In this article, we use the concept of proximity spaces to prove common fixed point results for mappings satisfying generalized (ψ, β)-Geraghty contraction type mapping in partially ordered proximity spaces. Finally, we investigate an application to endorse our results.
Citation: Demet Binbaşıoǧlu. Some fixed pointresults for nonlinear contractive conditions in ordered proximity spaceswith an application[J]. AIMS Mathematics, 2023, 8(6): 14541-14557. doi: 10.3934/math.2023743
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In this article, we use the concept of proximity spaces to prove common fixed point results for mappings satisfying generalized (ψ, β)-Geraghty contraction type mapping in partially ordered proximity spaces. Finally, we investigate an application to endorse our results.
Frigyes Riesz [1] first discovered the fundamental ideas of proximity spaces in 1908. Later, in 1934, Efremovich [2] resurrected and axiomatized this theory, which was later printed in 1951. Numerous studies on proximity spaces have been conducted over the years [3,4,5,6]. Smirnov [5] explained the relationship between proximities and uniformities, as well as the relationship between the proximity relation and topological spaces.
Inspired by [7], Kostic [8] introduced fixed point theory and defined the concepts of w-distance and w0-distance in proximity space. For more details on w-distance, see [9]. Naimpally et al. and Sharma [10,11] have also done studies on proximity spaces and their examples. Qasim et al. [12] give the theorems of Matkowski and Boyd-Wong in proximity spaces.
Geraghty [13] established a class of functions in 1973, which he designated as the set of functions. Khan et al. [14] introduced the idea of an altering distance function. Alsamir et al. [15] gave some common fixed point teorems in partially ordered metric-like spaces.
Moreover an important development is reported in fixed point theory via some applications. Hammad et al. [16,17,18] utilized some fixed point techniques to solve differential and integral equations.
In this study, we give some common fixed point results via generalized (ψ, β)-Geraghty contraction mappings in proximity spaces and an application of the existence of a unique solution of an integral equation.
In this section, we will include the basic definitions and theorems that will be necessary in the following parts of our work.
Definition 2.1. [5] Suppose Ω is a set and δ is a relation on the set 2Ω. If the following hold, the pair (Ω,δ) is considered to be in a proximity space: for any A,B,C∈2Ω, where 2Ω is the power set of Ω.
(p1) AδB⇒BδA,
(p2) AδB⇒A,B≠∅,
(p3) Aδ(B∪C)⇔AδB or AδC,
(p4) A∩B≠∅⇒AδB,
(p5) For all γ⊆Ω, Aδγ or Bδ(Ω−γ) implies AδB.
We shall write all ξ∈Ω and A⊆Ω as ξδA and Aδξ rather than {ξ}δA and Aδ{ξ}, respectively. If ξδμ means that ξ=μ for every ξ,μ∈Ω, then the proximity space (Ω,δ) is said to be separated. Generalizations of uniform features are used to describe the characteristics of proximity spaces, metric and topological continuity qualities, respectively.
Any proximity relation on a non-empty set Ω induces a topology τδ through the Kuratowski closure operator. When applied to all A⊆Ω, the Kuratowski closure operator can be described as cl(A)={ξ∈Ω:ξδA}. The topology τδ in this situation is always completely regular and if (Ω,δ) is separated, it is Tychonoff.
If (Ω,τ) is a topological space and δ is a proximity on Ω such that τδ=τ, it is said that τ and δ are compatible. Every completely regular topology on a nonempty set Ω, has a compatible proximity. Also, we obtain ξδ{ξn} if a sequence {ξn} converges to a point ξ∈Ω with regard to the induced topology τδ. Additionally, each uniform space (Ω,U) is associated with a proximity structure that is described by for all A,B⊆Ω, AδB if (A×B)∩C≠∅ for all C∈U. See [10,11] for more information.
Example 2.1. [12] Give us a metric space (Ω,p). Take into account the relation δ on 2Ω,
AδB⇔p(A,B)=0 and p(A,B)=inf{p(u,v):u∈A,v∈B}. |
δ is thus a proximity on Ω. Additionally, the metric topologies τp and δ are compatible.
In order to get the proximity space version of the Banach fixed-point theorem, Kostic [8] defined the concepts of w-distance and w0 -distance, which were inspired by [7].
Definition 2.2. [8] Let w:Ω×Ω→[0,∞) be a function and (Ω,δ) be a proximity space. Then w is a w -distance on Ω, if the axiom below is true:
(w1) if w(η,A)=0 and w(η,B)=0 imply AδB for all η∈Ω and A,B⊆Ω, when w(η,A)=inf{w(η,ξ):ξ∈A}.
Definition 2.3. [8] A w-distance on a proximity space (Ω,δ) is also referred to as a w0-distance if the axioms below are true:
(w2) For any ξ,μ,η∈Ω, w(ξ,μ)≤w(ξ,η)+w(η,μ),
(w3) Since w is lower semicontinuous in both variables with regard to τδ, we get
w(ξ,μ)≤liminfξ∣→ξw(ξ∣,μ)=supB∈Uξinfξ∣∈Bw(ξ∣,μ),andw(μ,ξ)≤liminfξ∣→ξw(μ,ξ∣)=supB∈Uξinfξ∣∈Bw(μ,ξ∣), |
where Uξ is a base of neighborhoods of the point ξ∈Ω.
Remark 2.1. [12] It is evident that for every sequence {ξn} convergent to ξ with respect to τδ, w(ξ,μ)≤ liminfn→∞w(ξn,μ) and w(μ,ξ)≤ liminfn→∞w(μ,ξn) exist. This is true if w is lower semicontinuous in both variables with respect to τδ.
Example 2.2. [12] Let Ω=R possess the usual metric as well as the proximity δ specified in Example 1. Definition of w1, w2:Ω×Ω→[0,∞) by
w1(ξ,μ)=max{|ξ|,|μ|} and w2(ξ,μ)=|ξ|+|μ|2, |
both w1 and w2 are w0-distance on Ω.
Lemma 2.1. [7,8] Let (Ω,δ) be a space of proximity with w-distance w. The following properties are then true:
(i) If (Ω,δ) is separated, then w(η,ξ)=0 and w(η,μ)=0 imply ξ=μ,
(ii) If w(η,ξ)=0 and w(η,ξn)→0 as n→∞, then {ξn} subsequently converges to ξ with respect to τδ.
Definition 2.4. [15] Let (Ω,⪯) be a partially ordered set and g,h:Ω→Ω be two mappings. Then
(i) The elements ξ,μ∈Ω are called comparable if ξ⪯μ or μ⪯ξ holds,
(ii) g is called nondecreasing i.e., if ξ⪯μ implies gξ⪯gμ,
(iii) The pair (g,h) is weakly increasing if gξ⪯hgξ and hξ⪯ghξ for all ξ∈Ω,
(iv) The mapping g is weakly increasing if the pair (g,I) is weakly increasing, where I is denoted to the identity mapping on Ω.
Definition 2.5. [15] Let (Ω,⪯) be a partially ordered set. Ω is called regular, if whenever {ηn} is a nondecreasing sequence in Ω w.r.t. ⪯ such that ηn→η, then ηn⪯η for ∀n∈N.
Definition 2.6. [13] If {xn} is a sequence in [0,∞) with α(xn)→1, then xn→0. The set of functions α:[0,∞)→[0,1) which holds the condition is denoted with a class of functions Π.
Definition 2.7. [14] If the circumstances below are true;
(i) ψ is continuous and non-decreasing,
(ii) ψ(x)=0⇔x=0,
afterward, the function ψ:[0,∞)→[0,∞) is referred to as an altering distance function.
The following lemma is introduced at the beginning of this section and will be used to prove our main results.
Lemma 3.1. Let (Ω,δ) be a separated proximity space with w0−distance w and {ηn} be a sequence in Ω such that limn→+∞w(ηn,ηn+1)=0. If limn,m→+∞w(ηn,ηm)≠0, then there exist ε>0 and two sequences {nk} and {mk} of positive integers with nk>mk>k such that following three sequences {w(ξ2nk,ξ2mk)}, {w(ξ2nk−1,ξ2mk)}, {w(ξ2nk,ξ2mk+1)} converge to r+ when k→∞.
Proof. Let {ηn}⊆Ω be a sequence such that
limn→+∞w(ηn,ηn+1)=0andlimn,m→+∞w(ηn,ηm)≠0. |
Then there exist r>0 and two sequences {nk}, {mk} of positive integers such that the lowest positive integer, nk, for which nk>mk>k, w(η2nk,η2mk)≥r. This means that w(η2nk−2,η2mk)<r. The triangular inequality implies that
r≤w(η2nk,η2mk)≤w(η2nk,η2nk−1)+w(η2nk−1,η2nk−2)+w(η2nk−2,η2mk)<w(η2nk,η2nk−1)+w(η2nk−1,η2nk−2)+r. |
Letting k→∞ in the above inequalities, implies that
limn→+∞w(ξ2nk,ξ2mk)=r+. |
Once more, we may determine that
|w(ξ2nk,ξ2mk+1)−w(ξ2nk,ξ2mk)|≤w(ξ2mk,ξ2mk+1) |
from the triangular inequality. In the inequality above, if we let k→∞ go, we get
limk→+∞w(ξ2nk,ξ2mk+1)=r+. |
Similarly, one can easily show that
limk→+∞w(ξ2nk−1,ξ2mk)=r+. |
Definition 3.1. Let (Ω,⪯) be a partially ordered set, (Ω,δ) be a separated proximity space with w0−distance w and g,h:Ω→Ω be two mappings. If α∈Π, ψ∈Ψ and a continuous function β:[0,∞)→[0,∞) exists with β(t)≤ ψ(t) for all t>0 such that
ψ(w(gξ,hμ))≤α(Kξ,μ)β(Kξ,μ), | (3.1) |
holds for all comparable elements ξ,μ∈Ω, where
Kξ,μ=max{w(ξ,μ),w(gξ,ξ),w(μ,hμ)}, |
we may then state that the pair (g,h) is of the generalized (ψ, β)-Geraghty contraction type.
Theorem 3.2. Let (Ω,⪯) be a partially ordered set, (Ω,δ) be a separated proximity space with w0−distance w and g,h:Ω→Ω be two mappings that meet the requirements listed below:
(i) The pair (g,h) is weakly increasing,
(ii) The pair (g,h) is generalized (ψ, β)-Geraghty contraction type,
(iii) Either g or h is continuous,
(iv) For all η∈Ω, any iterative sequences {gnη} and {hnη} have convergent subsequences with respect to τδ.
Then g and h have a common fixed point ν∈ Ω with w(ν,ν)=0. Furthermore, assume that if η, u∈Ω such w(η,η)=w(u,u)=0 implies that η and u are comparable then the common fixed point of g and h is unique.
Proof. Let ξ0∈Ω, ξ1=gξ0 and ξ2=hξ1. By continuing in this manner, we create a sequence {ξn}⊆Ω defined by ξ2n+1=gξ2n and ξ2n+2=hξ2n+1. Since the pair (g,h) is weakly increasing
ξ1=gξ0⪯hgξ0=ξ2=gξ1⪯...⪯hgξ2n=ξ2n+2⪯.... |
Thus ξn⪯ξn+1 for all n∈N. If there exists some l∈N such that w(ξ2nl,ξ2nl+1)=0. Hence ξ2l=ξ2l+1 and gξ2l=ξ2l. To show that hξ2l=ξ2l it is enough to show that ξ2l=ξ2l+1=ξ2l+2. Assume
w(ξ2l+1,ξ2l+2)≠0 and w(ξ2l+2,ξ2l+1)≠0. |
Since ξ2l⪯ξ2l+1, then by (3.1) we have
ψ(w(ξ2l+1,ξ2l+2))=ψ(w(gξ2l,hξ2l+1))≤α(Kξ2l,ξ2l+1)β(Kξ2l,ξ2l+1)=α(max{w(ξ2l,ξ2l+1),w(ξ2l,gξ2l),w(ξ2l+1,hξ2l+1)})β(max{w(ξ2l,ξ2l+1),w(ξ2l,gξ2l),w(ξ2l+1,hξ2l+1)}}=α(max{w(ξ2l,ξ2l+1),w(ξ2l,ξ2l+1),w(ξ2l+1,ξ2l+2)}) β(max{w(ξ2l,ξ2l+1),w(ξ2l,ξ2l+1),w(ξ2l+1,ξ2l+2)})=α(w(ξ2l+1,ξ2l+2))β(w(ξ2l+1,ξ2l+2))<β(w(ξ2l+1,ξ2l+2))≤ψ(w(ξ2l+1,ξ2l+2)), |
which is a contradiction. So w(ξ2l+1,ξ2l+2)=0 and similarly w(ξ2l+2,ξ2l+1)=0. That is ξ2l=ξ2l+1=ξ2l+2. Thus ξ2l is a common fixed point for g and h. We now presume that
w(ξn,ξn+1)≠0 and w(ξn+1,ξn)≠0 |
for all n∈N. When n is even, n=2t follows some t∈N
ψ(w(ξn,ξn+1))=ψ(w(ξ2t,ξ2t+1)) =ψ(w(gξ2t,hξ2t−1))≤α(max{w(ξ2t−1,ξ2t),w(ξ2t−1,hξ2t−1),w(ξ2t,gξ2t)})β(max{w(ξ2t−1,ξ2t),w(ξ2t−1,hξ2t−1),w(ξ2t,gξ2t)})=α(max{w(ξ2t−1,ξ2t),w(ξ2t,ξ2t+1)})β(max{w(ξ2t−1,ξ2t),w(ξ2t,ξ2t+1)})<β(max{w(ξ2t−1,ξ2t),w(ξ2t,ξ2t+1)}). | (3.2) |
Assume
max{w(ξ2t−1,ξ2t),w(ξ2t,ξ2t+1)}=w(ξ2t,ξ2t+1). |
By (3.2), we get
ψ(w(ξ2t,ξ2t+1))<ψ(w(ξ2t,ξ2t+1)), |
which is a contradiction. Thus,
max{w(ξ2t−1,ξ2t),w(ξ2t,ξ2t+1)}=w(ξ2t−1,ξ2t). |
Therefore
ψ(w(ξ2n,ξ2n+1))<ψ(w(ξ2n−1,ξ2n)). | (3.3) |
Because ψ is an altering distance function, we draw the conclusion that
w(ξ2n,ξ2n+1)<w(ξ2n−1,ξ2n) |
is true for every n∈N. n=2t+1 for some t∈N if n is odd. By (3.1) we have
ψ(w(ξn,ξn+1))=ψ(w(ξ2t+1,ξ2t+2))=ψ(w(gξ2t,hξ2t+1))≤α(max{w(ξ2t,ξ2t+1),w(ξ2t,gξ2t),w(ξ2t+1,hξ2t+1)})β(max{w(ξ2t,ξ2t+1),w(ξ2t,gξ2t),w(ξ2t+1,hξ2t+1)})=α(max{w(ξ2t,ξ2t+1),w(ξ2t+1,ξ2t+2)}) β(max{w(ξ2t,ξ2t+1),w(ξ2t+1,ξ2t+2)})<β(max{w(ξ2t,ξ2t+1),w(ξ2t+1,ξ2t+2)}). | (3.4) |
Assume that
max{w(ξ2t,ξ2t+1),w(ξ2t+1,ξ2t+2)}=w(ξ2t+1,ξ2t+2). |
By (3.4) we get
ψ(w(ξ2t+1,ξ2t+2))<ψ(w(ξ2t+1,ξ2t+2)), |
which is a contradiction. Then,
max{w(ξ2t,ξ2t+1),w(ξ2t+1,ξ2t+2)}=w(ξ2t,ξ2t+1). |
Thus,
ψ(w(ξn,ξn+1))<ψ(w(ξn−1,ξn)). |
We conclude that
w(ξ2n+1,ξ2n+2)≤w(ξ2n,ξ2n+1) | (3.5) |
holds for all n∈N since ψ is an altering distance function. The result of combining (3.3) and (3.5) is that
w(ξn,ξn+1)≤w(ξn−1,ξn) |
holds for all n∈N. The sequence {w(ξn,ξn+1)} is hence a decreasing sequence. Therefore, ν≥0 exists such that limn→∞w(ξn,ξn+1)=ν and the sequence {w(ξn,ξn+1)} is a decreasing sequence.
We now have proof that ν=0. Consider the contrary, which is ν>0. We get
ψ(w(ξn,ξn+1))≤α(w(ξn−1,ξn))β(w(ξn−1,ξn)) |
from (3.2) and (3.4). The inequality above indicates that ψ(ν)<β(ν)≤ψ(ν) if the limsup is taken. This is a contradiction. Therefore, ν=0. This implies that
w(ξn,ξn+1)→0 as n→∞. |
In a similar way, we can get
w(ξn+1,ξn)→0 as n→∞. |
We can now prove that
limn,m→∞w(ξn,ξm)=0 and limn,m→∞w(ξm,ξn)=0. |
Assume that
limn,m→∞w(ξn,ξm)≠0. |
Using Lemma 3.1, there exist r>0, two sequences {ξnk} and {ξmk} of {ξn} with 2nk>2mk≥k such that the three sequences {w(ξ2nk,ξ2mk)}, {w(ξ2nk−1,ξ2mk)}, {w(ξ2nk,ξ2mk+1)} converge to r+ when k→∞. From (3.1) we have
ψ(w(ξ2nk,ξ2mk+1))=ψ(w(gξ2mk,hξ2nk−1))≤α(Kξ2mk,ξ2nk−1)β(Kξ2mk,ξ2nk−1), | (3.6) |
where
Kξ2mk,ξ2nk−1=max{w(ξ2nk−1,ξ2mk),w(ξ2nk−1,hξ2nk−1),w(ξ2mk,gξ2mk)}=max{w(ξ2nk−1,ξ2mk),w(ξ2nk−1,ξ2nk),w(ξ2mk,ξ2mk+1)}. |
Letting k→∞ in (3.6) and using the properties of ψ, α and β, we deduce that
ψ(r)≤α(r)β(r)<β(r)≤ψ(r), |
a contradiction. Therefore
limn,m→∞w(ξn,ξm)=0. |
We can get limn,m→∞w(ξm,ξn)=0 in a similar way. According to (iv), if the sequence {ξn} has a subsequence {ξnl} that is convergent with regard to τδ to some η∈Ω, then we get
w(η,ξnl)≤limk→∞infw(ξnk,ξnl)=0 | (3.7) |
and symmetrically, we obtain
w(ξnl,η)≤limk→∞infw(ξnl,ξnk)=0. |
Since g or h is continuous, we get
limn→∞w(ξn+1,hη)=limn→∞w(gξn,hη)=w(gη,hη), | (3.8) |
limn→∞w(gη,ξn+1)=limn→∞w(gη,hξn)=w(gη,hη). | (3.9) |
Thus,
limn→∞w(ξn+1,hη)=w(η,hη), | (3.10) |
limn→∞w(gη,ξn+1)=w(gη,η). | (3.11) |
Combining (3.8) and (3.10), we conclude that w(η,hη)=w(gη,hη). Also, by (3.9) and (3.10) we deduce that w(gη,η)=w(gη,hη). So,
w(η,hη)=w(gη,η)=w(gη,hη). | (3.12) |
We now demonstrate how w(gη,η)=0 and w(η,hη)=0. Suppose that, on the contrary, is w(η,hη)>0 and w(gη,η)>0. Thus, we get
ψ(w(η,hη))=ψ(w(gη,hη))≤α(Kη,η)β(Kη,η)<β(Kη,η)≤ψ(Kη,η), | (3.13) |
and
ψ(w(gη,η))=ψ(w(gη,hη))≤α(Kη,η)β(Kη,η)<β(Kη,η)≤ψ(Kη,η), | (3.14) |
where
Kη,η=max{w(η,η),w(gη,η),w(η,hη)}=max{w(gη,η),w(η,hη)}, |
which is a contradiction. Thus we obtain w(gη,η)=0 and w(η,hη)=0. Hence gη=η, η=hη. So, η is a common fixed point of g, h.
We assume that u is yet another fixed point of g and h in order to demonstrate the uniqueness of the common fixed point.
We now show that w(u,u)=0. On the contrary, suppose that is w(u,u)>0.
ψ(w(u,u))=ψ(w(gu,hu))≤α(w(u,u))β(w(u,u))<β(w(u,u))≤ψ(w(u,u)) |
is a contradiction because of u⪯u. Hence w(u,u)=0.
So get the conclusion that η and u are comparable on the additional requirements on Ω.
We suppose that w(η,u)≠0.
ψ(w(η,u))=ψ(w(gη,hu))≤α(w(η,u))β(w(η,u))<β(w(η,u))≤ψ(w(η,u)), |
and
ψ(w(u,η))=ψ(w(gu,hη))≤α(w(u,η))β(w(u,η))<β(w(u,η)) ≤ψ(w(u,η)), |
this is a contradiction. Thus w(η,u)=0 and w(u,η)=0. Hence u=η. Thus g and h have a unique common fixed point.
Theorem 3.3. Let (Ω,⪯) be a partially ordered set, (Ω,δ) be a separated proximity space with w0−distance w and g,h:Ω→Ω be two mappings that meet the requirements listed below:
(i) The pair (g,h) is weakly increasing,
(ii) The pair (g,h) is generalized (ψ, β)-Geraghty contraction type,
(iii) Ω is regular,
(iv) For all η∈Ω, any iterative sequences {gnη} and {hnη} have convergent subsequences with respect to τδ.
Then g and h have a common fixed point ν∈ Ω with w(ν,ν)=0. Furthermore, assume that if η, u∈Ω such w(η,η)=w(u,u)=0 implies that if η and u are comparable, then the common fixed point of g and h is unique.
Proof. After proving Theorem 3.2 we create a a sequence {ξn}⊆Ω such that
ξn→ν∈Ω with w(ν,ν)=0. |
Since Ω is regular, ξn⪯ν for all n∈N.
Therefore, the elements ξn and ν are comparable for any n∈N.
We now show that w(v,hv)=0.
Suppose to the contrary, that is
w(v,hv)>0. |
By (3.1), we have
ψ(w(ξ2n+1,hν))=ψ(w(gξ2n,hν))≤α(max{w(ξ2n,ν),w(ξ2n,gξ2n),w(ν,hν)})β(max{w(ξ2n,ν),w(ξ2n,gξ2n),w(ν,hν)})=α(max{w(ξ2n,ν),w(ξ2n,ξ2n+1),w(ν,hν)})β(max{w(ξ2n,ν),w(ξ2n,ξ2n+1),w(ν,hν)}). |
Letting n→∞ in above inequalities, as a result, we say
ψ(w(ν,hν))≤α(w(ν,hν))β(w(ν,hν)). |
Utilizing the properties of ψ, α and β,
ψ(w(ν,hν))<ψ(w(ν,hν)), |
a contradiction.
Thus, w(ν,hν)=0 that is ν is a fixed point of h.
We can prove that is ν is a fixed point of g by using arguments similar to those used above. Similar arguments to those used in the proof of Theorem 3.2 are used to establish the uniqueness of the common fixed point of g and h.
Corollary 3.1. Let (Ω,⪯) be a partially ordered set, (Ω,δ) be a separated proximity space with w0−distance w and g:Ω→Ω be a mapping that meet the requirements listed below:
(i) There exist α∈Π, ψ∈Ψ and a continuous function β:[0,∞)→[0,∞) with β(t)≤ψ(t) for all t>0 such that
ψ(w(gμ,gξ))≤α(max{w(μ,ξ),w(ξ,gξ),w(gμ,μ)})β(max{w(μ,ξ),w(ξ,gξ),w(gμ,μ)}), |
holds for all comparable elements μ,ξ∈Ω,
(ii) gξ⪯g(gξ) for all ξ∈Ω,
(iii) g is continuous,
(iv) For all η∈Ω, any iterative sequence {gnη} has convergent subsequences with respect to τδ.
Then g has a fixed point ν∈ Ω with w(ν,ν)=0. Furthermore, assume that if η,u∈Ω such w(η,η)=w(u,u)=0 implies that η and u are comparable then the fixed point of g is unique.
Proof. Theorem 3.2 implies that by inserting h=g.
Corollary 3.2. Let we take that
(iii) Ω is regular,
Instead of (iii) in Corollary 3.1, again we can have the same result.
Proof. Theorem 3.3 implies that by inserting h=g.
Corollary 3.3. Let (Ω,⪯) be a partially ordered set, (Ω,δ) be a separated proximity space with w0−distance w and g,h:Ω→Ω be two mappings that meet the requirements listed below:
(i) There exist α∈Π, ψ∈Ψ and a continuous function β:[0,∞)→[0,∞) with β(t)≤ψ(t) for all t>0 such that
ψ(w(gμ,hξ))≤α(w(μ,ξ))β(w(μ,ξ)) |
holds for all comparable elements μ,ξ∈Ω,
(ii) The pair (g,h) is weakly increasing,
(iii) Either g or h is continuous,
(iv) For all η∈Ω, any iterative sequences {gnη} and {hnη} have convergent subsequences with respect to τδ.
Then g and h have a common fixed point ν∈ Ω with w(ν,ν)=0. Furthermore, assume that if η, u∈Ω such w(η,η)=w(u,u)=0 implies that η and u are comparable then the common fixed point of g and h is unique.
Corollary 3.4. Let (Ω,⪯) be a partially ordered set, (Ω,δ) be a separated proximity space with w0−distance w and g,h:Ω→Ω be two mappings that meet the requirements listed below:
(i) There exist α∈Π, ψ∈Ψ and a continuous function β:[0,∞)→[0,∞) with β(t)≤ψ(t) for all t>0 such that
ψ(w(gμ,hξ))≤α(w(μ,ξ))β(w(μ,ξ)) |
holds for all comparable elements μ,ξ∈Ω,
(ii) The pair (g,h) is weakly increasing,
(iii) Ω is regular,
(iv) For all η∈Ω, any iterative sequences {gnη} and {hnη} have convergent subsequences with respect to τδ.
Then g and h have a common fixed point ν∈ Ω with w(ν,ν)=0. Furthermore, assume that if η, u∈Ω such w(η,η)=w(u,u)=0 implies that η and u are comparable then the common fixed point of g and h is unique.
Corollary 3.5. Let (Ω,⪯) be a partially ordered set, (Ω,δ) be a separated proximity space with w0−distance w and g:Ω→Ω be a mapping that meet the requirements listed below:
(i) There exist α∈Π, ψ∈Ψ and a continuous function β:[0,∞)→[0,∞) with β(t)≤ψ(t) for all t>0 such that
ψ(w(gμ,gξ))≤α(w(μ,ξ))β(w(μ,ξ)) |
holds for all comparable elements μ,ξ∈Ω,
(ii) gξ⪯g(gξ) for all ξ∈Ω,
(iii) g is continuous,
(iv) For all η∈Ω, any iterative sequences {gnη} has convergent subsequences with respect to τδ.
Then g has a fixed point ν∈ Ω with w(ν,ν)=0. Furthermore, assume that if η, u∈Ω such w(η,η)=w(u,u)=0 implies that η and u are comparable then the fixed point of g is unique.
Corollary 3.6. Let (Ω,⪯) be a partially ordered set, (Ω,δ) be a separated proximity space with w0−distance w and g:Ω→Ω be a mapping that meet the requirements listed below:
(i) There exist α∈Π, ψ∈Ψ and a continuous function β:[0,∞)→[0,∞) with β(t)≤ψ(t) for all t>0 such that
ψ(w(gμ,gξ))≤α(w(μ,ξ))β(w(μ,ξ)) |
holds for all comparable elements μ,ξ∈Ω,
(ii) gξ⪯g(gξ) for all ξ∈Ω,
(iii) Ω is regular,
(iv) For all η∈Ω, any iterative sequence {gnη} has convergent subsequences with respect to τδ.
Then g has a fixed point ν∈ Ω with w(ν,ν)=0. Furthermore, assume that if η, u∈Ω such w(η,η)=w(u,u)=0 implies that η and u are comparable then the fixed point of g is unique.
Example 3.1. Let Ω={0,1,2} be equipped with the following partial order ⪯,
⪯:={(0,0),(1,1),(2,2),(1,0)}. |
Also, let Ω be endowed with the usual metric and the proximity δ on 2Ω as
AδB⇔p(A,B)=0,where p(A,B)=min{p(u,v):u∈A,v∈B}. |
Define w:Ω×Ω→[0,∞) by
w(ξ,μ)=max{|ξ|,|μ|}, |
w(0,0)=0, w(1,1)=1, w(2,2)=2, w(0,1)=w(1,0)=1, w(0,2)=w(2,0)=2,w(1,2)=w(2,1)=2.
It is easy to see that (Ω,δ) be a separated proximity space with w0 distance w.
Also define g,h:Ω→Ω with g(0)=0, g(1)=0, g(2)=1 and h(0)=0, h(1)=1, h(2)=0. It is simple to observe that g and h are continuous and that the pair (g,h) is weakly increasing with respect to ⪯.
Define α(t)=e−t16, β(t)=1011et, ψ(t)=1et if t>0 and α(0)=0.
We next verify that the functions (g,h) satisfies the inequality
ψ(w(gξ,hμ))≤α(Kξ,μ)β(Kξ,μ). |
For that, given ξ,μ∈Ω with ξ⪯μ.
Then we have the following cases:
Case i ξ=0 and μ=0. Then
ψ(w(g0,h0))=ψ(w(0,0))=0≤α(K0,0)β(K0,0). |
Case ii ξ=1 and μ=1. Then
ψ(w(g1,h1))=ψ(w(0,1))=ψ(1)=1e |
and
K1,1=max{w(1,1),w(g1,1),w(1,h1)}=max{1,1,1}=1. |
So,
α(K1,1)=α(1)=e−116 |
β(K1,1)=β(1)=1011e. |
Hence
ψ(w(g1,h1))≤α(K1,1)β(K1,1). |
Case iii ξ=2 and μ=2. Then
ψ(w(g2,h2))=ψ(w(1,0))=ψ(1)=1e |
and
K2,2=max{w(2,2),w(g2,2),w(2,h2)}=max{2,2,2}=2. |
So,
α(K2,2)=α(2)=e−216=e−18 |
β(K2,2)=β(2)=2011e. |
Hence
ψ(w(g2,h2))≤α(K2,2)β(K2,2). |
Case iv 1⪯0. Then we have two subcases:
Subcase i ξ=1 and μ=0. Then
ψ(w(g1,h0))=ψ(w(0,0))=0 |
and
K1,0=max{w(1,0),w(g1,0),w(1,h0)}=max{1,0,1}=1. |
So,
α(K1,0)=α(1)=e−116 |
β(K1,0)=β(1)=1011e. |
Hence
ψ(w(g1,h0))≤α(K1,0)β(K1,0). |
Subcase ii ξ=0 and μ=1. Then
ψ(w(g0,h1))=ψ(w(0,1))=ψ(1)=1e |
and
K0,1=max{w(0,1),w(g0,1),w(0,h1)}=max{1,1,1}=1. |
So,
α(K0,1)=α(1)=e−116 |
β(K0,1)=β(1)=1011e. |
Hence
ψ(w(g0,h1))≤α(K0,1)β(K0,1). |
Therefore, requirements of Theorem3.3 are all satisfied and so g and h have a common fixed point (0 is a common fixed point of g and h).
Let (Ω,δ) be the proximity space, where Ω=C[0,1] and δ is induced by the uniform metric p∞(ξ,μ)=sup{|ξ(t)−μ(t)|:t∈[0,1]}. In this case (Ω,δ) is separated proximity space. Consider the following w0−distance w on Ω defined by
w(ξ,μ)=sup{e−t|ξ(t)−μ(t)|:t∈[0,1]}. |
Now, consider the integral equation
ξ(t)=G(t)+1∫0S(t,s)F(s,ξ(s))ds,t∈[0,1] | (4.1) |
where F:[0,1]×R→R, G:[0,1]→R, S:[0,1]×[0,1]→[0,∞). By utilizing the outcome from Corollary 1, the objective of this section is to provide an existence answer to (4.1). We give Ω the partial order "⪯" provided by:
ξ⪯μ⇔ξ(t)⪯μ(t) |
for all t∈[0,1].
Theorem 4.1. Suppose that the following conditions are satisfied:
(i) There exists α:[0,∞)→[0,1] such that for all s∈[0,1] and for all ξ,μ∈Ω
0≤|F(s,ξ(s))−F(s,μ(s))|≤α(e−s|ξ(s)−μ(s)|) |
and
α(tn)→1⇒tn→0, |
(ii) 1∫0S(t,s)ds≤|ξ(t)−μ(t)|.
Then the integral equation (4.1) has a solution in Ω.
Proof. Consider the mapping g:Ω→Ω defined by
gξ(t)=1∫0S(t,s)F(s,ξ(s))ds, |
for all ξ∈Ω and t∈[0,1]. Then the (4.1) is equivalent to finding a fixed point of g.
Now, let ξ,μ∈Ω. We have:
|gξ(t)−gμ(t)|=|1∫0S(t,s)[F(s,ξ(s))−F(s,μ(s))]ds|≤1∫0S(t,s)|F(s,ξ(s))−F(s,μ(s))|ds≤1∫0S(t,s)α(e−s|ξ(s)−μ(s)|)ds ≤|ξ(t)−μ(t)|α(e−t|ξ−μ|)≤p∞(ξ,μ)α(e−t|ξ−μ|)≤p∞(ξ,μ)α(w(ξ,μ)) |
and then we obtain
e−t|gξ(t)−gμ(t)|≤w(ξ,μ)α(w(ξ,μ)) |
i.e.,
w(gξ,gμ)≤w(ξ,μ)α(w(ξ,μ)) |
for all ξ,μ∈Ω.
Now, let γ∈C[0,1] be an arbitrary function. Define a sequence of functions {ξn} as gnξ=ξn. Since e−tp∞(ξ,μ)≤w(ξ,μ)≤p∞(ξ,μ) for all ξ,μ∈Ω, we have p∞(ξn,ξm)→0 as m,n→∞. That is the sequence {ξn} is Cauchy and so has a convergent subsequence with respect to p∞ since (Ω,p∞) is complete. Consequently, there exists a unique ξ∈Ω which is a fixed point of the operator g, moreover w(ξ,ξ)=0. Hence the integral equation (4.1) has a unique solution in Ω.
The author is thankful to the referees and editor for making valuable suggestions leading to the better presentations of the paper.
The author declares no conflict of interest.
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