# Multidimensional Fixed-Point Theorems in Partially Ordered Complete Partial Metric Spaces under (ψ,φ)-Contractivity Conditions

Authors: A. Roldán; J. Martínez-Moreno; C. Roldán; E. Karapınar
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the existence and uniqueness of coincidence point for nonlinear mappings of any number of arguments under a weak (ψ,φ)-contractivity condition in partial metric spaces. The results we obtain generalize, extend, and unify several classical and very recent related results in the literature in metric spaces (see Aydi et al. (2011), Berinde and Borcut (2011), Gnana Bhaskar and Lakshmikantham (2006), Berzig and Samet (2012), Borcut and Berinde (2012), Choudhury et al. (2011),  Karapınar and Luong (2012), Lakshmikantham and Ćirić (2009), Luong and Thuan (2011), and Roldán et al. (2012)) and in partial metric spaces (see Shatanawi et al. (2012)).


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## Body

## 1. Introduction

The notion ofcoupled fixed point was introduced by Guo and Lakshmikantham [1] in 1987. In a recent paper, Gnana Bhaskar and Lakshmikantham [2] introduced the concept mixed monotone property for contractive operators of the form F:X×X→X, where X is a partially ordered metric space, and then established some coupled fixed-point theorems. After that, many results appeared on coupled fixed-point theory in different contexts (see, e.g., [3–6]). Later, Berinde and Borcut [7] introduced the concept of tripled fixed point and proved tripled fixed-point theorems using mixed monotone mappings (see also [8–10]).Very recently, Roldán et al. [11] proposed the notion of coincidence point between mappings in any number of variables and showed some existence and uniqueness theorems that extended the mentioned previous results for this kind of nonlinear mappings, not necessarily permuted or ordered, in the framework of partially ordered complete metric spaces, using a weaker contraction condition, that also generalized other works by Berzig and Samet [12], Karapınar and Berinde [13].Partial metric spaces were firstly introduced by Matthews in [14] as an attempt to generalize the metric spaces by establishing the condition that the distance between a point to itself (which is not necessarily zero) is less or equal than the distance between that point and another point of the space. In the mentioned papers, Matthews studied topological properties of partial metric spaces and stated a modified version of a Banach contraction mapping principle on this kind of spaces. After Matthews' pioneering work, the theory of partial metric spaces and particularly the field of fixed-point theorems have expansively been developed due to the increasing interest in this area and motivated by its possible applications (see [15, 16] and references therein).In this paper, our main aim is to study a weaker contractivity condition for nonlinear mappings of any number of arguments. This condition can be particularized in a variety of forms that let us extend the previously mentioned results and other recent ones in this field (see [2, 5, 7, 9, 11, 12, 16–20]). We also notice that our results cannot be obtained by the very recent paper of Haghi et al. [21] (for more details see Remark 26).


## 2. Preliminaries

Preliminaries and notation about coincidence points can also be found in [11]. Let n be a positive integer. Henceforth, X will denote a nonempty set, and Xn will denote the product space Xn=X×X×…n×X. Throughout this paper, m and k will denote nonnegative integers and i,j,s∈{1,2,…,n}. Unless otherwise stated, “for all m” will mean “for all m≥0”, and “for all i” will mean “for all i∈{1,2,…,n}”. Let ℝ0+=[0,∞[.Ametric onX is a mapping d:X×X→ℝ satisfying, for all x,y,z∈X: (i)
d(x,y)=0 if, and only if, x=y;(ii)
d(x,y)≤d(z,x)+d(z,y).From these properties, we can easily deduce thatd(x,y)≥0 and d(y,x)=d(x,y) for all x,y∈X. The last requirement is called the triangle inequality. If d is a metric on X, we say that (X,d) is a metric space (for short, an MS).Definition 1 (see [22]).
A triple(X,d,≤) is called a partially ordered metric space if (X,d) is a MS and ≤ is a partial order on X.Definition 2 (see [2]).
An ordered MS(X,d,≤) is said to have the sequentialg-monotone property if it verifies(i)
if{xm} is a nondecreasing sequence and {xm}→dx, then gxm≤gx for all m;(ii)
if{ym} is a nonincreasing sequence and {ym}→dy, then gym≥gy for all m.Ifg is the identity mapping, then X is said to have the sequential monotone property.Henceforth, fix a partition{A,B} of two non-empty subsets of Λn={1,2,…,n}; that is, A∪B=Λn and A∩B=∅. We will denote
(1)ΩA,B={σ:Λn→Λn:σ(A)⊆Aandσ(B)⊆B},ΩA,B′={σ:Λn→Λn:σ(A)⊆Bandσ(B)⊆A}.
If (X,≤) is a partially ordered space, x,y∈X, and i∈Λn, we will use the following notation:
(2)x≤iy⟺{x≤y,ifi∈A,x≥y,ifi∈B.LetF:Xn→X and g:X→X be two mappings.Definition 3 (see [11]).
One says thatF and g are commuting if gF(x1,…,xn)=F(gx1,…,gxn) for all x1,…,xn∈X.Definition 4 (see [11]).
Let(X,≤) be a partially ordered space. One says that F has the mixedg-monotone property (with respect to {A,B}) if F is g-monotone nondecreasing in arguments of A and g-monotone nonincreasing in arguments of B; that is, for all x1,x2,…,xn,y,z∈X and all i,
(3)gy≤gz⟹F(x1,…,xi-1,y,xi+1,…,xn)≤iF(x1,…,xi-1,z,xi+1,…,xn).
Henceforth, letσ1,σ2,…,σn,τ:Λn→Λn be n+1 mappings from Λn into itself, and let Φ be the (n+1)-tuple (σ1,σ2,…,σn,τ).Definition 5 (see [11]).
A point(x1,x2,…,xn)∈Xn is called a Φ-coincidence point of the mappingsFandg if
(4)F(xσi(1),xσi(2),…,xσi(n))=gxτ(i)∀i.
If g is the identity mapping on X, then (x1,x2,…,xn)∈Xn is called a Φ-fixed point of the mappingF.Remark 6.
IfF and g are commuting and (x1,x2,…,xn)∈Xn is a Φ-coincidence point of F and g, then (gx1,gx2,…,gxn) also is a Φ-coincidence point of F and g.Definition 7 (see [14]).
Apartial metric onX is a mapping p:X×X→ℝ0+ verifying, for all x,y,z∈X:(P1)  p(x,x)≤p(x,y);(P2)  p(x,x)=p(x,y)=p(y,y)⇒x=y;(P3)  p(x,y)=p(y,x);(P4)  p(x,z)+p(y,y)≤p(x,y)+p(y,z).
In this case,(X,p) is a partial metric space (for short, a PMS).Example 8 (see, e.g., [14]).
LetX=ℝ0+, and define p on X by p(x,y)=max{x,y} for all x,y∈X. Then, (X,p) is a partial metric space.Example 9 (see [14]).
LetX={[a,b]:a,b∈ℝ,a≤b}, and define p([a,b],[c,d]) = max{b,d}-min{a,c}. Then, (X,p) is a partial metric space.Example 10 (see [14]).
LetX=[0,1]∪[2,3], and define p:X×X→ℝ0+ by
(5)p(x,y)={max{x,y},if{x,y}∩[2,3]≠∅,|x-y|,if{x,y}⊂[0,1].
Then, (X,p) is a partial metric space.Example 11 (see, e.g., [23, 24]).
Let(X,d) and (X,p) be a metric space and a partial metric space, respectively. Functions ρi:X×X→ℝ0+(i∈{1,2,3}) given by
(6)ρ1(x,y)=d(x,y)+p(x,y),ρ2(x,y)=d(x,y)+max{u(x),u(y)},ρ3(x,y)=d(x,y)+a,
define partial metrics on X, where u:X→ℝ0+ is an arbitrary function and a≥0.
Obviously, if(X,d) is a MS and we define p=d, then (X,p) is a PMS. Indeed, a partial metric p on X verifies(i)
p(x,y)=0⇒x=y;(ii)
p(x,y)=p(y,x);(iii)
p(x,z)≤p(x,y)+p(y,z),
but the conditionp(x,x)=0 does not necessarily hold. For a partial metric p on X, the mappings dp,dm:X×X→ℝ0+ given by
(7)dp(x,y)=2p(x,y)-p(x,x)-p(y,y),dm(x,y)=max{p(x,y)-p(x,x),p(y,y)-p(y,y)}=p(x,y)-min{p(x,x),p(y,y)},
for all x,y∈X, are (usual) metrics on X. On a PMS, the concepts of convergence, Cauchy sequences, completeness, and continuity are defined as follows.Definition 12 (see [14, 25, 26]).
Let{xm} be a sequence on a PMS (X,p).(i)
{xm}  p-converges to x∈X (and one will write {xm}→px) if p(x,x)=limm→∞p(x,xm).(ii)
{xm} is called p-Cauchy if limm,m′→∞p(xm,xm′) exists (and it is finite).(iii)
(X,p) is said to be p-complete if every p-Cauchy sequence {xm} in X  p-converges to a point x∈X such that p(x,x)=limm,m′→∞p(xm,xm′).(iv)
A mappingf:X→X is said to be p-continuous atx0∈X if, for every ε>0, there exists δ>0 such that f(Bp(x0,δ))⊆Bp(f(x0),ε).We have used the previous notation because we need to distinguish betweenp-convergence and dp-convergence on X and usual convergence for real sequences.Lemma 13 (see [14, 25, 26]).
Let{xm} be a sequence on a PMS (X,p).(1)
{xm} is p-Cauchy if, and only if, it is dp-Cauchy.(2)
{xm}→dpx if, and only if, {xm}→px and p(x,x)=limm,m′→∞p(xm,xm′); that is,
(8){dp(xm,x)}⟶0⟺p(x,x)=limm→∞p(x,xm)=limm,m′→∞p(xm,xm′).(3)
(X,p) is complete if, and only if, the MS (X,dp) is complete.(4)
If{xm}→px and p(x,x)=0, then limm→∞p(xm,y)=p(x,y) for all y∈X.


## 3. Auxiliary Results

We will use the following results about real sequences in the proof of our main theorems.Lemma 14.
Let{am1}m∈ℕ,…,{amn}m∈ℕ be n real lower bounded sequences such that {max(am1,…,amn)}m∈ℕ→δ. Then, there exists i0∈{1,2,…,n} and a subsequence {am(k)i0}k∈ℕ such that {am(k)i0}k∈ℕ→δ.Proof.
Letbm=max(am1,am2,…,amn) for all m. As {bm} is convergent, it is bounded. As ami≤bm for all m and i, then every {ami} is bounded. As {am1}m∈ℕ is a real bounded sequence, it has a convergent subsequence {aσ1(m)1}m∈ℕ→a1. Consider the subsequences {aσ1(m)2}m∈ℕ,{aσ1(m)3}m∈ℕ,…,{aσ1(m)n}m∈ℕ; that are n-1 real bounded sequences and the sequence {bσ1(m)}m∈ℕ that also converges to δ. As {aσ1(m)2}m∈ℕ is a real bounded sequence, it has a convergent subsequence {aσ2σ1(m)2}m∈ℕ→a2. Then, the sequences {aσ2σ1(m)3}m∈ℕ, {aσ2σ1(m)4}m∈ℕ,…, {aσ2σ1(m)n}m∈ℕ also are n-2 real bounded sequences, {aσ2σ1(m)1}m∈ℕ→a1, and {bσ2σ1(m)}m∈ℕ→δ. Repeating this process n times, we can find n subsequences {aσ(m)1}m∈ℕ, {aσ(m)2}m∈ℕ,…, {aσ(m)n}m∈ℕ (where σ=σn⋯σ1) such that {aσ(m)i}m∈ℕ→ai for all i. And {bσ(m)}m∈ℕ→δ. But
(9){bσ(m)}m∈ℕ={max(aσ(m)n,…,aσ(m)n)}m∈ℕ⟶max(a1,…,an),
so δ=max(a1,…,an), and there exists i0∈{1,2,…,n} such that ai0=δ. Therefore, there exists i0∈{1,2,…,n} and a subsequence {aσ(m)i0}m∈ℕ such that {aσ(m)i0}m∈ℕ→ai0=δ.Lemma 15.
Let{am}m∈ℕ be a sequence of nonnegative real numbers which has not any subsequence converging to zero. Then, for all ε>0, there exist δ∈]0,ε[ and m0∈ℕ such that am≥δ for all m≥m0.Proof.
Suppose that the conclusion is not true. Then, there existsε0>0 such that, for all δ∈]0,ε0[, there exists m0∈ℕ verifying am0<δ. Let k0∈ℕ be such that 1/k0<ε0. For all k∈ℕ, take δk=1/(k+k0)∈]0,ε0[. Then, there exists m(k)∈ℕ verifying 0≤am(k)<δk=1/(k+k0). Taking limit when k→∞, we deduce that limk→∞am(k)=0. Then, {am} has a subsequence converging to zero (maybe, reordering {am(k)}), but this is a contradiction.Lemma 16.
If{xm}m∈ℕ is a sequence in a MS (X,d) that is not Cauchy, then there exist ε0>0 and two subsequences {xm(k)}k∈ℕ and {xn(k)}k∈ℕ such that, for all k∈ℕ,
(10)k<m(k)<n(k)<m(k+1),d(xm(k),xn(k))≥ε0,d(xm(k),xn(k)-1)<ε0.Proof.
We know that(11){xm}isCauchy⟺[∀ε>0,∃n0∈ℕ:(m,n≥n0⟹d(xm,xn)<ε)].
If this condition is not true, then
(12)∃ε0>0:(∀n0∈ℕ,∃m,n≥n0suchthatd(xm,xn)≥ε0).
Let n0=2. Then, there exists m1,n1∈ℕ such that m1,n1≥n0 and d(xm1,xn1)≥ε0. Let m(1)=min(m1,n1)≥n0=2>1, and consider the numbers
(13)d(xm(1),xm(1)+1),d(xm(1),xm(1)+2),…,d(xm(1),xmax(m1,n1)).
Since d(xm(1),xmax(m1,n1))=d(xm1,xn1)≥ε0, between the previous numbers there exists a first nonnegative integer n(1)∈{m(1)+1,m(1)+2,…,max(m1,n1)} such that d(xm(1),xn(1))≥ε0 but d(xm(1),xj)<ε0 for all j∈{m(1),m(1)+1,…,n(1)-1}. In particular, d(xm(1),xn(1)-1)<ε0.
Now, letn0=n(1)+1. Then, there exists m2,n2∈ℕ such that m2,n2≥n(1)+1 and d(xm2,xn2)≥ε0. Let m(2)=min(m2,n2)≥n0=n(1)+1>n(1), and consider the numbers
(14)d(xm(2),xm(2)+1),d(xm(2),xm(2)+2),…,d(xm(2),xmax(m2,n2)).
Since d(xm(2),xmax(m2,n2))=d(xm2,xn2)≥ε0, between the previous numbers there exists a first nonnegative integer n(2)∈{m(2)+1,m(2)+2,…,max(m2,n2)} such that d(xm(2),xn(2))≥ε0 but d(xm(2),xj)<ε0 for all j∈{m(2),m(2)+1,…,n(2)-1}. In particular, d(xm(2),xn(2)-1)<ε0.
Repeating this process, we can find two subsequences{xm(k)} and {xn(k)} such that, for all k∈ℕ:
(15)k<m(k)<n(k)<m(k+1),d(xm(k),xn(k))≥ε0,d(xm(k),xn(k)-1)<ε0.Definition 17.
LetΨ be the family of all continuous, nondecreasing mappings ψ:ℝ0+→ℝ0+ such that ψ(t)=0 if, and only if, t=0.These mappings are known asaltering distance functions (see [27]). Note that every selected ψ∈Ψ commutes with max; that is, ψ(max(s1,s2,…,sN))=max(ψ(s1),ψ(s2),…,ψ(sN)) for all s1,s2,…,sN∈[0,∞).Lemma 18.
Ifψ∈Ψ and limm→∞ψ(am)=0, then limm→∞am=0.Proof.
As there existsψ(am), then am∈domψ=[0,∞[. If the conclusion is not true, there exists ε0>0 such that, for all n0∈ℕ, there exists n≥n0 verifying an≥ε0. This means that {am} has a subsequence {am(k)}k such that am(k)≥ε0. As ψ is nondecreasing, ψ(ε0)≤ψ(am(k)) for all k∈ℕ. Therefore, {ψ(am)}m has a subsequence {ψ(am(k))}k lower bounded by ψ(ε0)>0, but this is impossible since limm→∞ψ(am)=0.With regards to coincidence points, it is possible to consider the following simplification. Ifτ is a permutation of Λn, and we reorder (4), then we deduce that every coincidence point may be seen as a coincidence point associated to the identity mapping on Λn (see, for instance, [28]).Lemma 19.
Letτ be a permutation of Λn, and let Φ=(σ1,σ2,…,σn,τ) and Φ′ = (στ-1(1),στ-1(2),…,στ-1(n),IΛn). Then, a point (x1,x2,…,xn)∈Xn is a Φ-coincidence point of the mappings F and g if, and only if, (x1,x2,…,xn) is a Φ′-coincidence point of the mappings F and g.Therefore, in the sequel, without loss of generality, we will only considerΥ-coincidence points where Υ=(σ1,σ2,…,σn), that is, that verify F(xσi(1),xσi(2),…,xσi(n))=gxi for all i. We also show some preliminary results on PMS.Lemma 20.
Let{xm} be a sequence on a PMS (X,p), and let x∈X.(1)
If{xm}→px and p(x,x)=0, then {xm}→dpx, {dp(xm,y)}→dp(x,y) and {p(xm,y)}→p(x,y) for all y∈X.(2)
If{xm}→dpx and {p(xm,xm)}→0, then p(x,x)=0.Proof.
(
1
) Since 0≤p(xm,xm)≤p(x,xm) and limm→∞p(x,xm)=p(x,x)=0, then limm→∞p(xm,xm) = 0. Therefore, limm→∞dp(x,xm) = limm→∞(2p(x,xm)-p(x,x)-p(xm,xm)) = 0, so {xm}→dpx. Since dp is continuous, then {dp(xm,y)}→dp(x,y) for all y∈X, and item 4 of Lemma 13 implies that {p(xm,y)}→p(x,y).
(
2
) Item 2 of Lemma 13 shows that p(x,x)=limm,m′→∞p(xm,xm′)=limm→∞p(xm,xm)=0.Remark 21.
Although the limit in a MS is unique, thep-limit in a PMS is not necessarily unique. For instance, let (X,p) as in Example 10. Then, (X,p) is a complete PMS (see [14]). Consider xm=2.5-1/(2m) for all m∈ℕ. Then, {xm}→dp2.5 but {xm}→px0 whenever x0∈[2.5,3].Definition 22.
LetN∈ℕ, let (X,p) be a PMS, let G:XN→X be a mapping, and let Y0=(x1,x2,…,xN)∈XN. We will say that G is αp-continuous at Y0 if, for all sequences {xm1},{xm2},…,{xmN} on X such that {xmi}→pxi for all i∈{1,2,…,N}, p(xi,xi)=0 for all i∈{1,2,…,N} and {p(G(xm1,xm2,…,xmN),G(xm1,xm2,…,xmN))}→0, we have that {G(xm1,xm2,…,xmN)}→pG(Y0) and p(G(Y0),G(Y0))=0. One will say that G is αp-continuous if it is continuous at every point Y0∈XN.Lemma 23.
If(X,p) is a PMS, and G:XN→X is dp-continuous at Y0∈XN, then G is αp-continuous at Y0.Proof.
Let{xm1},{xm2},…,{xmN} sequences on X such that {xmi}→pxi for all i∈{1,2,…,N}, p(xi,xi)=0 for all i∈{1,2,…,N}, and {p(G(xm1,xm2,…,xmN),G(xm1,xm2,…,xmN))}→0. Item 1 of Lemma 20 implies that {xmi}→dpxi for all i∈{1,2,…,N}. Since G is dp-continuous at Y0=(x1,x2,…,xN), then {G(xm1,xm2,…,xmN)}→dpG(x1,x2,…,xN). Item 2 of Lemma 13 assures us that {G(xm1,…,xmN)}→pG(x1,…,xN) and
(16)p(G(x1,x2,…,xN),G(x1,x2,…,xN))=limm,m′→∞p(G(xm1,xm2,…,xmN),G(xm′1,xm′2,…,xm′N))=limm→∞p(G(xm1,xm2,…,xmN),G(xm1,xm2,…,xmN))=0.
Then, G is αp-continuous at Y0.


## 4. Main Results

In the following result, we show sufficient conditions to ensure the existence ofΥ-coincidence points, where Υ=(σ1,σ2,…,σn).Theorem 24.
Let(X,p) be a complete PMS, and let ≤ a partial order on X. Let Υ=(σ1,σ2,…,σn) be an n-tuple of mappings from {1,2,…,n} into itself verifying σi∈ΩA,B if i∈A and σi∈ΩA,B′ if i∈B. Let F:Xn→X and g:X→X be two mappings such that F has the mixed g-monotone property on X, F(Xn)⊆g(X) and g is αp-continuous and commuting with F. Assume that there exist ψ,φ∈Ψ such that
(17)ψ(p(F(x1,x2,…,xn),F(y1,y2,…,yn)))≤ψ(max1≤i≤np(gxi,gyi))-φ(max1≤i≤np(gxi,gyi)),
for which gxi≤igyi for all i. Suppose either F is αp-continuous or (X,dp,≤) has the sequential g-monotone property. If there exist x01,x02,…,x0n∈X verifying gx0i≤iF(x0σi(1),x0σi(2),…,x0σi(n)) for all i, then F and g have, at least, one Υ-coincidence point.Proof.
The proof is divided into seven steps. The first two steps are the same as in the proof of Theorem 9 in [11], since the contractivity condition does not play any role in these parts of the proof.
Step  1. There exist n sequences {xm1}m≥0,{xm2}m≥0,…,{xmn}m≥0 such that gxm+1i = F(xmσi(1),xmσi(2),…,xmσi(n)) for all m and all i.
Step  2. gxmi≤igxm+1i for allm and all i.
Step  3. We claim that {p(gxmi,gxm+1i)}m≥0→0 for all i (i.e., {max1≤j≤np(gxmj,gxm+1j)}m≥0→0).
Indeed, defineδm=max1≤j≤np(gxmj,gxm+1j) for all m. As gxmi≤igxm+1ifor all m and all i, then condition (17) implies that, for all m≥1 and all i:(18)ψ(p(gxmi,gxm+1i))=ψ(p(F(xm-1σi(1),xm-1σi(2),…,xm-1σi(n)),F(xmσi(1),xmσi(2),…,xmσi(n))))≤ψ(max1≤j≤np(gxm-1σi(j),gxmσi(j)))-φ(max1≤j≤np(gxm-1σi(j),gxmσi(j)))≤ψ(max1≤j≤np(gxm-1j,gxmj))=ψ(δm-1).Therefore, for all m≥1, ψ(δm)=ψ(max1≤i≤np(gxmi,gxm+1i))=max1≤i≤nψ(p(gxmi,gxm+1i))≤ψ(δm-1). This means that the sequence {ψ(δm)}m≥1 is nonincreasing and lower bounded. Hence, it is convergent; that is, there exists Δ≥0 such that {ψ(δm)}m≥1→Δ. We are going to show that Δ=0. Since
(19){max1≤i≤nψ(p(gxmi,gxm+1i))}m={ψ(max1≤i≤np(gxmi,gxm+1i))}m={ψ(δm)}m⟶Δ,
Lemma 14 assures that there exist i0∈{1,2,…,n} and a subsequence {ψ(p(gxm(k)i0,gxm(k)+1i0))}k such that {ψ(p(gxm(k)i0,gxm(k)+1i0))}k→Δ. Repeating (18), for all k≥1,
(20)ψ(p(gxm(k)i0,gxm(k)+1i0))≤ψ(max1≤j≤np(gxm(k)-1σi0(j),gxm(k)σi0(j)))-φ(max1≤j≤np(gxm(k)-1σi0(j),gxm(k)σi0(j))).
Consider the sequence
(21){max1≤j≤np(gxm(k)-1σi0(j),gxm(k)σi0(j))}k≥1.
Suppose that this sequence has no subsequence converging to zero. Using ε=1, Lemma 15 assures us that there exists δ′∈]0,1[ and k0∈ℕ such that max1≤j≤np(gxm(k)-1σi0(j),gxm(k)σi0(j))≥δ′ for all k≥k0. It follows that
(22)-φ(max1≤j≤np(gxm(k)-1σi0(j),gxm(k)σi0(j)))≤-φ(δ′)∀k≥k0.
Then, (20) says to us
(23)ψ(p(gxm(k)i0,gxm(k)+1i0))≤ψ(max1≤j≤np(gxm(k)-1σi0(j),gxm(k)σi0(j)))-φ(max1≤j≤np(gxm(k)-1σi0(j),gxm(k)σi0(j)))≤ψ(max1≤j≤np(gxm(k)-1σi0(j),gxm(k)σi0(j)))-φ(δ′)≤ψ(δm(k)-1)-φ(δ′).
Taking limit in k, we deduce that Δ≤Δ-φ(δ′)<Δ, which is impossible. Therefore, the sequence in (21) must have a subsequence {max1≤j≤np(gxm′(k)-1σi0(j),gxm′(k)σi0(j))}k≥1 converging to zero. Since ψ and φ are continuous, taking limit when k→∞ in (20) using this subsequence, we deduce that 0≤Δ≤ψ(0)-φ(0)=0, so Δ=0. Then, we have just proved that Δ=0. Therefore, {ψ(δm)}m≥1→Δ=0, and Lemma 18 assures that {δm}m≥1→0, which means that {p(gxmj,gxm+1j)}→0 for all j since 0≤p(gxmj,gxm+1j)≤δm for all m and all j.
Step  4.
{
p
(
g
x
m
i
,
g
x
m
i
)
}
m
≥
0
→
0 for all i (i.e., {max1≤j≤np(gxmj,gxmj)}m≥0→0). It is the same proof of Step  3.
Sincedp(gxmi,gxm+1i)=2p(gxmi,gxm+1i)-p(gxmi,gxmi)-p(gxm+1i,gxm+1i) for all m and i, joining Steps 3 and 4, it follows that
(24){dp(gxmi,gxm+1i)}⟶0∀i.Step  5. Every sequence{gxmi}m≥0isdp-Cauchy. We reason by contradiction. Suppose that {gxmi1}m≥0,…,{gxmis}m≥0 are not dp-Cauchy (s≥1) and {gxmis+1}m≥0,…,{gxmin}m≥0 are dp-Cauchy, being {i1,…,in}={1,…,n}. By Lemma 16, for all r∈{1,2,…,s}, there exists εr>0 and subsequences {gxmr(k)ir}k∈ℕ and {gxnr(k)ir}k∈ℕ such that
(25)k<mr(k)<nr(k),dp(gxmr(k)ir,gxnr(k)ir)≥εr,dp(gxmr(k)ir,gxnr(k)-1ir)<εr,∀k∈ℕ.
Now, let ε0=max(ε1,…,εs)>0 and ε0′=min(ε1,…,εs)>0. Since {gxmis+1}m≥0,…,{gxmin}m≥0 are dp-Cauchy, for all j∈{is+1,…,in}, there exists n1j∈ℕ such that if m,m′≥n1j, then dp(gxmj,gxm′j)<ε0′/8. Since {p(gxmj,gxmj)}→0 by Step  4, there exists n2j∈ℕ such that if m≥n2j, then p(gxmj,gxmj)<ε0′/8. Define n0=maxj∈{is+1,…,in}(n1j,n2j). If m,m′≥n0, then
(26)0≤p(gxmj,gxm′j)=dp(gxmj,gxm′j)+p(gxmj,gxmj)+p(gxm′j,gxm′j)2<ε0′/8+ε0′/8+ε0′/82=3ε0′16<ε0′4.
Therefore, we have proved that there exists n0∈ℕ such that if m,m′≥n0, then
(27)dp(gxmj,gxm′j)<ε0′4,p(gxmj,gxm′j)<ε0′4,∀j∈{is+1,…,in}.
Next, letq∈{1,2,…,s} such that εq=ε0=max(ε1,…,εs). Let k1∈ℕ such that n0<mq′(k1), and define m(1)=mq(k1). Consider the numbers m(1)+1,m(1)+2,…,nq(k1) until finding the first positive integer n(1)>m(1) verifying
(28)max1≤r≤sdp(gxm(1)ir,gxn(1)ir)≥ε0,dp(gxm(1)ij,gxn(1)-1ij)<ε0,∀j∈{1,2,…,s}.
Now let k2∈ℕ such that n(1)<mq(k2), and define m(2)=mq(k2). Consider the numbers m(2)+1,m(2)+2,…,nq(k2) until finding the first positive integer n(2)>m(2) verifying
(29)max1≤r≤sdp(gxm(2)ir,gxn(2)ir)≥ε0,dp(gxm(2)ij,gxn(2)-1ij)<ε0,∀j∈{1,2,…,s}.
Repeating this process, we can find sequences such that, for all k≥1,
(30)n0<m(k)<n(k)<m(k+1),max1≤r≤sdp(gxm(k)ir,gxn(k)ir)≥ε0,dp(gxm(k)ij,gxn(k)-1ij)<ε0,∀j∈{1,2,…,s}.
Note that by (27), dp(gxm(k)ir,gxn(k)ir),dp(gxm(k)ir,gxn(k)-1ir)<ε0′/4<ε0/2 for all r∈{s+1,s+2,…,n}, so
(31)max1≤j≤ndp(gxm(k)j,gxn(k)j)=max1≤r≤sdp(gxm(k)ir,gxn(k)ir)≥ε0,dp(gxm(k)i,gxn(k)-1i)<ε0,
for all i∈{1,2,…,n} and all k≥1. Furthermore, for all j,
(32)2p(gxm(k)-1j,gxn(k)-1j)-p(gxm(k)-1j,gxm(k)-1j)-p(gxn(k)-1j,gxn(k)-1j)=dp(gxm(k)-1j,gxn(k)-1j)≤dp(gxm(k)-1j,gxm(k)j)+dp(gxm(k)j,gxn(k)-1j)≤dp(gxm(k)-1j,gxm(k)j)+ε0.
Therefore, for all j and all k,
(33)p(gxm(k)-1j,gxn(k)-1j)≤(ε0+dp(gxm(k)-1j,gxm(k)j)+p(gxm(k)-1j,gxm(k)-1j)+p(gxn(k)-1j,gxn(k)-1j))×2-1.
Next, for all k, let i(k)∈{1,2,…,s} be an index such that
(34)dp(gxm(k)i(k),gxn(k)i(k))=max1≤r≤sdp(gxm(k)ir,gxn(k)ir)=max1≤j≤ndp(gxm(k)j,gxn(k)j)≥ε0.
Then, for all k,
(35)p(gxm(k)i(k),gxn(k)i(k))=(dp(gxm(k)i(k),gxn(k)i(k))+p(gxm(k)i(k),gxm(k)i(k))+p(gxn(k)i(k),gxn(k)i(k)))×2-1≥ε02.
Applying the contractivity condition (17), it follows, for all k,
(36)0<ψ(ε02)≤ψ(p(gxm(k)i(k),gxn(k)i(k)))≤ψ(max1≤j≤np(gxm(k)-1σi(k)(j),gxn(k)-1σi(k)(j)))-φ(max1≤j≤np(gxm(k)-1σi(k)(j),gxn(k)-1σi(k)(j))).
Consider the sequence:
(37){max1≤j≤np(gxm(k)-1σi(k)(j),gxn(k)-1σi(k)(j))}k≥1.
If this sequence has a subsequence that converges to zero, then we can take limit when k→∞ in (36) using this subsequence, so that we would have 0<ψ(ε0/2)≤ψ(0)-φ(0)=0, which is impossible since ε0>0. Therefore, the sequence (37) has no subsequence converging to zero. In this case, taking ε0>0 in Lemma 15, there exist δ∈]0,ε0[ and k0∈ℕ such that max1≤j≤np(gxm(k)-1σi(k)(j),gxn(k)-1σi(k)(j))≥δ, for all k≥k0. It follows that, for all k≥k0, -φ(max1≤j≤np(gxm(k)-1σi(k)(j),gxn(k)-1σi(k)(j)))≤-φ(δ). Thus, by (36),
(38)0<ψ(ε02)≤ψ(max1≤j≤np(gxm(k)-1σi(k)(j),gxn(k)-1σi(k)(j)))-φ(max1≤j≤np(gxm(k)-1σi(k)(j),gxn(k)-1σi(k)(j)))≤ψ(max1≤j≤np(gxm(k)-1σi(k)(j),gxn(k)-1σi(k)(j)))-φ(δ)≤ψ(max1≤j≤np(gxm(k)-1j,gxn(k)-1j))-φ(δ).
Fix any γ>0 and we are going to prove that ψ(ε0/2)+φ(δ)≤ψ(ε0/2+γ). Indeed, by Step  3 and (24), since
(39){max1≤i≤np(gxm(k)-1i,gxm(k)-1i)},{max1≤i≤np(gxn(k)-1i,gxn(k)-1i)},{max1≤i≤ndp(gxm(k)-1i,gxm(k)i)}
are sequences converging to zero, we can find m1∈ℕ such that if m(k)≥m1, then
(40)max1≤i≤np(gxm(k)-1i,gxm(k)-1i)≤γ2,max1≤i≤np(gxn(k)-1i,gxn(k)-1i)≤γ2,max1≤i≤ndp(gxm(k)-1i,gxm(k)i)≤γ2.
Therefore, (33) implies that, for all j and for all k such that m(k)>m1,
(41)p(gxm(k)-1j,gxn(k)-1j)≤(ε0+dp(gxm(k)-1j,gxm(k)j)+p(gxm(k)-1j,gxm(k)-1j)+p(gxn(k)-1j,gxn(k)-1j))×2-1≤ε0+γ/2+γ/2+γ/22=ε02+3γ4<ε02+γ.
Then, (38) guarantees that 0<ψ(ε0/2)≤ψ(max1≤j≤np(gxm(k)-1j,gxn(k)-1j))-φ(δ)≤ψ(ε0/2+γ)-φ(δ). This means that ψ(ε0/2)+φ(δ)≤ψ(ε0/2+γ) for all γ>0. If we take γ=1/m>0 (where m∈ℕ), we deduce that ψ(ε0/2)+φ(δ)≤ψ(ε0/2+1/m) for all m∈ℕ. Since ψ is continuous, we have that ψ(ε0/2)+φ(δ)≤ψ(ε0/2), which is impossible since φ(δ)>0. This contradiction finally proves that every sequence {gxmi}m≥0 is dp-Cauchy.
SinceX is p-complete, then X is dp-complete (item 3 of Lemma 13). Then, there exist x1,x2,…,xn∈X such that {gxmi}→dpxi for all i. Furthermore, p(xi,xi) = limm,m′→∞p(gxmi,gxm′i) = limm→∞p(gxmi,gxmi) = 0 for all i. Since g is αp-continuous, then {ggxmi}→pgxi and p(gxi,gxi)=0 for all i. Item 1 of Lemma 20 shows that {ggxmi}→dpgxi for all i. Therefore, for all i, limm→∞p(ggxm+1i,ggxm+1i)=limm,m′→∞p(ggxm+1i,ggxm′+1i)=p(gxi,gxi)=0. Moreover, for all m and all i, ggxm+1i=gF(xmσi(1),xmσi(2),…,xmσn(n))=F(gxmσi(1),gxmσi(2),…,gxmσi(n)).
Step  6. Suppose that
F
is
α
p
-continuous. In this case, we know that {gxmi}→pxi and p(xi,xi)=0 for all i and
(42){p(F(gxmσi(1),gxmσi(2),…,gxmσi(n)),F(gxmσi(1),gxmσi(2),…,gxmσi(n)))}={p(ggxm+1i,ggxm+1i)}⟶0,
which implies that {F(gxmσi(1),gxmσi(2),…,gxmσi(n))}→pF(xσi(1),xσi(2),…,xσi(n)) and p(F(xσi(1),…,xσi(n)), F(xσi(1),…,xσi(n)))=0 for all i. Item 1 of Lemma 20 assures us that, for all i,
(43){ggxm+1i}={F(gxmσi(1),gxmσi(2),…,gxmσi(n))}⟶dpF(xσi(1),xσi(2),…,xσi(n)).
Since the limit in a MS is unique, we deduce that F(xσi(1),xσi(2),…,xσi(n))=gxi for all i, so (x1,x2,…,xn) is a Υ-coincidence point of F and g.
Step  7. Suppose that(X,dp,≤)  has the sequentialg-monotone property.  In this case, by Step  2, we know that gxmi≤igxm+1i for all m and all i. This means that the sequence {gxmi}m≥0 is monotone. As {gxmi}→dpxi, we deduce that ggxmi≤igxi for all mand alli. This condition implies that, for all m and all j,
(44)either[ggxmσj(i)≤igxσj(i)∀i]or[gxσj(i)≤iggxmσj(i)∀i]
(the first case occurs when j∈A and the second one when j∈B). Then, by (17), for all j,
(45)ψ(p(ggxm+1j,F(xσj(1),xσj(2),…,xσj(n))))=ψ(p(F(gxmσj(1),gxmσj(2),…,gxmσj(n)),F(xσj(1),xσj(2),…,xσj(n)))(F(gxmσj(1),gxmσj(2),…,gxmσj(n)),)≤ψ(max1≤i≤np(ggxmσj(i),gxσj(i)))-φ(max1≤i≤np(ggxmσj(i),gxσj(i)))≤ψ(max1≤i≤np(ggxmi,gxi)).
Since {ggxmi}→dpgxi for all i, then
(46)limm→∞p(ggxmi,gxi)=p(gxi,gxi)=0∀i.
Therefore, limm→∞(max1≤i≤np(ggxmi,gxi))=0. Taking limit when m→∞ in (45), we deduce that limm→∞ψ(p(ggxm+1j,F(xσj(1),xσj(2),…,xσj(n))))=0 for all j. As ψ∈Ψ, Lemma 18 guarantees that
(47)limm→∞p(ggxm+1j,F(xσj(1),xσj(2),…,xσj(n)))=0∀j.
Finally, for all j,
(48)dp(gxj,F(xσj(1),xσj(2),…,xσj(n)))=2p(gxj,F(xσj(1),xσj(2),…,xσj(n)))-p(gxj,gxj)-p(F(xσj(1),xσj(2),…,xσj(n)),F(xσj(1),xσj(2),…,xσj(n)))≤2p(gxj,F(xσj(1),xσj(2),…,xσj(n)))≤2[p(ggxmi,F(xσj(1),xσj(2),…,xσj(n)))p(gxj,ggxmi)+p(ggxmi,F(xσj(1),xσj(2),…,xσj(n)))].
Using (46) and (47), we conclude that dp(gxj,F(xσj(1),xσj(2),…,xσj(n)))=0 for all j.Remark 25.
In the previous theorem, if the imageImd of the metric d is not the whole set [0,∞[, then ψ and φ can only be defined on Imd, and we can consider a wider range of mappings since it is only necessary to impose that they are continuous and nondecreasing on Imd.Remark 26.
We notice also that our paper cannot be deduced from the recent interesting paper of Haghi et al. [21] on partial metric space. In fact, we use a partial order ≤. Then, we only suppose (17) for which gxi≤igyi for all i (not necessarily on points which are not comparable). Further, we use a self-map g:X→X which implies that
(49)P(A,B)=max1≤i≤np(gai,gbi),A=(a1,a2,…,an),B=(b1,b2,…,bn)∈Xn
is not necessarily a partial metric on Xn. For instance, let X=ℝ0+=[0,∞) provided with its usual partial order and the partial metric p(x,y)=max(x,y). Consider
(50)gx={0,if0≤x≤1,x-1,ifx>1.
Then, g is continuous, but
(51)P((0,0,…,0),(0,0,…,0))=P((1,1,…,1),(0,0,…,0))=P((1,1,…,1),(1,1,…,1))=0,
but (0,0,…,0)≠(1,1,…,1). Then, P does not verify the axiom p(x,x)=p(x,y)=p(y,y)⇒x=y. Therefore, we cannot apply Theorem 2.4 on Haghi et al. [21].
As a result, we cannot use Theorem 2.7 in [21] since T has an influence in -φ(max{p(x,y),p(y,Ty)}), and our mapping F has not a role in the left side of (17).


## 5. Consequences

Remark 27.
Theorem 9 in [11] is an easy consequence of Theorem 24 if we take p=d, ψ(t)=t, and φ(t)=(1-k)t for all t∈ℝ0+.
In the next result, letΓ0 be the family of all nondecreasing on each argument, continuous mappings ϕ:[0,∞[n→ℝ0+ verifying ϕ(x1,x2,…,xn)=0 if, and only if, x1=x2=⋯=xn=0. Examples of such mappings are the following, where k>0, αi>0 and ni∈ℕ for all i. (i)
ϕ(x1,…,xn)=kmax1≤i≤nxi.(ii)
ϕ(x1,…,xn)=∑i=1nαixini.(iii)
ϕ(x1,…,xn)=α1x12+⋯+αnxn2m.Lemma 28.
Letϕ∈Γ0, and define φ:ℝ0+→ℝ0+ as φ(t)=min(ϕ(te1),ϕ(te2),…,ϕ(ten)) for all t≥0, where {e1,e2,…,en} is the usual basis of ℝn. Then, φ∈Ψ and φ(max1≤i≤nxi)≤ϕ(x1,x2,…,xn) for all x1,x2,…,xn∈ℝ0+.Proof.
First part is clear. Ifxi0=max1≤i≤nxi, then ϕ(xi0ei0) = ϕ(0,0,…,0,xi0,0,…,0) ≤ ϕ(x1,x2,…,xi0-1,xi0,xi0+1,…,xn) = ϕ(x1,x2,…,xn). Therefore, φ(max1≤i≤nxi) = φ(xi0)≤ϕ(xi0ei0)≤ϕ(x1,x2,…,xn).Corollary 29.
Thesis of Theorem24 also holds if one replaces the contractivity condition (17) by any of the following list (for which gxi≤igyi for all i).(A)
This condition can be found in [11] and [12], there exist ψ∈Ψ and ϕ∈Γ0 such that
(52)ψ(p(F(x1,x2,…,xn),F(y1,y2,…,yn)))≤ψ(max1≤i≤np(gxi,gyi))-ϕ(p(gx1,gy1),…,p(gxn,gyn)).(B)
In  [17], there exist ψ,φ∈Ψ and β1,β2,…,βn∈[0,1] such that β1+β2+⋯+βn≤1 and
(53)ψ(p(F(x1,x2,…,xn),p(y1,y2,…,yn)))≤ψ(∑i=1n‍βip(gxi,gyi))-φ(max1≤i≤np(gxi,gyi)).(C)
There existψ,φ∈Ψ and α1,α2,…,αn>0 such that
(54)ψ(p(F(x1,x2,…,xn),F(y1,y2,…,yn)))≤ψ(max1≤i≤np(gxi,gyi))-φ(∑i=1n‍αip(gxi,gyi)).(D)
In  [2, 7, 9], there exist ψ∈Ψ, α1,α2,…,αn>0, and β1,β2,…,βn≥0 such that β1+β2+⋯+βn≤1 and
(55)ψ(p(F(x1,x2,…,xn),F(y1,y2,…,yn)))≤ψ(∑i=1n‍βip(gxi,gyi))-∑i=1n‍αip(gxi,gyi).(E)
In  [5], there exist ψ∈Ψ and α1,α2,…,αn>0 such that
(56)ψ(p(F(x1,x2,…,xn),F(y1,y2,…,yn)))≤ψ(1n∑i=1n‍p(gxi,gyi))-∑i=1n‍αip(gxi,gyi).(F)
In  [19, 20], there exist ψ,φ∈Ψ such that ψ is subadditive (ψ(s+t)≤ψ(s)+ψ(t) for all t,s∈[0,∞)) and
(57)ψ(p(F(x1,x2,…,xn),F(y1,y2,…,yn)))≤1nψ(∑i=1n‍p(gxi,gyi))-φ(max1≤i≤np(gxi,gyi)).Of course, it is also interesting to particularize all the previous items to the following cases:ψ(t)=λt (where λ>0), φ(t)=μt (where μ>0), or gx=x for all x∈X.Proof.
(A) By Lemma28, there exists φ∈Ψ such that -ϕ(x1,x2,…,xn)≤-φ(max1≤i≤nxi) for all x1,x2,…,xn∈[0,∞[, so (52) implies (17).  (B) It is obvious that ∑i=1n‍βip(gxi,gyi) ≤ (∑i=1n‍βi)max1≤j≤np(gxj,gyj)≤max1≤j≤np(gxj,gyj), so (53) implies (17).  (C) We only take ϕ(x1,x2,…,xn)=φ(∑i=1n‍αixi) in item (A).  (D) It is a mixture of (B) and (C).  (E) It is a particular case of (D) where βi=1/n for all i.  (F) If ψ is subadditive, then (1/n)ψ(t)≤ψ(t/n) for all t≥0, so we may choose βi=1/n for all i in (B).


## 6. Uniqueness ofΥ-Coincidence Points

Consider on the product spaceXn the following partial order: for (x1,x2,…,xn),(y1,y2,…,yn)∈Xn,
(58)(x1,x2,…,xn)≤(y1,y2,…,yn)⟺xi≤iyi,∀i.
We say that (x1,x2,…,xn) and (y1,y2,…,yn) are comparable if (x1,x2,…,xn)≤(y1,y2,…,yn) or (x1,x2,…,xn)≥(y1,y2,…,yn).Theorem 30.
Under the hypothesis of Theorem24, assume that for all Υ-coincidence points (x1,x2,…,xn), (y1,y2,…,yn)∈Xn of F and g there exists (u1,u2,…,un)∈Xn such that (gu1,gu2,…,gun) is comparable, at the same time, to (gx1,gx2,…,gxn) and to (gy1,gy2,…,gyn).
Then,F and g have a unique Υ-coincidence point (z1,z2,…,zn)∈Xn such that gzi=zi for all i.Proof.
From Theorem24, the set of Υ-coincidence points of F and g is nonempty. The proof is divided into two steps.
Step  1. We claim that if (x1,x2,…,xn),(y1,y2,…,yn)∈Xn are two Υ-coincidence points of F and g, then
(59)gxi=gyi∀i.
Let (x1,x2,…,xn),(y1,y2,…,yn)∈Xn be two Υ-coincidence points of F and g, and let (u1,u2,…,un)∈Xn be a point such that (gu1,gu2,…,gun) is comparable, at the same time, to (gx1,gx2,…,gxn) and to (gy1,gy2,…,gyn). Using (u1,u2,…,un), define the following sequences. Let u0i=ui for all i. Reasoning as in Theorem 24, we can determine sequences {um1}m≥0, {um2}m≥0,…,{umn}m≥0 such that gum+1i=F(umσi(1),umσi(2),…,umσi(n)) for all m and all i. We are going to prove that gxi=limm→0dpgumi=gyi for all i, so (59) will be true.
Firstly, we reason with(gu1,gu2,…,gun) and (gx1,gx2,…,gxn), and the same argument holds for (gu1,gu2,…,gun) and (gy1,gy2,…,gyn). As (gu1,gu2,…,gun) and (gx1,gx2,…,gxn) are comparable, we can suppose that (gu1,gu2,…,gun)≤(gx1,gx2,…,gxn) (the other case is similar); that is, gu0i=gui≤igxi for all i. Using that F has the mixed g-monotone property and reasoning as in Theorem 24, it is possible to prove that gumi≤igxi for all m≥1 and all i. This condition implies that, for all j and all m≥1(60)either[gumσj(i)≤igxσj(i)∀i]or[gxσj(i)≤igumσj(i)∀i].
Define βm=max1≤i≤np(gumi,gxi) for all m. Reasoning as in Theorem 24, it is not difficult to prove that {βm}m≥1→0 which means that limm→∞βm=limm→∞(max1≤i≤np(gumi,gxi))=0. As 0≤p(gumi,gxi)≤βm for all m and all i, we deduce that {p(gumi,gxi)}m≥1→0=p(gxi,gxi) for all i; that is, {gumi}→pgxi for all i. Item 1 of Lemma 20 shows that
(61){gumi}⟶dpgxi∀i.
If we had supposed that (gx1,gx2,…,gxn)≤(gu1,gu2,…,gun), we would have obtained the same property (61). And as (gu1,gu2,…,gun) also is comparable to (gy1,gy2,…,gyn), we can reason in the same way to prove that {gumi}→dpgyi for all i. Since the limit in a MS is unique, gxi=gyi for all i.
Let(x1,x2,…,xn)∈Xn be a Υ-coincidence point of F and g, and define zi=gxi for all i. As (z1,z2,…,zn)=(gx1,gx2,…,gxn), Remark 6 assures us that (z1,z2,…,zn) also is a Υ-coincidence point of F and g.
Step  2. We claim that (z1,z2,…,zn) is the unique Υ-coincidence point of F and g such that gzi=zi for all i. It is similar to Step 2 in Theorem 11 in [11].It is natural to say thatg is injective on the set of all Υ-coincidence points of F and g when gxi=gyi for all i implies xi=yi for all i when (x1,x2,…,xn),(y1,y2,…,yn)∈Xn are two Υ-coincidence points of F and g. For example, this is true is g is injective on X.Corollary 31.
In addition to the hypotheses of Theorem30, suppose that g is injective on the set of all Υ-coincidence points of F and g. Then, F and g have a unique Υ-coincidence point.Proof.
If(x1,x2,…,xn) and (y1,y2,…,yn) are two Υ-coincidence points of F and g, we have proved in (59) that gxi=gyi for all i. As g is injective on these points, then, xi=yi for all i.Corollary 32.
In addition to the hypotheses of Theorem30, suppose that (zσi(1),zσi(2),…,zσi(n)) is comparable to (zσj(1),zσj(2),…,zσj(n)) for all i,j. Then, z1=z2=⋯=zn.
In particular, there exists a uniquez∈X such that F(z,z,…,z)=z, which verifies gz=z.Proof.
LetM=max1≤i,j≤np(zi,zj), let j0,s0∈{1,2,…,n} such that p(zj0,zs0)=M, and let
(62)Λ=max1≤i≤np(zσj0(i),zσs0(i))≤M.
Fix j,s∈{1,2,…,n}. As (zσj(1),zσj(2),…,zσj(n)) is comparable to (zσs(1),zσs(2),…,zσs(n)), then either zσj(i)≤izσs(i) for all i or zσs(i)≤izσj(i) for all i. Since gzi=zi for all i, we know that either gzσj(i)≤igzσs(i) for all i or gzσs(i)≤igzσj(i) for all i. In any case, applying (17),
(63)ψ(M)=ψ(p(zj0,zs0))=ψ(p(gzj0,gzs0))=ψ(p(F(zσj0(1),zσj0(2),…,zσj0(n)),F(zσs0(1),zσs0(2),…,zσs0(n))))≤ψ(max1≤i≤np(gzσj0(i),gzσs0(i)))-φ(max1≤i≤np(gzσj0(i),gzσs0(i)))=ψ(Λ)-φ(Λ)≤ψ(M)-φ(Λ).
If Λ>0, then φ(Λ)>0, so ψ(M)≤ψ(M)-φ(Λ)<ψ(M), which is impossible. Then, Λ=0, and (63) implies that ψ(M)≤ψ(Λ)-φ(Λ)=ψ(0)-φ(0)=0, so ψ(M)=0. Therefore, p(zi,zj)=0 for all i and j.Example 33.
LetX=ℝ provided with its usual partial order ≤ and the partial metric p(x,y)=max(|x|,|y|). Let n∈ℕ, and let a1,a2,…,an∈ℝ∖{0} real numbers such that there exist i0,j0∈{1,2,…,n} verifying ai0<0<aj0. Let N>|a1|+|a2|+⋯+|an|, and consider F(x1,x2,…,xn)=(a1x1+a2x2+⋯+anxn)/N and gx=x, for all x,x1,x2,…,xn∈X. Then, F is monotone nondecreasing in those arguments for which ai>0 and monotone nonincreasing in those arguments for which ai<0. Furthermore, taking k=(|a1|+|a2|+⋯+|an|)/N∈(0,1), it follows that
(64)|F(x1,x2,…,xn)|≤|a1||x1|+|a2||x2|+⋯+|an||xn|N≤|a1|+|a2|+⋯+|an|Nmax(|x1|,|x2|,…,|xn|)=kmax(|x1|,|x2|,…,|xn|).
Therefore,
(65)p(F(x1,x2,…,xn),F(y1,y2,…,yn))=max(|F(x1,x2,…,xn)|,|F(y1,y2,…,yn)|)≤max(kmax(|x1|,|x2|,…,|xn|),≤maxkmax(|y1|,|y2|,…,|yn|))=kmaxi(p(xi,yi)).
If ψ(t)=t and φ(t)=(1-k)t, all conditions of Theorems 24 and 30 (and Corollaries 31 and 32) are satisfied. Indeed, it is clear that (0,0,…,0) is the unique fixed point of F.The following example is based on Examples 1.9 and 2.2 in [29].Example 34.
LetX={0,1,2,3,4}, and let p be the partial metric on X given by p(x,y)=max(x,y) for all x,y∈X. Then, (X,p) is complete, and p generates the discrete topology on X (indeed, dp is the Euclidean metric on X). Consider on X the following partial order:
(66)x,y∈X,x≤y⟺x=yor(x,y)=(0,2).
Consider F:Xn→X and g:X→X defined by
(67)F(x1,x2,…,xn)={0,ifx1,x2,…,xn∈{0,1,2},1,otherwise,gx={0,ifx=0,2,ifx∈{0.5,1},3,ifx∈{1.5,2}.
It is not difficult to prove the following statements.(1)
F and g are αp-continuous mappings (since dp generates the discrete topology on X).(2)
F and g are commuting.(3)
Ify,z∈X verify gy≤gz, then either y,z∈{0,1,2} or y,z∈{3,4}. Then, F has the mixed (g,≤)-monotone property on X.(4)
Ifx1,x2,…,xn,y1,y2,…,yn∈X verify gxi≤igyi for all i, then F(x1,x2,…,xn) = F(y1,y2,…,yn). In particular, (17) holds (whatever ψ and φ; for instance, ψ(t)=2t and φ(t)=log(t+1) for all t≥0).For simplicity, henceforth, suppose thatn is even, and let A (resp., B) be the set of all odd (resp., even) numbers in {1,2,…,n}.(5)
For a mappingσ:Λn→Λn, we use the notation σ≡(σ(1),σ(2),…,σ(n)) and consider
(68)σi≡(i,i+1,…,n-1,n,1,2,…,i-1)∀i.Then,σi∈ΩA,B if i is odd, and σi∈ΩA,B′ if i is even. Let Υ=(σ1,σ2,…,σn).(6)
Takex0i=0 if i is odd and x0i=2 if i is even. Then, gx0i≤iF(x0σi(1),x0σi(2),…,x0σi(n)) for all i.(7)
(X,dp,≤) has the sequential g-monotone property.
Therefore, we can apply Theorems24 and 30, and Corollaries 31 and 32, to conclude that F and g have a unique Υ-coincidence point, which is (0,0,…,0).



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