Skip to main content

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

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 m0’ 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: xx for all xX.

  • Transitivity: If x,y,zX verify xy and yz, then xz.

Henceforth, let {A,B} be a partition of Λ n ={1,2,,n}, that is, AB= Λ n and AB= 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 iA and σ i Ω A , B if iB.

If (X,) is a partially preordered space, x,yX 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 ,

XY 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 XY or XY.

Proposition 2 If XY 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 iA, 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 iB, 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 iA, 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 iB, 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,zX and all i,

yzF( 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 n1 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 n2 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 kN, 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 ψΨ 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 kN. 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,φΨ 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 Δ0 be such that {ψ( c m )}Δ and Δψ( 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,

0ψ ( a m ( k ) i 0 ) (ψφ) ( 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 k0 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 ΔΔφ(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,φΨ and { a m },{ b m }[0,) verify ψ( a m + 1 )(ψφ)( b m ) and ψ( b m )ψ( a m ) for all m, then { a m }0.

Corollary 10 If ψ,φΨ 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,zX:

  • ( G 1 ) G(x,x,x)=0.

  • ( G 2 ) G(x,x,y)>0 if xy.

  • ( G 3 ) G(x,x,y)G(x,y,z) if yz.

  • ( 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 xX 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 xX, let β x be a non-empty family of subsets of X verifying:

  1. 1.

    xN 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 xB(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 yx, 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 zB(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 zB(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 xX.

Next, let us show that τ G is Hausdorff. Let x,yX be two points such that xy. 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 zB(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 AX is G-open if for all xA 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 xX. 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 xX.

  • { 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 xX. 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 xX. 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 )=2G( 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)2max ( 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,aX.

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 )=2G( 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 XY 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 YZ, 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:XX 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,φΨ such that, for all x,yX with xy,

    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 zX 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 m1. Since T is non-decreasing (w.r.t. ), then x m x m + 1 for all m0. 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 andG( x n ( k ) , x m ( k ) 1 , x n ( k ) + 1 ) ε 0 for all k1.
(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 ) )δfor 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 zX 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:XX 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 φΨ such that, for all x,yX with xy,

    G ( T x , T y , T 2 x ) G(x,y,Tx)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 zX such that z 1 z and z 2 z, we obtain uniqueness of the fixed point.

If take φ(t)=(1k)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:XX 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,yX with xy,

    G ( T x , T y , T 2 x ) kG(x,y,Tx).

Then T has a fixed point. Furthermore, if for all z 1 , z 2 X fixed points of T there exists zX 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 ) ) (ψφ) ( G n ( X , Y , F ϒ X ) ) for all X,Y X n  such that XY

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 iA and σ i Ω A , B if iB. Let F: X n X be a mapping verifying the mixed monotone property on X. Assume that there exist ψ,φΨ 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 iA and σ i Ω A , B if iB. Let F: X n X be a mapping verifying the mixed monotone property on X. Assume that there exist ψ,φΨ 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 ) } 0for 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 (s1) 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 kN,

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 /4for 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 k1,

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