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On coupled fixed point theorems on partially ordered G-metric spaces

Abstract

In this manuscript, we extend, generalize and enrich some recent coupled fixed point theorems in the framework of partially ordered G-metric spaces in a way that is essentially more natural.

MSC:46N40, 47H10, 54H25, 46T99.

1 Introduction and preliminaries

In [1] Aydi et al. established coupled coincidence and coupled common fixed point results for a mixed g-monotone mapping satisfying nonlinear contractions in partially ordered G-metric spaces. These results generalize those of Choudhury and Maity [2].

Here we generalize, improve, enrich and extend the above mentioned coupled fixed point results of Aydi et al.

Throughout this paper, let N denote the set of nonnegative integers, and N be the set of positive integers.

Definition 1.1 (See [3])

Let X be a non-empty set, and G:X×X×X R + be a function satisfying the following properties:

(G1) G(x,y,z)=0 if x=y=z,

(G2) 0<G(x,x,y) for all x,yX with xy,

(G3) G(x,x,y)G(x,y,z) for all x,y,zX with yz,

(G4) G(x,y,z)=G(x,z,y)=G(y,z,x)= (symmetry in all three variables),

(G5) G(x,y,z)G(x,a,a)+G(a,y,z) for all x,y,z,aX (rectangle inequality).

Then the function G is called a generalized metric or, more specially, a G-metric on X, and the pair (X,G) is called a G-metric space.

Every G-metric on X defines a metric d G on X by

d G (x,y)=G(x,y,y)+G(y,x,x),for all x,yX.
(1.1)

Example 1.2 Let (X,d) be a metric space. The function G:X×X×X[0,+), defined by

G(x,y,z)=max { d ( x , y ) , d ( y , z ) , d ( z , x ) } ,

or

G(x,y,z)=d(x,y)+d(y,z)+d(z,x),

for all x,y,zX, is a G-metric on X.

Definition 1.3 (See [3])

Let (X,G) be a G-metric space, and let { x n } be a sequence of points of X. We say that ( x n ) is G-convergent to xX if lim n , m + G(x, x n , x m )=0, that is, for any ε>0, there exists NN such that G(x, x n , x m )<ε, for all n,mN. We call x the limit of the sequence and write x n x or lim n + x n =x.

Proposition 1.4 (See [3])

Let (X,G) be a G-metric space. The following are equivalent:

  1. (1)

    { x n } is G-convergent to x,

  2. (2)

    G( x n , x n ,x)0 as n+,

  3. (3)

    G( x n ,x,x)0 as n+,

  4. (4)

    G( x n , x m ,x)0 as n,m+.

Definition 1.5 (See [3])

Let (X,G) be a G-metric space. A sequence { x n } is called a G-Cauchy sequence if, for any ε>0, there is NN such that G( x n , x m , x l )<ε for all m,n,lN, that is, G( x n , x m , x l )0 as n,m,l+.

Proposition 1.6 (See [3])

Let (X,G) be a G-metric space. Then the following are equivalent:

  1. (1)

    the sequence { x n } is G-Cauchy,

  2. (2)

    for any ε>0, there exists NN such that G( x n , x m , x m )<ε, for all m,nN.

Proposition 1.7 (See [3])

Let (X,G) be a G-metric space. A mapping f:XX is G-continuous at x 0 if and only if it is G-sequentially continuous at x 0 , that is, whenever ( x n ) is G-convergent to x 0 , the sequence (f( x n )) is G-convergent to f( x 0 ).

Definition 1.8 (See [3])

A G-metric space (X,G) is called G-complete if every G-Cauchy sequence is G-convergent in (X,G).

Definition 1.9 (See [2])

Let (X,G) be a G-metric space. A mapping F:X×XX is said to be continuous if for any two G-convergent sequences { x n } and { y n } converging to x, y respectively, {F( x n , y n )} is G-convergent to F(x,y).

Let (X,) be a partially ordered set and (X,G) be a G-metric space, g:XX be a mapping. A partially ordered G-metric space, (X,G,), is called g-ordered complete if for each convergent sequence { x n } n = 0 X, the following conditions hold:

(O C 1 ) if { x n } is a non-increasing sequence in X such that x n x implies g x g x n , nN,

(O C 2 ) if { y n } is a non-decreasing sequence in X such that y n y implies g y g y n , nN.

Moreover, a partially ordered G-metric space, (X,G,), is called ordered complete when g is equal to identity mapping in the above conditions (O C 1 ) and (O C 2 ).

Definition 1.10 (See [4])

An element (x,y)X×X is said to be a coupled fixed point of the mapping F:X×XX if

F(x,y)=xandF(y,x)=y.

Definition 1.11 (See [5])

An element (x,y)X×X is called a coupled coincidence point of a mapping F:X×XX and g:XX if

F(x,y)=g(x),F(y,x)=g(y).

Moreover, (x,y)X×X is called a common coupled coincidence point of F and g if

F(x,y)=g(x)=x,F(y,x)=g(y)=y.

Definition 1.12 Let F:X×XX and g:XX be mappings. The mappings F and g are said to commute if

g ( F ( x , y ) ) =F ( g ( x ) , g ( y ) ) ,for all x,yX.

Definition 1.13 (See [4])

Let (X,) be a partially ordered set and F:X×XX be a mapping. Then F is said to have mixed monotone property if F(x,y) is monotone non-decreasing in x and is monotone non-increasing in y, that is, for any x,yX,

x 1 x 2 F( x 1 ,y)F( x 2 ,y),for  x 1 , x 2 X,

and

y 1 y 2 F(x, y 2 )F(x, y 1 ),for  y 1 , y 2 X.

Definition 1.14 (See [5])

Let (X,) be a partially ordered set and F:X×XX and g:XX be two mappings. Then F is said to have mixed g-monotone property if F(x,y) is monotone g-non-decreasing in x and is monotone g-non-increasing in y, that is, for any x,yX,

g( x 1 )g( x 2 )F( x 1 ,y)F( x 2 ,y),for  x 1 , x 2 X,
(1.2)

and

g( y 1 )g( y 2 )F(x, y 2 )F(x, y 1 ),for  y 1 , y 2 X.
(1.3)

Let Φ denote the set of functions ϕ:[0,)[0,) satisfying

  1. (a)

    ϕ 1 ({0})={0},

  2. (b)

    ϕ(t)<t for all t>0,

  3. (c)

    lim r t + ϕ(r)<t for all t>0.

Lemma 1.15 (See [5])

Let ϕΦ. For all t>0, we have lim n ϕ n (t)=0.

Aydi et al. [1] proved the following theorems.

Theorem 1.16 Let (X,) be a partially ordered set and G be a G-metric on X such that (X,G) is a complete G-metric space. Suppose that there exist ϕΦ, F:X×XX and g:XX such that

G ( F ( x , y ) , F ( u , v ) , F ( w , z ) ) ϕ ( G ( g x , g u , g w ) + G ( g y , g v , g z ) 2 )
(1.4)

for all x,y,u,v,w,zX with gwgugx and gygvgz. Suppose also that F is continuous and has the mixed g-monotone property, F(X×X)g(X) and g is continuous and commutes with F. If there exist x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and F( y 0 , x 0 )g y 0 , then F and g have a coupled coincidence point, that is, there exists (x,y)X×X such that gx=F(x,y) and gy=F(y,x).

Theorem 1.17 Let (X,) be a partially ordered set and G be a G-metric on X such that (X,G,) is regular. Suppose that there exist ϕΦ and mappings F:X×XX and g:XX such that

G ( F ( x , y ) , F ( u , v ) , F ( w , z ) ) ϕ ( G ( g x , g u , g w ) + G ( g y , g v , g z ) 2 )
(1.5)

for all x,y,u,v,w,zX with gwgugx and gygvgz. Suppose also that (g(X),G) is complete, F has the mixed g-monotone property and F(X×X)g(X). If there exist x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and F( y 0 , x 0 )g y 0 , then F and g have a coupled coincidence point.

In this manuscript, we generalize, improve, enrich and extend the above coupled fixed point results. We also state some examples to illustrate our results. This paper can be considered as a continuation of the remarkable works of Berinde [6, 7].

2 Main results

We begin with an example to illustrate the weakness of Theorem 1.16 and Theorem 1.17 above.

Example 2.1 Let X=R. Define G:X×X×X[0,) by

G(x,y,z)=|xy|+|xz|+|yz|

for all x,y,zX. Then (X,G) is a G-metric space. Define a map F:X×XX by F(x,y)= 1 12 x+ 7 12 y and g:XX by g(x)= x 2 for all x,yX. Suppose x=u=z

(2.1)

and

G ( g x , g u , g z ) + G ( g y , g v , g w ) = G ( x 2 , u 2 , z 2 ) + G ( y 2 , v 2 , w 2 ) = 1 2 [ | y v | + | y w | + | v w | ] .
(2.2)

It is clear that there is no ϕΦ that provides the statement (1.4) of Theorem 1.16.

Notice that (0,0) is the unique common coincidence point of F and g. In fact, F(0,0)=g(0)=0.

For some coupled fixed point and coupled coincidence point theorems, we refer the reader to [834].

We now state our first result which successively guarantees a coupled fixed point.

Theorem 2.2 Let (X,) be a partially ordered set and G be a G-metric on X such that (X,G) is a complete G-metric space. Suppose that there exist ϕΦ, F:X×XX and g:XX such that

(2.3)

for all x,y,u,v,w,zX with gwgugx and gygvgz. Suppose also that F is continuous and has the mixed g-monotone property, F(X×X)g(X) and g is continuous and commutes with F. If there exist x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and F( y 0 , x 0 )g y 0 , then F and g have a coupled coincidence point, that is, there exists (x,y)X×X such that gx=F(x,y) and gy=F(y,x).

Proof Given x 0 , y 0 X satisfying g x 0 F( x 0 , y 0 ) and F( y 0 , x 0 )g y 0 , we shall construct iterative sequences ( x n ) and ( y n ) in the following way: Since F(X×X)g(X), we can choose x 1 , y 1 X such that g x 1 =F( x 0 , y 0 ) and g y 1 =F( y 0 , x 0 ). Analogously, we choose x 2 , y 2 X such that g x 2 =F( x 1 , y 1 ) and g y 2 =F( y 1 , x 1 ) due to the same reasoning. Since F has the mixed g-monotone property, we conclude that g x 0 g x 1 g x 2 and g y 2 g y 1 g y 0 . By repeating this process, we derive the iterative sequence

g x n =F( x n 1 , y n 1 )g x n + 1 =F( x n , y n )

and

g y n + 1 =F( y n , x n )g y n =F( y n 1 , x n 1 ).

If for some n 0 we have (g x n 0 + 1 ,g y n 0 + 1 )=(g x n 0 ,g y n 0 ), then F( x n 0 , y n 0 )=g x n 0 and F=( y n 0 , x n 0 )=g y n 0 , that is, F and g have a coincidence point. So, we assume that (g x n + 1 ,g y n + 1 )(g x n ,g y n ) for all nN. Thus, we have either g x n + 1 =F( x n , y n )g x n or g y n + 1 =F( y n , x n )g y n . We set

t n =G(g x n + 1 ,g x n + 1 ,g x n )+G(g y n + 1 ,g y n + 1 ,g y n )
(2.4)

for all nN. Due to the property (G2), we have t n >0 for all nN. By using inequality (2.3), we obtain

(2.5)

Taking (2.4) into account, (2.5) becomes

t n ϕ( t n 1 ).
(2.6)

Since ϕ(t)<t for all t>0, it follows that t n is monotone decreasing. Therefore, there is some L0 such that lim n + t n =L.

Now, we assert that L=0. Suppose, on the contrary, that L>0. Letting n+ in (2.6) and using the properties of the map ϕ, we get

L= lim n + t n lim n + ϕ( t n 1 )<L,

which is contradiction. Thus L=0. Hence

lim n + t n = lim n + G(g x n + 1 ,g x n + 1 ,g x n )+G(g y n ,g y n ,g y n 1 )=0.
(2.7)

Next, we prove that (g x n ) and (g y n ) are Cauchy sequences in the G-metric space (X,G). Suppose, on the contrary, that at least one of (g x n ) and (g y n ) is not a Cauchy sequence in (X,G). Then there exist ε>0 and sequences of natural numbers (m(k)) and (l(k)) such that for every natural number k, m(k)>l(k)k and

r k =G(g x m ( k ) ,g x m ( k ) ,g x l ( k ) )+G(g y m ( k ) ,g y m ( k ) ,g y l ( k ) )ε.
(2.8)

Now, corresponding to l(k), we choose m(k) to be the smallest for which (2.8) holds. Hence

G(g x m ( k ) 1 ,g x m ( k ) 1 ,g x l ( k ) )+G(g y m ( k ) 1 ,g y m ( k ) 1 ,g y l ( k ) )<ε.

Using the rectangle inequality (property (G5)), we get

ε r k G ( g x m ( k ) , g x m ( k ) , g x m ( k ) 1 ) + G ( g x m ( k ) 1 , g x m ( k ) 1 , g x l ( k ) ) + G ( g y m ( k ) , g y m ( k ) , g y m ( k ) 1 ) + G ( g y m ( k ) 1 , g y m ( k ) 1 , g y l ( k ) ) = G ( g x m ( k ) 1 , g x m ( k ) 1 , g x l ( k ) ) + G ( g y m ( k ) 1 , g y m ( k ) 1 , g y l ( k ) ) + t m ( k ) 1 < ε + t m ( k ) 1 .
(2.9)

Letting k+ in the above inequality and using (2.7) yields

lim k + r k = ε + .
(2.10)

Again, by the rectangle inequality, we have

r k = G ( g x m ( k ) , g x m ( k ) , g x l ( k ) ) + G ( g y m ( k ) , g y m ( k ) , g y l ( k ) ) G ( g x m ( k ) , g x m ( k ) , g x m ( k ) + 1 ) + G ( g x m ( k ) + 1 , g x m ( k ) + 1 , g x l ( k ) + 1 ) + G ( g x l ( k ) + 1 , g x l ( k ) + 1 , g x l ( k ) ) + G ( g y m ( k ) , g y m ( k ) , g y m ( k ) + 1 ) + G ( g y m ( k ) + 1 , g y m ( k ) + 1 , g y l ( k ) + 1 ) + G ( g y l ( k ) + 1 , g y l ( k ) + 1 , g y l ( k ) ) = t l ( k ) + G ( g x m ( k ) , g x m ( k ) , g x m ( k ) + 1 ) + G ( g y m ( k ) , g y m ( k ) , g y m ( k ) + 1 ) + G ( g x m ( k ) + 1 , g x m ( k ) + 1 , g x l ( k ) + 1 ) + G ( g y m ( k ) + 1 , g y m ( k ) + 1 , g y l ( k ) + 1 ) .

Using the fact that G(x,x,y)2G(x,y,y) for any x,yX, we obtain from properties (G2)-(G4)

r k t l ( k ) + 2 G ( g x m ( k ) , g x m ( k ) , g x m ( k ) + 1 ) + 2 G ( g y m ( k ) , g y m ( k ) , g y m ( k ) + 1 ) + G ( g x m ( k ) + 1 , g x m ( k ) + 1 , g x l ( k ) + 1 ) + G ( g y m ( k ) + 1 , g y m ( k ) + 1 , g y l ( k ) + 1 ) = t l ( k ) + 2 t m ( k ) + G ( g x m ( k ) + 1 , g x m ( k ) + 1 , g x l ( k ) + 1 ) + G ( g y m ( k ) + 1 , g y m ( k ) + 1 , g y l ( k ) + 1 ) .

Next, using inequality (2.3), we have

(2.11)

Now, using (2.7),(2.10), the properties of the function ϕ, and letting k+ in (2.11), we get

ε lim k + ϕ( r k )= lim t ( ε ) + ϕ(t)<ε,

which is a contradiction. Thus, we have proven that (g x n ) and (g y n ) are Cauchy sequences in the G-metric space (X,G). Now, since (X,G) is complete, there are x,yX such that (g x n ) and (g y n ) are respectively G-convergent to x and y. That is from Proposition 1.4, we have

lim n + G ( g x n , g x n , x ) = lim n + G ( g x n , x , x ) = 0 , lim n + G ( g y n , g y n , y ) = lim n + G ( g y n , y , y ) = 0 .

Using the continuity of g, we get from Proposition 1.7

lim n + G ( g ( g x n ) , g ( g x n ) , g x ) = lim n + G ( g ( g x n ) , g x , g x ) = 0 , lim n + G ( g ( g y n ) , g ( g y n ) , g y ) = lim n + G ( g ( g y n ) , g y , g y ) = 0 .
(2.12)

Since g x n + 1 =F( x n , y n ) and g y n + 1 =F( y n , x n ), employing the commutativity of F and g yields

g ( g x n + 1 ) = g ( F ( x n , y n ) ) = F ( g x n , g y n ) , g ( g y n + 1 ) = g ( F ( y n , x n ) ) = F ( g y n , g x n ) .
(2.13)

Now, we shall show that F(x,y)=gx and F(y,x)=gy.

The mapping F is continuous, and since the sequences (g x n ) and (g y n ) are respectively G-convergent to x and y, using Definition 1.9, the sequence (F(g x n ,g y n )) is G-convergent to F(x,y). Therefore, from (2.13), (g(g x n + 1 )) is G-convergent to F(x,y). By uniqueness of the limit and using (2.12), we have F(x,y)=gx. Similarly, we can show that F(y,x)=gy. Hence, (x,y) is a coupled coincidence point of F and g. This completes the proof. □

The following example illustrates that Theorem 2.2 is an extension of Theorem 1.16.

Example 2.3 Let us reconsider Example 2.1. Define a map F:X×XX by

F(x,y)= 1 12 x+ 7 12 y

and g:XX by g(x)= x 2 for all x,yX. Then F(X×X)=[0,)=g(X)=X. We observe that

G ( F ( x , y ) , F ( u , v ) , F ( z , w ) ) + G ( F ( y , x ) , F ( v , u ) , F ( w , z ) ) = G ( 1 12 x + 7 12 y , 1 12 u + 7 12 v , 1 12 z + 7 12 w ) + G ( 1 12 y + 7 12 x , 1 12 v + 7 12 u , 1 12 w + 7 12 z ) = 7 12 | v y | + 7 12 | w y | + 7 12 | w v | + 1 12 | u x | + 1 12 | z x | + 1 12 | z u | + 1 12 | v y | + 1 12 | w y | + 1 12 | w v | + 7 12 | u x | + 7 12 | z x | + 7 12 | z u | = 8 12 ( | v y | + | w y | + | w v | ) + 8 12 ( | u x | + | z x | + | z u | )

and

G ( g x , g u , g z ) + G ( g y , g v , g w ) = G ( 1 2 x , 1 2 u , 1 2 z ) + G ( 1 2 y , 1 2 v , 1 2 w ) = 1 2 [ ( | x u | + | x z | + | u z | ) + ( | y v | + | y w | + | v w | ) ] .
(2.14)

Then, the statement (2.3) of Theorem 2.2 is satisfied for ϕ(t)= 2 3 t and (0,0) is the desired coupled coincidence point.

In the next theorem, we omit the continuity hypothesis of F.

Theorem 2.4 Let (X,) be a partially ordered set and G be a G-metric on X such that (X,G,) is g-ordered complete. Suppose that there exist ϕΦ and mappings F:X×XX and g:XX such that

(2.15)

for all x,y,u,v,w,zX with gwgugx and gygvgz. Suppose also that (g(X),G) is complete, F has the mixed g-monotone property and F(X×X)g(X). If there exist x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and F( y 0 , x 0 )g y 0 , then F and g have a coupled coincidence point.

Proof Proceeding exactly as in Theorem 2.2, we have that (g x n ) and (g y n ) are Cauchy sequences in the complete G-metric space (g(X),G). Then, there exist x,yX such that g x n gx and g y n gy. Since (g x n ) is non-decreasing and (g y n ) is non-increasing, using the regularity of (X,G,), we have g x n gx and gyg y n for all n0. If g x n =gx and g y n =gy for some n0, then gx=g x n g x n + 1 gx=g x n and gyg y n + 1 g y n =gy, which implies that g x n =g x n + 1 =F( x n , y n ) and g y n =g y n + 1 =F( y n , x n ), that is, ( x n , y n ) is a coupled coincidence point of F and g. Then, we suppose that (g x n ,g y n )(gx,gy) for all n0. Using the rectangle inequality, (2.15) and property ϕ(t)<t for all t>0, we get

G ( F ( x , y ) , g x , g x ) + G ( F ( y , x ) , g y , g y ) G ( F ( x , y ) , g x n + 1 , g x n + 1 ) + G ( g x n + 1 , g x , g x ) + G ( F ( y , x ) , g y n + 1 , g y n + 1 ) + G ( g y n + 1 , g y , g y ) = G ( F ( x , y ) , F ( x n , y n ) , F ( x n , y n ) ) + G ( g x n + 1 , g x , g x ) + G ( F ( y , x ) , F ( y n , x n ) , F ( y n , x n ) ) + G ( g y n + 1 , g y , g y ) ϕ ( G ( g x , g x n , g x n ) + G ( g y , g y n , g y n ) ) + G ( g x n + 1 , g x , g x ) + G ( g y n + 1 , g y , g y ) < G ( g x , g x n , g x n ) + G ( g y , g y n , g y n ) + G ( g x n + 1 , g x , g x ) + G ( g y n + 1 , g y , g y ) .

Letting n+ in the above inequality, we obtain

G ( F ( x , y ) , g x , g x ) +G ( F ( y , x ) , g y , g y ) =0,

which implies that gx=F(x,y) and gy=F(y,x). Thus we proved that (x,y) is a coupled coincidence point of F and g. □

Corollary 2.5 Let (X,) be a partially ordered set and G be a G-metric on X such that (X,G) is a complete G-metric space. Suppose that there exist k[0,1), F:X×XX and g:XX such that

[ G ( F ( x , y ) , F ( u , v ) , F ( w , z ) ) + G ( F ( y , x ) , F ( v , u ) , F ( z , w ) ) ] k [ G ( g x , g u , g w ) + G ( g y , g v , g z ) ]

for all x,y,u,v,w,zX with gwgugx and gygvgz. Suppose also that F is continuous, has the mixed g-monotone property, F(X×X)g(X) and g is continuous and commutes with F. If there exist x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and F( y 0 , x 0 )g y 0 , then F and g have a coupled coincidence point.

Proof Taking ϕ(t)=kt with k[0,1) in Theorem 2.4, we obtain Corollary 2.5. □

Corollary 2.6 Let (X,) be a partially ordered set and G be a G-metric on X such that (X,G,) is g-ordered complete. Suppose that there exist k[0,1), F:X×XX and g:XX such that

[ G ( F ( x , y ) , F ( u , v ) , F ( w , z ) ) + G ( F ( y , x ) , F ( v , u ) , F ( z , w ) ) ] k [ G ( g x , g u , g w ) + G ( g y , g v , g z ) ]

for all x,y,u,v,w,zX with gwgugx and gygvgz. Suppose also that (g(X),G) is complete, F has the mixed g-monotone property, F(X×X)g(X). If there exist x 0 , y 0 X such that g x 0 F( x 0 , y 0 ) and F( y 0 , x 0 )g y 0 , then F and g have a coupled coincidence point.

Proof Taking ϕ(t)=kt with k[0,1) in Theorem 2.4, we obtain Corollary 2.6 □

Remark 2.7 Taking g= l x (the identity mapping) in Corollary 2.5, we obtain [[2], Theorem 3.1]. Taking g= I x in Corollary 2.6, we obtain [[2], Theorem 3.2].

Now we shall prove the existence and uniqueness theorem of a coupled common fixed point. If (X,) is a partially ordered set, we endow the product set X×X with the partial order defined by

(x,y)(u,v)xu,vy.

Theorem 2.8 In addition to the hypothesis of Theorem  2.2, suppose that for all (x,y), ( x , y )(X×X), there exists (u,v)X×X such that (F(x,y),F(u,v)) is comparable with (F(x,y),F(y,x)) and (F( x , y ),F( y , x )). Suppose also that ϕ is a non-decreasing function. Then F and g have a unique coupled common fixed point, that is, there exists a unique (x,y)X×X such that

x=gx=F(x,y)andy=gy=F(y,x).

Proof From Theorem 2.2, the set of coupled coincidences is non-empty. We shall show that if (x,y) and ( x , y ) are coupled coincidence points, that is, if gx=F(x,y), g(y)=F(y,x), g x =F( x , y ) and g y =F( y , x ), then

gx=g x andgy=g y .
(2.16)

By assumption, there exists (u,v)X×X such that (F(u,v),F(v,u)) is comparable with (F(x,y),F(y,x)) and (F( x , y ),F( y , x )). Without loss of generality, we can assume that

( F ( x , y ) , F ( y , x ) ) ( F ( u , v ) , F ( v , u ) )

and

( F ( x , y ) , F ( y , x ) ) ( F ( u , v ) , F ( v , u ) ) .

Put u 0 =u, v 0 =v and choose u 1 , v 1 X such that g u 1 =F( u 0 , v 0 ) and g v 1 =F( v 0 , u 0 ). Then, similarly as in the proof of Theorem 2.2, we can inductively define sequences (g u n ) and (g v n ) in X by g u n + 1 =F( u n , v n ) and g v n + 1 =F( v n , u n ).

Further, set x 0 =x, y 0 =y, x 0 = x , y 0 = y and, in the same way, define the sequences (g x n ), (g y n ), (g x n ) and (g y n ). Since

( F ( x , y ) , F ( y , x ) ) =(g x 1 ,g y 1 )=(gx,gy) ( F ( u , v ) , F ( v , u ) ) =(g u 1 ,g v 1 ),

then gxg u 1 and g v 1 gy. Using that F is a mixed g-monotone mapping, one can show easily that gxg u n and g v n gy for all n1. Thus from (2.15), we get

G ( g u n + 1 , g x , g x ) + G ( g y , g y , g v n + 1 ) = G ( F ( u n , v n ) , F ( x , y ) , F ( x , y ) ) + G ( F ( y , x ) , F ( y , x ) , F ( v n , u n ) ) ϕ ( G ( g u n , g x , g x ) + G ( g v n , g y , g y ) ) .

Without loss of generality, we can suppose that (g u n ,g v n )(gx,gy) for all n1. Since ϕ is non-decreasing, from the previous inequality, we get

G(g u n + 1 ,gx,gx)+G(g v n + 1 ,gy,gy) ϕ n ( G ( g u 1 , g x , g x ) + G ( g v 1 , g y , g y ) )

for each n1. Letting n+ in the above inequality and using Lemma 1.15, we obtain

lim n + G(g u n + 1 ,gx,gx)=0and lim n + G(g v n + 1 ,gy,gy)=0.
(2.17)

Analogously, we derive that

lim n + G ( g u n + 1 , g x , g x ) =0and lim n + G ( g v n + 1 , g y , g y ) =0.
(2.18)

Hence, from (2.17), (2.18) and the uniqueness of the limit, we get gx=g x and gy=g y . Hence the equalities in (2.16) are satisfied. Since gx=F(x,y) and gy=F(y,x), by commutativity of F and g, we have

g(gx)=g ( F ( x , y ) ) =F(gx,gy)andg(gy)=g ( F ( y , x ) ) =F(gy,gx).
(2.19)

Denote gx=z and gy=w, then by (2.19), we get

gz=F(z,w)andgw=F(w,z).
(2.20)

Thus, (z,w) is a coincidence point. Then, from (2.16) with x =z and y =w, we have gx=gz and gy=gw, that is,

gz=zandgw=w.
(2.21)

From (2.20), (2.21), we get

z=gz=F(z,w)andw=gw=F(w,z).

Then, (z,w) is a coupled common fixed point of F and g.

To prove the uniqueness, assume that (p,q) is another coupled common fixed point. Then by (2.16), we have p=gp=gz=z and q=gq=gw=w. □

Theorem 2.9 Let (X,) be a partially ordered set and G be a G-metric on X such that (X,G) is a complete G-metric space and (X,G,) is regular. Suppose that there exist ϕΦ and F:X×XX having the mixed monotone property such that

G ( F ( x , y ) , F ( u , v ) , F ( w , z ) ) +G ( F ( y , x ) , F ( v , u ) , F ( z , w ) ) ϕ ( G ( x , u , w ) + G ( y , v , z ) ) ,

for all x,y,u,v,w,zX with wux and yvz. If there exist x 0 , y 0 X such that x 0 F( x 0 , y 0 ) and F( y 0 , x 0 ) y 0 , then F has a coupled fixed point. Furthermore, if y 0 x 0 , then x=y, that is, x=F(x,x).

Proof Following the proof of Theorem 2.4 with g= I x , we have only to show that x=F(x,x). Since y 0 x 0 , we get y y n y 1 y 0 x 0 x 1 x n x.

Thus, we have yx. Suppose that G(x,x,y)>0. Using inequality (2.15), we have

G ( x , x , y ) + G ( x , y , y ) = G ( F ( x , y ) , F ( x , y ) , F ( y , x ) ) + G ( F ( x , y ) , F ( y , x ) , F ( y , x ) ) ϕ ( G ( x , x , y ) + G ( y , y , x ) ) < G ( x , x , y ) + G ( x , y , y ) ,

a contradiction. Thus, G(x,x,y)=0 and x=y=F(x,x). □

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Correspondence to Billûr Kaymakçalan.

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Karapınar, E., Kaymakçalan, B. & Taş, K. On coupled fixed point theorems on partially ordered G-metric spaces. J Inequal Appl 2012, 200 (2012). https://doi.org/10.1186/1029-242X-2012-200

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Keywords

  • coupled fixed point
  • coincidence point
  • mixed g-monotone property
  • ordered set
  • G-metric space