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

([14]). A mappingσ : X × X → R⁺whereXis a nonempty set, is said to be a metric-like mapping onXiffor anyx,y, z ∈ Xthe following three conditions hold true: (σ1)σ(x,y)=0⇒x=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 τ_σ on X whose base is the family of open σ-balls

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

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

([19, 20]). Let (X, σ) be a metric-like space andUbe a subset of X. We say thatUis aσ-open subset ofX, iffor allx ∈ Xthere existss > 0 such thatB_σ(x,s) ⊆ U. Also, V ⊆ Xis aσ-closed subset ofXifX∖Vis aσ-open subset ofX.

([21]). Let (X, σ) be a metric-like space andVbe aσ-closed subset of X. Let {x_n} be a sequence inVsuch thatx_n → xasn → ∞. Thenx ∈ V.

([21]). Let (X, σ) be a metric-like space and {x_n} be a sequence inXsuch thatx_n → xasn → ∞andσ(x,x) = 0. Thenlimn→∞σ(xn,y)=σ(x,y)forally∈X.

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

1. ifσ(x, y) = 0, thenσ(x, x) = σ(y, y) = 0;

2. if {x_n} is a sequence such thatlimn→∞σ(xn,xn+1)=0,Then

   limn→∞σ(xn,xn)=limn→∞σ(xn+1,xn+1)=0;

3. ifx ≠ y, thenσ(x, y) > 0;

4. σ(x,x)≤2n∑i=1nσ(x,xi)holds for allx_i, x ∈ Xwhere 1 ≤ i ≤ n.

([21]). Let (X, σ) be a metric-like space and let {x_n} be a sequence inXsuch that

Iflimn→∞σ(xn,xm)≠0,then there existϵ > 0 and two sequences {m(k)} and {n(k)} of positive integers such thatn(k) > m(k)>kand the following four sequences tend toϵwhenk → ∞:

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 toϵwhenk → ∞, wheres ∈ 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]). LetF : X → Xandα : X × X →[0,∞). The mappingFis said to beα-admissible if for allx, y ∈ X, we have

([25]). Let (X, σ) be a metric-like space. Letα : X × X→[\mathrm{O},∞$) andF : X → X. We say thatFisα-continuous on (X, σ) , if

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 functionρ:[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]). Letζ∈Θ1.DefineΦ0(t) = ∫0tζ(t)dt.Then Φ₀is an altering distance

Let (X, σ) be a metric-like space, r ∈ N, A₁, A₂ , …, A_rbeσ-closed nonempty subsets of XY=⋃i=1rAiandα:Y×Y→[0,∞)be a mapping. We say thatF : Y → Yis a cyclic (ψ,ϕ,φ)_α-contractive mapping if

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

2. for anyx∈Aiandy∈Ai+1,i=1,2,…,r,whereAr+1=A1andα(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)

   whereρis an altering distance function, φ, ∈ Θ₂, ψ ∈ Θ₃, ϕ ∈ Θ₄and

   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 takeX = A_i, i = 1, 2, …; r, then we say thatFis (ψ,ϕ,φ)_α-contractive mapping. We denote the set of all fixed points ofFby Fix (F) , i.e., Fix (F) = {x ∈ X : Fx = x}.

Assume thatF : X → Xis a cyclic (ψ,ϕ,φ)_α-contractive mapping, x ∈ Fix(F)andα (x,x) ≥ 1. Then we must haveσ(x,x) = 0. Now, we prove this results. Suppose thatx ∈ Fix(F), α(x,x) ≥ 1 andσ(x,x) > 0. Then we haveM_σ(x,x) = σ(x,x). Moreover by (4) It follows that

which is a contradiction. Henceσ(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. Letrbe a positive integerA₁, A₂, …, A_rbeσ-closed nonempty subsets ofX,Y=⋃i=1rAiandα:Y×Y→[0,∞)be a mapping. Assume thatF : Y → Yis a cyclic (ψ,ϕ,φ)_α-contractive mapping satisying the following conditions:

1. Fis anα-admissible mapping;

2. there exists elementx₀inYsuch that, α(x₀,Fx₀) ≥ 1;

3. 1. Fisα-continuous, or;

   2. if {x_n} is a sequence inXsuch thatα(x_n, x_n+1) ≥ 1 for alln ∈ Nandx_n → xasn → ∞, thenα(x,Fx) ≥ 1.

   ThenFhas a fixed pointx∈∩i=1pAi.Moreover if

4. for allx ∈ Fix (F) we haveα(x,x) ≥ 1.

   ThenFhas a unique fixed pointx∈∩i=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 x₀ ∈ A_i0. Now by (3), F(Ai0)⊆Ai0+1 implies that Fx0∈Ai0+1. Thus there exists x₁ in A_i0+1 such that Fx₀ = x₁. Similarly, Fx_n = x_n+1, where x_n ∈ A_in. Hence for n ≥ 0, there exists i_n ∈ {1, 2, …, r such that x_n ∈ A_in and Fx_n = x_n+1. On the other hand, F is an α-admissible mapping, hence by Definition 1.8, we get

Again, since F is an α-admissible mapping, we have

Continuing this process, we can construct a sequence {x_n} in A_jn such that

and so

If for some n₀ = 0,1,2, …, we have x_n0 = x_n0+1, then Fx_n0 = x_n0, that is, x_n0 is a fixed point of F. So from now on, we assume x_n ≠ x_n+1 for all n ∈ N. Hence, by Lemma 1.5 (C) we have σ (x_n, x_n+1) > 0 for all n ∈ N. Now, we want to show that σ(xn,xn+1)<σ(xn−1,xn). If σ(xn,xn+1)≥σ(xn−1,xn), then by (4) and (5), we have

where

From Lemma 1.5 (D), we can get

and by (σ3) in Definition 1.1, we have

That is

Thus we have

Therefore, from (7) and (8), we get

By (6) and (9), we have

If max {σ(xn−1,xn),σ(xn,xn+1)}=σ(xn,xn+1), then by (10), we have

which contradicts to the condition (iii) of Θ₃ and Θ₄. Hence

Since σ(xn,xn+1)≥σ(xn−1,xn), by the properties of ψ, φ and Lebesgue integral function, we have

Hence, from (11) and (12), we can get

which is a contradiction. Therefore σ(xn,xn+1)<σ(xn−1,xn) holds for all n ∈ N and there exists δ ≥ 0 such that limn→∞σn=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

which is a contradiction. Thus, limn→∞σ(xn,xn+1)=0.

Next, we want to prove that limn→∞σ(xn,xm)=0, If limn→∞σ(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 limn→∞σ(xm(k),xn(k)+s)=ϵ, where s ∈ N. Chooses, s, l ∈ N such that s=m(k)+ls+1−n(k) then we have xm(k)∈Aiandxn(k)+s∈Ai+1 for some i ∈ {1, 2, …, r}. Therefore, using the contractive condition (4) with x=xm(k),y=xn(k)+s, we obtain

where

Letting k → ∞ in (13) and (14), then we have

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

On the one hand, we assume that (a) of (iii) holds, that is, F is α-continuous. Then it is obvious that

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 limn→∞xn=xandF(Aj)⊆Aj+1,j=1,2,…,r, where A_n+1 = A₁, the sequence {x_n} has infinitely many terms in each A_i for i ∈ {1, 2, …, r}. Suppose that x ∈ A_i. Then Tx ∈ A_i+1 and we take a subsequence {x_n(k)} of {x_n} with x_n(k) ∈ A_i-1 (the existence of this subsequence is guaranteed by the above mentioned comment). By using the contractive condition (4), we obtain

where

In (15) and (16), letting k → ∞ and using the lower semi-continuity of ϕ, we can get

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 x∈∪i=1rAi.

At last, we shall prove the uniqueness of the fixed point. Now, we assume that x,y∈∪i=1rAi are two fixed points of F and x ≠ y. 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

where

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

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 chooseϕ(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.

ConsiderX = {0, 1, 2}. Letσ : X × X→ [0,∞) be a mapping defined by

It is clear that (X, σ) is a complete metric-like space. Suppose thatA₁ = {0, 1} andA₂ = {1, 2} andY = A₁ ∪ A₂.

DefineF : Y → Yandα : Y × Y → [0, +∞) byFx = 1 for allx ∈ Yandα(x,y) = x² + 2 for all x, y ∈ Y.

Then we haveα(x,Fx)α(y,Fy)≥1for allx,y ∈ YandT(A1)⊆A2,F(A2)⊆A1.Thusx ∈A₁andy ∈A₂.

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

Case 1: Assume thatx = 0, y = 1. On the one hand, Fx = 1 andFy = 1, that is, σ(Fx,Fy) = σ(1,1) = 0. On the other hand, we have

Hence, we have

SoFis a cyclic (ψ,ϕ,φ)_α-contractive mapping.

Case 2: Assume thatx = 0, y = 2. Similarly as in the proof of Case 1, we can getFis a cyclic (ψ,ϕ,φ)_α-contractive mapping.

Case 3: Assume thatx = 1, y = 1. Similarly as in the proof of Case 1, we can getFis a cyclic (ψ,ϕ,φ)_α-contractive mapping.

Case 4: Assume thatx = 1, y = 2. Similarly as in the proof of Case 1, we can getFis a cyclic (ψ,ϕ,φ)_α-integral type contractive mapping.

ThusFis a cyclic (ψ,ϕ,φ)_α-contractive mapping. Clearlyα(0,F0)=α(0,1)≥1and so the condition (ii) of 2.1 is satisfied. Ifα(x,y) ≥ 1, thenx,y ∈ {0, 1, 2} which implies thatα(Fx,Fy) = 3 ≥ 1, that is, Fis anα-admissible mapping. Let {x_n} be a sequence inXsuch that, α(x_n, Fx_n) ≥ 1 andx_n → xasn → ∞. Then, wemust havex_n ∈ {0, 1, 2} and so, x ∈ {0, 1, 2} that is, α(x,Fx) ≥ 1. Hence, all the conditions of Theorem 2.4 hold andFhas a fixed pointx=1∈A1∩A2.

ConsiderX = {0, 1, 2}. Letσ : X × X → [0,∞) be a mapping defined by

It is clear that (X, σ) is a complete metric-like space. Suppose thatA₁ = {0, 2} andA₂ = {1,2} andY = A₁ ∪ A₂. DefineF : Y → Yandα : Y × Y →[0, +∞) by

for allx ∈ Yand

ThenF(A₁) ⊆ A₂, F(A₂) ⊆ A₁. Thusx ∈A₁andy ∈A₂. Now, ifx ∈ {0} ory ∈ {0}, thenα(x,Fx) = 0 orα(y,Fy) = 0. That is, α(x,Fx)α(y,Fy) = 0 ≤ 1, which is contradiction. Hencex ∈ {2} andy ∈ {1, 2}.

Also defineψ,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 thatx = 2, y = 1. On the one hand, Fx = 0 andFy = 1, that is, σ(Fx,Fy) = σ(0,1) = 3/2. On the other hand, we have

Hence, we have

SoFis a cyclic (ψ,ϕ,φ)_α-contractive mapping.

Case 2: Assume thatx = 2, y = 2. On the one hand, Fx = 0 andFy = 0, that is, σ(Fx,Fy) = σ (0,0) = 0. On the other hand, we have

Hence, we have