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Some multidimensional fixed point theorems on partially preordered G -metric spaces under ( ψ , φ ) -contractivity conditions

Fixed Point Theory and Applications20132013:158

https://doi.org/10.1186/1687-1812-2013-158

Received: 4 April 2013

Accepted: 24 May 2013

Published: 18 June 2013

Abstract

In this paper we present some (unidimensional and) multidimensional fixed point results under ( ψ , φ ) -contractivity conditions in the framework of G -metric spaces, which are spaces that result from G-metric spaces (in the sense of Mustafa and Sims) omitting one of their axioms. We prove that these spaces let us consider easily the product of G -metrics. Our result clarifies and improves some recent results on this topic because, among other different reasons, we will not need a partial order on the underlying space. Furthermore, the way in which several contractivity conditions are proposed imply that our theorems cannot be reduced to metric spaces.

MSC: 46T99, 47H10, 47H09, 54H25.

1 Introduction

In the sixties, inspired by the mapping that associated the area of a triangle to its three vertices, Gähler [1, 2] introduced the concept of 2-metric spaces. Gähler believed that 2-metric spaces can be interpreted as a generalization of usual metric spaces. However, some authors demonstrated that there is no clear relationship between these notions. For instance, Ha et al. [3] showed that a 2-metric does not have to be a continuous function of its three variables. Later, inspired by the perimeter of a triangle rather than the area, Dhage [4] changed the axioms and presented the concept of D-metric. Different topological structures (see [57]) were considered in such spaces and, subsequently, several fixed point results were established. Unfortunately, most of their properties turned out to be false (see [810]). These considerations led to the concept of G-metric space introduced by Mustafa and Sims [11]. Since then, this theory has been expansively developed, paying a special attention to fixed point theorems (see, for instance, [1228] and references therein).

The main aim of the present paper is to prove new unidimensional and multidimensional fixed point results in the framework of the G-metric spaces provided with a partial preorder (not necessarily a partial order). However, we need to overcome the well-known fact that the usual product of G-metrics is not necessarily a G-metric unless it comes from classical metrics (see [11], Section 4). Hence, we will omit one of the axioms that define a G-metric and we consider a new class of metrics, called G -metrics. As a consequence, our main results are valid in the context of G-metric spaces.

2 Preliminaries

Let n be a positive integer. Henceforth, X will denote a non-empty set and X n will denote the product space X × X × n × X . Throughout this manuscript, m and k will denote non-negative integers and i , j , s { 1 , 2 , , n } . Unless otherwise stated, ‘for all m’ will mean ‘for all m 0 ’ and ‘for all i’ will mean ‘for all i { 1 , 2 , , n } ’. Let R 0 + = [ 0 , ) .

Definition 1 We will say that is a partial preorder on X (or ( X , ) is a preordered set or ( X , ) is a partially preordered space) if the following properties hold.

  • Reflexivity: x x for all x X .

  • Transitivity: If x , y , z X verify x y and y z , then x z .

Henceforth, let { A , B } be a partition of Λ n = { 1 , 2 , , n } , that is, A B = Λ n and A B = such that A and B are non-empty sets. In the sequel, we will denote
Ω A , B = { σ : Λ n Λ n : σ ( A ) A  and  σ ( B ) B } and Ω A , B = { σ : Λ n Λ n : σ ( A ) B  and  σ ( B ) A } .

From now on, let ϒ = ( σ 1 , σ 2 , , σ n ) be an n-tuple of mappings from { 1 , 2 , , n } into itself verifying σ i Ω A , B if i A and σ i Ω A , B if i B .

If ( X , ) is a partially preordered space, x , y X and i Λ n , we will use the following notation:
x i y { x y , if  i A , x y , if  i B .
Consider on the product space X n the following partial preorder: for X = ( x 1 , x 2 , , x n ) , Y = ( y 1 , y 2 , , y n ) X n ,
X Y x i i y i for all  i .
(1)

Notice that depends on A and B. We say that two points X and Y are -comparable if X Y or X Y .

Proposition 2 If X Y and σ Ω A , B Ω A , B , then ( x σ ( 1 ) , x σ ( 2 ) , , x σ ( n ) ) and ( y σ ( 1 ) , y σ ( 2 ) , , y σ ( n ) ) are -comparable. In particular,
( x σ ( 1 ) , x σ ( 2 ) , , x σ ( n ) ) ( y σ ( 1 ) , y σ ( 2 ) , , y σ ( n ) ) if σ Ω A , B , ( x σ ( 1 ) , x σ ( 2 ) , , x σ ( n ) ) ( y σ ( 1 ) , y σ ( 2 ) , , y σ ( n ) ) if σ Ω A , B .

Proof Suppose that x i i y i for all i. Hence x σ ( i ) σ ( i ) y σ ( i ) for all i. Fix σ Ω A , B . If i A , then σ ( i ) A , so x σ ( i ) σ ( i ) y σ ( i ) implies that x σ ( i ) y σ ( i ) , which means that x σ ( i ) i y σ ( i ) . If i B , then σ ( i ) B , so x σ ( i ) σ ( i ) y σ ( i ) implies that x σ ( i ) y σ ( i ) , which means that x σ ( i ) i y σ ( i ) . In any case, if σ Ω A , B , then x σ ( i ) i y σ ( i ) for all i. It follows that ( x σ ( 1 ) , x σ ( 2 ) , , x σ ( n ) ) ( y σ ( 1 ) , y σ ( 2 ) , , y σ ( n ) ) .

Now fix σ Ω A , B . If i A , then σ ( i ) B , so x σ ( i ) σ ( i ) y σ ( i ) implies that x σ ( i ) y σ ( i ) , which means that x σ ( i ) i y σ ( i ) . If i B , then σ ( i ) A , so x σ ( i ) σ ( i ) y σ ( i ) implies that x σ ( i ) y σ ( i ) , which means that x σ ( i ) i y σ ( i ) . □

Let F : X n X be a mapping.

Definition 3 (Roldán et al. [20])

A point ( x 1 , x 2 , , x n ) X n is called an ϒ-fixed point of the mapping F if
F ( x σ i ( 1 ) , x σ i ( 2 ) , , x σ i ( n ) ) = x i for all  i .
(2)

Definition 4 (Roldán et al. [20])

Let ( X , ) be a partially preordered space. We say that F has the mixed monotone property (w.r.t. { A , B } ) if F is monotone non-decreasing in the arguments of A and monotone non-increasing in the arguments of B, i.e., for all x 1 , x 2 , , x n , y , z X and all i,
y z F ( x 1 , , x i 1 , y , x i + 1 , , x n ) i F ( x 1 , , x i 1 , z , x i + 1 , , x n ) .

We will use the following results about real sequences in the proof of our main theorems.

Lemma 5 Let { a m 1 } m N , , { a m n } m N be n real lower bounded sequences such that { max ( a m 1 , , a m n ) } m N δ . Then there exist i 0 { 1 , 2 , , n } and a subsequence { a m ( k ) i 0 } k N such that { a m ( k ) i 0 } k N δ .

Proof Let b m = max ( a m 1 , a m 2 , , a m n ) for all m. As { b m } is convergent, it is bounded. As a m i b m for all m and i, then every { a m i } is bounded. As { a m 1 } m N is a real bounded sequence, it has a convergent subsequence { a σ 1 ( m ) 1 } m N a 1 . Consider the subsequences { a σ 1 ( m ) 2 } m N , { a σ 1 ( m ) 3 } m N , , { a σ 1 ( m ) n } m N , that are n 1 real bounded sequences, and the sequence { b σ 1 ( m ) } m N that also converges to δ. As { a σ 1 ( m ) 2 } m N is a real bounded sequence, it has a convergent subsequence { a σ 2 σ 1 ( m ) 2 } m N a 2 . Then the sequences { a σ 2 σ 1 ( m ) 3 } m N , { a σ 2 σ 1 ( m ) 4 } m N , , { a σ 2 σ 1 ( m ) n } m N also are n 2 real bounded sequences and { a σ 2 σ 1 ( m ) 1 } m N a 1 and { b σ 2 σ 1 ( m ) } m N δ . Repeating this process n times, we can find n subsequences { a σ ( m ) 1 } m N , { a σ ( m ) 2 } m N , , { a σ ( m ) n } m N (where σ = σ n σ 1 ) such that { a σ ( m ) i } m N a i for all i. And { b σ ( m ) } m N δ . But
{ b σ ( m ) } m N = { max ( a σ ( m ) n , , a σ ( m ) n ) } m N max ( a 1 , , a n ) ,

so δ = max ( a 1 , , a n ) and there exists i 0 { 1 , 2 , , n } such that a i 0 = δ . Therefore, there exist i 0 { 1 , 2 , , n } and a subsequence { a σ ( m ) i 0 } m N such that { a σ ( m ) i 0 } m N a i 0 = δ . □

Lemma 6 Let { a m } m N be a sequence of non-negative real numbers which has not any subsequence converging to zero. Then, for all ε > 0 , there exist δ ] 0 , ε [ and m 0 N such that a m δ for all m m 0 .

Proof Suppose that the conclusion is not true. Then there exists ε 0 > 0 such that, for all δ ] 0 , ε 0 [ , there exists m 0 N verifying a m 0 < δ . Let k 0 N be such that 1 / k 0 < ε 0 . For all k N , take δ k = 1 / ( k + k 0 ) ] 0 , ε 0 [ . Then there exists m ( k ) N verifying 0 a m ( k ) < δ k = 1 / ( k + k 0 ) . Taking limit when k , we deduce that lim k a m ( k ) = 0 . Then { a m } has a subsequence converging to zero (maybe, reordering { a m ( k ) } ), but this is a contradiction. □

Let
Ψ = { ϕ : [ 0 , ) [ 0 , ) : ϕ  is continuous, non-decreasing and  ϕ 1 ( { 0 } ) = { 0 } } .

Lemma 7 If ψ Ψ and { a m } [ 0 , ) verifies { ψ ( a m ) } 0 , then { a m } 0 .

Proof If the conclusion does not hold, there exists ε 0 > 0 such that, for all m 0 N , there exists m m 0 verifying a m ε 0 . This means that { a m } has a partial subsequence { a m ( k ) } k such that a m ( k ) ε 0 . As ψ is non-decreasing, ψ ( ε 0 ) ψ ( a m ( k ) ) for all k N . Therefore, { ψ ( a m ) } m has a subsequence { ψ ( a m ( k ) ) } k lower bounded by ψ ( ε 0 ) > 0 , but this is impossible since lim m ψ ( a m ) = 0 . □

Lemma 8 Let { a m 1 } , { a m 2 } , , { a m n } , { b m 1 } , { b m 2 } , , { b m n } [ 0 , ) be 2n sequences of non-negative real numbers and suppose that there exist ψ , φ Ψ such that
ψ ( a m + 1 i ) ( ψ φ ) ( b m i ) for all i and all m , and ψ ( max 1 i n b m i ) ψ ( max 1 i n a m i ) for all m .

Then { a m i } 0 for all i.

Proof Let c m = max 1 i n a m i for all m. Then, for all m,
ψ ( c m + 1 ) = ψ ( max 1 i n a m + 1 i ) = max 1 i n ψ ( a m + 1 i ) max 1 i n [ ( ψ φ ) ( b m i ) ] max 1 i n ψ ( b m i ) = ψ ( max 1 i n b m i ) ψ ( max 1 i n a m i ) = ψ ( c m ) .
Therefore, { ψ ( c m ) } is a non-increasing, bounded below sequence. Then it is convergent. Let Δ 0 be such that { ψ ( c m ) } Δ and Δ ψ ( c m ) . Let us show that Δ = 0 . Since
{ max 1 i n ψ ( a m i ) } = { ψ ( max 1 i n a m i ) } = { ψ ( c m ) } Δ ,
Lemma 5 guarantees that there exist i 0 { 1 , 2 , , n } and a partial subsequence { a m ( k ) i 0 } k N such that { ψ ( a m ( k ) i 0 ) } Δ . Moreover,
0 ψ ( a m ( k ) i 0 ) ( ψ φ ) ( b m ( k ) 1 i 0 ) for all  k .
(3)
Consider the sequence { b m ( k ) 1 i 0 } k N . If this sequence has a partial subsequence converging to zero, then we can take limit in (3) when k 0 using that partial subsequence, and we deduce Δ = 0 . On the contrary, if { b m ( k ) 1 i 0 } k N has not any partial subsequence converging to zero, Lemma 6 assures us that there exist δ ] 0 , 1 [ and k 0 N such that b m ( k ) 1 i 0 δ for all k k 0 . Since φ is non-decreasing, φ ( b m ( k ) 1 i 0 ) φ ( δ ) < 0 . Then, by (3), for all k k 0 ,
0 ψ ( a m ( k ) i 0 ) ( ψ φ ) ( b m ( k ) 1 i 0 ) = ψ ( b m ( k ) 1 i 0 ) φ ( b m ( k ) 1 i 0 ) ψ ( b m ( k ) 1 i 0 ) φ ( δ ) ψ ( max 1 i n b m ( k ) 1 i ) φ ( δ ) ψ ( max 1 i n a m ( k ) 1 i ) φ ( δ ) = ψ ( c m ( k ) 1 ) φ ( δ ) .

Taking limit as k , we deduce Δ Δ φ ( δ ) , which is impossible. This proves that Δ = 0 . Since { ψ ( c m ) } Δ = 0 , Lemma 7 implies that { c m } 0 , which is equivalent to { a m i } 0 for all i. □

Corollary 9 If ψ , φ Ψ and { a m } , { b m } [ 0 , ) verify ψ ( a m + 1 ) ( ψ φ ) ( b m ) and ψ ( b m ) ψ ( a m ) for all m, then { a m } 0 .

Corollary 10 If ψ , φ Ψ and { a m } [ 0 , ) verifies ψ ( a m + 1 ) ψ ( a m ) φ ( a m ) for all m, then { a m } 0 .

Definition 11 (Mustafa and Sims [11])

A generalized metric (or a G-metric) on X is a mapping G : X 3 R 0 + verifying, for all x , y , z X :

  • ( G 1 ) G ( x , x , x ) = 0 .

  • ( G 2 ) G ( x , x , y ) > 0 if x y .

  • ( G 3 ) G ( x , x , y ) G ( x , y , z ) if y z .

  • ( G 4 ) G ( x , y , z ) = G ( x , z , y ) = G ( y , z , x ) = (symmetry in all three variables).

  • ( G 5 ) G ( x , y , z ) G ( x , a , a ) + G ( a , y , z ) (rectangle inequality).

Let { ( X i , G i ) } i = 1 n be a family of G-metric spaces, consider the product space X = X 1 × X 2 × × X n and define G m and G s on X 3 by
G m ( X , Y , Z ) = max 1 i n G i ( x i , y i , z i ) and G s ( X , Y , Z ) = i = 1 n G i ( x i , y i , z i )

for all X = ( x 1 , x 2 , , x n ) , Y = ( y 1 , y 2 , , y n ) , Z = ( z 1 , z 2 , , z n ) X .

A classical example of G-metric comes from a metric space ( X , d ) , where G ( x , y , z ) = d x y + d y z + d z x measures the perimeter of a triangle. In this case, property ( G 3 ) has an obvious geometric interpretation: the length of an edge of a triangle is less than or equal to its semiperimeter, that is, 2 d x y d x y + d y z + d z x . However, property ( G 3 ) implies that, in general, the major structures G m and G s are not necessarily G-metrics on X 1 × X 2 × × X n . Only when each G i is symmetric (that is, G ( x , x , y ) = G ( y , y , x ) for all x, y), the product is also a G-metric (see [11]). But in this case, symmetric G-metrics can be reduced to usual metrics, which limits the interest in this kind of spaces.

In order to prove our main results, that are also valid in G-metric spaces, we will not need property ( G 3 ) . Omitting this property, we consider a class of spaces for which G m and G s have the same initial metric structure. Then we present the following spaces.

3 G -metric spaces

Definition 12 A G -metric on X is a mapping G : X 3 R 0 + verifying ( G 1 ) , ( G 2 ) , ( G 4 ) and ( G 5 ) .

The open ball B ( x , r ) of center x X and radius r > 0 in a G -metric space ( X , G ) is
B ( x , r ) = { y X : G ( x , x , y ) < r } .

The following lemma is a characterization of the topology generated by a neighborhood system at each point.

Lemma 13 Let X be a set and, for all x X , let β x be a non-empty family of subsets of X verifying:
  1. 1.

    x N for all N β x .

     
  2. 2.

    For all N 1 , N 2 β x , there exists N 3 β x such that N 3 N 1 N 2 .

     
  3. 3.

    For all N β x , there exists N β x such that for all y N , there exists N β y verifying N N .

     

Then there exists a unique topology τ on X such that β x is a neighborhood system at x.

Let ( X , G ) be a G -metric space and consider the family β x = { B ( x , r ) : r > 0 } . It is clear that x B ( x , r ) (by ( G 1 ) , G ( x , x , x ) = 0 ) and N 3 = B ( x , min ( r , s ) ) B ( x , r ) B ( x , s ) . Next, let N = N = B ( x , r ) β x and let y N = B ( x , r ) . We have to prove that there exists s > 0 such that N = B ( y , s ) B ( x , r ) = N . Indeed, if y = x , then we can take s = r > 0 . On the contrary, if y x , then 0 < G ( x , x , y ) < r by ( G 2 ) . Let r ] G ( x , x , y ) , r [ arbitrary and let s = r r > 0 (that is, r + s = r ). Now we prove that B ( y , s ) B ( x , r ) . Let z B ( y , s ) . Then, using ( G 4 ) and ( G 5 ) ,
G ( x , x , z ) = G ( z , x , x ) a = y G ( z , y , y ) + G ( y , x , x ) = G ( x , x , y ) + G ( y , y , z ) < r + s = r .

Then z B ( x , r ) and, as a consequence, B ( y , s ) B ( x , r ) . Lemma 13 guarantees that there exists a unique topology τ G on X such that β x = { B ( x , r ) : r > 0 } is a neighborhood system at each x X .

Next, let us show that τ G is Hausdorff. Let x , y X be two points such that x y . By ( G 2 ) , r = G ( x , x , y ) > 0 . We claim that B ( x , r / 4 ) B ( y , r / 4 ) = . We reason by contradiction. Let z B ( x , r / 4 ) B ( y , r / 4 ) , that is, G ( x , x , z ) < r / 4 and G ( y , y , z ) < r / 4 . Using ( G 4 ) and ( G 5 ) twice
0 < r = G ( x , x , y ) = G ( y , x , x ) G ( y , z , z ) + G ( z , x , x ) = G ( z , z , y ) + G ( x , x , z ) G ( z , y , y ) + G ( y , z , y ) + G ( x , x , z ) = G ( y , y , z ) + G ( y , y , z ) + G ( x , x , z ) < r 4 + r 4 + r 4 = 3 r 4 < r ,

which is impossible. Then B ( x , r / 4 ) B ( y , r / 4 ) = and τ G is Hausdorff.

A subset A X is G-open if for all x A there exists r > 0 such that B ( x , r ) A . Following classic techniques, it is possible to prove that there exists a unique topology τ G on X such that β x = { B ( x , r ) : r > 0 } is a neighborhood system at each x X . Furthermore, τ G is a Hausdorff topology. In this topology, we characterize the notions of convergent sequence and Cauchy sequence in the following way. Let ( X , G ) be a G -metric space, let { x m } X be a sequence and let x X .

  • { x m } G-converges to x, and we will write { x m } G x if lim m , m G ( x m , x m , x ) = 0 , that is, for all ε > 0 , there exists m 0 N verifying that G ( x m , x m , x ) < ε for all m , m N such that m , m m 0 .

  • { x m } is G-Cauchy if lim m , m , m G ( x m , x m , x m ) = 0 , that is, for all ε > 0 , there exists m 0 N verifying that G ( x m , x m , x m ) < ε for all m , m , m N such that m , m , m m 0 .

Lemma 14 Let ( X , G ) be a G -metric space, let { x m } X be a sequence and let x X . Then the following conditions are equivalent.
  1. (a)

    { x m } G-converges to x.

     
  2. (b)

    lim m G ( x , x , x m ) = 0 .

     
  3. (c)

    lim m G ( x m , x m , x ) = 0 .

     
  4. (d)

    lim m G ( x m , x m , x ) = 0 and lim m G ( x m , x m + 1 , x ) = 0 .

     
  5. (e)

    lim m G ( x , x , x m ) = 0 and lim m G ( x m , x m + 1 , x ) = 0 .

     

Notice that the condition lim m G ( x m , x m + 1 , x ) = 0 is not strong enough to prove that { x m } G-converges to x.

Proposition 15 The limit of a G-convergent sequence in a G -metric space is unique.

Lemma 16 If ( X , G ) is a G -metric space and { x m } X is a sequence, then the following conditions are equivalent.
  1. (a)

    { x m } is G-Cauchy.

     
  2. (b)

    lim m , m G ( x m , x m , x m ) = 0 .

     
  3. (c)

    lim m , m G ( x m , x m + 1 , x m ) = 0 .

     

Remark 17 As a consequence, a sequence { x m } X is not G-Cauchy if and only if there exist ε 0 > 0 and two partial subsequences { x n ( k ) } k N and { x m ( k ) } k N such that k < n ( k ) < m ( k ) < n ( k + 1 ) , G ( x n ( k ) , x n ( k ) + 1 , x m ( k ) ) ε 0 and G ( x n ( k ) , x n ( k ) + 1 , x m ( k ) 1 ) < ε 0 for all k.

Definition 18 Let ( X , G ) be a G -metric space and let be a preorder on X. We will say that ( X , G , ) is regular non-decreasing (respectively, regular non-increasing) if for all -monotone non-decreasing (respectively, non-increasing) sequence { x m } such that { x m } G z 0 , we have that x m z 0 (respectively, x m z 0 ) for all m. We will say that ( X , G , ) is regular if it is both regular non-decreasing and regular non-increasing.

Some authors said that ( X , G , ) verifies the sequential monotone property if ( X , G , ) is regular (see [20]). The notion of G-continuous mapping F : X n X follows considering on X the topology τ G and in X n the product topology.

Definition 19 If ( X , G ) is a G -metric space, we will say that a mapping F : X n X is G-continuous if for all n sequences { a m 1 } , { a m 2 } , , { a m n } X such that { a m i } G a i X for all i, we have that { F ( a m 1 , a m 2 , , a m n ) } G F ( a 1 , a 2 , , a n ) .

In this topology, the notion of convergence is the following.
{ x m } G x [ B ( x , r ) , m 0 N : ( m m 0 x m B ( x , r ) ) ] [ ε > 0 , m 0 N : ( m m 0 G ( x , x , x m ) < ε ) ] [ lim m G ( x , x , x m ) = 0 ] .

This property can be characterized as follows.

Lemma 20 Let ( X , G ) be a G -metric space, let { x m } X be a sequence and let x X . Then the following conditions are equivalent.
  1. (a)

    { x m } G-converges to x (that is, lim m , m G ( x m , x m , x ) = 0 , which means that for all ε > 0 , there exists n 0 N such that G ( x m , x m , x ) for all m , m m 0 ).

     
  2. (b)

    lim m G ( x , x , x m ) = 0 .

     
  3. (c)

    lim m G ( x m , x m , x ) = 0 .

     
  4. (d)

    lim m G ( x m , x m , x ) = 0 and lim m G ( x m , x m + 1 , x ) = 0 .

     
  5. (e)

    lim m G ( x , x , x m ) = 0 and lim m G ( x m , x m + 1 , x ) = 0 .

     

Proof [(a) (c)] It is apparent using m = m .

  • [(c) (b)] Using ( G 5 ) , G ( x , x , x m ) G ( x , x m , x m ) + G ( x m , x , x m ) = 2 G ( x m , x m , x ) .

  • [(b) (a)] Using ( G 4 ) and ( G 5 ) ,
    G ( x m , x m , x ) G ( x m , x , x ) + G ( x , x m , x ) 2 max ( G ( x , x , x m ) , G ( x , x , x m ) ) .
  • [(a) (d),(e)] It is apparent using m = m and m = m + 1 .

  • [(d) (c)] It is evident.

  • [(e) (b)] It is evident. □

Corollary 21 If ( X , G ) is a G-metric space, then { x m } G x if and only if lim m G ( x m , x m + 1 , x ) = 0 .

Proof We only need to prove that the condition is sufficient. Suppose that lim m G ( x m , x m + 1 , x ) = 0 . In a G-metric space, the following property holds (see [11]):
G ( x , y , z ) G ( x , a , z ) + G ( a , y , z ) for all  x , y , z , a X .
Then, using a = x m + 1 ,
G ( x , x , x m ) = G ( x , x m + 1 , x m ) + G ( x m + 1 , x , x m ) = 2 G ( x m , x m + 1 , x ) .

This proves (b) in the previous lemma. □

Proposition 22 The limit of a G-convergent sequence in a G -metric space is unique.

Proof Suppose that { x m } G x and { x m } G y . Then
G ( x , x , y ) = G ( y , x , x ) G ( y , x m , x m ) + G ( x m , x , x ) .

By items (a) and (c) of Lemma 20, we deduce that G ( x , x , y ) = 0 , which means that x = y by ( G 2 ) . □

In the topology τ G , the notion of Cauchy sequence is the following.
{ x m }  is  G -Cauchy [ ε > 0 , m 0 N : ( m , m , m m 0 G ( x m , x m , x m ) < ε ) ] .

This definition can be characterized as follows.

Lemma 23 If ( X , G ) is a G -metric space and { x m } X is a sequence, then the following conditions are equivalent.
  1. (a)

    { x m } is G-Cauchy.

     
  2. (b)

    lim m , m G ( x m , x m , x m ) = 0 .

     
  3. (c)

    lim m , m G ( x m , x m + 1 , x m ) = 0 .

     

Proof [(b) (a)] Using ( G 5 ) , G ( x m , x m , x m ) G ( x m , x m , x m ) + G ( x m , x m , x m ) .

  • [(a) (c)] It is apparent using m = m + 1 .

  • [(c) (b)] Let ε > 0 and let m 0 N be such that G ( x m , x m + 1 , x m ) < ε / 2 for all m , m m 0 . Then
    m , m m 0 G ( x m , x m + 1 , x m ) < ε / 2 , m , m + 1 m 0 G ( x m , x m + 1 , x m + 1 ) < ε / 2 .
Therefore, using ( G 4 ) and ( G 5 ) ,
G ( x m , x m , x m ) = G ( x m , x m , x m ) G ( x m , x m + 1 , x m + 1 ) + G ( x m + 1 , x m , x m ) < ε / 2 + ε / 2 = ε .

Therefore, lim m , m G ( x m , x m , x m ) = 0 . □

4 Product of G -metric spaces

Lemma 24 Let { ( X i , G i ) } i = 1 n be a family of G -metric spaces, consider the product space X = X 1 × X 2 × × X n and define G n max and G n sum on X 3 by
G n max ( X , Y , Z ) = max 1 i n G i ( x i , y i , z i ) and G n sum ( X , Y , Z ) = i = 1 n G i ( x i , y i , z i )
for all X = ( x 1 , x 2 , , x n ) , Y = ( y 1 , y 2 , , y n ) , Z = ( z 1 , z 2 , , z n ) X . Then the following statements hold.
  1. 1.

    G n max and G n sum are G -metrics on X .

     
  2. 2.

    If A m = ( a m 1 , a m 2 , , a m n ) X for all m and A = ( a 1 , a 2 , , a n ) X , then { A m } G n max -converges (respectively, G n sum -converges) to A if and only if each { a m i } G i -converges to a i .

     
  3. 3.

    { A m } is G n max -Cauchy if and only if each { a m i } is G i -Cauchy.

     
  4. 4.

    ( X , G n max ) (respectively, ( X , G n sum ) ) is complete if and only if every ( X i , G i ) is complete.

     
  5. 5.

    For all i, let i be a preorder on X i and define X Y if and only if x i i y i for all i. Then ( X , G n max , ) is regular (respectively, regular non-decreasing, regular non-increasing) if and only if each factor ( X i , G i ) is also regular (respectively, regular non-decreasing, regular non-increasing).

     

Proof Let us denote G = G n max . Taking into account that G n max G n sum n G n max , we will only develop the proof using G.

(1) It is a straightforward exercise to prove the following statements.

  • G ( X , X , X ) = max 1 i n G i ( x i , x i , x i ) = max 1 i n 0 = 0 .

  • If Y Z , there exists j { 1 , 2 , , n } such that y j z j . Then G ( X , Y , Z ) = max 1 i n G i ( x i , y i , z i ) G j ( x j , y j , z j ) > 0 .

  • Symmetry in all three variables of G follows from symmetry in all three variables of each G i .

  • We have that
    G ( X , Y , Z ) = max 1 i n G i ( x i , y i , z i ) max 1 i n [ G i ( x i , a i , a i ) + G i ( a i , y i , z i ) ] max 1 i n G i ( x i , a i , a i ) + max 1 i n G i ( a i , y i , z i ) = G ( X , A , A ) + G ( A , Y , Z ) .
(2) We use Lemma 20. Suppose that { A m } G-converges to A and let ε > 0 . Then, for all j { 1 , 2 , , n } and all m,
G j ( a j , a j , a m j ) max 1 i n G i ( a i , a i , a m i ) = G ( A , A , A m ) .

Therefore, { a m j } G j -converges to a j . Conversely, assume that each { a m i } G i -converges to a i . Let ε > 0 and let m i N be such that if m m i , then G i ( a i , a i , a m i ) < ε . If m 0 = max ( m 1 , m 2 , , m n ) and m , m m 0 , then G ( A , A , A m ) = max 1 i n G i ( a i , a i , a m i ) < ε , so { A m } G-converges to A.

(3) We use Lemma 23. Suppose that { A m } is G-Cauchy and let ε > 0 . Then, for all j { 1 , 2 , , n } and all m, m ,
G j ( a m j , a m j , a m j ) max 1 i n G i ( a m i , a m i , a m i ) = G ( A m , A m , A m ) .

Therefore, { a m j } is G j -Cauchy. Conversely, assume that each { a m i } is G i -Cauchy. Let ε > 0 and let m i N be such that if m , m m i , then G i ( a m j , a m j , a m j ) < ε . If m 0 = max ( m 1 , m 2 , , m n ) and m , m m 0 , then G ( A m , A m , A m ) = max 1 i n G i ( a m i , a m i , a m i ) < ε , so { A m } is G-Cauchy.

(4) It is an easy consequence of items 2 and 3 since
{ A m } G -Cauchy each  { a m i } G -Cauchy each  { a m i } G -convergent { A m } G -convergent.

(5) A sequence { A m } on X is -monotone non-decreasing if and only if each sequence { a m i } is -monotone non-decreasing. Moreover, { A m } G-converges to A = ( a 1 , a 2 , , a n ) X if and only if each { a m i } G i -converges to a i . Finally, A m A if and only if a m i i a i for all i. Therefore, ( X , G n max , ) is regular non-decreasing if and only if each factor ( X i , G i ) is also regular non-decreasing. Other statements may be proved similarly. □

Taking ( X i , G i ) = ( X , G ) for all i, we derive the following result.

Corollary 25 Let ( X , G ) be a G -metric space and consider on the product space X n the mappings G n and G n defined by
G n ( X , Y , Z ) = max 1 i n G ( x i , y i , z i ) and G n ( X , Y , Z ) = i = 1 n G ( x i , y i , z i )

for all X = ( x 1 , x 2 , , x n ) , Y = ( y 1 , y 2 , , y n ) , Z = ( z 1 , z 2 , , z n ) X n .

  1. 1.

    G n and G n are G -metrics on X n .

     
  2. 2.

    If A m = ( a m 1 , a m 2 , , a m n ) X n for all m and A = ( a 1 , a 2 , , a n ) X n , then { A m } G n -converges (respectively, G n -converges) to A if and only if each { a m i } G-converges to a i .

     
  3. 3.

    { A m } is G n -Cauchy (respectively, G n -Cauchy) if and only if each { a m i } is G-Cauchy.

     
  4. 4.

    ( X , G n ) (respectively, ( X n , G n ) ) is complete if and only if ( X , G ) is complete.

     
  5. 5.

    If ( X , G ) is -regular, then ( X n , G n ) is -regular.

     

5 Unidimensional fixed point result in partially preordered G -metric spaces

Theorem 26 Let ( X , ) be a preordered set endowed with a G -metric G and let T : X X be a given mapping. Suppose that the following conditions hold:
  1. (a)

    ( X , G ) is complete.

     
  2. (b)

    T is non-decreasing (w.r.t. ).

     
  3. (c)

    Either T is G-continuous or ( X , G , ) is regular non-decreasing.

     
  4. (d)

    There exists x 0 X such that x 0 T x 0 .

     
  5. (e)
    There exist two mappings ψ , φ Ψ such that, for all x , y X with x y ,
    ψ ( G ( T x , T y , T 2 x ) ) ψ ( G ( x , y , T x ) ) φ ( G ( x , y , T x ) ) .
     

Then T has a fixed point. Furthermore, if for all z 1 , z 2 X fixed points of T there exists z X such that z 1 z and z 2 z , we obtain uniqueness of the fixed point.

Proof Define x m = T m x 0 for all m 1 . Since T is non-decreasing (w.r.t. ), then x m x m + 1 for all m 0 . Then
ψ ( G ( x m + 1 , x m + 2 , x m + 2 ) ) = ψ ( G ( T x m , T x m + 1 , T 2 x m ) ) ψ ( G ( x m , x m + 1 , T x m ) ) φ ( G ( x m , x m + 1 , T x m ) ) = ψ ( G ( x m , x m + 1 , x m + 1 ) ) φ ( G ( x m , x m + 1 , x m + 1 ) ) .
Applying Lemma 10, { G ( x m , x m + 1 , x m + 1 ) } 0 . Let us show that { x m } is G-Cauchy. Reasoning by contradiction, if { x m } is not G-Cauchy, by Remark 17, there exist ε 0 > 0 and two partial subsequences { x n ( k ) } and { x m ( k ) } verifying k < n ( k ) < m ( k ) < n ( k + 1 ) ,
G ( x n ( k ) , x m ( k ) , x n ( k ) + 1 ) > ε 0 and G ( x n ( k ) , x m ( k ) 1 , x n ( k ) + 1 ) ε 0 for all  k 1 .
(4)
Therefore
0 < ψ ( ε 0 ) ψ ( G ( x n ( k ) , x m ( k ) , x n ( k ) + 1 ) ) = ψ ( G ( T x n ( k ) 1 , T x m ( k ) 1 , T 2 x n ( k ) 1 ) ) ψ ( G ( x n ( k ) 1 , x m ( k ) 1 , T x n ( k ) 1 ) ) φ ( G ( x n ( k ) 1 , x m ( k ) 1 , T x n ( k ) 1 ) ) = ψ ( G ( x n ( k ) 1 , x m ( k ) 1 , x n ( k ) ) ) φ ( G ( x n ( k ) 1 , x m ( k ) 1 , x n ( k ) ) ) .
(5)
Consider the sequence of non-negative real numbers { G ( x n ( k ) 1 , x m ( k ) 1 , x n ( k ) ) } . If this sequence has a partial subsequence converging to zero, then we can take the limit in (5) using this partial subsequence and we would deduce 0 < ψ ( ε 0 ) 0 , which is impossible. Then { G ( x n ( k ) 1 , x m ( k ) 1 , x n ( k ) ) } cannot have a partial subsequence converging to zero. This means that there exist δ > 0 and k 0 N such that
G ( x n ( k ) 1 , x m ( k ) 1 , x n ( k ) ) δ for all  k k 0 .
Since φ is non-decreasing, φ ( G ( x n ( k ) 1 , x m ( k ) 1 , x n ( k ) ) φ ( δ ) < 0 . By (G5) and (4),
G ( x n ( k ) 1 , x m ( k ) 1 , x n ( k ) ) = G ( x n ( k ) 1 , x n ( k ) , x m ( k ) 1 ) [ x = x n ( k ) 1 , y = x n ( k ) , z = x m ( k ) 1 , a = x n ( k ) + 1 ] G ( x n ( k ) 1 , x n ( k ) + 1 , x n ( k ) + 1 ) + G ( x n ( k ) + 1 , x n ( k ) , x m ( k ) 1 ) = G ( x n ( k ) 1 , x n ( k ) + 1 , x n ( k ) + 1 ) + G ( x n ( k ) , x n ( k ) + 1 , x m ( k ) 1 ) [ x = x n ( k ) 1 , y = z = x n ( k ) + 1 , a = x n ( k ) ] G ( x n ( k ) 1 , x n ( k ) , x n ( k ) ) + G ( x n ( k ) , x n ( k ) + 1 , x n ( k ) + 1 ) + G ( x n ( k ) , x n ( k ) + 1 , x m ( k ) 1 ) G ( x n ( k ) 1 , x n ( k ) , x n ( k ) ) + G ( x n ( k ) , x n ( k ) + 1 , x n ( k ) + 1 ) + ε 0 .
Since ψ is non-decreasing, it follows from (5) that
0 < ψ ( ε 0 ) ψ ( G ( x n ( k ) 1 , x m ( k ) 1 , x n ( k ) ) ) φ ( G ( x n ( k ) 1 , x m ( k ) 1 , x n ( k ) ) ) ψ ( G ( x n ( k ) 1 , x m ( k ) 1 , x n ( k ) ) ) φ ( δ ) ψ ( G ( x n ( k ) 1 , x n ( k ) , x n ( k ) ) + G ( x n ( k ) , x n ( k ) + 1 , x n ( k ) + 1 ) + ε 0 ) φ ( δ ) .

Taking limit when k , we deduce that 0 < ψ ( ε 0 ) ψ ( ε 0 ) φ ( δ ) < ψ ( ε 0 ) , which is impossible. This contradiction finally proves that { x m } is G-Cauchy. Since ( X , G ) is complete, there exists z 0 X such that { x m } G z 0 .

Now suppose that T is G-continuous. Then { x m + 1 } = { T x m } G T z 0 . By the unicity of the limit, T z 0 = z 0 and z 0 is a fixed point of T.

On the contrary, suppose that ( X , G , ) is regular non-decreasing. Since { x m } G z 0 and { x m } is monotone non-decreasing (w.r.t. ), it follows that x m z 0 for all m. Hence
ψ ( G ( x m + 1 , T z 0 , x m + 2 ) ) = ψ ( G ( T x m , T z 0 , T 2 x m ) ) ψ ( G ( x m , z 0 , T x m ) ) φ ( G ( x m , z 0 , T x m ) ) = ψ ( G ( x m , x m + 1 , z 0 ) ) φ ( G ( x m , x m + 1 , z 0 ) ) .

Since { x m } G z 0 , then { G ( x m , x m + 1 , z 0 ) } 0 . Taking limit when k , we deduce that { ψ ( G ( x m + 1 , T z 0 , x m + 2 ) ) } 0 . By Lemma 7, { G ( x m + 1 , x m + 2 , T z 0 ) } 0 , so { x m } G T z 0 and we also conclude that z 0 is a fixed point of T.

To prove the uniqueness, let z 1 , z 2 X be two fixed points of T. By hypothesis, there exists z X such that z 1 z and z 2 z . Let us show that { T m z } G z 1 . Indeed,
ψ ( G ( z 1 , z 1 , T m + 1 z ) ) = ψ ( G ( T z 1 , T T m z , T 2 z 1 ) ) ψ ( G ( z 1 , T m z , T z 1 ) ) φ ( G ( z 1 , T m z , T z 1 ) ) = ψ ( G ( z 1 , z 1 , T m z ) ) φ ( G ( z 1 , z 1 , T m z ) ) .

By Lemma 10, we deduce { G ( z 1 , z 1 , T m z ) } 0 , that is, { T m z } G z 1 . The same reasoning proves that { T m z } G z 2 , so z 1 = z 2 . □

We particularize the previous theorem in two cases. If take ψ ( t ) = t in Theorem 26, then we get the following results.

Corollary 27 Let ( X , ) be a preordered set endowed with a G -metric G and let T : X X be a given mapping. Suppose that the following conditions hold:
  1. (a)

    ( X , G ) is complete.

     
  2. (b)

    T is non-decreasing (w.r.t. ).

     
  3. (c)

    Either T is G-continuous or ( X , G , ) is regular non-decreasing.

     
  4. (d)

    There exists x 0 X such that x 0 T x 0 .

     
  5. (e)
    There exists a mapping φ Ψ such that, for all x , y X with x y ,
    G ( T x , T y , T 2 x ) G ( x , y , T x ) φ ( G ( x , y , T x ) ) .
     

Then T has a fixed point. Furthermore, if for all z 1 , z 2 X fixed points of T there exists z X such that z 1 z and z 2 z , we obtain uniqueness of the fixed point.

If take φ ( t ) = ( 1 k ) t with k [ 0 , 1 ) in Corollary 27, then we derive the following result.

Corollary 28 Let ( X , ) be a preordered set endowed with a G -metric G and let T : X X be a given mapping. Suppose that the following conditions hold:
  1. (a)

    ( X , G ) is complete.

     
  2. (b)

    T is non-decreasing (w.r.t. ).

     
  3. (c)

    Either T is G-continuous or ( X , G , ) is regular non-decreasing.

     
  4. (d)

    There exists x 0 X such that x 0 T x 0 .

     
  5. (e)
    There exists a constant k [ 0 , 1 ) such that, for all x , y X with x y ,
    G ( T x , T y , T 2 x ) k G ( x , y , T x ) .
     

Then T has a fixed point. Furthermore, if for all z 1 , z 2 X fixed points of T there exists z X such that z 1 z and z 2 z , we obtain uniqueness of the fixed point.

6 Multidimensional ϒ-fixed point results in partially preordered G -metric spaces

In this section we extend Theorem 26 to an arbitrary number of variables. To do that, it is necessary to introduce the following notation. Given a mapping F : X n X , we define F ϒ : X n X n by
F ϒ ( x 1 , x 2 , , x n ) = ( F ( x σ 1 ( 1 ) , x σ 1 ( 2 ) , , x σ 1 ( n ) ) , F ( x σ 2 ( 1 ) , x σ 2 ( 2 ) , , x σ 2 ( n ) ) , , F ( x σ n ( 1 ) , x σ n ( 2 ) , , x σ n ( n ) ) ) ,
and F ϒ 2 = F F ϒ : X n X will be
F ϒ 2 ( x 1 , x 2 , , x n ) = F ( F ( x σ 1 ( 1 ) , x σ 1 ( 2 ) , , x σ 1 ( n ) ) , F ( x σ 2 ( 1 ) , x σ 2 ( 2 ) , , x σ 2 ( n ) ) , , F ( x σ n ( 1 ) , x σ n ( 2 ) , , x σ n ( n ) ) )

for all X = ( x 1 , x 2 , , x n ) X n .

Lemma 29

  1. 1.

    Z X n is a ϒ-fixed point of F if and only if Z is a fixed point of F ϒ (that is, F ϒ Z = Z ).

     
  2. 2.

    If F has the mixed monotone property, then F ϒ is -monotone non-decreasing on X n .

     
  3. 3.

    If ( X , G ) is a G -metric space and F is G-continuous, then F ϒ : X n X n is G n -continuous and F ϒ 2 = F F ϒ : X n X is G-continuous.

     

6.1 A first multidimensional contractivity result

In this subsection we apply Theorem 26 considering T = F ϒ defined on ( X n , G n , ) . In order to do that, we notice that joining some of the previous results, we obtain the following consequences.
  • If ( X , G ) is complete, it follows from Corollary 25 that ( X n , G n ) is also complete.

  • By item 2 of Lemma 29, if F has the mixed monotone property, then F ϒ is -monotone non-decreasing on X n .

  • By item 3 of Lemma 29, if F is G-continuous, then F ϒ : X n X n is G n -continuous and F ϒ 2 = F F ϒ : X n X is G-continuous.

  • If ( X , G , ) is regular, it follows from Corollary 25 that ( X n , G n , ) is also regular.

  • If x 0 1 , x 0 2 , , x 0 n X are such that x 0 i i F ( x 0 σ i ( 1 ) , x 0 σ i ( 2 ) , , x 0 σ i ( n ) ) for all i, then X 0 = ( x 0 1 , x 0 2 , , x 0 n ) X n verifies X 0 F ϒ ( X 0 ) .

We study how the contractivity condition
ψ ( G n ( F ϒ X , F ϒ Y , F ϒ 2 X ) ) ( ψ φ ) ( G n ( X , Y , F ϒ X ) ) for all  X , Y X n  such that  X Y
may be equivalently established. Let X = ( x 1 , x 2 , , x n ) X n and let z i = F ( x σ i ( 1 ) , x σ i ( 2 ) , , x σ i ( n ) ) X for all i. Then
F ϒ 2 X = F ϒ ( F ( x σ 1 ( 1 ) , x σ 1 ( 2 ) , , x σ 1 ( n ) ) , F ( x σ 2 ( 1 ) , x σ 2 ( 2 ) , , x σ 2 ( n ) ) , , F ϒ 2 X = F ( x σ n ( 1 ) , x σ n ( 2 ) , , x σ n ( n ) ) ) F ϒ 2 X = F ϒ ( z 1 , z 2 , , z n ) F ϒ 2 X = ( F ( z σ 1 ( 1 ) , z σ 1 ( 2 ) , , z σ 1 ( n ) ) , F ( z σ 2 ( 1 ) , z σ 2 ( 2 ) , , z σ 2 ( n ) ) , , F ( z σ n ( 1 ) , z σ n ( 2 ) , , z σ n ( n ) ) ) F ϒ 2 X = ( F ( F ( x σ σ 1 ( 1 ) ( 1 ) , , x σ σ 1 ( 1 ) ( n ) ) , F ( x σ σ 1 ( 2 ) ( 1 ) , , x σ σ 1 ( 2 ) ( n ) ) , , F ( x σ σ 1 ( n ) ( 1 ) , , x σ σ 1 ( n ) ( n ) ) ) , F ( F ( x σ σ 2 ( 1 ) ( 1 ) , , x σ σ 2 ( 1 ) ( n ) ) , F ( x σ σ 2 ( 2 ) ( 1 ) , , x σ σ 2 ( 2 ) ( n ) ) , , F ( x σ σ 2 ( n ) ( 1 ) , , x σ σ n ( n ) ( n ) ) ) , , F ( F ( x σ σ n ( 1 ) ( 1 ) , , x σ σ n ( 1 ) ( n ) ) , F ( x σ σ n ( 2 ) ( 1 ) , , x σ σ n ( 2 ) ( n ) ) , , F ( x σ σ n ( n ) ( 1 ) , , x σ σ n ( n ) ( n ) ) ) ) F ϒ 2 X = ( F ϒ 2 ( x σ 1 ( 1 ) , x σ 1 ( 2 ) , x σ 1 ( n ) ) , F ϒ 2 ( x σ 2 ( 1 ) , x σ 2 ( 2 ) , x σ 2 ( n ) ) , , F ϒ 2 X = F ϒ 2 ( x σ n ( 1 ) , x σ n ( 2 ) , x σ n ( n ) ) ) .
It follows that
G n ( X , Y , F ϒ X ) = max 1 i n G ( x i , y i , F ( x σ i ( 1 ) , x σ i ( 2 ) , , x σ i ( n ) ) ) and G n ( F ϒ X , F ϒ Y , F ϒ 2 X ) = max 1 i n G ( F ( x σ i ( 1 ) , x σ i ( 2 ) , , x σ i ( n ) ) , F ( y σ i ( 1 ) , y σ i ( 2 ) , , y σ i ( n ) ) , G n ( F ϒ X , F ϒ Y , F ϒ 2 X ) = F ϒ 2 ( x σ i ( 1 ) , x σ i ( 2 ) , x σ i ( n ) ) ) .

Therefore, a possible version of Theorem 26 applied to ( X n , G n , ) taking T = F ϒ is the following.

Theorem 30 Let ( X , G ) be a complete G -metric space and let be a partial preorder on X. Let ϒ = ( σ 1 , σ 2 , , σ n ) be an n-tuple of mappings from { 1 , 2 , , n } into itself verifying σ i Ω A , B if i A and σ i Ω A , B if i B . Let F : X n X be a mapping verifying the mixed monotone property on X. Assume that there exist ψ , φ Ψ such that
max 1 i n ψ ( G ( F ( x σ i ( 1 ) , x σ i ( 2 ) , , x σ i ( n ) ) , F ( y σ i ( 1 ) , y σ i ( 2 ) , , y σ i ( n ) ) , F ϒ 2 ( x σ i ( 1 ) , x σ i ( 2 ) , x σ i ( n ) ) ) ) ( ψ φ ) ( max 1 i n G ( x i , y i , F ( x σ i ( 1 ) , x σ i ( 2 ) , , x σ i ( n ) ) ) )
(6)

for which x i i y i for all i. Suppose either F is continuous or ( X , G , ) is regular. If there exist x 0 1 , x 0 2 , , x 0 n X verifying x 0 i i F ( x 0 σ i ( 1 ) , x 0 σ i ( 2 ) , , x 0 σ i ( n ) ) for all i, then F has, at least, one ϒ-fixed point.

6.2 A second multidimensional contractivity result

In this section we introduce a slightly different contractivity condition that cannot be directly deduced applying Theorem 26 to ( X , G n , ) taking T = F ϒ , because the contractivity condition is weaker. Then we need to show a classical proof.

Theorem 31 Let ( X , G ) be a complete G -metric space and let be a partial preorder on X. Let ϒ = ( σ 1 , σ 2 , , σ n ) be an n-tuple of mappings from { 1 , 2 , , n } into itself verifying σ i Ω A , B if i A and σ i Ω A , B if i B . Let F : X n X be a mapping verifying the mixed monotone property on X. Assume that there exist ψ , φ Ψ such that
ψ ( G ( F ( x 1 , x 2 , , x n ) , F ( y 1 , y 2 , , y n ) , F ϒ 2 ( x 1 , x 2 , , x n ) ) ) ( ψ φ ) ( max 1 i n G ( x i , y i , F ( x σ i ( 1 ) , x σ i ( 2 ) , , x σ i ( n ) ) ) )
(7)

for which ( x 1 , x 2 , , x n ) , ( y 1 , y 2 , , y n ) X n are -comparable. Suppose either F is continuous or ( X , G , ) is regular. If there exist x 0 1 , x 0 2 , , x 0 n X verifying x 0 i i F ( x 0 σ i ( 1 ) , x 0 σ i ( 2 ) , , x 0 σ i ( n ) ) for all i, then F has, at least, one ϒ-fixed point.

Notice that (6) and (7) are very different contractivity conditions. For instance, (6) would be simpler if the image of all σ i are sets with a few points.

Proof Define X 0 = ( x 0 1 , x 0 2 , , x 0 n ) and let x 1 i = F ( x 0 σ i ( 1 ) , x 0 σ i ( 2 ) , , x 0 σ i ( n ) ) for all i. If X 1 = ( x 1 1 , x 1 2 , , x 1 n ) , then x 0 i i x 1 i for all i is equivalent to X 0 X 1 = F ϒ ( X 0 ) . By recurrence, define x m + 1 i = F ( x m σ i ( 1 ) , x m σ i ( 2 ) , , x m σ i ( n ) ) for all i and all m, and we have that X m X m + 1 = F ϒ ( X m ) . This means that the sequence { X m + 1 = F ϒ ( X m ) } is -monotone non-decreasing. Since ( X n , G n , ) is complete, it is only necessary to prove that { X m } is G n -Cauchy in order to deduce that it is G n -convergent. By item 3 of Lemma 24, it will be sufficient to prove that each sequence { x m i } is G-Cauchy. Firstly, notice that X m + 1 = F ϒ ( X m ) means that
x m + 1 i = F ( x m σ i ( 1 ) , x m σ i ( 2 ) , , x m σ i ( n ) ) for all  i  and all  m .
Hence
x m + 2 i = F ( x m + 1 σ i ( 1 ) , x m + 1 σ i ( 2 ) , , x m + 1 σ i ( n ) ) x m + 2 i = F ( F ( x m σ σ i ( 1 ) ( 1 ) , x m σ σ i ( 1 ) ( 2 ) , , x m σ σ i ( 1 ) ( n ) ) , F ( x m σ σ i ( 2 ) ( 1 ) , x m σ σ i ( 2 ) ( 2 ) , , x m σ σ i ( 2 ) ( n ) ) , , x m + 2 i = F ( x m σ σ i ( n ) ( 1 ) , x m σ σ i ( n ) ( 2 ) , , x m σ σ i ( n ) ( n ) ) ) = F ϒ 2 ( x m σ i ( 1 ) , x m σ i ( 2 ) , , x m σ i ( n ) ) .
Furthermore, for all m,
F ϒ 2 ( X m ) = F ϒ 2 ( x m 1 , x m 2 , , x m n ) = F ( F ( x m σ 1 ( 1 ) , x m σ 1 ( 2 ) , , x m σ 1 ( n ) ) , F ( x m σ 2 ( 1 ) , x m σ 2 ( 2 ) , , x m σ 2 ( n ) ) , , F ( x m σ n ( 1 ) , x m σ n ( 2 ) , , x m σ n ( n ) ) ) = F ( x m + 1 1 , x m + 1 2 , , x m + 1 n ) = F ( X m + 1 ) .
(8)
Therefore, for all i and all m,
ψ ( G ( x m + 1 i , x m + 2 i , x m + 2 i ) ) = ψ ( G ( F ( x m σ i ( 1 ) , x m σ i ( 2 ) , , x m σ i ( n ) ) , F ( x m + 1 σ i ( 1 ) , x m + 1 σ i ( 2 ) , , x m + 1 σ i ( n ) ) , F ϒ 2 ( x m σ i ( 1 ) , x m σ i ( 2 ) , , x m σ i ( n ) ) ) ( ψ φ ) ( max 1 j n G ( x m σ i ( j ) , x m + 1 σ i ( j ) , F ( x m σ σ i ( j ) ( 1 ) , x m σ σ i ( j ) ( 2 ) , , x m σ σ i ( j ) ( n ) ) ) ) = ( ψ φ ) ( max 1 j n G ( x m σ i ( j ) , x m + 1 σ i ( j ) , x m + 1 σ i ( j ) ) ) .
Since ψ is non-decreasing, for all i and all m,
ψ ( max 1 j n G ( x m σ i ( j ) , x m + 1 σ i ( j ) , x m + 1 σ i ( j ) ) ) ψ ( max 1 j n G ( x m j , x m + 1 j , x m + 1 j ) ) .
Applying Lemma 8 using
a m i = G ( x m i , x m + 1 i , x m + 1 i ) and b m i = max 1 j n G ( x m σ i ( j ) , x m + 1 σ i ( j ) , x m + 1 σ i ( j ) )
for all i and all m, we deduce that
{ G ( x m i , x m + 1 i , x m + 1 i ) } 0 for all  i , that is , { G n ( X m , X m + 1 , X m + 1 ) } 0 .
(9)
Next, we prove that every sequence { x m i } is G-Cauchy reasoning by contradiction. Suppose that { x m i 1 } m 0 , , { x m i s } m 0 are not G-Cauchy ( s 1 ) and { x m i s + 1 } m 0 , , { x m i n } m 0 are G-Cauchy, being { i 1 , , i n } = { 1 , , n } . From Proposition 2, for all r { 1 , 2 , , s } , there exist ε r > 0 and subsequences { x n r ( k ) i r } k N and { x m r ( k ) i r } k N such that, for all k N ,
k < n r ( k ) < m r ( k ) < n r ( k + 1 ) , G ( x n r ( k ) i r , x n r ( k ) + 1 i r , x m r ( k ) i r ) ε r , G ( x n r ( k ) i r , x n r ( k ) + 1 i r , x m r ( k ) 1 i r ) < ε r .
Now, let ε 0 = max ( ε 1 , , ε s ) > 0 and ε 0 = min ( ε 1 , , ε s ) > 0 . Since { x m i s + 1 } m 0 , , { x m i n } m 0 are G-Cauchy, for all j { i s + 1 , , i n } , there exists n j N such that if m , m n j , then G ( x m j , x m + 1 j , x m j ) < ε 0 / 8 . Define n 0 = max j { i s + 1 , , i n } ( n j ) . Therefore, we have proved that there exists n 0 N such that if m , m n 0 then
G ( x m j , x m + 1 j , x m j ) < ε 0 / 4 for all  j { i s + 1 , , i n } .
(10)
Next, let q { 1 , 2 , , s } be such that ε q = ε 0 = max ( ε 1 , , ε s ) . Let k 1 N be such that n 0 < n q ( k 1 ) and define n ( 1 ) = n q ( k 1 ) . Consider the numbers n ( 1 ) + 1 , n ( 1 ) + 2 , , m q ( k 1 ) until finding the first positive integer m ( 1 ) > n ( 1 ) verifying
max 1 r s G ( x n ( 1 ) i r , x n ( 1 ) + 1 i r , x m ( 1 ) i r ) ε 0 , G ( x n ( 1 ) i j , x n ( 1 ) + 1 i j , x m ( 1 ) 1 i j ) < ε 0 for all  j { 1 , 2 , , s } .
Now let k 2 N be such that m ( 1 ) < n q ( k 2 ) and define n ( 2 ) = n q ( k 2 ) . Consider the numbers n ( 2 ) + 1 , n ( 2 ) + 2 , , m q ( k 2 ) until finding the first positive integer m ( 2 ) > n ( 2 ) verifying
max 1 r s G ( x n ( 2 ) i r , x n ( 2 ) + 1 i r , x m ( 2 ) i r ) ε 0 , G ( x n ( 2 ) i j , x n ( 2 ) + 1 i j , x m ( 2 ) 1 i j ) < ε 0 for all  j { 1 , 2 , , s } .
Repeating this process, we can find sequences such that, for all k 1 ,
n 0 < n ( k ) < m ( k ) < n ( k + 1 ) , max 1 r s G ( x n ( k ) i r , x n ( k ) + 1 i r , x m ( k ) i r ) ε 0 , G ( x n ( k ) i j , x n ( k ) + 1 i j , x m ( k ) 1 i j ) < ε 0 for all  j { 1 , 2 , , s } .
Note that by (10), G ( x n ( k ) i r , x n ( k ) + 1 i r , x m ( k ) i r ) , G ( x n ( k ) i r , x n ( k ) + 1 i r , x m ( k ) 1 i r ) < ε 0 / 4 < ε 0 / 2 for all r { s + 1 , s + 2 , , n } , so
max 1 j n G ( x n ( k ) j , x n ( k ) + 1 j , x m ( k ) j ) = max 1 r s G ( x n ( k ) i r , x n ( k ) + 1 i r , x m ( k ) i r ) ε 0 and G ( x n ( k ) i , x n ( k ) + 1 i , g x m ( k ) 1 i ) < ε 0
(11)
for all i { 1 , 2 , , n } and all k 1 . Next, for all k, let i ( k ) { 1 , 2 , , s } be an index such that
G ( x n ( k ) i ( k ) , x n ( k ) + 1 i ( k ) , x m ( k ) i ( k ) ) = max 1 r s G ( x n ( k ) i r , x n ( k ) + 1 i r , x m ( k ) i r ) = max 1 j n G ( x n ( k ) j , x n ( k ) + 1 j , x m ( k ) j ) ε 0 .
Notice that, applying (G5) twice and (11), for all k and all j,
G ( x n ( k ) 1 j , x n ( k ) j , x m ( k ) 1 j ) G ( x n ( k ) 1 j , x n ( k ) j , x n ( k ) j ) + G ( x n ( k ) j , x n ( k ) j , x m ( k ) 1 j ) G ( x n ( k ) 1 j , x n ( k ) j , x n ( k ) j ) + G ( x n ( k ) j , x n ( k ) + 1 j , x n ( k ) + 1 j ) + G ( x n ( k ) + 1 j , x n ( k ) j , x m ( k ) 1 j ) G ( x n ( k ) 1 j , x n ( k ) j , x n ( k ) j ) + G ( x n ( k ) j , x n ( k ) + 1 j , x n ( k ) + 1 j ) + ε 0 .
(12)
Applying Proposition 2 to guarantee that the following points are -comparable, the contractivity condition (7) assures us for all k
0 < ψ ( ε 0 ) ψ ( G ( x n ( k ) i ( k ) , x n ( k ) + 1 i ( k ) , x m ( k ) i ( k ) ) ) = ψ ( G ( x n ( k ) i ( k ) , x m ( k ) i ( k ) , x n ( k ) + 1 i ( k ) ) ) = ψ ( G ( F ( x n ( k ) 1 σ i ( k ) ( 1 ) , x n ( k ) 1 σ i ( k ) ( 2 ) , , x n ( k ) 1 σ i ( k ) ( n ) ) , F ( x m ( k ) 1 σ i ( k ) ( 1 ) , x m ( k ) 1 σ i ( k ) ( 2 ) , , x m ( k ) 1 σ i ( k ) ( n ) ) , F ϒ 2 ( x n ( k ) 1 σ i ( k ) ( 1 ) , x n ( k ) 1 σ i ( k ) ( 2 ) , , x n ( k ) 1 σ i ( k ) ( n ) ) ) ) ( ψ φ ) ( max 1 j n G ( x n ( k ) 1 σ i ( k ) ( j ) , x m ( k ) 1 σ i ( k )