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Some fixed/coincidence point theorems under (ψ,φ)-contractivity conditions without an underlying metric structure

Abstract

In this paper we prove a coincidence point result in a space which does not have to satisfy any of the classical axioms that define a metric space. Furthermore, the ambient space need not be ordered and does not have to be complete. Then, this result may be applied in a wide range of different settings (metric spaces, quasi-metric spaces, pseudo-metric spaces, semi-metric spaces, pseudo-quasi-metric spaces, partial metric spaces, G-spaces, etc.). Finally, we illustrate how this result clarifies and improves some well-known, recent results on this topic.

1 Introduction

Fixed point theory plays a crucial role in nonlinear functional analysis since, among other reasons, fixed point results are used to prove the existence (and also uniqueness) of solution when solving various types of equations. The Banach contraction principle is considered to be the pioneering result of the fixed point theory, and it is the most celebrated result in this field. The simplicity of its proof and the possibility of attaining the fixed point by using successive approximations let this theorem become a very useful tool in analysis and in applied mathematics. The great significance of Banach’s principle, and the reason it is possibly one of the most frequently cited fixed point theorems in all of analysis, lies in the fact that its proof contains elements of fundamental importance to the theoretical and practical treatment of mathematical equations. After the appearance of this result in Banach’s thesis in 1922, a great number of extensions (in many occasions, as well-known as the original result, such as those by Krasnoselskii and Zabreiko, Edelstein, Browder, Schauder, Göhde, Kirk, and Caristi; a comprehensive study can be found in [1]) have been proved in various different frameworks (see [2, 3] in partial metric spaces, [47] in G-metric spaces, [8, 9] in fuzzy metric spaces, [10, 11] in intuitionistic fuzzy metric spaces, [12, 13] in probabilistic metric spaces and [14, 15] in Menger spaces).

In recent times, one of the most attractive research topics in fixed point theory is to prove the existence of a fixed point in metric spaces endowed with partial orders. An initial result in this direction was given by Turinici [16] in 1986. Following this line of research, Ran and Reurings [17] (and later Nieto and Rodríguez-López [18]) used a partial order on the ambient metric space to introduce a slightly different contractivity condition, which must be only verified by comparable points. Thus, they reported two versions of the Banach contraction principle in partially ordered sets and applied them to the study of some applications to matrix equations. Their proofs involved combining the ideas of the iterative technique in the contractive mapping principle with those in the monotone technique. This approach led to a very recent branch of this field, with applications to matrix equations and ordinary differential equations. The literature on this topic has exponentially risen in recent years. To mention some advances on this topic, we highlight the following ones. Firstly, in order to guarantee the existence and uniqueness of a solution of periodic boundary value problems, Gnana-Bhaskar and Lakshmikantham [19] (and, subsequently, Lakshmikantham and Ćirić [20]) proved, in 2006, the existence and uniqueness of a coupled fixed point (a notion introduced by Guo and Lakshmikantham in [21]) in the setting of partially ordered metric spaces by introducing the notion of mixed monotone property. Later, the notions of tripled fixed point, quadruple fixed point and multidimensional fixed point were introduced by Berinde and Borcut [22], by Karapinar and Luong [23] and by Berzig and Samet [24] (see also [25]), respectively.

But the two main ingredients of all extensions are, basically, the same that we can find in the Banach contraction principle: a complete metric space and a self-mapping verifying a contractive condition. Although modern versions use, in many cases, different kind of mappings, the more intensively studied condition is based on the idea that the distance between the images of any two points (comparable or not) is upper bound by the product of a constant (small enough) and the distance between those points. The main aim of this manuscript is to provide a result powerful enough to guarantee that a nonlinear operator T has, at least, a fixed point, even when we consider that a measure mapping does not have to be an underlying metric structure on the ambient space X and the binary relationship is not necessarily a partial order on X. To do this, we present a result which can be applied in the following adverse conditions: the framework is a set X provided with a preorder and a measure mapping d:X×XR that does not necessarily verify any of the four classical properties of a metric space (in fact, it need not be one of the following metric structures: a metric, a pseudo-metric, a quasi-metric, a pseudo-quasi-metric or a semi-metric). Furthermore, d has not to be symmetric and the triangular inequality must only be verified by a kind of comparable points. Even if d would verify some of the classical properties of a metric, (X,d) would not be a complete space. In this setting, none of the theorems proved until now can be applied to guarantee that a nonlinear operator (even if it is a contractive mapping) has, at least, a fixed point. We illustrate our results with a particular example. Finally, we show that they extend and improve some well-known fixed point theorems.

2 Preliminaries

Preliminaries and notation about coincidence points can also be found in [25]. Let n be a positive integer. Henceforth, X will denote a nonempty set and X n will denote the product space X×X× n ×X. Throughout this manuscript, m and k will denote nonnegative 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}’. In the sequel, let F: X N X and T,g:XX be three mappings. For brevity, T(x) will be denoted by Tx.

Definition 2.1 A binary relation on X is a nonempty subset of X×X. For simplicity, we will write xy if (x,y)R, and we will say that is the binary relation. We will write xy when xy and xy, and we will write yx when xy. We will say that x and y are -comparable if xy or yx.

A binary relation on X is transitive if xz for all x,y,zX such that xy and yz. A preorder (or a quasi-order) on X is a binary relation on X that is reflexive (i.e., xx for all xX) and transitive. In such a case, we say that (X,) is a preordered space (or a preordered set). If a preorder is also antisymmetric (xy and yx implies x=y), then is called a partial order, and (X,) is a partially ordered space.

All partial orders and equivalence relations are preorders, but preorders are more general. From now on, (X,) will always denote a preordered space.

Definition 2.2 A metric on X is a mapping d:X×X[0,[ satisfying

( M 1 ) d ( x , x ) = 0 ; ( M 2 ) d ( x , y ) = 0 x = y ; ( M 3 ) d ( y , x ) = d ( x , y ) ; ( M 4 ) d ( x , y ) d ( z , x ) + d ( z , y )

for all x,y,zX. If d is a metric on X, we say that (X,d) is a metric space.

The function d is a premetric if it satisfies ( M 1 ); a pseudo-metric if it satisfies ( M 1 ), ( M 3 ) and ( M 4 ); a quasi-metric (or a nonsymmetric metric) if it satisfies ( M 1 ), ( M 2 ) and ( M 4 ); a quasi-pseudo-metric if it satisfies ( M 1 ) and ( M 4 ); and a semi-metric if it satisfies ( M 1 ), ( M 2 ) and ( M 3 ).

Remark 2.1 We point out that there exist different notions of premetric that are not universally accepted. For instance, Kasahara [26] used the term premetric to refer to a quasi-pseudo-metric defined on a subset of X×X. However, Kim [27], in the same issue as Kasahara, preferred using the term quasi-pseudo-metric (see also Reilly et al. [28]). For our purposes and for the sake of clarity, we prefer using the previous definitions because we consider that it is a more modern nomenclature.

Definition 2.3 A fixed point of a self-mapping T:XX is a point xX such that T(x)=x. A coincidence point between two mappings T,g:XY is a point xX such that T(x)=g(x). A common fixed point of T,g:XX is a point xX such that T(x)=g(x)=x.

Remark 2.2 If T,g:XX are commuting and x 0 X is a coincidence point of T and g, then T x 0 is also a coincidence point of T and g.

Definition 2.4 If (X,) is a preordered space and T,g:XX are two mappings, we will say that T is a (g,) -nondecreasing mapping if TxTy for all x,yX such that gxgy. If g is the identity mapping on X, T is nondecreasing (w.r.t. ).

3 An illustrative example

Let I be the real interval ]2,1] and let X=I{2,3,4,5,6,7} provided with the following binary relation:

for x,yX,xy { either x = y , or { x , y } = { 2 , 3 } , or ( x , y I  and  x y ) .

Define d:X×X[0,[, for all x,yX, by

d(x,y)= { 0 , if  { x , y } = { 2 , 3 }  or  { x , y } = { 4 , 5 } , 0.5 , if  x = y = 5 , 1 , if  ( x , y ) = ( 6 , 7 ) , 2 , if  ( x , y ) = ( 7 , 6 ) , | x y | , otherwise .

Let g the identity mapping on X. Define T:XX by

Tx= { x 2 , if  x I , 0 , if  x { 2 , 3 , 6 , 7 } , 1 , if  x = 4 , 1 , if  x = 5 , for all xX.

Then the following statements hold. Proofs can be found in Appendix 2.

  1. 1.

    The binary relation is a preorder on X, but it is not a partial order on X.

  2. 2.

    The measure mapping d does not hold any of the four classical properties ( M 1 )-( M 4 ) that define a metric space. Indeed, it is not a metric on X, neither a premetric nor any of the following: a pseudo-metric, a quasi-metric, a pseudo-quasi-metric, a semi-metric or a partial metric.

  3. 3.

    Even if d would verify some of the metric properties of Definition 2.2, (X,d) would not be a complete space.

  4. 4.

    T is not a d-contraction (that is, there is no k[0,1) such that d(Tx,Ty)kd(x,y) for all x,yX) because d(T4,T5)=d(1,1)=2, but d(4,5)=1.

Therefore, none of the theorems proved until now can be applied to the quadruple (X,,d,T) in order to guarantee that T has a fixed point.

4 Test functions

One of the most important ingredients of a contractivity condition is the kind of involved functions. Recently, many classes of families have been introduced, like altering distance functions, comparison functions, (c)-comparison functions, Geraghty functions, etc. In this section, we present the kind of functions we will use and we show how other classes can be seen as particular cases.

Definition 4.1 (Agarwal et al. [29])

We will denote by the family of all pairs (ψ,φ), where ψ,φ:[0,)[0,) are functions, verifying the following three conditions.

  • ( F 1 ) ψ is nondecreasing.

  • ( F 2 ) If there exists t 0 [0,) such that φ( t 0 )=0, then t 0 =0 and ψ 1 (0)={0}.

  • ( F 3 ) If { a k },{ b k }[0,) are sequences such that { a k }L, { b k }L and verifying L< b k and ψ( b k )(ψφ)( a k ) for all k, then L=0.

Example 4.1 If k[0,1) and we define ψ k (t)=t and φ k (t)=(1k)t for all t0, then ( ψ k , φ k )F. Furthermore, ψ k (t) φ k (t)=kt for all t0.

Notice that axiom ( F 2 ) does not necessarily imply the well known condition ψ(t)=0t=0φ(t)=0. Furthermore, we do not impose any continuity condition neither on ψ nor on φ. In order to prove that the family is very general, next we will show a variety of pairs of functions in that have been previously considered by other authors in the past.

A function ϕ:[0,)[0,) is lower semi-continuous if ϕ(t) lim inf n ϕ( t n ) for all sequence { t n }[0,) such that { t n }t. Similarly, ϕ is upper semi-continuous if, in the same conditions, ϕ(t) lim sup n ϕ( t n ).

Definition 4.2 (Khan et al. [30])

An altering distance function is a continuous, nondecreasing function ϕ:[0,)[0,) such that ϕ(t)=0 if and only if t=0.

Proposition 4.1 If ϕ is an altering distance function and { a m }[0,) verifies {ϕ( a m )}0, then { a m }0.

The following lemma shows some examples of pairs in .

Lemma 4.1 (see [29])

Let ψ,φ:[0,)[0,) be two functions such that ψ is an altering distance function.

  1. 1.

    If φ is lower semi-continuous and φ 1 ({0})={0}, then (ψ,φ)F.

  2. 2.

    If φ is continuous and verifies φ 1 ({0})={0}, then (ψ,φ)F.

  3. 3.

    If ψ and φ are altering distance functions, then (ψ,φ)F.

Notice that the condition φψ is not necessary.

Proof We prove item (1). Conditions ( F 1 ) and ( F 2 ) are obvious. Next, assume that { a k },{ b k }[0,) are sequences such that { a k }L, { b k }L and verify L< b k and ψ( b k )(ψφ)( a k ) for all k. Therefore, ψ( b k )(ψφ)( a k )=ψ( a k )φ( a k )ψ( a k ). Hence 0φ( a k )ψ( a k )ψ( b k ) for all k. Letting k and taking into account that ψ is continuous, we deduce that lim k φ( a k )=0. As { a k }L and φ is lower semi-continuous, we deduce that φ(L)lim inf t L φ(t) lim k φ( a k )=0. Hence L=0. The other two items immediately follow from item 1. □

Example 4.2 (see [29])

  1. 1.

    If a,b>0 and we define ψ(t)=at and φ(t)=bt for all t0, then (ψ,φ)F. The case ab is usually included in other papers, but the case a<b is new.

  2. 2.

    If ψ(t)=φ(t)=t+1 for all t0, then (ψ,φ)F. Notice that, in this case, ( F 3 ) holds because it is impossible to find such kind of sequences since 11+ b k =ψ( b k )(ψφ)( a k )=0. In this case, the condition ψ(t)=0t=0 does not hold.

Corollary 4.1 Let ψ,ϕ:[0,)[0,) be two functions such that ψ is an altering distance function and ϕ is upper semi-continuous verifying ϕ 1 ({0})={0} and ϕ(t)<ψ(t) for all t>0. Then (ψ,φ)F, where φ=ψϕ.

Proof It follows from item 1 of Lemma 4.1 because φ=ψϕ is lower semi-continuous and verifies φ 1 ({0})={0}. □

A function α:[0,)[0,1) is a Geraghty function if the condition {α( t n )}1 implies that { t n }0.

Lemma 4.2 (Agarwal et al. [29])

If α is a Geraghty function and we define ψ(t)=t and φ(t)=(1α(t))t for all t0, then (ψ,φ)F.

In [31], the authors used a contractivity condition as follows:

ψ ( d ( T x , T y ) ) β ( ψ ( d ( x , y ) ) ) ψ ( d ( x , y ) ) ,

where ψ is an altering distance function and β is a Geraghty function.

Lemma 4.3 If ψ is an altering distance function and β is a Geraghty function, then (ψ,(1βψ)ψ)F.

Proof Let φ=(1βψ)ψ, that is, φ(t)=(1β(ψ(t)))ψ(t) for all t0. Notice that the image of βψ is contained in the image of β, which is in [0,1). Therefore, β(ψ(s))ψ(s)ψ(s) for all s0 (if ψ(s)=0, both members are equal, and if ψ(s)>0, then β(ψ(s))ψ(s)<ϕ(s) since β(ψ(s))<1).

( F 1 ) Since ψ is an altering distance function, then it is nondecreasing.

( F 2 ) Assume that there exists t 0 [0,) such that φ( t 0 )=0. Then (1β(ψ( t 0 )))ψ( t 0 )=0. Since 1β(ψ( t 0 ))>0, then ψ( t 0 )=0, which means that t 0 =0. In such a case, ψ 1 (0)={0} because it is an altering distance function.

( F 3 ) Let { a k },{ b k }[0,) be sequences such that { a k }L, { b k }L and verify L< b k and ψ( b k )(ψφ)( a k ) for all k. Since ψ is continuous, lim k ψ( a k )= lim k ψ( b k )=ψ(L). Moreover,

ψ( b k )ψ( a k )φ( a k )=ψ( a k ) [ ψ ( a k ) β ( ψ ( a k ) ) ψ ( a k ) ] =β ( ψ ( a k ) ) ψ( a k ).

Let us show that L=0 reasoning by contradiction. Suppose that L>0. Since ψ is nondecreasing,

0<ψ(L)ψ( b k )β ( ψ ( a k ) ) ψ( a k )ψ( a k ).

In particular, ψ( a k )0 and

ψ ( b k ) ψ ( a k ) β ( ψ ( a k ) ) 1for all k.

Letting n, we deduce that {β(ψ( a k ))}1. Since β is a Geraghty function, ψ(L)= lim n ψ( a k )=0, which contradicts that ψ(L)>0 because L>0 and ψ is an altering distance function. □

In [32], the authors used a contractivity condition as follows:

ψ ( d ( T x , T y ) ) ψ ( N ( x , y ) ) φ ( N ( x , y ) ) , where  N ( x , y ) = max ( d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) + d ( y , T x ) 2 ) ,
(1)

ψ is continuous and φ verifies that {φ( t n )}0 implies that { t n }0.

Lemma 4.4 Let ψ,φ:[0,)[0,) be two functions such that ψ is an altering distance function and φ verifies the following condition:

if { t n }[0,) and  { φ ( t n ) } 0, then { t n }0.
(2)

Then (ψ,φ)F.

Proof ( F 1 ) Since ψ is an altering distance function, then it is nondecreasing.

( F 2 ) Assume that there exists t 0 [0,) such that φ( t 0 )=0. Letting t n = t 0 for all n1 and applying (2), we deduce that t 0 =0. In such a case, ψ 1 (0)={0} because it is an altering distance function.

( F 3 ) Let { a k },{ b k }[0,) be sequences such that { a k }L, { b k }L and verify L< b k and ψ( b k )(ψφ)( a k ) for all k. Hence 0φ( a k )ψ( a k )ψ( b k ) for all k. Letting k and taking into account that ψ is continuous, we deduce that lim k φ( a k )=0. By condition (2), L= lim k a k =0. □

A comparison function is a nondecreasing function ϕ:[0,)[0,) such that lim n ϕ n (t)=0 for all t>0.

Lemma 4.5 If ϕ is a continuous comparison function, and we define ψ(t)=t and φ(t)=tϕ(t) for all t0, then (ψ,φ)F.

Proof It is clear that every comparison function ϕ verifies ϕ(t)<t for all t>0. In such a case, if ϕ is continuous, then ϕ(0)=0, so ϕ(t)t for all t0. Moreover, if ϕ(t)=t, then t=0.

( F 1 ) Since ψ is an altering distance function, then it is nondecreasing.

( F 2 ) Assume that there exists t 0 [0,) such that φ( t 0 )=0. Then ϕ( t 0 )= t 0 , so t 0 =0. In such a case, ψ 1 (0)={0} because it is an altering distance function.

( F 3 ) Let { a k },{ b k }[0,) be sequences such that { a k }L, { b k }L and verify L< b k and ψ( b k )(ψφ)( a k ) for all k. Therefore

b k =ψ( b k )(ψφ)( a k )=ϕ( a k )< a k .

Letting k, we deduce that lim k ϕ( a k )=L. Therefore, as ϕ is continuous, ϕ(L)= lim k ϕ( a k )=L, which is only possible when L=0. □

In [33], Berzig et al. introduced the notion of pair of generalized altering distance functions, which is a pair (ψ,ϕ), where ψ,ϕ:[0,)[0,), verifying the following conditions:

  1. (a1)

    ψ is continuous;

  2. (a2)

    ψ is nondecreasing;

  3. (a3)

    (a3) lim n ϕ( t n )=0 lim n t n =0.

The condition (a3) was introduced by Popescu in [34] and Moradi and Farajzadeh in [35]. Notice that the above conditions do not determine the values ψ(0) and ϕ(0). If ϕ<ψ in (0,), then (ψ,φ=ψϕ)F. This is the case of Lemma 4.4.

After we have shown many different contexts in which some pairs of appear, we present some of their useful properties.

Lemma 4.6 (Agarwal et al. [29])

Let (ψ,φ)F.

  1. 1.

    If t,s[0,) and ψ(t)(ψφ)(s), then either t<s or t=s=0. In any case, ts.

  2. 2.

    If t[0,) and ψ(t)(ψφ)(t), then t=0.

  3. 3.

    If { a k },{ b k }[0,) are such that ψ( a k )(ψφ)( b k ) for all k and { b k }0, then { a k }0.

  4. 4.

    If { a k }[0,) and ψ( a k + 1 )(ψφ)( a k ) for all k, then { a k }0.

Remark 4.1 In [36], Berzig introduced the class of shifting distance functions, which are pairs of functions ψ,ϕ:[0,)R verifying the following two conditions:

  1. (i)

    for u,v[0,), if ψ(u)ϕ(v), then uv;

  2. (ii)

    for { u n },{ v n }[0,) with lim n u n = lim n v n =ω, if ψ( u n )ϕ( v n ) for all nN, then ω=0.

Pairs of functions in are intimately related with the class F shi of pairs of shifting distance functions, but they are different. On the one hand, pairs in F shi can take values in , but pairs in take values in [0,). On the other hand, if a pair (ψ,φ) verifies ( F 1 ) and ( F 2 ), then the pair (ψ,ϕ=ψφ) satisfies (i). Furthermore, if (ψ,ϕ=ψφ) satisfies (ii), then (ψ,φ) satisfies ( F 3 ).

5 A fixed point theorem without an underlying metric structure

The main aim of this section is to show sufficient conditions in order to ensure that T and g (given in Section 3) have a coincidence point. To set the framework, throughout this section, let (X,) be a preordered space, and let d:X×XR and T,g:XX be three mappings. The following definitions are usually considered when X has a metric structure. However, we do not suppose, a priori, any condition on the mapping d. Indeed, we will only be able to prove that d takes nonnegative values as a consequence of a particular version of the triangular inequality. However, in general, we do not consider necessary to assume this sign constraint.

Definition 5.1 We will say that a sequence { x m }X:

  • d-converges to x 0 X (and we will write { x m } d x 0 or simply { x m } x 0 ) if for all ε>0 there exists m 0 N such that d( x m , x 0 )ε for all m m 0 ;

  • is d-Cauchy if for all ε>0 there exists m 0 N such that d( x m , x m )ε for all m m m 0 .

We will say that (X,d) is complete if every d-Cauchy sequence in X is d-convergent in X.

With respect to the previous notions, the following remarks must be done.

Remark 5.1

  • When the distance measure d is not symmetric (that is, it does not verify axiom ( M 3 )), the definition of convergence or Cauchyness of sequences usually depends on the side, because d( x n , x m ) and d( x m , x n ) can be different. This is the case, for instance, of quasi-metric spaces. In such cases, the previous definitions correspond to the idea of right-convergence and right-Cauchyness (see Jleli and Samet [37]) because the more advanced term (which is nearer to the limit) is placed at the right argument of d. Similarly, it can be defined the notions of left-convergence (using d( x 0 , x m )) and left-Cauchyness (using d( x m , x m )) of sequences. Notice that some of the concepts we will present can also be introduced by the right side or by the left side. However, in order not to complicate the notation, we prefer avoiding the term right- in all definitions and theorems.

  • In [[38], Section 3], the authors did a complete study (completion, topology and powerdomains) of spaces verifying axioms ( M 1 ) and ( M 4 ), which they called generalized metric spaces. They solved the previous discussion using the terms forward convergent sequences (for right-convergent sequences) and backward convergent sequences (for left-convergent sequences). However, as we will not assume ( M 1 ) nor ( M 4 ), we also prefer avoiding these prefixes.

  • Notice that if d does not verify ( M 2 ), then the limit of a sequence, if there exists, might not be unique.

  • And if d only takes nonpositive values, then all sequences converge to all points.

Definition 5.2 We will say that a subset AX is (d,) -nondecreasing-closed if any d-limit of any -nondecreasing sequence of points of A is also in A.

[ { x m } A , { x m } d x X , x m x m + 1 , m ] xA.

Similarly can be defined the concepts of (d,) -nonincreasing-closed set and (d,) -monotone-closed set, and, more generally, a d-closed set, when any d-limit of any convergent sequence of points of A is also in A.

Definition 5.3 A mapping T:XX is (d,) -nondecreasing-continuous at x 0 X if we have that {T x m } d-converges to T x 0 for all -nondecreasing sequence { x m } d-convergent to x 0 .

In a similar way, the concepts of (d,) -nonincreasing-continuous mapping and (d,) -monotone-continuous mapping may be considered and, more generally, a d-continuous mapping, when {T x m } d-converges to T x 0 for all sequence { x m } d-convergent to x 0 .

Definition 5.4 We will say that a point xX is a d-precoincidence point of T and g if d(Tx,gx)=d(gx,Tx)=0.

In the following results, we will assume some of the following conditions.

  1. (a)

    T(X)g(X).

  2. (b)

    T is (g,)-nondecreasing.

  3. (c)

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

  4. (d)

    There exists (ψ,φ)F such that

    ψ ( d ( T x , T y ) ) (ψφ) ( d ( g x , g y ) ) for all x,yX for which gxgy.
    (3)
  5. (e)

    d(x,z)d(x,y)+d(y,z) for all x,y,zX such that xyz.

  6. (f)

    d(x,y)d(x,z)+d(y,z) for all x,y,zX such that xyz.

  7. (g)

    Every -nondecreasing, d-Cauchy sequence in X is d-convergent in X.

  8. (h)

    If { x m } is a nondecreasing sequence and { x m } d-converges to xX, then x m x for all m.

  9. (i)

    d(x,z)d(y,x)+d(y,z) for all x,y,zX such that yx and yz.

  10. (j)

    Every d-precoincidence point of T and g is a coincidence point of T and g (that is, if d(Tx,gx)=d(gx,Tx)=0, then Tx=gx).

As we have shown in Section 3, notice that a mapping d verifying (a)-(i) does not have to verify any of the conditions that define a metric on X.

Remark 5.2 A priori, d can take negative values in . However, condition (f) lets us prove some constraints about the sign of d. Indeed, if we take x=y=z in condition (f), we deduce that d(x,x)0 for all xX. Furthermore, if y=x in (f), it follows that d(x,z)0 for all x,zX such that xz. This does not mean that d is nonnegative because d could take negative values when zx or x and z are not -comparable.

Remark 5.3 As we shall show in the proofs, the mapping d could only be considered on the set Ω={(x,y) X 2 :xy}, that is, we will only use d | Ω :ΩR. In this case, the previous remark shows that, as usual, d(Ω)[0,[.

Theorem 5.1 Let (X,) be a preordered space and let T,g:XX be two mappings verifying (a)-(c). Then there exists a sequence { x m } m 0 such that g x m + 1 =T x m T x m + 1 =g x m + 2 for all m0. In particular,

g x 0 g x 1 g x 2 g x m 1 g x m g x m + 1 .

Proof Since T x 0 T(X)g(X), there exists x 1 X such that T x 0 =g x 1 . Then g x 0 T x 0 =g x 1 . Since T is (g,)-nondecreasing, T x 0 T x 1 . Now T x 1 T(X)g(X), so there exists x 2 X such that T x 1 =g x 2 . Then g x 1 =T x 0 T x 1 =g x 2 . Since T is (g,)-nondecreasing, T x 1 T x 2 . Repeating this argument, there exists a sequence { x m } m 0 such that g x m + 1 =T x m T x m + 1 =g x m + 2 for all m0. □

Theorem 5.2 Let (X,) be a preordered space and let d:X×XR and T,g:XX be three mappings verifying (a)-(e). Then any sequence { g x m } m 0 such that T x m =g x m + 1 for all m0, is d-Cauchy ({ x m } is given as in Theorem  5.1).

Proof Since g x m + 1 g x m + 2 for all m0, it follows from (d) that, for all m1,

ψ ( d ( g x m + 1 , g x m + 2 ) ) =ψ ( d ( T x m , T x m + 1 ) ) (ψφ) ( d ( g x m , g x m + 1 ) ) ψ ( d ( g x m , g x m + 1 ) ) .

By item 4 of Lemma 4.6, the sequence {d(g x m + 1 ,g x m + 2 )} d-converges to zero. Using the same reasoning, since g x m + 1 g x m + 1 for all m0, it follows that {d(g x m + 1 ,g x m + 1 )} also d-converges to zero. Therefore

{ d ( g x m , g x m ) } m 0 0and { d ( g x m , g x m + 1 ) } m 0 0.
(4)

Let us show that {g x m } is d-Cauchy reasoning by contradiction. Suppose that {g x m } is not d-Cauchy. Then there exist ε 0 >0 and partial subsequences {g x n ( k ) } and {g x m ( k ) } verifying

k < n ( k ) < m ( k ) < n ( k + 1 ) and d ( g x n ( k ) , g x m ( k ) 1 ) ε 0 < d ( g x n ( k ) , g x m ( k ) ) for all  k 1
(5)

(m(k) is the least integer number, greater than n(k), such that d(g x n ( k ) ,g x m ( k ) )> ε 0 ). Since n(k)m(k)1<m(k), we have g x n ( k ) g x m ( k ) 1 g x m ( k ) . By (e),

ε 0 <d(g x n ( k ) ,g x m ( k ) )d(g x n ( k ) ,g x m ( k ) 1 )+d(g x m ( k ) 1 ,g x m ( k ) ) ε 0 +d(g x m ( k ) 1 ,g x m ( k ) ).

Therefore, the sequence { b k = d ( g x n ( k ) , g x m ( k ) ) } k 1 satisfies

lim k b k = ε 0 and ε 0 < b k for all k.
(6)

Now, let us apply condition (e) and (5) to g x n ( k ) 1 g x n ( k ) g x m ( k ) 1 , and we deduce, for all k,

d(g x n ( k ) 1 ,g x m ( k ) 1 )d(g x n ( k ) 1 ,g x n ( k ) )+d(g x n ( k ) ,g x m ( k ) 1 )d(g x n ( k ) 1 ,g x n ( k ) )+ ε 0 .
(7)

By (d), we also have, for all k,

ψ ( d ( g x n ( k ) , g x m ( k ) ) ) =ψ ( d ( T x n ( k ) 1 , T x m ( k ) 1 ) ) (ψφ) ( d ( g x n ( k ) 1 , g x m ( k ) 1 ) ) .
(8)

By item 1 of Lemma 4.6, it follows that

d(g x n ( k ) ,g x m ( k ) )d(g x n ( k ) 1 ,g x m ( k ) 1 )for all k.

Joining this inequality and (7), we deduce that, for all k,

d(g x n ( k ) ,g x m ( k ) )d(g x n ( k ) 1 ,g x m ( k ) 1 )d(g x n ( k ) 1 ,g x n ( k ) )+ ε 0 .

Letting k and using (4) and (6), we deduce that the sequence { a k = d ( g x n ( k ) 1 , g x m ( k ) 1 ) } k 1 also verifies { a k } ε 0 , and by (8), we have that ψ( b k )(ψφ)( a k ) for all k. Since (ψ,φ)F, axiom ( F 3 ) guarantees that ε 0 =0, which is a contradiction with the fact that ε 0 >0. This contradiction shows that {g x m } is a d-Cauchy sequence. □

After the previous technical results, we give the main results of this manuscript.

Theorem 5.3 Let (X,) be a preordered space and let d:X×XR and T,g:XX be three mappings which fulfil conditions (a)-(h). Assume that the following condition holds.

(p) g(X) is (d,)-nondecreasing-closed.

Then there exists zX such that the sequence {g x m } (defined in Theorem  5.2) d-converges to gz and to Tz. Furthermore, if (i) holds, then z is a d-precoincidence point of T and g.

Notice that, in the previous result, g and T need not be continuous.

Proof Theorem 5.2 guarantees that {g x m } is d-Cauchy. Since it is -nondecreasing, condition (g) implies that there exists yX such that {g x m } d-converges to y (that is, {d(g x m ,y)}0). Moreover, since g(X) is (d,)-nondecreasing-closed, yg(X), so there exists zX such that y=gz. Applying (h), g x m + 1 y=gz for all m and, hence,

ψ ( d ( g x m + 2 , T z ) ) =ψ ( d ( T x m + 1 , T z ) ) (ψφ) ( d ( g x m + 1 , g z ) ) =(ψφ) ( d ( g x m + 1 , y ) ) .

Since {d(g x m ,y)}0, item 3 of Lemma 4.6 guarantees that {d(g x m + 2 ,Tz)}0, that is, {g x m + 2 } d-converges to Tz.

Now suppose that (i) holds. By (h), g x m + 2 Tz for all m and (i) implies that

d ( g z , T z ) d ( g x m + 2 , T z ) + d ( g x m + 2 , g z ) and d ( T z , g z ) d ( g x m + 2 , g z ) + d ( g x m + 2 , T z ) .
(9)

Therefore d(gz,Tz)=d(Tz,gz)=0. □

Theorem 5.4 Let (X,) be a preordered space and let d:X×XR and T,g:XX be three mappings which fulfil conditions (a)-(h). Suppose also:

  1. (p′)

    T and g are (d,)-nondecreasing-continuous and commuting and, at least, one of the following conditions holds:

    • ( p 1 ) T is a -nondecreasing mapping.

    • ( p 2 ) g is a -nondecreasing mapping.

    • ( p 3 ) If z,ωX and { z m }X is a -nondecreasing sequence such that {g z m } d-converges to z and to ω at the same time, then d(z,ω)=d(ω,z)=0.

Then there exists yX such that the sequence {g y m } (defined in Theorem  5.2) d-converges to gy and to Ty at the same time. Furthermore, if (i) holds, then y is a d-precoincidence point of T and g.

In any case, T and g have, at least, a d-precoincidence point.

Proof Theorem 5.2 guarantees that {g x m + 1 } is d-Cauchy. Since it is -nondecreasing, condition (g) implies that there exists yX such that {g x m } d-converges to y. Thus, taking into account that g and T are (d,)-nondecreasing-continuous, {gg x m } d-converges to gy and {Tg x m } d-converges to Ty. Furthermore, since T and g are commuting, Tg x m + 1 =gT x m + 1 =gg x m + 1 for all m, which means that {gg x m } d-converges, at the same time, to gy and to Ty.

Next, assume that (i) holds, and we claim that y is a d-precoincidence point of T and g. Firstly, if T (or g) is a -nondecreasing mapping, then the sequence {gg x m }={Tg x m 1 } is -nondecreasing. Since it d-converges to gy and to Ty, property (h) implies that gg x m gy and gg x m Ty for all m. Reasoning as in (9), we conclude that y is a d-precoincidence point of T and g. Secondly, if T and g are not necessarily -nondecreasing mappings, we could apply ( p 3 ) to the sequence {g x m } in order to deduce that d(gy,Ty)=d(Ty,gy)=0, that is, y is a d-precoincidence point of T and g. □

We summarize and improve all previous results in the following theorem.

Theorem 5.5 Let (X,) be a preordered space and let d:X×XR and T,g:XX be three mappings verifying the following properties.

  1. (a)

    T(X)g(X).

  2. (b)

    T is (g,)-nondecreasing.

  3. (c)

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

  4. (d)

    There exist (ψ,φ)F such that

    ψ ( d ( T x , T y ) ) (ψφ) ( d ( g x , g y ) ) for all x,yX for which gxgy.
  5. (e)

    d(x,z)d(x,y)+d(y,z) for all x,y,zX such that xyz.

  6. (f)

    d(x,y)d(x,z)+d(y,z) for all x,y,zX such that xyz.

  7. (g)

    Every -nondecreasing, d-Cauchy sequence in X is d-convergent in X.

  8. (h)

    If { x m } is a -nondecreasing sequence and { x m } d-converges to xX, then x m x for all m.

  9. (i)

    d(x,z)d(y,x)+d(y,z) for all x,y,zX such that yx and yz.

  10. (j)

    Every d-precoincidence point of T and g is a coincidence point of T and g.

Assume also either

  1. (p)

    g(X) is (d,)-nondecreasing-closed, or

(p′) T and g are (d,)-nondecreasing-continuous and commuting and, at least, one of the following conditions holds:

  • ( p 1 ) T is a -nondecreasing mapping.

  • ( p 2 ) g is a -nondecreasing mapping.

  • ( p 3 ) If z,ωX and { z m }X is a -nondecreasing sequence such that {g z m } d-converges to z and to ω at the same time, then d(z,ω)=d(ω,z)=0.

Then T and g have, at least, a coincidence point.

Notice that condition (j) does not mean that d(x,y)=0 implies x=y.

Remark 5.4 If g is the identity mapping on the ambient space, then the quadruple (X,,d,T) introduced in Section 3 verifies all conditions (a)-(j) and (p)-( p ) (see more details in Appendix 2).

Remark 5.5 Obviously, similar results can be stated changing the following hypothesis.

( b ˜ ) T is (g,)-nonincreasing.

( c ˜ ) There exists x 0 X such that g x 0 T x 0 .

( g ˜ ) Every nonincreasing, d-Cauchy sequence in X is d-convergent in X.

( h ˜ ) If { x m } is a nonincreasing sequence and { x m } d-converges to xX, then x m x for all m.

( p ˜ ) g(X) is (d,)-nonincreasing-closed.

( p ˜ ) T and g are (d,)-nonincreasing-continuous and commuting and, at least, one of the following conditions holds:

( p ˜ 1 ) T is a -nonincreasing mapping.

( p ˜ 2 ) g is a -nonincreasing mapping.

( p ˜ 3 ) If z,ωX and { z m }X is a -nonincreasing sequence such that {g z m } d-converges to z and to ω at the same time, then d(z,ω)=d(ω,z)=0.

The unicity of the coincidence point cannot be guaranteed unless additional conditions are imposed. A result in this direction is the following.

Theorem 5.6 Under the hypothesis of Theorem  5.5, let x,yX be two coincidence points of T and g verifying that there exists uX such that gugx and gugy. Then d(Tx,Ty)=d(Ty,Tx)=d(gx,gy)=d(gy,gx)=0.

Proof Define u 0 =u. Since T u 0 T(X)g(X), there exists u 1 X such that g u 1 =T u 0 . Repeating this process, there exists a sequence { u m } m 0 such that g u m + 1 =T u m for all m0. We claim that {g u m } d gx and {g u m } d gy. Firstly, we reason using x, but the same argument is valid for y.

Indeed, notice that g u 0 =gugx. As T is (g,)-nondecreasing, then T u 0 Tx=gx, which means that g u 1 gx. Again, g u 1 gx implies T u 1 Tx=gx, which means that g u 2 gx. By induction, it is possible to prove that g u m gx for all m0. Using condition (d), it follows that ψ(d(g u m + 1 ,gx))=ψ(d(T u m ,Tx))(ψφ)(d(g u m ,gx)) for all m0. Thus, by item 4 of Lemma 4.6, {d(g u m ,gx)}0. The same argument proves that g u m gy for all m0 and {d(g u m ,gy)}0. As a consequence, by (i), d(Tx,Ty)=d(gx,gy)d(g u m ,gx)+d(g u m ,gy) for all m, which lets us conclude that d(Tx,Ty)=d(gx,gy)=0. □

Corollary 5.1 Under the hypothesis of Theorem  5.5, assume the following conditions.

  • For all coincidence points x,yX of T and g, there exists uX such that gugx and gugy.

  • g is injective on the set of all coincidence points of T and g.

  • If z,ωT(X) verify d(z,ω)=d(ω,z)=0, then z=ω.

Then T and g have a unique coincidence point. Furthermore, if T and g are commuting, it is a common fixed point of T and g.

Proof Let x,yX be two coincidence points of T and g. Then gx=TxT(X) and gy=TyT(X). By Theorem 5.6, d(gx,gy)=d(gy,gx)=0. Therefore gx=gy. As g is injective on the set of all coincidence points of T and g, we conclude that x=y.

Now let xX be a coincidence point of T and g, and let z=Tx. By Remark 2.2, z is also a coincidence point of T and g. Then x=z=Tx=gx, so x is a common fixed point of T and g. □

Taking ψ(t)=t and φ(t)=(1k)t for all t0 in the previous results, we obtain the following particular case.

Corollary 5.2 Theorems 5.1, 5.2, 5.3, 5.4, 5.5, 5.6 and Corollary  5.1 also hold if we replace condition (d) by the following one.

( d ) There exists k[0,1) such that d(Tx,Ty)kd(gx,gy) for all x,yX for which gxgy.

6 Consequences

This section is devoted to show how to apply Theorem 5.5 in many different contexts, and how to deduce unidimensional, coupled, tripled, quadruple and multidimensional fixed point theorems (for completeness, they are included in Appendix 1).

6.1 Fixed/coincidence point theorems in partially ordered metric spaces

In this subsection we show that different results in partially ordered metric spaces, including unidimensional, coupled, tripled, quadruple and multidimensional fixed point theorems, can be seen as simple consequences of Theorem 5.5.

Corollary 6.1 Theorems A.1, A.2 and A.3 follow from Theorem  5.5.

However, Theorems A.1 and A.2 cannot be applied to the example of Section 3.

Corollary 6.2 Theorem  A.4 follows from Theorem  5.5.

Proof Let Y= X 2 provided with the partial order (x,y)(u,v) if and only if xu and yv, and the metric δ:Y×Y R 0 + given by δ((x,y),(u,v))=max(d(x,u),d(y,v)) for all (x,y),(u,v)Y. Define T F :YY by T F (x,y)=(F(x,y),F(y,x)) for all (x,y)Y, and let G be the identity mapping on Y. Let X 0 =( x 0 , y 0 )Y. Then the hypothesis of Theorem A.4 implies the hypothesis of Theorem 5.5 (for instance, T F is (,G)-nondecreasing because F has the mixed -monotone property). The contractivity condition holds since, if (x,y)(u,v),

δ ( T F ( x , y ) , T F ( u , v ) ) = δ ( ( F ( x , y ) , F ( y , x ) ) , ( F ( u , v ) , F ( v , u ) ) ) = max ( d ( F ( x , y ) , F ( u , v ) ) , d ( F ( y , x ) , F ( v , u ) ) ) = max ( d ( F ( x , y ) , F ( u , v ) ) , d ( F ( v , u ) , F ( y , x ) ) ) max ( k 2 ( d ( x , u ) + d ( y , v ) ) , k 2 ( d ( v , y ) + d ( u , x ) ) ) k 2 2 max ( d ( x , u ) , d ( y , v ) ) = k δ ( ( x , y ) , ( u , v ) ) .

Theorem 5.5 assures us that T F and G have a coincidence point, that is, F has a coupled fixed point. □

Tripled, quadruple and multidimensional theorems can be proved similarly using X 3 , X 4 and X n , obtaining the following result.

Corollary 6.3 Theorems A.5, A.6 and A.7 follow from Theorem  5.5.

We remark that the techniques used in this paper might be applied in order to prove other coupled, tripled, quadruple, n-tupled fixed point theorems in the framework of various abstract spaces, e.g., partial metric spaces, cone metric spaces, fuzzy metric spaces, b-metric spaces, etc.

6.2 Fixed/coincidence point theorems in quasi-metric spaces

Recall that a mapping q:X×X[0,) is a quasi-metric on X if it satisfies ( M 1 ), ( M 2 ) and ( M 4 ), that is, if it verifies, for all x,y,zX:

  • ( q 1 ) q(x,y)=0 if and only if, x=y,

  • ( q 2 ) q(x,y)q(x,z)+q(z,y).

In such a case, the pair (X,q) is called a quasi-metric space. Some preliminaries about convergence, Cauchy sequences and completeness in quasi-metric spaces can be found in [29, 37].

Theorem 6.1 Let (X,q) be a complete quasi-metric space and let T,g:XX be given mappings. Suppose that T(X)g(X) and that there exists (ψ,φ)F such that

ψ ( q ( T x , T y ) ) ψ ( q ( g x , g y ) ) φ ( q ( g x , g y ) ) for all x,yX.

Then T and g have a unique coincidence point.

Proof Assume that is the preorder in X given by xy for all x,yX, that is, all points are -comparable. Then all conditions of Theorem 5.5 hold. Moreover, as g(X) is (d,)-nondecreasing-closed, we deduce that T and g have, at least, a coincidence point. Furthermore, if u and v are two distinct coincidence points of T and g, then

ψ ( q ( g u , g v ) ) =ψ ( q ( T u , T v ) ) ψ ( q ( g u , g v ) ) φ ( q ( g y , g v ) ) <ψ ( q ( g u , g v ) ) ,

which is a contradiction. □

If g is the identity mapping on X, we have the following statement.

Corollary 6.4 Let (X,q) be a complete quasi-metric space and let T:XX be a given mapping. Suppose that there exists (ψ,φ)F such that

ψ ( q ( T x , T y ) ) ψ ( q ( x , y ) ) φ ( q ( x , y ) ) for all x,yX.

Then T has a unique fixed point.

If ψ(t)=t for all t0, we have the following particular case.

Corollary 6.5 (Jleli and Samet [37], Theorem 3.2)

Let (X,q) be a complete quasi-metric space and let T:XX be a mapping satisfying

q(Tx,Ty)q(x,y)φ ( q ( x , y ) ) for all x,yX,

where φ:[0,)[0,) is continuous with φ 1 (0)={0}. Then T has a unique fixed point.

6.3 Fixed/coincidence point theorems in G-metric spaces

Following [7, 39], recall that a generalized metric on X (or, more specifically, a G-metric on X) is a mapping G:X×X×X[0,) verifying the following properties.

Definition 6.1

( G 1 ) G(x,y,z)=0 if x=y=z;

( G 2 ) 0<G(x,x,y) for all x,yX with xy;

( G 3 ) G(x,x,y)G(x,y,z) for all x,y,zX with 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) for all x,y,z,aX.

In such a case, the pair (X,G) is called a G-metric space. Some preliminaries about convergence, Cauchy sequences and completeness in quasi-metric spaces can be found in [7, 29, 37, 39].

Lemma 6.1 (see, e.g., [29, 37])

Let (X,G) be a G-metric space and let us define q G , q G : X 2 [0,) by

q G (x,y)=G(x,y,y)and q G (x,y)=G(x,x,y)for all x,yX.

Then the following properties hold.

  1. 1.

    q G and q G are quasi-metrics on X. Moreover,

    q G (x,y)2 q G (x,y)4 q G (x,y)for all x,yX.
    (10)
  2. 2.

    In (X, q G ) and in (X, q G ), a sequence is right-convergent (respectively, left-convergent) if and only if it is convergent. In such a case, its right-limit, its left-limit and its limit coincide.

  3. 3.

    In (X, q G ) and in (X, q G ), a sequence is right-Cauchy (respectively, left-Cauchy) if and only if it is Cauchy.

  4. 4.

    In (X, q G ) and in (X, q G ), every right-convergent (respectively, left-convergent) sequence has a unique right-limit (respectively, left-limit).

  5. 5.

    If { x n }X and xX, then { x n } G x{ x n } q G x{ x n } q G x.

  6. 6.

    If { x n }X, then { x n } is G-Cauchy { x n } is q G -Cauchy { x n } is q G -Cauchy.

  7. 7.

    (X,G) is complete (X, q G ) is complete (X, q G ) is complete.

Theorem 6.2 Let (X,G) be a complete G-metric space and let T,g:XX be given mappings. Suppose that T(X)g(X) and that there exists (ψ,φ)F such that

ψ ( G ( T x , T y , T y ) ) ψ ( G ( g x , g y , g y ) ) φ ( G ( g x , g y , g y ) ) for all x,yX.

Then T and g have a unique coincidence point.

Proof It follows from Theorem 6.1 applied to the quasi-metric q G (x,y)=G(x,y,y) for all x,yX (as in Lemma 6.1). □

To conclude the paper, we include two appendices: in the first one, we recall some celebrated theorems that can be seem as particular cases of our main results; in the second one, we prove the statements announced in Section 3 and why our results can be applied.

Appendix 1: Some recent results we generalize

The following statements are well-known fixed point theorems in partially ordered metric spaces.

Theorem A.1 (Ran and Reurings [17])

Let (X,) be an ordered set endowed with a metric d and T:XX be a given mapping. Suppose that the following conditions hold:

  1. (a)

    (X,d) is complete.

  2. (b)

    T is nondecreasing (w.r.t. ).

  3. (c)

    T is continuous.

  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 d(Tx,Ty)kd(x,y) for all x,yX with xy.

Then T has a fixed point. Moreover, if for all (x,y) X 2 there exists zX such that xz and yz, we obtain uniqueness of the fixed point.

Nieto and Rodríguez-López [18] slightly modified the hypothesis of the previous result obtaining the following theorem.

Theorem A.2 (Nieto and Rodríguez-López [18])

Let (X,) be an ordered set endowed with a metric d and T:XX be a given mapping. Suppose that the following conditions hold:

  1. (a)

    (X,d) is complete.

  2. (b)

    T is nondecreasing (w.r.t. ).

  3. (c)

    If a nondecreasing sequence { x m } in X converges to some point xX, then x m x for all m.

  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 d(Tx,Ty)kd(x,y) for all x,yX with xy.

Then T has a fixed point. Moreover, if for all (x,y) X 2 there exists zX such that xz and yz, we obtain uniqueness of the fixed point.

Theorem A.3 (Harjani and Sadarangani [40])

Let (X,) be a partially ordered set and suppose that there exists a metric d in X such that (X,d) is a complete metric space. Let T:XX be a nondecreasing mapping such that

ψ ( d ( T x , T y ) ) ψ ( d ( x , y ) ) ϕ ( d ( x , y ) ) for all xy,

where ψ and ϕ are altering distance functions. Also assume that, at least, one of the following conditions holds.

  1. (i)

    T is continuous, or

  2. (ii)

    if a nondecreasing sequence { x n } in X converges to some point xX, then x n x for all n.

If there exists x 0 X with x 0 T x 0 , then T has a fixed point.

Theorem A.4 (Bhaskar and Lakshmikantham [19])

Let (X,) be a partially ordered set endowed with a metric d. Let F:X×XX be a given mapping. Suppose that the following conditions hold:

  1. (i)

    (X,d) is complete;

  2. (ii)

    F has the mixed monotone property;

  3. (iii)

    F is continuous or X has the following properties:

    • ( X 1 ) if a nondecreasing sequence { x n } in X converges to some point xX, then x n x for all n,

    • ( X 2 ) if a decreasing sequence { y n } in X converges to some point yX, then y n y for all n;

  4. (iv)

    there exist x 0 , y 0 X such that x 0 F( x 0 , y 0 ) and y 0 F( y 0 , x 0 );

  5. (v)

    there exists a constant k(0,1) such that for all (x,y),(u,v)X×X with xu and yv,

    d ( F ( x , y ) , F ( u , v ) ) k 2 [ d ( x , u ) + d ( y , v ) ] .

Then F has a coupled fixed point ( x , y )X×X. Moreover, if for all (x,y),(u,v)X×X there exists ( z 1 , z 2 )X×X such that (x,y) 2 ( z 1 , z 2 ) and (u,v) 2 ( z 1 , z 2 ), we have uniqueness of the coupled fixed point and x = y .

Theorem A.5 (Berinde and Borcut [22])

Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let F:X×X×XX be a mapping having the mixed g-monotone property. Assume that there exist constants j,k,[0,1) with j+k+<1 such that

d ( F ( x , y , z ) , F ( u , v , w ) ) jd(x,u)+kd(y,v)+d(z,w)

for all x,y,z,u,v,wX with xu, yv, zw. Suppose that either F is continuous or (X,d,) has the following properties:

  1. (a)

    if a nondecreasing sequence { x m }x, then x m x for all m;

  2. (b)

    if a nondecreasing sequence { y m }y, then y m y for all m.

If there exist x 0 , y 0 , z 0 X such that

x 0 F( x 0 , y 0 , z 0 ), y 0 F( y 0 , x 0 , y 0 )and z 0 F( z 0 , y 0 , x 0 ),

then there exist x,y,zX such that

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

A quadruple version was obtained by Karapınar and Luong in [23].

Theorem A.6 (Karapınar and Luong [23])

Let (X,) be a partially ordered set and (X,d) be a complete metric space. Let F:X×X×X×XX be a mapping having the mixed monotone property. Assume that there exists a constant k[0,1) such that

d ( F ( x , y , z , w ) , F ( u , v , r , t ) ) k 4 [ d ( x , u ) + d ( y , v ) + d ( z , r ) + d ( w , t ) ]

for all x,y,z,u,v,wX with xu, yv, zr and wt. Suppose that there exist x 0 , y 0 , z 0 , w 0 X such that

x 0 F ( x 0 , y 0 , z 0 , w 0 ) , y 0 F ( y 0 , z 0 , w 0 , x 0 ) , z 0 F ( z 0 , w 0 , x 0 , y 0 ) and w 0 F ( w 0 , x 0 , y 0 , z 0 ) .

Suppose that either F is continuous or (X,d,) has the following properties:

  1. (a)

    if a nondecreasing sequence { x m }x, then x m x for all m;

  2. (b)

    if a nondecreasing sequence { y m }y, then y m y for all m.

Then there exist x,y,z,wX such that

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

Later, Berzig and Samet extended the previous result to the multidimensional case in the following way.

Theorem A.7 (Berzig and Samet [24])

Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. For N, m positive integers, N2, 1m<N, let F: X N X be a continuous mapping having the m-mixed monotone property. Assume that there exist the constants δ i [0,1) with i = 1 N δ i <1 for which

d ( F ( U ) , F ( V ) ) i = 1 N δ i d( x i , y i )

for all U=( x 1 ,, x N ),V=( y 1 ,, y N ) X N such that

x 1 y 1 ,, x m y m , x m + 1 y m + 1 ,, x N y N .

If there exists U ( 0 ) =( x 1 ( 0 ) ,, x N ( 0 ) ) X N such that

x 1 ( 0 ) F ( x ( 0 ) [ φ 1 ( 1 : m ) ] , x ( 0 ) [ ψ 1 ( m + 1 : N ) ] ) , x m ( 0 ) F ( x ( 0 ) [ φ m ( 1 : m ) ] , x ( 0 ) [ ψ m ( m + 1 : N ) ] ) , x m + 1 ( 0 ) F ( x ( 0 ) [ φ m + 1 ( 1 : m ) ] , x ( 0 ) [ ψ m + 1 ( m + 1 : N ) ] ) , x N ( 0 ) F ( x ( 0 ) [ φ N ( 1 : m ) ] , x ( 0 ) [ ψ N ( m + 1 : N ) ] ) ,

where φ 1 ,, φ m :{1,,m}{1,,m}, ψ 1 ,, ψ m :{m+1,,N}{m+1,,N}, φ m + 1 ,, φ N :{1,,m}{m+1,,N}, and ψ m + 1 ,, ψ N :{m+1,,N}{1,,m}, then there exists ( x 1 , x 2 ,, x N ) X N satisfying

x 1 = F ( x [ φ 1 ( 1 : m ) ] , x [ ψ 1 ( m + 1 : N ) ] ) , x m = F ( x [ φ m ( 1 : m ) ] , x [ ψ m ( m + 1 : N ) ] ) , x m + 1 = F ( x [ φ m + 1 ( 1 : m ) ] , x [ ψ m + 1 ( m + 1 : N ) ] ) , x N = F ( x [ φ N ( 1 : m ) ] , x [ ψ N ( m + 1 : N ) ] ) .

Theorem A.8 (Dutta and Choudhury [41])

Let (X,d) be a complete metric space and let T:XX be a self-mapping satisfying the inequality

ψ ( d ( T x , T y ) ) ψ ( d ( x , y ) ) φ ( d ( x , y ) )

for all x,yX, where ψ,φ:[0,[ are both continuous and monotone nondecreasing functions with ψ(t)=φ(t)=0 if and only if t=0. Then T has a unique fixed point.

Theorem A.9 (Luong and Thuan [42])

Let (X,) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX be a mapping having the mixed monotone property on X such that there exist two elements x 0 , y 0 X with x 0 F( x 0 , y 0 ) and y 0 F( y 0 , x 0 ). Suppose that there exist ψΨ and φΦ such that

ψ ( d ( F ( x , y ) , F ( u , v ) ) ) 1 2 ψ ( d ( x , u ) + d ( y , v ) ) φ ( d ( x , u ) + d ( y , v ) 2 )

for all x,y,u,vX with xu and yv. Suppose either

  1. (a)

    F is continuous, or

  2. (b)

    X has the following properties:

    1. (i)

      if a nondecreasing sequence { x m }x, then x m x for all m,

    2. (ii)

      if a nonincreasing sequence { y m }y, then y m y for all m.

Then there exist x,yX such that x=F(x,y) and y=F(y,x), that is, F has a coupled fixed point in X.

Appendix 2: Proof of statements of Section 3 and Remark 5.4

Recall that I=]2,1], X=I{2,3,4,5,6,7},

xy { either x = y , or { x , y } = { 2 , 3 } , or ( x , y I  and  x y ) ,

and d:X×X[0,[ and T:XX are defined by

d(x,y)= { 0 , if  { x , y } = { 2 , 3 }  or  { x , y } = { 4 , 5 } , 0.5 , if  x = y = 5 , 1 , if  ( x , y ) = ( 6 , 7 ) , 2 , if  ( x , y ) = ( 7 , 6 ) , | x y | , otherwise . Tx= { x 2 , if  x I , 0 , if  x { 2 , 3 , 6 , 7 } , 1 , if  x = 4 , 1 , if  x = 5 .

We firstly prove some properties of this space.

( P 1 ) The binary relation is a preorder on X (reflexive and transitive), but it is not a partial order since 23, 32 and 23. Furthermore, if x,yX and xy, then

xIyI;x{2,3}y{2,3};x{4,5,6,7}y=x.

If xyz, then either x,y,zI or x,y,z{2,3} or x=y=z{4,5,6,7}. The same is true if yx and yz.

( P 2 ) d does not verify ( M 1 ) because d(5,5)=0.5.

( P 3 ) d does not verify ( M 2 ) since d(2,3)=0 but 23.

( P 4 ) d does not verify ( M 3 ) since d(6,7)=1 and d(7,6)=2.

( P 5 ) d does not verify ( M 4 ): if x=2, y=4 and z=3, then d(x,y)=2 and d(z,x)+d(z,y)=1.

( P 6 ) (X,d) is not complete. Since I=]2,1]X and d | I × I is the Euclidean metric on I, then the sequence { x m = 2 + 1 / m } m N is d-Cauchy but it is not d-convergent in X.

( P 7 ) If { x m } is a -nondecreasing sequence in X, then one, and only one, of the following cases holds.

x m I for all m.

x m {2,3} for all m.

There exists z 0 {4,5,6,7} such that x m = z 0 for all m (that is, x m is a constant sequence).

It follows from that fact that points of I (respectively, {2,3}, {4,5,6,7}) are only -related with points of I (respectively, {2,3}, themselves), and x 1 x m for all mN.

( P 8 ) If { x m } is a -nondecreasing, d-convergent sequence in X, then one, and only one, of the following cases holds.

x m I for all m. In this case, its d-limit is also in I.

x m {2,3} for all m. In this case, its d-limit is also in {2,3}.

There exists z 0 {4,6,7} such that x m = z 0 for all m. In this case, its d-limit is z 0 .

It follows from ( P 7 ). Notice that z 0 5 since d(5,5)=0.5>0.

( P 9 ) T is not a d-contraction (that is, there is no k[0,1) such that d(Tx,Ty)kd(x,y) for all x,yX) because d(T4,T5)=d(1,1)=2 but d(4,5)=1.

Now we prove assertions (a)-(j), (p) and (p′) taking into account that g is the identity mapping on X.

  1. (a)

    T(X)X. It is obvious.

  2. (b)

    T is -nondecreasing. Let x,yX be such that xy and xy. By ( P 1 ), if xI, then yI and xy. Then Tx=x/2y/2=Ty, being Tx,TyI, so TxTy. If x{2,3}, then y{2,3}, so Tx=Ty. The case x{4,5,6,7} is impossible since yx.

  3. (c)

    There exists x 0 X such that x 0 T x 0 . If x 0 =1, then x 0 =11/2=T x 0 .

  4. (d)

    There exists k[0,1[ such that d(Tx,Ty)kd(x,y) for all x,yX for which xy. Let k=1/2 and assume that x,yX are such that xy. By ( P 1 ), if xI, then y,Tx,TyI and d(Tx,Ty)=|x/2y/2|=|xy|/2=(1/2)d(x,y). If x{2,3}, then y{2,3} and d(Tx,Ty)=0=d(x,y). If x{4,5,6,7}, then y=x and Tx=Ty{1,0,1}I. Therefore d(Tx,Ty)=0.

  5. (e)

    d(x,z)d(x,y)+d(y,z) for all x,y,zX such that xyz. By ( P 1 ), there are only three cases. If x,y,zI, the triangular inequality holds because it is true in provided with the Euclidean metric. If x,y,z{2,3}, then all distances are zero. Finally, if x=y=z{4,5,6,7}, all distances are either zero (if x=y=z{4,6,7}) or 0.5 (if x=y=z=5).

  6. (f)

    d(x,y)d(x,z)+d(y,z) for all x,y,zX such that xyz. It is similar to the previous one using ( P 1 ).

  7. (g)

    Every d-Cauchy, -nondecreasing sequence in X is d-convergent. It follows from ( P 7 ).

  8. (h)

    If { x m } is a -nondecreasing sequence and { x m } d-converges to xX, then x m x for all m. It also follows from ( P 7 ).

  9. (i)

    d(x,z)d(y,x)+d(y,z) for all x,y,zX such that yx and yz. It is similar to (e) using ( P 1 ).

  10. (j)

    Every d-precoincidence point of T and g is a coincidence point of T and g. Assume that d(Tx,gx)=d(gx,Tx)=0. Since TxT(X)I, it is impossible Tx,gx{2,3} or Tx,gx{4,5} (the only cases, far from I, in which the distance between them can be zero). Then d(Tx,gx)=0 implies that Tx,gxI and 0=d(Tx,gx)=|Txgx|, so Tx=gx.

  11. (p)

    g(X)=X is -nondecreasing d-closed. It is immediate: indeed, g(X)=X is d-closed.

(p′) T and g are (d,) -nondecreasing-continuous and commuting, and g is -nondecreasing. It is only necessary to prove that T is (d,)-nondecreasing-continuous. Actually, it follows from ( P 8 ) since there are only three cases: if x m I for all m, its d-limit x 0 is also in I, and {T x m = x m /2} d-converges to T x 0 = x 0 /2; if x m {2,3} for all m, then x 0 {2,3} and T x m =0=T x 0 for all m; finally, if there exists z 0 {4,6,7} such that x m = z 0 for all m, then x 0 = z 0 = x m for all m.

We point out that T is not d-continuous: if x m =4 and x 0 =5, then { x m } x 0 but {T x m =T4=1} does not d-converge to T5=1.

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Acknowledgements

This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The second author acknowledges with thanks DSR for financial support. The first author has been partially supported by Junta de Andalucía by project FQM-268 of the Andalusian CICYE.

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Roldán-López-de-Hierro, AF., Shahzad, N. Some fixed/coincidence point theorems under (ψ,φ)-contractivity conditions without an underlying metric structure. Fixed Point Theory Appl 2014, 218 (2014). https://doi.org/10.1186/1687-1812-2014-218

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Keywords

  • Binary Relation
  • Fixed Point Theorem
  • Common Fixed Point
  • Coincidence Point
  • Unique Fixed Point