AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 484857 10.1155/2014/484857 484857 Research Article Coupled Coincidence Points for Mixed Monotone Random Operators in Partially Ordered Metric Spaces Jiang Binghua 1 Xu Shaoyuan 2 Shi Lu 1 Park Sehie 1 School of Mathematics and Statistics Hubei Normal University Huangshi 435002 China hbnu.edu.cn 2 Department of Mathematics and Statistics Hanshan Normal University Chaozhou 521041 China hstc.edu.cn 2014 2942014 2014 17 12 2013 08 04 2014 29 4 2014 2014 Copyright © 2014 Binghua Jiang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The aim of this work is to prove some coupled random coincidence theorems for a pair of compatible mixed monotone random operators satisfying weak contractive conditions. These results are some random versions and extensions of results of Karapınar et al. (2012). Our results generalize the results of Shatanawi and Mustafa (2012).

1. Introduction

Random coincidence point theorems are stochastic generalizations of classical coincidence point theorems and play an important role in the theory of random differential and integral equations. Random fixed point theorems for contractive mapping on complete separable metric space have been proved by several authors (see ). Fixed point theorems for monotone operators in ordered Banach spaces have been investigated and found various applications. Since then, fixed point theorems for mixed monotone mappings in partially ordered metric spaces are of great importance and have been utilized for matrix equations, ordinary differential equations, and the existence and uniqueness of solutions for some boundary value problems (see ).

Ćirić and Lakshmikantham  and Zhu and Xiao  proved some coupled random fixed point and coupled random coincidence results in partially ordered complete metric spaces. Moreover coupled random coincidence results in partially ordered complete metric spaces were considered in . Following Karapınar et al.  and Shatanawi and Mustafa , we improve these results for a pair of compatible mixed monotone random mappings F:Ω×(X×X)X and g:Ω×XX, where F and g satisfy some weak contractive conditions. Presented results are also referred to the extensions and improve the corresponding results in [19, 21] and many other authors’ work.

2. Preliminaries

Let (X,) be a partially ordered set. The concept of a mixed monotone property of the mappings F:X×XX and g:XX has been introduced by Lakshmikantham and Ćirić in .

Definition 1 (see [<xref ref-type="bibr" rid="B16">16</xref>]).

Let (X,d) be a partially ordered set and F:X×XX a mapping. Then the map F is said to have mixed g-monotone property if F(x,y) is monotone g-nondecreasing in x and is monotone g-nonincreasing in y; that is, for any x,yX, (1)gx1gx2  implies  F(x1,y)F(x2,y),gy1gy2  implies  F(x,y2)F(x,y1).

Definition 2 (see [<xref ref-type="bibr" rid="B16">16</xref>]).

An element (x,y)X×X is called a coupled coincidence point of the mapping F:X×XX and g:XX if F(x,y)=gx and F(y,x)=gy.

Definition 3 (see [<xref ref-type="bibr" rid="B22">22</xref>]).

The mappings F:X×XX and g:XX are said to be compatible if (2)limnd(gF(xn,yn),F(gxn,gyn))=0,limnd(gF(yn,xn),F(gyn,gxn))=0, where {xn} and {yn} are sequences in X such that limnF(xn,yn)=limngxn=x and limnF(yn,xn)=limngyn=y for all x,yX being satisfied.

Theorem 4 (see [<xref ref-type="bibr" rid="B17">17</xref>]).

Let (X,) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let F:X×XX and g:XX be two mappings such that F has the mixed g-monotone property and satisfies (3)d(F(x,y),F(u,v))φ(max{d(gx,gu),d(gy,gv)})+Lmin{d(F(x,y),gu),d(F(u,v),gx),iiiiiiiiiiiiiiiiiiid(F(x,y),gx),d(F(u,v),gu)}, for all x,y,u,vX with gxgu and gygv, where φΦ and L0. Let and F(X×X)g(X), F, g be continuous and let F and g be compatible mappings. If there exist x0,y0X such that gx0F(x0,y0) and gy0F(y0,x0), then F and g have a coupled coincidence point in X.

Denote Φ as the set of functions φ:[0,][0,] satisfying the following:

φ  is continuous,

φ(t)<t for all t>0 and φ(t)=0 if and only if t=0.

Let (Ω,Σ) be a measurable space with Σ sigma algebra of subsets of Ω and let (X,d) be a metric space. A mapping T:ΩX is called Σ-measurable if, for any open subset U of X, T-1(U)={ω:T(ω)U}Σ. In what follows, when we speak of measurability, we will mean Σ-measurability. A mapping T:Ω×XX is called a random operator if, for any xX,  T(·,x) is measurable. A measurable mapping ξ:ΩX is called a random fixed point of a random function T:Ω×XX, if ξ(ω)=T(ω,ξ(ω)), for every ωΩ. A measurable mapping ξ:ΩX is called a random coincidence of T:Ω×XX and g:Ω×XX if g(ω,ξ(ω))=T(ω,ξ(ω)) for each ωΩ.

Definition 5 (see [<xref ref-type="bibr" rid="B22">22</xref>]).

Let (X,d) be a separable metric space and (Ω,Σ) a measurable space. Then F:Ω×(X×X)X and g:Ω×XX are said to be compatible random operators if (4)limnd(g(ω,F(ω,(xn,yn))),iiiiiiiiiiF(ω,(g(ω,xn),g(ω,yn))))=0,limnd(g(ω,F(ω,(yn,xn))),iiiiiiiiiF(ω,(g(ω,yn),g(ω,xn))))=0, where {xn} and {yn} are sequences in X such that limnF(ω,(xn,yn))=limng(ω,xn)=x and limnF(ω,(yn,xn))=limng(ω,yn)=y for all ωΩ and for all x,yX being satisfied.

Theorem 6 (see [<xref ref-type="bibr" rid="B21">21</xref>]).

Let (X,) be a partially ordered set, (X,d) a complete separable metric space, and (Ω,Σ) a measurable space. Let F:Ω×(X×X)X and g:Ω×XX be mappings such that there are two nonnegative real numbers α and β with α+β<1 such that (5)d(F(ω,(x,y)),F(ω,(u,v)))αd(g(ω,x),g(ω,u))+βd(g(ω,y),g(ω,v)), for all x,y,u,vX with g(ω,x)g(ω,u) and g(ω,y)g(ω,v) for all ωΩ. Assume that F and g satisfy the following conditions:

F(ω,·),  g(ω,·) are continuous, for all ωΩ,

F(·,v),  g(·,x) are measurable, for all vX×X and xX, respectively,

F(ω×(X×X))X, for each ωΩ,

g is continuous and commutes with F and also suppose that either

F is continuous or

X has the following properties:

if a nondecreasing sequence xnx, then xnx, for all n,

if a nonincreasing sequence xnx, then xxn, for all n.

If there exist measurable mappings η0,θ0X such that g(ω,η0(ω))F(ω,(η0(ω),θ0(ω))) and F(ω,(θ0(ω),η0(ω)))g(ω,θ0(ω)), then there are measurable mappings η,θ:ΩX such that F(ω,(η(ω),θ(ω)))=g(ω,η(ω)) and F(ω,(θ(ω),η(ω)))=g(ω,θ(ω)) for all ωΩ; that is, F and g have a coupled random coincidence.

Now, we state our main results as follows.

3. Main Results

In this section, we study coupled random coincidence and coupled random fixed point theorems for a pair of random mappings F:Ω×(X×X)X and g:Ω×XX. Then we will prove some results for random mixed monotone mappings, which are the extensions of corresponding results for deterministic mixed monotone mappings of Karapınar et al. .

Theorem 7.

Let (X,) be a partially ordered set, (X,d) a complete separable metric space, (Ω,Σ) a measurable space, and F:Ω×(X×X)X and g:Ω×XX mappings such that (6)d(F(ω,(x,y)),F(ω,(u,v)))φ(max{d(g(ω,x),g(ω,u)),d(g(ω,y),g(ω,v))}+Lmin{d(F(ω,(x,y)),g(ω,u)),iiiiiiiiiiiiiiiiiiiiid(F(ω,(u,v)),g(ω,x)),iiiiiiiiiiiiiiiiiiiiid(F(ω,(x,y)),g(ω,x)),iiiiiiiiiiiiiiiiiiiiid(F(ω,(u,v)),g(ω,u))}, for all x,y,u,vX with g(ω,x)g(ω,u) and g(ω,y)g(ω,v) for all ωΩ, where φΦ and L0. Assume that F and g satisfy the following conditions:

g(ω,·) are continuous, for all ωΩ,

F(·,v), g(·,x) are measurable, for all vX×X and xX, respectively,

F(ω×(X×X))g(ω×X), for each ωΩ,

g is continuous and commutes with F and also suppose that either

F is continuous or

X has the following properties:

if a nondecreasing sequence xnx, then gxngx, for all n,

if a nonincreasing sequence yny, then gygyn, for all n.

If there exist measurable mappings η0,θ0X such that g(ω,η0(ω))F(ω,(η0(ω),θ0(ω))) and F(ω,(θ0(ω),η0(ω)))g(ω,θ0(ω)), then there are measurable mappings η,  θ:ΩX such that F(ω,(η(ω),θ(ω)))=g(ω,η(ω)) and F(ω,(θ(ω),η(ω)))=g(ω,θ(ω)), for all ωΩ; that is, F and g have a coupled random coincidence.

Proof.

Let Θ={η:ΩX} be a family of measurable mappings. Define a function h:Ω×X+ as h(ω,x)=d(x,g(ω,x)). Since xg(ω,x) is continuous, for all ωΩ, we conclude that h(ω,·) is continuous, for all ωΩ. Also, since ωg(ω,x) is measurable, for all xX, we conclude that h(·,x) is measurable, for all xX (see [23, page 868]). Thus, h(ω,x) is the Caratheodory function. Thus, if η:ΩX is measurable mapping, then ωh(ω,η(ω)) is also measurable (see ). Also, for each θΘ, the function η:ΩX defined by η(ω)=g(ω,θ(ω)) is measurable; that is, ηΘ.

Now we are going to construct two sequences of measurable mappings {ξn} and {ηn} in Θ and two sequences {g(ω,ξn(ω))} and {g(ω,ηn(ω))} in X as follows. Let ξ0,η0Θ be such that g(ω,ξ0(ω))F(ω,(ξ0(ω),η0(ω))) and g(ω,η0(ω))F(ω,(η0(ω),ξ0(ω))), for all ωΩ. Since F(ω,(ξ0(ω),η0(ω)))F(ω×(X×X))g(ω×X), by a sort of Filippov measurable implicit function theorem (see [25, 26]), there is ξ1Θ such that g(ω,ξ1(ω))=F(ω,(ξ0(ω),η0(ω))). Similarly, as F(ω,(η0(ω),ξ0(ω)))g(ω×X), there is η1Θ such that g(ω,η1(ω))=F(ω,(η0(ω),ξ0(ω))). Thus F(ω,(ξ0(ω),η0(ω))) and F(ω,(η0(ω),ξ0(ω))) are well defined now. Again, since (7)F(ω,(ξ1(ω),η1(ω))),F(ω,(η1(ω),ξ1(ω)))g(ω×X), there are ξ2,η2Θ such that (8)g(ω,ξ2(ω))=F(ω,(ξ1(ω),η1(ω))),g(ω,η2(ω))=F(ω,(η1(ω),ξ1(ω))). Continuing this process we can construct sequences {ξn(ω)} and {ηn(ω)} in X such that (9)g(ω,ξn+1(ω))=F(ω,(ξn(ω),ηn(ω))),g(ω,ηn+1(ω))=F(ω,(ηn(ω),ξn(ω))), for all n. Now, we use mathematical induction to prove that (10)g(ω,ξn(ω))g(ω,ξn+1(ω)),g(ω,ηn(ω))g(ω,ηn+1(ω)), for all n. Let n=0, and by assumption we have (11)g(ω,ξ0(ω))F(ω,(ξ0(ω),η0(ω))),g(ω,η0(ω))F(ω,(η0(ω),ξ0(ω))). Since (12)g(ω,ξ1(ω))=F(ω,(ξ0(ω),η0(ω))),g(ω,η1(ω))=F(ω,(η0(ω),ξ0(ω))), we have (13)g(ω,ξ0(ω))g(ω,ξ1(ω)),g(ω,η0(ω))g(ω,η1(ω)). Therefore, (10) holds for n=0. Suppose (10) holds for some fixed number n0. Then, since (14)g(ω,ξn(ω))g(ω,ξn+1(ω)),g(ω,ηn(ω))g(ω,ηn+1(ω)) and F is monotone g-nondecreasing in its first argument, we have (15)F(ω,(ξn(ω),ηn(ω)))F(ω,(ξn+1(ω),ηn(ω))),F(ω,(ηn(ω),ξn(ω)))F(ω,(ηn+1(ω),ξn(ω))). Also, since g(ω,ξn(ω))g(ω,ξn+1(ω)) and g(ω,ηn(ω))g(ω,ηn+1(ω)). and F is monotone g-nonincreasing in its second argument, we have (16)F(ω,(ξn+1(ω),ηn+1(ω)))F(ω,(ξn+1(ω),ηn(ω)))F(ω,(ηn+1(ω),ξn(ω)))F(ω,(ηn+1(ω),ξn+1(ω))). Thus, from (9), we get (17)g(ω,ξn+1(ω))g(ω,ξn+2(ω)),g(ω,ηn+1(ω))g(ω,ηn+2(ω)). Thus, by mathematical induction, we conclude that (10) holds for all n. Now, we prove that {g(ω,ξn(ω))} and {g(ω,ηn(ω))} are Cauchy sequences. Let n, and, by (6)–(10), we have (18)d(F(ω,(ξn(ω),ηn(ω))),F(ω,(ξn-1(ω),ηn-1(ω))))φ(max{d(g(ω,ξn(ω)),g(ω,ξn-1(ω))),iiiiiiiiiiiiiiiiiid(g(ω,ηn(ω)),g(ω,ηn-1(ω)))})+Lmin{d(F(ω,(ξn(ω),ηn(ω))),g(ω,ξn-1(ω))),iiiiiiiiiiiiiiiiiiid(F(ω,(ξn-1(ω),ηn-1(ω))),g(ω,ξn(ω))),iiiiiiiiiiiiiiiiiiid(F(ω,(ξn(ω),ηn(ω))),g(ω,ξn(ω))),iiiiiiiiiiiiiiiiiiid(f(ω,(ξn(ω),ηn(ω))),g(ω,ξn-1(ω)))}, which implies that (19)d(g(ω,ξn+1(ω)),g(ω,ξn(ω)))iiiiφ(max{d(g(ω,ξn(ω)),g(ω,ξn-1(ω))),iiiiiiiiiiiiiiiiiiiid(g(ω,ηn(ω)),g(ω,ηn-1(ω)))}). Similarly, we have (20)d(F(ω,(ηn(ω),ξn(ω))),F(ω,(ηn-1(ω),ξn-1(ω))))φ(max{d(g(ω,ηn(ω)),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξn-1(ω)))})+Lmin{d(F(ω,(ηn(ω),ξn(ω))),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiiiiid(F(ω,(ηn-1(ω),ξn-1(ω))),g(ω,ηn(ω))),iiiiiiiiiiiiiiiiid(F(ω,(ηn(ω),ξn(ω))),g(ω,ηn(ω))),iiiiiiiiiiiiiiiid(F(ω,(ηn-1(ω),ξn-1(ω))),g(ω,ηn-1(ω)))}, which implies that (21)d(g(ω,ηn+1(ω)),g(ω,ηn(ω)))φ(max{d(g(ω,ηn(ω)),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξn-1(ω)))}). From (19) and (21), we get that (22)max{d(g(ω,ξn+1(ω)),g(ω,ξn(ω))),iiiiiiiiid(g(ω,ηn+1(ω)),g(ω,ηn(ω)))}φ(max{d(g(ω,ηn(ω)),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξn-1(ω)))}). Since φ(t)<t, for all t>0, by (22), we have (23)max{d(g(ω,ξn+1(ω)),g(ω,ξn(ω))),iiiiiiiid(g(ω,ηn+1(ω)),g(ω,ηn(ω)))}<max{d(g(ω,ηn(ω)),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξn-1(ω)))}. Set dn=max{d(g(ω,ξn+1(ω)),g(ω,ξn(ω))),d(g(ω,ηn+1(ω)),g(ω,ηn(ω)))}, then {dn} is a nonincreasing sequence of positive real numbers. Thus, there is d0 such that (24)limndn=d.

Suppose that d>0; letting n in two sides of (22) and using the properties of φ, we have (25)d=limndniilimnφ(max{d(g(ω,ηn(ω)),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξn-1(ω)))})iiφ(d)<d, which is a contradiction. Hence d=0; that is, (26)d=limnmax{d(g(ω,ξn+1(ω)),g(ω,ξn(ω))),iiiiiiiiiiiiiiiiiiiiid(g(ω,ηn+1(ω)),g(ω,ηn(ω)))}=0. We will show that {g(ω,ηn(ω))} and {g(ω,ξn(ω))} are Cauchy sequences. Suppose, to the contrary, that at least one of {g(ω,ηn(ω))} or {g(ω,ξn(ω))} is not a Cauchy sequence. This means that there exists an ɛ>0 for which we can find subsequences {g(ω,ηn(k)(ω))} of {g(ω,ηn(ω))} and {g(ω,ξn(k)(ω))} of {g(ω,ξn(ω))} with n(k)>m(k)k  (k=1,2,) such that (27)max{d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))),iiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))}ɛ. Further, corresponding to m(k), we can choose n(k) in such a way that it is the smallest integer with n(k)>m(k)k and satisfies (27). Then, (28)max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)(ω))),iiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)(ω)))}<ɛ.

Using the triangle inequality and (28), we have (29)d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω)))d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω)))+d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)(ω)))d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω)))+ɛ,d(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))d(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω))+d(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)(ω)))d(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))+ɛ. By (27) and (29), we obtain (30)ɛmax{d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))),iiiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))}imax{d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω))),iiiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))}+ɛ. Letting k, in the inequalities above, we get (31)limkmax{d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))),iiiiiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))}=ɛ. By the triangle inequalities, we have (32)d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω)))d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω)))+d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω)))+d(g(ω,ξm(k)-1(ω)),g(ω,ξm(k)(ω))),d(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))d(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))+d(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))+d(g(ω,ηm(k)-1(ω)),g(ω,ηm(k)(ω))).

By the above inequalities and (27), we have (33)ɛmax{d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))),iiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))}imax{d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω))),iiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))}+max{d(g(ω,ξn(k)-1(ω)),iiiiiiiiiiiiiiiiiiig(ω,ξm(k)-1(ω))  d(g(ω,ηn(k)-1(ω)),iiiiiiiiiiiiiiiiiiig(ω,ηm(k)-1(ω)))}+max{d(g(ω,ξm(k)-1(ω)),g(ω,ξm(k)(ω))),iiiiiiiiiiiiiiiiid(g(ω,ηm(k)-1(ω)),g(ω,ηm(k)(ω)))}. Again, by the triangle inequality, we obtain (34)d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω)))d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)(ω)))+d(g(ω,ξm(k)(ω)),g(ω,ξm(k)-1(ω)))d(g(ω,ξm(k)(ω)),g(ω,ξm(k)-1(ω)))+ɛ,d(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))d(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)(ω)))+d(g(ω,ηm(k)(ω)),g(ω,ηm(k)-1(ω)))d(g(ω,ηm(k)(ω)),g(ω,ηm(k)-1(ω)))+ɛ. Therefore, (35)max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))}iiiiiiiimax{d(g(ω,ξm(k)(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiiiiiiiid(g(ω,ηm(k)(ω)),g(ω,ηm(k)-1(ω)))}+ɛ. Taking k in (33) and (35), we have (36)limnmax{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))}=ɛ. Since n(k)>m(k), g(ω,ξn(k)-1(ω))g(ω,ξm(k)-1(ω)) and g(ω,ηn(k)-1(ω))g(ω,ηm(k)-1(ω)). Then, from (6)–(10), we get (37)d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω)))=d(F(ω,(ξn(k)-1(ω),ηn(k)-1(ω))),iiiiiiiiiF(ω,(ξm(k)-1(ω),ηm(k)-1(ω))))φ(max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))})+Lmin{d(F(ω,(ξn(k)-1(ω),ηn(k)-1(ω))),iiiiiiiiiiiiiiiiiig(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiiiiiid(F(ω,(ξm(k)-1(ω),ηm(k)-1(ω))),iiiiiiiiiiiiiiiiiig(ω,ξn(k)-1(ω))),iiiiiiiiiiiiiiiiiid(F(ω,(ξn(k)-1(ω),ηn(k)-1(ω))),iiiiiiiiiiiiiiiiiig(ω,ξn(k)-1(ω))),iiiiiiiiiiiiiiiiiid(F(ω,(ξm(k)-1(ω),ηm(k)-1(ω))),iiiiiiiiiiiiiiiiiig(ω,ξm(k)-1(ω)))}φ(max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))})+Lmin{d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ξm(k)(ω)),g(ω,ξm(k)-1(ω)))}. Similarly, (38)d(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))φ(max{d(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω))),iiiiiiiiiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))})+Lmin{d(g(ω,ηm(k)(ω)),g(ω,ηm(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))}. From (37) and (38), we arrive at (39)max{d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))),iiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))}φ(max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))})+Lmin{d(g(ω,ξm(k)(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω)))}+Lmin{d(g(ω,ηm(k)(ω)),g(ω,ηm(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))}. Letting n in the above inequality and using (26), (27), and the properties of φ, we have (40)ɛφ(ɛ)+2Lmin{0,0}<ɛ, which is a contradiction. This means that {g(ω,ξn(ω))} and {g(ω,ηn(ω))} are Cauchy sequences.

Since X is complete, for all ωΩ, there exist the functions ζ(ω) and θ(ω) such that (41)limng(ω,ξn(ω))=ζ(ω),limng(ω,ηn(ω))=θ(ω). Thus, (42)limnF(ω,(ξn(ω),ηn(ω)))=limng(ω,ξn(ω))=ζ(ω),limnF(ω,(ηn(ω),ξn(ω)))=limng(ω,ηn(ω))=θ(ω).

Since F and g are compatible mappings, we have (43)limnd(g(ω,F(ω,(ξn(ω),ηn(ω)))),iiiiiiiiiiiiF(ω,(g(ω,ξn(ω)),g(ω,ηn(ω)))))=0,limnd(g(ω,F(ω,(ηn(ω),ξn(ω)))),iiiiiiiiiiiiF(ω,(g(ω,ηn(ω)),g(ω,ξn(ω)))))=0.

Suppose at first that assumption (a) holds. Taking the limit as n in the following inequalities (44)d(g(ω,ζ(ω)),F(ω,(g(ω,ξn(ω)),g(ω,ηn(ω)))))d(g(ω,ζ(ω)),g(ω,F(ω,(ξn(ω),ηn(ω)))))+d(g(ω,F(ω,(ξn(ω),ηn(ω)))),iiiiiiiiiiiiiF(ω,(g(ω,ξn(ω)),g(ω,ηn(ω))))) and using (9) and the continuity of F, g, we get (45)d(g(ω,ζ(ω)),F(ω,(ζ(ω),θ(ω))))=θ. This implies g(ω,ζ(ω))=F(ω,(ζ(ω),θ(ω))). Similarly, we can show that g(ω,θ(ω))=F(ω,(θ(ω),ζ(ω))) for each ωΩ. The proof is complete.

Suppose now that (b) holds. From (9), we have (46)limnF(ω,g(ω,ξn(ω)),g(ω,ηn(ω)))=limng(ω,F(ω,(ξn(ω),ηn(ω))))=limng(ω,g(ω,ξn+1(ω)))=g(ω,ζ(ω)),(47)limnF(ω,g(ω,ηn(ω)),g(ω,ξn(ω)))=limng(ω,F(ω,(ηn(ω),ξn(ω))))=limng(ω,g(ω,ηn+1(ω)))=g(ω,θ(ω)). Since g(ω,g(ω,ξn(ω)))g(ω,ζ(ω)) and g(ω,g(ω,ηn(ω)))g(ω,θ(ω)), we have (48)d(g(ω,ζ(ω)),F(ω,(ζ(ω),θ(ω))))d(g(ω,ζ(ω)),F(ω,(g(ω,ξn(ω)),g(ω,ηn(ω)))))+d(F(ω,(g(ω,ξn(ω)),g(ω,ηn(ω)))),iiiiiiiiiiiiF(ω,(ζ(ω),θ(ω)))((ξn(ω))))d(g(ω,ζ(ω)),F(ω,(g(ω,ξn(ω)),g(ω,ηn(ω)))))+φ(max{d(g(ω,g(ω,ξn(ω))),g(ω,ζ(ω))),iiiiiiiiiiiiiiiiiid(g(ω,g(ω,ηn(ω))),g(ω,θ(ω)))})+Lmin{d(g(ω,g(ω,ξn(ω))),F(ω,(ζ(ω),θ(ω)))),iiiiiiiiiiiiiiiiiid(g(ω,ζ(ω)),iiiiiiiiiiiiiiiiiF(ω,(g(ω,ξn(ω),g(ω,ηn(ω))))),iiiiiiiiiiiiiiiiid(g(ω,ζ(ω)),F(ω,(ζ(ω),θ(ω)))),iiiiiiiiiiiiiiiiid(g(ω,g(ω,ξn(ω))),iiiiiiiiiiiiiiiiiF(ω,(g(ω,ξn(ω),g(ω,ηn(ω)))))}. Taking n in the above inequality and using (46) and the properties of φ, we have (49)d(g(ω,ζ(ω)),F(ω,(ζ(ω),θ(ω))))φ(max{0,0})+Lmin{d(g(ω,ζ(ω)),F(ω,(ζ(ω),θ(ω)))),0}=0. Hence g(ω,ζ(ω))=F(ω,(ζ(ω),θ(ω))).

Similarly, one can show that g(ω,θ(ω))=F(ω,(θ(ω),ζ(ω))).

The proof is complete.

Remark 8.

Taking L=0, for all x,y,u,vX, α,β0, and α+β<1, we have (50)αd(g(w,x),g(w,u))+βd(g(w,y),g(w,v))(α+β)max{g(w,x),g(w,u),g(w,y),g(w,v)}=φ(max{g(w,x),g(w,u),g(w,y),g(w,v)}), where φ(t)=(α+β)t  (t0). Obviously, φ(t)Φ. Moreover, the conditions that

if a nondecreasing sequence xnx, then gxngx, for all n,

if a nonincreasing sequence yny, then gygyn, for all n, are weaker than the conditions that g is monotone mapping and

if a nondecreasing sequence xnx, then xnx, for all n,

if a nonincreasing sequence yny, then yyn for all n. Therefore, Theorem 7 generalizes Theorem 6 and [18, Theorem 2.2] and the following corollary is obtained.

Corollary 9.

Let (X,) be a partially ordered set, (X,d) a complete separable metric space, (Ω,Σ) a measurable space, and F:Ω×(X×X)X and g:Ω×XX mappings such that

g(ω,·) is continuous, for all ωΩ,

F(·,v),  g(·,x) are measurable for all vX×X and xX, respectively,

F(ω,·) has the mixed g(ω,·)-monotone property for each ωΩ and (51)d(F(ω,(x,y)),F(ω,(u,v)))φ(max{d(g(ω,x),g(ω,u)),d(g(ω,y),g(ω,v))})+Lmin{d(F(ω,(x,y)),g(ω,u)),iiiiiiiiiiiiiid(F(ω,(x,y)),g(ω,x)),iiiiiiiiiiiiiid(F(ω,(u,v)),g(ω,x)),iiiiiiiiiiiiiid(F(ω,(u,v)),g(ω,u))},

for all x,y,u,vX with g(ω,x)g(ω,u) and g(ω,y)g(ω,v) for all ωΩ, where φΦ and L0. Suppose that F(ω×(X×X))g(ω×X) for each ωΩ, g is monotone, and F and g are compatible random operators. Also suppose that X has the following property:

if a nondecreasing sequence xnx, then xnx, for all n,

if a nonincreasing sequence yny, then yyn, for all n.

If there exist measurable mappings ξ0,η0:ΩX such that (52)g(ω,ξ0(ω))F(ω,(ξ0(ω),η0(ω))),F(ω,(η0(ω),ξ0(ω)))g(ω,η0(ω)), then there are measurable mappings ζ,θ:ΩX such that (53)F(ω,(ζ(ω),θ(ω)))=g(ω,ζ(ω)),F(ω,(θ(ω),ζ(ω)))=g(ω,θ(ω)) for all ωΩ; that is, F and g have a coupled random coincidence.

Remark 10.

Comparing with [21, Theorem 2.6], we find that the monotone of g is essential. Also the condition that X=g(ω×X) is unnecessary and the proof of case (2) in [21, Theorem 2.6] was irrational. So our Corollary 9 generalizes and improves [21, Theorem 2.6].

Theorem 11.

Let (X,) be a partially ordered set, (X,d) a separable metric space, (Ω,Σ) a measurable space, and F:Ω×(X×X)X and g:Ω×XX mappings such that

F(·,v),  g(·,x) are measurable, for all vX×X and xX, respectively;

F(ω,·) has the mixed g(ω,·)-monotone property for each ωΩ and (54)d(F(ω,(x,y)),F(ω,(u,v)))φ(max{d(g(ω,x),g(ω,u)),d(g(ω,y),g(ω,v))})+Lmin{d(F(ω,(x,y)),g(ω,u)),iiiiiiiiiiiiiiiiiiiiid(F(ω,(u,v)),g(ω,x)),iiiiiiiiiiiiiiiiiiiiid(F(ω,(x,y)),g(ω,x)),iiiiiiiiiiiiiiiiiiiiid(F(ω,(u,v)),g(ω,u))},

for all x,y,u,vX with g(ω,x)g(ω,u) and g(ω,y)g(ω,v) for all ωΩ, where φΦ and L0. Suppose that F(ω×(X×X))g(ω×X) and g(ω×X) is complete subspace of X for each ωΩ. Also suppose that X has the following property:

if a nondecreasing sequence xnx, then xnx, for all n,

if a nonincreasing sequence yny, then yyn, for all n.

If there exist measurable mappings ξ0,η0:ΩX such that (55)g(ω,ξ0(ω))F(ω,(ξ0(ω),η0(ω))),F(ω,(η0(ω),ξ0(ω)))g(ω,η0(ω)), then there are measurable mappings ξ,θ:ΩX such that (56)F(ω,(ξ(ω),θ(ω)))=g(ω,ξ(ω)),F(ω,(θ(ω),ξ(ω)))=g(ω,θ(ω)), for all ωΩ; that is, F and g have a coupled random coincidence.

Proof.

Construct two sequences {ξn(ω)} and {ηn(ω)} as in Theorem 7. According to the proof of Theorem 7, {g(ω,ξn(ω))} and {g(ω,ηn(ω))} are Cauchy sequences. Since g(ω×X) is complete, there exist ξ(ω), θ(ω)Θ such that (57)limng(ω,ξn(ω))=g(ω,ξ(ω)),limng(ω,ηn(ω))=g(ω,θ(ω)). Since {g(ω,ξn(ω))} is nondecreasing sequence and g(ω,ξn(ω))g(ω,ξ(ω)) and {g(ω,ηn(ω))} is nonincreasing sequence and g(ω,ηn(ω))g(ω,θ(ω)), by the assumption, we have g(ω,ξn(ω))g(ω,ξ(ω)) and g(ω,ηn(ω))g(ω,θ(ω)) such that (58)d(g(ω,θ(ω)),F(ω,(θ(ω),ξ(ω))))d(g(ω,θ(ω)),g(ω,ηn+1(ω)))+d(g(ω,ηn+1(ω)),F(ω,(θ(ω),ξ(ω))))d(g(ω,θ(ω)),g(ω,ηn+1(ω)))+φ(max{d(g(ω,ηn(ω)),g(ω,θ(ω))),iiiiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξ(ω)))})+Lmin{d(g(ω,ηn(ω)),F(ω,(θ(ω),ξ(ω)))),iiiiiiiiiiiiiiiiiid(g(ω,θ(ω)),F(ω,(ηn(ω),ξn(ω)))),iiiiiiiiiiiiiiiiiid(g(ω,ηn(ω)),F(ω,(ηn(ω),ξn(ω)))),iiiiiiiiiiiiiiiiiid(g(ω,θ(ω)),F(ω,(ξ(ω),θ(ω))))}=d(g(ω,θ(ω)),g(ω,ηn+1(ω)))+φ(max{d(g(ω,ηn(ω)),g(ω,θ(ω))),iiiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξ(ω)))})+Lmin{d(g(ω,ηn(ω)),F(ω,(θ(ω),ξ(ω)))),iiiiiiiiiiiiiiiid(g(ω,θ(ω)),g(ω,ηn+1(ω))),iiiiiiiiiiiiiiiid(g(ω,ηn(ω)),g(ω,ηn+1(ω))),iiiiiiiiiiiiiiiid(g(ω,θ(ω)),F(ω,(ξ(ω),θ(ω))))}. On taking n in the above inequality and using (57), we obtain (59)d(g(ω,θ(ω)),F(ω,(θ(ω),ξ(ω))))φ(0)+Lmin{d(F(ω,(θ(ω),ξ(ω)))g(ω,θ(ω)),iiiiiiiiiiiiiiiiiiiiiiiiiiiiiF(ω,(θ(ω),ξ(ω)))),0(F(ω,(θ(ω),ξ(ω)))g(ω,θ(ω))}=0. This means that g(ω,θ(ω))=F(ω,(θ(ω),ξ(ω))). Similarly, it can be shown that g(ω,ξ(ω))=F(ω,(ξ(ω),θ(ω))). Thus, F and g have a coupled coincidence point in X.

The proof is complete.

Remark 12.

Following Theorem 7 and Corollary 9, we replace the continuity and monotone of g, the compatibility of F and g, and the completeness of X by assuming that g(X) is a complete subspace of X. Moreover, by the measurable space, our random fixed point theorems generalize the main results in .

Conflict of Interests

The authors declare that they have no conflict of interests.

Acknowledgment

The research is partially supported by Doctoral Initial Foundation of Hanshan Normal University, China (no. QD20110920).

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