N-Order Fixed Point Theory for N-Order Generalized Meir-Keeler Type Contraction in Partially Ordered Metric Spaces

Abstract

This paper concerns N-order fixed point theory in partially ordered metric spaces. For the sake of simplicity, we start our investigations with the tripled case. We define tripled generalized Meir-Keeler type contraction which extends the definition of [Bessem Samet, Coupled fixed point theorems for a generalized Meir-Keeler contraction in partially ordered metric spaces, Nonlinear Anal. 72 (2010), 4508-4517]. We then discuss the existence and uniqueness of tripled fixed point theorems in partially ordered metric spaces. For general cases, we generalized our results to the N-order case. The results will promote the study of N-order fixed point theory.

Share and Cite:

Wang, S. and Zhang, J. (2019) N-Order Fixed Point Theory for N-Order Generalized Meir-Keeler Type Contraction in Partially Ordered Metric Spaces. Journal of Applied Mathematics and Physics, 7, 1174-1184. doi: 10.4236/jamp.2019.75078.

1. Introduction and Preliminaries

Banach contraction principle [1] is classical and powerful in fixed point theory. It has been widely generalized (see [2] [3] [4] and others). Recently, fixed point theory in partially ordered metric spaces has been presented by many scholars: Ran and Reurings [5] , Agarwal et al. [6] , Bhsakar and Lakshmikantham [7] , Samet [8] , Berinde and Borcut [9] , Amini-Harandi [10] , etc., considered some coupled and tripled fixed point theorems. For more fixed point theorems in partially ordered metric spaces, one can refer to [11] [12] [13] and others.

This paper focuses on the tripled and N-order fixed point theory. For convenience, we denote N + = { 1,2, , n , } . Let ( X , , d ) denote a partially ordered set ( X , ) endowed a metric d (i.e., ( X , d ) is a metric space). Our work is carried out on the following two preliminaries: a result about fixed point in partially ordered metric space in [6] and a definition of generally Meir-Keeler type function for the case of coupled fixed points in [8] .

Lemma 1.1 ( [6] ). Let ( X , , d ) be a partially ordered metric space and suppose the metric space ( X , d ) is complete. Assume there is a nondecreasing function φ : [ 0, ) [ 0, ) with φ n ( t ) 0 as n for each t > 0 . If f : X X is a nondecreasing mapping with

d ( f ( x ) , f ( y ) ) φ ( d ( x , y ) ) , x y .

Assume that either

1) f is continuous or,

2) If a nondecreasing sequence x n x , then x n x , n N + .

If x 0 X with x 0 f ( x 0 ) then f has a fixed point. If for each x , y X , there exists z X which is comparable to x and y, then the fixed point of f is unique.

Definition 1 ( [8] ) Let ( X , , d ) be a partially ordered metric space and F : X × X X be a mapping. F is called generalized Meir-Keeler type function if for all ε > 0 there exists δ ( ε ) > 0 such that

x u , y v , ε 1 2 ( d ( x , u ) + d ( y , v ) ) < ε + δ ( ε ) d ( F ( x , y ) , F ( u , v ) ) < ε . (1.1)

Let ( X , ) be a partially ordered set with a metric d on X, M = X × X × X and F : M X be a given mapping. Let be the partial order on M : ( x , y , z ) ( u , v , w ) x u , y v , z w . We employ the notion of tripled fixed point introduced by Samet and Vetro which is investigated by Amini-Harandi [10] .

Definition 2 ( [11] ) An element x , y , z X is called a tripled fixed point of F : M X if

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

In this paper, we first define N-order generalized Meir-Keeler type contraction by adding some parameters (see Definition 3 and Definition 5), which is an extension of Definition 1. Then we use a simple approach introduced by [10] to discuss N-order fixed point theorems. We start our discussions with the tripled case. Section 2 devotes to tripled fixed point theorems. Section 3 devotes to N-order fixed point theory. Section 4 gives two examples to illustrate the results obtained in Section 2.

2. Tripled Fixed Point Theory

Recalling that ( X , , d ) is a partially ordered set with a metric d on X and M = X × X × X . Let ρ be the metric and be the partially order on M . For each ( x , y , z ) , ( u , v , w ) M , we define

ρ ( ( x , y , z ) , ( u , v , w ) ) = d ( x , u ) + d ( y , v ) + d ( z , w )

( x , y , z ) ( u , v , w ) x u , y v , z w

and

( x , y , z ) ( u , v , w ) at least one of the inequalities x < u , y < v and z < w hold .

Now, we define tripled generalized Meir-Keeler type contraction which is a useful tool for the following theorems in this section.

Definition 3 Let ( X , , d ) be a partially ordered metric space and F : M X be a mapping. F is called a tripled generalized Meir-Keeler type contraction if for all ε > 0 there exists δ ( ε ) > 0 such that

( x , y , z ) ( u , v , w ) , ε 1 3 [ l d ( x , u ) + k d ( y , v ) + j d ( z , w ) ] < ε + δ ( ε ) d ( F ( x , y , z ) , F ( u , v , w ) ) < ε (2.1)

where l , k , j are constants with 0 < l + k + j < 3 .

Theorem 2.1 Let ( X , , d ) be a partially ordered metric space. Let l , k , j be the given constants with 0 < l + k + j < 3 . If F : M X is a tripled generalized Meir-Keeler contraction mapping, then

d ( F ( x , y , z ) , F ( u , v , w ) ) < 1 3 [ l d ( x , u ) + k d ( y , v ) + j d ( z , w ) ]

for all ( x , y , z ) ( u , v , w ) .

Proof. Let ( x , y , z ) , ( u , v , w ) M such that ( x , y , z ) ( u , v , w ) . Then it follows that

1 3 [ l d ( x , u ) + k d ( y , v ) + j d ( z , w ) ] > 0.

Setting

ε = 1 3 [ l d ( x , u ) + k d ( y , v ) + j d ( z , w ) ] ,

we have

0 < ε = 1 3 [ l d ( x , u ) + k d ( y , v ) + j d ( z , w ) ] < ε + δ ( ε ) .

By F : M X being a tripled generalized Meir-Keeler type contraction, then

d ( F ( x , y , z ) , F ( u , v , w ) ) < ε = 1 3 [ l d ( x , u ) + k d ( y , v ) + j d ( z , w ) ] . □

Let F : M X be a mapping. We say F is nondecreasing in each of its variables if

x 1 , x 2 X , x 1 < x 2 F ( x 1 , y , z ) < F ( x 2 , y , z ) , y , z X ,

y 1 , y 2 X , y 1 < y 2 F ( x , y 1 , z ) < F ( x , y 2 , z ) , x , z X ,

and

z 1 , z 2 X , z 1 < z 2 F ( x , y , z 1 ) < F ( x , y , z 2 ) , x , y X .

By the monotone property of F, we can get

( x , y , z ) , ( u , v , w ) M , ( x , y , z ) ( u , v , w ) F ( x , y , z ) < F ( u , v , w ) . (2.2)

For all n N + , n > 1 , we define:

F n ( x , y , z ) = F ( F n 1 ( x , y , z ) , F n 1 ( y , z , x ) , F n 1 ( z , x , y ) ) (2.3)

with F 1 = F .

In order to investigate the tripled fixed point of F, we introduce a mapping T : M M which is defined by

T ( x , y , z ) = ( F ( x , y , z ) , F ( y , z , x ) , F ( z , x , y ) ) . (2.4)

Obviously, by the definition of ρ , we have

ρ ( T ( x , y , z ) , T ( u , v , w ) ) = d ( F ( x , y , z ) , F ( u , v , w ) ) + d ( F ( y , z , x ) , F ( v , w , u ) ) + d ( F ( z , x , y ) , F ( w , u , v ) ) . (2.5)

Simultaneously, by (2.3) and (2.4), we have

T n ( x , y , z ) = ( F n ( x , y , z ) , F n ( y , z , x ) , F n ( z , x , y ) )

with T 1 = T , and we have

F n ( x , y , z ) = F ( T n 1 ( x , y , z ) ) .

Theorem 2.2 Let ( X , , d ) be a partially ordered metric space and l , k , j be the given constants with 0 < l + k + j < 3 . Let F : M X be nondecreasing in each of its variables and be a tripled generalized Meir-Keeler type contraction. There exist ( x , y , z ) , ( u , v , w ) M with ( x , y , z ) ( u , v , w ) . Then, for n N + , we have

1) T n ( x , y , z ) T n ( u , v , w ) ;

2) ρ ( T n + 1 ( x , y , z ) , T n + 1 ( u , v , w ) ) < ρ ( T n ( x , y , z ) , T n ( u , v , w ) ) ;

3) ρ ( T n ( x , y , z ) , T n ( u , v , w ) ) 0, n .

Proof. We first prove 1). Since ( x , y , z ) ( u , v , w ) , due to the monotone property of F and (2.2), we have F ( x , y , z ) < F ( u , v , w ) , F ( y , z , x ) < F ( v , w , u ) and F ( z , x , y ) < F ( w , u , v ) . By T 1 = T and (2.4), 1) holds for n = 1 . Now we assume 1) holds for n N + , i.e.

( F n ( x , y , z ) , F n ( y , z , x ) , F n ( z , x , y ) ) = T n ( x , y , z ) T n ( u , v , w ) = ( F n ( u , v , w ) , F n ( v , w , u ) , F n ( w , u , v ) ) .

Then, we obtain

F ( F n ( x , y , z ) , F n ( y , z , x ) , F n ( z , x , y ) ) < F ( F n ( u , v , w ) , F n ( v , w , u ) , F n ( w , u , v ) )

which means F n + 1 ( x , y , z ) < F n + 1 ( u , v , w ) . Using the same strategy, we have F n + 1 ( y , z , x ) < F n + 1 ( v , w , u ) and F n + 1 ( z , x , y ) < F n + 1 ( w , u , v ) . Hence we have T n + 1 ( x , y , z ) T n + 1 ( u , v , w ) , that is, 1) holds for n + 1 . Simultaneously, we can also obtain that T n ( y , z , x ) T n ( v , w , u ) and T n ( z , x , y ) T n ( w , u , v ) .

Now, we prove 2). We consider

ρ ( T n + 1 ( x , y , z ) , T n + 1 ( u , v , w ) ) = d ( F n + 1 ( x , y , z ) , F n + 1 ( u , v , w ) ) + d ( F n + 1 ( y , z , x ) , F n + 1 ( v , w , u ) ) + d ( F n + 1 ( z , x , y ) , F n + 1 ( w , u , v ) ) .

It follows from Theorem 2.1 and 1) that

d ( F n + 1 ( x , y , z ) , F n + 1 ( u , v , w ) ) = d ( F ( T n ( x , y , z ) ) , F ( T n ( u , v , w ) ) ) < 1 3 [ l d ( F n ( x , y , z ) , F n ( u , v , w ) ) + k d ( F n ( y , z , x ) , F n ( v , w , u ) ) + j d ( F n ( z , x , y ) , F n ( w , u , v ) ) ] ,

d ( F n + 1 ( y , z , x ) , F n + 1 ( v , w , u ) ) = d ( F ( T n ( y , z , x ) ) , F ( T n ( v , w , u ) ) ) < 1 3 [ l d ( F n ( y , z , x ) , F n ( v , w , u ) ) + k d ( F n ( z , x , y ) , F n ( w , u , v ) ) + j d ( F n ( x , y , z ) , F n ( u , v , w ) ) ]

and

d ( F n + 1 ( z , x , y ) , F n + 1 ( w , u , v ) ) = d ( F ( T n ( z , x , y ) ) , F ( T n ( w , u , v ) ) ) < 1 3 [ l d ( F n ( z , x , y ) , F n ( w , u , v ) ) + k d ( F n ( x , y , z ) , F n ( u , v , w ) ) + j d ( F n ( y , z , x ) , F n ( v , w , u ) ) ] .

Thus,

ρ ( T n + 1 ( x , y , z ) , T n + 1 ( u , v , w ) ) < 1 3 ( l + k + j ) [ d ( F n ( x , y , z ) , F n ( u , v , w ) ) + d ( F n ( y , z , x ) , F n ( v , w , u ) ) + d ( F n ( z , x , y ) , F n ( w , u , v ) ) ] < ρ ( T n ( x , y , z ) , T n ( u , v , w ) ) .

Last, we prove 3). From 2), we know that lim n ρ ( T n ( x , y , z ) , T n ( u , v , w ) ) exists. If lim n ρ ( T n ( x , y , z ) , T n ( u , v , w ) ) 0 , we suppose that

lim n 1 3 ρ ( T n ( x , y , z ) , T n ( u , v , w ) ) = ε > 0. (2.6)

Then it follows that

1 3 ρ ( T n ( x , y , z ) , T n ( u , v , w ) ) ε , n N + .

By (2.6), we have

lim n 1 3 [ l d ( F n ( x , y , z ) , F n ( u , v , w ) ) + k d ( F n ( y , z , x ) , F n ( v , w , u ) ) + j d ( F n ( z , x , y ) , F n ( w , u , v ) ) ] = ε

which implies that there exists m 0 N + such that

ε 1 3 [ l d ( F m 0 ( x , y , z ) , F m 0 ( u , v , w ) ) + k d ( F m 0 ( y , z , x ) , F m 0 ( v , w , u ) ) + j d ( F m 0 ( z , x , y ) , F m 0 ( w , u , v ) ) ] < ε + δ ( ε ) . (2.7)

Since F is a tripled generalized Meir-Keeler type contraction, we get

ε > d ( F ( F m 0 ( x , y , z ) , F m 0 ( y , z , x ) , F m 0 ( z , x , y ) ) , F ( F m 0 ( u , v , w ) , F m 0 ( v , w , u ) , F m 0 ( w , u , v ) ) ) = d ( F m 0 + 1 ( x , y , z ) , F m 0 + 1 ( u , v , w ) ) . (2.8)

By (2.7), we also have

ε 1 3 [ k d ( F m 0 ( y , z , x ) , F m 0 ( v , w , u ) ) + j d ( F m 0 ( z , x , y ) , F m 0 ( w , u , v ) ) + l d ( F m 0 ( x , y , z ) , F m 0 ( u , v , w ) ) ] < ε + δ ( ε ) ,

and

ε 1 3 [ j d ( F m 0 ( z , x , y ) , F m 0 ( w , u , v ) ) + l d ( F m 0 ( x , y , z ) , F m 0 ( u , v , w ) ) + k d ( F m 0 ( y , z , x ) , F m 0 ( v , w , u ) ) ] < ε + δ ( ε ) .

Then, we get

ε > d ( F m 0 + 1 ( y , z , x ) , F m 0 + 1 ( v , w , u ) ) (2.9)

and

ε > d ( F m 0 + 1 ( z , x , y ) , F m 0 + 1 ( w , u , v ) ) . (2.10)

From (2.8)-(2.10), we get

1 3 ρ ( T m 0 + 1 ( x , y , z ) , T m 0 + 1 ( u , v , w ) ) = 1 3 [ d ( F m 0 + 1 ( x , y , z ) , F m 0 + 1 ( u , v , w ) ) + d ( F m 0 + 1 ( y , z , x ) , F m 0 + 1 ( v , w , u ) ) + d ( F m 0 + 1 ( z , x , y ) , F m 0 + 1 ( w , u , v ) ) ] < ε .

This is a contradiction. The proof is completed.

From the definition of T, we observe that the fixed point of T is exactly the tripled fixed point of F, that is,

( x , y , z ) = T ( x , y , z ) x = F ( x , y , z ) , y = F ( y , z , x ) , z = F ( z , x , y ) .

We will obtain the tripled fixed point theorems by investigating the fixed point of T.

Theorem 2.3 Let ( X , , d ) be a partially ordered metric space and ( X , d ) is a complete metric space. Let l , k , j be the given constants with 0 < l + k + j < 3 . Let F : M X be nondecreasing in each of its variables and be a tripled generalized Meir-Keeler contraction. T : M M be a mapping defined as (2.4) satisfying that there exists ( x 0 , y 0 , z 0 ) M with ( x 0 , y 0 , z 0 ) T ( x 0 , y 0 , z 0 ) . Then, there exists ( x * , y * , z * ) M which is a tripled fixed point of F, if either

1) F is continuous or

2) a nondecreasing sequence ( x n , y n , z n ) ( x , y , z ) , then ( x n , y n , z n ) ( x , y , z ) , n N + .

Furthermore, if

3) for ( x , y , z ) , ( u , v , w ) M , there exists ( a , b , c ) M that is comparable to ( x , y , z ) and ( u , v , w ) , we get the uniqueness of tripled fixed point of F and x * = y * = z * .

Proof. Since ( X , d ) is a complete metric space, it is obvious that the metric space ( M , ρ ) is complete. By Theorem 2.2, T is non-decreasing. Meanwhile, by Theorem 2.1 and (2.5), for each ( x , y , z ) , ( u , v , w ) M with ( x , y , z ) ( u , v , w ) , we have

ρ ( T ( x , y , z ) , T ( u , v , w ) ) = d ( F ( x , y , z ) , F ( u , v , w ) ) + d ( F ( y , z , x ) , F ( v , w , u ) ) + d ( F ( z , x , y ) , F ( w , u , v ) ) < 1 3 ( l + k + j ) ρ ( ( x , y , z ) , ( u , v , w ) ) .

By Lemma 1.1, we deduce that T has a unique fixed point denoted by ( x * , y * , z * ) , then ( x * , y * , z * ) is the unique tripled fixed point of F.

However, we can check that ( y * , z * , x * ) is also a tripled fixed point of F. In fact, since ( x * , y * , z * ) is the tripled fixed point of F, i.e., x * = F ( x * , y * , z * ) , y * = F ( y * , z * , x * ) , z * = F ( z * , x * , y * ) , we have

y * = F ( y * , z * , x * ) , z * = F ( z * , x * , y * ) , x * = F ( x * , y * , z * )

which implies that ( y * , z * , x * ) is also a tripled fixed point of F. By the uniqueness, we get x * = y * = z * . □

Corollary 1 Suppose that all the hypotheses of Theorem 2.3 are satisfied, then the tripled fixed point ( x * , y * , z * ) can be deduced by

F n ( x 0 , y 0 , z 0 ) x * , F n ( y 0 , z 0 , x 0 ) y * , F n ( z 0 , x 0 , y 0 ) z * , as n . (2.11)

Proof. By examining the proof of Theorem 2.3, ( x * , y * , z * ) is actually the fixed point of T on M . According to the proof of Lemma 1.1 in [6] , we have

lim n T n ( x 0 , y 0 , z 0 ) = ( x * , y * , z * ) .

By the definition of T n , we can easily get (2.11). □

Theorem 2.4 In addition to the hypotheses of Theorem 2.3 except (3), we have x * = y * = z * by adding the hypotheses (3*): x 0 , y 0 , z 0 in X are comparable.

Proof. Without the restriction of the generality, we assume that x 0 y 0 z 0 . Setting ( x 1 , y 1 , z 1 ) = ( x 0 , y 0 , z 0 ) and ( u 1 , v 1 , w 1 ) = ( y 0 , z 0 , z 0 ) , it’s easy to see that ( x 1 , y 1 , z 1 ) ( u 1 , v 1 , w 1 ) . From Theorem 1.1, we have ρ ( T n ( x 1 , y 1 , z 1 ) , T n ( u 1 , v 1 , w 1 ) ) 0 as n , which implies that

d ( F n ( x 1 , y 1 , z 1 ) , F n ( u 1 , v 1 , w 1 ) ) 0, n ,

i.e.,

d ( F n ( x 0 , y 0 , z 0 ) , F n ( y 0 , z 0 , z 0 ) ) 0, n . (2.12)

By the similar strategy, setting ( x 2 , y 2 , z 2 ) = ( y 0 , z 0 , x 0 ) and ( u 2 , v 2 , w 2 ) = ( y 0 , z 0 , z 0 ) , we can get

d ( F n ( y 0 , z 0 , x 0 ) , F n ( y 0 , z 0 , z 0 ) ) 0, n . (2.13)

It follows from the triangular inequality that

d ( x * , y * ) d ( x * , F n ( x 0 , y 0 , z 0 ) ) + d ( F n ( x 0 , y 0 , z 0 ) , F n ( y 0 , z 0 , z 0 ) ) + d ( F n ( y 0 , z 0 , z 0 ) , F n ( y 0 , z 0 , x 0 ) ) + d ( F n ( y 0 , z 0 , x 0 ) , y * ) .

Taking the limit as n , by (2.11), (2.12) and (2.13), we get x * = y * .

Similarly, by setting

( x 3 , y 3 , z 3 ) = ( y 0 , z 0 , x 0 ) , ( u 3 , v 3 , w 3 ) = ( z 0 , z 0 , y 0 )

and

( x 4 , y 4 , z 4 ) = ( z 0 , x 0 , y 0 ) , ( u 4 , v 4 , w 4 ) = ( z 0 , z 0 , y 0 ) ,

we can get two equalities,

d ( F n ( y 0 , z 0 , x 0 ) , F n ( z 0 , z 0 , y 0 ) ) 0, n (2.14)

and

d ( F n ( z 0 , x 0 , y 0 ) , F n ( z 0 , z 0 , y 0 ) ) 0, n (2.15)

respectively. Then it follows from (2.11), (2.14) and (2.15) that

d ( y * , z * ) d ( y * , F n ( y 0 , z 0 , x 0 ) ) + d ( F n ( y 0 , z 0 , x 0 ) , F n ( z 0 , z 0 , y 0 ) ) + d ( F n ( z 0 , z 0 , y 0 ) , F n ( z 0 , x 0 , y 0 ) ) + d ( F n ( z 0 , x 0 , y 0 ) , z * ) 0.

We get y * = z * . Hence we have x * = y * = z * . □

3. N-Order Fixed Point Theorems

Let ( X , , d ) be a partially ordered set with a metric d on X. Let K = X N , η be the metric on K and be the partially order. For each x = ( x 1 , , x N ) , y = ( y 1 , , y N ) K , we define

η ( x , y ) = d ( x 1 , y 1 ) + + d ( x N , y N )

x y x 1 y 1 , , x N y N

and

x y there exists 1 i N , such that x i < y i .

Definition 4 [11] Let X be a non-empty set and F : K X be a given mapping. An element x K is called a N-order fixed point of F if

x 1 = F ( x 1 , , x N ) , x 2 = F ( x 2 , , x N , x 1 ) , , x N = F ( x N , x 1 , , x N 1 ) .

We introduce generally N-order generalized Meir-Keeler type contraction.

Definition 5 Let ( X , , d ) be a partially ordered metric space and F : K X be a mapping. F is called a N-order generalized Meir-Keeler contraction if for all ε > 0 there exists δ ( ε ) > 0 such that for x , y K

x y , ε k 1 d ( x 1 , y 1 ) + + k N d ( x N , y N ) N < ε + δ ( ε ) d ( F ( x ) , F ( y ) ) < ε (3.16)

where k 1 , , k N are constants with 0 < k 1 + + k N < N .

Substituting the tripled case with N-order case in the discussions of Section 3, by the similar strategy, we can obtain the same results with Theorem 2.1, Theorem 2.2, Theorem 2.3, Corollary 1 and Theorem 2.4.

4. The Examples

This section provides two examples to illustrate Theorem 2.3 and Theorem 2.4.

Example 1 This example is aroused by [13] . Let X = R , d ( x , y ) = | x y | and F : M X , defined by

F = 4 x 4 y + 3 z + 1 15 .

It is easy to check that F satisfies all the hypotheses of Theorem 2.3 with

l = 1 , k = 1 , j = 3 4 , δ ( ε ) = 1 4 ε

and ( x * , y * , z * ) = ( 1 12 , 1 12 , 1 12 ) is the unique tripled fixed point of F.

Example 2 Let

X = { ( 0 , 1 ) , ( 0 , 2 ) , ( 1 , 3 ) , ( 1 , 0 ) , ( 2 , 0 ) , ( 3 , 0 ) } .

For x = ( x 1 , x 2 ) , y = ( y 1 , y 2 ) X , d ( x , y ) = | x 1 y 1 | + | x 2 y 2 | and x y x 1 y 1 , x 2 y 2 . F : M X is defined by

F ( x , y , z ) = { ( 0 , 3 ) , x , y , z { ( 0 , 1 ) , ( 0 , 2 ) , ( 0 , 3 ) } ( 3 , 3 ) , x , y , z { ( 1 , 1 ) , ( 2 , 0 ) , ( 3 , 0 ) } (4.1)

It is easy to check that:

1) F is continues on M ;

2) F is a tripled generally Meir-Keeler type contraction. In fact, we can deduce that

d ( F ( x , y , z ) , F ( u , v , w ) ) = 0 for each ( x , y , z ) ( u , v , w ) ;

3) Setting x 0 = y 0 = z 0 = ( 0 , 1 ) , then we have F ( x 0 , y 0 , z 0 ) = F ( y 0 , z 0 , x 0 ) = F ( z 0 , x 0 , y 0 ) = ( 0 , 3 ) . Clearly, we have ( x 0 , y 0 , z 0 ) T ( x 0 , y 0 , z 0 ) ;

4) Setting ( x , y , z ) = ( 0 , 1 , 0 , 2 , 0 , 3 ) , ( u , v , w ) = ( 1 , 0 , 2 , 0 , 3 , 0 ) , there are no elements in M which are comparable to ( x , y , z ) and ( u , v , w ) .

The above 4) implies that F doesn’t satisfy all the hypotheses of Theorem 2.3. However, the above 1)-3) imply that F satisfies all the hypotheses of Theorem 2.4, then F has the unique tripled fixed point ( x * , y * , z * ) with x * = y * = z * = ( 0 , 3 ) .

5. Conclusion

In this paper, we extend the definition generalized Meir-Keeler type contraction to N-ordered case. And we use it to discuss N-order fixed point theorems. In future work, we will study N-ordered fixed point theory with invariant set.

Acknowledgements

This work is supported by the National Natural Science Foundation of China (Grant No. 11701390).

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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