Open Access

Tripled coincidence point theorems for weak φ-contractions in partially ordered metric spaces

Fixed Point Theory and Applications20122012:44

DOI: 10.1186/1687-1812-2012-44

Received: 13 November 2011

Accepted: 21 March 2012

Published: 21 March 2012

Abstract

In this article, we present tripled coincidence point theorems for F: X3X and g: XX satisfying weak φ-contractions in partially ordered metric spaces. We also provide nontrivial examples to illustrate our results and new concepts presented herein. Our results unify, generalize and complement various known comparable results from the current literature, Berinde and Borcut and Abbas et al.

1 Introduction

Fixed point theory has fascinated hundreds of researchers since 1922 with the celebrated Banach's fixed point theorem. This theorem provides a technique for solving a variety of applied problems in mathematical sciences and engineering. There exists a last literature on the topic and this is a very active field of research at present. There are great number of generalizations of the Banach contraction principle. Bhaskar and Lakshmikantham [1] introduced the notion of coupled fixed point and proved some coupled fixed point results under certain conditions, in a complete metric space endowed with a partial order. Later, Lakshmikantham and Ćirić [2] extended these results by defining the mixed g-monotone property. More accurately, they proved coupled coincidence and coupled common fixed point theorems for a mixed g-monotone mapping in a complete metric space endowed with a partial order. Karapınar [3, 4] generalized these results on a complete cone metric space endowed with a partial order. For other results on coupled fixed point theory, we address the readers to [513].

To make our exposition self contained, in this section we recall some previous notations and known results.

For simplicity, we denote from now on X × X X × X k terms by X k , where k and X be a non-empty set.

Let (X, ≤) be a partially ordered set. According to [1], the mapping F: X2X 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, y X,
x 1 x 2 F ( x 1 , y ) F ( x 2 , y ) , for x 1 , x 2 X , y 1 y 2 F ( x , y 2 ) F ( x , y 1 ) , for y 1 , y 2 X .
An element (x, y) X2 is said to be a coupled fixed point of the mapping F: X2X if
F ( x , y ) = x and F ( y , x ) = y .
Theorem 1.1. ([1]) Let (X, ≤) be an ordered set such that there exists a metric d on X such that (X, d) is complete. Let F: X2X be a continuous mapping having the mixed monotone property on X. Assume that there exists k [0, 1) with
d ( F ( x , y ) , F ( u , v ) ) k 2 [ d ( x , u ) + d ( y , v ) ] , f o r a l l u x , y u .
(1.1)

If there exist x0, y0 X such that x0F(x0, y0) and F(y0, x0) ≤ y0, then, there exist x, y X such that x = F(x, y) and y = F(y, x).

Recently, Samet and Vetro [14] introduced the notion of fixed point of N-order as natural extension of that of coupled fixed point and established some new coupled fixed point theorems in complete metric spaces, using a new concept of F-invariant set. Later, Berinde and Borcut [15] obtained existence and uniqueness of triple fixed point results in a complete metric space, endowed with a partial order.

Again, let (X, ≤) be a partially ordered set. In accordance with [15], the mapping F: X3X is said to have the mixed monotone property if for any x, y, z X
x 1 , x 2 X , x 1 x 2 F ( x 1 , y , z ) F ( x 2 , y , z ) , y 1 , y 2 X , y 1 y 2 F ( x , y 1 , z ) F ( x , y 2 , z ) , z 1 , z 2 X , z 1 z 2 F ( x , y , z 1 ) F ( x , y , z 2 ) .
An element (x, y, z) X3 is called a tripled fixed point of F if
F ( x , y , z ) = x , F ( y , x , y ) = y and F ( z , y , x ) = z .

Berinde and Borcut [15] proved the following theorem.

Theorem 1.2. ([15]) Let (X, ≤) be a partially ordered set and (X, d) be a complete metric space. Let F: X3X be a mapping having the mixed monotone property on X. Assume that there exist constants a, b, c [0, 1) such that a + b + c < 1 for which
d ( F ( x , y , z ) , F ( u , v , w ) ) a d ( x , u ) + b d ( y , v ) + c d ( z , w )
(1.2)

for all xu, yυ, zw. Assume either

(I) F is continuous, or

(II) X has the following properties:

(i) if non-decreasing sequence x n → x, then x n x for all n,

(ii) if non-increasing sequence y n y, then y n y for all n.

If there exist x0, y0, z0 X such that
x 0 F ( x 0 , y 0 , z 0 ) , y 0 F ( y 0 , x 0 , y 0 ) , a n d z 0 F ( x 0 , y 0 , z 0 )
then there exist x, y, z X such that
F ( x , y , z ) = x , F ( y , x , y ) = y , a n d F ( z , y , x ) = z .

In this article, we establish tripled coincidence point theorems for F: X3X and g: XX satisfying nonlinear contractive conditions, in partially ordered metric spaces. The presented theorems extend and improve some results in litterature.

2 Main results

We shall start this section by recalling the following basic notions, introduced by [Abbas, Aydi and Karapınar, Tripled common fixed point in partially ordered metric spaces, submitted]. In this respect, let us consider (X, ≤) a partially ordered set, F: X3X and g: X → X two mappings. The mapping F is said to have the mixed g-monotone property if for any x, y, z X
x 1 , x 2 X , g x 1 g x 2 F ( x 1 , y , z ) F ( x 2 , y , z ) , y 1 , y 2 X , g y 1 g y 2 F ( x , y 1 , z ) F ( x , y 2 , z ) , z 1 , z 2 X , g z 1 g z 2 F ( x , y , z 1 ) F ( x , y , z 2 ) .
An element (x, y, z) is called a tripled coincidence point of F and g if
F ( x , y , z ) = g x , F ( y , x , y ) = g y , and F ( z , y , x ) = g z ,
while (gx, gy, gz) is said a tripled point of coincidence of mappings F and g. Moreover, (x, y, z) is called a tripled common fixed point of F and g if
F ( x , y , z ) = g x = x , F ( y , x , y ) = g y = y , and F ( z , y , x ) = g z = z .
At last, mappings F and g are called commutative if
g ( F ( x , y , z ) ) = F ( g x , g y , g z ) , x , y , z X .

In the same paper, they proved the following result.

Theorem 2.1. Let (X, ≤) be a partially ordered set and suppose there is a metric d on X such that (X, d) is a complete metric space. Assume there is a function φ: [0, +∞) → [0, +∞) such that φ(t) < t for each t > 0. Also suppose F: X3X and g: XX are such that F has the mixed g-monotone property and suppose there exist p, q, r [0, 1) with p + 2q + r < 1 such that
d ( F ( x , y , z ) , F ( u , v , w ) ) φ p d ( g x , g u ) + q d ( g y , g v ) + r d ( g z , g w ) ,
(2.1)

for any x, y, z X for which gx > gu, gυgy and gz ≥ gw.

Suppose F(X3) g(X), g is continuous and commutes with F. Suppose either

(a) F is continuous, or

(b) X has the following properties:

(i) if non-decreasing sequence gx n x (respectively, gz n z), then gx n x (respectively, gz n z) for all n,

(ii) if non-increasing sequence gy n y, then gy n y for all n.

If there exist x0, y0, z0 X such that gx0F(x0, y0, z0), gy0F(y0, x0, y0) and gz0F(z0, y0, x0), then there exist x, y, z X such that
F ( x , y , z ) = g x , F ( y , x , y ) = g y , a n d F ( z , y , x ) = g z ,

that is, F and g have a tripled coincidence point.

Before starting to introduce our results, let us consider the set of functions
Φ = φ : [ 0 , + ) [ 0 , + ) | φ ( t ) < t and lim r t + φ ( r ) < t , t > 0 .

Our first main result is the following:

Theorem 2.2. Let (X, ≤) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose F: X3X and g: XX are such that F has the mixed g-monotone property and F(X3) g(X). Assume there is a function φ Φ such that
d ( F ( x , y , z ) , F ( u , v , w ) ) + d ( F ( y , x , y ) , F ( v , u , v ) ) + d ( F ( z , y , x ) , F ( w , v , u ) ) 3 φ d ( g x , g u ) + d ( g y , g v ) + d ( g z , g w ) 3 ,
(2.2)
for any x, y, z, u, υ, w X for which gxgu, gυgy and gzgw. Assume that F is continuous, g is continuous and commutes with F. If there exist x0, y0, z0 X such that
g x 0 F ( x 0 , y 0 , z 0 ) , g y 0 F ( y 0 , x 0 , y 0 ) a n d g z 0 F ( z 0 , y 0 , x 0 ) ,
(2.3)
then there exist x, y, z X such that
F ( x , y , z ) = g x , F ( y , x , y ) = g y , a n d F ( z , y , x ) = g z ,

that is, F and g have a tripled coincidence point.

Proof. Let x0, y0, z0 X be such that gx0F(x0, y0, z0), gy0F(y0, x0, y0) and gz0F(z0, y0, x0). We can choose x1, y1, z1 X such that
g x 1 = F ( x 0 , y 0 , z 0 ) , g y 1 = F ( y 0 , x 0 , y 0 ) and g z 1 = F ( z 0 , y 0 , x 0 ) .
(2.4)
This can be done because F(X3) g(X). Continuing this process, we construct sequences {x n }, {y n }, and {z n } in X such that
g x n + 1 = F ( x n , y n , z n ) , g y n + 1 = F ( y n , x n , z n ) , and g z n + 1 = F ( z n , y n , x n ) .
(2.5)
By induction, we will prove that
g x n g x n + 1 , g y n + 1 g y n , and g z n g z n + 1 .
(2.6)
Since gx0F(x0, y0, z0), gy0F(y0, x0, y0), and gz0F(z0, y0, x0), therefore by (2.4) we have
g x 0 g x 1 , g y 1 g y 0 , and g z 0 g z 1 .
Thus (2.6) is true for n = 0. We suppose that (2.6) is true for some n > 0. Since F has the mixed g-monotone property, by gx n gx n +1, gy n +1gy n , and gz n gz n +1, we have that
g x n + 1 = F ( x n , y n , z n ) F ( x n + 1 , y n , z n ) F ( x n + 1 , y n , z n + 1 ) F ( x n + 1 , y n + 1 , z n + 1 ) = g x n + 2 ,
g y n + 2 = F ( y n + 1 , x n + 1 , y n + 1 ) F ( y n + 1 , x n , y n + 1 ) F ( y n , x n , y n + 1 ) F ( y n , x n , y n ) = g y n + 1 ,
and
g z n + 1 = F ( z n , y n , x n ) F ( z n + 1 , y n , x n ) F ( z n + 1 , y n + 1 , x n ) F ( z n + 1 , y n + 1 , x n + 1 ) = g z n + 2 .
That is, (2.6) is true for any n . If for some k
g x k = g x k + 1 , g y k = g y k + 1 , and g z k = g z k + 1 ,
then, by (2.5), (x k , y k , z k ) is a tripled coincidence point of F and g. From now on, we assume that at least
g x n g x n + 1 or g y n g y n + 1 or g z n g z n + 1
(2.7)
for any n . From (2.6) and the inequality (2.2)
d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) + d ( g z n + 1 , g z n ) = d ( F ( x n , y n , z n ) , F ( x n 1 , y n 1 , z n 1 ) ) + d ( F ( y n , x n , y n ) , F ( y n 1 , x n 1 , y n 1 ) ) + d ( F ( z n , y n , x n ) , F ( z n 1 , y n 1 , x n 1 ) ) 3 φ ( 1 3 ( d ( g x n , g x n 1 ) + d ( g y n , g y n 1 ) + d ( g z n , g z n 1 ) ) .
For each n ≥ 1, take
δ n : = 1 3 ( d ( g x n , g x n - 1 ) + d ( g y n , g y n - 1 ) + d ( g z n , g z n - 1 ) ) .
(2.8)
One can write
δ n + 1 φ ( δ n ) n 1 .
(2.9)
By (2.7), we have δ n > 0. Having in mind φ(t) < t for each t > 0, so we have φ(δ n ) < δ n . From (2.9), we get
δ n + 1 < δ n n 1 ,
that is, the sequence {δ n } is non-negative and decreasing. Therefore, there exists some δ ≥ 0 such that
lim n + δ n = lim n + 1 3 d ( g x n , g x n - 1 ) + d ( g y n , g y n - 1 ) + d ( g z n , g z n - 1 ) = δ + .
(2.10)
We shall prove that δ = 0. Assume, on the contrary, that δ > 0. Then by letting n → +∞ in (2.9) we have
0 < δ = lim n + δ n + 1 lim n + φ ( δ n ) = lim r δ + δ ( r ) < δ ,
which is a contradiction. Thus, δ = 0, and by (2.10), we get
lim n + δ n = 0 .
(2.11)

We now prove that {gx n }, {gy n }, and {gz n } are Cauchy sequences in (X,d).

Suppose, on the contrary, that at least one of {gx n }, {gy n }, and {gz n } is not a Cauchy sequence. So, there exists ε > 0 for which we can find subsequences {gx n (k)}, {gx m (k)} of {gx n }, {gy n (k)}, {gy m (k)} of {gy n }, and {gz n (k)}, {gz m (k)} of {gz n } with n(k) > m(k) ≥ k such that
d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) + d ( g z n ( k ) , g z m ( k ) ) ε .
(2.12)
Additionally, corresponding to m(k), we may choose n(k) such that it is the smallest integer satisfying (2.12) and n(k) > m(k) ≥ k. Thus,
d ( g x n ( x ) - 1 , g x m ( k ) ) + d ( g y n ( k ) - 1 , g y m ( k ) ) + d ( d z n ( k ) - 1 , g z m ( k ) ) < ε .
(2.13)
By using triangle inequality and having in mind (2.12) and (2.13)
ε t k = d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) + d ( g z n ( k ) , g z m ( k ) ) d ( g x n ( k ) , g x n ( k ) - 1 ) + d ( g x n ( k ) - 1 , g x m ( k ) ) + d ( g y n ( k ) , g y n ( k ) - 1 ) + d ( g y n ( k ) - 1 , g y m ( k ) ) + d ( g z n ( k ) , g z n ( k ) - 1 ) + d ( g z n ( k ) - 1 , g z m ( k ) ) < d ( g x n ( k ) , g x n ( k ) - 1 ) + d ( g y n ( k ) , g y n ( k ) - 1 ) + d ( g z n ( k ) , g z n ( k ) - 1 ) + ε .
(2.14)
Letting k → ∞ in (2.14) and using (2.11)
lim k t k = lim k d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) + d ( g z n ( k ) , g z m ( k ) ) = ε .
(2.15)
Again by triangle inequality,
t k = d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) + d ( g z n ( k ) , g z m ( k ) ) d ( g x n ( k ) , g x n ( k ) + 1 ) + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g x m ( k ) + 1 , g x m ( k ) ) + d ( g y n ( k ) , g y n ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g y m ( k ) + 1 , g y m ( k ) ) + d ( g z n ( k ) , g z n ( k ) + 1 ) + d ( g z n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g z m ( k ) + 1 , g z m ( k ) ) δ n ( k ) + 1 + δ m ( k ) + 1 + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g z n ( k ) + 1 , g z m ( k ) + 1 ) .
(2.16)
Since n(k) > m(k), then
g x n ( k ) g x m ( k ) , g y n ( k ) g y m ( k ) , g z n ( k ) g z m ( k ) .
(2.17)
Take (2.17) in (2.2) to get
d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g z n ( k ) + 1 , g z m ( k ) + 1 ) = d ( F ( x n ( k ) , y n ( k ) , z n ( k ) ) , F ( x m ( k ) , y m ( k ) , z m ( k ) ) + d ( F ( y n ( k ) , x n ( k ) , y n ( k ) ) , F ( y m ( k ) , x m ( k ) , y m ( k ) ) + d ( F ( z n ( k ) , y n ( k ) , x n ( k ) ) , F ( z m ( k ) , y m ( k ) , x m ( k ) ) ) 3 φ ( 1 3 [ d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) + d ( g z n ( k ) , g z m ( k ) ) ] ) = 3 φ ( t k 3 ) .
Combining this in (2.16), we obtain that
t k δ n ( k ) + 1 + δ m ( k ) + 1 + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g z n ( k ) + 1 , g z m ( k ) + 1 ) δ n ( k ) + 1 + δ m ( k ) + 1 + 3 φ t k 3 .
Letting k → ∞ and having in mind (2.11) and (2.15), we get
ε 3 lim k + φ 1 3 t k = 3 lim r 1 3 t + ϕ ( r ) < 3 1 3 ε = ε ,

which is a contradiction. This shows that {gx n }, {gy n }, and {gz n } are Cauchy sequences in (X, d).

Since X is complete, there exist x, y, z X such that
lim n + g x n = x , lim n + g y n = y and lim n + g z n = z .
(2.18)
From (2.18) and the continuity of g.
lim n + g ( g x n ) = g x , lim n + g ( g y n ) = g y , and lim n + g ( g z n ) = g z .
(2.19)
From the commutativity of F and g, we have
g ( g x n + 1 ) = g ( F ( x n , y n , z n ) ) = F ( g x n , g y n , g z n ) , g ( g y n + 1 ) = g ( F ( y n , x n , y n ) ) = F ( g y n , g x n , g y n ) , g ( g z n + 1 ) = g ( F ( z n , y n , x n ) ) = F ( g z n , g y n , g x n ) .
(2.20)

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

Suppose that F is continuous. Letting n → +∞ in (2.20), therefore by (2.18) and (2.19), we obtain
g x = lim n + g ( g x n + 1 ) = lim n + F ( g x n , g y n , g z n ) = F lim n + g x n , lim n + g y n , lim n + g z n = F ( x , y , z ) ,
g y = lim n + g ( g y n + 1 ) = lim n + F ( g y n , g x n , g y n ) = F lim n + g y n , lim n + g x n , lim n + g y n = F ( y , x , y ) ,
and
g z = lim n + g ( g z n + 1 ) = lim n + F ( g z n , g y n , g x n ) = F lim n + g z n , lim n + g y n , lim n + g x n = F ( z , y , x ) .

We have proved that F and g have a tripled coincidence point.

Corollary 2.3. Let (X, ≤) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose F: X3X and g: XX are such that F has the mixed g-monotone property and F(X3) g(X). Assume there exists α [0,1) such that
d ( F ( x , y , z ) , F ( u , v , w ) ) + d ( F ( y , x , y ) , F ( v , u , v ) ) + d ( F ( z , y , x ) , F ( w , v , u ) ) α ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g w ) ) ,
for any x, y, z, u, υ, w X for which gxgu, gυgy, and gzgw. Assume that F is continuous, g is continuous and commutes with F. If there exist x0, y0, z0 X such that
g x 0 F ( x 0 , y 0 , z 0 ) , g y 0 F ( y 0 , x 0 , y 0 ) , and g z 0 F ( z 0 , y 0 , x 0 ) ,
then there exist x, y, z X such that
F ( x , y , z ) = g x , F ( y , x , y ) = g y , a n d F ( z , y , x ) = g z ,

that is, F and g have a tripled coincidence point.

Proof. It follows by taking φ(t) = αt in Theorem 2.2.

In the following theorem, we omit the continuity hypothesis of F. We need the following definition.

Definition 2.1. Let (X, ≤) be a partially ordered set and d be a metric on X. We say that (X, d,≤) is regular if the following conditions hold:
  1. (i)

    if a non-decreasing sequence (x n ) is such that x n x, then x n x for all n,

     
  2. (ii)

    if a non-increasing sequence (y n ) is such that y n y, then yy n for all n.

     
Theorem 2.4. Let (X, ≤) be a partially ordered set and d be a metric on X such that (X, d, ≤) is regular. Suppose that there exist φ Φ and mappings F: X3X and g: XX such that (2.2) holds for any x, y, z, u, υ, w X for which gxgu, gυgy and gzgw. Suppose also that (g(X), d) is complete, F has the mixed g-monotone property and F(X3) g(X). If there exist x0, y0, z0 X such that gx0F(x0, y0, z0), gy0F(y0, x0, y0), and gz0F(z0, y0, x0), then there exist x, y, z X such that
F ( x , y , z ) = g x , F ( y , x , y ) = g y , a n d F ( z , y , x ) = g z ,

that is, F and g have a tripled coincidence point.

Proof. Proceeding exactly as in Theorem 2.2, we have that (gx n ), (gy n ), and (gz n ) are Cauchy sequences in the complete metric space (g(X), d). Then, there exist x, y, z X such that gx n gx, gy n gy, and gz n gz. Since (gx n ) and (gz n ) are non-decreasing and (gy n ) is non-increasing, using the regularity of (X, d, ≤), we have gx n gx, gz n gz, and gygy n for all n ≥ 0. If gx n = gx, gy n = gy, and gz n = gz for some n ≥ 0, then gx = gx n gx n +1gx = gx n , gz = gz n gz n +1gz = gz n , and gygy n +1gy n = gy, which implies that gx n = gx n +1 = F(x n , y n , z n ), gy n = gy n +1 = F(y n , x n , y n ), and gz n = gz n +1 = F(z n , y n , x n ), that is, (x n , y n , z n ) is a tripled coincidence point of F and g. Then, we suppose that (gx n , gy n , gz n ) ≠ (gx, gy, gz) for all n ≥ 0. Using the triangle inequality, (2.2) and the property φ(t) < t for all t > 0,
d ( g x , F ( x , y , z ) ) d ( g x , g x n + 1 ) + d ( g x n + 1 , F ( x , y , z ) ) = d ( g x , g x n + 1 ) + d ( F ( x , y , z ) , F ( x n , y n , z n ) ) d ( g x , g x n + 1 ) + 3 φ 1 3 [ d ( g x n , g x ) + d ( g y n , g y ) + d ( d z n , g z ) ] < d ( g x , g x n + 1 ) + d ( g x n , g x ) + d ( g y n , g y ) + d ( g z n , g z ) .
(2.21)

Taking n → ∞ in the above inequality we obtain that d(gx,F(x, y, z)) = 0, so gx = F(x, y, z).

Analogously, we find that
F ( y , x , y ) = g y , F ( z , y , x ) = g z ,

thus, we have proved that F and g have a tripled coincidence point.

Corollary 2.5. Let (X, ≤) be a partially ordered set and suppose there is a metric d on X such that (X, ≤,d) is regular. Suppose F: X3X and g: XX are such that F has the mixed g-monotone property and F(X3) g(X). Assume there exists α [0,1) such that
d ( F ( x , y , z ) , F ( u , v , w ) ) + d ( F ( y , x , y ) , F ( v , u , v ) ) + d ( F ( z , y , x ) , F ( w , v , u ) ) α ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g w ) ) ,
for any x, y, z, u, υ, w X for which gxgu, gυgy, and gzgw. Suppose also that (g(X), d) is complete. If there exist x0, y0, z0 X such that
g x 0 F ( x 0 , y 0 , z 0 ) , g y 0 F ( y 0 , x 0 , y 0 ) and g z 0 F ( z 0 , y 0 , x 0 ) ,
then there exist x, y, z X such that
F ( x , y , z ) = g x , F ( y , x , y ) = g y , a n d F ( z , y , x ) = g z ,

that is, F and g have a tripled coincidence point.

Proof. It follows by taking φ(t) = αt in Theorem 2.4.

Now, we shall prove the existence and the uniqueness of a tripled common fixed point theorem. For a product X3 = X × X × X of a partial ordered set (X, ≤), we define a partial ordering in the following way: For all (x, y, z), (u, υ, r) X3
( x , y , z ) ( u , v , r ) x u , y v and z r .
(2.22)
We say that (x, y, z) and (u, υ, w) are comparable if
( x , y , z ) ( u , v , r ) or ( u , v , r ) ( x , y , z ) .

Also, we say that (x, y, z) is equal to (u, υ, r) if and only if x = u, y = υ and z = r.

Theorem 2.6. In addition to hypothesis of Theorem 2.2, suppose that for all (x, y, z) and (u, υ, r) in X3, there exists (a, b, c) in X3 such that (F(a, b, c), F(b, a, b), F(c, b, a)) is comparable to (F(x, y, z), F(y, x, y), F(z, y, x)) and (F(u, υ, r), F(υ, u, υ), F(r, υ, u)). Also, assume that φ is non-decreasing. Then, F and g have a unique tripled common fixed point (x, y, z), that is
x = g x = F ( x , y , z ) , y = g y = F ( y , x , y ) , a n d z = g z = F ( z , y , x ) .
Proof. Due to Theorem 2.2, the set of tripled coincidence points of F and g is not empty. Assume now, that (x, y, z) and (u,υ,r) are two tripled coincidence points of F and g, that is,
F ( x , y , z ) = g x , F ( y , x , y ) = g y , and F ( z , y , x ) = g z , F ( u , v , r ) = g u , F ( v , u , v ) = g v , and F ( r , v , u ) = g r .

We shall show that (gx, gy, gz) and (gu, gυ, gr) are equal.

By assumption, there is (a, b, c) in X3 such that (F(a, b, c), F(b, a, b), F(c, b, a)) is comparable to (F(x, y, z), F(y, x, y), F(z, y, x)) and (F(u, υ, r), F(υ, u, v), F(r, υ, u)).

Define the sequences {ga n },{gb n }, and {gc n } such that a = a0, b = b0, c = c0 and
g a n = F ( a n - 1 , b n - 1 , c n - 1 ) , g b n = F ( b n - 1 , a n - 1 , b n - 1 ) , g c n = F ( c n - 1 , b n - 1 , a n - 1 ) ,
for all n. Further, set x0 = x, y0 = y, z0 = z and u0 = u, υ0 = υ, r0 = r, and similar define the sequences {gx n },{gy n }, {gz n } and {gu n },{ n }, {gr n }. Then,
g x n = F ( x , y , z ) , g u n = F ( u , v , r ) , g y n = F ( y , x , y , ) , g v n = F ( v , u , v ) , g z n = F ( z , y , x ) , g r n = F ( r , v , u , ) ,
(2.23)
for all n ≥ 1. Since (F(x, y, z), F(y, x, y), F(z, y, x)) = (gx1, gy1, gz1) = (gx, gy, gz) is comparable to (F(a, b, c), F(b, a, b), F(c, b, a)) = (ga1, gb1, gc1), then it is easy to show that (gx, gy, gz) ≥ (ga1, gb1, gc1). Recursively, we get that
( g x , g y , g z ) ( g a n , g b n , g c n ) for all n 0 .
(2.24)
By (2.24) and (2.2), we have
d ( g x , g a n + 1 ) + d ( g b n + 1 , g y ) + d ( g z , g c n + 1 ) = d ( F ( x , y , z ) , F ( a n , b n , c n ) ) + d ( F ( b n , a n b n ) , F ( y , x , y ) ) + d ( F ( z , y , x ) , F ( c n , b n , a n ) 3 φ ( d ( g x , g a n ) + d ( g y , g b n ) + d ( g z , g c n ) 3 ) .
(2.25)
Set
γ n = d ( g x , g a n ) + d ( g y , g b n ) + d ( g z , g c n ) 3
From (2.25), we deduce that γ n +1φ(γ n ). Since φ is non-decreasing, it follows
γ n φ n ( γ 0 ) .
From the definition of Φ, we get lim n + φ n ( t ) = 0 . Then, we have lim n + γ n = 0 . Thus,
lim n d ( g x , g a n ) = 0 , lim n d ( g y , g b n ) = 0 , lim n d ( g z , g c n ) = 0 .
(2.26)
By analogy, we show that
lim n d ( g u , g a n ) = 0 , lim n d ( g v , g b n ) = 0 , lim n d ( g r , g c n ) = 0 .
(2.27)

Combining (2.26) and (2.27) yields that (gx, gy, gz) and (gu, gυ, gr) are equal.

Since gx = F(x, y, z), gy = F(y, x, y), and gz = F(z, y, x), by the commutativity of F and g, we have
g ( g x ) = g ( F ( x , y , z ) ) = F ( g x , g y , g z ) , g ( g y ) = g ( F ( y , x , y ) ) = F ( g y , g x , g y ) , g ( g z ) = g ( F ( z , y , x ) ) = F ( g z , g y , g x ) .
Denote gx = x', gy = y', and gz = z'. From the precedent identities,
g x = F ( x , y , z ) , g y = F ( y , x , y ) , and g z = F ( z , y , x ) ,

that is, (x', y', z') is a tripled coincidence point of F and g. Consequently, (gx', gy', gz') and (gx, gy, gz) are equal, that is, gx = gx', gy = gy', and gz = gz'.

We deduce gx' = gx = x', gy' = gy = y', and gz' = gz = z'. Therefore, (x', y', z') is a tripled common fixed of F and g. Its uniqueness follows from Theorem 2.2.

3 Examples

Remark that Theorem 2.2 is more general than Theorem 2.1, since the contractive condition (2.2) is weaker than (2.1), a fact which is clearly illustrated by the following example.

Example 3.1. Let X = with d(x, y) = |x - y| and natural ordering and let g: XX, F: X3X be given by
g ( x ) = n + 1 n x , n = 1 , 2 , , x X ; F ( x , y , z ) = x , ( x , y , z ) X 3 .

It is clear that F is continuous and has the mixed g-monotone property. We now take φ ( t ) = n n + 1 t . We shall show that (2.2) holds for all gxgu, gy, and gzgw.

Let x, y, z, u, υ, and w such that gxgu, gy, and gzgw, and by definition of g, it means that xu, yυ and zw, so we have
d ( F ( x , y , z ) , F ( u , v , w ) ) + d ( F ( y , x , y ) , F ( v , u , v ) ) + d ( F ( z , y , x ) , F ( w , v , u ) ) = x - u + y - v + z - w = 3 φ d ( g x , g u ) + d ( g y , g v ) + d ( g z , g w ) 3 .

which is the contractive condition (2.2). On the other hand, x0 = 0, y0 = 0, z0 = 0 satisfy (2.3). All the hypotheses of Theorem 2.2 are verified, and (0,0,0) is a tripled coincidence point of F and g.

On the other hand, assume that (2.1) holds. Then, there exist p,q,r ≥ 0 such that p + 2q + r < 1 and φ satisfying φ(t) < t for each t > 0. If x > u, z = w and y = υ, we have
0 < x - u = d ( F ( x , y , z ) , F ( u , v , w ) ) φ ( p d ( g x , g u ) + q d ( g y , g v ) + r d ( g z , g w ) ) = φ n + 1 n p x - u < n + 1 n p x - u ,

which implies p > n n + 1 for any n ≥ 1, and letting n → +∞, we get p ≥ 1, that is a contradiction. Thus, Theorem 2.1 is not applicable in this case.

Following example shows that Theorem 2.2 is more general than Theorem 1.2.

Example 3.2. Let X = be endowed with the usual ordering and the usual metric. Consider g: XX and F: X3X be given by the formulas
g ( x ) = x , F ( x , y , z ) = 3 x - 6 y + 3 z 16 , f o r a l l x , y , z X

Take φ: [0, ∞) → [0, ∞) be given by φ ( t ) = 3 t 4 for all t [0, ∞).

It is clear that all conditions of Theorem 2.2 are satisfied. Moreover, (0,0,0) is a tripled coincidence point (also a common fixed point) of F and g.

Now, for x = u, z = w and υ > y, we have
d ( F ( x , y , z ) , F ( u , v , w ) = 3 8 ( v y ) > 1 3 ( v y ) k 3 [ d ( x , u ) + d ( y , v ) d ( z , w ) ] ,

for any k [0,1), that is the result of Berinde and Borcut [15] given by Theorem 1.2 is not applicable ( f o r a = b = c = k 3 ) .

Declarations

Authors’ Affiliations

(1)
Institut Supérieur d'Informatique et des Technologies de Communication de Hammam Sousse, Université de Sousse
(2)
Department of Mathematics, Atılım University
(3)
Faculty of Applies Sciences, University Politehnica of Bucharest

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© Aydi et al; licensee Springer. 2012

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