Ismat Beg a , Mujahid Abbas a , Akbar Azam b
3. Periodic and fixed points of rotative random op- op-erators
In this section, we first show an existence of a random fixed point of a nonexpansive rotative random operator which not only provides a random analogue of theorem 17.1 of [11] (see also [12]) but also improves theorem 2.1 of [17] in the sense that it does not require the boundedness of T(ω, F)for anyω ∈Ω. Moreover we replace continuous condensing random operator by nonexpansive rotative random operator.
Periodic point problems were systematically studied since the beginning of fifties (see [9, 10, 13, 14, 19, 20]). We show some results on the existence of random periodic points of random single valuedǫ- contractive operator in the setting of a separable metric space.
Periodic fixed points of random operators 43 Theorem 3.1. Let F be a nonempty closed and convex subset of a separable Ba-nach space X and T: Ω×F → F be a nonexpansive rotative random operator.
Then T has a random fixed point.
Proof. Letξ: Ω→F be any fixed measurable mapping. For0 < α <1and any arbitrary measurable mappingη: Ω→F,define Tα: Ω×F →F as,
Tα(ω, η(ω)) = (1−α)ξ(ω) +αT(ω, η(ω)).
Note that for eachα, the random operatorTα has Lipschitz constantα. we may apply [8] to obtain the sequence of random operators Fα: Ω×F → F such that Tα(ω, Fα(ω, ξ(ω))) =Fα(ω, ξ(ω)),for everyω∈Ω.Consequently, we have
Fα(ω, ξ(ω)) = (1−α)ξ(ω) +αT(ω, Fα(ω, ξ(ω))).
It can be verified that each Fα is nonexpansive random operator. By iteratingFα
we obtain
Fαk(ω, ξ(ω)) = (1−α)Fαk−1(ω, ξ(ω)) +αT(ω, Fαk(ω, ξ(ω))), k∈N. (3.1) Note that,
(1−α)Fα(ω, ξ(ω))
= (1−α)ξ(ω) +αT(ω, Fα(ω, ξ(ω)))−αFα(ω, ξ(ω))
= (1−α)ξ(ω) +α(T(ω, Fα(ω, ξ(ω)))−Fα(ω, ξ(ω))).
Thus for each ω∈Ω
(1−α)(ξ(ω)−Fα(ω, ξ(ω)))
=α(Fα(ω, ξ(ω))−T(ω, Fα(ω, ξ(ω)))). (3.2) Now suppose T is a (a, n)-rotative random operator, that is
kξ(ω)−Tn(ω, ξ(ω))k6akξ(ω)−T(ω, ξ(ω))k, for everyω∈Ω.Now,
Fα(ω, ξ(ω))−Fα2(ω, ξ(ω))
=
(1−α)ξ(ω) +αT(ω, Fα(ω, ξ(ω)))−(1−α)Fα(ω, ξ(ω))
−αT(ω, Fα2(ω, ξ(ω)))
=
(1−α)(ξ(ω)−Fα(ω, ξ(ω))) +α(T(ω, Fα(ω, ξ(ω)))
−αT(ω, Fα2(ω, ξ(ω)))
=
α(Fα(ω, ξ(ω))−T(ω, Fα(ω, ξ(ω)))) +α(T(ω, Fα(ω, ξ(ω)))
−αT(ω, Fα2(ω, ξ(ω)))
=αFα(ω, ξ(ω))−T(ω, Fα2(ω, ξ(ω)))
44 I. Beg, M. Abbas, A. Azam 6αkFα(ω, ξ(ω))−Tn(ω, Fα(ω, ξ(ω)))k
+αTn(ω, Fα(ω, ξ(ω)))−T(ω, Fα2(ω, ξ(ω))) 6αakFα(ω, ξ(ω))−T(ω, Fα(ω, ξ(ω)))k
+αTn−1(ω, Fα(ω, ξ(ω)))−Fα2(ω, ξ(ω))
= (1−α)akFα(ω, ξ(ω))−ξ(ω)k
+αTn−1(ω, Fα(ω, ξ(ω)))−Fα2(ω, ξ(ω)),
for everyω∈Ω.Now we claim that the following inequality holds for everyω∈Ω andm>2.
αTm−1(ω, Fα(ω, ξ(ω)))−Fα2(ω, ξ(ω)) 6(m−1)−mα+αmkξ(ω)−Fα(ω, ξ(ω))k
+αmFα(ω, ξ(ω))−Fα2(ω, ξ(ω)). (3.3) For this consider,
αT(ω, Fα(ω, ξ(ω)))−Fα2(ω, ξ(ω))
=αT(ω, Fα(ω, ξ(ω)))−(1−α)Fα(ω, ξ(ω))−αT(ω, Fα2(ω, ξ(ω)))
=α
(1−α)(T(ω, Fα(ω, ξ(ω)))−Fα(ω, ξ(ω)))−α(T(ω, Fα2(ω, ξ(ω)))
−T(ω, Fα(ω, ξ(ω))))
6(1−α)kα(T(ω, Fα(ω, ξ(ω)))−Fα(ω, ξ(ω)))k
+α2T(ω, Fα2(ω, ξ(ω)))−T(ω, Fα(ω, ξ(ω)))
= (1−α)2kξ(ω)−Fα(ω, ξ(ω))k+α2T(ω, Fα2(ω, ξ(ω)))−T(ω, Fα(ω, ξ(ω))) 6(1−α)2kξ(ω)−Fα(ω, ξ(ω))k+α2Fα2(ω, ξ(ω))−Fα(ω, ξ(ω)).
So (3.3) is valid for m= 2and for anyω∈Ω.
Assuming the validity of (3.3) form=j and for anyω∈Ω,consider αTj(ω, Fα(ω, ξ(ω)))−Fα2(ω, ξ(ω))
=αTj(ω, Fα(ω, ξ(ω)))−(1−α)Fα(ω, ξ(ω))−αT(ω, Fα2(ω, ξ(ω)))
=α
(1−α)(Tj(ω, Fα(ω, ξ(ω)))−Fα(ω, ξ(ω))) +α(Tj(ω, Fα(ω, ξ(ω)))
−T(ω, Fα2(ω, ξ(ω))))
6α(1−α)Tj(ω, Fα(ω, ξ(ω)))−Fα(ω, ξ(ω))
+α2Tj(ω, Fα(ω, ξ(ω)))−T(ω, Fα2(ω, ξ(ω))) 6jα(1−α)kFα(ω, ξ(ω))−T(ω, Fα(ω, ξ(ω)))k
+α2Tj−1(ω, Fα(ω, ξ(ω)))−Fα2(ω, ξ(ω)) 6jα(1−α)kFα(ω, ξ(ω))−T(ω, Fα(ω, ξ(ω)))k
+α[(j−1)−jα+αj]kξ(ω)−Fα(ω, ξ(ω))k +αj+1Fα(ω, ξ(ω))−Fα2(ω, ξ(ω))
6j(1−α)2+α2[(j−1)−jα+αj]kξ(ω)−Fα(ω, ξ(ω))k
Periodic fixed points of random operators 45 +αj+1Fα(ω, ξ(ω))−Fα2(ω, ξ(ω))
6[j−(j+ 1)α+αj+1]kξ(ω)−Fα(ω, ξ(ω))k +αj+1Fα(ω, ξ(ω))−Fα2(ω, ξ(ω)).
So by induction inequality(3.3)is valid for everyω∈Ωand m>2.
Now consider, forω∈Ω
Fα(ω, ξ(ω))−Fα2(ω, ξ(ω)) 6(1−α)akFα(ω, ξ(ω))−ξ(ω)k
+αTn−1(ω, Fα(ω, ξ(ω)))−Fα2(ω, ξ(ω)) 6(1−α)akFα(ω, ξ(ω))−ξ(ω)k
+ [(n−1)−nα+αn]kξ(ω)−Fα(ω, ξ(ω))k +αnFα(ω, ξ(ω))−Fα2(ω, ξ(ω)).
It further implies that
(1−αn)Fα(ω, ξ(ω))−Fα2(ω, ξ(ω))
6[(1−α)a+ (n−1)−nα+αn]kξ(ω)−Fα(ω, ξ(ω))k, for everyω∈Ω.Now we arrive at
Fα(ω, ξ(ω))−Fα2(ω, ξ(ω))
6(1−αn)−1[(1−α)a+ (n−1)−nα+αn]kξ(ω)−Fα(ω, ξ(ω))k 6(a+n)(1−α)(1−αn)−1−1kξ(ω)−Fα(ω, ξ(ω))k
= [(a+n)(
nX−1 i=0
αi)−1−1]kξ(ω)−Fα(ω, ξ(ω))k
=g(α)kξ(ω)−Fα(ω, ξ(ω))k,
for every ω ∈ Ω, where g(α) = [(a+n)(Pn−1
i=0 αi)−1−1]. Since g is continuous and decreasing for α∈(0,1]with g(1) = an < 1, there exists b ∈(0,1]such that g(1)<1 for α∈(b,1]. For suchα, the sequence of measurable mappings defined byηn(ω) =Fαn(ω, ξ(ω))→η(ω), for eachω∈Ω, η: Ω→F,being the limit of the sequence of measurable functions, is also measurable (see remark 2.6).From (3.1)
it follows that η is a random fixed point ofT.
Example 3.2. LetΩ = [0,1]andΣbe the sigma algebra of Lebesgue’s measurable subsets of Ω. Take X = R with d(x, y) = |x−y|, for x, y ∈ R. Define random operatorT from Ω×X toX as,T(ω, x) =ω−x.
Define a fixed measurable mapping ξ: Ω→ X as ξ(ω) = ω3, for every ω ∈Ω.
Note that T is nonexpansive random operator. Since random operator equation T2(ω, ξ(ω)) = ξ(ω)holds for every ω ∈ Ω, therefore it is (2,1)−rotative random operator. Thus the conditions of Theorem 3.1 are satisfied. Moreover a measurable mappingη: Ω→X defined asη(ω) =ω2,for everyω∈Ω,serve as a unique random fixed point ofT.
46 I. Beg, M. Abbas, A. Azam Theorem 3.3. Let X be a separable metric space and T: Ω×X → X be a ǫ-contractive random operator. Letξ0: Ω→X be any measurable mapping such that a sequence {Tn(ω, ξ0(ω))} has a point wise convergent subsequence of measurable mappings. Then T has a random periodic point.
Proof. Let{Tni(ω, ξ0(ω))}be a subsequence of{Tn(ω, ξ0(ω))}such thatTni(ω, ξ0
(ω))→ξ(ω)for eachω∈Ωasni → ∞where{ni}is a strictly increasing sequence of positive integers. The mapping ξ: Ω → X being point wise limit of sequence of measurable mappings is measurable. Define sequence of measurable mappings ξi: Ω →X as ξi(ω) =Tni(ω, ξ0(ω)).Given ǫ >0, there exists an integern0 such that
d(ξi(ω), ξ(ω))< ǫ
4, fori>n0 andω∈Ω.
Put k=ni+1−ni.Consider,
d(ξi+1(ω), Tk(ω, ξ(ω))) =d(Tk(ω, ξi(ω)), Tk(ω, ξ(ω)))
< d(ξi(ω), ξ(ω))< ǫ
4, for eachω∈Ω.
Now,
d(ξ(ω), Tk(ω, ξ(ω)))
6d(ξi+1(ω), Tk(ω, ξ(ω))) +d(ξi+1(ω), ξ(ω))
< ǫ 4+ ǫ
4 = ǫ
2, for everyω∈Ω.
Now we claim that ξ is a random periodic point ofT. To prove this, assume that η: Ω→X be any measurable mapping such thatη(ω) =Tk(ω, ξ(ω))but
η(ω)6=ξ(ω), for some ω∈Ω. (3.4) Which implies that0< d(η(ω), ξ(ω))< ǫ. AsT is aǫ- contractive random operator therefore forω∈Ωfor which (3.4) holds, we have
d(T(ω, ξ(ω)), T(ω, η(ω)))< d(ξ(ω), η(ω)).
Define h: Ω×X2 →R as,h(ω, x(ω), y(ω)) = d(T(ω,x(ω)),T(ω,y(ω)))
d(x(ω),y(ω)) ,where x(ω)6= y(ω) ∈ X for each ω ∈ Ω. Now h(ω, ., .) is continuous at (ξ(ω), η(ω)) for every ω∈Ωfor which (3.4) is valid.
Take 0 < α < 1, then there exists δ > 0 such that x(ω) ∈ B(ξ(ω), δ) and y(ω)∈B(η(ω), δ)gives
d(T(ω, x(ω)), T(ω, y(ω)))< αd(x(ω), y(ω)).
As, lim
r→∞Tk(ω, ξr(ω)) = Tk(ω, ξ(ω)) = η(ω), for every ω ∈ Ω. So there exists n1>n0 such that
d(ξr(ω), ξ(ω))< δ
Periodic fixed points of random operators 47 and
d(Tk(ω, ξr(ω)), η(ω))< δ, forr>n1 andω∈Ω.Hence we have
d(T(ω, ξr(ω)), T(ω, Tk(ω, ξr(ω))))< αd(ξr(ω), Tk(ω, ξr(ω))). (3.5) Consider,
d(ξr(ω), Tk(ω, ξr(ω)))
6d(ξr(ω), ξ(ω)) +d(ξ(ω), Tk(ω, ξ(ω))) +d(Tk(ω, ξ(ω)), Tk(ω, ξr(ω)))
< ǫ 4 +ǫ
2+ ǫ
4 =ǫ, (3.6)
for r>n1 >n0 andω ∈Ωfor which (3.4) holds. Now using (3.5) and (3.6), we have
d(T(ω, ξr(ω)), T(ω, Tk(ω, ξr(ω))))
< αd(ξr(ω), Tk(ω, ξr(ω)))< d(ξr(ω), Tk(ω, ξr(ω)))< ǫ,
for r>n1. Since T is a ǫ- contractive random operator so forr >n1 andq >0, we have
d(Tq(ω, ξr(ω)), Tq(ω, Tk(ω, ξr(ω))))
< d(ξr(ω), Tk(ω, ξr(ω)))< ǫ α.
Put q=nr+1−nr,we haved(ξr+1(ω), Tk(ω, ξr+1(ω)))<αǫ.Hence, d(ξs(ω), Tk(ω, ξs(ω)))< ǫαs−r.
Now,
d(ξ(ω), η(ω)) 6 d(ξ(ω), ξs(ω)) +d(ξs(ω), Tk(ω, ξs(ω))) +d(Tk(ω, ξs(ω)), η(ω))→0, ass→ ∞.
for thoseω∈Ωfor which (3.4) holds. This contradiction concludes the result.
Corollary 3.4. If in theorem 3.2, the random periodic pointξ(say) ofT satisfies d(ξ(ω), T(ω, ξ(ω)))< ǫ, for every ω∈Ω. (3.7) Then ξ is a random fixed point ofT.
Proof. Letkbe the positive integer such thatTk(ω, ξ(ω)) =ξ(ω),for everyω∈Ω.
Ifξis not a random fixed point ofT,thenξ(ω)6=T(ω, ξ(ω)for someω∈Ω.Since T isǫ- contractive random operator, using (3.7) we have
d(ξ(ω), T(ω, ξ(ω))) =d(Tk(ω, ξ(ω)), Tk+1(ω, ξ(ω)))
< d(ξ(ω), T(ω, ξ(ω))).
This contradiction concludes the proof.
48 I. Beg, M. Abbas, A. Azam Remark 3.5. IfX is a separable compact metric space andT: Ω×X →X is an ǫ- contractive random operator. Then applying theorem 3.3, we conclude that T has a random periodic point.
Theorem 3.6. Let X be a separable compact metric space and T: Ω×X → X be an ǫ- contractive random operator. Then T has finitely many random periodic points.
Proof. Let ξ, ζ: Ω → X be two random periodic points ofT with ξ(ω) 6=ζ(ω) and d(ξ(ω), ζ(ω)) < ǫ for some ω ∈ Ω. Let m, n > 1 be two integers such that Tm(ω, ξ(ω)) = ξ(ω) and Tn(ω, ζ(ω)) = ζ(ω) for every ω ∈ Ω. Obviously Tmn(ω, ξ(ω)) =ξ(ω)andTmn(ω, ζ(ω)) =ζ(ω)for eachω ∈Ω. Now consider,
d(ξ(ω), ζ(ω)) =d(Tmn(ω, ξ(ω)), Tmn(ω, ζ(ω)))
< d(ξ(ω), ζ(ω)),
which is contradiction. Therefore any two random periodic point of T must be at leastǫ- apart. Compactness ofX prevents us defining infinitely many random
periodic points from Ω×X to X.
Acknowledgement. The authors are thankful to referee for precise remarks to improve the presentation of the paper.
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Ismat Beg, Mujahid Abbas
Center for Advanced Studies in Mathematics, Lahore University of Management Sciences, Lahore-54792, Pakistan
Phone: 0092-42-35608229 Fax: 0092-42-35722591 e-mail: ibeg@lums.edu.pk
mujahid@lums.edu.pk Akbar Azam
Department of Mathematics,
COMSATS Institute of Information Technology, Islamabad, Pakistan
Annales Mathematicae et Informaticae 37(2010) pp. 51–75
http://ami.ektf.hu