RANDOM OPERATORS
ISMAT BEG AND MUJAHID ABBASReceived 2 June 2005; Revised 12 April 2006; Accepted 25 April 2006
We construct a random iteration scheme and study necessary conditions for its conver-gence to a common random fixed point of two pairs of compatible random operators satisfying Meir-Keeler type conditions in Polish spaces. Some random fixed point theo-rems for weakly compatible random operators under generalized contractive conditions in the framework of symmetric spaces are also proved.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
1. Introduction and preliminaries
The study of random fixed point theory was initiated by the Prague school of proba-bilists in the 1950s [12,13,26]. Random fixed point theorems are stochastic general-ization of classical fixed point theorems. The survey article by Bharucha-Reid [10] at-tracted the attention of several mathematicians and gave wings to this theory. Itoh [16] extended Spacek’s and Hans’s theorem to multivalued contraction mappings. Now this theory has become the full fledged research area and various ideas associated with ran-dom fixed point theory are used to obtain the solution of nonlinear ranran-dom system (see [9,19,25,27]). Papageorgiou [23], Beg [3,4], and Beg and Shahzad [6,8] studied the structure of common random fixed points and random coincidence points of a pair of compatible random operators and proved fixed point theorems for contractive random operators in Polish spaces. Recently Beg and Shahzad [7,8] had used different iteration processes to obtain common random fixed points. The aim of this paper is to study the necessary conditions for the convergence of random iteration scheme to common ran-dom fixed points of two pairs of compatible ranran-dom operators satisfying Meir-Keeler-[18] type conditions in Polish spaces. Also, in Section 3, we establish the existence of unique common random fixed points of random operators under generalized contrac-tive conditions. We first review the following concepts which are essential for our study in this paper.
Throughout this paper, (Ω,Σ) denotes a measurable space (Σ—sigma algebra). A sym-metric on a set X is a nonnegative real-valued function d on X×X such that for all
x,y∈Xwe have
Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 23486, Pages1–15
(a)d(x,y)=0 if and only ifx=y, and (b)d(x,y)=d(y,x).
Letdbe a symmetric on a set X. Forε >0 andx∈X,B(x,ε) denotes the spherical ball centred atxwith radiusε, defined as the set{y∈X:d(y,x)< ε}. A topologyt(d) onXis given byU∈t(d) if and only if for eachx∈U,B(x,ε)⊂Ufor someε >0. Note that limn→∞d(xn,x)=0 if and only ifxn→xin the topologyt(d). LetFbe a subset of X. A mappingξ:Ω→X is measurable ifξ−1(U)∈Σfor each open subsetUofX. The mappingT:Ω×F→Fis a random map if and only if for each fixedx∈F, the mapping
T(·,x) :Ω→Fis measurable. The mappingTis continuous if for eachω∈Ω, the map-pingT(ω,·) :F→Xis continuous. A measurable mapping ξ:Ω→X is a random fixed point of random mapT:Ω×F→Xif and only ifT(ω,ξ(ω))=ξ(ω) for eachω∈Ω. We denote the set of random fixed points of a random mapT byRF(T) and the set of all measurable mappings fromΩinto a symmetric spaceXbyM(Ω,X). We denote thenth iterate T(ω,T(ω,T(ω,. . .,T(ω,x))),. . .) ofTbyTn(ω,x). The letterIdenotes the random mappingI:Ω×F→Fdefined byI(ω,x)=xandT0=I. Letφ:R+→R+be a function satisfying the condition 0< φ(t)< t, for eacht >0.
Definition 1.1. LetXbe a Polish space, that is, a separable complete metric space. Map-pings f,g :X →X are compatible if limn→∞d(f g(xn),g f(xn))=0, provided that limn→∞f(xn) and limn→∞g(xn) exist in X and limn→∞f(xn)= limn→∞g(xn). Random operatorsS,T:Ω×X→X are compatible ifS(ω,·) andT(ω,·) are compatible for each
ω∈Ω. (See Beg and Shahzad [6].)
Definition 1.2. LetX be a Polish space. Random operatorsS,T:Ω×X→Xare weakly compatible if T(ω,ξ(ω))=S(ω,ξ(ω)), for some ξ∈M(Ω,X), then T(ω,S(ω,ξ(ω)))=
S(ω,T(ω,ξ(ω))) for everyω∈Ω.
Definition 1.3 [29]. Let{xn}and{yn}be two sequences in symmetric space (X,d) and x,y∈X. The space is said to satisfy the following axioms.
(w.1) limn→∞d(xn,x)=limn→∞d(xn,y)=0 implies thatx=y.
(w.2) limn→∞d(xn,x)=limn→∞d(xn,yn)=0 implies that limn→∞d(yn,x)=0.
IfX=Ris taken, define the symmetric function asd(x,y)=e|x−y|−1 with sequences {xn} = {1 + 1/n},{yn} = {1−1/n}, andx=1 inX, then it is easy to verify that axiom (w.2) holds.
Definition 1.4. Let{xn}and{yn}be two sequences in a symmetric space (X,d) andx∈X. The spaceXis said to satisfy axiom (HE); if limn→∞d(xn,x)=limn→∞d(yn,x)=0 implies that limn→∞d(xn,yn)=0.
For example, let X=[0,∞) with the symmetric function d(x,y)=e|x−y|−1, then (X,d) satisfies the axiom (HE).
Definition 1.5. Letdbe a symmetric function onX. Two random mappingsSandTfrom Ω×XtoXare said to satisfy property (I) if there exists a sequence{ξn}inM(Ω,X) such that for someξ∈M(Ω,X),
lim n→∞d
Tω,ξn(ω),ξ(ω)=nlim→∞dSω,ξn(ω),ξ(ω)=0 (1.1)
Example 1.6. LetΩ=[0, 1] and letΣbe the sigma algebra of Lebesgue’s measurable sub-sets of [0, 1]. TakeX=Rwithd(x,y)=e|x−y|−1, obviouslydis symmetric onR. Define two random mappingsTandSfromΩ×XtoXas
T(ω,x)= ⎧ ⎨ ⎩
1 + 2x−ω2 if (ω,x)∈[0, 1]×[0, 1], 31−ω2 otherwise,
S(ω,x)= ⎧ ⎨ ⎩
ω2+ 4x−1 if (ω,x)∈[0, 1]×[0, 1], 31−ω2 otherwise.
(1.2)
Also, the sequence of measurable mappingsξn:Ω→Xis given byξn(ω)=((n−1)/n)− ω2for everyω∈Ωandn∈N. Define measurable functionξ:Ω→Xasξ(ω)=3(1−ω2) for everyω∈Ω. Consider the case, whenξn(ω)∈[0, 1] for everyn∈N, andω∈Ω, then we have
lim n→∞d
Tω,ξn(ω)
,ξ(ω)=lim n→∞e
|T(ω,ξn(ω))−ξ(ω)|−1
=lim n→∞e
|−2/n|−1=0 for everyω∈Ω,
lim n→∞d
Sω,ξn(ω)
,ξ(ω)=lim n→∞e
|S(ω,ξn(ω))−ξ(ω)|−1
=lim n→∞e
|−4/n|−1=0 for everyω∈Ω.
(1.3)
Whenξn(ω)∈/ [0, 1] forn∈Nandω∈Ω, then obviously limn→∞d(T(ω,ξn(ω)),ξ(ω))= limn→∞d(S(ω,ξn(ω)),ξ(ω))=0 for everyω∈Ω. SoTandSsatisfy property (I).
Definition 1.7. LetX be a Polish space and letA,B,S,T:Ω×X→X be four random operators withA(ω,X)⊂T(ω,X) andB(ω,X)⊂S(ω,X) for everyω∈Ω. Consider the sequence of functions{ξn}and{ηn}fromΩtoXdefined by
ξ2n−1(ω)=Tω,η2n−1(ω)=Aω,η2n−2(ω),
ξ2n(ω)=Sω,η2n(ω)=Bω,η2n−1(ω),
(1.4)
whereξ0:Ω→Xis an arbitrary fixed map, defineM:X×X×Ω→R+as
M(x,y,ω)=max
dS(ω,x),T(ω,y),dS(ω,x),A(ω,x),dT(ω,y),B(ω,y),1 2
dS(ω,x),B(ω,y), +dT(ω,y),A(ω,x) .
Remark 1.8. LetFbe a closed subset of a Polish spaceXand let the sequence of measur-able functions{ξn}defined fromΩtoFbe pointwise convergent, that is,ξn(ω)→q:= ξ(ω), for eachω∈Ω. Now the closedness ofFimplies thatξis a mapping fromΩtoF, andξbeing the limit of the sequence of measurable functions is also measurable. SinceF
is a subset of a separable metric spaceX, so ifTis a continuous random operator, then by [2, Lemma 8.2.3], the mapω→Tn(ω,f(ω)) is a measurable function for any measurable function f fromΩtoX(see also [15,28]).
2. Common random fixed points of random operators
Meir and Keeler [18] obtained a remarkable generalization of the Banach contraction principle (see [1,11]). Park and Bae [24] extended Meir and Keelar’s theorem for two commuting maps following Jungck’s method [17]. Recent work of Pant [22] contains common fixed point theorem for four maps satisfying certain contractive conditions (also see [5,18,20,21] and references mentioned therein). Beg and Shahzad [7] studied ran-dom fixed point theorems for contractive type ranran-dom operators on Banach spaces. In continuation of these results now we obtain common random fixed point for four ran-dom operators under a (ε,δ) contractive condition which was initiated by Meir and Keeler [18]. Our result is a stochastic analog of a result of Pant [22]. To establish our main result, we first prove the following lemma.
Lemma 2.1. LetXbe a Polish space and letA,B,S,T:Ω×X→Xbe four random operators withA(ω,X)⊂T(ω,X) andB(ω,X)⊂S(ω,X) for everyω∈Ω. If for any givenε >0, there exists aδ >0 such that for allx,y∈X,
ε≤M(x,y,ω)< δ+ε=⇒dA(ω,x),B(ω,y)< ε for eachω∈Ω, (2.1)
then, for the sequence of function{ξn}, as taken inDefinition 1.7, (a) limn→∞d(ξn(ω),ξn+1(ω))=0, for eachω∈Ω;
(b) there existsn0∈Nsuch thatp,q≥n0, wherepandqare of opposite parity, give the following implication:
ε≤dξp(ω),ξq(ω)
< ε+r=⇒dξp+1(ω),ξq+1(ω)
< ε (2.2)
for eachω∈Ω, and wherer=min{ε/2,δ/2};
(c) the sequence{ξn(ω)}is a Cauchy sequence for eachω∈Ω.
Proof (part (a)). It is obvious that if for everyω∈Ωandx,y∈X, we takeM(x,y,ω)=0, thend(A(ω,x),B(ω,y))=0 for everyω∈Ωandx,y∈X.
Now,M(x,y,ω)>0=⇒d(A(ω,x),B(ω,y))< M(x,y,ω) for eachω∈Ωandx,y∈X. So
Consider
dξ2n(ω),ξ2n+1(ω)
=dBω,η2n−1(ω),Aω,η2n(ω)
=dAω,η2n(ω),Bω,η2n−1(ω) ≤Mη2n(ω),η2n−1(ω),ω
=max
dSω,η2n(ω),Tω,η2n−1(ω),dSω,η2n(ω), Aω,η2n(ω)
,dTω,η2n−1(ω),Bω,η2n−1(ω), 1
2
dSω,η2n(ω),Bω,η2n−1(ω)+dTω,η2n−1(ω),
Aω,η2n(ω)
=max
dξ2n(ω),ξ2n−1(ω),dξ2n(ω),ξ2n+1(ω)
,
dξ2n−1(ω),ξ2n(ω)
,1 2
dξ2n(ω),ξ2n(ω)
+dξ2n−1(ω),ξ2n+1(ω)
≤max
dξ2n(ω),ξ2n−1(ω),dξ2n(ω),ξ2n+1(ω)
, 1
2
dξ2n−1(ω),ξ2n(ω)+dξ2n(ω),ξ2n+1(ω) ≤maxd(ξ2n(ω),ξ2n−1(ω),dξ2n(ω),ξ2n+1(ω)
(2.4)
for eachω∈Ω. Now, ifM(η2n(ω),η2n−1(ω),ω)=0, thend(ξ2n(ω),ξ2n+1(ω))=d(ξ2n(ω), ξ2n−1(ω))=0 for eachω∈Ω. But ifM(η2n(ω),η2n−1(ω),ω)>0 for everyω∈Ω, then
dξ2n(ω),ξ2n+1(ω)< Mη2n(ω),η2n−1(ω),ω ≤maxdξ2n(ω),ξ2n−1(ω)
,dξ2n(ω),ξ2n+1(ω)
(2.5)
for eachω∈Ω. That is,d(ξ2n(ω),ξ2n+1(ω))< d(ξ2n(ω),ξ2n−1(ω)) for eachω∈Ωandn∈
N. Thus in any case we have
dξ2n(ω),ξ2n+1(ω)
≤Mη2n(ω),η2n−1(ω),ω
≤dξ2n(ω),ξ2n−1(ω) for each ω∈Ω.
(2.6)
Similarly,
dξ2n+1(ω),ξ2n+2(ω)
≤Mη2n+1(ω),η2n(ω),ω
≤dξ2n(ω),ξ2n+1(ω)
for each ω∈Ω. (2.7)
then for somem∈N, we have
L≤dξ2m+1(ω),ξ2m+2(ω)
≤Mη2m+1(ω),η2m(ω),ω
≤dξ2m(ω),ξ2m+1(ω)
< L+δ for eachω∈Ω. (2.8)
So, we have
dξ2m(ω),ξ2m+1(ω)
=dBω,η2m−1(ω)
,Aω,η2m(ω)
< L for somem∈N. (2.9)
Which contradicts the choice ofLand this completes the proof of part (a). Proof (part (b)). Since limn→∞d(ξn(ω),ξn+1(ω))=0 for each ω∈Ω, so, there exists a positive integern0such that d(ξn(ω),ξn+1(ω))< r/2 for all n≥n0 for everyω∈Ω. Let
p,q∈Nsuch thatp,q≥n0withp=2n,q=2m−1. Suppose for eachω∈Ω,
ε≤dξp(ω),ξq(ω)=dξ2n(ω),ξ2m−1(ω)< ε+r. (2.10)
Consider
ε≤dξp(ω),ξq(ω)
=dSω,η2n(ω),Tω,η2m−1(ω)≤Mη2n(ω),η2m−1(ω),ω
=max
dSω,η2n(ω),Tω,η2m−1(ω),dSω,η2n(ω),Aω,η2n(ω), dTω,η2m−1(ω),Bω,η2m−1(ω),
1 2
dSω,η2n(ω),Bω,η2m−1(ω)
+dTω,η2m−1(ω)
,Aω,η2n(ω)
=max
dξp(ω),ξq(ω),dξp(ω),ξp+1(ω)
,dξq(ω),ξq+1(ω)
, 1
2
dξp(ω),ξq+1(ω)
+dξq(ω),ξp+1(ω)
≤max
dξp(ω),ξq(ω),1 2
2dξp(ω),ξq(ω)+dξq(ω),ξq+1(ω)
+dξp(ω),ξp+1(ω)
≤dξp(ω),ξq(ω)
+r 2 < ε+
3r
2 < ε+ 2r≤δ+ε for eachω∈Ω.
(2.11)
Thus,
dξp+1(ω),ξq+1(ω)
=dξ2n+1(ω),ξ2m(ω)=dAω,η2n(ω),Bω,η2m−1(ω)
< ε for eachω∈Ω, (2.12)
Proof (part (c)). Takeα=2ε, part (b) assures the existence of positive integern1such that wheneverp,qare some positive integers of opposite parity andp,q > n1, then
dξp+1(ω),ξq+1(ω)
< ε ifε≤dξp(ω),ξq(ω)
< ε+r (2.13)
for each ω∈Ω. Since limn→∞d(ξn(ω),ξn+1(ω))=0 for eachω∈Ω, choose a positive integern0such that n0> n1and
dξm(ω),ξm+1(ω)
< r
6 for everym≥n0and for eachω∈Ω. (2.14) Selectq > p≥n0 so that (2.13) and (2.14) hold. Now we show that for eachω∈Ω, we haved(ξp(ω),ξq(ω))< α. If it is not true then for someω∈Ωwe have
dξp(ω),ξq(ω)
≥α=2ε. (2.15)
We first want to choosem > psuch that for all thoseω∈Ωfor which (2.15) holds, we have the following inequality:
ε+r 3< d
ξp(ω),ξm(ω)
< ε+r, (2.16)
with pandmof opposite parity. Letkbe the smallest integer greater than p such that
d(ξp(ω),ξk(ω)> ε+r/2. Sincer < ε, the integerkexists from (2.15). Moreover, we have
dξp(ω),ξk(ω)< ε+2r
3. (2.17)
For otherwise,
ε+2r 3 ≤d
ξp(ω),ξk(ω)
≤dξp(ω),ξk−1(ω)+dξk−1(ω),ξk(ω)
< dξp(ω),ξk−1(ω)+r 6,
(2.18)
which givesε+r/2< d(ξp(ω),ξk−1(ω)). Sincek−1> p≥n0> n1, so this contradicts the choice ofk. Thus we have
ε+r 3 < d
ξp(ω),ξm(ω)
< ε+r. (2.19)
Ifpandkare of opposite parity, we takem=kin (2.19) to obtain (2.16). Ifpandkare of like parity, thenpandk+ 1 will be of opposite parity. Sinced(ξk(ω),ξk+1(ω))< r/6, now, using triangle inequality, we have
ε+r 3< d
ξp(ω),ξk+1(ω)
< ε+5r
in this case, takem=k+ 1. So we can choosemso that (2.16) holds. Now forp,m≥n0, we get
ε+r 3< d
ξp(ω),ξm(ω)
≤dξp(ω),ξp+1(ω)
+dξp+1(ω),ξm+1(ω)
+dξm+1(ω),ξm(ω)
, (2.21)
which givesε+r/3< r/3 +d(ξp+1(ω),ξm+1(ω))< ε+r/3. This contradiction concludes the
required result.
Theorem 2.2. LetXbe a Polish space and let (A,S) and (B,T) be two pairs of compatible random operators from Ω×X toX withA(ω,X)⊂T(ω,X) andB(ω,X)⊂S(ω,X) for everyω∈Ω, and
(1) for any givenε >0, there existsδ >0 such that for allx,y∈X,
ε≤maxdS(ω,x),T(ω,y),dA(ω,x),S(ω,x),dB(ω,y),T(ω,y)
< δ+ε=⇒dA(ω,x),B(ω,y)< ε, (2.22)
for eachω∈Ω;
(2) alsod(A(ω,x),B(ω,y))≤k(ω)d(S(ω,x),T(ω,y)) for eachω∈Ωandx,y∈X, where
k:Ω→[1,∞) is a measurable mapping.
If one of the random operatorsA,B,S, orTis continuous thenA,B,S, andThave unique common random fixed point.
Proof. Consider, the sequence of function{ξn}as taken inDefinition 1.7. Now contrac-tive condition (1) in this theorem gives the similar contraccontrac-tive conditions as given in Lemma 2.1, so{ξn(ω)}is a Cauchy sequence for eachω∈Ω. Therefore,ξn(ω)→ξ(ω), whereξ:Ω→X, being the limit of the sequence of measurable mappings, is a measur-able mapping. Now
ξ2n−1(ω)=Tω,η2n−1(ω)=Aω,η2n−2(ω)−→ξ(ω),
ξ2n(ω)=B
ω,η2n−1(ω)=Sω,η2n(ω)
−→ξ(ω) (2.23)
for eachω∈Ω. Suppose thatSis a continuous random operator, then
Sω,Sω,η2n(ω)−→Sω,ξ(ω), Sω,A(ω,η2n−2(ω)−→Sω,ξ(ω) (2.24)
for everyω∈Ω. SinceAandSare compatible random operators, this gives
Aω,Sω,η2n−2(ω)
−→Sω,ξ(ω)=h(ω) for eachω∈Ω, (2.25)
whenn→ ∞. SinceXis a complete separable metric space, so for any continuous random operatorS, a mapping h:Ω→X defined by h(ω)=S(ω,ξ(ω)) is measurable [15,28]. Since A(ω,X)⊂T(ω,X) for eachω∈Ω, define mappings ζ2n−2:Ω→X asA(ω,S(ω,
η2n−2(ω)))=T(ω,ζ2n−2(ω)) for eachω∈Ω. Thus
Tω,ζ2n−2(ω)−→h(ω), Sω,Sω,η2n(ω)
for everyω∈Ω. Since
dAω,Sω,η2n−2(ω),Bω,ζ2n−2(ω) ≤kω)dSω,Sω,η2n(ω)
,Tω,ζ2n−2(ω)
(2.27)
for eachω∈Ω, on taking limitn→ ∞, we haveB(ω,ζ2n−2(ω)))→h(ω) for eachω∈Ω. Also
dAω,ξ(ω),Bω,ζ2n−2(ω)
≤k(ω)dSω,ξ(ω),Tω,ζ2n−2(ω).
(2.28)
Applying limitn→ ∞, we haved(A(ω,ξ(ω)),h(ω))≤k(ω)d(S(ω,ξ(ω)),h(ω)). SoA(ω,
ξ(ω))=h(ω) for eachω∈Ω. NowA(ω,X)⊂T(ω,X) andA(ω,ξ(ω))∈A(ω,X) for each
ω∈Ωallow to define the measurable mappingρ0:Ω→X asA(ω,ξ(ω))=T(ω,ρ0(ω)). Using (1) we haveA(ω,ξ(ω))=B(ω,ρ0(ω)) for eachω∈Ω. So
h(ω)=Aω,ξ(ω)=Bω,ρ0(ω)
=Tω,ρ0(ω)
=Sω,ξ(ω) (2.29)
for everyω∈Ω. ThusA(ω,S(ω,ξ(ω)))=S(ω,A(ω,ξ(ω))) gives
Aω,h(ω)=Sω,h(ω) for everyω∈Ω. (2.30)
Moreover
Aω,Aω,ξ(ω)=Aω,Sω,ξ(ω)=Sω,Aω,ξ(ω)=Sω,Sω,ξ(ω) (2.31)
for eachω∈Ω. Also
Bω,Bω,ρ0(ω)
=Bω,Tω,ρ0(ω)
=Tω,Bω,ρ0(ω)
=Tω,Tω,ρ0(ω)
(2.32)
for eachω∈Ω. Using (2.30) we have
Aω,h(ω)=Aω,Aω,ξ(ω)=Aω,h(ω)=h(ω),
Bω,ρ0(ω)
=Bω,Bω,ρ0(ω)
=Bω,h(ω)=h(ω), (2.33)
for eachω∈Ω. So,his a random fixed point ofA,B,S, andT. Uniqueness of random
fixed point follows from (1).
3. Common random fixed points in symmetric spaces
Theorem 3.1. Let (X,d) be a separable symmetric space that satisfies (w.1) and (HE). Let T,S:Ω×X→X be two weakly compatible random operators satisfying the property (I). Moreover, for allx,y∈Xwe have
dT(ω,x),T(ω,y)
≤φmaxdS(ω,x),S(ω,y),dS(ω,x),T(ω,y),dT(ω,y),S(ω,y) (3.1)
for everyω∈Ω. IfT(ω,X)⊂S(ω,X) and one ofT(ω,X) orS(ω,X) is a complete subspace ofXfor everyω∈Ω, thenTandShave unique random fixed point.
Proof. Since random operatorsTandSsatisfy the property (I), so there exists a sequence {ξn}inM(Ω,X) such that
lim n→∞d
Tω,ξn(ω)
,ξ(ω)=lim n→∞d
Sω,ξn(ω)
,ξ(ω)=0 (3.2)
for every ω∈Ω, for some ξ ∈M(Ω,X). Therefore by property (HE), we have limn→∞d(T(ω,ξn(ω)),S(ω,ξn(ω)))=0 for every ω∈Ω. SupposeS(ω,X) is a complete subspace ofXfor everyω∈Ω. Letξ1:Ω→Xbe the limit of the sequence of measurable mappings{S(ω,ξn(ω))}andS(ω,ξn(ω))∈S(ω,X) for everyω∈Ωandn∈N. Now since Xis separable, thereforeξ1∈M(Ω,X). Moreoverξ1(ω)∈S(ω,X) for everyω∈Ω.This allows obtaining the measurable mappingξ:Ω→Xsuch thatξ(ω)=S(ω,ξ(ω)). Now we show thatT(ω,ξ(ω))=S(ω,ξ(ω)) for everyω∈Ω. Consider
dTω,ξ(ω),Tω,ξn(ω)
≤φmaxdSω,ξ(ω),Sω,ξn(ω)
,dSω,ξ(ω),Tω,ξn(ω)
,
dSω,ξn(ω)
,Tω,ξn(ω)
<maxdξ(ω),Sω,ξn(ω)
,dξ(ω),Tω,ξn(ω)
dSω,ξn(ω)
,Tω,ξn(ω)
for everyω∈Ω.
(3.3)
Takingn→ ∞, we haved(T(ω,ξ(ω)),T(ω,ξn(ω)))→0 for everyω∈Ω. Thus by (w.1) we haveT(ω,ξ(ω))=S(ω,ξ(ω)) for everyω∈Ω. The weak compatibility of random map-pingsT andSimplies thatT(ω,S(ω,ξ(ω)))=S(ω,T(ω,ξ(ω))), thenT(ω,T(ω,ξ(ω)))=
T(ω,S(ω,ξ(ω)))=S(ω,T(ω,ξ(ω)))=S(ω,S(ω,ξ(ω))) for everyω∈Ω. Let us show that
T(ω,T(ω,ξ(ω)))=T(ω,ξ(ω)) for eachω∈Ω. If not, then for someω∈Ω, consider
dTω,ξ(ω),Tω,ξ(ω)
≤φmaxdSω,ξ(ω),Sω,ξ(ω),dS(ω,ξ(ω),Tω,ξ(ω),
dSω,ξ(ω),Tω,ξ(ω)
≤φmaxdT(ω,ξ(ω),Tω,ξ(ω),dTω,ξ(ω),Tω,ξ(ω) ≤φdTω,ξ(ω),Tω,ξ(ω)
< dTω,ξ(ω),Tω,ξ(ω),
which is a contradiction, soT(ω,ξ(ω)) is a random fixed point ofT. NowT(ω,ξ(ω))=
T(ω,T(ω,ξ(ω)))=S(ω,T(ω,ξ(ω))) for everyω∈Ω. ThereforeT(ω,ξ(ω)) is a common random fixed point ofT andS. The proof is similar whenT(ω,X) is supposed to be a complete subspace ofXfor everyω∈Ω, asT(ω,X)⊂S(ω,X) for eachω∈Ω. To prove the uniqueness of common random fixed point, letη,η:Ω→X be two common ran-dom fixed points of ranran-dom operatorsTandSsuch thatη(ω) =η(ω) for someω∈Ω. Consider
dη(ω),η(ω)=dTω,η(ω),Tω,η(ω)
≤φmaxdSω,η(ω),Sω,η(ω),dSω,η(ω),Tω,η(ω),
dSω,η(ω),Tω,η(ω) ≤φdη(ω),η(ω)< dη(ω),η(ω).
(3.5)
This contradiction showsη(ω)=η(ω) for everyω∈Ω.
Theorem 3.2. Let (X,d) be a separable symmetric space that satisfies (w.1), (w.2), and (HE). Let (A,T) and (B,S) be two pairs of weakly compatible random operators fromΩ×X
toXsuch that one of the pairs (A,S) or (B,T) satisfies the property (I). Moreover
dA(ω,x),B(ω,y)
≤φmaxdS(ω,x),T(ω,y),dS(ω,x),B(ω,y),dT(ω,y),B(ω,y) (3.6)
for everyω∈Ω. If B(ω,X)⊂S(ω,X), A(ω,X)⊂T(ω,X), and one of T(ω,X), S(ω,X),
B(ω,X), orS(ω,X) is complete subspace ofX for everyω∈Ω, thenA,B,T, and Shave unique common random fixed point.
Proof. Suppose the pair (B,T) of random mappings satisfies the property (I). So there exists a sequence{ξn}inM(Ω,X) such that
lim n→∞d
Bω,ξn(ω)
,ξ(ω)=lim n→∞d
Tω,ξn(ω)
,ξ(ω)=0 (3.7)
for everyω∈Ωfor someξ∈M(Ω,X). As {B(ω,ξn(ω))}is a sequence of measurable mappings andB(ω,ξn(ω))∈B(ω,X) for everyω∈Ωandn∈N, now the factB(ω,X)⊂ S(ω,X) allows obtaining the sequence of measurable mappings ηn:Ω→X such that B(ω,ξn(ω))=S(ω,ηn(ω)) for everyω∈Ω. Hence, limn→∞d(S(ω,ηn(ω)),ξ(ω))=0 for everyω∈Ω. Now we show that limn→∞d(T(ω,ηn(ω)),ξ(ω))=0 for everyω∈Ω. For this consider
dAω,ηn(ω)
,Bω,ξn(ω)
≤φmaxdSω,ηn(ω)
,Tω,ξn(ω)
,dSω,ηn(ω)
,Bω,ξn(ω)
,
dTω,ξn(ω)
,Bω,ξn(ω)
≤φmaxdBω,ξn(ω)
,Tω,ξn(ω)
,dT(ω,ξn(ω)
,Bω,ξn(ω)
≤φdBω,ξn(ω)
,Tω,ξn(ω)
< dBω,ξn(ω)
,Tω,ξn(ω)
for everyω∈Ω. Therefore by property (HE), we have
lim n→∞d
Bω,ξn(ω)
,Tω,ξn(ω)
=0 for everyω∈Ω. (3.9)
Hence,
lim n→∞d
Bω,ξn(ω)
,Aω,ηn(ω)
=0 for everyω∈Ω. (3.10)
By (w.2), we deduce that limn→∞d(A(ω,ηn(ω)),ξ(ω))=0, for everyω∈Ω. Suppose for everyω∈Ω,S(ω,X) is a complete subspace of X. Now{S(ω,ηn(ω)}is a sequence of measurable mappings andS(ω,ηn(ω))∈S(ω,X) for everyω∈Ω. Letξ1:Ω→Xbe the limit of the sequence of measurable mappings{S(ω,ξn(ω))}. SinceXis separable, there-foreξ1∈M(Ω,X). Moreoverξ1(ω)∈S(ω,X) for everyω∈Ω. This allows obtaining the measurable mappingξ:Ω→Xsuch thatξ(ω)=S(ω,ξ(ω)). Now consider
lim n→∞d
Aω,ηn(ω)
,Sω,ξ(ω)=lim n→∞d
Bω,ξn(ω)
,Sω,ξ(ω) =lim
n→∞d
Tω,ξn(ω),Sω,ξ(ω)
=lim n→∞d
Sω,ηn(ω)
,Sω,ξ(ω)=0
(3.11)
for everyω∈Ω. Thus
dAω,ξ(ω),Bω,ξn(ω)
≤φmaxdSω,ξ(ω),Tω,ξn(ω)
,dSω,ξ(ω),Bω,ξn(ω)
dTω,ξn(ω)
,Bω,ξn(ω)
,
(3.12)
for eachω∈Ω. This immediately gives
lim n→∞d
Aω,ξ(ω),Bω,ξn(ω)
=0 for everyω∈Ω. (3.13)
By (w.1), we haveA(ω,ξ(ω))=S(ω,ξ(ω)) for everyω∈Ω. The weak compatibility of random operatorsAandSimplies thatS(ω,A(ω,ξ(ω)))=A(ω,S(ω,ξ(ω))) for everyω∈ Ω. Now
Aω,Aω,ξ(ω)=Aω,Sω,ξ(ω)=Sω,Aω,ξ(ω)=Sω,Sω,ξ(ω) (3.14)
A(ω,ξ(ω))=T(ω,ξ1(ω)) for everyω∈Ω. We now show that for everyω∈Ω,B(ω,ξ1(ω)) =T(ω,ξ1(ω)). If not, then for someω∈Ω, consider
dAω,ξ(ω),Bω,ξ1(ω)
≤φmaxdSω,ξ(ω),Tω,ξ1(ω)
,dSω,ξ(ω),Bω,ξ1(ω)
,
dTω,ξ1(ω)
,Bω,ξ1(ω)
≤φmaxdAω,ξ(ω),Bω,ξ1(ω)
,dAω,ξ(ω),Bω,ξ1(ω)
≤φdAω,ξ(ω),Bω,ξ1(ω)
< dAω,ξ(ω),Bω,ξ1(ω)
,
(3.15)
which is a contradiction. Hence,
Bω,ξ1(ω)
=Tω,ξ1(ω)
=Aω,ξ(ω)=Sω,ξ(ω) for everyω∈Ω. (3.16)
The weak compatibility of random operatorsBandTimplies that
Bω,Tω,ξ1(ω)
=Tω,Bω,ξ1(ω)
for everyω∈Ω,
Tω,Tω,ξ1(ω)
=Tω,Bω,ξ1(ω)
=Bω,Tω,ξ1(ω)
=Bω,Bω,ξ1(ω)
(3.17)
for eachω∈Ω. Let us show thatA(ω,A(ω,ξ(ω)))=A(ω,ξ(ω)) for eachω∈Ω. If not, then for someω∈Ω, consider
dAω,ξ(ω),Aω,Aω,ξ(ω) =dAω,Aω,ξ(ω),Bω,ξ1(ω)
≤φmaxdS(ω,Aω,ξ(ω),Tω,ξ1(ω)
,dSω,Aω,ξ(ω),Bω,ξ1(ω)
,
dTω,ξ1(ω)
,Bω,ξ1(ω)
≤φmaxdAω,Aω,ξ(ω),Aω,ξ(ω),Aω,Aω,ξ(ω),Aω,ξ(ω) ≤φdAω,Aω,ξ(ω),Aω,ξ(ω)
< dAω,Aω,ξ(ω),Aω,ξ(ω),
(3.18)
which is a contradiction. Therefore
Aω,Aω,ξ(ω)=Aω,ξ(ω)=Sω,Aω,ξ(ω) for everyω∈Ω. (3.19)
SoA(ω,ξ(ω)) is common random fixed point of random operatorsAandS. Similarly,
B(ω,ξ1(ω)) is common random fixed point of random operatorsBandT. SinceA(ω,
is a complete subspace of X, sinceA(ω,X)⊂T(ω,X) andB(ω,X)⊂S(ω,X) for every
ω∈Ω. To establish the uniqueness of common random fixed point of random operators, let ξ andη be two common random fixed points of the random operators such that
ξ(ω) =η(ω) for someω∈Ω. Consider
dξ(ω),η(ω)=dAω,ξ(ω),Bω,η(ω)
≤φmaxdSω,ξ(ω),Tω,η(ω),dSω,ξ(ω),Bω,η(ω),
dTω,η(ω),Bω,η(ω) ≤φdξ(ω),η(ω)< dξ(ω),η(ω),
(3.20)
which is a contradiction. So the result follows.
Acknowledgment
Authors are thankful to the referee for his critical remarks to improve the presentation of the paper.
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Ismat Beg: Department of Mathematics and Center for Advanced Studies in Mathematics, Lahore University of Management Sciences, 54792 Lahore, Pakistan
E-mail address:[email protected]
Mujahid Abbas: Department of Mathematics and Center for Advanced Studies in Mathematics, Lahore University of Management Sciences, 54792 Lahore, Pakistan
