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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 14, Issue 1 (Jan 2016)

Issues

Fixed point theorems for cyclic contractive mappings via altering distance functions in metric-like spaces

Jianhua Chen
  • Corresponding author
  • School of Mathematics and Statistics, Central South University, Changsha, 410083, Hunan, China
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/ Xianjiu Huang / Shengjun Li
Published Online: 2016-11-16 | DOI: https://doi.org/10.1515/math-2016-0080

Abstract

In this paper, we introduce two different contractive conditions and prove some new fixed point theorems for cyclic (ψ,ϕ,φ)α-contractive mappings and α-(κ,φ)g-contractive mappings in complete metric-like spaces via altering distance functions. Our results generalize and extend some existing results. Moreover, some examples are given to support the obtained results.

Keywords: α-admissible; (ψ,ϕ,φ)α-contractive; α-(κ,φ)g-contractive; Fixed point; Metric-like spaces

MSC 2010: 47H10; 54H25

1 Introduction and preliminaries

It is well known that functional analysis is made up of two main methods which are variational methods and fixed point methods. Variational methods are used to prove the existence of solutions for differential equations [19]. However, fixed point methods are studied by many scholars [10-13] in different spaces. Especially, in 2012, Amini-Harandi [14] introduced the notions of metric-like spaces which is considered to be an interesting generalization of metric spaces, partial metric spaces [15] and quasi-metric spaces [16]. At the same time, the author proved some fixed point theorems in complete metric-like spaces. Based on the work of Amini-Harandi [14], P Salimi et al. [17] proved some interesting fixed point theorems for generalized cyclic contractive mappings in complete metric-like spaces. Motivated by [17, 18], the purpose of this paper is to introduce two different integral type contractive conditions and prove some new fixed point theorems for cyclic (ψ,ϕ,φ)α-contractive mappings and α-(κ,φ)g-contractive mappings in complete metric-like spaces via altering distance function. Our results generalize and extend some existing results. Moreover, we give some examples and an application to support our results.

Throughout this paper, Let R = (−∞, +∞), R+ = [0, +∞), N be the set of all positive integers, N0 = ∪N}, and

Θ1={ζ|ζ:R+R+ is Lebesgue integral, summable on each compact subset of R+andl0ϵζ(t)>0for eachϵ>0}

Θ2={φ|φ:R+R+ is nondecreasing and continuous function with φ(0) = 0 if and only if t = 0}.

Let Θ3 and Θ4 be the set of the classes of all functions ψ, φ : [0, ∞) → [0, ∞) (respectively) satisfying the following conditions:

  1. ψ is nondecreasing and continuous;

  2. ϕ is lower semicontinuous;

  3. ψ(t)ψ(s)+ϕ(s)>0 for all t > 0 and s = t or s = 0.

Θ5={κ|κ:R+R+ is nondecreasing and continuous function with n=1κn(t)<}.

By the definition of Θ5, it is an easy matter to show that κ(t) < t for all t ∈ (0, +∞), κ(0) = 0 and limnκn(t)=0.

Now, we recall some basic concepts, notations and known results of metric-like spaces which will be used in the sequel.

([14]). A mapping σ : X × XR+ where X is a nonempty set, is said to be a metric-like mapping on X iffor any x,y, z ∈ X the following three conditions hold true: (σ1)σ(x,y)=0x=y;

(σ2)σ(x,y)=σ(y,x);

(σ2)σ(x,z)<_σ(x,y)+σ(y,z).

The pair (X, σ) is then called a metric-like space.

In [14], the author have shown that the metric-like space induces a Hausdorff topology, and the converge described is as follows.

A metric-like on X satisfies all of the conditions of a metric except that σ(x,x) may be positive for x ∈ X. Each metric-like σ on X generates a topology τσ on X whose base is the family of open σ-balls Bσ(x,ϵ)={yX|σ(x,y)σ(x,x)<ϵ}

for all x ∈ X and ϵ > 0. Then a sequence {xn} in the metric-like space (X, σ) converges to a point x ∈ X if and only if limx(nn,x)=(x,x).

A sequence {xn} of elements of X is called σ-Cauchy if the limit limnσ(xm,xn) exists and is finite. The metric-like space (X, σ) is called complete if for each σ-Cauchy sequence {xn}, there is some x ∈ X such that limnσ(xn,xm)=σ(x,x)=σ(xn,x).

([19, 20]). Let (X, σ) be a metric-like space and U be a subset of X. We say that U is a σ-open subset of X, iffor all x ∈ X there exists s > 0 such that Bσ(x,s)U. Also, V ⊆ X is a σ-closed subset of X if XV is a σ-open subset of X.

([21]). Let (X, σ) be a metric-like space and V be a σ-closed subset of X. Let {xn} be a sequence in V such that xn → x as n → ∞. Then x ∈ V.

([21]). Let (X, σ) be a metric-like space and {xn} be a sequence in X such that xn → x as n → ∞ and σ(x,x) = 0. Then limnσ(xn,y)=σ(x,y)forallyX.

([21]). Let (X, σ) be a metric-like space. Then,

  1. if σ(x, y) = 0, then σ(x, x) = σ(y, y) = 0;

  2. if {xn} is a sequence such that limnσ(xn,xn+1)=0, Then limnσ(xn,xn)=limnσ(xn+1,xn+1)=0;

  3. if xy, then σ(x, y) > 0;

  4. σ(x,x)2ni=1nσ(x,xi) holds for all xi, x ∈ X where 1 ≤ in.

([21]). Let (X, σ) be a metric-like space and let {xn} be a sequence in X such that limnσ(xn+1,xn)=0.

If limnσ(xn,xm)0, then there exist ϵ > 0 and two sequences {m(k)} and {n(k)} of positive integers such that n(k) > m(k)>k and the following four sequences tend to ϵ when k → ∞: {σ(xm(k),xn(k))},{σ(xm(k),xn(k)+1)},{σ(xm(k)1,xn(k))},{σ(xm(k)1,xn(k)+1)}.(1)

In [17], the authors give the following remark which is very useful in this paper.

If the condition of the above lemma is satisfied, then the sequences {σ(xm(k);xn(k)+s))} and {σ(xm(k)+1,xn(k)+s))} also converge to ϵ when k → ∞, where s ∈ N.

The notion of α-admissibility was defined by [18] and many authors [22-24] applied α-admissibility to some various metric spaces so that they can get some generalized fixed point theorems.

([18]). Let F : XX and α : X × X →[0,∞). The mapping F is said to be α-admissible if for all x, y ∈ X, we have α(x,y)1α(Fx,Fy)1.(2)

([25]). Let (X, σ) be a metric-like space. Let α : X × X→[\mathrm{O},∞$) and F : XX. We say that F is α-continuous on (X, σ) , if xnxasn,σ(xn,xn+1)1forallnNTxnTx.

In order to prove our results, we need to give the following definition of altering distance functions which was introduced in [26].

In order to prove our results, we need to give the following definition of altering distance functions which was introduced in [26].

([26]). An altering distance is a function ρ:[0,+)[0,+) which is increasing, continuous and vanishes only at the origin.

Fixed point problems involving altering distances have also been studied in [29-32]. In [30], the following lemma shows that contractive conditions of integral type can be interpreted as contractive conditions involving an altering distance.

([30]). Let ζΘ1. Define Φ0(t) = 0tζ(t)dt. Then Φ0 is an altering distance

2 Fixed point theorems for cyclic (ψ,ϕ,φ)α-contractive mappings

In this section, we present some new fixed point theorems for cyclic (ψ,ϕ,φ)α-contractive mappings in metric-like spaces. Before stating our main results, we need to give the following definition.

Let (X, σ) be a metric-like space, r ∈ N, A1, A2 , …, Ar be σ-closed nonempty subsets of X Y=i=1rAiandα:Y×Y[0,) be a mapping. We say that F : YY is a cyclic (ψ,ϕ,φ)α-contractive mapping if

  1. F(Aj)Aj+1,i=1,2,,rwhereAr+1=A1.(3)

  2. for any xAiandyAi+1,i=1,2,,r,whereAr+1=A1 and α(x,Fx)α(y,Fy) ≥ 1, we have ψ(ρ!(σ(Fx,Fy)+φ(σ(Fx,Fy))))<_ψ(ρ(Mσ(x,y)+φ(Mσ(x,y))))dGn+(1)(f)ϕ(ρ(Mσ(x,y)+ψ(Mσ(x,y))))(4)

    where ρ is an altering distance function, φ, ∈ Θ2, ψ ∈ Θ3, ϕ ∈ Θ4 and Mσ(x,y)=max{σ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy)+σ(y,Fx)4}.

If in Definition 2.1, we take X = Ai, i = 1, 2, …; r, then we say that F is (ψ,ϕ,φ)α-contractive mapping. We denote the set of all fixed points of F by Fix (F) , i.e., Fix (F) = {x ∈ X : Fx = x}.

Assume that F : XX is a cyclic (ψ,ϕ,φ)α-contractive mapping, x ∈ Fix(F) and α (x,x) ≥ 1. Then we must have σ(x,x) = 0. Now, we prove this results. Suppose that x ∈ Fix(F), α(x,x) ≥ 1 and σ(x,x) > 0. Then we have Mσ(x,x) = σ(x,x). Moreover by (4) It follows that ψ(ρ!(σ(x,x)+,(σ(x,x))))=ψ(ρ(σ(Fx,Fx)+,(σ(Fx,Fx))))ψ(ρ(σ(x,x)+,(σ(x,x))))ϕ(ρ(σ(x,x)+,(σ(x,x)))),

which is a contradiction. Hence σ(x,x) = 0.

Now, we want to state and prove our main result of this section.

Let (X, σ) be a complete metric-like space. Let r be a positive integer A1, A2, …, Ar be σ-closed nonempty subsets of X,Y=i=1rAiandα:Y×Y[0,) be a mapping. Assume that F : YY is a cyclic (ψ,ϕ,φ)α-contractive mapping satisying the following conditions:

  1. F is an α-admissible mapping;

  2. there exists element x0 in Y such that, α(x0,Fx0) ≥ 1;

    1. F is α-continuous, or;

    2. if {xn} is a sequence in X such that α(xn, xn+1) ≥ 1 for all nN and xn → x as n → ∞, then α(x,Fx) ≥ 1.

    Then F has a fixed point xi=1pAi. Moreover if

  3. for all x ∈ Fix (F) we have α(x,x) ≥ 1.

    Then F has a unique fixed point xi=1rAi.

By the condition (ii), let x0 be an arbitrary point of Y such that α(x0,Fx0) ≥ 1. Then there exists some i0 such that x0Ai0. Now by (3), F(Ai0)Ai0+1 implies that Fx0Ai0+1. Thus there exists x1 in Ai0+1 such that Fx0 = x1. Similarly, Fxn = xn+1, where xnAin. Hence for n ≥ 0, there exists in {1, 2, …, r such that xn ∈ Ain and Fxn = xn+1. On the other hand, F is an α-admissible mapping, hence by Definition 1.8, we get α(x0,Fx0)1α(x1,Fx1)=α(Fx0,F(Fx0))1.

Again, since F is an α-admissible mapping, we have α(x1,Fx1)1α(x2,Fx2)=α(Fx1,F(Fx1))1.

Continuing this process, we can construct a sequence {xn} in Ajn such that α(xn1,Fxn1)1α(xn,Fxn)1forallnN0

and so α(xn,Fxn)α(xn1,Fxn1)1forallnN.(5)

If for some n0 = 0,1,2, …, we have xn0 = xn0+1, then Fxn0 = xn0, that is, xn0 is a fixed point of F. So from now on, we assume xnxn+1 for all n ∈ N. Hence, by Lemma 1.5 (C) we have σ (xn, xn+1) > 0 for all n ∈ N. Now, we want to show that σ(xn,xn+1)<σ(xn1,xn). If σ(xn,xn+1)σ(xn1,xn), then by (4) and (5), we have (ρ!(σ(xn,xn+1)+φ(σ(xn,xn+1))))=ψ(ρ(σ(Fxn1,Fxn)+φ(σ(Fxn1,Fxn))))ψ(ρ(Mσ(xn1,xn)+φ(Mσ(xn1,xn))))ϕ(ρ(Mσ(xn1,xn)+φ(Mσ(xn1,xn)))),(6)

where Mσ(xn1,xn)=maxσ(xn1,xn),σ(xn1,Fxn1),σ(xn,Fxn),σ(xn1,Fxn)+σ(xn,Fxn1)4=maxσ(xn1,xn),σ(xn1,xn),σ(xn,xn+1),σ(xn1,xn+1)+σ(xn,xn)4=maxσ(xn1,xn),σ(xn,xn+1),σ(xn1,xn+1)+σ(xn,xn)4(7)

From Lemma 1.5 (D), we can get σ(xn,xn)σ(xn1,xn)+σ(xn,xn+1)

and by (σ3) in Definition 1.1, we have σ(xn1,xn+1)σ(xn1,xn)+σ(xn,xn+1).

That is max{σ(xn1,xn),σ(xn,xn+1)}maxσ(xn1,xn),σ(xn,xn+1),σ(xn1,xn+1)+σ(xn,xn)4maxσ(xn1,xn),σ(xn,xn+1),σ(xn1,xn+1)+σ(xn,xn+1)2=max{σ(xn1,xn),σ(xn,xn+1)}.

Thus we have maxσ(xn1,xn),σ(xn,xn+1),σ(xn1,xn+1)+σ(xn,xn)4=max{σ(xn1,xn),σ(xn,xn+1)}.(8)

Therefore, from (7) and (8), we get Mσ(xn1,xn)=max{σ(xn1,xn),σ(xn,xn+1)}.(9)

By (6) and (9), we have ψ(ρ!(σ(xn,xn+1)Z+φ(σ(xn,xn+1))))ψ(ρ(max{σ(xn1,xn),σ(xn,xn+1)}+φ(max{σ(xn1,xn),σ(xn,xn+1)})))ϕ(ρ(max{σ(xn1,xn),σ(xn,xn+1)}+φ(max{σ(xn1,xn),σ(xn,xn+1)}))).(10)

If max {σ(xn1,xn),σ(xn,xn+1)}=σ(xn,xn+1), then by (10), we have ψ(ρ(σ(xn,xn+1)+φ(σ(xn,xn+1))))ψ(ρ(σ(xn,xn+1)+φ(σ(xn,xn+1))))ϕ(ρ(σ(xn,xn+1),+φ(σ(xn,xn+1)))),

which contradicts to the condition (iii) of Θ3 and Θ4. Hence ψ(ρ(σ(xn,xn+1)+φ(σ(xn,xn+1))))ψ(ρ(σ(xn1,xn)+φ(σ(xn1,xn))))ϕ(ρ(σ(xn1,xn),+φ(σ(xn1,xn)))).(11)

Since σ(xn,xn+1)σ(xn1,xn), by the properties of ψ, φ and Lebesgue integral function, we have ψ(ρ(σ(xn1,xn)+φ(σ(xn1,xn))))ψ(ρ(σ(xn,xn+1)+φ(σ(xn,xn+1)))).(12)

Hence, from (11) and (12), we can get ψ(ρ(σ(xn1,xn)+φ(σ(xn1,xn))))ψ(ρ(σ(xn1,xn)+φ(σ(xn1,xn))))ϕ(ρ(σ(xn1,xn),+φ(σ(xn1,xn)))),

which is a contradiction. Therefore σ(xn,xn+1)<σ(xn1,xn) holds for all n ∈ N and there exists δ ≥ 0 such that limnσn=δ. Next, we shall show that δ = 0. In order to prove δ = 0, we assume that δ > 0. By (11), together with the properties of ψ,ϕ,φ, we have ψ(ρ!(δ+φ(δ)))=limnsupψ(ρ(σ(xn,xn+1)+φ(σ(xn,xn+1))))limnsup[ψ(ρ(σ(xn1,xn)+φ(σ(xn1,xn))))ϕ(ρ(σ(xn1,xn)+φ(σ(xn1,xn))))]ψ(ρ(δ+φ(δ)))ϕ(ρ(δ+φ(δ))),

which is a contradiction. Thus, limnσ(xn,xn+1)=0.

Next, we want to prove that limnσ(xn,xm)=0, If limnσ(xn,xm)0, then by Remark 1.7, there exists ϵ > 0 and two sequences {m(k)} and {n(k)} of positive integers such that n(k) > m(k)>k and limnσ(xm(k),xn(k)+s)=ϵ, where s ∈ N. Chooses, s, l ∈ N such that s=m(k)+ls+1n(k) then we have xm(k)Aiandxn(k)+sAi+1 for some i ∈ {1, 2, …, r}. Therefore, using the contractive condition (4) with x=xm(k),y=xn(k)+s, we obtain ψ(ρ(σ(xm(k)+1,xn(k)+s+1)+φ(σ(xm(k)+1,xn(k)+s+1))))ψ(ρ(Mσ(xm(k),xn(k)+s)+φ(Mσ(xm(k),xn(k)+s))))ϕ(ρ(Mσ(xm(k),xn(k)+s)+φ(Mσ(xm(k),xn(k)+s)))),(13)

where Mσ(xm(k),xn(k)+s)=maxσ(xm(k),xn(k)+s),σ(xm(k),xm(k)+1),σ(xn(k)+s,xn(k)+s+1),σ(xm(k),xn(k)+s+1)+σ(xn(k)+s,xm(k)+1)4maxσ(xm(k),xn(k)+s),σ(xm(k),xm(k)+1),σ(xn(k)+s,xn(k)+s+1),2σ(xm(k),xn(k)+s)+σ(xn(k)+s+1,xn(k)+s)+σ(xm(k),xm(k)+1)4.(14)

Letting k → ∞ in (13) and (14), then we have ψ(ρ(ϵ+φ(ϵ)))ψ(ρ(ϵ+φ(ϵ))))ϕ(ρ(ϵ+φ(ϵ))),

which is a contradiction. Thus we can conclude that limnσ(xn,xm)=0, that is, the sequence {xn} is a σ-Cauchy sequence. Since Y is σ-closed in (X, σ) , there exists xY=i=1rAi such that limnxn=xin(Y,σ); equivalently, σ(x,x)=limnσ(x,xn)=limnσ(xn,xm)=0.

On the one hand, we assume that (a) of (iii) holds, that is, F is α-continuous. Then it is obvious that x=limnxn+1=limnFxn=Fx.

On the other hand, we assume that (b) of (iii) holds and α(xn(k),xn(k)+1)1. Then we have α(x,Fx) ≥ 1 and so α(x,Fx)α(xn(k),Fxn(k))1.

Next, we prove that x is a fixed point of F. Since limnxn=xandF(Aj)Aj+1,j=1,2,,r, where An+1 = A1, the sequence {xn} has infinitely many terms in each Ai for i ∈ {1, 2, …, r}. Suppose that xAi. Then TxAi+1 and we take a subsequence {xn(k)} of {xn} with xn(k)Ai-1 (the existence of this subsequence is guaranteed by the above mentioned comment). By using the contractive condition (4), we obtain ψ(ρ(σ(Fx,Fxn(k))+φ(σ(Fx,Fxn(k)))))ψ(ρ(Mσ(x,xn(k))+φ(Mσ(x,xn(k)))))ϕ(ρ(Mσ(x,xn(k)),+φ(Mσ(x,xn(k))))),(15)

where Mσ(x,xn(k))=maxσ(x,xn(k)),σ(x,Fx),σ(xn(k),Fxn(k)),σ(x,Fxn(k))+σ(xn(k),Fx)4=maxσ(x,xn(k)),σ(x,Fx),σ(xn(k),xn(k)+1),σ(x,xn(k)+1)+σ(xn(k),Fx)4.(16)

In (15) and (16), letting k → ∞ and using the lower semi-continuity of ϕ, we can get ψ(ρ(σ(Fx,x)+φ(σ(Fx,x))))ψ(ρ(σ(x,Fx)+φ(σ(x,Fx))))ϕ(ρ(σ(x,Fx)+φ(σ(x,Fx)))),

which is a contradiction. Hence, σ(Fx,x) = 0, that is, Fx = x. Therefore, x is a fixed point of F. The cyclic character of F and the fact that x ∈ X is a fixed point of F, imply that xi=1rAi.

At last, we shall prove the uniqueness of the fixed point. Now, we assume that x,yi=1rAi are two fixed points of F and xy. Suppose that condition (iv) holds. Then we have σ(x,y)>0,α(x,x)1,α(y,y)1 and then by (4), we obtain ψ(ρ!(σ(x,y)+φ(σ(x,y))))=ψ(ρ(σ(Fx,Fy)+φ(σ(Fx,Fy))))ψ(ρ(Mσ(x,y)+φ(Mσ(x,y))))ϕ(ρ(Mσ(x,y),+φ(Mσ(x,y)))),

where Mσ(x,y)=maxσ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy)+σ(y,Fx)4=maxσ(x,y),σ(x,x),σ(y,y),σ(x,y)+σ(y,x)4.

Using Remark 2.2 we have σ(x, x) = σ(y,y) = 0, therefore it follows from the above inequality that ψ(ρ(σ(x,y)+φ(σ(x,y))))ψ(ρ(σ(x,y)+φ(σ(x,y))))ϕ(ρ(σ(x,y),+φ(σ(x,y)))),

which is a contradiction. Hence, σ(x,y) = 0, that is to say that x = y. The proof is completed.

If in Theorem 2.4, we choose ϕ(t)=0andρ(t)=0tζ(t)dt, then we can get Theorem 2.1 of [17].

Next, we give the following examples to illustrate the usefulness of Theorem 2.4.

Consider X = {0, 1, 2}. Let σ : X × X→ [0,∞) be a mapping defined by σ(0,0)=σ(1,1)=0,σ(2,2)=52;

σ(0,2)=σ(2,0)=2,σ(1,2)=σ(2,1)=2,

σ(0,1)=σ(1,0)=3.

It is clear that (X, σ) is a complete metric-like space. Suppose that A1 = {0, 1} and A2 = {1, 2} and Y = A1A2.

Define F : YY and α : Y × Y → [0, +∞) by Fx = 1 for all x ∈ Y and α(x,y) = x2 + 2 for all x, y ∈ Y.

Then we have α(x,Fx)α(y,Fy)1 for all x,y ∈ Y and T(A1)A2,F(A2)A1. Thus x ∈ A1 and y ∈ A2.

Also define ψ,ϕ:[0,)[0,)byψ(t)=tandϕ(t)=12t. Next, our process of the proof is divided into four steps as follows:

Case 1: Assume that x = 0, y = 1. On the one hand, Fx = 1 and Fy = 1, that is, σ(Fx,Fy) = σ(1,1) = 0. On the other hand, we have Mσ(x,y)=maxσ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy)+σ(y,Fx)4=maxσ(0,1),σ(0,1),σ(1,1),σ(0,1)+σ(1,1)4=max3,3,0,3+04=3.

Hence, we have ψ(ρ(σ(Tx,Ty)+,(σ(Tx,Ty))))=ψ(ρ(0+,(0)))=012(ρ(3+,(3)))=ψ(ρ(3+,(3)))ϕ(ρ(3+,(3)))=ψ(ρ(Mσ(x,y)+,(Mσ(x,y))))ϕ(ρ(Mσ(x,y)+,(Mσ(x,y)))).

So F is a cyclic (ψ,ϕ,φ)α-contractive mapping.

Case 2: Assume that x = 0, y = 2. Similarly as in the proof of Case 1, we can get F is a cyclic (ψ,ϕ,φ)α-contractive mapping.

Case 3: Assume that x = 1, y = 1. Similarly as in the proof of Case 1, we can get F is a cyclic (ψ,ϕ,φ)α-contractive mapping.

Case 4: Assume that x = 1, y = 2. Similarly as in the proof of Case 1, we can get F is a cyclic (ψ,ϕ,φ)α-integral type contractive mapping.

Thus F is a cyclic (ψ,ϕ,φ)α-contractive mapping. Clearly α(0,F0)=α(0,1)1 and so the condition (ii) of 2.1 is satisfied. If α(x,y) ≥ 1, then x,y ∈ {0, 1, 2} which implies that α(Fx,Fy) = 3 ≥ 1, that is, F is an α-admissible mapping. Let {xn} be a sequence in X such that, α(xn, Fxn) ≥ 1 and xn → x as n → ∞. Then, we must have xn {0, 1, 2} and so, x ∈ {0, 1, 2} that is, α(x,Fx) ≥ 1. Hence, all the conditions of Theorem 2.4 hold and F has a fixed point x=1A1A2.

Consider X = {0, 1, 2}. Let σ : X × X → [0,∞) be a mapping defined by σ(0,0)=σ(1,1)=0,σ(2,2)=52;

σ(0,2)=σ(2,0)=2,σ(1,2)=σ(2,1)=3,

σ(0,1)=σ(1,0)=32.

It is clear that (X, σ) is a complete metric-like space. Suppose that A1 = {0, 2} and A2 = {1,2} and Y = A1A2. Define F : YY and α : Y × Y →[0, +∞) by Fx=2,ifx=01,ifx=10,ifx=2.

for all x ∈ Y and α(x,y)=x2+3,ifx,y{1,2}0,ifx,y{0}.

Then F(A1) ⊆ A2, F(A2) ⊆ A1. Thus x ∈ A1 and y ∈ A2. Now, if x ∈ {0} or y ∈ {0}, then α(x,Fx) = 0 or α(y,Fy) = 0. That is, α(x,Fx)α(y,Fy) = 0 ≤ 1, which is contradiction. Hence x ∈ {2} and y ∈ {1, 2}.

Also define ψ,ϕ:[0,)[0,)byψ(t)=t,ϕ(t)=12t,ρ(t)=0tζ(t)dt,φ(t)=tandζ(t)=1. Next, our process of the proof is divided into two steps as follows:

Case 1: Assume that x = 2, y = 1. On the one hand, Fx = 0 and Fy = 1, that is, σ(Fx,Fy) = σ(0,1) = 3/2. On the other hand, we have Mσ(x,y)=maxσ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy)+σ(y,Fx)4=maxσ(2,1),σ(2,0),σ(1,1),σ(2,1)+σ(1,0)4=max3,2,0,3+04=3.

Hence, we have ψ(ρ!(σ(Tx,Ty)+φ(σ(Tx,Ty))))=ψ(032+φ(32)1dt)=12(03+φ(3)1dt)=ψ(03+φ(3)ζ(t)dt)ϕ(03+φ(3)ζ(t)dt)=ψ(0Mσ(x,y)+φ(Mσ(x,y))ζ(t)dt)ϕ(0Mσ(x,y)+φ(Mσ(x,y))ζ(t)dt).

So F is a cyclic (ψ,ϕ,φ)α-contractive mapping.

Case 2: Assume that x = 2, y = 2. On the one hand, Fx = 0 and Fy = 0, that is, σ(Fx,Fy) = σ (0,0) = 0. On the other hand, we have Mσ(x,y)=maxσ(x,y),σ(x,Fx),σ(y,Ty),σ(x,Ty)+σ(y,Fx)4=maxσ(2,1),σ(2,0),σ(1,1),σ(2,1)+σ(1,0)4=max3,2,0,3+04=3.

Hence, we have ψ(ρ!(σ(Fx,Fy)+φ(σ(Fx,Fy))))=ψ(00+φ(0)1dt)=012(03+φ(3)1dt)=ψ(03+φ(3)ζ(t)dt)ϕ(03+φ(3)ζ(t)dt)=ψ(0Mσ(x,y)+φ(Mσ(x,y))ζ(t)dt)ϕ(0Mσ(x,y)+φ(Mσ(x,y))ζ(t)dt).=ψ(ρ(Mσ(x,y)+φ(Mσ(x,y))))ϕ(ρ(Mσ(x,y)+φ(Mσ(x,y)))).

Thus F is a cyclic (ψ,ϕ,φ)α-contractive mapping. Clearly α(0,F0)=α(0,1)=321 and so the condition (ii) of 2.1 is satisfied. If α(x,y) ≥ 1, then x,y ∈ {1,2} which implies that α (Fx,Fy) ≥ 1, that is, F is an α-admissible mapping. Let {xn} be a sequence in X such that, α(xn,Fxn) ≥ 1 and xn → x as n → ∞. Then, we must have xn {1,2} and so, x ∈ {1,2} that is, α(x,Fx) ≥ 1. Hence, all the conditions of Theorem 2.4 hold and F has a fixed point x = 2 A1A2.

If in Theorem 2.4, we take ψ(t) = t, then we can get the following corollary.

Corollary 2.8. Let (X, σ) be a complete metric-like space. Let r be a positive integer A1, A2, …, Ar be σ-closed nonempty subsets of X,Y=i=1rAiandα:Y×Y[0,) be a mapping. Assume that F : YY is a mapping satisying the following conditions:

  1. F(Aj)Aj+1,j=1,2,;rwhereAr+1=A1;

  2. F is an α-admissible mapping;

  3. there exists element x0 in Y such that, α(x0,Fx0) ≥ 1;

    1. F is α-continuous, or;

    2. if {xn} is a sequence in X such that α(xn,xn+1) ≥ 1 for all n ∈ N and xn → x as n → ∞, then

      (v) there exist, φ ∈ Θ2 and ϕ ∈ Θ4 such that α(x,Fx)α(y,Fy)1ρ((Fx,Fy)+φ(σ(Fx,Fy)))ρ(Mσ(x,y)+φ(Mσ(x,y)))ϕ(ρ(Mσ(x,y)+φ(Mσ(x,y)))),

      where Mσ(x,y)=maxσ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy)+σ(y,Fx)4

      for all xAi,yAi+1,i=1,2,,r,whereAr+1=A1 and ρ is an altering distance function.

      Then F has a fixed point xi=1rAi. Moreover if

  4. for all x ∈ Fix(F) we have α(x,x) ≥ 1.

    Then F has a unique fixed point xi=1rAi.

    If in Corollary 2.8, we take X = Ai, i = 1, 2, …, r, then we can get the following results.

Let (X, σ) be a complete metric-like space. α : X × X → [0, ∞) be a mapping. Assume that F : XX is a mapping satisying the following conditions:

  1. F is an α-admissible mapping;

  2. there exists element x0 in Y such that, α(x0,Fx0) ≥ 1;

    1. F is α-continuous, or;

    2. if {xn} is a sequence in X such that α(xn, xn+1) ≥ 1 for all n ∈ N and xn → x as n → ∞, then α(x,Fx) ≥ 1.

  3. there exists, Θ2 and ϕ ∈ Θ4 such that α(x,Fx)α(y,Fy)1ρ(σ(Fx,Fy)+φ(σ(Fx,Fy)))ρ(Mσ(x,y)+φ(Mσ(x,y)))ϕ(ρ(Mσ(x,y)+φ(Mσ(x,y)))),

    where Mσ(x,y)=maxσ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy)+σ(y,Fx)4

    for all x, y ∈ X, where ρ is an altering distance function.

    Then F has a fixed point. Moreover if

  4. for all x ∈ Fix(F) we have α (x,x) ≥ 1.

    Then F has a unique fixed point.

It is obvious that Corollary 2.9 improves Theorem 2.7 of [14].

If in Theorem 2.4, we choose ψ(t)=t,ϕ(t)=(1k)t(0k<1),φ(t)=t then we can get the following corollary.

Let (X,σ) be a complete metric-like space. Let r be a positive integer A1, A2,…,Ar be σ-closed nonempty subsets of X,Y=i=1rAiandα:Y×Y[0,) be a mapping. Assume that F : YY is a mapping satisying the following conditions:

  1. F(Aj)Aj+1,j=1,2,,rwhereAr+1=A1;

  2. F is an α-admissible mapping;

  3. there exists element x0 in Y such that, α (x0,Fx0)≥ 1;

    1. F is α -continuous, or;

    2. if {xn} is a sequence in X such that α(xn,xn+1)>_1forallnNandxnxasn,then α(x,Fx)≥ 1.

  4. there exist k ∈[0,1)and φ ∈ Θ2 such that α(x,Fx)α(y,Fy)>_1ρ(σ(Fx,Fy)+φ(σ(Fx,Fy)))<_kρ(Mσ(x,y)+φ(Mσ(x,y))),

    where Mσ(x,y)=max{σ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy)+σ(y,Fx)4}

    for all xAi,yAi+1,i=1,2,,r,whereAr+1=A1and ρ is an altering distance function. Then F has a fixed point xi=1rAi.Moreover if

  5. for all xFix(F) we have α(x,x)≥1.

Then F has a unique fixed point xi=1rAi.

If in Theorem 2.4, we take X=Ai, i=1,2,… n, then we can get the following results.

Let (X σ) be a complete metric-like space. α : X × X →[0,蜴) be a mapping. Assume that F : XX is a(ψ φ,ϕ)α-contractive mapping satisying the following conditions:

  1. F is an α-admissible mapping;

  2. there exists element x0 in Y such that, α(x0, Fx0)≥ 1;

    1. F is α-continuous, or;

    2. if {xn} is a sequence in X such that α(xn,xn+1)>_1for all nN and xnx as n →蜴 , then α(x,Fx)≥ 1.

    Then F has a fixed point. Moreover if

  3. for all xFix(F) we have α(x,x)≥1.

Then F has a unique fixed point.

In what follows, we give a cyclic (ψ φ ϕ)α-contractive mapping in metric spaces.

Let (X,d) be a metric space, rN, A1, A2,…,Ar be closed nonempty subsets of X,Y=ri=1andα:Y×Y[0,)be a mapping. We say that F : YY is a cyclic (ψ φ ϕ)α-contractive mapping if

  1. F(Aj)Aj+1,i=1,2,,rwhereAr+1=A1.

  2. for any xAiandyAi+1,i=1,2,,r,whereAr+1=A1andα(x,Fx)α(y,Fy)>_1,we have ψ(ρ(d(Fx,Fy)+φ(d(Fx,Fy)))<_ψ(ρ(Md(x,y)+φ(Md(x,y))))ϕ(ρ(Md(x,y)+φ(Md(x,y))))where ρ is an altering distance function, φ ∈ Θ2, ψ ∈ Θ3, ϕ ∈ Θ4and Md(x,y)=max{d(x,y),d(x,Fx),d(y,Fy),d(x,Fy)+d(y,Fx)4}.

Now, we have the following result in metric spaces. Since a metric space must be a metric-like space, we omit the proof of the following theorem in here.

Let (X, d) be a complete metric space. Let r be a positive integer A1, A2, …, Ar be σ-closed nonempty subsets of X,Y=i=1rAiandα:Y×Y[0,)be a mapping. Assume that F : YY is a cyclic (ψ φ ϕ)α-contractive mapping satisying the following assertions:

  1. F is an α-admissi ble mapping;

  2. there exists element x0 in Y such that, α(x0,Fx0) ≥ 1;

    1. F is α-continuous, or;

    2. if {xn} is a sequence in X such that α(xn,xn+1)>_1forallnNandxnxasn,thenα(x,Fx)>_1.

    Then F has a fixed point xi=1rAi.Moreover, if

  3. for all xFix (F) we have α(x,x)≥ 1.

Then F has a unique fixed point xi=1rAi.

3 Fixed point theorems for cyclic α-(κ,φ)g-contractive mappings

In this section, firstly, motivated by [27], we want to give the definition of G-distance function. Afterwards, we define a cyclic α-(κ,φ)g-contractive mapping via G-distance functions and altering distance functions in metric-like space. At last, we prove some fixed point theorems for such mapping in complete metric-like space. Before stating our main results, we present the following definitions.

A function g is said to be a G-distance function if g: [0,蜴)5 →[0,蜴) is a continuous function for which the following properties hold:

  1. g is increasing in each co-ordinate variable;

  2. g(t,t,t,at,bt)<_tforeveryt[0,),wherea+b=4.

Let g(x1,x2,x3,x4,x5)=max{x1,x2,x3,x4+x54}forallx1,x2,x3,x4,x5[0,)then g is a G-distance function.

Let g(x1,x2,x3,x4,x5)=[17x12+17x2x3+17x1x5+17x2x4+17x1x3]12forallx1,x2,x3,x4,x5[0,),then g is a G-distance function.

Let (X,σ) be a metric-like space, pN, A1, A2, …, Ar be σ-closed nonempty subsets of X, Y=i=1pandα:Y×Y[0,)be a mapping. Let g be a G-distance function. The mapping F : YY is called to be a cyclic α(κ,φ)g-contractive mapping if

  1. F(Aj)Aj+1,i=1,2,,rwhereAr+1=A1.(17)

  2. for any xAiandyAi+1,i=1,2,,rwhereAr+1=A1andα(x,Fx)α(y,Fy)>_1we have ρ(σ(Fx,Fy)+φ(σ(Fx,Fy))<_κ(ρ(Mσ(x,y)+φ(Mσ(x,y))))(18)

where ρ is an altering distance function, φ ∈ Θ2, κ ∈ Θ5 and Mσ(x,y)=g(σ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy),σ(y,Fx)).

If in Definition 3.4, we take X=Ai, i=1,2,… r, then we say that F is a modified α-(κ , φ)g-contractive mapping. By Fix (F) we denote the set of all fixed points of F, that is, Fix(F)={xX : Fx = x}.

Assume that F : XX is a cyclic α(κ,φ)g-contractive mapping, xFix(F) and α(x,x)≥1. Then we must have σ(x,x)=0. Indeed, suppose that xFix(F),α(x,x)>_1andσ(x,x)>0.Then Mσ(x,x)<_σ(x,x).Moreover by (18), we have ρ(σ(x,x)+φ(σ(x,x)))=ρ(σ(Fx,Fx)+φ(σ(Fx,Fx)))<_κ(ρ(Mσ(x,x)+φ(Mσ(x,x))))<_κ(ρ(σ(x,x)+φ(σ(x,x))))<ρ(σ(x,x)+φ(σ(x,x))),

which is a contradiction. Hence σ(x,x)=0.

Now, we are ready to state and prove our main result of this section.

Let (X;σ) be a complete metric-like space. Let r be a positive integer A1, A2,…,Ar be σ-closed nonempty subsets of X,Y=i=1rAiandα:Y×Y[0,)be a mapping. Assume that F : YY is a cyclic α(κ,φ)g -contractive mapping satisying the following conditions:

  1. F is an α-admissible mapping;

  2. there exists element x0 in Y such that α(x0 , Fx0)≥ 1;

    1. F is α-continuous, or;

    2. if {xn} is a sequence in X such that α(xn,xn+1)>_1 for all nN and xnx as n→蜴, then α(x,Fx)≥ 1.

    Then F has a fixed point xi=1rAiMoreover if

  3. for all xFix(F) we have α(x,x)≥ 1.

Then F has a unique fixed point xi=1rAi

By the condition (ii), let x0 be an arbitrary point of Y such that α(x0, Fx0)≥ 1. Then there exists some i0 such that x0Ai0. Now, by (17), F(Ai0)Ai0+1implies that Fx0Ai0+1Thus there exists x1 in Ai0+1 such that Fx0=x1. Similarly, Fxn=xn+1, where xnAin. Hence for n ≥ 0, there exists in ∈ {1,2,…,r} such that xnA{in and Fxn=xn+1. On the other hand, F is an α-admissible mapping, hence by Definition 3.1, we get α(x0,Fx0)>_1α(x1,Fx1)=α(Fx0,F(Fx0))>_1.

Again, since F is a n α-admissible mapping, we have α(x1,Fx1)>_1α(x2,Fx2)=α(Fx1,F(Fx1))>_1.

Continuing this process, we can construct a sequence {xn} in Ain such that α(xn1,Fxn1)>_1α(xn,Fxn)>_1forallnN0

and so α(xn,Fxn)α(xn1,Fxn1)>_1forallnN.(19)

If for some n0=0,1,2,… , we have xn0=xn0+1, then Fxn0=xn0, that is, xn0 is a fixed point of F. So from now on, we assume xnxn+1 for all nN. Hence, by Lemma 1.5 (C) we have σ(xn,xn+1) > 0 for all nN. Then by (18) and (19), we have ρ(σ(xn,xn+1)+φ(σ(xn,xn+1)))=ρ(σ(Fxn1,Fxn)+φ(σ(Fxn1,FXn)))<_κ(ρ(Mσ(xn1,xn)+φ(Mσ(xn1,xn)))),(20)

where Mσ(xn1,xn)=g(σ(xn1,xn),σ(xn1,Fxn1),σ(xn,Fxn),σ(xn1,Fxn),σ(xn,Fxn1))=g(σ(xn1,xn),σ(xn1,xn),σ(xn,xn+1),σ(xn1,xn+1),σ(xn,xn)).(21)

From Lemma 1.5 (D), we can get σ(xn,xn)<_σ(xn1,xn)+σ(xn,xn+1)

and by (σ3) , we have σ(xn1,xn+1)<_σ(xn1,xn)+σ(xn,xn+1).

By (21) and the above two inequalities, we have Mσ(xn1,xn)=g(σ(xn1,xn),σ(xn1,xn),σ(xn,xn+1),σ(xn1,xn+1),σ(xn,xn))<_g(σ(xn1,xn),σ(xn1,xn),σ(xn,xn+1),σ(xn1,xn)+σ(xn,xn+1),σ(xn1,xn)+σ(xn,xn+1))(22)

Now, we want to show that σ(xn,xn+1)<σ(xn1,xn).Ifσ(xn,xn+1)>_σ(xn1,xn) ,by (22) and the properties of g, then we have Mσ(xn1,xn)=g(σ(xn1,xn),σ(xn1,xn),σ(xn,xn+1),σ(xn1;xn+1),σ(xn,xn))<_g(σ(xn1,xn),σ(xn1,xn),σ(xn,xn+1),2σ(xn,xn+1),2σ(xn,xn+1))<_g(σ(xn,xn+1),σ(xn,xn+1)),σ(xn,xn+1),2σ(xn,xn+1),2σ(xn,xn+1))<_σ(xn,xn+1).(23)

Using the properties of κ φ and (20) together with (23), we have ρ(σ(xn,xn+1)+φ(σ(xn,xn+1)))=ρ(σ(Fxn1,Fxn)+φ(σ(Fxn1,FXn)))<_κ(ρ(Mσ(xn1,xn)+φ(Mσ(xn1,xn))))<_κ(ρ(σ(xn,xn+1)+φ(σ(xn,xn+1))))<ρ(σ(xn,xn+1)+φ(σ(xn,xn+1))),

which is a contradiction. Therefore, σ(xn,xn+1)<σ(xn1,xn)holds for all nN and there exists δ ≥ 0 such that limnσ(xn,xn+1)=ı. Next, we shall show that δ = 0. In order to prove δ =0, we assume that δ >0. By (20) and (22) and together with the properties of κ φ , we have ρ(ı+φ(ı))=limnρ(σ(xn,xn+1)+φ(σ(xn,xn+1)))<_limnκ(ρ(σ(xn1,xn)+φ(σ(xn1,xn))))=κ(limnρ(σ(xn1,xn)+φ(σ(xn1,xn))))=κ(ρ(ı+φı)))<ρ(ı+φ(ı)),

which is a contradiction. Thus limnσ(xn,xn+1)=0.

Next, we want to prove that limnσ(xn,xm)=0,Iflimnσ(xn,xm)0then by Remark 1.7, there exist ∈ > 0 and two sequences {m(k)\} and {n(k)} of positive integers such that n(k) > m(k) >k and limnσ(xm(k),xn(k)+s)=ϵwhere sN. Choosing, s,lNsuch that s=m(k)+lr+1n(k) then we have xm(k)Aiandxn(k)+sAi+1for somei{1,2,,r} Therefore, using the contractive condition (18) with x=xm(k),y=xn(k)+s

we obtain ρ(σ(xm(k)+1,xn(k)+s+1)+φ(σ(xm(k)+1,xn(k)+s+1)))<_κ(ρ(Mσ(xm(k),xn(k)+s)+φ(Mσ(xm(k),xn(k)+s)))),(24)

where Mσ(xm(k),xn(k)+s)=g(σ(xm(k),xn(k)+s),σ(xm(k),xm(k)+1),σ(xn(k)+s,xn(k)+s+1),σ(xm(k),xn(k)+s+1),σ(xn(k)+s,xm(k)+1)).(25)

If we let k → ∞ in (25), then we have Mσ(xm(k),xn(k)+s)g(ϵ,0,0,ϵ,ϵ)g(ϵ,ϵ,ϵ,2ϵ,2ϵ)ϵ.

Letting k → ∞ in (24) and (25) we have ρ(ϵ+φ(ϵ))κ(ρ(ϵ+φ(ϵ))))<ρ(ϵ+φ(ϵ))),

which is a contradiction. Thus we can conclude that limnσ(xn,xm)=0, that is, the sequence {xn} is a σ-Cauchy sequence. Since Y is σ-closed in (X, σ), there exists xY=i=1rAi such that limnxn=xin(Y,σ); equivalently, σ(x,x)=limnσ(x,xn)=limnσ(xn,xm)=0.

On the one hand, we assume that (a) of (iii) holds, that is, F is α-continuous. Then, it is obviously that x=limnxn+1=limnFxn=Fx.

On the other hand, we assume that (b) of (iii) holds and α(xn,xn+1) ≥ 1. Then, we have α(x,Fx) ≥ 1 and so α(x,Fx)α(xn(k),Fxn(k))1.

Next, we prove that x is a fixed point of F. Since limnxn=xandF(Aj)Aj+1,j=1,2,,r,whereAr+1=A1, the sequence {xn} has infinitely many terms in each Ai for i ∈ {1, 2, …, r}. Suppose that x ∈ Ai. Then Fx ∈ Ai+1 and we take a subsequence {xn(k)}} of {xn} with xn(k)Ai-1 (the existence of this subsequence is guaranteed by the above mentioned comment). By using the contractive condition (18), we obtain ρ(σ(Fx,Fxn(k))+,(σ(Fx,Fxn(k))))κ(ρ(Mσ(x,xn(k))+,(Mσ(x,xn(k))))),(26)

where Mσ(x,xn(k))=g(σ(x,xn(k)),σ(x,Fx),σ(xn(k),Fxn(k)),σ(x,Fxn(k)),σ(xn(k),Fx))=g(σ(x,xn(k)),σ(x,Fx),σ(xn(k),xn(k)+1),σ(x,xn(k)+1),σ(xn(k),Fx)).(27)

In (27), letting k → ∞, using the properties of g, we have limn→!Mσ(x,xn(k))=g(O,σ(x,Fx),O,O,σ(x,Fx))<_g(σ(x,Fx),σ(x,Fx),σ(x,Fx),2σ(x,Fx),2σ(x,Fx))<_σ(x,Fx).(28)

In (26), letting k → ∞ and using the continuity of k, φ and (28), we can get ρ(σ(Fx,x)+φ(σ(Fx,x)))<_κ(ρ(σ(xFx)+φ,(σ(x,Fx))))<ρ(σ(x,Fx)+φ(σ(x,Fx))),

which is a contradiction. Hence σ(Fx,x) = 0, that is, Fx = x. Therefore, x is a fixed point of F. The cyclic character of F and the fact that x ∈ X is a fixed point of F, imply thatxi=1Air.

At last, we shall prove the uniqueness of the fixed point point. Now we assume that x,yi=1rAi are two fixed points of F and x ≠ y. Suppose that the condition (iv) holds. Then we have σ(x,y)>0,α(x,x)1,α(y,y)1 and then by (18), we obtain ρ(σ(Fx,Fy)+φ(σ(Fx,Fy)))κ(ρ(Mσ(x,y)+φ(Mσ(x,y)))),(29)

whereMσ(x,y)=g(σ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy),σ(y,Fx))=g(σ(x,y),σ(x,x),σ(y,y),σ(x,y),σ(y,x))g(σ(x,y),σ(x,x),σ(y,y),2σ(x,y),2σ(y,x)). Using Remark 2.2 we have σ(x,x) = σ(y,y)= 0, therefore it follows from the above inequality and the properties of g thatMσ(x,y)g(σ(x,y),0,0,2σ(x,y),2σ(y,x))g(σ(x,y),σ(x,y),σ(x,y),2σ(x,y),2σ(y,x))σ(x,y). From the above inequality and (29), we have ρ(σ(x,y)+φ(σ(x,y)))ψ(ρ(σ(x,y)+φ(σ(x,y))))<ρ(σ(x,y)+φ(σ(x,y))), which is a contradiction. Hence σ(x,y)=0, that is to say that x=y. The proof is completed.

  1. If we take g,k,κ for different values in Theorem 3.7, then we obtain some various corollaries. In here, we want to omit these corollaries for the sake of simplicity

  2. Similarly as in Section 2, we can get the corresponding conclusion in complete metric spaces. In here, we want to omit these corollaries for simplicity

4 Consequences

In [17], the authors state the following notions consistent with Jachymski [28]. Let (X, σ) be a metric-like space and Δ denote the diagonal of the Cartesian product ΔX × X. Consider a directed graph H such that the setV(H) of its vertices coincides with X, and the set E(H) of its edges contains all loops, i.e., E(H) ⊂ Δ. We assume that H has no parallel edges, so we can identify H with the pair (V(H),E(H)) . Moreover, we may treat H as a weighted graph (see [28]) by assigning to each edge the distance between its vertices.

By H-1 we denote the conversion of a graph H, that is, the graph obtained from H by reversing the direction of edges. Thus we haveE(H1)={(x,y)X×X:(y;x)E(H)}.

The letterH~ denotes the obtained from H by ignoring the direction of edges. Actually, it will be more convenient for us to treat H~H as a directed graph for which the set of its edges is symmetric. Under this convention, E(H~)=E(H)E(H1) If x and y are vertices in a graph H, then a path in H from x to y of length N (N ∈ N) is a sequence {xi}i=0N=0ofN+1 vertices such that x0=x,xN=yand(xn1,xn)E(H)fori=1,,N.. A graph H is connected if there is a path between any two vertices. H is weakly connected ifH~ is connected.

([28]). We say that a mapping F : XX is a Banach H-contraction or simply H-contraction if F preserves edges of H, i.e.,x,yX:(x,y)E(H)(F(x),F(y))E(H) and F decreases weights of edges of H in the following way:k(0,1),x,yX:(x,y)E(H)d(F(x),F(y))kd(x,y).

([28]). A mapping F : XX is called orbitally continuous, if given $x∈ X$ and any sequence $\{k_{n}\}$ of positive integersFknxyXasnimpliesF(Fknx)Fyasn.

([28]). A mapping F : XX is called H-continuous, if given $x∈ X$ and any sequence {kn} of positive integersxnxXasnand(xn+1,xn)E(H)forallnNimpliesFxnFx.

([28]). A mapping F : XX is called orbitally H-continuous, if given x, y ∈ X and any sequence kn of positive integersFknxyXandFkn,Fkn+1x)E(H), for all n ∈ N implyF(Fkn)xFyasn.

Let (X, σ) be a metric-like space endowed with a graph H. Let p ∈ N, A1, A2, …, Ar be σ-closed nonempty subsets ofX,Y=i=1rAi.. We say that F : YY is a cyclic (ψ,ϕ,φ)-graphic contractive mapping if

  1. F(Aj)Aj+1,i=1,2,,rwhereAr+1=A1(30)

  2. x,yY2(x,y)E(H)(F(x),F(y))E(H).

  3. for anyxAiandyAi+1,i=1,2,,r,whereAr+1=A1and(x,Fx),(y,Fy)E(H) we have ψ(ρ(σ(Fx,Fy)+φ(σ(Fx,Fy)))ψ(ρ(Mσ(x,y)+φ(Mσ(x,y)))ϕ(ρ(Mσ(x,y)+φ(Mσ(x,y))))

where ρ is an altering distance function, φ∈Θ2, ψ ∈ Θ3, ϕ∈Θ4 andMσ(x,y)=max{σ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy)+σ(y,Fx)4}.

If in Definition 4.5 we take Ai = X where i = 1,2,… r, then we say that F is a (ψ,ϕ,φ)-graphic contractive mapping. Now, we give an application by using Theorem 2.4.

Let (X, σ) be a complete metric-like space endowed with a graph H. Let r be a positive integer A1, A2,…,Ar be σ-closed nonempty subsets ofX,Y=i=1rAi. Assume that F : YY is a cyclic (ψ,ϕ,φ)-graphic contractive mapping satisying the following conditions:

  1. there exists element x0 in Y such that, (x0Fx0)∈ E(H);

    1. F is orbitally H-continuous on (X, σ), or;

    2. if {xn} is a sequence in X such that xnxasn,and(xn+1,xn)E(H)we have (x,Fx)∈E(H).

    Then F has a fixed point xi=1rAi. Moreover if

  2. for all x ∈ Fix(F) we have (x,x)∈E(H).

    Then F has a unique fixed point xi=1rAi.

Define α : Y × Y →[0,+∞) byα(x,y)=1,if(x,y)E(H)0,others:

Firstly, we prove condition (i) of Theorem 2.4, that is, F is an α-admissible. Since α(x,y)≥1, then (x,y)∈ E(H) . As F is a cyclic (ψ,ϕ,φ)-graphic contraction mapping, we have (Fx,Fy)∈ E(H), that is, α(Fx,Fy)≥1. So F is an α-admissible mapping.

Secondly, it is obvious that condition (ii) of Theorem 2.4 holds by Definition 1.8.

Thirdly, we prove condition (iii) of Theorem 2.4 holds. Let F be H-continuous on (X, σ). That is,xnxasnand(xnxn+1)E(H)forallnNimplyFxnFx.This impliesxnxasnandα(xn,xn+1)1forallnNimplyFxnFx.

which implies that F is α-continuous on (X, σ). From (i) there exists x0 ∈ X such that (x0 Fx0∈ E(H). That is, α(x0Fx0)≥1. Let xn ⊂ X be a sequence such thatxnxasnandα(xn+1,xn)1. Therefore, (xn+1,xn)E(H) and then from (ii) we have (x,Fx)∈ E(H. That is, α(x,Fx)≥1. Thus the condition (iii) of Theorem 2.4 holds.

Let 2xAiandyAi+1whereα(x,Fx)α(y,Fy)1. Then x∈ Ai and y∈ Ai+1 where (x,Fx)∈ E(H) and (y,Fy)∈ E(H). Since F is a cyclic (ψ,ϕ,φ)-graphic contractive mapping, we haveψ(ρ(σ(Fx,Fy)+φ(σ(Fx,Fy)))ψ(ρ(Mσ(x,y)+φ(Mσ(x,y))))ϕ(ρ(Mσ(x,y)+,(Mσ(x,y))))

where ρ is a altering distance function, ∈Θ2 ψ∈Θ3, ϕ∈Θ4 and Mσ(x,y)=max{σ(x,y),σ(x,Fx),σ(y,Fy),σ(x,Fy)+σ(y,Fx)4}.

That is, F is a cyclic (ψ,ϕ)-graphic contractive mapping. Thus the conditions (i),(ii) and (iii) of Theorem 2.4 are satisfied and F has a fixed point in Y=i=1rAi.

Finally, we prove the uniqueness of the fixed point. Since (x,x)∈ E(H) for all x∈ Fix(F), then α(x,x)≥1. Hence it coincides with the condition (iv) of Theorem 2.4. Then F has a unique fixed pointxi=1rAi.

  1. If in Theorem 4.6, we choose ϕ(t) = 0 and ρ(t)=t, then we can get Theorem 3.1 of [17].

  2. Similarly to Corollaries 3.1-3.3 and Theorem 3.2 in [17], we can get the corresponding results which are more general than the results obtained by [17]. Here, we want to omit these results because they are tedious.

Acknowledgement

The authors thank the editor and the referees for their valuable comments and suggestions. This work was supported by National Natural Science Foundation of China (Grant No. 11461016 and 11461043) and supported partly by the Provincial Natural Science Foundation of Jiangxi, China (20114BAB201003 and 20142BAB201005) and the Science and Technology Project of Educational Commission of Jiangxi Province, China (GJJ11346) and Hunan Provincial Innovation Foundation for Postgraduates (No. CX2016B037).

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About the article

Received: 2016-04-06

Accepted: 2016-08-22

Published Online: 2016-11-16

Published in Print: 2016-01-01


Citation Information: Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0080.

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