# L-Fuzzy Fixed Points Theorems for L-Fuzzy Mappings via βℱL-Admissible Pair

Authors: Maliha Rashid; Akbar Azam; Nayyar Mehmood
Journal: The Scientific World Journal
 (2014)

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## Abstract

We define the concept ofβℱL-admissible for a pair of L-fuzzy mappings and establish the existence of common L-fuzzy fixed point theorem. Our result generalizes some useful results in the literature. We provide an example to support our result.


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## Body

## 1. Introduction

A large variety of the problems of analysis and applied mathematics relate to finding solutions of nonlinear functional equations which can be formulated in terms of finding the fixed points of nonlinear mappings. Heilpern [1] first introduced the concept of fuzzy mappings and established a fixed point theorem for fuzzy contraction mappings in complete metric linear spaces, which is a fuzzy extension of Banach contraction principle and Nadler’s [2] fixed point theorem. Subsequently several other authors [3–17] generalized this result and studied the existence of fixed points and common fixed points of fuzzy mappings satisfying a contractive type condition.Zadeh published his important paper “Fuzzy sets” [18], after that Goguen published the paper “L-Fuzzy sets” [19]. The concept of L-fuzzy sets is a generalization of the concept of fuzzy sets. Fuzzy set is a special case of L-fuzzy set when L=[0,1]. There are basically two understandings of the meaning of L, one is when L is a complete lattice equipped with a multiplication * operator satisfying certain conditions as shown in the initial paper [19] and the second understanding of the meaning of L is that L is a completely distributive complete lattice with an order-reversing involution (see, e.g., [20–22], etc.).In 2012, Samet et al. [23] introduced the concept of β-admissible mapping and established fixed point theorems via β-admissible and also showed that these results can be utilized to derive fixed point theorems in partially ordered spaces and coupled fixed point theorems. Moreover, they applied the main results to ordinary differential equations. Afterwards, Asl et al. [24] extended the concept of β-admissible for single valued mappings to multivalued mappings. Recently, Mohammadi et al. [25] introduced the concept of β-admissible for multivalued mappings which is different from the notion of β*-admissible which has been provided in [24] and Azam and Beg [26] obtained a common α-fuzzy fixed point of a pair of fuzzy mappings on a complete metric space under a generalized contractive condition for α-level sets via Hausdorff metric for fuzzy sets.In this paper we introduce the concept ofβℱL-admissible for a pair of L-fuzzy mappings and establish the existence of common L-fuzzy fixed point theorem. We also have given an example to support our main theorem.


## 2. Preliminaries

Let(X,d) be a metric space, and denoteCB(X)={A:A is nonempty closed and bounded subset of X},C(X)={A:A is nonempty compact subset of X}.Forϵ>0 and the sets A,B∈CB(X) define
(1)d(x,A)=infy∈Ad(x,y),d(A,B)=infx∈A,y∈Bd(x,y),N(ϵ,A)={x∈X:d(x,a)<ϵ,forsomea∈A},EA,B={ϵ>0:A⊆N(ϵ,B),B⊆N(ϵ,A)}.Then the Hausdorff metricdH on CB(X) induced by d is defined as
(2)dH(A,B)=infEA,B.Lemma 1 (see [2]).
Let(X,d) be a metric space and A,B∈CB(X); then for each a∈A(3)d(a,B)≤H(A,B).Lemma 2 (see [2]).
Let(X,d) be a metric space and A,B∈CB(X); then for each a∈A, ϵ>0, there exists an element b∈B such that
(4)d(a,b)≤H(A,B)+ϵ.Definition 3 (see [19]).
A partially ordered set(L,≾L) is called(i)
a lattice, ifa∨b∈L,  a∧b∈L for any a,b∈L;(ii)
a complete lattice, if∨A∈L,  ∧A∈L for any A⊆L;(iii)
distributive ifa∨(b∧c)=(a∨b)∧(a∨c),  a∧(b∨c)=(a∧b)∨(a∧c) for any a,b,c∈L.Definition 4 (see [19]).
LetL be a lattice with top element 1L and bottom element 0L and let a,b∈L. Then b is called a complement of a, if a∨b=1L, and a∧b=0L. If a∈L has a complement element, then it is unique. It is denoted by a´.Definition 5 (see [19]).
AnL-fuzzy set A on a nonempty set X is a function A:X→L, where L is complete distributive lattice with 1L and 0L.Remark 6.
The class ofL-fuzzy sets is larger than the class of fuzzy sets as an L-fuzzy set is a fuzzy set if L=[0,1].TheαL-level set of L-fuzzy set A is denoted by AαL and is defined as follows:
(5)AαL={x:αL≾LA(x)}ifαL∈L∖{0L},A0L=cl({x:0L≾LA(x)}).Herecl(B) denotes the closure of the set B.We denote and define the characteristic functionχLA of an L-fuzzy set A as follows:
(6)χLA:={0Lifx∉A,1Lifx∈A.Definition 7.
LetX be an arbitrary set and Y a metric space. A mapping T is called L-fuzzy mapping if T is a mapping from X into ℱL(Y). An L-fuzzy mapping T is an L-fuzzy subset on X×Y with membership function T(x)(y). The function T(x)(y) is the grade of membership of y in T(x).Definition 8.
Let(X,d) be a metric space and S,TL-fuzzy mappings from X into ℱL(X). A point z∈X is called an L-fuzzy fixed point of T if z∈[Tz]αL, where αL∈L∖{0L}. The point z∈X is called a common L-fuzzy fixed point of S and T if z∈[Sz]αL∩[Tz]αL.Definition 9 (see [23]).
LetX be a nonempty set, T:X→X, and β:X×X→[0,∞). We say that T is β-admissible if for all x,y∈X we have
(7)β(x,y)≥1⟹β(Tx,Ty)≥1.Definition 10 (see [24]).
LetX be a nonempty set, T:X→2X, where 2X is a collection of subset of X, β:X×X→[0,∞) and β*:2X×2X→[0,∞). We say that T is β*-admissible if for all x,y∈X we have
(8)β(x,y)≥1⟹β*(Tx,Ty)≥1.Definition 11 (see [25]).
LetX be a nonempty set, T:X→2X, where 2X is a collection of subset of X and β:X×X→[0,∞). We say that T is β-admissible whenever for each x∈X and y∈Tx with β(x,y)≥1 we have β(y,z)≥1 for all z∈Ty.


## 3. Main Result

In this section, we introduce a new concept ofβℱL-admissible for a pair of L-fuzzy mappings and establish the existence of common L-fuzzy fixed point theorem.Definition 12.
Let(X,d) be a metric space, β:X×X→[0,∞), and S,TL-fuzzy mappings from X into ℱL(X). The order pair (S,T) is said to be βℱL-admissible if it satisfies the following conditions:(i)
for eachx∈X and y∈[Sx]αL(x), where αL(x)∈L∖{0L}, with β(x,y)≥1, we have β(y,z)≥1 for all z∈[Ty]αL(y)≠ϕ, where αL(y)∈L∖{0L};(ii)
for eachx∈X and y∈[Tx]αL(x), where αL(x)∈L∖{0L}, with β(x,y)≥1, we have β(y,z)≥1 for all z∈[Sy]αL(y)≠ϕ, where αL(y)∈L∖{0L}.
IfS=T then T is called βℱL-admissible.Remark 13.
It is easy to see that if(S,T) is βℱL-admissible, then (T,S) is also βℱL-admissible.Next, we give a commonL-fuzzy fixed point theorem for βF-admissible pair.Theorem 14.
Let(X,d) be a complete metric space, β:X×X→[0,∞), and S,TL-fuzzy mappings from X into ℱL(X) satisfying the following conditions.(a)
For eachx∈X, there exists αL(x)∈L∖{0L} such that [Sx]αL(x), [Tx]αL(x) are nonempty closed bounded subsets of X and for x0∈X, there exists x1∈[Sx0]αL(x0) with β(x0,x1)≥1.(b)
For allx,y∈X, we have
(9)max{β(x,y),β(y,x)}[H([Sx]αL(x),[Ty]αL(y))]≤a1(dx,[Sx]αL(x))+a2(dy,[Ty]αL(y))+a3(dx,[Ty]αL(y))+a4(dy,[Sx]αL(x))+a5d(x,y),wherea1, a2, a3, a4, and a5 are nonnegative real numbers and ∑i=15ai<1 and either a1=a2 or a3=a4.(c)
(S,T) is βℱL-admissible pair.(d)
If{xn}∈X, such that β(xn,xn+1)≥1 and xn→x, then β(xn,x)≥1.
Then there existsz∈X such that z∈[Sz]αL(z)∩[Tz]αL(z).Proof.
We will prove the above result by considering the following three cases:(1)
a1+a3+a5=0,(2)
a2+a4+a5=0,(3)
a1+a3+a5≠0 and a2+a4+a5≠0.
Case 1. For x0∈X in condition (a), there exist αL(x0)∈L∖{0L} and x1∈[Sx0]αL(x0) such that β(x0,x1)≥1 and also there exists αL(x1)∈L∖{0L} such that [Sx0]αL(x0) and [Tx1]αL(x1) are nonempty closed bounded subsets of X. From Lemma 1, we obtain that
(10)d(x1,[Tx1]αL(x1))≤H([Sx0]αL(x0),[Tx1]αL(x1))≤β(x0,x1)[H([Sx0]αL(x0),[Tx1]αL(x1))]≤max{β(x0,x1),β(x1,x0)}×[H([Sx0]αL(x0),[Tx1]αL(x1))].
Now, inequality (9) implies that
(11)d(x1,[Tx1]αL(x1))≤a1d(x0,[Sx0]αL(x0))+a2d(x1,[Tx1]αL(x1))+a3d(x0,[Tx1]αL(x1))+a4d(x1,[Sx0]αL(x0))+a5d(x0,x1).
Usinga1+a3+a5=0 together with the fact that d(x1,[Sx0]αL(x0))=0, we get
(12)d(x1,[Tx1]αL(x1))≤a2d(x1,[Tx1]αL(x1)).
It follows thatx1∈[Tx1]αL(x1), which further implies that
(13)d(x1,[Sx1]αL(x1))≤H([Tx1]αL(x1),[Sx1]αL(x1)).
By condition (c), forx0∈X and x1∈[Sx0]αL(x0) such that β(x0,x1)≥1, we have β(x1,z)≥1 for all z∈[Tx1]αL(x1). Since x1∈[Tx1]αL(x1), therefore β(x1,x1)≥1 and hence
(14)d(x1,[Sx1]αL(x1))≤β(x1,x1)[H([Sx1]αL(x1),[Tx1]αL(x1))].
Again, inequality (9) implies that
(15)d(x1,[Sx1]αL(x1))≤a1d(x1,[Sx1]αL(x1))+a2d(x1,[Tx1]αL(x1))+a3d(x1,[Tx1]αL(x1))+a4d(x1,[Sx1]αL(x1))+a5d(x1,x1).
Sincea1+a3+a5=0 and d(x1,[Tx1]αL(x1))=0, we get
(16)d(x1,[Sx1]αL(x1))≤a4d(x1,[Sx1]αL(x1)),
which implies that x1∈[Sx1]αL(x1) and hence
(17)x1∈[Sx1]αL(x1)∩[Tx1]αL(x1).
Case 2. For x0∈X in condition (a), there exist αL(x0)∈L∖{0L} and x1∈[Sx0]αL(x0) such that β(x0,x1)≥1 and also there exists αL(x1)∈L∖{0L} such that [Sx0]αL(x0) and [Tx1]αL(x1) are nonempty closed bounded subsets of X. By condition (c), we have β(x1,x2)≥1 for all x2∈[Tx1]αL(x1). From Lemma 1, we obtain that
(18)d(x2,[Sx2]αL(x2))≤H([Tx1]αL(x1),[Sx2]αL(x2))≤β(x1,x2)[H([Sx2]αL(x2),[Tx1]αL(x1))]≤max{β(x1,x2),β(x2,x1)}×[H([Sx2]αL(x2),[Tx1]αL(x1))]≤a1d(x2,[Sx2]αL(x2))+a2d(x1,[Tx1]αL(x1))+a3d(x2,[Tx1]αL(x1))+a4d(x1,[Sx2]αL(x2))+a5d(x2,x1).
Usinga2+a4+a5=0 together with the fact that d(x2,[Tx1]αL(x1))=0, we get
(19)d(x2,[Sx2]αL(x2))≤a1d(x2,[Sx2]αL(x2)).
It follows thatx2∈[Sx2]αL(x2), which further implies that
(20)d(x2,[Tx2]αL(x2))≤H([Sx2]αL(x2),[Tx2]αL(x2)).
By condition (c), we haveβ(x2,x2)≥1, and hence
(21)d(x2,[Tx2]αL(x2))≤β(x2,x2)[H([Sx2]αL(x2),[Tx2]αL(x2))].
Again, inequality (9) implies that
(22)d(x2,[Tx2]αL(x2))≤a1d(x2,[Sx2]αL(x2))+a2d(x2,[Tx2]αL(x2))+a3d(x2,[Tx2]αL(x2))+a4d(x2,[Sx2]αL(x2))+a5d(x2,x2).
Sincea2+a4+a5=0 and d(x2,[Sx2]αL(x2))=0, we get
(23)d(x2,[Tx2]αL(x2))≤a3d(x2,[Tx2]αL(x2)),
which implies that x2∈[Tx2]αL(x2) and hence
(24)x2∈[Sx2]αL(x2)∩[Tx2]αL(x2).
Case 3. Let λ=((a1+a3+a5)/(1-a2-a3)) and μ=((a2+a4+a5)/(1-a1-a4)). Next, we show that if a1=a2 or a3=a4, then 0<λμ<1.
Ifa3=a4, then λ,μ<1 and so 0<λμ<1. Now if a1=a2, then
(25)0<λμ=(a1+a3+a51-a2-a3)(a2+a4+a51-a1-a4)=(a1+a3+a51-a1-a3)(a1+a4+a51-a1-a4)=(a1+a3+a51-a1-a4)(a1+a4+a51-a1-a3)<1.
By condition (a), for x1∈X, there exists αL(x1)∈L∖{0L} such that [Tx1]αL(x1) is a nonempty closed bounded subset of X. Since a1+a3+a5>0, by Lemma 2, there exists x2∈[Tx1]αL(x1) such that
(26)d(x1,x2)≤H([Sx0]αL(x0),[Tx1]αL(x1))+a1+a3+a5≤β(x0,x1)[H([Sx0]αL(x0),[Tx1]αL(x1))]+a1+a3+a5≤max{β(x0,x1),β(x1,x0)}×[H([Sx0]αL(x0),[Tx1]αL(x1))]+a1+a3+a5≤a1d(x0,[Sx0]αL(x0))+a2d(x1,[Tx1]αL(x1))+a3d(x0,[Tx1]αL(x1))+a4d(x1,[Sx0]αL(x0))+a5d(x0,x1)+a1+a3+a5≤(a1+a5)d(x0,x1)+a2d(x1,x2)+a3d(x0,x2)+a1+a3+a5≤(a1+a3+a5)d(x0,x1)+(a2+a3)d(x1,x2)+a1+a3+a5.
This implies that
(27)d(x1,x2)≤λd(x0,x1)+λ.
By the same argument, for x2∈X, there exists αL(x2)∈L∖{0L} such that [Sx2]αL(x2) is a nonempty closed bounded subset of X. Since a2+a4+a5>0, by Lemma 2, there exists x3∈[Sx2]αL(x2) such that
(28)d(x2,x3)≤H([Tx1]αL(x1),[Sx2]αL(x2))+λ(a2+a4+a5).
By condition (c), for x0∈X and x1∈[Sx0]αL(x0) such that β(x0,x1)≥1, we have β(x1,x2)≥1 for x2∈[Tx1]αL(x1). So we have
(29)d(x2,x3)≤H([Tx1]αL(x1),[Sx2]αL(x2))+λ(a2+a4+a5)=H([Sx2]αL(x2),[Tx1]αL(x1))+λ(a2+a4+a5)≤β(x1,x2)[H([Sx2]αL(x2),[Tx1]αL(x1))]+λ(a2+a4+a5)≤max{β(x1,x2),β(x2,x1)}×[H([Sx2]αL(x2),[Tx1]αL(x1))]+λ(a2+a4+a5)≤a1d(x2,[Sx2]αL(x2))+a2d(x1,[Tx1]αL(x1))+a3d(x2,[Tx1]αL(x1))+a4d(x1,[Sx2]αL(x2))+a5d(x2,x1)+λ(a2+a4+a5)≤a1d(x2,x3)+(a2+a5)d(x1,x2)+a4d(x1,x3)+λ(a2+a4+a5)≤(a2+a4+a5)d(x1,x2)+(a1+a4)d(x2,x3)+λ(a2+a4+a5).
This implies that
(30)d(x2,x3)≤μd(x1,x2)+λμ.
By repeating the above process, for x3∈X, there exists αL(x3)∈L∖{0L} such that [Tx3]αL(x3) is a nonempty closed bounded subset of X. From Lemma 2, there exists x4∈[Tx3]αL(x3) such that
(31)d(x3,x4)≤H([Sx2]αL(x2),[Tx3]αL(x3))+λμ(a1+a3+a5).
By condition (c), for x1∈X and x2∈[Tx1]αL(x1) such that β(x1,x2)≥1, we have β(x2,x3)≥1 for x3∈[Sx2]αL(x2). So we have
(32)d(x3,x4)≤H([Sx2]αL(x2),[Tx3]αL(x3))+λμ(a1+a3+a5)≤β(x2,x3)[H([Sx2]αL(x2),[Tx3]αL(x3))]+λμ(a1+a3+a5)≤max{β(x2,x3),β(x3,x2)}×[H([Sx2]αL(x2),[Tx3]αL(x3))]+λμ(a1+a3+a5)≤a1d(x2,[Sx2]αL(x2))+a2d(x3,[Tx3]αL(x3))+a3d(x2,[Tx3]αL(x3))+a4d(x3,[Sx2]αL(x2))+a5d(x2,x3)+λμ(a1+a3+a5)≤(a1+a5)d(x2,x3)+a2d(x3,x4)+a3d(x2,x4)+λμ(a1+a3+a5)≤(a1+a3+a5)d(x2,x3)+(a2+a3)d(x3,x4)+λμ(a1+a3+a5).
This implies that
(33)d(x3,x4)≤λd(x2,x3)+λ(λμ).
By induction, we produce a sequence {xn} in X such that
(34)x2k+1∈[Sx2k]αL(x2k),x2k+2∈[Tx2k+1]αL(x2k+1),k=0,1,2,…,β(xn-1,xn)≥1,∀n∈ℕ.
Now, we have
(35)d(x2k+1,x2k+2)≤H([Sx2k]αL(x2k),[Tx2k+1]αL(x2k+1))+(λμ)k(a1+a3+a5)≤β(x2k,x2k+1)×[H([Sx2k]αL(x2k),[Tx2k+1]αL(x2k+1))]+(λμ)k(a1+a3+a5)≤max{β(x2k,x2k+1),β(x2k+1,x2k)}×[H([Sx2k]αL(x2k),[Tx2k+1]αL(x2k+1))]+(λμ)k(a1+a3+a5)≤a1d(x2k,[Sx2k]αL(x2k))+a2d(x2k+1,[Tx2k+1]αL(x2k+1))+a3d(x2k,[Tx2k+1]αL(x2k+1))+a4d(x2k+1,[Sx2k]αL(x2k))+a5d(x2k,x2k+1)+(λμ)k(a1+a3+a5)≤(a1+a5)d(x2k,x2k+1)+a2d(x2k+1,x2k+2)+a3d(x2k,x2k+2)+(λμ)k(a1+a3+a5)≤(a1+a3+a5)d(x2k,x2k+1)+(a2+a3)d(x2k+1,x2k+2)+(λμ)k(a1+a3+a5).
This implies that
(36)d(x2k+1,x2k+2)≤λd(x2k,x2k+1)+λ(λμ)k.
Similarly,
(37)d(x2k+2,x2k+3)≤H([Sx2k+2]αL(x2k+2),[Tx2k+1]αL(x2k+1))+(λμ)kλ(a2+a4+a5)≤β(x2k+1,x2k+2)×[H([Sx2k+2]αL(x2k+2),[Tx2k+1]αL(x2k+1))]+(λμ)kλ(a2+a4+a5)≤max{β(x2k+1,x2k+2),β(x2k+2,x2k+1)}×[H([Sx2k+2]αL(x2k+2),[Tx2k+1]αL(x2k+1))]+(λμ)kλ(a2+a4+a5)≤a1d(x2k+2,[Sx2k+2]αL(x2k+2))+a2d(x2k+1,[Tx2k+1]αL(x2k+1))+a3d(x2k+2,[Tx2k+1]αL(x2k+1))+a4d(x2k+1,[Sx2k+2]αL(x2k+2))+a5d(x2k+2,x2k+1)+(λμ)kλ(a2+a4+a5)≤a1d(x2k+2,x2k+3)+(a2+a5)d(x2k+1,x2k+2)+a4d(x2k+1,x2k+3)+(λμ)kλ(a2+a4+a5)≤(a2+a4+a5)d(x2k+1,x2k+2)+(a1+a4)d(x2k+2,x2k+3)+(λμ)kλ(a2+a4+a5).
This implies that
(38)d(x2k+2,x2k+3)≤μd(x2k+1,x2k+2)+(λμ)k+1.
From (36) and (38), it follows that, for each k=0,1,2,…,
(39)d(x2k+1,x2k+2)≤λd(x2k,x2k+1)+λ(λμ)k≤λ[μd(x2k-1,x2k)+(λμ)k]+λ(λμ)k=(λμ)d(x2k-1,x2k)+2λ(λμ)k≤(λμ)[λd(x2k-2,x2k-1)+λ(λμ)k-1]+2λ(λμ)k=(λμ)λd(x2k-2,x2k-1)+3λ(λμ)k⋮≤λ(λμ)kd(x0,x1)+(2k+1)λ(λμ)k,d(x2k+2,x2k+3)≤μd(x2k+1,x2k+2)+(λμ)k+1⋮≤(λμ)k+1d(x0,x1)+(2k+2)(λμ)k+1.
Then for m<n, we have
(40)d(x2m+1,x2n+1)≤d(x2m+1,x2m+2)+d(x2m+2,x2m+3)+d(x2m+3,x2m+4)+⋯+d(x2n,x2n+1)≤[λ∑i=mn-1‍(λμ)i+∑i=m+1n‍(λμ)i]d(x0,x1)+λ∑i=mn-1‍(2i+1)(λμ)i+∑i=m+1n‍2i(λμ)i.
Similarly, we obtain that
(41)d(x2m,x2n+1)≤[∑i=mn‍(λμ)i+λ∑i=mn-1‍(λμ)i]d(x0,x1)≤+∑i=mn‍2i(λμ)i+λ∑i=mn-1‍(2i+1)(λμ)i,d(x2m,x2n)≤[∑i=mn-1‍(λμ)i+λ∑i=mn-1‍(λμ)i]d(x0,x1)≤+∑i=mn-1‍2i(λμ)i+λ∑i=mn-1‍(2i+1)(λμ)i,d(x2m+1,x2n)≤[λ∑i=mn-1‍(λμ)i+∑i=m+1n-1‍(λμ)i]d(x0,x1)≤+λ∑i=mn-1‍(2i+1)(λμ)i+∑i=m+1n-1‍2i(λμ)i.
Since 0<λμ<1, so by Cauchy’s root test, we get ∑‍(2i+1)(λμ)i and ∑‍2i(λμ)i are convergent series. Therefore, {xn} is a Cauchy sequence in X. Now, from the completeness of X, there exists z∈X such that xn→z as n→∞. By condition (d), we have β(xn-1,z)≥1 for all n∈ℕ. Now, we have
(42)d(x2n,[Sz]α(z))≤H([Tx2n-1]αL(x2n-1),[Sz]α(z))=H([Sz]α(z),[Tx2n-1]αL(x2n-1))≤β(x2n-1,z)H([Sz]α(z),[Tx2n-1]αL(x2n-1))≤max{β(x2n-1,z),β(z,x2n-1)}×H([Sz]α(z),[Tx2n-1]αL(x2n-1))≤a1d(z,[Sz]αL(z))+a2d(x2n-1,[Tx2n-1]αL(x2n-1))+a3d(z,[Tx2n-1]αL(x2n-1))+a4d(x2n-1,[Sz]αL(z))+a5d(z,x2n-1)≤a1d(z,[Sz]αL(z))+a2d(x2n-1,x2n)+a3d(z,x2n)+a4d(x2n-1,[Sz]αL(z))+a5d(z,x2n-1).
Since
(43)d(z,[Sz]αL(z))≤d(z,x2n)+d(x2n,[Sz]αL(z)),
so we get
(44)d(z,[Sz]αL(z))≤d(z,x2n)+a1d(z,[Sz]αL(z))+a2d(x2n-1,x2n)+a3d(z,x2n)+a4d(x2n-1,[Sz]αL(z))+a5d(z,x2n-1)≤(1+a3)d(z,x2n)+(a4+a5)d(z,x2n-1)+a2d(x2n-1,x2n)+(a1+a4)d(z,[Sz]αL(z)).
This implies that
(45)d(z,[Sz]αL(z))≤(1+a31-a1-a4)d(z,x2n)+(a4+a51-a1-a4)d(z,x2n-1)+(a21-a1-a4)d(x2n-1,x2n).
Letting n→∞, we have d(z,[Sz]αL(z))=0. It implies that z∈[Sz]αL(z). Similarly, by using
(46)d(z,[Tz]αL(z))≤d(z,x2n+1)+d(x2n+1,[Tz]αL(z)),
we can show that z∈[Tz]αL(z). Therefore, z∈[Sz]αL(z)∩[Tz]αL(z). This completes the proof.Next, we give an example to support the validity of our result.Example 15.
LetX=[0,1], d(x,y)=|x-y|, whenever x,y∈X; then (X,d) is a complete metric space. Let L={δ,ω,τ,κ} with δ≾Lω≾Lκ, δ≾Lτ≾Lκ, ω and τ are not comparable; then (L,≾L) is a complete distributive lattice. Define a pair of mappings S,T:X→ℱL(X)as follows:
(47)S(x)(t)={κ,if0≤t≤x6,ω,ifx6<t≤x3,τ,ifx3<t≤x2,δ,ifx2<t≤1,T(x)(t)={κ,if0≤t≤x12,δ,ifx12<t≤x8,ω,ifx8<t≤x4,τ,ifx4<t≤1.
Define β:X×X→[0,∞) as follows:
(48)β(x,y)={1|x-y|,x≠y,1,x=y.
For all x∈X, there exists αL(x)=κ, such that
(49)[Sx]κ=[0,x6],[Tx]κ=[0,x12],
and all conditions of the above theorem are satisfied. Hence, there exists 0∈X, such that 0∈[S0]αL(0)∩[T0]αL(0).Corollary 16.
Let(X,d) be a complete metric space, β:X×X→[0,∞), and S, T fuzzy mappings from X into ℱ(X) satisfying the following conditions.(a)
For eachx∈X, there exists α(x)∈(0,1] such that [Sx]α(x), [Tx]α(x) are nonempty closed bounded subsets of X and for x0∈X, there exists x1∈[Sx0]α(x0) with β(x0,x1)≥1.(b)
For allx,y∈X, we have
(50)max{β(x,y),β(y,x)}[H([Sx]α(x),[Ty]α(y))]≤a1(dx,[Sx]α(x))+a2(dy,[Ty]α(y))+a3(dx,[Ty]α(y))+a4(dy,[Sx]α(x))+a5d(x,y),wherea1, a2, a3, a4, and a5 are nonnegative real numbers and ∑i=15ai<1 and either a1=a2 or a3=a4.(c)
(S,T) is βℱ-admissible pair.(d)
If{xn}∈X, such that β(xn,xn+1)≥1 and xn→x then β(xn,x)≥1.
Then there existsz∈X such that z∈[Sz]α(z)∩[Tz]α(z).Proof.
Consider anL-fuzzy mapping A:X→ℱL(X) defined by
(51)Ax=χLTx.
Then for αL∈L∖{0L}, we have
(52)[Ax]αL=Tx.
Hence by Theorem 14, we follow the result.
If we setβ(x,y)=1 for all x,y∈X in Corollary 16, we get the following result.Corollary 17 (see [26]).
Let(X,d) be a complete metric space and S, T fuzzy mappings from X into ℱ(X) satisfying the following conditions:(a)
for eachx∈X, there exists α(x)∈(0,1] such that [Sx]α(x), [Tx]α(x) are nonempty closed bounded subsets of X;(b)
for allx,y∈X, we have
(53)H([Sx]α(x),[Ty]α(y))≤a1d(x,[Sx]α(x))+a2d(y,[Ty]α(y))+a3d(x,[Ty]α(y))+a4d(y,[Sx]α(x))+a5d(x,y),wherea1, a2, a3, a4, and a5 are nonnegative real numbers and ∑i=15ai<1 and either a1=a2 or a3=a4.
Then there existsz∈X such that z∈[Sz]α(z)∩[Tz]α(z).



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*Source: 853032-2014-02-05.xml*