# Extension of Kirk-Saliga Fixed Point Theorem in a Metric Space with a Reflexive Digraph

Authors: Karim Chaira; Abderrahim Eladraoui; Mustapha Kabil; Samih Lazaiz
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

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## Abstract

We extend the result of Kirk-Saliga and we generalize Alfuraidan and Khamsi theorem for reflexive graphs. As a consequence, we obtain the ordered version of Caristi’s fixed point theorem. Some concrete examples are given to support the obtained results.


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## Body

## 1. Introduction

Fixed point theory is one of the most useful tools in mathematics; it is used to solve many existence problems such as differential equations, control theory, optimization, and several other branches (for the literature see [1]). The most well-known fixed point result is Banach contraction principle [2]; it is famous for its applications, proving the existence of solution of integral equations by converting the problem to fixed point problem (see [3]). Recall that a point x∈X is called a fixed point for a map T:X→X if Tx=x. Due to its importance, this theorem found a number of generalizations and extensions in many directions; for more details see [4] and the references therein. In 1976, Caristi (see [5]) gave an elegant generalization of Banach contraction principle, where the assumption that “T:X→X is continuous” is dropped and replaced by a weak assumption. Since then, various proofs, extensions, and generalizations are given by many authors (see [6–8]). It is worth mentioning that Caristi’s fixed point theorem is equivalent to the Ekeland variational principle [8]. Also, it characterizes the completeness of the metric space as showed by Kirk in [9]. Among those generalizations, there is Kirk-Saliga fixed point theorem (see [10]) which states that any map T:X→X has a fixed point provided that X is complete metric space and there exist an integer p∈N and a lower semicontinuous function φ:X→0,∞ such that (1)dx,Tx≤φx-φTpxand φTx≤φx for any x∈X. For more on the latter result, one can consult [11].Recently, Ran and Reurings [12] extend the Banach contraction principle in the context of partially ordered set where the contraction is restricted to the comparable elements which allowed them to give a meaningful application to linear and nonlinear matrix equations. Moreover, Nieto and Rodríguez-López in [13] have weakened the continuity assumption using a more suitable condition where the order is combined with the topological properties. For more details, one can consult [14, 15]. Also, in [16] Alfuraidan and Khamsi gave an analogue version of Caristi’s fixed point theorem in the setting of partially ordered metric space where the inequality holds only for comparable elements. However, the new approach in their work is mixing the concept of the reflexive acyclic digraph with fixed point results. In this article, we discuss an extension of Kirk-Saliga result and we generalize Alfuraidan and Khamsi theorem for reflexive graphs. As a corollary, we obtain the ordered version of Caristi’s fixed point theorem. Some concrete examples are given to support the obtained results. Throughout this paper we denote by N the set of all integers and by N∗ the set of all positive integers.


## 2. Preliminaries

We start by recalling some basic notions on graphs borrowed from [17].Definition 1.
LetV be an arbitrary set.(i)
A directed graph, or digraph, is a pairG=(V,E), where E is a subset of the Cartesian product V×V. The elements of V are called vertices or nodes of G and the elements of E are the edges also called oriented edges or arcs of G. An edge of the form (v,v) is a loop on v. Another way to express that E is a subset of V×V is to say that E is a binary relation over V. Given a digraph G, the set of vertices (of edges) of G is denoted by V(G) (E(G)).(ii)
The digraphG=(V,E) is said to be transitive if whenever (x,y)∈E and (y,z)∈E, (x,z)∈E.Definition 2.
A digraphG=(V,E) is said to be reflexive if Δ≔v,v∣v∈V is a subset of E. Otherwise, every vertex has a loop.Definition 3.
LetG=(V,E) be a digraph. (i)
A vertexx is said to be isolated if for all vertex y≠x, we have neither (x,y)∈E nor (y,x)∈E.(ii)
Two verticesx,y∈V. A path in G, from (or joining) x to y, is a sequence of vertices p=ai0≤i≤n, n∈N∗ such that a0=x, an=y and (ai,ai+1)∈E, for all i∈0,1,…,n-1. The integer n is the length of the path p. If x=y and n>1, the path p is called a directed cycle. An acyclic digraph is a digraph which has no directed cycle.(iii)
We denote byy∈xG the fact that y can be reached from x by means of a path in G.A metric space(X,d) endowed with a digraph G such that V(G)=X is denoted by (X,d,G). The following notion of regularity is borrowed from Alfuraidan and Khamsi in [16] that considered it for posets.Definition 4.
Let(X,d,⪯) be a partially ordered metric space. We say that X satisfies the condition (OSC) if for any decreasing sequence xn⊆X that is convergent to x∈X, x=inf⁡xn:n∈N.In the setting of digraphs, the analogue of the infimum of chain may be stated as follows.Definition 5.
LetX,d,G be metric space endowed with a digraph. We say that X satisfies the condition (OSCL) if for any sequence xn⊆X that is convergent to x∈X and for all n∈N, xn+1∈xnG, x∈xnG for all n∈N and if there exists y∈X such that y∈xnG, for all n∈N, then y∈xG.Remark 6.
LetX,d,⪯ be a partially ordered metric space. Let G⪯ be the digraph associated with the order ⪯ (see [16]). One can see that (2)x⪯y⟺y,x∈EG⪯⟺x∈yG⪯.Under the above observations, the (OSCL) property is reduced to the (OSC) condition.Letω be the first transfinite ordinal and let Ω be the first uncountable transfinite ordinal. ω is the order type of N “the set of integers” and Ω is the order type of R the set of real numbers. Note that, for each ξ<Ω, ξ is countable.Proposition 7 (see [11]).
The following is valid:(i)
The ordinalΩ cannot be attained via sequential limits of countable ordinals. That is if αn is an ascending sequence of countable ordinals, then the ordinal (3)α=sup⁡αn=lim⁡αnis countable too.(ii)
Each second kind countable ordinal is attainable via such sequences. In other words: ifα<Ω is of second kind (ordinal limit), then there exists a strictly ascending sequence αn of countable ordinals with property (3).The following result is needed throughout this work; for the proof see [18, Proposition  A.6, pp. 284].Proposition 8.
Suppose that a sequencexαα∈Ω⊆R is bounded and either nonincreasing or nondecreasing. Then there exists β∈Ω such that xα=xβ for all Ω>α≥β.We conclude this section by the following useful definitions.Definition 9.
Let(X,d,G) be metric space endowed with a digraph, p∈N and φ:X→[0,+∞[ a lower semicontinuous function. Let T:X→X be a self-mapping. We say the following: (1)
T is a G-monotone if for all (x,y)∈X2, (4)x∈yG⟹Tx∈TyG.(2)
T is a G-Caristi mapping if for all x∈X, (5)Tx∈xG⟹dTx,x≤φx-φTx.(3)
T is a G-Kirk-Saliga mapping if for all x∈X, (6)Tx∈xG⟹KS1:dTx,x≤φx-φTpx;KS2:φTx≤φx.


## 3. Main Results

Theorem 10.
Let(X,d,G) be a complete metric space endowed with a reflexive digraph satisfying the (OSCL) condition. Let T:X→X be a G-monotone and G-Kirk-Saliga mapping. If there exists an element x0∈X such that Tx0∈x0G, then T admits a fixed point in X.Proof.
Ifp=0 then Tx=x, for all x∈X such that Tx∈xG. Assume that p≥1 and consider the function ϕ defined from X into [0,+∞[ by (7)ϕx=∑i=0p-1φTix,∀x∈X.The idea of the proof is to construct a transfinite orbit xαα∈Ω, where Ω is the first uncountable ordinal satisfying, for each α∈Ω, A(α): Txα=xα+1;B(α): xα=limλ→α-⁡xλ, whenever α is an ordinal limit;C(α): xα∈xμG, whenever μ≺α;D(α): dxα,xμ≤ϕxμ-ϕxα, whenever μ⪯α.
Consider the sequencexn defined for each n∈N by xn+1=Txn. Since Tx0∈x0G and using the monotony of T, we obtain xn+1∈xnG for each n∈N. According to (KS2), the nonnegative sequence ϕxn is decreasing and then converges. From (KS1), we get that for all integers n>m(8)dxm,xn≤∑i=mn-1dxi,xi+1≤∑i=mn-1ϕxi-ϕxi+1≤ϕxm-ϕxn.Hence, xn is a Cauchy sequence and then converges to xω∈X. Let us put xω+1=Txω. Clearly the properties A(α)–D(α) are satisfied for each α≤ω. Let β∈Ω. Assume that the orbit xαα<β has been defined. We need to define xβ and show that the four properties Aβ–Dβ hold. For that, we have to distinguish two cases, when β is an immediate successor or β is an ordinal limit. Clearly Aβ and Bβ are satisfied; let us focus on Cβ and Dβ.
Claim  1 (Cβ holds)
Case  1. Assume that β is an ordinal limit; that is, there exists a strictly ascending sequence βnn of ordinals in Ω such that β=supβn;n∈N and βm⪯βn≺β whenever m≤n. Since Dα holds for all α≺β, we get (9)dxβn,xβm≤ϕxβm-ϕxβn,which implies that ϕxβnn is decreasing sequence in 0,∞ and hence it is convergent. Then xβn is Cauchy sequence, so it converges in X. Set xβ=limn→∞⁡xβn. By (OSCL) property, we obtain xβ∈xβnG for all n∈N. Let α≺β. There exists n0∈N such that for each n≥n0 we have (10)α⪯βn≺β,and thus for each n≥n0, (11)xβn∈xαG,xβ∈xβnG⟹xβ∈xαG.Since α is taken arbitrary, we obtain Cβ.
Case  2. Assume that β is an immediate successor; there exists α≺β such that β=α+1. (i)
Ifα is an immediate successor, there exists an ordinal μ such α=μ+1. From C(α), we have xα∈xμG and using the G-monotonicy of T it follows that xβ∈xαG and so C(β) holds.(ii)
Ifα is an ordinal limit, from Proposition 7, there exists an ascending sequence {αn}⊂Ω such that α=supαn:n∈N. From B(α) we have xα=limn→+∞⁡xαn. Using the (OSCL) condition, we have xα∈xαnG. Since T is G-monotone, xβ∈[xαn+1]G and as xαn+1∈[xαn]G, we get xβ∈[xαn]G. Again, (OSCL) insures that xβ∈xαG. Then C(β) holds.
Claim  2 (D(β) holds)
Case  1. Assume that β is ordinal limit. Let α≺β. There exists n0∈N such that for each n≥n0 we have (12)α⪯βn≺β.Then we get for each n≥n0 that (13)dxα,xβn≤ϕxα-ϕxβn,and for all i∈0,1,…,p-1(14)limn→∞⁡Tixβn=limn→∞⁡xβn+i=xβ.Since φ is lower semicontinuous, we get (15)φxβ≤liminfn→∞⁡φTixβn.From C(β), we have xβ∈xβnG for all n∈N. Using the same argument as above, we get Txβ∈xβnG and (OSCL) insures that Txβ∈xβG. Hence, for all i∈0,1,…,p-1, Ti+1xβ∈TixβG. This implies that (16)φTpxβ≤φTp-1xβ≤⋯≤φTxβ≤φxβ.By passing to limit superior in inequality (13), it follows that (17)dxα,xβ≤ϕxα-liminfn→+∞⁡ϕxβn≤ϕxα-∑i=0p-1liminfn→+∞⁡φTixβn≤ϕxα-pφxβ≤ϕxα-ϕxβ.Hence, Dβ holds.
Case  2. Assume that β=α+1 is an immediate successor; we have shown above that Cβ holds. Then Txα=xβ∈xαG and by assumption we get (18)dxα,xβ≤ϕxα-ϕxβ,and for all γ⪯α, we have (19)dxγ,xα≤ϕxγ-ϕxα.The triangle inequality implies that (20)dxγ,xβ≤ϕxγ-ϕxβ,for each γ⪯β, which completes the proof of Dβ in both cases.
Thus, the orbitxαα∈Ω is well constructed. Since ϕxα is nonincreasing on xα and Ω is uncountable, there must exist α0∈Ω such that ϕ(xα) is constant for all α⪰α0. From Dα0+1, we get (21)dxα0+1,xα0≤ϕxα0-ϕxα0+1=0.Hence, Txα0=xα0+1=xα0.We support our result by giving an example of a mapping which isG-Kirk-Saliga mapping, for some integer p>1, but not G-Caristi.Example 11.
Consider the metric space(X,d), where X=[0,1] and d(x,y)=x-y, for all x,y∈X. Endow X with the directed graph G=(X,E) represented in Figure 1, where (22)E=Δ∪12,0,1,0,12n,0,12n,12n+1:n∈N∗.
Consider the functionφ:X→[0,+∞[ defined by (23)φx=x,ifx∈0,1;0,ifx=1,and the mapping T:X→X defined by Tx=x2, if x∈0,1; T1=1/2.
One can see thatT1∉1G, T1/2∉1/2G and (24)XG≔x∈X:Tx∈xG=0,12n:n∈N∗.We verify the following assertions: (i)
X,d is complete and T is G-monotone obviously.(ii)
G satisfies the (OSCL) property. Indeed, let xn be a sequence in X such that {xn} converges to some x∈X and xn+1∈xnG, for all n∈N. Two cases to distinguish are as follows:
(1)
There existsn0∈N such that xn=xn0, for all n≥n0. Then for all n≥n0, xn=x. If x is an isolated vertex, the (OSCL) is obviously satisfied. If not, y∈xnG for all n∈N implies y∈xG. Thus, (OSCL) is satisfied.(2)
For allk∈N, there exists mk∈N such xmk≠xk. Then xn⊆1/2n:n∈N; that is, there exists a nondecreasing function ψ:N→N such that xn=1/2ψ(n) for all n∈N, and x=0. If y∈xnG for all n∈N, then y=0. Thus, (OSCL) is satisfied.(iii)
T is G-kirk-Saliga mapping in X with p=3. Indeed, for all x∈XG, (25)x-x2≤x-x4⟺dx,Tx≤φx-φT3x,butT is not G-Caristi mapping, since (26)d12,T12>φ12-φT12.(iv)
T0∈0G.
andT admits a fixed point in X which is 0.Figure 1
GraphG (the loops and the isolated vertices are not represented).If we remove the (OSCL) property, we are not certain that the fixed point will be obtained. Let us illustrate that by this counterexample.Example 12.
Replace in the above example the digraphG by the digraph G′ represented in Figure 2, where (27)E′=Δ∪12n,0,12n,1,12n,12n+1:n∈N∗,and we consider the mapping T:X→X defined as follows: (28)T0=1;Tx=x2,ifx∈0,1;T1=0.One can see that G′ satisfies (OSC) property but does not satisfy the (OSCL), since 1/2n→0 and for all n∈N, 0∈1/2nG′ and 1∈1/2nG′ but 1∉0G′. The mapping T satisfies all others conditions of Theorem 10 but has no fixed point in X.Figure 2
GraphG′ (the loops and the isolated vertices are not represented).Corollary 13.
Let(X,d,G) be a complete metric space endowed with a reflexive digraph which satisfies the (OSCL) property. Let n=∏i=0kpiαi, where pi is prime integer and (k,αi)∈N×N∗, for each i∈0,1,…,k, and let T:X→X be a G-monotone mapping such that there exists j∈{0,1,…,k}, for all x∈X, (29)Tpjx∈xG⟹max⁡dx,Tpix:i∈0,1,…,k≤φx-φTnx,φTpjx≤φx.Then T admits a fixed point in X provided that there exists an element x0∈X such that Tx0∈x0G.Proof.
Clearly,Tpj satisfies all conditions of Theorem 10; then there exists x¯∈X such that Tpjx¯=x¯ and so, Tnx¯=x¯. Since Tpjx¯∈x¯G, we get from (29) that(30)dx¯,Tpix¯≤φx¯-φTnx¯=0,∀i∈0,1,…,k\j.Now, let i∈{0,1,…,k}∖{j}; then Tpix¯=x¯. By Bezout identity, there exists (u,v)∈Z2 such that uv≤0 and pju+piv=1. Without loss of generality, we suppose that u<0 and v>0. Since Tpjx¯=x¯ and Tpix¯=x¯, then T1-pjux¯=Tx¯ and Tpivx¯=x¯. Since piv=1-pju, then Tx¯=x¯.We conclude this work by a discussion about preordered sets.Let(X,⩽) be a preordered set; that is, the binary relation “⩽” is reflexive and transitive.Given a reflexive digraphG=(X,E), we can always define a preorder ⩽G on X as follows: (31)x⩽Gy⟺x∈yG.Conversely, if (X,⩽) is a preordered set, we define the reflexive digraph G⩽ as follows: two vertices x,y∈X are connected by an arc from x to y if x⩽y. Note that G⩽ is transitive (i.e., if (x,y)∈E(G⩽) and (y,z)∈E(G⩽); then (x,z)∈E(G⩽)), so x∈[y]G⩽⇔(x,y)∈E(G⩽). These remarks lead to the following definition.Definition 14.
Let(X,⩽) be a preordered set. We say that (X,⩽) satisfies the (OSCL) condition if and only if G⩽ satisfies the (OSCL) condition.We shall say thatT:X→X is ⩽-monotone (resp., ⩽-Kirk-Saliga) mapping if G⩽-monotone (resp., G⩽-Kirk-Saliga) mapping.An analogue version of Theorem10 in the setting of the preordered metric spaces may be stated as follows.Theorem 15.
Let(X,d,⩽) be a preordered complete metric space satisfying the (OSCL) condition. Let T:X→X be a ⩽-monotone and ⩽-Kirk-Saliga mapping. If there exists an element x0∈X such that Tx0⩽x0, then T admits a fixed point in X.Remark 16.
If moreover the above binary relation⩽ is antisymmetric (i.e., (x⩽y and y⩽x) imply x=y), we obtain, from Remark 6, the following result established by Alfuraidan and Khamsi.Corollary 17 (see [16, Theorem  5]).
Let (X,d,⩽) be a complete partially ordered metric space satisfying the property (OSC). Let T:X→X be a ⩽-monotone and ⩽-Kirk-Saliga mapping with p=1. If there exists an element x0∈X such that Tx0⩽x0, then T admits a fixed point in X.



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