ d ( x 0 , x 1 ) + d ( x 1 , x 2 ) ] ,

which gives

d( x 1 , x 2 ) κ 1 κ d( x 0 , x 1 ).
(3.3)

Next, for this x 2 , there exists x 3 F 3 x 2 with x 2 x 3 such that

d( x 2 , x 3 )κ [ d ( x 1 , x 2 ) + d ( x 2 , x 3 ) ] .

Using (3.3), we obtain

d( x 2 , x 3 ) [ κ 1 κ ] 2 d( x 0 , x 1 ).

Continuing in this way, we obtain a sequence x n whose consecutive terms are comparable and

d( x n , x n + 1 ) [ κ 1 κ ] n d( x 0 , x 1 ).

Take 0h= κ 1 κ <1, then we have

d( x n , x n + 1 ) h n d( x 0 , x 1 ).
(3.4)

Therefore, { x n } is a Cauchy sequence in a complete metric space X, so x n x. By using assumption 2, x n x for all n.

Further, since M is closed, xM. Let F m be any arbitrary member of F n . Now, since x n F n x n 1 with x n 1 x n . Also, x n x for all n. By using (B), there exists u n F m x such that

d( x n , u n )κ [ d ( x n 1 , x n ) + d ( x , u n ) ] .
(3.5)

Now

d(x, u n )d(x, x n )+d( x n , u n ),

which gives

d(x, u n ) 1 1 κ [ d ( x , x n ) + κ d ( x n 1 , x n ) ] .

Letting n, we have u n x.

As u n F m x and F m x is closed, so x F m x, i.e., x F n x. □

Remark 3.5 Theorems 3.2 and 3.4 improve/extend [[2], Theorems 4.1 and 4.2].

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Acknowledgements

The present version of the paper owes much to the precise and kind remarks of the learned referees.

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Correspondence to Ismat Beg.

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The authors declare that they have no competing interests.

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IB gave the idea and ARB wrote the initial draft. Both authors read and agreed upon the draft and finalized the manuscript. Correspondence was mainly done by IB. All authors read and approved the final manuscript.

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Beg, I., Butt, A.R. Common fixed point and coincidence point of generalized contractions in ordered metric spaces. Fixed Point Theory Appl 2012, 229 (2012). https://doi.org/10.1186/1687-1812-2012-229

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Keywords

  • fixed point
  • coincidence point
  • partially ordered set
  • metric space
  • set-valued mapping